Higher Algebra, Barnard & Child.pdf

DAMAGE BOOK

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_1 60295

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HIGHER ALGEBRA BY S.

BARNARD, M.A.

FORMERLY ASSISTANT MASTER AT RUGBY SCHOOL, LATE FELLOW AND LECTURER AT EMMANUEL COLLEGE, CAMBRIDGE

AND J.

M. CHILD,

B.A., B.Sc.

FORMERLY LECTURER IN MATHEMATICS IN THE UNIVERSITY OF' MANCHESTER LATE HEAD OF MATHEMATICAL DEPARTMENT, TECHNICAL COLLEGE, DERBY FORMERLY SCHOLAR AT JESUS COLLEGE, CAMBRIDGE

LON-DON

MACMILLAN

fcf'CO

LTD

*v

NEW YORK

ST MARTIN *S PRESS

1959

This book

is copyright in all countries which are signatories to the Berne Convention

First Edition 1936

Reprinted 1947, ^949> I952> *955, 1959

MACMILLAN AND COMPANY LIMITED London Bombay Calcutta Madras Melbourne

THE MACMILLAN COMPANY OF CANADA LIMITED Toronto ST MARTIN'S PRESS INC

New York

PRINTED IN GREAT BRITAIN BY

LOWE AND BRYDONE (PRINTERS) LIMITED, LONDON, N.W.IO

CONTENTS

ix

IjHAPTER

EXEKCISE

XV

(128).

Minors, Expansion in Terms of Second Minors (132, 133). Product of Two Iteterminants (134). Rectangular Arrays (135). Reciprocal Two Methods of Expansion (136, 137). Use of Double Deteyrrtlilnts,

Symmetric and Skew-symmetric Determinants, Pfaffian (138-

Suffix,

143),

ExERtad XVI

(143)

X. SYSTEMS OF EQUATIONS. Systems (149, 150). Linear Equations in Line at Infinity (150-152). Linear Equations in Three Unknowns, Equation to a Plane, Plane at Infinity (153-157). Definitions, Equivalent

Two Unknowns, EXEKCISE XVII

(158).

Systems of Equations of any Degree, Methods of Solution for Special Types (160-164).

EXERCISE XVIII

XL

(164).

RECIPROCAL AND BINOMIAL EQUATIONS. The Equation Reduction of Reciprocal Equations (168-170). xn - 1=0, Special Roots (170, 171). The Equation xn -A =0 (172). The Equation a 17 - 1 ==0, Regular 17-sided Polygon (173-176). EXERCISE

XIX

(177).

AND BIQUADRATIC EQUATIONS. The Cubic Equation

(roots a, jS, y), Equation whose Roots are Value of J, Character of Roots (179, 180). Cardan's Solution, Trigonometrical Solution, the Functions a -f eo/? -f-\>V> a-f a> 2 4- a>y (180, 181). Cubic as Sum of Two Cubes, the Hessftfh (182, 183). Tschirnhausen's Transformation (186). (

-y)

2

,

etc.,

EXERCISE XX (184). The Biquadratic Equation

A=y + aS,

(roots a,

,

y, 8) (186).

The Functions

the Functions /, J, J, Reducing Cubic, Character of Roots (187-189). Ferrari's Solution and Deductions (189-191). Descartes' Solution (191). Conditions for Four Real Roots (192-ty). Transformation into Reciprocal Form (194). Tschirnhausen's Transetc.,

formation (195).

EXERCISE

XXI

(197).

OP IRRATIONALS. Sections of the System of Rationals, Dedekind's Definition (200, 201). Equality and Inequality (202). Use of Sequences in defining

a Real Number, Endless Decimals (203, 204). The Fundamental Operations of Arithmetic, Powers, Roots and Surds (204-209). Irrational Indices, Logarithms (209, 210). Definitions, Interval, Steadily Increasing Functions (210). Sections of the System of^Real Numbers, the Continuum (211, 212). Ratio and Proportion, Euclid's Definition (212, 213).

EXERCISE XXII

(214).

CONTENTS

x CHAPTER

XIV/INEQUALITIES. Elementary Methods (210, 217)

Weierstrass' Inequalities (216).

For n Numbers a l9 a 2 a

n (a* -!)/*

(a" -I)/*,,

xa x ~ l (a-b)$a x -b x

(l+x)

n

^l+nx,

\*JACJJ

>

n

n

(219).

^ xb x

~l

(a

- 6),

(219).

(220).

Arithmetic and Geometric Means (221, 222). -

-

^

V

n and Extension

(223, 224).

(224).

XV. SEQUENCES AND Definitions,

Maxima and Minima

(223).

EXERCISE XXIII

LIMITS.

Monotone

Theorems,

Sequences

(228-232).

E*

ponential Inequalities and Limits, /

l\m

/

i\n

and

1) >(!+-) m/ n/ \ /

1 \

n

lim (1-fnj n_ >00 V

EXERCISE

XXIV

l\-m /

/

=e,

nj

(1)

if

m>n,

(232,233).

(233).

General Principle of Convergence (235-237). Limits of Inde termination (237-240).

Theorems

~n

l\"w

=lim(l--) \

1 \

<(l--) nj \

(1--) mj \

Bounds of a Sequent

:

Increasing Sequence (u n ), where

u n -u n ^ l
^

1

u n >0 and u n+l lu n -*l, then u nn -*L u n l, then lim If lim (Ui+u 2 + ...

(2) If (3)

n (4) If

lim

+u n )jn

a n ~a, and lim

n-~>oo

lim (a n 6 1 n >ao

fe

w

= 6,

then

n~>oo

+ a w _ 1 6 1 + ...H-a 1 6 n )/n=o6,

(240-243).

Complex Sequences, General Principle of Convergence EXERCISE

XXV

I.

n->oo

>oo

(244).

(243, 244).

CONTENTS XVI. \CONVERGENCE OP SERIES

xi

(1).

Definitions, Elementary Theorems, Geometric Series (247, 248). Introduction and Removal of Brackets, Series of Positive Terms. p D'Alembert's of Order Terms, Comparison Tests, 271 /w Changing ,

and Cauchy's Tests

XXVI

EXERCISE

(248-254).

(254).

Terms alternately Positive and Negative (256). Absolute Convergence, with Terms Positive or Negative.

Series with Series

General Condition for Convergence, Pringsheim's Theorem, Introduction and Removal of Brackets, Rearrangement of Terms, Approximate Sum, Rapidity of Convergence or Divergence (256-261). Series of Complex Terms. Condition of Convergence, Absolute Conn n vergence, Geometric Series, Zr cos nd, Sr sin n6. If u n /u n+l = l+a n /n, where a n ->a>0, then u n -*Q. Convergence of Binomial Series (261-263).

XXVII

EXERCISE

(264).

^

XVII^CoNTiNuous VARIABLE.

<*> Theorems Meaning of Continuous Variation, Limit, Tending to on Limits and Polynomials (266-268) Continuous and Discontinuous Functions (269, 270). Continuity of Sums, Products, etc., Function Fundaof a Function, lirn {f(x)}, Rational Functions, xn (271). ,

.

mental Theorems

Derivatives, Tangent to a Curve, Notation (272). of the Calculus, Rules of Differentiation (273-277). Continuity of of {f(x)} and xn (278, 279). Meaning of Sign {f(x)}, Derivatives of J'(x) (279). Complex Functions, Higher Derivatives (2*9, 280). Maxima and Minima, Points of Inflexion (280-282).

EXERCISE XXVIII

(282).

Inverse Functions, Bounds of a Function, Rolle's Theorem, MeanValue Theorem (284-288). Integration (289). Taylor's Theorem, Lagrange's Form of Remainder (290, 291). Function of a Complex Variable, Continuity (291, 292).

XXIX

EXEBCISE

(293).

XVIIL, THEORY OF EQUATIONS TIONS

(2),

POLYNOMIALS

RATIONAL FRAC-

(2),

(1).

Multiple Roots, Rolle's Theorem, Position of Real Roots of/(&)=0 Newton's Theorem on Sums of Powers of the Roots of f(x) =0 (297). Order and Weight of Symmetric Functions (298, 299). (296, 296).

Partial Derivatives, Taylor's x, y,

...

Euler's

.

(299-302).

Theorem

for Polynomials in

for Polynomials, J

x

A Theorem on Partial Fractions (302). lf

EXERCISE

Theorem

XXX

(304).

+ ...=0,

(303).

dx

+y

x and in

+ ...~nu dy

The Equation

CONTENTS

xii

CHAPTER

XIX. EXPONENTIAL AND LOGARITHMIC FUNCTIONS AND

SERIES.

The Exponential Continuity, Inequalities and Limits (306, 307). x Theorem, Series for a (307, 308). Meaning of an Irrational Index, Derivatives of a x log x and x n (309). Inequalities and Limits, the x way in which e and log x tend to oo Euler's Constant y, Series for 2 The Exponential Function E(z), Complex Index (310-312). log ,

,

(312, 313).

Series for sinx, cos x arid Exponential Values (313). in Summing Series (314).

Use of Exponential Theorem

EXERCISE

XXXI

(315).

Logarithmic Series and their Use in Summation of Series, Calculation of Logarithms (315-319). The Hyperbolic Functions (319-321).

EXERCISE

XXXII

XX. CONVERGENCE

(321).

(2).

Series of Positive Terms.

2

-

'

Cauchy's Condensation Test, Test Series

Rummer's, Raabc's and Gauss's Tests

(325).

Binomial and Hyper-geometric Series (328, 329). %

(326-328).

De Morgan and

Bertrand's Tests (330).

Terms Positive or Negative. Theorem, Abel's Inequality, and Abel's Tests (330, 331). Power Series, Interval and Radius of Convergence, Criterion for Identity of Power Series (332Binomial Series l-f?iz4-... when z is complex (334). Multi334). plication of Series, Merten's and Abel's Theorems (335-338). Series with

Dirichlct's

EXERCISE XXXIII

(338).

XXI. JBlNOMIAL AND MULTINOMIAL THEOREMS. Statement,

Vandermonde's Theorem

Binomial Theorem.

(340).

Euler's Proof, Second Proof, Particular Ins tancesj 34 1-345).

Num-

";

1 + x (345-348). erically Greatest Term, Approximate Values of EXERCISE XXXIV (349). Use of Binomial Theorem in Summing Series, ^Multinomia^Theorem

(351-355).

EXERCISE

XXXV

(355).

XXII. RATIONAL FRACTIONS

(2),

PR,ECURRING)SERIES

AND DIFFERENCE

EQUATIONS. Expansion of a Rational Fraction (357-359).

EXERCISE

XXXVI

(359).

Expansions of cos nd and sin nOjsin 9 in Powers of cos 6 (360). Recurring Series, Scale of Relation, Convergence, Generating FuncLinear Difference Equations with Constant tion, Sum (360-363). Coefficients (363-365).

EXERCISE

XXXVII

(365).

Difference Equations, General

EXERCISE

XXXVIII

(370).

and Particular Solutions

(367-370).

CONTENTS

xiii

CHAPTER

XXIII. THE OPERATORS 4> E, D. The Operators J, E, Terms of u l9 Au^ A 2 u l9

INTERPOLATION. A ru x u x+r

Series for ...

,

;

U1 + u2 + u3 +

...

in

and

Interpolation, Lagrange's

(373-379).

d \n (uv) ( j- ) Vcte/ /

Bessel's

The Operator

Formulae (379-382).

f) 9

Value of

(382, 383).

EXERCISE

XXXIX

(384).

XXIV. CONTINUED FRACTIONS

(1).

Definitions, ForniationjoConvergents, Infinite Continued Fractions

(388-391).

EXERCISE

Simple and ^eeuTrirSg; Continued Fractions (391-394).

XL

(394).

Simple Continued Fractions, Properties of the Convergents, an Irrational as a Simple Continued Fraction (396-401). Approximations, Miscellaneous Theorems (402-406). Symmetric Continued Fractions, Application to Theory of Numbers (406-409). Recurring Con tinned Fractions (409-411).

EXERCISE XLI

(411).

XXV. INDETERMINATE EQUATIONS

OF THE FIRST DEGREE.

axby axbycz...=k

Solutions of the Equation

EXERCISE XLII

Simple

c (414-417).

Two

Equations in

x, y, z

;

(417-419).

(419).

XXVI. THEORY OF NUMBERS

(2).

Congruence, Numbers less than and prime to n, Value of q)(n) E
'

(424-427).

EXERCISE XL1II

(427).

Roots of a Congruence, the Linear Congruence, Simultaneous Congruences (428-431). Theorem on Fractions (431). The General Congruence, Division (mod n) 9 Congruences to a Prime Modulus (432-434).

EXERCISE

XLIV

(434).

XXVII. RESIDUES OF POWERS OF A NUMBER, RECURRING DECIMALS. Residues of a, ag ag* 9

(436, 437).

9

...

n) the Congruence gr*s 1 (mod n) 9 Modulus, Primitive Roots (438, 439).

(mod

An Odd Prime

9

Decimal Equivalent of m/n Number of Figures in the Period, Short Methods of Reckoning, Prime Factors of 10* - 1, 9

(

EXERCISE

XLV

= 1,2,...

10), (439-444).

(442).

The Congruence xn -sa (mod p) Methods of Solution (444 EXERCISE XLVI (445). 9

45).

CONTENTS

xiv CHAPTER

EXERCISE XLVII

(447).

Separation of the Roots, Sturm's Theorem, Fourier's Theorem (447-456).

EXERCISE XLVIII (456). Newton's Method of Approximating to a Root, Fourier's Rule, Nearly Equal Roots (456-460). Horner's Method (461-465). EXERCISE XLIX (466).

FUNCTIONS, CURVE TRACING. Implicit Functions, Rule for Approximations (467-469). for Roots of a Cubic Equation (when all Real) (470).

EXERCISE

L

Series

(471).

Tangents, Asymptotes, Intersection of a; Straight Line and Curve (473). Curve Tracing, Newton's Parallelogram (474-480).

EXERCISE LI

XXX.

(480).

INFINITE PRODUCTS. Convergence, Absolute Convergence, Derangement of Factors, Convergence discussed by Use of

Expansion as a Series (485-488). Logarithms (490). EXERCISE LII (491).

XXXI.yPERMUTATIONS, COMBINATIONS AND DISTRIBUTIONS. Combinations with Repetitions, Things not all Different (493). Arrangement in Groups, Distribution in Parcels (494-497). Derangements, General Theorem (498, 499). Partition of Numbers, Table of p Partitions of n, Euler's Use of Series (500Distributions,

504).

XXXII.

EXERCISE LI1I

(504).

EXERCISE TJV

(506).

^PROBABILITY. First Principles, Exclusive Events, Independent and Interdependent Events (508-513). Probability estimated by Frequency, Expectation, Successive Events (513-518). Probability of Causes, Value of Testimony, Appli cation of Geometry (518-522).

EXERCISE

LV

(523).

XXXIII. CONTINUED FRACTIONS

(2).

Expression of a Quadratic Surd as a Simple Continued Fraction, the Form The form V~AjB, Cycle of l )lr l (527-530). 2 2 Quotients, the 6 and r Cycles, Solution in Integers of Bx -Ay 2= z x The Form of Solutions 9 ^/N ~Ny (530-534). Integral (534-538). The Cycle belonging to (^/N + 6)/r found by the G.C.M. Process (538).

(VNb

=M

M

EXERCISE LVI

(540).

MISCELLANEOUS EXERCISES

A

(543),

B

(555).

HIGHER ALGEBRA CHAPTER

I

THEORY OF NUMBERS In *

'

this section

number

we

discuss properties peculiar to whole numbers.

'

fs

(1)

taken to

mean

l

The word

whole number,' and, unless otherwise stated^

positive whole number*

Division.

1.

(1)

Let b be any positive whole number, and consider

the sequence ...-36,

-26, -6,

0, 6, 26, 36,...

continued indefinitely both ways. Any whole number a (positive, negative or zero) is either a term of this sequence, or it lies between two consecutive

Thus two numbers q and

terms.

r

can be determined uniquely so that

and

a-=bq + r

To

divide a

ditions

is

6

is

to find the

called the quotient

numbers

and

r

q

and

r

which satisfy these con-

the remainder.

= 0, we

say that a is divisible by 6 or is a multiple of 6, and that a divisor or a factor of a. Among the divisors of a number we count

If r

6 is

q

;

by

0^r<6 ........................... (A)

number itself and 1. Whatever 6 may be,

the

= 6.0 4-0; hence

(2) If r

= 6~/, we have a = 6-fl-r'

if

and

R

then /
r

>|6

Hence

and

it is

and Ex.

0,

1.

For

if

1,

2,

must be regarded as

by every whole number.

divisible

also

zero

Every number

is

any number 5-2, 5-1.

is

always possible to find numbers

Q

JR<6.

of one of the forms 5n, 5n

1,

5n

divided by 5, the remainder

2. is

one of the numbers

DIVISION

2 Ex.

Every square number

2.

The square

these are divided by

and 5n -

l,

Ex.

For (2k +

by

2

l)

is of

one of the forms 5n,

odd number

form Sn + 1. and either k or k + l must be even, so that k(k +

-4k(k + l) + l,

(1

// both a and b are divisible by

(2)If

of the

l) is

r is the

Division.

remainder when a

so also is

c,

divided by

is

ma 6,

nb.

then cr is the remainder

ca is divided by cb.

For

a = 6
if

then

ca

(3)

(cb)q

and

+ cr

and

// both a and b are divisible by

divided by b, then is divided by b/c.

For also

is

2.

Theorems on

when

.

,

2. )

5wL

is

1.

Tlte square of every

3.

divisible

number

of one of the forms (5w) 2 (Sw^l) 2 , (5m 2) 2 If 5, the remainders are 0, 1, 4 ; and, since 4 = 5-1, the forms are

of every

if

is r,

(i)

r is divisible

c, and r is the remainder when a is and (ii) r/c is the remainder when a/c

a~bq + r and 0^r<&, then which

is

equal to a a/c

3.

by

c

-

bq.

since a and Thus we have

= (b/c)q + r/c where

6 are divisible

by

c,

so

O^r/c <6/c.

Theorems on the Greatest Common

Divisor.

is divided by b, the common divisors of (1) If a and b are the same as those of b and r. For, since a = bq + r, every common divisor of b and r is a divisor of a

r is the

remainder when a

;

and, since r

= a-bq,

common

divisor of a

every This proves the statement in question.

and

b

is

a divisor of

r.

(2) If a and b are any two numbers, there exists a number g, and one only, such that the common divisors of a and b are the same as the divisors of g.

For numbers (q v rx ), and uniquely, so that

(q 2 , r 2 ), etc.,

a = 6y 1 -f-r1

,

can always be found in succession,

6=r1 ? 2 + r2

,

6>r1 >r2 >r3

where

rx ...

= r2

3

+ r3

,

etc., ............... (A)

^0.

number of positive integers less than b remainder must occur and, supposing that rn ^ =0, the Since the

;

r -2

= r n-l?n + rn>

rn-l

=

is

limited, a zero

process terminates

GREATEST COMMON DIVISOR divisors

numbers have the same common

(A), the following pairs of

Hence, by

3

:

(a, 6)

(6,

rj,

(r

v

r2 ) y

...

(rn ^ v r n )

,

;

^

e common divisors of r n _ 1 and r n are the divisors and, since fn-i VZn+i* Hence the number g exists, and its value is r n of r n .

.

Since g

common

is

divisible

numbers and

divisor of these

as the greatest (3)

by any common

common measure

divisor of a

known

is

(G.C.M.) of a

and

the greatest in elementary arithmetic

and

it is

ft,

ft.*

If a and b are each multiplied by any number m, or are divided by a divisor m, then g, the greatest common divisor of a and ft, is multiplied

common

or divided by m.

For each (4) If a,

of the r's in the proof of (2) ...

c,

ft,

and

divisor of a

ft,

,

is

multiplied or divided by m.

and g l

are several numbers,

g 2 that of g 1 and

c,

ft,

d,

c,

. . .

),

(gv

ft,

c,

d,

. . .

common (# 2 ,

),

c, d,

Hence we arrive eventually at a number g, whose divisors of a,

the greatest 4.

ft,

and

d,

divisors . . .

),

common

so on, then :

etc.

divisors are the

common

uniquely determined. This number g is divisor , or the greatest common measure of a, 6, c, ...

c, ...

common

the greatest

g3 that of g 2 and

the following sets of numbers have the same (a,

is

;

also g

is

.

Numbers Prime

Two numbers,

to each other.

a and

6,

are

said to be prime to each other when their greatest common divisor is 1, so that they have no common divisor except 1. This is often expressed

by saying that a is prime to theorems are fundamental.

ft,

or that

ft

is

prime to

// the product ab is divisible by a number m, and factor a, then m is a divisor of the other factor ft. (1)

a.

m

The following

is

to

prime

one

For, if a is prime to m, the greatest common divisor of a and m is 1 hence the greatest common divisor of aft and mb is ft. But, by hypothesis, m is a divisor of aft, and is therefore a common divisor of aft and mb ;

;

hence

m is equal to,

// a is prime a divisor of N.

(2) is

or

to 6,

is

a divisor

of,

ft.

and each of these numbers

is

a divisor of

2V,

then ab

Suppose that N = aq, then ft is a divisor of aq, and since ft is prime to one factor a, it must be a divisor of the other factor q. Let q = wft, then

N = waft, and *

The

aft is

a divisor of N.

process, here stated algebraically,

is

that used in Elementary Arithmetic.

PRIME NUMBERS

4 If a

(3)

is

prime

to b, positive integers x,

ax-by For

it

y can be found such

that

1.

follows from equations (A) of Art. 3, (2), that TI

= a - bq v

r2 -

-

aq z + b (1

+ qfa),

Continuing thus, every remainder can be expressed in the form where x, y are positive integers. If

a

prime to

is

6,

the last remainder

and the theorem

is 1,

Theorems on Prime Numbers.

5.

A number

(ax-~by),

follows.

which has no

divisors except itself and 1 is called a prime number, or simply a prime. Numbers which are not prime are said to be composite.

A

(1)

prime number, p,

is

prime

to

every

number which

not a multiple

is

For, if a is any such number, q and r can be found such that a~pq + r where 0
therefore to a. (2)

least

If a prime p is a divisor of a product abed one of the factors a, 6, ... k.

...

hk,

it is

a divisor of at

the prime p is not a divisor of a, by Theorem (1) it is prime to a, hence it is a divisor of bed ... k. If, in addition, p is not a divisor of 6, it must be a divisor of cd ... k. Continuing thus, it can be shown that if p is

For

if

not a divisor of any of the numbers

a, b, c,

...

h, it

must be a

6.

Theorems on Numbers, Prime or Composite.

(1)

Every composite number

N

divisor of k.

N has at least one prime divisor. N

not a prime, it has a divisor, n, different from and from 1, which is not greater than any other divisor. Further, this divisor must be a prime for, otherwise, it would have a divisor less than itself and greater than 1 and this latter would be a divisor of N. This contraFor, since

is

;

,

dicts the hypothesis that It follows that every

n

not greater than any ot,her divisor. composite number, N, can be expressed as the is

product of prime factors. has at least one prime factor, p, we have N~pa, For, since where l<.p
N

factor, q

Thus,

;

and a = qb, where l
N =*papqb

;

COMPOSITE NUMBERS But the numbers

N

the set

pressed in the form,

N are limited

than

must

...

a, b,

y

less

finally

N=pqr

...

5

and N>a>b>...

;

end in a prime. Hence, where p, q, r, ... u are

u,

,

,

therefore

N

can be ex-

all

primes, not

all different.

necessarily That is to say,

any composite number, N, can be expressed

N=p where p,

q, r, ...

u are

,

a

b

.q .r

all different

c

u

as

s ,

primes.

(2) A composite number can be expressed as the product of prime factors in one way only. a b c = P A Q B R C where p, g, r, For suppose that P, Q, R, ^=p q r a b c are primes then, since the prime P is a divisor of the product p q r ...

N

. .

. . .

. . .

.

. . .

,

,

,

;

a divisor of one of the factors p, q,r, ... and is therefore equal to one of them. In the same way each of the set P, Q, R, ... is equal to one of it is

,

and no prime factor can occur in one of the expressions q, r, ... which does not occur in the other. Suppose then that N^p a qr c ...^p A q Br c ...

the set p, for

N

,

.

a --A one of them must be the greater. Let A>a> and suppose that + e, then q b r c .... t m ^p c qB r .... t M but this is impossible,

If

9

Aa

.

.

.

;

as the left-hand side of the equality right-hand side is divisible by p.

Hence, a must be equal to thus the two expressions for

A

;

a number prime to p, and the

is

and

similarly, 6

= B,

. . .

m~M

,

and

;

N are identical.

The' above theorem

is

one of the most important in the Theory of

Numbers, and the following propositions are immediately deducible. (3)

//

m

product ab (4)

// a

is ...

is

A

each of the numbers, a,

i,

...

,

k, it is

prime

to the

then a n is prime to b n , where

n

is

any

integer

;

and

*-*""*'

number

greater than 1

For, if 2V

to

to b,

prime

conversely. (5)

prime k.

N

and

is

a prime, if

less than, or

it

is

equal

to,

= a&, where b^a, then

not divisible by

any prime number

v^V.

N^a

2 ;

that

The sequence of primes is endless. For, if p is any prime, the number p 4-

is,

a^.*jN.

(6)

1 is

j

divisible

or by any smaller prime.

by p must have a prime

than

p

exists.

divisor greater than p,

If

greater than

then

and in

p+1

is

p and

is

not

not a prime,

either case a

it

prime greater

NUMBER

DIVISORS OF A

6

which are

...

N-

,

than

less

problem of discovering whether a large number one of great difficulty. Ex.

If n

1.

is

any number, prove

that

n(n + I)(n + 2)

Of the two consecutive numbers, n and n +

be obtained in order by

The process consists in writing 1 and erasing all multiples

a process called tlie(Sieve of Eratosthenes). down in order all the numbers from 1 to of the primes 2, 3, 5, 7,

N can

number

All the primes less than a given

1,

one

%/JV.

Nevertheless, the

prime or composite

is

is divisible

by

6.

by 2

divisible

is

is

and one of

;

the three consecutive numbers, n, n + 1, n + 2, is divisible by 3. Hence the product and, since 2 is prime to 3, n(/i-f-l)(n + 2) is n(tt + l)(tt-f-2) is divisible by 2 and by 3 ;

divisible

Ex.

We

G.

by

Prove that 3 2n+1

2.

have

3

+1

2r2

=3

.

+ 2"+ 2

is

divisible

7.

by

= 3 (7 + 2) n = Ik + 3

9W

.

2 n,

by the Binomial theorem,

and

7.

The

where p,

q,

the divisors

Then,

Divisors of a Given Number N. Let N=p a are primes ; and let n be the number, and s r, ... N, including 1 and N. of

.

,

n = (a + _1

(1)

pa+l

l)(6 yft+l

p-l For the divisors of

+ l)(c + l) _ l r c+l _

...

For,

if

N

the

rc

.

...

,

sum, of

;

r-l

q-l

N are the terms in the expansion of

N

and

1, is

%(n +

the product of

l) or %n, according as

a perfect square, each of the numbers, a,

is

.

I

and the expressions for n and s follow immediately. (2) The number of ways in which N can be expressed as two factors, including not, a perfect square.

b

q

N

6, c, ...

,

is,

is

or is

even,

and therefore n is odd but if N is not a perfect square, at least one of these numbers is odd, and therefore n is even. Further, if ^( = 1), d2 rf 3 ... d n _ 2 d n _ v d n (=N), are the divisors of A' '

;

,

,

,

in ascending order, the different

two

and

factors are

did n

,

d^dn

,

N

ways

of expressing

N

as the product of

:

d 2d n _ v

c^ n _

d 2dn , l9

dsd n _ 2

is,

or

is

...

2>

Hence, the number of ways according as

,

,

d xd x

d yd y+l

,

is

either

when

,

x or

not, a perfect square.

,

when

y,-i.e.

n=*2a?-l,

= 2y.

either |(w

+ l)

or \n,

PRODUCT OF CONSECUTIVE INTEGERS

7

The number of ways in which N can be expressed as the product of two m ~ 1 where m is the number mctors, which are prime to one another, is 2 of N. Different prime factors of (3)

,

For, such factors are the terms in the expansion of

~~ and their number is 2 W Hence the number of pairs is 2 m 1 For example, if # = 2 2 3 3 5 = 540, n = (2 + l)(3 + l)(l + 1) = 24, .

.

.

.

8.

The Symbol

l[x/y].

If

a

is

m==3

and

'

a fraction or an irrational number, the

symbol / (a) will be used to denote the integral part of where 0
// n v n 2 n 3 ,

,

...

are

,

any

>

integers,

and

Thus

a.

s is their

if

x = qy + r

sum and a

is

any

number, then

= aq^ + r l9 n 2 = aq% + r2 w 3 = ay3 -f r3

Let n

,

Hence,

etc.

;

then

/[/a]

and, since Ji^/f^i/a],

The

(2)

,

highest

r

52

==

^[ n 2/a ]>

>

^e resu

power of a prime p which

is

lt

follows.

n

contained in

is

\

For, of the numbers from 1 to n inclusive, there are I[n/p] which are 2 2 and so on hence the by p of these I[n/p ] are divisible by p

divisible

;

;

;

result follows.

9.

by

Theorems.

(1)

The product of any n consecutive

integers is divisible

\-n.

For (m + l)(w + 2) last expression is

occurs in

[m

n^

...

(m-f n)/\ n =

\

m + nj\ m

w,

and to show that the

it is sufficient to show that any prime p which Thus we occurs to at least as high a power in -f n.

an integer

m

j

have to show that /[ (m

+ n)/p] -f /[ (m + n)/p*] + 1[ (m -f n)/p 3] +

Now

/[(m + n)/p]>/[w/p]-f /[n/p], and the same is true by p p*, ... in succession hence the result in question. 2

ft

. . .

,

,

:

if

we

replace

NUMBERS IN ARITHMETICAL PROGRESSION

8 (2)

is

Ifn

a prime,

is divisible

C?

by n.

n(n - l)(n 2)

For by the preceding since n is a prime and r r is

Hence,

...

-r+

(n

1) is divisible

supposed to be less than n, r a divisor of (rc-l)(n-2) ... (n-r + 1) is

is

by

[/% aijid

prime to

;n.

|

|

and Thus

a prime,

t*

ft

t/

except the first

and

last,

text-books about

'

divisible

is

l)/[r

by

n.

in the expansion of (l+x)

all the coefficients

n ,

are divisible by n.

supposed to be acquainted with what is said in elementary for permutations and combinations and the binomial theorem

The reader

NOTE.

(n-r +

...

^(ft-l)

is

*

'

a positive integral index.' In what follows, F$ denotes the number of permutations, and C the number of combinations, of n things taken r at a time. Ex.

Find

1.

We have

the highest

power of 5 contained in

7[158/5]-31,

158. |

2

/[158/5 ]=/[31/5J-:6,

therefore the required power has an index

1

/|158/f>J=y|<$/5J

:

= 31 +6 + 1 =38.

an odd prime, the integral part of (x/5 +2) n -2 W+1 is divisible by 20n. Let (V5 + 2) #+/ where 0/5 - 2)n =/'. Then, since 0< V5 -2< 1, we have also 0
If n

2.

is

n=

;

N =2(Cf Moreover, since n

by 20w

;

which

is

.

is

2

Sty- 1 ) + C%

a prime,

C^

is

.

23

.

5*( n

divisible

~3

)

by

+ ... + C n,

.

2n

~z .

and therefore

5

+ 2n ).

N -2 n +

1

is

divisible

the required result.

10.

Numbers

(1)

Let a be prime

in to

Arithmetical Progression. n, then if the first n terms of the arithmetical

+ a, x + 2a, ... are divided by n, ... n - 1, taken in a certain order.

sion x, x 0, 1, 2,

.

,

the

progres-

remainders are the numbers

,

we suppose

that two of the terms as x + ma, x

+ m'a leave equal remainders, then their difference (m~m')a would be divisible by n. This is impossible, for a is prime to n and m-m' \
|

,

//

the progression is continued

beyond the n-th term the remainders recur in >

same order. For the terms x + ma, x + m'a leave the same remainder

the

If a and n are not prime to one another and g is divisor, the remainders recur in a cycle of n/g numbers. (2)

For

let

a~ga', n=gri, so that

a' is

prime to

n'.

if

m = m' + qn\.

their greatest

The terms

common

METHOD OF INDUCTION

9

if, and only if, a(w-w') is divisible by a! (m m') is divisible by ri. Since o! is prime to w', this can when m - m' is divisible by n'. Thus the first ri terms leave

x f m'a leave the same remainder that

is, if

o ly happen different remainders and, after that, they recur in order.

Method of Induction.

11.

Many theorems

to whole

relating

numbers can be proved by a process known as mathematical induction. In some cases this is the only method available. The method may be described as follows.

-

x

'

Let/(w) be a function of an integral variable n. Suppose that a certain statement S, relating to/(w), is true when na. Further, suppose we can prove that, true when n = + 1.

if

8

is

true

when n = m,

it is

also

m

Then

S

since

in succession, that

Ex.

is,

- 3w -

that 2 2n

Show

1.

2n -3n-l, Let/(tt) = 2 for n = l.

Also 2 2n

-

Hence

/(n)

if

=(3 + l)

divisible

is

-f(n)

1)

=2 2 w +1 (

= 3fc, where 9, so

by

fc

by

1)

;

so the theorem

9,

- 2 2W - 3 = 3(2^ -

an

is

is/(n +

divisible

is

)

a-f 3,

...

9.

by

=0 and

when n = a + l, a + 2,

true

it is

1 is divisible

then /(I)

f(n + n-1

Again, 1

when n = a, when

true

is

is

true

1).

integer,

and, since /(I) is so divisible, it follows that is, the theorem holds for ;

in succession that /(2), /(3), /(4), etc., are so divisible all values of n.

Or more Ex.

2.

using the method of (9) Ex. 2,

easily,

If

n

is

a

1

i

i

~Zi

n+2

n+1

1,111 __ 2n-l 23

111

i

i

j

i

T

i

7^

i*

^|*

1

1

.

1_

_

n Again,

= (3 + l) n - 3n - 1,

etc.

positive integer, prove that

11 ___T__

n

f(ri)

vn~^

if

t^en

Vti ^ f

1 -_

2n

= i>l + l

" ...+

23

1

1 I

i

~T~

i

4

2n -

,

1

1 -J_

2n + 1

*

"*

'

~~

I

"""rki"**

2n-l

2n

2n + l

1

^n+i-^wHence

if

succession,

un -vn then ,

it is

true

t*

n+ i=t;n+l

forn=2,

3, 4,

Now

.

...

,

that

the statement holds for n = l is,

for

any value of

n.

;

hence, in

ELEMENTARY THEORY OF NUMBERS

10

EXERCISE q

is

I

the quotient and r the remainder when a is divided by the quotient when a is divided by b + 1> provided that r^q.

1.

If q

2. If

is

a and b are prime to each other, show that - 6 have no common factor other than 2 (i) a + b and a 2 a* ab a +6 have no common factor other than b and + (ii)

6,

show

th*|

id

;

3.

and b are prime to each

If a

(a

have no 4. If 5.

If

common a

is

a+b

factor, unless

prime to b and

y,

and n is a prime, prove + )l(a b) and a + 6

that

other,

n + bn

and

is

b

is

3.

a multiple of n.

prime to

x,

then ax + by

ab' -a'b = l,

X=ax + by and 7 a'z-H&'t/, where X and y is the same as that of x and y.

prime to

is

ab.

common

the greatest

divisor of

If 2>

is

7.

a prime, and j9=a 2 -6 2 then a=%(p + 2 b 2 in two ways. Express 55 in the form a

8.

If

takes the values

6.

a;

,

(i)

1, 2, 3, ...

z 2 + x+

17,

(ii)

,

l) 9 b

\(p-

1).

in the expressions

2z 2 + 29,

(iii)

x*

+ a; + 41,

the resulting values of the expressions are primes, provided that in

(i)

#<16,

in

x<29, and in (iii) #<40. Verify for x = 15, 28, 39 respectively.

(ii)

9. Iff(x) is a polynomial, it cannot represent primes only. [Let uf(x) and v~f(x + ku), where k is any integer. Prove that u factor of v.]

is

a

10. For the values 2, 3, 4, ... 10 of x, the number 2 3 5 7 + x is composite. Hence write down nine consecutive numbers none of which is a prime. 11. If n is any odd number, then n(n 2 - 1) is divisible by 24 and if n is an odd prime greater than 3, then n 2 - 1 is divisible by 24. .

.

.

'

;

12.

that 2 n

Show

13. If

n

is

of the form

+1

prime to

or 2 n -

5,

1 is divisible

then n

2

+1

or

n z

by

3,

according as

1 is divisible

5m -f 1.

by

5,

n

is

and

odd or even.

therefore

n4

is

14. If n is prime to 5, then n 6 - n is divisible by 30 hence the fifth power of any number has the same right-hand digit as the number itself. 15. Show that 2 -f 1 or 2^ - 1 is divisible by 5, according as n is odd or even. 2n 2n 16. Show that 5 1 is divisible by 13, according as n is odd or -f 1 or 5 ;

even.

- 1 is divided by 13, show that the remainder is either 0, 2, or 8, as n is of the form 3m, 3m -f 1, or 3m - 1 ; hence prove that 3 W - 1 or according 3 2n -f 3 n + 1 is divisible by 13, according as n is, or is not, a multiple of 3. 18. If 2n -f 1 is a prime, then n must be a power of 2. 17v If 3 n

19.

Show

that 7 2n - 48n -

1 is divisible by 2304. 20. Show that 7*" + 16n 1 is divisible by 64. 21. Prove that 2 2W + 1 - 9?i 2 -f 3n - 2 is divisible by 54. 22. Find the number of divisors of 2000, and their sum.

1*

DIVISORS Let

23. is

s be the

called

a

perfect Show that, if 2 n *nd the three least '

,

24. If

<

1

;

8 the products, of the divisors of L and M are prime to one another,

numbers, and F, R,

a are the

r,

ft,

M

N

sum of the divisors of N, excluding itself; if s=N, then number. n~1 n - 1 is a and 1) is a perfect number (2 prime, then 2 numbers given by this formula.

NLM, and

L, respectively, where rove that (i)n=r.0, (ii)P=R s

''J7,

n

$ 25. If

the number, and

is

11

P

.Sr

.

the product, of the divisors of

N

prove that

9

=lfn.

N

26. If the product of the divisors of is

N

N

9 excluding the product of two primes or the cube of a prime.

itself, is

equal to

N

9

then

27. If N has 16 divisors, it cannot have more than 4 prime factors, a, 6, c, d, and it must be of one of the forms abed, a 36 3 a 36c, 7 6, a 15 Hence find the smallest number having 16 divisors. .

,

28.

Find the smallest number with 24

29.

Prove by induction that w(n + l)(w + 2)

30.

Find the highest power of the prime p (i)p

31. If less

N

is

r

divisors.

-l
...

in

(u)p

+r -

(n JV,

r

1) is divisible

by

\r.

when

-p
.

expressed as a polynomial in a prime p, with each of the coefficients s is the sum of these coefficients, prove that the power of p con-

than p, and

tained in |

32.

when 33.

N

is

(N - s)i(p -

1)

Show

N

is

that the index of the highest power of 2 contained in f - r when a power of 2, and is equal to 2 - 1.

N

Show

that

N

|2tt-l/(|n|w-l)is odd or even according

as

n

N

is

N-1

or

is

not,

|

is,

a

power of 2. 34.

Prove that

2n

is

divisible

[

35.

Prove that,

is divisible

by

m

by n

if g is

the greatest

1.

|

common divisor of m and n -f 1, then g

m 4- n

. \

n -f 1.

. \

|

Prove that the power of 2 in 2 in n + 2. n+1 [n 36.

.

n -f

.

|

3n

I

is

greater than or equal to the power Jf

.

|

37.

Prove that, ifn

38. If a, 6,

is

an integer 39.

c, ...

,

are

divisible

41 4f 2.

1

greater than 2, then

numbers whose sum

by

,

The

-f

1

is

by|n.|n + l.|n-t-2.

a prime, p then 9

that

wn

is

integer next greater than (3

integer next greater than (s/7

If 2"

3nis divisible |

p.

If tt n = (3 + v^5) n + (3 - V5) n show

Hence prove that the 40.

is

=&2/ prove that x

1

+ \^3) an

and y

-*

1

an

integer,

+ */5) n

is

is

divisible

and that

divisible 2n by 2

are divisible

n by 2

.

.

by the same power B.C.A.

CHAPTER

II

RATIONALS AND IRRATIONALS 1. Rational Numbers. number system is enlarged by

the introduction of fractions,

(i)

possible

in.

order that division

may be

always

;

the introduction of negative numbers and zero, so that subtraction

(ii)

may

Starting with the natural numbers, the

be always possible.

We

thus obtain the system of rational numbers or rationale, consisting of positive and negative integers and fractions, with the number zero. This system can be arranged in a definite order so as to form the rational scale.

With c

reference to

equal to

'

and

any two

'

less

than

'

rationals x

and

y,

the terms

*

greater than/

are defined as referring to their relative

positions on the rational scale. '

To say that x>y or y
To say

The Fundamental Laws of Order are as follows (l)Ifa=6 = 6 and 6 = c, then a = c. (3) If a^b and 6>c, or if (2) If a a>6 and 6^c, then a>c, (4) If a>b then -a< - 6. 2.

:

then 6 = a.

Hence we deduce the following rules for where it is assumed that zero is not used as a (5) If

b

+ x,

a-x = b-x

(6) If

a = b and x = y, then

(7) If

a>b

inequalities,

:

)

ax

bx

9

a/x

b/x.

then

and

ax^bx and a/x>b/x

(8) If

divisor

and

a= a+x

Also

equalities

a>b and x>y,

according as x

then a + x>b

both positive, then ax>by.

a-x>b-x. is

positive or negative.

+ y, and

if

a and y or b and x

ai!

REPRESENTATION OF NUMBERS BY POINTS Fundamental Laws of Arithmetic.

3.

Any two

13

rationals can

of addition, subtraction, multiplication and the result in each case /jvision, being a definite rational number, excepting lat zero cannot be used as a divisor. This is what is meant when it is

combined by the operations

j8

iid that the

system of rationals

The fundamental laws

;

f

(1)

(3)

4.

an

and multiplication are

+ 6 = 6 + a, ab = ba,

+ 6) + c = a + (6 + c), - ac + be, (a + b)c (a

(2)

(4)

(5)

The

closed for these operations.

a

a nd

fifth

is

of addition

= a(bc). (ab)c

and third constitute the Commutative Law, the second and

first

the Associative Law, the fourth the Distributive Law.

Theorem

integer

n

of Eudoxus.*

If a and b are any two positive rationals, nb>a. This simply amounts to saying that an greater than a/6.

exists such that

integer exists which

is

Representation of Numbers by Points on a Line. Take a

5.

OX

take a point 1 so that the segment 01 X'OX as axis in contains the unit of length. To find the point a which is to represent any Divide the segment 01 into n equal parts positive rational a, let a = m/n.

straight line

and

;

set off a length Oa, along

which represents - a

is

OX, equal

OX', at the

in

m

i

FIG.

The point

of these parts.

Ola

i

-a

X'

to

same distance from

as the point a.

I

i

X

1.

Points constructed thus represent the rational numbers in the following and one only. respects (i) For every number there is one point :

(ii)

The point point a

X

is generally taken to the right of to the right of the point 6.

Absolute Values.

6.

- a,

is

The points occur in the order in which the corresponding numbers stand on the rational scale.

according as a

|a-6|=|6-a|.

is

It

The

positive is

f

a

i is

or negative, and

x>a

is

denoted by

a>6, the

a

is

-fa or

.

Thus

a |

\

obvious that

positive, to say that

x |>a, then

if

absolute or numerical value of

+ 6, j

so that,

or

|

x<

x |
is

a+6

>

a - 6

the same as saying that

Sometimes ascribed to Archimedes,

.

-a<#
NOMENCLATURE

14

Large and Small Numbers.

Whether a number is regarded small depends on the purpose to which numbers are applied. A

7.

large or error of 6 inches

i

would be large in measuring a

but small in

table,

settir

out a mile course. If for purposes of comparison we choose some positive number c whuih we regard as small, then any number x is said to be small if x <. Any number x is said to be large if x >N where N is some previously \

|

\

|

chosen positive number, which is regarded as large. We say that x is large or small compared with y when x/y is large or small. If x/y is neither large nor small, x and y are said to be large or small

numbers of

We

the

same

order. '

use the abbreviation

order as If

'

y

is

0(x)

to indicate that y

is

same

of the

x.

is

small,

numbers which are

same order

of the

2

as

e,

,

e

3

...

,

are

orders respectively. called small numbers of the first, second, third, 2 2V 3 , ... are called If JV is large, numbers of the same order as N, , . . .

N

large If

orders respectively. numbers of the first, second, third, x - y is small, we say that x and y are nearly equal. . . .

Meaning of

8.

'

that x varies in such a less

To say that x

Tends.'

way

that

its

small.

is

If

from If

expressed by writing x-> 0. zero, where a is constant,

we say

or/rom the x tends to a and

above,

is

left,

always

less

than

a,

we say that x tends

)

is

great that number may be. In this case we also say that

from above, l/x tends to

;

oo

Any

,

we may

- x tends to - 00 Thus, and - l/x tends to - oo

collection of

.

number

as x tends to zero

.

numbers

is

the numbers themselves are the elements of the the

have what

is

called

an aggregate or

set.

^

of

called

to a

x becomes and choose, however

elements exceeds any positive integer however choose, great, we say that their number is infinite^ & If

and

to say that

remains greater than any positive number that

set

to a,

and we write x-> a -0.

tends to infinity (x -> oo

Aggregate.

that x tends

is

below, or from the

9.

how

always greater than a, we say that x tends to a This is expressed by writing x -> a -f 0. right.

x tends to a and

from To say that x

to say

choose, no matter

may

x - a tends to we write x -> a. If

is

numerical value becomes and remains

than any positive number that we This

tends to zero

an

infinite set.

^^

APPROXIMATE VALUES 10.

System Everywhere Dense.

system of rationals

many

infinitely

For

if

is

An

important property of the that between any two rationals a and b there are

is

rationals.

a
This fact

15

k

is

any

positive rational,

it is

sometimes expressed by saying that

easily seen that

the system of rationals is

everywhere dense. 1 1

.

A

Sequence.

according. to

some

succession of

definite rule,

Such a

denoted by (u n ).

numbers u ly u 2 u 3 ,

called a sequence,

is

rule defines

,

...

un

...

,

which

is

u n as a function of

,

formed

generally

the positive

integral variable n.

The rule may be quite arbitrary, and it is unnecessary that we should be able to express u n in terms of n by an algebraical formula. For instance, u n may denote the nth prime number, or the integral part of Jn.

A

sequence in which each term

is

followed by another term

is

called

an

infinite sequence.

12. Approximate Values. Suppose that the object of an experiment is to determine a certain number represented by A. No matter what care may be taken to ensure accuracy, it is unlikely that the exact value of A can be found. All that can be done in most cases is to find two numbers a and a' between which A must lie. If we find that a
is an approximate value of A, A -a is called the and Error, (A a)/A the Relative Error, in taking a to

Definition. (Absolute)

If

a

represent A. If

a
in defect

and

say that a and

a!

are approximate values of A, the first

the second in excess with errors numerically less than a'

Observe that the error

A - a cannot exceed a' - a

;

therefore a

upper limit to the error in taking a to represent A. Again, if a is positive, we have l/A
-a

;

1

- a.

-a

is

an

hence

(A-a)/A<( a'- a )/a; and

(a'

-

a)/a

is

an upper limit to the

We

relative error.

estimate the comparative accuracy of different experiments by comthe smaller the relative error, the greater is the paring the relative errors :

(degree

of accuracy.

FUNDAMENTAL THEOREMS

16

In writing down decimal approximations of a number, we adopt the followinjg

scheme (i)

If

:

we

write

7r

= 3-1416

(approx.),

we

mean

shall

that

between 3-14155 and 3-14165, and we say that 3-1416 significant figures, or to

than 0-00005, or 2i

(ii)

If

we

write

10

TT

by 3-1416

is

numerically

.

3-14159

-n

lyinig

correct to 15

TT

4 places of decimals.

Since 3-14155<7r< 3-14165, the error in representing less

some number

is

TT

the value of

is.

...

,

we

shall

mean

that the figures

3, 1, 4,

...

9 are those

which actually occur in the decimal representation of TT, and that TT is some number lying between 3-14159 and 3-14160. The error in representing TT by 3*14159 is positive and is less than 1/105 .

Fundamental Theorems.

In theoretical and practical work, the following theorems are of fundamental importance. If a, a', b b' are given numbers and A, B are any numbers such that 13.

9

a
and

(i)

b
and

a + b
;

if all the letters denote positive numbers,

ab
(iii)

The truth Proof of

of

(a

Similarly

Proof of

(i)

and

Since

(ii).

Subtracting

B+V

(iii)

(iv) a/6'

;

follows from Art, 2,

a
and 5<6',

a+

.*.

from each member

it

can be shown that Since

(iv).

a
f

b'.

a~b'
/.

);

and B
Ab'

oJB

Similarly

,

B
a,

A, B,

V

are positive,

aB
Bo

1.

(8).

A - B
Dividing each side of (A) by Bb', which

Ex.

< A/B < a'/b.

of the last inequality,

+ B)-(B + b')<(A+b')-(B + b

therefore

lir

,

a - V
(ii)

;

f

it

Show

Wehave

12

35

f<

. '

Bb''

is

positive,

A

a b

f< B

m

can be shown that A/B
1234567

13

< 3456789 < 34'

., Also

12

is less

than 0-05.

12

by oO

^

-.

^=0-34...

^|-^< 0-39 therefore the error

1234567

,0400709

- is less

,

and

13

00

^=0-38

-0-34 =0-05;

than 0-05.

....

Hence,

IRRATIONAL NUMBERS Ex.

7/0< x<

2.

division

By

we

find that

an approximate value of

is

-

-

-

=

1

- 2x + -

//" .4

-

---

an

is

by hypothesis, x cannot be an integer. Then (p/q) n

For,

so

Need for New Numbers.

15.

with an error in

<

1

- 2x + 5z 2

;

integer which is not a perfect

= A.

is

x

p/q,

whore

a fraction in lowest terms,

cannot be equal to the integer A.

it

rational

I/ (I -fa:)

If possible, let

a fraction in lowest terms.

is

~-

exists such that x n

no rational x

n-th power, then

and

-2x

Non-perfect 'Powers.

14.

p/q

then I

1,

5a? 2 .

than

defect less

17 2

numbers are inadequate

There are

for

purposes for which there is no rational

many

(i) example whose square equals 7 (ii) if the diameter of a circle is 1 inch and we denote the length of the circumference by x inches, it can be shown that x is not a rational number. ;

:

.

We

extend our number-system by inventing a new

class of

numbers

called Irrationals.

An

defined by some rule which gives the rational numbers. Such a rule is the following. irrational

Meaning of

is

We

"/A.

number

consider a positive

ordinal rank with

it

A

which

is

not a

perfect nth power, so that no rational exists whose nth power is A. The symbol *IA (the nth root of A) is used to denote the irrational

whose place among the numbers on the rational scale is determined by the following rule \IA is to follow every positive rational whose n-th power is less than A, and to precede every rational whose n-th power is greater than :

Thus

A.

if a, a'

are positive ratio rials such that

a n
then

and

a, a'

,

are approximate values of ^JA with errors, respectively in defect

excess, less

than

a'

- a.

16. Representation of a Number by an Endless Decimal. Take the following instance. No rational exists whose square is 7. Hence 7 lies between two consecutive terms of the sequence I 2 2 2 3 2 4 2 .... ,

Now

22

<7<32

2

(2-1)

2 ,

(2-2)

,

,

...

therefore 7 2

(3-0)

.

We

between consecutive terms 2

,

,

between two consecutive terms of

,

2

(2-0)

,

find that (2-6) 2 <7<(2-7) 2 , therefore 7 lies

of (2-60) 2

2 ,

(2-61)

,

...

2

(2-70)

,

and we

find that

2

<7<(2-65) However far this

(2-64)

otherwise

lies

.

it

process

is

carried on,

would be possible to

it

can be continued further

find a rational

whose square

is 7.

;

for

IRRATIONALS IN PRACTICAL RECKONING

18

The process such that 7 22

and

an endless decimal 2-6457 between the two classes of numbers

therefore gives rise to

lies

2

2

(2-6)

,

,J1 lies

2

and

...

^.(2-646)2, (2 .65)

is

which

2 ,

(2-7)

,

32

is

,

numbers

classes of

and

...2-646, 2-65, 2-7, 3.

meant when we write /7 = 2-6457

is

Observe that ^/7 2/3

2

2-6, 2-64, 2-645...

what

is

(2-645)

,

between the two 2,

This

(2-64)

,

...

related to the decimal 2-6457

....

...

in the

same way that

related to 0-6666 ....

is

For if we say that 2/3 is equal to the endless decimal 0-6666 ... this is only an abbreviation for saying that 2/3 lies between the two classes of numbers ,

0-6, 0-66, 0-666,

and

...

...

0-667, 0-67, 0-7.

It should further be noticed that the decimal equivalent of recur, for

if it

did, ^/7

^7 cannot

would be a rational number.

a process similar to that just described we can find the decimal but, of course, in practice representation of any root of a given number

By

;

other methods would be used.

Real Numbers.

17.

The

form the system of Real Numbers. and inequalities and the definitions tion

to

and

extended so as to apply

numbers.

Theoretically

it is essential

to

make

for purposes of practical reckoning,

Ex.

of addition, subtraction, multiplica-

division, already given for rationals, are

all real

may

and irrational numbers together In Ch. XIII the rules for equalities

rational

this extension,

but

where we replace any

occur by a sufficiently close rational approximation. 1.

where dn

Given that

is the

$2 = 1-259921

decimal continued

Letd' n =dw + l/10 therefore

n ,

then 2 -rf

to

n

,

prove that

places.

d%<2
<

Now

w rf'

rf

dn
It follows that

...

*l +

Wn + *l< 2

Hence we are justified in writing

- d* --* (

%/2 )

8

as

=2

.

2

(1'3)

n->

oc

.3
.

it is

unnecessary

irrational

which

EXPONENTIAL FUNCTION Ex.

We

^5 =2-236 ... , ^8 = 1-817 an upper limit to the error.

Given that

2.

Find

s/5 / x/6.

also

have

2-236 / 1-818<

Now 2-236/1-818 = 1-229

...

and

...

^5 f /6<

,

find

19

a decimal approximation

to

2-237/ 1-817.

2-237/1-817 = 1-231

...

;

l-229< N/5/^/8< 1-232. an approximate value of
/.

1-229

is

3. If a is positive and less than 1, show that 1 -a/2 2 -a) with an error in excess less than a /2. Applying the square root process to 1 a, we have

Ex.

less

than 0-003.

an approximate value of

is

-a -a-f^a

fa

2

2

-

fa

3

-

\a* =

N-

(

-

1

\a

-

|a

Now |a -|a -ia =^a (3-2a-a )=4a (3+a)(l-a); and /. a*(3 + )(!- a) >0 3

2

4

2

2

2

2 2 )

.

since

0
and

The Function

18.

a*.

(1)

Ifa>l and n

is

a positive

integer, then

a n >l-f-n(a-l). For,

if

a=

n by the Binomial theorem, a = (1

I -h 6,

-f

n 6)

>l 4- nb.

n n 7/a>0, ^en a ^l and a ^l according as a^l. For if a>l, then a n >l. If a1

(2)

Again, Similarly, (3)

if if

a>l, then a n >l

x=p/q, (4) //a>0 and For a /a y = a ~ (5)

m

x

ax

if

a?

for otherwise

we should have a = (an ) w
a
//a>0 and

For

;

and

aj

a

Ifa>l and

>l


>a y according as agl.

according as

a>l.

N is any positive number,

can be found such that

For we can choose

according as

^

>y, I/

>l

x positive rational, then a p 1, according as a^.1. %Ja

ts

=

an

>N

for

however great, a positive integer

n^m.

m so that m(a - 1)>N - 1,

and then by

(1),

APPROXIMATE VALUES

20 // a

(6)

is

If

a
and not equal

positive

awarding as a>\. First let a>l.

Then

in (5),

if

an is

any

a>l,

If

positive

as in

(5),

m

ax >am >N, and so ax -+.

z>w,

1

|<

oo

6*

,

~>

oo

and a = 1/6* -> 0. 35

n^m,

for

number, however small. can be chosen so that

and

a
If

as

or

m can be found such that

|

where

a^-xx)

then

1,

a = 1/6. Then 6>1, and as x ->

let

//"a>0, a positive integer

(7)

to

m

(1 -f e)

m >a. Hence

a"-l<. can be chosen so that

f-

(

>

that

m (l-c)
is,

Hence,

if

n>m,

n

(1

-e) <(l -e)

if

1

JL

m <, and

\m 1

/

'1

>-, d

therefore

A

l-a n <. If a = 1,

then a7 -

If a

is positive

(8)

*

First, let #-->

part of If

(7)

;

1

=0

and x > 0, ZAen ax >

then

if

x
m as x 1 -a
values, let

1

.

If a>l, choose I
through positive values.

a
and Thus

for all values of n.

ax

-

in the second part of (7)

;

both cases, ax 1->0 and a*->l. If a = 1/6 and x = - y. Then ax = 1 /6 V ->1

then

if

m as in the first x nt #a

x->0 through

negative

.

EXERCISE

II

APPROXIMATE VALUES

a
x

a, a are known numbers, and any number c lying chosen as an approximate value of A, the numerical value of the error in representing A by c cannot exceed the greater of the numbers 1.

If

between a and

where

of is

(c-^a), (a'-c).

Hence show that, if J(a-f-a') is chosen as an approximate value of A, the greatest numerical value of the error is less than it would be for any other approximation. 2. If

f(x)

= 2 + 3x - 4x 2 - 5x 3 + 8x* and x

is

small, find

an upper

numerical value of the error in representing f(x) by 2 -f- 3x. -4x*-5x* + 8x* < 4* 2 -f 5x 3 [If e is the error, \e\ = \

when

|

x

\

<

\

\

1

|

41

1

limit to the

8* 4

|

<

17* 2 ,

1.]

The length of a man's step is approximately 30 in., this value being correct to the nearest inch. If the number of steps which he takes per mile is calculated on the assumption that he steps exactly 30 in., show that the error in the result 3.

may

be as

much

as 35, but cannot exceed 36.

APPROXIMATE VALUES 4. If

21

0-2
(i)0-3<2/-z<0-5;

x

2<

(ii)

are all positive

1

y-x

<4

;

(iii)

and x
5.

If

6.

cannot If the numerical value of the error in taking a to represent a, and the numerical value of the error in taking a' to represent a cannot a', then a -fa' is an upper limit of the error in taking a! for A. [i.e. given

9

y, z

A

exceed exceed that

a-a
(x.

and a'-a'
it is

required to show that

+ (a + a').]

:a'

A

and J5, subject to errors 7. If a, b are observed values of two numbers (positive or negative), whose numerical values cannot exceed a and j3 respectively, show that an upper limit to the numerical value of the error in taking (i) (ii)

a-f /?

;

a -f ft

;

(iii)

(iv)

ajb

to

represent

a

is

small,

less

0
If

is afc-f

i/(

a

;

!/(!

+ a) 3

is

0
less

than

error in taking ---h -

o

x

-

x> 100,

x

is

if

By

o

the error in taking - as the value of -f

3 in descending powers of

show that

-

-

(i) ~r~

is

OX

i

o j, -

3

-h

x.]

nearly equal to

1

+ (a - b)x - b (a - b) x 2 is less

;

than 0-01.

,

,

,

R

v

+

R = (b + c -

By means of Ex. 1+ (a~b-c)x.

13.

Show is less

than 0-0007.

is less

4#

the division transformation, show that

I+ax

to

is

o -f" Q.X

-

4

small,

(1

where

Also show

x as the value of -

y

a = 5, b = 2 and #<0*1, the error in this approximation

12.

positive

.

A (ii)

any

o

[Divide 3# + 2 by 4# 11. If

is

2

o 10. If

6'

nearly equal to l-3a. error < 0-0007.

i

numerically

where

~l~)"0K

ft.

show that the

9

(a-f a)

is

yl/7?

than b

show that if

that, 9.

is

A - B is

a& to represent ^4J5

number 8. If

A +B

a + b to represent a - b to represent

also that if

12,

show that

a = 2, 6^3,

if

x

is

small,

c

=4 and x < 0-01,

r~r/l + ox)(l+cx) "\

(l

^ nearly equal

the error in this approximation

than 0003.

The focal length (/) of a lens is given by the formula l//=l/t> - l/u. The values of u and v may be in error by as much as 2 per cent, of the corresponding true values. If the true values of u and v are 20 and 13, show that the value of/ '

14.

as calculated from the observed values of its true value.

may

be in error as

much

as 9-9 per cent,

APPROXIMATE VALUES

22

The weight (w grammes) of water displaced by a solid is given by the w=w l -w 2 where w l9 w 2 are the weights (in grammes) of the solid (i) in

15.

formula

,

In determining the values of w i9 w 2 the error in each case per cent, of the true value. If the true values of w lf w 2 are 13 and 10, show that the value of w calculated from the observed values of w ti> a may be in error by as much as 7 per cent, of its true value. [The greatest and least possible values of w 1 are 13-13 and 12-87, those of w z are 10-1 and 9-9, hence those of w l -to* are 3-23 and 2-77.]

vacuo,

may

in water.

(ii)

be as

much

as

,

1

l
16. Given that the endless decimal corresponding to /123 is 2-618060 6 prove that 123 -(2-618060) < 0-0003. [Proceed as in Ex. 1 of Art. 17, using the identity a' 6 - a 5 = (a' - a) (a' 4 + a' 3a -f a' 2 a 2 -f a'a 3 + a 4 ).] 17. If x and y are positive and x>y, then x-y*j(2x) of s/(# 2 - i/ 2 ) with an error in excess less than t/ 4 /(2a; 3 ). 2 [In Ex. 3 of Art. 17, put a = y /x*.]

a and b are nearly equal, show that - (a + 6)

18. If

_ value of Va6 with an error in excess

less

-

than

(

a

x

,

an approximate value

-

is -j~

_M4 r

is

...

4(d -}- uj

an approximate

[Use Ex.

3

17.]

19. Given that ^14 = 3-741..., 4/3-1-442..., for each of the following obtain a decimal approximation with an error in defect, finding also an upper limit to the error :

(i)

20. If

^14

0
x 4/3

s/14/4/3

(ii)

;

then a -

(iii)

;

V A <---.

W

14 4- 4/3.

.

o(^JA)

[Put

y-ZJA

21. If

x

is

in the identity

a 8 - y 3 = (a - y) (a 2 + ay + 1/ 2 ).]

small, prove that

N/T

(i)

+x

nearly equal to 1-f rr/3

is

positive, the error in taking 1 +a:/3 to represent Z/l

cally less

;

+ xis negative, and

(ii)

is

if

a;

is

numeri-

than # 2 /8. 10# 2

-

-

(x\* o/ 1 H-

(

)

1 4-

and use the x)<---2tl ,

last example.]

Approximate values of A, B are a = 3083, 6 = 8377, each correct to the Using Ex. 7, show that an upper limit to the error in taking 4 i a/6 for A IB is Ii 10 1 22.

nearest digit.

.

[The error

<

(0-5

+ 04 xO5)<

.]

23. In the last example, suppose a/b expressed as a decimal to n places, to the nearest digit. Show that the numerical value of the error in taking this as the value of A /B is less than

111

W*2 "iO"' Hence show that

it is

useless to carry

on the division to more than four places

of decimals.

B = 1 -414213 by using Ex. 7 find how many B to find an approximate value of AB with an error

24. If figures

^=3-141592,.., must be kept in A,

...

,

s

numerically less than I/ 10 [Choose a, b so that a6-|-/3(a .

take

a,

We may

such that

2
therefore take

+ a)
4<1/103

.

This

.

Now 6<2 is

a = 3- 1416, 6 = 1-4142.]

and a + a<4,

satisfied if

a<^6

1(

^

,

/.

we may

<-

6

^. 1U ^

CHAPTER

III

POLYNOMIALS 1

An

Definitions.

.

a xn

where a

,

a 1?

expression of the form

+ a^"- 1 4- a 2z n ~ 2 +

a n are independent of

...

(1)

. . .

+ aw

x, is called

,

a polynomial in x of the

n-th degree.

A ax m

y

A

polynomial in x and y n ,

where a

any number

sometimes called a rational

y,

and m, n are

of terms of "the

form

positive integers.

and

is

degree

is

of variables is similarly defined,

integral function of the variables

:

its

its

the

sum

instance,

atfP is

}

number

of a

highest term. of the indices of the variables in each term of a polynomial a constant number, the polynomial is said to be homogeneous. For

If is

sum

the

independent of x

is

polynomial in

that of

is

+ a^-^y + a2 x n ~ 2y 2 +

. . .

+ a nyn

,

a homogeneous polynomial in x, y of degree n. constant may be regarded as a polynomial of degree zero. Polynomials of the first, second, third, fourth, degrees are

A

. . .

linear, quadratic, cubic, quartic,

...

homogeneous polynomial

as

functions respectively.

In some parts of Algebra, a homogeneous polynomial

A

known

in x, y of degree

n

is

is

called a quantic.

often written in the

form n atfc

where binomial

+ na^-iy 4-

*!

-

a 2x n

~2

2 y +

. . .

+ a nyn

coefficients are introduced for convenience.

denoted by (00,^,02,

..,

a n \x,y) n

d$x,

I)

d e\x,

I)

(a, 6, c, (a, 6, c,

y

may

3

=aa?3

4

= ax4 + 4&z3 4- 6cx2 4- idx 4- e.

-h

This

is

shortly

.

Using this notation, cubic and quartic functions of x

(

,

be denoted by

36x2 4- 3cx + d,

If A, 5, C are polynomials and A=BC, then JB and C are called factors ol divisors of 4, and .4 is said to be (exactly) divisible by J? of by C.

SYNTHETIC DIVISION

24 Let

Division.

2.

A

and

B

be polynomials in

A

x,

being of higher

degree than J5, or of the same degree. To divide A by B is to find an identity of the form '

'

are polynomials (or in special cases constants) and R is of the result is Such division is always possible lower degree than B. and is called the division transformation, Q being the quotient and

where

Q and R

:

unique R the remainder.

In dividing ax3

Synthetic Division.

3.

reckoning

be arranged as on the

may

+p + q + r

a

-

^

fcz

+

2

ex-}-

z d(ax +px + q

2

+ cx px *-phx

>

>

r

= qh +d 2

the quotient = ax 2

,

the

ax* - akx 2

where rp = ah + b, cr=ph r + c, 2

and then

+ ex -f d by x-A,

:

7

ah + yA + g

bx 2

x-h)ax* +

+d

a +b + c 7

left

-f

-f

:

qx-qn

px + q,

r

the remainder = r.

and

This process is known as synthetic division. parison with the reckoning on the right.

It is justified

by a com-

evident that the method can be used to divide any polynomia 1 in x by x-h. If any powers of x are missing, these must be supplied with zero coefficients. It

is

Ex.

We

Divide 3* 4 - x* f 2x 2 - 2x x- divide as follows 1.

by

(

2)

l-(-2)

.'.

Quotient

1

by x +

2.

:

3

-1

+2

- 2

-

3

-0 -7

+14 +16

-32 -34

+68 +67

= 3z 3 - 7z 2 + Ux - 34

;

remainder

1

= 67.

To divide a polynomial A by ax- 6, we may use the following rule Find the quotient Q and the remainder R in the division of A by x- b/a, *

:

then Q/a

A

by

and

ax- b.

R

are respectively the quotient

and remainder in

the division of

APPLICATIONS

25

Dividex* + 2x ~3x~4:by2x-l. as on the it is Dividing by x Ex.

2

2.

l

right,

J,

seen that

+2

-3

- 4

+t

+4

-

the quotient = a; 2

+ # -J and the remainder =

/.

the required quotient = J (a; 2

7/ the divisor to divide 4#4

+

+ fa; - J)

1

^

y-

5

^. ;

remainder = - ^.

of the second or higher degree, the process is as follows 3# + x-l by 2 -2z + 3, write the divisor in the form is

:

3

x 2 - (2x - 3) and proceed thus, 4

+3 +0 +1 -1 -12-33-30 8

+ 22 + 20

(b)

10-12-31 The quotient term in

line

8

in (a);

is

(c)

llo;

where

the remainder

-12z-31.

The

will

be understood

+ 3#3 + x - I

if it is

in the

e are independent of x. a, 6, Divide f(x) by x*-2x + 3 as above. 2 quotient 4# + 1 Ix + 10.

compared with ordinary

division.

form

. . .

The remainder

is

-12rc-31 and the 4

Divide 4z 2 is 19a;

+ 1 Ix + 10 by x 2 - 2x + 3.

- 2 and the quotient

4.

The remainder

+ 11 + 10 -12

Hence

8

~ 2

- 12* -31.

4. Applications of Synthetic Division. formations are often required.

Let/(x) be a polynomial in (I)

first

;

Express f(x)= 4x*

3.

+ 10 and

(c)

4; 4 (2 -3) = 8 -12; put 8 in line (6) and -12 ll(2-3) = 22-33 put 22 in (6) and -33 in (a);

+ 3 = 11;

and so on. The reckoning Ex.

4z 2 +

is

(a)

To

f(x)=a (x-h) +a l (x-h) is

an _ 2

is

x-h.

n -l

Suppose that

+ ...+a n _l (x-h) + an

.

the remainder when/(#) is divided by x - h. ItQ is the quotient, the remainder when Q is divided by x - h. If Q' is the quotient, the remainder when Q' is divided by x-h. Continuing thus, we

Then ar an-1

trans-

x.

express f(x) as a polynomial in n

The following

can find

is

all

the coefficients.

The reckoning

is

arranged as in Ex.

1, overleaf.

REMAINDER THEOREM

26 Ex.

1.

z

Express 2x + x

5x - 3 as a polynomial in x-2.

2

2

+13

(2) To expand f(x + h) in powers of x, we express f(x) as a polynomial x - h, and then change x into x + h.

Ex.

in

2.

Proceed as in Ex.

1

and substitute x + 2

for

x

in the result, thus

2 f(x + 2) = 2x* + 13* + 23* + 7.

(3)

Iff(n)

is

a polynomial in n of degree

aQ + a^n + a 2n(n +

1)

r, to

+ a%n(n + 1) (n + 2) +

This can be done by dividing by w, w-hl, n 2fo. 3.

4 Express n

Divide n4

and

+ 3n 2 + 2

+3n 2 + 2 by

+ 2,

...

in succession.

in the form

by n + 1, the quotient thus obtained by n + 2, by n + 3. The reckoning is shown below, and the required

numbers in thick type.

1+1

1+0+3+0

1+2

1

1+3

1

-1+1-4 _i +4-4 -2+6

1 /.

. . .

w, the quotient

finally the last quotient

coefficients are the

express f(n) in the form

-3 -3

+10

-6

n4 + 3n 2 + 2 = 2 - 4n + 10( + 1) - n(n + !)( + 2)

The Remainder Theorem. Iff(x) is a polynomial, then f(h) remainder when f(x) is divided by x - h. This follows on substituting J^ fof xirTtfre* identity

5.

is the

where of f(x)

Q and J? are respectively the quotient and by x-h; for R is independent of #.

remainder in the division

IMPORTANT THEOREMS Corollary. Iff(h) = 0, then For, in the preceding,

x~h

is

27

a factor off(x).

R

Ex.

Show

1.

n

Since

is

that, if

-

odd,

(

l)

n

n

is odd,

= - 1, an

hence

2.

///(a;) is

f(0-2)

a factor of xn + 1. - 1, x

1 is if

+ x = ( _ ! jn f ! _ _ X + X _

.*.

##.

x+

x 4-

1 is

a factor of x n

= 3** + 2** - 6x - 4, ./md

*fce

-+

1

.

.

w/we off (0-2).

the remainder in the division of f(x) by

a;

- 0-2.

Performing the division

:

1-0-2

Theorem.

6. values

a 1? a 3

,

...

// f(x) is a polynomial which vanishes when x has a n no two of which are equal, then the product (x-oc 1 )(x-(x 2 )(x-oc 3 )

is

the

,

...

(z-a w

)

a factor off(x). For since /(a^^O,

by the Remainder theorem f(x) = (x~oc l ) .fl (x) where /x (x) is a polynomial. Therefore /(a 2 ) = (a 2 - aj)^ (a 2 ). Now /(a 2 ) = and aa-aj^O, therefore /i(a 2 )=0, and consequently x-oe. 2 is a factor oif^x). Hence

y

.

where /2 (x)

is

and

(x^-(x-^)(x

a polynomial.

Proceeding in this way, we can show that (x-oc 1 )(x-oc 2 )...(x-oc n )

is

a

factor oif(x).

7.

A

Theorem.

polynomial f(x) of the n-th degree cannot vanish for

more than n values of x unless all its coefficients are zero. For otherwise (by Art. 6) the product of more than n expressions of the form x-cx. would be a factor of the polynomial. That is to say, a polynomial of the nih degree would have a factor of degree higher than n,

which

is

Thus it is

impossible.

the coefficients of f(x) are zero, f(x)

If all

is

said to vanish identically.

-a)(b-c) + (x- b) (c - a) -f (x c) (a - b) vanishes identically the first degree in x, and it vanishes for the values a, 6, c of

(x

of

Corollary.

;

An

equation of the n-th degree cannot have more than

n

for x.

roots.

it will be shown that every equation of the nth degree has n not exactly roots, necessarily all different, which may be real or imaginary.

NOTE. Later c

B.C.A.

UNDETERMINED COEFFICIENTS

28

Theorem.

8.

ax n then a

= a',

6

If for more than n values of x, ~ " bx n 1 -t+ hx + k = a'x n + b'x n l -f

-f

.

= 6',

...

. .

h = h', & = &'

,

-f

. . .

h'x

that is to say the

;

+ k',

polynomials are

identically equal.

For by the given conditions the polynomial n - 6> n ~ 1 + ... -f (h - h')x + (a a')x -h (6 vanishes for more than n values of x

-

k')

by the preceding

hence,

;

(k

article, it

follows that 6

a^a', 9.

Polynomials

(1)

///(x, y)

is

= 6',

Two

in

...

h

,

=h

f

k^=k'.

,

More Variables.

or

a polynomial in x and y which vanishes for

all

values

ofx

and y then all the coefficients off(x, y) are zero. The following method applies in all cases. Let the polynomial be y

u = ax 2 + bxy which

is

supposed to vanish for

all

-f

2

+ dx + ey -h/,

cy

values of x and y

u = ax 2 + y(by + d) +

and

for

any given value

vanishes for

all

a = 0,

/.

(2)

If for

all

ey +f)

by

6 = 0,

values of x

is

then

,

a polynomial in x, which

;

+ d = Q,

Also the relations (A) hold for

a = 0,

-f

of y, the last expression

values of x

hence

2

(cy

;

all

rf

and y

2

cy

+ e?/+/=0

values of y

= 0,

c = 0,

(A)

;

e-0,

/=0.

the 'polynomials f(x, y)

and F(x,

y)

have

equal values, then the coefficients of like terms in the polynomials are equal. This follows immediately from (1).

The fact that two polynomials, -writing u^v.

NOTE.

by

10.

The Method of Undetermined

necessary to enquire specified 4

u, v, are identically equal is often expressed

farm.

Undetermined

In

Coefficients.

It is

often

a given function can be expressed in a certain cases of this kind we may use the method of if

Coefficients/

which

may

be described as follows.

Assume that the function is expressed in the given form, where some of the coefficients are unknown. In order that the identity involved in this assumption may be true, the. unknown coefficients must satisfy certain equations. If these equations have a solution, the function can be expressed in the specified form.

QUADRATIC FUNCTION Ex.

Show

1.

that values of a,

c exist

6,

23

such that

and find these values. The right-hand side ^ a (a; 2 -x -2) + b(x 2 ~ 3z + 2) +c(z 2 This

equal to the left-hand side for

is

such that

a

~ b-~<- ~'2,

-3A"5.

u

values of x,

all

-a -

- c

2/y

if

1)

values for

a, 6, c

These equations have one solution, namely, a~2, 6 = 1, c= -1. In a question of this kind, i/ t has been shown that an identity of the exists, the values of the constants

may

be

can be found

1.

found by giving special valuer

specified

Thus, assuming that an identity of the specified form exists, wo can find

by putting a:!, -1,2 2.

a ti^(2x + y+p)(x ~3?/-f q) =2x~ - 5xy 3y + a?(^

Equating

coefficients,

we have

p

From

the

iirst

two,

er,

b, c,

in succession.

Search for factors of #=3 2x~ - 5r// - 3?/ 2 -x + 10// Since 2x 2 - 5xy - 3?/ 2 (2x + y) (x - 3y) we assume that

Ex.

form

to the variables.

2q

-\-

y>-^

tluj three

-

--

-3,

(/

1,

-

conditions 3/> 4

= !, and

7

.---

3.

+y(q -3^) +y^.

-i-2q)

:

pq

10,

:--

-

3.

these values hajtpcii to satisfy the third

condition.

11.

Quadratic Functions of x and

function of x,

y, z of

the second degree

ax 2

When z^l

-f 6//

2

is

of the

Every homogeneous

y.

form

+ cz* 4- 2///2 -f 2(jzx + 2hxy ...................... (A)

this reduces to

az*

+ 2hxy + by z + 2gx + 2fy + c,

........................ (B)

the general form of a quadratic function of x and y. 2 of Art. 10 it appears that the cases in which an expression of the form (B) can be resolved into factors are exceptional. The condition

which

is

From Ex.

that this

12.

may

he possible

Theorem.

is

investigated in the next article.

The necessary and

sufficient condition that the quadratic

s ^ aj? + 2hxy + by* + 2gx + 2fy + c

faction can be expressed as

the product of two linear factors

A

-=

abc

+ 2fgh

-

2

af

-

Itf

u

- ch* - 0.

Assume that S = (px + qy + r) (p'x -f q'y -f r') S=^0 this can be written

and consider the equation

;

ax 2 + 2x (hy +g) + giving

x

(hy+g)

(bif

+ 2fy -f c) - 0,

(hy+gf-a(by* + 2fy + c)

............ (C)

DISCRIMINANT

30

These values of x must be the same as are therefore rational functions of

root sign in (C)

is

this

may

+ r) and

be a perfect square

The following cases must now be considered // a

(i)

not zero,

is

//a =

(iii)

// both

fails

r ')

;

they

is

= 0.

:

but, except

;

solving for

by

-f

2 2fgh-af-bg*-ch =0 ......................... (D)

the reasoning

to the condition (D)

(q'y

follows that

it

abc + (ii)

f

Hence the expression under the

y.

a (abc + 2fyh - a/2 - bg 2 - ch?)

or

-

Now

a perfect square.

and the condition that

(qy

a and b are

zero,

when

6 = 0,

y and proceeding as

we

shall be led

before.

and A^O, then

S = 2hxy + 2gx + 2fy + c

A = 2fgh-ch 2 and only if, A =0.

for in this case

factors

S

//a = 6 = A = 0, S

Hence S can be resolved

not a quadratic function. can be resolved into linear factors, then J=0.

(iv) if

if,

.

is

into linear

Hence, in

all cases,

Conversely, if A 0, then S can be resolved into linear factors. This is evident in all cases, from the preceding.

NOTE.

The expression abc + 2fgh-afz

called the discriminant of S.

Ex.

1.

//,

Observe that p,

in the above, p, q,

r,

The square

;

and

real

if,

root of (hy

and only

,

ca-g\

+ g) 2 - a (by 2 + 2fy + c),

denoted by J, and is

are not necessarily all real.

prove that

ab-h\

2

or

-ab)+2y(hg-af)+g*-ac,

if

h - ab 2

Similarly,

2

is generally

q, r, p', q' 9 r' t

vice versa.

y2(h is

2

p', q' t r' are all real,

6c-/ are all negative

bg*-ch

by solving

S =0

and

g

2

-ac

as a quadratic in y t

are positive.

we find that/ 2 - be must

be positive.

DIVISION

AND FACTORS

EXERCISE DIVISION

31

III

AND FACTORS

In Exx. 1-6 find by synthetic division the quotient and remainder expression is divided by the second.

when the

first

3.

+ 6x + 2; x + 4. x*-4x* + 6x z -4x + I

5.

3x*-5x*-llx* + x-I; x*-2x~2.

6.

3z 5 + 6z 4 ~2o; 3 -:e 2

5x*

1.

3x-2.

;

4.

a;

5

- 5x* -f Ix* -

-2;*; + 4; x* + 2x-I. 3 2 If 4z 6x + 1 is divisible by 2x - 1, (a l)x -f ax 3 Express x + 2x* + x + 80 as a polynomial in (i) x + 5

8.

If

9.

giving

*/

x

given by the equation 3#~- 1 i/- a; -2 (ii) y

is

when

(i)

;

I

x-2.

;

-!; 2x + l.

4

7.

10.

2. x*

find the value of a. (ii)

;

# - a 2 - 8# + 3 = 0, = 3x + 4. (iii) y 3

x -f

find

1

(iii)

;

the

x 4- J.

equation

;

4 Express n in the form

a + bn + cn(n + 11. Iff(x)---Gx*

l)

+ dn(n + l)(n + 2) + en(n + l)(n + 2)(n + 3).

+ 4x* + 3x + 2,

find the values of /(

-!),/(-f), /(J).

Factorise the expressions in Exx. 12, 13. 12.

x*-2x*-lx* + 8x+l2.

13.

2z 4 - 3x 3 2/ - 6z 2 */ 2 -

3

8xi/

- 3y 4

.

14. Find a polynomial in x of the third degree which shall vanish when x = 1 and when x -2, and shall have the values 4 and 28 when x~ -1 and # = 2 [The polynomial is of the form (x - 1) (x 2) (ax + b).] respectively. -f-

15. Find a quadratic function of x which shall vanish the values 24 and 62 when x = 3 and x ~ 4 respectively. [The function is of the form (7x + 3)(ax + b).]

when x

-f and have

16. Find a homogeneous function of x and y of the second degree which shall vanish when x~y and also when # 4 and y~3, and have the value 2 when x 2 and y = 1 [The function is of the form (x y)(ax + by).] .

17. If

18. [It

19.

a

n

is

m odd, show that x

+1

is

a factor of xmn

-f-

1.

15 Express # + 1 as the product of four factors. follows from Ex. 17 that cc 3 + 1 and x 5 + 1 are factors.]

Prove that the condition that ax 2 + bx + c and a'x 2 + b'x + c'

common

linear factor

may have

is

(ca'

- c'a) 2 = (be' - b'c) (aV - a'b). x

be a common factor if a 2 -f6a + c=0 and a a 2 -i-6'a-hc'=0. Eliminating a between these equations, we obtain the required condition.] [x-
20.

will

Find the condition that ax3 + bx -f c and a'x3 + b'x -f c'

may have

linear factor. 21. If

ft

is

z any number, show that n can be expressed

rt(rc-l

and

find the values of a, 6,

c.

in the

form

a

common

DIVISION

32

AND FACTORS

22. If p, q, r, s are four consecutive numbers, show that p 2 q 2 , nected by a linear equation, and find this equation. [This follows from Ex. 21.] ,

r 2, s 2

are con-

23. If x y, z, w are four consecutive terms of an arithmetical progression, 2 2 2 2 prove that x t/ z w are connected by a linear equation with constant coefficients. Find this equation. + q, ivp + 3q, and find 6, c, d such that [Write x p-3q, y~p-q, t

,

,

,

zp

Show

24.

and

that constants a,

find the values of a, 6,

6, c

can be found such that

c.

In Exx. 25, 26 show that values of a, b can be found for which the sion

is

25.

divisible

by the second, and

x 5 + 2x 3 + ax 2 + b

;

a?

+

expres-

l.

[Find the remainder in the division of the

make

first

find these values.

first

expression by the second, and

this identically zero.]

+ bx*-l2x 2 + 21x-5; 2x 2 + 3x-l. [Arrange in ascending powers of x before dividing.]

26. ax*

Resolve the expressions in Exx. 27, 28 into factors. 27.

29.

2 28. 3x 2 + 2xy - Sy 2 - 1x IGy 6. 3x~i-xy~4y + 8x + 13?/-3. Show that x 6 ~ 6z 5 -f 24z 4 - 56z 3 -f 96x 2 - 96x + 64 is a perfect cube, and f-

find

the cube root. 30. If x3

show that

-f

3x 2 - 9x -h c

c is either

is the product of three factors, two of which are identical, 5 or - 27 and resolve the given expression into factors in ;

each case. 31.

(i)

Find the remainder in the division of x 3 + 3px-fg

(ii)

(iii)

32. If

by

x 2 -2ax + a 2

.

+ 3px + q has a factor of the form (x-a) 2 show that # a + 4/> 3 = 0. If the equation x 3 +3px + q has two equal roots, what is the relation connecting p and q ?

If

atP

,

x*+px* + qx + r can be expressed in the form 3 -: 2 8^p -21q* and p = 12r.

(x

3 -a) (x -b), show that

In Exx. 33-34 find the value, or the values, of A for which the given expression can be resolved into factors, and for each value of A resolve the expression into factors.

33.

x z -xy + 3y-2 + \(x 2 -y 2 ).

34.

x z + y 2 -6y + 4 + \(x 2 -3

ax 2 -f 2Lxy + by 2 + 2gx + 2fy + c

is a perfect square, show that /= *Jbc, h= that not the does an odd Va6, Vca, g provided negative sign precede number of the radicals.

35. If

36. If 5x 2 4xy -f y* - 24x - Wy -f 24 = 0, where x and y are real numbers, show that x must lie between 2 - ^5 and 2+^5 inclusive, and y between - 4 and -h 6 inclusive. [Solve for x and y in succession.] -f-

EXPANSION OF PRODUCTS 2

-f

2 xy-y + 2x~y+ 1

real,

show that x

lie

38. If lie

33

2# -f 4xy + y - I2x -8?/-f 15 = 0, where x and y are between I Jl and y cannot lie between 1 and 3.

37. If

cannot

2

a;

2

and -

between

|, biit

0, where x and y are real, show "that y cannot x can have any real value whatever.

39. If f(x) is a rational integral function of a; and the remainder in the division of /(#) by (x - a)(x -- b)

,

6 are unequal,

show that

is

[(x-a)f(b)-(x-b)f(a)]/(b-a). [The remainder

is

a linear function of

x,

and may therefore be assumed

to be

A(x-a)+B(x-b).] 40.

(i)

(ii)

If ax 5

+ bx + c has a

factor of the form x 2

In this case, prove that ax 3 quadratic factor.

[For (ii), in the identity ax* 3 multiply each side by # .] 41.

(i)

If ax*

+ bx 2 -\-c has

(a

13.

-

+ bx + c -(x*+ px +

c

2

2 )

In this case, prove that quadratic factor.

(a <7#

5

~

c

2

-f

c

show that a 2 - c*=ab.

l)(ax

6

and

+ a have a common

+ c) put

+px + 1,

I/a;

for

x and

prove that

8 .

ca;

5

4-

6x 3

+ a have

a

common

To expand the product

Expansion of Products. (a

+ 6c)=a

2 4 6x

8

1,

ex* i-bx 2

+ c and

a factor of the form x 2 2

(ii)

bx

+ px -f

+ b)(p + q + r)(x + y + z + w),

........................

(A)

we then by each term of (p + q + r) the products so formed by each term of (x + y + z + w),

we multiply each term

of (a

-f

b)

;

multiply each of and add the results.

Hence the expression (A) is equal to the sum of all the products which can be formed by choosing any terra out of each of the factors and multiplying them together. In general, the product of any number of polynomials is equal to the sum of all the

products which can be formed by choosing any term out of each poly-

nomial and multiplying them

14. i>

Theorem.

#2 a s>

For

(x

?

+ ax

//

an> token

)

(x

together.

p l9 p z p 39 ,

one, two, three,

+ a 2 ) (x + a 3 )

...

(x

4-

denote the

...

at

...

an )

i

sums of

the products of

a time, respectively, then

equal to the

products which can be formed by choosing a term out factors and multiplying these terms together,

sum of

of all

the

each of the

BINOMIAL THEOREM

34

n Choosing x out of each factor, we obtain the term x of the expansion. Choosing x out of any (n 1) of the factors and the a out of the remain-

we obtain

ing factor,

xn

~l

+ a 2 + d3 +

(a l

...

f an )

or

p^x*-

1 .

Choosing x out of any (n 2) of the factors and an a from each of the two remaining factors, we obtain xn

~


(a^i -h a^ 3

-f

4-

. . .

a 2a 3

-f

.

. .

p 2x

or

)

n ~2 .

Choosing x out of any (n r) of the factors and an a from each ~ the r remaining factors, we obtain,,jo r x n r

of

.

Finally, choosing

+ aj (x 4- a 2

(x

.'.

In particular, (x

which

a)

-f-

if

n

)

(x

4-

a3 )

. . .

a 2 = a3

a

(x

+ a n =x n + p^" 1 + p 2x n ~ 2 -f )

Use

1.

.

.

.C"xn

-f

x)

2n

= (l +2x+x 2

)

.

~r

ar

4-

4-

. . .

an

,

n to prove that

---3)

_

|2?

UH

TUT have

coefficient of

x n in

(1

+ x) zn

is

w <7^

n the coefficient of x n in (x z +2x) n is 2 , ~1 n n 2 the coefficient of x in C^(a: + 2a;) is CJ n n 2 -* the coefficient of z in is +

C"(x A

The

so on.

The use

result follows

by equating

Here

Hence,

1

.

C J" 1

(7J

.

(7^~

.

2 .

2n

the coefficient of

+ Sa^

r

= (1 +a? + x* + x*)

= 2m,

if

r

if

r=

x r in

then

(l

,

,

coefficients in the identity.

2.

x + x 2 4- x3 ) 11 n n +x) (l +x*)

(1 -f

n=

~2

~ 2W 4

of the Multinomial theorem, explained in Art. 16

Find

2.

,

22

2x)

page, can sometimes be avoided, as in Ex.

and,

pn

an

can be obtained by expanding a function of x in two

the identity (1

The

Ex.

-f

. . .

...

ways by the Binomial theorem and equating coefficients. -

and

a^a^a^

= a n = a, we have

. . .

- z n -f C%x n - la + C%x n - 2a* 4-

identities

Many

We

we obtain

of each of the factors,

the Binomial Theorem for a positive integral index.

is

different

Ex.

an a out

.

on the opposite

MULTINOMIAL THEOREM Expansion of f (x + h)

15.

in

Powers of x. Here /(x)

to be a polynomial in x. When the coefficients in f(x) are division, as in Art. 3.

The following general theorem in Art. 1 of this chapter.

is

f(x)

f(x + h)

then

=a

/

:

supposed synthetic

the notation is explained

1

r\

xn

= a^c n + nA^ ' 1 + ----'- A^n ^ + ...+A n 1

A = aQh + a

where

is

known numbers, we use

required later Ai

//

35

,

A 2 ^ a h 2 + 2%/fc -f a 2 A 3 = a h 3 + Sa^h 2 + 3a 2 h + a 3 a \h, l) r ^ r -(a ,a ,a 2

ly

,

1

...

,

,

.

r

For by the Binomial theorem,

ft

'YW

\JvfvJU

I

~L*

r*n\

~T~ A/-J i*/

\

j/>

/y7t

2

"T~ /i'o*/

I i

~ n/y nffl~~T _1

jf

. I

I

"^

<^

{/

tt

^" *" 1 r A;r = a C J?A -f + agC^^Y^^ 2 + a!^ iC^-Ti A

where

7

7

to r +

1

terms.

n

Now

C*!>C*rI%

|r-j)|n-r

Thus

kr ^C^(aQ

which proves the theorem.

Multinomial Theorem for a Positive Integral Index. n The product -f k) Expansion of (a + 6 -f c +

16. (1)

. . .

(a is

the

sum

.

+ 6 + c + ... +k)(a + b + c + ...

-f

k)

...

to

w

factors

which can be formed by choosing a term out and multiplying these terms together. therefore the sum of a number of terms of the form

of all the products

of each of the

n

factors

The expansion is ... k, where each index may have any

aab^cy

of the values 0,

1, 2, ...

,

w,

subject to the condition ...-fic

= tt

(A)

MULTINOMIAL THEOREM

36

Choose any positive integral or zero values for

To obtain the

satisfy this condition.

a,

coefficient of the

...

y,

jS,

term a

K which

,

a

b^cv

...

k" y

K factors, respecdivide the n factors into groups containing a, )S, y, of of factor the first a out each b out of each factor of Take group, tively. . . .

,

the second group, and so on.

The number

of

ways

in

which

this

can be done

is

In

and

which

of times

7

a a b^cy

this is the coefficient of

term

this particular

(B)

,

a ...

k" in the expansion

.aWcv where a,

from

y,

jS,

...

K are

,

numbers

the

to

have ...

0, 1, 2,

,

all possible sets

n,

subject to the

Jt,

(C)

of values which can be chosen condition

Theorem

called the Multinomial

is

...

+K = n.

a + j8-f y-f... This result

number

the

(i.e.

Hence

occurs).

for a positive integral

index.

Thus

(2)

in the expansion of (a

If there are

m numbers a,

+ b 4- c -f

sion of (a

. . .

-f

- b - c -f d) 6 the

b, c, ... k,

coefficient of a*b 2 c is

the greatest coefficient in the

expan-

n k) is

and r the remainder when n is divided by m. n I a ^ y The coefficient \K has its greatest value when a j8 has its least value. Denote this value by v, then v is the least value * if a and j3 alone vary and y, K are constant. a j8 where q

is the quotient

. . .

. . .

l/c

|

|

|

\

|

|

|

Thus

a

C

j8

must have

value subject to the condition a

its least

|

|

where

is

4-

j8= C,

a constant.

Let wa =|a|/J = |a|C-a, then ing as

of

. . .

. . .

|

|

a^C~a + l,

when a=j8=4<7

if

C

=

^~

and w a

agKC-fl). Hence ua even, and when a = i(C + l),

i.e.

is

^~

as

has

=

t/

a _1

accord-

its least

= i(C-l) j8

value

if

C

is

odd.

Thus a and /? must be equal or pair of the numbers a, 6, ... k.

differ

by

1,

and the same

is

true for

any

EXAMPLES

37

We

conclude that some (possibly all) of the set a, certain number q, each of the rest being equal to q +

Let r of the numbers be equal to q +

0
and

Hence

is

q

+

r(q

and

1

+ (m-r)q = n, that

l)

are equal to a

6, ... 1.

m - r of them equal to j, then

qm + r = n.

is

when n

the quotient and r the remainder

divided by m,

is

which proves the theorem. // there are

(3)

m

numbers

k, the coefficient

...

a, 6, c,

+ bx + ex 2 +

. . .

+ kx m

n

Z

is

)

-

n

I

sion of (a

ro~f^

r of x in the expan-

&*&&

r~~

.

. .

k",

where

_

a,

j8,

.../<:

y,

Aave any o/^Ae values

This follows from

by putting

(1)

w subject

...

0, 1, 2,

2 6x, ex ,

...

to the

&x m for

conditions

6, c, ...

A in equation

(C). j^o:.

Find

1.

the coefficient of x* in the

The where

a,

jS,

coefficient

y have any of the values

expansion of

=2

0, 1, 2,

=5

5

-

la

20 3v,

T

-^

j_a

j8

...

5 such that

.

|

.

.

.

y a

V

^

and

The solutions of the second equation are (4, 0), (2, 1), (0, and the possible values of a, /?, y are shown in the margin

2

3

2),

;

'

5

5

the coefficient

/.

=

| '

,

;

24

.

5 |

|

.

3

|4|J)

+ ^- L rr-r 2 2 3 + <--: |3 |2 2 1 '

.

r

'

.

2

.

3

= 80 + 360 + 90 = 530. Ex.

2.

Find

the greatest coefficient in the

expansion of

7 (a + 6+c)

Here

m=3

and 7^-2

.

3 .'.

+

1,

so that

.

q^2, r~l

greatest coefficient

=

;

2 |

7/(| 2)

3.

. |

EXERCISE IV

BINOMIAL COEFFICIENTS If (1 +ic)

n

~c + c

1 o;

+ c 2x 2 +...

r

n

x" prove the statements in Exx. 1-12. 9

COEFFICIENTS IN EXPANSIONS

38

+ 2c 1 x + 3c 2x z + ...+(n + l)c n xn = {I + (n +

3. c

+ 3c 2 -...+(-l

4. c (>-2c l 5. c a

+ 2c 3 + 3c4 + ...+(ri-l)cn = 1271

.-.^

6. C

+

7. c c r -f C!C r+1 -f c 2 c r+2

. . .

|2n

+ c w _ rc n =

n-r

...

"

c n-i 9.

n

is

that

c!5

-

c?

1 ) <5

=

of the products of

c

+ c^ -

. . .

-f

(

n

~ ,

or

,

according

zero,

Show

sum

that the

2n equal to 2

11.

,

c l9 c 2 ,

...

,

cn

taken two together

\2n

~1 -

.

[The required sum^JffCo + ^-HCjj-f

Using the result of Ex.

...

2

H-cn ) -(c^-f c^-f Cg-f-...

-f

c^)

show that

1,

2n ~ l n ~ l in the equal to the coefficient of x expansion of n(l +x) that the sum of this series is 2n n- 2

is

l/(

j

12.

is

as

even or odd.

10.

is

Show

Using the result of Ex.

3,

I)

|

Hence show

.

.

show that

n equal to the coefficient of x in the expansion of

(n

Hence show that the sum of this

+ 2) 2n-l

series is

13.

Prove that

n- I

n I

|

n> 1,

if

1

1

1

1

'

n-l|l 14. If the

terms of the

are expanded of xr

L-

is ,

I

\L\^l.

series

and the whole

In r

(3

w ~r

[n

is

arranged in powers of

x,

show that the

coefficient

-

l

15.

Show thatZ ;

[Using Ex. 11, the 2 2n ~ 2

sion of

n (l^-x)

sum of the series

.]

is

n equal to the coefficient of x in the expan-

THEOREM

IDENTITIES BY BINOMIAL In Ex. 16-18 find the

coefficients of the

products stated

expansion of (a + b + c+d + e)

16. abcde in the

39

:

5 .

17. a 6 c in the expansion of (bc + ca + ab)*. x z 5 [Write a*b'c = (bc) (ca)v(ab) and find x, y, z.] 3

4 5

18. a*b 2 cd in the expansion of (ab 19. If (l+x + x ) n =c (i)

(iii) /.

C

= C 2n

CQ

-f

x

(iv)c

C2

f

2

,

-f

c4 4-

Cr

C 2n-i>

. . .

4-

c 2w

n(n-l) v

-

|0

c2

-~

tt(n-l)(rc-2)-7

r^

l-.

.

v

term of

last

(i)

20. If

put

(i)

nc

is

or

is

,

is

w

,

to

equal

not a multiple of

- -

'

C tC z

I/a;

when n

this series

+ C 2C4

...

.

3.

,

x

write

down

A

.

the

is

even and when n

is

odd.

+ C 2n _2c 2n ~ cn+l

x and multiply by x m

for

2 3 n (l+# + a; + ;r ) :=c

(ii)

-l)

P~l2*

(vi) C C 2

[In

... -f (

= l + n(n~l) + n(n-l)(n-2)(n-3) +..., and ,

,

(v)c n

,.

,

c 3 -f

!-

In ""-- or zero, according as n

.

that

,

t

ci

-nc 1 +

+ ac + a d + bc + bd + cd)*.

+ c l x + c< x + ...+CrXr +...+c 2nx 2n prove

z

In

.

(ii)

put

-x

for

a:,

and observe that

c^ -f c 2a: 2 4- + C 3nx 3n Find the value of c -f Cj 4- c 2 + -f c 3n Prove that C + c 2 + c4 -f c e + ...= Ci + Ca + Cs + Prove that C = c 3n Ci^Cg^.j, C 2 =c 8n _ 2 etc. -f

. . .

:

. . .

.

v

(iii)

21.

,

,

In the expansion of

(i) (ii)

What

is

the term containing the highest power of x

Find the sum of the

?

coefficients.

that the sum of the coefficients of the even terms that of the odd terms.

(iii)

Show

(iv)

Show

is

equal to

that the coefficients of the terms equidistant from the beginning and end are equal.

22. In the expansion of

w

n

(l+z) (l+ y) (l+ z)

n ,

show that the sum of the

371

==

|

coefficients of the

terms of degree r

is

-.

|r

23.

By

.

3n-r

expanding the two sides of the identity

and equating the

coefficients of

xn show that ,

Also write down the last term of the series on the is odd.

when n

left (i)

when n

is

even,

(ii)

HIGHEST COMMON FACTOR

40 24.

By expanding

{(1 4-z)

fi2m

~^9

2

- 2x} m in two different ways, prove that,

/->m/nr2w-2 L

L/

-

"^92

/-TfW/t2

u

i

-

-~ 4

__/nfW ----

if

m>n,

*

25. If

prove that

(i)

a r = 2 r n {C* n

+ CjCj n C"r

the last term in the bracket being or odd.

"2

or (^

+ C^Cf

at

(iii)

=4n

= \V-

[We have

17.

(1

l l ii(l+3C- + 5C%- + ...

--2

n

n(n -]){!. 3

4-2x4- 2x

Highest

2

4-3

2

5C?"

.

accorc^ n g as r

4-5

7f7>

.

is

cven

n terms).

to

l

~2 4- ...

2

to (n - 1) terms}.

)^

Common

in a single variable x.

+ ...},

+ l)^7(H-i)

In particular, show that 2 2-n (ii)a 2

~4

Factor (H.C.F.). These will be denoted by

We consider polynomials capital letters.

where A, B, Q, R are polynomials, the common factors B are the same as those of B and R. For since A BQ 4- R, every common factor of B and R is a factor of A and, since R~A BQ, every common factor of A and B is a factor (1)

of A

of

// A = BQ + R,

and

7?.

(2)

(Q 2

,

If

R2

),

B

is

not of higher degree than A, polynomials or constants (Q }9 Rj), ... can be found such that

($ 3 J? 3 ) ,

where the degree of any one of the

set

B,

R R2 l

,

,

is less

...

than that of the

preceding.

from the

It follows (a) if

factor, this is a (h) if

theorem that

last

R2

one of the pairs (A, B), (B, RJ, (R v

common

one pair have no

factor of every pair

common

),

(R2

,

R3

)

have a

common

;

factor, the same

true for every pair.

is

Moreover, the process must terminate and that in one of two ways the last remainder, say R n must vanish identically, or be independent of x. :

,

Rn

If

Q,

then

R n _ 2 = Rnnl Q n

,

so that

Rn ^

is

a .common factor of

(7?n _ 2 ^n-i)> an d therefore also of (A, B). Moreover, every common factor of (A, B) is a factor of J? n _j, which is therefore the H.C.F. of A, B.

If

Rn is independent of x, then

the same (3)

is

(J? n _ 2 ,

Rn -i)

have no

common

factor,

and

true of (A, B).

The H.C.F. Process.

by successive

Equations

(1)

of the last section are

divisions, as represented below.

found

H.C.F.

In practice, the reckoning

B) A

is

PROCESS

41

generally arranged as on the right.

(Q,

A

B

BQi

(Q 2

Qi

BQ, etc.

etc.

The process may often be shortened by noticing that

:

multiply or divide any of the set A, J5, R l9 R 2 ... by any constant or by any polynomial which is not a factor of the preceding member of the set. (i)

We may

(ii)

If

we

,

arrive at a remainder

Rn

which can be completely factorised,

the process need not be continued. For factors is a factor of R n __ v we can write

(iii)

we find which, if any, of these down all the common factors of

if

We may use the following theorem X^lA + mB and Y ^I'A + m'B, where :

m, I', m' are constants Z, the H.C.F. of X and Y and such that then Im'-l'm^Q, different from is the same as that of A and B. For every common factor of A and B is a common factor of X and Y. //

zero

Moreover,

A(lm'-Vm) = m'X-mY therefore every Ex.

is

1.

Find

common factor

the H.C.F. of

Paying attention to the as follows

of

and

(i)

r

X and Y is a common factor of A and B.

A = 3#3 +# + 4

remark

B(lm -l'm) = lY ~-VX,

and

B = 2x*

and using detached

coefficients, the

reckoning

:

3+0+ 1+

4

(a)

Hence the H.C.F. =x + l. But it is unnecessary to go beyond the step marked (a), which shows that E t = 3x 2 + 2x - 1 = (3z - 1 (x + 1 ). Now x + 1 is, and 3x - 1 is not, a factor of B. Therefore x + 1 is the H.C.F. of B, R 19 and consequently that of A and B. )

PRIME AND COMPOSITE FUNCTIONS

42 Ex.

Find

2.

A and B where, ^8*5 + 5* + 12. A = 12x5 + 5x* + S and 2 A - 3B = 10*3 - 15z 2 - 20 - 5 (2* 3 - 3* 2 - 4), 3 A -2B = 20x 5 + 15*3 - Wx* = 5x*(4:X* -f 3x - 2).

the H.C.F. of

2

We have

is

3 2 3 Putting C = 2x -So; -4 and Z> = 4# the same as that of C and D.

-

2, it

A and B

follows that the H.C.F. of

C - 2D - 6z 2 + 3z + 6 = 3 (2z 2 + # + 2), 2(7 - D = 6x* + 3* 2 + 6z = 3a? (2x 2 + x -f 2).

Further,

of

+ 3x

Hence the H.C.F. of C -2D and 2C C and D and also that of A and 5.

-D

is

2x 2 + x + 2

:

this is therefore the H.C.F.

t

Prime and Composite Functions.

18.

factors except itself (and constants)

it is

a polynomial has no said to be prime otherwise it is If

:

said to be composite.

Thus x 2 + 3x -f 2 is a composite function whose prime factors are x 4- 1 and x + 2. Polynomials which have no common factor (except constants) are said to be prime to each other. Such expressions have no H.C.F. Thus 2(x + 1) and 4(cc 2 -f 1) are prime to each other.

We

can prove theorems for polynomials analogous to those of Ch. and 5, relating to whole numbers.

I,

Arts. 3, 4

Remembering the

between arithmetical and algebraical

distinction

primeness, the reader should have no difficulty in making the necessary verbal alterations. Thus, the theorem corresponding to that of 4 (i) is as follows

:

M

M

are polynomials and If A, JB, is a factor of B.

For

if

H.C.F. of

common

M

is

prime

is

a factor of AB and prime

prime to A, these two have no

common

to

A, then

Hence the

factor.

MB and AB B. But M a factor of AB, and therefore a MB and AB. Hence M equal to, or a factor of, B. is

is

is

factor of

A

is

is

B

and are polynomials in x, then polynomials If each other, can be found such that

Theorem. to

M

AX + BY = l according as

A

is or is not

prime

to

or

AX + BY = G,

B G 9

X, Y,

being the H.CJF. of

A

and

B

in the

latter case.

For

if

Q 19 Q2

,

...

are the quotients

process of searching for the H.C.F. of

and B l9 J?2 A and B,

,

...

the remainders in the

etc.

IMPORTANT USE OF we obtain

Therefore

where

in succession the following equations

we can

Continuing thus,

the last remainder

Hence we can

express any remainder in the form

find

Y

X,

AX + BY,

is

either G, the H.C.F.

In either case

X+

.

-f,

common

X

Y = 1 and G

factor of

or

it is

a constant

c.

or

prime to Y, for the

is

;

such that

AX+BY^G .

:

X and Y are polynomials.

Now

-^

43

H.C.F.

is

a factor of

X and Y would

A

and

first

of

equation

J3.;

may

be written

thus in either case any

be a factor of a constant.

In the second case, the polynomials in question are X/c and Y/c. NOTE.

This theorem

is

fundamental

in the

Theory of Partial Fractions, which

are dealt with in Ch. VII.

EXERCISE V H.C.F.

Find the

AND

H.C.F. of the functions in

ITS

Exx.

USE

1-6.

2. 3.

2z 4 - 13z 2 + x + 5

15,

3z 4 - 2z 3 - 11 x 2 + I2x + 9.

4.

z -f5* -2, 2# -5o; 3 -fl.

6.

12z 3 + 2* 2

In Exx.

5

2

5.

2x*-5x 2 + 3, 3z 5 - 5x3 -f 2.

-2LK-4, 6*3 -t-7* 2 -14a;-8, 2Lr 4 -28a; 3 -f 9* 2 -

7-9, find

16.

polynomials X and Y for which the statements are identically

true. 7.

(*-l)

2

Z-(* + l) 2 F=:l.

8.

9.

10.

Use the

and hence

find

H.C.F. process to obtain

A, B, C, D, such that

Ax + B

Cx + D

[Multiply the first identity by A, the second by /*, and add that (7A + 10 Li)/(llA + 28^)=2/(-3); and thus obtain

:

find A

and

p,

such

/

B.C.A.

CHAPTER

IV

SYMMETRIC AND ALTERNATING FUNCTIONS, SUBSTITUTIONS

A

Symmetric Functions.

1.

any two

interchange of

symmetric with regard

function which

of the variables

which

it

is

unaltered by the is said to be

contains

to these variables.

2 2 2 2 2 y + y z 4- z x) (x z -f y x + z y) are symmetric with regard to x, y, z. (In the second expression, the interchange of any two letters transforms one factor into the other.)

Thus yz + zx + xy and

2

(x

The interchange of any two letters, x, y, is called the transposition (xy). Terms of an expression which are such that one can be changed into the other by one or more transpositions are said to be of the same type. Thus all the terms of x 2y -f x 2 z -f y 2z + y2 x + z2 x -h z2 y are of the same type, and the expression is symmetric with regard to x, y, z. symmetric function which is the sum of a number of terms of the

A

same type

is

often written in an abbreviated form thus

of the terms and place the

x

-f

Again, (x It

y+z

is

letter

Z (sigma)

Hx and

represented by

For instance,

it.

before

yz

Choose any one

:

+ zx + xy by Zxy.

+ y -f z) 2 = x 2 + f + z 2 + 2yz + 2zx + Zxy = Zx2 + 2Zyz.

obvious that

is

If a term of some particular type occurs in a symmetric function, then

(i)

terms of this type occur.

all the

The sum,

(ii)

difference, product

and

quotient of two symmetric functions

are also symmetric functions.

Considerations of

symmetry greatly facilitate many algebraical

processes,

as in the following examples. Ex.

1

.

Expand

(y

-f

z

- x)

(z

+x

-

y) (x

-f

y

- z).

symmetric, homogeneous, and of the third degree in - x) + x - (x + ~ = a Zx3 + b therefore assume that (y + z y) y z) (z

This expression

may

is

.

where

a, 6, c, are independent of In this assumed identity,

.

x,

y

>

z.

We

2x zy + cxyz,

x, y, z.

then -1 a; 0, z (i) put x~ 1, y = 2a + 26, .'. (ii)puts = l y = l,z = 0; then a? = l, z 1 then l=3a + 66+c, ; 2/-1, (iii) put ;

f

Hence the required product

is

- a:3 - y3 -

z

3

-f

y-z

6^1; /. -f

-2.

c

yz-

-f

z~x

-f

zx 1

-f

x 2y

4-

xy

2

-

2xyz.

ALTERNATING FUNCTIONS Ex.

Expand

2.

(a

+ b+c+d)(ab + ac+ad + bc + bd + cd).

45 Test the result by putting

a~b=c~d = \. The product is the sum of all the products which can be obtained by multiplying any term of the first expression by any term of the second. Hence the terms in the product are of one of the types a?b abc. The coefficient of a z b in the product is t

1

and ab, and in no other way. The coefficient of abc is 3 for this term

;

for this

term

is

obtained as the product

of a

;

is

obtained in each of the three ways a(bc),

b(ac) 9 c(ab).

a 2 b + 32abc. Hence, the required product is The number of terms of the type a-b is

Test.

type abc

4

is

hence,

;

a

if

Ea Sab ^ 4 .

so that the test

Ex. 3

c

G

= 24

12,

and the number of terms of the

,

and

Za-b

\-

%abc = 12 + 3

.

4 - 24,

is satisfied.

Factor ise

.

.

~d = 1

b

(x

-f

y

-f

- x5 - 5 - 2 5 y

5

z)

.

Denote the given expression by E. Since E-0 when x - - y, it follows that x + y The remaining factor is a is a factor of E; similarly, y + z and z + x are factors. homogeneous symmetric function of #, y, z, of the second degree. We therefore assume that

where a and

b arc

independent of put x~l,

?/--!,

put x^\, y

2.

and proceed thus then 2a + b = 15. z-0

x,

//,

z,

:

;

= l, z^l

Alternating Functions.

;

/.

,

a ^-5

6=5

a + b^-lQ.i

then

If a function

of x, y,

z, ... is

trans-

formed into - E by the interchange of any two of the set x, y, z, ... then E is called an alternating function of x, y, z, ... n for the interchange of z n (x Such a function is x n ,

.

(y-z)+ y (z-x)+

any two

letters,

say x and

y,

n

transforms n

y)

it

n

y (x~z)+x (z-y)+z (y-x)

Observe that

the product

and

;

into

= -E.

the quotient of two alternating functions are

symmetric functions. 2

Thus

{x

3 ~ (y-z) + y (z-x)+ z* (x y)}/ (y -z)(z- x} (x y)

with regard to

is

symmetric

x, y, z.

Ex.

1. Factorisex*(y-z)+y*(z-x)+ z*(x-y). Denote the expression by E. Since E ~0 when x = y, Similarly y -z and z -x are factors, thus

it

follows that x

-y

is

a

factor.

E = (y-z)(z-x)(x-y).F, where

F

where k

is

symmetric, homogeneous and of the

is

independent of - 1, thus

find that k

x, y, z.

E=

first

Equating the

degree in x,

y, z.

coefficients of x*y

-(y-z)(z-x)(x-y)(x+y + z).

Hence

on each

side,

we

IMPORTANT IDENTITIES

46 3.

An

expression is said to be cyclic with k, arranged in this order, when it is un-

Cyclic Expressions.

regard to the letters a, 6, c, of, ... h, altered by changing a into b, b into

c into d,

c,

. . .

,

h into

k,

and k into

a.

This interchange of the letters is called the cyclic substitution (abc ... k). Thus a 2 b + 62 c -h c2d -f d2 a is cyclic with regard to a, b, c, c? (in this order), for the cyclic substitution (abed) changes the first term into the second, the second into the third, It is obvious that

. . .

,

and the

last into the first.

If a term of some particular type occurs in a cyclic expression, then the term which can be derived from this by the cyclic interchange, must also occur ; (i)

and

the coefficients of these terms

(ii)

The sum,

must be

difference, product,

and

equal.

quotient, of two cyclic expressions are

also cyclic.

In writing a cyclic expression, Thus X2 (y-z) + y2 (z-x)+z 2 (x-y)

unnecessary to write the whole. 2 is often denoted by Zx (y-z), where the meaning of 27 must not be confused with that in Art. 1. 2 2 2 Again, we sometimes denote x (y -z) + y (z-x)+z (x-y) by is

it

x2 (y-z)

+

.

..

+

...,

being understood that the second term is to be obtained from the and the third from the second by cyclic interchange.

it

The student should be

familiar with the following important identities.

2.

a(&-

3.

a 2 (b-c)+b 2 (c-a) + c2 (a-b)= -(b-c)(c-a)(a-b).

4.

bc(b-c)-\- ca(c-a)

5. 6.

a(b

2

2

-c ) + b(c

3

a (i-c)

2

first

+ ab(a-b) = -(b-c)(c-a)(a-b). -a + c(a 2 -b 2 = (b-c)(c-a)(a-b). 2

)

)

+ 6 3 (c-a) + c 3 (a-6)= -(b-c)(c~a)(a-b)(a + b + c).

7.

8. 9.

10. 11.

a+&+

It will be (i)

(ii)

Any

c& + c-ac + a-ba + b-c)=

proved later

that

symmetric function of a,

Any symmetric

27a, Za/J,

(pp. 95, 96)

Za/?y and

aj

-a* -b* -

function of a,

j8,

j8,

y can be expressed y, 8

in terms of

can be expressed in terms of

SUBSTITUTIONS mode

This

functions,

of expression is

and

47

extremely useful in factorising symmetric

in proving identities.

-6 2 )(1 -c 2 )+ 6(1 -c 2 )(l -a 2 )+ c(l -a 2 )(l -6 2 )-4a6c. Denoting the given expression by E we have Ex.

1.

Factorise

a(l

t

= Sa - Zab* + abcZab - 4abc. Za6 2 = Za Zab - 3abc E Za - Za Zab - abc + abcZab

Now

.

4. Substitutions.

ment

;

.

.*.

= (1

-be -ca-a&)(a + b+c-abc).

(1)

We

of a set of elements

may

consider processes

by which one arrange-

be transformed into another.

Taking the permutations cdba, bdac of a, 6, c, d, the first is changed into the second by replacing a by c, 6 by a, c by 6 and leaving d unaltered. This process

is

represented by the operator

/abc\

fabcd\ , \cabdJ f

or

)

,

)

.

(abc\Icaoa = bdac. , \cabj

and we write

,

(

\cab/

Such a process and

As previously

also the operator

which

stated, the interchange of

effects it is called a substitution.

two elements

a, 6 is called the

transposition (ab).

Also a substitution such as

fabcd\ _

(

in

),

which each

the one immediately following it and the last by the substitution or cycle, and is denoted by (abed). If is

letter is replaced first, is

called a cyclic

two operators are connected by the sign = the meaning ,

equivalent to the other, thus (abcd)

is

Let

that one

= (bcda).

Two

or more substitutions may be applied successively. as indicated follows, the order of operations being from right to left. (2)

by

This

is

S = (a&), T = (6c), then ^ Sacbd = bcad, and om = fabcd\ ST

mi

Thus

(bcad)>

This process substitution

is

is

TS = (abcd ,

called multiplication of substitutions,

and the

resulting

called the product.

Multiplication of this kind substitutions have

no common

is

not necessarily commutative, but if the

letter, it is

commutative.

The operation indicated by (aft) (aft), in which (ab) is performed twice, produces no change in the order of the letters, and is called an identical substitution.

SUBSTITUTIONS

48

substitution is cyclic, or is the product of two or

Any

(3)

cyclic sub-

which have no common element.

stitutions

As an

more

instance, consider the substitution

fabcdefghk\

s

\chfbgaedk/

Here a

c, c to /, / to a, thus completing the cycle (ac/). to h to d, d to 6, making the cycle (bhd). Next, e is h, changed to and to The element k is unchanged, e> giving the cycle (eg). g changed g and we write

Also b

is

changed to

is

S = (acf)(bhd)(eg).

or

S~(acf)(bhd)(eg)(k)

This expression for S is unique, and the order of the factors is indifMoreover, the method applies universally, for in effecting any substitution we must arrive at a stage where some letter is replaced by the ferent.

thus completing a cycle. The same argument applies to the set of not contained in this cycle.

first,

letters (4)

A

cyclic substitution of

product

ofn-1

transpositions.

(abo)

have equalities such as

also

(ae) (ad) (ac) (ab) (5)

is the

= (ab) (be), = (abc) (cd) = (ab) (be) (cd), (abed) = (abed) (de) = (ab) (be) (cd) (de), and so on. (abcde)

For

We

n elements

A

= (abcde)

(ab) (ac) (ad) (ae)

,

n

substitution which deranges

cycles is equivalent to

n-r

letters

= (edcba)

and which

is the

.

product of r

transpositions.

This follows at once from

(3)

and

(4)

.

Thus

if

S=(

^

a

ef^ h

]

\chfbgaed/

S = (ac/) (bhd) (eg) = (ac) (c/) (bh) (hd) (eg)

then

we

introduce the product (a6)(a&), transpositions is increased by 2. If

Thus,

a given substitution

if

number j

is

not unique.

j=n This

(6)

and

unaltered and the

number

of

equivalent to j transpositions, the prove that

is

shall

- r + 2s where

is

s is

a positive integer or

a very important theorem, and to prove

is

notion of

We

S

.

zero.

it

we introduce the

*

inversions.'

Taking the elements a, 6, c, d, the normal arrangement.

e,

choose some arrangement, as abcde,

call it

Consider the arrangement bdeac. Here 6 precedes a, but follows it in the normal arrangement. On this account we say that the pair ba con-

INVERSIONS stitutes

an

inversion.

Thus bdeac contains ba,

da,

dc,

49

five inversions,

ea,

namely,

ec.

Theorem 1. If i is the number of inversions which are introduced or removed by a substitution which is equivalent to j transpositions, then i and j are both even or both odd.

For consider the

a single transposition (fg). If f, g are consecutive elements, the transposition (fg) does not alter the position of/ or of g relative to the other elements. It therefore introduces effect of

or removes a single inversion due to the interchange of /, g. Iff, g are separated by n elements p, q, r, ... #,- then/ can be

the place occupied by g by n + 1 interchanges of consecutive elements, and then g can be moved to the place originally occupied by / by

n such

interchanges.

moved

to

fpq - x
(The steps are shown in

...xgf... ...

...gpq...xf

the margin.) Thus the transposition (fg) can be effected by 2n + 1 interchanges of consecutive elements. Therefore any transposition introduces or removes

an odd number

of inversions,

and the theorem

ever value j may have, it odd. Hence the following.

Theorem

2.

into another

B

follows.

a fixed number, and therefore whatmust be even or odd, according as i is even or

Again, for a given substitution,

i is

A

of a given set of elements is changed then j is always even or always odd. by j transpositions,

If one arrangement

In other words

:

The number of transpositions which are equivalent is not

to

a given substitution

unique, but is always even or always odd.

Keferring to (5), the minimum value of j theorem stated at the end of that section.

is

Thus substitutions may be divided into two that a substitution

is

even or odd according as

n-r, and

so

we have the

distinct classes.

it is

equivalent

to

We

say

an even or

an odd number of transpositions. Rule. To determine the class of a substitution S we may express it as the product of cycles, and count the number of cycles with an even number then S is even or odd according as this number is even or of elements :

odd.

Or we can

settle the question

by counting the number

of inversions, but

this generally takes longer.

Thus we inversions.

see at once that the substitution in (5)

is

odd.

Also

it

has 17

EXPANSIONS AND IDENTITIES

50

EXERCISE VI

SYMMETRIC AND CYCLIC FUNCTIONS 1.

Show

that (be -ad)(ca - bd)(ab-cd)

2.

Show

that the following expressions are cyclic with regard to a,

taken in this order

symmetric with regard to

(ii)

;

the expressions in Exx.

3, 4.

+ z-2x)(z + x-2y)(x + y-2z).

4. (x

+ y + z)* + (y + z~x)* + (z + x-y)* + (x+y-z) 3

Y

5.

(ft

6.

(a

6, c, d,

(a-b)(c-d) + (b-c)(d-a).

2

3. (y

Prove the

a, 6, c, d.

:

-d)

Expand

is

Exx. 5-14.

identities in

+ y 2" 2 +

2

.

+ ft + y) =

2 ) (

j

7.

8. 9.

10.

2 ( j3

2 2 2 2 y -f 0y + y a + ya -f
-f

a0

2 )

(a

4-

+ y) =27a3 + 227a 2

2

+ 2ajSyZ'a.

a3

11.

12. (a 2 13. (6

2

-f

6 2 -f c 2 ) (x*

+ y 2 + z 2 ) = (ax + by + cz) 2 + (bz - cy) 2 + (ex - az) 2 + (ay -

- ca) (c - ab) + (c 2 - aft) (a 2 - 6c) -f (a 2 - 6c) (6 2 - ca) 2

= 14. (a

2

6 8c 3

-f

6, c is

the geometric

c 8a 8 -f a 36 8

16. If the numbers x 9 y, z, taken in progression, use Ex. 3 to show that

(i)

-6c~ca

- 6c) (6 2 - ca) (c 2 - ab) = a6c (a 8 + 6 3 + c 3 ) - (6 3c 3 + c 3a s + a 3 6 3 ).

15. If one of the numbers a, use Ex. 14 to show that

17.

2 2 a -(6c-f ca-f a6)(a -h6 4-c

=a6c(a

3

4-

mean between

the other two,

6 3 -f c 8 ).

some order or

other,

form an arithmetical

Express 2(a-6)(a~c)-f2(6-c)(6-a)4-2(c-a)(c-6) as the

sum

of

three squares. (ii)

- c) (c - a) + (c - a) (a - b) 4- (a - b) (b - c) except when a 6 = c.

Hence show that

all real

[Put

values of a,

6, c,

b-cx, c-a

18. If

a?

(6

y

9

a-b

+ y+z=0, show 2

(i)2yz=o; ~i/

(in)

negative for

and notice that

that

-2 2

;

2 2 2 - 2 z ) + Sx*y*z* = ; y ) (x -ft/ ax 2 + by 2 + cz 2 + 2fyz + 2gzx + 2hxy can be expressed in the form 2 2 2 px + qy +rz ; and find p, q, r in terms of o, 6, c,/, g, h.

2

(ii)

a

z,

is

(y

-f

z2

- x 2 ) (z 2 -f x 2 -

FACTORS OP CYCLIC EXPRESSIONS

51

Prove the identities in Exx. 19-23, where 27a, 27aj9, etc., denote symmetric functions of a, 0, y, 8 : also verify each by putting a =/3 = y = 8 = l. 19.

+ y + 8) 0y8 -f y 8a -f 8a + ay) =27a 20y + 4aj3y8. 2 2 2 2 ()9y8 + y8a + 8aj3 -f ajSy) =:Za j8 y + 2ajSy827aj8.

20. (a 421.

(

22. (aj8 + ay

+ a8 + jSy -f j88 + y8) 2 =27a 2 jS 2 -f 227a20y -f 6aj8y8.

23.

Simplify the expressions in Exx. 24-28. 24. (6- 1

+ c- 1

)

(6

+ c - a) + (c^ 1 + a" 1

)

(c

+ a - b) -f (cr 1 4- ft- 1

)

(a

+ b - c).

(x-a)(x-b) (a-6)(a-c)

(6-c)(6-a)

(a-b)(a-c)

(b-c)(b-a)

c-a

ft-c

14-ca

(c-a)(c~b)

-6 l+a6

a(6~c) "" *

1+ca

l+bc

Factorise the expressions in Exx, 29-35.

(6-c) (64-c-2a) + (c~a) (c+a-26)-f(a-6) (a + 6-2c). [Put 6-c=a;, c-a=y, a-bz, noting that b + c-2a=y-z.] 2

29.

30.

2

2

(6~

31. (6[Put 6-f

c-a=, c + a -&=i/, a + 6-c=z.]

32.

[Put 33.

34.

[Put 64-

cx

c+a=y, a + bz.}

9

35.

(l-

36.

Express the following substitutions as the product of transpositions

^ <

l>

/123456X

.... ;

(654321 J

(U)

/123456X

..... ;

(246135J

(m)

/123456X '

( 641235J

:

CHAPTER V COMPLEX NUMBERS 1.

Preliminary. In order that the operation of taking the square number may be always possible, we extend the number system to include numbers of a new class known as imaginary numbers.

root of a so as

Consider the equation

x*-2x 4-5 = 0. Proceeding in the usual way,

we obtain the formal

a?=l%/^T,

x=

or

solution,

l2j~^l,

which at present has no meaning. Suppose (i)

the

symbol

possess the following properties

to

i

It combines with itself

and with

real

:

numbers according

to the

laws of

algebra. (ii)

We may substitute

(

-I) for

t

2 ,

wherever this occurs.

The reader can

verify for himself that (at any rate so far as addition, subtraction, multiplication and division are concerned) reckoning. under

these rules will not lead to results which are mutually inconsistent. Ex.

We

1.

Show

that if

x = l+2i, then x* x a = l + 4t + 4i a = l + 4i-4 = -3+4i;

have .*.

x 2 -2z + 5=

Immediate consequences

-3 + 4t-2(l+2t)+5=0.

of the supposition are

(i)

(x

+ ly) + (x + iy') = (x + x

(ii)

(x

+

1

f

1

+ itf

f

(x

)

+ iy') = (x-x') + i(y-y ). + iy') = xx' -yy' + i (xy + x'y). f

iy)-(x'

(x

(iii)

f

4-

+

iy) (x

f

iy') (x

f

-

' f

iy

)

x'*

+ y'*

x'*

+ y'*

This reckoning is at present meaningless, and can only be justified by of the number system.

an extension

THE FUNDAMENTAL OPERATIONS

A Complex Number

2.

53

represented by an expression of the form

is

The sign + does not indicate addition iy, where x, y are real numbers. as hitherto understood, nor does the symbol i (at present) denote a number. These things are parts of the scheme used to express numbers of a new x+

class,

a

and they

signify that the pair of real

single complex

The complex number x + number x, and we write

and

numbers

(x, y)

are united to form

number. be regarded as identical with the real

lO is to

iQ = x

x+

in particular

(A.)

=

tQ

-f

(B)

Thus the system of complex numbers includes all the together with new numbers which are said to be imaginary. For shortness we write _ Q

real

numbers,

+

and

+

in particular

tl

/QX

= *,

(D)

an abbreviation for + il, the symbol i denotes a complex number. Thus two real numbers are required to express a single complex number. This may be compared with the fact that two whole numbers are required

so that as

to express a fraction. In order that expressions of the form x

numbers, we have

to say

what

+ ly may be regarded as denoting meant by equality and to define the *

is

'

fundamental operations.

We

3. Definition of Equality.

x+

ly

= x' + iy'

if

say that

and only

if

x = x' and y = y'.

In applying this definition, we are said to equate real and imaginary parts. It should be noticed that the terms greater than and less than -have no significance in connection with imaginary numbers. '

4.

by

The Fundamental Operations.

as defining

'

'

In defining these we are guided

and the equations (i)-(iv) are taken addition, subtraction, multiplication and division.

the formal reckoning of Art.

5.

'

The sum

Addition.

1,

of the

complex numbers x-f

iy

and

x'

+ iy'

is

defined as (x

+

')

+

i(y

+ y')

and we write (x

+

iy)

+ (v' + iy') = (x+x') + i(y+y')

(E)

It follows that

x+ iy = (x + 40) -f (0+ iy), so that x

+ iy

is

the

sum

of the

numbers denoted by x and

(P) iy.

MULTIPLICATION AND DIVISION

54

Subtraction

6.

is

defined by the equation

Hence subtraction is the inverse of addition, for x-x' + i(y-y') number to which, if x' + iy is added, the result is x + ly. The meaning of - (x + ty) is defined by the equation

is

the

r

Hence

= (0 - a + i (0-y) aB (- a? + 4(-y) ................ (H) = 0, we have

.

)

In particular, putting

)

.)

-y=*(-y) 7. Multiplication

is

by the equation

defined

x'y) .................. (J)

It follows that multiplication

Putting

#'

= &,

2/'

=

is

commutative.

in the last equation,

number,

we

see that

if

k

is

a real

+ iy)k = xk + iyk = k(x + iy) ......................... (K)

(x

we put x=x = and y = y' = l, we have 2 or i 2 =-l .......................... (L) (0 + il) =-l f

if

Again,

8.

Division

is

the inverse of multiplication, that

is

to say, the

equations

x +

mean

and ly

From

the same thing.

Equating

real

and imaginary

the last equation

parts,

Xx'-Yy' = x mi

and

we have

and

Xy'+Yx' = y.

v x *-

f

Therefore unless x'

we have

y'

are both zero.

Thus

the equation defining division is

x + iy _xx' + yy' x'

+ iy'

excepting the case in which x'

Putting x'=k, y' =0, we

x'*

yx'-xy'

+ y'*

x'*

+ y'*

'

...................... (

'

+ iy' =0.

see that

if A; is

,)=

a real number different from zero,

+i

................................. (N)

ZERO PRODUCTS

55

It is easy to see that addition, subtraction, multiplication (as defined

and

division

above) are subject to the same laws as the corresponding opera-

tions for real numbers.

We have therefore justified, and attached a definite meaning of reckoning described in Art.

to,

the process

1.

Zero Products.

9.

is zero, then

where

For complex, as for real numbers, if a product one of the factors of the product is zero. For let?

Then

numbers.

x, y, x' , y' are real

1

xx -yy'

+ i (xy + x'y) = 0. f

Hence, by the rule of equality,

.\ 2

xx'-yy'~Q and 2 2 and z(*' + 2/' ) =

2

a;'

If x' 2

10. Ex.

1.

y(x'*

+ x'y

Q;

+ y'*)=Q.

x = 0,

y = 0, and therefore x + ty = 0.

real

they must both vanish, and

+ y' is not zero, it follows that + ?/' 2 = 0, then, since x', y' are therefore z' + 4j/' = 0. If

zy'

Examples. Express

(1

+ 24) 2 /(2 +

(2+4)

2

2

in the farm

= = ~4+4i-l""

X + iY.

3+4i~~ (3+40(3~4t) 24 7

-9 + 16 + 244 Ex.

If

2.

Solve the equation

x3 = 1

.

The equation can be written (#-l)(a; 2 + a; + l)=0, giving # = 1 or 2 a; + a; + 1 =0, we have in the usual way x=

Hence Ex.

3.

the

numbers

Fen/y

that

1,

J(

J(

-

1

11.

2

V3)

- 1 + K/3)

are ca//ed JAe ^reg cw6e roofe of unity.

M a roo CD,

3 o/ x

= 1.

we have

= |{ - 1 +34^3 -3(tN/3) 2 + (^3) 3} = |( - 1+34^3 + 9-34^3) =

Geometrical

+ a: + 1=0.

-

Denoting the given number by to 8

a;

Representation

of

1.

Complex Numbers.

In naming line-segments and angles the order of the letters will denote the direction of measurement. Thus AB denotes the distance travelled by a point in moving from A to B, and we write (1)

AB + BA-0,

and

EA--AB.

ARGAND DIAGRAMS

56 Again, we use L BAG must turn in order that it '

'

to denote the least angle through which

may lie along AC. Thus L CAB = - L BACy and we shall suppose

AB

that

-7T<

= x + iy, we

say that the number z is represented by the point whose coordinates referred to rectangular axes are x, y. (2) If z

When

there

is

no

risk of confusion, this will be called the point

z.

A

diagram showing points which represent complex numbers is called an Argand diagram.

Let point

be the polar coordinates of the

6)

(r,

then

z,

= x + iy = r(cos + 1 sin 5), z = rcos#, y = rsin0, tan = y/x. r = J (x 2 + y 2)

z

where

and hence Here

r,

FIG. 2.

,

which

denoted by

is

or

z \

|

is

essentially positive,

called the

by mod z. Thus modz = z\ = r = J(x 2 + y 2 |

It follows that if

The angle 6

is

then z

\

= 0,

z,

and

is

z

and

is

= 0.

denoted by

values differing by multiples of

many

z

).

y = 0, and consequently

called the amplitude of

angle has infinitely of 6 such that

is

=

z \

modulus of

2rr.

am z.

This

The value

called the principal value of the amplitude.

With the convention is L XOz.

of

(1), it will

be seen that the principal value of

am z

Unless otherwise stated,

amplitude of

amz

will

mean

the principal value of the

z.

and Subtraction. = z x + ty, z' = x' + iy' Draw

12. Addition (1)

Let

'.

parallelogram Ozsz', then

For then

if

s represents z

(X, Y) are the coordinates of

the

+ z'. 5,

X ==sum of projections of Oz, zs on OX = sum of = + X'. 05

projections of Oz, Oz' on

OX

FIG. 3.

ADDITION AND SUBTRACTION

Y = y + y',

Similarly that

therefore s represents z

+ z' = the

z

length Os,

|

|

Now Os<0z + 0z', the sign Therefore i.e. if am z = am z'. To prove

this algebraically,

where each root (x

is

and

+ z'.

am (z +

')

57

It is to be observed

= L XOs.

of equality occurring

we have

to

if

LXOz =

L.XOz'

show that

to have its positive value.

This will be the case

+ z') 2 + (y + y')*/(x2 + y2

2 )

(x'

if

+ j/ /2 ),

i.e. if

or

if

(x

or

if

(xy'

(2)

Draw

the

- x'j/) 2 >0, which parallelogram

^en

is

the case.

Oz'zd

d represents z-z'

(Fig.

4),

Y

.

by the last construction, the number = represented by d + z' z. For,

Observe that

= the length am (z - z') = z. Jf Od. - 2'

2

|

and (3) it is

= // s n Zt

|

+ z2 -f ...

4-z w ,

FIG. 4.

wAere zl5

required to find the point s n

~O

z'z,

2J

2>

z n are given complex numbers,

.

FIG. 5.

Draw in succession z^, sense as 022 Oz3 ,

2j

+ 22 + 23, This

is

,

Oz4

,

...

,

s 2s3 , 5 35 4>

then

s 2 , s3 ,

equal to, parallel to, and in the same ... are the points representing 2 1 + 2a ,

etc.

a continued application of the

first

construction of this article.

PRODUCTS AND QUOTIENTS

58

The modulus of the sum of any number of complex numbers

(4)

or equal to the

sum

For in the Ozl9 21$2> S2S29

an(^

the moduli of z v z 2

^at

Osn Therefore |

f sn

< Oz *n

1s

Os n

|
z2

+|

l

eng^s

Also

.

+ 2^2 + s2s3 +

l

^e

are

2s

>

. . .

4-... |

v

+ sn _ r *n

+|

|.

The sign of equality is to be taken if, and only if, che points O, the same straight line and occur in this order, that is, if z l9 z t ...

XOTE. are in

than

of the moduli of the numbers.

last figure, >

is less

,

z l9 * 2

zn

tS

have the

same amplitude.

To prove

this algebraically, |

and

sn

|

^

|

sn

^

+ |

we have by

\z n

I

<

I

(1) of this article,

s n _2

z n _i

-h 1

I

-h I

\z n

, I

so on.

13. (1)

Products and Quotients. Let z = x + iy =r(cos^H-tsin0), z'

By

= x' + iy' =r'(cos 0' + 1 sin 0').

the definition of multiplication,

==

rr'{cos

cos

0'

- sin

sin

& + t(cos

sin 0'

+ cos 0' sin 0)}

;

^^rr'tcos^ + ^-ftsin^ + fl')} ........................................ (A)

/.

Since division

is

the inverse of multiplication,

4 = ^{cos(0-0') + 'sin(0-0')} ........................ (B) For

this last is the

number which when

multiplied

by

z'

produces

z.

Hence

also )

z l *=x l

If

with similar notation for z2 (A)

,

+ iy^ = fj (cos 23 ,

...

zn ,

l -h i

.............. (C)

sin 0j),

by continued application

of equation

we have

^

2

---n = ^2---M COS (^l^^+---+^n)

In (D) put

Z1

+ ^sin(01 4-02 +...+0n )}. = = =z2 ==...~2 w z r(cos0 + tsin0), then zn

where n

rn (cos

n0 +

1

sin n0),

...(D)

........................... (E)

is

any positive integer. Similarly, from (C) it follows that ) ........................... (P)

DE MOIVRE'S THEOREM Show

Ex.

1.

We

have

few 1 -

that

is

nearly equal to

59

.

+ i)=*/26(cos0 + i sin 0), where tan = 1/5, 4 = + i) 676(cos 40 + 1 sin 40), by (E). (5 2 = 4 = 476 + 480* hence we have (5 + 1) (24 + 10t) cos 40 = 476/676, sin 40 = 480/676, and tan 40 = 1, (5

and therefore

But

;

(2)

Modulus and Amplitude. From (D) we complex numbers zl9 z2 ... zw

of the

see that

|Z|=V2 ...rw = KN*2 |...|*n|, axnZ

but note that the

---

last

if

Z

the product

is

,

,

,

and

nearly.

40=rr/4 approximately.

.*.

..................... (0)

+ amz n

(H)

equation does not give the principal value of

unless

am Z

-7T<01 +02 +. ..+0n < 77.

Again, from (B)

we have

M am =

and

am

0'

21

.(I)

am z'

(J)

,

z

although the last equation does not give the principal value of amz/s' unless

-7r<0-0'<7T.

important to notice that the amplitude of z/z' through which Oz' must be turned in order that it may and that, with the convention of Art. 11, (1), It

is

is

the angle

lie

along Oz;

ft

the principal value of

Again,

if z,

a, a' are

am - = L z'Oz. any numbers,

z-a

am

and

(3)

the length az the length a'z

:

z-a = L aza. z-a ,

.

De Moivre's Theorem.

FIQ. 6.

From

(E)

and

(F) it follows that if

n

is

a

positive or negative integer (cos

This, with

+ i sin

n 0)

an extension to be given

= cos n9 + 1 sin n0. later (Art. 16), is

known

as

De

Moivre's

Theorem. B.C.A.

COMPLEX ROOTS OF EQUATIONS

60 14.

Conjugate Numbers.

z=x + iy and

//

(1)

= x-iy

z'

9

then

are called conjugate numbers.

z, z'

Their product is a? 2 + y2 and polar coordinates of the point ,

point

z'

are r

- 0.

,

6 are the

if

r,

z,

those of the

Hence

O

am(x + ty) = = -am(x-iy). (2)

a polynomial in

is

// /(z)

with

z

real

= X + 1 F, where f(z) =f(x + iy) then f(x - iy) = X - 4 Y.

and

coefficients,

Y are real, We have /(# 4- iy) = a + a

X,

t

(x

+ iy) + a 2 (x + iy) 2 -f

FIG.

. . .

where a

7.

ax

,

,

are

. . .

real.

We

can therefore obtain expansions of the form f(x +

iy)

= 2axm (iy) n

f(x

)

-

ty)

= Zaxm - iy) n (

,

is real and m, n are positive integers or zero. Every term of f(x + iy) in which n is even or zero is real, and occurs with the same sign in/(x - iy). Every term of f(x + iy) in which n is odd is imaginary, and occurs with

where every a

the opposite sign inf(x-ty). Hence the result follows. (3) // an equation with real coefficients has complex roots, then these occur in conjugate pairs.

Let a +

tjS

be a root of f(x)

the polynomial /(x), are

Let /(a +

ij8)

A = 0, B = 0.

= A + iB,

= 0, where

a,

j8,

as well as the coefficients of

real.

where

A B 9

are real.

Since /(a +

1/?)

==

0,

we have

Hence

Therefore a -

ijS

is

also a root of f(x)

= 0.

EXERCISE VII

COMPLEX NUMBERS 1.

Find the modulus and amplitude of

2.

The condition that

3.

Find

4.

Express in the form JC

2 so that 2(3

(i)

(a + t6)/(a'

+ 1 7, iJ

+ 1, V3 - 1, -
may

-f i4) =2 -f 3*.

f

2~+3i

4- ib')

s/3

(iii)

be real

is

ofe'-a'6=0.

.

COMPLEX NUMBEEb x 4- iy> express

If 2

5.

form

in the

X + iY 9

Find the moduli of

7.

What (i)

If

(ii)

If

8.

2 |

|

2

z

(i)

if

(ii)

if

(2

+ 3i) 2 (3-40 3

2

= =

|

z' |

|

= 1 and am z =

|

- am 2', then

am 2, am z'

and

z' |

x + ty where

differ

by

z'

a

z describes

11.

Prove that

12.

|

|

22 cos a 4-2* |

22 cos a

4-

|<1 22

<2

1

z |

cos a

|

4-

Given that

1

+ 24

is

14.

the formula cos nO + deduce that cos w0 = cosn sin

< 2r 4- r < 1 2

2 |

if

___

one root of the equation, x 4 - 3a;3 -h 8x 2 -7a? + 5=0, .^

From

integer,

.

that

..

13.

2 |

z 1

then ,,,

,

/

|2|
if

Ifz=x + iy and Z = Jf + 1 7, show

find the other three roots.

'

circle.

Prove that |24-2'| 2 4-|2-2'| 2 =2|2| 2 4-2|2

[Let

1

-C\ cos

n0=(7? cos*- 1 1

sin

n^ = (cos 9 + 1 sin

n ~ 2 ^ sin 2

sin

-C?" cos

cos n

+ <7 n~3

tf

0)

~4

n where ,

^ sin 4

+ ...

sin 3 ^

-

a positive

ri is

...

,

,

(

tannf.

15.

By

putting

2=r (cos

=

-f i

sin 0) in the identity

prove that 1 -f t

coM+r

2

cos 26 +

2 r sin 04- r sin 26

?

;

10.

|

6

.

Thus, in either case, the point

=r, then

50)

2

|,

|

sin

.

.

then z'= -z.

TT,

2|z-l| = |z-2|, then 3 (z -f y ) = 4z 22- 1 = 2-2 then # 2 4- y 2 = 1 |

=-

1

4-

+ 4t) 3

Prove that

y are variables.

x,

2

|

2

(2-f 3t) /(3

(ii)

;

the principal value of the amplitude of (cos 50

is

|

Let

9.

(i)

*

2~ 1

Z 6.

61

...

+rn ~ l cos (n - 1)0 I-rco&O - rn cos n6 -f rn+1

+ ... -t-r - 1 sin 11

(rc

~ 1)0

cos

'

rsin0-rn 8inn0-f rn

" " t

1

sin

(n~

(TI

-

1

)

ROOTS OF COMPLEX NUMBERS

62

= cos

16. If z

and n

sin

4- 1

zn

is

+ z -n = 2

a positive integer, prove that

- z~n = 2i

zn

cos nd,

sin n0..

Hence show that ~ 2n 1 cosn 6 = cos n0 + C%coa(n -2)6 + 0% cos(?i-4)04-

and

if

n

is

...

,

even n

-

and

if

n

2"- 1 sinn

1)2

(

odd n-l n-l (-1) 2 2

= cos ?i0 -Of cos (n -2)0 + 0% cos (n -4)0 -

is

Binn

0==8mn0-C%8w(n similarly for

...;

17.

By

(*-ar

n

l )

-l

the method of Ex. 16, prove that

16 cos 8 18. If cos a

+ cos

0= - cos

sin*

50 - cos 30 4- 2 cos

0.

+ cos y=0 and sin a 4- sin + sin y=0, prove cos 3 + cos 3jS 4- cos 3y = 3 cos (a 4- ]8 4- y)

/?

that

8in34-sin3)54-8in3y = 3sin(aH-)S4-y).

and

[Put a = cosa4-t sin

and a 8 4- 6 8 + c3 = 3a6c, (19)

.

From

a,

b

= cos j3 4-

1

sin

c

j5,

= cos y4-t sin y,

then

etc.]

the identity

(x-b)(x-c)

(x-c)(x-a)

"

t

t

(a-6)(a~c)"

(ft-cXft-a)"*"

(x-a)(x-b) (c-a)(c-6)

deduce the identities

-

sin (0

sin (a

B) sin (0

j5)

-

sin (a

y)

y)

[Put x = cos 20 4- i sin 20, a = cos 2a 4- 1 sin 2a,

Roots of Complex Numbers. number whose nth power is equal to any 15.

z = r (cos

Let

where

r>0 and

root of

r.

is

(1) If n is a positive integer, z is called afc n-th root of z.

sin 0),

Let ^r denote the positive arithmetical nth

-7r<0<7r.

Consider the values of

>

where &

4- 1

etc.]

s

any integer or zero. un = r {cos (0

from which

it

follows that

u

By -I-

is

+ L sin

n

Art. 13

2for)

+ 1 sin

I

n

)

we have (0 4- Zkn)}

an nth root of

z.

z

;

PRINCIPAL VALUES

63

and only if, the corresponding That this may be the case, the must differ a and of 2ir the corresponding values of k angles by multiple must differ by a multiple of n. Therefore u has n distinct values, namely Again, two values of u will be equal angles have the same sine and cosine.

if,

y

those given

Thus

by

& = 0, 1,2,

the n-th roots

cos-

ofz = r(cos 6 + -

n

lvalue of

(2)

,

n .

J

/n + cos-

.

_ 1, 2, ...,

, where &=A 0,

, n-1.

,

0\ i

sin

~i

is

called

the

principal

|

^z,

To find

the points representing the

find the length represented

Find

1.

representing

where

%Jz,

we must be

able to

V#

into

Q lt Q2

the points values of

the

n values of

by $r

and to divide the angle n equal parts. Ex.

n-1.

sin 6) are the values of \

+ 1 sin

v^. v

,,^ v ^,,,

i

...,

Q^

,

/z,

z

Oz=r=\/5 + 3~= ^Oz =tan 8 ^=

Here and

tan

Also fyr= */2* an(* so Ci ^2 ^3 are and points on the circle with centre radius */2, such that

FIG. 8.

General Form of De Moivre's Theorem. If x w one o/ the values of (cos 6 + isin 6) x cos x0 + M*n

16. then

The

^

is rational,

.

case in which

2

is

a positive or negative integer has been considered

in Art. 13.

Let x~p/q where p, q are integers and q (cos 0/q

is

positive, then since

+ 1 sin 0/q) q = cosO + t sin 0,

therefore

i

cos 0/q also, since

j? is

an

+ 1 sin 0fq

is

a value of

(cos

+

t

sin 0)*

;

integer,

(cos 0/q +

1

sin 0/q) p

= cos pO/q + 1 sin j?0/#

therefore

;

,,

cos p0/q 4-

1

sin jo0/j

is

a value of (cos

+ 1 sin 0)

.

ROOTS AND FACTORS

64

The n-th Roots

17.

+1

then since cos

of Unity.

= 1,

sin

2k-rr

r-1 and 0=0,

let

the n-th roots of unity are the values of

2kn -hsm~ n n

cos

In Art. 15,

where #7 = 0,

.

,

1

n

1, 2, ...

1.

The principal n-th root, given by A = 0, is 1. w is even, the root 1 is given by k \n. -1 n = 1 (x 1 Since x + x n ~ 2 + + 1 ), it follows that (x If

71

. . .

)

other than

of x"~-

1,

71

The

-

+1)77 (2* VL-

.

sm

-ft

-

(

(2A

.n

If

n

-

n-th root of

odd, the root

is

+ 1)7T T where k = n Q, n ,

-

(

1) is

(

1) is

cos

root,

given by

1

O

1, 2,

...

1 n~l.

6 = 77.

l,

to k = 0. n corresponding & = -|(n - 1).

-+ n

Factors of x n -1 and x n +

19.

any

These are the values of

1 ).

This follows from Art. 15 by putting r =

The principal

is

+an ~ 2 +...+<* + 1=0.

n-th Roots of

cos

a

1=0, then a -1

18.

if

1

sin

,

1.

n

(1)

The n factors ofx -l are x - {cos 2r7r/n + 1 sin 2rrr/n} where

r

= 0,

1, 2, ...

n-1.

This follows from Art. 15.

= 0, (i) If n is even, the factors x-1 J.nd x + 1 are given by r The remaining (n - 2) factors can be grouped in pairs as follows

r

= n/2.

:

Since

2r7r/n

+ 2(n-r)7r/n = 27r,

the

factors

corresponding to r and

n - r are

x-

f {

cos

2rn --

[

And

f-

n

sm 2f7rl .

*

the product of these

'/

j

x-

f {

I

cos

2rn -n

1

--

2f7T - cos 2f7T\ = x^ - 2x cos 2f7T + sm^2 n n / n .

a;

.

and

sm 2r
.

]

is 2

/ \

n

)

h 1

;

therefore

n

/

where 77 denotes the product of the factors as indicated. - 1 is = (ii) // n is oddy the factor x given by r 0. The remaining factors can be grouped in pairs, and

(A)

n-1

(B)

CUBE ROOTS OF UNITY

x-{cos // n

(i)

15

(2r

//"

n

+

l)77/n

l

=

+ ism

+1

(2r

are

+ l)7r/n},

where

'r

= 0,

1, 2,

...(n-1).

even

x + (ii)

factors of xn

n

the

(2) Similarly,

65

is

n^

n - 2)

C (x*-2x \

os&-^ + l) ............. n

(0)

/

ocW,

(D)

20. The Imaginary Cube Roots of Unity. of x 2 4- x -f 1 = 0. Their actual values are

.277

277

cos

i

l

cube roots of

1,

co,

the other

+w+

2

=

^3). for (a> 2 ) 3

o> 2 ,

is

= (6t> 3 2 = l. )

.................................. (A)

and

o>

2r

are the imaginary

so l-f-o/4-

The following

co

.

/Q

-1,

r is a multiple of 3, co r

when and

(

o

Denoting either of these by Moreover,

Again, except

-,

or

sin

j

These are the roots

o> 2r

=

.................................. (B)

identities are important.

x*-y* = (x-y)(x-o)y)(x-aj

2

y) ....................... (C)

+ y* = (x + y)(x + ajy)(x + a> 2y) ....................... (D) x2 + y2 + z 2 - yz - zx - xy = (x + coy 4- o> 2 z) (x + o> 2 y 4- ojz) ......... (E) z?

x 3 + y3 + z3 - Sxyz = Ex.

Prove *Aa*

1.

(x*

(x

-f-

y + z) (a; +

coy

+

+y* + z* -3xyz)* = X*+Y* + Z*

X=x

z

+2yz,

2

co z)

(x

+

co

-3XYZ,

2

y -h

coz) ....... (F)

where

Y=

This follows from (F) on observing that

and Ex.

2.

If (l+x)

n =c

Q

+ c 1x + ctf(;* + ...+cnxn and

o/ the series being continued as far as possible, show that the values of are

|(2 \

n

+ 2 cos ~

Putting

*

1, o>, o>

2

) ,

where r = n,

n - 2, n + 2,

Slt #2

,

3

respectively.

/

for #, in succession,

(l-fl)

n -c

+ Cl -f...-fc r 4-...-fcn ..................................... (A) n

...-hc n a>

,

4-...+c n cu

........................... (B)

2n ..........................

(C)

COMPLEX ROOTS

66

r 3, 1-f co

not a multiple of

If r is

we have

Now

-i

1

hence, adding (A), (B), (C) together, n 2 n +o>) + (l+o> )

+c3 +c 6 4-..0=2n + (l

3(c

XT

+ o>2r =0;

+co = l i

2rr.27r

.

7r/7r.

TT\ + cos -^- + 1 sin -5- =2 cos- cos^+tsmO /), O O O O rt

\

and, since

AI

i

Also

1

cos ~

=i, we have

01l 4-cos 27T -i sin

+aj 2 =

.

O

Again, multiply (B) by

2-7T

co

a ,

,

,

O

by

(C)

1 -f o>

= cos - + 1 sin -

,,

and

Ttrr

+o> 2 ) n ,.,

(1

cos

.

-tsui

o

and add to (A)

co,

;

Al

a>

277 mn / (l+eo) =( cos -^--4

2 /i

.

sm

27T\/

(n-2)ir -

If

.

tJ

then

;

n 2 n n 2 8(^+04+07 + ...)=2 +co (l4-a>) +o)(l+a> )

Also

nrr

TlTT

cos-

.

-

W7T\

+tsm-~

1

(n-2)ir -^~

.

^

^

-

,

and 3

Finally, multiply (B)

by

o>,

(C)

by

o>

2 ,

and add

EXERCISE

to (A)

;

it will

be found that

VIII

COMPLEX ROOTS 1.

Prove that the values of

^-1

are

~^(1

v^

2.

Find the values of ^(1 + 1) and *J(l-i).

3.

Use

Art. 19 to

t).

show that 1),

) 4.

Prove that

=a; 2 (a;

where

a,

j3

are the roots of z* + z- 1=0.]

'

(#-

2x

cos^-H

l).

COMPLEX ROOTS

67

Give a geometrical construction to find the points

5.

the values of

z l9 z 2 corresponding to

*Jz.

Oz l bisects the angle XOz, and is a mean proportional z-f) produced, so that Oz a =2 1 O.]

[If 01 is the unit of length,

to

01 and Oz ; and

2 2 is

on

If a, b are complex numbers, show that

6.

a [Let 2 1 =a + */a 2 I

*i

I

+

-6 2

,

2 I

*$J

a-*Ja*-b 2 then by Exercise VII, Ex. 10, =i *!+** 2 + i *i -*2 2 =2 a )t + 2 a*-6> z2

,

!

I

I

I

I

. |

|

|

a -6

-f |

|

a -6

;

2 |

}

= {|a + 6| + |a-6|}M Solve the equation

7.

f

+ 2(l+2i)z -(11 + 20=0.

- (11 + 20. Verify that the sum of the roots is -2(1 + 20 and the product [Put z~x + iy, equate real and imaginary parts to zero and solve for x, y.] Prove that, with regard to the quadratic

8.

z2 (i) if

(ii) if

+ (p + ip')z

the equation has one real root, then

the equation has two equal roots, then

and

p*-p'*=:4q If a

[(i)

Eliminate

In

(ii)

is

a

real root,

by

a.

this case (p

+ ip')*=4:(q + i,q'),

If z ~x + ly =r (cos 6 +

9.

1

etc.]

sin 0), prove that

Jz=z--={'Jr + x + i*/r-x} according as y

Also

if

is

y>Q

-

-f l

sin

J

then 0<^
cos 6 = -

and

if

the roots of zn = (z +

them

{Vr -f x

or

- i\lr - x}

9

positive or negative.

*/r ( cos

[\/z

pp'ty'*

the rule of equality a 2

y<0 we

l)

n ,

;

have -7r<8<0

9

etc.]

and show that the points which represent

are collinear.

[The roots are -J( points

lie

on the

line

l

+i

cot

x + i=0.]

J,

where r=0,

1, 2, ...

n-1. The corresponding

COMPLEX FACTORS

68

Show r

= 0,

that the roots of (1-f 2) n = (l -z) n are the values of n - \ but omitting n/2 if n is even.

1, 2, ...

12.

tan-^, where

n

Prove that

* 2W -2xn cos

(i)

194- 1

*n + ar* - 2 cos

(ii)

cos

(iii)

[(i)

i

,

1 2 /Trio" (* - 2x cos

=

"*

=

- cos nS = 2n - 1 77^o

rz<

~*

x m - 2#n cos 6 + 1 ~~ {xn - (cos 4 t sin (iii) Put x =cos + 1 sin ^, and for

n

is

odd and not a multiple of

+ l) n -xn -l. - 1, co, where [Put x=Q,

cos

(a?

cos

put

3,

- cos

<

n-

0)} {z

Art. 15.

13. If

+ ar * - 2

/T^I JJ

+ -1

(cos

~

sin 0)}.

Now

use

n0.]

prove that x(x + l)(x* + x +

1) is

a

factor of (x

14. If

(

1

+ x + x2

)

n=a

-f

a; is

an imaginary cube root of unity.]

a^ -f a 2 + a:

2

. . .

-f

2

n^

2n >

prove that

aQ + a 3 + a 6 +

u

15. If

v=x+y

+ z + a(z

w=x + y + z + a(x + y-2z), 2

prove thUt 21 a (x

(a

where

3

4-

y

3

-f

z

-

3

Sa^z)

= w 3 -f v 3

3

- Suvw.

,

r

a>

and w are the imaginary cube roots of unity, prove that

[IT] If 2^

4-

2 2 2 4- z 3 2

[For

18. If

o>=J( then

-

2;

2 2;3

x 4- a>2:2

-

1 4-t\/3),

4-

x)~

l

-f (c 4-

z)-

1

4-

(d

4 x)- 1 = 2ar 1 .]

- Zfa - 2^2 = 0, prove that *-

^^s = 0,

/.

ZB

- zt =

being an imaginary cube root of unity, and

+ oA+
a, 6, c

are

2


where

1^

+ w')- 1 -f (6 -f co')- 1 + (c + co')" 1 + (d + w')- 1 - 2ft/- 1

1 [Consider the equation (a -f a)"" -f (6

real,

-f-

J{>

D = s/(

provided that 6>c. If 6Jx denotes the positive square root.]

PRODUCTS AND QUOTIENTS 21. Points representing the Product Let z = r(cos0 + *sin0), Given Numbers. Construction for

(1)

the

point

69

and Quotient of z' = r'

(cos 0'

Two

+ 1 sin 0').

representing

the product zz'.

be the point on OX which represents Draw the triangle Oz'P unity, so that 01 =1. similar to the directly triangle Olz. 1

L^t

P

Then

represents the product zz'

z

.

For by similar triangles

i

FIG. 9.

OP

Oz -,

OP

.

that

r

Lz'OP^ LlOz = 0,

also

But

zz'

Therefore

P

= rr

f

{cos (0

represents

Construction for

(2)

..OP = rr

r-,

IB,

-f i

;

LlOP =

:.

+ 0')

,

sin (0

+ 0')}.

zz'.

the

point

representing

the quotient z/z'.

Draw

the triangle

the triangle Oz'l

directly similar to

OzQ

.

i

Then

Q

represents the quotient

by the

For,

FIG. 10.

z/z'.

last construction,

(number represented by Q) // k

(3)

is constant,

z-a z-a

and

.z'

z.

z varies so that

"k,

then the point z describes a circle of which a, a! are inverse points; unless k~I, in

which case z describes the perpendicular bisector of

For at d,

let

d

f .

aa

f .

FIG. 11.

the bisectors of L aza' meet aa

Then, by Art. 13,

Therefore

ri,

(2),

az

:

d' divide aa' internally

r

a'z

=k

:

1.

and externally

and are fixed points. Therefore z describes the circle on dd' as diameter.

>

in the ratio

k

:

1,

DISPLACEMENTS AND VECTORS

70 Also a,

f

,

we have

ca.ca'

= cd2

;

hence

are inverse points with regard to

a'

the

the mid-point of dd

c is

if

circle.

If

&=

1,

then az = a'z\ and

on

z lies

the perpendicular bisector of aa'. if

Conversely,

which

circle of

the point z describes a a, a' are inverse points, Z ~Q>

then

z-a

For, since ca

.

ca'

the same ratio, k to

Hence,

:

a'z

=cd 2 then ,

where



on

is

dd'

is

divided internally and externally in

say.

1,

=k

:

1 .*

If z varies so that

(4)

circle

az

FIG. 11.

K.

,

am

z-a

a constant angle, then the point

aa',

containing an angle

z describes

an arc of a segment of a

.

\a

FIG. 12.

FIG. 13.

For by Art. 13, La'za = on which the segment lies.

.

The Thus

sign of

is



determines the side of aa'

positive in Fig. 12

and negative

in Fig. 13.

22. Displacements

and Vectors.

(1) In connection with the geometrical representation of complex numbers, we introduce the notions of displacement and directed length or

vector.

Displacements in a given Plane. Let P, Q be two points in the plane OXY. The change of position which a point undergoes in moving from P to Q is called the displacement PQ. (2)

* See Elements of Geometry,

Barnard and Child,

p.

316

' :

Circle of Apollonius.'

ADDITION OP VECTORS If

any

sense as

71

drawn equal to, parallel PQ, the displacements PQ, P'Q' are straight line P'Q'

is

and

to,

in the

same

said to be equal.

To

specify completely a

we must know its

(i)

magnitude, direction

(ii) its

its

(iii)

i.e.

denoted by the

sense,

;

;

we draw OL equal and

If

PQ

the length

the letters, and arrow.

that L

PQ

displacement

:

if

order of

-

necessary by an

FIG. 14.

PQ and in the same makes with OX.

parallel to

XOL is the angle which PQ

sense,

we say

This angle determines the direction and sense of

An

(3) Vectors.

with reference to line

PQ) used to denote a line-segment and sense, the actual position of the

expression (such as length, direction

its

being indifferent,

is

called a vector.

Quantities which can be represented by lines used in this way are called Velocities and accelerations are vector quantities.

vector quantities.

A

force can be represented action of the force.

by a vector

'

'

localised

to

lie

in the line of

Quantities (such as mass) which do not involve the idea of direction are called scalar. (4)

Connection with Complex Number.

If

z=x + t,y and P

the point

is

(x, y), a one-to-one correspondence exists between the number z of the following (ii) the displacement OP (i) the point P :

and any

;

;

(iii)

the

vector (or directed length) OP.

Any to be

one of these three things

represented by

may

therefore be said to

represent

%*

z.

23. Addition of Displacements

and of Vectors.

If (1) Let P, Q 9 R be any three points. a point moves from P to Q and then from Q to R, the resulting change of position is the

same as to

R.

if

We

it

had moved

therefore

displacements as follows *

Some

from

P

addition

of

directly

define

the

Flo

:

writers use

an underline instead

of

an

overline.

15

z,

or

ZERO AND NEGATIVE VECTORS

72

The

result of

adding

QR

expressed by writing

This equation

PQ, QR,

If

to

PQ

is

defined to be

PR\ and

this is

PQ+QR=PR .................................. (A)

also taken as defining the addition of vectors.

is

RS

are

any three displacements or vectors

(Fig. 17),

Q FIG.

FIG. 16.

17.

The Commutative and Associative Laws hold for placements and vectors. (2)

In Fig. 16, complete the parallelogram

(i)

and

PQRS\

the addition of dis-

then by Arts. 23,

(1),

22,

Hence the commutative law (ii)

holds,

and

PR is called the sum of PQ and

QR.

In Fig. 17,

therefore

and the NOTE.

associative law holds.

In Fig. 16,

placements and

PQ + PS = PR

vectors are

a fact which is expressed by saying that t added by the parallelogram law.

dis-

24. Zero and Negative Displacements and Vectors. If after two or more displacements the moving point returns to its initial position, we say that the resulting displacement is zero. Thus we write

This equation the meaning

of

is

also written in the

form

-PQ = QP,

which

defines

- PQ.

These equations are also taken as defining the meaning of zero and negative vectors.

DISTRIBUTIVE

LAW

73

For displacements and vectors the meaning of

25. Subtraction.

PQ-QR is defined by PQ-QR - PQ + (-QR) = PQ + RQThus OP-OQ -OP + QO = QO + OP = QP.

multiply a displacement or a vector PQ by a positive

Number. To number k is to multiply

The

resulting displacement

26. Definition of Multiplication by a Real its

length or vector

by is

A,

its

direction unaltered.

denoted by JcPQ or by

we

Further,

leaving

define

PQ

.

k.

- k)PQ by the equation

(

In particular, (

So that

to

multiply a vector by

27. The Distributive number then

(

1) is to

Law.

turn

We

it

through two right angles.

shall prove

that if

k

is

a real

,

Let k be positive. Along PQ set off PQ' = kPQ. in R'. By similar triangles, QR to meet

(i)

to

Draw

Q'R' parallel

PR

and

and

Q'R'

= kQR\

kPQ + kQR^PQ'

Hence the theorem holds

for

positive

-k),

we have

numbers. (ii)

For a negative number

and

(

IG *

(

hence the theorem holds for negative numbers.

Thus

the distributive

vectors by real

NOTE.

law holds for the multiplication of displacements and

numbers.

The diagram of

Fig. 18

is

drawn

for a value of k greater

than unity

student should see that the same result follows from a diagram in which k unity.

is less

:

the

than

VECTORS AND COMPLEX NUMBERS

74

Complex Numbers represented by Vectors.

28.

It will

now

be seen that, so far as addition, subtraction and multiplication by real numbers are concerned, complex numbers are subject to the same laws as the vectors which represent them. This fact is fundamental in theory and

very useful in practice. It should be noticed that

if

a

|

|

Theorem

1.

If

C

divides

z is represented by a vector AB, the angle which the directed line AB

number

AB

and am z then z is the length makes with the directed line Ox.

is

AB in

the ratio

n

:

m and

is

any point, then

(m + n)OC - mOA + nOB. For

mOC^mOA+mAC,

Also

mAC = nCB = - nBC

whence the

Theorem in the ratio

result follows

2.

If

n:m,

by

;

O FIG. 19.

addition.

z z point which divides the straight line joining v 2 then the corresponding numbers are connected by the

z is the

relation

This follows from Theorem

1, in

accordance with the principle stated in

this article.

In particular, if z

mid-point of

z^ then

z

= %(z l + z 2 ).

that // a, b are complex numbers, prove geometrically

Ex.

1.

Let

A B

Bisect

is the

9

be the points which represent C, then

a, 6.

AB at

OA+OB=20C and Therefore

OA-OB^'BA^Z'CA. a +6

200"and2CJ; Now, since C

and a -b are represented by

hence

is

mid-point of base

A B,

SYMBOLS OF OPERATION Ex.

2.

//

OA, OB, OC

are connected by the relation

= 0,

?

Men A, B, C [This

a -f b

where

-f

c

= 0,

are collinear.

the converse of

is

75

(a

Theorem

We

of this article.

1

+ c)(W==a.OA+c.OC'; hence

The Symbol

a

have

AB^c

.

.

BC.]

as an Operator.*

Along two straight lines at consecutively, equal lengths OP, OQ, OP', OQ' in the

29.

i

right angles set off, positive direction of rotation.

Let the symbol

i

applied

through a

it

of turning

operation

a vector denote the

to

angle

right

in

the positive direction of rotation.

To bring our language into conformity with that of algebra, we say that to multiply a vector by i is to turn it through a right angle in the

FIG. 21.

positive sense.

Thus

in Fig. 21

OQ=i OP

,

OP' =

Therefore

where i*OP

Thus

i

2

an abbreviation

is

1(1

OP) -

1

2

OP,

Hence

for i(iOP).

t

and -1 denote the same operation, and OQ' =

OQ' = iOP' = i(-l)OP and

Again,

Either of these results to

OP' = i OQ.

and

multiply a vector by (-

is i)

turn

it

OP -

(

- 1)OP.

in this sense

-OQ =

written in the form

is to

2

(

1)

(

we

write

-l)iOP.

OP^OQ',

so that

through a right angle in the negative

sense.

Again,

Ex. z, z'

1

.

if i

//a,

3 .

i

2

OP

is

taken to mean

t

3

2

(i

OP),

it is

obvious that

b are complex numbers, find numbers z, z' and a, 6 may be opposite

so that the points

corners of a square.

Let

c

be the mid-point of a6, then

Oc + icb

Similarly, *

2'=!- (a

;

+ &)+* (a- 6).

For the moment, the reader should forget

FIG. 22. his conception of

I

as denoting a number.

B.C.A.

PRODUCT OF COMPLEX NUMBERS

76

30. The Operator cos0-MSin0. Draw two equal straight lines OP, OP' inclined at an angle 0. Draw P'N perpendicular to OP. Along NP set off NQ equal to NP'. Then

OF = ON + NP' = ON + iNQ. ON = ^ n OP - cos

Also

OF - cos

/.

.

OP +

*

.

OP,

sin

OP,

.

which we write in the form

OF-(cos0+ism0)OP, and we say that

to

multiply a vector by cos

+

i

sin

is to

turn

it

through

the angle 0.

31. Multiplication

Number. say that

to

and Division of a Vector by a Complex

In accordance with Arts. 26, 29 and 30 of this chapter, multiply a vector OP by the complex number r (cos -h i sin 0)

we is

length by r and turn the resulting vector through the angle 0. the stretching factor and cos -ft sin the turning factor. These are independent of each other, and the order in which they are

to

multiply

Here

applied If z,

OQ

we

its

r is

indifferent.

is is

the vector obtained by multiplying

OP

by the complex number

write

OQ^zOP

arid

OQ/OP^zi

also say that the ratio of OQ to OP is the number z. Division is the inverse of multiplication, so that if OQ = zOP then the result of dividing OQ by z.

we

y

is

Therefore

to divide

a vector

OQ

resulting vector through the angle

The

result

is

by

(

z is to divide its length

by r and tnr^ the

0).

the same as that obtained

32. Product of

OP

by multiplying

OQ by

I/z.

Complex Numbers. OQ^z'OP and OR^zOQ,

Let then we write

where

zz'

applied to

OP

denotes that the operators

in succession, in this order.

z',

z are to

be applied

AEGAND DIAGRAMS

77

Since the stretching and turning factors may be applied in any order, OP may be transformed into OR by multiplying its length by rr' and turning the resulting vector through the angle (6 + 6').

Hence the operations denoted by and rr' {cos z'z zz',

(0

+ 0') +

i

sin (0 + 0')}

are equivalent.

Again,

if

we take

(cos 0-h

i

sin 0) n to

+1

(cos is

to be applied

n times, the

result

is

mean that

the operation

sin 0)

the same as that given by the operator

cos n0

-f i

sin 0) n

= cos n0 +

sin n0.

In this sense then (cos

+

i

i

sin w#.

It will be seen that complex numbers used as operators on vectors conform to the

laws of algebra.

EXERCISE IX

ARGAND DIAGRAMS VECTORS :

1. If z = 3 + 2i, z' 1 -f i, mark the points z, z' in an Argand diagram, and by geometrical construction, the points representing

Z

+ Z',

Z-Z',

ZZ\

find

Z/z'.

be complex numbers of which a, b are constant and z varies. If Z is given in terms of z by one of the following equations, it is required to find the point Z corresponding to a given point z. Explain the constructions indicated in the diagrams, 01 being the unit of length. 2.

Let

z,

a, b

FIG. 24.

(ii) (iii)

Z=

(iv)

Z=tz where

Z~az+b.

t is

real,

CENTROIDS

78

find tho point Z corresponding to a given point z t the point Z is on the y-axis.

Z=(l + z)/(l-z),

3. If

and show that

|

|

= 1,

numbers and

are given complex

b

4. If a,

z

if

J^ 6^,

find the point z corresponding to

value

of

-oo to + 00,

z

passes through a, b, to the values from

[Along the

from which

line

the segment ah corresponding to 1 of t.

ab set off the length az = t.ab,

line

Oz=0a+az~0a + t

then

varies

i

the entire

describes

real

any given

and prove that as

t,

.

ab;

:.

z=a + (b-a)t,

etc.]

5. If z=a(l + it) where t is a real number, prove that as tho line through the point a perpendicular to Oa.

6. If

c,

a are given numbers, a being

real,

and,

find the point z corresponding to a given value of cf>

show that as t varies from with centre c and radius a.

is real,

t

once the

circle

7. If

A, B, C,

that this theorem

8. If

G

is

is

is

.

any point, then (m l +

m m w at A A A& = m m OA + + ...)OG + msOA + maOA

29

z

3

. . .

all

9. If z is the centroid of particles of

mass

of the

Any

three coplanar

Moreover,

2,

m w m

s , ...

and non-parallel

pOA +qOB + rOC = Q,

OA,

vectors

where

p

9

:

we may suppose acting at A l9 A% ... .

at z l9 z 29 z a , . . .

OB OC 9

...

,

then

.

are connected by a

q 9 r are real numbers.

p:q:r=& OBC A OCA A OAB :

...

3 4- ...

sign,

m^ + m 2z 2 + w 3z3 +

...)s

relation of the form

same

m m 2,

29

2

l9

l9

19

l

1

are not

3 , ...

z,

19

to be the centre of parallel forces, proportional to

10.

the point z describes

,

= tan J<^.]

t

.

the centroid of particles of mass

l9

G

+ oo

[Put

an immediate consequence of the identity

m To include cases where m m O

GO to

D are any four points in a plane, then AD BC^BD AC + CD AB. .

and

;

z=c + a(l + it) I (I - it),

(ii) if

Show

varies z describes

z = c+a(cos< + isin<),

(i) if

where

t

9

where the signs of the area# are determined by the usual rule. Also the points A, B C are collinear if p + q + and conversely.

rQ

9

[For

and

p

9

q,

let

G

p:q:r= &OBC AOCA :

:

9

&OAB,

be the centre of parallel forces, acting at A, B, r then G coincides with O, and pOA

let 9

(7,

and proportional to

.

VECTORS AND COMPLEX NUMBERS 11. (i) //a, of the form

j8

are non-parallel vectors

involves the two equations

If a,

(ii)

,

p=p'

andp,

q,

79

p' 9 q' are real numbers, adequation

q=q'.

9

y are non-parallel coplanar vectors connected by the equations = Q and p'a + q'fi r'y = 0, p
where p

9

q, r,

9

is

[These theorems follow at once from Ex. 10.] 12.

three complex

Any

numbers

z l9 z 2 , z 3

are connected

by a

relation of the

form where p,

q, r

are real numbers.

p:q:r~ A0z 2z 3

Moreover,

:

the signs of the areas being determined as usual. Also the points z l9 z 2 z a are collinear if p + q + r Prove this algebraically, and deduce Ex. 11.

Q,

,

[If z l

=x + iy l

whence the

l9

z2

=z 2 + iyl9

zs

=z 3 +

B C are collinear, and is any point, OA EC + OB CA + OC AB^O. Theorem 1, m n m + n=CB AC AB.] A,

.

(i)

:

Zj, z 2 , z a

then

9

.

[In Art. 28,

Let

we have

results follow immediately.]

13. If the points

14.

it/ 3 ,

and conversely,

.

:

:

:

be complex numbers, no two of which are equal, then

If the points z l9 z 2 z 3 are collinear, ,

l

\

2|2~2l

Z 2~*3\

!

*8hl-2l|

=-

Also, if the point z l lies between z 2 and z 3 , the ambiguous signs are both minus. (ii) If the above equation holds, then either z l9 z 2 , z 3 are collinear, or else O is the centre of a circle which touches the sides of the triangle Z 1 z 8z 3 . 15. If A i^4 2^4 a is an equilateral triangle, the vertices occurring in the positive direction of rotation, prove that s

where

~

4- 1

sin

an imaginary cube root of unity. numbers corresponding to A l9 3 X -f wz 2 + cj*z 3 = 0, and consequently 2 2 z^ 4- z a + z 3 - z^ - Z& - Z& = 0. o> is

Also, if z l9 z 2 , z 3 are the

16. If

numbers

A X YA'X'Y' a, a' 9

is a regular hexagon and then the numbers represented by

where 6 has the values

~,

3

~ 3

-

.

A

X

9

9

A t9 A

B9

prove that

A' represent given complex X' Y Y' are given by 9

9

TRANSFORMATIONS

80 The

17.

triads of points

A B C and X, 9

9

triangles if the corresponding by the relation

x(b

Y,

Z are the vertices of directly

complex numbers

- c) +

y(c

[The triangles are directly similar

and

a, b, c

x, y, z are

similar

connected

- a) + z(a - b) =0.

if

= AC XZ

i.e*.

,

if

-

c-a

=.] z-x

18. If ABC is a triangle and triangles BCX, CA Y, ABZ are drawn on BC, CA, AB, directly similar to one another, the centroids of XYZ &ndABC coincide.

x

[By 1 J Ex.

c

17. i

b-c

-

a

= y--- = z br. Hence show that# + yv + z=a + & + c.lJ c-a a-b

drawn on the sides of a given triangle ABC, all inwards. Prove that their centroids form an equilateral triangle. be the centroids of the triangles drawn outwards on BC, CA, AB.

19. Equilateral triangles are

outwards or

all

[Let P, Q,

R

Prove that

_ _

I

QA = CA

(cos

30 + L sin 30),

__ __ A R =AB

i .

\/O

30 -

(cos

L

QR \CB + ^ (CA -AB), and that RP has show that EQ = EP (cos 60 + sin 60), and use Ex. 15.]

Hence show that

Hence

sin 30).

*Jd

a similar value.

L

'

TRANSFORMATIONS

'

20. If Z, z are connected by any of the relations in Ex. 2 and given curve s, then Z will describe a curve S. Explain the following, where a, b are given complex numbers. = z -f a, S can be obtained from 8 by a translation. (i) If Z

if z

describes a

Z=tz where t is real, the curves s, S are similar and similarly situated, the centre of similitude. In this case we say that S is a magnification being (ii)

of

If

s. (iii)

If

Z = (cos a-f

S

tsin a) z,

can be obtained from s by a rotation about

through L a. (iv) If

Zaz

-f 6,

S

can be obtained from s by a

Z = l/z, S

is

the reflection in

a magnification and a

rotation,

translation. (v) If

OX

of the inverse of

s,

O

being the centre

of inversion. 21.

Show

22.

that each of the substitutions in Ex. 20 converts a circle c into a

except that in

circle C,

Show

(v), if c

passes through O, then

[Z L or

a

circle into

= ~> H--a

is

a straight

line.

that the substitution

a'z

converts

C

a

more of those

?

a

circle or, in

-; az + br?

in Ex. 20.]

9

an

+ b'

exceptional case, into a straight line.

therefore the transformation

is

to one equivalent ^

CHAPTER

VI

THEORY OF EQUATIONS 1

Roots of Equations.

.

type f(x) =0

of the

an equation

Under this heading we consider equations a polynomial, and equation will mean '

(

where /(x)

is

of this kind.

The general equation forms

(1)

nth degree

of the

be written in one of the

will

:

xn ~ l + pfln - 2 + ~ a xn + c^z"- 1 + a2xn 2 + M (yt 1 \ ~ aQxn + na^- 1 + a zx n 2 +

xn

+p

l

-~

or,

where binomial briefly in the

= 0, = + an

... -f j0 n . . .

;

+ a n = 0,

-

coefficients are introduced.

. . .

The

last

equation

is

written

form (a0>

a 1? a 2

,

...

a n $x,

l)

n

= 0.

For the present, we assume that every equation has one root. This is the fundamental theorem of the Theory of Equations, and will be proved in another volume. (1) It follows that every equation of Ike n-th degree has exactly n roots. ~ For let f(x) = xn +p l x n l + ...+p n and let a be a root of/(x)=0. By the Remainder theorem, f(x) is divisible by x a; we may therefore assume that

Let

may

/J

be a root of

=0

(x)

;

as before,

(x)

is

divisible

by x /J, and we

assume that

Proceeding in this way,

we can show

where there are n linear factors on the A, and no others. a, /?, y, ... ,

that

right.

Hence f(x) =

has n roots

ROOTS AND COEFFICIENTS

82

Imaginary Boots. Let the coefficients of f(x) be real then, is a - t/J (Ch. V, 14). Therefore/(cr) is divisible by

(2) is

;

if

a -Mj8

a root, so

that

is,

2 by (x-a) +

2 .

Thus a polynomial in x with

can be resolved into factors

real coefficients

which are linear or quadratic functions of x with real coefficients. Multiple Roots.

(3)

by

re

-a, a

When

is

r

If

f(x)^(x-a)

.

where

(f>(x)

(x)

not divisible

is

an r-nndtiple

called

we say that a

2,

r

root otf(x)=^Q. a double root.

is

Relations connecting the Roots and Coefficients of an Equation. .

Theorem.

then the

//a 1? a 2

...

,

a n are

,

the roots of the equation

sums of the products o/a t a 2 a 3 ,

,

...

,

taken one, two, three,


,

...

,

n

at a time, are respectively equal to

Pv

-Pi>

For

xn

+piX

n -l

n

+pzX -*+

-P*>

...

(~ 1 )>rr

>

+ p M ^(ff-a,)(x-a 2 )...(#--a,

- x - Zaj

xn

11

.

~l

l

)

+ 27a,a 2 x n ~ 2 + .

...

n

,

...

........... (A)

and, equating coefficients,

we have .a n = (-l) wy n ............. (B)

Conversely, if ot^ a 2

,

...

xn

an

satisfy the equations (B), ffoiy are the roots of

-\-p l x

n ~l

4-y 2 x

n ~a

+ ...

+7> n =^0.

For under these circumstances the identity (A) holds. It follows that the result of eliminating a 2 a 3 ... a n from equations (B) ,

is

a

-f

j^a;. 1.

a~ x -f // a,

j3,

...

y

+t) w

arc

Mo roote a + /?

Write the equation

3.

in the

3

o/ 2a; -f-

y,

form x3

+ x* - 2x - 1 =0, write dQwn ^y + ya + a/J, ajSy.

-f

2 -|# 4-

(

l).r 4-

Transformation of Equations. = 0, and suppose that we require

of /(z)

),

...

,

= 0.

where

(x)

is

(

-|)

=0

Let a,

the valuer of

;

j8,

y,

...

be the roots

the equation whose roots are a given function of x.

TRANSFORMATION OF EQUATIONS Let y =

and suppose that from

(x)

83

we can

this equation

single-valued function of y, which we denote by ~ (y). Transforming the equation /(a?) =0 by the Substitution

find

a;

as a

l

obtain /{^"^(yJJ

A -a

O,

case in which x

Ex.

1

which

1.

//

a,

,

9

j-

*

Let

^

~p

1

y~ l---x

y

)9,

x^-x-lQ,

are the roots of

Hence

.

ivrite

down

-

then x

,

-

and,

;

(y) 9

we

#3 - x -

if

1

is

given in Ex.

2.

find the equation wJiose roots are

the value of 27(1

y

-

l

the equation required.

is

not a single-valued function of y

is

x=$~

0,

y H- 1

-f

a)/(l -a).

then y

is

given by

y-l

/y-l\

UTIJ which This

is

is

equivalent to

Hence

the required equation.

'

also

i+v + i+P.^+Y^-l 1-a 1-J8 1-y Ex.

2.

roots are

If a,

a3

3 ,

jS

,

y

j9,

y

fere

Me

2

.

Let ,v=rc 3 then a;-?/3 where ;/3 denotes a?^y cube root of therefore obtained by rationalising ,

is

)-y+ p for the result will be the

equation,

let

y+p*

l,

pM* ~m,

Hence the required equation

is

the

same s

V

l

+ p^

y

4 p,

=

The required equation

y.

;

same whichever cube root y* stands

-

which

equation whose

off(x)x? 4-piX +p zx+p3~Q, find the,

roofs

3

Z

n,

pdj* 3

f w*

for.

To

rationalise the

then Z-f-m + n = 0, and therefore

+ n8

is

as

+ (Pi3 - 3piP2 + 3p3 y2 + (p 23 - 3p!p2 p3 4- 3p 3 a y + p a 3 - 0. )

)

4. Special Cases, The following transformations are often required. Let a, jS, y, be the roots of f(x) = 0, then . . .

(1)

the equation whose roots are -a, ~j8, -y,

(2)

the equation whose roots are I/a,

(3)

the equation whose roots are

This transformation (4)

We

is

called

The equation whose find f(x

+ h)

as in

*

diminishing the roots

1//J,

1/y,

is/( -x)==0

...

...

is

kot, A*j8,.Ary, ... is

/(1/rr)

multiplying the roots off(x)

a - A,

ft

- h,

y

-A

is

/(x

-f

Q by A)

6y

A.'

k.'

= 0.

Ch. Ill, 4, and the transtormation

off(x) =0

;

f(x/k)=0.

*

roots are

=0

;

is

called

SPECIAL CASES OF TRANSFORMATION

84 (5)

The second term of f(x)^a&n +ai$n "~l + aiXn -* +

removed by diminishing the roots by

For

f(x

-f

...

+a n =

can be

ajna^.

= a^(x + h) n + a t (x + A) - 1 -f h) 11

~ so that the coefficient of xn l

iuf(x + h)

is

na

. .

.

,

A+a1? and

this is zero

when

To transform /(#)=0 into an equation in which the coefficient of the term is 1 and the other coefficients are the least possible integers proceed first as follows (6)

,

:

Ex.

Consider the equation of

1.

Write this Putting x

ar>

-

fz

and multiplying by &

yjk

2

3

+

& which

least value of

will

-!*

make every

3 2 y - 9y f 90y - 168

Transform x* 6

2.

^ =0. =0.

coefficient

an integer

2

+ 5z + 12 =

= into

i_2)

1

lacking the second term. diminish the roots by 6/3 = 2.

The reckoning on the the resulting equation o:

which

is

jr.

3.

3

-7a;

right

"J

shows that

// a,

)8,

2/S + 3,

y

is

\

+ 6^0,

are

x

4-5

-8

-6

ITJ

173

_^_

^_2

-4

T2 +2

ITy

+12 g

1

Me

roots of 2x*

+ 3x* - a: - 1 -0,

find the equation whose roots

2y-i-3.

The equation whose 3.3

-6 +2 -

the one required.

are2a + 3,

so that the

y = 12a;.

where

an equation

Wo

is 12,

is

required equation

j&a:.

7=0.

-

the equation becomes

,

f-\W +^y The

fa:

-

2

roots are 2a,

x

2.~'+3--j-g-l=0,

or

2j3,

2y

is

1

- (

3) )

^ + 3^-2^-4=0.

1

3 - 2 - 4 - 3 + + 6

-h

1+0-2 -3

Increasing the roots of this by 3 (as in the margin, where divide successively by x + 3), the required equation is

"j

we

_ 3

+ +

+ 2

9 7

^

are connected 6y i^e relation ft~ we can J?*. 4. // two roots, a and j5 of f(x) ^S(a) generally find them as follows: The equations /(se)=0 and /{$()} =0 have a common root, namely a. Therefore x - a is a common factor of f(x) and

and may bo found by the

H.C.F. process.

If however f(x) and/{0(a:)} are identical, then the method

fails.

CUBIC AND BIQUADRATIC Ex. o.sfiolve 2x* + a;8 - Ix - 6 =0, given

Let the roots

be

a + 3,

a,

that the difference between

+ + +

2

Find the

j8.

equation whose roots are a -3, a, j8~3. The reckoning on the right shows that this equation

is

a

common

~ 6

6

+ 21

+ 42

+53

^

is

3

wsa^ + 36z

2

# + 1, therefore

and the third remaining root

;

We

The Cubic Equation.

""^"19

this proves to be

Again, the product of the three roots

If

7

+ 13

Find the H.C.F. of these expressions; and hence - 1 and 2 are roots.

5.

is 3.

~

?

factor of

and 2a^ + & 2 - 7* - 6.

2^ + 19z 2 + 53z + 36

two of ike roots

1

-

.

Hence, x - a

86

is

= -1 -

;

3/2.

take as the standard form

+ 3cz + d!:==0 ........................... (A)

x = y + h, this becomes

B=

where If

h=

-b/a, then

JB

= 0, and

^+ H = ac-b

where Or,

if

we

write

z

]8,

^

,

= ay = ax + b,

y are the roots

+ 03

.................................

the equation

+ 3ffz +
+ 6, ay +

a]8

is

a + 6/a,

We

The function u

is

called a quartic.

2

If

6#

.

+ 6/a, y + 6/a, and

take as the standard form

w = ax + 46x + 6ca; + 4da; + e = 3

/J

6.

The Biquadratic Equation. 4

(B)

.................................. (C)

of (A), those of (B) are

the roots of (C) are aa + 6, 6.

= 0,

is

G - a*d - 3abc + 2R

2

z3

If a,

the equation

x^y-bja

....................... (A)

the equation becomes

K

40

(B)

where

T

ac - i2

6

1

,

= a2d - 3a6c + 26s 3

If z

= ay = ax + 6,

the equation s4

If

(as for the cubic),

K = a e - 4a 6d + 6a6 c - 36

and


2

2

4 .

is

+ 6#z2 + 43 + #=0 ............................. (C)

y, 8 are the roots of (A), those of (6) are

of (C) are aa+*6, etc,

a + 6/a,

etc.,

and those

EQUATIONS WITH CONNECTED ROOTS

86

EXERCISE X

TRANSFORMATION OF EQUATIONS 1. If a, jS, y are the roots of 2ar* 4- 3# 2 - x- 1=0, find the equations whose roots are (i) l/2a, 1/2)3, l/2y ; (ii) a- 1, 0- 1, y- 1 ; (iii) !/(!-), l/(l-0), 2 l/(l-y); (iv)a + 2,0 + 2, y + 2; (v) a , [(iii) should be deduced from (ii).]

2.

If a,

roots are

[For 3.

,

y\

2 y are the roots of 80^ 4# -f 6# -1=0, find the equations whose + i, y + i; (ii) 2a-f 1, 20 + 1, 2y + l.

)8,

(i)

(ii)

2

a + i,

use the result of

(i).]

x*-2x*-3x 2 + 4x-~l=:Q, given

Solve

unity. 1 [If a, a"" are these roots,

(a;

- a)(x - or 1 )

x4 - 2x* - 3x* + 4x -

and

1

that the product of two roots

is

a

x*

common

factor of

- 4x 3 + 3z 2 + 2x -

1.]

Solve 4s4 - 4x 3 - 13# 2 + 9# + 9 =0, given that the sum of two roots - a are these roots, (x ~-oc)(x + a) is a common factor of [If a, 4.

4x*-4x*-13x* + 9x + 9 5.

is

zero.

4* 4 + 4a?- 13z 2 - 9s + 9.]

and

Solve 4# 3 -24o; 2 -f 23^+18=0, given that the roots are in arithmetical

progression. [Let the roots be a 6.

is

a+

S, a,

8,

.'.

3a =^.]

Solve 6x - Ilx 2 -3x4-2=0, given that the roots are 3

gressions

[Solve 2#

- 3t/ 2 -

3

lit/

+ 6 = 0, whose

roots are in A.P.]

- 8# 3 -f 14# 2 ^ 80; - 15 =0, whose roots are 7. Solve [Let the roots be -38, a 8, a-f 8, a-f-38.] a;

in harmonical pro-

4

in A. P.

Solve 2# 4 - 15# 8 + 35o: 2 - 3Qx + 8=0, whose roots are in G.P. 3 3 1 [Let the roots be a/>~ a/)" , ap, a/? .] 8.

,

9. Solve

a?

3

-7x 2 + 36=:0,

given that the difference between two of the roots

is 5.

10. Solve 4x 4 -4a? 3 -25aj 2 the roots is unity.

+ a:-h6=0, given that the

difference

11.

Transform into equations lacking the second term

12.

Transform into equations with integral

(u)

13, If

n

Un

is

coefficients

between two of

:

:

^-

odd, prove that

n0 = (-l)Kn-i)tan0tan(V-Wn^

\n/

\n/ V

[Regard the equation of Exercise VII,

14, as giving

n/

\

n

)

tan 6 in terms of tan n$.]

GENERAL THEOREM

87

Character and Position of the Roots of an Equation. We say that the character of the roots is known when we know how many are real and how many are imaginary. 7.

The position (jf a real root is its position on the scale of real numbers, and is determined roughly for a non-integral root by finding two conFor a complete discussion, secutive integers between which the root lies. we require Sturm's Theorem (Ch. XXVIII), but a good deal of information can be derived from the elementary theorems which follow.

Some

8.

general Theorems.

= n f(x) X +p where

p l9 p 2 + im is a ,

...

Xn 1

~l

Here we suppose that

+ p2Xn

~~ 2

+ ...

-f

pn

,

are real numbers.

root of f(x) = 0, then l-im is also a root (Ch. V, 14) and 2 - = - 2 has the f(x) quadratic factor, {x (1 4- im)} {x (I iw)} (x 1) + m If

l

.

The last expression is positive for all real values of x. are the real roots of f(x) = 0, these being not necessarily f(x)

Thus

if

a,

/?,

. . .

K

all different,

= (x-*)(x-p)...(x-K).(x),

where (x) is positive for all real values of x. factors is unique.

Also this expression in

Hence, if x varies, the sign of f(x) can change only when x passes through a real root of /(x) = 0, and we draw the following conclusions. (1)

case

If x

when

is greater

than any of the roots, f(x)

all the roots

Whence

is positive.

This

is also the

are imaginary.

pllows^tbat for sufficiently large values of x, any polynomial

it f

s its highest term. (2)

Let a and b be any

real

numbers

then

:

///(a) andf(b) have like signs, an even number of roots o//(z) between a and b, or else there is no root between a and b. (i)

(ii)

lie

=

///(a) and f(b) have unlike signs, an odd number of roots off(x)

between a

and

=

lie

=Q

b.

an equation of odd degree, it has at least one real root. (ii) Iff(x)~Q is of even degree and p n is negative, the equation has at least one positive root and at least one negative root. For, by (1), a positive number a can be found so large that /(a) is of ( - l) n positive and/( -a) has the (3) (i)

///(#)

is

sign

Hence lies

n

.

odd, /(a) and/( -a) have unlike signs and at least one root between a and - a. If n is even- and p n negative, /(a) and /( - a) if

is

are positive and /(O) is negative, so that at and a and at least one between and -a.

least

one root

lies

between

NUMBER OF REAL ROOTS

88

Descartes' Rule of Signs. In what follows, every equation is supposed to have a term independent of x, so that zero roots do not occur. In considering the signs of the terms of a polynomial taken in order from 9.

we say

that a continuation, or a change, of sign occurs at any particular term according as that term has the same sign as the preceding, or the opposite sign.

left

to right,

Thus x1 - 2x? - 3xt - 4X3 + 5x2 - 6x + 7 has 2 continuations, and 4 changes of sign, the continuations occurring at the terms -Sz4 -4Z3 and the ,

,

-2Z6 +5z2 -6z, 4-7. The equation a xn + a l xn " 1 ^-a^c n ~ 2 +...-{ a n =0 is said to be complete when no coefficient is zero. If ar = we say that the corresponding term changes at

is

,

,

missing.

In a complete equation : (i) Ifx is changed into -x,a change of sign becomes a continuation and vice versa. If

(ii)

p

number of changes, and p!

is the

of sign, then

p -f p,' = n,

Descartes' Rule

is

where n

the

number of continuations,

is the degree of the equation.

as follows

The equation f(x) = cannot have more more negative roots than f( x)

:

positive roots thanf(x) has changed of sign, or

has changes of sign.

To prove the first part, we shall show that if u is any polynomial and v~u(x-a), where a is positive, then v, when expanded, has at least one more change of sign than u. First suppose that no term is missing in

instance

:

+

Signs of terms of u,

---

Signs of terms of v,

where

indicates that the sign

is

+ + - + + + 4---- + -+ -

+-+-

term

u and consider the following

may

be

-f

or

- or that the corresponding ,

zero.

In the diagram of corresponding signs, observe that (i) If the rth sign of u is a continuation, the rth sign of v (ii)

is

ambiguous.

Unlike signs precede and follow a single ambiguity or a group of

ambiguities. (iii)

On

A

change of sign

account of

(i)

is

and

introduced at the end of

(ii),

v has at least as

even in the most unfavourable case in which tinuations

:

sign than u.

and on account

of

(iii)

v.

changes of sign as u, the ambiguities are con-

many all

v has certainly one more change of

NUMBER OF IMAGINARY ROOTS

89

That no changes of sign are lost on account of any terms which may be missing from u appears on considering such instances as

+00-

+

-+00+-00 -+-0-

--00+ -

-

+

+

Thus v has at least one more change Next let f(x) = (x) (x - a) (x - j8) .

of

.

of sign . .

than

where

a,

/(z)=0.

u.

are the positive roots

j8, ...

If (x) is multiplied in succession

+

+

by

a?

-a, x-fl,

multiplication introduces at least one change of sign. Hence /(x) has at least as many changes of sign as/(x)

=

...

,

each

has positive

roots.

Again, the negative roots of/(x) = are the positive roots of/(-x)=0, with their signs changed. Hence the second part of the theorem follows

from the

first.

Corollaries.

10.

Let n be the degree of/(x), and

number of changes of sign in/(x), the number of changes of sign in/( -x), m the number of positive roots of /(#)=0, m' the number of negative roots of f(x) =0 [JL

the

f

fji

then

(i)

roots

ifp,

+//
the equation /(#)

=0

has at

least

;

n - (^+/A') imaginary

; (ii)

if all the roots

For by Descartes'

o//(x)=0 are

rule,

m^p, and

real, then ra=ju,

m'^p,'

'.

and m'

=///'.

Hence

m + m'^fi +fjL'^n,

(A)

and the number of imaginary roots = n - (m + m')^n - (p. +/u/). Again, if all the roots are real m + m' = n, therefore by (A) m + m

Now

and, if w^ , which m^fji Therefore w=/u, and m'=///. Ex.

;

1.

//

q, r, s,

is

f

=/z +fi'.

impossible.

are positive, show that the equation f(x)

-x* 4- qx 2 +rx - s =0

has one positive, one negative and two imaginary roots. By Art. 8, (3), /(a;)=0 has a positive and a negative root.

Also

/*

1

and // = !,

therefore these are the only real roots.

Ex*

2.

Here

Show

= 2, /u

real roots,

that x?

= ^' 1

;

-2#2 +7 =0 has therefore

at least

3. /Lt-|'ft-'~

two imaginary

roots.

Hence there cannot be more than three

namely, two positive and one negative.

Therefore the equation has at least two imaginary roots.

EFFECT OF MISSING TERMS

90

DeGua'sRule.

11.

If a group of r consecutive terms

if r is even, the equation has at least r

(i)

if r is odd, there are at least r

(ii)

+1

imaginary

roots

-

or at least r

1

is

missing from

:

imaginary

roots,

according as the terms which immediately precede and follow the group have J like or unlike signs. "

~ ~ terms between hxm and kx m r l are missing from/(x). m ~r - l + = m and fl (x) = t/j(x) + (f)(x), (x), f(x) hx + kx

Suppose that the Let

1 m ~ 2 4m ^ (x) = hx + c^x- + c2x

where

none

r

of the set, c l3 c 2 ,

...

cr ,

,

The number of changes in 0(-z) is r + 1.

= number = number /*'

Let

/x

...

c rx

m ~ r + kx m ~ r - 1

,

being zero.

of sign in

ifj

(x)

of changes of sign in

+ the number ~ r ~l

hxm + kxm

of changes of sign in h(

~x)

,

m+

k( -x)

(-x)

m and

m

"~ r

~l .

and /( - x)

Thus the total number the number in/^x) and/x ( -x) by r + 1 -(/u,-f/x'). Hence f(x) = has at least r + 1 - (/x + /z') imaginary of changes of sign in f(x)

First suppose that r is even, then

of changes of sign

is less

than

roots.

m ~ r ~ have opposite l

(-~x)

and

signs,

(i) if

h,

(ii) if h,

k have the same signs, k have unlike signs,

/x

=

and

is odd,

(i) if

h,

(ii) if

h,

(-x)

=1

;

= 1 and p =0, and /x

~1 m and have ( -#)-' t

If r

//

in both cases

the same, sign, and

= r + 1. k have the same sign, p = 0, /// = and r + 1 - (p, = 1 and r + 1 - (/x +/z') =r - 1. k have unlike signs, ft = 1, -\-JJL')

/it'

This completes the proof. .

1.

If

H = ac - 6 > 0, 2


(a,6,c,

/wo imaginary

For by the substitution

by De Gua's

12.

w ...%,l) -0

roots.

x~y~ 6/a,

-^

the equation becomes #n

rule this equation has at least

is

said about the roots of

Also, the equation

is

;

an equation

.-

a2

two imaginary roots

Limits to the Roots of an Equation.

whatever

+

if

(1)

xn ~ 2 -f

...

=0, and

H > 0. In this

article,

refers to the real roots only.

supposed to be written with

its first

term

positive.

UPPER AND LOWER LIMITS

91

In searching for the roots of an equation, it is advisable to begin by finding two numbers between which the real roots lie. If k, I are two such numbers and A>Z, then h is called an upper limit and I a lower limit to the roots.

Upper Limits. Any number h is an upper limit to the roots of f(x) = 0, provided that/(z)>0 when-x^h. Method of grouping terms. The process consists in arranging the terms (2)

=

of the equation /(x)

we can

in groups, so that

find

by inspection a

number h such that the sum of the terms in each group is ^0 for x^h. To distribute the terms conveniently, it is often advisable to multiply j(x) by some positive integer. With a little ingenuity, this method can be made to yield quite close limits, as in the examples at the end of this article. Other methods are given in the next exercise, and in Ch. XXVIII.

Lower Limits.

(3)

If

(i)

an upper

is

li

limit to the roots of f(

0, then

x)

li

a lower

is

limit to the roots <>//(#)= 0.

For (

a

if

the least root of /(x)=0, then -a -a, and therefore -h'
is

Hence, h

-a;)=0.

f

the greatest root of

is

>

(ii) If h" is an upper limit to the positive roots 0//(l/x)-=0, then I/A" a lower limit to the positive roots off(x) 0.

For

if

a' is

the least positive root of

Ex.

Here

Qf(x) f(x)

>0

= 6#3 - 60s; 2 - 6frr - 600 ^

for

x^l& and

Again, putting x =

Now The

greater root of 16?/

Hence (y)>0

Ex.

We The

l/y,

15

10-6

is

for y"^

is

4x~(x

-

2

+ Hty -

2

~~

>

i-e-

(20*/ 1

15)

the greatest

+ x(x* - 66) + (x3 ~ 600).

an upper limit to the

the equation becomes

<(*/) =5*/

Thus

I/a' is

so l/A"
Find an upper and a lower limit to the positive roots of 3 2 /(z)=z - 10* - llx - 100=0.

1.

Thus

then

/()= 0,

Hence A">l/a' and

positive root of /(!/#) =0.

is

-0

-



(y)

roots.

= lOOy3 -f lly 2

-f-

IQy

-10.

- 1). 1) + (16r + 10y 1-4

^^

is

lo

for x = -

a lower limit to the positive

;

^v-p

and

if

ie. for

y has this value,

x

20t/

-

1

> 0.

10-6.

roots.

Find an upper and a lower limit to the roots of 8 f(x) =3x*~ 61s -f 127#* + 220a; 520 = 0. 2 2 have f(x) =z (3a; Qlx + 127) (220* 520).

2.

-I-

greater root of 3

2

-61o;4-127=0

is

17-9 ....

Thus 18

is

an upper limit to

the roots.

- z) = (3s4 - 520) 4- x(6lx* -f 127a: - 220), and each group 4 is a lower limit to the roots of /(a;)=0. Hence

Again, /( if

.0^4.

G

of terms

is

positive B.C.A.

RATIONAL ROOTS

92

To

13.

roots which

px~q

is

and

:

+ bx n ~ l +

p, q are integers

. . .

prime

. .

to

one another, then

.

p

is

a factor of

a,

q is a factor of k.

For, denoting the polynomial 0, and therefore

aq

Hence aq n =p x an to

such

Any

can be found by using the following theorem If 4- hx -f k, where a, 6, h, k are integers

exist

may

a factor of axn

and

or zero

Roots of an Equation.

find the Rational


therefore

divisible

Hence

by

p

n

by

~

+ bqn lp +

px-q

if

f(x),

...+ hqp n

~l

is

a factor of f(x), then

+ kp n = 0.

n integer, so that aq is divisible by p. n Hence a is divisible by p. prime to q

Now p

is

is

prime

Similarly k

.

is

q.

the rational roots of

axn + bxn

~l

+

. . .

+k=

must

be included

among

q/p, where p is prime to q, p is a factor of a and q a factor of k. We can test the values of q/p by synthetic division, or we may use the method of divisors given below. the values of

Ex.

We

Search for rational roots of f(x) = 2x 3 - 5x 2 + 5# - 3 = 0. have f(x) ~2x 2 (x --f) +5(.r - |), therefore % is an upper limit of the roots. 1

.

Also

has no positive roots, therefore f(x) has no negative roots. If qjp is a root, p \ or 2, q 1 or 3, and the only values of q/p which lie between and | are 1, i, -2. Testing these by synthetic division, we find that | is a root and

/(

-

x)

;

it is

the only rational root.

Newton's Method of Divisors.

14.

Let the given equation be

transformed into n n ~l + p 2x n f(x)=x + p l x

where p^ p 2

~2

+

...

+p n = Q,

..................

... are the least (See Art. 4, possible integers. the last theorem, if a is a rational root of (A), then a

By and p n

is

,

divisible

by

(6).) is

an integer

a.

Let h be any factor of p n by x h, the usual reckoning .

1+ft

I-/*)

(A)

-h

a factor off(x), and

If

x

is

as follows

is

we

divide f(x)

:

+y 2 +...+?V_ 2

+p n -l

+Pn

h 9l

Where

q^pt

+ h,

q

Newton's method consists in performing these operations in the reverse order, thus

Pn+Pn-l -gn-l

+Pn-2 +...+ft -92 -gn-2

+Pl " gl

+1 "1

NEWTON'S METHOD OF DIVISORS

93

this is added to divided by A, the quotient is -~q n -i giving ~q n _ 2 h. Dividing this by h, we get ~g n _ 2 which is added to

Here p n

is

p n -.i,

;

;p n

>

and

If any q in the process is not an integer, f(x) the reckoning need not be continued. Herein lies the advantage of Newton's method.

number

Further, the last

To

lessen the

Ex.

Find

1.

is

,

0.

4/(

a

1.

lie

a rational root, it is possible values of h are therefore 2,

1,

A number of these so that

1

h

4,

3,

-7,

6,

=2 3 .3.7.

-8

(A)

can be excluded at once, for

5 /(I)- -160- ~2

if

is

2 off(x)^x* -39z + 46z 168==0.

between - 9 and 7. h is a factor of 168 an integer and

is

The

x-h

if

2 - 30) + 2 (23* - 84), f(x) = x (x* 3 = -x) 2x*(x* 78) + (x* -672) + x (x - 184),

follows that all the real roots

Also

must be

.

and h

line

not divisible by x - h and

trials, choose any number a, then divisible by a -h. We generally take a

the rational roots

Since

If

bottom

in the

is

number of

factor of/(x),/(a)

it

_2

so on.

.

5

;

/(

- 1)-

-252- -2 2

32

.

.

7,

are not roots. is

a root,

1

-h

is

Again, -I -h must be a

a factor of 160, thus we can exclude - 2, factor of 252, and so we can exclude 4.

4,

-

~

6,

8.

The remaining

numbers are -7. +3, -3, +6, that 6 and - 7 are the only find Newton's we method, by 2,

Testing these as below rational roots.

1-3) -168

+46 -56 ~^To*

1-6) -168

4-46

^28 Tl8

-39 4-3 -36

4-0

_-6

-JL

-1J

~~6

EXERCISE XI

REAL ROOTS 1.

Use Descartes'

(i)

If q

(ii)

x*

is

positive,

- 3#4 + 2a?3 *

rule of signs to sc?

+ qx + r = Q

-1=0

show that

:

has only one real root,

has at least four imaginary roots.

The reckoning stops at these stages because 10 and 73 are not

divisible

by

4-1

3.

REAL ROOTS

94:

Show

2.

and by

that the roots of (a - x) (b

- x) - h* =

are real

and are separated by a

6.

[Denoting the left-hand side by/(#), we have

By grouping terms, Exx.

in

find

an upper and a lower

limit to the roots of the equations

3-9.

3. x*

x 5 - 10** - 5a?3 + 6* 2 - 11* - 350 =0. - IQx - 5) + (6z 2 - llx thus

- 3z3 - 2* 2 + Ix + 3 -0.

4.

[For an upper limit in Ex. 4, group x*(x* For a lower limit, write x = -y and group thus :

350).

:

- llx*

-9z- 50=0.

5.

x* + 4x3

7.

2x*~llx*-I()x-l=Q.

6.

x*-

[Group thus: x(2x*~ llx-ll) + (x-l).] z54-z 4

-6*3 -8o,' 2 --15a;-10:-0. - 2 - 15) + (z 4 - Sx 2 [Group thus x(x* 6z 8.

:

9.

a;

5

- 3z 4 - 24z 3 + 95z 2 - 46x - 101

= 0.

For the equation f(x) =xn +p 1 xn

10,

-l

greatest negative coefficient is equal to roots.

[Let

sol, then/(z)>0

This holds if xn

>p

if

xn

.

----,

i.e.

xn >p(xn if x -

1

10).]

+p

~l

p

2

xn

9

~2

+ ... +p n =0, if the p + 1 is an upper

then

+ xn -* +

...

+ I),

i.e.

if

numerically limit to the

xn

x n >p

1 .

.

x

I

> p.]

For the equation f(x) ~xn +p l x n l +piX n 2 + ... +p n = Q, if the numerically greatest negative coefficient is equal to -p and the first negative coefficient is p r then Zjp 4- 1 is an upper limit to the~ roots.- n n r X H T I + ... + + l), i.e. if l, then/(z)>0if x >p(x -

11.

,

This holds

if

x n >^

-r

,

i.e.

if

(x-l)x

r~ l

>p, which

holds

if

(*-

r

I)

>p,

i.e.

12. If the rule in Ex. 1 1 is applied to the equations in Exx. 3-7, show that - 51 - 1 ; (5) 7, - 6. the limits are (1) 4, - 4; (2) 351, - 5 (4) 6, (3) 9, ;

;

13.

By

1 100# 2 - 237=0. using Ex. 11, find an upper limit to the roots of x

Find the rational roots of the equations in Exx. 14-19. 14. * 3 -9z* + 22o;-24:=0. 15. a? 3 -5* 2 - 18*4-72=0. 16.

3* 3 - 2* 2 - 6*

4-

4=0.

17.

4# 4 i

18.

6s

4

-25^4-26^ + 4s-8=0.

19.

2

6a;H53^-95a; -25ar+42=0.

ELEMENTARY SYMMETRIC FUNCTIONS

95

Symmetric Functions of the Roots of an Equation.

15.

Let a,

y, 8, c,

/J,

then by Art.

2,

be the roots of ^ xn +p 1 xn 1 +p2x n - 2 +

...

27a=

-p

27aj9=]9 29

l9

... -f

pw = 0,

27a/?y

= - j>3

...................... (A)

etc .............. (B)

,

be shown that these equations can be used to express any symmetric function of the roots in terms of the coefficients. It will

Functions of the type ZoPfPy*

...

where

,

a, 6, c

...

are positive integers,

These can be calculated in order by a of and process any symmetric function of the roots can be multiplication, expressed in terms of them.

will be called elementary functions.

Ex. (i)

For equation (A) find

1.

The product

Sat,

27aj9yS consists of terms of the types

.

of these occurs once, in

product of any one of Hoc

a,

a88 =27a 2

.

Za 3

yS and S**ffy. a 2/?y8, a/?yf The first the product. The second occurs five times, namely as the c and a term from 27a/?y8. Therefore j5, y, 8, the values of

8 +527
.

and

2 2 (ii) Consider the product 27aj8 y, ZajSy. This consists of terms of the types a The first of these occurs once. The term a 2/?y8 occurs three times, a s)8y8, a/tySe. The term ajSySc namely, as each of the products (a/?)(ay8), (ay)(a/?8), (a8)(ay). occurs ten times, for we can select two out of the five, a, j, y, 8, c, in 10 way?. .

Therefore

^aft

Using the

NOTE. becomes

.

^y -

27a 2 j8y

and equations

last result

These results can

+ 3Za a]8y8 + lOZ^ySe ........................ (C)

(B),

we

find that

Z^Fr^PiPi-ptPa-Sps .................................. (D) be tested by putting a=]8 = y ...=1. In this way (C)

CJ C? = 3(7? + 12(72 + 10(7?. .

For reference, we give the following

results,

indicating briefly the

-" 1

"" 2

process of reckoning.

// a,

j3,

y

,

. . .

are ZAe roots of

- (27

2 .

aj8)

j8,

y,

...

+ p2;rn

+

. . .

+ y n == 0,

-

Here the elementary functions in a,

xn + fto;71

and fourth degrees and obviously the process can be

of the second, third

are calculated in order,

continued so as to find any function of this kind. It is important to notice that the degree n of the equation does not occur in the reckoning.

DIFFERENCES OF ROOT3

96

Symmetric Functions involving only the Differences

16.

of the Roots of f(x)=0. These are unaltered

the roots are diminished

if

by any number

they may be calculated by using the equation chosen so as to remove the second term.

A useful check is provided by the theorem in Ex. in

Exx. Ex.

2,

h.

Hence

f(y + h) = 0, where 1

below.

h

is

The functions

3 are important.

Let v be any symmetric function of the differences of the roots of

1.

(o c ,a 1 ,fl a ,...an $a:,l)

then if v is expressed in terms of a

,

a l9 a 2

...

,

n =0

the

f

sum

of the numerical coefficients is

zero.

The sum of the numerical coefficients is obtained by putting a^ a^ a 2 But in this case the given equation becomes (x + l) n every root of which since t; involves only the differences of the roots, in this case, v=0. ,

Ex.

2.

// a,

jS,

y

+ 3bx* + 3cx + d =0, find

are the roots of ax3

1 is

*

n

v

-

-1, and

the values of

have a (x - a) (x - jS) (x - y ) = ax 3 + 3bx- + 3cx + d. - 1 for x in this identity, we find that Substituting 4- 1 and (i)

We

a 2 (a 2 +

therefore

(ii)27(

Ex.

3.

j

1

2 )

(j8

+

1

+ 1 - (a - 3c) 2 + (36 - d) 2

2 )

(y

)

.

8-y)(y-a)-^y-Za2 -3 // a,

/3,

y, o are the roots of ................................ (A)

0,

find the values of f


(ii)r(a-j8)VS + a-/3-8)(a+/3-y-8).

-/3)

.

a)

(ii)

6

and

(iii)

its

The equation whose d,

roots are -

2

,

-i

8Za|8 ,

^ y

thus

-,

-^ is

fi 8

2

(6

-

ac).

obtained by interchanging o and

Denote the function by v. Since v is a function of the may be found by using the equation

differences of a,

j3,

e,

y, 8,

value

,

obtained by the substitution

xy-b/a.

+

4#

K

,

y+i-O,

................................... (B)

SYMMETRIC FUNCTIONS OF ROOTS Let

a',

',

97

Then since Za'=0, + a-8=jB' /-a'-8' = -2(8' -fa ), -0/8 = (8'ta')(S'+jB')(8' + y')

be the roots of (B).

y', 8'

7

and

- 8'3 + S' 2 (a' + j8' -f y') + 8' (j8

=

.'.

#*.

7/a + jS + y + 8 = 0, prove

4.

-

827a'j8'y'

y + yV +

a,

j3,

!=<), 5 2

-2^p 2 , 5 3

Hence by

tions.

tftaf

2

Also a 5

.

+p

.

= 0, and 4 2

2
let s r ~ZoL r then by Art. 15, +pi

addition,

The following are

y.

j8,

= -3p 3

& +p %* +P& +jp

Equations whose Roots are Symmetric Functions of

17. a,

y, 8 be the roots of

+ a'j8'y'

= 320/a*.

a 5 + )3 5 + y 5 + 8 5 - - SjScejB Let

'')

If a,

^J&f'l. 2

(j8-y)

(y-a)

,

y

/?,

typical examples.

+ qx + r=Q,

are JAe roote of x?

2

find the equation whose roots are

2

(a-jS)

,

.

Let 2~(j8 -y) 2 then since 2?a = ,

z

and a/?y = -r, = (jg + y) 2 - 4j9y =a 2 f 4r/ a

.

Hence the required equation can be found by eliminating a from a3 +ga + r=0,

By

subtraction,

a 3 -za-f4r=0.

(z -f q)oc

Substituting for

a in the

first

= 3r.

equation and simplifying, we have

which on expansion becomes -f

The -

Ex.

artifice

2.

(i)

employed

Find

in the next

the condition that the

4g

3

+ 27r 2 =0.

example

sum

is

often useful.

of two roots a,

xt+p^+p^+paX+p^Q may

ft

of

................................. (A)

be zero.

Use

(ii)

a, j8 are

the result to find the equation

any two

whose roots are the six values of |~(a-f /?), where

roots of

a^ + 46a? + 6cz2 + 4da; + e=0 ................................. (B) (i)

Since a and

-a

are roots of (A),

a4 + P&* + p 2a2 + P3a + Pi - 0, a4 4

.*.

Now a 9^0 which

is

unless

a

p4 =0,

a

-2>i
-fj 2 a

2

+^4=0

therefore

the required condition.

+ #2 2

2>1

2

~^3 +^4 =0 and a^a 2 ^ p a )=0. ;

P4~2>1#2JP3+2?3 2=: 0>

............................. (C)

ROOTS OF THE CUBIC EQUATION

98 Let

(ii)

If then

z

-\ (a + j8),

we have

then

we diminish the

- z) =0. - z) + (j8

(a

roots of (B)

by

z,

B=az + b, D-(a,

For equation

b,

x~y + z,

0,2*

I)

,

sum

of

two roots of

.............................. (D)

+ 2bz + c,

E = (a, b,

3

c,<%,

the

is

+ 4By* + GCy* + 4Dy + E=Q,

ay*

where

writing

This equation

the resulting equation in y will be zero.

c,

d,e$z, I)

4.

(D), the condition (C) gives

an equation of the sixth degree

in

2,

whose roots are the values of J(a

EXERCISE XII

SYMMETRIC FUNCTIONS OF ROOTS If a,

j3,

2 y are the roots of ax* + 36# 4- 3cx + d=0 and H = ac-b 2 G = a*d - 3a6c + 2b\ ,

prove that

:

a^a3 ^

1.

2.*

a^V 2

2

-3(a' d-9a&c-f9& 3

= 3 (^

3 ).

a - 9^6 -h 9c 3 ).

8 2 2 2 3 2 ^y + ya*)(ay + j8a -h yj8 ) = a*(3a ]8V = 9 (aW - 6a6aZ + 3ac 3 -f 36 8rf).

3.

a*(a^

4.

a 2 {(j8-y) 2 -f(y-a) 2 4-(a-j8) 2 }-18(6 2 -a^).

5.*

-f

a f {a a (j3-y) 8 + j8 8 (y-) 8 + y 2 (a-j8)} = 18(c 8 -c).

*Explain

how

to derive (2) from (1)

and

(5)

from

a {a(j3~

7.

a 4 {(j8-y

8.

a3 (^4-y

9.

a 3 (2a-~y)(2j3-y-a)(2y-a--/3)^ -27(9. (i) (ii) (iii)

[(i)

The condition that a, 0, y are in A.P. is ^=^0. The condition that they are in O.P. is oc 3 =6 3d. The condition that they are in H.P. is rf 2a - 3dcb + 2c3 =0.

If

2a = /M-y,

3

ax + d=0. 11.

(iii)

then

a=

-b/a.

(ii)

Eliminate x between

Put x~l/y and use condition

The condition that

aC 3 =5 3D, where 12.

(4).

2

6.

10.

is

2

a,

y

/?,

may

be connected by the equation

B=

The equation whose

roots are

jgy-a

2

yq-j8

-

2 )8

2j8' is

Show that

(i).]

a(aa:

2

-h 26a;

this reduces to

y

2

a-f j8-2y

+ c) 3 = (aa^ + 36x 2 4- 3ca; + d) (ox -f

a cubic equation.

[Use Ex.

11.]

u=

and

SYMMETRIC FUNCTIONS OF ROOTS y are the roots of

13. If a, 0,

rootsare -

ft

[Let 2 =

14. If a,

-

4-

+-

prove that the equation whose

is

a

Prove that a 8 -f 2ga - rz = 0.

.

Eliminate a between this equation

any two roots of x* + qx + r=Q, the equation whose roots are the

are

]8

~

- + -, a y

+ -,

y

&+qx + r =0,

99

six values of a/0 is

[Let z = -

a = /fe, hence the required equation

/.

;

^ 15. If

[Let a,

+g0-fr=0

6, c

last

example to show that 5 x5 - a 5 5ax(# -f a) (# 2 -f a.r -f a 2 ), (x -f a) 7 x* - a 1 = lax(x-{-a)(x z -f aa; + a 2 ) 2 (# -fa)

17. Solve(x-fa-f-6)

18. If a,

5

-x -a -6 -f -x) + 6

5

(

y, 8 are the roots of

,

5

(

j8,

-) +

-f

26 3 , prove that

from

a*Z(* a 32;(a -

-6)~0, and use Ex.

(ii)^

7

-

4, p. 97.J

prove that

~lp*(p**-pt).

+ 4bx*-\- 6cx 2 + 4dx + e=Q and //~ac~6 2

,

20.* a 2

= I2(ad~2bc).

*Note that

.

:

a 2 a 2 =4(4& 2 -3ac).

21. a*i;a z p

(

x*+p&*+pjc+pt=0,

8 are the roots of ax*

y,

Ga*d- 3abc

23.

j8

-:0.

(1)^ = 2(^-2^),

24.

got by eliminating

;

[Observe that x -f a -f 6

19.

is

+ qzp + r = 0.]

8

+ b b + c 6 =5abc(bc + ca + ab) (ii) a 1 + b 7 + c 7 ~labc(bc + ca + ab)*. be the roots of a: 3 -f qx + r =0.]

a*

Use the

If a,

z3

and

a + & + c=0 show that (i)

16.

3

22.* a 2

from

(20) follows

(19)

and

(22)

from

(21).

= 144(c 2 - bd). 8) (y + y8 + S - 32(7. y)(a 8) = ]8)(a 2

2

2

)

J

25. If a, 0, y,

...

are the roots of xn +p l xn

-1

+p xn

-2

2

+ ...+p n = 0,

prove that

Ev* j3 2 = - ptf

Show

that the squares of the roots of n1 2 4a
a^-

V

n

- M*"" 1 +

a^-

M

n~2

-

. . .

-i-

(

... 4(

-

-

l)

na

n =0

1)

n6

w = 0,

where

= 2 - 2a 2 = oS * b r =ar 2a r _ a r+ ! -f 2a r ^ &o

ft

1

27. If the equation

is

2

ar+2

r+8

. . .

,

. . .

.

whose roots are the squares of the roots of the cubic

identical with this cubic, prove that either

the roots of 3 a

a a %<*>ia* + 2a a4 - 2ar ^30 +

^2 ==

i

i

a =6=0 or a =6 =3, or

a, b are

CHAPTER

VII

PARTIAL FRACTIONS 1

.

and Q are polynomials

in x,

P

An

Rational Fractions.

expression of the form P/Q, where called a rational fraction. If P is of lower

is

degree than Q, P/Q is called a proper fraction. than Q P/Q is called an improper fraction.

P

If

is

not of lower degree

By means

y

of the division

transformation, an improper fraction can be expressed as the

**

If

!

X' A' + -^-,

A+^

,

where the

f

X'

A A

Also

X

therefore

Hence

^7

and X' are respectively

YX'-XY'

A - A' =0;

is

2.

x

and X/YsX'/Y'. *

-y.

of lower degrees

of lower degree

than

Y

and

Y',

and

than YY'.

for otherwise, the polynomial is

A - A'

impossible.

would be identiThus Az=A' and y

X/Y^X'/Y'. To resolve a given fraction into partial the sum of two or more simpler fractions.

is

to express

it

as

Fundamental Theorem. A,

denote polynomials in

Partial Fractions.

fractions

B

A^A'

X YX'-XY' yy

cally equal to a proper fraction, which

consequently

letters

#

and X/Y, X'/Y are proper fractions, then r or

of an

2

,

ix

X

Theorem.

+

sum

Thus

integral function and a proper fraction.

are prime

to

//

such that

For since

C/AB

CX A

is

is

a proper fraction and the factors X/A and Y/B can be found

each other, proper fractions

prime to

B

}

Y

polynomials X',

BX'+AT = 1

j^

* and therefore

-

C = CX> CT + -~j-

Y

r

can be found so that (Ch. Ill, 18.)

9

.

FORMS OF PARTIAL FRACTIONS

A

and CY'/B are not proper polynomials Q, Q', X, Y such that If CX.' I

fractions,

by

101

division,

we can

find

9

X

CX' where

X/A and Y/B C

Now X and 7 are

BX + AY

is

fraction

so also

:

~

Q

X Y

_

Y +>

and then

are proper fractions,

_

_,

A and B, therefore Thus (BX + A Y)/A B is a proper

respectively of lower degrees than

than AB.

of lower degree is

CT = n

,

and

C/AB,

therefore

C + tf-O and

To

resolve a Proper Fraction P/Q into set of Partial Fractions. We shall prove that 3.

(i)

To a

non-repeated factor

x-a

of

Q

its

there corresponds

simplest

a fraction of

the

A form

x-a

(ii)

To a factor

- b) n

(x

,

To a non-repeated .

-

.

4

there corresponds J

J

x-b (lii)

Q

a group of the form

---o?_T\O'/-- --'***',

-i

1

of /

(x-b)

2

2

1\*>

L

-L.

(x-b)

(x-b)*

2 quadratic factor x

+px + q

of

n

Q

*

there corresponds

a

.

fraction of the form

(iv)

2

To a factor

(x

+px + q) +-

n

of

Q

there corresponds

'"

2

(x

Here A, Proof

B is

B B2 I}

(i).

Let

prime to

Cv (72 ... are independent = $ (#-a) S, then since x-a ,

...

,

,

.

x - a, and

a group of the form

+px + q)

n

'

of

x.

is

a non-repeated factor,

so

P/Q = */(*- a) +7/B, where the fractions on the right are proper

fractions.

Hence

X

is

a

constant. (ii)

that

Let JS is

B. It is assumed that x - b n and (x 6) consequently

Q = (x - b) n prime to

.

is

,

where the fractions on the right are proper

fractions.

not a factor of

JB,

so

TYPICAL EXAMPLES

102

X is of degree n - 1

Hence

Bi(x-b) where

51

+ B2 (x-b) ~*+...+ Bn

,

(Ch. Ill, 4.)

are constants, which proves the statement in question.

2

,

at most, and it can be put in the form ~ n l n

The proof is similar to that of (i), but X is of the form Cx + D. (iv) The proof is similar to that of (ii), but each of the set J91 B2 ... the form Cx + D. (iii)

,

is

,

of

Various ways of finding the constants are explained in the following examples. X*

Ex.

1.

Express

as the

~ j^-*

'

sum

of

an

integral function of

x and

three

proper fractions.

We have Hence is

- a) (x - 6) (x - c) = x* - a:2

(x

it is

2a + xZab - abc.

.

4 easy to see that the quotient in the division of x by (x a)(x ~b)(x~ c) We may therefore assume that

x + a -f b + c.

re

5 C 4 x-a x-b x-c*

4

(x~a)(x-b)(x-c)

To

find A, multiply each side of (A)

by

(A)

-a, and put x=a.

a;

=

'

(a-b)(a-cY The

values of

B

and

C

can be found in the same way

;

a4

(a-b)(a-c)'

(x-a)(x-b)(x-c) Ex.

Resolve into the simplest possible partial fractions

2.

t

6

Multiply each side by x 2, and then put x

a?

8

a;

.

c

2

;

-5

- -

m Therefore ,

x-a

-

2

-5 =

r^rr s

(#-l) (a;-2)

21- -43-

-x-l

-f

,-

=-^ 2

;

+T

(*~1)

rr* 8

+

(a;-!)

x -2^

Second method. Find d as before. Next, multiply each side of (A) by (x then put

a?

=l

;

C

Next, multiply each side of (A) by

a?,

r~

l'-5 ^~

and ;

Finally,

put

a:

=0 in

(A), /.

--= 2i

~a-f 6

let /.

.

4.

#->oo

a=

~c~JL

,

;

-2. .'.

6

= 1.

8

1)

,

and

METHODS OF QUADRATIC DENOMINATORS

103

In this way two equations connecting the coefficients can always be written down. The steps are equivalent to the following Multiply each side of (A) by 3 3 and the sc coefficients of (x~-l) (x-2) equate (the highest power of x) and the absolute terms on each side. Particular attention

is

called to the last

two

steps.

:

W ...

33? - 2x 2 - 1

T

Multiplying by (x

z

ax + b

+ x -fl) (x* - x + 1), we

cx

+d

...

find that

3x*~2x2 -l=(a and, equating coefficients,

= -2, a-6 + c + d=0, a=2, 6 = 1, c = l, d- ~2. J

Whence

6

+ d=-l ............ (B)

Note that the first and last equations in (B) can be found by (i) multiplying (A) by x and letting #->oo and (ii) putting x=0. Two other equations may be quickly obtained by putting x~l and x- - 1. ,

The following method

is

sometimes

3s - 2x 2 x 2 -x -fl 3

whence by

The

a +c =3

= 2,

6

= 1, c = l,

and 9

b

1

also the fractions are identically equal,

find e

and /.

2c

-2c=

+ 2d= -2,

and so

-2,

rf= -2.

"

__

first

+ d) x 2 - x 4-

+ 2c + d = l,

1

(m)

We

ca^-f (c

_ ~~ aX

(A),

division,

integral functions

giving a

1

From

useful.

a

6

c

^ + ^l>

+

Multiply both sides of (A) by x

2

c?

+

-f

(^^ + ^

ea;+/

x + 1, and then

.

*" (

}

let

This gives

D

We have &2 =

- a? ~

1

= * B+/

'

when

and #3 = 1, whence we

find that

and It follows that

l=3( /.

3(/-2c)*-3(e+/) = l.

This linear relation holds for two values of x (namely, the roots of x 2 It

is

therefore

an

identity,

/-2e=0 and

4-

x + 1=0).

and so

l=-3(e+/),

giving

e=-l/9 and /=

-2/9.

104 Next, multiply (A)

PARTIAL FRACTIONS by x + 1, and then let x = 1 ;

" Multiply (A) by (x -I)

Multiply (A)

In

(A), let

by

x,

xQ,

3

and

,

a

and then

1

#=1

let

-> oo;

let

1

3 ~(-2) .l~ ;

= a -f 6 -f e,

/.

-l=a~& + e-d+/,

.'.

8*

-i

giving c

= ^ + ^ = ^-|.

6

giving ;

1 '

(x-l)(x>- !)(**-!) 1

1

17

1

!**

*-l

1

1

1

s + 1'72

1

_

/ 2r 6

1 \9

/__

^

4

(*-l)

1

1

_

i

8

'

(x~l)

3

9

a;

EXERCISE

PARTIAL FRACTIONS Resolve into partial fractions 3( *~ 6)

J

2

I.

3.

.

.

a: "

'

'

--g

'

10

11

'

I

r

:

v

2

as the

-fl)

sum

(l-ax)(l~bx) Hence show that if 1 -o&x 2

is divided quotient of (n 4- 1) terms, this quotient is

and

a;

12

'

(a:

-

Express r

2

H-a: -f 1

3

- I9x - 15

'

)(a? -f2)-

15.

3

'

'

2

2

(a;

of a constant

by

*

-f-2)"

1

and two proper r r

fractions.

-(a + b)x + dbx\ so as to obtain a

find the corresponding remainder.

16.

Express

fractions

-- -- -

~ =~
;

sum

of a constant

as the

sum

of a constant and three proper p

x3

17.

Express

fractions 18.

;

r--

-

(-a)(*-6)(a-c)

^

as the

and three ^ proper

^

Hence show that

_ ""

IDENTITIES 19.

Express r

FROM PARTIAL FRACTIONS

3.4

~

.

rr-,

sum

as the

(x-a)(x-b)(x-c)

function of x and of an integral 6

three proper fractions. Hence show that

(a-b)(a-c)(a-d)

(b-c)(b-d)(b-a)

__ (c-d)(c-a)(c~b)

(d-a)(d-b)(d-c) is

equal toa-ffc-fc-fd. 20. If ( 1

+ x) n

c

+ c x + c 2x 2 + ...+ cn xn show ,

v

x+ 1

x [Assume that 21.

...

,

=^ +

Hence show that

v; 2 _ 22.

"

that

ct

_c

Use Ex. 21 and (\\

v;

34 _

___

;

714-2

similar identities to prove that

_^___fL_4__^__ " 1.2 2.3 3.4 _

,

3.4

vlvy

V

-4_f_l\n V ; (

_. 4.5 ;

v

1.2.3

2.3.4

2.3.4

370 T '"'

(n

+ l)( + 2)(n + 3)

23. Prove that

24.

Use Ex. 23 to prove that

[In Ex. 23, write

n -f 1 !

for n,

and put an =w, ...

= !, a,=2,

,

105

and

a?=0.]

2(n-f3)

CHAPTER

VIII

SUMMATION OF SERIES 1

.

Meaning of Summation.

integral variable,

and

Let u n be a function

of the positive

let

possesses this peculiarity it is the sum of n terms, these terms cannot be added up, unless the value of n is specified.

The function Sometimes

sn

it is

possible to express s n as the

number which does not depend on

and the number is

of

n.

sum

of

p

terms, where

p

and is

a

For example,

terms on the right (namely two)

independent of n. To sum a series to n terms or to find the sum of the first n terms of a series but this is not always possible. to express s n in the form just described is

;

Among the series capable of summation are arithmetic and geometric a harmonic series cannot be summed. series ;

The sum

of the first

n terms

of the series

ui + is

often denoted

2.

by

H

r

r

^iU r and sometimes by 2u n or simply by ,

Method of Differences. v n - vn _

form to n terms. For,

1,

where v n

by hypothesis,

whence by addition

is

// we are able

some function of

n, then

to

express

we can sum

sn

.

u n in

the

the series

METHOD OF DIFFERENCES Ex.

1.

Sum

the series

I

'

and since v

Q,

2

.

u n ~n(n +

Here

.

3 +2

.

3

4

.

-f

3

4

.

.

5

to

...

-f-

107

n terms.

l)(n-t-2)

u n= vn~ vn-

where

vw

=-i(

we have by the preceding, J )

(n

4-

+ 3).

2) (n

3. Series in which u n is the Product of terms of an A. P., beginning with the n-th.

The

last

example

an instance

is

of this type,

be applied to the general case, which It is required to

sum

to

n terms

is

r

successive

and the same method can

as follows

:

which

the series in

"

wn = where

We

(a

+ w&)(a + n + I

.

6)...(a

+ w + r-

1

n-

un

.

/>),

a, b y r are constants.

__ _

have __u n {(a

_

u n (a -f n + r

Therefore w n = v n - v n _^

where

and consequently Here v is independent the following rule of the A. P.

;

:

+ n + r .b)-

On

b)

.

vn

5W

-

-f

(a

1

.

b)

n -/

= vn - % sum

of n, hence the

the right of

n terms may be found by

to

u n introduce as a factor the next term

number of factors and add a constant. found by putting w = l,

divide by the

so increased

and by

the

common

difference of the A. p.

The constant substituting

is

for

n

in v n

as in the next

1. Sum to n term* 1 3 5 + 3 5 7 -f 5 7 9 f Here u n ~ (2n- l)(2w-f l)(2w + 3), and applying the

Ex.

To

.

find the constant C,

put

.

?i

.

=1

.

.

.

The expression

which follows

H

. . .

;

or

by

in the bracket (} is formed (n-l-2) in the A.P. 1, 2, 3, ... .

3)(2

.

rule,

in (A), noting that s l

-f

*

example

.

=I

.

3

.

5

;

+ 5) + 15}.

by subtracting the term which precedes n from that B.C. A.

SUM OF INTEGRAL FUNCTIONS

108

which u n

4. Series in

In Ch. Ill, of

n

a Rational Integral Function of n. has been shown that a rational .integral function

4, (3), it

is

can be expressed in the form

of degree r

o + bn + cn(n + l)-f dn(n + l)(n + 2)-f

where

a, 6,

division.

e, ...

We

are constants whose values

I.

Sum

to

may

can therefore use the rule of the

this type, as in the

Ex.

...

(r

+ 1)

terms,

be found by synthetic

last article to

sum

series of

next example.

the series

2.34-3.6 + 4. Here

ll-K..-f(tt

+ l)(n 2 -f2).

= (n + l)(n' + 2)

tt

1

+ 1 )*

+2

+2

+0 -2

Dividing by n, n + 1, n + 2 in succession, as on the right, the reckoning shows that tt

+1

1

-2

n = n(

therefore

3n 8 + Wn* + 2ln + 38), the

'

constant

'

clearly being zero.

which u w is the Reciprocal of the Product of r successive terms of an A. P., beginning with the n-th. A 5. Series in

simple instance

the following

is

:

Ex.l.

111

Here

n

4

(2/i-l)(2w

+ l)

4

1

where

t>

n

The general in which u n is

case

is

as follows

*

:

It is required to

sum

to

n terms

the series

the reciprocal of

(a-f n6)(a

where

--

+w+ 1

.

6)...(a-hn-t-r~l

.

6),

a, 6, r are constants.

The numerator

is

the difference between the

first

and

last factors of the

denominator.

SUM OF RECIPROCALS

109

Proceeding as in the last example, we have 1

_

(a

u n ^v n _ l -v n

Therefore

+ n + r-1 6)-(a-f nb) .

and

,

sn

= vQ -vn

1

1 .

Now

where

,

b)(a + n + 2

.

b)

...

(a

+ n + r-

1

.6)

n, hence the sum to n terms may be found by Remove the left-hand factor from the denominator of u n ; divide by the number of factors so diminished and by the common difference of the A.P. and subtract the result from a constant.

v

independent of

is

the following rule

The constant in v n

is

:

found either by putting n = 1, or by substituting

,,

Ex.

for

n

Thus, in the last example, by the rule

.

2.

Sum

the series

4-

l.Tc

n

n(n + 3)

J*

n(?i

,

+

O

+ ^-^ + O.U

l)(n

. . .

+

W

.,

(

7i "f

C== _L,

1

+ 2)(n

1

*

To

find the constant C,

"

put w = l in

_L^ 1.4

(A), observing that

2

x

3

.

We

have

l

n .

8

=

*

,

x

11

1

4

5 1

u 1

^ ""18

which

is

11

" c^ 18*

3.2.3.4*

Alternative method, by Partial Fractions.

s^

the same result as before, in a different form.

*

1

TWO IMPORTANT TYPES

10

The

last

an instance of a

is

example

whose n-th term

scries

P is a polynomial in n of degree m. m
is

where If

a If

w=r-

1,

H-

% (a + n&) + a

2 (a 4-

the series cannot be

n+

4-

w6) (a

1/>) 4-

do

for to

summed,

the form . . .

.

this

we should have

sum

a harmonical progression. If w>r~l, the nth term consists of an integral and a fractional part which can be dealt with separately.

to

Another important Type.

6.

a

+ 6

a(a + b(b

+

a (a +

\)

1) (a

If u n

+ 2)

a(a+

"

term of

is the n-th 1)

...

(a-f

n-

1)

1

K

sn

=

rn

= u n (a + n)/ (a - b + I

AT Also

w,

=w

Therefore

sn

= vn

then

-

-

+ *) -

(

}

For where

t

.

(a-fl)-fc v / - ~= r T a-b+1

a

The

Series U 4-u 1 x-f-u 2 x 2 nomial in n of Degree m. 7.

If x

= l,

this series

then

s (1

where

^i

can be

r

7

=

a-6-hl

v 2 ^u z

a f

r

1 ).

_-

a~fe+l 1 =

r

a-6 + 1

as in Art. 4.

- x) = w + v& -f t'^2 +

= %-^o>

>

{w n ^ * (a

.

+ n) - a}. '

+ u n x n where u n

4-...

summed

(n

)

-u lt

. . .

If

x ^1,

+ vw a:n - u nx n+l

a Poly-

let

,

= w n -w n . r

... t? n

Now v n is a polynomial in n of degree m - 1

is

thus by repeatedly multiply- x (m the problem of finding in we can reduce all), ing by multiplications that of a 8 to summing geometric series. ;

1

It follows that s = P/(l ~x) mfl where Ex.

1.

Sum

Denoting the

sum by s(I

Let

+ 2 2z + 3 2za + we have

the series I 2 s,

. . .

- x) = 1 + So: H- 5x* + 5'

=

P

+ n*xn ~ l

. . ,

a polynomial in

is

.

+ (2 H -

1

)

xn - 1 -

x.

SUM OF POWERS then

a' (I

- x) = 1 + 2x + 2x* -f

. . .

111

+ 2a;w - 1 -

n (2n l)x

.__

The

8.

Series

1

r

+ 2 r + 3 r -f

and

cr

...-f

nr

= n(n-f-l),

We

.

shall write

a'

Theorem. // r is a positive integer, S r can be expressed as a polynomial in n of which the highest term is n r + l/(r + 1 ). For we can find a l9 a z ... a r independent of n, such that (1)

,

^a l n + a 2 n(n-l) + ... +a r n(n-l)(n~2) ... (n-r + 1), ....... (A) = obviously a r l. Writing n- 1, n-2, ... 1 in succession for n and n

where

,

r

adding, by Art.

3,

we have

+ The right-hand highest term

is

Show

#*.

1.

We

have

side,

-a r (w + l)w(tt-l)

when expanded,

n r+1 a r/(r +

that

.

1),

that

a polynomial in n of which the

is

n r+l /(r

is,

-f-

1).

S^ln(n + l)(2n + l) S3 ^4n 2 (n + l) 2 9

7i

2

-n-f W(TI~

(n-r + 1) ............. (B)

...

1)

.

;

Also

(2) If /S r fe

= 61 n + 62n 2 -ffe3 n 3 + ...-f6r4 1 n r .

"

-

f

1

=/(n), the values of & r41 , 6r ,

may be found as follows. = l r + 2 r + ...+nr and /(n-l)-l r -f 2 r -f ... + (n-l) r /(tt)

r _j, ...

,

w =/(n) -/(n - 1 )

therefore

so that

n r = 6r41 {w r+1 -

(n

Expanding and equating J

7,

6r+1

,

r

""'

A

r

=

1 '

-

r+1 l)

}

,

+ b r {n r - (n - l) r} +

coefficients, it will

h - = ftr 1

and so on. If r is large, b lf 62 and we use other methods.

,

...

r '

h 6r -

2= n

'

. . .

+ 6r

be found that h ftr 8=

r(r-l)(r-2)

cannot be found conveniently in this way,

ODD AND EVEN POWERS

112

The values

(3)

of

ing the function

8^ S3 S5> ... may be found in v n = n r (n + l) r -(n-l) rn r ,

,

succession

by

consider-

.

We have

v1 + v 2 + ...+vn =n (n +

r

r

l)

,

and, expanding by the Binomial theorem,

Putting n

n - 2,

1,

1 in

...

n and adding,

succession for

_5 +

a- n(n + 1). The

where

last

term

is

rSr+l or Sr

,

...

..................

,

according as r

is

(A)

even or

odd.

When r = l,

2, 3,

this

...

formula gives

=4

$3

=

-87

Thus if r is odd and >1 and Sr =f(n), then n2 (n + 1) 2 is a factor of/(n), and the remaining factor is a rational integral function of n(n + l). Of z 2 course, it does not follow that S r is arithmetically divisible by n (n + 1) .

(4)

S2 /S4 S6 ... may be found as follows. = wn nr (n + l) r (2n + 1) - (n - l) r r (2n - 1)

The values

Let

of

,

,

;

W1 + w2 + ... +wn =ri

(n

+ l)

and

cr'

r

then

a=n(n + l)

where

On

,

,

r

r

(2n + l)=a a,

expansion, we have

*2r-4

Putting n

1,

- 2,

in succession for

...

1

Sr

or

(r + 2)S r+1

...

this

.

n and adding, (B)

the last term being

When r 1, 2, 3, r=l Jour' =3S2 = r 2

,

according as r

is

even or odd.

formula gives

r-3 r=5

' |

(3o*

- lOa8 + ITo2 - 15a + 5).

BERNOULLI'S NUMBERS

113

Thus if r is even and $ r =/(n), then n(n + l)(2n + 1) and the remaining factor is a rational integral function It will presently

from (A)

this

$ 3 $5 5 7 ,

,

. . .

,

number

,

//r>l,

all

...

of

/(ft),

1).

immediately deducible

Sr

as the coefficient of n in

,

contain the factor n (n-f I)

2 .

for

S 1 =ln(n + l) and

The numbers

B -B4 BQ9 2>

,

Bv B2 B3 ,

is

;

the following.

+ l)Bar + iC5(2r-l)B^ 2 + iC5(2r-3)fi2r ^ + ...=0, ...... (C) on the left being B r or (r + 2) Br+1 according as r is even or odd. ,

the zero

B2m

on

the right of (C) is to be replaced by |.

the coefficient of

is

n

in

the left-hand side of (C)

(4),

r

%n (n +

as Bernoulli's numbers.

...

,

formulae can be given for calculating these numbers

then

(2r

For

Br

2

many

the last term

Art. 8,

n(n +

are the famous numbers of Bernoulli, in terms of which a large of functions can be expanded. For convenience we shall refer to

great

= 1,

define

B

one of the best

If r

We

expressed as a polynomial in n. = 2, and 7?3 , B^ ly ... are all zero;

the whole set

A

of

is

^

Thus

is

a factor of

of the last section.

Bernoulli's Numbers.

(5)

when

- #8

appear that the formula (B)

is

r

l)

(2n

Putting r=l,

+

is

52w

.

Hence by the formula

(B) of

the coefficient of n in the expansion

l).

2, 3,

...

in (C),

,

5B4

.B 2

=

1

30"

Continuing thus, we find that

"" A

691

7

M "6*

p ~ 16

43867

3617

510'

1

798

'

20

_ ~

174611 *

330~

few more of the extraordinary properties possessed by the sequence

Bv B2 B39

... are given below. the Binomial theorem, ,

By

Putting

n-1, n-2,

Equating the

...

1

coefficients of

in succession for

n on each

side,

n and adding,

we have (D)

SYMBOLIC NOTATION

114

from the equation

Similarly,

it

follows that

n r+l

=ci

and consequently 1

From

and

(D)

(E)

+

...

+ (-ir~ 1 Cj

{1

5

1

+ (--ir==0 ......... (E)

by addition and subtraction,

............ (G)

),

where each

B

series continues

B

or Q till 1, appears. 19 are to (F), (G) equivalent equations 10', 11' in Chrystal's

Formulae

They may be used for calculating 5 2 B^ twice the amount of labour required by (C).

Algebra, Vol. II, p. 231.

but their use requires Symbolic notation. cally thus

,

Equations

etc.,

can be written symboli-

(D), (E), (F), (G)

:

(D) (E) (F)

(G)

where, after expansion, each index is to be changed into a suffix be written for

Bm

Theorem.

(6)

9

i.e.

Bm

is to

.

//

8f =

l

r

+

2r

+ ...+n r where

r

is

a positive

integer,

then

Sr - rS r _i + Br

(i)

where

B

r

is


independent of n and

defined in Art. 8,

is,

(")

S r = r [s r ^dn + nBr

in fact, the

Bernoullian number as

(5).

For S r can be expressed as a polynomial in n this polynomial by /(?*), we have

of degree r

-f 1.

Denoting

* r -/(*)-/(*-!) This relation involves only positive integral powers of n up to n r y and holds for all positive integral values of n. Hence it is an identity, and we

may

differentiate

both sides with regard to r -l rn

Writing n-1,

ft

-2,

...

/'(n)f{n where

B

r is

r-1

1 in

Therefore

=f(n)~f'(n-l). and adding,

succession for n,

+ (n-l) r

independent of n.

n.

-l

+ ...H-2^ 1 + r

BERNOULLI'S THEOREM

Br is the

Moreover, lian

number

No

first.

Ex.

1.

Putting

coefficient of

defined in Art. 8,

(5).

constant of integration

Deduce formula (B) of Art. r + 1 for r in (A), we have

115

n in/(n), and is therefore the BernoulThe second statement follows from the

is

added, for

8, (4),

Sr

is divisible

from (A) of Art.

n.

by

8, (3).

Differentiating with regard to n,

i(r

No

+ l)aV = C[ fl (2r4a)S2r +

constant

r

Dividing by (7) I'sin^

/S'

...

2 , N.j

C^

added, for every term is divisible by + 1, we obtain the formula (B).

is

the relation

^V-M^'r-i^ +

1

.)

we can

//#,.,

S = n,

Thus

find the values of/S'j.

(At any stage, the constant

by successive integrations.

found by putting n =

n.

B r can

be

=-

xS A

2.;

a

==

Continuing the process,

we

arrive at the equation 7i

which

3

r-1

2

'

r-2

|3

'"

k

!

be written

may

+ I)S r = n r + l + B 1 C[' n n r + B2 C 2r +l n r - l + ...+B r ^

(r

and expressed symbolically, as above,

(H)

form

in the

is known as Bernoulli's Theorem. = l, we have equation (D) of Art. n Putting

This result

Ex.

1.

8, (5).

Verify the following, either by substitution in (H),

n3

n4

by successive

or, starting

with

nz

integration.

n* ~

=

5

+

n1 ~2

+

n3 n 3^~30

^

8

n8

" 10

9

2n 7

7?i

3

"15

5

2w 3

"

9

2 9

_n ^ w + 3 "TO

2

e

8

4"

"9

__7

"T(J

+

11

_

30

w*

n2

2

"20 '

Tl 8

7

n In* _w s + + "8

"2

T2

In 4

w2

""24"*" 12

2

66

SUMS OF INTEGRAL FUNCTIONS

116

EXERCISE XIV

SUMMATION OF SERIES

A Sum

n terms the

to

(Arts. 3, 4.)

Exx.

series in

1-9.

1.

1.2.3.4 + 2.3.4.5 + 3.4.5.6 + ....

2,

2.5.8 + 5.8.11 + 8.11.14 +

4.

1.5 + 2.6 + 3.7 + a

a

....

6.

1.2 + 2.3 + 3.4 +

8.

1.2.3 + 4.5.6 + 7.8.9 +

Find the value 10. 12.

a

...,

to

P+3 +5 2

2

2+2

2

2

+ 3

...

.

+ 3* .4+

5.

I

7.

1.2.4 + 2.3.5 + 3.4.6 +

9.*

....

.

.

...

.

.

2.1+5.3 + 8.5 + ....

of

lP + J2 2 + 13 2 + ...+2P. lP + 12 3 + 13 3 + ...+2P.

Sum

3.

....

n terms

11.

1P-12 2 + 13 2 -... ~20 2 + 2

13.

20 3 -19 3

+ 18 3 -...+2 3 ~P.

the series in Exx. 14-16.

14. ! 16. l

17.

Show (i)

(ii)

18.

Show

that whether n

P-2

is

odd or even, to (n-1) terms- 1( to (n - 1) terms = |( -

+ 3 2 -4 2 + ... 1* - 2< + 3 4 - 4 4 + .. 2

.

that the

natural numbers

is

sum

n l)

l)

n(n- 1). n(n -l)(n*-n~

n

1).

of the products, taken two together, of the

2 -^n(n -

first

n

l)(3?i

19. Find (i) the sum of the squares, (ii) the together of the numbers 1, 4, 7, ... (3n -2).

sum

of the products taken two

20. Find the number of shot which can be arranged in a pyramidal pile triangular base, each side of the base containing 10 shot. 21. Find the number of shot in a pile of 8 courses side of the base containing 12 shot.

22. Show that the number of shot in a pile of side of the base containing 2n shot, is

n

on a

on a triangular base, each

courses on a square base, each

23. 'Find the number of shot in a pile of a rectangular base, the sides of the base containing 16 and 10 shot respectively, and the top course consisting of a single row of shot.

A

24. pile of shot of n courses stands on a rectangular base, and the top course consists of a single row of x shot. Show that the number of shot in the pile is

*

Here the nth term

3, 3, 5,

t

...

is

the product of the nth terms of the arithmetic series, 2,

.

Here w f =r(w

r

l)=nr

(r

l)f.

5, 8,

...

,

and

SUMS OF OTHER TYPES 25. Show that or zero, then l*

26.

Show

n

if

of the form

is

+ 2 4 + 3 4 + ...+ n*is

that

if

n

5

is

B to

n terms the

,111 __ __ 1

2.3

2 2 2 by P-f 2 -h3 4-...+7i

divisible

by

1

3 .

3.4.5

I

i

3.6

6.

1

.

4

1

i

1.3

1

3

3

1

o O

1

1

F + 2 + 3 + ...4- n

1-24.

i

2.3.4

.

(Arts. 5, 6, 7.)

3.4

1

1.2.3

m is a natural number

5m -f 3, where

then

____

j

i

1.2

3.

3m -f 1,

Exx.

series in

or

divisible

of the form

is

.+w

Sum

5m -f 1

11?

7*""

1

L

I

f

2.5.8

6.8.

11

.

8. 11

.

14

1

5.

5 6.

8.

__ 2.3.6 __ __ 1.2.4

g

07II + -+ ~-f ~ +

+

,1 +

2.4

*

5\"9

JLL 2. 4.

.

_LL*L*. + "

6^2. 4^6.8

+

12.

...

32

14. 1

9

3.4.6

' * *

77

F75

52

13.

.

a

72

+- + 2i + ~ 1 ~

h

(!+*)(! +2*) 16.

,

3. l.| 1-f 2.|2-f |3

1*

+ ...

1-

.

3

2

n.]

(2rn-l)-l

(n-fl)

1

3. 9.

1

1

* * 11. 137. 13. 15 115.

""

on

L

"

n ~ l = a;* " 1 - a:n .] (a: + n) [Observe that nx 1

21.

[The value of s n (l -x) can be found by Ex.

11.]

2

-n 2

BERNOULLI'S NUMBERS

118 22 "'

n

17273

-.1

24

-[3 25.

5

+

+ +

il

n~

l

27374 11

+

23 M

475

3~.

-+w

a

l '

273

+22 +

2

+ 2*

'37I

'

+ re-l

[5

Prove that

l+n + n(n v

Sum 26.

n- 2 +

+

to

+ l) ft

'

r^

-f

2

+ ... xtor

terms =

n terms

+

--

-

-

+

+-

.... 3* 2

1

1

/r*

1

aa (1

C 1.

If

[Put -Ti + 2.

If

3.

-n + 2,

1,

...

/(-n) = (-ir+y(n-l). -1 for n in the identity f(n)-f(n- l)~nr .]

Sr ^ b^n + b 2 n 2 + b a n 3 +...+ b r+l nr + show that 64-61 4- 6 8 -6 4 + -...=0. 26 2 - 36 3 + 46 4 - 56 5 + ... =-0, 3.26 3 -4.36 4 -h5.46 5 -...^0. 1

,

[Use Art.

8, (2).]

Given that

value of

(Art. 8.)

F-f2 r + 3 r -f ...+n r =,/(n), show that

Ss by

9

= -^a 2 (2a 3 - 5a 2 + 6or -

3)

where a

n (n +

1

),

deduce the

differentiation.

4.

Continue the reckoning in Art.

5.

Show

8, (5), to find

that formula (C) of Art. 8,

form

(5),

Bm is to

be written for

B 12

1),

B.

6.

Find the value of B^ by using formulae (F) and (G) of Art.

7.

Show

8, (5).

that

C\+WBr + C$+W-*Br_ [Expand (2n +

l)

r+l

-(2n -

!)'+>.]

z 4-

and

J514 .

be written symbolically in the

- \Y (2B -

'

where after expansion

may

the values of

C rb +*2r -'BT _ 4 +

. . .

=r+ 1

.

CHAPTEK IX DETERMINANTS Notation.

complicated expressions can be easily handled if they are expressed as determinants/ The theory which we are about to explain arises in connection with linear equations. 1

.

Many '

= equations a^x -f 6j 0, a^c -f b 2 = of x, then 0. The expression If the

a^ a^

of the second order,

and

is

are satisfied

a^

-

aj) v

is

by the same value

called a determinant

denoted by or

by

(a^).

Next consider the system a i x + ^\y +
a 2x

-f

b2y

+ c 2 = 0,

= 0.

equations are satisfied by the same values of x and to show that If these

The expression on the is

y, it is

easy

a determinant of the third order, and

left is called

denoted by or

This determinant in

its

a 1 6 2C3

expanded form

C&ifyjCg 4-

In considering the form of

by

is

^ 2 63 C 1

this- expression,

observe that

:

Disregarding the signs, the terms can be obtained by writing the letters a, 6, c in their natural order and arranging the suffixes in all possible (i)

'

The number

ways. (ii)

As is

arrangements

is

six terms. [3, giving

any particular term depends on the order of the suffixes. Ch. IV, 4, (1), the interchange of two suffixes, 2, 3, for example,

The sign in

of

of

called the transposition (2 3).

The term o 1 62^3> i n which the the sign

-f

.

suffixes occur in their natural order, has

DEFINITION OF DETERMINANT

120

The

sign of any other the suffixes in that term

or by an odd third terms,

Also

1

in this

2. a, 6,

number 1

2 3

is

term is

+

-

according as the arrangement of derived from the arrangement 1 2 3 by an even is

or

,

Thus, considering the second and

of transpositions.

changed to 1 3 2

by the

single transposition (2 3).

changed to 2 3 1 by the two transpositions These ideas will now be generalised. order.

23

is

of

Definition c, ... I, and n

Determinant.

a

suffixes 1, 2, 3,

an

The determinant

of the

...

bn

n

bn

c n ...l n

of

performed

n

letters

2 represent n numbers thus,

.

nth order, denoted by ,

i

By means

we can

n,

(1 2), (1 3)

cn

or

by

(^i^2 c3

ln)>

... l n

sum of all the products, each with the sign determined the rule below, which can be formed by writing the letters a, 6, c, ... I by in their natural order and arranging the suffixes in all possible orders.

is

defined as the

Rule of Signs. The sign of the term a 1 6 2c 3 ... l n where the suffixes The sign of any other term is -f or occur in their natural order, is + ,

.

,

according as the arrangement of suffixes in that arrangement 1 2 3 ... n by an even or by an odd (that by an even or an odd substitution). This rule is justified by Theorem 2 of Ch. IV,

term

derived from the

is

number

of transpositions

is,

Each

of the

2

n numbers a l9

6X

...

a 2 62 ,

,

4. ^s

called

an element

of the

determinant.

The diagonal through the left-hand top elements a l9 6 2 c3 ,

a i&2c3

l

n

i fi

,

... l n ,

is

corner, which contains the

called the leading or principal diagonal,

and

called the leading term.

The expanded form of the determinant has n terms half of these have the sign + and half the sign For suppose that there are p terms of the first sort and q of the second. Since the interchange of the two suffixes transforms a positive into a negative term, p^q* Similarly q*^p, and therefore p = q. Every term contains one element and one only from each row, and one element and one only from each column. We can find the sign of any particular term by the rule at the end of Art. 4, Ch. IV. :

.

FUNDAMENTAL THEOREMS Ex.

Jn the expanded form of (a 1 6 2c 3^4 e 6)> find Me signs of (i) ajb^c^e^

1.

= (I 2 4 X 3 5 ) , ? t Q) f
(i)

121

=

(

l

2 H 2 4 H 3 5 )>

and tho

si

Sn

is

(ii)

ajb^c^d^.

*

~ (

The

last steP is

JL

unnecessary

if

we use the

we may make

Or,

rule referred to.)

single transpositions, as follows,

24513, showing that

three,

() (\ II I ^}=( l Or

as

if,

14523,

of these are required,

5 >( 24 )' and the

some text-books

advise,

12345, is

-

number

of

so the sign

.

+

si s n is

wo count

the

*

inversions

'

(Ch. IV, 4)

and (ii), these are 5 and 10 respectively the odd and the second even, giving the same results. But this takes much longer.

in the arrangements of the suffixes in first

12543,

and

3.

Theorem. A

columns and

its

(i)

determinant

is

unaltered by changing

its

rows into

columns into rows.

A

Let

The leading term of each determinant is a^c^. The remaining terms of A are derived from a 1 6 2 ca by keeping the letters in their natural order and arranging the suffixes in all possible ways, an interchange of two suffixes producing a change of sign. The remaining terms of A are derived from a^b^ by keeping the suffixes in their natural order and arranging the letters in all possible ways, an interchange of two letters producing a change of sign. The results in the two cases are identical. The same argument applies f

to a determinant of

4.

any

Theorem. The

order.

interchange of two rows, or of two columns, changes

the sign of a determinant without altering its numerical value.

For the interchange of two rows is equivalent to the interchange of the suffixes, and the interchange of two columns is equivalent to the interchange of two letters. Hence in either case the sign of every term of the determinant 5.

is

changed.

Theorem.

A

determinant in which two rows or two columns are

identical is equal to zero. If

of

A

two rows or two columns is

But

of a

determinant

A

are identical, the value

unaltered by the interchange of these rows or columns. this interchange transforms

A

into

- J,

J=~J, and J = 0.

therefore

EXPANSION BY ROW OR COLUMN

122 6.

Expansion of a Determinant

Row

of any

in

Terms

of the Elements

or Column.

A

Let

Every term in the expansion of A contains one element and one only from each row, and one element and one only from each column. Hence,

A can

be expressed in any of the six forms

a 1 A l + b l B l -f CjCp 4-

A

a 1 u4 1

b%B 2 4- CgCg,

:

+ a2 A 2

b}

Bl -f b2 B2

cx

Cl

+ c3 C 3

,

contains no element from the row or column containing a j? a similar remark applying to every A, B and C.

where

l

To determine

A^

every term of

A which

contains a 1 can be obtained

from a^^s by keeping the letters in the natural order, retaining 1 as the suffix of a and arranging the other suffixes in all possible orders, the interchange of two suffixes producing a change of sign. Hence A l is equal to the determinant (6 2 c 3 ), obtained by erasing the

row and column in A which contain a v To determine any other coefficient, say C 3 consecutive rows, bring the row containing of

interchange to the extreme

consecutive

bring

c3

the

to

the

interchange

the top.

of

the

By

column containing

c3

left.

Thus we have

in succession *3

and now

columns,

By

.

c3 occupies the

&3

c.

top left-hand corner, and the relative positions

of the elements not belonging to the

row and column containing

c3 is

unaltered.

Every interchange of rows or of columns introduces a change of sign. <__ The number of interchanges is equal to the number t of vertical and horizontal moves (indicated by arrows) left-hand which in c to the to top corner, required bring 3 '

^

this case is 4.

Hence,

<73 is

c8

equal to the determinant obtained by erasing the row and

column containing

c3 , multiplied

by

(

4

1)

.

Similar reasoning applies to a determinant of any order.

MINORS AND COFACTORS Ex.

1.

A~(a 1 b zc B

)

with reference to the element*

in the

results are -

2.

aa

J--

(ii)

Ex.

(i)

in the second column.

first roiv, (ii)

The

the determinant

Expand

123

62

-63

Prove that

a

h

g \^abc

h

b

f

9

f

c\

Expanding with reference

+ 2fgh-af 2 -bg 2 -ch z

to the

first

row,

b

f-h

h

b

f

*

g

f

A=a

.

\

= a(bc -/ 2 - h(hc -fg)+g(hf-bg) = abc 4- 2fgh - or/ 2 - bg 2 - ch*. )

Minors and Cofactors.

7.

For the determinant

the minor of any element is the determinant obtained by omitting from the row and column containing the element.

The

(H r )

cofactor

of

J

an element (h r ) is the coefficient of that element in J, and is therefore equal to the corresponding minor

the expanded form of with the proper sign prefixed.

The sign

is

of consecutive

determined as follows rows and columns

:

Count the number of interchanges of vertical and horizontal moves ') '

(i.e.

required to bring the element to the left-hand top corner. The sign is + or according as this number is even or odd. The rule may be stated thus. For the element h r which is in the r-th row arid the s-th column, ,

H

the cofactor

For the number is

of

*

r

= (- l) r + s

.

(the

minor of h r ).

'

moves required to bring h r to the left-hand top corner

(r-l)-f (.9-1). 1. Find the cofactors of d 2 and c 4 in the determinant A "(a^b^d^). Omitting the row and column which contain d 2 it will be seen that

Ex.

,

the minor of ^ 2 = (a 1 6 3 c 4 ).

Denote the cofactors

Similarly

O4 =

(

The student who

1)

in question

4 ^" 3

is

(a 1 6 2 c? 3 )

=

by

-

Z) 2 , (74

;

by the

rule given

above

(a^b

acquainted with

*

partial differentiation

'

should observe that

B.C.A.

MULTIPLICATION OF DETERMINANT

124 8.

For the determinant

Identities.

Important

A = (aj)&) we have a number (i)

of identities of the following types

a 2 A + b2 B l + c2 C l = 0, l

(ii)

61

4 1 -f b 2 A 2 + b z A^ = 0.

where

A B ly

...

where

A B

C^ are independent of a l9 b l9 c v

19

19

19

are the cof actors of a l9 b l9

:

...

.

Hence the expression

the result of substituting a 2 6 2 c 2 for a 19 b l9 c 1 in A, and equal to a determinant with two rows identical, therefore

is

a 2 A^

The second

identity

is

+ b2 B l ~f c 2 C l = 0.

obtained by putting b l9 62 b 3 for a l9 a 2 a3 in the ,

,

A = a^! + a 2 ^ 2 -f a3 ^ 3

e q ualit 7

therefore

is

,

,

.

Similar identities hold for determinants of any order. 9.

Theorem

minant

is

// every element in any row, or in any column, of a determultiplied by a number k, (hen the determinant is multiplied by k. .

For every term in the expansion of a determinant A contains one element and one only out of each row, and one element and one only out of each column.

Thus ka

Conversely,

common

'2

^2

'3

^3

if all

a2

the elements in any row, or in any column, have a is a factor of the determinant, and can be taken

factor k, then k

outside.

10.

Theorem. A

determinant can be expressed as the

sum

minants by expressing every element in any row or column as numbers.

For example, we

shall

of two detersum of two

the

prove that

=

a.

a2

o2

c

oc

63

c.

3

\

Denote the determinants on the right by A A' and the one on the by A". 9

9

left

ADDITION OF BOWS OR COLUMNS

We

A

have

+ a2A 2

a^A^

-f-

a^A,3

A^ A 2 A 3 are independent of a-,, a 2 a 3 Now A is obtained from J by substituting oq,

where

,

.

,

1

And A" a 19 a 2 a3 ,

is

obtained from

J by

A

f

and

Similar reasoning applies in

(or

A

Theorem.

.

,

3

.

+ a 2 a 3+ a 3 ,

+ a 2 ^ 2 + a3 ^ 3

J +

a +a

,

a

+

a^

all cases.

unaltered by adding to the elements of (or column) k times the corresponding elements of any other raw

column), where k

For example, we

is

determinant

is

any given number.

shall

prove that

=

e,

3

+ kb3

63

a

c3

Denoting the determinant on the right by

we have

/r

=

a

c

Note that by a second application X

-I-

A and

^

6X

cx

?

62

c2

2

left

by

A',

-0.

of the above,

/c x

+ I U3

ha

that on the

+

\

=*

Now

NOTE.

,

substituting a^+v.^ & 2

=oc l A l

J" = a+a-4-f

any row

a 2 a 3 for a l9 &2

.

Hence

1 1

125

C3

In simplifying determinants the following points should be carefully

observed.

one which (i) When, as in the last part of Art. 11, a column (or row) is replaced by contains the elements of that column (or row) all multiplied by a number h, then the determinant is multiplied by h. For instance, if the second column is subtracted

from the minant (ii)

is

first,

and the remainders are put

in the second

column, the sign of the deter-

changed.

When more than

this kind, at least oni

one column

column

(or

(or

row)

row) must be

is

changed by a succession of changes of unaltered, as in Art. 13, Ex. 1.

left

APPLICATION OF REMAINDER THEOREM

126

Theorem.

If the elements of a determinant A are rational integral functions of x and two rows (or columns) become identical when x~a, then x-a is a factor of A

12.

.

A

can be expressed as a polynomial in x, and so that the result follows by the Remainder theorem.

For

(is

Again, ifr rows become identical when a a factor of

J

For

K)/

J=0

when x = a;

is substituted for x, then

(x

- a) r ~ l

A.)

rows can be replaced by rows in which the elements hence are the differences of corresponding elements in the original rows x - a is a factor of all the elements in each of these (r - \ ) substituted r

of these

1

;

rows

and the

;

13.

result follows.

Examples.

The determinant

in question

Expanding with reference to the bottom row, J =19( - 19 13 + 23 31) -f 5(19 .

[In the

first

step, the 3rd

is

column

taken

denoted by A.

+2 31)^10209. from the 2nd to form a new 2nd column, .

11

.

taken from the 3rd to form a new 3rd column, and twice the 1st taken from the 4th to form a new 4th column. In the second step, six times

the 1st column

column

.

is

is

the 3rd column

is

is

taken from the

1st,

the 3rd

is

subtracted from the 2nd, and the

4th added to the 3rd.

Ex.

2.

Prove that 1

If

a/?, two columns

Similarly

ft

y, etc.,

1

in

-)(-)(

1

A

-8) (0-8)(y -8).

are identical, and consequently a-/?

are factors.

is

a factor of J.

Hence

Since J is of the sixth degree in a, /?, ... , k is independent of th^se quantities. Now the term j5y 2 S 3 occurs in J, and on the right-hand side the coefficient of this term is - k, therefore k ~ - 1.

METHODS OF EVALUATION Ex.

3.

//

a cube

oj is

root of unity, prove that

a

b

c

b

c

a

a -f&eo +c
2

127 is

a factor of

cab Hence show

To the

that the determinant is equal to

first

column add

A

CD

times the second and

4 bat + co>

wz

times the third, therefore

2

b

c

1

a

c

6-f cco-foco 2

c

a

o2

b

a

4 6o> 2

ct

6

a |

C 4- #cu

jab

j

a4&o>4co> 2

.*.

The

Ex.

2.

This method applies

to

is derived

are factors.

a determinant of any order in which the second row the third from the second, and so on, by a cyclic

the first,

from

and a

a factor of J, similarly

is

rest follows as in

substitution.

Ex.

Prove that

4.

(6-fc)

2

(c+a)*

(&4c)

a 2 -(&4c) 2

2

c2

a2

(a

Subtracting the

sum

of the

second and third rows from the

-6

be

2(a-f6+c)

b*

2

and taking 2 outside,

first

6c bc

+ ba

a+b -

In the

last step the first

column

is

ac

added to b times the second_ajid to

third, which is equivalent to multiplying by be

Alternatively.

If

we

substitute

+ bc

c times the

;

for a,

=0;

hence a

is

Again,

a factor

if

-

;

and

(6 4- c) is

similarly b

and

(-a) 62 c2

and

in this there are three identical

factor of J.

c

are factors.

substituted for a, 2

a2 (-6) c2

a2 a

62 (

- c) 2

columns; hence, by Art.

12,

(a

4-

b -he) 2

is

a

RULE OF SARRUS

128 Hence, since

A

is

of the sixth degree,

the remaining factor necessarily being of the first degree and symmetrical, since the cyclic substitution (a, 6, c) gives a determinant equal to A.

Putting o =

whence

N

l, b

= 1, c~l, we have 4

1

1

1

4

1

1

1

4

= 27iV,

2.

EXERCISE XV A. 1.

The

SIMPLE EVALUATIONS

positive terms in the expansion of (aj}.^) are

fliVa*

Hence the

rule of Sarrus,

which

is

btCMz, as follows

Negative terms

Cjtta&a. :

'2->

V*/

^r

a2 b,c 3j.--'

a3.'3 ' C a3 "3. 3^

The determinant (!^ 2 C 3)

-

D=P-N

is

the

sum

of the products of the three elements lying

on each of the six lines in the above diagram on the left, the sign of the product - according as the line which determines it is drawn downwards or being -f or upwards, proceeding from left to right. A more convenient way of using the rule in numerical cases is shown on the right. 2.

3.

Apply the

Show

rule of Ex. 1 to

a

h

g

h

b

f

g

f

c

show that

= abc -f 2fgh - a/2 - bg* - ch 2

that, in the expansion of (^i

.

following terms occur

:

-f c

and that the other

positive terms are obtained

from these by the substitution

SIMPLE EVALUATIONS

T

129

any term in the expansion of (a^Ca ...) and another term T' is T by a cyclic substitution involving r letters, the order of the suffixes being unchanged, then T and T' have the same sign, or opposite signs, according as r is odd or even. [Exx. 1 and 3 are instances.] 4. If

is

derived from

5.

In the determinant (a 1 6 a c 8 Z ), the ^i> sign of the term a w 6 n _ 1c w _ 2 M second diagonal has the sign -f if and only if n is of the form 4k . . .

forming the or 4& + 1. 6.

'

'

show that the terms

In the expansion of (a

-f

aj

occur. 7.

show that the terms

In the expansion of

4-

a 26 4c 6^ 1 e 8/

occur. 8,

10.

Expand

the following

:

Prove that

m

is

-^-M^-f m +nz )-t(L l

equal to

where 11.

L lf

l

2

Jlf2 ,

^

8

are the cofactors of

1 19

m^,

n3

in

Prove that I

m

a;

n a;

y values whatever. w have Z, m, any [From the first, second and third columns

where

fourth column.]

respectively take a,

/?,

y times the

ADDITION OF ROWS OR COLUMNS

130 12.

abe

Prove that

d

-6 -c

a

-d

d

a

b

-d

-c

b

a

[Multiply the rows by a, -6, -e, 13.

=

(

c

-d

new top

to form a

row.]

Prove that

+

1

,

1

I

i

1i

^ i

i

"*"

a

a

1

i

i

'

a

a

i

1

and that a

similar result holds for a determinant of the nth order

and of

this

form.

[Subtract the top row from each of the others and expand with reference to the top row.]

and u ll =

u

14. If

o 2 [From the fourth column subtract x times the and the second, third.]

15. If

u=ax*

y*

2

.'y

ax + by

2

-xy

x

b

c

b'

c'

a'x

prove that

,

bx + cy

+ b'y

b'x

column, 2x times the

first

1

+ c'y

ax + by

y

a'x \-b'y

'

[Add x times the second column column to y times the second.]

to

y times the third

:

then add x times the

first

16.

Prove that a

b

ax 4- by

b

c

bx + cy

ax + by

bx + cy

[Multiply the 17.

first

column by

the second

-x b

equal to

(x

by

z )

2 (ax 4 2bxy

+ cy 2

c

-x

d

a

c

d

a~-x

b

d

a

b

c-x

-a -6 -c -d)(a? -a 4-6 - c +d){x* -

).

and subtract from the

y,

bed

Prove that the determinant a

is

x,

= - (ac - b

(a

-

c)

f

-

(6

-

third.]

APPLICATION OF REMAINDER THEOREM

METHOD OF FACTORS

bed

B. 18.

x+a

Prove that

19. If 25 ~ a

x+b

c

a

b

x+c

d

a

b

c

x+d

+ b 4- c, prove

(s-c) 2

[Show that s' and find

factor,

2

(.s-a) 6

(*~6)

5

--

6,

by putting a

A'

-a)

(s

2

c

are factors

2,

1

1

:

deduce that

Ns is

the remaining

&~0, c0.]

20. Express the following in factors (i)

2 2

2

- a,

s

,

d

that

3

(a

-x*(x + a + b+c + d).

a

a - a 6)

131

:

(ii)

(iii)

y 1

A-h/x 21.

Prove that

[Divide the '6'

/x,

22. If

c/c'

bc'

ca

caf 4- c'a

c'af

ab

ab' + a'b

a'6'

row by

first

b'c',

= y, and use the

A-~y4-aS, I

fA

I

A

=

b'c'

the second by example.]

c'a',

the third by

a'6'.

last

,

prove that

I

V

fJL

^

A2

+ b'c

be

va

Hence show that the following determinant has the same value

23. Prove that

x3 a

y

By

2

x

I

a2

a

1

a;

3

3

y

2

y

i

equating the coefficients of x on each side, show that 1

au

y

3

a2

1

2

i

y

:

Put

a/a'

A,

COMPLEMENTARY MINORS

132 24. If

w

is

a root of #

4

= 1, show

that a

4- baj

abed b

c

d

a

c

d

a

b

d

a

b

c

+ cw z + dw 3

Hence show that the determinant is equal to - (a + b + c 4- d) (a - b + c - { (a - c) 2 d)

14.

Minors of a Determinant.

-f-

(b

is

a factor of

- d) 2 }.

The determinant obtained by

omitting any number of rows and the same number of columns from a given determinant A is called a minor of A. If

we omit two rows and two columns, the resulting determinant is and so on. from A we omit the rows and columns which contain every element

called a second minor, If

of a given minor,

to the

first.

we

Thus

get another minor which is said to be complementary (a 1 6 2 ) an d (c3 eZ4 e5 ) are complementary minors of

the interchange of rows and columns, any minor of A can be put into the left-hand top corner, and then the complementary minor can be

By

seen at a glance.

15.

M

i

Expansion of a Determinant by means of

n o rs

its

Second

Consider the determinant

.

In the expansion of A, the

sum

of the terms

which contain a v hz

is

and the sum of those which contain a 2 b l is - a^b^c^d^) Hence every term in the product (^^(c^d^) occurs in the expansion. Again, every term in A contains one element out of the first row and one out of the second, and these elements have different suffixes. .

Hence A

is

sign, of the

where the rows

equal to the

sum

of all the products, each with its proper

form

first

is any minor formed from elements of the first two two columns), and the second is the complementary

factor

(or of the first

minor.

The sign

of a

product

is

determined as follows. Take, for example,

EXPANSION BY MEANS OF MINORS In the expansion of this product, the term suffixes in the

same

133

occurs.

Keeping the

order, observe that

abcde Therefore, in the expansion of J, this term occurs with a negative sign, sign of the product is

and the

More

.

generally,

if

any two rows

of a

ar

br

cr

df

br

cr '

r

...

A

determinant

hr

... rif'

kr

...

.

.

.

kr

'

...

. .

'.

are

,

,

where hr hr belong to the sth column and kr kr to the s'th column, then A -ZX - I) r + r '+*+*'(h r kr >) (the complementary minor of (h r kr >)). >

*

,

,

.

The summation is to include all the minors which can be formed from The sign of the product may be determined as above, or as in Art. 7 by counting the number of moves required to bring h r to the place occupied by a x and k r to the place occupied by 6 2

the two rows.

'

>

-

Thus, in the instance given above, the sign of

In a similar

way we can expand

(b 1 d 2 )(a 3 c4 e^) is

given by

a determinant in terms of the rth

minors which can be formed from any r rows (or columns).

NOTE.

In the

c

double-suffix

following convenient form: is

sign

Ex. It

is

1.

'

In

notation of Art. 20, the rule takes the a rs ...) (complementary minor), the (a>j> q

1 VP+^'t" *"+*+

(

Expand A = (a 1 6 ac3rf4

)

in terms of the minors of the rows containing al and a,.

easy to verify that

+(A)(to) Ex.

2.

Prove that if aiui

<>i

c8 c*

m

Pi Pz

P*

l

nt

m^

rc 2

m2 Pa

Considering the minors of J, which can be formed from the elements of the first three rows, the only o^e not equal to zero is (a 1 6 ac a ), and the complementary minor is Moreover, the term a^c^^m^n^ occurs in the expansion of J, therefore

Again, A second and

is

transformed into A' by the interchange of the first and fourth, the and the third and sixth columns. Whence the second result follows.

fifth

MULTIPLICATION OF DETERMINANTS

134 16.

Product

As a typical case

of Two Determinants of the we shall prove that bl

4-

Denote the determinants on the a,

left

61

m

by A and

a^

1

A

;

4-

bl

same Order.

m

then we have

al

6,

a2

Cinvn

-1

-1

-1

-1 where the second determinant the third column add fourth column

2

l

times the

v

add L times the

Hence, as in the

last

JJ'H-1)

obtained from the

is

first

first

and

m

and

m

l

as follows

first

times the second

:

To

to the

:

times the second.

example,

2 . I

+v

aj, 2

4-

-1

bfl

-1 whence the

result.

The product of two determinants of (he n-th order (n>2) can be expressed in a similar form, and by the same method. Second method.

It

is

aJi

4-

b^wii

a3Z1

4-

b3

m

The determinant A

required to show that

+ c^n-^

l 4-

is

c^ the

^A, where

al 2 4- b^w

4-

61

m3

a3

4-

63

m

sum

2 4-

b^i

of 27 determinants,

two

4-

cx

3

of

which are

and

wliich are the

same

as ^i

ai

c\

b2

a2

c2

and b3

The second

of these is zero,

and the

first is

b3

equal to

With regard to the sign, observe that Iim 2n 3 is transformed into by the same number of interchanges of letters as those required to change a 1 62c3 into b^c^. This number is odd, and therefore the sign is .

Of the 27 determinants, all are equal to zero except the six of the first type, in which a's are in one column, 6 s in another and c's in the third. J

RECTANGULAR ARRAYS

135

Moreover, for reasons similar to the above, (a^c^) is a factor of each of the six, and the other factor is a term (with the proper sign) of (Iitn 2 n$), there-

J = (a

fore

A

6 c

m

(I

)

n

).

similar result holds for determinants of the nth order (n

= 4,

5

...).

17. Rectangular Arrays. A number of symbols arranged in rows and columns is called a rectangular array. If the number of rows is not equal to the number of columns, the array is generally enclosed by double lines, and is sometimes called a matrix. Let the number of columns exceed the number of rows.

(1)

Consider the arrays

i

!;

Proceeding as in the

al

a2

li

"2

7

C2

last article,

fro

m

l

n^

Wl I lit)

W tin

\

we form the determinant

J = al + bm + cn !

By

reasoning similar to that in Art. 16

This process

is

it will

be seen that

called multiplying the arrays.

determinant formed thus from two such arrays is equal to the sum of the products obtained by multiplying every determinant which can be formed from one array by the corresponding determinant formed from the

In general,

the

other array. (2) Let the If

number of rows exceed

number of columns.

the

from the arrays a2

b2

1

2

Z

3

I

a3

we form

||

the determinant

^2^3 -f

it will

63

be seen that

J =0. J

=

b 2 m^

+ bm 22

This follows, as in Art. 16, or from the fact that &x

.

I

^

62

1

2

b,

Z

3

m m2 w3 l

In general, the determinant formed by multiplying two such arrays

is zero.

RECIPROCAL DETERMINANTS

136 Ex.

// *r = a

1.

r

r

+

+ yr + 3r

,

prove that

^2

S3

The

first

54

two

55

^3

&A

S6

results are obtained

by squaring the arrays

1111

1111,

aj8y&

a

The

last is

Ex.

2.

8

y 2 y

j3

a2

2

S2

got by squaring the determinant in Ex. 2 of Art. 13.

Prove

that

2(a t - a z ) (o a - a 3 ) (a 3 - ^5(6! - 6 2 ) (6 2 - 6 3 )

This result

is

obtained by forming the product a,

Reciprocal Determinants.

18.

where

^ij,

B

l9

...

2

-2b 1 -26 2

I

Let

are the cofactors of a l9 b v

...

6j

622

^ = (^63

in A.

...

k n ) and

For reasons connected

B

are sometimes called inverse with linear transformations, A v 19 ... elements, and A' is called the reciprocal determinant. (1)

//

A

is

a determinant of

minant, then

The .following identities of the

A'

method type

=A

.

applies in all cases.

al A l

is the reciprocal deter-

n ~~ l

Let A==(a i b2 c3 ), then using

+ fej JBj + c l C l == A

aA% -f 6^82 + it will

and A'

the n-th order

c i@2

,

" 0>

be seen that

A

J A whence

it

follows that

J'= J

2 .

SPECIAL METHODS OF EXPANSION

Any minor

(2)

of A' of order r

is

137

equal to the complement of the corre-

r ~l

provided that A^O. sponding minor of A multiplied by A As a typical instance, let J = (a 3 62 c3d4 ). We shall prove that, ,

then .

We

A

and

(A,B<>C) =
if

2 .

have

A A

c2

A

C^

7?

1

64

4

000

C4

1

ca

d1

c2

d2

c3

d3

c,

d.

therefore

and The second identity

is

obtained by multiplying

11

jLjL-\

1

JLJ-\

juL

2

-^2

Oi

/y.

2

2

1

D B 0001

AS

Two Methods

19.

C3

3

of Expansion.

reference to the elements of a

With

(1)

3

A by

and

Let

row and column.

^= m n

m Let ^j,

jBj, ...

be the cofactors of a v bv

...

r

in A.

In the expansion of K, the sum of the terms containing r is Ar every other term contains one of the three I, m, n and one of the three V m', n'. :

>

and

Again,

are complementary minors of

r hence, coefficient of IV in

K=

coefficient of a^r in

= - coefficient and

of a x

in

K J = - Al

;

similarly, coefficient of

mri

m K=

- coefficient

= - coefficient Thus

it will

K

of c2r in

K

of c2

J = - C2

in

.

be seen that

Ar - {AJl' + B2 mm' + Cz nn -f

f

4-

C2mn'

B3m'n + A 3nl' + C^ril + BJm' + A zl'm}.

A

;

DOUBLE-SUFFIX NOTATION

138 (2)

With reference

Let

a diagonal.

to

c.

^

/? 2

C3

,

D

,

4

x2

-f

is

+ #3^64 -f x^

# B2 -f

03 +

J

Note that the expression

x)

in brackets is the

sum

of

x^x^a^ x

terms derived from these by the cyclic transpositions

The 20. in the

verification

d2

c2

be the cofactors of the elements in the leading diagonal

Cr^efy? 4

7^4

a2

2

of J, then the value of A'

-f

K=

&3

3

Let

and

dl

Cx

&!_

a2

is left

xj) 3 c 2

and

to the reader.

A

Definitions.

}

(a&c), (123).

determinant of the nth order

is

often written

form

n

lw

Denoting any element by a rs the determinant is said to be symmetric a sr = a rs a sr = -a rs the determinant is skew-symmetric: it is If ,

if

.

,

implied that

the elements in the leading diagonal are zero.

all

For example,

if

a h

first

b

g

1

,

L

z

~

f m

g

f

c

I

m

n

n

z

-y

K

is symmetric and L determinant is bordered by

the determinant

that the

h

is I,

y

Ox -x

skew-symmetric. m, n.

We

also say

Symmetric Determinants.

21.

// A rs A 8r are the cofactors of the elements a rs> a sr of a symmetric determinant J, then A r8 = A sr (1)

,

.

For

A r8

Thus

is

transformed into

A sr

by changing rows into columns.

^ = (011

if

au

a 12

a31

a32

a*>.

OM), Z

a21

32*

SYMMETRIC DETERMINANTS (2)

For the determinants

J=

a

h

h

b

9

f

K=

139

a

h

g

f

h

b

f m

c

9

f

c

m

n

g

,

I

we have

the following important relations (i)

EC - J 2 - Aa, GH-AF = Af,

where A, B, ... are the cofactors of These follow from Arts. 18, 19.

:

etc.

;

a, 6,

...

in J.

Consider the following symmetric determinants

Let A, B, if J=0,

n

K = - (Al* 4- Bm* + Cn*+ 2Fmn 4- 2Gnl + 2TZm),

(ii)

(3)

I

be the cofactors of

...

then

AK==Z -(Al +

Because A = 0,

...

a, b,

:'

We

in J.

shall

prove that,

Hm + Gn)*.

the coefficient of r in the expansion of

K

is

zero, therefore

- AK = A (Al* + J3m2 + Cn2 4- 2*W -h 2flW 4- 2ff Iro).

^J5-ff2 = Jc,

C^-(? 2 -J6, GH-AF=Af, etc. Hence AB = H2 AC^ G 2 AF = G^F, etc. and therefore - AK = ^ 2 Z 2 + #2m 2 + 2 n 2 + 2GHmn + 2^GwZ -f 2AHlm = (Al + Hm + Grip. Moreover,

9

,

,

way we can prove that - BK = (TZ 4- Bm 4- JFn) 2 and

In the same

More generally, using

the notation of Art. 20,

A n = (a n a22

...

,

...

and

let

A U) A 12

inA n r Then, ifA n _ l A nA n (A u a ln + A 12^2n + .,

.

let

^ n -i==(a n a 22

a nn ),

the determinants being symmetric,

a 12

- CX - (67 H- J?m 4- Cn) 2 a n-i> ,

...

n-i)>

be the cofactors of a n

,

)

with similar values for

A 22A n A^A n ,

,

*

A!, n _ 1 a w _ 1 n ) ,

,

....

asr = ar8 A 8r ~A r8 and (1), since expanding J w -i == 0, we have A n - - (^ n a ln 2 + ^ 22a 2n 2 + + 2^ 12 m2n + Now, since A n _l = 0, every minor of the reciprocal determinant as

For,

in

Art.

19,

,

)

(-^11^22 is

zero.

Hence, by

Using such

K

^n-l> n-l)

and A^A^^A^A^. (2), A n A 99 ^A^ follows that A n A n has the value stated.

Art. 18,

identities, it

B.C.A.

SKEW-SYMMETRIC DETERMINANTS

140

The following

(4)

result is of great importance.

Consider the symmetric determinants

a-x

h

a-x

h

h

b-x

h

b-x f

9

c

g

a-x

f

h

~x

g

h

m

I

J 2 J 3 J4

Denote these by

J 2 ^3

Jj,

Then

above type.

n

d-x

n

a-x, and suppose

Let A^

.

I

m

the sequence

be continued indefinitely by constructing determinants of the

t

>

,

,

g

b-x f f c-x

the roots of the equations

4=0, J 2 =o, J 8 =o,..., are all real, and the roots of any one of them are separated by those of the

preceding equation.

A 1? A 2 (A 1
If

cise

XI,

x has the values -

oo

,

/x2 ,

/u, 3 ,

if

,

we have

/x t

x has the values

J4

are

so

Exer-

1


= 3

^1

2


then A^A^

+,-,+>->

+

fa,

are real

negative. -f

-

/^2 ,

-f

Hence .

in ascending

.

oo

,

is

are

is

negative; hence,

,

;

and are separated by those

of

Zl 3

= 0;

Skew-symmetric Determinants.

For

cofactors

For

(see

on without end.

22. (1)

J4 =

if

zJ 3

and denoting them


/^3 ,

oo

therefore the roots of

and

A x
then A^A^

the signs of

,

are real,


Again, as in the last section,

the signs of

J 2 = 0,

A 1? A2 oo

J^O

Therefore the roots of order by fa,

if

then

2).

Also, as in the last section, if

J 2 = 0,

are the roots of

skew-symmetric determinant (^u - n ~l A rs ofa rs asr then A sr = ( l) the

,

A r8

22 , ...

a nn ), if

A rs A 8r ,

are the

.

,

A sr

by changing the sign of every element and then changing rows into columns and since A rs is a determinant of (n-1) rows, the result follows. Thus in the determinant is

transformed into

;

b

c

-c -6 the cofactors of c and

c are

-c -b and one

of these

is

(

a

-a

2

1)

a

and

-

times the other.

-a

THE PFAFFIAN

141

Every skew-symmetric determinant of odd order is equal to zero. For if A is a skew-symmetric determinant of odd order n, and A' (2)

is

the

determinant obtained by changing the sign of every element in A, A'

But A' can

= (-\} nA=-A. from

also be obtained

thatJ'=J: whence

A by

changing rows into columns, so For example,

the result follows.

c

-c -6 (3)

-a

Every skew-symmetric determinant of even order

(i)

A n = (a u a 22

Let

. . .

J n ^ = (a n a 22

a nn ),

.

.

a perfect square.

is

a^,

.

n _ x ),

where

n

is

For n = 2,

even and the determinants are skew-symmetric.

n

2

12

.

*12

Let

^4

U

^4 12 ,

,

...

,

A n _l =

by the preceding that a sr =

-a rs

,

... in J n _!. n = and A sr (-l) ~ 2A rs = A rs

be the cofactors of a n a 12

exactly as in Art. 21,

A ll A n = (A ll a ln + A

l

(3),

Since n

,

.

is

even,

Remembering

we can show that

^ 2n + ...+A lin _

a n _ ltn )* ................. (A)

l

A n is a perfect square, so also is A n moreover, A u is a skewdeterminant of order n - 2 and it has been shown that such a symmetric determinant of order two is a perfect square, therefore the same is true for Hence

if

;

;

those of orders four, six, etc. (ii)

If

n

is

even and

J n = S n2

the value of 8 n *

,

is

given by the following

rule.

Write down all the arrangements in pairs of the numbers Denote any such arrangement by (M), (Im), ... (uv), then %n = Z(

a hlfllm

where the positive or negative sign ... uv is derived from 1234

hklm

The term a l2a u

transpositions. (a)

where

The

rule is suggested

A n A 12 A 13 ,

,

is to be ...

...

a n _ lt n

,

which agrees with the *

is to

or by

have the sign

,

J3

find that

rule.

Named

the Pfafflan,

an odd number of

when n = 4. By

Whence we

n.

taken according as the arrangement

are the cofactors of a n a 12 a 13 in

A 12 = -

...

a uv)>

n by an even

by the case

2, 3,

1,

by Cayley,

after J. F. Pfaff.

,

+

.

equation (A),

so that

VALUE OF THE PFAFFIAN

142 (j3)

n-2> we shall holds when n = 4,

that the rule holds for determinants of order

Assuming

prove that it holds for those of order n. Then, since it will hold for w = 6, 8, etc. By equation (A),

Now J n _ 1 = 0,

therefore

it

Art. 18, (2),

by

Hence

etc.

by the rule. We shall show that the signs on the right are alternately positive and negative. The values of JA n and Consider the product P ^m^^n a vn\JA J>v are rule from the sets the obtained by JAgp

where the values

of *JA n ,

.JA^,

etc.,

are given

.

2, 3, 4,

n-1

...

and

1,2,3, ...y-l,p

+ I,

...

n-1.

Moreover, every term in the expansion of P, with the proper sign, must occur in J n .

For clearness, the element a rs If p

the product

is even,

P

will

be denoted by

contains a term

M,

(rs),

so that (sr)

which

is

= - (rs).

the product of

and

By changing the order of the numbers in each of the J(n - p) 4- 1 pairs which are underlined, it will be seen that ( - !)*(*-*)+* jf j g the product of elements (rs) in which corresponding values of r s are as below. y

1,

2,

2,

3,

3,

...p-2, p-1, p, p +

4,

...y~l, p + l,

The numbers those in the

n,

p + 2, 2> + 3,...w-2, n-1, p + 3, p + 2,...n-l, n-2,

l,

p,

in the second line can be written in the

first

by %(n+p)

transpositions,

and

n, 1.

same order as

since

%(n-p) + l+i(n + j?) = n + l, an odd number, it

follows that

value of 8 n

//

p

is

1)

numbers

is n,

It

of

<

is

a term in

Jn

,

so that

-#,> nNMj>j> occurs in the

odd, the corresponding process requires changes of order in - 1) transpositions. Since the sum of these pairs, and i(n + p

i(n -p +

rule.

M

.

which

is

even,

+a pn *JA p9

occurs in 8 n

;

thus

remains to be shown that the value of S n can be written down by the It is obviously necessary only to consider the first term in the value

MISCELLANEOUS METHODS This

143

(pn)(l2)(34:)...(p-l,p +

is

or

according as

p

is

Now

even or odd.

the arrangement

+ l,

p, n, 1,2,3, ..._p-l,p

...

n-l

n by ^ + n-l transpositions. The rule asserts that the sign of the term is + or - according as p + n - 1 is even or odd, that is according as p is odd or even, which agrees with what has can be derived from

...

1, 2, 3,

been proved.

EXERCISE XVI

MISCELLANEOUS METHODS

A. 1.

If a

+6

-I-

c

=0, solve the equation

a-x c

b

b

a

0.

c-x

a

b 2.

c

-x

Prove that (&iC 2 ) (#^2) "f ( c i a z) ftA) -f

(0iW (^1^/2) =0.

[Use the identities tti(6 lc 2 )

3. If (a l b 2 c 3

+ 6i(c

1

o2 )

)=0 and

+c

a 2 (6 1 c.2 ) 4-62(0^2)

(o 1 6 2 )=0,

1

(^62^3)

-f

=0, prove that

(& 2c 3 )

4.

Prove that c a -fa 2

5.

aa

Prove that

- a2

l/(a +

6. If

ca

)

!/(&

-

+ *)

l/(c

+ a) ''Q'

where

Q

is

the product of the denominators, prove that

03(^62) =0.]

MISCELLANEOUS METHODS

144 7.

Show

that c

8.

is

b

a

a

b

Prove that the

'

6

a

a

b

-(a + 64-c)(6-hc-a)(c+a-6)(a + 6-

c c

skew

*

determinant

x

c

c

x

b I

b

I

-a

x

m

n

n x

am

equal to 9.

Prove that

6

c

a

c

10.

is

=

6

a

a

6

c

6-c c -a a -6 a + 6-f-c

=2(a + 6 + c){a6c

-

(6

-

c) (c

- a) (a -

6)}.

Prove that

equal to (a + l)(6-

1 lj

[This follows from Ex. XV, 13, by the same method.]

by putting

1

+a ~ l

-

I/a, 1

+a

2

or, directly,

11.

Prove the following (i)

(ii)

(Hi)

(iv)

:

1

=

- ax

by

cz.

- (ax + by + 02) (ax' + 6t/' + cz').

-6)(a + 6-c).

where 2*=

-

1/6, etc.

;

PRODUCTS OF DETERMINANTS 12.

145

Denoting the squares of the differences ft - y, y - a, a - j8 by - 8, - 8 by I', m' n' 8, /? y respectively, prove that

and those of a -

/,

m, n

9

prove that

determinant A, when x= a ~ the same, then (x -a) p l is a factor of J rows become (i) p - a) p + ~~ a (ii) p rows become the same and q columns become the same, then (x a factor, provided that the common elements are zero or are divisible by (x - o) 2 14. If in a

:

;

ff

is

.

B. 15.

PRODUCTS

Write down as a determinant the product b

is

c

y x

a

Z

:

z

y X

Hence show that the product of two expressions of the form a 3 + 6 3 -f c 3 - 3abc of the same form. 16.

Prove that

[Consider the product

17.

By

forming the product on the

a*

where J 1 =a 1 c 1 -6 1

1 ,

c,

-26,

a,

c.j

- 2&2

Oj

I^^

left,

prove the identity on the right,

-^b^

etc.

24, 7 12

2J

/31

/

/,

2J 3

PRODUCTS OF ARRAYS

146 18.

forming the product

By

a+

ib

c+ld

-c +

id

a - ib

b'+la' -d'+lc'

+

d'

ic'

V - la'

prove that 2

(a

H-

62

+ cz + d2

where l=(6c'), 19.

(a

)

/2

-f 6'

2

+ c'* + d'*)

m = (ca'), n = (ab')

9

V = (ad'), m' = (M'), n' = (ci').

2 2 2 2 Express (P + 2 4-3 + 5 ) as the sum of four squares by taking c, rf and 2, 5, 1,3 for a', &', c', d'.

1, 2, 3,

5

for a, 6,

20.

By

multiplying the arrays on the

left,

prove that the determinant A

V m'

V

I

2ir

m

m'

m

lm' -f I'm

n

n'

n'

n

nl'

I

Hence prove that

if

+ l'm 2mm' mn' + m'n Im'

+ n'l

ax*2hxy+ by* +-2gX'*r2fy + c

\

nl'

21.

h

g

h

b

f

g

f

c

+ n'l

2nn'

the product of two linear

=0.

Prove that =0.

[Multiply the two arrays of four rows ||

al

a^

1

1

etc.

Prove that (i)

x

1

1

...

1*1... 1

1

X

...

the determinant being of the nth order. (ii)

V

etc.

MISCELLANEOUS THEOREMS AND EXAMPLES

C.

22.

-26 t

a

ab

a + nb

a

a + b.

a + nb

a

a + (n-

a+b

1)6

a + 2b

a

+ 2b

a + nb

a. (n-

zero.

mn' + m'n

factors, then

a

is

APPLICATION TO GEOMETRY

147

23. If

prove that u n ^a nu n

[Expand with reference to the bottom row.

Such determinants are

called

continuants,,]

(ii)

sin

Hence show (j3

y) sin (y

f sin (y

(iii)

that, If a,

Use the

- a)

-a)

sin

,

(a

sin (y

last result to

-

y, 8 are

any

+ sin

-

/?)

8) sin (a

(/?

-

show that

angles,

y) sin

8) -fsin (a

if a, 6, c,

(j3

/?)

d and

8) sin

sin (a

7 ,

b', c', d',

where

Z

[(ii)

In

(i)

put

m = (ca'), ~ cos z

(iii)

Let

a, 6, c,

circuits a, b,

c,

n=(a6

x = cos 2a 4- 1

2y + 1 sin

a', &', c', d'

sin(a-8)__

x

),

sin 2a,

d make angles

d and

8) sin (ft

6, c,

-

8)

= 0.

are the lengths

d are parallel to

LMN + Lmn + Mnl + NlmQ,

respectively, then

= (6c'),

-

8)

a', b', c', d'

of the sides of two quadrilaterals such that the sides a,

a

-

(y

2y,

a,

on

_

/?,

L = (ad')> y = cos 2

y, $

M=

2)9 -f

= cos 28 +

(bd') 9

t

sin

1

sin 28.

with a fixed

N = (cd').

2/J,

line.

By projecting

lines perpendicular to d, a, 6,

_sin(ft-

c,

the

prove that

APPLICATION TO GEOMETRY

148

Prove that the equations (i), (ii) are respectively the equation to the joining (x^y^, (x t y t ) and the equation to the plane through 26.

x

x

(ii)

*

2/2

Show

-0.

1

y

also that the last equation '>__'* A/

W

X/j

^2

"~

_

'i

Vi

*i

^2

2/2

^2

'3

2/3

23

be written in the form

may 41

line


Mj

~

#1

2/1

2/2

fi, C, Z> are four points in a plane and the lengths BC, CA, AB, AD, are denoted by a, b, c, x, y 9 z, prove that these are connected by the

27. If ^4,

BD,

CD

equation c2

62

z2

1

2

2

1

a b2

y

a2

x*

z*

z2

I

Q

1

-0.

11110

[Let (x l9

y^

(x 2 ,

y*

be the coordinates of the points.

y 2 ),

X2

1

l/ 2

1

!

1

-2z

Multiply the arrays

000 28. In a tetrahedron

by

a, 6, c, d, e,

sphere

is

y

...

,

;

the volume

is

F, the radius of the circumscribing

E.

Assuming equation

A B

A BCD, the lengths BC, CA,AB,AD, BD, CD are denoted

/ respectively (i),

prove equation

where (x v y lt

Zj),

(x 2 ,

y2

,

z 2 ),

...

are the coordinates of

(ii).

and use the equations z.i

*

Note the value of

this

determinant given in Ex.

)^c>, 11, (iv).

etc.]

CHAPTER X SYSTEMS OF EQUATIONS 1.

Systems of Equations.

variables x, is

t/, ...

,

which

If

is satisfied

w=

is

an equation connecting the ... then (xj, j/j, ...)

when x = x ly y = y ly

,

u = 0.

called a solution of

= 0, If the be a system of equations connecting x, y, solution are said to be have a common con(x l9 y l9 ...), they equations sistent, and (x l9 y v ...) is called a solution of the system u^Q, v~0 ... Let u =

0, v

. . .

. . .

.

9

.

Equations which have no common solution are said to be inconsistent. Equations which are homogeneous in x,y, ... have the common solution (0, 0, ...), in

which each

of the variables is zero

some other common solution

be consistent unless

:

but they are not said to

exists.

one equation of a system is satisfied by all the values of x, y, ... which satisfy the others, the equations are said to be not independent. To eliminate a variable x from two equations is to obtain from these If

two an equation

free

from

Any

x.

result so obtained

must hold

if

the

equations are consistent.

To exist).

solve a

This

system of equations is to obtain done by a process of elimination.

all

the solutions

(if

any

is

Illustrations.

The equations x + y = 1, x + y = 2, which represent

planes), are inconsistent. The equations u 0, v

So also are the equations

= 0,

lu

+ tnv-0

u~ 0, v = 0,

parallel lines (or

2% + 3v = 4.

are not independent.

Equivalent Systems. (1) Two systems of equations are said to be equivalent when every solution of either system is a solution of the other. In such cases we may say that one system is the same as the other. 2.

(2)

// a v a 2

,

are constants

...

MI is equivalent to the

= O,

and a l ^0 w 2 =0,

...

)

the system of equations

w n = 0,

system

For any solution

a solution of the second and, since a l is a constant (not zero), any solution of the second system is a solution of the first. of the first

system

is

SIMULTANEOUS LINEAR EQUATIONS

150

m, V m' are constants such that Zw'-Z'ra^O, then = is equivalent to the pair equations w 0, v If

(3)

I,

9

lu

+ mv

Vu + m'v = 0.

0,

For any values of the variables

the pair of

z, y,

which

...

satisfy

w = 0, v =

also

satisfy the second pair.

Moreover, any values of

x, y,

satisfy

m'(lu

-

which

...

+ mv) - m (Vu + m'v) = 0,

+ mv)+

l(l'u

- I'm) u - 0,

(Im

l'(lu

satisfy the second pair also

+ m'v)~0,

which are the same as f

(lm

and

since

It

Irri

is left

-Z'w^O,

these are the

1

same as w = 0,

Any

(5)

the two pairs

M2 = V 2

1/^2-^2, But

:

solution of the system

^1=^1,

W2 = v2

.

the second pair of equations is equivalent to the two pairs

w1 = t;1

,

u%

= v2 and w2 = 0,

t;

2

= 0.

Systems of Linear Equations.

3.

the chief difficulty (1)

is

formal proof

(2)

to include

The equation ax + by + c

6=0, e^O, in which is

all

In considering this subject, the special cases which may arise.

has infinitely

many

solutions, unless

case there is no solution.

hardly required.

Consider the equations

excluding the case in which either (i)

= 0.

to the reader to supply proofs of the following

a solution of

A

t>

The pair of equations uv = 0, w==0 is equivalent to and v = 0, w = 0. w=0, w =

(4)

is

- I'm) v = 0,

is

of the

form Ox + Oy + c = 0.

If (a 1 62 )^0, there is an unique solution. 2, (3), the equations are equivalent to

For by Art.

Wj62

which are the same as

-w

2 &!

=0,

- w xa2 + u

<&

= 0,

INCONSISTENT EQUATIONS (ii)

//

(a 1 62 )

= 0,

if

(ai& 2 )

= 0, we

For

151

the equations are inconsistent unless also

have w x 62 - u2 b

(b^) =0 and

s

Hence the equations cannot have a common solution unless (b^) and c a ( i %) are

both

zero.

If (a 1 62 ), (ft^), (c^) are

(iii)

same, and there are infinitely For in this case

all zero, the

many

two equations are one and the

solutions.

and

u^a^u^av Hence any solution of one equation is a solution of the other, or else an a is zero and a 6 is zero. The latter alternative cannot occur. For example, suppose that a^O; then by hypothesis fc^O and a 2 61 = a 1 62 = 0, therefore a2 = and b2 ^0. Hence the equations are one and the same. uj) 2 =zuj)i

The case in which

(iv)

the equations are inconsistent

may

be regarded from

another point of view.

Suppose the (6^2)

-

bja

We

&2/a

finite

;

y to vary in such a way that (a^-^O while then #-> y->oo and the fractions x/y, ,

equality.

= say that in the limiting case when (a^) 0, the equations are - fe a l or - 62 a2 x by infinite values of x, y which are in the ratio

shall

satisfied

(3)

coefficients of x,

and (c^) remain l9 2 tend to

:

:

Ifa^x + b l y =

and a 2x + b<$ = 0, then

either

x = 0,y = Qor

else

.

(a^) = 0.

Conversely, if (afi^^Q these equations have solutions other than (0,0), they are consistent.

i.e.

This

is

a particular case of the preceding.

The following Ex.

1.

=0

=0 or

and (a x c2 ) ^=0, then (6 x c 2 ) (o^) satisfy the equations a x 6 2 aj} { =0,

//

For a lf a 2

results are required later. else

a x =0, a a =0.

a^ - afa ~0.

are not all zero and Ex. 2. // (a 1 6 8 )=0, (6 1 cs )=0 and (c l a 2 )=0, where a lf b lt #2 ^2> c a are not all zero, then of the pairs (a l9 a 2 ), (b l9 6 2 ), (c lf c2 ) there is at least one in

q

which neither quantity is zerb. For suppose that fl^O, then 6 t =0,

if a a =0 we should have o^ =0, a^ =0 which the contradicts =0, Cg hypothesis. Hence, if a^O, then

;

Ex. 3. Let A =(0^3) and l& A\> B oe the cofactors of a lt 6j ..... A =0 and CB = Q, then either Cl =0, C2 =0 or else A B =0, B3 =0. For we have a l C l 4 a2 C 2 = 0, &!
and therefore

Prove that if

^=0, C2 =0, or else (6 1 c 2 )=0, (c 1 a 8 )=0, i.e. ^4 3 =0, B^-0. Or thus: Using identities of the type (^^3)= Ja t it will be seen that (OjBa), (C 2 A 9 ), (C 2 B3 ) are all zero. Whence the result follows. therefore either

,

(Ci-4 t ),

LINE AT INFINITY

152 4.

Three Linear Equations.

Consider the equations 0,

Let ^^(a^Cg), and

let

.4^

B

...

l9

be the cofactors of a x b l9 ,

(1) //"
For on account of the

.

J = 0.

identities

+ a 2 C 2 -f o3 C3 = 0,

a l Cl

C^ +

we have

...

Hence, if the equations are must have

(7 2

w2

satisfied

by the same values

of x, y,

we

AQ.

= If A

Ae equations are consistent and have an unique solution except when every C is zero ; in which case each of the equations (assumed to be distinct) is inconsistent with either of the others. (2)

For by the preceding, u l C l + u 2 C 2 + w3 <73 = 0. (i)

// no

C

is zero,

the equations are not independent.

= and w 2 = 0, w3 = are (ii) // CjL-0, C 2 7^0, C 3 =Q, then u 2 C 2 4- u 3 C 3 the same equation. In both of these cases, since C3 ^0, the equations w 1 = 0, w 2 = have an unique solution, which is the solution of the system.

= 0, C2 = 0, C3 9^0, then w 3 ==0. This case is excluded. (iv) // every C is zero and the equations are distinct, each is inconsistent with either of the others. In this case we may say that the equations are (iii)

7/C'1

satisfied

5.

by infinite values of

The Line

say that the

x,

y which are in

the ratio

For the sake

at Infinity.

of

- 6X a r :

complete generality we

'

equation/

Ox + 0y-hc = represents the line at infinity. w = meets the line at infinity

The is

where

c^O,

ideal point in

which a straight line on that line.

called the point at infinity

Parallel straight lines are to be regarded as meeting on the line at infinity.

For the

lines

w = 0, u + k = Q

intersect

on the

line

Ox-f Oy

+ k = 0.

These

conventions enable us to include various special cases under one heading. Illustration.

// u 1 =0, w 2

and A =0.

then the necessary at infinity) is

0,

u3 =0

(as above) are the equations to three straight lines,

sufficient condition that they

may

meet at a point (which

may

be

SYMMETRICAL FORM 6.

Two

Equations

in

Three Unknowns.

Consider the equations

^O,

........................... (A)

= 0,

........................... (B )

excluding the case in which a l9 b l9 (1)

Of the

three

153

(0^), (^2)5

(

or a 2 , 6 2 , Cg are all zero,

Cj

c i a a)>

that at least one,

oppose

say (a l b 2 )

>

is

not zero, then the equations are equivalent to

- %a 2 + u2a l = 0,

Wj&jj

~ w^i ^

which are the same as

-(aA)y + (Cia2 )z=(a ld2 ), -(b l c 2 )z +

........................... (C)

(a l b2 )x^(b l d 2 ) ............................ (D)

z, there are definite values of x and y, and (c^) are not both zero. If (6 1 c 2 = and (c 1 a 2 ) = 0, then x and y have definite values and z may have any value. (In fact, we have ^ = 0, c2 = 0.) Thus in every case, by Art. 3, Ex. 1, there are infinitely many solutions.

Corresponding to any value of

provided that

c (fc 1 2 )

)

(2)

(a v

Suppose 2 ),

(b ly

quantity

is

that

(a^), (b^), (c^) are

all

(See Art. 3, Ex. 2.)

zero.

Also

Then

zero.

6 2 )j (Cj, c 2 ) there is at least one, say (a l9

of the pairs

a2 ), in which neither

we have

t^aj-wg^ss -((*&) ............................... (E) Thus the equations are of the form 1^ = 0, + ^ = 0, where k~(a l d2)/a 2 and they are inconsistent unless k = 0.

^

(3)

In particular

it

follows that the equations

a i x + b^j are always consistent are equivalent to

y

(i.e.

-f

cz = 0,

a 2x

+ 6gt/ 4- c^ = 0,

they have solutions other than

(0, 0, 0)).

They

^/(b^) =y/(cl a 2 )=z/(a 1 b 2 ),

provided that no denominator

is

zero.

=0, they are equivalent to x = 0, y/(Cia 2 )=z/(a l b2 ). = 0, then x = 0, y 0, and z may have any value. (6^2) =0 and (0^2)

If (6^2)

If

(4) If (a,

j3,

y)

is

any solution

of (A), (B), these

may

be written

which are equivalent to

(x-a)/(V2 ) = 2/-^)/(c1 a2 ) = (z-y)/(aA) (

provided that no denominator

is

=0, they are equivalent to

If

=0 and

(0^2)

= 0,

................. (F)

zero.

If (6^2)

(6^)

)

then x=oc,

y==)8,

and

z

may have any

value.

INTERSECTION OF TWO PLANES

154 NOTE.

Much

Equations

labour can be saved by the following considerations. unchanged by the simultaneous cyclic substitutions (Oi6 1 c 1 ), If then we make these changes in any equation derived from (A) and

(A), (B) are

(a*6 2 c2 ), (xyz). (B),

the resulting equation will also hold.

from

For example, equation (D) can be derived

(C) in this way.

The Equation

7.

to a Plane.

The equation

(1)

represents a plane unless a, 6, c are all zero. For the sake of complete generality we say that the

Ox + 0y + 0z + d = represents an (2)

to

Two

'

*

equation

d^O

where

ideal plane called the plane at infinity.

= 0, w2 = be the equations Eeferring to Art. 6, let In general these intersect in a line, and we say that the

^

Planes.

two

planes. equations to the line are 1^=0, u2 Q. The equations may be written in one of the forms described in Art. 6, (4), where (a, /S, y) is any point on the line. The equation to any plane through the line u 1 ==0, w2 = is of the form

u l + ku 2

t

where k

Parallel Planes.

is

a constant.

(a^), (b^), (c^) are

If

zero, the planes

all

^ = 0,

u2 =

do not intersect, that is to say they are parallel. Hence the equation to any plane parallel to the plane ^==0 form ul + k^0, where A; is a constant. All this follows from Art.

We

is

of the

6, (2).

therefore say that parallel planes intersect in a line on the plane at

infinity.

Ex.

1.

The condition

whose equations are

that the line

(x-a)ll = (y-fllm = (z-y)ln

may

be parallel to the plane

is la

+ mb + nc == 0.

If the line

// also au + bp+cy + d=Q, then the line is in the plane. meets the plane at (x, y, z) and each fraction in (A) is denoted by

where

2.

The condition

~~

may and If

meet in a point

if

(x,

y

lll

-a ) = 0, /

m /

y

then

~~

n

z)

= m/m and

which

satisfies

value.

the last equation unless

Hence the

results follow.

whose equations are

~

'

m'

I'

- a') (mn' - m'ri)

(

the lines are parallel if

the lines meet at

r exists

may have any

thai the lines

I

r,

+ mb + nc)-faa

r(la

If /a-fra6 + wc=0, no value of o + bp -f cy + d = 0, in which case r

Ex.

............................... (A)

+(f$- fi') (nl'

~ n'

'

- n'l) + (y y') (lm' I'm) =0,

/

n/n'. f

(x -a)/J=r, (x-
mr~wV

/

/

4-(j8~^ )=0,

,

we have

nr-nV-f (y -/)=0.

CRITERION FOR UNIQUE SOLUTION These equations are

by the same

satisfied

a -a'

P~p' y-y'

(finite)

J

I'

m

m'

n

ri

values of

155

r, r' if

=0,

provided that (mri), (nl') (lm') are not all zero : but if these three are all zero, the values of r, r' are infinite. In the latter case equations (A) are satisfied by infinite values of x, y, z and the lines are parallel. 9

Three

8.

Unknowns.

Three

in

Equations

equations i2/

~*~

ciz

Consider the

+ d-i = .(A)

Let A==(a 1 b 2c3 ) and Also

A

A l ^(d 1 b2c,)

let

On

let

B

19

ly

...

be the cofactors of a l9 b l9

A 2 ^(a 1 d2c3

y

)

)

...

in A.

J 8 -(

account of the identities

.

we have and and

Suppose

(1) it is

u l B l -i-u 2 B2 + u^B3 ^ Ay + A 2y

similarly,

that

given by

We

........................ (C)

A^O. The last identities show that if a solution exists 3= -4/J, y= -'A^/A, z=-AJA .................... (D)

,

can show by actual substitution that these values of #, y, z satisfy Or we may use the following method, in which every step

the equations.

is reversible, so that verification is unnecessary.

Because

Suppose that

zero.

Because

that

2 (A^C^) =A ^0,

is

A^^Q,

to

A^gC^^O,

so that no one of

U2 = o,

w3 = 0,

u^ = 0,

* It

L

2>

not

Ax - - A r

^ = 0, Jy=~J

essential to show that suffixes p, q r exist, This step has been omitted in the text-books. is

19

is

zero.*

B2 ^0 and C3 ^ 0, the equations are equivalent

Hence the equations have the unique zero.

(A^B^C^

A B C3 is

the equations are equivalent to

In the same way, because to each of the sets,

therefore at least one term of

t

2

;

%==0, w2 = 0,

solution,

(

Az= - J 3

- A-JA, -

such that none of A$ t J5a See Chrystal, Algebra, Vol. I, p. 360.

all different,

.

,

Cr

B.CJU

is

THREE PLANES

156 (2)

= and A v A 2 A 3

If A

,

are not all zero, the equations are inconsistent.

For example, suppose that J 1 ^0, then

since

UiAi + UiAz + UsA^Av it

(E)

follows that the equations have no common solution. If we suppose the coefficients to vary so that J->0,

A v J2 J3

and at

least

one of

tends to infinity. d in with the d. d large terms, we see that in Neglecting l9 2 3 comparison general the following ratios tend to equality ,

remains

then at least one of

finite,

x, y, z

,

:

A :B :C

x:y:z,

1

l

A2

lJ

:

J5 2

:

C2

,

A^

:

B3 C3 :

.

In the limiting case when A=0, we say that (he equations have a solution (x, y, z) such that at least one of x, y, z is infinite and

x:y: z = A l B l

C^A

:

:

:

2

2

:

C a = 4 3 #3 C8 :

:

.

A l9 J 2 A 3

are all zero, the equations are not all independent and there are infinitely many solutions ; or else each of the equations (assumed to (3)

If A,

,

be distinct) is inconsistent with either of the others.

u^A^ + u 2 A 2 + u^A 3 =

For we have

A

If no

(i)

A =

//

(ii)

l

9

(F)

the equations are not independent,

is zero,

A^O, ^

3

7^0, then u 2 A 2 + u B A 3

= Q; and w 2 = 0, w3 =

same equation.

are the

the equations M 1 = 0, w 2 = solutions which are solutions of the system.

In both cases, since

many

A.^Q,

have

infinitely

A = 0, A 2 =-Q, A^Q, we have w 3 sO, which case is excluded. If, in the above, every A is replaced by the corresponding B or If

(iii)

(iv)

L

C,

Therefore the only remaining case is that in which minor is A zero, and then every two of the equations are inconevery of or they are one and the same. sistent similar results follow.

;

Ex.

We

1.

// a,

6,

c

are unequal, solve the equations

have 1

1

1

1

1

1

a

b

c

d

b

c

a2

62

c2


b*

c

x

and the values

9.

of y, z are obtained

Three Planes.

by the

(i)

(ii)

which

If

A^O

(d-b)(d-c) (a-6)(a-c)'

2

cyclic substitutions (a6c), (xyz).

Referring to 8, let

equations to three planes, then

x+y+z~I,

^=0,

w 2 = 0, Wg=0 be the

:

they meet in the point given by the equations in Art.

// A =0 and A l9 A 2 J 3 ,

8, (1).

are not all zero, two of the planes meet in a line

is parallel to the third plane.

FOUR PLANES (iii)

// A,

J x J 2 ^s >

,

157

are a M zero the planes have a >

common

line of inter-

section or they are all parallel.

Proof of

Suppose that

(ii).

Jx ^

We

0.

have

J 1 ........................... (A) A l ............................. (B) ,

On account of (A), at least one A is not zero. Let A l = w2 0, w 3 = are not parallel and, on account of (B), section does not meet the plane % = 0. Observe that

which

all the

planes are parallel

is parallel to the line,

The truth

of

(iii)

follows

Hence we conclude

0,

then the planes

their line of inter-

to the line

w 2 = 0, w3 = 0. from Art.

that, if

or else they hare a

infinity,

^

8, (3).

J=0,

common

a point at which may be

the planes either meet in line of intersection,

at infinity.

Four

10.

Equations

in

Three

Unknowns.

Consider the

equations

u^a^x + b^j + c x z -f d x =0, = 0, = 0. Let

A = (^i^a^) an ^

^ ^i>

BI>

^ e ^he cof actors

If the equations are consistent, then For, as in Art. 4, (1), we have

(1)

of a l9 b

...

.

A =0.

u l D 1 + w 2 D2 + w 3 J53 -f ^4^4 = d (2)

//

A =0,

^Ae equations are consistent

when every D is

and have an unique

solution except

zero.

This follows from the identity == 0.

D

is zero, the equations are inconsistent, but they are // every as having a common infinite solution, unless every minor of regarded

(3)

to

be

A

is

which case there are infinitely many solutions. explained with reference to four planes in Exercise XVII, 17. The methods already described apply to systems of linear equations con-

zero, in

This

is

necting more than three variables,

and lead to

similar results.

SIMULTANEOUS EQUATIONS

158

EXERCISE XVII

SIMULTANEOUS LINEAR EQUATIONS 1.

Solve by means of determinants

:

x-y + z~Q, 2x 4- 3y - 5z = 7, 3x - 4y - 2z = - 1.

(i)

(ii)

y+z ~

x

-

w

1,

3y- 22 4-3^ 2.

With regard

to the equations

4x 4- 7y - 142 = 10,

show that the

first

l(2x

and

find

3.

I

:

m

:

n.

2# + 32/-4z= -4,

can be expressed in the form 4-

3y - 4z) + m (x + y + z)

= n,

Hence show that the equations are

inconsistent.

With regard to the equations 4x + ly- I4z = -24, 2x + 3y-4z~ -4, first can be put in the form

x + y + z6,

show that the and

find

I

:

m.

4. If x, y, z

Hence show that the equations are not independent. are not

ax

all

-f-

zero

where

-f-

63

+ c 3 H- 3a&c ==

Having given the equations x cy + bz, y x, y, z are

not

az + cx,

-a 2 )^i/ 2 /(l -b 2 )^z 2 /(l -c 2

x*j(l

Having given the equations a = b cos C c cos J5, 6 c cos -f-

deduce the following

A -f-cos 26c cos A 6

and 8.

2

Having given that =1 a? 4- y 4-

prove that

.4

+a

cos

C

f

c

,

).

=a

cos

b cos

jE? -f-

:

cos 2

7.

bx + ay

z

prove that

all zero,

and 6.

bx -f cy -f- az ~cx -}~ ay + bz = 0,

a3

prove that 5.

and

by + cz

2

jftH-cos -he 2

2

- a2

(74-2 cos ,

^4

a/sin J.

cos

B cos (7

=6/sin

J5

1,

=c/sin 6

ax + by + cz=d, a zx 4- 6 2t/ 4- C 2z = cZ 2 a 3 x 4- 6 3 ?/ 4- c*zd 3 -(d-a)(d- b) (d - c) a*x + b*y + c*z= d* ~(d -a)(d -b)(d -c)( ,

If the equations

cy

4-

bz

6~c

__

az c

-f

ex _ bx + ay _ ax

a

4-

6y 4-

a -b

are consistent, prove that either

a + b + c~Q .or (b -c)(c -a)(a -b)=abc. [Put each fraction equal to k and eliminate x, y, z, &.]

f

,

.

^4,

APPLICATION TO GEOMETRY 9. If the three

that

equations of Art. 8 are distinct and

J 2 =0 and J 3

[Since therefore

unless b 1 :

are consistent

if

Cj

=6 2

:

c2

=63

:

A =0 and Ji=0, show

if

ca .

d=Q, A l A^: A 3 =B l Bz B3 and since d l A l +d< J 2 = d l Bl + dpB2 + d3 Bz =Q unless ^4 2 ^ 3 are all zero.] :

:

10. If the equations

Also

:

.159

^

,

xby + cz + du,

t

,

y~cz + ax + du,

uax + by + cz z=ax-\-bydu, and no one of a, 6, c, d is equal to - 1, then

a= -

1,

then one of 6,

c,

d

is

equal to

1.

[Ex.

XVI,

10.]

Using the notation of Art. 8, show that if t* l =0, u 2 =0, w 8 =0 are the equations to three straight lines, the coordinates of the vertices of the triangle formed by 11.

them

are

triangle 12.

(4, |),

if

13.

(# 4 ,

2/4,

and that the area of the

|), (^, |),

i JV^A^-

is

Prove that

if

c^O,

=

valent to

and

,

(

^=0,

the equations

w 2 = 0, u^=0 of Art. 8 are equi-

X

A^

deduce the solution in the form already given.

Show

that the condition that the points (x l9 y l9 2 X ), (x 2 , y29 z z ) be in the same plane is

24 )

9

,(x^ 9

t/ s ,

may

i

y\

~ "i

9

2/3

23

1

4

2/4

Z4

1

.

r

n ^*

i

*

14. Using the notation of Art. 10, consider the lines whose equations are U!=Q, 1*2=0 and w a =0, w 4 =0. It is required to find the equation to a plane which passes through either of these and is parallel to the other. therefore the required planes are [We have u 1 D 1 +u 2 D 2 -\-u 3 D 3 -^-u t ~\=0 and

D^A

15. Consider the lines

whose equations are

X-aS

S-ql/-j8Sm n Show

9

m

that the equation to the plane which contains the

to the second

first line

and

is

parallel

is

Deduce the condition that the [The required equation and aV + bm' + en' =0.]

is

may meet in a point or be parallel. + b (y - ft) + c (z - y ) = where al + bm + en =

lines

a (x - a)

16. Using the notation of Art. 10, show that if ^=0, % 8 =0, t* 8 =0, w4 =0 are the equations to four planes, the vertices of the tetrahedron bounded by them /'A

are the points

B

f^r> y^

the tetrahedron

C

\

jr)

>

is **-A

/A

B

(jr*

ff

z

ID

>

C

\

77

)

>

e ^c

>

an(^ * na ^ tn

volume of

DDD

shown in works on Coordinate Qeom&try that the volume of the tetrahedron whose vertices are (x l9 y l9 Zj), etc., is one-sixth of the determinant in Ex. 13.] [It is

EQUATIONS OTHER THAN LINEAR

160 17.

be the equaUsing the notation of Art. ]0, let t^ 0, w 2 0, w 3 0, w 4 and suppose that J = 0, then If at least one Z>, say D 19 is not zero, the planes meet at the point

tions to four planes, (i)

(ii)

Suppose that every

D is zero,

then

(a) If all the fninors of A are not zero, the four planes are parallel to a certain line, that is to say, they have a common point at infinity. (/?) If every minor of A is zero, the planes have a common line of intersection or they are all parallel.

For example, if and we consider the planes u l9 u 2 i/ 3 Since D 4 =0, two of these, say u l9 u 29 meet in a line which is parallel to the third. Also, considering the planes, u l9 u. u^, since /) 3 =0 the plane w 4 contains or is parallel to the line u l ~0, M 2 =0. Hence all the planes are parallel to the line

A^O

.

,

9

x\ (6^2)

Given the equations a 2 c 3 + a 3 c 2 - 26 2 6 3 = 0,

18.

a 3c l

= yj (Cjda) - zf (a^). -f

iC 3

-

26^3

= 0,

a tc 2

+ a^ - 2bJ) 2 = 0,

prove that

gl

[Prove that

L lf

20. If

Jf ,,

a;,

~ 2&1

^ 2 (i

Cl

=

c i-^i 2 )

-.

X= N

prove that P

where

=

etc.,

are the cofactors of

y are connected with J,

1'

N X-N Y=L -M

given by

t

l

1

l lt

m

lt

etc., in

the determinant

by the equations of Ex. #,

considered in detail in another volume.

The question of Here we illustrate

various methods of dealing with systems of special types. (1)

Systems which are symmetrical with regard

Ex.

1.

Find

to

x

9

y, ....

the rational solutions of

x *+ y * +558=50,

(y+ 2 )(z +*)(*+#) = 14,

3 ir'+^+z - 32^2 = 146.

If

x=p, Syz~q x\jzr t

2 Z>

y

the given equations are equivalent to

-2g = 50,

are

^2 .

Systems of Equations of any Degree. is

Y

y are given by

= m v -f

elimination

and X,

t,

prove that the corresponding values of

11.

19,

pq-r = U,

3

p -3^-146;

SPECIAL TYPES OP EQUATIONS whence we If

-

- 23,

r

find that

p =2, then

Thus

3

x, y, z

q

~

p

I50p -f 292 =0, - 60 and x,

~

2

p

giving

-

or

161 1

^147.

the roots of

y, z are

0*- 20 2 -230 + 60=0. may have the values 3, 4, - 5 taken in any

order,

and there

is

no other

rational solution.

(2)

A

solution

sometimes be discovered by considering the properties of

may

a particular determinant. Ex.

Solve

2.

x 2 - yz = a,

y

2

zx = b,

z2

- xy

c.

Consider the determinant

x

z

z

y x

y

The

cofactors of x 9 y,

if (x, y, z) is

be

.

z

are yz - x", zx ~ y 2 , xy - z 2

in the top line,

z,

/.

Hence

3 - 3 - 3 z 3xyz - x y

y x

2

2

(zx~y )(xy-z )-(yz-x

z 2 )

;

^Ax.

a solution,

-a 2

Ax, and similarly, ca

-b 2 =Ay

a, b, c, and adding, 3abc - a 3 - 6 s - c8 = A (ax

+ by+cz)

f

ab

-c 2 ~Az.

Multiplying these by

- A2

.

Therefore the only possible solutions are given by ~~

a 2 -be ~~b 2 -ca "c 2 -ab It is easily verified

in

which there

(3)

A

Ex. 3.

is

no solution unless

+ 6s -f c3 - 3a6c)

*

are solutions unless a 3

by substitution that these

a, b, c are all zero.

change of the variables. Solve

x(x-a)=yz, Let x

3

*J (a

l/X, y

1/7, z

= l/Z,

X -YZ 2

a

z(z-e)=xy.

y(y-b)=zx,

then the equations are equivalent to

Y*-2 o

c

Therefore, as in Ex. 2,

where the

A

first

a 2 -be,

B=b 2 -ca, 0=c 2 -ab,

and k

is

to be determined.

Substituting in

equation \ k_(k__ a

&_

A\A

k =0, giving the solution

therefore, either

or

k (BC

~A

with similar values for

2

8 )= a ABC, giving k (a

y, z.

(0, 0, 0),

+ 63 -f c 8 - 3a6c) = - A EC

;

SOLUTION BY SUBSTITUTION

162

In

(4)

In equations involving

useful.

Ex.

the case of two variables x,

4.

x, y, z,

we may

solution

t'x.

f

is

+ bt + c2) -. x {i + mj),

x^Q,

if

the solutions are given

x

x 2 (a'+b't + c'P) -x(l' + m't).

by

+ mt)/(a + bt + d 2 ),

(l

2 (a + bt + ct

where

)

(V

y

tx,

2 m't) = (' + 6'e + c'J (Z + mi). = become cy* my, c'y 2 m'y,

-f

)

If x~Q and y^O, the equations - y m/c. In this one value of < is infinite.

solution

7/a

(5)

sometimes

+ bxy -f cy2 = lx + my, a'x2 + b xy + c'ya = I'x -f m'y. = tx the equations become (0, 0), and if y

x *( a

Hence

write

is

Solve

ax*

One

y = tx = y tz, z

the substitution

y

x = x 1 ,y = y l

,

/.

cm'-c /m =

and

can be guessed, then a suitable substitution

...

is "V

x ~~~ Ju\ *&r. 5.

One

i

-A-

"V

y ~~ y\

)

*

i

j

/S'o/ve

solution

is (6 -t-c, c -fa, ar

a

+ 6), and

if

we

write

= 6+c + JT, y=c+o-l-7,

=

2

the equations become

Hence

if

none of the

three, 1

F

X, F, Z,

+ |__1 Z""

a

I 9

Z

i# zero,

+ !___!

X~

Special cases.

If

This

JT=0,

gi ^es

7=0

I

L_^I

1

X+Y~~

9

;

c

1

2_1_1 X~a b

'

with similar equations.

we have

c'

the only solution, beside the obvious one.

and

Z^O,

then a=0,

6=0 and Z may

have any

value. If

JC=0, 7^,0,

Z^Q,

then 6=0,

c=Q and

7,

^ may

have any values such that

7Z + a(7+Z)=0. Ex.

6.

The

eight solutions of

x*

+ 2yz=a,

y*

+ 2zx=b,

z*+2xy=c,

.......................... (A)

are given by

where

-X"

and

By

=2a-6~c2v/(a2 + 6 2 -fc2 -bc-ca-ab)

k=>J(a + b+c). addition,

From where

2

(+i/ + z) 2 =a + 6+c,

/.

x + y + z-k.

the second and third of the given equations, by subtraction,

X~k~Zx.

SIMULTANEOUS QUADRATICS Adding the second and third

163

and subtracting the

of equations (A)

first,

we

easily

find that

^V

A/ -

3

and solving

for

X

z ,

we obtain the values

Another solution.

.

=6+c-a;

stated in the question.

Equations (A) are equivalent to

-A

(say),

2 2 2 (x + w y + ojz) ~ a + o>6 + co c = B,

(x+uy +o> 2z) 2 =a + oj 26 +o>c = (7, where

o> is

an imaginary cube root

of

1.

Therefore the eight solutions are given by

x+w zy+a)Z=*/B, root of A and so for *JB,

x + y + z=JA,

where *JA

From

is

either square

z

x+a)y+a) z=,JC

9

............... (B)

*JC.

equations (B), by addition, 3a?= N /^+Z,

where X=*JB+>JC, and therefore

Also,

from the second and third of equations

(B),

%c

and

(

These Ex.

7.

^^

results agree

Show

)

with the preceding

that, if the

if

we change the sign of X, which is permissible.

equations

(A) are consistent, then

abc

From

+ 2fgh-af*-bg*-ch*=Q ................................ (B)

equations (A), by multiplication,

ax *(b*y* + c 2 z*) + ...+...+2abcxzy*z*=Sfghx*y*z* ..................... (C)

6V + c z = (4/ 2 4

Also

2

- 26c) y*z2

,

with two similar equations. Substituting in (C) and dividing by x*y*z* (assumed not to be zero), the result (B) follows. Special cases. (i)

If

Considering the original equations

z=0, y^O, 3^0,

then

C2 a =0,

it will

be seen that

and c=0;

hence the equations become, by* = 2fyz, &y*=0, and

we have 6=0, and/=0.

=0, y=0 and 2^0, the equations reduce to the single equation c 8 =0; (ii) and thus do not constitute a system of equations. Thus, if the equations (A) form a system of equations with a solution other than (0, 0, 0), then the condition (B) holds. If

COMMON ROOT OF TWO QUADRATICS

164 Ex.

Prove

8.

that, if

x 2 + y 2 + z 2 - 2yz - 2zx - 2xy =. 0,

and no two of x

then, if

(b-c)

a, b, c are equal, z

y

2

xyz

"

(c- a)

2

(a~b

2

+ b 2y 2 -f c 2z 2 - 2bcyz - 2cazx - 2abxy

a^x 2

)

'

The given equations can be written

-ax + by + cz)

x(

+y-

Multiplying these by (x find that

-by + cz)

-\-y(ax

+ by - cz)

(ax

z),

-f

+ by -cz)=Q,

z(ax

respectively

and subtracting, wo

xy{y(a-b)+z(c-a)}=xy{z(b-c)+x(a-b)}; similar equations can be obtained. is zero, we may write

and two z, y, z,

y(a-b)+z(c Hence,

if

-a)k,

no two of a,

+x(a-b)=k,

z(b -c)

x~

are equal,

b, c

Hence, assuming

-

-~-

-

a) (a

A

~ 2

2

-a(b~c) +b(c-a) +c(a-b)* :.

none of

the three,

x(

k(b -c)

(c

whence also

that

j~

,

0)

ax -by - cz

(b+c)(c-a)(a~b)'

2 2 2 Zax(ax-by-cz)^X (b-c)(c-a)(a-b) 2a(b -c .

)

-A 2 (6-c)2(c-a) 2 (a-6 and the result in question follows. The consideration of special cases

is left

to the reader.

EXERCISE XVIII 1.

If (a,

ft), (a', j8')

are the roots of

u prove that

a^^^-^^ and consequently

common

root is

the necessary

E=0,

and

R = (ac'- a c) f

2.

Eliminate

x,

sufficient condition that the equations 2

- (ab' - a'b) (be' - b'c).

y from the equations

and show that the (a, 0), (a', j3')

roots of the resulting equation in z are aa', are the roots of

ax 2 + bx + c-0,

Show

have a

where

that this equation

a'x 2

a/3', /?
where

+ b'x + c'Q.

is

f a 2a' zz* - aa'bb'z* + (b z a'c + b /2ac - 2ocaV) z 2 - bcb'c'z + c 2c /2 = 0.

Prove also that the equation whose roots are a/a', from the last equation by interchanging a' and c'.

x

a//T,

/a

,

//T is

obtained

SYSTEMS OF EQUATIONS

165

Solve the equations in Exx. 3-28. 3.

* 3 + 3^ = 8,

4.

X 2y + xy 2 + x + y^3,

5.

x(l-y*)=2y,

6.

x + y + z = $,

7. (y

*2 8

'

xy+(x + y)=2.

+y

l)

y(I-x*)=:2x.

x 2 + y 2 + z 2 ^ 29,

+ z)(z + x)(x + y)=*, 9

[Put

2(x - l)(y -

(x

a;- 1

+

25 8

"

'

9.

10.

-=l6-xy,

^--^Q-xy. x

y

12 l^

f

^ ---

15. [(

a>

2

a z --y~

+

4.

i y -. _ -a z -x- a ?

l-p, and prove that

2 2 ==a

+ 2/"f z,

2 2:

2

4-a;

6-i-

- x 2 = 6 s - 7/ 2 yz + zx {-xy= a 2

+ y)(a? + 2)=o

c

2

1

X

,

[Put 18, ayz

V

y

b

-X(a-b) + a-b

z-f re,

- 22

.

etc.]

1

1

1_ d

A2

a

16.

1

B

+ (l-2/) 2 -(l-o:)(l-2/)(l-a: y),

a;l-A, y

[Put

i/

,

/,

1

13. (l~rr) 2

14.

x

-----4. y2 a y a a?

V

,

1 -

O

<2

,

z

11 C

x

c

~~

.

a

x= + by + cz= bzx + cz -f ax ~

19.

20.

_

= -_ = ..2

+ 0;

X4-7/-1

21.

22.

x(bc-xy)=y(xy~ac), xy(ay+bx - xy) ~abc(x + y - c).

-x)=by (z-x)(z-y)=cz. 2 xy(c-z)cz 2 zx(b-y)=by

23. (x-i/)(2:-2)=aa;,

24.

yz(a-x)=ax

2 ,

(y ~-z)(y

t

.

,

25. 26. 27.

28.

'

'

x+y=x' + y'=r,

112 112 = --.

-

x

y'

-,

-

y

y

\^e + 1 + *Jy~-2 = *Jx -1 + N/jMh2 = 3.

x'

x

and

ELIMINATIONS

166 29. If

3

5a?

5t/

-2

pr0ve

,

8

y+\=x,

that

and

solve

the

equations. 30. If

a

-}-^ o

=~+ x

=a + b, show

that either

+ y=a + b.

(X+Y + I)(aX + bY -a-b)=Q.]

[Puta^aJf, y=bY, zndshov? 31. Eliminate #, y, z

or x

-4-^4-1=0 b

a

y

--2)

from

---

-

and ax + by + cz-Q.

C

CL

x3 -y*~ y*-x*=a* and aM-y= -a, and a:, y are unequal, prove that 2 8 2 2 2 ~ (iii) 3a (s 4-t/ )= (ii) 3a -3a + 3a- 1 =0, xy = a*(a-l)l(2a-l), 4 4 2 3 3 = and (v) 3a(a; + t/ 4- l) = 2+4a 3a(z 4-y ) 2-a,

32. If (i)

(iv)

1,

.

yV + (y + z)

33. If

2

2

z

+

2

2 2 a: !/ 4-

then will

z

(z-fir)

(x

Ayz -

+ +

2

+ i/) 2

1

+ A,

A2^ = l4-A and Aa:y

-f-

x^y,

= 1 -f A.

111111111 __ __ __

34. If x, y, z satisfy the equations

__

__

i

x

y-\~z

__ y

_,

9

a

_._

i

z

_

__

_

i

9

+x

b

z

x-\-y

9

c

+ c a)=y(c-\-a-b)z(a + b c): and hence solve the equations. + by* -f cz 2 = 0, ayz bzx + cxy = and x3 + y3 -f z 3 -f Xxyz = 0, prove that

prove that x(b 35. If ax*

36. If a

-f-

+ 6-fc=0 and x + y + z = Q, prove

that

2 2 [Use the identities (6y + cz ax) = (bz + cy) ,

37.

Given that

a 2o: 2

and Sh

etc.]

-f-

+ c zz 2 - 2bcyz - 2cazx - 2a6a:y = 0,

b 2y 2

W 38. Eliminate x, y, z

from (

showing that the result

(x

+ y-z)(y+z~x)~bzx,

(y

+ z-x)(z+x-y)=;cxy,

is

abc~(a + b + c-4) 39. Eliminate x, y, z

41. If

a;

2

2

a?

4-t/ -f2

.

from

x + y-z=a, 40. Eliminate

2

a:

2

+

a t/

-2 2 ==6 2

,

x8 + y8 -2 3 =c 3

,

xyzd*.

and y from 2

= l,

+ 2y 4-32=3, a;

3o;

+ y4-2 = 3,

-8y-f52=

3.

prove that

CONSISTENT EQUATIONS Show

42.

167

that the equations

are consistent,

and

x y

find

:

:

z.

and x 2 - ayz ~y 2 - bzx = z 2 - on/ f (a; 2 + 1/ 2 + z 2 ), a 2 -f & 2 4-c 2 = 2a&c + l and z a /(l -a 2 )^2/ 2 /(l -6 2 )=z 2 /(l -c 2 ).

43. If x, y, z are not all zero

prove that

Show

44.

that

z=c is a solution of 3 3 px* + #y -f rz = &M/Z,

#

if

a, ?/=&,

then

xa(qb* ~rc 3 ), is

i/

= &(rc 3 -pa 3 ), z=c(pa 3 -gfc 8

)

another solution. 45. If

3/

7V

-(y + z-x)

*-

2

+ x-y) = -

(z

(x

+ y-z) and J-fw + n=0, show

46. Eliminate x, y from (fc~#)(c-t/)=a 2 , (c-a;)(a-i/)=& 2 , (a -x)(b

that

-y)=c

2 .

47. If the four expressions

u2 +

xz+zu), are equal, prove that each 48. If x lt x 2 ,

... Xf,

2 2 z 2 equal to %(x + y + z + u ).

is

are all different

and 2

prove that 49.

(i)

2

2

2

_

,

x1 2 + xs 2 ~(k*- 4k 2 + 2) a:^. from the equations x ax + by + cz + eft = 0, a a; 6'y -I- c'z -f rf^ =0,

Eliminate

t

x, y, z,

-I-

a

-

x showing that the result

&

-

-f-

c -f

-

d

y

- + 6' - + c- d' T =0, x y z

+ -=0,

z

x

a'

.

.

-f-

t

t

is

L MN + Lmn + Mnl + Nlm = 0,

m = (ca

f

where

l-(bc

)9

L = (ad'), (ii)

Use

this to prove Exercise

n = (ab'),

/

),

N=(cd').

M=^(bd'),

XVI,

25,

(iii).

L

I

[(i) LV }

Prove that

...

...

2

2

(x*-t )yz the unwritten fractions being obtained

(LMN). Now put x\

2 t/

,

z\

t

2

for x, y, z 9

...

.

v. ,

(y -z*)xt

by the 1

cyclic substitutions (xyz), (Imn), in Ex. XVI, 25, (i).

Let # = cos a-f-tsina, y = cos -f t sin j5, etc., where a, j8, y, 8 are the angles which the sides a, 6, c, d make with a fixed line. By projecting along and perpendicular to this line, prove that the four given equations hold.] (ii)

CHAPTER XI RECIPROCAL AND BINOMIAL EQUATIONS

A

Reciprocal Equations.

1.

reciprocal equation

which possesses the following property a root. Let

a,

a xn

then

+ a^"-1 4-

a nx n

. . .

+ art^x*" 1 +

a reciprocal equation

is

(A)

any

root, then I/a is also

+ a n _yX + a n = 0,

.....................

(A)

are the roots of

...

I/a, l/)3,

Hence

is

an equation

be the roots of

...

/J,

a.

If

:

is

Denoting each

of these fractions

+ a x x + a = 0.

. .

.

if

by

and only

if

we have

k,

an a so that

We

&=

1.

say that Reciprocal Equations are of the first or second types +1 or k= -1, /&aZ i5 according as a r = a n _ r or a r = ~a n _ r according as k shall

,

= /or r 0, 1, 2, Theorem

1

...

n.

TAe

.

reciprocal equation of the first

Let f(x)

=a

x n + ajX**" 1

of the first type,

xnf (

\3//J

-f

.

.

Therefore/(z)

(z is

2m + 1

.

-f

an =

be a reciprocal equation.

that of

// this

2m + 1. Then

+ 1) 4-a 1 x(a;

divisible

by x -h

2m ~ 1 1,

+ 1) +

and

if

may be m +a mx (x + 1) = 0.

the equation ...

(x)

+x =

is

is

is

written

the quotient,

x+

X hence $(x)

a

=/()

Let n be odd and equal to

= f(x) a

any reciprocal equation depends on type and of even degree.

solution of

a reciprocal equation of the

first

type of degree 2m.

TYPES OF RECIPROCAL EQUATIONS Iff(x)

=

Let n = 2w +

xn

of the second type,

is

-

/( x/

)

=

169

-/(#)

Grouping the terms as before, we can show that /(x)

1.

2m (-} = by x 1, and that, if <(x) is the quotient, then x with the same result as before. = ~a m Again, if n = 2m, since a r ~ -a 2 m-r> ^ follows that a m therefore a m = Q. Hence the equation may be written divisible

<

/(x)

=a

-

2

(x

Therefore /(x)

1)

+ a 1 x(x 2 - 2 -

divisible

is

equation

We the

of the first

if

(x);

,

and

+a m _ 1 m -1 (x 2 - 1) = 0. tf

<^(x) is

the quotient,

~

in every case the solution of /(x)

= (x)

...

2 by x -!, and

/

Thus

+

1)

is

type and

say that a reciprocal equation type and of even degree.

=

depends on that of a reciprocal even degree.

of

of the standard

is

form when

it is of

first

Theorem

The

2.

solution of a reciprocal equation of the first type

and of

2m

degree depends on that of an equation of degree m. Let the equation be -1 a x2m 4- fljX 2 4 ax + a = 0. 4. . .

Dividing by

xm

,

this

may

be written

1 \

Let x f

-^=z

x

and x r +

-

xr

/

_

=t/ r so that ,

1\2

1

,

u*=z

+ -) -2 =

~

\

2

;

, A .(A) .

then

-2.

x/

l

x+

Also

r

= 3,

and so on. Thus equation z 1 is

= ^ + -r ) + (x*~* -f -i-A + -^ ^xv V x 1 /) f\ xr 2 / 7

"-

wr = 2w

therefore

Putting

l

} (x'~ x/\

4,

...

in succession,

(A) can be expressed as an equation of degree

one of its roots, the corresponding values of xare given by

a?

2

m in z, and

if

- z^x + 1=0.

THE BINOMIAL EQUATION

170

The reader may

verify that

#*.

Solve 6s5

1.

One

root

-

is

w 8 = z8 - 8z6 + 20s4 - 16z2 + 2.

- 7z,

14z 3

1*4 - 33** - 33z 2

4- 1

Dividing by x +

1.

+ 1 1* + 6 = 0. we have 6x4 + 5z3 - 38x2 + 5x + 6 =0.

1,

2 Dividing by # and grouping the terms,

If 2

= 3; + -,

this

x

2

.".

Whence we

becomes

=

find

a;

2"

6(z

2

an d x

or ~^sr

= 2,

|,

-3

that

-2)+5z-38=0, i8

gi y ^ n

hy

x-\

The Binomial Equation x n -1=0.

(1)

If a

(2)

//

m

common For

a

is

root of

xn -

prime

to n,

1

ar

so aZso is

=0,

=|- or

Thus the roots are -1,

or ~J.

2.

is

is

,

i^Aere r is

2, |,

any

-3,

integer.

m -l=0 and xn -l=0 have no

the equations

root except 1.

if

a

a

is

common

positive integers.

can be found so that If n n roots are (3)

then a pm = l and afln = l, where p, q are any pm ~qn = 1, and since is prime to n, j9 and 5

root,

Therefore

m

(x.

pm-qn=

(See Ch.

1.

a prime number and a

is

1, a,

a

2 ,

...

1

a*

is

I, 4, (3).)

any imaginary

Hence a = l.

root of

xn -

1

=0, $e

"" 1 .

For every one of this set is a root and no two of them are equal. For a a~ = l, and the suppose that the two a and of (a>b) are equal. Then a n b = =0 and x 1 o^~ l have a common This is root. equations imaginary - 6. is a for and n is thus n to a l
;

If

(4)

of

xn

-

n=pqr

1 =

...

are the

wAere p, q,r,

...

are primes or powers of primes, the roots

n terms of the product

a root o/x p -l=0, jS o/a?-l=0, yo/xr -l=0, ete. Take the case of three factors p q, r similar reasoning applying in

where a

is

9

;

cases.

Any term

of the product, for instance

(a

and

6

similarly

(/J )

n

=* 1

)

and (y )*

1

= 1.

5

/?}/ , is

= (a*) a *r = l,

a n 5

a

a root.

For

all

SPECIAL ROOTS any two terms are equal,

If

w hich

-

=0

1

and

;

since

aj8 y

--htsin -

where

,

n

xn -

of

.

.

n

prime to

is

p

have no common root except 1. (5) As explained in Ch. V, 17, the roots cos

if

instance,

c

=a

&/ '^9

c/

y

impossible, for jg*-*y-' is a root of z

is

^ft-fty-c'^a'-a^ and oca '~ a is a root of x p

for

171 6

= 0, /x

r

1

qr,

=0

>

r

-l=0

these equations

are

i

w-1. -i

1, 2, ...

The imaginary roots are therefore

2r7T

cos

2riT

.

4

.

sin

= l, .,

where

,

r

according as n is odd or even. If x n - 1 is divided by x - 1 or x* the quotient, then (x)Q

is (x)

or

...

2,

according as n is odd or even and a reciprocal equation, and if this is

1

is

transformed by the substitution z = x + -, the values of x o 2 cos

r = l, 2,

If r is less (1) v '

= 0,

1

same

but

to than n and prime r i

/

.-

I or, 1

a

is not

type.

TH

-.

&

&

-n

,,

.

h

i

sm

-

2rn\ m

n

)

=

cos

Therefore divisible

by

rm

is

n.

This

where

1,

/

2rm7T 2rm?r ---sm ---- = 1 h

Definition.

is

by

and since n

n,

of

xn -

1

The

roots of special r

prime to

number

#w -

1=0

is

n

and of

degree,

the

m
.

is

prime to

r,

m

must be

is

not a root of an equation of

called a special root of the equation.

are cos

--hi sin n

n

,

where

r
-l=0 has (n) special roots, where (n) numbers less than n and prime to n, including unity. and prime to n then n -- r is prime to n, and therefore n.

n Thus, x

is

the

of

r
2f7T

cos

n

.

db

i

sin

2r7T

n

are special roots. These can be arranged in pairs, fa, -j, so that they are the roots of a reciprocal equation.

M

a root of

m
=0 which

the same type and of lower degree

2rir -is .

sin

n

impossible, for

Any root

t

,

i

n

divisible

n

.

h

,

,

.

then

-2rrr

then cos

n,

any equation of lower

root of

2r7r

cos

if

\

If

or

-

-

...

z are

n Special Roots of x -1 -0.

3.

xn -

10

v where

n

M-

71-1

rt

fj8,

^J, ^

etc.,

B.C.A.

SOLUTION OF BINOMIAL EQUATION

172

a special root of x n - 1 =0, the n roots are 1, a, a 2 ... a 71 1 For every one of the set is a root. And if two of them are equal, for ~ a b then a a b = l, so that a is a root of x~ b = l. This is instance, if a =a ""

If QL

(2)

is

.

,

>

impossible, for the last equation

a

6

,

c, ...

,

,

1

=0.

xn -

I =0, the complete set of special roots is are the numbers less than n and prime to it,

root of

//a is any special a ac ... where a, 6,

(3)

a

than xn -

of lower degree

is

including unity.

a is any number less than n arid prime to it, the remainders when 3a ... (n-l)a are divided a, 2a, by n are the numbers 1,2,3, ... w-1, taken in some order or other (Oh. I, 10). And since a n 1, it follows ~ that every number in the set oc a a 2a a 3a ... a (n 1)a occurs somewhere in the set a, a 2 a 3 ... a n-1 Therefore a a is a special root.

For

if

,

,

,

4.

The Equation x n -A^O.

(1)

Let

A =a(cos a + ( \

.

are

^/a

.

i

For by De its n values are

,

a

.

4

2r7rl

-h

_

,

where

J-,

n

then the roots of x n -

|,

|

-ft sin

w

I

,

where a =

sin a),

a + 2r7i cos

,

.

r

= 0,

_ 1, 2, ...

^

o,.

it

cos

1.

J

----

.

4- i

a

=

,

------

-

sin

cos

n

n

n values of

^A may

a\

.

-

-f i

n

\

sm -

cos

j

n/

\

-n

r?r

.

sin

h t

he found by multiplying

n

roots of xn

+ 1 =0

are

This

is

-

---------i-ivsm

Solve x 5 -

Ex.

1.

real root is 1.

1

~0.

Deduce

where

,

n

obtained by putting a^=l, a

The

= A(),

,

.

cos

The values

of

then

a;

22

-fz-1^0,

are

#2

is,

-

54

TT

.'.

cos -

=coa 0JO 36

s/5 ^-- +

.

1.

,+ai-f

-f

t

--

+ 1 =0.

a:

2-|(-ls/5).

giving

cos - o

Therefore the values of z are 2 cos

n ~

30 and cos 72.

or-

2=3 + -

1, 2, ...

1,

x*+x*+x* + z + 1 ~0, that Let

,

in (J).

77

the values of co#

Dividing by x

r

sin

-

(

r

=1

,

2)

.

3

--

5 l

,

and 2 cos ,

,

and

Now

55

-

54 .

5

2?r

cos --

72 =cos ,,^0

cos

=

V "-! = ^-

\

),

/

any value of

by the n-th roots of unity.

The

and

r,

all different.

follows that the

(2)

=

t

n

Moivre's theorem, this expression is a root for every

bmce

^4

- cos -;

5

.

SOLUTION BY QUADRATICS Ex.

2.

the roots.

173

What are the special roots of a; 15 -1=0. Find the equation of which these are Show that this can be reduced to an equation of the fourth degree. What are its

roots?

The numbers

< 15 and prime to 1,

The

2, 4, 7,

special roots are

cos

15 are

15-7, 15-4, 15-2, 15-1.

-~

i

sin

The non -special roots are the The L.C.M. of x3 - 1 and x* - 1

(r

ID

15

roots of z 3 is

2

4-

(x

x+

1

= 1,

1 5

)

(x

-

2, 4, 7).

and xb 1) and

Therefore the equation whose roots are the special roots

x - x1 + ar - x* 8

1

+#

- a; +

3

1

1

-0.

is

=0,-

thatis, If

ZX + -, this becomes

that

z*

is,

of which the roots are 2 cos

2f7T (r

-r-^-

15

5.

- z3 - 4z 2 + 4z + 1 =0,

= l,

2, 4, 7).

The Equation x 17 - 1 =0.

this equation

Gauss discovered that

the solution of

depends on that of four quadratic equations.

This can be

proved as follows.

The imaginary roots

of

2f7T

cos

-=y

x17 .

L

1

=0

2f77

sm -y=-,

are

= 1, ^

,

where

r

rk

o

o

2, 3,

...

8.

Let yr = cos then we have --y ,

yn-r = yr Also the

sum

of the

and

2yrys - y f+J 4- y r .s ...................... (A)

imaginary roots

is

-

1,

therefore

Equations (A) lead to an arrangement in pairs of y l9 y 2 ... yg which has a remarkable cyclic property. Starting with the pair y^y^ we have ,

On

trial, it will

like result.

Let then

appear that no y except y4 taken with ,

t/j,

,

will lead to a

EXPRESSION IN SURD FORM

174 It follows that

relation connecting a,

any

j8,

y, 8 continues to hold after

the cyclic substitution (a/3yS).

Such

By

relations are the following.

a + j3 + y +

(B)

8= -i ................................... (E)

Again, using equations (A),

ay -i

to* 2/1^8+

2/42/2

+w^^ ay = - -^ ...................................... (F)

therefore

Consequently

j88

we can show

Similarly

=

-^

...................................... (G)

that

16aj8=-l+4a-4j9, .............................. (H) 4a 2 =l+2 8 + y ................................... (I) j

Then by

+ S = 2/x ............................. (J)

a + y = 2A,

Finally let

A+^= ~i

(E)

Also

4A/x,

=
Hence by (H) and three equations derived from stitution (a/?yS), we find that

2/2>2/8

+ j3 = y ~2fy + y -0 2 + 8 -0 2/ ~2yy

2/6,2/7

y

^/*

2/

2 y l9 y^ are the roots of y

Thus

this

by the

cyclic sub-

2ocy

2

2/3.2/5

Since cos quadratic

2

2

(K)

+ 82/-iV =

decreases as 6 increases from

is that

which

to

77,

the greater root of each

is stated first.

Thus, starting with the last equation and working upwards, the values of y l9 y 2 ... ys can be found, and then the roots of x 17 - 1 =0 are given by ,

a

-2^3 + 1=0, Ex.

1.

etc.

Express cos

as a surd.

and

= Also

and by

(I),

~

(34 -2^17),

g^

(34 +2^/1 7).

REGULAR therefor.

cos --

=

f

{

-

17-GON

175

,

1

J^{17 +3^17 ^(34 -2^17) -2^(34 + 2^17)}.

NOTE,

The

(i)

cosines of

~,

9

...

-^=-

can be expressed in a similar form; the

surd values of the sines of these angles involve one more radical sign. (ii)

It

is

(Ex.

XIX,

22.)

length determined by a quadratic equation can be found geometrically. therefore possible, by the use of ruler and compasses only, to effect the following

Any

construction.

To

inscribe a regular 17-sided Polygon in a Circle. be the centre. Take the radius of the circle as unit of length. Let Draw x'Ox, y'Oy at right angles, cutting the circle in A, A' and JB, B'. 6.

Let the vertices of a regular inscribed 17-gon be marked 0, 1, 2, ... 17, the The ordinates y l9 y 2 ... ys of the points point coinciding with B. ,

To solve these geomet8 are given by the equations of Art. 5. 1, 2, 2 = 6 are the ordinates of the observe that of + the roots 2ay rically, y 2 2 = cuts the circle x + y 2ay + 6 = points where the axis x Thus to find the points (0, t/j), (0, y 2 ), ... (0, j/8 ) (two of which are ...

marked

The

t/ 3 ,

y5 in Fig.

27),

we have

centres of the circles are

to

draw the

marked

o>,

circles in the following table.

A,

/z,

The To draw the

a, etc., in Fig. 27.

p are easily drawn, giving the points a, last four circles notice that circles co, A,

j9,

y, 8.

:

(i)

The

circles a,

j3,

y, S meet

y=0

at the

as diameters, respectively. For, since the equation to the circle on

By,

same points as

the circles

on

J5jS,

J58, J5a

_ 2

which

is satisfied

by

x=

*J

- f$,

J5j8

as diameter

is

~ /

.y=0,

2

we can draw

the circles

j8

und

y.

REGULAR

176 (ii)

The diameters of the

to the circles

j8,

circles a,

/3,

17-GON y, S are equal to the tangents

from

B

y, 8, a, respectively.

For (diameter of tangent from JB to

- j3) = (2jS + y + 1) - 4j8, and the square Hence we can draw the circles a, 2/2 -f y.

Q a) = 4 (a Q = 2

j8

1

2

of 8.

14,

Construction.

r

Along Ox take OJ^^OA. Along Oy take and radius oo7 draw a circle cutting yOy in A, f

With centre co With p. centre A and radius A J draw a circle cutting yOy in a, y. With centre ft and radius /i7 draw a circle cutting yOy' in jS, 8. Draw the circle on By as diameter cutting Ox in G. With centre jS and radius $6 draw a circle f

cutting yOy' in

t/3 , j/ 5 .

Through y 3 y 5 draw parallels to x'Ox. These lines meet the given circle and 5, 12 of the required 17-gon, which can be comthe arc 35 at 4 and setting off arcs equal to 34. pleted by bisecting Or we may draw the circles a, y, 8 as explained above. ,

in the vertices 3, 14

RECIPROCAL AND BINOMIAL EQUATIONS

177

EXERCISE XIX

RECIPROCAL AND BINOMIAL EQUATIONS Solve the equations in Exx. 1-9

:

1.

x'-Sx* + Ilx*-Sx+ 1=0.

2. x*

3.

2# 4 + 3z 3 + 5* 2 + 3a; + 2=0.

4.

2x* + x* -I7x*

5.

2x* +

5x*-5x-2=0.

6.

3x 5 - 10* 4 - 3* 3 - 3s a - 10* + 3=0.

7.

2z 6

8.

2z 7 -a;-32; 4 - 3^-0; + 2=0.

9.

*

8

-z 5 -2z3 ~a; + 2=0.

-f

l+(s + l) =2(* + :r + 2

8

1 [Write this x* + x-* + (x + x~

+ Qx*-5x* + 6x + l=Q. + x + 2^0.

4 .

I)

+ 2)*~2(x + x~ l + 1) 4 and show ,

that this reduces

to 10. If sn

~l+u + u + ...+un

where u n

2

1

un+l -Un

j

and

*n=-T-9-2 *j

xn +x- n and

z=x + x~

l

prove that

,

*n= wn-i- 5 n-.

Hence show that

+sn = 1

[1

:

A

11. If a is 2

n= xn

+ a5

,

as

12. If 13.

15 the

+

C

C

JL

-10 depends.

l

and

(1 -ar)(l

-ar')=2-z.]

an imaginary root of # 7 - 1 =0, the equation whose roots are a -fa 6 , 8 a is *,=3 + z -22J - 1 =0.

+ a4

n

is

The

a prime, the special roots of x 2n

special roots of 2T7T

The

~

i

sin

special roots of

values are

d-

(

x9

-1=0

2f7T

values are cos 14.

n+ l equation on which the solution of x*

x~ n

1

cos -

b

i

sin -

where

,

x 12 )

,

r

by 12 [To find the equation, divide a; 6 L.C.M. of a; the x* i.e. by 1, L] \

15. If a,

/?,

are the roots of x n

are the roots of x*

~ 1,

-1=0 that

-1^0

1

+ x* + 1~0, and

=0. their

2, 4.

are the roots of

is,

-f 1

g-(N/3

by the

# 4 -a; a -f

=0, and their

1

t).

L.C.M. of

x2 -

1,

#3 -

1,

x*

-

1,

x9 -

1,

3 2 y are the roots of o# -f bx + cx + d=0, prove that the equation

1 l l * whose roots are a-f-, 5 + -, y-f a ^ y

is

+ (ac + bd)z* + (ab + bc + cd- Bad) z + (a - c) 2 + (6 - d) 2 = 0. 2 3 2 [Prove that adIJ{x -a;(a-fa- + l} = {(aa; + bx* + cx + d) (dx* + ex + bx + a)}, and write (x* + l)fx=z.] adz*

1

)

that there is a value of k for which x* - I5x 3 -&r 2 + 2 find this value. and 1, that x*-15x*-Sx* + 2=0 has two roots a, p such that [Show 16.

x*

4-

Show

is

divisible

by

kx -f

a/?

= l. Then

CONSTRUCTION FOR REGULAR PENTAGON

178 17.

For the equation x*+px* + qx* + j34-l=0, prove that (i)

and only if, 2 4(g-2)

the roots are

all real

if,

2

;

(ii)

least

;

(iii)

two roots are

x + x~

[If z

l t

then

z*

-f-

and two imaginary

real

pz + q 2 = 0.

Let

z l9 z 2

or

i(z 2

&=i(i>/z7^4)

if (fg-f

I)*
2 .

be the values of z, then N/z a 2

-4).

any value of a: is real, one and therefore both values of z must be real, and so p*>4(q-2). The conditions in question can now be obtained by considering the equation whose roots are z^ - 4, z a 2 - 4.] If

18.

// in

the last

all the roots

example

values of z are real, prove that

\

[For then 4(q 2)
-

l)*=a(x* -

1)

of the given equation are imaginary and the

p <4 and q<6.

\

and

etc.]

x^ 1, prove that '

x~

2(1 -a)'

Also if 0 16, the given equation has three real and two imaginary roots, if a <0 or a> 16, the only real root is 1.

but

20.

Show

that, if r

is

any

integer, the roots of

(4r+l)7r

are the values of tan

2fTT

21. If

yr = cos-~ show that o

,

and deduce the construction indicated figure

a

for

inscribing

a

regular

in the

pentagon in

circle.

In the figure

OH=\QA, and

22. Prove that the values of sin

and

cos-^ are respectively

-^^ and

A

(I

-v'17-f N /(34-2 N /17)}--i N /{174-3 N /17

[Referring to Art. 5, Ex.

1,

+ N/(34-2 x/17) + 2

sin=V(i-~iy2) and

{!

+A-

CHAPTER

XII

CUBIC AND BIQUADRATIC EQUATIONS

TheCubic Equation.

1.

The standard form of the cubic equation

is

w = ra + 3ta + 3cz + rf-0 ............................ (A) 3

If

x

y

bja this equation

where

= ay = ax + 6,

becomes

= ac-6 2

//

If 2

If a,

j8,

y are the roots

2.

3

is

+ 3//z +

6-'

=

................................... (C)

a -ft/a,

of (A), those of (B) are

fl

+ b/a, y + ft/a,

a/3 + 6, ay + b.

+ 6,

Equation whose Roots are the Squares of the Differ-

ences of the Roots. equations (A) and

(B).

The differences of the roots are the same Hence the equation whose roots are

(-y) 2 is

G=^a*d

,

the equation z

those of (C) are aa

2

obtained by putting

The

(y-) 2

>

= 3H{a 2 q

resulting equation

,

,

r = G/a*

(<*-)

for

2 ,

in Ch. VI, 17.

is

aV + 18a4^

2

-h81a 2flr2 2;-f27(G?2 -h4Jy 3 )==0,

...............

(D)

whence the important equality a6 (]3-y) 2 (y-a) 2 (a-j8) 2 = - 27(

2

4-4# 3 ) ................ (E)

It is easily verified that 2

.................................. (F)

A = a?d2 - 6a6crf + 4ac 3 + 463rf - 36 2c 2

where

4

2

2

,

.................... (0)

2

a (^-~y) (y~a) (a-j3) =->27J ...................... (H)

so that

The function A is

+ 4#3 = a 2J,

is

called the discriminant of the cubic u,

and

the necessary and sufficient condition that the equation

have two equal

roots.

its t/

vanishing should

=

CARDAN'S SOLUTION

180

Excluding the case in which two

Character of the Roots.

3.

roots are equal, possibilities are (i)

and remembering that imaginary roots occur

All the roots

may

One root may

and then by

be real,

and two imaginary

(H),

in pairs, the

J<0.

and, denoting the roots by a, A IIJL, we find that the product of the squares of the differences of the roots - A) 2 + /it 2 } 2 so that A >0. is equal to 4p,*{ (a (ii)

be real

;

,

Hence if J<0, all the roots are real and unequal ; ifA>0, one root is real and two are imaginary. Also it follows from equation (D) that J7 = 0, 6r=0 are the necessary and sufficient conditions for three equal roots.

Cardan's Solution.

4.

z=az + 6,

If

the equation

az + 3&z -h3cz-hd = 3

2

becomes

z z = w*~

Let

Comparing

(B)

3

+ 3#z + #=0 .............................. (B) z3

+ w*; then

and

(C),

.............................. (A)

- 3rwn*z - (m + n) =

m*n* = -H, m + n=

-G.

m and n are the roots of 2 + <#-#3 =0, and we may take m = J - G + >/(r2 + 4/73). Therefore

t

................ (C)

........................

(D)

(

If

Q

denotes any one of the three values of

the three values of

m*

are Q, ajQ,

<*)

2

Q, where

CD is

an imaginary cube root

of unity.

Also, because

m

3

w* = -/?, the corresponding values of

-H/Q, Hence the values

of

z,

that

is

ri*

are

-<*HIQ, of

ax 4- 6, are

Q-H/Q, O>Q+4H 8 <0, all the roots

NOTE. If C? 2 of the cubic are real, but Cardan's solution them in an imaginary form, which is very unsuitable for practical purposes.* In this case a solution can be easily obtained as follows. gives

5.

Trigonometrical Solution when

G 2 +4H 3 <0.

Taking the

equation z3

when

6r

2

3

+ 4Jff <0, a

+ 3Hz 4-0=0,

solution can be obtained

This

is

sometimes called the

................................. (A)

by using the equation

irreducible cote.

FUNCTIONS OF THE ROOTS If cos

30

is

known, cos

determined by

is

0-i cos 30 =

cos 3 0-f cos z = q cos 0,

Let

181

(B)

so that

cos 3

Equations (B) and

3ff

G

q

q*

+ ^-cos0+ 2

=

-^

(C) ' v

(C) are identical if

$

4

ff

= 2V-# and cos30-~-~= 3

2>/-/P

2

Now

G?2


equation.

ifif

a

is

3 ,

and so a

(D)

,___..

can be found to satisfy the last

real value of

any such value, the roots of (A) are ^27T

jcosl ^x.

1.

reduced

+ 3&# 2 + 3cx + d =0

3 // all the roots of ax

are reai, a/iow

tJiat

the equation

can be

to

3-j-f^=0,

2

where

27/x

<4,

xp

+ qt, where p and q are by a substitution of the form If z = ax + 6, the equation becomes z 3 - 7b + Q = 0,

A= -3#=3(6 2 -oc).

where

-aj

real.

2

Since aU the roots are real,

+4# 3 <0,

/.

#<0

and

h>0. Writing

the equation becomes

z=*Jh.t,

substitution

6

x~ NOTE. reduction.

The

XXIX,

a

t

a

M =a-f

2

co y,

an imaginary cube root

a> is

~- .t. + >Jh

3

1

=1,

of unity,

+ CO + C0 = 0, 2

TAe interchange of any two of the

M

CO

3

(0

2

a> /J-f coy,

and observe that

-^=1^3.

letters a,

into the other, functions L , Thus the transposition (a, j8) changes

/J,

y transforms

either of the

3

+ coa +

a>

2

y)

3 ==

(cua

+

i3 into

3 co ^8

+

2

co y)

3

= coW 3 = M3

.

It is the existence of functions possessing this property which the solution of a cubic equation depend on that of a quadratic. (2)

the

3.)

.L=a-hco/J-f

CU

(1)

= GI*Jh\ and

Important Functions of the Roots..

Let

where

jj.

roots of the cubic can be expressed as convergent series by using this

(See Ch.

Two

6.

where

t*~t+p.~

is

L and

M are functions of

the differences

1-f co-f

they are unaltered by writing a + A,

j8

a>

2

o/a>

j8,

y.

For since

=0,

+ h, y + h

for

a,

j3,

y.

makes

SUM OF TWO CUBES

182 (3)

We

have the following equalities

wM = 2j8 ~ y - a,

orL + a>

+

cu

2

M = 2y - a - 0,

by Ex. XII,

Also,

a>

2

:

L - wM =

a>L -

'

cu

W

-

(

(

a>

-

co

-

a>

oj

2 )

2

(y

- a)

- j8) ) (a

,

.

9,

Z 3 + M 3 = -27/a 3

Therefore

,

..............................

(A)

i3 -M 3 =-3(
iJf = 2k 2 - 2?j8y = -9#/a2 ......................... (C)

Moreover, (4)

From

equations (A) and (C)

follows that (-|a) 3 , (-JaM) 3 are

it

Ae

roofs of

which 7.

is

the auxiliary quadratic in Cardan's solution.

The Cubic

Sum

as the

of

Two

3 2 // u == ax -h 36x -f 3cx + rf, ^Aew constants

A

Cubes. ,

B,

A, p,

u^A(x-\Y + B(x-tf\ provided that u has no square factor

H

s=

(ac

Also

Eliminating A,

B

from the

equations,

coefficients it will be seen that the identity

\A + ^.B^-

first

-

2

a( A

Assume that

are the roots of

)

A+B = a

(A)holdsif

A, /x

........................... (A)

- 6 2 x 2 + (acZ - be) x + (M - c2 ) = 0.

By expanding and equating

of

.

can be found such that

2

A/u

jit

)

three

and from the

last three of these

+ 6( A2 -p?) + c(A -/x) -0, - A 2 ) = 0.

and that A^/x, and divide the these equations by A-/x and the second by Aju,(A~/t), then = 0, and aA/Lt + o(A+jLt)4-c

Eliminating

neither A nor

/u,

is

zero

first

/^,

Therefore A and, by symmetry,

^ are the roots of - (bx + c) 2 = 0, (ax + b) (ex + d)

which

is

the same as

H(ac~62 )x Also A,

B

are given

by

+ (od~6c)x + (6d!-c2 )==0 ................... (C) A + B =a, A-4 +fJ3 = - 6. 2

THE HESSIAN

183

// A = 0, then by equations (B), TJ 2 =A &i-c

so that also in this case A

// A

jit,

two equal

the reasoning roots.

i and

c o
and p are the roots

Ex.

w=

if

and only

)

A

has two equal roots, so also has

H=

UG the preceding

1.

if,

Let

2x*

to solve the

)

2x*

equation

A+B = 2,

and

A+2B=;-l,

2 -(#-7) =0,

so that

vice versa.

giving

A

= l,/i=2;

.4=5, 7?^ -3.

.

,

and the roots

.

_ 5"^ -2A;3^ _ (5Jfc3*)

,

has

+ 3#2 - 2lx + 19 =0.

Hence the given equation may be written 5(x- 1) 3 3(#-2) 3 2 by -^/5 (x - I ) = k /3 (a: - 2), where k 1, co, co

are given

=1

w=0

+ 3x*-2lx + l$ = A(x-X)* + B(x-iJL)* ........................ (A)

Then A, //, are the roots of (2* + 1)( -lx + 19) and, equating the coefficients of x 3 and # 2 in (A),

Since ^3

if,

For using equation (G) of Art. 2, we find that 2 - 6 2 (bd - c2 = 4 (ad 6c) (ao

Hence,

H = 0.

of

This case will arise

fails.

,

(

Hence the roots are |(

1

-

/75 -4/45),

The Hessian.

8. is

-

called the Hessian,

The roots

of

H=

will

hQ =

where

(1)

and

is

of great

H = ac- 6

2 ,

A,

ft,

and we

h^ad- be,

shall write

h2 = bd-

c2 ,

A 1 2 ~4A A 2 = A .................................. (A)

Using the identities of Exercise XII,

~18H = a (3)

importance in the theory of the cubic.

be denoted by

so that

(2)

H which occurs in the last article

The function

We

2

2

2

2

{(o;--a) ()8~y) 4-(a:--j3)

can find the values of

A, p, in

4, 5, 6,

we

find that 2

2

(y-a) -f(x~ y ) (a^j9)

terms of a,

/J,

y as

2

} .......

follows.

(B)

Let

M =a + eo j8 + coy, y, 2 = Z/ j8y + coya 4- a> aj8, M' = /Jy -f co 2 ya 4- a>aj8. L =

Then,

and

if

2

2

P = (z-a)(j8~y), Q = (a?-j8)(y-a), R = (x~y)(a-j8), P +
we have

AUXILIARY QUADRATIC

184

from (B) that one root A

It follows

that

2 A(> -

is,

(4) It is

by

+ (co 2 - w)L' =0,

^ = -M'/Jtf ...................... (C)

A = -L'/l and similarly

so that

easy to verify that the substitution

reduces the equation h^x?

which

<*>)L

H=0 is given

of

+ hfl + h2 =

to

the auxiliary quadratic in Cardan's solution. Thus the roots of the auxiliary quadratic are is

EXERCISE

XX

THE CUBIC (i)

Find to five places of decimals the real root of z s + 293-97=0, (ii) z 8 + 6a; 2 H-27a;-26^0, (iii)

2.

Find to

1.

five places of (i)

3.

decimals the real roots of

x* -3a;-Hl=0,

(ii)

- 3qx +r =0 and k = y are the roots of a: (0-; 3

If a, 0, (i)

>/3(4g

3

-r 2 ), show

tl

t

08 -;

(ii)

The equation whose

(iii)

roots are

j9

- y, y - a, a - j3

is

+ y + y a=|(r + Jfc), a y-f ^ a + y ^ = f (r-t). a 8 8 = 3 a y-hj5 3 a + y 3 )8= -9g 2 a (v) j8 + ^ y-f-y a If the are all roots real and a> j8> y, the difference between any two (vi) of the roots cannot exceed 2 N /(3g), and the difference between the greatest and least must exceed aa

(iv)

2

a

2

2

2

.

4. If a, 0, (i)

y are the roots the equation whose roots are

j8

y,

y

- a, a - jS )

(ii)

=

is

;

the equation whose roots are a 2 j8 + ]8 2 y -4r y 2 a, a 2 y + 2 a -f y 2 )9 a4* 2 - 3a 2 (acZ - She) x -f 9 (a 2d 2 - fabcd + 3ac 3 + 36 8d ) = 0.

is

In &e following examples ike letters L, M, L', M' 9 h , h l9 h 2 have the meanings assigned in Art. 8, and unless otherwise stated a, j9, y are the roots of

.5.

Show

that

L t + "Jf=3.M', and M* + a

a

L=3L'.

TYPES OF SUBSTITUTIONS If

6.

-

A (x - A)

u

A)

3

-f

B(x -

l jB

A/B=

k=

s/(

;[

y are the roots of u =0, assume that

^=

(1, 1, 1), (1, o>, o>

2

),

(1, o>

2

- JUT '/ Jtf

w),

,

(A)

.

add and use Ex.

5.]

+ i )>

^(*i

and

a 2 y + j3 2 a + y 2

+

3^

=

~

(/H-fc)^

Find the values of I, m, n such that

2^0^-^ + ^,

proving that

That

is

to say,

'

homographic ^ ^

which

jS,

A = - L'/L,

- L 9 /M\

y + J/? + ray -fr&=0,

9.

a,

- JJ), prove that

+ /8V4V + 8 =

8.

and

2

Hence prove that

If

/x)

185

+*(a - M )=a>4*(|3 - A) + J3*(j8 - ,x)=a) ^*(y - A) -f JS*(y - ft)=0.

[Multiply equations (A) by 7.

3

'

yoc

mh

2h

any iwo roots

+ Jy + ma-fn-0,

(a,

l

-k,

h

a/?

n=h

+ Za + w/? + tt=0 where &

29

o/ i^e cubic equation

j8)

u =0

;

iV( -JJ).

are connected by the

relation

Except when J =0, there are two substitutions of the form will transform the cubic equation

u=Q into

itself,

and these

are

(

where 3& 2 =4A ^ 2 ~^i 2 and ^ ^i> Explain why this fails if A ~0. >

[This

is

merely another

way

^2 are the coefficients of the Hessian.

of stating the last part of Ex.

m

8.]

f

10. Find substitutions of the form x~(ly + m}l(l'y + transform ) which will the equation x 3 - 3# 2 + 3=0 into itself, showing that these are

* = (2y-3)/(y-l) 11.

and

* = (y-

Find substitutions of the form x~(ly + m)l(l'y + m') which will transform 2 60; -f 1 ^0 into itself, showing that these are 4- 3#

the equation z 3

12. Show that any pair of roots a, j3 of the equation x 3 -3ic 2 4- 3s- 2 = are connected by a relation of the form pa. -\-qfi + r=- where p q, r are the same numbers whatever pair is chosen, proving that the relation is 9

13. If a,

j3,

y are the roots of x* -30#-f r=0, find Za

proving that,

where

2

+ ma 4- ft

if a,

j8,

y are

=

I,

m, n such that

n = y, Z /j9 + mj5 + real and a> > y, then 2

/?,

jfc

Hence show that with these values of I, m, n the substitution transforms the equation x 3 - 3qx

+ r=Q

hi to

itself.

1S6

BIQUADRATIC EQUATION

14. If

either

,

2 y are the roots of a?-21a; + 35=0, show that a + 2a- 14

jS,

15. If a,

y=x* + 2x~

(In other words, the substitution

or y. equation into ft

is equal to 14 transforms the

itself.)

w=0 and = = k, <**+pk, y

y are the roots of the cubic

/?,

2

^

* prove that

where A

p=

36

a

A=

if

Further,

^=

tt=0

Prove that the equation y x* + px + q,

16.

6

3i=Jf(A-,*) =

-Jf'/Jf, then

2 8 (1, 1, 1), (1, ), (1, to , o>),

by

[Multiply

-L'/A

,

g= a hA-a

H =0.

a root of the Hessian

is

2c

,

-fA,

27/2

and add.] transformed into

is

36

.-,

p= a

if

where A

is

H=0

a root of

2c

N

6

N g= a + A-, a

+A,

and k

determined as in Ex.

is

Find substitutions of the form y ~x 2 + px -f q which

17.

by the

y*=k?

substitution *

15.

will

transform

=0 into the form y 3 = ^ 3 , giving the value of k in each case. 3 - 3 =0 and ( L) 3 , [A, p, are the roote of x* + 2x (JJif ) are the roots of

i.e.

of

t*

=4 3

If

.

we take A = 1,

fi

=

- 3, we must take

found that each of the substitutions, reduce tfc=0 to 18.

In Ex.

Explain

9.

J^^S

2

why it is not to

x = y - 6/a,

If 2= ay = ax + 6,

have

6/a,

j5,

3

-f

i/

= 8,

giving

ic

a

-f

6H 4G .OA +_ _^ !f+?

* This is

is

(A)

m

iC

2

G =a2d~3a6o + 26s

=a3e-4a26d + 6a62c-364

,

,

(B)

f

.

the equation becomes

+ &Hz* + Gz + K=Q

(C)

y, S are the roots of (A), those of (B) are

and those

4x + 3=0.

6c#2 + 4dz-f e=0

s

,

z*

>

It

becomes

this

iTac~62

where

If

will

The Standard Form of the Biquadratic Equation

-

-f

thus

6.

be expected that both roots of this equation satisfy u =0.

t ay +

8

is

y=# 2 -f-7,

.]

u=*ax* + 46x If

M=

and

3

y-x + 4x + 11, we may

17, if

L = - 6,

y~x + 4a;-f 11 2

of (C) are ooc *

+ 6,

a/J 4- 6,

ay

-f 6,

a + 6/a,

/?

+ &/#, y + 6/a,

aS + 6.

a case of Techirnhausen't Transformation.'

(See Art. 17.)

REDUCING CUBIC

Some Important Functions

10.

A = /Jy-fa8,

Any rearrangement

(1)

functions

A,

/i,

of the Roots.

= ya-f-/JS,

/A

of the letters a,

v into itself or into

187

j8,

Let

i>=a/J-fyS. y, 8 transforms

any one

of the

one of the other two.

found that the existence of functions having this property makes the solution of (A) depend on that of a cubic equation. It will be

(2)

We

The

cyclic substitution (a/Jy) changes

have

A to

/it,

/z

to

v,

v to A.

(j8-y)(a-S)=i>-/z

(y-a)(j3-SHA-v Hence if two

of a,

/J,

three of a,

j8,

y, 8 are equal,

then two of

........................... (D)

A,

//,,

v are equal,

and vice

versa.

Also (3)

if

The functions

A,

then A=/i = *>, and vice versa.

y, 8 are equal,

/x,

v are Ae roote

o/*

a*y*-&a cf + 4a(bd-ae)y-8(2ad* + 2eb*-3ace) = Q ......... (E) 2

For

2;A = 2:aj8 = 6c/a, 7a

.

ZajSy

+ aj3y 827a

- 4aj8y8 = 4 (4W - ae)/a2

,

2

~ (4d2 - See) + ^3 (4fc2 - Sac) '

a2

The second term

a

of this equation

can be removed by the substitution

and the equation becomes where

4^3_/^ + j = o, = / ae 4bd + Sc2

(G)

(H)

,

Expressed as a determinant a

&

c

c

d

e

bed

It will be is

shown that the

solution of (A) depends

on equation

(G),

which

called the reducing cubic.

The functions 7 and J are

of great importance in the theory of the

biquadratic.

K

B.C.A.

THE DISCRIMINANT

188

Boots of the Reducing Cubic.

(4)

(Q) corresponding to A,

Let t^t 2

^a = A-^-A-i(A+ a

/

t

s

be the roots of equation

Then by

v respectively.

/x,

,

(F)

+ v)=i{A- /t -(v-A)};

i

whence, by (A), t^^a{(y - a )(j8-8)-(a-jB)(y-8)}, with similar values for 2 2 tz If real,

y, 8 are all real, or if they are all imaginary, then

/?,

and

of the three

or

if

A,

a,

Also

t

/z,

jS,

t% is real,

a,

y, 8 are real

and

v

t%

v,

and therefore

y

y, 8 are of the

,

if

j8

are real

and two are imaginary

t

v

Z

3,

2,

lim,

forms

and y = / -f ^m,

are

V 8

all real if a,

im'

The Functions /?,

t

2,

t

3

are all

I,

Since

J.

,

then only one

t

l9

t%,

3

jS,

y, 8 are all real,

.

m,

== /

imaginary. statements are true. (5)

v

conversely.

also

Moreover, one of these cases must

of a,

t

conversely.

If two of the roots a,

For

.................... (L)

.

,

a,

............. (K)

is real and A, /x are and so the converse

then v arise,

are functions of the differences

y, 8, so also are /, J.

/ and

It follows that

J

therefore in equations (H)

same

are the

and

(I)

6 = 0, c-H/a,

a I= z

giving

K + 3H*

whence the important

for equations (A)

and

(B),

and

we may put d=

2

/a

e

,

= #/a 3

,

a*J^HK-G*-H*>

and

identities

= a 2/-3ff2 Another important equality

is

8-y)2(a-8)2 + (y-a)

()

G* + H* = a*(HI-aJ) .................. (M)

,

2 8-8) + (a-j3)2(y-3)2 = 24//a2 ....... (N)

2 (j

For

and

(6)

The Discriminant.

By

equations (D),

a(j8- y )(y -)( -/3)(-8)(

- <3 ) 2 (<3 -

= 256J by where

^ft -

< ) 2

2

by equation (K)

Art. 2,

............................................ (0)

J =/

-27,72 .................................... (P)

3

REDUCING CUBIC The function A ing

is

called the discriminant of the quartic w,

and

the necessary and sufficient condition that the equation

is

have two equal 11.

its

vanish-

w=

may

roots.

Character of the Roots.

J = 0.

this

If two roots are equal, then and remembering that if there are any

case

Excluding imaginary roots, then these occur in (1)

189

All the roots

may

be real,

pairs, the possible cases are

:

and then J>0.

Two roots may be real and two imaginary. Denoting the roots by a, /?, im, it is easily shown that the product of the squares of the differences of the roots is -4w 2 (a-0) 2 {(a - 1)* + !*}*{ (0 - 1) + 8 }*, and so J<0. (2)

I

w

(3)

All the roots

may

Denoting them by

be imaginary.

lim, /'iw',

we

find that the product of the squares of the differences of the roots

is

16w2m' 2 {(Z Hence

(4)

12.

- m') 2

+ (w + m')*}*{(l + Z') 2 +

2

}

,

and so

A>0.

// J<0, two roots are real and two imaginary.

(5) If

For the

2 Z')

J>0,

the roots are all real or all imaginary.

criterion distinguishing the

Solution

Ferrari's

two cases

of the

of (5), see Art. 15.

Biquadratic.

Writing

the

equation

u=ax* + we assume

bx*

+ &cx2 + 4dx + e^0,

(A)

that

au = (ax* + 2bx + s)*~(2mx + n)* Expanding and equating

coefficients,

2w = as + 262 ~3ac, 2

(B)

we have

wn = &s-ad, n2 = s 2 -ae

(C)

Eliminating m, n, 2

(s

- ae) (as + 2V* -Sac) = 2 (bs - ad) 2

,

which reduces to 53

-3c5 2 + (46d-ae)5 + (3ace-2ad2 -2662 ) =

(D)

The second term can be removed by the substitution

= 2t + c

(E)

&3 ~/* + J = 0,

(F)

s

and equation (D) becomes which

is

the

c

reducing cubic.'

i

(See Art. 10.)

= 2bt + bc-ad

Equations

I

(C)

become

(G)

FERRARI'S SOLUTION

190 Thus, if fj

a root of

is

m the equation

its

n t = (26^ + 6c - ad)/m v

b 2 - ac), l =\/ (at v +

u^=0 can

be put into the form

(ax

and

and

(F)

2

+ 26z + c + 2^) 2 - (2m 1 a? + n x ) 2 - 0,

roots are the roots of the quadratics

ax 2 + 2bx + c + 2^ =

(2m1x + 14).

It should be noticed tha^ the three roots of (F) correspond to the three of expressing

ways

u as the product

of

two quadratic

=x* + 3x 3 + x* - 2 = 0. u = (x 2 +px + s) 2 - (mx + n) 2 and equating coefficients, wo have Expanding

Ex.

Solve n

1.

Let

The

may

last

equation

is

a

and

u

[x*

-

if s -~

satisfied

take

and then

J-,

+ (p

-f-

77^) a;

?yr

-

-1+4, inn-

4.

Thus we

+ 5 -f n} {a; 2 + (p - ?/i) a; + s - n}

^(x*+2x-2)(x*+x + l). -1 >/3, w, a> 2 \vhero w ,

is

an imaginary cube root of

1.

Deductions from Ferrari's Solution.

13.

then

.

=-|,

=i,

=--j,

Therefore the roots are

Let

factors.

]8,

y be the

ax 2 + 2bx + c + 2^ + (2m 1 x + 7i 1 )=0,

roots of

2

ax + 26x + c +

a, S are the roots of

Further,

let (y, a), (a,

/?)

(H) by changing the suffix

1

2/ x

-

(2^0; + n x = )

...........

(H)

............. (I)

be the roots of the equations obtained from into 2 and 3 respectively.

= c + 2t l + n v a(3y

Now

a (/?y 4- aS)

therefore

which agrees with equation

(F) of Art. 10.

a(j8y-a8) = 2w

Also

= 2c + 4^,

1

,

a(]8-f

y-a-8)== -4m 1}

with similar equations given by the cyclic substitutions

................ (J)

(a,

j8,

y)

and

(1, 2, 3) of the suffixes.

Hence

(i)

the values of

m are

In connection with these functions

Ex.

3, (iii), p.

it

should be noted that by Ch. VI, 16,

96, ............ (K)

DESCARTES' SOLUTION by

Also,

This

(F)

and

The values of n are

(ii)

The values

(iii)

whose roots are

(G), the equation

an important equation, obtained

is

of

J^ZY^

___

and from

y + oc ~ p

o

Of.

- yS).

r(

ya-)8S 3+y

la(oc^

are

~

,

2m 92

in Art. 15.

way

- j8S),

|a(ya

-~- 2 >

2ml

another

in

\a(fiy -aS),

-~-

191

S

ot

+p

o

y

that these functions are the values of z found by

(F), (G) it follows

eliminating tfrom Z

+ bc~ad

2bt

1 __ _ ~

,, (M)

and

~2' 7

Biquadratic as the Product of Quadratic Factors.

14.

9

Let u

Descartes method.

ax*

4-

4foe 3

u = a (x + 2Zx -f 2

Expanding and equating l

+l =2

--

a

,

?/i

2 )

(x

coefficients,

m+m =6

a

~ 4a

-f

6cx 2

-!-

4c?o; -f e,

and assume that

+

we have k

d

:

,

a

,

,

-m?w

=-e

.

a

Now 1

1

I

l

1

m

1

l

m'

m'

I

Substituting the above values for

row by

If

we

a/2,

-f

Z

m 1

m + m'

2

|-

f

f

V

211'

m + m' lm +l'm m + m', etc., and + 1',

-0.

+ I'm 2mm'

lm'

multiplying each

we have a

b

3c-2all

b

all'

d

3c-2all'

d

e

f

=0.

\

write

the equation becomes

c

which when expanded

I,

+ 2t

a

b

b

c-t

d

d

e

+ 2t

c

=

0,

is

Corresponding to any root ^ of this equation, we can m, V, m' which will satisfy the conditions.

now find values

of

POUR REAL ROOTS

192

2 = given in the form u^-ofl + px + qx + r Q, we proceed as in Art. 15, using I instead of 2Z, to obtain the cubic in the form 2 2= 2 irl 2 0, which reduces to an equation lacking the second P(l +p) y

Again,

if

the equation

term on substituting

z

is

- 2p/3

for

I

2 .

In numerical work, unless this cubic has rational roots, the solution becomes by the methods described in this chapter the student can convince

NOTE.

very laborious himself of this

;

by verifying the steps of Cardan's method for solving the equation, 2 8 - 1 \z - 15 =0, which is obtained as above for the biquadratic, x* - 3# 2 - x -f 2 = 0.* Later, by one of the methods described in Ch. XXVII, it will be found that one root can be found approximately with very little trouble. This is all that it is only necessary to have one way of breaking up u into quadratic

15.

Four Real Roots.

In Art.

11, it

is

required, since

factors.

has been shown that

necessary condition that the four roots should be real rule (Ch. VI, 11, Ex. 1), if #>0, at least two of the roots

Hence, J>0, 77 <0 are necessary conditions that real but these two alone are not sufficient.

;

J>0 is a

also, by De Gua's must be imaginary.

the roots should be

all

;

The complete

may be found by using Sturm's theorem, the usual course This, given adopted in text-books, involves rather tedious reckoning the conditions can be found much more easily by the set of conditions

later.

;

use of either Ferrari's or Descartes' solution of the biquadratic. /

The necessary and sufficient conditions u == ax* -f 46x3 -f 6cx2 + kdx -f e =

Theorem.

aw

all real i*

(i)

or their equivalents

Let

z = ax

+b

;

v = z*

(ii)

J>0, J>(),

then by Art.

that the roots of

//<0,

and

12# 2 >

H<(),

and

2Hl>3uJ.

9, (C),

and Art.

+ 6Hz* + 4:Gz + K = O

t

2

/,

10, (M), z is given

K = a I-3H 2

where

by

2 .

v= has the same number of real roots as w = 0. For v = 0, using the method of Descartes, and supposing that - 2lz + v 55 2 + 2lz + m) m')

Now,

(z*

(z

we

get,

by equating

Eliminating

Hence, since

m

9

coefficients,

and m' from these equations,

3

(2Z -f

3HI)

2

- G2 = 1 2K, or

W + l2Hl* + (9H*-K)l*-G*=Q.

K=a*I -3H2

,

the values of

I

2

are the roots of 2

which, with the substitution, y +

H = at,

=0;

reduces to 4J

3

..................... (A)

-

*Thus, tt'+v^lS, ut?ll/3, t*-t?= 5-27397, ti=-2-16423, t>=l-0943, which ttS(*H2-42042s+l-63580)(s-2-42042a:+l-22265)=0.

-i+t>- 3-85845, from

FOUR IMAGINARY ROOTS Now

the three values of

I

2

correspond to the three ways of expressing

v as the product of two quadratic factors.

and

the roots of v =

If

(i)

i.e.

positive,

are

all real,

Hence,

the three values of

the three roots of (A) are

the roots of v =

If

(ii)

193

I

2

must be

all real

all positive.

imaginary, then, since their sum is zero, 2 Ait/i, -A-ti/i/, and thus the values of I are 2 -i(/A-/i') i.e. the roots of (A) are all real, but only one are

all

we may denote them by A2 is

2

-4(/4+/z')

,

positive If

(iii)

>

,

and conversely.

;

two

=

of the roots of v

are real

and two imaginary, we may denote

and then the values of Z2 A, JJL, A'i/*', where A-Fju-h2A' = 2 2 i.e. the roots of of which one is real and the other (A), are A' J(A -f X' t///) two imaginary and conversely.

them by

,

;

,

,

;

It follows that the roots of v if,

and only

if,

Now, the is

the case

the roots of (A) are

roots of (A) are real if,

and only

if,

Descartes' rule (Ch. VI,

negative and theorem.

if

12H 2 -a 2I

is

I2H

2

J=/ -27/

2

2

is

2

First observe that,

p + pq + q

2

but /

is

hence,

2

;

so that

,

J>0,

/>0

hence, since

;

//<0,

if

p and
q are

:

q according as

|

2HI\>\ 3aJ\;

any

2HI<3aJ.

and real

numbers, positive or negative,

hence, since

;

i.e.

,

2HI<3aJ.

2 3 3 2 p -q = (p- q) (p + pq + q ),

it

p g q*

we have 8/P/ 3 <27a3 J 3
2HI<3aJ, and J>0; and therefore

G 2 + I2H*
addition,

l2H 3 <.a 2HI, and

;

;

dividing by H, which Thus, the two sets of conditions are equivalent.

therefore

suppose

i2H 2 >a 2I.

4# 2/ 2 >9a 2
ff<0,

3

positive,

by

;

we have

>0,

= (p + %q) 2 + f 2 >0

follows that p* Then, since

this

also, the roots of (A) are, by and positive, if, and only if, H is which proves the first part of the

and

negative, therefore

Next, suppose that 2

and

;

HI<0

P>a P>27a J

HI

and, since

are real

of the second set of conditions,

//<0> 3

Then, since

all real,

all real

positive

that

Also,

and positive. those of 4 3 - It + J =

u = 0, are

9, 10), all real

follows that

it

therefore those of

A = / 3 -27e/ 2 >0

To prove the equivalence first

= 0, and

is

negative,

12H 2 >a 2 I.

~l * In 1 general, if n is odd, =*(p-~q)(pn- +pn-*q+ ... +gn ), where the last factor is the qn according product of pairs of complex conjugate factors, and is therefore positive hence p n

pnqn

^

;

&spr^q. But,

if

n

is

even,

and not as p :g q.

pnqn^ty

q)(p+q) (a positive factor), hence

pn^^n

as

|p|<|ff|,

EECIPEOCAL TRANSFORMATION

194

Alternatively, using Ferrari's method, the equation

whose roots are

s

which, when expanded, is equation (A) on p. 192. There are three possibilities. (i)

AH

the roots of u =

all three real (ii)

and

may be imaginary = /? A-
(iii)

Two

take a,

j8

;

I/JL',

of (A) are

-iaV-fO

three real, but owe owfa/

all

in that case the roots of (A) are

;

u=

a = A + i/z,

are

be real

positive.

All the roots of

and the roots

may

of the roots of

,

8 = A'

-iW)

I/A'

2

2

1

>

we may take ;

(A-

2

A')

,

which

is positive.

w=

to be the real roots

2

in that case

may

be real arid two imaginary 8 = A - t/i and then ;

and y = A + t/i,

we may

;

are the roots of (A), of which one only is real. The proof proceeds as before.

Another method Exercise

XXI,

of solving the biquadratic,

due to Euler,

is

given in

17.

Transformation into the Reciprocal Form. The substitution x = py + q transforms the equation w = into apY + *BrPy* + Wp*y* + 4:Dpy + E = Q, .................. (A) B = ag + b, C = aq 2 + 2bq + c, where D = (a, 6, c, <%, 1)3, E-(a, 6, c, d, efa I) 4 16.

.

= if ap* E, Bp*==Dp, 2 = 2^, p* = D/B .......................... (B) by aZ>

This will be a reciprocal equation that

is, if q,

p

are given

Suppose that the conditions (B) are satisfied, and that the roots of (A) where J/22/3 == J/iJ/4 :== l* Let a, j8, y, S be the corresponding !/i> 2/2> J/3> 2/4

are

roots of

w = 0, so that

It follows that

? H~?)(y-?) = (a~?)(S-?), 2

and

...................... (C)

therefore

Hence by

Art. 13,

(iii),

the values of q are the roots of the equation in z found

-

by eliminating tfrom

S' 2

i

LO

an d

TSCHIRNHAUSEN'S TRANSFORMATION

qp

The points corresponding to q and respectively the centre C and the foci NOTE.

F

F' of the involution determined by the

y

"p/

9

as given

COtfl

points a, j8, y, 8, in which (j8, y) and (a, 8) are corresponding pairs of points. That is

Find a

1.

substitution of the

form

x*

Use

into the reciprocal form.

o = l,

Here

4* 3

The equation One

and by

solution

is

(C), are

Y

fi"

F

C j8.Cy = Ca.C8=CJ2 ==JF C a /

f

xpy+q which will change the

+ x*-2x-l=Q

.

equation

....................................... (A)

this to solve the equation.

6

= J,

-/* + Jr=0

$=-J,

by

FIG 29

to say, the line segments are such that Ex.

195

and by

is

c

= 0,

4e 3

d=

c=-l;

-^,

-f^-^=0,

that

8

is

(4*)

+2 (40 -3=0.

(E) of the text, the corresponding value of q is

(B) of the text, the corresponding value of

p

2

is

-i + |~|_

*

Taking p = 1, one substitution of the kind required is * =y 1. 2 4 ~ becomes the + in 3t/ -f 1 =0. %!? 3y y equation (A), Substituting 1 2 Dividing by y and putting z=y-f-y"" we find that ,

2 1

Also

y -#s /.

17.

+ l=0,

a

- 3* + 1 = 0,

giving

z

= | (3

/5).

2/=i(2V2 -4), and aj=J(z~2db^ a ~4) a

.*.

*=i(- 1+^5^-2 + 6^5),

or

i( -1

;

-^^^2 -6^6).

Tschirnhausen's Transformation.

If

we

eliminate

x

between ~ + a^-1 + a2x n 2 + + a n = r "" 2 1 y=xr + p^" + ^2^ + + Pr

/(x)

and

we

shall obtain

=a

o;

n

an equation

. . .

of the

form

Thus for a single value of y corresponds to each of the n values of x. can be chosen so that r of the ... p r theoretically, in general p v p 2 , ^n are z ^ro.

coefficients bl9 62 >

Transformation

:

it

This process

is

called Tschirrihausen's

has been applied to the cubic in Exercise

XX,

16,

TSCHIRNHAUSEN'S TRANSFORMATION

196

Let

Case of the biquadratic.

a,

y, 8

j8,

be the roots of

u = ax 4 -f 46x3 -f 6cz2 + 4dz + e = 0.

We shall

prove that in general three if x is eliminated between

such that

u=

sets

of values of

p and

q can be

found

2 y = x + pz + q,

and

the resulting equation is of the form 4

Further, the values of p

and

is

+ =

.................................. (A)

<7

q are given by

ap=-Ab + where z

2

+/
2/

aq = 3c + 2bz,

2az,

......................... (B)

one of the three

aff-yS a + jS-y-S

ya-/3S

j8y~
'

'

y+a-jS^

J3Ty-a-S

*

the resultof y corresponds to each of the four values of x Let p, q be chosen so ing equation in y is therefore of the fourth degree. 3 and let are and of that the coefficients zero, y 2 be the values y l9 y y

One value

of y.

;

Then we may take 4-j>y

+ ?=

-~2/2

whence by addition and subtraction,

y-a-8) = 0, ................... (D) s 2 +^5 1 + 4y = 0, ................................. (E)

and where

sl

= 2oL,

sz

= Z
2

Now therefore

(j3H-y)

-(

where

-p^s.-Zz, *

2

we

a/

j8y-a8

=

p=

Hence Substituting in (E),

have, from (D),

46

+ 2z ..................................... (F)

find that

a

a

\

3c

a 2

giving

(

Moreover, is

of the

if

form

p, q (A).

have these values,

it is

}

obvious that the equation in y

SUBSTITUTIONS As

197

be seen in the next example, each value of y gives one and only

will

one value of x which* satisfies the equation w = 0. Ex.

1

.

Find a

4

to the

form y Here a = 1, we have

2

+/?/ b

substitution of the

+ g ~0. Find

=0,

c

=0, d = 3,

e

of

t

is

1,

-

5, so

that

where

4* 3

z~f and

giving

y=xz +px + q

the values off, g,

=

z^ One value

form

which

and use

will reduce the equation

the result to solve the equation.

7=-6, J=-9; and using Art.

13,

(iii),

+5*-9=0.

then

One such

p=3, q=0.

substitution

is

therefore

y

To

eliminate x from (A)

and

(B),

x2

O

flH\

\D j

H~ oiC.

we have

*

- 3x) =a;4 - 9x 2 - - 9x2 - I2x -H 5,

y(x*

xz + 43#

But

y

frcttn

(A)

;

;

which reduces to

- 200-0.

The

values of y are

Hence we

10*,

\/2,

and the corresponding values of x are given by

find that the roots of (A) are

-lN/2,

EXERCISE XXI

THE BIQUADRATIC Unless otherwise stated

1.

If

2.

If

+ y = a + 8, ( j8

-H

If

j8,

y, 8 are the roots of

y)8 = (-*- 8) j8y, show

[Deduce from Ex. 3.

a,

u^ax* + 4bx* + 6co; 2 4- 4d(a: + e =0. show that a zd + 26 3 - 3ofo = 0. that e 2 6

-f

2d 3 - 3edc = 0.

1.]

2 2 0y=a8, show that od =6 e.

L

4. If

show that the interchange of any two of the four 3 and into L 3

M

a,

j8,

* In the case of the general equation, tt=0, and the substitution, obtaining the second quadratic (G) is

y [ao? +(46 ~ap)x] * (

and

thus,

y, 8

changes

L3

into J/ 8

,

y**&+px+q,

the

method

for

FERRARI'S METHOD

198 5.

Show

that the equation whose roots are j8y

- a8

ya

/94-y-a-o'

-

-

a/?

/?S

yS

-/

is

where

B-az + b, D-(a,

[Dimmish the roots of 6. If a,

ft,

3

b, c,

u~Q by

y, 3 are the roots

I) #-(a, 6, and use Ex. 3.]

d$z, 2

,

of w==o; 4 -fg# 2

-f

d, e$z, I)

c,

ra-f s ~0,

4 .

show that the equation

whose roots are z3 -

is

and that

if 2 X is

2


- 4sz -f 4g# - r 2 = 0,

a root of this equation, then

where A ls A 2 are the roots of

A^Az^-s^O. 7,

Show

that the equation whose roots are 3

s

[Use equations 8.

Solve x*

23

(D), (G) of Art. 10,

- 2x* -\-a(2x -

1)

^0 by

and Exercise XX, putting

it

3, (Hi).]

form

in the

and choosing

s so that the right-hand side is a perfect square. the equation has two and only two real roots unless a = 1 or 0.

Hence show that

.

Solve by Ferrari's method 9.

:

* 4 + 12:r-5=0.

11. x*

- 3z 2 - 4x ~ 3 ~0.

13.

* 4 -f 12x3 4-54^ 2 -f 96x4- 40 = 0.

14.

4 4x 3 + lx z Express #

6a: -h

x*~2x- 1 ^0.

10.

a4

12.

x 4 - 4* 3

4

4- 5.u

4 2

= 0.

3 as the product of quadratic factors in three

different ways. 15. If a,

jS,

y, S are

the roots of x*

-\-

3x 3 4-# 2

-20,

prove that the equation

whose roots are

(j34-y-a-S) z

is

[Use equation (L) of Art. 16.

Show

and two

3

2 ,

(y 4-a

ft

- S) 2

,

(a

+ j8-y-S) 2

- 19z 2 -f 243z - 225 = 0.

13.]

that the equation whose roots are

similar expressions

is

3 2 4(*4-L) -e /(z4-L)4-eV=0

where L=ce-d*. [This follows from equation reciprocals.]

(L) of Art. 13

by changing

a, 0, y, 8 into their

EULER'S SOLUTION If

17. Euler's Solution.

and

= *Jl + \/m -f \/n,

z

(i)

(ii)

then

Z,

which (iii)

z

- 223

4

.

z2

Jl,

*Jm, *Jn denote either of the square roots of

is

.z

+

the same as

The complete

4(s-f//)

and(K)

6//z 2

4-

4s

3

-f

+ 4Gz + A" = 0,

127/s 2

~-K)s- G

2

4-

(9//

is

given by

2

- 0,

-a 2/(s + //)+a3
3

solution of the biquadratic u

aa -f b = -

*Jl + sjm + >Jn,

ay + 6 =

^Z + Jin -

Observe that the values of

18.

z4

identical with

a8

v/Z

.

-h

6

s/w

m, n are i\a

I,

=

af3 + b~

*Jn,

the square roots being chosen so that

(L)

m, n

l~- 22!tom =0.

m, n are the values of a given by is

/,

prove that

- &*Jl*Jm*/n

If this equation

199

.

=

*Jl- *Jm

-

Jl-~

*Jm -

(j3

>/n,

same

s/n has the

2

+ -s/n,

+ y-a-8)

2 ,

sign as G.

etc., see

equations

of Art. 13.

Find a substitution of the form

x=py + q

which

transform the

will

equation into the reciprocal form. to solve the equation. 19, In

Ex.

1

Show

of Art. 17,

if

that one such substitution

each value of y which y

is

substituted for y in y

*

is

satisfies

x = 2y -

1.

Use

this

the equation

+ 9% - 200^0 2

x + 3x, we get eight values of 2

x,

four of which satisfy

Without solving any equations, show that the other four values of x are the roots of x* f 12.r 3

[The left-hand side

+ 54x 2 + 96x

f

40 = 0.

(Cf.

Ex.

13.)

is

5H(z 4 + 12* - 5).] Find a substitution of the form y=x 2 +px + q which 3 2x - 1 to the form y4 -f fy 2 + g = 0. equation x* + x 20.

will

reduce the

that one such substitution is y =x 2 - x - 1, this reducing the equation 2 Use this to solve the equation. (Cf. Ex. 10. ) (2y) + 26 (2y) -11^0.

Show to

4

21.

Find a substitution of the form

y=x 2 +px + q

which

equation to the form

2 y*+fy + g=Q, showing that one such substitution

this reducing the equation to

4(1

is

will reduce

the

CHAPTER

XIII


Lower

class

A

Upper

O

class

A

P FIG. 30.

The system

of rationals is thus divided into

A

called the lower class

the rationals to the

two

parts,

which

will

be

upper dass A'. The class A contains all and the class A' contains all those to the of P,

and

left

the

A

is less tlutn any number in A'. Two right of P. Thus, any number in or does distinct cases arise, according as the point not, represent a does,

P

rational number.

P

to represent some rational, 3 for example. The Suppose contains every rational less than 3, and the class A' contains every rational greater than 3. The number 3 may be assigned to either dass. If First case.

class

A

3 belongs to A,

number For

it is

the greatest

number

of this class,

and there

is

no

least

in A'.

if a' is

A

supposed to be the least number in yationals exist which and greater than 3, and which therefore belong to the ',

are less than a' class

A

.

Similarly if 3 is assigned to -4', it has no greatest number. the class

is

the least

number

of this class,

and

A

Thus corresponding the

dass

A

to

has a greatest

any rational, the classification is such that number or the dass A has a least number. 1

either

DEDEKIND'S DEFINITION ticcond caw.

number.

201

P does not represent a rational be equal to the side of a square whose area

Suppose that the point

For example,

let

OP

square units. Classify the rationale according to the following rule The lower class A is to contain all the negative numbers, zero and every 7

is

:

number whose square

positive

The upper greater than

The

A'

is

is less

than

7.

to contain every positive

number whose square

is

7.

such that

classification is

Every

(i)

class

rational is included in one or other of two classes

For

A, A'.

exists whose square is equal to 7. number in A is less than any number in A. For if a is any (ii) Any number A and a' is any number in A', we have a 2 <7
no rational

,

a
so

The

(iii)

number.

and we

class

A

has no greatest number and the class

have to find b (greater than case

A

has no least

2 For, suppose that a is the greatest number of A, then a <7, 2 can find a rational b such that 6>a and 6 <7. To do this, we

that & 2

a), so

if

7

-a 2 <7

- <-s -a 2

7

6-flK-r

a2

.

This will be the

-a 2 .

2a

b -fa

Thus, 6 belongs to the class A, and a is not the greatest number in A. Similarly it can be shown that A' has no least number. This classification

by is

-s/7,

is

and we assign

said to define the irrational number which we denote a place on the scale with rationals as follows ,J1

it

to follow or to precede

:

any

positive rational a, according as a

iJJKDedekind's Definition.

2

<7

or

Suppose that a certain rule enables

us to divide the whole system of rationals into two classes, a lower class A and an upper class A', so that any number in is less than any num-

A

ber in A'

.

Two

cases arise

:

A

has a greatest number or if A has a least number, the tion defines a rational number, namely the greatest number in (i)

least (ii)

If

'

number

A

A

or the

.

no greatest number in A and no least number in A', the defines an irrational number which is to follow all the numbers

If there is

classification

in

in

classifica-

A and to

precede

all

those in

A.

number

is called a real number^ and the and is irrationals known as the system of real numbers. aggregate of rationals

Any

rational or irrational

EQUALITY AND INEQUALITY

202

The

be called the classification or the section (A, A'), and the real number defined by it may be known as the classification just described will

number (A, A'). The real number

zero is defined

A

by the section (A, A), where

A

contains

the negative rationals and contains all the positive rationals. A contains some positive rationals, the section (A, A') defines a

all

If

positive real number.

A' contains some negative rationals, (A, A') defines a negative

If

real

number.

We we

'

'

are not justified in regarding a real number as a have given fresh definitions of equality, inequality

'

'

number until and the fundadefinitions must be in

mental operations of arithmetic. Moreover, these agreement with those already given for the system of rationals.

equal If

Equality and Inequality. Two real numbers are said when they are defined by the same classification of rationals.

a and

precede

/J,

// a and

/J

are real numbers,

we say that a j8

is less

and some of the than j3 and that j8

are unequal real numbers, infinitely

to be

rationals

which follow a

is greater

than a.

many

rationals lie between

them.

For (B,

B

a
let

f

)

rational r lies between r belongs to

a and

jS.

each of the classes

Then r follows a and precedes jS, therefore A' and B. If r is the only rational between A'

ex

B Fio. 31.

a and

,

then r

Art. 2,

is

the least

by number r, Thus at least two, and andjS.

(ii)

in

A and the greatest in B.

therefore infinitely

(J5,

Hence,

B') defines the

many,

Each

Let

A and A'

be two classes of rationals such that

class contains at least one

Any number in

same

impossible, for a
is

(See Ch. II, 10.)

Theorem. (i)

number

(i), each of the classifications (A, A), and we should have
number.

A is less than any number in A

1 .

ENDLESS DECIMALS

203 .

(iii)

Two numbers a and

can be chosen from

a'

*

**

A

and A'

respectively, so

that a'

-a<,

is any positive number we may choose, however Then there is one and only one real number a such any number in A and a' is any number in A'.

where

is

a
that

where a

Divide the system of rationals into two classes according to the The lower class is to contain every rational less than any A The upper class is to contain every rational greater than

Proof.

following rule number a' in

any number a It will be tion.

small.

For

:

f

.

A.

in

proved that not more than one rational can escape classificatwo rationals Z and V (l
if

This contradicts the hypothesis, for a'

so that either a or a' lies between

we can choose a and

a! so

that

-a
and

I'.

one rational which escapes classification, then it is to be the upper or the lower class. to either Thus, by Dedekind's assigned a number defines real the a which satisfies the classification definition, If there is

Conditions stated above.

This theorem

may

also be stated as follows

:

If (a n ) and (a n ') are sequences of rationals such that (i)

and

(ii) it

a 1
is possible to

<

3 ...


e is

number 5.


...


find n such that <*n

where

...

-<**<>

any positive number we may choose, however small) then one and only one, exists such that # n ^a
real

'

a,

Endless Decimals.

Using the ordinary notation, a -a 1o 2 ...a n ...

let

be an endless decimal, where the successive figures are formed according to

some Let

definite rule.

dn = a

a^

The sequences (d n ) and theorem, and therefore a

We o

. . .

aw

and

dn

'

= d n + 1/10W

.

f

obviously satisfy the conditions of the last 8 is defined by real (d n

)

number

say that the decimal represents or

is

equal to this real

number

S.

B.O.A.

FUNDAMENTAL OPERATIONS

204

Conversely, any real number 8 can be represented by a decimal. For, using the same notation as in the preceding, if 8 is known, then for any

we can find d n so that d n <8
suffix n,

to any

That

'.

Of course, a terminating decimal

to say,

is

represents

we can

calculate

S.

be regarded as an endless decimal

may

in which, after a certain stage, all the figures are zeros.

Since

any

by a decimal which terminates or that a non-recurring endless decimal represents an irrational

rational can be represented

recurs, it follows

number, and conversely.

The fundamental Operations of Arithmetic. follows, a Greek letter denotes a real number, a small

In what

Roman

letter

Real numbers will be defined by using the theorem

represents a rational. of Art. 4, and when

we say that a

number a

real

is

defined

by a^a^a',

presumed that a, a' are any numbers in classes A, A! of rationals which satisfy the conditions of the theorem.

it is

(1)

Addition.

then a -f j8

is

If a,

j3

numbers defined by a
by a -f 6
defined

This classification defines a real number, for (i)

(ii) (iii)

There

Any

We

a

is

at least one a

-f

b

< any a' +

can choose

e is

and one

a'

+ b'.

b'.

that

a, a', 6, b' so

a<^e

a'

where

6

-f

and

b'

6
,

any positive number, however small, and then - (a 6) = (a - a) + (V - b)<. (a -f b') f

f

-f-

We

define

a -f /? + y as meaning

(a

-f jS)

+ y, and

it is

easy to show that

the commutative and associative laws hold good. .

It

is

Show

1.

obvious that a

and

j3

= a. b< 0<

-Ha are defined

defined

by

positive rational. Let a be defined

by

is

a< a<

&',

a',

since every 6

where

-f

6<

classification of rationals.

is

is

any negative rational and

defined

+ 0<

a'

b' is

any

by

-f 6',

X

is

negative and every 6 positive,

a + 6<

Thus a +

6

then a + a

and

by the same

Prove that a +

x. 2.

Zero

-f ft

a< a' -f 6'.

and a are defined by the same

classification of rationals

and are equal.

MULTIPLICATION OF IRRATIONALS If

.

3.

a

is

Prove that defined

by

205

a + ( - a) =0.

a< a< a',

then a +

(

a - a'< a

- a) 4- (

defined

is

by

- a)< a' - a.

every a< any a', therefore a a'< 0< a' a. Hence a + ( -a) and are defined by the same classification, and

Now

We

(2) Subtraction.

If a,

(3) Multiplication.

/?

are positive real

and

a^oc^a' aj3 is

Whence

define
follows that

it

for

(a-j8)+j8=a,

then

are equal.

defined

numbers defined by

6
These conditions define a real number,

by afe
for (i)

(ii) (i\\)

There

is

at least one ab

Every a6
a'b'

We can choose a,

6, a', &'

For

positive number.

and one

a'b'

.

.

a'b'

so that a'b'

-ab<,

-ab = a'(b' -b) + b(a'

a

a

6'

>

/}

where

is

any assigned

-a).

a'

'//

k

FIG. 32.

Choose rationals

a'6'

We

can

now

choose

For

a'6'

zero

- ab
a, 6, a', 6', so

6'

and then

-

^/7

Show

4.

that a'

-a
= = 0. a,

a(-j8H-aj8 = (~a) and (-)( -J8)=aj8. It is easy now to show is taken to mean (
that

defined by

that the

and associative laws hold good.

^1.^1 = 7.

a< N/7<

a',

where

rt,

a2

Hence *y7 ^7 is defined by a2 < 7< .

a'
negative factors, the definitions are

commutative, distributive

is

if

b)

and

b
a.

^ Ex.

and &>/J, then + k(a' - a).

- ab<.

and

Further, a/Sy

A>a

that

h, k, so

a' 2 ,

a' are

any

< 7< a'

and this

positive rationals such that

2 .

classification also defines the

number

7,

POWERS AND ROOTS

206 (4) Division.

If

a

is

a positive real number defined by

a^a^a', then

is defined by l/a'
I/a

There

(i)

We

(iii)

and one

at least one I/a'

is

can choose

Every l/a'< any I/a. so that l/a-l/a'<, where e is any assigned

a, a'

I/a.

(ii)

-

positive number.

To prove

this,

we have

show that

to

"i

a, a'

7*

can be chosen so that

a

;

^

p IG f

-a
choose a, a

First choose h so thaia'

"~~l

'

so that a>/i

and

a
h
NOTE.

Tliis

Consequently

JSfo.

it

5.

meaning

is

follows that

// a,

and

a'-a
reasoning depends on the existence of positive rationals

rio

j9

l/(

(a/j3)

.

less

than

a.

to 1/0.

assigned

Further definitions are

whence

h 2
we have

= - I/a and a/j8=a l/j8 for (a I//?) jB =a(l/jB /? =a -a)

.

.

;

are positive real numbers defined by

.

;

.

jB)

a< a< a' arw2 b< j8< &',

=a.
a/<

a' ft. defined by a/6'< This follows from the definition of

7.

Powers and Roots.

and multiplication.

l/j3

a positive integer and a a real numdefined by a n ==a a a to n factors. n m n m + follows that a .a =a and (aw ) w =amn

n

If

n ber, the nth power of a (written a )

is

is

.

.

. . .

m, n are positive integers, it That these formulae may hold for zero and negative values we must have a = 1 and a~ n = l/an

If

.

of

m and n,

.

Theorem

1

.

(i)

// a

x, x' positive rationals

is

a positive real number, n a positive integer and

such that

n


,

then positive rationals

xl9 #/

exist such that

For

if

We

can choose ^ 1 >x, so that

x
then

x x w - xn
The existence

of a

number x

~#<(a -xn )fnx' n

xx ,

so that

such that

~l :

and then

x^
a
can be proved in a

similar way. (ii)

//

is

any

positive number, however small, then

For as in the preceding, x/ n

~x1 w <(x1

/

-ajj)

.

n n x/ ~x 1 <, provided

nx^- 1

.

PRINCIPAL ROOTS Theorem number

,

207

a positive real number, there and one only, whose n-th power is equal to a. 2.

//a

If a = z n where #

is

is

a positive rational, there

is

is

one positive real

only one such number

x,

for the nth If

powers of unequal positive rationals are unequal. no such x exists, divide the system of rationals into a lower

class

A

and an upper class A' according to the following rule The class A is to contain every negative number, zero, and every positive :

',

number x such that x n a. Then (i) no rational escapes classification for no rational x exists such ;

that x n =a. (ii)

Every #
(iii)

There

is

x',

for

sc

n


no greatest x and no

n .

least

x

f .

This follows from the last

theorem. Therefore the classification defines a real number Again, for every

x and every

.

x'',

xn
xn
and

.

Also every rational is an x or an x', therefore these may be chosen so x is as small as we like. Hence, as in Theorem 1, they may be where e is any positive number, however small. chosen so that x' n -xn

that x'

<,

n are defined

Thus both a and

by the same

classification of rationals,

a classification which satisfies the conditions of the theorem of Art. 4 and n

therefore

=a.

a positive real number, the principal n-th root of a defined as the positive real number whose nth power is equal to a. This Definitions.

is is

If

written J^a, and

a

is

is

generally called the n-th root of n

Thus

(;/a)

a..

=a.

It follows that

ya.y]8-y(aj8) If

n

is

odd,

we have (-

When n

is

WH^a^ y(.

and

an odd

simply the n-th

n==

(-

1 )n

)

w== -a.

we therefore define the -a as - ^a thus

integer,

root, of

(^a

principal n-th root, or

;

even, the 7^th power of every real number is positive, and therefore no real number exists which is the nth root of a negative number. If

n

is

SURDS

208 Indices

which are Rational Fractions.

If p, q are positive integers,

?

aq

is

p defined as *]a* or (!ja) This definition assigns no meaning to a x is

.

considered in Art.

Surds.

8.

If

a

when x

is

This case

irrational.

9.

is

not a perfect nth power, ^Ja

is

called a surd of the

nth order.

Two rational

surds of the same order are like surds

when

their quotient is

otherwise they are unlike surds.

:

The following theorems and examples depend on the number cannot be equal to an irrational.

Theorem

1.

If

x+Jy^a + jb

fact that a rational

x, y, a, b are rationals, then

where

x=a

and y = b,or else

y and b are the squares ofrationals. For suppose that x-^a and let x = a 4- z,. then z + Jy

Jb,

and by squaring,

Therefore ,Jy is rational, and from the given equation the other hand, if x = a, then y 6.

is

Jb

rational.

On

Ex.

1.

// a

+ b^/p + c*jq

where

a, 6, c are rationals

and Jp

t

^/q are unlike surds,

thena = Q, 6=0, c=0. By transposing and squaring we can show that 2 c*q ~a b~p. 2ab^/p

If

ab^Q, the left-hand side would be irrational and the right-hand side would

be rational ; which

is

impossible.

Therefore a6=0, and consequently a

or

6~0.

a=0, then b*Jp + CsJq~Q; and, if 6^0, then Vp/\A?~ -c/6; and which is not the case. */p *Jq would be like surds Therefore 6=0 and c = 0. //

so

that

;

7/6 = Ex.

2.

then a // a

-{-

~ c^q ~ 0, and therefore a

+ 6^/p + c^ 2 =0

wAere

a, 6, c,

and

#

c

= 0.

are rationals and

p

is

not

a perfect cube,

then a, 6, c are all zero.

Multiplying the given equation by %Jp

cp +a?Jp

and eliminating the terms containing Since ^/p

is irrational, it

therefore

c^Q we should have p 2 =(

fy

= ac and c 2p 3 2 2 2 c*p = a 6 = a c.

-

) \ c/

Hence c=0, and

+ 6Ap 2 =0,

follows that b2

If

we have

therefore also

,

so that ^/p 2

a=0

ab,

would be

and 6=0.

rational,

which

is

not the case.

IRRATIONAL INDICES Theorem

2.

Suppose

square, then if

p + *Jq

is

For we have

Since

Jq

is

Putting

a polynomial with

2 &ndf(x) = Q{(x-p) -q}

9.

Irrational Indices.

9

if

f(x)

is

identity of the form

x=p + Jq, we

irrational it follows that

=

a perfect

a root off(x)

we obtain an

are rationals.

is

0//(#)=0 where f(x)

is also

divided by (x-p)*-q,

S

root

that q is not

= 0. p-*Jq 2 = + {x-(p Jq)}{x-(p-Jq)} (x-p) -q, and

rational coefficients,

where R,

and

that p, q are rationals

a

209

have

Rp -f S = and R = 0.

p-Jq

therefore

The theorems

Consequently

a root of/(z)

is

of Art. 7 hold

= 0.

when a

is

a

number, the indices being rationals, and lead to the following Definition. Let a be a real number greater than unity, and let be an

positive real

:

irrational defined

by x
f

x belong to classes of rationals

x,

Then

satisfying the conditions stated in Art. 4.

a*

is

defined

by

These conditions define a real number, for (i) (iii)

there

at least one a x

is

'

and one a x>

we dan choose x and

x every a*
(ii)

,

' x' so that a x

-

ax

<

where

is

any assigned

number, however small.

positive

To prove

this,

ax Next, by Ch.

'

then a x
&>,

choose a number

-a x = a x (a xl -x - l)
II, 18, (7),

find a positive integer

n such that

l

an

NOTE.

The

'

-a x
At present

7/0
a*

is

defined

show that ax .

hold for

all real

it

way, is called its from other values which will be found

of will stand for its principal value.

by the

defined as (I/b)*, where 6 = I/a. able to assign a meaning to a* It is easy to

;

real positive value of a*, defined in this

principal value, to distinguish later.

.

x<.l/n, and then

Finally, choose x, x' so that x'

ax

-l
if

classification

We

ax '
define 1* as

when a

is

1.

negative.

a>0, the index laws

ay

= a*+* and

values of x and y.

(a

x y )

= axy

At

*

Or

present,

it

we

may

be

are not

OPEN AND CLOSED INTERVALS

210 10.

and

a^l,

a single real number

there exists

N

If a and

Theorem.

Logarithms.

are positive real numbers

such that a*

= N.

N
x and first let Suppose that no rational x exists such that a = x x '. n>l. Then, by Art. 7, rationals x, x' exist such that a Divide the system of rationals into a lower class A and an upper class A'

Proof.

by the following rule The class A is to contain every rational x such that ax N. Then (i) no rational escapes classification for it is assumed that no for a x
;

;

There

(iii)

is

no greatest x

the last article,

a*
we can

Similarly

;

for

find x2

if

>x

x

ly

supposed to be the greatest, as in so that a^ aPK^N i.e. so that is

-a^

can be shown that there

it

classification therefore defines a real

Again, for every

x and every

is

no

least x'.

number

.

#',

a x
and

a*
;

and since every rational is an x or an x', these may be chosen so that x -x is as small as we like. Hence, as in Art. 9, they may be chosen so that a x -a x <, where e is any positive number, however small. Thus both N and a* are defined by the same classification of rationals, f

'

a classification which If

a
satisfies

the conditions of Art.

find f so that (l/a)*

= l/N, and

4,

and so

a^

= 2V.

then a* = N.

any positive real number a is chosen as base and N is any positive real number, the number given by the relation a*~N is called the logarithm of N to the base a, and we write f = log a N. Definition.

Definitions. We say that between a and 6 forms the (open)

1 1

lie

.

(ii)

The

real values of

or the range (iii)

If

closed at '

a^x^b form

(x)

which

a
the closed interval

(a, 6)

the real numbers x form an interval

(a, 6)

open

at

a and

or the range (a <#<&). '

'

generally means open interval/ number x in the interval (a, 6) is often called a point in the interval.

Interval

Any

x such that

the set of real numbers

(i)

interval (a, 6) or the range

a
a
'

'

If

x l9 x2 are any numbers in the interval (a, 6) and/(x) is a function x such that f(xl )^f(x 2 ) when x 1
of

case/(x)

,

is

constant in the interval.

DEDEKIND'S THEOREM )

when x^x^

then

211

said to increase steadily in the

/(a?) is

stricter sense.

f(x l )^f(x 2 ) when #T <#2 f(x) decreases steadily; and

If

if

,

when x l <.x2 ,f(x)

f(x 1 )>f(x^)

decreases steadily in the stricter sense. 9

'

is the set of a real number, the neighbourhood of a real numbers in the interval (a-e, a + e), where e is as small as we like.

a

If

(v)

is

12. Sections of the

System of Real Numbers.

Dedekind's Theorem.

vx(i)

// the system of real numbers

is

divided into

A

and A' such that (i) each class contains at least one number, number belongs to one or other of the classes, (iii) any number (ii) every real in A is less than any number in A' then there is a single real number a, such that all numbers less than a belong to' A and all numbers greater than a belong The number a may be regarded as belonging to either of the classes. to A' two classes

9

.

Consider the rationals in

Proof.

and AI, which form a section

number

a

If

(1)

Two

a. is

A

of the

and A.

system

These form two classes

and define a

of rationals

A

l

real

cases arise.

rational,

it is

the greatest

number

in

Al

the greatest number in A v it is also the greatest suppose that there is a number ft in A greater than a.

or the least in A^.

number

in A. For Between a and /? there are rationals greater than a which therefore belong to A and also to A'. Hence j8 must belong to A', which is not the case. Similarly if a is the least number in A^, it is also the least in A. If

a

is

If

(ii)

a

is irrational, it is

greater than

all

the numbers in

Al

and

less

Also, a must belong to A or to A, and just as in the show we can that it is the greatest number in A or the least in A' preceding,

than

all

those in A^.

The theorem

just proved

section of the system of real

is

of great importance.

numbers

defines

It

shows that any

a real number.

Thus the con-

sideration of sections of this kind leads to no further generalisation of our idea of number. All this is sometimes expressed by saying that the system

of real numbers is closed, or the aggregate of real numbers is perfect. On the other hand, the rational numbers do not form a closed system, or in other words, the aggregate of rational numbers is not perfect, for a section of the system of rationals does riot always define a rational. (2) // the numbers in the interval (a, b) are divided into two A' as in (1), the section determines a real number a.

classes

A

and

9

For

if

we take with

A 1

all

all

the real numbers x such that

the real numbers x such that x'^-b,

of real

numbers.

we obtain a

x^a

and with A'

section of the system

ARITHMETIC CONTINUUM

212

1 that if we take any we in a straight line as origin and any length may choose as unit point of length, then to any point P in the line there corresponds a real number p.

The Continuum.

13.

from Art.

It follows

choosing suitable axioms regarding the characteristics of a straight line we can ensure the truth of the converse of this statement, namely, that to every real number p there corresponds a point P in the line. Thus

By

the correspondence between the points of a straight line and the

numbers If

that

p p

is

complete. the real number corresponding to the point P of the line, we say Hence the system of real is the measure of the length of OP.

is

numbers This

real

is

adequate for the measurement of the length of a straight

an instance of what

is

The aggregate

of real

is

known

numbers

as a continuous magnitude. called the arithmetic continuum,

is

of the points of a straight line is called the linear

the aggregate

line.

and

continuum.

RATIO AND PROPORTION NOTE. small

In

Arts. 14-16 capital letters will denote concrete magnitudes

and not numbers

;

denote positive integers.

letters will

Equality and

Inequality of Concrete Magnitudes. In how a magnitude or a quantity of any kind is to be measured, considering the at outset we must define what is meant by saying that 'A is equal to 14.

A

and B are quantities of the greater than or less than J5/ where of these statements depends on the particular kind of quantity we are considering. For instance, let A and B denote segments

J5,'

'A

is

The exact meaning

kind.

if A can be made to coincide with J5, we say A = B can be made to coincide with a part of B, we say and B>A. The student should reflect on the meaning of equal angles, equal

of straight lines if

;

;

A
A

velocities,

equal forces, equal quantities of heat, etc.

We

15. Ratio.

number

said to contain

This

is

B

able,

C

exactly

Definition.

(that

B

can be divided into any

Denoting any one of these parts by C, B is n times, and C is called the nth part of B. (exactly)

expressed shortly by writing

nth part of

a and

assume that a magnitude

(n) of equal parts.

we denote

C = ^B.

this

If.

by writing

A A

contains the

B.

A and B are magnitudes of the same kind, and if integers such that A contains the fcth part of B exactly a times

A~j BJ,

and the

times,

or

If

b exist is, if

m

B^nC

ratio of

the magnitudes

A

to

B

(written

A

A

and :

B)

B is

are said to be commensur-

defined as the

number

a/b.

RATIO AND PROPORTION If

no integers a and

6 exist such that

213

A = ^B, the magnitudes A

and

B

are said to be incommensurable.

Axiom

of Archimedes.

If

A

no matter how small

A may

always be found such that

mA>B

kind,

16.

Two

Ratio of

B

are magnitudes of the

same

be compared with 5, an integer

m can

and

or

^B
Incommensurables.

A

If

and

B

are incom-

mensurable magnitudes, the ratio of A to B (written A B) is defined as the irrational number determined by the following classification of the :

entire

the the

system of rationals

Lower

Class Class

Upper

:

is

to contain every rational a/6 for which

is

to contain every rational a'/b' for

This classification defines an irrational (1)

every rational

hypothesis, there

(2)

(3)

every a/6 there

is

hypothesis

is

falls

than any

no greatest a/6

bl

A>a

l

;

because

into one or other of the

no rational a/b

is less

number

;

for

a'/6'

;

which bA for

bA>aB,

which b'A
two

= aB

classes;

for,

by

:

^B
:

for suppose ajb^ to be the greatest, then

by

B.

Now by

the axiom of Archimedes, a multiple of a magnitude (however small the magnitude may be) can be found which exceeds any magnitude of the same kind, however Therefore an integer r can be found such great.

^-a

JS)>a 1 J5, and therefore rb l A>(r + 1)a 1 B. Let r61==62 and (r-fl)a 1 -a2 then b^A>o 2 B and a^-a^r-f l)lrb{> ajbl9 so that be shown that there is no ajbi is not the greatest a/6. Similarly it can number. irrational an defines Hence the classification least that

r(6 1

1

,

a'/6'

Theorem.

If

A, B, C,

D

are four magnitudes of the

same kind, the

ratio of A to B is equal to that of C to D if mA> =oi=of
mC^nD and A/B = C/D = n/m. If no such rationals m, A, B and also C, D are incommensurables, and the ratios

n

exist,

then

A/B, CjD are therefore are and irrationals defined by the same classification equal. = C:D, A:B and kind same the of If A, B, C, D are magnitudes then A, B, C,

D are said to be in proportion. * This

is

Euclid's definition of the equality of

two

ratios,

SQUARE AND OTHER ROOTS

214

EXERCISE XXII

IRRATIONALS is

1. Show that no rational exists whose nth power is equal to ajb, where a/6 a positive fraction in its lowest terms unless a and 6 are perfect nth powers. /x \** a - where x/y is a positive fraction in its lowest terms. [Suppose that ( j =

Then bxn =ayn where k

is

Therefore

Since y

prime to a whole number. Then .

is

a~xn and byn Jx + *Jy = */a + >Jb ,

that

is

y

x,

to say,

then either 2. If 9 Ja, */b are all rational or all like surds.

Hence a and b are .

xa, y=b

and *Jxy = */ab, unless

[Squaring, x + y=a + b, 3.

n must be a divisor of

akxn

^fa

where

a, b are

in terms of

b

= kyn

x

for

b,

Va6 are

y=a,

,

6.

or *Jx, *Jy,

rational.]

not a perfect square.

Also find x and y

b.

[Proceed as in Ex. 4. If

is

Let

exist such that

x, y may + ^/b = *Jx + *Jy,

given rationals and 6

a and

6.

a is prime to perfect nth powers.] Jk

or

*fxy,

Find the condition that rationals

= l,

2,

and show that x

a \(a dbVa -6).]

a + b*Jp + c*Jq + d*JpqQ where *Jp

9

*Jq are unlike surds,

then

a, 6, c,

d

are all zero.

[We have Hence by

Art. 8, Ex.

1,

ac=bdp,

and consequently,

if c

^0 and a ^0,

bc=ad, c*=dp, then p=c*.]

5. If x + $y=a + $/b where x, y, a, b are rational cubes, then x=a and y 6.

[Put

z= a; -a.

rational.

Show

Now use

that 2 8 H-y~6

+ 32\/6y =

a$lp* + b$]pq + c$q*Q that a, 6, c are all zero.

and neither of

7.

cients,

not zero, since pjq

2

2

j? ^, j?^

_+ _

3 z 2 [Show that C*q*-a*p -b pq = 3db(ap\/pq

is

.*.

;

y, b are

not perfect

$y=mg/b 2 where

m

is

Art. 8, Ex. 2.]

6. If

in brackets

and

is

bq\/p

2

q),

is

a perfect cube, show

and that the expression

not a cube.]

Show

that *J2 and v/3 are cubic functions of J2 -f *J3 with rational and that ^2-^/6 + 3 is the ratio of two linear functions of *J2 -f \/3

coeffi-

=

8 [For the first part, let y s/2 + \^3 ; find t/ , and eliminate \^3 and V2 in turn : for the second part, find the value of the product ( \/2 - V6 + 3) (V2 -f N/3 + a:), and show that this is equal to y + 5 when x 1.]

=

8.

root

Find the equation of lowest degree with rational

is (ii)

[(i) (ii) (iii)

If * =

&2 + 3^/4;

(iii)

2 N^ + %/SL a: -l=2\a;, etc. 8 if y=#2 + 3#4, i/ = 110 + 18(4/2 + 8 =5 + 34/6.2, and (28 -5) 8 = if z=4/2 + 4/3, ;

coefficients of

4/2 + 4/3.

which one

RATIONAL COEFFICIENTS 9.

If all the solutions of

as 2 + 2hxy + 6t/ 2 = 1,

a'x 2 + Zh'xy + b'y* = 1

then (h -A') 2 - (a -a') (6 -&') and - a'6) 2 - 4 (aA' - a'A) (W - h'b) are the squares of rationals.

and

are rational (oft'

215

a, A, b, a', h' 9 b' are rational,

[Let (a lf j8i), {a 2 , are the roots of

j8 2 ), (

-

i,

-

ft) (

-

-

2>

^2)

be the solutions. Prove that

~

,

,. /

y

show that a^, a a j3 8 are the roots of - 4 (a*') A6')} + 2z { (a - a') ( A6') - (6 - 6') (aA')} + (a - a') (6 - 6') = 0, - a r&. Hence show that =ab' where (ab') Also,

z 2 { (oft') f

(

aiaa

/ .

whence the second 10. If a, 6, x,

prove that

i )

/

-4(aA

f

/

)(A6

)

result follows.]

y are rationals such that

eiJAer

a;

a,

yb,

[Put*-a = -S:,y-&==r; If

(o6

or I .'.

-ab and

1

-cry are the squares of rationals.

(aY -bX)* + 4XY=Q.

X X and 7 are not zero, solve for ~

Next observe that the given equation y and 6.]

is

,

and show that

1

-ab

is

a perfect square.

unaltered by the interchange of x and a,

CHAPTEK XIV INEQUALITIES IN

this chapter, various

of the results are of

Many

of dealing with inequalities are explained.

fundamental importance.

Weierstrass' Inequalities.

1. less

methods

than

1

whose sum

Ij a v a 2 denoted by s n then

is

,

a n are positive numbers

...

,

+sn ), -s n ), where, in the last inequality,

For

(1

and continuing

-a

x ) (1

thus,

it

supposed that s n
is

- a2 ) = 1 -

(a l

+ a 2 + a x a 2 > \-(a l + a 2 ), )

we can prove that

(l-o 1 )(l-o 8 )...(l In the same

way

Again, 1 -a 1
and

-a,)(l -o.)

if s n

(1

2.

-a^,

...

(1

-a n )
...

(1

H-o n )
) ...

Many

inequalities

depend on

-00(1 -o,)

...

Ex.

2.

For

Using

+a n )
7/a>0 and 6>0,

// a,

6, c

(a

then

are positive

and

(a

(1

-)
the fact that the square of a real

positive. 1.

0

+ 0l )(l +az)

Ex.

(1

for

+ b)>*Jab.

not all equal, then

(a + b + c) (be + ca + ab) >9abc. + 6 + c) (be + ca 4- ab) - 9a6c

number

is

TYPICAL METHODS 3.

A

Ex.

1.

We

have

217

special arrangement of terms or factors is sometimes useful.

n

Show

that

[n>w

(|^)

and r(n~r + l)>n

8==1

if

r2

2 .

n 2(n-l) 3(n-2)

...

.

.

-r(n + l) + n<0

that

;

r(n-r + l)

is if

...

n

.

1,

~n)<0

(r-l)(r

or

if

l
Therefore

j&e. 2.

are positive numbers

6, c

// a,

any two of which are

together greater than the

third, then

Now,

since

a(c

+ a-b)(a + b -c)>0, we have

a(a + 6 -c-f c-f a-6)>2(c~f-a ~6)(a

if

which

the case.

is

greater than

4. If a,

Similarly, l/(a

2/6, 2/c respectively

x are

6,

For since

6 (6

according as ab

+ x)/(b + x)^a/b,

that

+ c - a) + l/(c -f a - 6),

l/(6

result is obtained

(a -f x)/( b

positive,

+ bx^ab + ax,

o 2 >a 2 -(6-c) 2 ;

or

is,

by

are

addition.

according as

a$b.

+ x)^ a/b

according as

ax^bx

or a $6.

Prove that

r. 1.

^

positive, then (a

-f x) is

+ 6-c)

+ a-6) + l/(a + 6-c)>2/a,

+ 6 - c) + 1/(6 -f c - a), and the

;

l/(c

8

*

1

..

(2*-D

,

A then

'

/o

AI Also

(2*

4 6 >- -.

+ ix !)^-.-...

^
Therefore

-

*^

and

'

5. If both sides of an inequality are symmetric functions of a y no loss of generality in assuming that

6, c, ... A, k,

there is

Ex.

"

1.

6, c

// a,

are all positive

and

n^O or n^

-

1,

then

a n (a-b)
an

let

w>0.

n

n

^b ^c

therefore

whence the //

On

account of symmetry, we may assume that a^6>c. and a n (a-6)(a-c)>6n (a-6)(6 -c), also c n (c-a)(ca n (a-6)(a-c)+c n (c-a)(c-6)^6 n (a~6)(6-c),

Hence,

result in question. 2;a n (a-6)(a-c)=Z(a-6)(a-c)=27a2

n=0,

27a 2

and -l,

let

easily seen that

n= -w-1,

-

-.

2:6c

<

-276c

so that

Ean (a - 6) (a - c) = -r-

m^O. 2?a

Let a = l/a, 6 = l/jS,

m (a -

j8)

(a

-

y)

^

c

= l/y,

then

by the preceding.

it is

INEQUALITIES INVOLVING POWERS

218 6.

//!, a2 ((!&!

-f

,

a n and bl9 62

...

let

a 2 b2 +

A=

bn are two sets of real numbers, then

...

2

2

2

-f a n ) (6 X #tA) 2:^( a i + a 2 + = = occurring only when a^Jb^ a^jb^ -f

. . .

the sign of equality

For

,

. . .

B = 2arbr C = 2b rz then 2 Z(ar + A6 r - A + 2 A

a r2 ,

,

,

have a r -h \br =

for every

is

the

7. If a 19 a 2

. . .

= a n/b n

. . .

-f

6n2 ),

.

+ A2 C.

the sign of equality occurs,

we must

so

2

of

,

also follows

sum

from the identity

\n(n -

of the

1)

squares of the form (a^g

a n are n positive numbers, not

...

all

equal

according as x and y have the same or opposite signs. First suppose that x and y have the same sign. Let

the numbers

-f

-f

The theorem

where S

and

62 2

A + 2XB + A C = are equal and B 2 = A 2A$ + A2 C>0 for all values of A, and therefore B2
Hence the roots Otherwise

r,

If

-f

for all values of A,

)

Therefore ^4-2AJ5+ A 2 C^O.

2

1, 2,

negative or zero.

...

n,

then a rx -a sx and arv

Hence a

x

(a r

x+v

to

- a 2 ^i) 2

-

one another, then

r, s

be any two of

are both positive, both

-a/ -a8x )(ar y -a8 v )^0, and

x y v v -h a*+ ^*a r*a 8 + a s a r

There are 2^(^-1) relations of this sort, not occurs in n - 1 of them and a r xa s v in only one.

consequently

.

all equalities.

Also a rx + v

Hence by addition

(n- \)Za r Therefore

nEa

x +v

x +v

>Z:a rxasy where s^r.

> Sa x +v + Safa* = Za x r

If x and y have opposite signs, then (a r be substituted for in the preceding.

x

.

2a rv

.

- a,*) (a r y - a/)<0, and

< must

>

8. The inequality in Art. 7 can be written

n Hence to

it

follows that if

one another, and x,y,z,

a^ a 2

...

,

...

a n are n

positive numbers, not all equal

are all positive or all negative, then

27a r +y++...

n

n

n

Ea ^

n

Zgv Zaf ^

n

n

FUNDAMENTAL INEQUALITIES

**-

9. If a is any positive number except then First suppose that

we have (a

,

_

l)/p

_

(aP

_

= (a - l){pa~ l - (a p

p, q are positive integers.

_

V)/(p

~

If

1}

4-a*~ 2

l

y are positive rationals,*

x,

x>y.

if

x~p, y = q where _i

and

1,

219

+

... 4-

l)}/p(p

- 1).

Now, the expression in the large brackets can be written 1 a*~ 2 ) + (a*- 1 - a*- 3 ) + + (a~ l - 1 )} (a*. . .

{

and each term

of this is positive or negative according as a.^\. in either case, (a p -l)/p- (a** 1 - l)/(p - 1 )>0,

Hence, that

p,

(a

is,

p

l)/p decreases as

if

y are fractions, we

may

one or both of

q,

d are positive integers and p>q----

TT~ d

P/

p>q.

(a-l)/p>(a<*-l)/q If

cc,

and therefore

decreases,

j>

> --TT~~ d

x,

take x=p/d, y = q/d where have to show that

We A

>

.

that, is

9

P

?/

-

,

9

i

where b = a d NOTE.

6>0 and

This follows from the preceding, for

.

It has

been shown that

a>0

and ^1,

passes through positive rational x values, (a ~l)/x increases with x. Putting I/a for a, it follows that (a- ~l)/x increases. x and as decreases Further, putting increases, consequently (l-a-*)/* if

as

a;

x

i

i

we conclude that x(a x - 1)

Ifx for x,

decreases

~ Ex.

Ifa>l and n^l.then

I.

i

i

n(a

n

and x(l ~ a x ) 1

-l)>l

.

i

__ji

If

w>l, n(an -

If

n = \ the inequality becomes a -1>1 --, or a (a-

1)

=an .n(l-a n)> a^(l

- a~ 1 )>l - a- 1

9

1)

.

> a-1,

10. If a and b are positive and unequal and x except

unless

1,

then


Proof.

6)>a _

^-i (a ,

*

in which case

xa

x ~l

6x

>x6

x-i

(a-b)
P

x

(a

is

any

i.e.

2 (a-l) >0.

rational

number

_ 6)f

- bx
.

First observe that, in either of the cases, the truth of one of

the inequalities involves that of the other. * It

increases as x increases.

can be shown that

this is true

when

x,

v are positive

Thus, real

if

for all positive

numbers.

(See Ex.

XXIII,

and

37-39.)

B.CJU

FUNDAMENTAL INEQUALITIES

220

x ~l x x we may interchange (a-~b)>a -b unequal values of a and 6, xa l x - a x and therefore a x -b x a and 6, so that xb*~ (b a)>b >xb x ~ l (a-b). Thus it is only necessary to prove one inequality in each case. ,

',

Let a/b = k, then

x

(i) if

according as

x

(ii) if

is

- bx ^xbx ^ l (a - b) according as kx - I $x(k - 1), or x (k l)/xg(A 1)/1, that is according as xgl (See Art. 9.)

x positive a

is

;

negative and

equal to

-

y then xa

x ~l

y

- b) ^a x - b x according (a

-l

(k-l)>k-v-I, that is according as -y(k~l)^k-k y + v +l - 1 5 k (y + 1 ) - y - 1 or according as according as k (*+'-l)/(y + 1)2 (*-!)/!; -yk~ v

as

l ,

or

,

that

is

Now y>0, and

y + l^l, or t/^0.

(by Art. 9) according as

xa*- l (a-b)>a x

so

-bf*.

This completes the proof of the theorem. 1 1

.

When

6

= 1,

the second inequalities in Art. 10 become

ax

-I>x(a~I) -l
ax

Writing n for x and

and

1,

1 -f

x> -1 and 7^0,

x for

,

0
n

>l+nx

if

n<0

(l+x)
if

0
-x)

(l-x) Ex.

or

n

>l -nx

if

n<0

n


if

0
// m, n are positive and

1.

>1,

any

rational except

that

m
>1,

x
again, changing the sign of x, if (1

it

or

is possible to

>1,

find positive values of x such

(l+x)< this inequality holds

Now (1

if

or

follows that, if n is

it

n

For

x<0

JAen

(l+x)

Or

if

(1

+ x)- m >l -mx, or

+7ix)(l

~mx)>l,

Ex.

// sn = a + a 2 +

2.

1

Denote the left-hand

if

if

x>0

and

hence the inequality holds

x{(n-m) -mnx}>0, . . .

side

+ an

where a v a 2

by un then ,

r=n4- 5

HH-Un-^^-j*

if

,

or ...

if

if

1

-mx>l/(l +nx),

0
an are

all positive t

n>l,

*_

^(l+aja+o.)...

then

or

if

ARITHMETIC AND GEOMETRIC MEANS Now

s r

n

-^ n Li> ra n ^n r^Tl

1 >

when r>l

and when

(Art. 10),

r

221

= l,

this

becomes

an equality, therefore

un - u n _ l >an u n ^

therefore

Now

M!

>0

and

n=2,

for

n >(l

+an )u n _ >u n , 1

l

.

3, 4, etc.

Arithmetic and Geometric Means.

12. are

0, therefore w n

If

(1)

a,

k

c, ...

6,

n numbers and

A^ then

^4

and

+ c+...-h)

arithmetic

n G= J (abc

...

&),

mean and G the geometric mean

of

&.

Theorem.

13.

not all equal

For

the

called

is

...

a, 6, c,

-(a + 6

n

A

let

T/?e arithmetic mean of n positive numbers, which are one another, is greater than their geometric mean. he the arithmetic mean and G the geometric mean of the n

to

numbers

positive

a, 6, c, d,

Suppose that

a^any

other

number

...

A* ..................................

(A)

and that fc^any other

of the set

number, then Let

&'

na>a + b + c + ... +k>nb and a>A>b. =a + 6 - -4 so that ?/">0, and consider the set A,b',c,d,...k,

obtained by substituting A,

Thus

A + b' =a

-f

b

the same arithmetic

b'

for a, 6 in (A).

and Ab'>ab.

mean

.................................

We

(B)

have

Hence the numbers

of the set (B)

as those of (A), but a greater geometric

have

mean.

the numbers in (B) are not all equal, then A is neither the the least of the set. By repeating the process we can therenor greatest fore obtain a set of n positive numbers containing two A's, with the same

Again,

if

mean

as before, but a greater geometric mean. we shall finally arrive at a set of n numbers each equal of which the geometric mean is greater than (?, and since is the

arithmetic

Continuing thus, to

A

A

geometric

Thus

mean

for the

of these equal

numbers

10, 1, 2,

7;

The arithmetic mean

numbers,

it

follows that

A>G.

10, 1, 2, 7, the successive sets are 5,

of the

increasing from left to right.

6,

2,

7;

numbers

5,

6, 4,

5

;

5,

5,

5,

5.

in each set is 5, the geometric

mean

FUNDAMENTAL THEOREM

222

Taking the

Alternative proof.

set (A),

we have

Proceeding in this way, we can show that abc

k^

...

(1

U

{

(a

-f

V

6+

n

is

k)

}* >

if

...

a power of

2,

then

,

'/

so that

n

is

where

A

If

not a power of

2,

occurs r times

and n-f r

a b

consider the set is

a power of

c9

9

9

k

...

By

2.

A A A 9

9

y

9

...

9

the preceding,

i

abc

The right-hand

G^A,

...

n +r

A n+r

side is equal to

Gn A r ^A n + r

and therefore

,

when

the sign of equality only occurring

the numbers

all

,

so that

a, 6,

...

k

are equal. Shoiv that n n >l

J&r. 1.

We have

(1

Now

AV

>\

/

1.3.5... ...

For

...

(2n

-

1).

(2w !)}/> +2n - 1 =

&{!

(2-l)

and

.3.5

l

+

a^ (n

-

There are n inequalities

- a 2 ) + (0 - a 8 ) + - a 2 ) (5 1) ^ {(5

. . .

...

-

(2

(2*-l).

2 1)} -r/

n n >l .3 5 .

a n are positive and (n - 1)* =

(*

therefore

.

+3 +5 -f ...

// a x , a 2 , a 3

2.

3 5

+ 3 -f 5 + ... + 1

.'.

.

...

^+a

2

;

(2- 1).

+ ...

-f

an

,

+ (* - a w = a lf )

8 ) ...

(a

-a

n )}

(or equalities) of this kind,

l /(

n~ l \

and the

result follo^a

by

multiplication.

14.

Theorem,

//a,

6, ...

k and

x, y,

...

w

are two sets of

n

positive

numbers, those in the second set being rational, then

+ ku

For in Art. 13 we may take x^x'lg, y = y'lg, ... w = w'/g, where x' y ... w' and g are positive integers, and then it is sufficient to prove that 9

+ ''

ax'

Now is

fy'

+ ...*M/ '

the left-hand side

the geometric

mean

b occurs y' times, the last theorem.

...

is

...

,

'" k

the arithmetic

of a, a,

r

6, 6, ... k,

mean and k

...

,

the right-hand side where a occurs x' times,

and k occurs w' times, so that the

result follows

by

MAXIMA AND MINIMA .15. all

Theorem. to

equal

// a, 6, one another, and

k are n positive numbers which are not

c, ...

m is any rational except

n

.,

a

b

c

k

x

y

z

w

let

or 1, then

n

V

m does not or does Iw between

according as

For

223

/

and

1.

a + 6-f c-f

...

+&

n

then 1

fl m ^ l-a m m m -a -(a + & + c + ...+ k )^ l-(a + + c + .:.+

and

xm + ym -f zm +

according as If

m does not lie

xm + y m +

Hence

...

and therefore If

0
and

between

-f

w-n) = Q,

. . .

Art. 11, the sign

>

must be replaced by +wn
xm + i/m + ...

inequalities, so that

Art. 11,

+ wm -n>m(x4-y-h ... rw xm -f m -f -f t^ > n. y

by

+ wm $ n.

. . .

by

1,

,

<

in all these

This completes the proof of the theorem. 1

6.

Theorem

.

Ifa,b,...k and x,y, ...w are two

sets

...

+wk m

ofn

rationals

and

m being any rational,

those of the first set are not all equal to one another, then, .

J

m

does not or does lie between and 1. according as * This depends on Art. 15 in the same way that Art. 14 depends on Art. 13.

17. Application to Questions of Maxima and Minima Val ues. In the theorems of Arts. 13-15 the inequalities become equalities when all the num'bers a, 6, c, ... k are equal. Hence we draw the following ;

conclusions

:

Suppose that then: (1) If

x, y, z,

x+y+z+

...

. . .

w are n positive variables and that c is a constant,

+w = c,

the value of xyz

so that the greatest value of xyz (2) If

...

w

is (c/n)

xyz...w = c, the value of x + y +

...

...

+ y+

... -f

w

is

is

greatest

when

n.

w i

so that the least value of x

w

ncn

.

is

least

when x = y=...

=w?,

APPLICATIONS

224 if

Again,

ra is

rational except

any

or

from Aft. 3 we conclude that

1,

:

between

x + y + z + ...+w = c then, according as m does not or does lie and 1, the least or the greatest value of xm +ym + ... +w m occurs

when x

y...w,

(3) If

the value in question being n l

~m

cm

.

.

m m m = c, then -f w (4) If x + y + according as m does not or does lie between and 1, the greatest or the least value of x + y + ...+w occurs when x = y=...=w, the value in question being n 1 " 1 /. c1 /. . . .

Ex.

1.

Find

the greatest value of

xyz for positive valuer of

x, y, z, subject to the con-

dition

Since yz + zx + xy~l2, the value of (yz)(zx)(xy) is greatest when yzzx^xy, that when x~ y~z-2. Hence the greatest value of (yz) (zx) (xy) is 2 8 and the greatest value of xyz is 8.

is

,

Ex.

2.

// the

sum of the

sides of

w given, prove that

a triangle

when

the area is greatest

the triangle is equilateral.

Let If

a, b, c

A

is

be the sides of the triangle, and let 2 - b) zJ s(s - a)

the area, then

(s

-

Now

(.?

Hence the value

of (s

Ex.

3.

Since

Find the x 2 y3 6.

o%3 = 6,

Therefore A#

if

-f (s

- a) (s - b)

Therefore the value of

condition

a)

A

is

/x

are

6)

(s

-I-

- c)

(s is

-

(s

-f

+ c ~2s.

a constant,

greatest b

6

- c).

s

c)

when a

greatest

when

s

-a~s -b~s -c.

= c.

3x + 4y for positive values of x and

least value of

A,

-

a

any constants, we have

+ Xx+^y -f/xy+^y

is

least

when 2

(

\x) (Xx) (^y) (py) (/z//)

A#=/u,?/

= (6A 2/z 3

)

+ 4t/

of 3z

Find the least value of x~* + y~ l + z~ l for positive values of x, the condition x + y + z = c. In Art. 17, (3), putting w= - 1, n~3, since m does not lie between

+ y~ l + z' 1

is

SV" 1

.

.

4.

value of x~ l

= 6A 2/x 3

y

3

Hence the least value of 2Az -f 3pty is 5(6A /i )^. Putting 2A = 3 and 3/x 4, it follows that the least value

Ex.

y, subject to the

is

y, z

which satisfy

and

1,

the least

or 9/c.

EXERCISE XXIII In

marked with an

the inequalities

*, all the

numbers concerned are supposed

to

be

positive. ... a 1. If ,, cr 2 n and b 19 b 2 ... b n are two sets of numbers and the second set are positive, then (a l -f a a + -f a n )/(6 1 -f 6 2 + + bn) lies between the greatest and least of a^/b^ an lb n ..., 2 /6 2 ,

,

. . .

.

.

,

2.

If a,

6, c

are (b

any

4-

c

-

real

a)

2

numbers, show that a - 6) 2 -f (a + 6 - c) 2 ^ 6c

-f (c -f

. .

-f

ca

-f

ab.

all

those in

TYPICAL EXAMPLES

225

3. If J + w -fn = l and J' + m' -f w' ^l, then 2 2 [E(mri - m'ny + (W + mm' -hnn') = El* Z7' .] 2

2

2

2

2

2

.

4. Show that a(x + y)* the quantities involved. 5. If any two of showthat

a>0

If

6.

+ bz + cz(x ~-y) cannot be 2

positive for all real values of

together greater than the third and x

a, 6, c are

and x>y, show that ax + a~ x >a y + a~ y

+ y + 2=0,

.

c, d are in harmonical progression, then a + d>b + c. [Let p 3g, jp f, 2>-l-#> ^>4-3^ be the reciprocals of a, b, c, c?.] ~ ~ 8.* If x^ty, then (axn + byn )l(axn l + byn l ) increases as n increases through

7.* If a, 6,

integral values. 9. If a, 6, c,

[2(ab

m* 10.*

-4-

dareall >1,

1)> (a +

+ 1),

!)(&

then

etc.]

--- + ^ a + b + c. c+a 2 2 [For, (& + c )/(& + c)^(& + c), etc.] 11. If x>0 or <-l, then r zn ' r ^ I +x [Show that x + + x -r &4-c

,

+

-

-

,

l

12.*

(i) (ii)

a(a-b)(a-c)+b(ba 3 -f 6 3 + c 3 + 3a5c ^ a 2 (6 + c)

[The second inequality 13. If a, 6, c

> 27a

15. If

[If

.

a>6, then

16. If

n>0

Za Ella^n* .

and not

p>l

Za 3

(ii)

;

all

.

El'.a'&nZa*.

equal, then

y are any two positive numbers such that

#,

\*

(^

-l
a?

+ y = l,

then

.]

anda?>l, then

> ~nx

x n ~x~- n 17. If

first.]

and Za(6-c) 2 >0.]

JEb"

a 9^6 and

+ a) -f c 2 (a 4- b).

merely another form of the

is

14. If a, b 9 c are positive

8

6 2 (c

are n positive numbers, then

... fc

(i)

[327a

-I-

and

r is

(p

any

4.

A

.

A

[Use Art.

ni 9.]

rational except zero, then

l)f+i

- r+i^( r + l)pr >-pr + l p

where the upper or lower signs are to be taken according as lie between and - 1.

Hence show that

r

(l

+ 2 a + 3 r 4-

...

+

r

)/w

r4

'

1

lies

r

does not, or does,

between l/(r-fl) and

[In Art. 10, let a;=r4-l. First put 0=73 + !, &=/>, Finally let p = l, 2, 3, ... n in succession, and add.]

and next a=|9, 6=p-l.

HARMONIC SERIES

226

19.* a*b

+ b*c + c*a^3abc.

21. If

x> I

22. If

any two of a,

and n

is

20.

a positive integer,

(#

n

-

1)1 (x

-

are together greater than the third, then

6, c

23. If all the factors are positive, then

abc^TI(b+c-a)

(i)

24.

The

and

25. If

x+y+z

of,

is

a, 6, c,

abcd^ TI(b + c+d -2a).

*

-ax -by -cz)

greatest value of xyz(d

factors are positive,

(ii)

;

4

4

provided that d are given positive numbers. is


/4 o&c,

all

the

6, c are positive rationals and x, y, z are positive variables such that a b constant, show that x y sf has its greatest value when x/a y/b zlc.

26. If x, y, 2 are positive

and x + y-f- z

then

1,

8xt/z

<( 1 -z)(l

-i/)(l

V

-z)<-2

[Observe that 27(1 -a:) = 2.] 27. If

*=!+-A + -+. ..+_ n o

and n>2, show that

[Consider the sequences \, \, ?,

^04:

...

!L^

and ?, ?,

*

1ZO

71

...

!tl

.]

71

H

28. Let a l9 a 2 , ... an be a sequence of positive numbers and let A,Q, respectively be the arithmetic, geometric and harmonic means of a^ and an , then n
29. (i) (ii)

,

an

a harmonical progression, nH

G

,

If a x a 2

...

.

is

n IfzjX&t ...xn =a where a

is

...

an <(? n

.

a constant, then

r The sum of the products of every r of the x's > CJ?a The least value of (x l -f (a?a + k) ...(xn + k) is (a 4- k) n .

ifc)

30. If a is

.

any positive number except 1, and x, y, z are rationals no two of which and which may be positive or negative, then a x (y - z) +a*(z - x) +a*(x - y)>0. let z be x, [First y, integers and let x> y> z (Art. 5). Consider the A.M. and the O.M. of the set of numbers a x a x a x ... a 2 a 2 a 2 ... where a x occurs (y -z) times and a? occurs (x y) times. If x, y, z are fractions, denote them by p/d, q/d, r/d where p, q, r, d are integers and d is positive. This case can then be at once deduced from the preceding.]

are equal

,

31.

,

,

,

Deduce the theorem of Art. 9 from Ex.

32. If

a l9 a 2

,

...

,

(a 1

[In Art. 15 put

a n are positive and + aJ + ...+an*)<

*

w = ?, a-a^, etc.]

p>q

,

,

30.

then

,

333

IMPORTANT GENERALISATIONS

-f-

16

a-f b

w= -1, n4, $(b+c+d)

[In Art. 15, put

227

+ c + d'

fora, etc.

Use Art. 16 to show that

34.*

a" 1 + b~ lc + cr l a

a+b+c

*)

35.*

[Use Ex. 34.] 36. If

a2

!,

...

,

OK and b l9 b 2 , then

...

two

bn are

sets of positive

numbers arranged

in descending order, (a l

+ a 2 + ...+an )(b + b 2 + ...b n ^n(a b +a 2 b 2 + ... +a n b n - 2a br Use the relation (a r ~a n )(b r ~bn )^0 r r r l

~na b

[Let u n that w n >w-n _!.] 37.

If a>

1

and p, q are a*

1

).

1

.

positive integers,

aa

-I

p [Let

)

.

~l

p-q, a 2

q

u^(a? - l)lp -(a*- - 1)1 (p ~ 1), 1

show

tfiat

.

,

to

2 -I) when p>q.

then by Art.

9,

... -f

-

p

show

1)};

~.

p(p-L) a p -l aq ~l

and

p 38. If

a>l and

-

q x,

y are

"

positive rationale, then

x

2

y

[Put x pfd, yqjd where p, Ex. 37, we find that

q,

d are positive

Let a lld

integers.

b

;

then, using

i

and by

Art. 9, Ex.

39. //

a> 1

ami

1,

x,

d(a

y are

d

- 1)> 1 -a- 1 .]

positive real numbers, then

(a*-l)lx>(av-l)ly

if

x>y.

[Let x' y' be rational approximations to x, y such that x'>x, y' x - 1/. Let /( # ) )/, and suppose that (a 9

then

We

/(*)

can choose x', / so that when *>!/. See Ch. XIII, 7.]

|

-c'
-a" 1

2 )

,

and so

/(a?)

-f(y)>0,

CHAPTER XV SEQUENCES AND LIMITS (Continued from Art. 11 on p. 15)

Limit of a Function of the Positive Integral Variable

1

When

n

is

large,

1 4-

-

enough, we can make This as

n tends to

infinity.

I

further,

;

by making n large

l

as nearly equal to

1 -f

roughly what

is

nearly equal to

is

n

n.

is

as

1

we

meant when we say that

like.

1 is

the limit of

In this explanation, the meaning of

1

+-

'

large

enough/

'

as nearly equal to 1 as we like,' is by no means clear. If u n is a function of the positive integral variable n> the n tends to infinity is usually defined as follows

i

limit of

u n as

'

:

that we (1) If corresponding to any positive number no matter how a there is m such that small, positive integer may choose, for every integer n greater than or equal to m Definitions.

where

is

a fixed number, then

is

expressed by writing

I

This

lim u n

=

I

called the limit

is

tends to infinity.

lim u n = L

or simply

/,

ofu n as n

n->oc

We also say that

un

tends to

I

as n tends

u n -~>l

as

to infinity

n-+vo

M

corresponding to any positive number how great, there is a positive integer matter no

then u n

this

by

that

m

y

we may

choose,

such that for every

n greater than or equal to m, u n >M, is

said to tend

to infinity

wn

>

as n tends or

oo

,

,

to

lim u n

w n ->oo and v n = - u n we say that If u n -->l we say that the sequence (u n ) If w n ->oo or - oo to the limit L the If

and express

.

(2) If,

integer

;

infinity)

= oo

.

v n tends to is

and we write

minus infinity (-<#). and that it converges

convergent,

sequence (un )

is

said to bo divergent*

THEOREMS ON LIMITS u n does not tend to a

If

oscillate

un

If

limit,

It

if|a|
a n ->

is

u 2 ),

-

or

oo

a=-l,

if

according as n

oo

an

is

This

u3 ),

(3,

...

=r

1

even or odd,

often useful to represent the

graphically. (2,

it is

that, for

first

a n ~> oo if a - 1, a n = 1 for and a n oscillates finitely; if and so a n oscillates 1,

;

infinitely.

few terms

of a

sequence (u n )

done by plotting points whose coordinates are

is

It

.

said to

\
does the function a n behave as n-> oo ? II, 18, (5), it will be seen that if a>

How

1.

1,

,

Otherwise, u n oscillates infinitely.

said to oscillate finitely.

is

every n; -

oo

n,

Referring to Ch.

a<

nor to -

M

\u n

Ex.

,

and the sequence (u n ) is called oscillatory. can be found such oscillates and a positive number

every positive integer

then u n

+ oo

nor to

229

(1,

u

}

),

usual to join points representing consecutive

is

terms. Ex.

2.

un = ( -

//

n l)

I

1

+-

represent

,

j

Jew terms of the sequence and examine the diameter of the sequence

graphically the first

(un ), as

n->

oo

.

As ?i~> oo u zn -> 1 and u 2n ,!-> Thus un oscillates finitely. ,

2.

1.

FIG. 34.

Fundamental Theorems on Limits.

(1)

Ifu n -*l and

v n -*l\

then

u n + v n ->l + l',

(i)

l/u n ->l/l, unless

(iv)

Proof. (i)

and

un

(ii)

-v n ->l-l',

= 0,

1

(v)

Choose a positive number (ii).

By

w n ^ n ->Zr,

(iii)

u n \v n ->l\l' unless ,

I'

= 0.

no matter how small.

,

the definition of a limit

we can choose MJ and

w

2

so

that

\u n -l\<.l

and If

\v n

we denote

will hold for

-l'\<\t

the greater of

n^m, and

and |

Wn

mv m2

for

n^m

for

n^m%.

by m, then each n we have

for such values of

_ Vn _

(J

_ J')

Hence, by the definition of a

|

sg |

Mn

limit,

u + v->l + l'

l

and

of these inequalities

MONOTONE SEQUENCES u n = I + a and vn = T + Then

230 Let

(iii)

j9.

Choose any positive number

L

greater than

V + j8

|

then

1,

\u nv n -ll'\<\fll\+\\.L. the definition of a limit

By

provided only that

n^m. Hence

Let

tt n

= Z+a,

Z

is ftoJ zero

1

we can choose a

n

Now by

IV \

<.

This follows from

(iii)

number L

positive

and

than

1

+a

|

|,

and

|ifX'

we can choose

m so

that

also

* and therefore

11

i_

c,

less

<

-|

n>m. Hence

<

a

-

the definition of a limit

provided only that

(v)

|

then

A // then

u nvn -

also

u nvn -+ll'.

Therefore (iv)

m so that

we can choose

w

>T

.

Z

for

(iv),

when r Ifu n ->l and v n ~w n ->0, l. For Vn -Z = Vn - Wn ) + (w n -Z), therefore

(2)

(

Now we

can choose

m

|w n

Hence

-J|<|

vn

-un

\

+\ u n -l

\.

and

|^ n

-w n |
for such values of n, |

3.

w

n^m both

so that for

-Z|
f> |

vn

-1 1

<

c,

Monotone Sequences.

said to increase steadily, and (u n ) If w n +i
and therefore w n +i^ M

If is

un

v n -> Z. for a11 values of n,

un

is

called a monotone increasing sequence. is

called a monotone decreasing sequence.

said to decrease steadily,

and

(w n )

is

EXISTENCE OF A LIMIT Theorem less

231

// every term of a monotone increasing sequence (u n ) is than a fixed number k, then the sequence converges to a limit equal to, or 1.

less than, k.

Divide the system of rational numbers into two classes as

Proof. follows :

of

The lower class is to contain every rational a such that, for some value n (and therefore for all greater values), w n >a. The upper class is to contain every rational a! such that u n
values of n.

No rational escapes classification, and every a the classification defines a real number A.

We

shall

is less

prove that the sequence converges to

First observe that

every rational

which

A

than any

a'.

Hence

as a limit.

u n cannot exceed A. For if u n >A, u n would exceed This is impossible, for such lies between u n and A.

a rational would belong to the upper class. Next, choose a positive number e, no matter

how

small,

and

let

a be a

rational such that

A-
0
therefore

is

exceeded by some term um of the

;

if

limu = A.

Hence

k^A,

Again

for otherwise k

would be exceeded by some term

of the

sequence, thus

It follows that

or to

For

u n
-f

oo

if

if

un

increases steadily as

n->oo

,

then u n tends to a limit

.

u n does not tend to +

oo

,

number k

a positive

exists such that

for all values of n.

Theorem 2. If every term of a monotone decreasing sequence (u n ) is greater than a fixed number k, then the sequence converges to a limit equal to or greater than k. This can be proved by an argument similar to the preceding, or thus Let v n = -u n then (vn ) is an increasing sequence of which every term is Therefore (v n ) converges to a limit less than - k. Hence less than - k. :

,

(un ) converges to a limit greater than k. Hence also if u n decreases steadily as n->oo

to -oo

.

,

then u n tends to a limit or

EXPONENTIAL INEQUALITIES

232 i

Ex. If

1.

x

un

//

~0

un

1,

--ii(x

n -

1)

where

for every n,

#>0, show

that

un tends

to

a limit as n-^&

.

and so lim u n ~0.

x>I, u n >0, and by XIV, 9, un decreases limit, which we shall denote by /(x), as ?i~> x If

Hence un tends

as n increases.

to a

.

If

0<#<1,

(1) ///or all values of n* u n is positive and u n a constant greater than unity, then ?/ n ->oo

Theorems.

4.

where k

is

.

For u n >ku n _{>k*u n _2 (2)

y>\ and

x~l/y, then

let

all

Iffor

>k

...

values of n,

un

n ~l

u1

Now &>1,

.

therefore u n -+co

and u n i
is positive

is

.

a positive

constant less than unity, then u n -0.

The proof (3)

// u n

For

if

is

similar to that of

is positive

Theorem

(1).

and lim u n n /u n = l then w n ->o> ?//>!, and u n -+Q .

t

/>! we can choose k so that >&>!, and then hy the definition we can find m such that u n+1 /u n ^>k for n^-m. Hence by

of a limit

Theorem

(1), w n ->oo l<] we can choose k so that lw. Hence, by Theorem (2), w n ->0.

If

Ex.

.

9

1.

If p

is

a given

integer, positive or negative,

* (i)

un =

If

and the 5.

(1)

results follow

p

xn -^

->oo

if

x>l u

then

n

/

=(

-

~->0

(ii)

;

sJww

\v -_

if

find

m so that

that

|*|<1.

'X>x

as

rt-> oo

,

)

from Theorem

(3).

Exponential Inequalities and Limits. If n

For,

is

a positive

integer,

f

1

+-

)

<3.

by the Binomial theorem, ...

to

n -f

12 '(3

*

Or,

it'

n

^ m where M

is

a fixed number,

i.e.

after a certain stape.

1

terms

EXPONENTIAL LIMITS (2)

m and n are positive integers

//

1+

1

m

\

>

)

w/

/

n

1

+

1

\

I

1

\n

+m/

)

>

1

and

n/

ml

+

m

n

m>n, /

\

-

m.

11,

f

and therefore

,

then

1-

\

For since m/n>\, by XIV,

m

and

233

+

1

(

\

m

1 \ 1

m/

l\ n

/

>l-f-j. \ n/ n

therefore

Moreover,

0
Hence

l-
'

'

7

by J XIV,

1\*

f 1

11,

--

<1

)

n/

\

n

1

m

n

.

and

n/

(3)

The number

andu n = (l--) \ n

Ifu n = (l+-) n/

e.

'

\

u

through positive integral values,

and the

same for

the limit is the

,thenasn-+oo

'

tends to a finite limit, so also does

71

This limit

both.

is

denoted by

and

e,

is

un

'

;

one of

most important numbers in mathematics.

For

(i)

every n.

has been shown that u n increases with n and that u n <3 for Therefore u n as n->co where e is a fixed number less than 3.

it

-e

,

\~ n (f\ -~

' '

(ii)

Again,

Wfl

-*- - u = u fu n ( n \u n

\ 1

)

/

= un

f

0
therefore

'

- un <

n

u n '-u n -+Q

therefore

(by

2

V

n 2 /)

{\

l
Nowifn>l,

1

\

<-. Now

-

->0,

n

n

and

u n '-+e.

It should be noticed that

u n
can be shown that 6-2-7182818284

f .

....

EXERCISE XXIV 1.

State

how

the following functions behave as n->

(ii)

n + (-l) n

n

(v)

n 2 -f(

->

as n->

(i)

2

+ (-l) n

(iv)

2

+ (~l) n -.

.

xn 2.

Prove that

n

QO

.

.

(iii)

XIV,

-

n

1

1

.

\ j

11),

EXAMPLES ON LIMITS

234

Represent graphically the

3.

say

how each behaves

||
n->oo

%n

un ~

If

6.

-

A*

I

[The limit

.

W

~~

Show

x

is 1, if

\

|

and -

> 1,

1

otherwise.]

-H---- _H-------o+...-fr-, prove that f
tends to a limit as w-*oc 7.

*

n

-

then

^.fl

Find lim

5.

few terms of the following sequences, and

first

.

--*

(Os.i. 2 o 4. If

as tt~>oo

that

.

-n< -^ + o- + 4- -f

vr
... -f

,4

Hence show that

and that un

1 H-

J

+

4-

. . .

H

J

--->< n

/I

1

[Group the terms thus: -+

1\

^+-

)

+

as ?i->oo.

/I

1\

1

1

-4- -4-

)

+

... .]

^-4-

1

8.

If lim

71(0;^

-

prove that

1) =/(a;),

z>0 and

if

1

Ex.

llli [See Art.

xn y n -

3,

=yn (xn -

1

For the

1.

n

-

1)

lim yn ^

and

,

1

Q 9

T If

tt ^v

[If

/>0, then -

5 2

I

n

vn ->oo

n ->0,

/v

2

10. If

4

wn -

1

;

_^_^ and

i<

w n i' n

<

1

-

according as x

,

^

2n

"

^^'s-srn:' 4

Now

l

_____

_

.

1).

1

1

1) -f (y

1

we have f(xy)-\im n(xn yn -

last part

5 y, 1

hence

v< 2-^n

and

6

2w

46

,

-

''*

-

2n + 2

N/2/Hrl, prove that

between l/\/2 and 1 as n->oo [Show that w n ^
lies

wn

1

tends to a limit which

.

,

"

/2" has shown that this limit = A/ - .] ' 7T

9,

wn *=un vn and

^


Wallis

GENERAL PRINCIPLE n ^ 3,

11. If

n+ prove that \/(n +


1)

/

[For the

first

also

steadily;

part use the inequality

w>4

(

/

(r + 1)^

(i)

between

lies

7J 71

and

1

(

1 -f

+ l>0,

lim u n ^\l(r f

l

r

-f

2r

-I-

r-f-i

1

-

)

nj

\ (ii)

that

lim {(a n-

-

3 r -f

r-,--l

Corresponding small,

.

decreases

. .

-f

.

+

r

1

)

4-

l ,

n +

(a

-f

r

2)

-f ... 4-

(a

The

and add.

+ tt) r j/tt f + 1 = l/(r + l).

^-00

to

any

positive

must be possible

it

n r )/ti r

.

The necessary and

cient condition for the convergence of the sequence (w w )

how

oo

1).

General Principle of Convergence.

6.

n ->

|

,

Show

as

s'w

[In Exercise XXIII, J7, put 1, 2, 3, ... n in succession for ?;, necessity for the condition r~hl>0 in (ii) should be explained.] 13.

1

If Z>1, for ?*>3 we C'^>1, therefore "?i->Z where Z^l. n But J n /n->oo when J>1 (XV, 4, Ex. 1).] l /n
any rational except zero and

12. If r is

If r

Thus

<3.

4 ---I

1

(

that */n ->

n

]

^n>L and

should have

show that

235

to

number

m

find

we may

choose, no matter

so that, for every positive integer p,

\u m + ,-u m 1(

that

suffi-

as follows.

is

\<

(A)

say, the difference between u m and any subsequent term must be numerically less than e, so that if the terms of the sequence are repre-

That

is to

sented graphically, all the terms beginning with u m lie in a strip of width 2e. This statement, called the general principle of convergence, is of the greatest importance, (i)

The condition

and may be proved as For

is necessary.

|

um + p - u m

< \

The condition is points Wj, u 2 ... on the (ii)

,

so that

all

|

um + P -

I \

un

we can

>l,

find

u m+p - u m - (u m

Now

for every positive integer p.

Hence

if

follows.

+ u m -l\

,

]1t

and |

\

m

-l)

um

+ (l- u m ).

^ -um \<

e.

Represent the terms u v w 2 ... by the The condition asserts that m can be found

sufficient.

x-axis.

so that

,

the points,

on a segment of length with its middle point at um

lie

2e

FlG

35

.

By like.

choosing e small enough, we can make this segment as short as Hence it seems reasonable to conclude that as n->oo the point ,

tends to a limiting position somewhere near to um If the student satisfied with this reasoning, he may omit the proof on the next page. .

Q

B.C. A.

we un is

NECESSARY AND SUFFICIENT CONDITION

23G Given

for every positive such that

m

integer

e,

that,

no matter how small,

un - um for every integer n greater than m,

a positive


required to prove that the sequence (u n )

is

it

|

there exists

is convergent.

Using the given condition, we can determine a positive integer w 2 >Wj, u mi> is within the interval

Proof.

m

so that, for every integer

l

('U mi

Denote

this interval

by

-,

U mv

+).

(a v b^.

w >w2

Again, we can determine m 2 so that, for every integer 3 within the interval (u m
W 3 >m

2

Thus the whole or part

.

Denote the common part This //

^Hi.2

is

of length

The

.

(w ?Wo

-

Je,

of these intervals

lies

-e,

of

entirely within

figure illustrates the case in

um^ +

,

um ^

is

for (c^, 6 T ),

within (a v

b^).

and contains u n

tor

\t)

is

(a 2 6 2 ).

by

,

/^),

(c/ 1?

which the intervals overlap.

FIG. 36.

repeating the process, we can find a succession of intervals (a^ such that (i) each contains all that follow ... (a,., b r ) ...,

By (a 2

,

6 2 ),

(ii)

(iii)

;

for sufficiently large values of n, b n -ff

a n
^

*

.>, t -->

tl

Hence by Ch. XIII, that

b^),

1,

there

is

and

for every n,

ns

uu H

is

~~*

within each of the intervals

^

;

'

one real number

I

and one

only, such

this is the limit of the sequence (u n ).

NOTE. The necessary and sufficient condition for the convergence of the sequence (u n ) is usually stated (less simply) as follows. Corresponding to any positive number e that we may choose, no matter how small,

it

must be possible

to

find \

if

>w,/w

m

so that

",

-

"


.................................

(B)

every positive integer p.

The condition (B), though apparently more general than (A), is equiTo prove this, we have only to show that if (A) is possible, valent to it. then (B) is possible when n]>m.

BOUNDS OF A SEQUENCE we can

the condition (A),

By

all

the terms w wfl

m so that

U^^-Un

I

and then

find

,

I


um+z> m+z

wf 2 .

237

He in the interval

Therefore the difference between any two terms which follow u m so that Hence we can find

than

is less

m

.

|w n + p -tt n |<

H>w.

if

Therefore the conditions (A) and (B) are one and the same.

Bounds

7.

of a Sequence.

If

(1)

there exists a

number

M

such

that, for all values of n,

the sequence (u n ), and also the function u w are said to be bounded above (or

on

the right). If

number

there exists a

JV

such that, for

all

values of n,

then (w n ), and also w n are said to be bounded below (or on the left). A sequence (u n ), or a function w n which is bounded above and below, ,

,

is

said to be bounded.

Examples.

Here

(un ) has a greatest term,

namely w 2 ="2

:

^ has

also a least term,

These are called Ae upper and lower bounds of (MW ), or of u n (FiG.

namely

%

- 2.

37).

-1

FIG. 38.

FIG. 37.

1

2.

Although there

wn

,

is

no greatest term, we say that

for the following reasons (i)

(ii)

No term

of (u n ) exceeds

Infinitely

many terms

1

is

the upper bound of (un ), or of

:

1.

exceed any number

For similar reasons, we say that -

1 is

less

than

1,

for

u.in

-+

1.

the lower bound of (u n ) of u n (FiG. 38).

UPPER AND LOWER BOUNDS

238 // u n

Theorem.

no term of (u n ) exceeds h

(i)

at least

(ii)

This

bounded above

is

number

function u n

a number h such

there exists

;

one term of (u n ) exceeds any number h

is

that

less

than

h.

called the upper bound of the sequence (u n ) or of the

.

Divide the system of rationals into two classes. The lower class is to contain every rational a such that for some value

Proof.

of

n u n ^a. The upper

of

class is to contain

every rational

a'

such that for

values

all

n

u n
is

classification,

defines a real

bounded above, both classes and every a is less than any

number

exist. a'.

No

rational escapes

Hence the

classification

h.

Moreover, no term of (u n ) exceeds h, for in that case it would exceed a rational a' greater than h. This is impossible, for a' belongs to the upper class.

Also

if e

is

any

positive

number, no matter how small, we can find a

rational a such that

h-
:

(i)

(ii)

no term of (u n ) at least

The number function u n NOTE. h and

is less

than

one term of (u n ) I

is

I

;

any number

is less tJian

called the lower

bound

of the

greater than

I.

sequence (u n ) or of the

.

// (u n ) JLOS no greatest term, infinitely

many

terms of the sequence

lie

between

h-.

For every term is less than h, and at least one term u m is greater than h - e. Since ia no greatest term, infinitely many terms are greater than u m and therefore also greater than h e. there

Similarly, if (un ) has no least term, infinitely

and

,

many

terms of the sequence

lie beticc.cn

I

LIMITS OF INDETERMINATION Terms selected from u l9 u 2 said to form a subsequence. Ex.

Explain how

1.

to select

t/

,

3,

...

239

according to some definite rule are

a subsequence which will converge

to the

upper bound h

If a 2 ^h, let a 3 be the lirst If Ui^h, let a, 2 be the first term of (un ) greater than v term of (un greater than a.2 Continuing in this way we form the required subsequence The formal proof is loft to the student. ^i a2
.

,

,

Lower Limits

Upper and

8.

.

.

)

(of

Indetermination) of a

Sequence. Let (u n ) be a hounded sequence and let h l9 h 29 A 3 ... and l v J2 ^> the upper and lower bounds of the sequences u 19 u 2 u% ..., u 2 u 3 w 4 (1)

,

,

,

,

u39 w4 u 5 ,

If

fi

...

,

...

,

etc., respectively.

then u v

l=sUl9

,

* >e

h^u

h^h^.

If

hl = h 2

Thus, in

not

is

ly

than any term of

less

then Aj

is

t/

2,

^h

%, ^4

,

...

u 2 u3 w4

the upper bound of

,

,

>A

,

;

therefore

...,

and so

and so on. 4 Similarly A 2 ^/fc 3 A 3 any case, h l 2 and Hence h l9 h 2 A3 ... is a steadily decreasing sequence every term is greater than or greater than l v Therefore this sequence tends to a limit .

.

,

,

,

,

H

equal to l v Similarly

than h v

I

l9

1

2,

Z

3,

Therefore this

The numbers

H and

an increasing sequence and every term is less sequence tends to a limit L less than or equal to h r is

...

L

are called the upper

and lower

limits (of indeter-

mination) of the sequence (u n ).

lim u n = //

It is usual to write

(2)

and

lim u n ~L.

Examples. Here h 1

= h 2 =h3 = ...=H.

(Fig. 39.)

FIG. 40.

FIG. 39.

1\

Here

^ = ^3-

;

h^h^u^;

=l z ~Uq-,

etc.

etc. .',

/.

//=lim u2n ^l.

L~tim w 8n+1 =

-1.

(Fig. 40.)

GENERAL PRINCIPLE OF CONVERGENCE

240 (3) It is all

easy to see that,

terms of u lt w 2

...,

,

wn _1

After a certain stage,

(i)

that

if

A n >Aw+1 then h n = u n and h v A 2 ,

As w->oo two cases

.

arise

the A's are equal, that

all

7> /> /> ~ "'m+2 ~ n m ~ Mrn+l

:

m exists

to say,

is

h n -\ are

...

,

so

H "

-

In this case, (u n ) has no greatest term. None of the terms from u m onwards - e. can exceed H, but infinitely many terms exceed

H

There

(ii)

no such stage as that described in (i). That is, no matter h n >h n+l be, we can find w, greater than m, so that

is

m may

how

great Therefore every h

.

is

a term of (u n ), and in

cases //

all

H

such that, after a

is

certain stage, every term of (u n ) is less than + e ; Hence in all cases, infinitely many terms of (u n ) are greater than a in other than are certain all the terms less + also, after stage of (u n ) .

H

H

words, only a finite number of terms exceed II

In a similar

way we can show

}

+ e.

that infinitely

many

terms of

un

are less

L+e

than

and, after a certain stage, all the terms are greater than L~-e; in other words, only a finite number of terms are less than L-c. It follows that, if

H=L=

then the sequence converges

l,

For, after a certain stage, all the terms of (u n ) //

-f

that

,

is

between I -

and

l

lie

to

I

as a limit.

between

L-e

and

+ e.

We can at once deduce a proof (4) General Principle of Convergence. that the condition in Art, 6 is sufficient for convergence. Given that

m

can be found so

that

\u n

-u m \
nl>m,

for

it

is

required to prove that (u n ) is convergent.

u n cannot tend to

First observe that

-f

co

or to

oo

.

Therefore

We can therefore choose m (u n ) has upper and lower limits // and L. u n - um and n (n> m) so that un 3-6 Je, Je, \u m L |

and since

H

H

-L

H

- u n + um

\

<

-L + u n - um

\

<

\

|

<

;

,

\H-L\^\H-u n \+\u n -L\+\u n - u m

|


H and L is H = L and

to say, the difference between the fixed numbers Therefore less than any positive number c, however small.

That is

(u n )

9.

is

convergent.

Theorems.

(1)

certain stage,

where k

is

But

un

if

In

// (u n )

u

a positive constant -u n ^ ;>w n - u n }

the first case,

,

is

an increasing sequence and

if,

after

a

u less ^.

2

,

than unity, then the sequence the sequence is divergent.

suppose that the conditions hold for

is convergent.

Then

if

therefore

IMPORTANT THEOREMS w>s, u m+l -um
241

M

.

Also

Hence by addition

Choose any positive number can find

,

no matter how small then, since &<1, we and since 'Umirn >um it follows that ;

m so that &m
tWn

,

- um

|


for every n.

Therefore the sequence (u n ) converges. In the second case, if n>s, u n+l - w n >w s

Now (2)

us

w s+1

is

^

t

~ w 6-

a fixed positive quantity, therefore obviously

// M M >0 /or a^

m^.9

o/ n awrf w n+1 /?/, n ~>/>0 where

I

w n ->oo

is

.

a fixed

i

n number, then u n ~+L Because Z>0, we can choose a positive number like. Then we can find m so that

e< and

as small as

we

and by multiplication,

Since

Z-e>0,

it

follows that

^u n

^u n therefore

,

(Z-)

n

<

J m <(Z + e) n

i

/

\

n

1

'

and

i (t

m

')

.

(I

/

-)
j V

m 1

Now

u m fl m

is

a fixed number, therefore (u ni /l m ) n

1

>]

and

?/

n n

>i!.

i

The converse of this theorem is not true. That is to say, if ^n n -> follow that w n4l /w n ->/. For consider as a special case the sequence 1,

NOTE.

/,

it

does not

2, 1, 2,

...

in

i

which the terms are alternately a limit.

1

and

2.

Here w n n ->

1,

but n n

fl /^ n

does not tend to

IMPORTANT THEOREMS

242 (3)

We

//

,
= u + uz +

w

.

.

.

l

can find

w

-f

u n and Urn u n = I,

so that for

any

Urn

then

n-*x

n

positive

>

-

oc

^

sn

=

/.

c,

Hence by addition, - m)(l - e)
for

and

Now let m and n lend to oo in such a way that w/m->oo we may suppose that m~>cc and n>m 2

(For example,

.

.)

We

show that under these circumstances s m /n~>0. Since u n -*l we can find a fixed number k such that u r and then m < WA', 8m < % + h/ 2 4- ... 4shall

|

r,

I

I

I

1

I

so that

m

<

I

.v

Thus

(4)

//

Let

like,

and

/

s

n

P n = a n & +a n _ 1 6 2 + a n _ 2 1

and

bn

';

3

+

= + f3 ni l>

---

+a

(3),

We

shall

+a 2 + ...+a n )->0 and

l(a 1

e

can betaken

j8 n

->0, then

4-a^). i(j8 1+

j8 a

+

...

+j8 n )-^0,

Te-

prove that $,,->0.

have |

and since a n->0 and

jS n

can also find |

m

as

|

A-m

Q n |< ~n T| r =i

->0 we can find a |a f

We

and

to zero

so that a n -~>0,

?Z

and we

/

n->(>.

A

Q = ~(^i+n-ift +

where

by

for every

therefore s n jn->l.

nfy^^a-fay,

Now


I

m/n and 5 m /?i tend

in the inequality (A),

we

1

7

A:

n

7i

as small as

I

|

|

1

\

and

|<*

fixed

I

,

number k such that for every s

|ft|
so that

<

/k

and |

f38

\

< c/A

for

Hence

|a r ]9 n _ r+]

and

COMPLEX SEQUENCES for n-r + l>m, |<*| j8 n-r+1 |< * r

n>2w,

10.

zn

If

(1)

i.e.

= xn +

corresponding to any positive such that integer

m

-z\
\z n

where

we

a fixed number, then z

z is

zn = z

lim

write

said

(z n ) is

is

a positive

n^m,

for

called the limit

is

limz n = z

or

however small, there

e,

r^m,

where x n and y n

iy n

are functions of a positive integral variable n, the sequence to be complex. If

for

B)

P n-~>ab.

|Q n |< e: consequently Q n->0 and IV

Complex Sequences.

(

l^r^n.

|

Therefore

(A)

r>m f^>r r
|r0n-r+l|<*| a r|<

the inequality (A) will hold for so that, with this^proviso, oc r n _ f3 r+l \
243

or

ofz n as n->oo

,

and

z n ->z.

n->oo

Under these circumstances the sequence limit

(z n ) is

said to converge to the

z.

In order

lim z n

that

z

=x+

it is

ty

limx n = x

and

necessary and sufficient that

limy n ~y. 2

K-H=-y{(z-z) + (2/--?/) 2

For Therefore |

zn

z \

^

|

First suppose that z n -*z.

and

x

xn

|

\

By

l^n" !"^^?

z \

|# n -#|

therefore

~z\-> 0,

All this is evident geometrically, for \z n

points z n

and

yn

|

and

tend to zero, so that xn +x and //>?/. Again, if x n ->x and y n ->y, from equation (A) \z n

(A)

}

^

~

y

. \

definition,

therefore

2

zn

-z\

\y n ~y\

it

follows that

zn

-> z.

is

the distance between the

z.

The theorems of Ch. XV, 2, regarding the limits of the sum, difference, product and quotient of two functions of n, continue to hold when the (2)

functions are complex. (3)

Limit of

zn

as n->oo

Otherwise z n does not tend to

For

let z

(i) if

(ii) if

= r(cos 6+ i

=l

to a limit.

then z n

z is

complex.

a limit except then z n = r ri

sin 0),

|z|=r
when

r n ->0

= l, and

and z

if |

zn

->0;

1

then

|>

7/|z[
then z n ->0.

when z = l, and then limzn = l. (cos n0+ sin nd), and i

r n ->oo

and

zn

does not tend

SUBSEQUENCES OF PROPER FRACTIONS

244

11. General

condition for the convergence of a complex sequence

sufficient

as follows

so that \

to

(z n ) is

Art. 6,

any

e,

positive

-z m \
zm + p

If

Proof.

and by

(z n )

is

:

Corresponding

m

The necessary and

Principle of Convergence.

however small,

convergent, then (x n )

we can therefore choose m \

\

N W

X ^m+y

__

Jr'm

it

must be possible

to

find

every positive integer p.

|<-1<: ^ 2 fc

and

(y n ) are

convergent sequences

;

so that for every positive integer p,

MTU] *iu-i

< 2f e Mm \^l

lay |

\

__,,/

.7^40

->.

-

\

htn + p-^ml

therefore

Hence the condition Next suppose

that \

Therefore (x m ) (y n )

is

As

any

,

^ - z m \
then |

convergent (Art. (z n )

is

6),

however small,

e,

and

x w +p - x m

|

it

similarly

<

z |

m

-

f

it

may

be stated thus.

must be possible

to

find

p

can be shown that

convergent, and the condition

in Art. 6, the condition

positive

if

necessary.

Hence

convergent.

NtfTE. to

is

is

zm

is

sufficient.

Corresponding

m

so that

for every positive integer p.

EXERCISE XXV 1. If u n is a decreasing function of n, then, after a certain stage, all the terms of u n have the same sign. - QO First let u n -> I. If / ^ 0, u n is positive f Kithor u n tends to a limit or to for all values of T?, for un tends to / from above. If I <0, we can choose so that .

m

l
n^m.

for

Therefore the terms of (u n ) beginning with u m are negative. If u n ->--<*), we can find so that u < - k for

m

.

2.

If u n

is

n^m.]

n

an increasing function, the statement

in

Ex.

3. Show that the positive proper fractions can be arranged in a definite order, so as to form a sequence. Represent the first few terms graphically indicate terms which form subsequences converging to the upper and lower bounds as a limit. [The sequence, omitting such fractions as |, is

1 is

true. ;

'

|

I

!

;

1

1

1

2'

3'

I

9

3

1

2

4'

5'

5' 5'

The required subsequences dotted

lines.]

3

5'

6'

6

arc indicated

by the

-Q

FIQ. 41.

GEOMETRICAL CONVERGENCE 4.

un >0

If

[See Ex. 5.

6.

n and un+l /un ->l then (nun

for all values of

XXIV,

and Art.

11,

}

\+1+ 2 n (\+ V 3

un ^

,

)

n -+l.

9, (2).]

Prove that lim -

If

245

. . .

-

+

)

-0.

[Art. 9, (3).]

n/

show that

->

and consequently

e,

(n

e.

4-

"n-l

7.

n

- .^

z~

,

"~l.n

5-

+

71

1

~w

-7i

prove that

1

1

oo

[Show that

r(7i

k>0

prove that

(ii) if

>

where

.

for

and

(j) if

.

+ -^T;,

-zr*+

prove that un -*

If

1

1

If

9.

,

i .

+l

1

8.

+

...

3(71-2)

a,

r

= l,

2, 3,

...

n.

Hence show that

and negative roots of x 2 and %>0, then ^ ->a; and v 1 j8.

-/? are the positive

= */ (k + u^) vn = */ vn

un

__ l )

(A:

-x-k

rt

Illustrate geometrically. [(i)

Here

t*

n

2

= ^ + w n -i ^n

-

1

~&+w

'

f2-2

wn 2 ~ wn - 1

~ u n~i -Un-zi

hence u n '^u n _ l9 according as u n _^'u n _ z therefore (u n ) is monotone. 2 2 u t ~ a) (u l + /?), J^ir5< suppose that u t > a, then u l >u 1 +.k for 'i^! - u t - k ( 2 = 2 a u and is /. /. u 2 2 u 2 + k .'. u 2 >oc. Similarly u n >
i/,

,

)

.

f

^n -> Z ^ a.

The

Thus wn

case in which

i^ n_!

and

->

follows that

u l
of y x and (ii) Draw graphs meeting at A. In Fig. 42, we have

in the

same way.

y~

PrPr ~ u r ~ u r+\*

and thus

The

it

figure illustrates the case in

which

Ui = ON = N p > a, l

1

l

and shows that (un ) is a decreasing sequence. Thus J\^rPr > Nr pr for every r, and so Nr is always to the right of N. Thus un -> I ^ a and

Hence

N

r

tends to

N

as a limiting position

^

and un -> a.]

NN r N 2 FIG. 42.

N,

CONVERGENCE TO ROOTS OF EQUATIONS

246

Wn=/( wn-i) where f(x) is such that (i) f(x)>0 and (iii) the equation x=f(x) has a single w that ->a. prove n [It is advisable to draw a figure.] 10. If

increases with x,

11.

Show

number

that as the

tend to the roots of x 2 - x - 7

=0

if

*>0,

positive

f(x) root a,

(ii)

of roots tends to infinity, the expressions

as limits.

3 = v7-v7+y, prove that 2/ ^( a; 2_7)2_7 last the Draw a graph of equation, and prove that 12. If

%

un ^

\7-V7-f \/7-N/7T...

%, w 3 u5 u Z9 u^ u$

then

,

to

if

2n

roots,

a decreasing sequence,

... is

... is an increasing sequence, and that both sequences converge to 2 as a limit.

Prove also that

\7-fv7- V 7 WT"^"... 13. If

to oo ^3.

u l9 v1 are given unequal numbers and

n^^

for

u n decreases, and vn increases as n increases, (ii) Each of the functions u n and vn tends to a limit, which is the same for both. This limit is called the arithmetico- geometric mean of u and i\ (Gauss). ~ v = i ( \^w_i - N/v -i ) 2 > u > v for ever ^ prove that

K

(i)

*

n

>

n

Un - I u n~i + vn-i) < Wn-i ...
.'

Thus v^^ Hence u.n tends

l

t;

Vn

.

t*

i

= s/(^n -i^n-i)> ^n-i-

.

and so does vn Also lim vn = lim n_ 1 = lim (2un - Mn^) = lim to a limit,

y

n

and

(

n .j

CHAPTEK XVI CONVERGENCE OF SERIES 1

.

Let u n be a function of n which has a definite value

Definitions.

An

for all positive integral values of n.

which every term

in

is

R Ui v

by

,

sn

expression of the form

followed by another term,

This series will be denoted by En^\u n

n terms by The sum

(1)

,

or

by

is

27w n

,

called

an

infinite series.

and the sum

of its first

.

of the

p terms immediately

following the nth term

is

denoted

so that **n, v

As n tends

~ w n-f 1 + u n+2

~^~

~^~

u n+j> ~ s n+v~~ s n-

to infinity, there are four distinct possibilities

to a finite limit, to infinity or to

minus

infinity, or it

:

sn

may

may do none

tend

of these

things. If s n

tends to a

called its

sum

finite limit s,

to infinity.

Thus

lim

s n = s,

the series

is

said to be convergent and s

is

defined by

s is

or briefly,

lim s n

= s.

n->oo

This

is

also expressed

w t + w 2 + w3 + If s n

tends to

oo

by writing

=s

to oo

...

or to -

oo

,

or

the series

27fw n = s, is

or

2u n = s.

said to be divergent.

tends to no limit, whether finite or infinite, the series is said to oscillate. We say that the series oscillates finitely or infinitely, according as s n oscillates between finite limits, or between + oo and - oo If s n

.

Series

which diverge or

The sum not a

sum

be non-convergent. essentially a limit, and

oscillate are often said to

(to infinity) of a

convergent series

is

in the sense described in the definition of addition.

We

is

are

therefore not justified in assuming that the sum of an infinite series is unaltered by changing the order of the terms or by the introduction or

removal of brackets. In fact, changes of this kind

may

alter the

sum, or they

may

a convergent series into one which diverges or oscillates. Thus the series (1 - 1) + (1 - 1) + (1 - 1) + ... is convergent and zero,

but the

series 1

1

+ 1-1 + ..

.

oscillates.

transform

its

sum

is

ADDITION OF SERIES

248

The Geometric

2.

converges

l^l^l and

if

oscillates finitely if x== - 1, For sn (l-z n )/(l-z) if

=

5 n~>oo

If

x^l,

If

x< - 1,

.

s n->oo

If

x=

or

-

Theorems.

3.

Uj

The

Series.

series

+x + x* + ... +xn - 1 -t-...

1

its

sum

and

oscillates infinitely if If |z|0

is

It

l/(l-x).

diverges if

x< - 1.

x^l. -1, sn = l

and

or 0, according as n oo according as n is odd or even.

(1)

m

If

+ Wg + tia-f

is

...

a given number, the

is

* n->l/(l

-x).

odd or even.

series

um+l

and

converge, both diverge or both oscillate.

6
For

let

Then

5n

n = sm 4. n -Sfn

(i) if

(ii) if

(2)

= w 1 + w 2 -f...+w n

sm+n

an d

tends to a

5

m

is

,

*n

a fixed number, hence

finite limit or to

oo

so also will t n

,

;

sm + n tends to no limit, whether finite- or infinite, neither will t n

If k

.

a fixed number and Eu n converges to a sum s, then Eku n is its sum is ks. Also if the first series diverges or oscillates, so

is

convergent and

does the second, unless & = 0.

-oo

if s n ->s, then ks n ~+ks. If k 9^0 and s n so. does ks n tends to no fixed limit, finite or infinite, neither does ks n

For $n

,

.

If

.

4.

Addition and Subtraction of Series. wx

converge and (i) (ii)

(Wj

4-

(MJ

their

Vj)

Vj)

+ (u 2 -f v 2 ) + -f-

(ii)

v2 )

(u 2

-f

NOTE.

-f

sums are

w2 -f

In the proof,

The proof

is

-f

. . .

all

w2 + s

. . .

. . .

and

- . .

and

t

vl

+ v2 +

If the

series

. . .

respectively, then

converges,

and

its

sum

is s

converges,

and

its

sum

is s

+ w n ) + Iim(v 1 -f v 2 4-

the series in brackets are

. . .

+1

;

t.

+ v n ) = s 4- 1.

finite.

similar to the preceding.

SERIES OF POSITIVE TERMS

We

first

heading we

consider series in which all the terms are positive. Under this include series in which all the terms are positive beginning with

some particular term. 5.

Introduction and Removal of Brackets. Let Zun be a series and let the terms be arranged in groups without altering

of positive terms,

SERIES OF POSITIVE TERMS Denote the sum

their prder. (i)

// Su n

For

converges

to

a sum

let

Then

m tends

where

Now

lim

s

n we can

terms in the nth group by vn then ,

so does

s,

2v n

.

=

s

for every

of the

find

m and p,

to infinity with n. therefore

m = lim sm + 9 = s,

so that

lim t n = s.

If Zv n converges to a sum s, so does Zu n For corresponding to every n we can find (ii)

.

m

But lim tm = lim tm+l = s,

y

so that

lim s n = s.

therefore

Changing the Order of Terms.

6.

249

// the terms of a convergent remains convergent and its

series of positive terms are rearranged, the series

sum

is unaltered.

Proof.

manner. is

a

Let

Eu n -be

Denote the

.

first

n terms

Further, suppose that the

(n +p) terms of

Zu n

of

first

Zu n

m

are,

among

the

first

n^m^n+p,

m tends to infinity with

Also

n. sn

< >m
.

lim 5 n ==lim 5^ +J> ==5, therefore lim m = s. Hence Zvn is convergent and has the same sum as

Now

NOTE.

The argument

fails for

MI

Zu n

.

such a derangement as

+ wa 4- u6 -f

.

..

+ uz + w4 + w e +

. . .

,

where Zun is broken up into two (or any finite number of) infix For here we cannot find m so that the first n terms of Zun m terms of Zvn Thus u2 does not occur till infinitely many of ti have been placed. The rule for the addition of aeries applies in cases of this kind, but Sun is broken up into infinitely many infinite series. Derangements of the last sort are considered in another volume. .

m

tsrms of

terms of 2v n are among the

.

Thus

and

,

Let

u.

Suppose that the

Zv n

convergent, and let the terms be rearranged in any v n so that every u is a v and every v series by

new

first

COMPARISON TESTS

250

Theorem.

7.

of positive terms cannot

series

and

oscillate,

than some fixed number k, the series converges and

less

always

A

if s n is

sum

its

is less

than k.

For

XV,

(Ch.

and therefore tends to a

increases with n,

sn

Hence the

3).

every n, then

s n ~>s,

s is

Comparison Tests. Ua n is known to be

8.

terms and

* (i)

(ii)

u n ^a n for

* or if or if

(iii)

Proof,

u n/a n

to

+

...

a + a2

If

(i)

u l + uz

and since For

(ii)

is

n

less

to oo

-r ...

be

Eu n and Za n are divergent, then Uu n will (ii) (iii)

3).

=

we have

,

...

a n
un

follows that

it

;

is

;

convergent.

Now

.

ka n

is

convergent,

convergent.

//

(i)

for

limit.

+ u n ^.a 1 +a 2 -f

This follows at once from

*

XV,

than some fixed positive number k

for

(ii),

if

lira

a positive constant k, such that u n /a n

;

we have u n
values of n,

all

than k (Ch.

//

a finite

a fixed number,

is

un

therefore (iii)

t

sn

if

Further,

less



2u n and Sa n are two series of positive convergent, then Zu n will be convergent if (1)

always

u n/a n tends

oscillate.

a fixed number

values of

all is

cannot

series

where

finite limit or to

u n /a n

all

we can

is finite,

find

values of n.

two series of positive terms and

Za n

is

known

to

be divergent if

^a

un n for all values of n ; * or if u n /a n is always greater than some fixed positive or if

u n/a n tends

to

a limit greater than

/. :

s

;

zero.

TV be a positive number, no matter (i) Let Proof, Ea n is divergent, m can be found such that

how

great.

a l + a2 +

. . .

-f

>N

provided that

n> m

t^-f w 2 +

...

+w n >JV

provided that

n>m,

an

number k

Since

;

divergent. values of n,

u n >ka n

% therefore Eun

is

.

Now Ea n

is

divergent

;

therefore

divergent.

at once from (ii), for if lim u n /a n is finite, a positive ound such that u n/a n >k for all values of n. .y these tests, we require certain series which are known or divergent. The geometric series and the series of the .hose which are most generally useful as test series. flcient

that these conditions should hold for

all

values of

n

greater than

some

METHOD OF COMPARISON Theorem.

9.

The

series

p>l and divergent if Denote the given series by convergence or divergence

any way we

in

First

let

is

+ H^ + O +

TJ

S

Group the terms

is

Now

the

(B)

unity, for

Next, if

thus:

as follows

:

1

1

,

2, 4, 8,

and the

terms, respectively. than the corresponding term in

is

/ 1

1 \

!+_+_ +

same as

1

1

1

!+_+

or

...,

a geometric series whose

common

+ ....... (B)

ratio 1/2 P-1

is less

than

.

last

Each term

,

terms in the brackets are 1/2 2

brackets contain

Now

etc.,

,

in (A) after the first is less

(

is

convergent if

p>l. Hence (B) is convergent, and therefore S is convergent. p = 1, the series S becomes 1+J-f^ + J-H... Group the terms i i+ i + i +i + ... .................. (C) i + + + i) +( i)

where the

which

^s

is

,

...

I

which

.

terms in the brackets are l/2 p l/2 2p l/2 3p

brackets contain

Each term

+

a series of positive terms, its not affected by grouping the terms in brackets Since

S.

1

first

.+-

please.

p>l.

where the

.

251

2, 4, 8,

of (C)

...

4

3 ,

1/2

,

1/2

,

and the

etc.,

terms, respectively.

greater than the corresponding term of

is

the same as

+i + f + f + ...,

1

l+ + i + J + .......... (D)

or

and therefore S is divergent. Lastly, ifp
is

divergent,

;

fore

S

is

10.

divergent.

Examples on the Comparison Tests.

1.

Is the series

i

- -f

~

= .

K

Q

1

Ex.

Z

.

3

j.*>.4: let

+ o~~l

an

r o 4 o .

=

+ ~*

converge^ or divergent

*

Then we have

=

Now Zan

is

convergent

;

un

therefore

*

is

?

.

2

.

convergent.

The nth tenn a w of a suitable test series is found thus Keeping only the highest powers of n in the numerator and denominator of u n we obtain 2n/n8 or 2/n*. Disregarding the numerical factor 2, we take !/ for e* n :

,

.

B

B.C.A.

D'ALEMBERT'S TEST

252 Ex.

Test for ro nvcrgence or divergence the series x xn ~ l 1 x2

2.

<1,

If

a;

n

l+#2

,

l+o: 3

->0 and vn lxn ~ L

->

a:~

n ->

and

w^ 71

#n

=1 1

a:

= 1, u n = ^ and

If

}

the series

is

+-

...

l+zn

+

Now Zxn ~

1.

convergent. If a;>l,

+

_

4.

^

1+z

+ xn

=1

l

...

where x>0.

,

convergent, therefore

is

1

1 --

x x~ n +

->

Hence

.

x

1

ZV-, n

is

divergent.

divergent.

TESTS DERIVED FROM THE GEOMETRIC SERIES

A

D'Alembert's Test.

11.

u n+l ju n
vergent if

9

where k

Zu n

series

is

of positive terms is eon may have an\

a fixed number and n

value.

The

series is divergent if

For

if

u n+} fu n ^l for

u n+l /u n
Thus w w
71

.

2^7/j

.

all

values of n.

n, then

"1

A-

convergent, therefore

is

27w n

ii

convergent.

Again,

if

u n+l/'u n ^l for

+u n ^nu

w t -f w 2 +

therefore

values of n, then

all

1

-*cc

In particular, if u n ^/u n tends

l<\ and

to

.

Hence Su n

a limit

is

divergent.

then 2!u n is convergent whei

Z,

when />!.

divergent

For if Z<1, choose k so that l^ where m is a fixed number. Hence

of a limit

?

Wi + is

convergent, and therefore

Again,

if

/>!, then

um+2 + NOTE,

(i)

is

m

Zu n

tt

is

m+2 +... convergent.

can be found so that u n+ ^>u n for

divergent, and therefore

It is sufficient that the conditions

un

is

n>m.

Henc<

divergent.

should hold for

all

values of

n

greate

than some fixed value.

Nothing is said as to the ease in which Urn un+l lun ~l. In this case the be convergent or it may be divergent, and we say that the test fails. may For example, consider the series (ii)

1111 + + + +

i

The If,

first is

2

3

divergent, the second

however, u n ^^u n tends to

the series

is

divergent.

The

_

"'

4

series

and

1

r

+

1

+

2i

1 i

+

1

4*

+

is

convergent, and in both

1

as a limit

from

1+2+3 + 4 +

...

seriei

--

we have lim un +Ju n = 1 u n+l >u n

above, so that always is an instance of this .

CAUCHY'S TEST 12.

A

Cauchy's Test.

Eu n

series

253

of positive terms

is

convergent if

i

w n n <&
is

a fixed number and n can have any value. i

The

series is divergent if

u n n ^l for

many

infinitely

values of n.

i

u n n
if

series

Zk n

many

values of n,

,

con-

is

i

w n w >l,

If

Zu n

then

is

and consequently w n >l,

for infinitely

obviously divergent, i

In particular, if u n n tends and divergent when l>l.

The proof

Zu n

a limit L then

to

is

convergent when l
similar to that in the last article.

is

It is sufficient that the conditions should hold for

n>m,

where

m

is

a

i

number.

fixed

but

Also nothing

the limit

if

is

Zu n

approached from above,

Comparing these

13.

said as to the case in which lim u n n

is

useful than Cauchy's.

concerned, u n+l ju n

is

is

divergent.

more generally with which we are

tests, that of D'Alembert

For

most

in

the series

of

\,

a simpler function than u n

is

.

On

the other hand, Cauchy's test is more general than D'Alembert's. That this is the case will be seen by noting that

Cauchy's condition for divergence

(i)

is

greater than a fixed value,

and

this

is

For

wider than D'Alembert's.

in D'Alembert's test the condition has to be satisfied

by

all

values of

n

not the case in Cauchy's. i

(ii)

It

has been shown in Ch.

XV,

9, (2),

that

if

u n+1 /u n->l then u nn ->l.

i

But

if

u nn ->l,

So we

does not follow that u n+l /u n tends to any limit. expect that Cauchy's test will sometimes succeed when it

may

D'Alembert's

fails,

as in Ex. 3 below. 2

Ex.

Show

1.

that the series

ic

+

+

#3

+

yA

+ ...

is

convergent if

0<#<1

gent if

un =

Here Hence, If

and

x

is

if

1,

xn -

n

,

un n+1 =

0<:r
~;

n+1

.*.

lim

n w n41 =lim T un n-fl

series is convergent,

D'Alembert's test

divergent.

xn+l

fails,

and

if

but in this case the

x>l, series

xx. it is

divergent.

becomes

and

diver-

EXAMPLES OF SLOW CONVERGENCE

254 Ex.

Show

2.

that

series

the.

+x + x2 /

1

+ x* f

12

+ ...

convergent for all positive

is

(3 "~~

values of x. />.n-i

TT

~n

7?

lim

II

-= 1' lim

iv

=i

-

series is convergent.

+ 6 + a 2 4- 6 2 + a 3 H- 63 -f

a

Test for convergence the series

3.

*

,,,

!

'

.

Hence the Ex.

=

=

Horo

. . .

.

-^

Here

according as n divergent

is

a^l,

if

test

is

odd or even, the or if b^\.

series is

0
if

convergent

w n n ->a a or 6 2 and 0<6<1, and

test, since

Using Cauchy's

inapplicable.

i

i

~

and D'Alerabert's

EXERCISE XXVI 1.

State the character of the series

showing that

for the

second series lim$o M ~-l and lim

2.

Show

that the scries x

3.

Show

that,

if

ci

l -\-

-?>

+ x ^ + x2

+ 3 + a t ) 4-

a2

(a,

ft

...

-\-

is

5 2n _

3

x~ 2 + x + x~* +

-}

)

I

3

(

4-^4)

Criticise the following 2(-J

+

i-

+

i

+

J-f-.-)

which 5.

(iii)

is

absurd, for

If

n

is

11

J

<1,

^-i-*-l-i

+ l-+-J +

J

1

<,

v

'

(ii); v

>

some

fixed

number m.

=(i + 4 +4 +

...)

+ (* + ! + i +

)

= n-i+4 + ...,

,..11111

1

first

1

million terms of the

. .

that

(m +

:

The sum lies

(iii)

is

....

4-

Given that 2 20 > 10 6 4- 1, show that the sum of the series I -f i f 1 4- j -f is less than 20.

Show

never convergent.

and so on.

1

.

6.

-..

...

a positive integer, show that 1

is

...

:

=i+ .-.

2.

1

convergent, so also

J (a 2 + a 3 +

Prove also that the converse is true if a n >0, for n [Consider the sum to n terms of the second series.] 4.

. |

between

Hence show 4-

m-fl

and

m+1 ,

. .

.,

must be taken.

with an error

-~

(m-fl) -f(ra

that, in order to find the

^ + 4~r +

o

-

to infinity J of the series

less

2

+

/rr2 + : ~^r2 + (w-f2) (m-l-3)

rri.

+ 1) 2 sum to

than

infinity of the series

001, at least 1000 terms

CONVERGENCE OF SERIES Determine whether the n

1

,

9.

-

a+6

1

+

+ 26

+

a + 36

3.4

.2

1

Exx. 7-14 are convergent or divergent.

series in

o.lll

1

a

+

.

10

.

. .

255

5.6

3

1

7

_5_

~ 13.

7

J_ + __ + _?_ + ....

-A-

H.

Test for convergence or divergence the series whose nth terms are 15.

19.

-^_.

is.

-JL_.

17.

/t

-. +a

20.

n-

21.

N/rc

2n

.

/_4.

is.

_______ 2 + l-n.

22.

_

V^-fl-s/n 8

1

.

|

For what positive values of x are the what values are they divergent ? 23.

25

'

l+2x + 3x* + 4x* + ... .

Z

1

29.

32.

terms

--

x

2

30.

.

-

-

31.

.

:

refer to series of positive terms.

Zun

is

34. If

Hun

is

35. If

Eun

is

-2! ^ -^t-

1

convergent and

convergent, so

un

vn

^t-1 ^ ^=^ vn un

divergent and

[After a certain stage

un z

is

un
for

,

37. If

2un

z

is

convergent, so

Find the sum to n ^ 1.3.5 ,

+

ww ->0

2

.

2.5.6

then Zt'

is

is

convergent.

divergent.

un 2
/.

;

).

+ ~~^-~ -f ... 5.7.9 4

3 i

1.4.5

-

n ^ m, then Zvn

Zun f(\ +un is Sun ln. [For ww /

infinity of the series in

3.5.7

for

.

36. If 27wn is convergent, so is

A 41)

2

Explain why D' Alembert's test cannot be stated thus A series of positive convergent if the ratio of each term to the preceding is less than unity.

33. If

-

a;

is

Exx. 33-37

38.

-

..... 26

i

3a

2a

1

7.8.9

Exx. 23-31 convergent, and for

+ */2 2 4- x 2 /

24. 1

.

,

4.5.6

1.2.3

series in

_i

i

3.6.7

.

Exx. 38-41. 39. -

-

1.4

-f :r~

2.5

3.6""

.,12

41

3

-i

1

.3

.

7

3 .6

9

u

--

i

.

5

.

7

.

11

ABSOLUTELY CONVERGENT SERIES

256

TERMS ALTERNATELY POSITIVE AND NEGATIVE

Theorem.

14.

alternately positive

The

and

u^-u^ + u^-

series

negative, is

...

,

in which the terms are

convergent if each term

is

numerically

less

than the preceding and lim u n =^0.

We

Proof.

and

have

can be written in the form

this

= w l-( w 2- w 3)-( w 4- w 5)----- w 2n

*2n

Now

Wj-i/a, w 2 -i/ 3?

From

(A)

it

3

-w 4

,

...

are

all

therefore follows that s 2n

positive, for

is

and increases with

positive

n,

(B) always less than the fixed number u v Hence, s.2n tends to a limit which is less than MJ (XV, 3).

and from

it is

seen that

2n is

$.

Again, S 2ni l ~s 2 n + u 2n+i an( l l* m U 2n+i same limit as s 2n The series is therefore convergent, its sum .

therefore

\

tends to the

s 2n+l

.

Ex.

Show

1.

that the series

1

-

^ 4- -3"--

4

+

...

is

positive

and

less

than u v

is convergent.

Here the terms arc alternately positive and negative, each term is numerically than the preceding and lim l/--0 hence the series is convergent.

less

;

SERIES WITH TERMS EITHER POSITIVE OR NEGATIVE

Absolutely Convergent Series. The series 27w n containing positive and negative terms, is said to be absolutely convergent if the series 15.

2

1

un

,

is

convergent.

|

Hu n

convergent and

E

u n divergent, then or conditionally convergent semi-convergent. If

is

Thus the series 1

-f

series

f+

\

;

+

\

l-| + i-i + --+

ally convergent, for

An

Theorem.

Zu n

Let

|

.

.

.

it

is is

|

Su n

said to be

is

absolutely convergent, because the

is

But, 1-^4-

convergent.

convergent and

1 4-

+

1

3 4-

3-

1

-$+... 4-

. . .

is

is

condition-

divergent.

absolutely convergent series is convergent.

be absolutely convergent

;

then by definition

Z

\

un

is

con-

|

vergent.

Now

i/

n

4j

un

=2u n

\

or 0, according as u n

positive or negative.

is

Therefore every term of the series (u n + u n \) is term of the series 272 u n |. corresponding convergent \

>0

and

is

< the

1

Hence Z(u n + u n \

convergent.

\)

is

convergent and consequently, by Art.

4,

Zu n

is

PRINGSHEIM'S THEOREM NOTE.

if

we say

is

absolutely convergent, and not that of 2u n itself.

,

\

Zu n

that

another series, un From Arts. 11 and 12, \

it

a fixed positive number

is

absolutely convergent


less

than unity.

Since s n

of the positive integral variable n,

Zu n

sufficient condition for the convergence of

Corresponding it

to

must be possible

//j/7/ HUM

any

to

-jo t-o,

j

Thus

Zu n

if

is

positive

m

find

7? it n>

Ex.

is

p 1.

to

tend

if

y

-4^

convergent and

allowed

Apply

Mft-fi

I

I

and in particular u n ^0. But the condition (A) is unless

so that

-4-

'*n~f 2

is

'

*

in

to oo

-U II

* *

'

<^~ ~-

f

ensure the convergence of

Hu n

any way whatever.

11 4 + '"~

2 "3

7f w

'

p ~

n+

1

f "'

h

4-

1

n+2

+...+

n+p

For any fixed value of p, p/(n H />) -> as ?i-> oo hence, Rn p cannot tend to zero, and so the series

>

n~\-p

But

.

is

t

Pringsheim's Theorem.

'

lp

/i'

1

1

Here

tive

I

n-f j?|

the general condition to the series

+

17.

and

:

any integer,

sufficient to

'not

a function

e that

?/

p

as follows

is

is

(B), the necessary

we may choose, however small, u^in^for every positive integer p,

number

'}/

j,

a

after

|uj<*,

or if

General Condition for Convergence. by Oh. XV, 6,

16.

if,

i_

,

?p is

Zu n

follows that the series

certain stage,

where k

statement of this theorem, he should we assert the convergence of

If the reader fails to see the point in the

note that,

257

// Su n

if

2

ri

p

,

then p/(n

-f

p) ~>

1

;

not convergent.

is

a convergent series of posi-

decreasing terms, then nu r ~>0.

For

if

p

is

any

positive integer

as W->OQ P u m+v< u m+l+ u m+2+ + u m+jT> = Let p m, therefore mu 2m >0 and 2mu 2m->0 i.e. nu n -*Q, when n = 2m. and Let p = m 4- 1 therefore m + 1 u 2m +l~-> ;

,

(2m +

1)

u 2m+l

(

)

= 2(m + 1) u 2m+l - u2m+l ~~>0

This proves the theorem, whether n Ex.

1.

Use

Hero nun and, since

this

n-

it is

;

-

n

theorem

to

show

=1, and does not tend to it

nu n

when n = 2m -f

1

.

Hence E- cannot be convergent n

;

i.e.

>Q,

odd or even.

is

that the series

a series of positive terms,

;

27

zero.

-

n

is divergent.

must be divergent.

REARRANGEMENT OF TERMS

258

of

Zu n

sum

Removal of Brackets.

Introduction and

18.

Denote the

be arranged in groups without altering their order. terms of the nth group by v n then

of the

,

2u n

If

(1)

Let the terms

is convergent, so also is

Sv n and

these series have the

,

same

sum.

For

Then

= !*! +

$

let

for

any value sn

of

= m

+ ...+M n

2

n we can find

or

t

,

sn

t

= v l +v% + ...+vn

n

.

m so that either

= tm + (ww4l + w m+2 +

.

..u m + 9 ),

where the terms in the bracket are included in the group vm+1 If

Zu n

is

convergent, fi

n,p

w n+1 + t w+2 +...ttw4 ->0 as n~oo, therefore s n - Z m-> 0. as n ~> oo

:=s

.

m~> QO

also

J>

,

Hence Svn converges to the same sum as Su n If

(2)

Sv n

is convergent,

Zv n

For, given that for all values of p.

Denote

(3)

as in

Ev^

.

|

un

is

Eu n

is

.

not necessarily so.

convergent,

we cannot conclude that

by u n \ and suppose that the terms of Zu n

\

f

Rnjtf->Q

are grouped

'

Let v n be the sum of the terms of the nth group. (1). is convergent, so also is u n (Art. 5) and, as n -> oo

Then, if

'

,

;

Therefore, as in

(1),

Zv n

Rearrangement

19.

converges to the

of Terms.

same sum as Zu n

.

The following example shows

rearrangement of the terms of a semi-convergent series may alter its sum. and a rearrangement of the signs may change the series into one that

it

which Ex.

is

1.

divergent.

With regard t 1

I

sum

Lft's be Ihe

to the, series I

I

oj the tern i -convergent series

-5-4*3-5

I

I

1 + 5-

/ri

i

1

1

1

i

1

^r^rr^r-

,

,

........... (B)

......... (C)

obtained by rearranging the terms and the tigns of (A) respectively, prove thai (B) is convergent with a

(i)Let

then

w

sum

- 1 -5

{s

and

(C)

w

divergent.

-r- ---,-... to M terms.

^

or

w

= l -~ --H-- -- -~ + ...

to

n terms

:

SUM ALTERED BY *. 1

1

1

--lim9 2n =-*. Hence, the second (ii)^t

n

*>(

then

converges and

series

<

Also

= i + l_l + l + l_l

-+-

-

}+( -+- -O

D

I/,

Now

the series

+- +-

1

t

sn +%

&U tend to

oo

-f

is

. . .

3

2i

Hence the

.

1

divergent

;

series (C) is

Theorem.

// the terms of an absolu remains convergent and its su?

the series

Denote the new

We And (u n

have un since E\

un

-f |

Let

E\u n

\

|),

-f

\

|

un

v n) so tha series by u n = 2u n or 0, accor is

\

a convergent se

for its terms are

<

the c

Z\u n \=s and Z(u n + \u n \)^ and (u n + \u n \) -are series o

unaltered by any rearrangement of tern

27(^ + 1^1) =

2\v n \=s> Ex.

Find a range

1.

^

of values of x far whicJ 2 z l+(x-* + (x-x*)

-

)

can be arranged as a convergent

series of ascendi

Expanding the terms of (A) and removing h

Now,

|a?-a^|
if

justify either the

However, by Arts. convergent

where

/

a?

;

16, 18 (1), 19,

and, by Art.

=| x

|,

is

convergent

is

NOTE.

be the case

that

;

x'<\(*J$-l) which

each of these

5, this will

t

or

is,

if x'

+

-tON/5-

the required range.

What

has been said does not preclude

can be made over a wider range.

ERROR

7

^arching for the sum of an infinite content with an approximate value.

and >,ries

approximate value of

i

Rn

sum, denoted by Thus we have (A). its

,

called

is

It should be

5.

s.

)er

limit to the numerical value of the

number which

o find a

|

Rn

cannot \

t series

w 4-...+(-l) W w +...

,

Here

ive terms.

n+ 3 Brackets

Hence lij less

is

positive

the

than w n

less

than u n ^

error in taking s n as _ _ i

ror in taking x

1

an

.

- x2

/2 as the

sum

of the

than x 3/3.

3ss

.n

and

which

erms are

all

the terms are positive or in the same sign. This is

all of

n often proceed as 3

--K"H 3

xn

n

+... where

follows.

0<^<1.

The

series is

(

n +1 numerically less than x / (n

le of s is

;ence or Divergence, lue of

Rn

+1

) (

1

- x).

For a con-

for a given v&lue of n, the

more

ge.

ilue of r y the

2

+

more rapid is the convergence

convergent, but in order to find mals, at least 1000 terms must be taken /4

. . .

is

fore say that the series converges slowly.

SERIES OF

-v"

Next, consider the of a million terms

series

is less

1 -f

|-

f

_

than 20

(st

therefore said to diverge slowly. In prat, the reckoning so that we have to deal

w

For instance, the series, 5 = 1 - J J By grouping the terms, we find '

1

This series converges more rapidly 8 5 *-2[t + i(i) + i(i) +.

find that

rapidly than a geometrical series '

22. Series of

Complex Zx n and Sy n c

that the series sn ,

r n be the sums to n

cr n ,

Ch.XV,10,

tei.

lim Sn

= lu

and we say that z n converges ti Thus Ez n is convergent if Zx n a

(2)ByCh. XV, U^t of Zz n it

is

as follows

must be possible \s n+v -s n (3)

to

Definition of

As

zn

find

tha

\<,

terms and 2\

Corrcs

:

P

is \

for real sei

convergent.

then 2\

zn

is c \

Consequently (4) It is

impo.

2x n and Zy n9 an The

first

Hence

if

part

Z \

c

xn

\

The theorems c brackets and the c

(5)

of

complex

series.

This

1

T

;

'

,

FOR SERIES

is

..+zn

complex, the series

~l

and when

1

convergent,

z

if |

+ ...

|>

0, (3), 3

sum

is

1, z

its

n does

sum

is 1/(1 -z).

not converge to

*->0 and

s n -*l/(l-z),

1/(1 -0).

0s2tf + tsin2fi) + ... to oo

re

- r cos

+

ir sin

1

- r cos 6

that these results are also true

1

1-5

FIG. 43.

then

w n->0.

then we can find

ULTIMATE SIGNS OF TERMS

Now

m m+I m+2 + +

the series

-

H

. . .

The Binomial -

1

is

~ 7 -~ + n(n-l)(n-2) When n

called the Binomial Series. its

sum

is (1

which we denote by

+ x)

n

In

.

+ ti 1 +

tt

2

+

all ...

stage, the terms are alternately positive

'

if

+ t< r +

...

.

1).

which

1, 2, ...

,

,

Convergence, etc. We shall prove that the series 1 ; and when x = l, the series converges if vergent if x (2)

\

\

Denote the

Proof.

n-r r+1

ur

and the

<

n = -1 and

finitely if

oscillates infinitely if

is

-.x,

r->oo

as

series is absolutely

;

convergent

x = l.

. . .

is absolutely

n> - 1,

con-

oscillates

then

,

therefore lim

x

if

greater than n, negative, or all

w< - 1.

by UQ -h u + u2 4-

series

a certain

negative, or they all have the

um+2 ... are alternately positive and have the same sign according as x^Q. ,

+

the beginning, or after

and

m is the least of the numbers 0,

the terms u m um+l

x

a positive integer, the series terother cases, it is an infinite series,

same sign according as x^O. For w r +i/ M r = (n-r) xj(r +

Hence

of n, the series

+ 1) (n-r *

...

is

From

Ultimate Signs of the Terms.

(1)

w n->0.

For any given value

Series.

9 + nx + n(n-l) -Vs ' ^ +

minates and

divergent, therefore

and

uju n-> 25.

is

263

|

\

<

*r+l

1.

^

and u r have opposite signs, so that, after a If r>n, ur Next let certain stage, the terms are alternately positive and negative. Also

r-n (i)

Also |

n>-l,

If

u r+ i

|

<

|

ur

(ii)

//

n= - 1

(iii)

//

n<-l,

as |.

-si

\

r

r~>oo,

a,,.->n

Hence, by

the series let

is

n + l=

Hence u r -> oo and the The case in which x~ - 1 |

+-T where a r = Hence, by Art. 24, w r-0.

+ l>0.

Art. 14, the series

1-1 + 1-1 +

-m

so that

..

.

is

and

w>0,

convergent. oscillates finitely,

then

series oscillates infinitely. is

considered in Ch.

XX,

6.

CONVERGENCE OF SERIES

264

EXERCISE XXVII

RANGE OF CONVERGENCE Show

3

. K

that the series in Exx. 1-5 are convergent.

L

_J___ a+ , |

1

a+2 a+3

1

I

L__JL

,

1

1

1 1 L _____

__ ____ __ 22 32 42 52 I

I

.

6.

Prove that the

7.

Show

V'2.4 3

"

,

+4

a

4

J

2

1

-

3

'

5

2.4.6

,

'

62

series

that the series

X -l+x

#2

-

1 4-

,

*

x + x2

j

2

+4-

X^

-

-

...

*

^3/ 3

4-

. . .

is

1

|

convergent 6

is

if


values

a divergent

series.

convergent for

of x t real or complex.

8.

9.

Show Show

that - +

a

-

-a+2

-

- -f

~

a-f-1

a

+3

+

-+ a-f5 -

a+4 ;

...

is

that the series

3

is

2

convergent if /n <4/27. [Consider separately the 10.

is

(i)

sum

odd and the sum of the even terms.]

of the

Find a range of values of x

for

which the

series

convergent.

(ii) Find a range for which the series can be arranged as a convergent series of ascending powers of x.

11. If

0
0-4,

show that the series 2 1 4- (2x cos 6 - x + (2x )

cos 6

- a: 2 ) 2 -f

. . .

can be arranged as a convergent series of ascending powers of x. 2x cos [It is enough to show that with the given condition

\

|

4-

x 2 < L]

12. Let the terms of Su n be arranged in groups, without altering their order, and denote the sum of the terms of the nth group by v n Given that 2vn is convergent, prove that Eun converges to the same sum as .

Evn

in the following cases

(i) (ii)

If the

:

number of terms

in every bracket

If all the terms in each bracket have the

[See Art. 18. In case (i), J?n , j,->0. v n tends to zero, and case (i) applies.]

In case

is finite

same (ii),

and w n->0.

sign.

v n -+Q

;

hence, every term in

SUMMATION OF SERIES Show that the series l/(l+i), and find its sum.

l+z + z~ + ...

13. 2

14. If z n

=

+-~i n

n

show that Ezn

L>

is

is

265 convergent

absolutely

when

convergent, but not absolutely con-

vergent.

Rn

Prove the results in Exx. 15-17, where the given series.

the remainder after n terms in

is

^

15.

For the

series 1

16.

For the

series

+ry-

1

+ 7-5 - - +

r

1

1

. . .

\R n \
,

o

I

that

I

X 17.

For the

series

1

Q
18

AnL--

27

+-J +

-.

20. If

2L

l_

1^:

Exx. 18-23

in

X^ i

2

that ~ x provided

--

tOQ0

^ ]

or

!^

.

as

TI~> ^cording

|

x \$l.

then

a(a

6-2>a>0,

Show

l

|n

+ l)

+

a(a-f l)(a

+ 2)

6-1 ^ ^^

f "

"

'

then

___ __

o

24.

r-

""' i

6-l>a>0,

6

+1

19 ii/

/-

+

tl/

n+

:

L -i- + -l- + _i- + ...to*=- T ? I-x* x-l

|*|>1,

a

23. If

Rn <

1

li + l-i4 + ri~;8 +

22. If

X^

!_

IJ

Prove the results stated JLU<

X^

+ r^ + |- -+...,

+r

6(6

+

that the

+

6(6

1)

sum

l)(6

cos I

^(6"-a-l)(6-a-2)'

n terms of the

to

is

+ 2)

{sin

series

+ cos 2^ + cos 30 + ... - 1}. (n + 0) cosec \Q

Hence show that the series oscillates of TT, in which case the series diverges.

finitely

except

when

6

is

an even multiple

n

[2 cos r6 sin

25.

of

TT,

26.

Show in

(r

+ 16) - sin (r - 10).]

that the series Z'sin nO oscillates finitely except it converges to zero.

when

6

is

a multiple

which case

Show

convergent, onvergent,

by

-sin

that the series z

when

z

= cos + -

1

+ |2 2 +

sin -

4

plotting the points 8 19 s 29

4 ...

$7

.

z

3

+ ...

is

convergent, but not absolutely

Illustrate the slowness slownc of the convergence

on an Argand diagram.

CHAPTER XVII CONTINUOUS VARIABLE 1

it

We

Definitions.

.

passes once through

cumstances x

is

say that x varies continuously from a to b when real values between a and b. Under these cir-

all

called a continuous real variable.

Limit of f(x).

Consider the function

When x = Q,f(x)

f(x)

= {(1

has no definite value, for

-

3 -f-

x)

l}/x.

then takes the form 0/0. In other words the function f(x) is undefined for the value zero of x. For 3 2 2 every value of x except zero, we have f(x) = (3x + 3x + x )/^ = 3 4- 3x -f x it

.

Thus, if x

is

nearly equal

to zero,

f(x) is nearly equal to 3 : further, by the difference between f(x) and 3 as

making x small enough, we can make small as we please. This is roughly limit off(x) as x tends to zero.

what

The matter may be put precisely number e, no matter how small. If x

|

f(x)

Thus, as x tends positive

number

-3 1


meant by saying that 3

as follows

|

therefore

is

\

< 1,

is the

Choose any positive

:

then

x

provided that

|

\

<

\
|/(z)-3| becomes and remains less than any we may choose, no matter how small. This is precisely

to zero,

that

meant by saying that 3

the limit off(x) as x tends to zero. The usual It should be observed that f(x) never attains the value 3.

what

is

definitions are as follows (1) //,

I

any

small, there exists I

where

:

to

corresponding

how

matter

~ f( x ) 1

1

is

<

positive

a number

number

rj

provided only that

and a are fixed numbers, then

I

c

which we

may

choose,

no

such that

x-a |

\

<

is called the limit

TJ,

off(x) as x tends

to a.

This

is

expressed by writing lim f(x)

=

I.

z-+a

Or we may say that /(#)-> I as x-^a, where it is understood that x may approach a from either side, i.e. x may tend toa+Oora-0. Observe that nothing is said as to the value of f(x) when # = a. Such a value

may or may not exist.

In any case, we are not concerned with

it.

THEOREMS ON POLYNOMIALS corresponding

(2) If,

exists

to

any

number

no matter how small,

c,

there

N such that - < provided only that x> N, f(x)

a positive number

1

|

where

positive

267

I

This

c,

1

off(x) as x tends

is

a fixed number, then

is

expressed by writing lim /(#)== Z or f(x)->l as x->oo

is called the limit

I

to infinity. .

'x-+

,(3)

If

any positive number number N such that

a positive

or

no matter how

and we write /(#)-> oo

that f(x) tends to infinity with x,

correctly) lim f(x)

(less

oo

great, there corresponds

x>N,

provided only that

f(x)>M, we say

M,

to

as

x->oo

.

3->oo

The reader should be able to frame for himself corresponding for the meaning of f(x) tends to minus infinity as X-+CG

definitions

'

.'

Fundamental Theorems.

2.

tend to the limits

lim (u v) = I + V, liml/w = l/Z unless

(iv)

infinity,

lim (u - v) = I - 1',

(ii)

(v)

Z--=0,

(iii)

Two Theorems

(1)

= aQ + ax -f a 2 x 2 4-

is

If

number

17

any

. . .

f(x)

a 1


and the

how

small,

provided only that |

Z'

make

for

said in this article coefficients are real

we can find a

positive

|

'
f(x) a Q

then

kx'f(I

\

<

al

|

|

/(k+*)

is

f(x)

\

|,

-f |

,

\a n

-a 1

f

),

<

that c,

<

77.

;

1

that

is if

x

if

a suitable value of

\

\

a 2 x' 2

-x')<, provided

kx'<(l-x Therefore

|a 2 x'

...

x \

k be the greatest of the

let

\

(a^,

then

= x' and

x

let

numbers

so

is

easily

= IV, = 0.

a n xn , then

positive number, no matter

For shortness,

Proof.

and

-f

lim uv

so that

|

If

What

on Polynomials.

holds good, no matter whether the variable x or complex.

Let f(x)

then

limufv = l/l' unless

Apart from some verbal changes which the reader can himself, the proofs are the same as those in Ch. XV, 2. 3.

x which

u, v be functions of

V as x tends to a or as x tends to

I,

-f-

(i)

Let

|

\

<

/(k

+ e)

;

77.

B.C. A.

QUOTIENT OF POLYNOMIALS

268 (2)

//

number

M

is

a positive number, no matter how

m so that

\f( x )

The

letters

\>M,

provided only that

having the same meaning as in

|/(*) !=*">.

great,

-

n -l

+n-2

we can find a

positive

\x\

we have

(1),

^2 1

i

1

X

Choose any number

e

and a n

between

|

1

-

X

For such values

of

x we have

= i|a w

if

theorem

#'>

is if

f(x)

\

> x' n

{|e&

.

n

\

- e}.

then

|,

K

|/(x)|>t|a B

Hence

last

<

that

-

|

In particular, put

Then by the

1

< K+

if

. \

if

x':

m is the greater of the two numbers and

then

f(x)>M

From Theorem

(1) it follows that

if

\x\>m.

lim f(x)

= aQ *

a;->0

= Again, if x y -fa, then y->0 as x-^a and /(a) Hence of y in the polynomial f(y + a). -

lim /(x)

.

1.

// f(x)

_3

=

the term independent

lim f(y -f a) =f(a).

Km /(a;) -3

is

;

(i)

0,

(ii)

,

(iii)

cw

a?-

3

^5' (ii)

Let

a;

= 1/y,

then y ->

*-~

lim

->

as

lim /(X) (iii)

*

Let x = 1

-f y,

then

as x -+

t/

The reader must understand that

a?

oo

->

1

and

and

this is not

a roundabout way of saying that f(x}^a Q when

CONTINUOUS FUNCTIONS 4.

269

Continuous and Discontinuous Functions.

We

confine

ourselves to one- valued functions of the real variable x.

We may say that/(x) if

is

a continuous function of

the curve whose equation

where

xa and x =

yf(x)

is

is

or

in the interval (a, 6)

continuous between the points

6.

This supposes that we know what is meant by a continuous curve/ representation of the curve in Fig. 44 can be drawn without allowing '

A

the pencil to leave the paper, and we say that it is continuous. In Fig. 45, the curve cannot be so drawn near the points where x = x and x~x 2> and

we say that

the curve

is

discontinuous at these points.

FIG. 44.

We curve

:

FIG. 45.

choose the following as the distinctive feature of a continuous As the point P (x, y) passes along the curve, any small change in the

value of x is accompanied by a small change (or by no change at value of y. is true everywhere in Fig. 44, but xl and x=*x% in Fig. 45.)

(This

x

More

precisely thus

P(x, y) any other point on c, no matter how small. values of |

Further,

X~XQ

PQ (xQ

let

:

on

it

|,

is

point on it. Thus the curve

AB is

(1)

(or at the point

XQ

,

for all sufficiently small

continuous

is

if it is

P

.

continuous at every

x=xt

and

x=x2

it is

discon-

.

single-valued function f(x) is continuous at x = x or for # ), if for any positive number e that we may

The

however small, a number provided that x XQ
\

.

\y-yQ \<

if

continuous in Fig. 44, but in Fig. 45

tinuous at the points where Definitions.

yQ ) be any point on the curve and Choose a positive number ofP

said to be continuous at

we say that a curve

in the

not true near the points where

is

either side

Then

the curve

,

all)

77

exists

such

that \

f(x) -f(xQ ) <<=, \

\

This is equivalent to the following f(x) is continuous at x iff(x)~*f(xQ ) as X-+XQ, where x may tend to XQ either from the left or from the right. :

TYPES OF DISCONTINUITY

270 It

important that the student should fully realise all that If /(#) is continuous at z then

is

the definition just given. (i)f(x) has a single

is

implied in

,

definite value

when

x~x

;

h tends to zero through positive values, then /(x + A)-> a number L 19 and/(x - A), a finite number L 2 (ii) if

finite

;

i x =/(# =

and we must have

(iii)

)

."#. 1. A simple case, in which f(xQ ), given by Goursat.

If f(x)

zero,

= x2 +

x2 ...

^-f 1 ~}-X"

and so

f(x)0

x2

+7

:

4-

(1

,^i T* ar)

+

...

2

-

L 19

to oo

,

L%

but are not

all

equal,

is

when #=0, every term

of f(x)

is

all exist,

then,

but, for all values of x except zero, f(x) = 1 -f x lira /(*) = 1, so that f(x) is discontinuous at

2.

;

Thus, /(O) =0, and

# = 0.

*->0

TAe function f(x)

(2)

is

continuous ih any interval if

every point of the interval. If the interval (a, b) is closed (a
f(x) as x-+b-Q.

continuous at

end points are included,

x->a + Q and

continuous at a if f(x)-~>f(a) as

is

it is

at b if f(x)->f(b)

b X

FIG. 45.

Types of Discontinuities, 7

C,

C

as

x->x l from the

,

then y-^y^ as x-^x l i.e.

right,

lim JC

(ii)

In Fig. 45 let y v y/ be the ordinates of from the left, i.e. as x-^xl -0 and y->y%

(i)

^ 7

'^^ 7 JL

~

^ fl/j

and

lim X

At the point where x = x2

Thus y

This

as x-^a^-hO.

f

2 example, 1/x and 1/x

expressed by writing

/

^ 3/1 ~H

supposed that y is infinite. For x = x1 and at # = #2 are discontinuous at x = 0. in Fig. 45, it

is

a discontinuous function of x at

is

is

v = Vo. 7a

,"

.

that f(x) has real values only on one side of the In such a case the curve y=f(x) stops abruptly and/(x) is discontinuous at # (iii)

It

may happen

= point x x

.

.

Taking x,

we

see

(as usual)

Jx

that v x - x

to denote the positive square root of the is

discontinuous at x

.

number

FUNCTION OF A FUNCTION

Theorems on Continuous Functions.

5.

(1) It follows

x=x

from the

called

a,

and product of these functions. = 0. except when (x Q )

// f(x) tends lim tinuous at x^lythen

y=

If

function of a function of

Theorem.

and

x,

is

is

{f(x)}.

and

as x->a,

I

then y

w=/(x),

denoted by

= cf>{lim {f(x)}

if

where

(f>(u)

a finite limit

to

= lim/(x) = Z, then /(z) Z = continuous at x I, therefore

For

are continuous at

(x)

is true for the quotient f(x)/
Function of a Function.

(2)

and

definition that iff(x)

so also are the sum, difference

,

The same

is

271

con-

is

(x)

f(x)}.

+ 7?, where

77

-->0 as

x->a;

also

(x)

lim

Continuity of Rational Functions. If/(x) is a polynomial, by Art. 3, lim/(x) =/(&), where a is any real number.

6. then,

x >a

Therefore the polynomial

continuous for

is

values of x. a polynomial or the quotient of

all

Hence also any rational function of x (i.e. two polynomials) is continuous for all values of x except for such as make the denominator vanish.

The Function x n

7.

// n

.

is

rational,

xn

is

continuous for

all

values of x for which x n has definite real values.

Suppose

(i)

that

x>0.

x

If

is

any

positive

number, we can choose

a, b

0
so that

;

Now

~ both x"- 1 and x n 1

greater of the last If

then

c is

(ii)

|

7/x<0,

At

<,

|

and

bn

let

~l ,

hence

|<&.|w|.|x~rX

provided that

continuous at x = x

x-x |

is

the

|.

small,


=-.

1

k

if

r

.

n I

I

.

x= -y, so that xn = (-l) nyn where it is supposed that Now yn is continuous for positive values of y> therefore

xn has a real value. xw is continuous for (iii)

x

w

~1

any assigned positive number, however xn - x n

is

xn

two numbers,

|

Hence xn

between an

lie

the point

negative values of

x = 0.

If

n

is

x.

positive,

limx n = 0.

If

n

is

negative,

s~>0

xw ->-oo or to tinuous at

x=

-oo, according as x->4-0 or to -0.

Hence xw

for positive, but not for negative values of n.

is

con-

CONTINUITY OF A FUNCTION

272 8.

Fundamental Theorems.

and f(a)^Q, then f(x) has the neighbourhood of

the

(1)

is

continuous at x=*a

same sign as f(a) for

that is to say,

a;

If f(x)

if

values of

all

a-r]
r]

where

x in 77

is

arbitrarily small.

For since /(a)^0, we can choose e so that f(a)-e and f(a)+e have and since f(x) is continuous at x a, we can find

the same sign as /(a) so that 77 f(a)

-

:


For such values

provided that

-f e,

of x, f(x) has the

\

<

77.

same sign as f(a).

// f(x) is continuous for the range the value a to the value 6, f(x) assumes at (2)

and

x-a |

then as x passes from least once every value between f(a)

a^x^b,

f(b).

From

the graphical point of view the truth of this statement

is

obvious.

Let A, 6,

and

We

let

B

be the points on the curve y=f(x) whose abscissae are a and k be any number between /(a) and f(b).

assume

that

any

straight line

which passes between the points A,

B

cuts the curve at least once. It follows that the line y

varies

=k

cuts the curve at least once.

from a to b,f(x) takes the value k at

Hence as x

least once.

y-k

x

a XQ

l

x

f

FIG. 46.

Proof.

Suppose that

tween /(a) and

f(b).

f(a)
of x in the interval (a, 6) for which f(x)&, there are- values of x in

there are values

because f(x) (a, 6) for

is

a
which f(x)>k.

DERIVATIVES Divide the real numbers in the interval

The lower class

A

is

to contain every

273

(a, b)

into

two

classes as follows

:

number x such that f()
any number

in the range a
A

(#!<#' <#2

some

) belongs to class A', for although f(x')
It has been

shown that both

(a, b) is

included,

Therefore the classification defines a real number. x

A

in

for

classes exist.

Also every number in the interval than any x'.

number

f()>k

or the least

number

A

in

,

and every x which

is

is less

either the

greatest It remains to prove that /(# ) = k. we can find If f(x )-k<0, by reason of the continuity of f(x) at x so that f(x)-k<.0 if # Thus f()
,

-<<#

<

Hence x +

belongs to class A, which is impossible, as it is greater than x If f(x )~k>0, we can find e so that/(x)-fc>0 if XQ Q + c. Thus XQ e belongs to class A', which is impossible, for it is less than x .

-^X^X

.

Hence it follows that/(# ) = case when /(#)

It

.

is

NOTE. If f(x) =k for more than one valuo mined as above, is the least of these values. 9. Derivatives.

If f(x) is

easy to modify the proof to suit the

of

x in tho interval

(a, b),

is

and

by the equation

denoted byf'(x).

o:

,

deter-

a function of x such that

tends to a limit as h tends to zero, this limit is

then

Thus/'(x)

is

defined

called the derivative off(x),

The function f'(x) is also called the differential coefficient of f(x). It is possible that for some values of x, this limit does not exist. If x is such a value, f(x) has no derivative at X = XQ This is certainly the case when .

discontinuous at X = XQ

For in order that {f(xQ + h)-f(xQ )}/h f(x) tend to a limit as h-+Q,f(x + h) must tend to/(x ). is

Ex.

1.

Wo

have

// f(x)

=xn

,

where n

f'(x)= lim

is

.

a positive

integer,

prove that f'(x)

may

GRADIENT OF TANGENT

274 10.

to a Curve.

Tangent

Q

be a point which

P

from either

Let

P

supposed to move

is

be any point on a curve and let along the curve, so as to approach

It is supposed that the curve continuous near P, so that Q may be as near

is

to

P

as

we

side.

like.

T'PT

If a straight line

approaches

PQ

P

exists such that, as

from either

makes with T'PT

tinues to approach P) remains less than

we may

Q

the angle which becomes and (as Q conside,

choose, however small, then

any angle

T'PT is

called

47.

the tangent to the curve at P.

This at

P

is

sometimes expressed by saying that

is the limiting

position of the chord

PQ

as

tangent to a

the

Q

curve

tends to coincidence

with P.

Notice that

P,

PQ

The curve

(ii)

is

(i)

Q

would cease

direction

at B,

supposed a chord,

in Fig. 48

thrown from A,

ball hits the

is not

to be

is

to

reach P, for if

Q

were

to

coincide with

supposed to represent the path of a ball which ground at B and proceeds to C. Just as the

strikes the

it is moving in the direction TB and, just after, in the We may, if we choose, say that the curve has two tangents BT and BT Strictly speaking, according to the definition,

ground,

BT'

namely

.

'.

the curve has no tangent at B.

This at

is

an instance

of a continuous curve which has

no

definite tangent

one point.

B FIG. 48.

Gradient of Tangent. on the curve whose equation

1 1

P

.

curve has a definite tangent

PT

Let

be the coordinates of a point is y=f(x). We shall suppose that the at P, and that this tangent is not parallel (x, y)

to the y-axis, as in Fig. 49. Let Q be the point on the curve whose coordinates are (x

+ A,

y-f k).

DIFFERENTIAL COEFFICIENT Draw

QM, and draw PR

the ordinates PiV,

R.

Then

and

since

P and

($ are

Now,

as

PT and h

y + k =/(x 4- h)

gradient of chord

.'.

Q tends

PQ = n\

J^^M

.

h

;

gradient of

is the

;

PQ tends to the limiting position

Pr =

lim^4^W'(*)^-~>;o

Hence /' (x)

OX to meet QM in

k =/(* + h) -f(x)

/.

;

to coincidence with P,

tends to zero /.

parallel to

on the curve, we have

and

y =f(x)

275

fl

gradient of the tangent to the curve y

f(x) at the point

fey).

P

PT

tends to is nearly parallel to OY, f'(x) will be large, and if If coincide with a point where the tangent is parallel to the y-axis, we may expect that/' (x) will tend to infinity. Ex.

1.

Find

the equation of the tangent to the curve,

point whose abscissa

whose equation

y = x?,

at the

is a.

If f(jc)~x^, we have /'(#) the gradient of the tangent at the point 3# 2 .*. 3a 2 , and the equation to the tangent is 3 2 -a) or y = 3a z 2a ;

is

is

(a,

a3 )

.

12.

Notation of the Differential Calculus. In the differential is given a different name and a different

calculus the derivative of f(x) notation is used.

Any number by

z-f 8x; and,

nearly equal to x is denoted if
responding to x thus we have

+ Sx

is

denoted by y + 8y

2/

This limit

is

denoted by

(with regard to x). to x,

In finding

and the process

We

often write

-=^.

,

is

of differentiating f(x)

:

is

-=-

we

called 2Ae differential coefficient of y

are said to differentiate y with regard

called differentiation.

-~ in

dx

and

ctx j

;

where ^~ denotes the operation the form ~7 -/*(o:), v ' *

it is


dx

also written shortly as -~-

ax

.

DIFFERENTIATION

276

Rules of Differentiation.

13.

(1) If u,

v are functions of

d

,..

(l)

.....

(Ill)

dx

,

v

v

'

(.

.

and

in (v)

Proof.

dv

iP\

dx

dx

dx

and

-=-

ax

have

(i) (iii)

and

v

du dv T"""T"> dx dx

,

'

I du 5 -Tu ax

d /1\

.

-

i

i

>

definite values, also that in (iv)

Let

-5-

ax

(ii).

8 (uv)

Sv> 8(uv), etc.,

Su.,

,

^~ are not ax These are

= (u + Sw)

be the increments in w,

x.

infinite,

left to

v, wv, etc.,

Sw and Sv tend to zero as 8x->0.

the student,

( v + Sv)

d

^+

lim(fF

dw dx

dv

dx

+ fc)lim

s

'

l^___Sw_^ du dx

A^V ^ v ^ dx\v/ dx dx 1

N

v

/

^

1

1N .

dx v/

du

v dx

u dv dv_l1 / du 2 v dx~^v*\ dx

fo J3 + l)2-(2-l)8_

'da:

.

w

N

v

d /^N /u\ dx\v/

(i) (

rf

(Sar+l)"

-l

dvN dv\

dx/

5

~(3a: + l)*'

.

du

1

wlim(u-fSw)

u(w-fSu)

We have

u

v^Q.

responding to an increment 8x in

Because

.

dx

ax-u/

du

-=-

~

.

^

..

d

T-""

'

(iV) -y-

,

/

1

assumed that .

,.. % ^

dx

.

.

dx\v/

j

dv -J-

d dv du T- (UV) = U -7- + V -7dx dx dx

d /wN

It is

du Tdx

then

x,

-_*._

2

cor-

POWER OF A FUNCTION Ex.

Slww

2.

277

a product and a quotient can

that the rules for the differentiation of

be written in the forms

du dv dp /!)___+_ p dx u dx v ax ...

where (iii)

1

1

1 dq = 1 du---1 dv u \_^ qdx u ax v ax

.... /

,

an
q = ujv.

p = uv and Deduce

1

that

2

if

= (u1 w2

un )l(vt v2

...

u

wtare

v m ),

...

u 29

t> , 2

..., t^,

...

are functions of x; then

\- Z zdx (i) ,

If

p

uv,

then

,

-V-T- + u-=-

~f-

ax

dx

and,

;

dx

each side

if

divided by

is

p(=uv) the t

result follows. (ii) (iii)

This follows in a similar manner.

Let

7

UjU2

...

^n V

v^

,

...

__...

/ Z_ \J

and by repeated application

I

'

= !^+! du*

u 3 u^

= Z ~ m(-~}; ^^ V dx ^\v dx 8

i ssc

Prove

that,

dx

I

rr U

7

,

*j^

j:

V dx

dx

.

2

un)

w3 55"

dx

... i*

n

and the

a function of

is

rs=1

dx

\ur

result follows.

J

s

3.

I dV dz _ _ dU ~_ _ _

z

'

!

dx

Ex.

1

of the rule for a product,

dU _ 1 d% 1 rf(w U dx ~u dx u2 uB ... un 1

SimUarly J

then

;

onH cHlU.

[/ r

/ /

vm

and -~

if

y

let

n be a positive integer

x,

then for

exists,

all

rational

values of n,

Let

w

y

n \

and . f t

(ii)

Let n

(i)

Idw == 1 dy 1 dy _. _^__^. w dx y dx y dx

-m, where

m

is

<

Hence, by^

wy = 1

.

and

m dy 1 ^w ~ y- +- dx = w dx y

,

,-r

(i),

dy __j____ y dx

a positive integer _

/.

,.

then

;

1

/

w~y

y

.

y^

.

n terms) = ndy T -

...

y (n factors)

;

.

y dx

;

then w=y~~ m ,

m Idw -~ 1 dy + m -|~ =0. w dx y dx - -=-

~1 rfw -r- =-

or

.

w dx

y dx

then

wy*

P (iii)

Let n=p/q where p and q are integers 'u

cr

Hence, by J

That

is,

we have

/-\ (i)

w or

/--v (11), ' v

in all cases

d2

^ c?y i-^^-fl-^ wdx zdx ydx

g

-

1

rft^

-=-

=-

*f oo?

NOTB.

;

~

;

* (

oo;

and

and

t)

9

say.

- dw ndy --. 1

w

dx

y dx

and multiplying each

^ )=ny

wQ =y =z

side

by w(=y

n ),

-i.*

It should be carefully observed that there

00; is

always a factor

y-,

unless

y=s.

FUNCTION OF A FUNCTION

278

Function of a Function.

14.

is

If w=/(x)

(1)

and y~(u)

a single-valued function of

x, continuous for

a single-valued function of u, continuous for the correvalues sponding range of of w, then yisa continuous function of xfor a^x^b. Let x be any value of x in the given range, and w , yQ the corresponding is

u and

values of

y.

Because y = (u) is continuous at w we can find a number 77 such that

for

,

any assigned positive number

e,

provided that \u-uQ \\< is continuous at x we can find r\ so that >

and because /(x)

,

and consequently

\u~UQ\
\y-yo\
if |

x-x

]
This proves the theorem.

If the derivatives

(2)

values of u,

and

if

f (x) and

a
(u) exist for

'

F(x) = (u) = {f(x)},

fW-f <) ./<>,

then

ttati,,

|=|.g.

Let 8w, Sy be the increments u and y corresponding to an increment 8x in x, then

Sw^0,

if

8y^8y Su Sx

du ax

8u

.

--->- and

As o#->0, fi

oo;

~8u'

du -r~ dx

since

and therefore

ou

w

Sx

.

is

.

a finite,

Sw->0

-+--. du

Consequently

dy = .. 8y _. 8w dv dw r ax lim/-lira^--^-ow ox du dx

,^ v (B)

/

It

may happen

that Su =

for

some value

_

In this case we must have 8y = 0, for otherwise 7

7

= 0.

hence

and then equation (A)

of Sx,

does not hold.

Also

-^-

= 0,

8w = 0.

for

~ would not be

finite

U/U

Thus the equation

dy ^dy du dx du dx is

true also in this case. Ex. (i)

/Y (11)

1

Find

.

Let

,

u=ax + b,

*

Let

^

ti

a

(i)

then

70.1

-f-

when y =

(ax

y=u n

,

1

ox% then y =u

+

and i

,

n fc)

and

,

dx <^V -

dx

(ii)

=-~ du /

=( \

when y = .

dx 1

=nu n~ l .a=na(ax+b} n~ l

\ )

u*J

'

v

~,

.

26x =

26a;

(a+bx*)

2

.

.

:

SIGN OF FIRST DERIVATIVE

//+,!, a 6

&.*.

We

279

'

2

2

regard y as a function of x, and differentiate both sides of the given equation

with regard to x thus 9

2k 2y dy ~ + 7? -5? =0. a j cr eta

15. Derivative of

xn

// y = x

.

n wfore

w

is

any

rational, then


dx x

If

Proof. of h,

is

positive, x

and by Ch. XIV,

Also

a;

71

"1

is

x

is

positive for all sufficiently small values ~ 1 -1 'and nhx n l . lies between

is

nh(x + A)*

a continuous function of

dy -ax If

+h

+ h) n - xn

10, (x

Therefore

x.

--

..

(x

hm

v

+ h) n -xn '

h

h-+Q

- u, then = ( y

= negative, let x

l)

n wn

and

r~

n n l when n is a Or,* having proved that the derivative of x is nx positive integer, we can prove that this is also true when n=p/q as follows.

p

j

Let y = a^, therefore y* = x p ,,

f

therefore

16.

qtf^

.dy 1 1 /- = px*>dx ^

.

Now _

.

and

Meaning of the Sign of

7

q

~T-y

= qy9 " 1 ~/-

-

~ dy = px p l = pdx qy- 1 q -/

.

,

xp ~l

Z^v

xq

Iff(x )>0

f'(x).

9

--*

= -p x ^ q

then for all values

^

x of x in the neighbourhood of x /(x) 0. ,

sufficiently small values of /

Similarly

t//

(x

A,

Hence

/(x + A) ~/(x

)<0, /or every x in z

.

S

JAe

a5

has the same sign as ) neighbourhood of x

for

A.

,

5x

.

NOTE. If /'(a;)>0 only for the single value # of #, it does not follow that f(x) increases steadily as x increases through XQ For if x t
that f(Xi)
17.

and f(x2 )
)t

and we cannot conclude that f(xi)
Complex Functions of a Real Variable is

See also Art. 13, Ex.

3.

+ i<(x), + i'(x). f'(x)

x. If y=/(x)

called a complex function of x and its derivative rules of Art. 13 obviously hold for such functions.

then y

The

)

is

MAXIMA AND MINIMA

280

Higher Derivatives.

18.

the derivative oif'(x) exists, it is called the second derivative or the second differential coefficient of f(x). If

this is written in

y =/(#)>

If

of the forms

any

So by n differentiations (when possible) we obtain the nth derivative or the nth differential coefficient of f(x) which is written in any of the 9

forms

The, following are important instances. (i)

If

(ii)

If

t/

= zn

-t = nxn - 1

,

,

= a^xn + na^*- 1 + - ~^~1

a 2xn ~* +

(

f(x)

4-

. . .

a n)

L?

then /' (a?)

=n

a^f

1

+ (n - l)a^- 2 + (n

Or, using the notation of Ch. Ill,

= n(a f'(x)

then

and by successive

,

a l9 a 2

,

...

2)

1)

^" f(x)

a 2^~ 3 +

= (a Qy a v

a n _Jx, I)*-

a2

,

...

. . .

+ an

a n $x,

w l)

,

...3.2. (a^-f a x ),

Maxima and Minima. We say

that/(z) has a

maximum

XQ when f(xQ )>f(x) for all values of x in the neighbourhood In other words, if a number exists such that f(xQ )>f(x) when

at X

then /(x replaced

)

is

by

a

<

maximum ,

,

1

differentiation

Hnfn- 1) 19.

1, if

"

value of /(x). If in the preceding, the sign then f(xQ ) is a minimum value of (fx).

Maxima and minima

value of

>

z

.

is

values of a function are sometimes called stationary

or turning values.

A

necessary condition that f(xQ )

may

be a

maximum

or a

value of f(x)

is that f'(xQ ) =0. This follows from Art. 16 but the condition ;

is

not

sufficient.

minimum

TEST FOR MAXIMUM OR MINIMUM If f(x

)

a maximum

is

281

then for sufficiently small values of

value,

A,

Hence by Art. 16, as a increases through x the sign of ). from -f to -. f'(x) changes Suppose now that /"(# ) exists and is. not zero. Then since f'(x) decreases as x increases through rc f"(xQ ) must be negative. /(z + A)
,

,

Similarly sign of f'(x)

be

minimum value, as x increases through z the f(xQ ) must change from - to + and if f"(xQ ) is not zero, it must a

is

if

,

,

positive. If the sign of

/

-

(x)

does not change as x passes through x

is

f(x)

,

neither a

maximum nor a minimum. The /" (x = will be considered ma 7 be a maximum, a minimum or neither of these. later. In this case /(# To illustrate this geometrically, in Fig 51, the tangents at A, B C to case in which

)

)

9

the curve y ~f(x) are parallel to

OX.

FIG. 51.

The

sign of f'(x) for different parts of the curve

has

f(x)

maximum values

Fig. 52 represents the

C and

graph of

t/

a

minimum

= x3 -f 1. Here

as indicated,

Thus y 1.

Here

Thus

-=^

= 3z2

dx

is

neither, a

Search for

maximum

nor a

,

so that

-~=0

ax

maxima and minima

/'(*)=6*

a

minimum when x=0. values of

-6*-12=6

/'()=0

if

*=-l

or

2.

As x increases through -1, the sign of f'(x) changes from increases through 2, the sign of /'(#) changes from - to 4Thus /( - 1) is a maximum and /(2) a minimum value.

+

to

-

.

Or

and

value at B.

dy =0, but as x passes through 0, the sign of -~ does not change. ax

when

Ex.

at A,

is

thus,

f"(x) = 12s

- 6 =6(2* -

so that f"( - 1)<0 and /"(2)>0, leading

to the

1),

same

results as before.

;

and as x

CALCULATION OF LIMITS

282

20. Points of Inflexion.

APE

In Fig. 53

is

supposed to be part

of a curve represented

by y =/(#). moves along the curve from A to B, the gradient increases along the arc AP and then decreases. If a point

Thus

~

maximum

has a

ax

of the tangent

value at P,

,

2

and therefore

-~=

at P.

dx* If

Q

is

the secant

a point on the curve near P, QP cuts the curve at another

point Q' near P, and we may regard the tangent TPT' as the limiting position of the line QPQ\ when the three points Q, P,

and Q' are very Definition.

minimum

is

At such a

Q FIG. 53.

close together.

A

point

(x, y)

on a curve at which j-

is

a

maximum or

a

called a point of inflexion.

dty -~

point,

dx*

and the curve

0,

crosses the tangent.

EXERCISE XXVIII Find the limits of the functions in Exx. 1-3 tends to oo . 1.

(i)

as x tends to zero

;

3.

'

(1

+*)- (l-

Find the values of 4. lir

Prove that

(i)

10.

Prove that

sin

[sin (x -4- h) -sin

a;

8.

Iim

1;

a;

hm ,.

6.

(ii)

Iim -

a:

= 1.

x and cos x are continuous functions of

x.

=2 sin^A cos

For what values of x

12.

Prove that

r

Iim

Iim cos

11.

L

5.

sin

9.

-7-

sin

is

tan x discontinuous

x = cos x 9

Bm(x + h) -sin a; = 2 sin _ %h

(ii)

-=-

,

.

cos x ,,

v

?

- sin x

y

CO8 (x -f \h) ~> cos x as

-=-

tan x = sec 2

x.

as x

DERIVATIVES AND TANGENTS 13. If

f(x)=z(x-a)

Hence, 14.

m (x-b) n

if

m, n are

prove that

,

positive, f'(x) vanishes for a value of

Prove the following d \r

283

x between a and

b.

:

(i)

dxj

dx (m) 15.

\

(d

n

_

1

_1 / jl l-x*-2\"((l-

(dx)

1

+ ~

}

'

l)

Find the derivatives of (iii)

n (iv) sin z. 16.

Find -%

n (v) cos #.

if (i)

(vi)

ax 2 + 2hxy + by* + 2gx + 2fy + c=Q,

tan n tf.

(ii)

if

x 3 + y* =

(Z-iC

,-

17.

_

2

cZ o;

.

12 =

Prove that

rfy

- d*y //dy\*

-^ / aa: / \^/ 2

.

(

/

^

2

18. If

y=A

19. If

y = cos x -f

20.

cos

Prove that

if

|Lia;

+ J5sin

sin

21. is

r7 y then T-O

"

M

2

2/-

then -~ = iv.

or,

dx

= tan-

[Consider the values of

IJLX,

1

l

/

a/a:,

(x

-

then

I /

10)

(x

+ ia).]

Find the equation to the tangent at the origin to the curve whose equation

y

on the parabola y ~px 2 + qx + r whose abscissae are a - h, a + k respectively, prove that the tangent at P is parallel to the

22. If P, Q, Q' are points a,

chord QQ'. 23. 24.

x(x25.

Prove that the function 2x 3 - 3x 2 + 6x - 5 always increases with Sketch roughly the graph of y = x(x~ 2 for values of a; between and 1.

2

I)

,

and

x.

find the greatest value of

I)

(i)

If f(x)

=(x-l)*(x- 2),

for

what values of x does f'(x) vanish

?

= to x = 2, and show that one (ii) Sketch roughly the graph of y=f(x) from # of the values of x found in (i) gives a minimum value of f(x) and that the other gives neither a maximum nor a minimum. 26. Prove that aa; 2 -f

T

x~

-b/2a

bx -f c according as a

^ 0.

gives

a

maximum

or

a

minimum

value B.C. A.

of

INVERSE FUNCTIONS

284 27.

When

(i)

does the function 3x* - 4# 3 -

36a; 2

-

1

increase with

x

?

Find the turning values of the function, and state the character of each.

(ii)

Show

28.

that the turning values of (x-l)(x -2) (2 -3) are approximately

0-38.

Four equal squares are cut off from the corners of a rectangular sheet of 8 inches long and 5 inches wide. The rest of the sheet is bent so as to form an open box on a rectangular base. Find the volume of the box of greatest capacity which can be formed in this way. 29.

tin,

30. The regulations for Parcel Post require that the sum of the length and girth of a parcel must not exceed 6 feet. Prove that the right circular cylinder of greatest volume which can be sent is 2 feet long and 4 feet in girth.

Show

31.

that the height of the right circular cylinder of greatest volume ball of radius r is 2r/\/3.

which can be cut out from a spherical

21.

Inverse Functions.

Let f(x) be a single-valued function of

x, continuous for the range a
shall

prove that

The equation y=f(x) determines x as a

(1)

function of y. This function

x=/~

1

(y)

denoted

is

mean the same

by/~

thing.

1

(j/),

single-valued

so that the statements

Also the function/"

1

is

continuous

y=f(x) and

called the inverse

of/.

Since f(x) is continuous, as x varies continuously from a to 6, Also f(x) takes f(x) takes every value from /(a) to f(b) at least once. Thus, correevery value once only, for f(x1 ) =f(x2 ) only when x 1 = x2 Proof.

.

sponding to any value yQ of y between /(a) and/(6), there XQ of x such that yg=f(xQ ). l For, taking Again, f~ (y) is continuous. the case in which y increases with x, let yQ be any value of y between /(a) and /(&) and

is

a single value

I

XQ the corresponding value of #, so that

For

+

sufficiently small values of c,

x -

and

Let yQ -k

X Q~6 X Q X^ X and y + k be the corresponding values of y, FIG. 54. and let 77 be the smaller of the two k, k'. As y varies from y -7? to y + 7?> x ^ es between # -e and XQ + C, for a; are within the interval

(a, 6).

f

increases with y.

x~x

Hence |

\

-1

and therefore x =/

<

(y) is

,

provided that

continuous at

j/

|

.

y

PRINCIPAL VALUES

285

- k' as the values of y y decreases as x increases, we take yQ + k and y ~ To show that/~ 1 (y) is continuous at and XQ + corresponding to x If

.

y=zf(a) and y =/(&), we consider the variation

(2)' ^

// J

For ~ax

if

-~- exists

dx

and

is

then ~-

a
dy

Sx and 8y are corresponding increments of x and

is finite,

as Sz->0,

8y

dx T dy

ox

,.

-

22.

never zero Jfor the range y

of y for the ranges

and

.

by

Sv

/,.

1

Si"

I

I

~-

.

dx

then, since

Therefore

vice versa.

Joi,

,.

^-

y,

=1

,

lim^ = l

Idy --

.

dx

The

value of

Inverse Circular Functions. The numerically smallest = y which has the same sign as x and is such that sin y x is called

the principal value of sin~ l x. A similar definition applies to tan^ 1 x. The smallest positive value of y such that cos y = xi$ called the principal value of cos~ l x.

Y

Y

Y

l

jy=s//i"

jr

FIG.

Thus the principal values ,

and that

of cos" 1

x

lies

of sin" 1

between

5^>.

x and tan" 1 x

and

lie

between - - and

n.

Unless otherwise stated, sin" 1 ^, cos" 1 ^ and tan"" 1 ^ will be used to represent the principal values of the functions for these, the derivatives of ;

sin*"

1

a;

and cos" 1 ^ are

positive

and negative

respectively, as can be seen

from Fig. 55. Ex.

I.

Show

Let y = tan~ 1

that

-=-

dx a;,

then

tan~ l x =

l+x*,

x=t*ny and

.

dx -=-=8

dy

'

dy te

=

1

r+?'

BOUNDED FUNCTIONS

286

Bounds of a Function.

23.

Let f(x) be a function which has

a definite value for every value of x in the interval (a, b). If a number exists such that/(z)
M

is

(a, 6),

then/(x)

said to be bounded above. If

a number

N exists such that/(x)>2V for every

a;

in (a, b),f(x) is said

to be bounded below.

A function

which

is

bounded above and below

said to be bounded.

is

Suppose that f(x) is bounded above in (a, 6). Let S denote the aggregate of the values of /(x) as x varies continuously from a to 6. applying Dedekind's theorem (Ch. XIII, 12) to the set S, exists such that

By

we can

show that a number h

no value of f(x) exceeds h

(i)

at least

(ii)

one value of f(x) exceeds any number

number h

This

;

is

than

h.

called the upper bound of f(x).

Similarly, if f(x) is bounded below, a (i)

less

no value of f(x)

is less

than

I

number

I

exists

such that

;

one value of f(x) is less than any number greater than (ii) This number I is called the lower bound of f(x). at least

Ex.

Consider the function f(x)

I.

=

lim

.

n-^-oo 1

If

+nx 2

3=0, f(x)=0.

If

I.

x^O.

Thus f(x) has a definite value for every value of x. But f(x) is not any interval including zero, for by making x small enough, we can make exceed any number we may choose.

f(x)~ljx.

bounded I/a;

in

Theorem bounded in

1.

If f(x)

is

continuous in the closed interval

(a, 6),

it

is

(a, b).

Suppose that f(x) is not bounded in (a, b). Because f(x) is continuous at x = a, for sufficiently small values of x - a, f(x) lies between f(a) - c and /(a)

x in

+ e, where

is

any small

such that f(x)

(a, 6)

is

positive

bounded

possible) not bounded in (x, b). Divide the real numbers in (a,

A' as follows f(x)

is

is

or

is

:

The number x

not bounded in

b)

is

number. in (a, x)

Hence there are values and (if our supposition

into a lower class

to be placed in

A

A

and an upper

of is

class

or in A' according as

(a, x).

According to the supposition, both classes exist and every number in A less than every number in A'. The classification therefore defines a real

number

a,

which

Now f(x)

is

the greatest

number

in

A

or the least in A'.

continuous at x=a, therefore for sufficiently small values of x-a, f(x) lies between /(a-e) and /(a-f e), where is any small is

ROLLE'S positive

belongs to A, which

Theorem

2.*

is

287

bounded in (a, a-fe), and so impossible, for a + >a. This proves the theorem.

Hence f(x)

number.

THEOREM

// f(x)

is

is

continuous in

lower bounds f(x) takes the values h 9

and

I

(a, 6)

and

h,

at least once as

I

are

its

x wries continuously

from atob. For if c is an assigned positive number, however small, there one value of x in (a, b) for which f(x)>h -, so that

h-f(x)<

and

1/{A

upper and

is

at least

-/(*)}> 1/c.

Hence l/{h-f(x)} is not bounded. But if h-f(x) does not vanish for some value of x in the interval, l/{h-f(x)} is continuous in (a, 6). Therefore h=f(x) for some value of x in (a, 6). = Similarly there is a value of x for which f(x) I.

Theorem. Suppose that f(x) is continuous in the closed and has a derivative f (x) for every x such that a
24. Rolle's interval (a, b)

If /(a)

=

a and 6.** This theorem is of the greatest importance. A strict proof (as given on the next page) depends on the rather difficult considerations of the last article, but from the graphical point of view its truth is obvious.

Let the curve y=f(x) cut the x-axis at A, B. Suppose that it is conA to B, and that at every point it has a definite and finite gradient. The theorem asserts that there is at least one point on the curve

tinuous from

between

A

and

B where the tangent is parallel to the x-axis.

FIG. 56.

FIG. 57.

= l -x$. Here f'(x)= -GO at j/ not zero for any value of x between - 1 and + 1.

Fig. 57 represents the curve

f'(x)

is

* This proof

is

and

taken from

** If f'(x) exists for a
must be continuous

in the

open interval

(a, &),

but not neces-

MEAN-VALUE THEOREM

288

Because f(x) is continuous in (a, b), it is bounded in (a, b) and attains at least once its upper bound h and its lower bound /. Either (i)/(z)=0 for all values of x in (a, 6), in which case /'(z) = Proof.

x

for every

;

or

some values

(ii)

of f(x) are positive or negative.

positive and unequal to /(a) or /(&). Moreover, f'(h)=*Q, for otherwise there would be values of x in the neighbourhood of h for which f(x)>h, which is impossible. If positive values of f(x) exist,

h

is

there are negative values of f(x), then I is negative, unequal to /(a) or /(6) and /'(Z)=0. For if f'(l)^Q, there would be values of x near I If

impossible. This proves the theorem. simpler proof, for the case in which f(x) is a polynomial, is given in

which

for which f(x)
A

is

the next chapter.

25.

The Mean-Value Theorem.

and f'(x) where

exists for

is

a<#<6,

m

-f(a)

= (b-a)f (),

some number between a and Let

Proof.



then

(x)

=/(&) -f(x)


is

Iff(x)

continuous for

a
then ........................... (A)

b.

-

--

= -f(x)

-/(a)},

{/(&) -/(a)}.

Also ^(a) = and (b)=Q, hence by Rollers theorem '(x) = Q for some value f of x between a and 6, which proves the result in question.

Let A, B be the points on Geometrically. the curve y~f(x) where x = a,b respectively. The theorem asserts that there is a point C on the curve between

A

and

B where the

If P(x, y) parallel to AB. the curve, and the ordinate is

NP

duced) meets

AB in

Q,

it is

tangent

any point on

is

easily

(or

NP

pro-

shown that Fia. 58.

Corollary. (i) (ii)

//,

throughout any given interval,

f'(x) is always positive, then f(x) increases with if f'(x)

is

(ill)

if

f

(x)

All this follows

two values NOTE.

of

;

always negative, f(x) decreases as x increases

f

is

x

always

zero, then f(x)

from equation

x in the

(A),

is

where

it is

supposed that

interval.

It has not been

assumed that /' (x)

is

;

constant throughout the interval.

continuous.

a, b are

any

AREA UNDER A GRAPH 26.

a given function of x, and that Such a function is '(x)~f(x).

Suppose that f(x) we try to find a function ^>(x) such that called

If

and

I

ntegration

an

and

(x)

t/j(x)

differ

Hence

if

(x) are

denoted by \f(x) dx.

is

that

'(x)

any value

of

(iii),

<}>(x)

is

(f>(x)

-i/f(z)

=a

+ C.

Thus

equations which Ex.

1

.

NOTE.

mean

^ dx

Since

xn

denoted by

j

Integration

Ex.

2.

//

n>0,

+ 1)

j-(x) =/(#),

and



(x)

=

r

f(x) dx

+ C,

are

two

the same thing.

^^

( v

n+

1

) '

x n , therefore (x n dx -

...

dx

is

*-

n+1

+ C.

to bo regarded as denoting the operation of integra-

These are inverse operations.

denotes that of differentiation. '

where

(x)

J

-j-

4

then

/(x)rfx, the general value

J

The symbol

tion, just as

= $'(x),

constant C.

/i

f(x) dx

\l(n

and

by a constant.

r is

or),

any two functions such

Art. 25, Cor.

by

therefore,

iff

is

.

integral of f(x) (with regard to

289

PN

can be applied to find the area of a curve as in the next example. ordinate of

is the

prove that the area

any point P(x

ONP

of the rectangle contained by

9

y)

on

bounded by ON,

the curve represented

NP

and

the arc

OP

by

y~kxn

is

equal to

ON, NP.

FIG. 59.

Denote the area by S and let #, y be the coordinates of P. Let Sy, 8/9 be the increments in y, S corresponding to the increment ox in x. If Q is the point (x -f oxt y + Sy) and QM its ordinate, then rect. PM
therefore

y<

JO

Of

-

< y + Sy

!f y rfa: =

Now S=Q when

a;

=

I

;

and

since Sy -> 0,

.*.

y=

-

.

11 "1" 1

A:a;

n


/b: = --

-j-

(7,

^0 and 5=

where

C

is

a constant.

TAYLOR'S SERIES

290 27. Taylor's Theorem. is

If f(x)

tives f'(x),

Theorem,

known

also

as the General

the first

n

values of x between a and

b,

continuous in the closed interval

f"(x) .../

(n)

(x) exist

for

all

=/() + (6 - a)f () + -(&-

f(b)

/" (a) +

.

.

.

1

i^

some number between a and is

then

"

|>*-1

The proof

deriva-

2 )

f/

is

and

(a, b)

1

where

Mean Value

b.

similar to that of Art. 25.

Let

and then

F' (x)

and

f (x)=

Now

= -

1

p

(b

- x)"-

-

= Q and

^(6)^=0, therefore by Rolle's theorem for some value ^ of x between a and 6, hence

which

(a)

is

Let b

'(#)

=

equivalent to equation (A).

= a + h,

then equation (A)

may

be written

..+

...(B) /^

where

J2 B

=

n

pA /<

w >(a

+ 0A)

and

i.

0<0<1.

For any number between a and a-h/* may be written in the form a + Oh where 0<0<1. In general, the value of depends on n. Taylor's Series.

(a-c, a + c),

and

Suppose that f(x) is continuous in the closed n derivatives f'(x), /"(#), .../(n) (^)

the

a/wi /Aa< |

h |
Then equation

(B) holds.

interval

FUNCTIONS OF A COMPLEX VARIABLE

Rn =

Further suppose that

-.-

h"f

w (a

-f

n-> oo

as

9h)->Q

291 then

,

vL "

For

sn

if

is

the

sum

/(a + h)

The :he

to

n terms

...to*>.

of the series,

s n = 7? n ->

~

so that

When

V(0) +

a

Rn

A) .

-4-

is

Lagrange's form of

equation (A) becomes

0,

AT

1

s n ->f(a

and

series in (C) is called Taylor's Series

remainder after n terms.

...(C)

/

(0)

+

...to

................

(D)

11 This expansion

is

known

as Maclaurin's Series.

THE COMPLEX VARIABLE

The Quantity

28.

x-f ty, which we denote by

z, is

called the

com-

Here x and y are real variables, independent of one variable. oo to +00. another and each capable of assuming any values from As z passes from a value z to a value z v we suppose the variation to be plex

By this we mean that the point z is to move along a continuous curve from the point z to the point z v Thus the variation of z is continuous if the variations of x and y are continuous. continuous.

29. Function of a

X

Y

and

Complex Variable

are functions of x

and

y,

z.

If

we say that Z

is

Z=X+

iY, where a complex function

of the real variables x and y. If z is known, so also are x and y. Consequently X, Y and Z are known. Hence it would be perfectly reasonable to call Z a function of z.

But

it is

usual to restrict the meaning of the term

'

function of z

'

as

follows.

We

use the equation Z=f(z)

to indicate that

Z

is the result

operations performed on z, the variables x and y not occurring we say that Z is a function of z.

Thus 2x -f 3iy

is

a function of z in the

first sense,

of definite

explicitly,

but not in the

and

restric-

ted sense.

For

if.

terms of

z is

known, so

z alone, i.e.

It follows that

is

2x + 3iy

;

but 2x

-t-

3iy cannot be expressed in

without the explicit use of x or

y.

if

one series of operations determines

X and Y in terms of x and y.

CONTINUITY OF RATIONAL FUNCTIONS

292

30. Definitions relating to Limits (1)

way

and Continuity.

To say that z tends to zero (z->0) is to say that z varies in such a that z becomes and remains less than any positive c we may |

\

choose, however small.

a

Thus, after a certain stage, the point z (on an Argand diagram) is within which may be as small as we like. with centre and radius

circle

,

To say that z tends to a (z->a), where a is a fixed number, is to say that 2-a->0. After a certain stage, the point z is within a circle with centre a and radius c, which may be as small as we like. (2)

We

say that /(z) tends to a limit I as z tends to a if, for any posithat we may choose, however small, 77 can be found so that

tive

|

/(z)

-1 1

<

provided only that

e,

z

-a

|

\

<

rj.

In other words, as z approaches a by any path whatever the distance from the fixed point I becomes and remains less than

of the point f(z)

any length e that we may choose, however small. = z or at (the point) (3) The function /(z) is continuous for z e that we however choose, small, we can find 77 may any positive z -z provided only that f( z ) ~f( z o) < <^. I

>

I

z

if

for

so that

1

|

the same as saying that as z tends to z along any path whatever, /(z) tends to /(z ) as a limit. It is assumed that /(z) has a definite value at every point in the

This

is

neighbourhood of the point

z

.

A

The function

if it is continuous /(z) is continuous in a region of the region. at every point Hence if z describes a continuous curve in the region of continuity of the function /(z), the point /(z) will also describe a continuous curve.

(4)

31. Continuity of Rational Functions of z. All that has been and 3, with regard to the real variable, holds for the complex

said in Arts. 2

In particular

variable. (i)

(ii)

(iii)

lim (o

z-o

:

+ az + a^

2

-f

. . .

+ a wzn ) = a

.

A polynomial in z is continuous for all values

of

A

values of z except for

rational function of z

those which

make

is

continuous for

the denominator zero.

all

z.

EXPANSIONS IN SERIES

293

EXERCISE X3TTX 1.

Prove that

(i) (ii)

(iii)

x-x always decreases with x. tan x-x increases with x in the intervals

sin

(-lU,

-

ITT),....

(71-,

|TT),

decreases as x increases from

x

to

TT.

2 3 2 1 2. If a, ax + b, ax + 2bx + c, ax + 3bx + 3cx + d and ax + 4bx* + bcx* + 4
F=^o w 4~4ii 1 w 3 + 3% 2 then

=0

-7-

,

ax

and

ax

=0, so that

U

and F are indepen-

dent of a;. 3.

/ -l-

cos,

If

re 2

and dun

dvn

^=- -

xi_ x prove that

^=

v

Hence show that un and vn are

M

'

positive or negative according as

n

is

odd or

even. [If

#=0, then wn =0 and dui -7

Now

= ~v >0,

.*.

= 0;

vn

%

is

^

also

= cos x- 1<0,

an increasing function,

v

= sin x - x < 0.

%>0;

.*.

rfv

and

^-=w

and so 4.

1

>0,

/.

Prove that for (l)

(ii)

[(i)

Let

t

3, for

all

r

an increasing function,

t;

2 ),

(w a v3 ), ,

sina;

/.x

... .]

y*

3 X* = x- X + -

QO;

to oo

.

r^

be the rth term of the series arid sr the x, s zn

x as

< cos x < s 2n+l and s

r -> oo

2

+i

- s zn

sum

tzn+i

to r terms. ~> as ?i ->- oo

Then by .

There-

.]

-

/

sin (x

rv =sm x + hT + h) -

cos

x

W- sin x - & cos x r^ o

2i

(ii)

= cos - A sin a: - h

cos (# -f- h)

a;

2 -

cos x

+

2i

6.

^!>0;

Use Taylor's theorem to prove that (i)

[(i)

/.

values of x

COSX = l-+r-... tO

every

fore sr -* cos 5.

is

on, in succession for (u 2 ,

X2

Ex.

vx

If f(x) = sin x,

|^n

|<-|^n |->0as

+ ...

;

a; -f- ...

.

h3 sin

3

n-*oo.]

Deduce the expansions of sin 2 and cos x

in

Ex. 4 from Taylor's theorem.

CASES OF BINOMIAL

294 7.

Prove, by means of Art. 22, Ex.

tan" 1

;*;

=# -

o

x--5 -f

x'

5

7

Sr =x-~ + ~-...+(-l) r ~i

{Let

o

o

/(0)=0, and/

/

(x)

0
that, if

1,

x*

THEOREM

;

tooo.

4-

^,

and

/(s^tan- 1 *-^;

1

JLf

= l/(l-fa: 2 )-(l-x 2 4-a:*~...4-(-l) r-1 x 2r ~ 2

so that

,

which is positive or negative according as r is even or odd. Hence, tan~*aj lies between Sr and Sr+1 and 8r - Sr+l so long as x is positive and not greater than unity.] ,

8. If f(x) is

a function whose

first

= [Proceed as in Ex. 9. If

1

|

derivative

show

is I/a;,

then

as r ->

-> 0,

that, if

oo

,

0
x3

x2

x-j + ---...

to QO.

7.]

0<0<7r/6, prove that tan

+ i0 3 and

between

6 lies

+ f0 8

= + a03 -tan

.

show that /(O), /'(O), ; [For the second /"(O), are all zero, and that /'"(0)^0 throughout the interval 0<0^7r/6 if 9a>8 hence, in succession, / // (0),/ (0),/(0), are all positive.] inequality, let /(0) /

;

The Binomial Theorem a positive integer, show that the expansion of (a+h) n by the Binomial theorem is given by Taylor's series. 10. If

11. If

n

n

is

is

any

rational

number and -

n(n-l)^ z 2 + ^

1

.

-

...+

-- -


(i)

show that

n(n-l) ...(n-r + 2) x^T-* '

,

-.

-,

IT

1

-

where (ii)

If

0
that

J r

->0 as

r -> QO

,

so that in this case ...

[For r>n, (l+0x) (iii)

-

If

n - r
and

"'

to oo ...................... (A)

xr ->0 as r~>oo.]

<#<(), explain why we are unable to say that + Ox) n ~r > 1 for r>n, and so far as we can tell, 1

ftr

-+Q.

Rr may not tend to the expansion (A) in this case, another form (named after Cauchy) of the remainder after n terms of Taylor's series must be used. See any treatise on the Calculus.'] [Here

zero.

(

1

To deduce 4

CHAPTER XVIII THEORY OF EQUATIONS

(2), POLYNOMIALS RATIONAL FRACTIONS (1)

Multiple Roots.

1. is

a polynomial, then

(x

For /(#) = (#- a) r

*

(x)

one real root of

Let

Proof.

If
Theorem

2. Rolle's at least

.

a,

j8

where

(x)

is

a polynomial and

for Polynomials.

f (x) =

lies

// f(x) is a polynomial, between any two real roots of f(x) = 0.

be consecutive real roots of f(x) = 0, so that

f(x)=(x-*r(x-p).(x) integers, and (x) is not

where m, n are positive

Then

#-/?.

(2),

is

(x)

divisible

of invariable sign for oc^a^jS

by x-at or

and

/'() = (x a) where

Now

0(a)

= m(a-j8)<(a) and

Therefore 0(a) and 0(jS) have unlike signs where continuous.

Hence

and consequently /' (x)

\fj(x)

between a and

3. (

1)

is

;

moreover,

ift(x)

is

every-

equal to zero for some value of x

/?.

Deductions from Rolle's Theorem.

= // all the roots of f(x)

are real, so also are those of

f (x) = 0,

and

the

roots of the latter equation separate those of the former.

For if f(x)

is

of degree n, /' (x) is of degree n-1, and a root of /'(x) n - 1 intervals between the n roots of f(x) = 0.

=

exists in each of the (2)

//

all the roots

f" (x) = 0,

= f'" (x) 0,

of f(x) . . .

,

and

those of the preceding equation.

This follows from

(1).

=

are real, so also are those of /'(x)=0,

the roots of

any one of these equations separate

DEDUCTIONS FROM ROLLE'S THEOREM

296

Not more than one

(3)

(ii)

be the real roots of

j8x , /J 2 , ... ]8 r

a x =a 2 then a x ,

If 0^:56 a 2 ,

Hence

by if

(i)

if (ii)

;

(iii) if

or

jS1

= 0, any

/'(o;)


Let

.

of which o^,

may

a2 be

real

= f(x) 0.

roots of

roots

off (x) = 0,

be greater than the greatest of them.

be multiple roots, and suppose that If

can

be less than the least of these, or

(iii)

let

=

between two consecutive roots

(i) lie

For

root of f(x)

j8 r

is

one of the set

=

j8 1? j3 2 ,

(Art. 1.)

j8 r .

between o^ and a2 cannot be consecutive

Rollers theorem, a root of f'(x) lies


j31

a 1

then

,

then

,

/J 2

^

cannot be the least root of f'(x)=Q cannot be the greatest.

then

,

and

j8 x

.

j8 r

Thus, not more than one root of f(x)

can

;

in any one of the open

lie

intervals

(4)

// /'(x)

=

Aas r razZ roote,
f(x)=0 cannot have more than

r

+1

has no multiple roots, none of the roots of f'(x) = is a root of /(x)=0, and the theorem follows from (3). If f(x) = has an in-multiple root, we regard this as the limiting case in

If /(#)

=0

which

m

(5)

If

f

(r

is

\x)

It follows

from

(4)

= f(x) that

true in

is

and

all cases.

(r)

the equation / (#)=0 has as many. has at least as many imaginary roots

derivative of f(x)

any

roots, then

imaginary

Thus the theorem

roots tend to equality.

has at

/(#)=0

least

as/'(z)=0. (6) If all

number of

A

the real roots real roots of

...

j8j, j8 2 ,

f(x)=Q by

Ex.

and 1.

have opposite

/(/J 2 )

Find

the character of the roots'of

* /(*)

and /(#)=0 has a

real root

f(x)

same

-1

+

-

1,

j3 x

and

/? 2 ,

faw a4

2.

a // 6

tecw*

oc<

0,

2

+

-

are

1

one between

(uw imaginary

For

and the equation

then the equation roots.

1, 1, 2.

00

+ + -1 and

oa; -f-26a;

= (a,

6, c,

See Ch. VI, 11, Ex.

/(->(*) = ?i(tt-l) 2

f(x)

...3.(a

according as

sign.

and two imaginary

1,

roots.

Ex.

....

= 3a;4 - &E3 - 6* 2 + 24x + 7 = 0.

roots of /'(*)-=

-00

<-

between

lies

signs, or the

Here f'(x) = 12(x*-l)(x-2). The

find the

considering the signs of /(&), /(/J2 ),

single root of /(x)=0, or no root,

f(Pi)

f'(x)=Q are known, we can

of

3i

-f26

+ c =0 has two imaginary

. . .

1.

+ c) roots.

k $ #,

1

w )

~

NEWTON'S THEOREM

Sums

Powers Newton's Theorem. 4.

Let a,

j3,

of

n

let s r ==a r

+ j8r -h... + K T

n ~l

(r

+pzxn

= 0,

-2

+

...

1, 2, ...),

+p

(fi)

*r

+Pi*r-i+Px*r-*+- +Pn*r-n=<>

(i)

sr _ 1

We

..................... (A)

= n,

have identically f(x)

- (a? - a) (x - j8) ... (x - /c),

and, assuming the rule for differentiating a product (Ex.

(B)

2, p. 277),

synthetic division, 2 - a) = xn ~ l + Atf"- -f

f(x) / (x

where the

A

l

coefficients are given

by

A2=a?+p
=
in succession

,

l

;

,

other quotients in (C) are expressed in a similar way,

If the

Q l - s l + np v

and so on

;

and

Q
- ^ 2 -f p^ L -f np2

Q n ^ - * n-1 -f jvn _ 2 -f

Now f'(x)nxn ~ + (w-l)p l

comparing

this

with (D),

-*

"-*

-*

'

1

xn

*~ 2

^ - (n -

on,

it will

be

. . .

+ ...+Q n

................. (D)

,

Q3 = s3 + p^ 2 + p^ -f np3

,

,

+ ;v-2*i + w ^n- 1-

-h(n-2)^2x 1

and so

~"

~*

seen that

where

.........

x~

x-oc

By

then

+p2 s r _ 2 + ...+slpr _ l +rp

sr

1

+;p n =0,

so that $

(i)

Proof,

Equation

be the roots of

... AC

f(x)^x +p lx and

Roots of an

the

of

297

n -3

4-...

+|> w _!

^^ Q 2 - (w - 2)^2

,

. . .

,

;

and

Q n =p n - V

Therefore

2

(ii)

+ --+^--

1

)Pn-l

=

Let r^zn, then multiplying each side of /(a) "- 1

a r -h fta 7

+ p 2 a *- 2 -f 7

. . .

-f-

=

r-w = 0, ^ a rt

and similar equations hold for each root. Hence the second result follows by addition.

(E) r by a

~n

we have

ORDER AND WEIGHT

298 Ex.

an

For equation (A) of

1.

the last article, find the value of

even number, in terms of S Q s l9 ,

If q, c 2 ,

(a

-

r x)

In this equation, substitute

= a - qxa r

a,

/?,

...

y,

in the expansion of (1 4-x) r ,

...

7"' 1

+ C 2z V~ ~...+x 2

is

then

r .

K in succession for x and add, therefore,

Similar equations can be deduced from the expansions of (/3-#) r ,

(y~ z)

r >

etc.

we have

hence, by addition,

5.

where r

,

....

s.2

are the coefficients of x, x 2 ,

...

r

27(a-/?)

Order and Weight of Symmetric Functions.

Leta,

/J,

y

...

be the roots of xln

+p

1

xn

-l

+p&n -*+...+p n = Q ...................... (A)

Consider any function of the coefficients

;

for example, let

Let ^ be expressed in terms of the roots by means of the equalities

Consider Jfo highest power of any root a which occurs in u, and denote it by a*.

Each of p 19 p 2 ... involves a to the first degree from the highest term Pi 2p% and /row. /i/s alowc ,

:

degree of this term, namely

The

order of a

this expression for

5 only, therefore a arises also the index s is the

3.

symmetric function of the roots of an equation

is

defined

as the index of the highest power of any root which occurs in the function. Thus the order of 27a 3/8 is 3, and that of Za/3y is 1.

From what function of

the coefficients

If u

is

has been said

it

will

be clear that

the roots is the degree of the

p l9 p^

...

pn

a symmetric function of order atfv

+ al

xn l

s

of the roots of

+ ...+a n = Q,

s

then a Q u is a homogeneous function of # a l9 a 2 For if the equation is written in the form (A), u ,

which the highest term

The in a,

is

of degree s

,

and Pi =

........................... (B) ...

is

of degree s. a function ofp l9

i/a

>

weight of a symmetric function (u) of the roots a, |8,

...

.

Thus the weight of Za3/?

a symmetric

.

n

in

the order of

function ivhen expressed in terms of

is 4,

and that of ZajSy

is 3.

?2 ==a 2/ a o j8,

...

is its

j> 2 ,

e^ c

...

-

degree

PARTIAL DIFFERENTIATION Theorem.

// a symmetric function

expressed in terms of the coefficients term,

when

299

(u) of the roots of equation (A) is

pv pz

,

sum

the

...

of the suffixes of every

written at full length,* is equal to the weight (w) of u.

Denote any term in u by Pi hp 2 kp^ and in equation (A) write n 1 n The equation becomes y -f A^y"" + A2j92 y n ~ 2 44- A p n ... 0, 1

Proof.

-

x = yl A.

. . .

and the roots of If then (i) (ii) (iii)

;

this are

Aa,

we multiply every

to multiply

A/3, Ay, ....

root by

\w

u by

the effect of this

is

;

to multiply piSpz, p^

to multiply

A,

pfp^pz

1

by ...

A2 A 3

A,

,

,

...

respectively;

by

/

which proves the theorem. down the literal part of Za 2/? 2 expressed as a function of the coefficients. of 2 and weight 4. The only terms of weight 4 are such as contain order p 2 z The first two of these are of too high an order. Pi*> Pi Pz> PiPv Pz Pv Ex.

2

oc

1.

2

Write

,

is

>

2 2 2 Thus we have 27a j8 = ap\p$ 4- bp 2 -f cp^, where a, 6, c are independent of p lt p 2 p 3 .

,

6. Partial

Derivatives.

Suppose that u=f(x,

y, z, ...)

is

a func-

tion of the independent variables x,y,z, ... The operation of differentiatu with on to the x alone varies, is denoted that x, ing respect supposition .

by the symbol OX ^~ respect to x.

,

and ^~ CfX

called the first partial derivative of

is

u with

-.

Sometimes the ordinary symbol -p

the context implying that the differentiation

_ 2

d u ___

,

used in the same sense,

is partial.

_3 fdu\ _ .

.

dxdy

is

I

1

-

We

also write

d*u

dy\dx/

with similar notation for higher derivatives.

For example,

if

u = xm y n

j.nereiore

,

(xx)

^ ^ ox oy

oy ox

It follows that equation (A) holds

when u

*

U

Thus the term

2

2>i j> 2

3

is

is

a polynomial in

x, y.

written PiPiPzPzPzB.C.A.

TAYLOR'S THEOREM

300 It is

proved in books on the

represented by (A) holds

'

when

'

that the commutative property the functions in question are continuous. Calculus

A strict proof is difficult, and here we are only concerned with polynomials. Again,

we

often regard

h^ + k-^-

as

an operator, and we write

du da ( d 9 \ 7 7 ^ h-~-+k-=- = (h-r-+k-^-}u = Du, ox oyj oy \ ox

^^=^^^--5-^

dx

if

h k are constants y

a result which

;

dx 2

\

dyJ

\

dxdyj

we may

dxdy

express as follows

:

u.

dxdy The value of

D n u can be written down in the same way

by the Binomial

theorem.

7.

Taylor's

Theorem

nomial of degree n in

for Polynomials.

//IN

2

7

-r^r

+

:

av

...

a n are independent of

A,

+ TT. a n

'

\^

and

if

X=x+h

Hence, by successive differentiation with regard to

f (x + h) = aj + ha

The

result follows

h

2

by putting h =

2

+ :-- a3 +

(A)

we may assume that

[L

,

a poly

...+~fM(x) n

This follows from Ch. XVII, or thus

f(x + h) a + hdi + where a

is

x, then

~ f(x + h)=f(x) + hf'(x) + f"(x) + & Proof.

// f(x)

(1)

. . .

+

h"-

h,

1 -

-^

we h&ve

an

in these equations

;

,

for

we

find that

EULER'S THEOREM If u z=f(x, y)

(2)

is

a polynomial of degree n in 1

f(x -f A, y + k)

For we

where a

may assume

301 x, y, then

_+*-Dn u

u + Di

(

that

... are independent of A, k. = k A we have a = u', and differentiating 0, Putting with regard to h and r times with regard to k, we find that ,

6

,

^T) /(# + A, y +

Then,

if

d

f

\

X = x + A, Y = y m -r

( d

-f

yfc,

k)=pr + terms

m-r

times

k

A,

(D)

equal to

is

\r

and putting A = 0, A; = 0, we have which proves the theorem. // u?Ef(x,

involving

the left-hand side of (D)

= \SV/ /(^ ^) /tn-r,r(^

^3T/

(3)

y, 2, ...) is

^)> sa 7 ^ e fm~r -

^ r =/m _ r

,

r

(x, y)

=^

a polynomial of degree n in

^

v^

x, y, z,

ZAen

...

1

1

(E)

n where

D denotes the operator

The proof

is

8. Euler's x,

B )*

dx and, in general,

+ l~- +

...

.

oz

similar to the above, using the Multinomial theorem.

Theorem.

y and of degree -~-

h -- + k^ox oy

Let

(1)

u

be a polynomial

homogeneous in

n, then

J ^~

dy

=,

^9 dx 2

y

^^-

dxdy

+ y-^-) w = n(n-l)(n~2),..(n-r=f (x^ \ ox oy/

Proof.

Let u=f(x,

y),

then by Taylor's theorem,

f(x + \K,y + Xy)

du

+

---

>

and

since

u

is

homogeneous and

of degree

n

6.

}

........................... (B)

The coefficients of the powers of A in the expressions on the (A) and (B) must be equal, whence the results follow, end of Art.

r

in x, y,

/(x + As,y+Ay) = (l + A)t

* See

/AX (

right of

PARTIAL FRACTIONS

302 // u=f(x,

(2)

a polynomial homogeneous in

is

y, z, ...)

x, y, z,

...

and

of degree n, then

du du dn z *x a" +y~*-+ y oz ox dy

D u = n(n-l)(n-2) r

wAere

D r tfawdx for J

(x \

^dx

4-

y J

^~ dy

9. Partial Fractions.

.

. .

...

Let

in

)

form. J expanded r

its

/

in equation (E) of Art. 7, the proof

is

be a rational proper fraction // x a is a non-repeated

f(x)l(f>(x)

let

x-a

the fraction corresponding to

(x),

(f)'(a)

(x) rv

= (x-a)^(x) /r

'x-a

^

and

is

.

=

x-a

^>(x)

From

l)w,

to be expressed in partial fractions.

is

factor of

For

4-

(n-r +

...

dz

= Putting h = Xx, k Xy, /-Az, similar to the preceding.

which

2 -5-

4-

+-^nu>

4-

the second equation f(x)

and (rom* the

first,

by

= At(x) + (x

differentiation,

Putting x = a and observing that 0(a)^0, we have

and

f(a)=Aif>(a) Ex.

and If

1.

^(a)=0(a),

therefore

A=f(a)/'(a).

m - 1 /(x n ~l) in partial fractions, where m, n are positive integers Express x

m^n. x-a is any

factor of x n

xn ~

The imaginary roots of xn cos

according as

n

Let

-

1,

then a n

na n ~ l

I

=

1

= 1 and

x-a

x-a'

are

sin^-^ where r = l,

i

n

n

n

2,

...

i(w-l) *

or ^(n-2),

odd or even.

is

a = cos

n

and denote the sum of the

+

*

sin

u

,

then a" 1

= cos

i

n

fractions corresponding to x - a

sin

and x - a"

the preceding,

wr = -

-

# cos rmoL

2

/.

.

a;

r 2

-

tr - 2a:

cos

r(m - l)a where a = +1 ,

-

cos ra

;

n

2?r *

1

by u r then by ,

SYMMETRIC FUNCTIONS OF ROOTS Hence, according as n 1

1

n

x-l

r-*i

Theorem.

10. 7/a,

...

j8,

are

x

is odd, or even,

+r =J(n-

m -l

l(x

n-

equal to

is

1)

303

1

1..J .(zl) n x-l

'

r'

n

+

!. a;

4-1

1--K-V T

r--i

The following theorem is often useful. n w==(a a p ... a n $#, l) =0 awd

roote of

f/te

,

t;s^(a ,a 1 ,...a n ) is

a symmetric function 0/a,

...

j3,

involving only differences of the roots,

,

then v satisfies the equation '

dv

dv

dv

For the substitution x = X + h transforms w =

A l = a l + a^i J 2 = a 2 + 2a

where are

9

a -A, ]8-

A,

...

,

1

A-f a A 2

etc.,

,

into

and

since the roots of [7

=

we have

v = (f)(a^

a l9

= (f>(a^ A^

an )

...

... ^[

n ) ................... (B)

If the right-hand side of (B) is expanded in powers of A, every coefficient must vanish, and by the extension of Taylor's theorem the coefficient of h is the same as that in

dv

dv

dv

a Qh

2 2 ^- + (2a x A -f a Qh ^~ + (3aji + 3a x A + ajh?) ^ )

dv

.'.

^or

J&x. 1.

^

dv

+

. . .

;

dv

a,Q^~ + 2a l ^~ + 3a 2 ^ +...=0. ^a x ca 3 aa2

biquadratic (a

,

a lf a 2 a 3 a$x, ,

,

4

1)

~0, find in terms of

the coefficients

the value of

This function

where p of

a,

j8,

9

q, r

. . .

,

is

of order 3 and weight

are independent of a

,

a lt

...

3.

.

We may And

therefore

since v

is

assume that

a function of the differences

we have l l5

+ 3a 2^

+ 4a 83 ^

=

;

=

;

4 2

Q

This

is

true for

all

values of a

,

a 1$ a 2 therefore ,

= 0,

giving

q-~3p,

r-2p;

The value of 3? may be found by giving special values to a, y, 8. Thus by taking = 1, y=-l, 8=4, we find that ^=32, which agrees with Ex. 3 of VI, 16. ,


APPLICATIONS OF ROLLE'S THEOREM

304

EXERCISE XXX 1.

Prove ating

The equation f(x)==x* + 3qx + r=:0 has two equal this (i) by applying the II.C.F. process to f(x) and x between /(#) and f'(x)=0.

The condition that the equation

2.

equal roots

Show

that this

may

Find the values of a and solve the equation in one

bx 2

and

by

elimin-

may have two

+ 2cx + d~Q.

which ax3 -9x 2 -f I2x -5 =

for

the

5. All are real. .

roots

Ii

!_?

[Use Ex. VII,

Show

if

-

of

roots

has equal roots,

case.

~ The equation xn -qxn m + r=Q has two equal

14.]

that the equation x 3 - Ix + 7

and one between - 3 7. If

(ii)

be obtained by eliminating x from

3.

6.

+ 3bx* + 3cx + d =

if ;

is

ax 2 + 2bx + c=Q

4.

ax*

roots f'(x)

=

has two roots between

1

and 2

- 4.

arid

~ ~ the equation x n +p 1 x n l +p 2 x n 2 show that a is a root of n~2 + n _ n txn-i + n _

+ ... +p n ~0 has

each

three roots,

equal to a,

n ~*

-f ... +Pn -i =0. p 2X 2 equation x -4a^-hlOx + 7a;-5-0 has one (

l)*p lX

2)

(

2

8. Show that the negative and two imaginary roots. 1

9.

The equation 3x 4 + Sx3 - 6# 2 - 24x 4- r = 7 13
has four real roots

positive,

one

if

-8,

real roots if - 8
two

if

r> 19.

The equation f(x)=(x - a) 3 + (x - 6) 3 + (x - c) 3 -0 has one real and two imaginary roots. 10.

11.

The equation x 6 -5a# + 4&=:0 has

three real roots or only one, according

as 12. The equation x 5 + 5o (i) if a>0 ; (ii) ing cases :

3

+b

if

a<0

has one and only one real root in the followand 6 2 -f 108a 5 >0.

13. The equation x 6 + 5ax 2 + b has one and only one real root in the following cases (i) if a, b have the same sign (ii) if a, b have opposite 5 signs and b(b* + 108a )>0. [Deduce from Ex. 12 by putting I/a? for a?.] :

14.

Show

and that

that

;

if

36 2
a>0

these values are all positive if and a <46 . 2 [x 9 y, z are the roots of an equation of the form 0*-a8* + b + 2 2 3 this reduces to one of the form +<(6 Ja ) + fc'=0.] B= + a/3, 2

<

2

#=0

and

if

FUNCTIONS OF ROOTS OF EQUATIONS 15. If

A2

Aj,,

A 3 are the values of A given by the equation

,

jc*^ +

^A prove that these are

all real

order, then

;

y &2

2

z2

^+

^A ^-A~

also, if

6,

,

'

and A lf A 2 A 3 are

c

,

...

+

a-n-i

+

x n- 2

!

ZV^O

show that

+ "- +_____

I

for

U

_i

'

h*~~

|2

|j.

r

= 2,

3, 4,

...

n.

17. If f(x)~0 is a cubic equation whose roots are harmonic mean of the roots of /'(a?) ~0, prove that a 2

The sum

19. If s n

a,

the

is

sum

and a

y,

/?,

is

the

/?y.

of the ninth powers of the roots of # 3 -|~3#4 9

of the nth powers of the roots of

that

x-

is 4

-x

2

4

zero. 1

-~0,

show

H7r s 2 tt-i^0,

a,

/?,

...

equation whose

m and n

according as n

is

* 2n -=-4cos

-. 3

are the roots of (a a l9 a 29 ... n ^/, l) w ---0, are ay-x, py-~x etc., is (a u l9 u 2 ,

roots

where ^^a^x-^a^, 21. If

descending

are the roots of the equation 'vn j______ _________i_

20. If

in

a 2 > A x > 6 2 > A2 > c2 > A3

16. If a, 0, y,

18.

305

U.2 ~~a

x 2 + 2a 1 xy + a 2 y*,

are positive integers

odd

or even,

and

,

show that the un $t, l) n ^0,

...

etc.

m^n,

prove that

where

__-to

#cos(2r-f l)wa-cos _ __ ___ // ~ _ ___ " r j

rc*

22. Let k be given

,

y

^^

__

cos (2r

(2 \_

+

by the equation A J'^ a-fc ^ 6 -

/

f7

If this equation has a rejx?ated root

^, prove that

k^a-gh/f^b- hf/g - c -fg/h, provided that none of another (h), and then

/,

&!

g,

h

is

zero.

= a and

(a

If one of these (g)

-b)(a- c)

/

is

zero, so also is

2 .

Under these circumstances, show that ax 2 + by 2 + C2 2 + 2/yz + 2gzx + 2hxy - k t (x 2 + y* + z 2 ) a perfect square. [Denoting the cofactors of the elements of J' by A' 9 F', etc., the equations are qiiadratics in ^, and the roots of any one of them A' =0, B' 0, C" = Hence if k~kn then A\ B', C' and consequently separate those of J' 0. See Ch. IX, 21, (4).] F', G', H' are all zero. is

CHAPTER XIX EXPONENTIAL AND LOGARITHMIC FUNCTIONS AND SERIFS Continuity of a* and log a

1.

tinuous for

all real

XVII,

(Ch.

2. Ch.

21.)

Exponential

XV,

(1)

The reader

log a x

and Limits.

where

and

(l--}~*-*e. xj

\

m is

a positive integer, then

11 + ~
1,

1

x

l

therefore

If

z->oo

(l

,

~

+

<(

}

1 -f

-

1+

\

7

1

/ /

t

\

XV,

m

5, (3),

in -f-1

"

*

=fl+

<(l +

}

xj

then m->oo and by Ch.

1\ m

/

\

/

1 A

i

-7-I1+-

7

,

m + lj

\

'

v

_

m+l

.

1+1) (\m m/ .(l+j^

e;

therefore

Next

let

x=y +

l,

then as

x->x

,

?/->co

and v

=(i+ii .fi+h V \ y/ yj It follows that if

x->x

or

-

x

,

= lim(l+^)*

,

i

and consequently

9.

(Continued

through real values,

m
lim

con-

continuous for positive values of

is

Inequalities

(l+-)%c \ xJ let

is

reminded that log x stands for log e

is

5.*

As x->oc

For

a>0, the function a x

If

This follows from Ch. XIII,

values of x.

Hence the inverse function

x.

1 -f x) (

x

6.

x.

x.

from

EXPONENTIAL THEOREM If z

(2)

is

any

real

number, z \x

/

= ez

lim (14--) X/ a^oo \

This

is

obviously true

or -

oo

z

if

= 0.

*Aew

\

a*_i

.

hm

is

= obviously true jf a

Now by

Ch. XVII,

=

1/2;,

then as

x

L

y

a^l,

If

ax = l+y, then as x->0,

let

5, (2), 1

lim log a

a;

= loga.

------

x-*Q

This

let

In either case, by the preceding,

y/

(3)// a>0,

.

z^O,

If

according as z.50.

,

307

(1

1

+ y)* - log a {lim

-

y~>o

(1

+ y) y } - loga e = r

a*

lim

1

X

ic->0

;

Aoga

i/->o

therefore

1

= log a.

Here x may tend to zero through positive or negative values, so that x 1 - a~ ax 1 and x x

-

--

each of the functions

tends to log a as a limit as #->0.

been shown in Ch. XIV, 9, that the first of these functions increases and the second decreases as x increases from zero. Hence It has

.,

if

A

x>0,

.

then

l-o-* ---

The Exponential Theorem.

3.

x

2

x

3

.

<


a*-l

all real

-

values of x,

xn

'

I

J

Proof.

The

series (A) is

function of x which

x, n

we

Convergent for all values of x,

shall denote 2

x

by E(x).

If

- --

n

is

where

sum

is

a

8

y

I

its

a positive integer,

, n(n~l)/x\ w(n-l)(n-2)/a\ ~ =l+n.~ + + (1+-) f-) \ 3 n --Vo2 ^(-) \n/ \n/ n/

/-

and

+...

ton + 1 terms

IRRATIONALITY OF

308

Now

jo r


since

E(x) are positive

1/n, 2/n, ...(n-l)/n

than unity, for the values

- 1)

2, 3, ..., (n

of r

we

numbers

have, by Ch.

less

XIV,

1,

n .;

Let

be the

sn

sum

0
n terms

to

I


is finite

let |

x =^x 1 \

I

2.3

Now

and

of (A),

3 -4

>-D

,.

for all values of

then

i.

TT a!l+ - +

+ ,

Xl

;

g Xl

n

^.\ j

x ly therefore

/

x\

Iim^ n4 1 = lim( 1-f

n

.

n '00^

>QO

71

but W-^OO

This result

is

called the Exponential Theorem.

In particular, when x = 6

1

we have

11 + +-+

- ii 1+1+

[2

from which Ex.

it

can be shown that

|3 e

1

5

+

= 2-7182818284

...

$A#tt> JAaJ e i$ irrational.

1.

Suppose that e=pjq

where p and q are integers

t

then

;

where ?

If the equality (A) holds,

account of

4. //

(B).

a>0

Hence

e

and x

we should have

|

is

integer,

which

is

impossible on

cannot be a rational number.

is

any

real

a* = e*

that

R y=an

+2

number. lo 8

a

.

= jE(zloga),

........................... (A)

to say 2

1

3 )

+

.................. (B)

GRAPH OF EXPONENTIAL FUNCTION

An

6.

When a>0 and x

Irrational Index.

309

is irrational,

we take

ax

~E(x log a) as defining the meaning of a*. This agrees with what has been said in Ch.SXIII, 9, for it will be shown later that the sum of the series denoted by E(x) is a continuous function. the equation

6. (1)

Derivatives of a, logx and xf. Suppose that a>0, then by Art. 2, (3), x

-j-a

r lim

=

ah -l

lim a x

r

ax

A

In particular,

NOTE.

e

= a x log a.

x .

Tx be observed that,

It should

a<0, ax has not been

if

defined

for all real values of x.

y = logx, then

(2) If

^ therefore

and

dx -j~

= ev = x

d

-,-

-=-

is,

dx

!

=log G x

i

.

x

has been shown that when n

(3) It

prove that this

is

true for

_d NOTE.

Since

is

y log a,

all real

d y,n

dx it

en log a;

_n __

.

if

rational,

en log a?

instance,

the curve

(1) that, for

TN

y

,

TN

P

:

i.e.

y=ax

until it

at B.

9

the gradient

OX

the subtangent is constant.

l,

A,/'

mark one of the rulings as the axis of y. On this take OA, equal to, say, 4 intervals, as the unit; and on OX, to the left of O, lake Oa OA. Draw

OY

To

:

on the graph meets

sheet of ruled exercise paper and draw a line perpendicular to the ruling as the axis of x ; and

Aa

.

--nx n

the tangent at a point

PN is the

xn = nxn ~' 1

x

= l/log a ordinate of P, then This fact gives a rapid construction for the graph of y=ax , with considerable accuracy. = ex where For for take a and

in T,

ax

we proceed thus

_n _ xn

x

dx

follows that,

is

values of n,

has -been shown in

it

9

ay

fy = 1 xi. * -; that dx x

x

dyjdx

x = ev

meets the first ruling to the left of Without' removing the pencil-point

da

O FIG. 60.

B9 rotate the ruler until it passes through 6, the foot of the ruling next to the left of or, and draw bB until it meets the ruling next to the left of A ; and so on. Find .points on the curve to the right of Y in

fr6m

the same way, and replace the broken line by a If,

in Fig. 60,

Also,

if

'

fair curve.'

OX and OY are interchanged,

the paper

is

turned through 90 in

its

the curve

is

then the graph

own plane, and then turned

over, the curve as seen through the back of the paper y = log x, with the axes in the usual position.

is

the graph of

EXPONENTIAL AND LOGARITHMIC LIMITS

310 7. (1)

For

For

all real

let

f(x)

so that /' (x)

from (2)

values of x other than zero

= ^-1 -x,

(3)

then

according as x 5:

:

//

- 1 and ^0,

x>

//

t/

For

x>0,

then

Hence, if x increases or decreases from zero, and is therefore positive.

0.

x>log

ix2/(l + x)
-l
-|x (1

2


i -*

^

if

Hence

0(#)

x

+x)
/(l

+ &).

T2

_

increases

$

according as x

= a;~log(l + x)-%-

from -1, and since /(0) = 0,

it

0.

x2 X

1 ~r

then

,

and

increases as x increases from -1,

(x)

,

2

I+x

Hence f(x) decreases as x

Again,

(1

+ x)<-|x2

,

1

$

(1

- |-x 2 then

+ x)

f'(r\ Ji vv**'/'

follows that f(x)

+ x).

(1

by taking logarithms.

(1)

then

e*>l +x.

and /'(z) = e*-l,

=

/(0)

zero, in either case f(x) increases

This follows from

and

and Limits (continued).

Inequalities

since <(0)

= 0,

it

^

0. according as x (x) This proves all the inequalities in question.

follows that

(4)

//

x>

greater of the

"2

-'1

^0

and

numbers

and

and

1

I,

1 -f x,

h are respectively the smaller and the then

x2

This

(5)

is

merely another

For any

x2

way

of stating the inequalities in (3).

positive value of n*

(i)

lim a;->oo

(i)

For

all

positive values of

logx _2__ xn Choose

m so

that

0
(ii)

<-

n - w ->ao

m

we have by Art. 1 _1n x__m -l__ ^ ______ x mxn ~ m m

_

and

(log x)/x

2, (3),

,

~>0.

and

= ~ (log y)/yn ->0. log x

lim (x n logx) x >o

*

n

J/->QO

(ii)

.

then a^ x~>cx)

Put x = l/y, then as x~>0, xw

-~- =0, X

= 0.

EULER'S CONSTANT (6)

For

This

is

all values

of

lim x

r,

obvious when r<0.

x

r

/e

= 0.

r>0,

x let e

y

\

If

311

= y,

then, putting -

=n,

'

e*

Now lence 8. >e

as it

x-^oo, y->oo;

and, by follows that x r /e x -+0.

The Manner

which e and log x tend

any positive number. First suppose that #-^oo

Now, by n ;

in

Art. 7,

however great n

,

xn

(6),

w = ->0, (logy)/yw ->0.

since

(5),

may

,

then

e x ->oo

Let n

to Infinity.

and logx->oo.

x

/e -+Q, therefore e* tends

to infinity

faster than

be.

n Again, by Art. 7, (5), (log x)/x ~>0, therefore Zo<7# tends n nore slowly than x no matter how small n may be.

to infinity

,

n If #->0, then -logx->ao., and by Art. 7, (5), (log x)/x~ ->0; hence n log x tends to oo more slowly than l/x , no matter how small n may be.

9.

// M n = l+-g- + J

Theorem.

...H----

+

logn, then as n->oo

,

w n->y

Tt'

y

a fixed number lying between

is

Art. 7, (2),

By

Proof.

n-l

1\

/i

i

log

1+-

)

w/

V

log f

log(n-f

, ,

that

n-

<1

1,

n - 2,

...

2, in

7, (2),

n+1

is,

log

n

1

<-w

n

,


succession for n, we. have

and by addition,

;

n

Art.

.

I

(

& \

.

\

Substituting

Also

,

I

n _l

Again, by

1\

/, < n <-log& (1-n/

6

n

that w n decreases as n increases.

1.

n

1

n

1

and

it

l)
follows that ;

i.e.,

log (w

But log(w + l)-logw>0, hence 00. Consequently ;

V>y NOTE. /

where y

is

a fixed

The number y

0-577215

....

It

is

is

number >0; and, known

as

since

Euler's Constant.

w w
can be shown tnat

often convenient to state the theorem as follows ,

where

->0 as n-^oo

:

,

EXPONENTIAL FUNCTION

312

10. Series for log 2.

Let

i_

l_-i

(-n-i!

,,11

.,

then

1 N

= (y + log 2n + Therefore

2B )

-

(y + log

the

z2

)

- log 2 + e2n - en

~s2n --0, hence

.

s 2n+1 ->log 2.

l+z +

series

r-

+-- +

r

jE

Proof.

It has

fv hen T

...

is

is

sufficiently

is

z.

real,

first of

these equalities

and exactly the same proof holds when

z is

complex.

II

+- = 1 +- -M - =p(cos n n n

For

sum

z n

(

3T

1

real

z,

Its

convergent.

been shown in Art. 3 that the

Let

values of

x (z)=lim (1+-) =e (cosy + 1 siny).

z is real, Z

all

denoted by E(z), or by exp

// z = x + ly where x and y are

Theorem.

For

!

and

called the Exponential Function,

(1)

z3

I

holds

1

w+

The Exponential Function.

or complex,

(2)

1

1 -+... to oo. log2 = l-| + |-... + (-l)"- 71

Thus,

11.

to r terms of the series, to oo

/I

Also s 2n+1

2n --log2.

sum

be the

sr

+



i

sin

),

where

p>0 and ~7r<^<7r.

cr

values

large

of

n,

1+->0; n

P>0,

hence, since

and />cos< = l+-, we have cos ^>0, and -\-n n

^ = y/(n +

Also tan

x \2

<- 0, and ^/tan ^ -> 1 = lim {ny/(n + x)} == y. <^)

)-> 0, therefore

lim 7^^ = lim y2

(n tan

x \2

/

n/ +^i=(l+-) nz \ ( 1+-) n/

w therefore p r /

and,

lim see"

since

we have

lim p n = e*

and

;

- lim

1

sin

(

1

+ tan 2

hence, JE(z) jE (z)

(3) It follows

nor cos y +



\n

/

sec 2 <; T

)*

n = lim

f

x = (l-f-j

+^

)

=

sec"^;

n/

\

w2\in

1

hence

;

JL 2n

2

1

(^

)

=1

>

= lim p n (cos n + i sin n^)

= e?(cos y -ft sin y).

that E(z) cannot vanish for any value of y can vanish.

z,

for neither e*

AND COSINE SERIES

SINE

313

The function E(z) is periodic, and its period is 2i7r. For if k is any integer, positive or negative, the values (4)

siny are unaltered by adding 2&7r to y. Therefore E(z) < adding 2ik7r to z, thus E(z + 2 ikn)^E (z).

is

and

of cos y

unaltered by

t

A Complex

Index. When z is real, e z is a multiple-valued The real positive value of e z is function, except when z is an integer. called its principal value, and it has been shown that 12.

-

E(z)= the

z principal value of e

.

We take this equation as defining the principal value ofe when z is complex. z

This implies that for imaginary (as for real) values of garded as a multiple-valued function.

For

the present,

Thus e

if z

z=x 4

we

and %' = x' 4-

iy,

= E(z) = ex e

thus,

z .

e

take e z as denoting

(cosy-f

i

f

iy

,

e z is to

be

re-

principal value defined by E(z). we write

its

and

sin y),

z,

ez

'

= ex

'

(cos y'

+

tsint/')

;

= ex e x (cos y + i sin y) (cos y' + i sin y) = ex + x> {cos (y + y') + i sin (y 4- y')} = e* +2

z'

'

.

'.

Hence

the

Again,

if

same index laws hold for imaginary as for real indices. a is real and positive and z is real, a z = ezlo ^ a = E(z log

In agreement with

E(z log

this, if

a) and, at present,

we

a>0, we

take a

z

a).

a z as

define the principal value of

as representing this value.

and Cosine. For all real values of x and y, it x E(x + ty) = e (cos y + i sin y). When x = 0, this gives

13. Series for Sine has been shown that cos y

Equating

real

+i

sin

and imaginary

-

14.

mean

yE(iy)

v

2

parts,

we have,

4

t/

-

for all real values of

-*y

.

5

3 ?/

-

Exponential Values of Sine and Cosine.

Using

exactly the same as E(z)> by the preceding, cos y

therefore

+i

sin

cos

y = E(iy) j/

=

(c

= &u

9

ty 4- e"~ l *0,

cos ?/-

t

sin ?/=

sin

y~E( - iy)~e~~ iy

(^ rJL~^li ""*"JWK*"

;

e*

to

SUMMATION BY EXPONENTIAL THEOREM

314

which can be summed

15. Series

by the

Exponential

Theorem. (1)

The

Zu n

series

xn

where u n

,

is

a polynomial in

n.

'-

We

can find a

un = a

av

,

-f

a 1n

...

ar

independent of n, so that

,

+ a 2n(n -l)-f

...

+a r n(n~ 1)

...

(n-r-f

1),

and then s^n

/v>tt

1

spf*

n=rr

= (a + a^x + a2x 2 4Ex.

1.

w

Here 2

in

s

-f

the value of

(n*

n-r

a rx r ) e?.

5)/ n. i

6=n(n-l)(n-2)-f 3n(n- l)+n-f5;

TAe

n 5en*65 2/M n x /(w

n and

a, 6,...

+ a)(nH-6)...(rn-A;)

A are unequal positive

Series of this kind can be Ex.

-f

-f

2

hence,

putting

a

5,

a8 = l, and * = 1 in the result above, we have

= 3,

(2)

Find

. . .

T**

2..

Find

summed

|n w;Aere

t/

n is

a polynomial

integers.

as in the following example.

the value of

Denote the sum by

$, then

Now where

a, 6, c,

d=l.

Thus

rf

can be found by division, their values being

XT Now

=->-'-*)> _ -"' 6

a=

-5, 6 = 5,

c=

-2,

LOGARITHMIC THEOREM

315

EXERCISE XXXI X

Show

1.

lim fl -

that

Find the sum to

}

= 1. Exx. 2-9

infinity of the series in

3579

0,234

3579 +

n

f 4 f ji" [6 "L?"

1*1 23 1

+

Prove that

33

1+2

"" + '-

7

'

s " 4 + ""

^

A

*

A

23

+2+3

ni~

38 4"

a;

,

1

48

4""

" "

1

""^" 2**

3a:

8

:

-=t(4S-16e).

11.

+

~ii"

1

1

:

12.

(

cos 6 ox

13. If

= tan" 1 -

where

16.

-f

^^ + ...4- wa:n + ...

//

Suppose then

(i)

yn = log

rz^

n

(

1

- !<#
+ x) - s n

^

^ 0,

0
that

dx

n terms of

and

1-hx*

" " l

s n~>log (1

is

even or odd.

and s^^ 1 /( w f i)"^ -

+x)

this case of the theorem.

in

to

9

n

The case

sum

x = 0,

when

1+a;

a;

sn

which proves

be the

sn

__ 1 -f

according as

therefore

Let

yn =

then

;

Hence log(l-fx) lies between n n But n+l-n = (~l) ^

Therefore y n

X

i

.

Theorem.

series,

(ii)

then

,

a

Proo/.

the

cosfea;=w

as (Of.

for every n.

as W-^QO,

n->oo Ex.

,

XXIX,

which x = 1 has been dealt with in Art.

8.)

10. B.C.A.

LOGARITHMIC SERIES

316

When x<0, changing

(iii)

the sign of

has to be shown that

x, it

3 Xn X ~ X + -x) = x + + -+...+ 2

0
if

Let y n = -log

therefore y n

>0

-log

(1

-x)-s n9 where

(1

yn =

then

series;

then

sn

x = 0,

when

n

and

~r

n _ x ~ =

(i

then

'

~5

i

xn

|

i-x

~dx

-x

)_ ~

'

(T7*)

r^Ti

2

~

/

0
Logarithmic Series.

^i

Changing the sign

X

and, since

0
s n~>-log (1

-

+# =

x

x

log (1

+ x) - log

/

- x),

(1

x3

./

1+x = n-f --1 1-x

-

,,

.

so that x

-

+

1

=^

4-

(A)

follows that

it

x5 5

\

+

-;

/nv

......................... (C)

,

7,

we nave

2n-fl

7^

.....

3

L

//

0
o: and sn ~x+ o

a;

4-

If

the,

Rn

value of log

"

is

is less 1

~~

x

than

...(D)

5

- +... to

1-fa;

2s n as

-x),

have

/-tv~,i** 71 - 1

3 .

;

of x,

1+x

Writing &

x

;

-...+ (-I)

1

1

("r-lo^

;

x

1

-l
If

X

log

--

'

thus y n->0 and

;

since

__ 1x =

x n+1

1

-

-------

the last expression ->0 as n->oc which proves the theorem in this case.

and

x

this

x" +1

1

17.

n terms of

when x = 0, and

=

2n

x n+1

1

*

Hence

z n <0.

therefore

to

for every n.

n+i x n+

ofe n

=^

1

cfx

Zn^yn-"*

Let

sum

the

is

....

-

2ii\>

n

terms,

---

-

-

+1

1

x

the remainder after n terms of the series,

.

show

that the err.or in taking

CALCULATION OF LOGARITHMS 18. Series

317

which can be summed by the Logarithmic

Series. Ex.

1.

Ex.2.

The

If k

If

is

a positive integer and x < 1, then n xk f x2 1 ^n=oo' x \

\x\
series is

\

find the value of

convergent when \x\

< 1.

*-*

~~~

Let

8

be the sum, then since 3

1

1

*"

1

wehave

TAe method of Ex. 2 can term

sum any

6e applied to

series of

which the general

is

where

P

is

a polynomial in n and

a, 6,

19. Calculation of Napierian

k are unequal positive integers.

...

Logarithms.

The number

e is

chosen as the base of the system of logarithms used in theoretical work. Such logarithms are called Napierian after Napier, the inventor of loga}

rithms.

In theoretical work, log^V means log e

reckoning, log^ means log ]0 jV. The method of applying the equations

Napierian logarithms Ex. T

x

Let

1.

N;

just as, in practical

of Art. 17 to the calculation of

exhibited in the following example.

is

Calculate Iog e 2 to seven places of decimals.

i+*

i

,

.-.=-;

=2;

rt

.-.log, 2

-2

Carrying the reckoning to nine places, we have 333 333 1/3 =0-333 1/3 1/3

8 5

1/3=0-333 3 8 ) =0-012

333

345

679

3 5 ) =0-000

823

045

=0-000

065

321

=0-000 3 11 ) =0-000

005

645

000

513

=0-037

037

037

l/(3

.

=0-004

115

226

l/(5

.

7

=0-000

457

247

I/ (7

.

9

=0-000

050

805

I/ (9

.

=0-000 18 1/3 =0-000

005

645

1/(11

.

000

627

I/ (13

.

000

070

1/3

1/3 1/3

11

15

1/3

=0-000

/.

log e 2

3') 3')

3 13 ) =0-000

1/(15.3

15 )

333

000

048

=0-000

OOP

005

=0-346

573

589

0-693

147

178

=0-693147178 nearly.

BORDA'S METHOD

318

The (i)

this

error in the value obtained for log e 2

We

have taken 2S 8

account

is less

for log e 2.

JL

.

The

Ex.

:

the error on

1 of Art. 17,

than

17 (ii)

By

due to two causes

is

1

>000 000 002.

9

2SQ

error in the calculated value of

is

numerically

2x8x0-000 000 000 5 = 0-000 000

than

less

008.

Adding the results of (i) and (ii), we see that 0-693 147 178 is an approximate value of log,, 2 with an error numerically less than 0-000 000 01. V. 0-693 147 168
can be found in succession by using formula of (D) Art. 17. The logarithms of successive prime numbers may be calculated by the following method, due to Borda, in which the series employed converge very rapidly.

Using the identities

x3 -3x + 2 = (x-l) 2 (x-f2) it will

and

z3

-3z-2

be seen that

(x

+ l) 2 (z - 2)

"

x3 - 3x - 2

~ 1

- 2/(z3 - 3z)

-f ...

Putting z = 5,

6, 7, 8,

log -|

it will

+ -

+ log 5

log 7

Hence

it

log 3

+i

log 5

+

=0-0181838220

...

,

=0-0062112599

...

,

= 0-0040983836

....

= 0-0101013536 log 7

-2 log 2 + 2 log 3 -i log 5 ~I

.

be found that

2-f log 3

log 2

'

log 7

can be shown that

= 0-693147 180 = 1-609437912 log 5 log 2

The logarithms

= 1-098612288 7 = 1-945910149

...

log 3

...

log

of successive primes 11, 13, etc., can etc., in the formula.

....

now be obtained by

putting z + 2 = ll, 13,

The method

of the

example which

follows, in

which logarithms to a base

10 are calculated without the use of the logarithmic interest, being one of the methods given by Briggs.

series, is of historical

HYPERBOLIC FUNCTIONS Ex. 2. Obtain 5 decimal places.

319

by the use of square-roots only,

the logarithm of 7 to the base 10,

to

In the reckoning below, when a square-root has been obtained which is greater than the square-root of the product of it and the last square-root obtained that was less than 7 is found next and vice versa. Thus we have 7,

;

= log ,4 =0-5, log */T6I=log 5-623413 ... = log J5=:0-75, =0-875, Jog ^105 = log 7498942 ... = log log >/]&;= log 6-493816 ... -log D = 0-8 125, ^0-84375, log s/'OD^log 6-978305 ... = log = 7-233941 ... ^ 0-859375, log */(7JE=log =log log

and so on

;

until

we

VT6-log 3-162277

arrive at

= 0-845100 = 0-845096

log 7-000032 log 6-399968 thus, Iog 10

...

7=0-84510 to

five places,

and probably

Common

20. Calculation of

...

is

j

equal to 0-845098 to six places.

We

Logarithms.

have

logjo.V-log.tf/log.10.

The expression of logarithms and

is

l/(log e 10) is

modulus of the

called the

denoted by /*.

It

common system

can be shown that, to

fifteen places

of decimals, /*

Thus Iog 10 tf =/*log

21

.

tf

0434294481903251. where

tf,

has the above value.

ju,

The Hyperbolic Functions.

hyperbolic cosine, sine and tangent and tanh x, are defined by

The functions known as the and denoted by coshx, sinhx

(1)

of x,

--* ,

The

reciprocals

secant, cosecant

(2)

x .

= .

cosh x

.

,

cosech WQViVAU.

x = -r-i */

-7-

ax

.

i

sinh x

,

T- cosh x = sinh x,

and we write

of x,

Hence the following equations cosh 2 x - sinh 2 x == 1 ax

,

cosh x, sinh x, tanh x are called the hyperbolic

of

and cotangent

sech x =-

*~

A tannx

,

sinh x

,

coth VVVJLX

x rf/

--

1

i

.

tanh x

:

1

- tanh 2 x = sech 2 x,

cosh

x,

-7-

ax

tanh x = sech 2 x.

ADDITION FORMULAE

320

cosh ( - x) = cosh x, sinh ( - x) = - sinh x, tanh ( - x) = - tanh x, so that cosh x is an even function, and sinh x, tanh x are odd functions of x. Graphs of these functions are shown below rough sketches of the Also,

:

two can be constructed quickly from the curves y = ex y^e~ x (see p. 309), by observing that, in Fig. 61, C is the middle point of AB, and ED = BC. The ordinate of y = tanh x is equal to the ratio ED/EC

first

,

;

this

can be found numerically or by a geometrical construction.

61.

(3)

We

Addition Formulae.

2 cosh x cosh y

have

= \(ex + e~ x -f-

)

(e

y

+ e~ y

)

e

= cosh (x -h y) -f cosh (x-y). Similarly,

= cosh (x + y) - cosh (x-y), 2 sinh x cosh y = sinh (x + y) + sinh (x - ^ 2 cosh x sinh y = sinh (x + y) - sinh (x

2 sinh x sinh y

-

Therefore

(xy) = cosh x cosh y sinh (x y) = sinh x cosh y

cosh

sinh x sinh y,

cosh x sinh

y,

and consequently A

,

tanh

,

.

(x v dby) y/

=^

tanhxtanht/ T

r

T

r^-

ltanhxtanhy

where in each equation the upper or lower sign

is

,

to be taken throughout.

INVERSE HYPERBOLIC FUNCTIONS (4)

321

Using the exponential values of sin x and cos x (Art.

14),

we have

cos x =

2

x=

sin

= ~ sinh

tan x=

tanh

ix,

ix,

i

cosh x = cos

and

tx,

sinh x = i sin *x, tanh x

Inverse Hyperbolic Functions. Let x e -x _ and e2x - 2e*z/ -f 1 e

22.

^ = ?/v/(2/

therefore

Thus

if

2

there are

#>1,

i/ ==--

2^

_j_

and

-l)

two

tan

i

= coshz,

positive

x~log{y,J(y

2

-l)}.

real values of x, equal in

which correspond to any given value value is denoted by cosh" 1 y, so that cosh" 1 y = log {y + J (y2 ~

Next,

let

?/

values of #,

= sinhx,

1)}

6

magnitude and

of y.

.

-2e /-l=0, and since we have eF~y + ,J(y 2 + l), so that # = log then

2a;

then

0,

of opposite signs,

The

ix.

a?

ex

{y

>0

for real

+ J(y2 + 1)},

and we write sinh- 1 y = log {y *

A

Again,

.

it

/

e2

then

*

=-

= tanhx ~

+ Jy2 + 1}.

e^-e-* =

and

Hence we write ,,

Thus cosh" 1

/

is

1

a single- valued and continuous function of y (see

sinh"" 1 y is a single-valued and conCh. XVII, 21) for all values of j/>l tanh"" 1 y is a single-valued tinuous function of y for all values of y and continuous function of y over the range -l
;

EXERCISE XXXII The notation of Art. 9 1.

Show

2.

Prove

that

that

is

used in Exx. 2 and

-~H--- -+

1 -f

t+

---

+ ...

-J-^71

-f

. . .

3.

-f ur-

2n

~"=log 1

-> log 2 as

n->oo.

+ Jy + J

n+

2

log

2n

~

n,

so

that

LOGARITHMIC SERIES

322

1-J-fJ-J-f... be deranged as follows. Write down the by the first q negative terms, then the next p positive terms followed by the next q negative terms, and so on. Prove that the series so deranged is convergent, and that its sum is 3.

first

Let the

series

positive terms followed

p

log 2

[Let ar be the sum of the

first r

+ | log plq.

groups of

p

positive followed

by q negative

terms, then 1

1

\

I/

1

1

1

= (log 2 + Jy + \

- i + log qr + c ) -+ log 2 + \ log plq. log pr + 2pr (y ar pr ) Also, if sn is the sum to n terms of the deranged series, r can be found so that Thus ar -sn is the sum of a all these terms are contained in oy, but not in a r i. Also r->-oo as w->oo, and finite number of terms each of which ->0. consequently sn -> log 2 4- i log p/q.] 4.

Use the (i) if

if

(ii)

[For

(ii)

series for log

(

1 -f x)

to

show that

:

0<#-log(l + x)<\x* 0<#-log(l +x)<\x*l(l + x). x= -y, then 0
0
then

let

;

then

...),

5.

etc.]

(i)Showthat (ii)

Hence

calculate Iog 10 7 to five places of decimals, given that

Iog 10 2 =0-301030, 6.

(i)

(ii)

7.

(i)

(ii)

Showthat

lo glo

1-1=2,1

|l

+i

g=2 M {^ +

that log,

-

=0-434294.

i+i

-

-

21-.+ ...} 11 to five places of decimals.

Given ^=0-434294, find Iog 10

Show

p,

1 .

^.

+

L

^+

.

...}

Given that Iog 10 2 =0-301030, Iog 10 3=0-477121,

^=0-434294, find

log lo 13 to five places of decimals. 8.

(i)

(ii)

x + z 2 ) = log e ( 1 - x9 ) - log e ( 1 - x). x x2 ) to six terms, and show that the coefficient of + Expand log c (l-f xn in the expansion is -2/n or Ijn according as n is or is not a

Show

that log e

(

1 -f

multiple of 3. 9.

Find the

Sum

coefficients of x*

m and

x* m+ l in the expansion of

to infinity the series in Exx. 10-14. 1

1

1

'

IT2

+

37*

+

6~6

APPROXIMATE FORMULAE 15.

323

In the expansion of (I +bx + cx*) log e (1 -fa;), (i) Find the coefficients of x* and a*, and determine the values of b and c so that each of these coefficients may vanish, Giving to 6 and c the values so determined, expand the given expression as far as the term containing x*.

(ii)

-

(iii)

Hence

an expression of the form

find

-

r-

the best possible approximate values of log e 16.

(6

( 6 J .

---

Denoting the

(ii)

--

(

by u v

series in brackets

un

(in- 1)

=

will give

+ x) when x is small.

(1

- (6x + 3x*)

+ 6x + x*) log e 1 -f x) 1.2 2.3 3.4 t2 x xH x 5.6. 7 [3.4.5 4.5.6

Prove that

(i)

which

-

}

... V

w a -fw 3 -

-

,

/' ...

prove that

,

2

*n-i (n-l)( + 4) If 0
(iii)

-i

(iv)

17.

Use Ex. 16

759

(iv) to

with an error in defect

show that Q0^AO

less

than 5/10

11

is

an approximate value of log e

Hence

.

find Iog 10 1-024

and

to ten places of decimals. 18. If

a-b=h, where h is a-6/1 l\h a-b\

small,

h2

show that h

2/a-6\ 3 _^

h*

" +

+

+

A8

i

(iii)*

Hence show

that, if

a

,

a

*

+ "'"

is^nearly equal to 6,

a-bfl

l

the error being approximately equal to (a -6) 3 /6a 3

Sum in

the series in Exx. 19, v"*

83*

00

n

~*~*

21. If 8n is the

on

~n

sum

to

n terms of the 1

1

1. 2.

20':

3*5.

6. 7

+

v WamQO

(

n + *-J

xn

series

1

1

9. 10. 11

13.14.15

'

prove that a

a

8^= Hence show that the sum to

1

1 i

,

..,

infinity of the series is

* This is Napier's formula.

,

|

J log

2.

.

1-024,

Iog 10 2, each

LOGARITHMIC SERIES

324 22. If sn

is

the

sum

to

n terms of the

series

1

2. 3.

1

45. 6. 78. 9. 10

(3w l)3n(3n +

1)

111^3+".+3^T- L 2^^+^

prove that

1

+

Hence show that the sum to

infinity of the series

is

(log

3-1).

23. Prove that

+++ 24.

2

By expanding (i ,

log(l-f-o;-fa;

n

as ,..,

,

is

of the form 3r or

n a (?i 2 -l 2 ;)

w2

1-^+-^ 1^ Li n(n +

,

g)

[3

cording as n ,...,

in various

i- + ^Ll)-(-f)(|J8

(u)

)

-

-

ways show that

+ ... =( -i r

-

.-ii

is

.

+

n

is

+1

1XM

=(-!)"

.

or

of the form 3r or

the form

6r,

6rl, 6r2,

[These are obtained from the identities

-, -

or

71

-

according as n

71

6r~3. :

2 -^)-21og(l -x ) + log(l -s)=log

identities

^-^ to show that

/

This formula

is

.

ac-

3rl.

(ffi

Use the

lmn - 1

(-|)

or

7Z-

25.

...

3rl.

of the form 3r or

-=0,

log(l

according

(n-

l)

according as

(iv)

(-ir-

l_z

2n

(ii)and

2 or

.

3rL

n 2 (n 2 -l 2 )(n 2 -2 2 ); i

-=

"- to

72

1

/

3

U

72 4

-

2/

+ ""

"

J

useful in finding the logarithms of large primes.

is

of

CHAPTER XX CONVERGENCE

(2)

SERIES OF POSITIVE TERMS (Continued)

Cauchy's Condensation Test.

1.

and a

creasing function of n, n and 2a n

f(a

Group the terms

is

any

// f(n)

positive integer

is

>1,

a (positive*) dethen the two series,

are both convergent or both divergent.

),

of Zf(ri) as follows

:

{/(I)

sum

v n denotes the

if

where,

v n =f(<*

The number

n -1

of the

+ 1) +f(a n ~ l + 2) + ... +/(a).

of these terms

function, n (a /.

- aw ~ 1

terms of the nth group,

is

n

)/(a

~ a n - a n l and since f(n) ,

< vn < (an - an ~ )f(an l

)

1

)

-(a-lJaYKXvn^^- 1 )^" /^'

a nf(a n )

verges (Ch.

XVI,

;

1 )-

therefore

2.

5).

vn and therefore Sf(n) conconvergent, so also is a nf(a n ) is divergent, so also is vn and therefore i

If

But a

An important p>l

gent if

a

;

terms must converge or diverge, n n are both and a f(a ) convergent or both divergent. f(n)

2f(n) diverges. .

is

a decreasing

1

a

Now, if

is

and

series of positive

Test Series.

The

-

2r

series

}

^

is conver-

divergent if

positive integer, this series is convergent or divergent according as the series Zv n is convergent or divergent, where If

is

any

n

~

a n (log an )*

~ (log

ay n p

"

Since (loga) p is constant, the given series is convergent or divergent according as 21/n p is convergent or divergent, that is according as p>l or *

In any case, the terms of the decreasing function /(n) are ultimately of the same sign (Exercise

A-A.V,

1}.

RAABE'S TEST

326

Kummer's

3.

and

terms

Test. that

suppose

Vn = dnUn/Un+i-dn+l ;

for

vn >k>0,

n^m, if v n

(ii)

Proof,

condition,

...

that

so that after

a certain

stage,

say

is convergent ;

n-1

,

the

s r is

if

sum

. . .

in succession for

n

+ u n < 7 (dm um - dn u n ) < 7 dm um

in the given

Su n

is

,

a fixed

convergent (Ch. XVI, -

dn

Since

.

to r terms of

m + dm u m

(ii)

suppose

is divergent.

m + 1,

Putting w,

found

Zu n

the series

um+l + ww + 2 +

therefore

also

divergent;

we have

whence

Thus

be

<0 for n^m, Zu n

(i)

is

2l/dn

be two series of positive

then

number k can

if a fixed

(i)

Su n and 21/dn

Let

dn4 1 <0 .

number

7).

n^m, we

for

;

have

"n+i d therefore

u n >dmum

divergent, therefore

Now

-5-.

an

2u n

is

d mw m

is

a fixed number and

l/dn

is

divergent.

NOTE. If t>n ->-0 we cannot find k so that t;n >&>0 for n^m, and the test fails. In part (i) of the test, S\jdn need not diverge in fact, dn may be any positive function of n. ;

Raabe's Test.

4. (i)

Let

Zun

be a series of positive terms, then

if after a certain stage, say for

where (ii)

"k

If

is

a fixed number, then

n(

~

Xu n+l

1)<1, '

then

Zu n

Zu n

is convergent.

is divergent.

This iollows at once from Kummer's test by taking d n = n, giving

GAUSS'S TEST In

This test

^u

1

n+l

may

TT

*,

un+l -~-^ (2-l)

D'Alembert's fails

test, 27wn

with Raabe's

;

-'

.

.-.

; '

2un

NOTE.

From 5. that

is

* -f

4 o

hm

n 2wn+1 x92

<

l

.3 x 5

5 ^2

.

\

=-

+

.

1 and diverges converges if x we have however, |

-r

4 o

if |

x >1.

x

If |

\

\

= 1,

test,

-n

3

.

convergent.

This series

is

the expansion of

he deduced the series for sin

it,

fails.

--

x+-

..

-~

*a

6n 2 therefore

when l> 1 and

converges

l

.

,.

-T5L

Here

By

Zu n

then

Z,

the convergence or divergence of

1.

the test

=

:

.,,-,.., Consider

Ex.

'J

but if Z = l, the test fails. be tried when D'Alembert's test

when l
diverges

-

n(

particular, if Urn

327

sin" 1 x,

and was discovered by Newton.

x.

Gauss's Test. Let Zu n be a series of positive terms and suppose u n/u n+l can be expressed in the form ^n-fi

where

and

p>l

\

limit as n->

\< a fixed number k or

bn

then

,

Zu n

First suppose that

Proof.

U n( n \u n+l n ,, |

bn


|

and diverges If

a=

in Art. 3

1,

bn

j=a+ n p^ --*r ,

a>l

and

diverges if

><

as

/

and p>l. Therefore, by Raabe's if

a finite

then

a^l,

l\ i

(

for

converges if

(in particular) b n tends to

test,

Su n

converges

if

a>l

a
Raabe's test

fails

Taking dn = n log n

and we apply Kummer's.

we have -

(n

+ l)

log (n

log

n-

(n

+ 1)

^n+l

=n( 1

Now and

(n-f-l)log

i^.6n->0,

Therefore v n-> divergent.

l

4-

~

for

-

+ -j J

= (w-f l)logf 1 jp>l (Ch. XIX,

+ 1)

log (n

+1

r)->-l

and, after a certain stage,

as

and

7, (5)),

t>

)

n <0.

|

Hence

,^

27w n is

BINOMIAL SERIES

328 In

cases

many

u n/u n+l can be expanded

and we have the following

rule

in a series of

:

b u n _ ^I j__a j___ i -r -r o ~r n n2 U

r* Tf

.,_

i

M

// J

,

valws of

the series being convergent for sufficiently large

convergent if

For

a>l and

sum

Zu n

n, then

is

a^l.

divergent if

in this case b n is the

powers of 1/n,

of a convergent series.

the most generally useful test for series of positive terms, and in cases where it applies, it is better to use it at once instead of beginning

This

is

with D'Alembert's and Raabe's Ex.

_.

By

P -

Discuss the convergence or divergence of

1.

_

tests.

.

.

.

division

n

we can show Athatx

where

bn

,

=^1

^

-r

1

\2

32.52 42 6 2

+

'

"

'

n n->oo.

aa

j

)/(l 4-^) is

+n

v- - i

i

32

l%

+ W~T* * 02

-

I/, + -i\ //,

-

Therefore by Gauss's test the series

Raabe both

-

=[

2

Here the

divergent.

tests of

D'Alembert and

fail.

CONTINUED FROM CH. XVI, 25 6.

Binomial Series.

When x=-l,

(n-l) v

2

the series

n(n-l)(n-2)

-

/

L

L converges if n>0 diverges if n<0. When x== -1, after a certain stage

and

+ Wj -f w 2 +

Denoting the series by u

M-

r

*

n-r

w r +i & r = n(n

where

Hence, by Gauss's

5 1,

+ 1)

test,

+1

_

l( I

the terms have the same sign. division that

all

. , .

we can show by

,

n-fl

.,

i

b~ *

i

t

~j

)

r

--)->n(n + l)

-

rz

r~>oo.

as

the series converges or diverges according as

^

0. according as n the that the test says series diverges when n Apparently the reasoning fails because u r/ ur+i * s f the form 0/0.

n+1

7.

that

is

The Hyper-geometric X+a

1+ rT^ is

known

Series.

The

0,

series

xi

+

"-

r.2.y(y+V

as the hyper-geometric series,

and

is

of great importance.

but here

HYPER-GEOMETRIC SERIES If none of the constants a, and we shall prove that (i)

j3,

is

y

329

a negative integer, the series

the series is absolutely convergent

|x|
if

is

is

endless,

divergent if

\x\>\; (ii)

when $

1,

(\\\)

when x

-1,

the series converges if y-f

Denoting the

Proof.

,

and not otherwise;

the series converges if y>a-f-/8,

by UQ + ^ +

series

. . .

-

=

l>a + /3,

-f

un +

. . .

and not

otherwise.

where

...

1.2...n.y(y + l)...(y + n-l) .

we have

Hence the

(i)

series

converges absolutely

if

|x|
if

|x|>l. (ii)

When x = l, w n /w n+1 ~>l.

have the same

Hence, after a certain stage, the terms

sign.

Applying Gauss's

we can show by

test,

division that

n

n+i

where n-

\

and 4,

5

are fixed numbers.

verges according as y-f

1

Thus

-a->l

y>a-f/J (iii)

If Aen

x= - 1,

or or

w

w/

\

b n -+A

and Z"w n converges or

<1, that

is

di-

according as

y^a-f/?.

after a certain stage, the terms are alternately posi-

and negative. Also |w n +il
tive

(a

+ n)(]8 + n)<(l+n)(y + n)

;

that

is,

if

(a + j8)
or

l+y~a-j8>(ajS~y)~.

if

This

is

true for sufficiently large values of

n

if

--!*_-.! +-5

y 4- l>a -f /?.

where a n->y-f 1 ~a~)3 as w n+1 n n->oo. Hence by Ch. XVI, 24, u n->0 if y-f I>a4-j8, but not otherwise. Therefore by Ch. XVI, 14, the series converges if y-f l>a+/}, but Again, as in the preceding,

not otherwise.

ABEL'S INEQUALITY

330

De Morgan's and Bertrand's

8.

positive terms,

and suppose

Test.

u n/u n+l can

that

Zun

Let

be

a

series of

be expressed in the form

~~

n

u n+l

a certain stage an >l + k where k

then

(i)

Zun

is convergent.

if after

// a n
(ii)

Zu n

Using Rummer's

Therefore v n ->a n diverges If

is

a fixed positive number,

is divergent.

test,

we take dn = n log w. Then,

v n ~n log

if

nlogn'

-l

n

u u n+l

.

as n->oo

(n

as in Art. 5,

+ 1 ) log (n + 1

)

Hence Zu n converges

.

if

an >l

+ k and

a n
a n = 1, vn <0, for log ( 1 \

-n

r

-f-

-n+

)<

=

1/

,

Zun

therefore

1

is

divergent.

SERIES WITH TERMS POSITIVE OR NEGATIVE 9.

// Eu n is a convergent series of positive terms and a sequence of real numbers, positive or negative, and all than some fixed number k, then 2a n u n is absolutely con-

Theorem.

av a2>

t5

39

numerically

less

vergent.

For |

Since 27wn

where

is

is

^

\

+ a m+2w 1

convergent,

we can

find

m

such that for

any positive number, as small as we choose. |

Hence

am ^um

am+lum+l

-f |

|

am+2um+2

-f ... -f 1

|

values of p,

Therefore

am+ ,um+ ,
27a n w n is absolutely convergent.

^a:. 1. // Eun is a convergent series of positive terms, are absolutely convergent for all values of 0.

10. Abel's

numbers

all

swh

for the values

Inequality.

// uv

ti

t^,

2un cosnO and Sun ainnB

8 , ... t*

a sequence of

real

that

nofr, and if al> aa a3 ... is a a j < Q,^ + a + + aj^ n < a

1, 2, 3, ...

of positive terms, then

,

^

m . .

,

decreasing sequence

^

DIRICHLETS TEST For

let

tf

n

~ u l + u^ 4-

-f

. . .

un

331

,

- a Wj 4- a 2 n 2 + + a n w w n = Then u l s v u^s z -s v W 3 = s 3 -.s 2 etc., therefore = 1 5 1 + 2 5 2 ~ S l) + + a n (*n ~ S n~l) * = (a l - a 2 }s 1 + (a 2 - a 3 )* 2 >+ (a n ^ l -o n n _ 1 + a n * n

and

t

.

.

.

.

3

,

' ' '

(

-t-

Now and

the factors a l

sum

their

Also

52 ,

sv

-a 2>

2

~a

.

av

...

5 n are all greater

.

).s

.

a n-i~~ a ni a n are positive or zero,

3>

is

.

than

and

I

than

less

h.

Therefore

Dirichlet's Test. // Zu n converges or oscillates between finite and a l9 a2 a 3 ... is a decreasing sequence- of positive terms which tends to zero as a limit, then Ea n u n is convergent. 11.

limits

,

,

Let

sn

= u l 4- w 2 +

+ u n-

we can

Since 27w n converges or oscillates finitely, and p, that, for all values of

find

numbers

A,

I

so

m

l that

is

to say,

l< um+I + w m+ 2 + by Abel's inequality

Therefore,

Now

m+1 ->0 as w->oo

w

// (an )

limit, both the series

which case

For

therefore for

,

^a nu n

hence j&ar. 1.

if

Ean is

+

cos

n6

*s

any

e

we can

find

m

so that

convergent.

a decreasing sequence of positive terms which tends to zero as a an sinn& are convergent unless 0=0 or 2for, in an cos nQ and diverges.

not a multiple of

2?r,

ScoBnO and

tho series

27 sin

n^

oscillate

between

(Exercise XXVII, 24, 25). the result follows by Dirichlet's test.

finite limits

Hence

12. Abel's Test.

n n converges and a 1? a 2 then Za nu n is convergent. sequence of positive terms, For as n tends to infinity, a n tends to some limit

Hence by

Dirichlet's test,

a n u n - lLu n

But Zu n y

If

Z(a n -l)u n

is

,

a3

I,

,

...

is

a decreasing

therefore

convergent; that

is

to aay,

is

is

convergent. convergent, therefore

a nu n

is

convergent. B.C.A.

CONVERGENCE OF POWER SERIES

832

POWER

A

series of the

x may be

real or

type

SERIES

Za n xn is called

a Power Series.

In Arts. 13-15,

complex.

Za nx n

converges (absolutely or conditionally) for x x x1 verges absolutely for all values of x such that

13. //

\

For

a n x^

since

convergent, a n a: 1

is

k exists such that |

a n xf

\
n ->0.

\

<

= x ly

con-

.

\

|

Therefore a positive

Hence

for every n.

it

x

if |

r

\

<

|

xl

number |,

In

4 Therefore every term of 27 a n

xn

is

less

\

1

than the corresponding term

of the convergent series 7

,

k+k

Z

Therefore

\

Za n x n

14. //

x such For

that

is

is \

7

i

H-An I

X T ^l

convergent and

non-convergent for

Ea nx n

x^x v

is

it is

absolutely convergent.

non-convergent for every

\x\>\x\. x 2 where I^l-H^il* by the last which is contrary to the hypothesis.

the series converge for x

if

theorem

an

xn

X Tl X

converges for

it

x^x^

and 15. With regard to the series Za n xn either (i) it converges for x = it all other or values no value x, (ii) converges absolutely for of x, for of a n x n converges absolutely or (iii) a positive number R exists such that ,

when

x |

\


and

is

non-convergent when

\x\>R.

Suppose that there is a value of x other than zero for which converges. Also, suppose that there is a value x' of x for which x' non-convergent. Choose a positive number a greater than

Proof.

2a nx n it

is

,

|

then,

by Art.

14, the series is

Divide the real numbers in the interval

The lower

class is to contain

(0, a)

every real

into

number

two

classes.

such that

r

Sa n x n

The upper class is to contain every real number r', does not converge if |x|=r'. Both classes exist, and follows from Arts. 13 and 14 that every r is less than any r'. Therefore the classification defines a real positive number R which

converges such that it

\

non-convergent for x = a.

if

|

x\ =r.

Za nxn

R

This number separates the classes and may be assigned to either class. such that a n xn converges absolutely if x
is

|

if |

x \>R.

Nothing

is

said as to

\

what happens when x = R. |

\

RADIUS OF CONVERGENCE

333

= = writing R Q, R in the exceptional cases tively, these may be included in (iii).

By

All that has

been said applies to any power

16. For a real series of convergence. according as x

Za nxn may

The is

Za nx n

,

the interval

series, real

(~R,R)

and

(i)

is

(ii)

respec-

or complex.

called the interval

converges absolutely or is non-convergent within or outside the interval. At either end point series

converge, diverge or oscillate.

Fdr a complex series Za n zn R is called the radius of convergence. The circle (0, R) with centre at the origin and radius R is called the circle of convergence. The series converges absolutely or is non-convergent according as the point z is inside or outside the circle. If z is on the circum,

ference, the character of the series

Sa n z n

17. Let

be any

not determined.

is

series, real or

complex, and suppose that

lim \a n+l /a n

Then and

it

\=l

= l\z\,

lim

R = l/l.

follows from D'Alembert's test that i

i

Again,

if

lim |

a n n = l, then lim \

|

anz n

n \

^l

z \

|,

and by Cauchy's

test

18. In considering the behaviour of a complex series at points on of convergence, the following theorem is often useful.

its

circle

// (a n ) is a decreasing sequence of positive numbers such that a n+l/a n->l and a n ->0, then, (i) if Za n converges, 2a n z n converges absolutely at every n point on its circle of convergence, (ii) if Za n diverges, Za n z converges = 1, (though not absolutely) at every point on the circle, except at the point 2 n n where it diverges, (iii) the same statements hold for the series (-I) a n z ,

except that

For

it

diverges at the point z

= lim 1/72 an

Moreover,

(i)

a n+l /a n = zn if

l,

= - 1, when Z( -

and at a point on the

l)

na

n is divergent.

circle of

convergence

= a n (cos + i sin d) n = a n (cos nO + 1 sin nO). Za n

converges, both

absolutely convergent (Art. 9,

Ex. 1);

Za n cos nO and Za n sin nO are therefore Za n z n is absolutely

a n cosnO and convergent when |z|=l. (ii) If Za n diverges, both = or 2&7r; in which case 2a n sinnd converge unless a n cosn0 Ex. the Hence statement follows. diverges (Art. 11, 1). (ii) (iii)

This follows from

(i)

and

(ii)

on substituting -z

for

z.

COMPLEX VARIABLE

334 19. If

Za n

n z

vergent when

when

converges

z |

|

<|

zx

z=*z lt

Denoting

|.

Art. 13,

by

sum by

its

/(z),

= a + a 1z + a 222 + --- + (a n + n)z n where /(z) z n = a n+l z n + l + a n +2* n+2 + We have >

|

As

|<| a n+l z+

T?Z

we can

in Art 13,

\

+ a n+2z+

2

find a fixed

number

+

1

1

as

|z|->0.

>

T?

and

absolutely conprove that

shall

|^| >0

f

l

it is

we

...

.

such that for every

k,

r,

20. Criterion for the Identity of Power Series. // Za nzn =Zb n z n = for every value of z whose modulus is less than some number /z, then a n b n for every value of n.

For by Art.

where

77

and

19, for

every

n,

tend to zero as

17'

z |-->0.

|

- 6
be seen that

z |-^0, it will

1

|

.

,

can divide by

z,

therefore

+ a n + 'V)^n ~ 1 = ^i +

a l + a2? +

Making z->0 as a n = 6n

21

.

(

before,

^+

'

+ (^n + ^)2;n "" 1

*

we have a l = bv and by continuing the

process,

for every n.

We

Binomial Series. (n-l)

consider the convergence of the series

2

n(n

...

1)

(n

r

+ 1)

i

where w

is real

Denoting the

and

z

series

complex.

by

...

n Therefore, (i)

it

,

we have

=1 r

follows from Art. 17 that

The series converges absolutely

if

z |

|

<1, and

is

non-convergent if

z>l.

MULTIPLICATION OF SERIES

When shall

|=1, the point

2; 1

z

on the

is

circle of

335

convergence, and

we

prove that

(ii)

// n>0, the

series converges absolutely at every point

on

its

circle

of convergence.

ln, (lii)

'

|

/

ur

n

r

4. i

"

r

u/ Therefore

If n

^~

u'r+l

1?

ur

=u

r ',

+

1

according as r

increases with

r,

or

is

n^r + l,

that

is

according as

constant and the series cannot be

convergent.

- 1, we If ?i> apply the theorem of Art. 18. After a certain stage, u r a decreasing sequence, u'r /u r '- >l and, as shown in Ch. XVI, 25, '

is

<->0. These are the conditions required in Art. 18, to apply the theorem of ur and we have to consider the convergence of '

this article,

'.

After a certain stage, the terms of Zu r are alternately positive and r - l) ru r negative, so that Zu r is convergent or divergent, according as Z( is convergent or divergent, that is according as n^O. (See Art. 6.)

Hence the statements

(ii)

and

(iii)

follow immediately from Art. 18.

MULTIPLICATION OF SERIES 22. //

t/j

sums being

... and V ^v 2 + v + ... are 1 3 convergent series of posiare absolutely convergent real or complex series, their respectively, then the series

+ Wg-f w3 +

tive terms, or if they

s

and

t

U l v l + ( U IV2 + U 2V l) + ( U 1 V3 + U 2V 2 + U 3 V l) + is absolutely

convergent

and

Multiply the terms of products as below.

its

sum

is st.

%-f w 2 -fw3 +

...

by vl9 v2

,

v3

,

...,

and arrange the

This array extends to infinity on the right and below, and we shall consider two ways of arranging the terms so as to form simply infinite series.

PRODUCTS OF SERIES

336 Let

an

is

sn ,

the

t

sums

n be the

sum

to

n terms

n terms

of the first

of 27w n

Evn

,

of the first

FIG. 62.

Draw

respectively

n rows

then

;

if

we have

of (A),

Fia. 63.

a set of squares, as in Fig. 62, where the terms are represented by cr Then the terms n

dots and the nth square just contains the terms in in (A) can be arranged to form the infinite series

u i vi + ( u i

2

.

+ W 2v 2 + uflj + (ttjVa + U 2v3 + u3Vz + U 3 v2 + u^) +

where the nth term

is

the

sum

of the terms of (A)

. . .

,

......

(B)

between the nth and

the (n - l)th squares.

The sum

to n terms of this series

and

is cr n ,

its

sum

to infinity

is

st.

Again, the terms of (A) can be arranged to form the infinite series

U V l + (M l Vt + U 2V 1

where the nth term

the

is

1 )

+ (tijVs + Wg^ + tVi)*...

sum

of the terms

(n-l)th diagonals, drawn as in Fig. 63. If the brackets are removed in (B) and differ

only in the

convergent and series

Ex.

The

(Chap. 1.

sum

and

= l +S +

+

5,

series equivalent to

~n

where

is st

6

XVI,

// E(x)

arrangement

its


= r- + \!L

it

we have two

The

series

first series is

which

absolutely

this is therefore also true for the second

;

19).

-

+ ...

,

prove

that,

for all value* of x

and

y,

E(x) and E(y) are absolutely convergent, therefore ~n~i

l^Lil II

In Arte. 23 and 24, the series dn

............... (C)

between the nth and the

(C),

of terms.

,

= uvn -f

i/

!

n ii2

(C) will

l?_

!

be denoted by

w s l -2 + 2 tv_i -^

should be noted that, in every term of d n the ,

Zdn where

+ w w Vi

,

;

sum of the suffixes is n + 1.

MERTENS' THEOREM

337

23. // Su n and Ev n are convergent with sums s and t respectively and one of these series, say Zu n is absolutely convergent, then Zd n is convergent ,

and

sum

its

Let

sn ,

is

tn ,

.st.

(Mertens.)

Dn

Considering the array (A) in Art. 22, 2

wl = v n

w2 -

,

3

l

t;

2

+ vn

w-1

4

wn -i = ^2 +

lastly

?;

+ ...+u n w n _ l

............... (B)

,

w3 = vn _ 2 + t;^ + vn

,

wm - vn _ m+1 + v n _m+2 +

and generally

3

,

,

be seen that

it will

-Dn = u w +u w +u w

sn tn

where

u n 2vn Edn respectively.

be the sums to n terms of

.

+ vn

;

+ vn-

+

s

. .

,

The series Zvn is convergent; and hence (i) we can v r +v r+l +v r+2 + ... number p such that v n |< for is an arbitrarily small number therefore and positive -f

|

find a positive

r>/x,

where

c

;

l^l'

l^il*

l

w'm|j

the sequence number k such that

is

(t n )

(ii)

t |

n

are

\
/x;

all

bounded

hence we can find a positive

;

for every

/?,

and consequently

We

therefore consider separately the sum of the first series in (B), and the sum of the remaining terms, where m

p=u

Let

chosen.*

^ + u#o + 2

+ u m+l w m

. . .

...(C)

m is

terms of the

to be properly

,

Q^W then

with the condition 2?|t/ n

|,

then

if $'

is

the

sum

of the convergent series

\P\
Q

Also

and

Hence,

(C).

since

27 1

that |

vm f 2

un H 1

is \

1

convergent,

u m +% +

. . .

|

-f |

nn

and then |

Q

we can

<,

|

find a positive

provided that

number

m

4-

pf such

2>//,

. .

.(D)


The conditions (C) and (D) are satisfied if n>2m and m> either of the - 2. Hence, as ?/?, and p 1 and consequently w, tend to infinity,

two,

/u,'

\Q\->0

|^|->0,

and

sn tn

-Dn ->0,

which proves the theorem. *

This artifice

is

often useful.

THEOREM

ABKIAS

338

24. // Eu n and Zv n are convergent with sums if

Sd n Let

it

sn,

convergent, then t

Dn

n,

sum

its

is st.

and

s

t

respectively,

and

(Abel.)

he the sums to n terms of

Zu ny ZV n Ed n ,

respectively,

then

+ D 2 + D3 +

Hence

D

Now

and /->/,

s n ~->s

i

/r Also,

if

D

the

is

sum

+ D n - ,Vi + n-Va ^ *n therefore by Ch. XV, 9, (4), . . .

s n t,1

+$ /<^1.L2 -f

of

ZWW which

-

...

is

,

convergent,

by Ch. XV,

9, (3),

1

D=

Therefore l/^. 1.

//

.s/.

-l<.r^l,

-prove,

(log

(

dn

that 1

+ x) 2 ^ rfrr2 - ^.r3 -f c?3 r4 -

=

"-, '/*

Under

4-

1

(

is

. .

+

:

-f

-

.,

^

+

...

4--

,

).

W- /

suitable conditions, the rule for the multiplication of series gives

easily

applies,

1

V

// -1<#<1, the series for log ( 1 + question holds by Art. 22. - + ~ + ~ 4 2 "a // x 1, the series 1 it

.

J

shown that dn
and the statement

is

a;)

is

absolutely convergent and the result in

and so is d l - d% + d3 is convergent for , and dn ->0 by Ch. XV, 9, (3J. Hence Art. 24

^

. . .

;

true in this case also.

EXERCISE XXXIII Prove that the .

t

!

1

+

1 H-

1.3 --

series

+

:

1.3.5 -

.

r T-^-f...

^

is

,.

divergent.

-...

la + 1.3

2

6

2T4

a(o + '

l)

is

convergent

1.3.5 a(oH-l)(o+2)

if

6-l>a>0,

its

sum

-

MULTIPLICATION OF SERIES un

rr If

A 4.

=

-

nP +
a - a' > 1 and divergent

series 27#n cos

The

5.

.

a - a' <

if

+

+k

. .

M '-*

,,

,

,

v~ z

.-

2/w n

is

if

convergent &

1.

Exn sin n0

n# and

show that

=-> ' >

339

are absolutely convergent

x

if |

\

< 1.

4

2n

6.

The

series

7.

The

series 27 - cos

a 2

and

cos nO

diverges

The

8.

sin

n6 are convergent excepting that the

71

when

an even multiple of

6 is

27 - cos nS

series

are absolutely convergent.

n2

nO and 27 -

71

first series

~ sin nO

27

and 27

-

IT.

n6 are not absolutely convergent

sin

71

71

excepting the last series, [Use the following

when

a multiple of TT.

is

:

|

cos

n6

|

>

cos 2 nB&%(l+ cos 2n0)

9. If u l9 u, u 3 , ... as a limit, the series

is

is

and

|

sin

nB

\

^

sin 2

^

nO

r

-J

(

1

- cos

2n0).]

a decreasing sequence of positive terms tending to zero

convergent.

[Deaote the *>i9

v z> ^3 ,

...

is

series by vl - v% -f v3 Using Ch. XV, 9, a decreasing sequence tending to zero as a limit.]

convergence of

10. Discuss the

If.

Show

11.

that, if

Q
Zun

where

1

1

show that

(3),

1

the rule for multiplication

fails

to

obtain a

convergent series for

/i__L_j__ [Show that the

series is 1

- d^

-f

d2 -

. . .

L_j_

where 1

n

12. If

/ an = ( 1

-f

^\ -

-l=-2

is

13. If

that

i)v

-~ e

x

J

absolutely convergent for

27l/n

_ r+

r p( n

r(n-r+l)^%(n + l ) 2 .]

and

[an

r=I

show that the

,

all

series

whose nth term

is

an -

1

is

values of x.

2 2 -t-r~-...; therefore n .|an -l|-*-|2;| as n->oo.

Also

convergent.]

+ FfrlKX^l+x *Y L

yJP(a,j8 > y,*)-(y-a)JP(,j8-f

Under what circumstances

is

1

1,

-^-y(V + y+

this true

1,

**^ 1

and \x\
)

*)=a(l -*)F(a-f

(i)

when z = l, and

1

9

p + l,y +

(ii)

l 9 x).

when a= -

1 ?

CHAPTER XXI BINOMIAL AND MULTINOMIAL THEOREMS

A

1.

the

sum

General Statement.

The Binomial theorem

asserts that

of the series

n(n-l) _J_^_J x 2+_

when convergent,

is

L\(w-r - +Jx

n(n-l)

+__V----

one of the values of

...

(1

-f

1)

n x)

.

^ r

+

mtm

(Of course, in a com-

must be specified.) and only if, n is a positive integer

plete statement, this value

The the

series

sum

terminates

is (1

+x)

if,

;

in which case

n .

This leads to the following theorem, which enables us to deal with the case in which

is

rational number.

any

Vandermonde's Theorem.

2.

used

n

;

we

The following notation

x r = x(x-l)(x-2) where x

is

often

is

write

any number whatever.

...

(x-r +

1),

Vapdermonde's theorem

asserts that

if m and n are any numbers whatever, then

Proof.

First suppose that

m and

n are

positive integers

1

;

then

+r n *+ro*+r Z

:

1

x r in the product of these series the coefficient of x r in (1 -f x) m + n that is to say,

The

coefficient of

is

therefore equal to

,

r

m

r ._

^r-1^1

^r-2^2

.^r

Multiplying by r, we obtain equation (A). This equation holds for all positive integral values of since each side

is

a polynomial in m, n

it is

true for

all

'

m

and n

;

values of m, n.

and

BINOMIAL THEOREM Case of a Rational Index. !, the sum of the series

3.

n(n-l) is the real positive

We

value of

// n

nln-l)

2

(1 -f x)

is

...

341 rational

any

(n-r + 1) '

number and

r

n .

give what is essentially Euler's proof, although in the we have to assume a property of infinite series which is not proved

The series and let

Eiderjs proof.

sum by

its

last step

first

f(n),

is

absolutely convergent

M fM n r ~n(n

1 \

Art

O\ &)

///.

i)\n

...

f

M

A*

r +

\n

I

x

if |

\

<1

till

;

later.

demote

1 \ i j,

so that

n2 For

values of

all

m

and

wr

2

r

n, the series

absolutely convergent, therefore

corresponding to f(m), f(n) are the rule for the multiplication of by

series,

f(m) f(n) = .

dr =

where

d vx

1-1-

-I-

dzx2 -f

. . .

-f-

A^x

r

-f

. . .

,

^+ -^l + ,-^A + - +^-

Hence, by Vandermonde's theorem,

and

/(/).

/"(/A)

/(m)./(n)=/(m + n) .............................. (A) ./(p) =/(m 4- n) .f(p) =/(m + n +p),

so that

Hence

=

/(m) ./(n)

a similar result holding for any number of factors. Positive index.

If

n=p/q where

jo,


are positive integers,

by the

preceding,

Hence

/(~j

by

(1 -fx)

,

we

With regard

is

a

q-th root of (I

+ x)

......................... (B)

the real positive gth root of (l+x) p ? " /p\ = (1 +x)* that is f(n) conclude that

Thus, x being ?

*

real, if

/(

to sign,

it will

a continuous function of x in

)

,

is

denoted (1 -f x)

be shown in another volume that f(n) the interval

-

n. is

BINOMIAL THEOREM

342

Now

f(n) does not vanish for any value of x in the interval, and consequently its sign is the same throughout the interval.

But when

cc

= 0,/(n) = l,

therefore the sign

to the real positive value of (1

-f

x)

is

positive

is

equal

.

Let n be a positive rational.

Negative index.

and/(n)

n

Putting

m~ -n

in

equation (B), ............................. (0)

l; therefore

/(

- n) =

,

= (1 +

NOTE, (i) Equations (A)-(C) hold when x is complex, provided that x |< 1. Hence x is complex, \'x\
if

.

a positive integer, /( -n)~ (1 +x)~ n x |< 1, it can be shown that f(n) is continuous for all n x are real and and (ii) // values of n. Hence if (l+x) n is defined as in Ch. XIII, 7, when n is irrational, we have (1 -Hx)n f(n). is

if n.

particular,

.

\

Second Proof for the Case of a Real Index.

4.

The following

proof depends on very elementary considerations and, although it is unsuitable for reproduction in toto, the reader will find it a very useful exercise to work through the details. We shall prove that if n and x are real

and \

x |<1, then

n(n-l)' v

-

n(n-l)

'

2

...

lo

T*

(1

+ x) n

Proof.

Let

where

Also

series.

has t

its real

...

,

positive value.

be the rth term and

nt r

-

(w-r + 1)

...

s nt r

sum

the

to r terms of the

let

Q
First suppose that 1/n,

Also

~j^

If for

= ny n ^

any assigned values

of invariable sign for sign as ny n _ lt r-1

r

according as

^

and y n

of

n and

0
,

r

n^O ................ (A)

when x =

=

................ (B)

can be shown that yn -\, r-i i Q (B), y nt r must have the same

r it

then by

.

Repeating this reasoning when n-\, n-2,

and observing

that,

by

(A),

y n - r +i

t

i

^e

^ as

...

are substituted for

same sign as n-r-f 1,

n it

follows that the three expressions

yn

,

r

,

n(n-l)...(n-r-f-2)2/n ^ r4 1>1

have the same

.

sign.

Therefore y n

,

After a certain stage, say for r

r

,

and

n(w-l)

has the same sign as

> some

fixed

(n-r + 1)

...

number

t

nt

k,

r+v the signs of

ALTERNATIVE PROOF ^n,

an(i consequently those of y nt

r+i>

Therefore (l+x) n lies

increases.

But

are alternately

r

between

since the series is convergent, s n> s W| r-> ( 1

Therefore

which proves Next

let

0
/i

(l~x)

Here so that

+ x)

n

343

and

s Wjt

.

r -> oo

as

* Wff+1

s n> r~^0 as

y^

4-

and - as for r>&. ,

r->oo

r

.

,

this case of the theorem.

x<0.

Changing the sign of

x, it

-

then

-

has to be shown that if

(n-r-f 1) r -^ = il~nx + n(n-l) x 29 ---+ (- 1 _ n(n-l) -x + ... ^T-* -~2~~ ^^ = (1 ~^) n ~n,n n y njl = (l -cc) -1^0 -according as wgO; ............... (C) x

)

r

also

-

,

-^-

= -^-i,*.-i and

y w ,r =

...

^=

when

.

............. (D)

Reasoning as in the previous case and observing that by (C), yn _r+l> and n -r ~h 1 have opposite signs, it follows that the three expressions

r

and

(~l) n(w-l)

have the same

...

(n~r-f

1)

Hence, as before,

sign.

yn

has the same sign as

r

^

t

r,

t

n>

r+l is invariable,

and

:

n

For every

r+1 ..................... (E)

nt

After a certain stage, say for r>k, the sign of

we proceed thus

t

n

r

,

,

r

-

-

n r+1

.

,

we have

+^- r Therefore

(l-x)* n>r = * n+1 '

Hence

(1

Hence by

(E), 2 Wt

r

,

r

-x

- x) z w

by

,

r

and

(1

-x)y n

,

r

= y n + lff + 2* n>r ....... (P)

= (1 - a?)y n r - tn> r+1 ^ 2/n+l, r + x^n, r ~ ^n, r-fl ,

has the same sign as

,

so that

ntr

t

n+lj r+2 .

Now

r+2

(E) z n>T

has the same sign as

(w-

+ l)y nif ................ (H)

PARTICULAR INSTANCES

344

n>

If

(i)

fore

lies

yn>r

follows that y nt

SL8

cession

n-h it

Hence

r-^

since

m> -1,

and

,

-x)

z nt

t

-x)

n ,

r

.(1 -x)""

r +i

have opposite 1

But

.

We shall prove that

if

n

where

by

case

from

(F) that

yn + m

(i),

,

a negative

,

21)

and

(compare equation

J

n>r ->0

as

r->o>

.

yn> r->0.

r->0;

Hence

^nd

to zero.

complex and

in suc-

z \

\

<1,

2

Using the same notation as before, we see that the

XX,

,

in all cases.

integer, z

n(n-l)

(Chap.

r+1 ->0 as r->oo

u positive integer. If

is

//*

a ll

is

,

.

r yn +m-2,r> 2/n,r and the theorem is proved ,

^n

There-

signs.

n

ntr->Q it follows

therefore,

follows that y n + m _ 1 s nt r->(l

*w

s Wjr ->(l

and

,.

-M-l
//)/<-!, then

Now

and

between

,

yn+i,r~*Q

5.

1 it

yn r->0 and

therefore (ii)

-

Also

(1

series is

+z)ynj r =

convergent

\

(F)).

Hence as before, if yn+l} r->0, then y n> r~>0. Now when n= -1,

Hence

in succession y n

,

r

~>0

for n==

-2, -3, -4,

...

.

So that

nir ->(l+z)n

when w is a negative integer. The theorem is obviously true when n is which n is a fraction is considered later.

a positive integer.

The case

in

6. The reader should be familiar with the following particular instances is assumed that x 1

in each case it

|

and

\

<

:

in a similar

way 2 (l-z)- = l+2z 3.4.5

3 4 -

n(n-f-l)

1.3

.

1.3.5

3.4.5.6

;

NUMERICALLY GREATEST TERM

345

2

Ex. 1. Expand l/(3 + 2a:) to three terms (i) in ascending powers of x; scending powers of x ; (iii) say for what values of x each expansion is valid.

(ii)

in de-

27

The

(iii)

first

The second

Ex.

To

see

if

valid

if |

-|ar

|

<1, that

is if |

x |
valid

if

this

that

1,

1

+ +

is if

~

can be done by the Binomial theorem, take some particular term, say

the fourth, and proceed thus

5.8.11 ._..

:

o f.f.^s o 3

/5.\3

__.

J !-! J

1 ^

O

.

_______

* \4 5.

".

'

sum

is

equal to 2n(n

of the series

;

+ l)(n + 2)(n + 3)x*/\

4,

where

treating every term as above,

Numerically Greatest + x) n where |x|<1.

7. (1

1.2.3.4

2.3.4

8.12.16 which

.

3 is

Sumtheterie*

2.

is

expansion

w~|

and

a;

= -|.

Let

5

be the

we have

Term

in

the

Expansion

of

Let u r be the rth term, and let accents indicate numerical values so that x' = x and u r = t/ r t/ n is positive, f

|

|

|

|

;

wr Let

ifc

be the positive integer such that

therefore ^4-1, greatest term If r^.k, .,

,

that

.

is

r

is

% +2j among

-

is

the

a decreasing sequence, k 4- 1 terms.

and the numerically

first

then wr _}_iw/

,. ^ -according & as r dr. -* 1

i
according as

(n-f 1

r)x' .-

r,

NUMERICALLY GREATEST TERM

346

-

~

If

(i)

r~ ~i+f, where

~\~

proper fraction, the

I

tth term

Next,

is let

~r~

(i

+ l)th term

= i, where

-

//

(ii)

X

/

is

a positive

numerically the greatest term.

is

a positive integer, then, numerically, the

t is

+ l)th, and each is greater than any other term. and equal to - m, then

equal to the

n

a positive integer and

i is

x

j

1

(i

be negative

l '

wr+1 ~w/ according

therefore

,,,.,. according 8 as

that

r

is

It follows that

(ii)

If

= i +/, where

i is

a positive integer or zero and

(m-l)x' = "*~~

X

to the (i-f l)th,

We

.

first

^

JL

7 '

r,

the

(i)

//

positive proper fraction, the (iii)

<- (m-l)x' ~ -:--

=

m^l,

:

If

(m + r-l)x'

as

f

(i

where

i,

and each

term

+ l)ih term

i is

is

is

the greatest,

is

greater than

any

other.

have thus proved the following important property n // |x|
series:

is

a

the greatest.

a positive integer, the iih term

of these

/

is

equal

.

of a binomial either decrease

numerically from the beginning of the series or else they increase numerically up to a greatest term (or to two equal terms) and then decrease numerically. ""

Ex,

when

Which

1.

numerically greatest term (or terms) in the expansion of (I

is the

+ a;)"2

x%?

Let ur denote the rth term

;

ur or

if

r

^

14.

5

r

First consider the terms for which

4r>54

54 - 4r

%^ -r + l 4

u r+i

uril jur

For such values of M r+i

(A)

5r is

negative.

This will be the case

if

r,

ur

Hence the

greatest term or terms must be included among the

first

1

4 terms of the

expansion. If r

^

13,

u r+l /u r

is

|

positive,

u r+1

rr 1

and from (A) \u r

that

Hence, the sixth term other term.

is

|

is

it

follows that

according as 54 4r according as r

~

^

5r,

6.

equal to the seventh, and each of these

is

greater than

any

CALCULATION OF ROOTS The Binomial

Approximate Values.

8.

tain approximate values, as follows Ex.

347

series

can be used to ob-

:

Calculate */2 to six places of decimals. perfect cubes are 1, 8, 27, 64, 125, .... Search for have roughly twice the other. 1.

The

two of these such that one

We

is

.

*

3* 5 s 1

L

1.2.3

1.2

59

+ ""/

L

(

Denoting

this series

by u t + u 2 u 3 + u t - u 5 +

. . .

'

we have

,

=1*25,

^ -

*.=

-0-01,

--

= 0-000

07

001

8

=

08

=,0-000 Jj

~= 4

0-000

OOP

02

0-000

080

02*

'

'2

Therefore,

07

001

1-260

= 1-260 = 1-259

001

07-0-000

921

05 nearly ......................................... (B)

02 nearly

080

possible error in this result, let s n be the sum to n terms of the series R and after n terms. The error arises from two causes the remainder (A) n The error caused (i) by taking s 5 as the value of '^2 is R 5 Now, in the series (A), the terms beginning with the second are alternately positive and negative, also each is therefore J? 5 is positive and less than w e numerically less than the preceding

To estimate the

:

.

.

;

u= -^5

Now

.'.

The

.~-u 5 = ~~u 5 < 0-000 000 000 4; 3

the error

5

5*

is less

than 0-000 000 000

4.

from the errors in the calculated (ii) values of w 4 and u 6 These cannot exceed 0-000 000 005, hence the error in the calculated value of s 5 cannot exceed 0-000 000 01. Adding the results of (i) and (ii), we see that the error in the result (B) is numerically less than 0-000 000 010 4. Thus, to six places of decimals, error in the calculated value of S B arises .

/2

= 1-259921. i

Ex.

2.

p + qx -,

,

We

and

If x

where p,

is

small and

q,

p' 9

q'

n>l,

obtain

an approximation

to

(l+x)

n in the form

are independent of x.

have

(B) B.C.A.

NEWTON'S APPROXIMATION

348 Choose a and

b so that 2

and

a-b-l/n

-&)&-=(&- l)/2w and a = (n + l)/2/i.

(a

b-^(n-l)l2n Substituting these values for a and 6 /.

n

-l)x

we have

in (B),

2

;

2

\n J both convergent, and since they

\nj

\

(C)' v

J

If x is small, the series (A) and (C) are as far as the terms involving x 2 it follows that

are identical

,

.

+x

(1

9.

Theorem.

n

)

^

_

----

-

,

- rf- nearly.

}

n>l and 0
// ^

~ n2 -

1

proximate value of

By it

(1 -f

n

x)

w/A an

,

1

error in defect less than -ys~3

^3

-

multiplying each side of equation (A) in the last example by

2n + (n- l)x,

_

can be shown that

__

i

-x-x .x)

-

X

2

on"5

Denoting the

< 1

--

-

series in brackets

MI

-h

u r-

we have

w r _!

=7

OX

_

.

4w .on r

+ (-

rn

l) '^r

r

1

-

--,

-

(r-f-2)n

-

-

r-1

+

. . .

i-

.

J

,

x.

(rn-l)r<(r-f2)(r-l)?t provided that r>2, for _ 1 r _ ( r + 2) (r - 1) n - - r ~ (r - 2) n<0 (

m

...

by

-^ + 2 ^3

,

Now

LX

4n

^

)

and since 0
;

.

and, since

n> 1

and x

is

Ex.

2

-l)rryi2n

v2

than

3 ,

that 63.5/504 is

an approximation

\2

to

with an error in defect less than

0-0000005.

Here

is less

;

result follows.

Show

1.

than unity

positive, the last expression

(n

whence the

less

*

1

(

+

J

j

henco, an approximation

5750-4-12 4

750

i

6

635 r

'

504'

--

... r * an error in defect with i

4

12.27

-

liM 3

10 8

10 7

is

KXAMPLES OF APPROXIMATION

349

EXEECISE XXXIV If a:

1.

is

positive,

which

is

the

first

negative term ih the expansion of I

x

2. If

is

(

1 -f

x)

*~ ?

9

terms in the expansion of (1 -x)~* beginning with a Prove this, and find which is the first of this series

positive, the

,

certain term, are all positive, of positive terms.

Which 3.

(1+x)*' when ar^J

5.

(l~x)~

7.

Between what numbers must x _

(I

the numerically greatest term or terms in the expansions of 35 _3

is

-x)

(l-x)

when x = ]l

6.

(l+x)~* when *---{??

2,

the third term

x> 0,

9.

If

0<#<1, show

(ii)

lie

order that, in the expansion of

in

3

If

than

when *-:&?

4.

8.

and

defect

2

?

less

which

than

~x~ \ ~\x

\/l

may the

is

be the greatest

first

that

(i)

?

negative term in the expansion of 1

Vl 4-x

1

+lx-{x-

nearly, the error being in

^x'^. -

.r-

nearly

*

the error being in excess and numerically less

-

x

10. If

is

large

and

positive, s/a;

3 -t

show that ~

\~x-\3

^2

the error being in defect arid less than 5/81 a11.

IfO<
1/(1 -ir)

2 ,

and show that

/? n

is

x5

9

nearly,

8 .

the remainder after n terms in the expansion of

0
0<#<1, and Rn

13. If

1/(1

-x)

3 ,

14. If

1/(1

-x)

(ii) 1

less

3

is

the remainder after n terms in the expansion of

show that

0<#<0-01,

show that

with an error

in defect less

-3#

f fix* is

(i)

l+3o; + 6x 2 is an approximate value of than 0-000011.

an approximate value of

1/(1 +x)*

with an error in excess

than 000001.

15. If

is

small,

show

that, neglecting cubes

and higher powers of e,

APPROXIMATION TO ROOTS

350

16. // we have tables giving the square roots and cube roots of numbers to f signicant figures, show that the roots can be found by division only, to 2f significant figures as follows.

If *Jx~a

to

f significant figures,

then *Jx

f significant figures,

then

-

2 (\a 1

If

$x-a

Show also

Given

Given 4/10

19.

Show

x

is

20. If

(i)

(iii)

1

\ to )

2f significant figures.

/

that

x

is

by

= 141 421356237

division, that

(approx.).

2-1544 (approx.), show, by division, that $ 10 = 2- 154434690 (approx.). if

(i)

x

is

small,

v/(z

large,

2

*J(x*

+ 4) - *J(x* + 1) ~ 1 - i* 2 + 3 /

s ~ +4) -V(x + 1)= 2

1

~

(

5

7 4 6 4~z

~

nearly

;

21 \

^ + ^)

nearly-

show that approximately

small,

V(4 - 3*)

(

/x

$x = o-(\a~ + 2a

1-41421 (approx.), show,

v/2

18.

if

2f significant figures.

use Ex. 15.]

>/2

(ii)

to

/

that in each case the error is in excess.

*Jx-, and

[Put a 17.

to

/x - + a\ )

=1

.

4-

# (8 + 3*) -r (9 - |z)

+ 4a; -f 10x 2

* )

= 1 -f 2x + 3z

--,

7

(1

-

-iVr)

;

2

(iv)

(v)

(

I -fa:

+ * + *)* = l+ix-h A* 8 + ^5**.

In each of the examples 21-26, (i) prove the given equality ; (ii) by expanding the right-hand side to four terms, deduce an approximate value of the given surd ; (iii) find an upper limit to the numerical value of the error. 21.

^1001 = 10(1+0-001)3.

22.

V2 =-J(l -0-02) "i

-f(l+ 0-024) "i

24.

#5 =(1+0-08)*.

26.

#4=f(l

23. 4/4 25.

*/3=i(l-0-01696)*.

27. If

a

is

nearly equal to

b,

show that ~

^ and that

if

a>6,

the error

is

in defect

and

is less

3 tt -! /a-6\ "12^ V~T/

than

2

[Assume the result of Art. 28. 10*. 3/10*.

Show

that

/543

9,

2709

^^ = 2706 ^TTO ^ 54 5

J

'

C

'

and put x = (a - b)/b.] '

nearly, the error being in defect

and

less

than

SUMMATION OF SERIES 29. If a -b=^h,

whore h

a

I

+

ft" 1

2 (iii)

h

6~a

2 ft~8 1

A2

4 ft" 8

8

1

+

small,

is

I h'1

7i

+ "

Hence show that

4

a

show that 5

I 3

T6

ft~

1

&3

16

ft"

h*

~T28 b** 1

&*

"

ft"

351

3

+

'*

""

"

""32 6*

nearly equal to 6, then I a +b = + nearly b a+b 4~'b'-' if

is

a

Va

'

the error being approximately equal to (a-ft) /128ft 4

Application of the Binomial

10.

4 .

Theorem

Sum-

to the

mation of Series. (1) Many series may

be summed by expanding some function of x as a convergent power series in two different ways and equating coefficients. Ex.

1.*

If

prove that

It will be easily seen that

w2 -|?.(g + l),

wl

therefore

and

+i/ 2 o;

+ w3a;2 -f-... /

-f

wn _!a;n<" 2 + ... =

-

%--f v

Hence

if

8

is

the

sum

of the series in question, of n 2 in (1

[g=coefft.

=coefft. of

(2)

|g (1

Theorem.

where k

// f(x)

-a

-f

xn ~* in

-*)-(-).

(1

(1 -a:)-<

a^ + a2x2 -f a3 x2 +

. . .

+ a nxn +

positive number, then a + aj-f a 2 the coefficient of x n in the expansion of f(x)/(l -x).

For

is

any

~

-f ...

. . .

+a n

for is

\

x
equal

to

convergent for sufficiently small values of

x.

fix)

z=

x

and the

-

x

series are absolutely * This equality

is

required in the Theory of Probabilities.

SUMMATION OF SERIES

352

In what follows, we suppose that x <1 except when n tive integer, so that all the series which occur are convergent. (3)

|

We

shall write

,,

w

then,

(1

+ x) n = c

/i ,n- * , 1 H- ;r) (

4-

4-

c^x

c 2x 2

a posi-

is

\

c3x

4-

3

-h

-f

. . .

c rx r

-f

. . .

n(n-l)^ x ~n(n-l)(w~2) --~~----- x2a + = w -f -^r 4-

l

;

. . .

r

!

= c x 4- 2c 2x

-f

3c 3 ,r 2

-f

+ rc^ r

. . .

'1

-f

. . .

,

and, by a succession of similar steps,*

=l

2

and so

.

+2

2c 2

3r3 :r

.

+

... -f

r(r

where u r

is

-

on, as far as required.

Hence we can sum

*_ Q u r c r x

the series

r

a polynomial in

For, dividing u r and successive quotients by r, r-1, we can find constants A Q A L A 2 A 3 ... such that

order

,

Z^^u

then,

Ex.

c

r r

Find

2.

x r ^At(l+jc) n +

Z*^

(r

Denoting the sum by

+1 *V

3 )

c

r

xr

,

A

v

S=-(l+x)

= (1 + (4)

n

-f

n -8 a;)

nx(l +x)

n ~l

+ A 2 n(n- })x 2 (l +x) n

'2

.

and proceeding as above, we find that +7/' + 6r(r - 1) +r(r - l)(r -2)

1nx(l +ar) }l -f (7

in this

...

,

,

3 (r + l) -a .'.

r-2,

r.

n -1

;

+ 6n(w-

l)x~(l +x)

n

~2

+ n(n-l)(n

4 3) a: -f (6/& 2 + 8w + 3) x 2 4 (w + 1)V}.

Again, by expanding

+x)

(1

1 (Ijfx^^^ " " "V" n+1 ?i + l ^

i

w+1

^n^

we can show

,

GI '

*^

r

2

2

that

i. ..i" _c, -i~X

l "r~

r-fl

and, by similar steps, "

(n

+

l)(n

+ 2)

(M

+

l)(

+ 2)'n +

n 3 (l+x) +

"

^,2

.

-,

Jv

~~~^,

i

(r 1

^.3

__

j

1.2.3 on.

i

\

1

.

and so

i

1.2

l

+

l)(r

+ 2)'

_

_ X+

?r

'

2(n + ,v.r+3

l) ,

"

These results can also be obtained by integration,

* Tlirse results

can also bo obtained by differentiating with regard to

process has not been justified in ease of an infinite series.

x,

although so far the

SPECIAL TYPES OF SERIES Hence, if u r

is

a polynomial in

we can sum

r,

353

the series

r r

and

For, dividing u r

we

in this order,

successive quotients by r -f k, r can find an equation of the form

A

ur r

where vr

is

^4^, -4fc_!,

...

Al

l

~~~

"*.

":;

i

r

r

+1

A% \ ^rr (r + l)(r + 2) '

..

/

The value

z r~~

'"

1,

. . .

r

+ 2,

sum

-f

1

Ak

.

(r +

l)(r + 2)

...

and

of the series is

-

r+1

r

r+2

+

x

Zv r c r x r

of

i~

+k-

are the remainders in the successive divisions

the final quotient, and then the

zvx rr +

/

'

is

found by

(3),

and the other sums are given by

the above equations.

Further, we can

sum

the series

k are unequal positive integers of which k is the greatest. can be reduced to the preceding by multiplying numerator and denominator of the fraction by such factors as will convert the where

a, b,

For

...

this

denominator into

Find

Ex.

2.

We

have

hence,

S = 2^1 -

r

(r -f 1 ) (r

-

+ 2) (r + 3)

. . .

(r 4- k) .

f -s


-

+3

*"-**

(MMT,

(7TIj(iFT2MF?S,

z2

n -Hi

n (i+x) +*-i

v.

d+*) n+l

i

i

a;

n-fl

2

r

i

*

x

MULTINOMIAL THEOREM

354 yiOO

^r^O 7+3

Wo

tr

""

'

~ =r -

have r

and

4-

3

H

3

r

+3

27(r

where

f,

If 6, c,

...

i are any

m

f(x) (i)

2.

numbers, n

= (1

&#

4-

and

is real,

4 ex 2 4-

fcc

... 4-

m

n )

,

/(x) can be expanded in the form 1

4 ux + u 2x* 4-

where the series terminates if n

is

urx r 4-

... 4-

a positive

. . .

The

u r of x r

coefficient

n(n -

~

1)

(

is the

sum

and in

other cases is

x.

of all terms of the form

+ 1)

H--2) .^(a

,

integer,

absolutely convergent for sufficiently small values of (ii)

the same as in Ex.

Multinomial Theorem for any Real Index.

11.

then

is

, /B

y

c

""

^

'

B'Ty/'-E where

/?,

y

,

. . .

K:

Aave any positive integral or zero values such that

and the corresponding values of a are given by a-f/3-fy-f-...-f/c = w then For let y = 6x + ex 2 -f 4- fcr by the Binomial theorem . . .

,

* provided that |y|
By

substituting bx

+ cx 2 +

...

for

(1),

we can

find a positive

y in (A) and expanding the terms, we

obtain an expansion of /(x) in powers of x which is convergent if x\
are

made

where

This will be the case

positive.

x', b', c', ...

stand for \x\, |fej, this is the case if

if

c |

number a such that

|

|,

+ k'x' m
b'x' ...

;

. . .

x
This concludes the proof of the statement (i). r Again, s being any positive integer, by Ch. Ill, 16, the coefficient of x 2 s m in is the sum of all terms of the form (bx + ex + ... 4 kx )

EXPANSIONS where

. . .

y,

j8,

ic

have any positive integral or zero values such that jS

and

+ 2y + ...+w/c = r, K

y-K..+

/?+

from (A) that the coefficient the sum of all terms of the form It follows

is

n(n-l)...(n-* + T-7f

where the values of of s

by

Ex.

1.

Here

j8,

y,

JPt'nrf f/ie coefficient

n~

--J,

j8,

y, 8

x r in the expansion of f(x)

of

l)

fl

c/

A;

. . .

,

j

of X* in the expansion of (\

r

^4, and the

( v

~i)( -i--l)( v -i ~2) I _JLI *_ -_ _z

coefficient is the ...

(a

sum 1

.

2#(

3

8

terms of the form

- 4) v ( - 2) 6 ,

have positive integral or zero values such that

and the corresponding values of a are given by

~~

+ 2x - 4# 2 -

of all

+ 1)

i

-

where

(B)

S ............................... (C)

c; i

..............................

are given by (B) and the corresponding values = n-s. (ii) follows on writing a

...

The statement

(C).

355

/J

+ 2y + 33 -4,

a+/?+y + 8=~-|-.

The possible values of a, /S, y, 8 are shown in the margin. The term corresponding to the first set of values is

a

y

ft

2>

>

-3i,

2,

1,

-4,

4,

0,

-2j,

0,

2,

S

'

j

Similarly it will be found that the other terms are ^, 6, so that the required coefficient is

15,

-3 + 15+4f + 6=22-|.

EXERCISE XXXV 1.

By

n considering the expansion of (1 -a:)~

A ~r H H

n(n +

n(n +

1)

,

show that

l

r

r^,:

2

L 2.

Show

that, if

z

= ~^, and x+y

2 |

|

<1, then n-f2);

3. If n, r are positive integers (including zero), show that n f| - 1 2 . t he expansion of (1 2r)2 x) /(l -a;) is (n

n+r-i

^

+

+

n - 2 [Expand {2 - ( 1 x)} l ( 1 a;) .] 4. If w is a positive integer and (1

show that

+ c*i + a, +

+x) n .

+ an_! = $n (n 4- 2) (n + 7) 2n

-4 .

the coefficient of

COMPARISON OF COEFFICIENTS

356 5. If

n

a positive integer, show that

is

4

3.4n(n-l)

= coefficient of z n 6.

If n

Prove that

Show

n

if

!?L 8.

{1

p

in

{2

-

-

(1

n x)} l(l

(n

+ l)(n + 2)

- x)*=(n z + In 4- 8)2W ~ 3

"~~ ""

1)

(

"

f

-# (2 -x)}~ 1 = (I -x)~ 2 .]

'

JL??_

that, for all values of

p 4-^ = 1 and

n(tt-l)(n~2)

+" ~

'

*

m and n

9

coefficient of

x r in the expansion of

n, r are positive integers (n>r),

put 1 q for p, and show that the hand side when expanded is

(1

+x)

n~m

.

show that

p

r

,

coefficient of q k

^coefficient of z fc in (1 +3;) n x (1 +a?)-
+0^+

value, find the

sum

2 r==Q mc x

of the series,

r

IL^J; 14. If

'

n

[Divide by

any

.

a positive integer,

is

n n

= the 9. If

5 n(n - l)(n -2)

a positive integer, show that

is

[From the expansion of 7.

.

12.

r 9

when

on the

left-

- r - fc+1 >.]

n may have

m is

,-,!-=,;

13.

ur =Ao + A lr + A s r(r-I) + ...+A h r(r-l)...(r-h + l) and ~ h~1 h 4- A l nx + A 2n(n- l)xh 2 -f + A h n(n ~I) ...(n-h + l),

(x)-A Qx

. . .

prove that

15.

Prove that

CfCfJlJI

[Consider the product 16.

Show

(1

+ Cf^Cfilf H-(7J +1 C;Si; + ... T +^ x (1 - X )~^- T \] -x)~(

that

3 2n + a

1

to 40 terms -6' JJ.

-3

2714-3*

the summation being taken for n. 9 r such that p + q + r [Consider the expansion of

all

positive integral values, including zero, of

p q y

(a 4-6

+c)

2n+3

- (6 *

-f

c

-a)

2n + 3

- (c 4-a - ^>) 2n+3 - (a 4-6 -c) 2n + 3 .]

Check the answer by putting

nl.

CHAPTER XXII RATIONAL FRACTIONS

RECURRING SERIES AND DIFFERENCE

(2),

EQUATIONS 1.

Expansion of a Rational Fraction.

fraction in x,

Any

rational proper

whose denominator does not contain # as a

factor,

may

be

written in the form

P

+a _ - - -+p

+ a 1 a; + a 2z2 + ... 2 l+piX+p^x + ...

a _

Q

mx rx

r

Various ways of expanding such a fraction in a series of the form

Zu n x n

are given below.

can be shown that the

obtained by any process is convergent for all values of x in any interval including zero, then by Ch. XX, 20, any other method of expansion yields identically the same series. If it

2.

Theorem

.

The

series

rational fraction

P/Q can

be

expanded in a convergent

series of the form

U + u l x + u 2x2 + if

and only

if

For, by the

x \

\

<

A |

method

of several terms of the

|,

where A

+u n xn + ...

...

is the root

of partial fractions,

of

,

$=

with the least modulus.

expressed as the sum a constant, r a positive

P/Q can be

form A(x-a,)~- r where ,

A

is

integer and a is any root, real or imaginary, of the equation Q Q. All of these terms can be expanded in a convergent series of ascending

powers of x if, and only the least modulus.

When

this

the various Ex.

1.

is

if,

x |

\

<

A |

|,

where A

the case, the expansion of

is

the root of

the fraction

be expanded in a series of ascending powers of x? Equating the denominator to zero, the roots are

-iV2

and t(i

and their moduli are -s/2il, >/5, ^5. Hence the expansion is possible if and only

with

P/Q can be obtained by adding

series.

For what values of x can

$=

if |

x \<*J~ -

1.

RECURRING SERIES

358

Methods of Expansion.

3.

where the Art.

1.

(1)

Suppose that

assumed to be convergent and P,

series is

are as described in

Q

Multiplying by Q,

Expanding the right-hand side and equating uQi u v u2 ... are determined by the equations

coefficients, tjie values of

,

Hence, after a certain stage, any r successive coefficients are connected by a linear relation. A series having this property is called a Recurring Series. (2)

The process

equivalent to synthetic division carried Ch. Ill, 3.) (See

just described

out as in the next example. Ex.

Show

1.

that,

The reckoning

is

for sufficiently small values of x 9

1

+ X + X 2 + X3

is

as follows.

1

-

-3

(2

4-

4)

1+1+1+1 + 4 + 12 + 16 + 16 + 32

-3-9-12-12-24 + 2+6 + 8+8 + 16 1+3 + 4 + 4+ 8 + 20 The

first

term

diagonally in

(c),

in the quotient (d)

2

(a);

(6),

+ 1-3,

is

1

;

(a) (b) (c)

(d)

2, -3, +4 3(2-3+4)^6-9 + 12;

1(2-3 + 4)^2-3+4; put

the next term in

(d)

;

- 9, + 12 6 - 3 + 1 =4, the next term in (d). diagonally in (c), (6), (a) do this example by the method in Art. 3, (1), he will see that the reckoning is essentially the same as that just described.

put

6,

;

If the reader will

(3)

The method of Partial Fractions.

Ex.2. If We have

where

Hence

Now hence

a,

j8

1/(1

-2ar+5x 2 )=w

a;

1

1

are the roots of i

+w1 + ...+Mwarn + ...

x*

find the value of un

_

= _1 /__?_ _ _J )~a-8 \1 -a* 1 -j -j8*)~a-j8

- 2x + 5 =0, so that we

may

take a = 1 + 2*,

i

l a=v/5(cos0 + isin0), 0=V5(cos0 -tsin0), where 0=tann +1 n ^ 1 =5*(n + 1 /. a ).2tsin(n + l)0; -/3

w n =i.5*(n fl )sin

.

-

(n

+ l)0, where 0=tan"-

!

2.

2,

ft

= 1 - 2i.

HOMOGENEOUS PRODUCTS

A

(4)

fraction can also

expanded by the Multinomial method and that -of Partial Fractions and equating

rational

Theorem.

359

Using this we can obtain

be

many important

coefficients,

results, as in

the following

exercise.

EXERCISE XXXVI Use synthetic division to expand

1.

w (l-a;)

v

2

'

1 (l-2*)' as far as the terms containing a; 6 Find also the .

Show

2.

that, if

p and

coefficient of xn in

each expansion.

2 q are real numbers, the condition that 1/(1 4- px + qx ) series of ascending powers of a; is as follows

can be expanded in a convergent 2

p >4#, then

(i) if

<

x

|

\

:

- Vp 2

\

(p'

*q)/2q

where p'-\p\\

|,

(ii)ifp*<4q, then \x\
x lies between a and show that the fraction l/(x -
|

|

|

|

. . .

[Let If

The

|a|
,

= (a- ]8){l/(a;- a) -!/(*- J8)}.

fraction

|a|<||<|j8|, prove that

and

H

[Hn i.e.

is

sum

of the homogeneous products of n dimensions which can show that dn +*(b-c) + b n +*(c-a)+cn +*(a-b)=-Hn (b--c)(c-a)(a--b). the coefficient of x n in the product

4. If n is the be formed with a,

6, c,

The

in the expansion of 1/(1 -ax)(l -bx)(l -ex). fractions.]

result

obtained by the

is

method of partial 5.

Prove that, for

sufficiently small values of x, 1

1

a n =p n

where

n

-(n-

""

2

l

-px + qx*~~

~

+

l)p ~*q

n

-p

n

~*q*

-

. . .

,

I

the (r+ l)th term being 6.

(

T

l)

r

C^- p

n - zr r q

.

use the identity

If a-f/?=jp, aj9=5 , f

/

to prove that

an + p" =pn - pr

n~2 j3

Li the (r + l)th term being

[Expand both

sides,

(

I)f

g

+

1

-
__ *>\ ,

p

f-

w~ 4

g

1

-

2

-

1

fix

. . .

-px + qx*

,

!j_

^~>-l)^-^-2) - (n-2r + l) yn _ ifgr

equate coefficients and use Ex.

5.]

SCALES OF RELATION

360 7.

+ p=p, 4=9,

If a

to prove that

-

a

p

the (r + l)th term being

[Expand both 8. If

%

is

sides

(

(-1?

^n

*~ r "

and use Ex.

~^

(r

(ii)

+ 1) th term

552* sin

-*V-- y

.

5.]

n 0)

-

~ (2 cos B) n ~

2

+

r

Li

the

"' (n

a positive integer, prove that

2 cos nB = (2 cos

(i)

-

use the identity

being

(2 cos

-

r

0r- -

-

the (r+ l)th term being

(

n n~r~ ^

^

2

I

n (2 cos 0)

~4 -

...

,

_

l)

(

^r~

(2 cos

1

(n

*) "' (*>-

^

~ T ~ l)(n

(

(2 CO8

"- 3

4)

y-

"'

(

-

(2 cos *)

12

~*~ 2)

)n -sr.

"~ 2r)

1)'

(2 cos

...

,

0)-->.

7"

[In Exx. 6, 7 put a 9.

(ii)

Show The

that

(i)

= 2, p = z~

l

where

z~ cos

0-fi sin

0.]

the coefficient of x m in the expansion of -~

coefficient of x* n in the

expansion of

-- -

--

-

(1

-x)(\

-x

;

2

)(l

- x4 )

1

+#

is

(n f

2 Recurring Series. (1) Let u Q + u 1 x + u 2x + ... be a series which any r + 1 successive coefficients are connected by the equation

4.

u n +PiU n -i+P2U n _ 2 +...+p r u n _ r = where r

is

a fixed number and

p l9 p 2

,

...

pr

)

is

-, ''

1 )

2 .

in

.................... (A)

are constants.

Such a

series is

called a Recurring Series of the r-ih order.

Some authors

equation (A) the Scale of Relation of the series 1 +p 1 x+p 2 x 2 + ... + p r x r others use this term to denote the polynomial We shall take it to mean either of these things. call

u l9

;

.

u r _^ are known, the subsequent coefficients can be found in succession by equation (A). Thus a recurring series of the rth order depends on 2r constants. Hence if the first 2r coefficients are given, in general the series can be continued as a recurring series of the rth order, and in one way only. Also If the coefficients

it

u

,

...

can be continued as a recurring series of the (r + l)th order in a doubly number of ways. For to do this we can give ur and ur+l any values.

infinite

2r terms belong to a recurring series of necessary that this may be so, and then the series can be continued as a recurring series of the (r - l)th order.

may happen that the first order r-1. Two conditions are It

CONVERGENCE OF RECURRING SERIES Ex.

361

Discuss the question of continuing 2un x n as a recurring series when the

1.

first

six terms are known.

un +pu n _ l + qu n _ 2 + run _ 3

Let the scale of relation be

Then

are given

q, r

p,

u3 4- pu2

4-

qu

0.

by

4- r u

w4 4- pu9 4- #1*2 4- r^j

0,

Now Sunx n

can be continued as a recurring definite values and rr^O. This will be the case

w e 4- pw 4 4- g?/ 3

0,

4-

ru 2

order

series of the third

if

0.

have

p, q, r

if

and

(A)

"3

In case either determinant then

is

zero,

suppose that the scale

%

u & +pu^ 4- gw 3 =0.

That these equations may be

one of the

u n ^-pu n _ l

W 4 4-^ 3 4-
first

+ p W2

4-

^ Wj = 0,

consistent, each of the determinants in (A)

vanish, so that two conditions are necessary that the second order. If it is

is

u2 4- p^ + qu = 0,

order, the scale being

Eun xn may

must

be a recurring series of

un +pu n ^ 1 =0, then

Ut+pUQ Q, +pu 1 = 0, ... uB +pu4 0, 2 conditions u l -u^u 2 = Q u 2 2 -u u 3 =0 w 3 2 -w 2 w 4 u2

so that the four

must be

(2)

9

9

1

u4 z

0,

-u 3 u5 =Q

satisfied.

Any

u n xn

recurring series

is

convergent for sufficiently small values

of x.

For

let

the scale be

u n +p l u n _ l +p 2 u n __ 2 +...+p r u n _ r = 0,

and let U Q Then by (A),

|,

|

|,

... |

pl

|,

p2

|

|,

be denoted by UQ ',

...

u,

p, p%',

...

(A) ....

'

the greater of the numbers Pi +p% + +p r an d 1, and un ^ the same way the greatest of the set u' ... In __ n 2 l9 UQ.

where # is

u |

..................

is

u^

where

w^

is

,

the greatest of u^_ s _ l) u'n _ 8 _ 2

Continuing thus, by multiplication

where

A

is

the greatest of Wn-i> Wn-2>

Thus |

,

...

u$.

we can show that |

...

2u n x n

\

u \

u nx n \^\gx\ n .A,

n convergent if x | for then w nx (w4-l)th term of a convergent series of positive terms.

and

un

is less

is

|

|

|

than the

GENERATING FUNCTION

362

n

when convergent, of any recurring series Zu n x a rational proper fraction, which is called the generating function of The sum

(3) is

to infinity

,

the series.

Let the scale be

U n +PlU>n-l+P2U n _ 2 +...+pr Un __ r = Q, and

..................

(A)

let S

= w + w 1 x + w 2 x2 -f ...

to QO ......................... (B)

p^, p2^> p rxr and adding, we find that + p& 4- p 2x2 4 4 p rx?) = a 4- djX 4 a 2x2 4 + a r _ 1 xr - 1

Multiplying (B) by s(1

where

. . .

a

=w

al

,

=w

. . .

4r

,

......

(C)

a 2 = t/ 2 4 jo^

^WG,

the coefficients of powers of x higher than x r

~l

vanishing on account

of equation (A).

Thus

s is

It follows

A

ti;Aer6

i

equal to a rational fraction, of which the denominator

from Art. 2 that

// w n

(4)

i

a recurring

is

Q=

iA6 root of

To prove

a

of which the scale

is

(1

-x)

n of r+l

is

r,

then

Zun xn

. . .

Let ,

1

so that v n is a polynomial in

,

r steps of this kind,

where &

degree

n 2 - x) = w 4 v^ 4 1 2^ 4 ... 4 v nx 4

s (1

where vn = u n -u n _ l

|,

.

we can proceed as in Ch. VIII, 7. s = u 4 ux 4 u%x 2 -f 4 u nx n . . .

then

if\x\<\ A

wjAicA Aas $Ae Zeas^ modulus.

positive integral function of

series

this,

2Ae recurring series is convergent

is

we

. . .

,

n of degree r-1.

By

find that

independent of

x,

and therefore

s(I-x)

r+ l

~u +(k-u Q )x,

which proves the theorem, Ex.

2.

Find

the

sum l

This

by

is

to infinity 2

of

42 2*43 2a: 2 44V4..., when |*|<1.

a convergent recurring

series,

and the

scale

is

(1 -a;)

8,

= =

-3

.

l*x

-3

.

2 2 *2 - 3

.

3.

and

S--

8 .

Denoting the

sum

FINITE DIFFERENCE EQUATIONS The sum

(5)

n terms of a recurring

to

similar to that in Ex.

Find

3.

the

to

can be found by a process

or in the last example.

(3),

sum

series

363

n terms

+ u^x 4- u 2x2

of w

+

. . .

,

where u n

+pu n _ t + qu n , z =

n>l.

for

=u + u lx+ u2 x 2 + ... + u n _ xn ~ 2 + pu n _ 2 xn - + pu n _^xn pas n = 2?tt # -f >Wi# +

Let

l

6n

l

1

then

. . .

Since w r +7>w r _ 1

+ ^w r _ 2 =0

for

::

*n (1

+ JP^ +

^

2 )

t

r>l, by addition,

~ uo + w i +P ?/o) x ~ u n xn (

which determines the value of sn unless a; is a root a of the equation 1 In this case the sum may be deduced from the general result by supposing that #,

Given the

(6)

uv

w n -i-^ 1 w n _ 1 +

scale

ur _ v we can

...

find the value of

4-j? r w n _ r

= 0, and

the values of w

,

u n by

(i) finding the generating function (ii) expressing this as the sum of partial fractions of the form ~ ~ m where a is a root of xr A(l-ocx)" +p 1 x r l + p 2xr 2 + ... -f 7? r = 0; and n (iii) expanding the partial fractions and collecting the coefficients of x ...

;

}

.

Examples

Examples of Finite Difference Equations.

5.

any

are given in the next article.

r

+1

consecutive terms of the sequence (u n )

u n +PiU n ~i+P2U n _ 2 +

equation

where p l9 p2 ...pr are constants. ,

is

From

...

Suppose that are connected by the

+;p r w n _ r = 0,

................. (A)

this point of view, equation (A)

called a Linear Finite Difference Equation with constant coefficients. So far as equation (A) is concerned, the first r terms of the sequence

may have any

values whatever, that

is

to say they are arbitrary constants.

Thus the process described in the last section enables us to find the general value of u n which satisfies equation (A), and shows that this value involves r arbitrary constants. The general value of u n is called the general solution, and the process of finding it is called solving the equation. The process leads to the following results. The general

(i)

(A

-f

wJ5)a

n

where a,

,

u n + pu n _ l + qu n _ 2 ^=Q is wn = ^4a n -f J3/?" or are the roots of x 2 +px + q = Q and A, B are arbitrary

solution of j8

constants, the first or second form being taken according as (ii)

The general

where a,

j3,

y

solution of

//a^^y,

then

2 +px + qx + r = Q and A,

no two of

the roots are equal. n a n -f

u n = (A+ nB) u n = (A+nB + n*C)oL n

If a = P 7^ y, the solution 2A

u n +pu n ^ l + qu n _ 2 + ru n _3 = Q

are the roots of x 3

constants, provided that

a^/?

is

Cy

B,

or

a = j8.

is

C

are arbitrary

.

.

B C A -

-

-

GENERAL SOLUTIONS

364 For any

(iii)

other value of r the general solution of equation (A) has

similar form.

We If

give the proof of

w

,

(ii).

%, u 2 have any values, then

+ u 2*x2* + + u^x l

UQ

where

a, 6, c

where

a,

j3,

. . .

+ u n xn +

are functions of ^

y are the roots

. . .

=

-a + bx + cx2

_

u v w2

,

~ 2

==

3

P

a

.

J 7: (say),

Q

Also

.

of

# 3 + px* 4- qx 4- r = 0. If

(i)

no two of

where A, B,

a,

j8,

are equal,

y

C

are arbitrary constants. the fractions on the right and equating coefficients, Expanding

(ii)

If

a = /? 9^ y, then

P/4 (iii)

If

w

JJ

u n = (4

and

f^

JD

^L

-f

nfi)a

n

+ Cy w

.

a = j3 = y, then

B

P__A__ ~ + T - ocx

Q

(^

+

(l^axp

(1-a;

u n = ^4o

and

where

JB',

C' are arbitrary constants.

The first term of a sequence is 1, the second is 2, of the two preceding tertrw. Find the n-th term.

j&r. 1.

sum

Denoting the sequence by u l9 w 2 , ... we have u -u and n -2^Q

Hence

Eliminating A,

,

B from these (a

-

equations,

a

-a

we

0K = (2 - )3)a" -'

Now a+0 = l, a/3=-l and 2

1,

other term is the

U 2 ^2.

2-Ao^ + B^, l=.4a + ^, where

1

Un^AoP+Bp

^=

n~i-u

n

and every

a,

j8

are the roots of

find that

-

(2

- )"-'.

a>/8, a- 8=v'5, Therefore -1=0), and similarly 2-a=/?

hence 2-y8

if

J

2

.

I

-d

-

= l+a=a 2

(for

GENERAL TERMS Ex.

Find

2.

a recurring

the n-th term of

which

series of

Find also the sum of the first n terms when x ~ 1 and Denote the series by w -f u^x + u 2 x 2 + . . .

,

365 the first

.

let

the scale be

7+2p+# = and 20 + 7^4-2^=0, giving p= -2, un -2u n ^ l -3u n _ z -Q. The roots of x*-2x-3^Q are 3, - 1, and therefore Then

is

un =A.3n + B(-l) n

A+B^\,

where

Thus the nth

-^2.

3.4

n

term-w^a;"- -i-[3 + ( 3

.*.

sum to n terms = -

1

l-3#

4

---h

--

1

Now the

sum

to

Ex.

3.

The

solution

n terms

are arbitrary constants.

The values

of a,

ft

Hence the values

~iin

a,

==

;

...

n-1 + u n _ 2 =0 ^w a rea/ /orm. are the roots of x*-x + \

and

j8

Changing the constants,

this

^3

.77

may

(7,

be written

are

an

of

,,!.. solution

;

.

1

-w

scale

n ~l

-2 + 7- 20 +

1

of un + />^3n where

1

and the

v\n

of the series

Can

un

]x

1+s

J'iwrf the general solution is

l)

!-(-*)" _

/

and the

-3,

,

iZL-^^ limLJL

lim

q=

B=^\. n ~l

--

l-3w z n

4

/.

A-%

Hence 1

four terms are

,

j3

n are cos

t

sin

-^~, o

o

AA

un

is

7T

r + B am -

cos u

and so

~u

EXERCISE XXXVII 1.

Find the wth term and the sum to n terms of the recurring (i) 1

2.

is

Show

+2+5+

Show

vergent

.

;

1+2 + 5 + 12 +

(ii)

x |

|

-

<

sum

to infinity

..

.

series

:

.

is 7J-.

un -2un _ 1 + 4un _ 2 -3un _ 3 = Q

that, if

if

infinity is

+ ..

that the recurring series

convergent, and that the 3.

14

;

and

if

the

first

the series

9

three terras are

1

2unxn

+ 2x + 3#

2 ,

** ( i

^ ~\

9

_1_ 1*2^i ~^~ fjjL,

//Ix

; i

_

f?l* 4Utfe

I

n^

4.'/*^ xC/ <

-

^f*^ O*<

!

3 2 [The moduli of the roots of x -2# + 4x-3=0 are

1, 1/^/3, l/>/3.]

the

is

con-

sum

to

EXAMPLES OF RECURRING SERIES

366

3un -lun_ l -{-5un _ 2 -~u n _ 3 =0 and

4. If

%=

u 3 = 17,

u 2 = 8,

!,

the

find

un and u l

values of 5.

Find the nth term of the recurring

6.

If u n

series

l-f2-f34-54-74-94-...

.

the nth term of the recurring series 14-24-34-84-134- 30 4-... , find the sum to infinity of the

is

show that un ={(3n-4)( -l) n -f 2n + 2 }, and 8eries u If w n

7.

-^~w _ n

un ain6~u 2 q

+ gw n _ 2 =:0, where ^ 2 <4g, and -^7r<6<^rr show

1

9

2

2

?L:J

sin (n

- 1)0

2

-u^

sin

tan#=

where

(n- 2)0,

that

\/(4

2 ) .

J

Zu n x n Er n xn

If

8.

and

1

,

+p'x + q'x

2

then

,

are recurring series of which the scales are 1+px + qx 2 2(un + vn )xn is a recurring series whose scale is (

ZW n ZV n

9. If

1 -\-px 4-

2 flo:

) (

1 4- 2>'z 4- q'x*)-

are recurring series of which the scales are and vn + p'u n 4- q'u n__ 2 = 0,

,

un + pu n _^ 4- qu n _i ^

^

7^ 4^', then Sun vn is a recurring series whose ^ 4<^ and un ~PP' un-i + (PV +P' 2<1 - 2^0^-2 -pp'OU'lln-z + '^4-g ~0 n n /n n /n n " n then tively, 4- ^?a /3 4- 6V + Dai' p Un r n ^Aoc where ^4, B, C, I) are constants. Hence the scale of, Zun vn xn is 1 - OLOL'X) ( 1 - *P'X) 1 - affix) I ~ OL'fi'x).] (

where

2

>'

^>

2

7

r/

scale

is

respec-

'

,

(

10. If 1

2un

- 2kx 4- A: 2x 2

,

xn

Evn

are recurring series whose scales are then Eun r n xn is a recurring series whose scale ,

Hence show that Ununxn

Eun

(

xn

xn

is

a recurring

series

whose

scale

l+px + qx 2 and is

is

(1

a recurring series whose scale is 1+px + qx 2 whose scale is [1 - (p 2 ~ 2q)x 4- g 2 # 2 ] ( 1 - qx). are the roots of z 2 4-pz4-?=-0, then u n * = A(ot*) n [If a, fi where A, B, C are constants. Hence the scale of Eu n 2 x n is 11. If

is

a recurring

12. If

-

is

-

l+px + qx

13.

,

then

Uun

2

xn

series

+ B(fi*) n + Cq n

,

(l-a 2^)(l-^)(l-^).] , 2

= ^r u xn n==0 n

Prove that each of the COS a sin

use Ex. 11 to show that

series

4- X COS

oc + x

,

sin

(0

4-

a)

4-

X 2 COS (20 4- a)

4- ...

,

2

(04-a)+# sin (204-a)4-... 2 being 1 2x cos 4- x in each ,

Show

also

and that the sum to n terms of the second series may be obtained from expression by changing cos to sin in the numerator.

this

a recurring series, the scale that the sum to n terms of the is

case.

first series is

cos a - x cos (a - 0) - x n cos (a -f nO) + xn + l cos (a _____._____

+n -

Id) 9

FORMATION OF DIFFERENCE EQUATIONS Finite Difference

6.

367

Let (u n )

be a sequence in u n ^ r are connected by

Equations.

all of the terms u n u n __ v u n _%, ... an equation which holds for all values of n^r, where r is a fixed number. Such an equation is called a Finite Difference Equation.

which some or

An

,

important class of these equations has been considered in Art.

5.

Here we give methods which apply to various cases which occur in ordinary algebra. The general theory belongs to the Calculus of Finite Differences. To solve a difference equation is to express u n as a function of n in the most general form. The result is called the general in Art. 5

it is

clear that this

and from what has been said solution must involve one or more arbitrary solution,

constants.

A

particular solution

obtained by giving these constants special

is

values.

un

a given function of n containing one or more arbitrary constants, a difference equation may be obtained as in the following examples. If

Ex.

is

// u n

1.

~-Gn-2

where

C

an arbitrary

is

constant, obtain the corresponding

difference equation.

The required equation

is

found by eliminating

un ~Cn-"l

- l)w (n n

giving

C'

from

^n-\~ V(n -

and

1)

-2,

-nw n- 1 = 2. .

n n where A, B are arbitrary constants and // un ~AoL +Bf$ numbers neither which is zero, prove that unequal of u - (* + $)u -=0. fl _

Ex.

2.

The

result is

n _i +aj8t*

n

found by eliminating A

Solution of

7.

,

a,

ft

are given

8

B from

some Elementary Types.

n // u n = au n _ l9 then u n = Ca where C is an arbitrary constant. For w n /w n _ 1 ==M n _. 1 /w w ._ 2 ==... =w 2 /w 1 = a, and u v may have any value. (1)

// u n = a n u n-i where a n

(2)

where

C

is

For *or

an

Wi/a i

n> then

arbitrary constant.

Un ~ l ~Un un-\ u n-Z *

un =

therefore

and

a given function of

is

mav have any

value

...

al

we

^-aa --a na n -i ul

a^

. . .

choose.

a ny

a

<*2i

w n = Ca xa2

...

an

SOLUTION OF ELEMENTARY TYPES

368 (3)

= bn

// u n a n u n _i

where a n and b n are given functions of n, the

general solution is

where

C

is

an

arbitrary constant.

Let vn = a 1 a 2 ...a n

The

result

,

and divide each

side of the given equation

by v n

.

is

Un/Vn~Un-lK-I = b n/ v n' In

this,

writing n

Whence by

n~

1,

2 in succession for w,

...

we have

addition,

wK This

2,

the

gives

^1/^1

= &iM + 6 2 /v 2 +

fe

the value of

question, for

in

revsult

+ & nK - iKu l lv l -bl jv l

is

arbitrary.

In examples of this type, it is better to apply the rather than to use the actual result. -l)u n - nu n _ 1 ~2. we have Dividing by n(n-l), u n \n - u n ^l(n - 1) =2/ Ex.

1.

1,

nn

-2,

...

2 in succession for

U.2

I

Hence un ~ Cn-2 where C. (See Art. 6, Ex. 1.) Ex.

2.

Find

(4)

we

2.3

C~ 2 + u v

the general solution of

(n

-

1).

n and adding,

(n-l)n) and

since u t

u n - nu n _ { =

I

is

an arbitrary constant, so

also

M.

and proceeding the equation becomes un j\n--u n _ l l\n-\ l, find that wn + (7) where C is an arbitrary constant.

Dividing by as before,

just described

Solve the equation (n

Writing n

is

method

n,

|

The general

~[n(n

solution of

^a n + c according as a and

?/

n

n

~aw n _ 1 = cjSn

+Y03--a)

is

or

are unequal or equal.

j3

For dividing by an we have ,

Writing n-1,

n-2,

...

2 in succession for n and adding,

a

LINEAR DIFFERENCE EQUATIONS u n = coc n

Therefore

=u

where B( Hence,

l

#

-

gn r

a

if

+ -^2 + [a a -

{

.

.

.

n

369

l

+ r-n \ a

j

an arbitrary constant.

is

/ot-cft/oi)

B

(B

2

_ an

-

n

n

,

/3-a which-

may

be written

where A(^B-cft/(ft~oc)) If a = ]8, then i/ n = cna n

The general

(5)

is

an arbitrary constant.

-

w n ~(a-f j3)w n _ 1

solution of n

u n ^AoL + Bft according as a and

/?

n

-faj3w n _ 2

=

u n ^(A + nB)oc.

or

are unequal or equal, where A,

i

n ,

B are arbitrary constants.

For the equation may be written

^-atV-i^K^-a^o). Hence by where C

is

u n ~ocu n

(1),

^^C^

t

an arbitrary constant, and therefore, by (4), w n ==^a w + Cj8 n+1 /()8-a) or (4+Cw)a n ,

according as a 7^)8 or

Cft/ (ft -of)

is

a = /J.

This establishes the result in question, for an arbitrary constant if a^/3. (See also Art. 5.)

Linear Difference Equations.

8.

au n + bu n _ l + cu n __<>+ is

With regard (i)

it

An

equation of the form

ku n , r = l

r is a fixed

n or constants.

to equation (A),

The general

... 4-

Here

called a linear difference equation.

k, I are given functions of are of this type.

(1)

.....................

number, and

(A)

a, 6,

...

All the equations in Art. 7

should be noticed that

u r+l9 wr4 2 >---

solution involves r arbitrary constants, for

can be found in succession in terms of u l9 u^

...

u r) which

,

may

have any

values. (ii)

// v n

is (he

general solution of

au n -f 6w n _ x -f cw n _ 2 +

and

wn

is

solution

any particular For it of (A).

. .

.

+ ku n , r = 0,

solution of equation (A), then v n is obviously a solution, also

+ wn it

is

is the

general

the general

solution, because v n involves r arbitrary constants. (iii)

The method

of dealing with such equations as

au n -f bu n ^ where

a,

fc,

...

k, r

-f

. . .

are constants and

trated in the next example.

-f

ln

ku n _ r = /, is

a given function of n,

is illus-

GENERAL TERM OF A SEQUENCE

370 Ex.

-Su^j-f 6w n _ 2 =n

Solve the equation u n

1.

The general

-5%^-f 6w n _ 2 =0

un

solution of

A

*^tl

Next search

Assume

w n =a + fru + en 2

a + bn+cn -5{a + b(n-l)+c(n-l)

powers of

coefficients of

6

-J-,

+ 2n

.

is

^

a

J,

-^

n,

we

then for

,

2

}

A

/

all

values of n,

+ 6{a

find that

2a

26-14c=0,

2c^l, c

5n

+/J.O

z

giving

-f

for a particular solution of

as a trial solution

Equating

2

~ n d so a solution

is

15) ...................................... (B)

Again, take the equation

and assume as a

a(5

Thus a solution

n

a 5n

un

trial solution

-5.5n " 1

.

,

.

then

+6.5n - a

)=5

n ,

? giving a = 6^.

is tS.

Finally, take the equation

Since 2

un

is

a root of x 2

an. 2 n ,

-5x + 6=0,

it is clear

we have a{n

- 2. giving a =

.

2n

-5(w -

Thus a solution

l)2

n -x

that a

.

2W

is

not a solution.

+ 6(n -2)2n ~ 2 } =2n

Trying

,

is

u n = -n.2n + l .......................................... (D)

The general

solution of the given equation

sides of equations (A)-(D),

and

is

obtained by adding the right-hand

is

EXERCISE XXXVIII The first term of a sequence is 1, and every other term the preceding terms. Find the nth term. 1.

2.

If

3.

The

2ttn

-tt n _ 1

first

the arithmetic

is

the

sum

of

all

=na, show that

is a, the second is 6, and every other term of the two preceding terms. Show that the nth terra is

term of a sequence

mean

thus proving that the nth term tends to J (a

+ 26)

as a limit as

n ->

oo

is

.

4. The first term of a sequence is 1, the second is a, and every other term the geometric mean of the two preceding terms. ~ Show that the nth term is ax where x ^i:{2 + ( - i) n 2 }.

is

DIFFERENCE EQUATIONS

371

5. If ww = />_! gw n _ a and un tends to a limit other than zero as w-> oo are 1 and -~q; also show show that |g|
^i

,

x

6.

i-

Show

+ qui

Ui

hm unn

that

.

,

1+q

that the equation

un - (a 4 ft + y) U n _i 4 (fty

4-

ya 4 aj8)

tt n _2

~

can be written in the form

Hence prove that

Un-oMn-i^Bp + Cy" fi^y or j9 y, where J5, C

OT

are arbitrary constants. according as as of the in Art. 5, (ii). solution equation given general 7.

If

un

nu n _ l + (-l) n and w ~l, show that un j\n-*e~* as ^->oo.

K/|n_= Wn_ /ln_-J-K-D 1

8.

If un = ?i(u n _ l

+ un _ 2

[Write the equation 9.

If

wn

(^

t*

n

)y

n

/[n..]

un

find

- (w 4-

in terms of

w and ^, and show that

u n _ l ~ - (^n-i ~ nu n-2)-]

1)

- l)(^ n ~i + ^n-2) an(i

V 3

j

l

M2 = l

= 0,

"

1

"

(

prove that

1)n 1

,

li

/'

L?

[Write the equation in the form

un - nun ^ = - [u^.! - (n 10.

Obtain a particular solution of

by assuming un =an + b. Hence find the general solution of the equation. Find the general solution of un -5u n _ l -i-Qu n _ 2 n [Assume un ~a .l as a trial solution.] 11.

12.

Find the general solution of

[Take as a trial solution 13.

wn -wW -i-Wn_2=ft 2 un ~ an 2 + bn + c.]

.

Find a particular solution of

un + u n-i sinna by assuming un a cos no, -f b sin noc. Hence show that the general solution is

~~ 2(1 14. If

Deduce the

u n un _i

=-

[nu n = (n l)?/n_ 2

un

in terms of

-f

cos

)

,

find

;

no^e that there are two forms.}

u^

ln

.

GENERAL SOLUTIONS

372 15.

Show

that the general solution of the equation

Vn-l + aun + bun-l + C = can be written in the form

where

A

an arbitrary constant and a, /3 are the roots of 2 - ab =0 a: - (a - b)x 4- c

is

;

unless

a.

=

fi,

in

which case

[Write the equation as ^(^ n _i4-a)4-&w n _ 1 4-c^0. v The equation then reduces to so that u n 4 a /

.

v n-l

vn

unless vn _j 16. If

4-

(6

- a)v n _ 1

4- (c

- ab)v n _ 2

= 0.]

Vn-i + 3ww - 4

w_ 1

-2=0, prove

that

>l.2 n 4-5 n ^

/I

17. If

<

wn w w _ 1

4- 5itw 4-

WTJ_I 4-

9

0,

2w

_,, ""

1

4-5 n

~1

<

then 2

,

18. If

tt

n+1

=-~

-'

2n-wn

and w l = a, prove that

a4-w(l-a)

and [Put

wn ^nvw .]

0,

Put

CHAPTER XXIII THE OPERATORS 1.

The Operator

variable n, (i)

J, E, D.

INTERPOLATION

Let u n be a function of the positive integral

A.

and consider the sequence u v u 2 u3 ,

The meaning

symbol A

of the

is

,

...,

un

...

,

.

defined by

^n^tVrt-f^, and A

regarded as denoting the operation which when applied to u n the Here n may have any value, so that difference u n+l - u n produces is

.

Jw 1 = w2 -^^ 1

Au 2 = u3 -u 2

,

etc.

Au n = v n *Aen Vj + Va + Va + .-.+Vn^Wn^-ti!. = w a - tij, = v n = w w+1 - ti w Vj *?_! w n !*__!, ...

(ii)//

,

For

,

whence the (iii)

,

result follows

by

addition.

d(u n + v n )=Au n + dv n A (u n + v n ) = (w w+1 + v w+1 ) - (u n + w n ) ~ *n) + (V i - V n =A
// u w v n are functions of n, then

For

(

In this connection

then As n

is

it

n-fl

w+

not equal to

.

)

should be noted that

.

if

Au v +Au% + ...+Au n

unless

w 1 = 0, for

and

Following the index notation of Algebra, the symbol that the operation A is to be repeated, and we write (iv)

,

and so on

;

thus

J2

denotes

TABLE OF DIFFERENCES

374

This process may be continued to any extent and, from the way in which the successive differences are formed, it is obvious that the numerical 3 coefficients in the expressions for J 2w n , J w n ... are the same as those ,

in the expansions of (1

-x)

2 ,

(1

-x)

3 ,

Hence we conclude that

....

4 ru n = u n+r -Clun+r _ 1 + C'2 un+r

(A) 2 -...+(-I)nu n in Art. 4. more concisely proved For the sequence uQ u l9 w2 ... a table of differences is conveniently

a result which

,

is

,

written thus:

,

UQ

u2

Ul

2 Thus, for the sequence O ,

u

I2,

2 2 , 32 ,

Au

...

W3

,

4

1

1

A*u A*u

(v) If 'u x is

_.

where u n 9

222

a function of any variable

n2 we write ,

16...

5

3

...

7

x, the

meaning

A applied to u x is defined by Aux ^u x ^-u x All that has been said about the symbol A still

of the'operation

.

successive differences of II u x>

I/

holds, for in finding the

u x we are merely concerned with the sequence

07 x+V ux+2>

Illustrations,

(i)

u n = n(n~ l)(n-2), then

If

J rw n -0

and (ii)

If

w n = n3

2. Special

,

since

for

r>3.

w 3 = n(n~l)(w-2)-f 3w(w-l)-hw, by the preceding,

Cases,

(i)

The following notation

xr =x(x-l)(x-2)...(a;-r-f 1) and 3_ r = If

x

that

is

is

variable

and J

is

-

often used

-^-^-

x(x + l)(x-f 2)

...

;

we

write

-

applies to x, then

Jx r = rx r _

1

........................................

(A)

THE OPERATOR & Similarly

we can show that Ax^ r

-

=

ra( f +D,

375 ...........................

(B)

being assumed that no denominator vanisJies. These formulae may be compared with

it

x r = rx r

-r--

ax Ex.

Shaw

1.

~l

and

-_-

ax

x~ r = - rx~ (r + 1 \

that

n(-l) + (n-l)(n-2) + ...+2.1 w(7i-l)(/i-2) + (n-l)(n-2)(n-3) + ... +3

.

2

.

=(/t + l)?i(n l=(n + l)n(w

-

1),

and so on.

8

If

and

is

sum

the

of the

Jn3 = 3wa by

since

>

first series,

Art.

1, (ii),

S=^ (n + 1) 3 +

vhere

C

is

a constant.

Putting n =2, we find that (7 = 0. Hence the result. Similarly for the second Of course, the rule in Ch. VIII, 3, might have been applied.

A

(ii)

by

Polynomial.

a?

x,

dent of

1,

#,

a? -2, ... such that

a polynomial in x of degree r, dividing in succession, we can find a a^ ... ar indepenIf

ux

is

,

ux = ao + a i^i + a 2 x2 +

J w^ = a l + 2a2 x x + 3a 3#2 -f

and then

and so

on.

series.

Finally

A ru x = a r

+ a rx r

,

. . .

J r ^wx = 0.

and

r

.

. . .

,

|

2.

?x.

For

//

A mu

x

w
as a factor.

\

The Operator

3.

.

when applied

to w n ,

is

E.

The

the operation denoted by E by unity. Thus for all values of n,

effect of

to increase

n

Eu n ^u n+l a+

.

a constant, the following equations define the meaning of E, Ea and a$, regarded as operators.

If

a

is

(E + a) u n = w n ^! + a?/ n = (a + J0) w n Further,

and so on

;

we

write

thus,

if

r is

y

a positive integer,

,

E + a,

LAWS OF OPERATIONS

376 Hence,

p, q are positive integers,

if

E*E*u n = E* (Eu n ) = Eu n ^ =

E*E*u n = E*>+
so that

Again,

.

a, 6 are constants,

if

(E + a)(E + b)u n = (E + a) (u n+l + bu n )

+ abu n Thus, so far as addition (subtraction) and multiplication are concerned, the operator E combines with itself and with constants according to the laws of Algebra.

The same

for A, for

is true

4* n = ti w+1 -w n = (^-l)ti n More defined

4.

x

9

generally,

if

ux

is

a function of any variable

x,

the operation

E

is

by

Fundamental Theorems.

and

.

r is

a positive

// u x

is

a function of the variable

integer, then

(i)^ux ^ux+r -C[ux+r ^^C 2 ux+r ^-...+(-lYu X) ......... (A) = ux + C[Aux + ClA 2 ux +...+A r u x ....................... (B) (ii) ux+r r

The first theorem Proofs. on writing x for n. More easily thus

follows from the reasoning in Art.

1, (iv),

:

(i)

J'.-(-l)X = {Er - CJ^- 1 + Cj'- -...+(- l) r}ux

Again,

(iii)

(ii)

If u x

u x+r =Eru x = (l +A) ru x and therefore ,

is

a polynomial in x of degree

For the left-hand (iv)

//

s n is the

side

sum

=4^ =

to

n(n-l)

l,

then

(Art. 2,(ii)).

n terms of the A

r

series

n(n-l)(n-2) '

u

-f

u 2 -f u3 -h

. . .

,

then

METHOD OF DIFFERENCES For w

=

n

l+^) ~X>

377

therefore

Similarly

Adding and using the Ex.

4

3,

Find a

1.

results of Art. 2,

n which has

cubic function of

Ex.

the result follows at once.

1,

the values

-3, -

1,

1,

13 when n

= \,

2,

respectively.

Denoting the function by u n

and A r un =

,

we have

for.r>3, since u n

Also

tt

is

a cubic function.

= ^ w -X =

-

_

+6

j&a;.

Sum

2.

the series

2.3+3.6 + 4. Here un ~ (n + l)(^ 2 +2). Writing down the as

on the

first

right,

This

four terms,

that

is

ll

+ ...+(

a cubic function of

we

un

find,

^=6, 4^ = 1

?i,

A r un

for

44

18

Aun A 2u

A^u^U, J 3 u 1 = 6.

so that

6

26

12

n>3. 90

46 20

14 6

Hence by

(iv),

i9 12

.

Ex.

3.

//

positive integers,

For by Art.

if

2,

A

(Cf.

Ch. VIII,

5 = *n -C7J(a: + l = (-l) n \n then 8 if r>n and S

applies

(ii).

,n(n-l)(n-2)(n-3)

1A 14

-

to

x,

S=^(l

4,

Ex.

1.)*

r,

if r

- E) rxn ^(-l) r A rx n f

are

= n. and

the

result

follows

BERNOULLI'S NUMBERS

378 Ex. 4.*

(Bernoulli's Numbers.)

For nr =tf n O r =(l+J)n O r and

If

Br

is

defined as in Ch. VIII, 8, (5), then

-l+$-...

+ (-ir


m>r;

if

i

}r

1)

...

......................... (A)

...

Li

II.

#r = l f + 2r + ... +nr we have f(n + l)n (n + l)n(*-l)

Hence

if

,

l)n(n-

(n-f

(*~r +

l

'

*'~\

LT+JL

[3

[2_

Now, by

definition,

Br

coefficient of

n

in

Sr

hence

,

Again, equating the coefficients of w in (B), A3

J "T + {AZ Hence

~- +( ~ 1)f "

Ar^

^ =i

and (-l) rJ5r =J5r Hence equation (A) holds for r>0.

5.

5.

J0r =l,

Apply equation

in

Operators '

'

following

"

r>l,

if

Moreover,

Ex.

r=

r}

proof

(A) of

for

r>l

Ex. 4 to show that B^

a Fractional Form.

of the

theorem in Art.

4, (iv)

(Ch. VIII, 8,

(5)).

^.

Many

writers give the

:

En -l n(n-l)

.

n(n-l)(n-2)

i

Here the operator is written in a fractional form in order to transform a polynomial in E into a polynomial in J. This result was obtained his collected works).

by Cayley by a much more

difficult process (Vol.

IX, pp. 259-262 of

LAGRANGE'S INTERPOLATION FORMULA The

when

result,

obtained, can be verified by multiplication

379 and addition ;

this fact justifies the process.

But

Thus the

at every stage the operator must be regarded as a whole.

E -l n

equation Sn

(E -

does not imply that

In fact the last statement

is

~~E^T Ul 1) s n

= (En - 1) %.

not true, for

(E-l)s n = s n+1 -s n = u n+l

(E -l)u l ^un ^ l -u l n

and

.

We

conclude that, in such transformations, the operator can be put in the form /(J)/^(J), where f (A) and (d) are polynomials in J, provided that

f(A)

is divisible

6.

by <(J), andf(A),

(d)

are not regardedas separate operators.

Suppose that y

Interpolation.

a function of x whose value

is

has been determined by experiment, or otherwise, for x = a,

The problem

of interpolation intermediate values of x.

The graphical method is to plot the points draw a smooth curve through the points.

(x, y) for

We

'

*

6, c, ...

.

to find approximate values of y for

is

x = a,

6, c, ...

assume that

,

and

this curve

represents as nearly as possible the graph of the function y. Let us suppose that the equation to the curve so drawn is y=f(x). It is natural to take for f(x) the simplest function of x which satisfies the conditions.

If

y

is

known

for

n values

we assume that

of x,

where the n constants p p^ ... are to be determined by making the curve go through the n points. If only two values of y are known, the curve is the straight line joining ,

the corresponding points, and we have the rule of proportional parts as used in working with tables of logarithms and trigonometric functions. 7. ively,

Lagrange's Formula. we assume that b)(x

(x

y=syi

For y

is

when x = a, Similarly

c)

(a-b)(a-c)

If

+y*

(x

y=y

y2 y$ for x=a,

-a)(x

6, c

,

l9

(x

c)

(b-a)(b-c)*

y3

a)(x

respect-

b) *

(c-a)(c-6)

a quadratic function of x of which the values are yl9 y& y3 6, c. if

y = j/1

y2

,

y s- y

>

-

^3

when x = a,

-^4

(x-b)(x-c)(x-d) \

'

_L^

'

b,

.

+ three

c, .

d we take ..

similar terms

;

*(a-b)(a-c)(a-~d)

with similar formulae when more than four values of y are given. B.O.A. 2B

INTERPOLATION

380

8

.

If

the values of y are

known

of applying Lagrange's formulae,

we

For example, suppose that y

Assume that y is

a positive integer,

at equal intervals, instead

use the operators

y&

y\,

a cubic function of x

is

x

for values of

;

y%,

then

E

for

j/3

J ry =

and A. x v x2

X

XQ

for

r^4, and

,

x$.

,

if

n

= (l+A) n y0)

n

y n ^f! y therefore

Hence the equation

represents a cubic curve which passes through the four points (# ?/ ) (x l9 T/J), ... and gives an approximate value of y corresponding to any value ,

of x within the specified range. Ex.

1.

Given

tJiat

sin 45 =0-7071,

sin 50 =0-7660,

in 55 =0-8192,

A-rn

60 =0-8660,

n62. Let

1*3

= 10 4 sin (45+5x).

Construct a table of differences as below.

Au

u

A 2u

A*u

589

-57

Mj-7660 532 "

7

-64 468

J^ = 589,

/.

Assuming that

r

J w

J 2 w = - 57,

-=

A92

7.

-

M x(x-l)(x-2) _j

x = 1-4 and

so that

u

=7071 D

= 1-4x589 MA

=

2

=

824-6

=

- 0-7 x 0-4 x 57

|

ILl-JpZJ

A 3 u Q ^ 1 -4

x 0-4 x 0-7

=

0-392

7895-992

I

15-96

u x = 7880-032 .'.

The

-

r>3, approximate values of w^are given by

for

x(x-l) -L_J

52=45 + 5z,

J 3w

sin 52 =: 0-7880 to four places.

correct seven-place value

is

0-7880108.

- 15-96

BESSEL'S INTERPOLATION If

wo only

FORMULA

381

use the values of sin 50 and sin 55, the rule of proportional parts would

give sin 52

=0-7660+f

x 0-0532 =0-7873.

If UQ and several other terms of the sequence U Q u v u%, ... are known and approximate values of the missing terms are required, we may proceed ,

as follows.

= 4-7046, w 3 = 5-6713, w 5 = 7-1154, find approximate w = 4-3315, and w 4 Four points on the graph of u x are known. We therefore assume that it can be 4 represented by a cubic function of x. This is the same as assuming that J wa = 0. Ex.

2.

^

//

values of u

.

.

Hence we have approximately 0,

/.

+ ^i) ^11-4619. = w 4 6-3199 (approx.).

w 2 ^5-1420,

NOTE. In this example w ~tan77, w 1 -tan78, w 3 tan 80, W 5 ~tan82, we have found that tan 79 ^5-1420, tan 81 ^6-3199 (approx.). The correct values are 5-1446 and 6-3138. It should be noticed that the

approximations just found are

given by )

9.

=5-1879,

much

closer

so

than those

w 4 = \ (u 3 + u 5 )= 6-3933.

Bessel's Formula.

second differences

In using Mathematical Tables, the effect of be easily and accurately allowed for by using a

may

formula due to Bessel.

Let u_ l9 u

,

u lt u

be four consecutive values of a function u XJ as

and suppose that values and 1.

given in the tables, of

x between

of

u x are required for values u -i

Denoting ences by a

9

on the

first 6, c

a

and second differand d e, as shown

,

b

9

right, Bessel's

formula

u*

is

f e

c

\

x(x-l) d + e which

is the

same as a) ............................ (B)

The values little

labour

is

of

uQ9

a, 6, c

can be taken from the tables, so that very

required in using the formula.

THE DIFFERENTIAL OPERATOR

382

if

To show that this gives a good approximation, and to estimate the error, we neglect differences of the fourth and higher orders, we have

Now

e-d^f; and

w -w_;i=a, b-a=^d,

x(x-l).

+ l)x(x-l),

(x

'

V

e-d\

x(x~l) fd + e so that

if

?'

x is

Hence the

(x

+ l)x(x-

1)

the approximation given by (A),

error

C/5^ to6/65

Ifo. 1.

therefore

is

approximately izx(x- l)(2x~

of square roots

to

I)/.

find

Using Barlow's tables, we take

045020= a -:

11 1-085553

045001=6 M1

= N /12350 = ............... 044983 -c.

^2-^/12360 = ................ (The differences

a, 6, c are

z^-4,

given in the tables,)

For >/12344, we have

c-a-^ -

-000037. -J*(3?-l)=--06, w = 111-085553

Ja?(a?

-

1

)

(c

-

6= a) -

-0180004

-00000224

~

N /12344=!ll-1035o7>

The

result is correct to the last figure.

10.

The Operator D. du

the symbol

-r-

is

When we d

dn

write

u/ d

\

n

regarded as an operator, which

is

sometimes denoted by

118

If

a+

a

is

a constant, the following equations define the meaning of

A Da and aD, regarded as operators.

D + a,

THEOREM

LEIBNIZ' Hence

(D + a) (D + 6) u =

a, b are constants,

if

383 -I-

(Z)

a) (Dw

-f

6w),

= Z)2 w -f 6Z)w -f aDw -f afrw = {D2 + (a + b)D + ab}u. (/) 4- a) (D -f 6) u

and

Thus, so far as addition, subtraction and multiplication are concerned, the operator D combines with itself and with constants according to the laws of Algebra.

The n-th Derivative of a Product.

11.

d

du

dv

dx

dx

dx'

x

of

D D

Let the symbols

l9

If w, v are functions

denote differentiation with regard to

2

x,

D

suppose that only operates on u and its differential coefficients, that 2 operates only on v and its differential coefficients, so that

and and

D

Dn

D

As

/

l

\

(uv)

=

ax

and addition and / ^

&u

t

D

dv

D 2 (uv) = u jr\

v,

i

\

ax

\dz/^ that

WV '"~'

is,

-j

1+

*'

(uv)

=^

This result

Ex.

is

(

1+11 -

uv >~\

known

If ij~u

+v

and

-

(uv)

n -f D n = (D l 2

.

)

uv.

where u m

=x + (xz - ^)

vm

,

=x -

-f

uv n

,

x.

- k) ",

/^

W = 0,

1, 2, ...

.

we have

Differentiating with regard to x,

_ i +x Wi

'

2

)>

often written in the

is

(

2)( u

2

with regard to

=

tn-i

-

. . .

suffixes indicate differentiation

ww

ax

21

2

as Leibniz' Theorem,

=

1.

~=

are independent of each other, they obey the laws of multiplication, therefore

form where the

d

,

2

n \

,

and hence

;

_ jt)"^ = tt w (a; 2 - Jt)"" (x z

;

hence (# a - k)^u l =u/m.

Differentiating again,

.'.

Similarly

~*

*-

*

(x

-

2

(#

2

k)

-

wa

!

-f-

arM!

k) v^-^-xvl

= (x* - kf*

- w/m 2

0.

- v/m?

0,

.

ujm*

;

- y/m 2 =0. whence by addition, k) y 2 -f ar^ (x Differentiating w times by Leibniz' theorem, 7 2 ^-i (* k) yn +* + n.2x. t/w+1 + 2

-

-

This example leads to a remarkable theorem on cubic equations, given

in

Ch.

XXIX.

USK OF OPERATOR A

384

EXERCISE XXXIX 1.

Find the polynomial of the third degree in n which has the values when w = l, 2, 3, 4 respectively. (Use Art. 4.)

2, 11,

32, 71, 2.

Use the method of Art.

P+3 +5 + 3

(i)

3

u n = xn and A

3.

If

4.

Show

(ii) I

;

apph'es to

/-l, 2 or

S be = u r (* n/) r

the

(ii)Lct

.

show that

0,

according as n

of the series .

(

r

~y)

n

n~2

or ?z->2.

n(n +

l),

I,

2

sum

n~ 3

if

n>2

of the series.

S=

then

',

?/

.

;t,

2

[In each case let

Also J

+ 22

(w-l) C?4-(n-3) r^ + (w-5) Ct4-...^2

(iii)

r?

.

xn -C'i(x-y) n

2

Let

22

2

that

(i)

(i)

sum to n terms 3 2 + 3 a 4 2 + ...

Ex. 2 to find the

4, ...

(

1

=

- E) n ii

-

1 )

(

nJ w

.

etc.

n, |

w r = (a:-r)(y-r), then S = ( -

l)

nA n

Also

u^.

w if n>2 Aut = (x-l)(y-\)-xy=-x-y+\ J M = 2, J w r = n w = r then S = i-{(^+ l) ~(E I) )w (iii) Let w r n n Also w = 0, Jw = l, ^ w = 2, and JX = 0if /. S-=l-{(2 + A) -A }u 9 8

9

2

,

;

2

.

5.

^x(x-l)(x~2)...(xr + l), show that - 1 )a:r_ 2 -h (x -h n) r = a- r + ( ?rar r _ l + CJr (r

If x r

the last term being r(r-

Verify that,

6.

If

if

n

1)

...

(r

3 and r

xr = x r -f o^^!

-}-

~n-\- l)x r _ n if

. . .

a^l-r(r-l),

[Equate to 7.

Show

,

,

if

prove that

6=A-(r-l)(r-2)(3r-5).

the coefficients of x r

JV+^fc+JrJ'rj-JL

(ii)

8.

.

~1

" and x r 2 on the

right.]

that

(i)

[Use Ex.

.

the identity becomes

2,

6a:r _2 -f

.

r>n and C^jr

J rx r + 2 =

J

[x

2

-f 7-a:

4-

;

^r (3r -f 1)]

. |

rjh

2.

6.]

Prove that xn -Ci(x+l) n + Cl(x + 2) n -...+(-\Y(x\-r) n

+\

= - Yl {& + r-r + iV(3r -f 1)}

and [Use Ex.

(

7.1

1

|

when

n~r-\

when n = r 4- 2. r+_2

1,

EXAMPLES OF IDENTITIES 9. If

385

none of the denominators vanish, prove that 1

x

(-ir\n

= __

Deduce that

4- C? ^-fC? ^-"'+ 1

+

(

1)n

1

1 -

n(n-l)

1

n

-

-

(x+l)(x + 2)

2

n(n-l)(n-2)

I

3 10.

Prove that 2

1 *

~r 71

r-

22

1.2

m+2

m-f-1

n.2.3 n ~ l

n

^3

m(w-f-l) 11. If u n

2W

yt(n-l) T

nt/^x/.^-.um(m + l)(m + 2)

,

'

/

i\n.

m(m-f 1)

...

(m-fn)

~\l(an + b), show that

(i)

A ru ~( -

r

l)

r

111

(an

+ b)(a .n 1

n ' n(n-l)

a '

n(n-l)(n~2) "

a2 '

"3

= l/a;(a: + l)(a; + 2)...(a: + r-l), show that + n)_ T = a?_ r -6 ?ra- r+1) +(7?r(r+l)a;_(r +8)-...

12. If a:_ r

I

(

n+ 1

to

(

13. If n is a positive integer vanishes, prove that l

+

n

n n ~l) (

and y

is

any number such that no denominator

n(n-l)(n-2)

n. ^n(n-l)

terms.

_

y+

l

n(n-l)(n-2)

/.x (in)

each

series being

continued to n 4-

[These results follow

1

terms.

from Ex. 12 by putting successively

r

= l, x=

-(t/4

1)

;

USE OF THE OPERATOR J

386 14.

Show

tfcat

ln + :

1

+ 2)... 1

V

/

,*//*.

i

I\/,M

i

* //

o\

IW/-M

i

-...to n +

-.

_

15. If

>

...

'

16. If

(a-n-f

a(a-

a-b a~6-fl

a-6-fr-l

tti=7

^2=7-77 6(6

b

_

^ rWn=

.

1)

...

(a-n-f 1)

a(a-

...

1)

(a-n-f

1)

e tc., as in Ex. 15, prove that

1)

a-6 a-6-fl a-6-f2 ^ T 6^1 6^2~

[Putting n = l in Ex. 15, find

17.

T;>

+2

prove that

66-16

1

ln '

1)

a-6 a-6 + 1

w

'

o\

..

___

__a~b ~~ a(a- 1) U==

i

terms=

1

--...-

^^

~w

o\/ M

i

Au 0t A*u

a-6-fr-l *

6-r-fl

etc., in

,

succession from the equations

Prove that fI

_? )

__

"*

a

M n

a 6

+

a

)(1

6

6/\

\

j

...

?L 6

Ti-

n(n - l)(n-2) a(a- l)(a-2)

n(n-l)a(a-l)

JT~

(1

6-2/"\

1/\

6(6-1)

6(6-l)(6-2)"

[3

to n-fl terms.

[Using Ex. 16, the left-hand side = ( 18.

Show

that the

sum

n J nw

n =(l -E) u

.]

of the products two together of

1

x is

l)

9

JL JL x + l'

i

x+2

equal to

n(n-l)

(

n-2

"--

I

~+ (n-2)(n-3)

1

"-

[2

[Equate the

coefficients of a* in

Ex. 17 and put

#=

-6.]

to

n ~ 1 term8

1

r

INTERPOLATIONS 19.

In Art.

4,

Ex.

387

has been shown that

4, it

Prove also that

and use each of these formulae to show that B^ = [Equation (D) of Ex. 4 can be written

and

(

^12500 = 1 1 1-803399, ^12510 = 111-848111, ^12520 = 111-892806,

20. Given that

^12530= 111-937483, ^12516 = 111-874929.

show that 21. Given that

show that

sec 89

4'

sec 89

5'

= 61-3911, = 62-5072,

sec 89

6'

=63-6646,

sec 89

7'

= 64-8657,

Bessel's formula gives

sec 89

22. If

Sn = l n +

on

on *

/

L

w

+rz+.-+ ^ 1 i

2 fl .l f

3n

40" = 63-2741.

5' i

i

\n

+

to oo

971

n

r

r

.2 4 .3 + WJ^^ +--+_ ,.,

A

IJL

J"5. = F-i +

or-i

+ LL

(iu) (iv)

(v)

and

. . .

= 5, _,.

J

Use

(iii)

(iv),

S19 SZ9

or (iv) to find 2,

4,

3,

)

S19

showing that for

6,

-T

B

+-

i

L_

(iii)

5,

Qn

rn = l n -"r? + r5-- + ('" I m

(u)

...

7,

^n /e=2, 5, 15, 52, 203, 877, 4140. we have SW =(J + l) n and Sn ^ = (E - l)n S on

23. If

+

L_

Sn = -Sn_ 1 SB = 5 ^ 1 n = l,

[(iii)

....

d

I

or~i (ii)

prove that

,

I

i

= Tn.!

Using

(ii),

show that

for

n=l,

rn .e=0,

2,

3, 4, 5,

-1, -1,

2, 9.

to

.]

<, prove that

CHAPTER XXIV CONTINUED FRACTIONS 1

Definitions.

.

Any

(1)

expression of the form .................................

i-: called a continued fraction,

is

The

and

-

Oj

may be supposed to be attached fraction may be expressed in the form

- Ui ~ F-a + JU

where the

and

a's

ai

^2/a 2

by

stopping at ,

b n /a n

.,

.

6's ...

&3

&2

~

"

is

,

convergents are

av +

,

1

is

.............. (B)

and so any

6's,

........... .......

called a convergent.

flu

. . .

~

..

........

(C) VI \

be positive or negative numbers. The quantities are called the elements of jF, and the fraction obtained

,

second, third,

3

may

any particular stage

,

bn

"...

1

--

to the

signs

continued

It

a

written

is

obvious that the fraction

is

6<> "

a:

,

H

6<>6q----

. . .

,

#2

unaltered

Thus the

first,

.

2^~ ^3

if

6n

,

an

,

b n+l

(u

= 2,

3, ...)

by any number k or if the signs of 6 n a n and 6n fl are Thus any continued fraction may be written in the form

are all multiplied all

(C)

changed. where a 2 a 3 ,

Ex.

1.

Prove

,

are

...

,

that, if

all positive.

JLJLA

2.

fraction

on the

left is

*

elements,

*

l

1 .

4+ 4+

4-f

The

n

each fraction contains

2+ 4+

to s equal ^

4+

2-h

1122 ---- -

-

2+4+4+44-

Formation of Convergents.

(1)

...

""

=x

1111 -

~

x ^ 2+4+2+4 +

Let u n denote the nth con-

vergent of the continued fraction, _.

62

60

a 2 + a3

We

shall write

+

bn

.

w=^) 1 /

a,

J

an + 1,

,

so that

w=^22

fi,=r 1 1

where

J a2

a2

THE RECURRENCE FORMULAE Observing that u3

389

be obtained from w 2 by changing a 2 into

may

a2 4- 63/03, we have u3 =

(

and so u$ = psjq& where y3 - a3j> 2 4- 63^ and Proceeding thus we can show that, if p n and values of n by the equations

? =

?-! + &nPn-*

Vn

?3

==

q n are defined for successive

= anqn -i + b n qn _ 2t

.................

(A)

the nth convergent of jP. p n/q n Equations (A) are called the recurrence formulae. It will be found convenient to write = 0, ............................. i and

then

is

yo==

and

it

will

Pz^^Pi^^Po an d

be seen that

important to state

For

Definition.

#Ae quantities

?2 ==a 29 i r

rather more precisely

all this

with the initial values

If F. convergent of

pQ = l 9

p n and

qo

=

Pi

9

= ^i>

q\

:

q n are defined in this way, then

,. .

.1

II

'

I

J.

I JL

I

............... (A)

,

=1

Assume that this holds for the values 2, 3, Proof. the nth convergent by u n , then

Now p

so

the continued fraction

p n and q n are defined by the equations = a n?n-l + 6wyn _2 Pn = ttfiPn-l + ^n-2 ?n

Theorem.

+ ^25 o> r

n^2.

equations (A) hold for (2) It is

(B)

?0

...

p n /qn r of n,

is the n-th

and denote

f,

also u r may q r _ z are independent of a r and b r be transformed into u r+1 by writing ar + br+1 / a r+i for ar therefore r

_v

gv_

j,

pr _z

,

;

,

(

. Ur+1

r

+

Hence

br+l

^ \

w^

,

,

Thus the theorem holds therefore for

w = 3,

4, 5,

TPr ~ Z

P '~ l+

...

=

-

_ 5^i (r^i

-

r +-'

for

n = r-fl;

in succession.

but

it

holds for

n = 2,

and

INFINITE CONTINUED FRACTIONS

390 (3)

For the fraction

p n and

q n arc defined

b

by equations

Again, the fraction

b2

ft

is

bn

b

(A) with the initial values

b3

obtained from the fraction in

by changing the

(2)

signs of the 6's, so

that the recurrence formulae are

Pn = fln^n-l

~ &nPn-2>

?n

= a n?n-l - 6 n?n-2'

In the following articles we shall call the reader's attention to certain types of continued fractions which will be considered in detail later. Ex.1.

Prove that (I-

x)-^l+~ |j 3^ y

Calculating the convergents of the continued fraction,

we have

2.1-3&.1 6-8o;~

which provea the statement in question.

Notice that no fractions are reduced

to

lower

terms, until the final stage.

3.

Infinite

Continued Fractions. ,_ If

JL'

/I t*i

I

T"

694

"

With regard

__6q?_

to the fraction

6n )

-

if

the

number

of elements is finite,

F

is

called a terminating continued

fraction.

We

may, however, suppose the b's and a's to be determined by a rule In this case F stands for an is no last element.

of such a kind that there infinite

continued fraction, which

regarded as defining the sequence of

is to be

convergents Pl/

?2/?2>

>

Pn/

>

and we say that

the continued fraction is convergent, divergent or oscillatory this as according sequence converges, diverges or oscillates. is defined as If the sequence converges, the value of the fraction

F

lim Pn/q n

,

and we write jF

= a 1! +

-

--

a2 +a3 +

...

SIMPLE CONTINUED FRACTIONS Let x stand for ^ 2

Illustrations.

after the first is

Again,

meaningless.

,

2

may

x=lJ2i,

leading to

where every element

be treated as a number,

a result which

is

clearly

x stands for

if

333

t000

2T2T2T"* and we assume that x

to oo

...

-=

^ 2

// we assume that x

-3/2.

we have # = 3/(2-x)

333

391

is

we have

a definite number,

x = 3/(2+se);

z2 + 2z-3 = 0;

/.

We

have thus proved that unity, but we have not shown

.'.

that

a value

z=*l

or

has a value,

the fraction

if

'

its

-3. value must be

exists.

In neither of these examples can the first step be justified without knowing that the continued fraction is convergent. 4.

An

Simple Continued Fractions.

_

1

a i T^___ i

_ _ 1

1

a 2 + a3

expression of the form

..."

+

>

an +

where every a is a positive integer, except that a l a simple continued fraction.

may

be zero,

is

called

Theorem. Any rational number can be expressed as a simple terminating continued fraction which can be arranged so as to have either an odd or an even number of quotients. Let

A/B

be the number,

of finding the G.C.M. of r2 ,

...

A

and B being positive integers. In the process and 5, let a v a2t ... be the quotients and rl9

A

the remainders, so that

A^a^B + r^ and therefore

IB

A= #1 + B l ^prrBr

-^

JS==r 1a 2

,

r

Again,

if

>!, we can

r1

,

1

=flo"f -

This process terminates, and

A B

+ r2

g,^

if

an

r,

7-

rr is

,

r

111

-^

write 1

=O+ r 3

etc.,

,

1

7-, etC.

the last quotient,

1

a n ~an -l+

andifa n

= r2a3 + r3

1

RECURRING CONTINUED FRACTIONS

392

Thus matters can always be arranged so that the continued either an odd, or an even, number of quotients, as required. Ex. 1. Express ^g$- as a simple continued fraction with add number of quotients. al

157

.

i.e.

(i)

fraction has

an even and

an

I

JL

=2 +

(ii)

-gg-

a4 =5

i

i

i

number of quotients can be made odd or even as required For instance, the penultimate convergents to 157/68 in the two forms above are 30/13 and 127/55; and it will be found that (127, 55) and (30, 13) are The

NOTE.

is

fact that the

of importance.

the least solutions in positive integers of

68* - 157y = + (See Art. 9

respectively.

Ex.

Use

2.

The

z -x)~ as a continued fraction.

is

giving

*

- x) 2

(1

^ l/2a?

-

1+

"^4/3+"

1/3

-

L

^9/2*+

1/3

2x

2s 3*

+

1

-3/2

1- 2+ -9/2* + l/3 x 2x

2x 3x

~ 1+ last three steps, the

+ - 4/3 + -9/2* +

??.

""

In the

1

3.)

the H.C.F. process to express (1

H.C.F. process

68* - 157y = -

and

1

and Ch. XXV,

ri2 + 3^T*

statement at the end of Art.

1

has been used.

Recurring Continued Fractions. If after a certain stage, the elements recur in the same order, we have a recurring continued fraction. 5.

,

'

The recurring elements form the the non-recurring elements,

'

'

*

recurring period

such exist, form the

if

or the

*

9

cycle,

*

'

acyclic part

and

of the

fraction.

The

cycle is usually denoted

last of the recurring elements, _____ _____ _ _ ~~~ _

.-

-

-

_____ _

-.

~.

1+2+3+4+3+4+ A

by putting

asterisks

under the

* . .

1Q AD

/l ATir'I'f'.Afl VA.\3JLlv Uv7vA

MY J

r4"\7

-

._,....

_.

.

_

1+2+3+4 *

Here 3-f

-

4+

is

first

thus

and the cycle J

-

1+2s

^e acyclic part. J *

*

.

and

SPECIAL TYPES Ex.

is

1.

equal

to

(a

n+1

are the roots of x* -00;

ft

-^n+l )l(
1

Hence show

(ii)

is the

// a,

(i)

(l

+ l/n)a

- ax + 6 of x 2

according as ot^fi or a=j5.

a 2 ^ 46, and that

value

Here pn and qn are given by

together with

p = l, p^^a, qQ

where A, B, A',

E

f

= aq
4n

^ = 1.

0,

a^jS, then by Ch. XXII,

First suppose that

p

(i),

B^a=a+^

=A +B

gvng

5,

Therefore

are constants.

p 1 =Aoc +

=1,

^l/= yw =

thus leading to the

Next

let

n+1

-]8

(

w+1

and

)/(-]8)

fc

n (

n -)8

)/(-^

= a=)S, so that a 2oc and 6=^2, then by Ch. XXII,

A~B~\

,

If

=

first result.

Proceeding as before, we find that n -1 n yn = na ;>n -(n + l)a a 2 >46, then

a,

j3

>

^n /gn = (l + l/n)a.

,

l-W

^

5, (i),

A' = Q, B' = l/a., hence

n

Let |a|>|/?|, then

are real.

qn

j8

/a

n ->0 as n->oo and

'

2

a 46, then a~/? and pn /qn (l+I/n)a-+oc. Hence, in both cases, the continued fraction is convergent and

If

Ex.

2.

//

pn lqn

is the

11 prove that

and

n

is

its

value

is

a.

n-th convergent of the recurring fraction 1

aT 6TaT

If

its

= 0.

Pn = <*Pn-

(ii)

prove that the n-th convergent of

that the continued fraction is convergent if

numerically greater root

(i)

or

)

+ 6=0,

393

even, the recurrence

1

'"

'

6-f

p n - (ab + 2)p n _ 2 qn -(ab + 2) qn formulae give

Pn

=t>Pn

Pn-2 = pn

whence eliminating p n -i and pn -3> we have n +2)pn -2

p

(ab

+Pn-t =0.

If n is odd, we have the same equations as before, except that a and b are interchanged, leading to the same result. And so for the g's.

BROUNCKER'S FRACTION

394 Ex.

3.

//

pn lqn

n4h

ta the

convergent of

3a

52

1+ 2+ 2+

2+

1

I

(2w-3)

1)

<

Hence show

this holds

and

that the fraction is convergent

If vn denotes either

if

n

Replacing the

we

n-

write

ft

1,

- 2,

-(2- lK_i=(- l) t;'s

by

p's

n ~1

'

2 ^Ti

that its value is w/4.

we have

or qn ,

pn

"

2+

Pn l3n = l ~ 3+5-- + ~

prove that

and

"

2 in succession for

...

n -1

(2-3)(2n-5)

and observing that

=

jt)

3

...

and

n, therefore

.

^=

!(,-,). 1, we have

^n-i

2n-l'

,

" where

^4 is

"" 4

independent of

Next, replacing the

v'a

n.

by

find that qn

Again, the series fraction converges

-

1

and

=1 1/3

its

n~ 1, we find

and observing that ?n

whence we

1

3

1

Putting g's

1/5

value

A =0.

that

an(^ 5 i

l

(fo

F

we

== l

-(2n- l)7n-i-0,

.3.5... (2n -f-

""'"" ,/. "*

.

,

1.3. ..(2-l)""

- ...

is

1),

leading to the

convergent and

its

first result.

sum

is Tr/4

;

therefore the

is Tr/4,

EXERCISE XL 1.

Find the

first

four convergent (unreduced) of

w JLJLJL* 2+3+4+5 r\

2.

W JLAJL! l-2-3-4' /-\

f

Without calculating the convergent^, prove that 2

3

2+ 3T

5_ 1 1 23 T 5^1+ 2T 3T 4

4

'

3. Express fff as a simple continued fraction having odd number of quotients.

4.

The reckoning on the 85

5.

3- 4-

52 57 5

an

85 104 19

I

this.

Express

(ii)

20

5*

^

in the

form a l

greater than unity. 6. Express (2a -a- l)/(2a*-3a) calculating the convergents. 2

by

an even and

right shows that

1_J_1

52""

Explain

(i)

a * ~ a* in the

...

,

where the

"""

form

-

1+^ /i + /a +

a's are integers

...

,

and verify

INFINITE CONTINUED FRACTIONS Show

7.

that the nth convergent of

JL_1__!

2-2-2- -

2 1 is

+ l)/n, and

(n 8.

Show

395

that the value of the infinite fraction

is

unity.

that the nth convergent of 1

1

1

1

rr^r^r^r is

n,

and that the

9.

infinite fraction is divergent.

There being (n +

1)

elements in

2~2- 210. If

show that

all,

1

'"

2^"

l_(n + l)a?-n nx-n + T

*

a?""

a z >46, the numerically smaller root of x 2 -ax + 6=0

A JL J A a

a

[This follows from Art. 5, Ex. 11.

For the fraction

b

is

equal to

a

#

1.]

b

b

b

a+ a+ a+

"

*

*

a

+

"

'

pn = - a -)/(- 0), gn -( n+1 - j3n+1 )/(a - ]3) where a, ^ are -a# -6-~-0. Hence show that if a, 6 are positive, the fraction is convergent and equal to the positive root of x~bj(a + x).

prove that

(

an

the roots of # 2

12. If

p n jqn

IL

the (unreduced) nth convergent of

and the number of terms tively,

then u n \vn

is

in

>

n,

qn

respectively are denoted

by un

,

v n respec-

the nth convergent of

Hence show that

13.

Prove that

,

x

l

fi\ v

14.

;

_z_ z = _ 1

(

_o/JLJL-i-

1^4^1-4^'

^

\2~2^2-"V"

Hence show the nth convergent of the

first

fraction reduces to 2n/(n

+ 1).

Prove that 1

showing that the 2o

9fy.

Prove that, each fraction containing n elements,

J_ J_ JL JL

15.

_

1- 1+ 1- 3+ 2-

I2

infinite

22

2

J3_

(^-1]

_1

continued fraction

is

11

1

divergent.

B.C.A.

SIMPLE CONTINUED FRACTIONS

396 Prove that

16.

~

' ' '

2a + 2n -I+5= k n a (a + 1) (a + 2) (a + n)

......

'

2a

if

showing incidentally that,

Use

show that the

this to

fraction in Ex. 15

Putting

a~ -m,

where

in Ex. 16,

m J-itJli

a

O

iftl

m is

~~ ju'tTl

~~

9

then

divergent. l)
a positive integer, show that 2

I

~~

x

(w--2)

jw-l) i.

is

!){?! -(a 4- w -

[Here gn -(a + H)g n _ 1 ^=(a + n17.

. .

.


2

Also prove this result by induction. 18.

For the fraction

-

+

1

Hence show that the n

...

prove that r

,

n+

^1_1_

fraction

12

is

13

~ convergent and equal to l/(e

1).

a positive integer, prove

is

n

by induction that 2 11 ~ n 1 "" ?T-~ "2+ w+~2 2"+ 1 + 2 n - -2

[

w"+ w - 1*+

-t-

'

For the fraction I

2

22

1_

r+ 7)

n

-

prove that

1+1+

1"*-

^

1

3a

11 + -

-

fraction

is

. . .

(

where the

,

...

l

_

,

...

3/3

>

.

2.

l *

*

'

~j

except that a x

1

1

then x2) ^3

"1 1 -

are called the (partial) quotients

X2 =
'

are of the form

+ ~~"~" r~T

a's are positive integers,

Here a l9 a 2

71

I)

""

convergent and equal to log

1

at

2

r+

+ -

Simple Continued Fractions xl

(?i-l)

"*

t>

Hence show that the 6.

...

\n-i I

n

20.

-

+

?n

^j

J>n_

in -hi

19. If

~^- _ 2+ 3

Gt

3

+

-

-

a4

be zero.

may and

if

11 ~ --

+ a5 +

are called the complete quotients.

Thus we have Xl

that

is

"~

JL ai f "

a2

_L_JL.

+ *"a n ^+ x n

to say, the continued fraction x l

is

l

obtained

its

from

by substituting the complete quotient x n for the quotient a n

.

n-th convergent

FUNDAMENTAL PROPERTIES

397

PROPERTIES OF THE CONVERGENTS

As

7.

with the

in Art. 2,

initial

p n and

values

mi

!,

)

U/

be seen that q n a2 ... an _ x

by the equations

p = a 1?

,

1

-

,

,

and so

the

*'or

the case of

an

From

9.

and

a n as p n ^

...

,

etc.,

is

.

,

We have p2 >p v and if n>3, p n ~ pn ^ = (a n -

8.

. ,

1

-4-

#o#3

same function of a 2 a3

is the

.

!

,

a2

1

=1

g^

a^a., ~*--3 ~f-o\ -ha, -

a^o-fl -2--

0i1 ,

= 0,


l

/1\ --

The convergents are

of

=1

jo

,

it will

q n are defined

since

y's,

thus

q^q^,

1

)p n

,

-f

pn - 2 and

j) w a/?d g n increase with n, .

infinite continued fraction, they tend

to infinity

in

with n.

we have

the recurrence formulae (A)

Pn ln-l -Pn-dn^fanPn-l +#n -2)^-1 ~ Pn-l (n?n-l +
= This holds

if

we

write n

therefore

Again,

(B)

-(Pn-i9'n-2-? n-27n-l)-

-

1,

n-

2,

.

Pnfr-1-.Pn

..2

in succession for n, also

-i7n

= (-1) w ............................ (B)

we have

P1n-z-Pn-&n =

therefore

From

and

(C)

^} n " ^n ........................ (0) l

(

we obtain the formulae

10. ^uer?/ convergent

pjq n

is

in

its

lowest terms, also

Pn-\ and q n to q n _ v For if the numbers belonging to any (q

7n-i)

have a

would divide 1 1

.

3)

>

From

common

pn ,

the recurrence formulae (A), -

n

prime

to

of the pairs (p n q n ), (p n p n ^) factor g, then on account of equation (B), g ,

1.

Pn-l

is

- --

Pn-l/Pn-2

we have

and

-

=

n

?n-l

The first of these holds if we write n - ] n ~ 2, The second of these holds if we write n - 1, n - 2, ,

---;

-


2 for n.

3 for

n.

.

9

ODD AND EVEN CONVERGENTS

398

1111

and

Also PI/PQ^^I

therefore

JaAh^^J

/n

-,

(E)

(F)

11,

It follows that if

pr/qr pr '/([r ,

a*2

are respectively the r-th convergent of

and

+

an

-r

an +

i*

11

'

a-, n-l +

a,l

where in the second the quotients occur in the reverse order, then

p n '/q n ~p n lpn -i and pn ^ 1/?n-1 = ? n/?n-i '

For by the preceding,

by

and

Art. 10 these fractions are in their lowest terms.

12. It has been shown that

if

p'q', p/q are consecutive convergents

of a simple continued fraction, then pq' ~-p'q =

1.

where q'^q, and suppose that p/q is Conversely, let as a fraction continued with an even or an odd number expressed simple

pq-p'q=l

of quotients according as the arbitrary sign

that p'jq'

For

is the

p" jq"

if

is

-f

or

-

.

We

can prove

convergent immediately preceding p/q.

is

the last convergent but one,

by hypothesis and equation

(B) of Art. 9, f

pq

-p q = pq" -p"q, and f

r

therefore

p(q

-q")q(p' -p")> .

Now p else

is

prime to

q,

for

pq -p'q=

1,

therefore q divides

q'

-q"

or

= q"q'

Also

5'<<7

sequently

13.

and q"^q, therefore q cannot divide q'-q", and con-

= q" and p'=p", which q'

proves the theorem.

Sequences of Odd and Even Convergents. Let w n

then, from equations (D)

of Art. 9, it follows that

Wn-w-i

and

u n -u n _ 2

Therefore u n lies between w n _rl and w n _2 between the two preceding convergents. say, every convergent

are of opposite sign.

>

that

is

to

lies

Now W!
consequently W1

on, therefore

Thus

the

odd convergents form an increasing sequence, the even convergents

form a decreasing sequence, and every odd convergent convergent.

is less

than any even

CLOSENESS OF APPROXIMATION

399

Consequently, as n tends to infinity, both w 2n _! and u2n tend to limits, and these limits are the same, for by equation (D) of Art. 9,

finite

W2 n 7 U*n-l == l/?2n?2n-l-> 0.

Hence un tends to a

limit x such that for every n,

In other words,

and

(ii) its

continued fraction is convergent, (i) every infinite simple value is greater than any odd convergent and (iii) less than any

even convergent.

4. Each convergent is a closer approximation fraction than the preceding. 1

For

let

so that

z'

is

the (n + l)th complete quotient, then

= a n+lPn + ^n-l)/( a n+l?n + ?n-i) z = (z'p n +p n -.i)/(z q n + ?n-])

Consequently

2;'

and

(zg^

and

>? n _i

- j> n ) = "

2

r

'

(

f

and therefore

gn

and 2'=a

...

Pn+l/
Now

value of the continued

1

1

1

= %-{-

z

to the

(2;<7

- y n _!)

n _1

=

2;'^!, therefore

z-that

is

to say,

is

nearer to

^an to

p n/
pn -\l
5. ulwy convergent p/q of a simple continued fraction is a closer approxito the value of the fraction than any rational fraction r/s with a denominator less than q. 1

mation

For

let

z

al

preceding p/j. to z than p'/q'. lies

-\

--

By

...

and

,

Art. 14,

if

let

Moreover, between y/g and p'jq', and

that

?

|

ry'

nearer to z than ^/j, it is also nearer between p/q and p'/gr' therefore r/s ;

1

P'i

r_ Therefore

be the convergent immediately

r/s is

z lies

\~8

p'jq'

i.e.

9

q'

- ^jp' <^ |

;

9'

and, since

ry'

- sp'

is

an integer>

it

follows

16. If

is the

numerical value of the error, in representing the value z

of the continued fraction, by

and within wider For

then

,

limits

u n =p n /q n

if

p n/q n

then by Art. 13,

,

w n> is

ERROR

LIMITS OF

400

U n+l9

^>

^n+2>

an increasing or a decreasing sequence, therefore ~ ~ t*n Un+2 <\ W n *
I

I

Also by equations (D), I

W n - W n +2

whence we have the

I

=nW?n7nf2 and

I

w n ~ w n-h

inequalities (H).

Again, g n + a = a n +29Wi

+ ?n>

therefore

^ ?n +2/a n+2 Also qn+l >q nJ and therefore l/9 n ?n Hence the inequalities (I) follow from (H).

17.

fraction,

Let then

and al

z

l/z 2

that in only

a3

is

Z2


,

one way.

+ -- where a l and Z2

where a 2

can be expressed as a simple continued

positive irrational z

Any

,

...

Z3

,

-a 3 -fl/Z4

z a >l,

it

Zn

...

,

,

be zero),

let

r,

...

,

11

1

may

l,

follows in succession that a 2 a 3

and we have

(which

= <*n+l/Zn+l>

are integers such that, for every

a r <2 r
z

Continuing the process,

z 2 '>l.

= a 2 4-l/03

the integral part of

are positive integers

The process can be continued indefinitely, and leads to the infinite continued fraction ] j =O I + --

F

S

0-2

Again, we have a 2 <2 2

,

# 3
,

-

an

etc.,

j and

al

+

a2

>z,

11

-f --------

a + a3

.

+

therefore 1

1 !

----- ...

...

+

a2

+-">2 2 a3

~ 1


a 2* +

-

a3

-

H-

,

etc.,

1

a4

<2o, -

etc,

EXPRESSION FOR A SURD Continuing thus, F, then

Now

it

can be shown that

ur

if

401 the rth convergent of

is

n tends to infinity, u 2n _ l and u 2n tend to the same limit, which must therefore be equal to z. That is to say, the fraction F converges to z as a limit, and we write as

JL '" JL

1

It remains to

show that

a 2 -h

this

+

an

mode

of expression is unique.

Suppose

that

_JLJ_ The

1

1

_A

1,1

integral parts of these continued fractions are equal, therefore

j and

= 11

a,

i

6,

------

a2

H

a3

+

.

.

= 62 + T

.

Hence, by similar reasoning, 1

a 2 -62

and

... ,

and so on

When

Thus the continued

indefinitely.

z is

=6t 3 + r

1 >

v:

C/4 -r

C&4 -r

fractions are identical.

expressed in this way, any convergent of the continued

fraction will be called a convergent to

18. The value of any number.

infinite

z.

simple continued fraction

For any rational number can be expressed as a

finite

is

an

irrational

simple continued

fraction. Ex.'l.

The

Express (/v/37+8)/9 a* a simple continued fraction.

integral part of this surd

N/37+8 9

"~ -- ~ 1 1 -f ^37-l_, iT "9"

and

is 1,

4

7m

~\

s/37

-

s/37-4

3

4

.

^2+

\/37~3 ~'

:

7 1

4

-

~ "f*

V37+3'

4 N/37~5_ o -r -- + 4 ___ud H--H -7=^ 3 3 V37+5 -i

7

,

"

1 H A

:

4 ------

>

^37-3 =2+-, 4

--

\/37

-7=

V37+4 V37+5

+1

N/37 ,

+r

_

tj

.

=l+eto.

-

7

v/37+3

The complete quotient (s/37+3)/7 has occurred

before,

and the subsequent work

consists of repetitions of the last three steps, thus, with the notation, of Art. 5

is/37+8 9

1

IH-

>

J_ JL_ 1+ 3+ x-

the continued fraction being a recurring fraction.

1

2' -ti-

:

INTERMEDIATE CONVERGENTS

402

a given number x is expressed as a which pr/qr is the rth convergent, then pn/q n is an approximate value of x in defect or excess, according as n is odd or even, and the error is numerically less than I/qn q n +v an(i a fortiori

19.

Approximations.

If

(1)

simple continued fraction of

less is

than l/q n 2

Now

.

2n+i

= 0n+i7n + ?n-i>

hence if a n+1

is large,

p n/qn

a specially good approximation.

Or again, if we require an approximation with an error numerically less than a given fraction I/a, we calculate the convergents until we find one p n/q n such that q n q n+ i ^a. This is the required approximation. Of course, the conditions are satisfied

if

q n ^*Ja.

Ex. 1. It is required to find good approximations to the value of vulgar fractions, given that 77

The convergents

_q

,

TF

in the form of

JL _JL _L _JL. L JL 7+ 15+ 1+ 292+ 1+ 1+""

are

3

22

333

356

103993

I*

T*

Toe*

m'

33ioT'""

The denominators 106, 33102 are large compared with the preceding ones, therefore 22/7 and 355/113 are specially good approximations, and they are both in excess. For 355/113, the error is less than 113x33102

3740000

20. The problem may be of such a kind that no suitable approximation can be found among the convergents (as in Art 22, Ex. 2). We may then proceed as follows Choose two convergents p/q, p'/q, :

one an odd and the other an even convergent. integers, the fraction

;

mq + m'q'

and p'/q'. Also x lies between these convergents, a closer approximation than p/q, and perhaps also than P/Q and it may be possible to choose m, m' to suit the conditions of

between

therefore p'/q'

m, m( are positive

p mp + m'p' Q

lies

If

p/q is

the problem in question.

21. The most important fractions of this kind are the so-called Let p n/q n be the nth convergent of

mediate convergents.

1

Consider the sequence

Pn

Pn+Pn+l

the (r-hl)th term being

PrjQ r

where

Pf

inter-

APPROXIMATIONS The terms

of this sequence (excluding the first

p n /q n and

intermediate convergents between

possesses the following properties (i)

It

or even, (ii)

(iii)

and

last) are called

Every

Any

PT jQr

fraction

fraction

which

is

the

The sequence

;p n +a/?n+a-

:

is an increasing or a decreasing sequence according as n and the denominators form an increasing sequence.

is-

odd

in its lowest terms.

lies

denominator greater than either (iv)

403

Pr/Qr

between

Qr

and

Pr+1 /$r+1

has a

Pr/Q

as the

or Qr +v

Limits to the numerical value of the error in taking

r

value of x are given by

For we have

P r+l Q

r

whence the statements

- PrQ

(i)-(iii)

PrIQr> is

either

r+l

=pn+l q n -p n q n+ - ~ l) w+1 i

follow

:

Pn+2/q n +2>

Now

x

2W]/?n+1>

>

an increasing or a decreasing sequence, therefore

r
Qr

(iii)

all

by writing down the odd cqnvergents

the intermediate convergents, then

an increasing sequence of fractions in their lowest terms, and a closer approximation in defect to x than the preceding. The denominators form an increasing sequence. This

is

each fraction (ii)

this proves the statement.

,

consider the sequence formed

and inserting (i)

,

and because

Qr Since

(

If

is

P/Q, P'/Q' are successive terms, any fraction which greater than Q or Q

lies

between

f

them has a denominator

.

Again, the sequence

formed by the even convergents and their intermediates, has the same properties as the preceding, except that it is a decreasing sequence in which each term is a closer approximation in excess to x than the preceding. 22.

Problem

denominators

less

It is required to find the fraction which, of all those with than a given number a, is the closest approximation (in .

defect or in excess) to

a given number

x.

APPLICATION TO CALENDAR

404

To do

we

express x as a simple continued fraction, and calculate the convergents p^q^, pdq& until we find p n/qn such that this

q n ~a, then

If If

we

q n <.a,

is the required fraction. consider approximations in defect.

p n/qn first

Taking the

sequence (A) formed by the odd convergents and their intermediates, then P/Q P/Q, P'jQ' such that Q^a
find successive terms

;

the closest approximation in defect of less

than

all

we is

fractions with denominators

a.

For if h/k is nearer to x than P/Q, then h/k must coincide with a term to the right of P/Q in the sequence, or else it must lie between two

and

suc.h terms,

in either case

k^Q'>a.

In the same way, by considering the sequence (B) formed by the even convergents and their intermediates, we can find the fraction with denominator

two

Ex.

As

less

than a which

fractions, the

Find

1.

Art

in

18,

the,

is

the closest approximation in excess. Of these is nearer to x is the number required.

one which

fraction with denominator less than 500 which is nearest to s/14.

Ex.

1,

we can show that

N /14

=3 +

--

-

-

- -

1+24-1+6

,

and we

<,

.

,

t

P!

,

find that

~ 333

q7

89

p B - 449 , '

q8

120

p-9 = 3027 -

,

qg

,

*

* .

809

For the approximation in defect, we insert intermediate convergents between p 1 lq1 and p 9 /q9 These are of the form Pr /Qr -= (333 + 449r)/ (89 + ] 20r). .

The conditions $ r ^500<(> rfl give

r---3, and 2\IQ 9 -- 1680/449. For the approximation in excess, we consider the convergents 8 /io/

,

?/& =449/120. -

2

Finally, we can show that 14 (1680/449) <(449/120) so that the required approximation is 1680/449.

Ex.

2.

The mean tropical year

convenient method of correcting

We have

is the,

-

14,

It is required to find a practically calendar, the year being taken as 365 days.

365-2422642 days.

*

<

2

1

'

'

and the convergents are 1

1*

4'

7

29*

j*

33'

39

j47 f

161"

194'

321

1325""

Take 1/4 as a first approximation. This, amounts to adding 1 day every 4 years, and is an over -correction. The fraction 7/29 is inconvenient but 8/33 suggests an addition of 24 days in every 99 years, i.e. make every fourth year a leap-year, except at the end of a century. This is also an over-correction. ;

CHARACTER OF APPROXIMATION

'

405

None of the other convergents give convenient approximations ; and the same is therefore proceed true for the intermediate convergents in the first few intervals. choose the convergents 1/4 and 39/161, the first an oven and the as in Art. 20.

We

We

second an odd convergent

if

then,

;

m,

m

f

are positive integers,

m. l4-m

1

x

.39

39

ra.4 + m'. 161

161

'

4

The values

m'^2

??i~19,

give the approximation

19.1+2.39

97 *

+ 2. Tel "400

19. 4

This gives one more day to be added in 400 years than the fraction 8/33 i.e. it makes every fourth year a leap-year, except those at the end of a century, with the further ;

correction that every fourth century is to be counted as a leap-year. The fact that this approximation, 97/400, can be obtained from the convergents 1/4 and 47/194, which are both in excess, shows that this also is an over-correction.

The

error

is less

than a day in 4000 years.

MISCELLANEOUS THEOREMS

23 Let p/q be a fraction in its lowest terms, which is an approximate value some quantity z. It may be important to know under what circumstances .

of

p/q

is

a convergent

The answer to

to z.

this is contained in the following

Let p/q be expressed as a simple continued fraction with

number of quotients according as p/q

~g

and

z,

an even

or

:

an odd

f

let

p

/q' be the last convergent

but one. !

z

If J

~P

I

\

<

q\ then p/q is

------

^

q(q + q)

a convergent

For suppose that

1

V

q

\

to z.

11

p= a l + ---

...

an

#2+

even or odd according as p/q

is

i

or a fortiori if J

9

q

where n

P -

1

"7

g

1

^a-i

,

z,

1

and

let z'

be determined by

1

a

prove that z'>l, for then z' can be expressed as a simple continued fraction in which the first quotient is not zero. We have It is sufficient .to

*=& +q

andthcrefore

Now

f

pq -p'q^I

f

q(qz

4- q')

is

positive

or

1,

according as p/q

qz'

q(qz L

z

;

+q)

therefore in both cases

and

____ q(qz'+q')

Hence

'-*-&-:** q

qz

\q

+ q'>q + q' and z'>l, which proves the theorem.

SYMMETRIC FRACTIONS

406 24. If

pr/
which the value

For

if

k

is

Therefore

is th6

r~th convergent of the simple continued fraction of

then z 2

is z,

+ Pn-l? ~PnPn-l (*? + ?-l) 2 ^ left is

25. // p/q is an even convergent, and continued fraction whose value is z, then according as p/q precedes or follows

X/Y.

ately precedes

than p/q or

else

X/Y

is

even or odd, for

let

X'

-

p~

p'/q' precedes p/q. with or coincides else X/Y

is it,

pX

9

be the convergent which immediately it occurs later than

let p'/q'

// p/q follows X/Y,

an odd convergent, and therefore

x _/

7 since p/q is

in the sequence of convergents.

an odd convergent immediately preceded by X'/Y',

Then

p

|<*
an even convergent immediately preceded by

p'/q',

and

z*>

qY

qq

26.

is

Then X'/Y'

XX'

and

n

X/Y an odd convergent, of a simple

X/Y

X

(ii)

0.

X'/Y' be the convergent which immediis an even convergent, and it occurs coincides with it, therefore

// p/q precedes X/Y,

since

even or odd.

and qn >qn -i, which proves the theorem.

Pn>p n -i,

and

is

equal to

this is positive or negative according as

later

n

according as

z^^n-i/g^n-i*

The expression on the

(i)

according as

the complete (n-f l)th quotient,

Jrfn-l ( kPn

and

5 PnPn-il^n-v

Symmetric Continued Fractions.

A

simple finite continued

which the quotients equidistant from the beginning and end are equal, is said to be symmetric. For instance, the fractions, fraction, in

77

^

.^:

4+l-h4+3'


132

are symmetric, the former having an odd latter

an even number.

_L__

4+1+IT4T3' number

of quotients,

and the

SUM OF TWO SQUARES The

is

following

407

an important property of a fraction of

with

this kind

an even number of quotients. /n Let r , P - = #!

H

(1)

j and

...

1

^=

ar + a r + a r _ +

q the fractions

P/Q

p/q,

being in their lowest terms,

the convergents immediately preceding p/q

P=p*+p'2, For by Art.

11,

therefore

P/P'=P/Q,

f

let

p'/q

,

P'lQ' be

and P/Q, then

= q* + q'* and

Q'

and

ax

Q* +

P ~Q,

l=

also

therefore

p

pq+p'z" /

ar + q

X=p* + p' 2

If

therefore

and

Y^pq+p'q',

any common factor

to one another, therefore

therefore

##.

1.

X/Y ,

PQ -P Q = (~l) 2r /

//

pr /qr

/

is the r-th


r

+ q'

we have

X and Y divides p and p', which are prime

of

is

P = X and Q = Y and

Moreover,

= pq+pq2

in its lowest terms.

ff

Q~

consequently

and

P/==Q,

therefore

1111 _____

i

"

._

at+'" ar + ar +'" a^

prove that the remainders in the process of finding the G.C.M. of Q,

P=p 2r

is

convergent of the continued fraction

P

For

So also

P are

Q=p* r -i

,

with sinaikr equations, the last being p 2 =a jp l + l. Hence the G.C.M. process is that shown on the right.

&r.

2.

Given that 218 2 -f

1

=25

.

1901, express 1901 ow fe

The reckoning shows that 1901

+ _LJL_L J_ JL J_I 1+ 2+ 1+ 1+2+ 1+ 8*

218 Hence,

if

1901

p r /qr

=^a

2

is

the rth convergent,

+z> 4

2

= 26 a + 35 2

.

(See Ex. 1

and Art.

26.)

*ww

of two squares.

P/Q,

SUM OF TWO SQUARES

408

27. Application to the

any factor of

positive integer,

Theory of Numbers. // Q Q 2 + 1 can be expressed as the sum

is

any

of two

squares.

For

let

Q 2 + l=PR

P>R,

where

p -=

l

Therefore

P>Q

is

P'fQ'

divisor of

p P= a 2r 4

Consequently ^ J 7;= P

Q

a 2r _ 1 = a 2

;

/A* (A)

the convergent immediately preceding P/Q,

P(R-Q') = Q(Q-P') Q-P' or else Q = P'.

.1

...

l

and P>P', therefore

n

..............................

and

l

0,+ ----

Q if

is

P>Q

Let

prime to Q.

then

P

then obviously

P

and, since

The

-1

a 2r _ t

...

11 --a 2 4- a l

4-

and the continued fraction

is

is

it

a

impossible, for

R = Q'.

and

Q==P'

prime to Q,

is

alternative

first

therefore and1*1.*

,

a 2r = a l9

is

symmetric. Denoting the mth convergent of (A) by p m /qm we have, by Art. 26, ,

etc.,

,

which proves the theorem.

28. Conversely integer

Q

--

if

x2 + y~ where x

can be found so that Q*

z 1 T Let - = a r 4 a y r-1 then,

P

if

9

pml^m

is

1 .

.

the

.

.

ax

-h

,

and

+1

X = a, --

4-

.

a r-1

.

.

ax

4-

=p

Using Art. 28, we have

2

?

= 46 2 4- 39 2

o9

Q

therefore

,

2

an(j

4-

i ^ ^ 3 54-14-14-

,

4-"

1027 -

is

LA

l

1 -f 1 4-

3637 -

f

3

we have 3637

,

x 200 - 1027 2 =

1,

10/27

(1027, 290).

the last two convergents of 3

290

and writing down the continued

,

<
and the required solution therefore

^=^2^.

find a solution in positive integers of the equation

= l4-r

of which the last two convergents are

;

= Pq^r-v

1

JL _ X -1 JL _L 1 1 4- 1 5+ 1 4- 1 + 5T

:

ar

2

fraction

Or thus

4-

to the last continued fraction,

Given ^Aa^ 3637

.

1

----

ar

mth convergent

Also ^L^r-i ~ Qpzr~\ == 1

,

111

. . .

a2 +

Q

r

1.

by P.

1

4------

y and x>y, then an

to

prime

is divisible

Z=y 4-^ i==a;2_h2/ Ex.

is

+ 1-

F

-

--

are TT ancl

and x = >/(3637 290- 1)=: .

i

o

Therefore

RECURRING CONTINUED FRACTIONS

409

29. If x is prime to y, any factor of x* + y* is the sum of two squares. For we can find an integer Q such that x2 -f y 2 is a factor of Q2 + 1 and every factor of

Q2

-f-

1

is

the

sum

two squares.

of

30. Simple Recurring Continued Fractions. grouped as follows (i) (ii) (iii)

These

may

be

:

Fractions which have no acyclic part.

Those with an acyclic part consisting of a single quotient. Those with an acyclic part consisting of more than one quotient.

The reasons

for this classification will

and we proceed

of fundamental importance,

later.

appear

The conclusions are

to consider these types in

order.

31.

No Acyclic

iT

Part. U-t

(X>

T~ 7

1111 .

7~~~

&2+ b3+

................

7

7

b n+

&1+

&2

the fraction having no acyclic part, then a with integral coefficients.

1

7

Moreover, if

j8

+

(**)

a root of a quadratic equation

is

second root of the equation ,

is the

then

where the order of the quotients

is reversed,

whence

it

follows that

-1
l/i

T" 7

'"

62

therefore

and

a

in the

is

~"

-fractions (A), (B)

by p r/q r and

'

.

+

T

b n +oc

the positive root of

same way the

fraction (B)

is

the positive root of

Jn'zMK-CJ*-^--!^ which by Art. 11

is

'

the same equation as

Now

equation (D) can be transformed into equation (C) by writing Therefore - l/j3 is the positive root of (D), and is conIfx for x. sequently equal to the fraction (B).

-

'Also, from (B),

/?

is

negative,

and

-^>1

;

from which

it

follows that

STANDARD TYPES

410

A

32.

single Acyclic Quotient. l


siw/Ze quotient

l

]

a l does not recur, then a

equation whose second root cannot

and -

between

1,

11

11

then x-a^^I/y, and

lie

a root of a quadratic

is

if

pr/gv

is

1

1

the rth convergent of the second fraction,

as in Art. 31,

^-(Pn-ffn-Jy-Pn-l-O,

(

Denoting the

a'^ + l/a, The root

a'

(B)

pn -i(x-<*i) + (Pn-q n -i)(x-<*i) -?n = ................. c ) roots of (B) by a, j3 where a is positive, and the correspondby a', /?', we have 2

and therefore

ing roots of (C)

........................

is

'

= 04 + 1/0.

the value of the fraction (A), and by Art. 31

where /is a positive proper fraction. Now a l and b n are unequal, for

if

we have

a l = 6 n then a a would belong to

the cycle, whiqh is contrary to the hypothesis. Consequently /?' = /--/ or -I -f where 7 is a positive integer, and therefore /J' cannot lie between

Oand -1.

1111111

More than one Quotient

33. // XI

OC

_ -n

l

Uf-l

"1

Acyclic.

in which at least two quotients do not recur > then a is a root of a quadratic equation with rational coefficients whose second root is positive.

- -111 +6 +6

,,11-

1

Let x = a t +a2 +

and

Pr/Q r

let

...

=

am

>

2

1

+

...

and

t/ y

= 0i +-

7

62

and p r/q r be the rth convergents

+

...

=

bn

of the first

+

...

,

and second

fractions respectively, then

2

and

?n!/

If a,

/J

are the roots of !

A ___ _ t'lii /.

i

1

__

11

_____ i

-(F-?-i)!/-yn-i =

......................... (C)

_

a being positive, then by Art. 31,

(C), O

j

Tlfi

<*l-l.vl

1 -.

^

i

A) t/

n "

~ i

1

_

7

.

. .

_11_ ,

r

...

.

I II l ...IJL/I v '

SUMMARY OF RESULTS

411

eliminate y from equations (B), (C) we obtain a quadratic in x with rational coefficients, whose roots a', /?' are found by substituting Thus a' is equal to the fraction (A) and a, ft respectively for y in (B).

Also

if

From

we

equation (D)

it

+ l/p=a m -bn -f, where / is a not equal to b n for if a m = bn

follows that a m

Now a m is positive proper fraction. then aw would belong to the cycle, which It follows that am + l/j8-/-/

where I

is

_ 1

1

.

and

since a /m __ l

fore also

/?',

is

is

or

contrary to the hypothesis.

-/-/,

Thus

a positive integer.

Um - i+

,

is

1

__

ft

m - i+ I

1 fYf or

________

ft

-1

a positive integer, the expression on the

NOTE. The argument depends on the existence of a m _ l9 and than one quotient in the acyclic part.

34.

left,

and

there-

positive in all cases.

Summary.

It has

fails if

there

is

not more

been proved that any simple recurring con-

tinued fraction is a root of a quadratic equation with rational coefficients. Also the second root (3 of the equation is restricted as follows : (i)

If there

is

no acyclic part, then

1


(ii)

If the acyclic part consists of

a single

(iii)

If the acyclic part contains at

least

Ex.

1.

Find

the value of 1

JL

~ o

-

+~f~

JL

~f"

quotient, then

/?<

-

1

or j8>0.

two quotients, then j3>0.

-

.

~}~

*

*

Denoting the value of the fraction by xy we have

^l + _. i

1

1

where

^ = 1+ ^

and y = (s/37

giving

+ 3)/7

and

__.. 1

1

1

;

y=

*=

EXERCISE XLI 1.

provided that 2

D

---

For the fraction

|

x

\

<|

(

...

- a +
,

prove that

(See Ch.

pn

XXII,

q n -i and

2.)

B.C.A.

COMPARISON OF PHYSICAL UNITS

412

----

For the fraction

2.

(ii)

3.

l>n?n-4

-S /iJPn-4 =

-

f

w l)

(

...

-JT4-1)^ CL '

1

prove that

~Mnn-ian.-2

4

2+

:

1

1

2a

+ 2+

1

-

...

~

.111

/-T\<2(
.

'

1

1

1

1

Prove that /

n

1

A

,~

2a -f 2a

va 2 ~-2 5.

,

Prove that

Deduce the following

4.

... ------

2a -f

-f

1

a-

1

1

Prove that (i)

(n)

6.

Ex.

Show

a -f

f

1111 r~ *T

&4-c-f6+2a

that 449/120 differs from

//

A/

V\

a

1\/

+7

(

c

a

6/\

^14 by

less

+ r~"7i

\

6c-{-2/

than 1/90,000.

(See Art. 22,

1.)

A

7. straight ruler is graduated in inches and centimetres, the ratio of an inch to a centimetre being N. If is expressed as a simple continued fraction and p/q is any convergent, show that the distance between the graduations p centimetres and q inches is less than the distance between any two preceding graduations corresponding to a convergent of the fraction.

N

4

'

'

'

8. Find the fractions which, of all those with denominators less than 500, are the closest approximations (i) in defect and (ii) in excess to 0-2422642. (See

Art. 22, Ex. 2.) 9. Given that 1 metre 3 -2809 ft., obtain the two approximations, 8 kilometres 5 miles, 103 kilometres 64 miles and find an upper limit of error in each case. ;

Given that 1 kilogramme = 2 '2046 pounds, show that 44 kilogrammes slightly greater than 97 pounds, the error being less than half a grain. Obtain also the usual approximation, 7 grammes 108 grains. 10,

11. If p/q,

p'lq',

is

p"lq" are consecutive convergent^ to a simple continued F lies between *Jpp'/qq' and *Jp'p"Iq'q"-

fraction F, prove that

SPECIAL TYPES

F=

For the fraction

12.

Also

-.

------

...

-i

a+ b+ a+

b

2 p are the roots of x - (ab 4- 2)x -f

if a,

ap 2n = bq 2n __ l ~abd n /d l

(iii)

e^a& + 2, and

(v) if

cc

2

1

(iv)

;

< |(

that prove r

,

+

413

~

and dn = a n - /_*", then

^ 2 n4i = _/2n

:r::

(^n-|.i

-dn )/d

1

;

-c-f \ 7c 2 h4), then

2

l+frr-o: _ - CX 2 -f iC 4

1

1 -f fix

[If w n

stands for

un =-p 2n

If

or for

y; 2n

we have

,

- ^2

.4 n-

>

2 n-i>

then, by Art. 5, Ex. 2,

B~ p ~ 0,

Ay.\Bfi~p 2 ~b,

Q

-4~ -B=6/c/ lf

giving

w~^? 2 n-i> since c^a-f-^, then Ja

If

^=(l--i)/^,

giving

Finally,

it is

and then the 13. If

p^n^d^d^

/.

B~-(l-fi-*)ld l9

results in (v)

# n /gn

is

can be obtained by equating

the ?ith convergent to ------

+

2

the numerator of the (n~5)th convergent to

rt

that

the

fraction

a

----a- a-

then

...

n

n

Prove

coefficients.]

--->

-f

Pw-i^s-i-gn-iPs-i

14.

1,

d fi-i)/ d ij) 2w -i-=(^n-

:.

easy to show that

i

is

Aa. ZjrBfP~p z -=c-

1,

\~lJf3=pi

-

+ a n-l +

----

a- x

,

n_2

+

....

which a

in

is

equal

- 1

and is repeated any number of times, must have one of three values, to and that if x satisfies the equation 2# 3 -f 3# 2 - 3# - 2 ~ 0, the fraction satisfies this equation. 15. If n L

H

--a2

. . .

-f

terms, prove that

---#r-i

$2 -

1

.

. .

+

a r + a r-i

+

is

divisible

by P.

2

+ a

i

P, Q, R are positive integers such that can be expressed in the form

16. If

P __ _ Q and

if

Verify

pm jq m

is

1 i

^,

x

a,+

'*'

111 _

__________

a r_t

=~ V

and P/Q

Q~-l=FR

...

+ a r + a r _ t -f

'"

11_ _.'

the rath convergent, then

^ =Pr (Pr +Pr-i)> G =.Prfr-i +Pr-i
R = V2r = 3r-JL

in its lowest

and JP>JR, then

_______

cu-f

is

(

CHAPTER XXV INDETERMINATE EQUATIONS OF THE FIRST DEGREE In

this chapter a,

6,

stand for positive integers.

c, ...

Underthis heading we

consider integral solutions, and more particularly positive integral solutions, of a single linear equation in two or more variables or of a system of linear equations in n variables (m
m

2.

that

A Single Equation of the Form we need only consider the forms ax

If either of

any common

by

by

have a

b

a,

division.

prime

to

common

Hence and we

b,

(i)

To find a

by c.

integral values of #,

Hence

if a, b have

t/,

then

a common

no solution in integers exists. divisor which divides c, this can be removed no

assume

a

of generality in supposing that

loss

is

this to be the case.

The Equation ax~by = 1

3.

obvious

c,

there is

shall

It is

ax -f by = c.

and

these equations is satisfied divisor of a, b must divide

divisor which does not divide If

=c

axby=c.

(a prime to b).

solution in positive integers, express

as a simple

a/6

continued fraction with an even number of quotients (Ch. XXIV, 4). The last convergent is a/6. Let p/q be the convergent immediately preceding (Ch.

a/6.

XXIV,

9),

Then since and therefore

is

a/6 (q,

p)

is

an even convergent,

(ii) To find the general solution in integers, suppose that solution in integers, then

ax - by = Therefore since a

where

prime to

is

t

a(x~~q)

is

This

is

6,

x-

q

Hence a(x-q) must be divisible by

zero.

y~p + at

is

(x, y)

any

=aq - bp.

= b(y-p).

an integer or

x = q + bt,

1

aq-bp~l

a solution.

is 6.

divisible

by

Consequently

Hence where

= 0,

called the general solution in integers.

1,

6,

2,

etc.

and

GENERAL SOLUTIONS Since

(iii)

that (q,p)

We

a convergent which precedes a/6, it follows that and from the form of the general solution we conclude is

p/q

and q
p
only solution in positive integers such that least solution in positive integers.

is the

shall call this the

Since

(iv)

integers of

'

p
and q
'

a(b-q)-b(a~-p)= 1, the ax -by = -1 is (b-q,a-p).

The Equation ax-by = c

4.

415

'

'

solution in positive

least

(a prime to b).

(i) To find a solution in positive integers, express a/b as a simple continued fraction with an even number of quotients, and let p/q be the

convergent immediately preceding

aq-bp = l Hence (ii)

(qc,

pc)

To find

is

Then

a/b.

aqc~bpc = c.

and

a solution.

The equation may be written prime to 6, x-qc must be divisible

the general solution in integers.

a(x-qc) = b(y-pc) and

since

9

a

is

Therefore (x qc)/b = (y-pc)/a by Hence the general solution is b.

x = qc + bt, Let

where

t,

y = pc + at

= 0,

(

is

t

an integer or

zero.

2, ...).

1,

m

be the integral part of the smaller of pc/a, qc/b 9 then the in positive integers is (a, /8) where a qc- 6m, ft =pc - am, the general solution in positive integers is

(iii)

least solution

and

x=oc + bt,

Thus Ex.

1.

the equation has infinitely

Find (i)

We

y~p + at many

(J

= 0,

1, 2, ...)

positive integral solutions.

the general solution in positive integers of

13.r~l7y = 5:

(ii)

13z-17y= -5

(iii)

;

13*-172/=996.

find that

l^o+J--^^*-!---^-!-! 17 1+3+4 1+3+3+1* The convergents

of the continued fractions are

f

1

3

13

i'

I'

n

13.4-17.3 = 1,

and

Hence for

the first equation

* = 20 +

The

1

P P

;

a solution 17*,

3

10

13

4'

Is'

IT

13.13-17.10=-!. is

The. general solution in integers is

(20, 15).

y = 15 + 13*

(*=0,

1,

2, ...).

-

solution in positive integers is x = 3 -f 17$, Similarly for the second equation a solution

1) is (3, 2), and the general by y = 2 + 13* (t 0, 1, 2, ). is given by # = 13. 5 = 65, y = 10.5=50,

and the general

is

least solution in positive integers (given

solution in positive integers

t

. . .

NUMBER OF SOLUTIONS

416 In case of

the third equation, the

be shortened thus

may

reckoning

:

Dividing 996

by 13 (the smaller of the coefficients), we find that 996 = 13 76 + 8 = 13 77 5. Hence the equation may be written 13 (# 77) ~\ly -5, and by the preceding the least positive integral solution is given by x - 77 = 14, y 11. .

Therefore the general solution in positive integers

s = 91+17*

is

y = ll + 13* (=0,

f

.

1,2,

...).

c can be made to depend on that NOTE. In this way the solution of ax - by of an equation of the same form in which c is less than or equal to the smaller of $a, \ b.

The Equation ax + by = c

5.

Let

General solution in integers.

(i)

prime to

(a

continued fraction with an even number last

Then aq-bp =

convergent but one.

b).

be expressed as a simple of quotients, and let pfq be the a/b

\

and the equation may be

9

written

ax + by = c(aq -bp) a

Since

prime to

is

a(cq-x) = b(y + cp).

or

cq-x must be

6,

t is an integer or zero. Hence the general solution in

divisible

by

6,

and therefore

where

x = qc-bt (ii)

integers

y^at-pc

9

(2

is

= 0,

The values

Positive integral solutions.

2, etc.).

1,

of

t

(if

such exist) for which

x and y are positive are given by the conditions, pc/a
then

y

x,

Hence

m+f,

qc/b~n+g,

be positive integers if ... n. is the number of solutions in positive integers,

will

if

N

b

Now

0
where

-1


,

a

ab

and therefore

ao

-..

ab

not divisible by ab, the number (N) of solutions in one of the integers next to c/ab 9 and is greater than or positive integers than less c/ab according as f^,g.

Hence, if

c

is

is

If

c

is divisible

by ab

9

/=0,

then pc/a and qc/b are integers

= 1 and N--.-L ao

;

therefore

THREE UNKNOWNS Ex.

Find

1.

13*

and the

+ 17^3000(13.

4

4 - 17

.

-17.

_

Moreover,

3

.

= 1, and

may be written = 13(12000 -*) 17 (y + 9000)

or

3)

the equation

:

x~ 12000 - lit, y~l3t- 9000.

by

_

9000

15

_. c =706

^,

3000

13

any

also

4

15

/=Yo LO

^"i^ Li

-693 +

1

13,

^ nat

so

= 093,

by

694,

...

These

705.

hence the number of

/<#

this result agrees with the preceding.

Equations with Three Unknowns.

may

exist,

1.

<14;

Since 705

13.

The general

(1)

Ex.

is

Two

6.

-

XO.JLV

solutions

if

<~'--

4

^692 ^

therefore the positive integral solutions are given solutions are (15, 165), (32, 152), ... (219, 9).

Now,

3000.

that there are 13 such solutions.

integral solutions are given

HT

+ 17/

13#

the positive integral solutions of

Use the rule just given, to show As in Art. 4, Ex. 1, we have 13

417

Find

solution in integers, and all the positive integral solutions, be found as in the next example.

all the positive integral solutions of the

2x -

= 7, Zij + 3z

pair of equations,

- 4# + ly + 3z = 1 9.

4 ; and tho general Eliminating one of the unknowns (z), we have 2x 3y = solution in integers of this equation is #~3w-2, 1, 2...). y--2m, (in~0, Substituting these values of x, y in one of the given equations, say in the first, we

-

find that

2m + 35 = 11 and the general solution in integers of this equation is z = l+2, and therefore a;^=3/-2 = 10-9, y~2w = 8-6*. ;

M = 4-3t,

Hence the general solution in integers is x 10 - 9J, y = 8 - 6^, 2 1+2^. and 1, therefore the only The only values of t for which x, y, z are positive are positive integral solutions are (10, 8, 1) and (1, 2, 3). (2)

//

(a,

j3,

y)

is

a solution in integers of the two equation

ax -f by + cz = d, it is

required

(i)

to

a'x

4-

b'y

f

the three

(be'),

the equations are equivalent to

(ab

(ca') y

/

is

a

common

be integers for

factor of

(ca') 9

t

all integral

(ab

by Oh. X,

6,

...............................

r

(bc') t

zero,

(4),

v

;

(aV)

Denote each member of (A) by t/f where and / is an integer to be chosen later, then

will

is

)

-=

(te'P(ca')

x, y, z

',

find the general solution in integers.

// none of

Thus

+ c'z = d'

).

may

be any integer or zero

values of

t

if

and only

if

SINGLE EQUATIONS

418

Now hence

common

every

and

is the

where g

any

The only

G.C.M. of

is the

l

A

Ex.

(ab

(ca'),

)

and

(

= 0,

easily seen that the general solution in integers

(ca'),

2,...

1,

.

exist)

Find

1.

are

...

a, 6, c,

may

all

is

(ab').

Form ax + b

Single Equation of the

(1) If (if

(be'),

If (6c')=0, the equations are equivalent to

it is

7.

G.C.M. of

a factor of their G.C.M.,

(ca') 1

where g

is

are given by

all integral solutions

(be')

(ii)

numbers

factor of these

then

positive,

be found as follows

the positive integral solutions

all

:

lx + 1 1?/ 4- 26z = 123. + 11 +2(>2< 123, hence z<4.

the positive integral solutions of

greatest coefficient is

2(>,

and

7

Therefore the

possibilities are 2

+ \\y~ 97, + llr/=:71, 7z + ll?/ = 45, 7# + !!?/:= 19, lx

1,

2-2,

= 3, = 2 4, 2

Thus there are two solutions Ex.

Find

2.

Here

Il:r4

~(3-f 4

+ 7),

possibilities,

are 28 solutions in

(2) If

(a,

namely

(0, 5),

(7,2),

no

therefore

x3 = 4,

one solution

one

no

all the positive integral solutions of

^67

is

in positive integers,

~ 2,

The other

of which there

7.r

3arj,

a:

4

x^

2

t

1,

(6, 5, 1)

and

3^ + 4x 2 -f 7# 3 + 1 1.r4

2

= 6, may

no solution. be treated in the same way.

^ y)

w

There

all.

is

an integral solution of ax + by + cz = d (where

xa + bw-cv,v,

07.

no solution,

are positive or negative integers), other solutions are given by

where u,

(7, 2, 2).

<5, and we may have

+ 4#2 = 9,

^xl + 4

namely

y = /3 + cu-aw

9

have any integral or zero values.

z

a, 6, c

GENERAL SOLUTIONS This

obviously true.

is

a, 6, c are

For

prime rr-a,

if

we can show that

Further,

each other, then every integral solution can be so expressed. y~fi, z-y are denoted by X, Y, Z respectively, the

a-Y

Let a be prime to

6,

Z may

+ ftY + cZ-O ................................. (A)

then integers

u, v

and

X, 7,

then,

since a

w

where

is

is

Z

can be found so that, whatever

have,

av-bu If,

if any two of the three

to

equation becomes

integral value

419

Z.

are integers which satisfy (A),

we must have X + cv == biv, Y - cu = - aw,

prime to

6,

some integer or

This proves the statement in question.

zero.

EXERCISE XLII Find the least solution

1.

in positive integers of

=l

(ii)

;

68s

In Exx. 2-7, find the general solution in integers and the least positive integral values of 2. 5.

3.

27a:-8y = 125.

4.

17a;-41y^l.

6.

12a?-7y = 211.

7.

Find 8.

11.

x, y.

8z--27?/:--125.

all

7* +

1.

the positive integral solutions of equations 8-10. 8//

= 50.

9.

lLr + 7y = 151.

10.

Express 68/77 as the sum of two proper fractions whose denominators

and

are 7

29* - 130 =

11.

12. Show that I/ (13x17) can be expressed in two ways as the difference between two proper fractions whose denominators are 13 and 17 ; and express

the given fraction in these ways.

AB

AB

CD

are two rods, each one foot in length. is scaled off into and 13. 13 equal parts, and CD into 17 equal parts. If the two rods are placed side by side, with the scales in contact, and the ends of the rods in alignment, show that and one of CD can never be less the distance between one scale division of than I/ (13 x 17) of a foot, and that there are two cases in which the distance has this value. Which are the pairs of divisions in these cases ?

AB

The sum of two positive integers is 100. If one is divided by 7 the remainder and if the other is divided by 9 the remainder is 7. Find the numbers.

14. is 1,

15. If c is increased

ax + by=c

is

by

kab, then the

increased by k.

[For c/ab

number

is

of positive integral solutions of increased by k and/, g are unaltered.]

SOLUTIONS IN POSITIVE INTEGERS

420 16.

The

least value of c for

which the equation ax + by

[This follows from Ex. 15, coupled with the fact that a of c for which the equation has one solution.] 17.

c

has

+b

is

N solutions

is

c^a + b + (N -l)ab.

given by

Find the

least positive integer, a, for

the least value

which the equation 3x + y

a has

six positive integral solutions. 18. Find the least value of c in order that the equation l%x + Yty have 13 solutions in positive integers, showing that the required value Verify by finding the number of solutions for c-2682, 2681. 19.

The number of

the integral part of

Find

all

20.

x + 3*/-4z-

solutions in positive integers of the equation

may 2682.

x + by~c

is

c/b.

the positive integral solutions of equations 20-23.

Ix + lly + 133 = 201,

21.

S,\

7z 22.

9x+I6y + 25z = IQ9.

24.

Find the

remainders

c is

23.

number which when divided by

least

1, 5, 9,

= 209.

respectively.

7,

11,

Find also the general form of

all

20,

leaves the

such numbers.

a collector has 25. A certain set of stamps is made up of four different sorts one of each sort and some duplicates. The values of the stamps are 2d., 3d., and the lot are worth 3s. 9d. Find all the different ways 9d., Is., respectively in which the batch could have been made up. :

;

A

man has in his pocket three half-crowns, seven other silver coins which 26. are either shillings or sixpences, and eightpence in coppers he paid a bill of 9s. 7d. Find the possible number of different ways of doing so. ;

27. Find the sum of money which is expressed in farthings by the same digits written in the same order as those which are required to express it in pounds, shillings

and pence.

How many

solutions are there to this problem

?

28. If (p l9 2h> P& P*) i g an integral solution of ax l that other solutions are given by

+ bw -cv du', x 2 =p 2 + cu - dv' - aw, ~ #3 ~p3 + dw' + av bu, XIPI + an' + bv' - cw'

+ bx z ^-cx^ + dx^

e9

show

x i ~P\

where u

9

29. If

v,

w, u' 9 v' 9 w'

ax + by + c~Q, a'x + b'y + c'

integer, unless both b

[For a(bc')

9

have any integral or zero values.

and

b'

Q and x

are divisible

+ b(ca') + c(ab') = Q

and

by a'(

an integer, prove that y - a'b.

is

ab'

is

an

CHAPTER XXVI THEORY OF NUMBERS 1

Congruence.

.

(1)

Let n be any positive whole number, which we any numbers, positive or negative, such

shall call the modulus.

If a, b are

that a

by

~~

b

is

divisible

respect to the modulus n,

This

is

expressed

(2)

then a and b are said to be congruent with and each is said to be a residue of the other. n,

= by writing a ^b (mod n) or a 6 an abbreviation for is congruent with.''

(mod

n),

When no where the symbol doubt exists as to the modulus in question, we simply write a E=&. Any statement of this kind is called a congruence. '

==

Every residue

is

a to the modulus n

of

is

form a -f qn, where q

of the

may be positive or negative. If r is the remainder when a a = r (mod n) and 0^r
is

divided by n,

;

thus

r is the least positive residue of a.

Again, a

= - (n - r) (mod n), and either r or n - r^.^n. Hence we can and |/|<^n, where r' may be r so that a = r'(modw)

== r

always find

This

positive or negative.

For example,

if

5

is

number

the modulus, 34

least residues are (2) (i)

1

and

^6 + 6'.

(ii)

4

s - 1 and - 34 = -4 = 1. Thus are 4

and

and the absolute

1,

1 respectively.

Fundamental Theorems.

a -fa'

==

-34

the least positive residues of 34 and

-

called the absolute least residue of a.

r' is

f

a^b(modn) and a ^b'(modn),

//

a -&==#' -6'.

(iii)

then

aa'^bb', and in particular uc^bc.

d, a number prime to n, then a/d = b/d. For by hypothesis a b and a' - b' are divisible by n, hence - (b + b') and (a - a') -(b-b ) are divisible (a + a) by n. = (a - 6)a' 4- (a' - 6') 6, aa' (iv)

// a and b are divisible by

f

W

Again,

therefore aa' -66'

divisible

by

- = Lastly, a 6 qn, where q

is

is

by d, so also is qn. Now number, and since a/d-b/d = q/d

divisible

n. This proves the theorems (i)-(iii). a whole number, then since a and 6 are d is prime to n, therefore q/d is a whole .

n>

it

follows that a/d = b/d (mod

ri).

In particular, if a == 6 (mod p) wAere p is a prime and if d is any common divisor of a and b, then afdz=b/d(modp), except when d^Q(modp).

Thus a close analogy and equalities/ *

exists

between

'

congruences to a prime modulus

'

TEST OF DIVISIBILITY

422

The following

(3) (i)

ab

...

f

then

,

(2)

:

same modulus

to the

.

(ii)

and

are immediate consequences of the theorems in

If any modulus as=a', 6 = 6',... k~k &==a'6' ... &', and in particular a m z=a' m to

// f(x\ y, 2, ...) is a polynomial in x, y, z, ... with integral coefficients x = x', y^-y', z^z, ... to any modulus, then to the same modulus

f(x,y,z, ...)=f(x,y',z, Ex.

1.

Give a

test

as

...).

to the divisibility

number by

of a

N may be written in the form N=a +a + a. + +an n where

11 or 13.

7,

Any number

l

t

i

t

2

t

...

f^lOOO

0^a r
and

s-a -a 1 + 2 -... + (-l) n an then we have + l =1001 =7 t~ -1 and r =(-l) r with regard to each of the moduli Let

rt

,

t

,

N~s

Consequently

J

(mod

7,

.

11

.

13;

11

7,

hence

and

13.

11 or 13).

Thus if 2V = 12345671, then 5 = 671-345 + 12 = 338. Now 338 is divisible by 13 but not by 7 or 11 therefore 12345671 is divisible by 13 but not by 7 or 11. ;

The Numbers

2.

less

than a Given Number and prime to

n is any number greater than 1, the number of positive integers than n and prime to it is denoted by
it.

less

:

value of <>(1) as 1. Here, it should be noted that 1 is to be regarded as prime to every other number. Thus the numbers less than 12 and prime to

it

(1)

are

and

1, 5, 7, 11,

If a

is

prime

For

if

to

n, so also

n

is
by

in a certain order (Ch. its

is

remainder.

gression are prime to

If

number of terms of the arithmetical progression x + 2a, ... x + (n-l)a x-i-a,

these numbers are divided

w-1, taken

(2)

= 4.

to n, the

x,

which are prime

gp(12)

I,

n, the remainders are 0, 1, 2, 10),

Consequently, as

n as there are numbers

m is prime to n,

and

then cp(mri)

(p(m)

.

less

if

a

number

many

...

prime to terms in the prois

than n and prime to

it.

(f>(n).

To enumerate the numbers less than mn and prime to it, we arrange the numbers 1, 2, 3, ... mn in n rows and m columns as below :

1

...

w-hl 2m-fl

...

...

(n-l)w-t-l

m

2

m-f2 2m + 2

...

k

...

...

m+k

...

...

2m + k

(n-l)w-f2...

....

(n-l)m+k...

m 2m 3m nm.

is prime to n, the numbers in question are those which are to both and n. Now member of the &th column is or is not prime every prime to m, according as k is or is not prime to m.

Since

m

VALUES OF

423

p(n)

Thus the numbers in question, being prime to w, are members columns headed by numbers prime to m. 99 (w) ft,

of the

Moreover, every column contains cp(n) numbers which are also prime to for the numbers in each column form an arithmetical progression of

n terms with a common

It follows that there are

m and n,

to both

(3)

//

that

m v m w3

cp

(m)

m

.

cp(ri)

are prime

r

prime to n. numbers less than is

and so cp(mn) =
to mn,

is

...

2,

m which

difference

.

mn and


one another, then

to

cp^m^m^ ... w r = 9?(%)
)

w2

,

therefore

and so on, (4)

If p

for is

it is

m m2

prime to

any number

)

.

)

.

)

)

and

1

.

prime

both m^

Hence

of steps.

a prime, then q>(pr )

p

r

(l-l/p).

when r = l,

The proposition is true If r>l, since p is a prime,

for cp(p)=p-I. numbers 1, 2, 3, ... p r those which ~ are not prime to p r are p, 2p, 3p, ... p r and their number is p r l ~ r l which All the rest are prime to p r and their number is p r -p of the

,

.

,

y

,

proves the theorem. (5)

If

n=pa

.


r

.

...

where p,

p/\ For pa (f, (3) and (4). ,

Find

Ex.

1.

We

have

r

q/

are prime to one another,

...

the

are the prime factors of n, then

... q, r,

number of

integers less than

and the

n and prime

to

result follows

it,

from

when

w = 1024,1025, 1026.

1024=2 10

,

1025 = 5*. 41 1026

= 2.

= 1024 (1 -|) =512. and 9? (1025) = 1025(1 -J)(l -^-)=800. 19 and
hence

33

.

(p (1024)

This illustrates the irregularity in the variation of the function

Ex.

2.

This

is

prime to

//

n^2,

the

sum of the when n

obviously true it,

so also

is

Hence the numbers each pair has a

sum

integers less than 2.

If

n>2

q>(ri).

n and prime to it is \nq>(n). is any number less than n and

and x

n - x. less

than n and prime to

it

can be arranged in pairs such that

n.

Thus
pairs is

fyp(ri)

and the sum of the

FERMATS THEOREM

424 (6)

Theorem.

number

// ^(

= 1),

d3

2,

dr (=-n)

...

,

n^p^r

+ y(d2 + 93(^3) + )

where p,

...

q

l)

then every divisor of n

0
...

)

form p x qVr z

of the

is

and

let


where

...

Q^x^a, Q^y^b,

so that

etc.,

S = Zcp(p*qyr*

S

have

...

x, y, z,

= 27{y(p*)

...)

.

99(7")

.


Hence

values subject to the above conditions.

all

equal to the product

is

Now

1

+ (jp(p) +

2

cp(p

-f

)

.

.

.

4-

a (jp(p

)

= l+(p-l) + (p a -ji) + ...+(^and

any

= n.

(p(d r )

are primes

r, ...

y

+

...

S =
where

are the divisors of

n, then


Let

rf

similarly for the other series in the brackets.

S-=p For instance, the divisors

~l to p, then a p

For

a

1

is divisible

p

)=

Therefore

.r ...=n.

.q*>

//

-1

p

of 24 are 1, 2, 3, 4, 6, 8, 12, 24

Fermat's Theorem.

3.

a

fl -

is

a prime and a

is

;

and

any number prime

by p.

-

l)a are divided by p, the a 1 in taken certain order. p Also the product of the numbers a, 2a, 3a, ... is congruent (mod p) ~ with the product of the remainders, therefore a p l \p-I = \p-I; and since

is

prime to p,

if

a, 2a, 3a,

remainders are

1, 2, 3, ...

since

prime to p, therefore

1

\p

is

...

(p

a^^

^mod^), which

proves the

theorem. Corollary 1. is divisible

If p

is

a prime and a

is

p any number whatever, then a -a

by p.

"" ~~ For a p -a = a(a p 1 -l), and if a is prime to p, a p l - 1 is divisible p - a is divisible by p which is a prime number. Hence in all cases a by p. y

Corollary 2.

If p

an odd prime and a

is

a i (11-1)52

For a therefore

number.

p -1

p

is

prime

is

a divisor of

I>

a^^~ 1^

p, then

l(modp).

-l = (a*
to

1

or of

a"- 1 -!

a^/>-i)

+1

?

is

divisible

since

p

is

by

p,

a prime

THEOREM

WILSON'S Ex.

The fifth power of any number N has is divisible by 5 and by 2.

1.

For A75

-N

Ex.

The ninth power of any number

If

2.

N

prime to 19, since

is

\r(19

425

same right-hand

the

N is of one of the forms Cor. 2

-1)=9, by

19w, 19m

Euler's Extension of Fermat's Theorem. and a is prime to n, then afW ^ 1 (morf n).

^( =

a2

1),

,

a3

than n and prime to

...

,

9

19).

If n

a
divided

is

any num-

be the numbers, in ascending order, less

and consider the products

it,

aa l9 If these are

1.

N ^l(mod

we have

4.

Let

N.

N = 19m.

Otherwise we must have

ber

digit as

by

aa 2

aa^

,

n, the

...

aa9(n}

remainders are

(A) all different.

For

if

we

suppose that two products as aa r aas (r>s) give the same remainder, then a(a r -as ) would be divisible by n. This is impossible, for a is prime to n and a r a s
Also the remainders are

all

prime to

n, for the factors of

any product

are both prime to n.

Hence the remainders are the numbers a v a 2 ... a 9 n taken in some Now the product of the numbers in the set (A) is con,

(

)

order or other.

gruent (mod n) with the product of the remainders, therefore a*( n >

.

and dividing by a x a 2

a xa 2 ...

...

a
a

^a^

which

is

...

a9(n) (mod

prime to n,

n),

we have the

result in

question.

Wilson's Theorem.

5.

divisible

1,

is

.

is

by p. a is any one of the numbers

a prime number, then

\p-l +1

is

p 1, and if the products the remainders are the numbers a, 2a, 3a, (p-l)a by p, >~1, taken in some order or other. Hence, for every a, there 2, 3, ... one number a' and one only, such that aa' =zl(modp).

For 1

If p

if

If

1, 2, 3, ...

are divided

...

2 a'=a, then a -!

must be

divisible

by

p,

and

since

p

is

a prime

\p

or a -1=0. and a
such that the product of each pair is congruent with 1. 2 3 ... (p~2)~l (mod p) and \p- 1 ==j?- 1 (mod^).

Therefore

Hence NOTE. the other.

\p

.

-

1

+ 1 ^0 (mod p), which

The numbers (Euler.)

a, a'

is

the result in question.

are called associated residues, each being the associate of

LAGRANGE'S THEOREM

426 Conversely, if

p-1

\

is divisible

4- 1

p would have

For otherwise

by

p

then

9

p

is t3

prime number.

a factor q which would divide

and

[p-1

[p-1 + 1. 28 4-233

Ex.

1.

We

have 899 = 29.31,

Prove that

by 899.

is divisible

|

233-1 (mod 29) and 233= 10 (mod 31). By Wilson's theorem, 28 + 1-0 (mod 29), therefore |j|8 + 233=0 (mod 29). 30 + 1 = 0, therefore 30.29. 28 4- 1 = 0, Again, to the modulus 31, we have also

|

1

|

and

so

(

-

1)(

28 4-32=0.

2)

Since 2

prime to

is

31, it follows that

[28

|

and therefore

[28

Hence 128 4-233

4-233=0.

is

divisible

by 29 and

31,

4-

16=0,

and therefore

by 899. 6.

than

Lagrange's Theorem. If p p-I, the sum of the products of

a prime and r is any number less numbers 1,2, 3, ..., p-1 taken

is tJie

r together is divisible by p.

Let f(x)

= (x 4~ 1

products of

(x

)

4-

...

1, 2, 3,

,

2)

p

.

.

.

-

p - 1)

then

4-

1

taken r together,

;

a r denotes the sum of the

if

(x

= tf>- +a l x*- 2 + a&*-* + ...+a P _ l ................... (A) = (x + I)f(x + l), that is, identity (x+p)f(x) l

f(x)

Also

we have

the

(B)

Equating

~ the coefficients of x p 2 ,

etc., are all all divisible

,

...

x,

we

find that

~2

1

a prime, Cf is divisible by p, for r

since

Now,

~ zp 3

p

is

>

;

,

by

which

p,

is

the theorem in question.

For another proof,

see Art. 11, (6).

We

NOTE. (i)

and is

can immediately deduce the theorems of Wilson and Fermat, thus

Equating the terms independent of

since

divisible

av a2

,

...

,

by p which t

re

a p _ 2 are divisible by is Wilson's theorem.

p

9

so also

is

a^^ + 1,

that

is

any number prime to p one of the numbers x 4-1, x 4- 2, ... by p, hence f(x)=0 (mod p). ~ ~ Also /(#)= x p l 4-ap^ x p l - 1 (mod p) therefore x3*- 1 -1=0 (mod p) (ii)

If

must be

x

is

:

in the identity (B),

9

\p

,

-1

4-1

x+p - 1

divisible

9

Format's theorem.

9

which

is

EXAMPLES OF DIVISIBILITY Ex.

If p

1.

is

greater than 3, prove that

awrime, '1

1

1

+ '"

^2 in the identity (A),

If,

(p

and

N6w,

p

since

divisible

Hence,

is

-

(2p ...

1)

a prime greater than

a 3) _ 2

.

divisible

is

If p

divisible

is

a prime, greater than

by p

m

~\

3,

mp

r

~l,

2, 3,

since 3

divisible

*s

prove

-

m

I

.

we have

-f ...

7>

;

and, by Lagrange's theorem,

prime to p,

is

by p

that, I

(

3

it

follows that

2 .

for all integral values of m,

p\m ~

.

Substituting rp for x in the identity (A),

where

2 .

subtraction,

a/p-s exists

3,

by p*\ and,

1

is

P+^-s

ap^ 2

j)~i-f

in succession,

by p.

3_p

2.

p and - 2p

+ l)~2>_i ~%_2 2; are identical, we have by

a^_ 2 Ex.

to

(p

since the left-hand sides

it is

p~-}

we put x equal

+ l)(p + 2)

(2p -l)(2p -2)

427

...

m - 1,

,

Hence,

|

(r

and, by giving r the values

and p -f-

1)

p

I

I

(

1, 2, ...

hence,

is

I

mp -

j

as in Ex.

prime and greater than

rp

.

I

/;)

m - 1,

,

we have,

ra(|

~r+1 it

(mod

1,

3.

3 )

jt?

;

follows that

p\ m =0 (mod

p

m +s ).

EXERCISE XLIII 1.

N

If

ao

+ a t + a 2 t* + l

...+a n t n

,

... -f (

-

I)

n tn

9

T

Q^a r
where Z~ 10000 and

S^aQ-^^a^ -

and

prove that J\ ~/S(mod 73 or 137). Hence state a rule for the divisibility of a number by 73 or 137. Apply it to 30414 and 81103. 2. If

and

p

is

a prime and x l9 x 2 , X Q

in particular (ax) p ~ax p

Prove that,

4.

Prove that,

if if

n

n

is

...

xa are any numbers, show that

(mod p), where x is any number. and if a is not divisible by p, a p~~ 1 == I (mod p)

p Consequently, a ^= a (mod p) which is Fermat's theorem. 3.

,

9

a prime greater than

is

then n 6 -

7,

1

is

a prime greater than 13, then n 12 -

divisible 1

is

by

9

504.

divisible

by

65520. 5.

If

p and

6.

If

p

is

q are different primes, then

a prime of the form

pq

~l

4m + 1, show

-f

q

p~ l

that

I

1

is

%(p -

+ 1 = (mod p). 2ms (4m) (4m -1) (4m -2) 1.2.3

divisible 1)

is

by pq.

a solution of

the congruence, x 2

[Show.that 2 E

.

.

...

(2m -f l)(mod^).j B.C.A,

ROOTS OF A CONGRUENCE

428

4m -

7. If p is a prime of the form the congruence, x 2 - 1 = (mod p).

the congruence,

8. If

-

(i)

9.

\p

Show

10. If

11. If

1 =

that

-

2

^

l)fc

(

118

a'

+1

=s

-

p ~^

is

1

divisible

-3 + 1

|

(modp), has no

and

;

show that

1,

(ii)

by

p

a prime, 2 \p

n

is

any odd number, show that

divisible

\n-\\ I+S + QT--- + 11 I

12.

Ad

I

Arrange the numbers aa ~\ (mod 17).

...

2, 3, 4,

,

p

-

1)

is

a solution of

show that of the form 4m -

solution, is

1.

437.

is

is

that

\(p

by

p,

T h=0(inod7i). l_j

15 in pairs, a and

a',

such that for each

f

pair,

13.

Let

n~a pb Qc r

...

,

where

a, 6, c,

...

and consider the

are different primes,

groups a, 2a, 3a,

...

7i/a

.

a

6, 26, 36,

...

n/6

.

b

1

a&, 2a6, 3a6, .

(

f

A

)

6c, 26c, 36c,

...

n/ab

...

n/bc

.

.

a6 be

1 f

(

B

)

together with groups (C), (D), etc., formed in a similar way from the combinations of a, 6, c, ... taken 3, 4, ... together. If s l9 s 2 5 3> denote the sums of the rth powers of the numbers in the groups (A), (B), (C), ... respectively, show that the sum of the rth powers of the numbers not greater than n and not prime to it is ,

14. If less

# 2 is

the

sum

of the squares, and it, show that

$ 3 the sum

of the cubes, of the numbers

than n and prime to

where

a, 6, c, ... are the different prime factors of n. [Use the last example to show that the sum of the squares of the numbers not greater than n and not prime to it is

U 3

7.

\

a~

ab

+ n +

+ '")

Roots of a Congruence.

degree in x, condition

with integral

f(x)=Q(modn)

2

6^

a~

a+ ~->-l

is a polynomial of the rth value of x which satisfies the any called a solution or root of the congruence

If f(x)

coefficients, is

f(x) =0, which is said to be of the rth degree. If xQ is a root of the congruence f(x) ^0(mod n), so also congruent with x to the modulus n.

is

any number

Two roots will be regarded as distinct only when they are incongruent with respect to the modulus, and when we say that a congruence has r roots we mean that it has r distinct roots.

THE LINEAR CONGRUENCE Ex.

from Format's theorem that, (modp) are 1, 2, 3, ... p 1.

It follows

1.

1 congruence x^"

1

if

is

p

429 a prime,

the roots of the

E\rery square number is of the form 5n. or 5wl> thus no square can be z congruent with 2 to the modulus 5. In other words, the congruence x ==2(mod5) Ex.

2-.

has no solution.

NOTE.

It

13x=

congruence

-

z=

1

obvious that, in dealing with the congruence /(#)=0 (mod n), we may by any number congruent with it to the modulus n. Thus the

is

coefficient

any

replace

(mod

17

(mod

5)

is

3#=2

identical with

or with

3a;=

-3

or

with

5).

8. The Linear Congruence. (1) If a is prime to n, the congruence ax H= b(mod n) has just one distinct root. For if the terms of the arithmetical progression 0, a, 2a, ... (n - 1) a are is prime to n, the remainders are the numbers 0, 1, some or other. Hence there is just one value of x order 2, n-1, such that ax^b(modn) and 0^x
divided by n, since a in

...

be the last convergent but one.

Then ax - nyQ = 1 and since a is prime to ,

a

If

so that n,

x^=bxQ

axQ == ,

1

(mod

which

is

n).

,

the required solution.

small, or the product of small primes,

is

ax E= 6 = abxQ

Therefore

it is

generally easier to

proceed as in Exx. 1-3 below. (2)

// a

is

not prime to n,

and g

is the greatest

numbers, there are g incongruent solutions of is divisible

by

common

ax^b(mod

n),

divisor of these

provided that b

If b is not divisible by g, there is no solution.

g.

a~ga' n=gri, so that a' is prime to ri. We require values of x such that ga'x - b is divisible by gn'. No such value exists unless 6 is divisible by g. If this condition is satisfied and b=gb', then a'x-b' is divisible by n' and the given congruence is equivalent to a'xs-fe'fmod n'). Since a' is prime to ri, the last congruence has one solution a
(3)

let

,

The expression 6/a(mod

n)

is

used to denote any solution of

ax = 6 (mod If

n).

common divisor which does not divide 6, then 6/a(mod n) no meaning in every other case it has infinitely many values. If a

a and n have a

has

:

if a is not prime to congruent (mod n) is the where congruent (mod n/g) g greatest common divisor

is

prime to

n, these values are all

n,

they are

all

of a

and

n.

:

>

SIMULTANEOUS CONGRUENCES

430

Such expressions possess many of the properties of ordinary and it is easy to prove the following

fractions,

:

a = #'

If

(i)

and

a7&'(modn) ak/bk(mod

(iii)

Ex. (i)

(ii)

equivalent to a/6 (mod n). equivalent to a/b (mod n), if k

am/bm(mod mn)

(ii)

and

a/6 (mod n)

is

is

n)

the expressions

bz=b'(modn), are equivalent.

is

1. Solve (i) 5ar=2 (wod 7) (ii) ISzssG (mod 21). Hero 6#==2-7== -5 (mod 7), therefore #~ - l==6(mod

prime to

n.

;

Since 3

is

a divisor of

so that

z==6(mod

Ex.

2.

Solve

We

have

and x~G,

7)

3.

We

have

?=

12*==

Sotoe -)-

1

20 (mod

13,

7z=

1

~ 2 4-

=43

.

- 50,

^|?== o 11

-

(mod -

-

2 - 17

.

5~

7).

equivalent to

as -Hf?=

therefore

5#=2(mod

7),

IA

^s ~ O

==

O

22 (mod 157).

43).

and the third convergent - 17

.

5(mod

Simultaneous Congruences,

have x = a +ocy where y

is

a + ocy = 6 (mod

and

43)

(i)

is

5/2, thus

&(mod 43 )~

~5

'

55= 31 (mod 43). find x so that

It is required to

and

x-a(moda)

We

is

21).

x= -\^- (mod 43)~ -

Therefore (4)

the congruence

36* = 7 (tnod 157). O

Ex.

15, 6, 21,

x==b(mod

/?).

given by

j8)

or

a2/

= fe~a(mod j8).

Let ^ be the greatest common divisor of a and j8. If 6 - a is not divisible by g there is no solution. Otherwise, there is just one value y l of y less than /?/
+ f}/g

.

t,

so that the general solution is

x = x l + afilg

.

t

x t = a + at/j.

where

and only if, 6 - a is divisible by g, the greatest common divisor of a, j8, and then the conare gruences equivalent to the single congruence x = x (mod I), where I is the L.C.M. of a and j8. Thus a solution

^

of the given congruences exists

It should be observed that solutions (ii)

To find x

so as to satisfy

xz=a(mod
always exist when a

a number of

x = b(mod

if,

j8),

relations of

tJie

x^c(mody),

is

prime to

/?.

form

...

replace the first two by x = x 1 (modi!) as in the last section. this with the third, the first three congruences are equivalent to

we

Taking

x = x2 (mod m),

where x2

is

any

solution (found as before)

and

m

is

the L.C.M. of

I

and

y,

THEOREM ON FRACTIONS and

therefore of a,

congruences are together equivalent

and L if

is the L.C.M.

A

<*>

Ex.

We

35x^4 (mod

Driven that

1.

x=~

have

oo

Hence

.

Thus y = l+4t and

a;

and

9),

~ (mod 12) - 1-- 2 (mod 12).

x=-

&o

3

= 5 4- 9 (1 + 4$)

+ 362, which

14

solution

any

12), find the general value of x.

1

12) =s--

y^-Wmod y

giving

12),

is

the given

Moreover, a solution always exists

and 55x=2(mod

5 (mod (mod 9)==-^r== 1

= 5 + 9ys 2 (mod

a;

9)

where

xz==g(mod L)

of a, /?, y, are prime to one another.

...

y,

to ...

be seen that

will

it

Continuing thus,

y.

j8,

431

o

is

(mod

4)

si (mod 4).

the general solution.

We

can apply these methods to solve a linear congruence when the modulus is a composite number, as in the next example. (5)

Ex.

2. Solve 19*== 1 (mod 140). and have 140 4 .5.7, and therefore 19#==1 for the moduli 4, 5 and 7 since these numbers are prime to one another, any value of x which satisfies these conditions is a solution. Thus we have

We

:

#==

17

=

-

(mod 4)==

1

- l(mod

x=

-

1

5),

7)

= -=-4 (mod 7).

- l(mod 5), so that y = 5z. - 4 (mod 7) z= and -f 202=

+4t/==

x= -

Therefore

(mod

1

8

z==y-(mod

Hence

:==-- (mod 5)= 1 t/

1

and

4),

1

7)= 3 (mod 20 (mod

-^r-

Thus

7).

and x = 59 In

this case the solution

_. = 7 + 2

1<7

Therefore

-

-

+

1

2*

~\r

4"

2

-f 140J, which is the general solution. can be found more easily by the method of

The fourth convergent

.

is

~ o

,

and

59

.

(3),

Ex.

19 - 140

3,

.

thus 8

= 1.

x~ ~ (mod 140) == 59 (mod 140). 19

A Theorem

9.

on Fractions.

If n

is the

one another, and Z, m/n can be expressed uniquely in the form

c 9 d,

which are prime

...

to

m = a- B y A + y + + ...+ 7 n

where a,

Let

a'

)8,

...

A,

b

c

is

product of factors a,

prime

6,

to n, the fraction

...

7

..............................

fc,

v

A)'

I

k are positive integers and a
etc.

and therefore integers x, x' can prime to a', for any common these two would divide m, which is prime to a'. Thus we have bed

then a

... I,

be found so that a'x factor of

a

m

is

prime to

a',

+ ax' = m. Moreover,

m _x

x

aa

a

f

a

'

. -I

1

x' is

OUJ.VA

w_x n

a

T

7

:i

7

bed ...I

THE GENERAL CONGRUENCE

432

Proceeding in this way, we can obtain a relation of the form

m-x

._

n where

u I

_. . .

I

j

a

w

z L ~___ J

__ _i-

b

c

w are integers, and then mfn x^a + y^/3 + kJ), etc.,

x, y, z,

by writing

j

,

I

frjcr,

can be put in the required form where a, 'j8, ... are the least

with regard to the moduli a, 6, ... respectively. we multiply each side of the equality (A) by n, it will be seen

positive residues of x,

Again, if that a he

...

.

?/,

...

Z^m(moda),

ac

.

j8

= w(mod

...

6),

etc.

Since

6c...

i!

is

a has just one positive value less than a. Similarly j8 has just one value less than 6, and so on. Hence the expression found for m/n is unique. prime to a,

Ex.

}

Express

.

^ e form

Yr~~7\

- + -+- + - + A% 4

O

/

J.

the fraction* being positive

1

proper

fractions and k an integer. Here x, y, z, u are determined by

5.7. liars

4.7.

1

101

(mod

= 101 (mod ly

1

4), 5),

(

4.5 Ilz=sl01(mod7), With

a?

.

5

.

= l,

lu

~ 101 (mod

y = 2, = 1,

it

1)

(

2

.

1

.

.

= 1,

== 1

= 1 (mod

a?

-4,

4);

y^2(mod5); ZEE! (mod 7);

s 1= 12,

w

mod

== 3(

i + f +y+^r = l^^;

find that

The General Congruence.

10.

y

4?t

11),

= 3, we

- l)x

1)(

(-3)(-2) .4^-3,

.

4

.

In this article

11).

so that jfe=-l.

/(re), 97(3),

...

will

denote polynomials with integral coefficients. (1) If

every coefficient of a polynomial /(x)

is divisible

by n

for all values of x,

gruent with zero to the modulus n.

/(x)


r=

divisible

by

n,

then /(x)

and we say that f(x) is identically This is expressed by writing

con-

said to be identically congruent with (p(x)(modri) 0(mod n), which is also written as /(x) H^ 99 (x) (mod n).

Again, /(x)

-

is

is

if

If /(x) is divided by
whore the remainder R'(x)

is of lower degree than
is called division

(mod

n),

f(x)

where R(x) If

72 (x)

is

and leads

= Q(x)

of lower degree

^0 (mod w),

.

to

an

identical congruence of the

form

(p(x)+R(x)(modn),

than

then f(x)

is


said to be divisible (mod n)

by

(p(x).

CONGRUENCES said

is

Ex.

divisible

is

// f(x)

have r roots equal

to

(mod

shown on the

,

the congruence

7)

right.

f(x)==Q(modn)

to a. 2

2, 4,

-6 by + 1. In

and

1

r

G are roots of 5x* + 3# - \x - 2===0(mod - 2 and x 4 in succession is x by In the third line 13 is replaced by

Prove that

1.

Division

-

(mod n) by (#-a)

433

the last line 19

is

replaced

0. The reckoning shows that + 3z2 - 4x - 2~ 5(x -2)(x -4) (x + l)(mod

by 5

7).

12

5+342

\ __4

5-1+1 + Q

10-2+2

and 21 by or*

when x ~2, 4

so that the congruence holds

or

20 + 9 7),

6.

Congruences to a Prime Modulus.

The analogy between a step further by the following

11.

congruences and equalities is carried theorems, in which the modulus p is a prime.

or is

N ^Q(mod p).

For since

MN

is

M

= Q(modp), a prime, then either or N divisible by the prime p, either is

M

so divisible.

If a

a root of f(x)z=Q(modp), then f(x) is divisible (mod p) by For by di vision (mod p) we have f(x) ^Q(x) (x a) -f R, where

(2)

x-a.

R

where p

MN-O(modp)

//

(1)

is

.

is

of

independent

therefore /(.r)

then

it

of x

- a and x

(mod

p)

divisible (mod p)

divisible (mod p)

is

/?

is

x=oc

Putting

divisible

is

is

Iff(x)

(3)

x.

x-oc.

by

r

and by (x-fly where a For any common divisor which is independent of x.

by

(x~a)

-

r

by (x
we have 72=

- ft)* (x /?

.

(4) If f(r) is a polynomial of degree n, the congruence f(x)^Q(rnod p) cannot have more than n roots, any root which occurs r times being counted

as r distinct

This follows at once from the preceding theorems.

roots.

// f(x)'^Q(modp) is a congruence of the n-th degree with n roots and a factor of'f(x) of degree r, then the congruence (p(x)^0(mod p) has
is

(6)

p

//

is

a prime,

p congruence x

Hence it

if

~l ~

(Art. 6)

Ex.

1.

;

it

follows

ar

a^"-

^0(mody)

also

I

p

-

1

+

sum

-a

I

2

from Fermafs theorem that the roots of 1, 2, 3, ... p - 1, and therefore

are

^0(mod p)

a r denotes the

follows that

Hence

1

of the products r together of

p -3 ar

for

4- ... 4-

a p _,

r
=0(mod

p),

the

,

9

x2

{

I

p

-

which

which

is

1 -f 1}

1

,

2, 3,

...

p-

1,

^ 0(mod p).

Lagrange's theorem Wilson's theorem. is

Solve

Here 4z 2 4-43;-f4==0, or (2^-fl) 2 ^

-3s 4;

and2.r-f IE=

2.

Hence x~ 2 or 4 (mod

7).

ROOTS OF CONGRUENCES

434

In treatises on the Theory of Numbers the notion of congruence is extended to complex numbers, thus completing the analogy referred to at the head of this

article.

Let /(a;)^0(mod p) be a connth degree to a prime modulus p. Let the roots of the congruence be oc v 2 ... a r where some of these may be multiple roots. By a modified H.C.F. process we can find a congruence R(x) ^0(mod p) of 12.

Reduction of a Congruence.

gruence of the

,

,

degree r of which the roots are ot v
,

1, 2, 3, ...

~ xp l

-I ==0(modjp). /(#)=EO(mod^) is a solution of ~ Let the H.C.F. process be carried out for the functions f(x) and x p l - 1 with the modification that any coefficient may be replaced by any number congruent with it (mod p) then the last divisor R(x) is the highest common solution of

.

p - 1, hence every

;

di visor ( mod p)

and x p ~ l -l.

of f(x)

The truth of this statement depends on the following. Let u, v be any two consecutive remainders in the process. If w is the next remainder, w au + bv where neither a nor 6 is divisible by p. Hence the common divisors (mod p) of u and v are the same as those of v and w. It follows that

Ex.

1.

SJunv that the congruence

distinct root

is

R(x)^0(modp)

and

solve

is

the congruence referred to above.

3# 4 -2x* ~4x 2 -x-l==Q(inod

7)

has only one

it.

6 Carrying out the modified H.C.F. process for the left-hand side and re -!, it the x 2 is the distinct found that last divisor is root. + 2, showing that only we find that 3z4 - 2#3 - 4z2 - x - 1 == 3 + 2 x~ division

(mod 7) By Now afe3{mod7)

(x

-2

has no solution, hence

is

2)

(

3) (mod 7).

the only distinct root.

EXERCISE XLIV 1.

Find the form of x, (i)

29x=

1

if

(mod

13)

;

(ii)

2:c==3

(mod

7)

and 3*^5 (mod

2.

(iii) 8z= 1 (mod 7), 3z=4 (mod 11), and 7z~3 (mod Solve (i) 78z= 1 (mod 179) (ii) 78a;= 13 (mod 179).

3.

Solve

16 ^^ 31

.

(mod 1217) (iii) 15*= 28 (mod 1009) #= a (mod 22) and x =: b (mod

(H)

;

4. If

20).

If

x~a (mod

b

16)

40),

(mod 5)==c (mod

30#=~31 (mod 1861)

If x==a

(mod

7)== 6

z==

;

(iv) 26#=35 (mod 1901). show that a -b is even and

11),

prove that

x- 385a + 1766 - 560c (mod 880). 6.

;

;

(i)

5.

11)

(mod 11) c (mod 13), prove that - 286a + 3646 - 77c (mod 1001).

BICYCLE-GEAR PROBLEM 7.

If x = a

8.

Express

integers 9.

and

Express

integers

435

(mod 3)^6 (mod 5)=== c (mod 7), prove that x== - 35 4- 216 4- 15c (mod 105). --

1

/

form

in the

-

OU

< 32,

?/

-

< 5,

2

< 11

k,

where

x, y, z,

k are positive

k,

where

x, y, z,

k are positive

11

-

form

in the

~

.

1001

and #<7,

4-

O

11

7

2<13.

2/

10. Find tfie form of numbers, of which the first, second and third powers are of the forms 3^4-1, 4^4-1 and 5>i4-l respectively. 2

-2

divisible

11. If

?i

12. If

# 2 4-4oH-2

is

13. If

n 2 and

+

is

(n

by

divisible 2

l)

and n 2 ~ 3

7,

is

divisible

23, find the

by

are of the forms

11, find the

by

form of

form of

n.

x.

llm 4- 4 and 12m, 4- 4,

find the

form

of n. 14.

Find four consecutive numbers

divisible

15. 'Prove that the congruence 3# - 4s - a; incongruent solutions, and find them. 3

4

by

2

4-

5, 7, 9, 11 respectively.

3# - 4^

(mod

16. Prove that the congruence incongruent solutions, and find them. 17.

7)

(mod

~0 (mod 29)

Prove that the congruence and find it.

has only two

11)

has

four

has only one distinct

solution,

18. Bicycle gear as

A

a revolution

counter.

bicycle with no free wheel has

,,-

19 teeth in the back chain- wheel, 62 teeth in the front chain- wheel, 121 links in the chain,

^

c

and the circumference of the back wheel of the bicycle is known to be almost exactly 7 feet. Prove the following rule to find

p IG>

^

the number of revolutions (R) of the back wheel (and so the distance travelled) in any journey of not more than 10 miles. Mark a tooth .4 in the back chain- wheel, a tooth in the front chain- wheel, and a link C in the chain. At the start count the teeth or links from A to B and from A to C in the direction of turning. At the finish count these again. Take the increase from A to B or 62 minus the decrease, and denote it by a. Take the increase from A to (7 or 121 minus the decrease, and denote it by 6. Then the number E is the remainder in the division (3025a 4- 21086) 7502. [If x is the number of revolutions of the back wheel,

B

19z=a(mod Hence show that and therefore

62)

#==49a(mod

and 62)

u;= 3025a

-f

19#==&(mod 121). and #=51&(mod 121),

21086(mod

7502).J

CHAPTER XXVII RESIDUES OF POWERS OF A NUMBER, RECURRING DECIMALS Residues of Powers of a Number. 1

// a and g are prime

.

to

any modulus

n, the least positive residues

successive terms of the geometrical progression

number of terms in

is the

t^cp(n) and

the recurring period,

and

recur ,

...

a, ag, ag~,

is

of

if

t

independent

of a.

Denote the residues by r r l9 r2 etc. Each of these is one of the
,

,

,

;

,

prime to n, hence ag

i+2

~ag

2 ,

etc.,

Moreover, e

<7

~l(mod

the

= l(modn).

^*

and so r rQ r t+l r^ f^^^* number of terms in the period is

rM

If

/A==v(mod

t

~a

t+l

^ag, where t^
ag

9

e ^ c ->

,

t

the

n).

in (i) No two residues = and 0(modJ), conversely, /z

It should be noticed that (ii)

that

It follows

= l, J),

then

:

and conversely,

(iv)

Thus for

the

(modn) with

modulus

~gr,

2

-g

,

,

0,1,2,3,

Residue.

1,2,4,8,16,13,7,14,9,18,17,15,11,

7,8,

= /v,

rM

then

n-l occurs w-r3 etc.; for

,

,

etc.

,

1 2 2, 2 2

Index.

5,6,

3

If

(v) If

n-rv n-r2

~g

19, the residues of

4,

(iii)

^^^^(mod n).

as a residue, the subsequent residues are

they are congruent

the period are equal,

,

...

are

9,10,11,12,13,14,15,16,17,18.... 3,

6,12,

5,10,

1

...

.

Having found r4 = 16, we have r 5 = 16 x 2 = 13 (mod 19). Again, r = 18 = 19 - 1, and the sul>sequent residues are 19-2, 19-4, etc. Here J = 18 =
For

the

modulus

13, the residues of

1 2 5, 5 5

Index.

0,1,

Residue.

1,5,12,8,1,5,12,8,1,5,12,

Here

2,3,4,5,

= 4, which

is

,

,

6,7,8,9,10,11,12. 8,

a divisor of 99(13)

1.

~

12.

...

are

RESIDUES OF POWERS 2. // g

is

g ^\(modn}

to

prime

i

then

9

For #o (n = l(mod )

If

ceding t^(p(n).

n and

t = (p(ri)

t

r

for

;

for

b
Hence,

=g mod

is

not the case, for r-s
S

denote the set of numbers

S

the numbers in

b of the set

the residues of bg, bg l bg 2 ,

,

6

if

included in the

"l 9

&A

g

,

than n and prime to where s
r~5

=1

,

than n and prime to t

it.

it.

If the set

q>(n).

does not occur in the set (A), consider

bg ,

less

then

9

S t

...

and

n)

(

9

less

a r ~as

if

s

Let

some number

numbers

q>(ri)

which

all

= 0,

9

g

These are

6

q>(n).

of these is one of the

all

+ &, where 0<6
and consequently

9

Also no two of them are equal

If

be equal to q>(n), and by the pre-

may

t

let

Let the least positive residues of g^ g l g 2 ... g*~ l be a t-i ............................... ( A ) a i> a 2> o>

Second proof.

(A) includes

index other than zero for which

or is a divisor of
n), so that

Therefore g b ^l(modn) t
Each

is the least

t

437

,

denoting them by

62 ,

set S, for

b

bt

...

is

^ ............................... (B)

prime to

n.

Also they are different

from one another and from those of the set (A). For if br = bs where s
9

9

)

which

is

contrary to supposition. and (B) include all the numbers in

If the sets (A)

/S,

then

2t

t

=
>

Otherwise 2t<(p(n)> and then some number c of the set S does not occur We then consider the residues of cg cgl cg 2 ... cgr*-1 ,

in (A) or (B).

denoting them by

As

be'fore,

CQ>

we can show that

^

9

^^

...

these

numbers are

9

9

................................ (C) all

included in

S and 9

that they are different from one another and from those in the sets (A)

and (B). Thus 3t^
(p(n).

Incidentally, this proof establishes the truth of Euler's generalisation of

FermatVtheorem.

PRIMITIVE ROOTS

438

An Odd Prime Modulus:

3.

prime, g any

g

t

s 1 (mod p),

number not then g

is

Primitive Roots. Let p be an odd

divisible

by

p,

and

t

the least index for which

said to belong to the index

t.

Since cp(p) = p-l, t is equal to or is a divisor of p-1. If tp-l, so that the period of residues includes every number less than p, the period is

said to be complete

is

a divisor of

and g

is

called a primitive root of the

the period

p-1,

is

modulus p.

incomplete and g may be

It

t

called a sub-

ordinate root.

Hence if g is a primitive root of p, every number prime to with some power of g.

p is

congruent

proved that, for any odd prime modulus, primitive roots exist for the present we assume this to be the case. When one has been found, It can be

;

the others can be found as in the following illustration. Illustration.

In the margin, for the modulus 13 the least positive U are written down round ... 2

residues of 2, 2 l 2 2 ,

,

the circumference of a

Every number

circle.

than 13 occurs, and so 2

loss

is

a

primitive root of 13. Hence, without further reckoning, we can write down the residues (mod 13) of powers of any number. 5 6 2 s2 10 etc., powers of 6, since 6-2 1 2 the residues of 6, 6 6 ... are found by starting with 1 and taking eveiyffth number. The residues are

Thus

for

,

,

,

,

1,

6,

10, 8, 9, 2, 12, 7, 3, 5, 4,

11.

Every number less than 13 appears, the reason being that 5 is prime to 12. Hence 6 is a primitive root; and, in the same way, if is any number less than 12 and prime to it, 2 M is congruent (mod 13) with a primitive JJL

root.

Since

with

2


2

5 ,

= 4, 27

,

there are four primitive roots of 13; they are congruent and are equal to 2, 6, 11, 7, the first, fifth,

2 11

,

seventh and eleventh

from

numbers

in

order

round the

counting

circle,

1.

9 and the residues of 5, 5 1 5 2 ... are obtained Again, we have 5 = 2 by counting every ninth number, or every third number in the reverse ,

,

,

order.

Since 3

is

four counts

Thus 5

is

the greatest

common

divisor of 9

we reach the number, a subordinate root

is 1, 5, 12, 8.

;

5,

it

and 12 and 12 = 3

from which we

.

4,

after

started.

belongs to the index 4, and the period

RECURRING DECIMALS

439

an odd prime, g any number >1 and not divisible by p, t the r v r^ ... the least positive least index for which #' == 1 (mod y) and r residues of g, g l g 2 ... then 4. If

p

is

fl ,

,

that

,

sum

to say, the

is

Thus

,

= 13 and jo

if

of the residues which form the period is divisibk by p. = 5, then 2 = 4 and
Again, suppose that e

with any residue

rm

and

= ef. Starting (not t itself) and taking every eih residue, we have the sequence is

a factor of

^m>

^m+e*

t

^m+2e>

These residues recur, and they form a sub-period containing f residues for g x ^l{modp) if x = ef, and for no smaller value of x. Also

^ + W.+W2

+

-.-+W(/-i)=^

Hence the sum of the residues forming Thus if p = 19 and # = 2 (See Art. 1,

5. (1)

Recurring Decimals. If n

is t

period of

prime

;

to 10,

figures, where

m/n t

Let

is

is

equal

the sub-period is divisible by p. (v)),

m
the least

then

*

= 18 and

and prime to

n, .then

a pure recurring decimal with a index other than zero for which

10* =

1 (mod n) ; so that t=*(p(n) or is a divisor of cp(n). For the remainders rv r 2 3 ... in the process of expressing m/n as a 2 decimal, are the least positive residues (mod n) of m, 10m, 10 m, .... Hence, by Arts. 1 and 2, the remainders recur, with a period of t figures. ,

Also all

m/n = O-a^

if

...

/*

dt) then

the fractions m/n, rjn, r^n, (i)

If

t

If

groups of

...

r8 /n = 0'ds+l as+2

...

a ta l

have the same period.

...

as

,

so that

It follows that

= y(n) and m is any number
m/n have the same (ii)

,

,

(jp(n)=et t

period.

where e>l, the fractions m/n can be arranged in e such that the periods are the same for fraction in

fractions,

the same group and different for those in different groups. This follows from the reasoning in the last part of Art. 2.

DECIMAL PERIODS

440

np

If

p

any prime except 2 or

according as 10 is or

p-l, r ~l

np,

//

(iii)

r

then

,


The case

(p-l).

is

t=p-I,

or

not a primitive root of p.* 1) and t is equal to or (p

n=p

2

a divisor of

is

t

r ~l

which

in

5, then

is

a divisor of

deserves further consideration.

5, and l/p has a period of t figures, then If p t l/p has a period of or pt figures. For let T be the number of figures in the period of l/p 2 and let

a prime, not 2 or

is

(iv) 2

,

Then F(t)^0(mod p),

lO'sl(mod^), and therefore

for

Hence T is pt, or a divisor where r is t, or a divisor of t.

r<, we

If

1

for 10

T

+ 10 + 102T +

...

+ lO(*-V* = (lQ(p-Ur _

a prime,

T=t

or pr,

Consequently r

T

1)/(10

-

1)

E=0(mod

p),

Hence F(T) = 10<*-V = l(modp). 10 PT - 1 - (10 T - l)F(r),

and therefore 10* T -1

JJL,

is

T

isnot si.

is

is

not ^O(modj) 2 ).

T=

not
The only known

NOTE.

p

have

T

Now

p =3

of pt, and, since

cases in which

t

or pt. Ijp

2

has a period of

t

figures are

when

or 487.

(2)

//

v,

then

.F

iyy\

=

<1, where

^

F

is

For 10*F

is

m is prime to

10,

and k

is the greater

of the two

a terminating decimal with k figures. an integer with not more than k digits, the last of which

is

not zero. (3) is

//

m/

F=

<1, where m, n and 10

a mixed recurring decimal, with a period of t figures, where

of figures in the period of l/n. For let k be the greater of the

prime

n and

to

integer of not

period of (4)

//

t

than

more than k

jit,

v,

t

is the

then IOkF = m'/n, where m'

Therefore IQ kF = I+f, where /

10*.

digits

is

is

an

and /is a pure recurring decimal with a

m

where p,

q, r, ...

are different primes other than 2

prime to p, q, r, ... , then if t, t', t" ... are the a figures in the periods of the decimal equivalents of F, l/p , respectively, *

number

figures.

F = -~j~

and 5 and

less

two

F

are prime to one another, then

The primes

t

is

is the

less

L.C.M. oft', t",

than 100 of which 10

is

numbers of b

l/q

,

etc.,

t'", etc.

a primitive root are

7, 17, 19, 23, 29, 47, 59, 61, 97.

USEFUL THEOREM For

n = p a (f

let

and, since

t'

Therefore

;

t

similarly, is

t

then

,

s 1 (mod n),

10*

a multiple of t",

is

common multiple of any common multiple

a

t' ,

Again, if r is each of the moduli p*, f, It follows that

t

the

is

...

,

least

and

since

common

10 e s t

a

1

(mod p ) must be a

t'" 9 etc.

t", etc.

of f,

p

therefore

10*' = 1 (mod p a ),

the least index such that

is

$'

multiple of

. . .

441

a ,

(f,

...

we have 10 T ^1

...

",

for

are prime to one another,

multiple of

t"', etc.

t',

Given a table showing the periods of decimals equivalent to fractions 1/n, where n is a prime or a power of a prime, we can easily express any fraction as a decimal by using the theorem of Ch. XXVI, 9. 5. In practice, the following theorem is useful. Let m/n=0' a^a^ where n is prime to 10 and . . .

r steps of division,

9-a

decimal are

For

let

m,

9-a 2 9-a3

x,

,

m m2 ly

...

,

n-m

remainder

tfie

r

a r+l

is

is

,

-m

-m2

l9

,

etc.,

etc.

a r+l = 7(10-0^ +/)

/

...

.

are congruent with ,

the greatest integer in

hence

where

the period contains 2r figures.

m (mod n)

n- m =

Hence the subsequent residues and are equal to n m l9 n w2 Moreover,

and

,

//, after

be the least positive residues of m, 10m, 10 2 w,

m

then

...

,

m prime to n.

occurs, the subsequent figures of the

10(n-m)/n;

= 9 - %,

Similarly a r+2

a positive proper fraction.

= 9-a 2

,

etc.

Ex. 1. To express 1/73 as a decimal, the division need not be carried beyond the stage shown in the reckoning, for since the remainder g \ ^AQ / .9136 72(=73-l) occurs, the subsequent remainders are 73-10,

73-27, etc., the corresponding 9-0, 9-1, etc. Thus

figures in the decimal being

270 ~510

-^Q

1/73-0-01369863. Again, the number of figures in the period is 8 and (p (73) 72. Hence for values m less than 73, the fractions ??f/73 can be arranged in 72/8=9 groups such that the period is the same for those in the same group. By the preceding, the complete of

set of remainders for

1/73 1,

is

10,

27,

51,

72,

63,

46,

22, .......................... (A)

and one group is obtained by giving m these values. To get a second group, choose any number loss than 73 not included in the set (A), say 2. The remainders for 2/73 are congruent (mod 73) with 2x1, 2 x 10, 2 x 27, etc., and are therefore 2,

A

20,

54,

29,

second group is obtained by giving in the same way by choosing a number (B)

and

so on.

m

71,

53,

19,

less

44 .......................... (B)

A

third group can be found than 73 and not included in the sets (A),

these values.

RESIDUES

442

EXERCISE XLV The prime

NOTE.

the following table

/(4)-3 /(7)-3

2

11

.

2

/(8) = 3

4649,

.

= 10*-1

for

{

= 1,

2, 3,

9

...

are given in

3 /(6)-3 .7.

2 /(5)-3 .41 .271,

101,

.

239

.

factors of /(0

:

2 .

11

73

.

101

.

/(9)^3

137,

.

4 .

37

11 .

.

13.37,

333667.

1 2 etc. given the least positive residues of 2, 2 2 2 2 1 1 are etc. What 7 7 of and write down those 10 10 7, 10, , etc., (Art. 1), the primitive roots of 19?

1.

For the modulus

19,

,

,

,

2.

For the modulus

are the primitive roots of 17 3.

If

p

is

the least positive residues of 3, 3 1 3 2 , etc.

17, find

,

(ii)

p

an odd prime, g belongs to the index

pccurs as a residue

1

v^o

rt

...

_^

when #^13, g^2, (iii)

If g

-

and only

if,

t,

and

r l9 r 29 etc., are

,

the

even.

is

if, t

according as

t

even or odd.

is

Verify

3, 5.

a primitive root of p, show that theorem

is

r

prove that

+1 (modp)

or

1

What

?

least positive residues of
,

,

(ii)

becomes Wilson's

theorem. [(i) (ii)

If

is

t

even, g ...

/wr.,

rt_ l

f

!2

~

-

I

(mod p).

s

^g (modp)

where

9

= + I +2 + ... +(t -

8

1).]

RECURRING DECIMALS Here n

is

(modn

etc.

4.

O-a^

etc.,

1

...

...

1

at

aA

t

a **./ ...

l/n= O'(iia 2

,

= 0'a 1aotf 3

a2

at

...

and

in the division,

...

,

^/(lO*

7^-1

r is

then

r

t

+

-

where

1)

. . .

a 1 a 2 ...a t

is

the

.]

J

= 0-6i6

n

....

.

+ ---^ +

and

r ...

...

a^a^

8uchthat

j.

6 rf

2 ...

...

y,

prove that j.

the remainder immediately preceding the

and a t are the

least positive integers

x y 9

lte-ny = l

and where the

5, t is the number of figures in the are the least positive residues of 10, 10 1 10 2 ,

p

digits are a l9

= a& 2

---O-atGU

V) n

If

a prime not 2 or

n

Prove that

remainder

is

r 19 r 29

,

or p), and we write - or ~ *

[The decimal

6.

r

;

number of which the

5. If

p

to 10,

prime

period of l/n or l/p

a^z

...

a's are digits of the

the figures in the period

[l/ft^O-a^ to be carried

...

a^!

when a l

may

...

is

,

at x

r

a^a^

. . .

a^_ l9

numbers

indicated. be found as in Ex. 8.

rln

Q-a t a l

multiplied by

...

r.]

a^

...

;

Having found explain

r

why no

and a t9

all

figure has

RECURRING DECIMALS 7. If

p

is

a prime and 1/p

=

a^

. . .

and give an example to show that such number. 8. (i)

arar+l

. . .

relations

a 2r

443 ,

prove that

do not hold

if

p

is

a composite

= Express -^ as a decimal by using the equality 10 x 4 13 x 3 1. Put down 3 (the last figure of the period) and multiply by 4 as follows

:

076923 4

If the process

4x3 =12, 4x2 + 1= 9, 4x9 =36,

put 2 to the

left

put 9 to the

left

put 6 to the

left

4 x 6 + 3 =27, 4 x 7 + 2 = 30,

put 7 to the

left

is

put

of 3 and carry of 2.

1.

of 9 and carry of 6 and carry

3. 2.

to the left of 7.

continued, the figures recur and tV=0'076923.

in Ex. (ii) Or thus Hence the following

a ta^a z ... a^_i/r=a 1a 2 ... a t Put down 3 and divide by 4 as below

7, :

.

:

"076923.

which is the first figure of the period. and 2 over. Put 7 to the right of 0. 4 into 27 is 6 and 3 over. Put 6 to the right of 7. 4 into 36 is 9. Put 9 to the right of 6. 4 into 9 is 2 and 1 over. Put 2 to the right of 9. 4 into 12 is 3. Put 3 to the right of 2. 4 into 3 is 0, and the figures recur. 4 into

3

4 into 30

is 0,

is

7

9. Cases in which l/p can be quickly expressed as a decimal by the methods of Ex. 8 are p = 17, 19, 23, 29, 43, 59, 79, 89.

10. Justify the following

:

A

If

C=

~0

9408

_

then

tV = -0588235294

where jB=4J, (7=4J5,

etc.

. . .

[The division shows that 10* =588 x 17 + 4

11.

By

the

;

method of Ex. 10 show that j& =0-032258064516129.

[After 6 steps of division, the remainder 2 occurs.] 2F

B.C.A.

BINOMIAL CONGRUENCES

444

12. Show that for values of m<41 the fractions r/i/41 can be arranged in 8 groups such that all the fractions in any group have the same period. What for the group to which 2/41 belongs ? are the values of

m

13. Prove that gV^O-0 12345670, and show that the values of m/81 has this period are given by m 1 (mod 9). 14. If n is any number prime to 10, prove that a that every digit in the product nn' is 1. Find n' when n~ 41 and when n~123.

~ or ~- as a decimal, according as n

[Express

fit

is

m

for

which

f

number n can be found such

or

not prime to

is

3.]

\jTt

15. If n is a prime or a power of a prime, prove that the values of n for which the decimal equivalent of 1/n has a period of t figures are as follows :

1

3

4

5

3,9

27,37

101

41,271

6

9

|

n

|

7,

13

239, 4649

73, 137

333667

[The values of n for which 1/n has a period of t figures are included among the divisors of 10* - 1. See note at the head of this Exercise.]

Show that the only prime p for which the decimal equivalent of l/p has a period of 10 figures is 9091, and (ii) a period of 12 figures is 9901. Also express 1/9091 and 1/9901 as decimals, explaining why only 5 and 6 16.

(i)

steps of division respectively are necessary.

Primitive Roots (continued). Theprocessof finding a primitive one of trial. The orthodox methods of shortening the reckoning given in another volume but in the example below we give a method,

6. root are

is

;

founded on the assumption of Art.

3,

which

is

convenience in solving the congruences in Ex. least primitive root,

= 2, = 3, 0-5, Ex.

Find

1.

g,for any prime modulus, p,

j0-3,

5,

usually successful. For we here give the

XL VI,

less

than 100.

11, 13, 19, 29, 37, 53, 59, 61, 67,

p = 7, 17, 31, 43, 79, 89; = 6, y = 23, 47, 73, 79;

p=

I

;

= 7,

83;

p = ll.

the primitive roots of 67.

converted into a decimal fraction, the reckoning shows that 10 is a subordinate root of modulus 67, with a period of 33 figures. Art. 3 suggests that this If 1/67

is

period consists of every alternate figure of the period of some primitive root, g 2 10(mod 67) will lead to a primitive root. consequently, that g

;

and,

\Ve find that a value of g is il2 and, converting 12/67 into a decimal fraction, obtain, as remainders in the process, the residues of the odd powers of 12. When ;

we

these are interpolated between the remainders in the reckoning for 1/67, we find that the numbers less than 67 are included ; thus 12 is a primitive root, and its period is

all

1,

From

12, 10, 53, 33, 61, 62, 7, 17, 3, 36, 30, 25, 32, 49,

this, as in Art. 3, all

U,

21, etc.

the other primitive roots can be immediately written down, together with their periods thus we find that other primitive roots are 61, 7, 32, 51, 41, 13, 2, etc., these being the residues of 12 fc where k is prime to 66. ;

,

OF PRIMITIVE ROOTS

tFSE

The Congruence x n ^a(mod

7.

It is

p).

445 supposed that

p

an

is

odd prime.

Ifn

(1)

if

For,

is prime to p- I, there xn z=a, then x nA = aA

is

3, a single

value of

A has this value, xnX = x (x v which is the only solution. If

~l

k

.

Ex.

x* ~ 21* (mod 23) z:EE21

// n

(2)

)

A, less

prime to j>-l, there than p-l, such that

= x (mod p

)

is

and

;

is,

x s= ax (mod

thus,

p),

Solve the congruence, x 13 == 21 (mod 23).

1.

Here

and, since n

;

by Chap. XXV,

a single solution.

is

17

and 13

;

17 E=

.

17

~(~2) ~(-2)

not prime to

.

1

(256)

(mod

a

22)

= (-2)

;

hence we have 2

.

(3)

s5(mod

23).

there are several solutions, or there is

p-l,

no

solution.

common

Let d be the highest

n and p - 1

divisor of

and- let g be a

;

Then, since the period of g contains all the

primitive root of p.

numbers

r

than y, we can find r, such that g z=a(modp). s r x^g (mod p); then g ns = a^g (mod p), and therefore

less

Let

ns^r(mod p-l) ................................ (A) The congruence solutions, or

no

and therefore

(A),

also the congruence

solution, according as r

is,

or

is

a factor of

is

by d

not, divisible

n and p 1 (Chap. XXVI, 8). n # E=a(mod p) are given by x^gs (modp), where

d

xn ^a, has d ;

for

Also, the solutions of 5

has any value which

satisfies (A).

Ex.

A

Solve the congruence, # 15 =31(wod 37).

2.

primitive root of 37

2 15S ==2 9 ,

is

and 15$~9(mod

and we

2;

36)

;

z^2 3

2 15

,

,

2 27

9

2 =;31. Hence, 27 (mod 36), and

find that

hence 5=3, ,

15,

i.e.

8,

if

x=2 s

,

then

23, 6.

EXERCISE XLVI Solve the congruences in Exx. 1-6 1.

z 3 =l(mod 11

19).

= 5 (mod 31).

:

2.

x 5 =2(mo&

5.

^=31(mod41).

17).

3.

*7 ==8(mod

6.

a:

21

a;

7.

Find the general solution

8.

Find three solutions in integers of # 8 = 73t/4-3.

9.

Show, without calculation, that a primitive root of 237

in integers of

least of the congruences, <7== 10, g a

= 10,

4

13).

= 2 (mod 31).

4.

# 2 = 19z/-f-5.

s 10.

will satisfy

one at

CHAPTER XXVIII NUMERICAL SOLUTION OF EQUATIONS 1

The Problem

.

mate values of

under consideration

the real roots of

is to

witli

find the actual or approxinumerical coefficients.

any equation This question is quite distinct from that of finding an algebraical soluIn fact, no algebraical solution of the general equation of the fifth tion. or higher degree has been discovered. If any rational or multiple roots exist they can be found and removed from the equation (Ch. XVIII, 1, and VI, 14). It may be convenient not to

is

remove the multiple roots. Thus we are only concerned with as follows

We

(i)

irrational roots.

The usual procedure

:

find

an interval which contains

all

the roots.

Ways

of doing

have been given in Ch. VI, 12. A method due to Newton which involves more calculation, but which yields closer limits, is given in the this

next

article.

We

separate the roots, that is to say a contains single root or a multiple root. (ii)

we

Taking any interval which contains a

(iii)

find intervals each of

which

by a process of which contain the

single root,

approximation, we find smaller and smaller intervals root.

Newton's Method of finding an Upper Limit to the

2.

This depends on the following theorem a number such that f(x) and all its derivatives are positive when then h is an upper limit to the roots of f(x) =0.

Roots. // h

x = h,

For

:

is

if

Hence that

is

NOTE.

power

of

n

is

if

the degree of f(x), then

f(h), f'(h),

to say

/(x)>0

...

for

fM (h)

are all positive, f(x + A)>0, for which x^h, proves the theorem.

In applying this theorem, we suppose that the x inf(x) is positive, so that

coefficient of the highest

SEPARATION OF ROOTS

We then

^ such that

find the least integer

of x does not

make

Continuing thus,

n f^

~^(x)>0

we can

9

we

find

by

f^

n

xh^

"^(x)>0 when

trial

find the least integer

447 If this value

a greater integer h2 such that

h which makes f(x) and

all its

derivatives

positive.

It should be observed that if

are

all positive,

above theorem, Ex.

1.

Find

we have found a number h r such that This follows from the positive for x>h r n (r+l) are (z), .../ -^(), f^(x). / f^(x)

then these functions are for the derivatives of

all

.

the integral part of the greatest root of

f(x)=x*-10x*-Ilx -100 = 0. Here

The

2 f'(x) -3rr

-20*-

11,

least integral values of x for

(Cf.

Ex.

1, p.

91.)

/"(a) =2 (3s -10),

/'"(z)=6.

whieh

and

f"(x), /'(a*)

f(x)

are positive are

respectively 4, 8 and 12. Thus 12 is the least integer which exceeds the greatest root, the integral part of which is therefore 1 1 .

EXERCISE XLVII In Exx. 1-5 find an upper limit to the roots by Newton's method. 3 2 2. z 3 ~2z 2 -51z- 110=0. 1. z -20z - Six + 1609^0. 3.

* 4 -4x3 -3* + 23-0.

4.

5.

Separation of the Roots. Rules for this purpose were given and by Sturm. Fourier's rule, though often convenient, is Fourier by incomplete, while that of Sturm definitely separates the roots, but its application may be very laborious. We require the following theorems. 3.

Subsidiary Theorems.

(1) If

a

is

a root of /(x)=0, as the variable x

increases through the value a, the functions /(x) and/'(#) have opposite signs just before x=
For

since

/(a)=0, we have 7>2

/(a + A)-A/'(a) +

7,r

/"(a)

a)

+ ... +

/W(a) + ...,

+ ...+ 1

/0(a) +

........... (A)

................. (B)

Suppose that /^(a) is the first term of the sequence /'(a), /"(a), ... which is not zero. For sufficiently small values of h, the signs of /(a -h h) and /'(a + A) are respectively the signs of the first terms in (A) and (B) r r) they are therefore the same as the signs of A /* (a) and h f^((x). Hence /(a + h) and /'(a -f h) have the same or opposite 0, which proves the theorem. signs according as h

which do not vanish

;

r ^l

^

STURM'S FUNCTIONS

448 If

(2)

an

is

a.

r-multiple root of

then, as the variable

f(x)=Q,

x increases

through the value a, the signs of the functions /(*),

/"(*), .../

/'(*).

- 1)*,

(1

/'>(*)

(C)

x = a, and all

- and

the functions + and -, or -f, just before have the same sign just after X QL. For, when z = a, all the terms of the sequence (C) except the last vanish, and the result follows by applying theorem (1) to every two consecutive It follows that as x increases through the value a, r changes of terms.

are alternately

in the sequence (C). important to observe that since

sign are It is

lost

#=a,

at

is

continuous

just before and just after X

derivative.

Let f(x) be a polynomial and fi(x) its Let the operation of finding the H.C.F. of f(x) and fi(x)

be performed with this alteration

The sign of each remainder the sign of the last

remainder

:

is to be

changed before

is also to be

it is

used as a divisor

\

known

:

changed.

then Denote the modified remainders by /2 (z), /3 (#), ... fr ( x ) fi( x )> fz( x )> "-fr( x ) are called Sturm's functions and f^x), f2 (x), ... are

OL.

Sturm's Functions.

4. first

/^(aJ^O and f^(x)

same sign

this function has the

/(#),

fr (x)

as the auxiliary functions.

In the ordinary H.C.F. process we can multiply (or divide) any remainder by any constant. In the modified process it is essential that such multipliers should be positive. Sturm's functions are connected

by the equations

.(A)

where q v q2

are the quotients in the process just described, these quotients being functions of x. It is important to notice that the relation between f(x) and fi(x) is ,

...

essentially different

from that connecting f^x) and/2 (x)

;

fz(x),f3 (x)

;

etc.

5. Sturm's Theorem. Iff(x) is a polynomial and a, b are any real numbers (a<6), the number of distinct roots of f(x) = which lie between a and b (any multiple root which may exist being counted once only) is equal to the

excess of the

number of changes of sign in

the sequence of

Sturm's

functions /<*),

when

x=a

over the

A(*),

MX),...

fr (x),

number of changes of sign in

the sequence

(S)

when x~b.

STURM'S THEOREM

449

which f(x)^0 has no repeated Equations (A) lead to the following conclusions. the case in

Proof for

(i)

factor, (ii)

Since /(x)

=

has no repeated root,/(z)

(iii)

If

any term

and follow

it

if

this

happened,

Thus

if

have opposite

signs.

if

(x)

x

(S)

have no

common

can vanish for the

the subsequent terms including

is

remark applies

similar

all

of (S) except the first

/.-i (a)

A

and/

and consequently fr (x) is independent of x. No two consecutive terms of the sequence

same value of x, for, fr (x) would vanish.

root.

terms which precede /s (^)==0, then zero, the

=-/,+i (a).

one of the

y's is zero.

Thus

if

<7 S

= 0,

then

/s-i(*H -/mfc). Suppose now that x increases continuously from x a to x b. As x varies, no one of Sturm's functions can change its sign unless x passes through a value which makes that function vanish, for these functions are polynomials.

Suppose that x passes through a value a which makes just one

of Sturm's

functions vanish. (i)

If

a

a root of /(x)

is

= 0, one change

of sign is lost in the

sequence

For f(x) and f^x) have opposite signs just before x=a, and they have the same signs just after x=a.

(S).

(ii)

is

If

a

a root of fm (x) where

is

gained or

lost.

For

m = l,

/rn (a)=0, and

2,

...

or

therefore

r-1, no change

of sign

fm -i(oc)=-- -/w+iM-

= l, /

Moreover, fm ^(x) and fm+ i(x) are (a) stands for /(a). continuous at x=a, so that each of these is of invariable sign near #=a. If

ra

Hence, just before and also just after #=a, the signs of /m _i(^),/m (x), - or + or h-f or 1-, showing that fm+i( x ) are either -f -f

no change

of sign is lost or gained.

If x passes through a value a which makes more than one of Sturm's functions vanish, no two of them can be consecutive functions. If f(x) is

one of them, then by the preceding, one change of sign

is

lost

and/1 (a?), and no change is gained or lost in the sequence ... /1 (x),/2 (x), fr (x). Thus one change is lost in the sequence (S). If f(x) is not one of them, no change of sign is gained or lost. between f(x)

Thus, as ar increases, a change of sign in the sequence (S) is lost whenever x passes through a root of f(x)=0, and under no other circumstance is ;i change gained or lost. This proves the theorem.

ESSENTIAL PROPERTIES

450

Remarks.

In applying Sturm's theorem, labour

be saved by the

may

following considerations. If there is

(i)

of

x and

its

no repeated

sign only

is

root, the last function

the last three functions are

then

a'"

and from (ii) If,

any

stage,

fr -i(x) = a'x 2

9

then

+ b'x + c',

we

fr (x)=

a!'x

= - (a' -^ - V \

this the sign of a'"

at

is

independent

required.

Let x have a value such that if

fr (x)

+ 6",

Thus,

-fr-zfa)a'",

+

may often be found

without

much

calculation.

arrive at a function fs (x) such that all the roots

/s (.r)=0 are imaginary, the H.C.F. process need not be continued, and we can use /(x), /j(x), ... fs (x), instead of the complete set of functions. of

For the

essential property of the last function

of invariable sign for all real values of x,

then, in counting the

a\

number

that

and fs (x) has

Suppose that one of Sturm's functions as

(iii)

x

is

it

this property.

fm (x)

may regard the sign of/rn (a) as either + /TO (a) = 0, then/TO-1 (a) and /m+1 (a) have opposite signs.

Ex*

1.

Find

the

number and

when

vanishes

of changes of sign in the sequence

/(),/i (), .../r (^)> we if

should remain

position, relative to

numbers on

or

-

.

For

the scale of integers,

of the real roots of

Here fl (x)

4x* - 9# 2

-4z + 7, and

the modified H.C.F. process

is

as follows

:

43

-99 2237

/(*) =

Thus /2 (aO-43a;2 -72:r-69, fs (x) = S3x-191, /4 (*)= +48074. The sign off^x) is found more easily by Remark (i). If /3 (#) =0, then/4 (s) = -

When x=-^> /2 (x)

is

negative.

Thetefore /4 (x)

is

positive.

MODEL SOLUTIONS The number values of x are

of changes of sign in the sequence of Sturm's functions for various

shown below. x

-oo

-2

-1

-f

4-

2

/i(*)

Number ,

o*

,

.

.

} changes of signj

-

451

4

-

+

+

3

Hence the equation f(x)=0 has one root between 1 and and two between 2 and 3. Ex.

(i)

Find

2.

the

number and position of

Taking f(x) ~x* - x -

Also,

1,

For various values of

x,

-f

-f

-2 and

-1,

one between

the real roots of

we have f^x) 5**-l

when /2 ()=0, fs (x)^

oo

2200

.

4

-f

3

5x* -

and the modified H.C.F. process is

1,

-fi(x), therefore

the signs of Sturm's functions are

-oo

2

1

:

oo

/(*)

Number

/>(*) of

"

'

changes of sign/

showing that there

is

one real root between

1

and

and that the other roots are

2,

imaginary.

x.

The

roots of

/2 () =0

are imaginary, Jfor 15 2 <4

.

8

24,

.

and we need not proceed

further.

Thus there

is

one real root between

other roots being imaginary.

- 2 and -

1,

another between

1

and

2,

the

ALL ROOTS REAL

452

Sturm's Theorem (Multiple Roots).

6.

the case in

which the equation f(x) = /(*)

where p,

and

a,

are

...

y,

/?,

Suppose that

-

= (*-)"(* -W*-y) r

are positive integers

...

q, r,

It remains to consider

has repeated roots.

all different (real

or imaginary), then

u = (x-a)*- l (x-p)*- l (x-y) r - 1 ...

where

Hence u

is

the H.C.F. of f(x) and /^z), and /(*)>

are divisible

u.

by

all of

Sturm's functions,

.......................... (S)

/a(*),-/r(*),

fl(*)>

.

Denote the quotients by

so that

(x)

= (x-
...

,

Also for any value of x, the number of and r (x) changes of sign in the sequence (S) is the same as that in the sequence (S'). Dividing equations (A) of Art. 4 by u, we have the identities is

tfj

independent of x.

(x)

=q

l i/J l

(x)-ifj 2 (x),

tl*i(x)

= q.j/i2 (x)-fa(x),

etc.

no change of sign is gained or lost as x a value makes which one of the set ^(x), $%(x}, ...iff r (x) passes through

Whence, as

in Art. 5, it follows that

vanish.

^ (a) =p(oc - /3)(a - y)

Again,

...

-^(a),

have the same sign when x = oc, and since these functions are continuous, they have the same signs for values of x near therefore

x=oc.

^(x) and



L (jr)

Hence, by Art.

change of sign

is

5,

as x increases through

lost in the

It follows that the

sequence

number

any root a

of

(x)

a

(S').

of real roots of

(f>(x)^0

in the interval

number of changes of sign a
is

equal to the excess of the

Conditions for the Reality of all the Roots. Let/(x) be a polynomial n with its leading coefficient positive. In general, the number of Sturm's functions is n -f 1.

of degree

The necessary and sufficient conditions that the roots of f(x) = Q may be and different are : (i) the number of Sturm' s functions must be n -f 1.

all real

and

(ii)

the leading coefficients of all these functions

For, as x passes from - oo to sequence of Sturm's functions if

+ oo n changes ,

and only

if

must be

positive.

of sign will be lost in the

both of these conditions hold.

THE BIQUADRATIC 7.

where

Application of Sturm's Theorem.

K=a2 I-3H

2

(Ch. XII, 9; 10

(6)),

453 For the biquadratic,

Sturm's functions are /() and

2 fz (z)=-3Hz ~3Gz-K, f3 (z)=-(2HI-3aJ)z-GI,

In proving this

we

require the identity (Ch. XII, 10, (6)) 2

Sturm's process

+ 4#3 = a2 (#/-aJ) ............................ (A)

is

-Hz G

Again,

if

f3 (z)=-Lz-GI where L = 2HI-3aJ. /3 (z) = 0, then /4 (z)= -/2 (z), therefore = 3H(GI) 2 - 3G(GI)L + KL2 )

= (a2/ - 3H2 (3aJ - 2HI) 2 + 3G2I (3aJ - HI) = 9a4 /J2 - 12a 3HI*J + a2H2 (U 3 - 27J2 )

)

+ 3I(G2 + 4:H3 )(3aJ-HI)* = a2 H2 (4/ 3 - 27 J2 ) - 3a2#2/3 Disregarding square factors,

we may

therefore take

3 2 /4 (z)=/ -27/ Hence if A is positive and both H and 2HI - 3a J .

are real

If A

is positive

and

are negative, all the roots

at least one of the two,

H and

2HI - 3aJ,

is

are imaginary. - oo to -f oo , in the first case four changes of sign as x from For, passes are lost in the sequence of Sturm's functions, and in the second case no

positive, all the roots

change of sign If A

is lost.

two roots are real and two imaginary. All this has been proved more easily in Ch. XII. is negative,

*

The

identity (A)

is

used in these steps.

FOURIER'S THEOREM

454

Fourier's Theorem.

8.

Let f(x) be a polynomial of the n-th degree

and let fi(x), /2 (x), ... fn ( x ) be its successive derivatives. Let R be the number of real roots of f(x)~0 which lie between a and b, where a, b are any real numbers, of which a is less than 6, and an r-multiple root is counted r times. Let N, N respectively denote the number of changes of sign f

in the sequence

AW,

/(*),

when x

a

and when x = 6,

/.(*),../(*)>

........................

(?)

then

N^N

f

R^N-N'i

and

(N - N') - R is an even number or zero. In particular if N - N' = 1 there is just one = no real root lies between a and b. if N N'

also

real root in the interval,

,

In this connection /(x) and

its

and

derivatives are called Fourier's functions.

Let x increase continuously from x = a to x = b. As x varies, no change of sign can he gained or lost in the sequence (F) unless x passes through a value which makes one of its terms vanish. Proof.

(1) (i)

Suppose that x passes through a root a of /(x) = 0. If a is an unrepeated root, a change of sign is lost between /(x) and This has been proved in Art. 3.

fi(x). (ii)

If

a

an r-multiple

is

f(x),fi(*)..:fr(*)-

changes of sign are lost in the sequence

Suppose that x passes through a value

(2)

of the functions /j(x), (i)

root, r

(See Art. 3.)

If

/m (a') = 0,

/2 (x),

a'

which makes one or more

.../n (x) vanish.

but neither

fm -i(
nor

fm +i(a)

is

zero, there is

no

gain or loss of changes of sign when /m _i(a') and fm+ i(at') have opposite but if these have the same sign, two changes are lost. signs :

For fm (x) and fm+ i(x) have opposite signs just before x=a', and they have the same sign just after x=a'. (ii)

all

Suppose that when x=a' the

vanish, but neither

Then as is

if

fm -i(x)

/m + r (x)

is

zero,

r-

odd there is a loss of 1 or r + 1 changes of sign according and fm + r (&) have opposite signs or the same sign, while if r

r is

/m -i(a')

nor

r functions

even there

is

a loss of r changes of sign in either case.

Thus, under this heading, no change of sign are lost, their

number is

even.

is

ever gained, and

if

any

DISADVANTAGES Ex.

Apply Fourier's method

1.

to

455

separate the roots of

2 ~ f(x) =2^ + Ix* 4Qx* 23* + 38* 4 -0. 3 2 /x (x) = 2 (5^ -f 14* - 60* - 23z + 19), 2 120* +42* ft (x) =2(20** -23),

Here

/4 (a?) =8 (30* + 21),

/6 (*) = 240. By Newton's method we find that

all

the roots

lie

in the interval

(

7, 4).

For

various values of x the signs of Fourier's functions are as below.

These results show that there is one root in each of the intervals ( - 7, - 6), ( - 2, - 1 ), (3, 4), and either two roots or none at all in the interval (0, 1). Subdividing the last interval, we find that when x =0-5, f(x) is positive. Hence there

a root in each of the intervals

is

It should be noticed that

for it lias

(0, 0-5)

and

we need not find the

(0-5, 1).

signs

oifl (x), /2 (#), e te when re =0-5, and 1. all, lie between

been shown that either two roots, or none at

Remarks.

In applying Fourier's theorem, as stated above, the what sign is to be attached to one of Fourier's functions

(1)

question arises as to

which

may happen

number

roots

vanish when

x=a

or

b.

met by observing that, for sufficiently small values of A, between a and b (a
This difficulty

the

to

is

of roots

and b -h instead

of

where A->0.

6,

Disadvantage of Fourier's Method. Suppose that (as in the last example) all the roots of an equation are accounted for except two, which, (2)

if

real,

must

it

may

require a large

lie

in a certain interval (a,

number

this the possibilities are (i)

The

roots

separated by

it

may

further

6).

If these roots are nearly equal,

of trials to separate them.

we

may

to do

be very nearly equal, in which case they can be for equal roots

The roots may be imaginary, and no matter how many

made, we

fail

trials.

if we begin by testing (ii) The roots may be equal, and would be better to use Sturm's process. (iii)

If

:

not be able to discover whether this

is

or

is

trials

are

not the case.

NEWTON'S METHOD

456 (3) If

N

so that is the

the roots of f(x) are diminished

is the

by

the transformed equation

a,

number of changes of sign in

is

Similarly N'

this equation.

number of changes of sign in the equation obtained by diminishing the 7 f(x)=0 by b. If N and A arc defined in this way, we obtain '

roots of

Sudan's statement of Fourier's theorem.

from -

oo to 4- oo n changes of sign If it is known that a are lost in the sequence of Fourier's functions. = no of root certain interval contains f(x) 0, and that, as x increases through

Imaginary Roots.

the interval,

than n - s

s

As x

increases

changes are

real roots,

lost,

,

then the equation cannot have more

and must therefore have at

least s

imaginary roots.

EXERCISE XLVIII 1.

From

Fourier's theorem deduce (i) Descartes' rule of signs of finding an upper limit to the roots.

method

In Exx. 2-11, find intervals

(a, b)

where

a, b are

;

(ii)

Newton's

consecutive integers which

contain the real roots of the following equations.

In Exx. 2-4 use Fourier's method; and in Exx. 5-11 use Sturm's method. 2.

a^-3a; J -4a:-f 11=0.

3.

z3 -20z 2 - 3 Ix +1609 = 0.

4.

2z3 + llz 2 - 10z + l=0.

5.

s3 + 2z 2 -51z + 110=0.

6.

2z 3 -15z 2 -8a; + 166=0.

7. a;*

8.

3z 4 -10z 2 -6a;+16=0.

9.

10.

12.

z - 3z - 24z 5

4

3

+ 95s - 46* - 101 =0. 8

11.

~4* 3 - 3x-f-23=0.

x*~x*-4x* + 4z + l=0. x 5 + 3x4 + 2s3 - 3x a - 2x - 2 =0.

Use Sturm's theorem to show that the equation xz + 3Hx + G=Q has and distinct roots if and only if G* + 4# 8 <0.

three real 13. If

real

e<0-55,

4 8 2 prove that the equation # -f 4# -f 6x

and two imaginary

9.

roots.

[/ = e

-4x + e=0 has two

+ 7, J = - 4.]

Newton's Method of approximating to a Root.

Suppose

equation f(x)~Q has a single unrepeated root in a small interval (a, jS), so that f'(x)^Q within the interval. Let a 4- h be the that

the

by the second mean value theorem, /(a + h) =/(a) + hf (a) + &*/" (a + 0h) - 6,

actual value of the root, then

where

0<0<1.

Hence

FOURIER'S RULE It follows, that

if

k

the greatest value of f"(x) in the interval, and

is

=a -/()//'(),

,

the value of the root

is

457

ocj -f

e

where

Thus, if /? - a is a small number of the first order, the error in taking o^ as the value of the root is of the second order of smallness, unless k/f'(
approximation can be carried to any

By repeating the

process, the of accuracy. required degree

is

large.

NOTE.

In the preceding, f(x) may be any function of are continuous in the interval (a, jS). /"(a?) i

x such that

f(x), f'(x)

and

.

Find an approximate value of a, such that a A/a = 3. Put a = x*; then f(x)=x 2x -3=0; and f'(x)-x zx (2 + 2 log e x). Also /(ar)=-2 when z = l, and f(x)- +0-375 when 3 = 1-5. Starting with a - 1 5, the work proceeds as follows.

Ex.

I.

/(a) =0-375, /'(a) =9-487, /(a)//' (a) =0-04, /( ai ) =0-0193, /'( ai )

/(,)= 0-000255,

/'(a a )

hence, the value of a required

This method

is

often

is

and

ai

/

= l-46;

and

/(a 1 )// (a 1 )--0-0023,


= l-4577;

=8-262, f(*jlf '(**)= 0-000031, and a 3 = l-457669; (1-457669)*

employed

= 2-12480.

in cases

where the interval

very small, important to know under is a closer approximation to the root than a.

and then

11

= 8-324,

(a,

is

jS)

not

what circumstances a t

it is

We shall assume that neither f'(x) nor f"(x) is zero in value x the interval (a, /?), so that each of these functions reof any tains the same sign for all values of x with which we are concerned. Now Fourier's Rule.

for

where

0<0 <1, 1

so that

*=-/()//'( + Hence

Now

if

oq

t

is

W

-/(a)//'(a), it follows that t has the certainly a better approximation than

and a + A, and

since

This condition

a1

=a +

/,

this is the case

1 |

7 '

is

if

same sign as h. a if o^ lies between a

satisfied if /(a)

and/

(a)

1<| h

|,

that

have the same

increases with x or decreases as x increases according as

is if

sign.

.For

/"()0,

f(x) and h 2^0 according as /(a) and /'(a) have different signs or the same sign. Thus we have Fourier's rule, which is as follows // neither f (x) nor f"(x) is zero near the point a, and /(a) and f"(a) have the same siyn, then :

ax

is

a

better

approximation than

a.

VARIOUS CASES

458

The various cases which can arise are illustrated in Fig. 66, (i)-(iv). It is to be noted that the tangent to the graph of f(x) at the point where #=a cuts the x-axis at the point a x .

In the figures on the left, /(a) and /"(<*) have the same sign; in those on the right they have opposite signs and Fourier's rule does not apply.

+./(*)-

a

a a

a cn+h

FIQ. 66.

We

now show

shall

that a a

is

a better approximation than a, provided

that the interval is sufficiently small.

The true value

of the root

approximation than a JLA2

if

= a 4- h =! +

|J<|A|,

/f"(\

that

or

and therefore

,

/" (x) ^t

such that

1 1

If neither

is

a

in the interval (a,

\<.\f"(
f (x)

better

+ 0h)

\,

nor f"(x)

^3),

x

is

a better

is if

if |

If


<2

we can

find a positive

number

and we draw the following conclusion is zero

in the interval (a,

approximation than

a.

/J)

and

:

I

EVALUATION OF ERROR This

is

certainly the case

if .|

j3

-a |

<

'

^

459

'

,

.

Moreover, in cases where Fourier's rule does not apply,

at

a closer approximation than a

is

that

,

|

h \<\

t\,

and

if

is, if

2

Z|/(a)|<2{/'(a)}

............. (C)

In this case, by (A),

J5te.

2.

Let

Find

/(x)r,x -2a;-5,

Since /(2)<0
/"(2)

Xs = 2x + 5

the real root of

3

to

nine significant figures.

th-n

f'(x)=3x* -2, /"() -60:. / the root lies between 2 and 3. Also /(2)= -1, / (2)=:10,

= 12.

Taking a

2 and applying Newton's method,

ai

=2

-=^-=

2-1.

Here /(2) and /"(2) have opposite signs, and Fourier's rule does not apply. /i<0-l, ;and the least value of |/"(#)| in the interval (2,3) is 12. Hence

But

t

and

KI is a better approximation than aMoreover, /"(#) increases with a:, and therefore

/"(a

,.,. Next, taking a -2-1,*

,

.

we have

Hence

ai ---2-l

/(2-1) =0-061,

-Pl -^?

/'(2-1) -11-23,

//

/ (2-l)-12-6.

=2-094569

....

in the first

approximation, so that

J jL'2ij

Also the value of h

is

equal to that of h |

and consequently

|

|

<~

|

< 0-007,

(0-007

)2^|~|<

0-00003.

Moreover, e is negative, and so the root lies between 2-09457 and 2-09453. the error in taking 2-09455 as the root is numerically less than 0-00002. Again, taking a

find that

.

Also

|

c is negative,

*

2o

we

/(a)- -0-000016557... and /'(a) = 11-1 1607, ai =a -/(a)//' (a) =2-094551483 ...

giving

Moreover, 2-09455148,

2-09455,

Hence

\
2

n ^<10~ 11

and BO the value

.

24.

of the root to nine significant figures

This approximation might have been obtained graphically. B.C. A.

is

NEARLY EQUAL ROOTS

460

Ex. 3. A donkey is tethered to a point A in the circumference of a circular field of radius a by a rope of length L Find the value of I in order that he may be able to graze over exactly half of the field.

In Fig. 67

AB

Thus

2a'

is

LAOB=x, LOAB^x'

the stretched rope,

+ *= w

measured

in radians.

J

,

Area donkey can graze over segment A BB' + segment CBB'

= a 2 (x - \ sin 2x) + l*(x' - \ is

JaV,

to

will

Equating

this

given by

f(x)~x- tan x -f r.

Hence,

/' (x)

it

r

( I

be found that x sec

x-

n

f/

f (x)~

} ,

(

\ **

and

0.

1)

x - tan x

- sec

tan x

sin 2x').

sec x(\

j

+ 2 tan 2 x)-2 tan x sec

a:.

In a question of this kind it is important to start with a really good approximation, to be found graphically or otherwise. draw the figure so that the shaded areas are nearly equal, and then by measurement

We

Z_^O# = 71^l-239 Taking a 1-239, we find and the next approximation ^=

i

If

x increases from

positive.

radians (nearly).

that, approximately,

/(a) ^0-018,

^

7

-/(a)// (a)

= 1-239

--

0-003

-=

1-236.

therefore /"(a) is f"(x) increases from ?r/2 to oo have the same sign and, by Fourier's rule, a z is a closer

to ir/2,

Hence /(a) and /

"

(a.)

;

approximation than a. Further, we can show that /"(a)<34 if x
Thus the following

therefore / '(x)<34

a

radians

Hence

/

;

|
results are correct to the last figure

L AGE = 1-236

/'(a) ^-5-578,

is

- 70 50

X ,

and

/7a/6, very

Ifa ^2

:

sin 35

25'

= 1-16.

nearly.

10. Nearly equal Roots. Suppose that the equation /(z) = has two nearly equal roots within the small interval (a, j8), and let oc + h be the true value of one of them, then

where

a=/(a),

6=/'(a),

(?-/"(),

etc,

Neglecting squares of /?, we have a + 6A==0, giving Newton's rule. But the roots are separated by a root of /' (x) ~ 0, so that /' (x) is zero for some 2 Consequently /'(a) is small, and the term ch cannot be neglected in comparison with bh, so that Newton's rule does not 2 apply. Neglecting cubes of h, we have a + 6A + cA = 0.

value of x in the interval.

HORNER'S METHOD This equation has nearly equal roots, and

same as

(a

+ |6A)

2

= 0.

This result

is

therefore approximately the

is

Hence we have 2a

b

T = "2c

i

value of

461

.

f

"

fcPP 1

*-)'

used in Homer's method (Art. 11, Ex. 3) to suggest a

when

trial

2a/b and

b/2c are nearly equal. only applicable Jn the same way, if the equation has three nearly equal roots in the interval (a, j8), the value of one of them being a-fA, neglecting A 4 h.

It

is

,

we have a -f bh -f ch 2 + dh 3 Hence,

1 1

.

;

3a

7

*

which

is

b

nearly the same as c

--y---- -35

.

.

,

(nearly).

Homer's Method. The

Horner,

is

process described in this article, due to the most convenient way of finding approximate values of the

Any

rational roots

which

may

=

where f(x) is a polyexist can also be found in the same

irrational roots of equations of the type /(z)

nomial.

(a

+ |M) 3 = 0.

way. One great merit of the method is the very concise and orderly way in which the reckoning can be arranged. The root is evolved as a decimal, the figures of which are obtained in succession. In what follows, we are only concerned with positive roots. To find the negative roots of f(x) = 0, we find the positive roots of /( x) =0.

Consider the equation

and suppose that

f(x)

= a x n -f a^~

this has a single root

a

v

4-

.

.

.

a n ^x

-h

in the interval

-f

(p

an y

p

(A)

0,

f

1

),

where

a positive integer. (The case of two nearly equal roots will be conp sidered later.) Let (x^p-qrs ... where q, r, s ... are the figures in the decimal part of the root then p can be found as in Ch. XXVIII, 2, is

:

or preferably figure by figure as in Ex. 2 below. Decrease the roots of (A) by p, and let the transformed equation be n n -l + ...+b n _ l x + b n = a

(B)

This has a single root between and 1, namely 0-qrs .... In finding the value of q, we avoid the use of decimals by multiplying the roots of (B)

by

10, obtaining the

equation

= aQx n + lOV? (x)

71

-1

* ...+I0 n - l b x + 1

W

n

bn

=Q

(C)

This has a single root between and 10, namely q-rs .... Moreover, is the number that such the roots of (C) are decreased by q the greatest q if sign of the last term of the transformed equation is the same as that of th>e last y

term of q and and

(C).

q+

This follows from the fact that

l, and there is no root between have the same signs. (q + I)

<(#)

=

and

has a root between q,

so that

<(0),

(q)

HORNER'S METHOD FOR CUBE ROOT

462

In practice the value of q given by Newton's method. For, regarding 0-qrs

p ...

is suggested

and, after a certain stage,

as an approximate value of a,

-&/&_!

-f(p)!f'(p}

Having found the correct value of

is definitely

we have

(approximately).

we

decrease the roots of (0) by q, multiply the roots of the resulting equation by 10, so obtaining an equation of which r-s ... is a root. y,

Continuing as above, we can find as decimal representation of a.

When

a certain

number

many

we choose

figures as

have been found, about as

of figures

in the

many more

can be found by a contracted process explained in the next example. Ex.

I.

Find

to ten significant figures the real root

/(*)=

3

of

- 111=0

(A)

The work is arranged below in a form suitable for explanation. In practice be arranged as in Ex. 2. The explanation is given on the opposite page. 1

0+ 0-111(4 4

47.

48

it

would

(a)

+ 16- 47

8+48

_ uO .Qy

12 (6)

120+4800-47000(8 128 + 5824408 136 + 6912 144 1440

+691200

-

408000(0

14400 +69120000 -408000000(5 14405 +69192025 - 62039875

(c)

(d)

14410 +69264075 14415

11=0-8,

144,15

+ 6926407,5- 62039875(8 6927560

-

()

6619395

6928713 1,44

+ 692871,3 692884

6619395

9 (

(/)

383439

692897 69289,7 6928,9 692,8 69,2 6,9

383439 (5 36990 5

(j

(

2345 267

3 (

3 (

59

8 (

4

Hence the required root

is

4-8058955338, with possibly an error in the last figure.

EXPLANATION OF PROCESS NOTE.

Denoting the root by

Also that for the values f(x)

=

4-80
4-8
4-8,

4,

-47,

the successive steps show that

a,

4
463

-0-408,

4-8058955338 of

...

-0-062

...

...

,

etc, x,

-0-000000004

....

EXPLANATION OF THE SUCCESSIVE STEPS (a)

Decreasing the roots by

the resulting equation

4,

is

(B) If

(b)

by Newton's method

a4-fA,

Multiplying the roots of (B) by 10, the resulting equation (f)

^o;

(a?)

3

is

+ 120a; +4800* -47000-0 ............................ (C) 2

Taking 9 as a trial figure, we find that this is too large, for decreasing the roots of (C) this shows that (0) and <^ (9) have opposite by 9 changes the sign of the last term ;

dgns,

and thus a root of

Trying

8,

we

(f>

(x) lies

find that this

resulting equation

is

and

between

9.

the correct figure, and decreasing the roots by

the

8,

is

s3

+ 144s 2 + 6912&- 408 .=

Denoting the root of (C) by 8

-f k,

................................. (D)

we have

Hence we may expect that the next two

and

figures of the root are

5,

and

this

proves to be correct. (c) Here the roots of (D) are multiplied by 10, and those of the resulting equation iecreasod by if they were decreased by 1, the sign of the last term would be changed, is the next figure of the root. 30 :

(d) is

The

roots of the equation obtained in

(c)

are multiplied by 10,

and we

find that 5

the next figure. (e)

The equation found ar

and the next

3

in (d)

is

+ 1441 5.T 2 + 69264075z - 62039875 = 0,

figure of the root is 8.

We

...

..................... (E)

have to multiply the roots

decrease those of the resulting equation by

of (E)

by 10 and

8.

In doing this the important figures will retain the same relative position if, instead of introducing O's, we write down the last term of (E) as it stands, cut off one figure from, the 3 2 coefficient of x, two figures from, tJiat of x , and neglect the coefficient of a: .

In

an equation of higher x3 and so on.

the case of

coefficient of

NOTE.

In

degree,

we should

cut off three figures

from

the

,

the multiplication allowance should be

(f)

This step

(g)

Only the

is

similar to

last

mod J 1

r the figures cut off.

(e).

two terms remain, and the process

is

simply contracted division.

TWO NEARLY EQUAL ROOTS

464 Ex.

Find

2.

to

nine significant figures the real root of

It is easily seen that the root lies

the roots by 10.

x*

- x* - x - 2000 =0.

between 10 and 20, and the

In practice, the work

is

arranged thus

first

step

is

to decrease

:

|f =0-7.

The required NOTE.

which

3.

lie

is

12-9686724, to nine significant figures.

It will be seen that

figure until

Ex.

root

we come

Find approximate

is

more than the correct

1

values of the two roots of

3s4

- 61X3 + 127z2 + 220*

as follows,

and the explanation

between 2 and

The reckoning

Newton's approximation gives

to the 8.

-520-0

3.

3-61 -55 -49 -43 -37

is

+127

+220-520 (2

+

+ 254+ 92

17

- 81 -167

3-370 -16700 -364 -17428 -358 -18144 -352 -18848 -346

on the opposite page. (a)

12

+92000 +57144 +20856

- 120000 - 5712

2 (

3-3460-1884800 + 20856000-57120000 (5

+ 1 1345875 -3430-1919175 + 1750000 - 3445 - 1902025

,.(6)

(c)

390625

-3415-1936250 -3400 -3,4

-19362,5 + 175000 -390625(4 -19376 + 97496 641

- 19390 -19404

+

19930

.(d)

SEPARATION OF ROOTS For the smaller root

For the greater root

:

-194,04 + 1993,6-641 (0

-641

199,3 193

- 62

18,7

- 62

+ + 187

465

(e)

(/)

(3

-149,9 +1582 (9 - 167 79 f

-1,9'

-

1,8

(3

(g)

6(3

...... (h)

:

-194,0 + 1993,6- 641(9 + 247 +1582 -1499

-18,4

79

-1,8

Thus the case

is

and 2-2549942..., where the

roots are 2-2540333...

(4

(')

....(/')

....(

5(2 last figure in

each

not to be relied upon.

EXPLANATION OF THE SUCCESSIVE STEPS (a)

Decreasing the roots by

2,

the resulting equation

3a^ - 37z - 167z 3

is

2

+ 92z -

By

the question, this has two roots between

(6)

Suspecting two nearly equal roots,

12 =0.

and

we apply

is

1.

the rule of Art. 10, and the figure 2

suggested by ...

Decreasing the roots by 2, the sign of the last term of the resulting equation is If the roots are decreased by 3, it will be found that again the sign of the last term .

-

So far, then, we have no assurance that 2 is the proper figure. But when the roots arc decreased by 2, the signs of the terms are + - - + so that two changes and when they are decreased by 3, the signs are of sign are lost. Hence, by Art. 8, the two roots lie between 2-2 and 2-3. is

.

,

+----,

(c)

and

(d).

Similar remarks apply to these steps.

~

2(-5712)

Thus the (e)

and

matter of

roots f

(e

).

trial.

of the terms are

lie

__

20856--'

figure 5 in

20856

.

and

The

(c) is

suggested by

nr5 -'

2rT8848T

between 2-254 and 2-255.

Here the

rule of Art. 10 fails,

If at (e)

+--

+

and the separation of the roots

is

a

we

decrease the roots by 1 (instead of by 0), the signs +, one change of sign being lost. The same is true if,

we

decrease the roots by 9. Thus one root lies between 2-2540 and 2-2541, (e'), and the other between 2-2549 and 2-255 ; and the roots are separated.

as at

Immediately after the roots have been separated, is suggested by Newton's rule. (/'), the proper figure

i.e.

beginning at the stages

(/),

APPROXIMATION TO ROOTS

466

EXERCISE XLIX Find by Newton's method to five significant figures 1.

The two

2.

The

positive roots of x*

3 positive root of 2x

:

- Ix -f 7 ^0.

- 3x - 6 =0.

Find the roots indicated below correct to seven places of decimals by Horner's method. 3

+ 29* - 97 = 0.

3.

The

4.

The cube root of 4129,

5.

The three

6.

The two roots of 2* + ll# - 10z-f

real root of

a;

real roots of x*

- 3z 4-

3

1 =- 0.

2

1=0

7.

The two roots of # -2(Xr - 3 1#+ 1609 =

8.

The two

9.

The

3

roots of 2x* -

greatest and

in the interval

2

Wx

z

(0,1).

in the interval

- Sx + 166 =0

in the interval

(13,15).

(5, 6).

least roots of

3s 4 - 6Lr 3 + 127* 2 -f 220* - 520 = 0.

The two

roots of 3z 4 - 10z 2 ~ 6x

+ 16 =

in the interval (1,2).

11.

The two

roots of 3x4 - 2x* - 34x + 40 =

in the interval (1,2).

12.

The two

roots of Ix* - llx 2 -32x-f -40 =

13.

The two roots of x* - 2x 3 - 13s 2

14.

The two

roots of

15.

The two

roots of 2z 4

16.

The two roots of

17.

The two

10.

3a;

x*

4

-4^ 3 ~

+ 4z

18.

The

positive root of 2z

19.

The

greatest root of the following (i) (ii)

(in)

-f 5a;

2 -f-

(4,5).

in the interval

f2, 3).

3x = 8002.

:

x + 4# - 2o;3 4- 10* 2 - 2x rr 962. x 5 + toe* - 10r3 - 1 12x 2 - 207* x5

in the interval

in the interval (2,3).

-27z 3 + 25z 2 -j-179:r-275---0 3

(3, 4).

12^ + 9^0 in the interval (1,2).

2

12o; -f

-16z 3 -17a a + 392- 782 =

4

(1,2).

llx + 133 --0 in the interval

+ 5$x*-388x + 511=0

roots of 2* 4

5

in the interval

4

4-

12z 4

4-

1 10,

59z 3 + 150* 2 -f 201* = 207.

x 20. If f(x) = e - 1 - 2#, show that the equation than other zero, the solution being a? =1-260 .... 1-260. Show also that f(x)<0 if

f(x)-Q has

just one solution

0
21. If

f(x)~e

x

-e~ A

positive solution given

-4x, show that the equation ,/(x)=0 has just one by x = 2- 177 ... and that/(z)<0 if 0
Find an approximate solution of (nearly), where the last figure is not to be 22.

of x, x*

a:*

=10,

relied on.

showing that

x -2-50616

Verify that for this value

9-9995 (approx.). l is given by f(x)~x~ ~log, ;r--0. By drawing the curves y~xr l and Five applit/=log 10#, it will be found that x 2-5 is a good approximation. cations of Newton's process give x 2-50616.1 -=

[Here x

CHAPTER XXIX IMPLICIT FUNCTIONS, CURVE TRACING 1.

Implicit Functions.

If

an equation represented by /(#,

y)

two variables x and y are connected by =0, we say that y is an implicit/unction

of x.

In particular, is

called

is

f(x, y}

an algebraic function

Taking

this gives

But

if

a polynomial in x, y

and

/(x,

y)=0, then y

of x.

as a simple case the equation

y=x

-

j(1i

t

a;

2 ),

and so defines

?/

as a two-valued function of x.

we

are unable to express y explicitly in if, terms of x, we are not justified in assuming without further enquiry that y is really a function of x in the sense hitherto understood. as generally happens,

The general question involves

difficult analysis,

and cannot be con-

Some exercise in tracing curves from their equations will, convince the student that, at any rate in a large number of however, does define y as a single- or multiple-valued cases, the equation f(x, y) =

sidered here.

function of

x.

The general form

of a curve represented by an algebraic equation can be found by pure algebra, and affords excellent practice in finding generally

approximate values. 2.

A

Rule for Approximations.

The following process

is

of

great practical use.

Suppose that

J/

= ^ + ^/(?/)

*~

x2 (y) +x?i/j(y)

+

...

,

small and/(y), <(?/), etc., do not involve approximate to the value of y, in ascending powers of x.

where x

Let

is

f(y + h)

=

f(y)

+

A/, (y)

.................... (A)

x.

It is required to

+

................................................. J

The first approximation

is

obtained by putting x=0, giving

y-

....................................... (0)

INVERSION OF SERIES

468

For the second approximation, y = a + xf(a+
2

y = a-fa where a

let

+ ...

<(a-fa)

= a-f x {/(a) -fa/! (a) +

+ x2 {(a) +

...}

0(x),* then

is

and neglecting small quantities of the second 2 etc., the second approximation is XOL, x

order,

+

...}

...

,

terms containing

i.e.

,

y = a + xf(a), so that

Next

oc

xf(a).

let

y = a-ha-fj8 where

j8

y = a + xf(a -fa -f j3) +

and neglecting small quantities

is 2

0(x

2 ),

...................................

then

<(a -ha + ]8)


Similarly,

we

if

y

is

0(x

3 ),

. . .



(a) ,

...........................

(E )

namely

2 jS-x {/(a) ./x (a) +

Thus

4-

of the third order,

2 y==a + xf(a + a) + x

which gives

(D)

^(a)}.

neglecting small quantities of the fourth order,

shall find that ,

which gives

the fourth

.................

(F)

approximation.

Hence the following Rule

Having found

:

r successive

approximations

to the value of y, to find the equation giving the next approximation, in equation (A) substitute the result of

the rth

approximation in xf(y),

the (r-l)th approximation in x 2 (y), the (r-2)th approximation in x 3 ifj(y)

the

first

r approximation in the term containing x

and neglect terms containing powers Ex.

1.

Newton used the expansion of sin~ l

x=x-

sin

?/,

of

x higher than x r

2 (1 ~a; )~i

a

1

8

-

g

Putting x

y

1.3

-f

3

^

,

.

to prove that

z -

5

.

+

(A)

he deduced that

1^,U'* This

means

'

of the

same order

as


a;.'

(Ch. II, 7.)

ROOTS OF A CUBIC To do

469

write equation (A) in the form

this,

7

If

y

is

small, so

is

sin y,

and the

first

2/- ......................... (C)

is

approximation

Substituting y for sin y in the second term of (C) and neglecting higher terms, the second approximation is

y-%y* ........................................... (E) Following the rule, the third approximation

is

given by

siny^y-^^ and neglecting the powers of y higher than (F)

The fourth approximation

is

given by

Expanding and neglecting powers

than y7 ,

of y higher

this gives

and so on. Ex.

2.

If

is

fj,

If fj,

is

small, so

small, find approximate values of

is

y(y

- l)(y + 1), therefore y

y which

satisfy the equation

nearly equal to

is

or

1.

Write the

equation (A)

(i)

If y is small, the first approximation is 2/ = /x. For the second, substitute p? for y3 in (A), giving

For the

third,

y = /x -f (p, + /r ) 1

3 ,

^+

For the fourth, y =fi + (p. 4-

giving 6 3

3/z

) ,

y=

//,

4-

2/=/

p? + 3/A

y -p +

giving

^ + 3^

5

+ 12/z 7

.

Further approximations can be found in the same way.

(ii)

If

The

y-

1

first

is

approximation

The second For the

small, put

then equation (A)

is

Y = - ^/LI. (

F=

-J/i-|( -i/x-f/x 1

(iii)

is,

Similarly

a 2 )

that

if

y+

1

is

zero.

3 ,

.

small, then approximately

This result also follows from the preceding, for the is

.

-^

2/

we can show

be written

-i(-^)

y^-iM-lf* = 1 - i^ - f M 2 - ift8

giving

that

may

y = - J/i -f - Jft)= - J/t -f /x 2

is

third,

Y ~y - 1,

sum

of the roots of equation (A)

ROOTS OF A CUBIC

470

Cubic Equation with Three Real Roots. If all the ax -h 36x2 -f 3cx + d Q are real, the equation can be reduced to 3.

roots of

3

y -t/+/x = 3

by a transformation (Ch. XII, 5,

Ex.

1).

last

/x

x=p + qy,

form

of the

The

2

where

2

//

/^

+

'"

the roots of

<4/27,

....................... (A)

where p, q are

real

numbers

example suggests the possibility of expressing

the roots of equation (A) as power series in

Theorem.

<4/27

y

//,.

-y+^ =

3

are

where

+

V2

\n

y*-y = 2x,

Write the equation in the form

where

x=

-/z/2

..............................

of y can be expressed as power series in values these are given by theorem, Taylor's

Assuming that the values

/

and ^)o

(^i)o

x,

by

xn

x2

where y n = 'r: ax

(A)

and x2
are the values of y, y l9

(^2)0*

j/ 2 ,

...

n

when cc = 0. (Of course, x is to be made zero after differentiation.) The assumption will be justified if the series is convergent. Let y = u + v, then y 3 - 3uvy = u* + v 3 with (A) if w3 + v 3 = 2x and

w = x + (x2 - fc)i, 8

Thus we may take un =

Writing

dn u -=-^

vn

,

=

dnv -7-^

and

,

This equation will be identical

.

3uv v

3

1.

= - (a; 2 - &)*, where a?

& = 1/27.

differentiating with regard to x,

we have therefore

2

(x

= u/ 3

- fc)

Differentiating again, 2

giving

(x

;

jfc)

Since

y *

An

u + v,

- &)% = - v/3.

- ^)^w + 2 xu^x* k)~% = Wj/3, = w2 4- x^i = (x2 - &)i 2

(x

.

(x it

(x

similarly

we have

2

Similarly,

2

and

i) v 2

follows that

alternative proof of this result

2

(x is

^/S

w/9.

+ xt;j = v/9.

-k)y2 + xyl ~y/

.................... (C)*

indicated in Ex. 6 of the next Exercise.

EXPRESSION AS POWER SERIES Differentiating

n times by Leibniz' theorem (Oh. XXIII,

471 11),

(^-k)y n+2 + (2n + l)xy n ^ + (n^^)yn ^O ................. (D)

#=0

Putting

The

possible values of (y)

(i)J/ (yJo

-

885

Hence from

and

= 1/27, we have for n = 0, 1, 2, ... = 3(3w + l)(3M -l)(yn ) ........................ (E) (yn +2 )o

and

1*

are 0,

yi* 2

since

/^

where

3

x

4.6

a

(n

, 3

such values of x as make the

Denoting the series

Hence the convergent //

u 2n

series

%

series convergent.

u z + u3 +

=l, then

(y)o

Since the

O

is

Putting

A

we have

,

consequently also

un

(y 1 )

=l-

i

sum

Hence the numerical value of (y n Q is + or - according as n is odd or even. )

-F(-x).

of the roots of (A) is zero, the root corresponding to

y = F(-x)-F(x). -/x/2, we have the results stated above.

given by

x=

more general theorem, which includes

given in Ex. L,

this, is

7.

EXERCISE L Having given that x = tan x - J tan3 x + -| tan 6 a: - ^ tan 7 as + tan x - x + fx3 + ^-a; 5 4- ^^x 7 + prove that x + ayn where x, y are both small, prove that approximately 2. If y ~ y=x + ax n + na 2x m l + f n (3n - 1 ) asa: 3n 2 1.

. . .

. . .

,

.

.

3.

If

are

,

x
y

ss

. . .

Eu 2n+v and

,

-

-

+ l)(n + 3)...(3n-3) ^

2

if

(y)

by

4-

the same as before, the sign being Therefore =l (iii)

we have (y^o^l.

1)

(E) in succession,

y=-l+F(x) ,

(ii)

-

n

if

Thus

for

1. 2

are both small, then .approximately y=x + ay* + by* and = x + a* 2 + (6 + 2a )x3 + 5a (b + a 2 z4 a;, ?/

2

?/

)

.

APPROXIMATIONS AS POWER SERIES

472

y=l+xav

4. If

and x

small, then approximately

is

z 2 3 2 y = 1 -f xa -f x a log a + fa^a (log a)

5.

If

are positive integers (>s>

8, t

and A

t)

is

.

small, the equation

y*-y* + (-OA = has a root nearly equal to

l-X-(-l+8 + t)-(-l+2s + t)(-l+s + 2t). ii

y*-y=2x, show

If

6.

2

(ii)y,=

where y1 ~-~, 7.

yt =

y=u + v

If

li

that

-l) = (32/ ~l) 2

24*,

-

3

12 (t/-

Deduce that

~.

v m ~x-*/(x 2 -k),

um =x + ^(x 2 -k),

where

m

being

a

positive integer, then (i)

x and y are connected by the equation '

~m

~

'

2

li (ii)

vhere

If

a;

(t/)

is

sufficiently small,

is

a value of y obtained by putting dii d 2y

are the corresponding values of (iii)

Using Ch. XXIII,

11,

,

u>x

Ex.

1,

-^~ dx

m is odd,

one value of

. . .

in (A),

and (y^,

(t/ 2 )

,

...

.

show that

Hence prove that equation (B) holds (iv) If

,

#=0

if

(y) Q is 0,

x 2
and then

L* + ?3 VP Mn-^16'F "^^ L- 'H

K

|

I

(v) If

.

m=5

and

A;

6

=^, equation (A) becomes 6 y -

and the values of (y) Q are

Thus

all

I

0,

A/f g

the roots are real, and they can be expressed as power series in x

! &' (

vi) If

m=7

and

Ff = |,

the equation

is

y

1

- y 5 -f fi/ 3 - ^i/ = 2x.

if

DOUBLE POINTS Tangents and Asymptotes.

4.

Q

a point on

of the secant of

Q If

is

P

P

a point on the curve and defined as the limiting position is

near P, the tangent at P is PQ as Q moves up to P. Thus as it

.

from the tangent at

curve

If

P

473

Q moves to

P, the distance P the

becomes indefinitely small, so that near

indefinitely close to the tangent.

moves to an

the tangent at

infinite distance

along a branch of the curve, and

P tends to a limiting position,

this limiting position is called

an asymptote to the curve. It may happen that an infinite branch has no asymptote. We then find the simplest equation which nearly

and say that this is the equation to a represents the branch at oo curvilinear asymptote. In the examples which follow, tangents and asymptotes are found by a ,

process of approximation.

Intersections of a Straight Line and Curve. equation to a curve of the nth degree and y =/*# -f v that 5.

line,

-+

the abscissae of the points of intersection are given ...-fa n

= 0,

w=0

If

is

the

of

any straight by an equation of

.....................

(A)

obtained by substituting px + v for y in u = 0.

*

(i) If this equation has two equal roots, the line meets the curve in two coincident points and, except in special cases (see (iii)), touches the '

three roots are equal, the line meets the coincident points. Except in special cases (see (iii)), the

curve.

If

curve in three

point of contact is a point of inflexion (Ch. XVII, 20). At such a point, the curve crosses the tangent.

If

a

= 0,

a

If

(ii)

and the

line

=

one root of equation (A) meets the curve in one point at

and

^ = 0,

is infinite, '

F

infinity.

68

the line meets the curve in

two points at infinity and, except in special cases (see

where a curve crosses

(iv)), is

an asymptote,

a double point, or node. At such a point there are two tangents, each meeting the curve in three consecutive points. Any other straight line through (iii)

Any

point- (D)

D meets the curve in two consecutive points. If

(Fig. 69.)

the double point is at infinity, the

parallel straight lines.

two

itself is called

tangents are So in this case the curve has

parallel asymptotes, each

meeting the curve in

three points at infinity. Any straight line parallel to the* asymptotes meets the curve in two points at infinity.

FIG. 69.

CURVE TRACING

474

tangents at a double point coincide, the point There are two kinds of cusp, as shown in Fig. 70. If the

called $ cusp.

is

FIG. 71.

In tracing a curve, it is important to be able to account for all the n points in which any straight line meets the curve, remembering that imaginary points occur in pairs. (iv) Fig. 71, (i), is

a sketch of the curve whose equation

is

= (x-a)*(x-b). y* The

axis of x meets the curve in

A (a, 0)

and B(b,

2

(x

a)

The equation

0).

to

any

AP through A is y = m(x - a) and, on eliminating y we have ~m = hence, AP meets the curve in two coincident points (x -b

straight line

t

;

2

)

;

at a point P, given by x = 6 -f m 2 Therefore, A is the only point on the curve to the left of the line y = b and thus may be regarded as a double point- at which the tangents are imaginary.

at

A and

.

,

A

;

A point on a curve, in its

such as A,

is

immediate neighbourhood

is

The conjugate point at A, in Fig. form, when a a', of the oval in curve whose equation is

Curve Tracing.

6.

To

called a conjugate point, if

no other point

on the curve. 71,

(i),

may be considered as the limiting

Fig. 71,

(ii),

which

is

a sketch of the

find the general shape of a curve represented

by a given equation, proceed as follows, (i) Consider any symmetry which the curve may possess, e.g. symmetry with regard to either axis, or in (ii) Mark any points on the curve which can be opposite quadrants, found easily, e.g. the points (if any) where it cuts the axes, (iii) Find the approximate form of the curve near some of these points. Near (0, 0) (if on the curve) x, y are small, but not necessarily of the same order. To find the form near any other point, move the origin to that point. The quickest way of finding the tangent at any point is to find the value of T-

But, without further different!

of the tangent the curve

may

possess.

lies,

or discover

So, for a/etu points,

mation, as in Ex.

1.

ition,

it is

any

we cannot say on which peculiarities

side

which the point

better to use a process of approxi-

APPROXIMATIONS TO FORM (iv\

Find any regions

Find the form

(v)

which there

in

is

no part

of the curve near infinity,

475

of the curve. i.e.

(Exx.

1, 2.)

where one or both of

x and y are large, but not necessarily of the same order. In doing this, we find the (linear) asymptotes (if there are any), and the approximate form of any parabolic branches. '

'

The reader should be able to draw the 'parabolas' represented by m and n are positive integers. For example,

m n where y =x ,

y x

x

f

O

O

y-

x

y FIG. 72.

By

(vi)

are parallel to the axes.

(vii) It

or again,

We

~

considering the value of

may be

,

dx

find the tangents

(if

any

exist)

which

70 JS)

d y dx 2

Notice that, at a point of inflexion,

''

possible to find y explicitly in terms of x or x in terms of y able to express x and y as functions of some variable /. ;

we may be

can then find any number of points on the curve and the gradients at '

these points.*

Examples on Curve Tracing.

7. Ex.

1.

Trace the curve

The equation

is

(

The points

is

= (x2 - 4) x

~x,

~y

y

1),

(A)

Hence there

for x, y.

(2,

0),

(2,

is

are

symmetry

in

on the curve. (A)

y

y~4=x.

the equation to the tangent at

2nd approx.,

1)

- x3 - y + x

x and y are small and of the same order.

1st approx.,

This

1 )y

(0, 0), (0, 3

Write the equation (0, 0),

-

unaltered by writing

opposite quadrants.

Near

if

~ 3 y 4x + 63^

x

(0, 0).

,

obtained by putting (4#) 3 for y3 in (A), showing that the curve is above the tangent in the first quadrant and below it

in the third quadrant.

Thus

(0, 0} is

a point of inflexion.

FIG. 73.

* The examples given in Art. 7 and in Exercise LI have been chosen so that one of these methods can be used. Thus in each case the curve can be plotted to any degree of accuracy required so that the student can verify the conclusions arrived at by the methods just described, and gain confidence in using them. However, when only the characteristic form of the curve is required, the advantages of the above method over the method of plotting, even when the latter can be done, will be obvious. [See Ex. 22 of Exercise LI.] '

'

;

2

H

B.C. A.

APPROXIMATE FORM AT INFINITY

476 Near

(0, 1

2 J = - 4x

1st approx.,

which

is

Y = y - 1,

putting

),

y=I

or

27

equation (A) becomes

- 2x,

the equation to the tangent at

+ 3 Y 2 + 7 s = (x* - 4) x.

y

(0, 1).

2 7 3 - 2#) 2 - - 4z, 2nd approx., giving - 2# - 6# 2 which shows that the curve is below the tangent. -f-

y=1

(

,

\

Near (2,0), putting X=x-2 we find that = - 8z - Gz 2 (approx.), showing that the t/ -S(x -2) and that the curve equation to the tangent is y

FIG. 74.

9

Again,

-

dyjdx

(3x

2

gradient of the tangent

and the tangent

is

2 - 4)/ (3?/

1

parallel to Oy,

3 2/

From

is

it.

1 ), the Therefore, at each of the points ( 2, is parallel to Ox where x=2/\/3=l-16;

).

where

y~

large

-a^~0,

dil/s/3-

and

of the

giving

0-58.

same order

;

and approximately

y~-x~Q.

y - x=

(A),

This shows that, as :r~>

which

below

The tangent

is 4.

Near (00,00), x and y are both

2nd approx.

is

oo

,

the curve becomes indefinitely close to the line y - x

0,

therefore an asymptote.

Again, on the curve that the curve

is

Thus the curve

y~x--

(approx.),

and on the asymptote y = x; showing

below the asymptote on the right and above is

as given in

-Fig.

it

on the

left.

75.

y

Putting y ~tx in equation (A), it will bo seen that any number of points on the curve can be found by giving any values to t in the equations y = tx, x 2 = (4 - *)/(! - 13 ). If

l
in the space

the corresponding point is imaginary, thus no part of the curve lies from the position y = x to the swept out by a line turning round

position y = 4x.

ASYMPTOTES PARALLEL TO AXES Ex.

z 2 (x-l)(x~2)y = 2x

Trace the curve

2.

477

-y

(A)

henco the curve for y symmetrical The equation is unaltered by writing hence no part of the curve lies with regard to Ox. Also y is imaginary if 1
;

;

* 2 )=0

N ear

the curve

(0, 0),

by

y

z

-x 2 =

is

(A)

approximately given

y~x.

or

Taking y =x as a first approximation, from (A) 2 (y - x) = (3*2/ 2 - re j(y 4- x)

V

;

)

giving as a second approximation

showing that y -x Similarly y + x~Q

Near

(1, GO

),

is

x-

1

3z 3

3

2

2x

4

FIG. 76.

.

is a tangent at (0, 0) and that the curve a tangent at (0, 0) which is a double point.

2^2 1

;

-..

^ear

(2, oo

),

9^2 "

r-2{i

-

.-,

(#-")?/

1} y3

~r~

-2)#

(1

O

-

^

92

~-

showing that x 1 =0 and x 2 =~0 are asymptotes, the curve approaching the first from the left and the second from the right. x 2 (y z -

Again, writing equation (A) in the form

near

We

3

'"

(oo

>s/2),

y

above the tangent.

I2

2 "

~

is

and near

(oo,

2)

FIG. 77.

=

-

I

-^2),

y

- 2y 2

,

it

+ N/2 ^ -

will be seen

that

3

can now sketch the curve, as in Fig. 78.

parallel to Ox can be found by equating to zero the coefficient of the power of y in (A) similarly for those parallel to Oy. Also, for all values of k 1, the line y - k meets the curve in two finite points and in two points at oo except The same is true for the line a; /, where /2.

NOTE. The asymptotes

highest

:

.

FOLIUM OF DESCARTES

478

8. In the next example we consider the possibility of finding the approximate form of a curve near the origin, or near infinity, by retaining only certain terms in its equation.

Ex.1.

Trace the curve

+ y*-3axy-Q

x*

(a>0) .............................. (A)

3 term, we have y -3axy=0. the we the equation factor consider y, Disregarding

Omitting the

If

y

is

first

small, x

is

small and 0(y 2 )

so that r*

;

t/

- Sax =

),

and can

2 6

is

0(?/

................... (B)

therefric be neglected in comparison with the other terms in (A). Hence (B) approximately represents the curve or part of the

curve near

(0, 0).

by omitting the second term, we find that part x 2 - Say = 0. (0, 0) is represented by 3 3 0. Omitting the third term, we get a; -)-?/

Similarly,

of

the curve near

O

FIG. 79. This represents the curve near (oo oo ) ; for if y is O(x), then 2 Saxy is O(x ), and can be neglected in comparison with the other terms of (A). Thus near ( oo, oo), as a first approximation ?/-ha: = 0. ,

For a second approximation,

A

third approximation

is

-f

x

3a

(the last step being obtained

-a

a3 4-

approaches

;

it

,

-

z

-

-

-

=

- x - a for

y,

*

f

~x(x-\-a)

by long

division).

showing that y -f x = - a

is

/V.2

3o -^-^

;

obtained by substituting

xy

~ y+x

_

SAJ

y

Hence the

- a. thus "2

third approximation

is

an asymptote, and that the curve

from above at both ends.

y

o

-a \

\

\ \

-a

FIG. 80.

The curve

is symmetrical with regard to the line y x, for its equation is unaltered x and?/ are interchanged. Further, if y tx, then s = 3a//(l -f J 8 ), so that any number of points can be found by giving different values to t e.g. < = 1, = 3a/2.

if

;

xy

ORDER OF MAGNITUDE OF TERMS Newton's Parallelogram.

9. t/ =

A

where u

is

a polynomial in

479

Consider the curve represented by

x, y.

Let v be the expression obtained by retaining only certain terms of u. rule which goes by the name of Newton's Parallelogram enables us to *

'

choose these terms so that v =

may

near the origin or near infinity,

when such

represent the curve,

or,

part of

it,

parts exist.

axyn

Let

cx r y s

1

bx^tf

,

,

,

...

be any terms

of u. Represent these by points A, J5, C, ... whose coordinates referred to any axes are

(m, n), (p,

The

q),

(r, s), ...

formed

figure so

.

H is

m n Suppose that ax y

,

'Newton's Diagram.' bxVy* are small or

K Fio. 81.

~m 0(x p ), and therefore y is O^te-wO/fr-fl)). Join AB, cutting Ox in H, and let the angle xHA=0; ~ ie and axmy n is 0(xm ncoie ). MHO, w-ncot0 = 0#; then y is 0(x) therefore the terms represented by A, B are 0(x H ). Draw CK parallel to AB to meet Ox in K then, for similar reasons, Thus the term C is of an the term cx r tj s represented by C is 0(xos ). order lower or higher than the terms A, B according as C is on the same side of AB as 0, or on the opposite side. Hence the following rule. large

numbers

of the

same order

;

then yn

~~

(l

is

;

Rule. Mark on Newton's diagram points A, B, C, ... representing all the terms of the polynomial u. Choose any two of these, say A, B. Let v be the expression whose terms are those represented by A, B and

any points which may happen to then v =

lie

on the

line

AB.

AB

remote from the origin 0, = (or part of it) near approximately represents the curve u

If all the other points lie

on the side of

on the curve).

(if

v=

AB as 0, then

the other points lie on the same side of sents the curve u Q (or part of it) near infinity. If all

NOTE. If the straight line AB passes through 0, the rule terms cannot be of the same order. In the following examples the points representing the the equation are marked 1, 2, 3, ... in the diagram. Ex.

fails

1st,

;

repre-

for the corresponding

2nd, 3rd,

...

terms of

1.

The terms

1,

3 and

2,

3 give the form of the curve near

the origin.

The terms Art.

8.)

1,

2 give the form near infinity.

(See Ex.

1,

Fio. 82.

EXAMPLES OF CURVE TRACING

480 Ex.

2.

- by)

(ax

2

= ax 2

ij

-f

4.

y

Write the equation in the form, a 2 x 2 - 2abxy + b 2 y 2 - ax 2 y -y*^0.

The form near the

by terms

origin is given 1, 4 and 4, 5.

by the terms

infinity

Thus near (0, 0), (ax - 6//) 2 =0. Near infinity, a z x 2 ~ax z v-a=0 and

yQ

ax 2 y-l-y*

and

and near

1, 2, 3,

~^

giving

0,

EXERCISE LI Trace the curves represented by the following equations and work out the accompanying details. (For hints for plotting, see Ex. 22.) 1.

y*-y + x=Q.

Near

(GO

,

2 ?/

3.

a;

4.

a;

3

3

Near 5.

(i).]

3

Near

.

oo

),

The tangent 2.

[Fig. 84,

y ^x + x y~ -x* -

Ne,ar (0, 0),

[Fig. 84,

1).

+ 2/ 3 ^a 2 x.

+ 1/ 3 - 3ax 2 3

(0,0),

t/

[Fig. 84,

= 3aa; 2

(a, oo

a

passes through

Near

(ii)].

(0,0),

Near

(iii)].

a

(a;

no

=

),

8.

x(y-x)

Near

Near

(a, oo

a

2nd approx.,

2

O

.

2

near

) ;

z y*=a x;

a),

closer

Ox where

Near

oo

2 ,

?/

^a; 3 .

near (00,00),

y=x*Jax + a.

dy -~

on the curve, but the curve

0,

(0, 0),

a (

x

O

is

y

*

a

conjugate point.'

-,

2-a(>/5

x*^a*y.

approximations being

[Fig. 84, (viii)].

.

y=

consequently If

[Fig. 84, (vii)].

x = a-{

:

\

-1.

2

),

!#

.

(0,0),

is

[Fig. 84, (vi)].

y~xa,

=;aij

(a; -f

(0, 0),

near (oo,

;

+ a).

Tangents are parallel to 2

^

y=

/

real point near

2 2 x(y-x) =a y.

Near (00,00),

.

near (00,00),

;

Near (00,00), y 7.

- \x - |z 2

at the points

[Fig. 84, (v)].

,

y (# ~a)~tf

1

1

Ox

y 6.

y ==

),

[Fig. 84, (iv)].

.

(x-a)y*=a*x. near

1

3~.

perpendicular to

is

^a; 2 (2;-f

$jc~

(0,

+ 1)

Near

I'D approx.

(0, oo

),

x=?* y

y=

3x.

Near

Near (00,00),

(0, 0),

a3

-

1st approx., la*

(A).

3rd approx., y

x

V^T"

This shows that the curve becomes indefinitely close to that represented by (A), to the parabola (y-x-a) 2 ~ax. This is therefore a parabolic asymptote.

i.e.

EXAMPLES OF CURVE TRACING

482 x*

9.

Near

- 3*y 2 + 2y* = 0. a;

(0,0),

3

-3i/

Near (00,00),

2

[Fig. 85,

^0

and

(i).]

t/^fz-fz

1st approx.,

# 4 -f2i/ 3

2nd approx.,

y-^

2 .

i.e.

0,

r ..................... (A)

y

2^ y3

-

T + &- ..................................... ( B )

03

-----

in the diagram ; it gives the general Equation (A) is represented by trend of the curve, but (B) shows that it is not a parabolic asymptote. If y tx then x = t*(3-2t). Mark points on the curve corresponding to < = 1,2, -A -1. %

10.

2 ax(y-x) = y*.

[Fig. 85,

Near

(ii).]

and

y-x=Q

(0, 0),

ax^y*.

3'

With y = #

as 1st approx., the

2nd

x is

z

?/

z .

^

Near (00,00), ax*-y*

0.

a2 a2 11. x'A

x'-a'sy-iy^O.

-a zy=0.

powers of

x,

[Fig 85,

Near (00,00), by solving show that approximately

The equations y =

(0,0) (0,

2nd approx.,

y~^2

beuig 3x 2?/

0,

Near (oo,3),

2nd approx.,

?/

3

and

-

-^

in descending

a

by -------

in the diagram.

...................................... (A)

[Fig. 85, (iv).]

are on the curve.

Near

O -

3x~2y^0.

(0,0),

xJ, showing that the point

O

is

a cusp, the tangent

2*^/2

12 -

.

x

The asymptote

2/

viz.

(

- f, -

3 cuts the curve again at

3x 2 + y*~Q (represented by

-----

)

(

-J).

-f

,

3).

................... (B)

y~ -3 s X s -

3# 2 ~ I) (y If 3z 2 + ?/ 3 ^0, then - f - f ). the ( 3

-f

y and expanding

which meets the curve at one other point,

Near (00,00), that

2)

O

for

a*x

are represented

(3x~2y)*-3x*y-y*^0.

The points

x3

y=-r$--i D

(0,0),

tt

x2

12.

Near

(iii).]

-4

I, showing that the curve (A) is below (B) and a parabolic asymptote. 2 (3.r -2*/) -0, showing that the curves (A), (B) touch at

is

point solving for x, show that ,

By lines

y=

-

1

and y

Fig. 85, (v), (vi), 16.

y< - 1 or 4 touch the curve.

(vii), (viii),

else

0
Also show that the

are curves whose equations are given in Exx. 13-

These are of the form xy*-2ky~px* + qx* + rx + $* (See Exx. 2

- I2y = x- 15x 2 4- MX - 85.

14. xy*

15. zz/ a

3 2 -242/^:c -10:r ~5a:+150.

16.

13. xt/

17, 18.)

- IQy^x* - 14z a + 60x -

104.

a 8 8 a^ -8y = x -8a; -3aJ-8.

17. Show that every curve whose equation is xy z - 2ky = px* -f- qx 2 -f rx + s has three asymptotes, or only one, according as p is positive or negative. * Curves having an equation of this form belong to the first of four classes into which divided cubic curves (Enumeratio Linearum tertii ordinis, 1706).

Newton

/

FIG, 85*

(viii)

POINTS OF INFLEXION

484 18.

Show

(i)

that, if

of x other than zero, y

xy

2 2ky px* + qx + rx + s, where p^Q, then, for values given by an equation of the form

2

is

(xy

-

2

k)

=p (x -a)(x- b) (x -c)(x~d).

(ii) In Exx. 13-16 employ the methods of the previous examples to find first and second approximations to the curve at infinity and, by the help of the points, A, J5, C, Z>, which correspond to x=a, b, c, d, and of the points where the curves ;

cut the asymptotes, trace the curves.

NOTE. Diagrams (v) and (vi) show that a cusp ia of a higher order of singularity than a node ; while (vii) and (viii) show the change of partners that generally '

*

occurs when two of the points, A, B, C t D, approach one another, coincide, and then separate again, owing to a change in the coefficients, p, q r, s. t

Points of Inflexion 19.

Show

that the points of inflexion of the cubic curve z

y(a'x lie

on the straight

line f

%y(
[From

(i),

by

+ 2b'x + c') = ax* + 3bx z + 3cx + d

whose equation is - 6"2 - 2bb' + ca') x + (be' (ac' -:=

)

2cb'

....................... (A)

+ da').

differentiation,

(B)

Differentiating again,

and putting

d 2 )/

~0, we have

ct>x

(C)

The coordinates of a point of

inflexion satisfy equations (A), (B),

and

(C).

Multiplying equations (A), (B), (C) respectively by a',

the result

is

-2(a'x + b')

9

a'x 2

+ 2b'x + c',

obtained by addition.

20. Show that the curve (p. 475), whose equation is (y*-l)y~(x 2 -4:)x 9 has three points of inflexion which very nearly lie on the line x 64y. 2 = 6 4 6 [If p=dy/dx 9 show that, at a point of inflexion, # (4p -p )/(^ -l), and

y* = (4p*-l)/(p*-l); so that

21. Prove that the points of inflexion on the curve y*~x z (x 2 + 2ax + b) are determined by the equation 2x*-'+ Qax 2 + 3 (a 2 -f b)x + 2ab =0.

[At a point of inflexion

22. Plot

For For For For

+ 3ax 2 + bx, and y~-~2x* dx

some of the curves

in

Examples

-^

( ) \flte/

1-12, as follows

= 6z + 6o# -f 6.1 2

:

Ex. 1, plot x~ -t/ 3 , and # = ?/, and add corresponding abscissae. Exx. 2, 5, 6, 7, 8, 11, express y in terms of x. Exx. 3, 4, 9, 10, use y=te, or x=ty. Ex. 12, express x in terms of y.

CHAPTER XXX INFINITE PRODUCTS a v a2

1. If

product a x a2

an

...

,

...

an

is

...

will

a sequence of numbers, real or complex, the

be denoted by

Pn

or

by II^ia r .

Definition.

n-> oo a 1 a 2 a3

we

,

Pn

If

tends to a

except when a factor of

...

is

convergent and

Pn

Thus we write

.

limit P, different

finite is

an

.

from

zero, as

we say that the infinite product

zero,

that its value is P, or that it converges to

P, and

write

P=a Pn-> oo

If

infinite If

Pn

or

product

is

oscillates,

a 2 a3

1

-

oo

...

,

to oo

or

= /7"~5a n

Pn->0

if

or simply

,

when no

=

factor

Uan of P n

.

is

zero, the

said to diverge or to be divergent.

the infinite product

is

said to oscillate.

The reader may ask why we say that the product diverges (to zero) when Pw ->0. The reason is as follows. According to the definition, if a a2> ... are real and positive, the infinite product Ua n and the infinite >

series

2 log a n

both converge, both diverge or both

For logPn^logaj-f log a 2 +

...

-flog

oo

,

an

,

and

or oscillates according as log , to zero oo or oscillates. to

to oo

If

Pn->

a

finite limit,

then

Pn

..

1

Pn

Pn

oscillate.

tends to a

tends to a

~> the same

limit,

and

finite limit,

finite limit, to

this limit is

if

not zero a n = Pn/Pfl _ 1~>l. Thus if Ua n is convergent, then a n->l. It is generally convenient to write an infinite product in the form

A necessary (but not sufficient)

condition for convergence

that w n->0,

is

and so, after a certain stage, un |<1. The character of P, as regards convergence, will not be affected by removConseing any number of factors from the beginning of the product. quently there will be no loss of generality in supposing that u n |<1 for |

|

every

n.

The simplest and at the same time the most important case the us have the same sign.

is

when

all

CONVERGENCE

486 2.

Theorem.

// u l9 u2 w3 ,

and

are positive

...

,

than unity, each

less

of the infinite products

is

convergent or divergent according as the series

The

Proof.

series

Suppose that found so that

Zu n

2u n is

is

convergent or divergent.

convergent and that

um + u m+l + um+2 + Also the omission of the its

first

m-l

sn

quently

By

Ch.

is


XIV,

1,

no

;

then

m

can be

factors of a product does not affect

Also

Pn

increases

1

)(l-fw 2 )...(l+w w )<

-U,)(l

-,)

...

Q n >0.

3.

T1

-<,

1

-

(!-)>

By

Ch.

:

1

XIV,

-*.

u n ) diverges to

to

any

so that 77(1

+u n

)

zero.

is

convergence of the

Corresponding

1,

and Q n ->0,

complex, none of which

sufficient condition for the

as follows

>

,

Pn->oo

Therefore

and 77(1

real or

1~6

6n

General Condition for Convergence.

numbers,

conse-

and Q n decreases as n increases. Therefore P n and and both of the products 77(1 -f w n )

and 77(1 ~u n ) are convergent. Next suppose that Zu n is divergent.

diverges to oo

s
for every n.

tend to positive limits as n->oo

Moreover,

assuming that

we have

g n = (l

is

is s

<1.

to oo

...

loss of generality in

Pn = (l+u

of

sum

its

convergency.

Hence there

Qn

Eu n is convergent or divergent.

positive

Let

zero.

infinite

number

(a n )

be a sequence

The necessary and product

6,

however small, an

integer m must exist such that for all values of p

that is to say,

Proof.

which

is

First suppose that

< P

is

for

^ = 1,2,3,

convergent.

not zero, and consequently a positive

(A)

Then P n tends to a limit number k exists such that

for every n.

ABSOLUTE CONVERGENCE by Ch. XV,

Also,

6,

we can

find

m such that y = l,2,3,....

for _rn+p

Therefore

487

fa

~i

so that condition (A) holds.

condition (A) holds, then

if

Conversely,

l-

and Taking e
I

I

I

~~

I

I

\

|

XV,

so that

|,

|P n |>some

\

IP m\ *

fc

\

I

\

is positive and as small as we choose. is convergent. zero and the product

P

6.)

Absolute Convergence.

numbers whose absolute the infinite product

is

convergent. Hence the product is

u l9 u 2 w3 ,

...

,

are real or complex

by w/, u 2 ', u3 ',

...

,

P = (l +*!)(! +^(1 +113)...

said to be absolutely convergent

Zu n

If

values, or moduli, are denoted

is

the series

|

IP * f m \< ^ m+p -IP

Pm+p Pm \ a limit different from

(See Ch.

Pm

.

Consequently Pn cannot tend to zero. Pm+p and Pm have the same sign, and therefore

IP L m * m+p -P Hence Hence

-e)

................ (B)

number when n>m.

Also by (B),

4.

Pm+J) |>(1

|


77(1

+u n

)

when

is

the product

absolutely convergent

if,

and only

if,

absolutely convergent.

Theorem. An absolutely convergent infinite product (P) is convergent. Let P and P' be defined as above and suppose that P' is convergent. Then, by Art. 3, we can find m so that

|(l+t4+i)(l+t4+a)-..(l+t4+p)-l|<6

Now

for

^ = 1,2,3,....

(1

This follows on expanding each side and remembering that for any

numbers

a, 6, c,

...

Therefore

|(l+t/m+1 )(l+wm ^2 )...(l+wm+P )-l|<

Hence, by Art.

3,

P is convergent.

for

2>

= 1,2,3,....

EXPANSION OF PRODUCT

488

We

Derangement of Factors.

6.

shall

prove that if the factors

of an absolutely convergent infinite product (P) are rearranged in any way, remains absolutely convergent and its value is unaltered.

the product

Consider the infinite products

where

P is absolutely convergent and Q is formed by rearranging the factors

of P, so that every v

Since

is

a u and every u

a

is

v.

P is absolutely convergent, so is the series Su n when

absolutely convergent

This series remains

.

Zv n

Therefore

terms are rearranged.

its

absolutely convergent, and consequently the product

Q

is

is

absolutely

convergent.

For any

Pm =a and

is

a2

...

w,

am

it is

q,

...

|

Pm

are

s

a p aa

all

>w.

follows

it

. . .

as -

1

Also, as

n

= 1 + vr

br

,

.

so that all the factors of

Q n = b^b2

the factors of

...

bn

,

a*

-

m tends to

oo

so does n,

,

and

since

from Art. 3 that

and therefore

1->0,

and Q n tend to the same

Expansion of an

6.

among

QJPm = a vaQ

convergent,

Thus

possible to find

are included

therefore

where p,

P

1

suffix

Let a r = 1 + u r

P = Q.

shown that

It remains to be

Qn /Pm -* 1

.

P = Q.

limit, that is to say,

Product as a Series.

Infinite

Consider

the infinite product

where ul3 w 2 w3

... and x are any numbers, real or complex. Let p r be the sum of the products of u ly u 2 ... u n taken r together, so ,

,

,

that

Pn = (l +xu We

(ii)

)(l+xu%)

prove that if

shall

(i)

1

As n->oo For

,

...

(1 -f

limit

so that j>/

Each

is

|

,

the

lr

+px + p2x* +

...

+pn xn

is absolutely convergent,

= for r

l, 2,

...

.

then

n.

of x,

P = (l +zw 1 )(l -f xuz )(l +xiiz) Let w/ = u r x' = x and |

Eu n

the series

pf-> a finite

all values

xun )^l

...

to oo

= 1 + ^x +

Z

2

x2

-f ...

to oo

.

,

|

sum

|

w2 ', ... w n ', taken r together. r occurs in the expansion of

of the products of W/,

of these products multiplied

by

1

GENERAL COEFFICIENT '

therefore

pr

.

Eu n

Since

occurs in the expansion, and

r

I

is

erefore

Also

If

pr '
Zu^

absolutely convergent,

r->oo

then

,

s'

r

/

I

follows that

it

converges to a

sum

s',

and

r

j\r.

increases with n, therefore, as n->oo

pr

489

r->0, and consequently

,

#>/-> a limit

Z

r

'

l

'->0, so that

such that

r

f

lr

is finite,

however great * may be. Thus as n->oo p r becomes an infinite series, which is convergent and pr becomes an absolutely convergent series. Hence pr -> a finite limit l r) and we may write Z r = (l +e r )Pr where e r->0. f

;

,

Let

where

Qn = l +

m

I

l

x+

I

2

x2 +

...

the greatest of

is

Q n -Pn ->0,

Therefore

+l n xn

|

and

l |,

then

,

|

c2

I?

e^c

-

Pn->P,

since

...

JE'ar.

1.

^e

^t'nc?

coefficient

In this identity put zx for

therefore

(1

+zx*)(l +zx?) +zxn+l

2,

therefore

... (1

)(l -\-p l

+z*n + 1

z+p 2 z 2 + ...

-f-pn 2"

coefficients of z r ,

Equating the

1

-xn - r+1 ^

Putting

r

~-

1

,

r

-

2,

...

2,

1

in succession for

n ~\-x Whence by

multiplication

r

r,

follows that, for

to oo.

r of z in the expansion of

Assume that

(1

it

all

values

LOGARITHMIC METHOD

490 Ex.

2.

//

|

x <1, show

that for all values of z,

|

...

where

= l+f

00

to

I

x 4- x 2 -f jr3 and denoting convergent,

The

where

series

Z

its

pr and pr

lim n *>

r

is

. . .

tend to zero as

absolutely convergent, therefore the infinite product value by P, by the theorem just proved,

is

the same as in Ex.

therefore

?i->oo,

Another Method.

Z

r

Also

1.

xn

xn ~\ ...xn

,

~r +l

is

all

has the value stated above.

When

the factors are real and positive, the theory of infinite products can be made to depend on that of infinite series by the use of logarithms. Let P w -(l + wi)(l +^)...(1 +u n ), 7.

where u l9 Art.

1,

t/

2 , ...

~I
are real and

Hence the

Pn = log

(1

+1^) + log

P

infinite

if

I

is

z>-l

+w2 + ... +log )

1 ~

um+l

x~

y

Putting

(1

ti^+i,

,

+w m+1 + log )

w m 4.2>

...

ww

4.

(1

^w_,jn ^- r

L is l+u n

(i)

the least and

_L 77

+ti ro+2 )

+

for x in (A),

2 "if

where

numbers

of the

H the

m>n

2

both

)

1

and

.........

,

(A)

1 -f x.

,

~ sm^ <%

... -f

log (1 +ti w )

by addition we

_ 4 m n ^-^ ?i5

2~L

>

+u n

7, (4), 1

m+I^m+2^~'^ U n

log

XIX,

in

).

27 log (1

r-
a*

+uM+2 + ...+u n = rm n

4-7/2

7/2

Ch.

By

+w n

(1

2

and h the greater

Let

infinite series

oscillate.

and ^0, then

tV> smaller

(1

and the

product converge, both diverge or both

...,

As explained Then

r.

the last assumption involves no loss of generality. log

where

for every

r

greatest of the

= Jmfn

.

find that

'

numbers

1

,

1 4-

uly

1 -f

w2

,

.

Suppose

that

u n2

convergence we can find

is convergent,

m

so that for

then by the general principle of all

values of n,

5 m>n <2Lc,

and

therefore

c is positive and Hence by the general

where

as small as

please.

principle of convergence,

diverges to 0, or oscillates, as

or oscillates.

we

un

P converges, diverges to oo

converges, diverges to oo

,

diverges to

-

oo

,

,

EXAMPLES OF INPINITE PRODUCTS (ii)

it

Zu^

that

Suppose

then since

diverges,

491

r,n,n~^,n^**Vn

follows that, for sufficiently large values of n,

for

any

T

A

positive value of

however

,

(a)

If

u n converges or

oscillates

(b)

If

u n diverges

+

then

P may

Hence

between finite limits, then

or oscillates so that

o>

P diverges

upper limit

its

is

to 0.

-f-

GO

,

converge.

possible cases, and when the logarithm of a complex has been defined and its properties investigated, the same method

This includes

number

to

great.

all

can be applied when the factors of

P

are complex numbers.

EXERCISE

LII

Obtain the results in Exx. 1-5 by considering the value of 1.

d+t)(H

2.

(i-

4.


5.

(l+i)(l

6.

Show

\P*ni =

l

Pn

in

each case.

and

that .

A

Hence

x

if |

\

(l+x)(l+x z )(l+x*) 7.

Show

x = 2n

sin

---

Remembering that Jim 0->o

^

~ cos

1,

that

Show

^v. A

C

+ a:) (1 +-)( 1 V 2 /^

|

cos

^

...

cos

^

.

x

sin

a;

diverges to oo or to 4-^ 3 /)...

according as

that the infinite products

are convergent for ~

(1

^

O. 9.

to oo

deduce that

xx

Show

...

that sin

8.

X

< 1,

all

values of x. B.C.A.

EXPANSION OP PRODUCTS

492 10. If

|a;|
_(* a ? '~

where CTe

[Proceed as in Art.

-*")(*

1.]

11. If n, r are positive integers,

exactly divisible by

is

12. If

(

1

)

(!-*)(! -*)...(l

Ex.

6,

~* +i - d

show that

- x) ( 1 - x 2 )

.

.

.

(

\x\<\ and \zx\<\, show ...

where

*

[Having explained

r

why

^zr/(l

~ xr ).

1

at that

~ 1 + *iZ +

to oo

2 -x)(l -a: )

...

such an expansion

is

* 2Z

2

+

-

H-

tf ^

r

+

. . .

to 00

,

r (1 -a; ).

possible,

put zx for

z

and show

that

(l-zz)(l+tl z + t 2z* +

Hence by equating the

...)

coefficients of z

= ! + tjzx +

t 2z

z

xz

-f

. .

.

.

r ,

(l-X^t^X.t^.] 13. If

(l+zx)(l+zx*)...(l+zx*

x

14. If

|

|

< 1,

show that

15. If

a; |

|

<1

and

zx |

5

2

l

\

< 1,

,

z,

)... ==1

--____

^

/

)^l+p z+p z*+...+pnzn show

for all values of

(l+zx)(l+zx*)(l+zx where

n- l

.

x r\

show that

,

where

(ii)

where

- - --

3 (1 -sa;)(l -2o; )(l

^

r

-zx 5 )

...

to oo

=l+^2 +

=a; r /(l -a: 2 )(l -a; 4 )

J

^2

2 2 -}-...tO 00 .

...

(1

-x* r ).

,

that

CHAPTER XXXI PERMUTATIONS, COMBINATIONS AND DISTRIBUTIONS 1.

Combinations with Repetitions. The number n things r together, when each may be taken as

tions of

is

please,

denoted by

H,

and

number

the same as the

is

of

of combina-

often as

we

homogeneous

products of degree r which can be formed with n letters a, 6, c, ... k. The sum of these products is the coefficient of x r in the expansion of i

When

a, 6,

...

k are

all

equal to unity,

we obtain

the number of the

Thus

products.

therefore

#? = coefft.

of

x r in

(1

+x+

= coefft.

of

x r in

(1

-x)~

//

-

2

-f ...)

n

n ;

"^

Combinations and Permutations of n Things, not all different. LetC be the number of combinations, and P the number of 2.

permutations, of n letters aa (1)

C

is

equal

...

to the coefficient

bb

...

cc

is the

ofVs, r

the

taken k together, then

k of x in the product

number of different letters, p number ofc's, and so on.

where f

...

is the

number of cCs,

For the sum of the combinations in question

is

q the

number

the coefficient of x k in

the product

...

and

their

number

is

(1

+cx + c 2 z2 -h ... + c rx r )

unity.

-h#

+

2

+...

2 +*)(! +Z-fZ -f

obtained from the preceding by putting Whence the result follows.

is

to/factors

;

the coefficient of x k in the expression

(l+X+X 2 + ... +X p )(l which

...

... -f

a, b, c,

Xr) ...

equal to

DISTRIBUTIONS

494 (2)

P

k \k times the coefficient of x in the product

is

x\ L

x2

x

/

For the number

x

x2

tf\

of permutations involving Aa's,

re's,

//,6's,

...

is

Hence where the summation

is

to include every set of values of A,

(with repetitions) from 0,

1, 2, 3,

A Therefore Ex.

1.

We

+v+

. . .

= k.

xk in the above product.

number of combinations (C) and

the

the letters of the ivord parallel taken

four

chosen

such that

\k is the coefficient of

P/

Find

-f/z

...

... /x, v,

number of permutations (P) of

the

together.

have by the preceding,

PI whence we

3. result (i) (ii)

1

C

coefficient of x* in (1

4

coefficient of x*

find that

C22, P

Distributions, is

Some

286.

If

of the things

5

i

n things are divided into

called a distribution.

The

~x)~

may

A number be

In the

first

of

classes, the

of different cases arise,

alike, or

they

order of the things in each class

considered.

r lots

may

may,

or

be

thus

:

all different,

may

case the classes are called groups

not, have to be ;

in the second,

they are called parcels. (iii)

When a

distribution has been

made,

it

may, or may not, be necessary

to consider the order in which the classes stand.

In the

first

case the classes are said to be different (meaning that they in the second case they are

have to be distinguished from one another)

;

said to be indifferent. (iv)

Blank

may, or may not, be admissible, a blank

which contains none

Ex.

1.

How many

lot

'

being a

lot

of the things.

For example, consider what

three masts ?

'

'

lots

is

different signals

implied in the following questions. can be made by flying

five flags a, 6, c,

d

y

e

on

ARRANGEMENT Denoting the masts by

(A)

495

the following arrangements give different signals.

1, 2, 3,

123

123

123 d

IN GROUPS

*

e

<

< {C ,


:

c

c

; c

Because such arrangements as (A) and (B) are

different, it is a case of distribution

in groups. (A), (C) are different, the groups are said to be different.

Because

Further blank

two masts.

lots

are admissible, for all the flags

(See Art. 4, Ex.

may

be flown on only one or only

1.)

Ex. 2. In how many ways can five books a, 6, <% rf, e be divided among three people denoted by 1, 2, 3 ? Such distributions as (A), (B) in Ex. 1 are identical, so it is a case of distribution in parcels. Also the distributions (A), (C) are different, and so the parcels are, said to be different.

Further,

it is

Ex.

implied that each person gets at least one book, and so blank lots are

(See Art. 5, Ex.

not allowed.

1.)

In how many ways can jive books

be tied up in three bundles ? Here the order of the books in a bundle does not matter, so the distribution is in Also the bundles are not to bo distinguished from one another, and so the parcels. 3.

parcels are indifferent.

4.

(See Art.

Arrangement

different things

in

5,

Ex.

2.)

Groups.

The number of ways in which n

can be arranged in r different groups r(r

+ l)(r + 2)...(r + n-l)

or

is

|wC*l},

according as blank groups are or are not admissible. Proof,

(i)

Let n

letters

k be written in

...

a, 6,

a,

row

in

any

All

order.

the arrangements of the letters in r groups, blank groups being admissible, - 1 marks of can be obtained thus partition, place among the letters r :

and arrange the n + r-l things possible orders.* Since ferent arrangements is

r

-

1

\n-\- r

(consisting of letters

and marks)

of the things are alike, the

-

I

/

\r -1 =r(r + l)(r + 2)

...

in all

number of (r + n- 1).

dif-

arrangements of the letters in r groups, none of the groups can be obtained as follows (a) Arrange the letters in all being blank, possible orders. This can be done in In ways, (b) In every such arrange(ii)

All the

:

ment, place (r-1) marks of partition in (r-l) out of the (n-1) spaces between the letters. This can be done in C"~\ ways.

Hence the required number Ex.

1.

How many

n C^l*.

is

different signals

|

can be made by flying

five different flags

on

three masts ?

The required number = 3.4.5.6.7= 2520. *

Thus one arrangement of

7 letters in 5 groups, the

second blank,

is

indicated by a

|j

bed \ef\g.

PARCELS OF UNLIKE THINGS

496

Distribution in Parcels.

5.

(1) The number of ways in which n different things can be distributed into r different parcels, blank lots being admissible, is r n For each of the n things can be assigned to any one of the r parcels. .

The number of ways in which n

(2)

r different parcels, there being

which

different things

no blank

can be distributed into

lots, is

In times the coefficient of xn in the expansion of (ex

is

-

In any distribution, denote the parcels by a r a 2 Proof. consider the distributions in which blanks are allowed. ,

The total number of these is The number in which a x is blank

number

Therefore the

not blank

Of these

in

which

rn is

(r

the

last,

number

in

,

1

,

blank

Therefore 2>

i>

which

in

-

1 )

(r-1

a.*,}

n > r

1

the

last,

,,

.

is

n .

n

-(r-2".

is

1W| -2ft/ r-l) + r-2 tt

blank

is

number

in

which]>

.

n .

(T

Z(r

1)

4-

A)

o)

(r

.

J

the

number k

blank

in

which^

is

J it is

obvious that

number of distributions in which no one of x assigned parcels r n _ (j*(r - l) n + Of (r - 2) n -... + (- l) x (r - x) n

is

.

When x = r,

A

.

J

is

a a are n

the

ft

,

This process can be continued as far as we like, and the coefficients are formed as in a binomial expansion.

Hence

and

J

! a 2 are not blank

a3

,

.

,

>

number

Therefore the

Of these

ar

...

o^ is]

which]

U11 blank is

is

,

.

is

.

a2

r

l)

this gives the result in question.

complete proof

is

as follows

:

Let ux denote the number of distributions in which no one of the parcels a t a2 ... a x is blank. Of these, the number in which ax+1 is blank is ,

,

obtained by changing r into r -

number

is

changing

represented by

r into

in

1

ux (which

E~ ux where E~ l

,

l

a function of

is

blank

is

number of distributions in which no one of
Also we have

This

denotes the operation of

r-1.

Therefore the is

r).

.

,

u { - r n - (r - l) n = (1 - E~ l )r n

.

PARCELS OF LIKE THINGS w x = (l

Hence

-E~

l )

xr n

~2 -

= (1 - Cf E- 1 + Cf and when x = r, Ex.

. . .

+ ( - 1 )*E- X

)

rn

this gives the result in question.

In how many ways can

1.

497

five different books be distributed

among

three persons,

if each person is to have at least one book ?

The required number ^3 5 - 3 Ex.

25

+3

In how many ways can five

2.

The required number ~ (3)

.

I

(3

= 150.

5

different books

- 3 2s + 3

5

O

.

.

.

1

-

--

5 )

be,

bundle

tied HJ> in three

?

= 2,~>.

v)

The number of ways in which n

1

tributed into r different parcels is

same

things of the

can be dis-

sort

or C^Ii, according as blank lots

C>i{~

are or are not admissible.

If there are no blank lots, any such distribution can be effected as follows place the n things in a row and put marks of partition in a 1 out of the n l This can be selection of r spaces between them. which is therefore the number of such distributions. done in C n ways, Proof.

:

~~^

// blank of

lots

are allowed, the

n + r things

of the

same

number

of distributions

sort into

r

parcels with

such a distribution can be effected thus

and

into each of the r parcels, parcels, blank /Tn-fr O r _i

lots

:

is

no blank

put one of the

Hence the number

lots.

n

h r

For

things

in

question

r is

i .

These theorems can also be proved as in Ex. 2 of the next (4)

as that

n things into

distribute the remaining

being allowed.

same

the

The number of ways in which n things of the same

into r different parcels, where no parcel is to contain there are

no blank (x

lots, is

the coefficient of

+ x 2 + x* +

...x k ) r

or of

sort

section.

can be distributed

more than k things and

x n in the expansion of

(l-x

k+l

r )

(l-x)-

r .

r For the product of r factors, each of which is x x 2 -f x 3 -f 4- x may x v A v x^x where X x^, x ... are any terms selected be denoted by 2x from x, x2 ... x k the summation to include every such selection. The coefficient of x n in the product is therefore the number of ways in -f

. .

,

which

r

numbers

.

,

,

.

.

.

,

,

,

A,

/x, ...

can be chosen from

being allowed, such that their

sum may be

n,

1,

2, 3,

and

...

this

repetitions the required

A:,

is

number. Distributions of things of the sidered in Art. 8.

same

sort into indifferent parcels are con-

DERANGEMENTS

498 Ex.

In hoiv many icays can

1.

total score

three persons, each throwing a single die once,

make a

1 1 ?

of

The required number

is

>3 2 11 equal to tho co. of x in (x + x + x f

^co. of x

11

3

-^)

3

in

u;

of x 8

in

(l-o:)- (l

^co. of* 8

in

(^*

^c',o.

(l

-a;)-

(1

.

..

+ a; 6

3 )

3

3

2

~*

+

8

6

)(l"3z

9

)

= 45 -18 -=27. Ex.

2.

Apply

this

method

to

The number of distributions r or in in (1 +x + x z -f ... to GO

prove the theorems of the la^t section. of n like things into r parcels is the coefficient of x n (x

}

-f

x2

-f a;

3 -|-

.

,

.

to oo

r )

according as blank lots are or

,

are not allowed.

These expressions are respectively equal to

Derangements.

6.

-

(1

in

Any change

x)"

r

and x r (\ -x)~ r

,

etc.

the order of the things in a

group is called a derangement. If n things are arranged in a row, the number of ways in which they can be deranged so that, no one of them occupies its original place is

Denote the group of things by a j? a 2 arrangements is \n\ and of these

Proof. of

The number

in

which

stands

a,l

first,] I

in situ,

i.e. is

not in situ

Of these .

.

is

last,

in

which a

l

number

1M-

is

n

In

1.

is

the

number

in

which a 2 l \

.

.

in situ

total

is

number

Therefore the

The

a n-

>

is

i

\n~\-

i

!tt-2.

I

J

number

the

Therefore

.

neither a l nor a 2

Of these is

last,

in situ

the

is

which] I

in situ

,

,

|w-2 74-1 +|tt-2.

is J

number in which

a3

*\

is

Therefore the

one

in

.

number

....which

,

or aj, a 2 , a 3 is

?,?i

in

no] y

,

\n--3

^ w-1+3 ln-2,

,

in-

,si^/ is

This process can be continued as far as we like, and formed as in a binomial expansion.

it is

obvious that the

coefficients are

Hence inaitu If

the

number of arrangements in which no one of x assigned

is

x = n, the

n-Cf JB-l+Cfjn -2 -...+(-!)* last

term

is

|n

letters is

-a.

(-1)", and we obtain the result in question.

THEOREM OF WHITWORTH A

499

complete proof is as follows Let Vy denote the number of arrangements in which no one of a ...

:

ax

Of these, the number into n - 1 in uj. (which This of

number

is

changing n into n in situ

vx

is

which a x ^ l

in

,

a%,

obtained by changing n

is in situ is

a function of n).

is

l represented by E~~ v x

where E~~ l denotes the operation

,

1.

number -E~*v x

Therefore *he is

}

in situ.

is

,

of arrangements in which

that

is

no one of a {9

fl

2,

...

ax

to say,

.* we have

Also, as in the preceding,

Hence

1 ,

it

follows that vx

=-

(\-E~iy\n = (i c 'f A'- + <7/?- 2 -... + (- 1 yE-*) n - j>i-Cf l'/&-l+0 |n--3-...+([)*~[n-.r. 1

-

|

=

If

NOTE,

by

n.

1

1

(ii)

??.,

(i)

It

is

the last term

is

(

Tho expression (A) easily shown that

Tho values

n

of

n

for

||

-I) is

1

77 ,

and we obtain the

sometimes called

n

is

1,

2, 3,

^n(>factorial n,

the integer nearest to \n

1

result in question.

can be calculated thus

...

and

,

I

a.

I

-1^0,

|

denoted

Begin with

:

and multiply successively by 1, 2, 3, ... increasing any even product by ing every odd product by 1 before the next multiplication. Thus ||

is

Ic.

1

1

and decreas-

2=0.2 + 1^-1,

j

3=3. |

I

-1^2,

1

[[4^4.24

-9.

A

7. General Theorem. In Arts. 5 and 6 we have instances of a * as follows theorem, stated by Whitworth :

If there are letter

N sets of letters and, if out of r assigned letters a, N sets, each combination of two letters occurs in N

occurs in

,

each

sets,

each

6, c, ...

t

combination of three

letters

occurs in

N%

and

sets,

so

on; and finally

all the

N sets, then the number of sets free from all the letters is N ~ C[N + CIN - GIN., + + - YNT Or more generally * If there are N events and r possible conditions such that every single condition is satisfied in N of the events, every two of the conditions are simultaneously satisfied in N of the events, and finally all r letters occur in

r

t

.

Z

.

.

]

(

.

:

l

...

2

the r conditions are simultaneously satisfied in

N

r

of the events, then the

number of events free from all the conditions is as stated above. The proof is exactly similar to that in the last article. *

Whitworth, Choice and Chance,

p. 73.

CONSTRUCTION OF PARTITION-TABLE

500

Numbers.

Partition of

8.

In this section we consider a num-

(1)

ber as formed by the addition of other numbers. Any selection of the numbers 1, 2, 3, 4, ... (with repetitions), of which if the selection contains p numbers, the sura is n, is called a partition of n ;

it is

called a partition of

n

into

p parts

or, shortly,

Thus 7=34-2 + 1+1, and we say that 3211 Ex.

Siij}posing that all the partitions of

1.

explain how to find all the partitions of Take first 6 then 5 followed by ;

a p-partition of n. a 4-partition of 7.

2, 3, 4,

5 have been written down,

6. 1

;

then

4,

followed by

all

the partitions of 2

then 2, followed by all the partitions of 3 (that is, 3, 21, 111) followed by the partitions of 4 which contain no part greater than 2 (that is, 22,211, 1111); lastly, 111111. The complete set is (that

11)

is, 2,

;

6,

(2)

//

Up

then

1,

is

3,

;

51, 42, 411, 33, 321, 3111, 222, 2211, 21111, is

the

111111.

number of p-partitions of n* then

n$=n$-i + n$To make the p -partitions

of

n

is

v

..............................

to distribute n elements into

p

(A)

indifferent

parcels, blanks being inadmissible. This can be done by placing one of the elements in each of the

and then distributing the remaining n

p

1

p parcels or 2 or 3 ... or

Therefore

parcels.

Changing n into n

and p

1

n

TTH-I

n -p

llp^l^lll whence the

p-1, we have

into

rjn-p -Til*

-r

rjn-v +

11$

-

rjn-pij + 11 p

-

result (A) follows.

If P>ln> then

II%= n"1\.

For

p^np,

(3)

~p

elements into

in this case

and therefore 77^^

Construction of a table of values of 77^. 77? =

1,

77S

= i(w-l)

or

0.

We

have

\n,

according as n is odd or even. Also 77^ = 1, 77^-1 = 1. In the table on page 501, denote the nth space of the mth row by (n, m) call the set of spaces (n, 1), (n + 1, 2), (n + 2, 3), ... the n-th diagonal.

We

can

mid using

up the first and second rows and the (2), we have

fill

n

i 77r -77 1

f2 ,

77?

Hence the numbers three,

...

^

l

77? +77'2

in the (n

,

first

and second diagonals,

77^= 77? + 77^+ 77*,

+ l)th diagonal are the sums

terms of the nth column, starting from the top. *

Hp

is

subsequently denoted by P(n, p,

:

*).

etc.

of one, two,

CLASSES OF PARTITIONS

1, 1,

The second column is 1, 1, 0, 2, 2, 2, .... The third column is and so on to any 2, 3, 3, 3, ... ;

0,

...

1, 1, 1,

therefore the third diagonal is and the fourth diagonal 0, 0, . . .

,

extent.

TABLE OF ^-PARTITIONS OF n

Various classes of partitions of

(4)

,

501

(i.e.

VALUES OF P(n

n may

y

p, *))

be considered

:

the

num-

ber of those in any particular class is denoted by a symbol of the form P(n, , ) where the number and the nature of the parts are respectively indicated in the first and second spaces following n.

In the

the first space,

that there are

p

parts

;

and


means that

number

In

and

p means

the <;<7

An

of parts does not exceed p. second space, q means that the greatest of the parts means that no part exceeds q.

asterisk

means that no

restriction

nature of the parts, as the case If the parts are

P(n, p,

q) is

to

may

be unequal, then

the

number

of which

is

is

q

;

placed on the number or the

be.

Q

is

used instead of P.

of partitions of

n into p

Thus

:

parts, the greatest

is q.

number of jo-partitions of n. P(n, ^p, *) is the number of partitions of n into p or any smaller number of parts. number of partitions of n into unequal parts, none the is Q(n, *, ^q)

P(n, p,

*) is the

of

which exceeds

q.

CONJUGATE PARTITIONS

502

The values

(5)

of

P(n, ^p,

For instance, the number is

sum

the

the 4th

That

can be found from the table on

of partitions of 10 into

of the first four

number

*)

numbers

and

therefore

is

23.

<4,*HP(14,4,*),

we have

and, in general,

P(n,
We

is

to say that

is

P(10,

(6)

not more than 4 parts

in the 10th column,

in the llth diagonal, that

p. 501.

= P(n+p,p,*) ......................... (C)

have the following relations

:

Q(n,p,q) = Q(n-q,p-l,
and

For consider any partition of n into p parts, of which the greatest is q. l parts, If we remove the part q, we have a partition of n q into p none of which exceeds q. Also from every partition of the latter kind we can derive one of the former. Consider a partition (7) Conjugate Partitions. which the greatest is 4, say 431. This

may

be written as in the margin, and

columns instead of the rows, we have 3 2 2 into 4 parts of which 3 is the greatest. Clearly

431

can be derived from 3 2 2

1

1,

if

of 8 into 3 parts of

we sum the

i

]

1

1

111

a partition of 8

1

by the same

process,

and these

are caljed conjugate partitions.

Thus

to

every partition of

n

into

p

parts of which the greatest

corresponds a partition of n into q parts of which the greatest

Consequently,

is q, there

is p.

we have P(n,p,q)^P(n,q,p),

............................. (E)

P(n,p,*) = P(n,*,p), P(n,
g)

= P(n,

q,


.......................... (F)

P(n,
5,

,*) ........................ (H) (8)

Euler's use of Series in the Enumeration of Partitions.

(i)

For

Q(n,

*,


is the coefficient

this coefficient

adding a selection of

n of x in the expansion of

number of ways in which n can be obtained by the numbers 1, 2, 3, ... q. is

the

EULER'S USE OF SERIES P(n,

(ii)

*,

n of x in the expansion of

is the coefficient

^q)

(1

503

- X )~ l (l -x*)- l (l -x 3 )- 1

...

(1

-a*)-

1 .

For consider the product

terms out of the

Any by

ce,

x2/3 x3y

...

,

a,

j8,

a+

y

2/3

...

factors


are

any we have

...

n is x ,

the product of these terms

If

second,

first,

x q& where

+ 3y 4-

. . .

-f

may

be represented

numbers

of the

0, 1, 2,

...

= tt.

qO

Hence the coefficient of x n in the product is the number of ways which n can be made up by adding any selection of the numbers 1, 2, ... Hence the result follows. q, repetitions being allowed.

Q(n

(iii)

p,

9

is the coefficient

^q)

(1

and

For the in

zpx

coefficient of

n

in 3,

p n in the expansion of of z x

3 +zx)(l +zx*)(l +Z3 )

...

(1

+z&),

n -iP(p-fi) in the

therefore the coefficient of

is

y.

in the first

expansion of

product

number

the

is

which n can be obtained by adding a selection of

of the

p

of ways numbers

1, 2, 3, ...q.

The

rest follows

follows that

(iv) It

of

expansion

Hence by

(i

L

,

9

3,

...

,

P(n, p,

0(17, total

4,

number

-x

3

1

)-

...

into

(1

i), *,

1),

p unequal

~x

p

~W+

l

>

in the

l

)~

.

),

p, *)

;

parts can be found from the

*) in (3). be

made up by adding four

of the

*)=P(17 -^4.3, 4,*)=P(11, of

of forming 17

numbers

ways

4, *)--=ll.

by adding four of the numbers

this

number we must subtract the number

1,

2,

of partitions with a part greater

9.

These partitions are

Or

the coefficient of x n

is 11.

From than

1.

?

havo

Thus the

is

Ex.

*)=P(n~^p(p-

In how many ways can 11

...

6,

- x )- l (l -x2 )~ l (l

Q(n,p,

(5),

table of values of

Wo

Q(n,p,*)

number of partitions of n

/Au^ ^Ae

Ex.

XXX,

Q(n,p, *)=P(n~l-p(p +

(ii),

whence by

1,2,3,

from Ch.

thus,

and similarly

10, 4, 2, 1

G(17,4,

;

11, 3, 2, 1

;

and

10) = C(7,3, <10)=0(7,

<2(17, 4, 11)

1.

their

3,

number

= P{4, *)

Hence the required number

is 2.

3, *)

is

= !,

11-2 = 9.

EXAMPLES OF SELECTIONS

504

The reader should have no

(v)

-zx)-

(1

l

(I

-zx2 )~l

...

difficulty in verifying the following (1

1 -2^)- = l + EP(n,

l

-Z)-

1

1~ZX~ \-ZX 2 )- 1 1

to oo

...

)~

...

tO 00

EXERCISE

=l

=

LIII

Find the number of ways of displaying 5 flags on 4 masts if more of the masts need not be used

1.

are to be used, and (i) one or masts are to be used.

Find the number of ways

2.

in

which n things, of which

all ;

the flags the

(ii) all

r are alike,

can be

arranged in circular order. 3. Show that a selection of 10 balls can be made from an unlimited number of red, white, blue and green balls in 286 different ways, and that 84 of these contain balls of all four colours.

4. Show that the number of made from five a's, four 6's, three

Show

5.

which n are sort, in

n

.

which can be

different selections of 5 letters c's,

two

and one

cf s

e is 71.

that a selection of (n - 1) things can be made from 3n things, of n are alike of another sort and n are alike of a third

alike of one sort, ~ 2 n 1 different

ways.

O

w where n is In the expansion of (a x +a 2 + + any positive integer not the that coefficient than ofany term in which none of the numbers greater p prove a lf a 2 ... a v occurs more than once is In. 6.

9

,

1.

in

In how

m different 8.

many ways can a bags, if one or

white

balls, 6

black balk and c red balls be put

more of the bags may remain empty

?

In an examination, the maximum mark for each of three papers is n Prove that the number of ways in which a candidate it is 2n.

;

for the fourth paper

can get 3n marks

is

i(w +

[Number = coefficient 9.

10.

of

z 3n in

8 + 10n + 6). - xn+l ) 3 1 - x 2n+1 )

l)(57i (

1

(

(

1

- x)~ 4 .]

In how

many ways can 5 rings be worn on the 4 fingers of one hand ? In how many ways can an examiner assign 30 marks to 8 questions,

giving not less than 2 marks to

any question ? 11. The number of ways hi which 2n things of one sort, 2n of another sort and 2n of a third sort can be divided between two persons so that each may have 3n things is 3n 2 + 3n+ 1.

[Number of selections of 3ft things = coefficient of x? in (1 -a;2n+1 (l -#)~ .] 12. The total number of permutations of n things taken 1, 2, 3, ... or n together n

3

)

is

the integer nearest to e in [Total L

number=e n I

L_

I

-

1.

/I [

\n +

x

1

+ l

n + l.n + 2

-f ...

)

/*

1 J

3

EXAMPLES OF ARRANGEMENTS

505

13. There are 5 letters and 5 directed envelopes, (i) In how many ways can all the letters be put into th wrong envelopes ? (ii) In how many ways can 2 letters be rightly placed and 3 letters wrongly placed ?

The number of arrangements of the n terms a l9

14.

of r sequences

a^,

2

a ija r+i occurs

a 3>


...

an

in

which no one

is

r r C [n-CT [w-l + r:; \n-2-. + (-l) r \n-r. number of arrangements is \n. Any one of ..

[For the total occurs in \n-l of these

Show

15.

that the

any two

;

number

none of the sequences

in

I

n-2

and so

;

the sequences

on.]

of arrangements of the letters abed which involve Verify by writing down the arrange-

ab, be, cd is 11.

ments.

The number

16. c,

d come together

Explain

this,

of arrangements of the letters abed in which neither a, b nor is

= 8. 3 + Cl .2* [4-CJ.2 [2 down the arrangements in question.

and write

A

party of 10 consists of 2 Englishmen, 2 Scotsmen, 2 Welshmen and of other nationalities (all different). In how many ways can they stand in a row so that no two men of the same nationality are next one another ? In how many ways can they sit at a round table ? 17.

4

men

18.

n are

The number of combinations n together ~~ and the rest unlike is (n 4- 2) 2 n 1

b

of 3n letters of which n are a

[Number == coefficient of x

n in

(1

~x n ^)

= coefficient of xn in 2 n (l

2

(l

+ x) n (l ~x)~\

-x)~ -n z

.

~ 2n 1

(l

or in (2 - F"-a-) w (l

~x)~\

nm

.

,

Show

also that the

number

-x)~

z

etc.]

Show that if a l9 a 2 ... a m arc distinct prime numbers other than number of solutions in integers (excluding unity) of the equation 19.

is

and

.

of solutions in which at least one x

unity, the

is

unity

is

[Solution follows from the general theorem of Art. 7.] 20. Given n pairs of gloves, in how many ways can each of n persons take a right-handed and a left-handed glove without taking a pair ? 21. (i) If n things are arranged in circular order, the number of ing three of the things no two of which are next each other is

ways of

select-

ira(n-4)(n-5). (ii)

22.

n things are arranged in a row, the number of such sets of three is -J(n-2)(n-3)(n-4). Find the number of positive integral solutions of x + y -f-z-h w>~20 under

If the

the following conditions (i) (ii)

:

Zero values of

x, y, z,

w

are included.

Zero values are excluded.

(iii)

No

(iv)

Each

(v) x,

y

variable

y

may

variable

z,

w

is

exceed 10

;

zero values excluded.

an odd number.

have different values (zero excluded).

PARTITIONS

506 23. Prove that the of the equation

number

of positive integral solutions (zero values excluded)

x 4- 2# + 3z~n i(*--l)("

is

where c~--H

or

tt-9, 10. [The required

24.

according as n

J

number

is co.

or

is

not a multiple of

is

3.

when

Verify

of xn in

The number of independent

solutions in unequal positive integers, zero

included, of

x is

4-7/4-z \-u-

the integer nearest to -&$p*(Zp ~ 25.

Show

that

TIlj'^S/*

2p

3).

2 .

26. A necklace is made up of 3 beads of one sort and tin of another, those of each sort bein# similar. Show that the total number of possible arrangements of the beads is 3n 2 -I 3/i -f 1. [Having put on the three beads, the number of arrangements of the CM, beads

between the three (which are

in the spaces

27.

Find the values of

28.

In how

integers 29. total

30.

/>(18, 5 *),

(ii)

P(12, <5,

ways can 20 be expressed as the

many

is

*),

sum

P(12,

(iii)

*, 4).

of four unequal positive

?

Show that the total number of number of partitions of 10. Show

Art. 8, 31.

(i)

indifferent)

3

that

7 (2ri-fr,

(3), to tind

P(25,

n

partitions of

+ r, *)=P(2n*

r?,

*).

n

is

Find the

P(2/i, n, *).

Hence

the table of

use

18, *).

Show that P(n, p, ^q)

is

cqiioi to the coefficient of

~ xn p in the expansion

of

Hence show that Q(*,P, 32.

Show

that

EXERCISE LIV * 1. There arc n points in a plane, no three being collinear except p of them, which are collinear. How man}7 triangles can be drawn with their vertices at

three of the points

?

There are n points in a plane, no three being collinear. number of n-sided polygons with their vertices at these points is

Show

2.

n-

-J j

if

n=4. *

The

results hi

Exx.

i>-12

were given by Dudeuey

~~

that the

1.

Verily

GEOMETRICAL PROBLEMS

507

m

points in a straight line are joined in all possible ways to n points in line, then, excluding the (ra-f n) points, the number of points of intersection is If

3.

another straight

^mn(m- l)(n- 1). n straight lines of indefinite length are drawn being parallel and no three meeting in a point, then 4. If

(i)

the

number of points

(n) the plane [(ii)

If

un

Show

5.

is

the

is

of intersection

+ n-f

divided into \ (/t 2

number

of parts,

that, in general,

show that

is

2)

in a plane,

^n(n

1)

no two of them

;

parts.

u n ~-u, _ l

<

l

n planes divide space into

4-

n.] 3

J-(?i

+ 5?j.

\-

0)

rerions.

What

are the exceptional cases ? [If no three planes have a common line of intersection and no four meet in a v n ~\ + u n-i where v n is the required number and u n is the same point, then v n as in Ex. 4.] 6.

lines

The number of squares formed on a piece of squared paper by and n vertical lines (m
%m(m -

m

l)(3>i

-i(a

according as a (ii)

is

+ l)(a-l)(a-3)

or

horizontal

l).

n are positive integers such that m^n and m + n (i) constant, prove that the greatest value of m(m ~ l)(3w- -m - 1) is If M,

7.

m

where a

a,

is

ia(a-l)(a-2),

odd or even.

If a straight rods of indefinite length are placed so as to number of squares, prove that this number is

form the greatest

possible

Ar(a according as a

is

+

l)(a

-])(- 3)

or

Aa(a-l)(a-2),

odd or even.

8. Show that the number of ways in which three numbers in arithmetical - I) 2 or \n(n - 2), according progression can be selected from 1, 2, 3, ... n is i(n as r? is odd or even.

The

9.

If c

sides of a triangle are a, b, c inches

is

given, according as c

where

a, b, c

are integers and

show that the number of different triangles is J (c + is odd or even.

1 )2

or

-Jc (c

+ 2),

10. Of the triangles in the last question, the number of those which are isosceles or equilateral is i(3c - 1) if c is odd and -J-(3c - 2) if c is even.

Each

side of a triangle is an integral number of inches, no side exceeding Prove that the number of different (i.e. non- congruent) triangles winch can be so formed is 11.

c inches.

or

according as

c is

odd or even.

12. Of the triangles in the last question, the number of those which are isosceles or equilateral is J(3c 2 -fl) or Jc 2 , according as c is odd or even.

2K

B.C.A.

CHAPTER XXXII PROBABILITY First Principles. In speaking of the probability of an event, doubt is implied as to whether it will or will not happen. The degree of 1.

doubt depends on our knowledge of the controlling conditions. knowledge

of these enables us to say that the event

or certain to

is

A complete

certain to

happen

fail.

Measurement of Probabilities. We are constantly forming rough estimates of probabilities, saying that some event is unlikely, likely, very This implies that (at any rate in certain likely or certain to happen. can be measured and can be compared with certainty, cases) probability which must be regarded as a degree of probability. Thus there is nothing to prevent us from choosing certainty as the unit of probability, and this is always done. In what follows the words probability and chance are synonymous, both of them being used to denote measure of prol)ability.

Suppose that n balls A, B, ... K> all of the same sort, are placed in a bag, and that one of them is drawn. There is nothing to favour the drawing

more than another,

of one ball

A, B,

...

Thus say 5,

or

if

is

x

K will is

the chance that

drawn

is

the chance that either

n

words

it

equally likely that

is

A

is

drawn, the chance that any other

ball,

also x.

Our notions with regard

so

in other

be drawn.

A

or

to probabilities therefore lead us to assert that

B is drawn is

2x,

and the chance that one

of the

drawn is nx. But it is certain that one ball will be drawn, and nx = 1 and x = l/n. Thus the probability of drawing A is 1/n. The following definition is a statement of Laplace's First Principle.

balls is

Definition,

If

an event can happen in a ways and

fail in b

ways, and

nothing to lead us to believe that one of these ways should occur rather than another, that is to say, all the ways are equally likely, then

there

is

the chance that the event fails is

+ 6), + 6/(a 6),

the odds in

a

the chance that the event

happens

is

favour of the event are as

the odds against the event are as

a/ (a

b

:

:

b,

a.

EXCLUSIVE EVENTS If b = 0, the event

a = 0,

certain to

it is

is

fail,

If a = 6, the event is there is an even chance of

509

certain to happen,

and

and the probability as likely to happen

of its

its

as to

probability

happening

1

is

;

if

is 0.

and we say that

fail,

its

happening. the respective chances of the happening and failing of an

If p, q are

event, then

Events

of a type to

which

this definition does not directly apply will be

considered later. Examples, (ii)

(iii)

(i)

// a coin

// a six-faced die // a ball

is

is tossed,

the chance of

thrown, the chance that

drawn from a bag containing 3

is

'

heads '

ace

;

It is easy to

Ex. toss

1.

A man

i.e. all

heads or

is

J-.

turns up

is

ace of spades

is

the chance that the card

is

a spade

is i

the chance that the card

is

an ace

a mistake as to the meaning of

much

of

-^>-

|

simultaneously,

is

an even

-J-.

the chance of

;

;

'

equally likely.'

practical experience in the laying of it

is

balls,

2.

the chance that the card

make

three pennies

:

'

and 2 black

white,

drawing white is J the odds against black are as 3 (iv) // a cord is drawn from a well-shuffled pack,

'

odds said

cha?ice that they

'

// you

:

will fall all alike,

all tails.

1

For they must fall all heads, or all tails, or 2 heads and a tail, or 2 tails and a head. So they fall all alike in 2 out of 4 possible cases.' As a matter of fact, the odds are 3 to 1 against their falling alike. For, if the pennies are denoted by a, 6, c, then a can fall head or tail, and so for b and c. Hence there are 8 equally likely cases of which 2 are favourable.

Events are said to be mutually exclusive 2. Exclusive Events. when the happening of any one of them precludes the happening of any other.

In the case of mutually exclusive events,

the ctiance that one

sum of the chances of the separate events. known as Laplace's second principle. It can

or

other of them occurs is the

This statement

is

also be

If an event can happen in several different ways, one only of which can occur on the same occasion, the chance of its happening is the stated as follows

:

sum

of the chances of its happening in the several different ways. For consider n events of which the chances are ajg, a,2/<7, . . .

,

a n /g respec-

denote integers. Out of g equally likely ways, tively, the events can happen in a ly a 2 ... a n ways respectively, and since the events cannot concur, one or other of them can happen in a l + a 2 -f -f a n

where the

letters

,

.

ways.

sum

The chance

of this

is

therefore

of the chances of the events.

(a4

-f

a2

-f

. . .

-f

a n )/g,

which

.

.

is

the

CALCULATION OF PROBABILITY

510 Ex. (ii)

1.

In a

Any

face of either die

may

These are made up of the thirty throws (1, 2), (2, 1), ...

We

count

what

single cast with two dice,

doublets, (ui) five-six (i.e. one die turns

(i)

two aces,

the other six) ?

turn up, so theie are 6x6 equally likely cases. six throws (1, 1), (2, 2), ... (6, 6), together with the (5, 6), (6, 5).

as the

(5, 6), (6, 5)

chance of throwing

the

is

up five and

same throw, and

aces^g

the chance of throwing doublets the chance of throwing

;

3%

five-six:

In a single cast with two dice,, what are two numbers of which the sum is 7 ?

Thus

call it five-six.

the chance of throwing two

;

-3%.

Ex.

2.

Out

of 36 possible cases, those in favour of the event are and the number of these is 6, therefore

odds against throwing

the

7, i.e.

against

(6, 1), (1, 6), (5, 2), (2, 5),

(4, 3), (3, 4),

the chance of throwing 7

and the odds against throwing Ex. (ii)

3.

eleven,

-/

J~,

1.

:

In a single cast with three dice, what is the chance of throwing (iii) less than eleven, (iv) more, than ten ?

The number (i)

7 are as 5

-

is

Among

of possible cases

is

63

1

3 times

;

3

(ii)

Let

s

of cases in which 11

is

-^ o

~Q^ t5u

thrown, then

s^coeflt. of x 11 in (x f x'2 +x*

+

+ :r 6 3 ^27

...

)

27 .*.

1

i

chance of throwing four-five -six

number

be the

four -five-six,

.

those four -jive-six occurs .'.

(i)

chance of throwing 11

-.

-

is

;

1

These cases are (iii) Let / be tho number of cases in which less than 11 is thrown. those in which exactly 3, 4, 5, ... 10 are thrown. Therefore t is the sum of the coeffi4 10 cients of 3 x~ a; in the expansion of }

,

,

.

,

.

.

(x

+ x- + x*

t- coefficient

hence,

= coefficient

of x of

+ #6

...

3 ;

)

x 3 (l -o; 6 ) 3 (l -x)~*

in

x 1 in

(1

-z)~

^coefficient of x 7 in ( 4x

4

(l -3rc

+ -, I

\

.

8 )

6V6

2

.

= 120-12^108; chance of throwing

.'.

(iv)

are

The

total

number

thrown and those .'.

(In the old

Ex.

4.

game

^4 parti/

in

of possible cases

which

less

less

is

made up

chance of throwing more than 10

of ten take, their seats at

B)

=-

is

^-

.

s>

of those in

than 11 are thrown

of passe-dix, a player stakes

specified ptrsons (A,

than 11

which more than 10

;

1

-J ==i.

on throwing more than

a round

table.

What

10.)

arc the odds against two

sitting together ?

A having taken his place, B has a choice of 9 places, 2 of which are next to A, hence the odds against B sitting next to .4 are as 7 2. :

INDEPENDENT EVENTS

511

Four cards are drawn from a pack of 52 cards. What is the chance that one of drawn ? Four cards can be selected from 52 in C% z ways and this is the number of possible One of each suit can be selected in 1.3* ways and this is the number of cases. Ex.

5.

each suit

is

;

;

favourable eases

;

28561

1.3*

Ex. is the

Five, balls are

6.

chance that 3

white.

drawn from a bag containing G white and 4 black and 2 black balls are drawn ?

What

balls.

Five balls can be selected out of ten in CJ ways. This is the number of possible Also three white balls and two black balls can bo selected in C\ C% ways,

cases.

which

.

the

is

number

of favourable cases, therefore

-iv

,i

the required chance

3.

C

T

^L

5

Events are said to be independent when

Independent Events.

the happening of any one of the others.

- 10 = C ^*'t ^p ">T-

them does not

affect the

happening of any of

The probability of the concurrence of several Laplace's third principle. independent events is the product of their separate probabilities. It is sufficient to consider

two independent events

A

A2

and

t

,

of

which

the probabilities are p l and p 2 respectively. Suppose that A L can happen in a L ways and fail in 6 t ways, all of which are equally likely. Also suppose that A 2 can happen in a 2 ways and fail in 6 2 ways, all equally likely.

Then

Pi

= ai/(<*>i + b

^ 2 -=a 2/(a 2

1 ),

f 6 2 ),

and there are the following

:

possibilities

A and A 2 may both happen, and this may occur in a^a* ways. A may happen and A 2 fail, and this may occur in a 6 2 ways. A may happen and A fail, and this may occur in aj> ways. A and A may both fail, and this may occur in 6 6 2 ways. The total number of ways in which both A and A 2 are concerned }

1

l

l

}

2

l

t

l

(a l

+b 1 )(a 2 -\-b.2), and

both

A

1

and

A2

occur

these are is

all

equally likely.

is

Hence the chance that

# 1 a 2 /(a 1 + 6 1 )(a 2 4-6 2 )j which

is

equal to Pip%.

It should also be noticed that

A that A 2

the chance that

the chance

l

happens and happens and

the chance that both

Hence

and

also the chance that one

PiO -ft) Also the chance that at fail) is

Al

1

-(1

A2

A 2 fails is p (l -p 2 A fails is p 2 (l -pi) t

)

1

fail is

(1

and only one

-

Pi)(l

-

p

;

;

2 ).

of the events

happens

is

+^ 2 (1 -PiH Pi +Pz- ZpiP*

least

one of the events happens

-^(l -pzl^

(i.e.

both do not

INTERDEPENDENT EVENTS

512 Ex.

What

I.

is the,

chance of throwing ace with a single die in two

trials ?

The chance

of not throwing ace in one trial is -|, the chance of not throwing ace in two trials is () a

the chance of throwing at least one ace

.'.

Or

1

-

;

2

(^ )

= ^Q.

thus,

Chance Chance Chance Chance Ex.

of success in both trials

3^. of success in the first trial and failure in the second

and success in the second one success yg + 2 ^ f J^~.

of failure in the of at least

2.

Referring

f

first

to the last

example, explain

The chance of throwing ace in the first is J, therefore the chance of ace in two

why

\

.

-|.

the following reasoning is incorrect

:

^ and the chance of ace in the second trial

trial is

trials is

-|.

The throwing

of ace in the first trial does not exclude the possibility of throwing ace in the second trial. Thus the events are not mutually exclusive and Art. 2 does

not apply. It should be noticed that, if the reasoning were correct, it would follow that the chance of throwing ace in 6 trials would be -f i.e. that ace would certainly occur in 6 trials which is obviously false. ,

;

Ex.

In how many throws with a

3.

single die will

it

be

an even chance

that ace turns

at least once ?

up The chance against ace turning up in n throws is (|) n and the chance that n If then n is the required number, we have up at least once is 1 - (|) ,

it

turns

.

l-() n =

/.

;

w (f)

=4;

/.

n = 3-8

nearly.

Hence it is less than an even chance that aco turns up once than an even chance that it turns up in 4 trials.

in 3 trials,

and more

Interdependent Events. If two events A v A 2 are such that p the probability of A ly and p 2 is the probability of A 2 on the supposition

4. is

A

that is

p p2

has happened, then the probability that both

l

A

and

A2

happen

.

}

Also

p3

if

is

the probability of a third event

happened, then the probability that A v A 2 and and so on for any number of dependent events.

For the reasoning Ex.

One bag contains

1.

.and 3 black

chance that

If a bag white ?

balls. it

is

The chance that the from

of the last article holds

it is .".

first

3 white balls is

A 3 after A l and A% have A 3 will happen is Pip^p^

good under these conditions.

and 2 black

balls

chosen at random and a ball

bag

is

chosen

is

i,

;

is

another contains 5 white

drawn from

and the chance

of

it,

ivhat is the

drawing a white

;

the chance of choosing the

first

bag and drawing a white

Similarly the chance of choosing the second bag .'.

is

\ \

and drawing a white is \ random is

the chance of drawing a white from a bag chosen at

.

f-

;

ball

FREQUENCY NOTE. its

Ex.

were put into one bag and a is not the same as before.

If all the balls

being white

A

2,

person draws a card from a pack, replaces

at least three trials,

The chance of

(ii)

success at

any particular

second caw, he must

What

balls.

is the

First method.

chance

It

In

trial is \.

the first case,

he will have

;

2

at the

first

(J)

~-fij.

and second

2 required chance = (2)

trials

and succeed at the third

;

~~Q.

are drawn, one by one, from a bag containing 6 white and 4 black

Five balls

3.

the pack. He conhe will have to make

and shuffle

it,

is the chance, that

required chance

fail

.'.

Ex.

were drawn, the chance of

exactly three trials ?

.'.

the

What

a spade.

to fail at the first anct second trials

In

ball

-^, which

is

tinues doing this until he draivs (i)

513

3 white

tliat

makes no

and 2 black

balls are

difference to the result

drawn

?

we suppose the

if

5 balls

drawn

Hence, as in Art. 2,

simultaneously.

the required chance -=C'*

Second method. Suppose the balls drawn chance that this occurs is easily seen to be -

_6_.

10

.

9

4 8

.

C*/C\ ~iy.

.

in the order indicated

4

.

7

by wwwbb.

The

3.

6-

Also the chance that 3 white and 2 black balls are

drawn

in

any other particular

obtained from the preceding by merely altering the order of the numerators. The number of orders in which this can occur is the number of permutations of the 10 5 / 3 2 letters wwwbb, and is therefore order

is

1

;

j

.*.

which

5.

is

the required chance

10

-^ f

-f

-f

'5,

equal to the result previously obtained.

Another

Way

of estimating Probabilities.

It

is

obvious

that the definition in Art. 1 only applies to events of a restricted type in particular, to games of chance. There are other cases in which probability ;

is

estimated by considering the frequency with which the event occurs, or believed to occur, in the long run/ For example '

is

:

(i)

It

is

an observed fact that about 51 per cent, of the children born in Hence we say that the probability that a child, about to

Europe are boys. be born, will be a

boy

is

about 0-51.

'

According to tables of mortality,' such as are used by Insurance Companies, out of 81,188 men 50 years old, 70,552 live to the age of 60. We therefore say that the chance that a man of age 50 will live another (ii)

10 years

is

70552/81188 = 0-869.

In general, trials, where

an event has happened in pN trials out of a total of N a large number, we say that p is the probability that happen on a fresh trial.

if

N

the event will

is

EXPECTATION

514

This notion of probability

is

not at variance with that contained in

Laplace's first principle (Art. 1). If a six-faced die is thrown a large

number

we expect one

of times,

face

up about as often as another, and this has been verified experimentally. Thus in a large number of throws we may expect ace to turn up in about J of the total number of trials, and from this point of view we

to turn

say that the chance of throwing ace

is |.

Expectation. Suppose that a person (A) has a a. lottery which gives him a chance p of a prize of 6.

ticket in

a

N

N

is a large number, we may the lottery were held times, where him to the about expect get pN times, receiving pNa. Thus we prize may say that on an average he receives pa for a single lottery, and this

If

is

called his expectation.

Next suppose that

him a chance p t

ticket gives

/1's

chance p 2 of receiving a 2 and so on. If the lottery were held times, where

of receiving

ap a

,

N

N

is

we may expect him

large,

pfl occasions, 2 on about p 2 N occasions, and so on. Altogether he may be expected to get about N(p a l -f p 2& 2 + ) f r a Thus we may say that on an average he gets ) (p l a l 4-jt) 2 tf 2 -f his from sum is sum of This the the single lottery. expectations arising a l on about

to receive

<2

l

chances of securing the separate sums, and

is

called his expectation.

The average value of a quantity P, subject to risk, is the value P assumes in the long run. This value is also called which average the expected value of P or the expectation with regard to P. Definitions.

If

that

P

a quantity can assume the values P l9 P2 P3 it has these values are respectively p v p 2 jo 3 ... ,

,

,

expected value of

This

is

P

merely a

,

^1^1+ j>2 P 2 + p 3 P3 ... generalisation of what has been is

...

and the chances the average or the

,

.

-f-

said in the case of a

lottery.

It should be observed that every p denotes the chance that corresponding value, so that events need not be independent.

P

has the

1. A man's expectation of life, is usually taken to mean tho average number of which men of his age survive. years If p r is the ehanre that he will survive r years, dying before the end of the next year,

Ex.

his expectation of life

is roughly given in another volume.

Ex.

2.

A

pl

.

1

+ p2

.

2

+p 3

.

3

4-...

.

A

closer estimate will be

person draws 2 balls from a bag containing 3 white and 5 black balls. If he for every white ball which he draws and Is. for every black ball, what is his

is to receive 10s.

expectation

?

The number

of

ways

in

which 2

balls

can be drawn

is

C\

28.

SUCCESSIVE EVENTS Of these the number of ways in which 2 white balls can be drawn 1

white and

2 black balls can be

Hence

expectation

= (/g

Successive Events.

7.

drawn

his chances of receiving 20s., lls. .*.

happening and of the

For the chance

Let

failing of

specified order is

of its r

drawn

2s.

3,

= 15, Cf = 10.

3

is

5

.

is

and

20 + |f

-

are respectively 3/28, 15/28, 10/28;

11

+ Jf

=88. 9d.

2)

chauces of the

p, q be the respective

an event at a single

The chance of its happening

(1)

C\

is

black ball can bo

1

515

exactly r times in

n

then

trials ix

times and failing

r

happening

trial,

n-

:

C"p r

n ~r

r

q

.

times in a

n ~r

p q Now the number of different orders in which these things can occur is the number of different arrangements of r things of one sort and n - r n - r, which is equal to C r things of another sort. This number is n/ \r \n .

:.

I

Thus the event in question can happen in C ways, which are mutually r n ~r exclusive, and the chance that it happens in any one of them is p q ;

therefore the required chance

We

(2)

n

exactly

have times,

C^ = C^ r

Cp q r

is

(n~l)

The greatest term

(3)

part of

-

that of

n-r

If

+ q) n =p n +

,

nq

is

exactly

is

and

of

C> ~V

(n

+

r failures

integer, then

r

])q.

Hence

where

r is

= nq and n

p

:

times,...

2)

are

.

.

.

.

which

the

r is

the integral

most probable case

is

the integral part of (n + 1) q. = np, so that in the most

r

probable case the ratio of the number of successes is

(n

...

in the expansion is that in

that

successes

an

times,

terms in the expansion n~l n + C^p q 4-

respectively the first, second, third,

(p

.

therefore the chances that the event happens

,

exactly

n ~r

to the

number of failures

q.

may be useful to state an important theorem due to Bernoulli chance of an event happening on a single trial is p and if a number of independent trials are made, the probability that the ratio of the number of successes to the number of trials differs from p by less than (4) It

:

If the

any assigned number, however small, can be made as near to certainty as we choose, by making the number of trials sufficiently great. The proof of this is beyond our present scope, and requires a use of approximate values of large factorials. For instance, we may use Stirling's theorem, which states that

PROBLEM OF POINTS

516 If unfT

(5)

we

is the

chance that the event happens at least r times in n

trials,

shall prove that (i)

u nt r = p n + C% p n ~ l

For if the event happens at least r times, then (i) it must happen exactly n times, or exactly (n-1) times, or exactly (n - 2) times, ... or exactly r times, and the chances that these things occur are the successive terms of the first expression. (ii) it

Or,

must happen exactly (r + 2) trials, ...

or in just

trials,

r

times in just

or in just

n

r trials, or in just (r

Now

trials.

happen exactly r times in just (r + s) trials, it must happen these and also in (r - 1) out of the preceding (r + s -I) trials. r 1 1 of this is p C and therefore j;+*- ^r-igS^Cy+*~ ^ r/

that

it

+ 1)

may

in the last of

The chance

f

;

Un,r

-f + Vr-lP'q + ^iP^ +

-

+ Cr~l

P^f^

the second of the expressions in question. That these two expressions are equal may be proved as in Exercise XXXV, Ex. 9.

which

Ex.

is

// a die

1.

3 timej,

at least

(ii)

is

thrown 5 times, what

3 times

is the,

chance that ace turns

up

(i)

exactly

?

Let x and y be the required chances, then

and or by the second formula,

46

fe. 2. A and B throw alternately with a single die, A having the first throw. The person who first throws ace. is to receive 1. What are their expectations ? The chances that the stake is won at the first, second, third, ... throws are i

5

6'

6*6'

i

5 \4

Hence, the expectations of

i

/sy !

*6'

+

-

..

A and 5

i

/r>y

\Qj

;

\Qj '6""

!

!

=

'

are

-ff

and

G

^B r,,

;

^-

,65 =

chance==1 i

-

respectively.

The Problem of Points.' A and B play a scries of game* which cannot be J^a;. 3. drawn, and p, q are their respective chances of winning a single game. What is the chance that A wins m games before B wins n games ? '

Here

A

must win at

least

m

games out

obtained by substituting these numbers for

of r

m + w-1, and n

and the required chance

in either of the formulae in (5).

is

CHANCE UNDER DIFFERENT CONDITIONS (6)

So

Several Alternatives. the event

alternatives

at each trial there have been

far,

succeed or

may

517

may

it

two

fail.

Consider the case of an experiment which at any trial must produce one of three results denoted by A, B, C. Let the chances that A, B, C occur at

B

be p,

trial

any

therefore

occurs

q,

P

One

r respectively.

p + q + r = l.

C

times and

of the three results has to occur,

Also the chance that in n trials occurs y times (where a

A

occurs a times,

+ p -f y = n)

is

Jl [*[(/ that

is

to say, the chance

For the chance that in a specified order

p

is

the term containing

is

A may a

qPr

Y

happen a times,

p

a

q^r

B

y in

the expansion of

times and

j8

C y

times

.

Now the number of orders in which these things can

occur

is

the

number

arrangements of a things of one sort (A), j8 things of another sort and y things of a third sort (C), placed in a row. of

This Ex. this is

number

4.

A

card

In/ la

is

I

Hence the

\y.

j8

(J5),

result follows.

the card is replaced and the pack shuffled. If chance that the cards drawn are 2 hearts, 2 diamonds and

drawn from a pack,

is

done six times, what

is the

2 black cards ?

At any

trial,

the chance that a heart or a diamond

chance that a black card

is

drawn

is

is drawn Hence by the preceding,

^.

is

^=4

;

also the

=

512* Ex.

5,

chance that

A it

'

hand

'

of six cards

from a pack in the ordinary way. diamonds and 2 black cards.

is dealt

consists of 2 hearts, 2

Find

the

The chance that the card dealt

is

a heart

is

second card dealt

is

a heart

is -^-f ,

card dealt

is

a diamond

is

fourth card dealt

is

a diamond

is

^f

fifth

card dealt

is

a black

is

ff,

sixth

card dealt

is

a black

is

f-f .

first

third

Hence the chance that 2 in any specified order

hearts, 2

,

,

,

diamonds and 2 black cards are dealt in

this or

is

j^3 5~2~

is

^f

The required chance therefore equal to

is

. *

1,

2

5 1"

*

3,

3

50

. *

12 40

*

26 48

, *

25 4 7"

obtained by multiplying this number by [6/[2 [2

1_2

;

and

INVERSE PROBABILITY

518

Ex. 6. Find the. chance that in a game of whist the dealer (A) has exactly two honours. The card turned up is, or is not, an honour and the chances of these possibilities are ;

respectively ^3

A

an d Y^.

In the first case A certainly has one honour. Of the remaining 51 cards, 12 belong to and 39 to the other players. Let p be the chance that, of the 3 remaining honours, 1 is among the 12 cards and 2

are

the 39, then

among

^_/T3 P ^I In

the second case,

2 honours belong to

.

12.

51

39.

.

where the card turned up

A and

49-

not an honour, we have

if

p'

is

the chance that

2 to the other players, '

P

__

p4 U

12.

.

.

51

2

1 1

3j9

.

50

the required chance^

/.

is

3J.

.

50

49

.

38

.

48

>

-$p +-sp = f

8. Probability of Causes. (1) Suppose that an event has happened which must have arisen from one of a certain number of causes C l9 C2 ... ,

What

This question Ex.

bay

is

Each of three bags A, B,

1.

are as follows

A

a specified cause, C l9 actually led said to be one of inverse probability.

is the probability that

is

C contains

v)hite

and black

.

to the event ?

balls, the

numbers of which

:

A

B

C

white

-

-

al

a2

as

black

-

-

bl

62

63

chosen at random, a ball

is

drawn from

It is required to find the probabilities

Q l9

it

and

is

found to be white. came from A, B,

Q%, Q% that the ball

C

respec-

tively.

If

the numbers of the balls are altered as below, the probabilities in question remain

ABC

unchanged:

when

#,

?/,

z

are

white

-

-

atx

a^y

a3 z

black

-

-

bvx

b^y

b^z

any numbers whatever. x(a l

Choose these so that

+ 6j) ^y(a 2

-t-

62 )

^ z(a 3 + 6 3

).

The three bags now contain the same number of balls, therefore any one of the a^y + a 3 z) white balls is as likely to be drawn as another. If the ball which was drawn came from A, then it belonged to the group a x x of

(a^

-|-

white

balls,

and so

for the other possibilities.

Therefore

Pi^ 1 /(a 1 + 6 1 ),

where

Now

Qi'.Q 2 :Q3

a white ball

is

It is to

:

a 3 z ^p l

'

'.p z

Ps>

P2 =

+ e 2 +
#2 Qv p is

be noticed that

thai the ball

az y

:

drawn, therefore

Ci with similar values for

= <*>&

l

and

the probability that, the event will occur

comes from A, and so for p^

pz

.

on

the supposition

PROBABILITY OF FUTURE EVENTS Ex.

The same as

2.

a x white and

m

z

such as B, and

from an

A

y

m3

such as C, as in Ex.

C

B,

and as

is

m

l

1.

containing

Q lt Q2t Q3

bag respectively.

Then any one

Alter the numbers of the balls as in the preceding.

white balls

A

Also

the preceding, except that there are ra x bags such as

b t black balls,

are the chances that the ball came

one of the

519

of the

as likely to be drawn as another, and Q l is the chance that balls, so also for (? 2 Q 3 , therefore groups of

ax

Ql + Qz -f
before,

therefore

with similar values for

it

comes from

,

Q Qs 2,

3#j)

.

A

NOTE. If PI is the chance estimated before the event that an and P2 P3 have similar meanings, we have

bag

will

be chosen

,

P2 PS^W! W 2 m3 l Q x Q 2 Q3 =Pl p l P2P 2 PsPa Qi^PiPiKP&i + PsP* + ?&*), etc. P

and Observe that (2)

A

:

:

:

:

:

:

;

*

therefore

Pp J

l

is

:

the chance that a white ball

drawn, and that from an

.4

bag.

Suppose that an event has occurred which

General Statement.

must have been due to one

is

of the causes,

Cl9 C 2

,

...

Cn

.

Pr

be the probability of the existence of the cause Cr estimated the event took place. before Let p r be the probability of the event on the assumption that the cause

Let

Cr

,

exists.

Then

the probability

Qr

of the existence of the cause

Cr

,

estimated after

the event has occurred, is given by

For an argument similar to that in the Qi

and

:

%

:

3

:

last

two examples shows that

=

since the event has happened,

Gl + 02

whence the NOTE.

result follows.

It is usual to call

the causes and

P P 19

2 , ...

Pn the

a priori probabilities of the existence of

Q l9 Q 2 ... Q n the a posteriori probabilities. The product Prp r is the antecedent probability that the ,

event

will occur,

and that

from the rth cause.

The argument depends on the assumption

P2Pz>

(1)

that

Qlf $2 >"'

ar ^ proportional to

which is justified in (1) PiPit In particular, if an event is due to one of two causes, the odds in favour of its having occurred from the first cause are as P1 p l P^Pf The way in which these principles are applied to determine the Probability of Future Events is illustrated in Ex. 4. :

A POSTERIORI PROBABILITY

520 Ex.

A

1.

A

white.

white ball

bag contains 5

ball is

drawn and

and of

balls, is

these

What

is the

only

-

5

i

.

1+2 + 3 + 4 + 5 = 1 ~5~~~" 2

1

g /.

the required chance

2 5,

p2

,

,

73

may

= Ql

P

l

pl

---

// ut
be

0, 1, 2, 3, 4, 5.

a notation similar to the

,

,

2.

5 are

1, 2, 3, 4,

chance that this

of white balls

C lt ... C5 and using - u, p l _,i5 nm p Q ^n ana

Y

1

number

the

;

Denoting these possibilities by 6 _i above, we have 2^ p ~~JJ _ _p ^

$#.

is the

?

There are 6 possible hypotheses

that

equally likdy that 0,

it is

to be white.

found

^

\

.

.

,

^5

_5.. 5

>

;

.

is replaced,

what

is the

chance

a second drawing

~6.5 2X Hence the required chance Ex.

3.

the

A

pack of cards

Two

missing.

cards are

is

The

event

'

is

counted, face downwards, and to be spades. ?

that two spades are drawn. (Cj)

(C 2 )

is found that one card is What are the odds against

it

drawn and are found

missing card being a spade 4

li-

The missing card The missing card

There are two possible hypotheses is is

:

a spade. not a spade.

The a priori probabilities P19 1\ of C19 C 2 are P l J, P z = %. The chances p l9 p 2 of the event under the hypotheses C\, C 2 are

Mil *

5TO>

1312 = 5~l 5~0 '

P'2.

The odds against the missing card being a spade

are as

m

balls which are either white or black, all possible numbers Ex. 4. A bag contains being equally likely. If p white and q black balls have been drawn in p + q successive trials without replacement, the chance that another drawing will give a white ball is

The

possible hypotheses, all equally likely, are that the

m Denote these by If

pL p2 ,

,

we have

.

. .

m

q,

C lf C 2

,

...

,

lf

...

Cm_q

m-g-rH-1,

If

A

is

r-r+l

p.

r(r

nq \-r-\\rm _/j ^ 7/ur,.vr> '^q l^p^q

+ l)(r + 2)

...

(r

independent of r. ... are the a posteriori probabilities of

Q19 Q&

where

...

of white balls are

_^ r

+ q- 1),

...

and

number

are the antecedent probabilities of the event under these hypotheses,

^ /?m Pr'-^p ur

where

q

Cr

...

a^irVr-i

\p \q

(n-r+p-l),

C C2 lt

,

...

,

then

Qr ur vT jS

\m-\-\

^

(by Ch.

XXI,

10.

Ex.

1.)

t

VALUE OF TESTIMONY If

pr

'

is

the chance that, with hypothesis , f

n^

thus

pn

l

r

-.

<7

521

_

another drawing gives a white

r,

m-q-p-r+l _ __ _ _. n-l -r m-q-p* m-q-p *

*

__

.

=0, and the chance that another drawing gives a white

r~r*QrPr' = S'l(m-q-p)S, where S' = 2 /S" can be obtained from S by writing n r

r r

Thus

T"T

ball,

^\(n-l ~r)(n-r)

-_

j

and

ball is

;

n and p

for

1

the chance

^-

for

...

1

;

(n-r +p). and hence

=

9. Value of Testimony. The theory of probability has been used to estimate the value of the testimony of witnesses. Such an application is open to adverse criticism. It rests on two assumptions which can hardly

be

namely (i) that to each witness there pertains a constant p credibility], which measures the average frequency with which he speaks

justified,

(his

:

the truth; (ii) that the statements of witnesses are independent of one another in the sense required in the theory of probability. If

we are prepared to make these assumptions,

the procedure

as follows.

is

probability that a statement made by A is true and p has a similar meaning for B, what are the odds in favour of the truth of a statement which A and concur in making ?

Ex.

1.

If

p

/

is the

B

4

The

the agreement of A and the statement is true

'

event

is

possibilities are

:

(i)

;

B

in

The

making a certain statement. false.

(ii) it is

chance that they both say it is true is pp'. the chance that they both say it is true is (1 -p)(l -p')> Thus the antecedent probabilities of the event on the hypotheses (i), (ii) are

If it is true, the If it is false,

pp' and (\-p)(\-p') and the odds in favour of the truth of the statement are t

as pp'

:

(1

-p)(l -p')

2. A bag contains n balls, one of which is ivhite. The probabilities that f speak the truth are p, p respectively. A ball is drawn from the bag, and A and assert tJiat it is white. What are the odds in favour of its being white ?

Ex.

B

The are

'

(i) it is

The a If

'

event

is

true,

the agreement of

(ii) it is

A

and

P P

priori probabilities 2 , of (i), (ii) are 19 are the chances of the event under (i)

In the case of

(ii),

(n

The

and both

possibilities

false.

pv p2

-

1)

balls

remain

P

1/n,

t

and

(ii),

in the bag,

(n-

P^

l)/n.

we have p ~pp'.

and one

of these

A

is

white.

The

should choose this ball and wrongly assert that it was drawn from the (lp)(n-l). The chance that B should do the same thing is (1 p') (n- 1) ;

chance that

bag is hence

B in making a statement.

A B

and the odds

Pt =(l-p)(l-p')(n-l), in favour of

a white

ball

having been drawn are as

RANDOM

POINTS TAKEN AT

522

axiomatic

that

:

If a

(i)

The following statements are

Geometrical Applications.

10.

taken at random on a given straight line AB, the chance on a particular segment PQ of the line is PQ/AB.

point

it falls

is

Or we may say that the total number of cases and the number of favourable cases by PQ. (ii)

a point

If

the chance that the point

L

Ex.

Given that x +

and 2a are equally

AB

Let

likely,

on a

falls

y-2a show

where a

that

a/S.

is

constant

and

an even chance

is

it

is

P

at

random

in

AB,

includes an area

that all values of

o-,

x between

that

and

be a diameter of a circle with centre

Take a point

radius a.

represented by

S which

taken at random on an area

is

is

AB.

AP = x, PB~y,

then x + y~2a, and all values and 2a are equally likely. Draw the ordinate PQ, then PQ 2 = AP PB^xy. Let

of x

between

.

If A',

B' are the mid-points of OA, OB, the ordinates

at these points are equal to a

.

^f

O P

.

Hence PQ>cuJ% if, and only if, P lies in A'B'. Hence the chance that z//>f a 2 is A'tt'/AB, that Ex.

Two

2.

The points

We may

Q PQ>b,

points P,

that the chance that

are taken at

random on a

where b
are as likely to

is

(a-b)

is

therefore suppose that

Q

is

B

\.

straight line

2

OA

of length a.

Show

z .

ja

the order 0, P, Q,

fall in

B'

FIG. 86.

A

as in the order 0, Q, P, A.

to the right

of P.

Draw OA

'

at right angles to

OA and

equal to

it.

Complete the figure as in the diagram, where

OL-^PR^b. If

8x

is

distance of is in

PA,

is

small, the

P from

number

lies

of cases in

which the

between x and x + $x and

Q

by the area

of

represented by

Sx.PA,

i.e.

the shaded rectangle.

Of these, the favourable cases are those in which lies in EA, and their number is represented by the upper part of the shaded rectangle cut off by LM.

Q

Hence the the total

total

number

number

of cases

.'.

represented by area of the triangle OAA', and by the area of the trianglq LMA' ;

is

of favourable cases

FIG. 87.

the required chance

=

A OAA'

CALCULATION OF CHANCE

523

EXERCISE LV 1.

balls,

two balls are drawn from a bag containing 2 white, 4 red and 5 black what is the chance that If

they are both red ? one is red and the other black

(i) (ii)

A

?

white and n black balls. If p+q balls are drawn bag without replacement, the chance that these consist of p white and q black balls a rim finisim+n is C p .^/C p+g 3. If four balls are drawn from a bag containing 3 white, 4 red and 5 black balls, what is the chance that exactly two of them are black ? (i.e. two are black and two are white or red.) 2.

contains ra

;

.

4. In the last question what drawn from the bag are black ?

is

the chance that at least three of the four balls

5. Out of 20 consecutive numbers, two are chosen at random chance that their sum is odd is yf 6.

alike,

In a throw with three dice what

two

(ii)

Explain

alike

the

why

and the third

sum

is

different,

;

prove that the

the chance that the dice (iii) all

of these probabilities

is

different

unity,

fall

(i)

all

?

and

verify the truth of

this statement. [(ii)

Any

there are 6

particular throw of this sort, as five-five-six, occurs in 3 cases, 5 throws of this kind.]

and

.

7.

What

8.

In a single cast with four dice what (ii) exactly 12, (iii) less than ^2 ?

is

the chance of throwing 10 in a single cast with three dice is

?

the chance of throwing

(i)

two

doublets, 9.

How many

times must a pair of dice be thrown that may occur at least once ?

it

may

be at least an

even chance that double six 10. If four cards are

king, queen,

knave

A letter is

11.

out of

i

are ace,

'

'

chosen at random out of assinine and one is chosen at random Show that the chance that the same letter is chosen on both

assassin.'

occasions

is |.

[The number

of possible cases

12. If the letters of

that

drawn from a pack, what is the chance that they same or of different suits) ?

(of the

(i) all

the

'

'

attempt

are together,

/'s

is

(ii)

7

.

Of these 14

8.

are written

no two

's

are favourable.]

down

at random, find the chance are together.

13. A party of n men of whom A, B are two, form single rank. What is the chance that (i) A B are next one another, (ii) that exactly m men are between them, (iii) that not more than m men are between them ? 9

m

balls arc drawn simultaneously and 14. From a bag containing a balls, then n balls are drawn. Show that the chance that exactly r balls replaced are common to the two drawings is 'C'^~/( '^. Prove also that this ex;

m

and n. by interchanging a the second There are C n possible cases. The favourable [Consider drawing. cases consist of selections of r balls out of the already drawn, with selections n-r out of not already drawn.] pression

is

inialtered

m

a-m

B.C. A.

INDEPENDENT EVENTS

524

15. A squad of 4w men form fours. are next one another in the same four. (16.}

two squares are chosen at random on a chess board, the chance that they have a side common is y\-

If

(i)

show that

;

the chance that they have contact at a corner

(ii)

[(i)

Find the chance that two specified men

Out of 64 x 63

possible cases, the

number

is

yj^.

of favourable cases

is

4.2 + 24.3 + 36.4.] 17. One bag contains a white balls and 6 black balls another contains a' white and b' black balls. Let p be the chance of drawing a white ball from a bag chosen at random, and let^/ be the chance of drawing a white ball from a bag containing (a + a') white f and (b + b ) black balls. r Find p, p', and show that p ~", p according as a-\-b ~a' ~b' and ab' a'b ;

have the same or opposite

signs.

18. If v4's chance of winning a single game against winning (i) two games (at least) out of the first three out of the first six.

B

f find /Ts chance of four games (at least)

is

,

(ii)

;

the card is 19. A draws a card from a pack of n cards marked 1, 2, 3, ... n replaced in the pack and then B draws a card. Find the chance that A draws (i) the same card as B (ii) a higher card than B (iii) a lower card than B. Verify that the sum of these chances is unity. ;

;

[For

n 112 + - + ...+

1

-

chance

(ii),

;

{--

n ^n

n

1.1 } J

n

bag contains n cards marked 1, 2, 3, ... n. If A is to draw all the cards and is to receive 1 shilling for every card which comes out in its proper order, prove that his expectation is 1 shilling. 20.

A

in succession

[The expectation arising from each ball

~ of a shilling, etc.

is

The

fact that

Ti

the probabilities, on which the several expectations depend, are not independent does not affect the argument.] 21.

A bag contains n tickets numbered

at once, and

drawn. 22. (ii)

to receive a

What

What

at least

[For

is

(ii),

is

is

his expectation

1, 2, 3, ... n.

A person draws two tickets

of shillings equal to the product of the

number

p3

pz< (i)

,

numbers

?

the chance that a hand of five cards contains

two aces

23. If p l9 chance that

number

(i)

exactly two aces

;

?

of favourable cases

8

-C\ C

-\-C\

.

.

6*2

+48.]

pi are the probabilities of four independent events, find the (ii) at least

two and not more than two of the events happen

;

two of the events happen. 24. Two persons each make a single throw with a pair of chance that the throws are equal ? [If

and

(x

+ x2 +

this is

. . .

+ x6 2 - a 2x 2 + ...+ a 12x 12 )

dice.

then the chance = Q A~2 ( a * 2 +

,

t>D

equal to

{coefficient of

a;

10

in

What

f\-

(

\

JL

X*

4 \

2

-

x

j

/

}/36

.]

is

the

+ an2

)

:

GAMES OF SKILL

525

25. The decimal part of the logarithms of two numbers taken at random are found to seven places. What is the chance that the second can be subtracted from the first without borrowing ? '

'

26. In a game of whist, find the chance that each of the four players should have an honour :

(i) if

the card turned

(ii) if this

is

up

is

an honour

not an honour

before the last card

(iii)

27.

card

;

;

turned up.

is

In a game of whist, find the chance that each party has two honours the card turned up is an honour (ii) if this card is not an honour (i) if

:

;

;

before the last card dealt

(iii)

g.

is

turned up.

Out of n persons

sitting at a round table, three, A, B, C, are chosen at Prove that the chance that no two of these three are sitting next one

random. another

(n-)(n ~

is

5)/(n ~l)(n-2). For 2 of t^em, C can [Relative to A, 13 can sit in ri-~3 places. for n - 5 of them, C can sit in n - 6 places, Hence, Ibhe places favourable cases = 2(n - 5} + (n - 5) (n - 6).]

sit in

n-5

number

;

of

29. Three cards are drawn from a bag containing n cards marked 1, 2, 3, ... n. Find the chance that (i) the three cards form a sequence (ii) they contain a (iii) they involve no sequence. sequence of two ;

;

between A and 7J, the chance that the server wins a game always p, the chance that he loses is q. The score is five all, they proceed to play deuce and vantage games and A has the service. Denoting /Ts chance of 30. In a set of tennis

is

2 winning the set by x, show that x=pq + (p + q*)x, whence and B have equal chances of winning the set.

31.

it

follows that

A

A and B

before

B

chance

is

play with two dice on the condition that A wins if he throws 6 throws 7 and B wins if he throws 7 before A throws 6. Show that A's to J5's chance as 30 31. (Huyghens.) :

Three players A B, C of equal skill engage in a match consisting of a games in which .-1 plays 13, 13 plays O, C plays A A plays B, and so on, in rotation. The player who first wins two consecutive games wins the match. Prove that the chances which A B, C respectively have of winning the match are as 12 20 17. [First show that if ^4 loses the first game, his chance of winning the match is ^ if he wins the first game, his chance is JJ.] ^2

,

f

series of

,

,

:

:

:

A

33.

white. (i) (ii)

34.

bag contains 3

A

ball is

What What

is

is

balls,

drawn and

and

is

it is

equally likely that

1,

2 or

all

of

them are

found to be white.

the chance that this

is the only white ball ? the chance that another drawing will give a white ball

The same question

as the last, except that each ball

is as.

?

likely as

not to

be white.

[Proceeding as in Ex. 3 expansion of (i + i)

.

1,

p. 520,

The

we

see that

rest proceeds as in

P P P ,

Ex.

l9

2

^3 are ^ ne terms in the

2, p. 520.]

GEOMETRICAL PROBLEMS

526

A

bag contains 5 balls, and it is not known how many of these are white. balls are drawn, and these are white. What is the chance that all are white ?

35.

Two

36. The chances that A , B, C speak the truth are respectively p, p', p". are the odds in favour of an event actually having happened which (i) all (ii)

A,

three assert to have happened ? assert to have happened and C denies

B

What

?

Geometrical

The sides of a rectangle are chosen at random, each than a given length a, all such lengths being equally Show that the chance that the diagonal is less than likely. a is 7T/4. [Draw a square, OACB, whose side is a and the arc AB. If is one of the rectangles, the favourable cases are 37.

less

;

ONPM

when 38.

P lies in the quadrant GAB.] A floor is paved with rectangular

a and breadth the

A

b.

Show

floor.

bricks, each of length circular disc of diameter c is thrown on

that

the

chance

that

it

falls

entirely

on

one

brick

is

(a-c)(b-c)lab. 39. A point P is taken at random in a straight line AB. Show that the chance that the greater of the parts AP, PB is at least k times the smaller is 2/(k+l).

40. If two points are taken at random on the circumference of a circle, the chance that their distance apart is greater than the radius of the circle is -|. [The unfavourable cases are represented by the thick arc in Fig. 89.]

FIG. 89.

random on the circumference of a the chance that they do not lie on the same semicircle is J. [Choose P. Choose Q on one semicircle on PP' E must lie on P'Q'. (Fig. 90.)

41. If three points, P, Q, R, are taken at circle,

;

If

PQs,

number

of favourable cases-

total

number

ic

.

c,

etc.]

Jo

42.

On

AB of length a + b, two segments PQ P'Q', of lengths measured at random, (i) If a>6, the chance that P'Q'

a straight line

a, 6 respectively, are

9

PQ is (a-b)la. (ii) If c is less than. a or of PQ, P'Q' is less than c is c*/a&. part Let AP==x. If [(ii) <2=c, ^g=rz, and P'Q' overlaps PQ towards B, P' must lie in LM. Hence,

lies

entirely within

the

common

the

number of favourable

cases

=

I

(c

- x)dx =-|c 2

b,

the chance that

kP

L

c

M

Q

FIG. 91.

;

Jo total

number

of favourable cases

c

2 .

Total

number

of cases

ab,

etc.]

CHAPTER XXXIII CONTINUED FRACTIONS

(2)

EXPRESSION OP A QUADRATIC SURD AS A SIMPLE CONTINUED FRACTION

Surds of the Form

1.

(JNb

l

)/r v

follows that

From

any simple recurring continued fraction surd. In other words, its value is of the form (JN are positive integers, except that 6 t

may

is

Oh.

XXIV,

33, it

a quadratic equal b^)/r v where N, b l9 r } to

N

be zero, and

is

not a perfect

square.

We

shall

prove that conversely any positive number of

UNb

1

)/r 1

can be expressed as a simple recurring continued fraction. In considering this theorem, it is to be observed that There

(i)

is

no

loss

the form.

of generality in assuming that

:

N - b^

is divisible

by r v

For

(JNb

1 )/r l

= (jNr

1

*b r

1 l )/r 1

*

and Nrf-farJ*

is

divisible

by

2

hence (^/N b 1 )/r 1 can always be replaced by an expression of the r^ same form for which the above condition holds. ,

We

(ii)

For

if

need only consider surds which are greater than unity. 2 where r2 is a positive integer, then ^V-^i 2

=^

It follows that every positive quadratic surd or its reciprocal is of ,four types considered in the next article.

2. (i) (ii)

(A)

Types of Quadratic Surds. The surd in question

N ~b^

is divisible

The type

it is

assumed that

greater than unity.

is

by

In every case

one of

rv

(JN + bJ/Tt

where

b l *
This

will

normal type, and is of special importance in the theory forms \/AjB and JN.

:

be it

called

the

includes the

SURDS OF NORMAL TYPE

528

To express

(^N + b^)/^

as a simple continued fraction,

we form

the

equations -

where a a

is

-a

_n f

,

1

U-1

W"i

the integral part of (^/N -i-b l )/r

shall first

Since

^

is

show that

62

are positive integers. + b l )/r l

r2

,

and

li

2 r I r 2 = N-b^ ......................... (B)

kj^'Vi-fri*

We

,

the integral part of (*JN

,

&!?!<*/ N + b^af.L +

and therefore

b2

we suppose that

If

62


<0,

it

2 4-

........................... (C)

r!,

r x .................................. (D)

follows that

N/A

r


1

,

but

61

< N/^

r >

and consequently 6 1
therefore 6 1
Again,

62

is

.

a positive

and consequently

r 2 is positive.

N-b2 and

since

N -b^

positive integer.

for

n = 2,

3,...

2

number

*

is

= N- (a

x

fj

divisible

less

than

Further,

is

N

62 2 >0,

we have

- ^) 2 = N-b*-r l (afa - 2a 1 61 ), by

rp

N -b

2

Hence we form the equations

so also

Continuing the process,

where a n

therefore

^/-ZV,

is

2

.

r2

is

a

an integer such that (F) .................... (G)

leading to

steps similar to the above, we can show that every a, 6, r is a positive For taking n = 2, by the preceding (*JN + b 2 )/r 2 >l, therefore a 2 integer.

By

is

it

N

- b^ is divisible a positive integer. Also by r 2 Reasoning as before, will be seen that 6 3 andr3 are positive integers, and so for n = 3, 4, 5, ... .

.

Finally, the fraction is periodic, for the

and

for every n, b n


y

therefore

rn

nth complete quotient

is

= (b n + b n+l )/a n <2JN.

Hence the fraction (JN + b n )/r n cannot have more than 22V distinct values, and one of the complete quotients must occur again. From this the stage, all the succeeding a's recur and in the same order for the fe's and r's. Thus the continued fraction is periodic. :

same

is

true

REDUCTION TO NORMAL TYPE

529

It will now be shown that in the process of expressing a positive quadratic surd of any other type as a simple continued fraction, a stage must occur at which the complete quotient is a surd of normal type. Whence it ivill follow that in every case the continued fraction is periodic. 2

The lype (*JN -ftj)'?*! where quotient is a surd of normal type.

/>

(B)

-

N/57-7

T.

X or instance,

/

r

~~p

The form

_

/>/57 + 7

,

_

__

b

i-t

.

where a x

a2

,

1

1

1

%

-1

form the equations

rz

__.,

are the integral parts of the fractions on the left and,

...

,

.1

--i

>N. We 0l+

1

3+l-t-l+l+l4-7

4

/

2

1

1

2

+ JN)/ri where

(bi

Here the second complete

.

/

s/57-7

2

(C)

T


=2- --- =

/

= -/ 1

dim and

1

for every n,

"

"w-f-l

As

in (A), every r

hypothesis

}

~~

n

a n rn>

r n r n+ 1

~ ^n 4-1 ~

an integer and every b

is

an integer or

is

zero,

and by

> S/JV.

b^

Suppose that are

*

6 15 6 2

,

6r

...

b l9

all positive, arid

b2

,

are

,

...

all

6 n41

^N

then r ly r 2 ... r n greater than is a decreasing sequence of integers. ,

;

Hence a stage must occur at which This being

and therefore Hence n is since also is

so,

n cannot

b n+l

odd,

JN

-

be even

;

for in that case

we should have

>JN. and

bn f t

+,JN>a n rn

bn

> 0,

it

x /-ZV

+ 6 n+l >0; and

Ar ~^ 4 >0, and

therefore r n+1

Consequently

.

follows that

.

1

negative.

Now

(n

+

being even, the

1)

(n

+ l)th complete

quotient

may

be

written

and

since

7

^w+i^^

according as 6 W a. 1 (D) The type

quotient

is

(b l

?

^i s

or

^^ ie

next quotient

is

a surd of normal type

0.

~JN)/r 1 where

a surd of type

(C).

6 1 2 >^V.

Here the second complete

MENTAL PROCESS OF RECKONING

530 3.

Theorem.

Let the surd

ing fraction, then

if

(i)

(JN + b)/r

b<>JN
be expressed as a simple recurr-

has no acyclic part;

the fraction

if ,JN>b + r, it has an acyclic part consisting of a single quotient; b > JN, it has an acyclic part of one or more quotients. (iii) if For if x = (JN + b)/r then (rx-b) 2 = N and the second root of this - /iV + hence, from Chap. XXIV, 33, it follows that quadratic is &)/r + (i) t/ -1<( N/2V 6)/r<0, the fraction has no acyclic part and these (ii)

,

(

;

conditions are equivalent to 6< N/2V<6-hr; (ii) i/" ( - N/A + 6)/r< - 1, that is, if and 6 -f r, the acyclic part consists of a single quotient that if contains one + the is, (iii) if b)fr>0, acyclic part b>JN, r

JN>

;

(~JN

or more quotients.

In particular, if

v A/B has a

A>B,

simple continued fraction equivalent

the

to

single non-recurring quotient.

4. Method of Reckoning. In practice we replace the written work involving surds by an easy mental process, as in the next example. Ex.

I.

Express (\/37

-f

as a simple continued fraction.

8)/9

Here a i integral part of the surd = l, also are now found in succession from the equations fe

n

=

fl

-&-n

n.-i r n-i

giving the table

rn

= N~ (

W

r n-i>

^ n

if

continued,

part of

a repetition of the part between the dotted

is

s/37+8 " 9

+

1111

lines,

and

i+I+3V2+"" Jf

The

= integral

various quantities

:

The reckoning,

NOTE.

^=9. The

8,

-X-

third complete quotient, (\/37+3)/7, marks the beginning of the 3<
recurring period, for

chapter that, from

5.

this stage, the above reckoning

can be replaced by a G.C.M. process.

>/A/B as a Continued Fraction.

(1) It is

supposed that A,

have A/ -= = \ />

-5 jD

=

l ,

B

are positive integers

where

N = AB,

6a

and that

= 0,

rl

= B.

A>B. We The surd

is

7*j

therefore of the type discussed in Art. 3, and it is expressed as a simple continued fraction by constructing equations of the type

,

for values of

n = l,

2, 3,

...

;

CYCLE OF QUOTIENTS where a n

is

an integer such that a n

an(i

b n+l

leading to

As

< (JN + bn )frn < an +

1,

= =

A/"S r is a positive integer and the continued Also, it has been shown in Art. 3 that a 1 is the only

in Art. 2, every a, 6

fraction

631

is

periodic.

and

non-recurring quotient. (2) is

The following

greater than

inequalities are required.

JN + bn >rn

therefore

1,

Every complete quotient (A)

.

= (# - bn*)/rn = (JN - b n (JN -f b n )/r n >JN- b w therefore ,/#<& + rnHl Again, if n>l, the continued fraction equivalent to (*JN + b n )/rn no acyclic part, therefore, as in Art. 3, 6n <\/^ <^n + rn n ^2) Also

r n _!

)

(B)

has

r

(

For any

suffix

w, 6m

(3)

hence from (B) and

bm <*JN,

-6 n
and

111

/-^

*

Now JNfr^-a^ is a root equation being V#Ai ~

Therefore by Ch.

Hence

ac

,

K -f

-

a 2"^" a c-fl

and therefore It has

recurring fraction in

+

from

2a l

;

(iii)

the beginning

etc.

for the rest of the

and end are

B

rl

l

_L_JL a 2 -i- a 3 -I-

"

*

_L

_J

a3

a2 4- 2a x

-f-

*

and we

,

cycle, the

equal.

be expressed by writing

M_N/A[ \

a3

<._!==

been shown that \I^AjB can be expressed as a simple (i) there is a single non-recurring element, a l ;

partial quotients equidistant

may

31,

which

the last partial quotient of the cycle is

All this

XXIV,

J

= o2a t -f

. . .

2a l9 a c = a%,

a c+l

Summary. (ii)

1

...

^c+

the other root of this

111

l

I

TT

*

r^ (x + a^^N,

of

i-

(D)

Let c be the number of elements in the

The Cycle of Quotients.

cycle, then

(C),

-6n <^n (n^2)

6m

(C)

shall call the sequence, a 2

cycle of quotients.

!_ +

*

"

'

* ,

a3

,

...

a 3 a 2 the reciprocal part of the ,

,

CALCULATION OF QUOTIENTS

532 (4)

The b and

Let

r Cycles.

cycle, so that

be the number of elements in the

c

jjf

\

j

*

*

The recurrence

due to the fact that the second complete quotient

is

From

appears again as the (c + 2) th. recur in the same order hence,

and therefore

bc + m

Hence, bc

63

,

r c +m

= 2a

complete quotients

ci

1,

rm

,

^^2.

f r

m = 0,

c _ m ^a 2+m for

1, 2, ...

,

we have

.

(i)

the cycle of

etc. ;

?'

=

c

The character is

Vs

6 2 , 63

is

,..

,

the cycle of r's is r v

(ii)

first term, i.e.

sequences

and

that a c+1

Kemembering

and so on

bm

this stage all the

(JN + b c+m )/r e+m = (JN 4- b m )/rm

;

2,

f c -i

of

the

/'

= r^

bc

r2

^ and is

,

i.e.

reciprocal,

and

rc ,

...

b c+l

= b2

,

is reciprocal after the

etc.

and the reciprocity of the three

recurrence

exhibited below, where the recurring periods are enclosed in

brackets.

Q

T

al

a2

&3

It should be noticed that the a

parts of the a

and

i

,

...

and

tt

,

a2

a

2

o

.

6 cycles correspond,

and the

reciprocal

r cycles correspond.

Calculation of the Quotients. In finding the values of 6 n r w a n teH w/ien middle the the b z/ of cycle has been reached, no further calculation is necessary. This matter is settled by the following theorem. (5)

,

,

,

we can

If b m r rn a m are respectively equal to are equidistant from the ends of the b cycle. ,

,

For

rn+i

and

:=

(6 n42

- bm _

the

l )/r

m

inequalities

__ l

less

rn ,

n+1 r n+1

Subtracting and using the hypothesis,

Now by

,

an

,

and

b n+l

+ 2 )/r n+l

and

then b m

= (N-b$ + i)lr n = (N-b m *)!rm = rm _ r

^m-i rm-i =b m _, l + bm

Also

b n+}

in

.

.

1

find that

(D),

(2),

are less than unity.

we

= 6 w+2 + 6 n4

both

(6 rn _ 1

- bn

Therefore the right-hand side of (A)

is

than unity, and consequently a w _i = a n+ i and bm _ l = b n+2

numerically We can show by a similar argument that 6 TO _ 2 tively equal to 6 n+3

,

r n4 2 , .

n+2

,

and

so on.

,

^m _ 2

-

a m-2 are respec-

CONVERGENTS

m=n

Putting

m=n+ 1

and

533

we have

in succession,

the following

:

If two consecutive b's, namely bm and bm + v are equal, then rm and a m are the mid-terms of the reciprocal parts of the r and a cycles. In this case (i)

c is even.

= // r n rn+l and a n =^a n+l

(ii)

two consecutive

(so that

likewise the corresponding a's), then

rn ,

and a n

rn+l

r's are

equal and

an+1 are

,

the

mid-

terms of the reciprocal parts of the r and a cycles, and bn+l is the mid-term of

In

the b cycle.

Ex.

Express ^/^i ana V^l as simple continued fractions.

1.

= V.

(i)

this case c is odd.

......

we construct the

N

Henco

.

A/T^J-

following table

^ = 11,

187,

al

by a mental process

[11

13

r

[11

6

3

22

a

1

[4

8

1

6

~l

11

and proceeding

t

as in Art. 4,

:

11

4

8

2].

Because 6 4 = 6 5 11, therefore a 4 is the inid-term of the reciprocal part of the a cycle. This cycle can therefore be completed, for its last term is 2a x = 2, thus 17

V

111111

+

II"

4~+

8+ r+

8"+

IT 2* *

x-

(ii)

For

/s/61

r

[I

60 a7

we have

= l,

=7 and

the

6, r, a,

table

is

[1431221341

12

4

3

a 7 and r 6

Seeing that a e

r7 ,

reciprocal part of the a cycle.

Therefore

ax

[757546 rj_

v

5

9

5

we conclude

Also the last quotient of the cycle

-

---

-

J+

+

j-

2+ 1+

where

*

Relations connecting the p's and q's.

N = AB, VIN -

rl

=

^B- and J

let

.-

-}-

^ B t

o-f

^i

p r /qr be

11~

1 05,-

an +

?i

zq n

Suppose that

the rth convergent, then

t.

wHere

z

= *PnPn=i

therefore

.

J+

* (6)

is 14.

11111111111 _ 2+ 4+

,

14].

that a 6 , a 7 are the mid-terms of the

+ ? w -i

and Equating rational and irrational parts,

21

==

7 V/A + h W-Hl ~~~

r^,.

i

CONTINUED FRACTION FOR

534

Solving for b n+1 and rn+1 and putting that

N = AB

and

r^B, we

find

=(l)rw+1 ........................ (B)

Bp n *-Aq*

2 2 Integral Solutions of Bx -Ay = M. If for some value of n it n happens that (- l) r n+l = M, then, by equation (B), (j? n q n ) is a solution

(7)

,

of

Ex.

1.

Find two

Expressing ^/^f- as a continued fraction

= 5/4, lqi 6.

^N

and therefore

1^4/^4=46/37,

In

(

(5),

(5,4),

Ex.

the cycle

2

3.

17?/

1),

we

find that r3

r6

= 3.

Also

(46,37) are solutions.

as a Continued Fraction.

Here

(1)

rx

= l and

the

all

Additional theorems are the following.

conclusions of the last article hold. (i)

-

z positive integral solutions of 11 x

o/Vs, one member and only one is equal to 1, namely r, r m = l, then, remembering that 6 2 = a 1 and using the

For suppose that

_r

ar

equations

,

+y

we have a m = integral part Also o

= &H-&,

Therefore

beginning of a (ii)

//

r n >2,

cycle of

^n

- 1V /V ~

(JN + 6J/r TO - % + &m = 6 2 + b m

of

/^H-6 m+ i)/ m ^i =

new

mrw+l

.

therefore

i

(N

f

--J

(

and

N /^-r62)/r 2>

marks

^tn^l

the

r's.

a n ^b n

.

For

in this case

n>l

and

so that

=

// r n 2, JAe number (c) of elements in the cycle belonging even. Also a n is the mid-term of the reciprocal part of the a a n = a x or a l - 1. (iii)

'

For a x Also

an

is is

cycle,

the integral part of 6 n therefore the integral part of (^i + b n )jr n and so if r n = 2 fl

.

,

therefore 6 n = 6 w+1

is

and

a^bn

,

,

a w >(a 1 H-6 )/2-l>6 n -l>6n But by the preceding, a n <6 n therefore a n = bn

Hence by

JN

to

3

.

we have

.

Also

2a n = 6n

+ 6n+1

,

.

even and a n is the mid-term of the reciprocal Finally a n is the integral part of (^i + n )/2, so that

(5), (i), c is

part of the a cycle. a n = (a 1 + a n )/2 or (a l ^ a n - 1)/2, and therefore a n ^a l or a^-l. (2)

Equations (A) and (B) at the top of the page become

^?n?n-l-^n^n-l = (~l)

p n *-Nq n

*

W

&4-l

....................... (A)

=(-l)rn+1 ......................... (B)

CALCULATION OF CONVERGENTS (3)

x2 -Ny 2 = M.

Integral Solutions of

(-l) n+1 = M, nr

found that

In particular,

if

c is

If,

for

535

some value

of w, it is

then (p n q n ) is a solution of x -Ny* = M. the number of elements in the cycle belonging to 2

,

the necessary and sufficient condition that r n+l may be equal to 1 that n = tc where = 1, 2, 3, ... hence, it follows that

JN, is

;

x -Ny = l,

(i) for the equation and- (PM C ?2fc) when c >

2

is

(ii)/or the equation x* c is

odd

2

odd

Find a

Ex.

1.

We

find that

Ny =

c

= 5.

>

18,

6

x2 -

2

13^/

-

1.

* q5

~ 5 and (18,5)

is

a solution.

The following formulae are

Calculation of Convergents.

(4)

we multiply equations (A), (B) firstly by and secondly by p n and p n _i> we obtain by addition

this connection.

q n -i>

even

when

solutions are (p(2*-i)c> ?(2*-i)c) solution when c is even.

positive integral solution of

Also

c is

I,

* so that

when

;

2

the method giving no

;

solutions are (p tc yjtc)

If

useful in

and

qn

Again, changing n into n 1 in equation (B) and using equation (A), find that p n ~i = r n q n ~b n+l q n and Nq n _ l ^rn p n ~-b n ^p n _ l ..... (D)

we

^

In general, the p's are considerably larger than the
The Convergent p c /q c

(5)

elements in the cycle belonging

.

(i)

11~ ?c

f

i

where

~

m + n = c,

where c

is the

number of


to

.

For

//

_..._ a

2+

111

m+ f

-- "

~

q n+l

therefore If

^

is

the numerator and

Y

the denominator of the last fraction,

m Xq m -Ypm = (-I)-iqn and Xft^- Yp m .^(- l) qn+l Hence any common factor of X and Y is a factor of q n and of

*

q n+l

,

and since these numbers are prime to one another, so also are X and Y. Thus the fractions pjqc and X/Y are in their lowest terms, and therefore pc = X and qc = Y, which are the results in question.

536

HALF-CYCLES In particular, if c

(ii)

odd and n = |(c -

is

l>c=P*!7n+;W5Wi // c

is

(iii)

//

even and n

m

w

equivalent

to the

m

-qm p c -(p m qm ^-q m p m ^)q n -=(-l)q , r

^

p m -Nq m ^(-lY"rm V-AV-f-lJ'VHi; mtn = c, we have r^^^r^^; and therefore ~ N
-Pn

remains to consider the sign of p mp c -

The convergent

m

.............. (H)

)

(PmPc It

),

2 2 %n?c) - # (PmVc ~ VmPc? = (?> w ~ ^?m2 (^ 2 ~ W)2

since

and,

~

(P*-JNqm ){p e + JNq e

two equations

For using equations (E), p m q c Also we have the identity

Now

= ? n 2 + ?JM-i ................. (F)

n = c,

-f

(PmPc

?c

\c,

Pn + JNqn = (-I) icA

and

1),

pjq^

precedes

pjq c

.

2 -

Nqm q c

.

First suppose that c

is

even.

and p m p c >Nq n q e If m is odd, it follows from Ch. XXIV, 25, that p m P c <^r Tlie samc results follow q mq C w in a similar way when c is odd, and thus pm p c = I) 7> n Nq m q c If

is

even,

JN

.

c

'

.

(

It should be noticed that these results hold if c is replaced by (iv)

N is a prime and c is even and n = ^c

//

>

tc.

then

Pc + JNq = l(p n +JNq n )*, ........................... (J) 2 2 and q c - p n q n p c = (p n + Nq n e

which

is

Also

a cycle

equivalent 2

jt;

n

-

is a^

to

-J-

= 2Vyn or a, - 1. (

For by equations

(I),

therefore /T) 2

and

M

2

_ JV0

2

2

1)

.

2,

.

)

and

the

mid-term of the reciprocal part of the

p np c - Nq nq e = - l)p n and p n q c ~ q n p c = - l) n qn = (Pn* + ......................... (K) Pc/Vc n
/-n

V

21 NqM>

(2s_^L) ^(A) -A^-,; V 2p w 7 n / ?c 2

,

(

hence

^2 * ~ Nq " \7

(

pw ?n

L

9

2\

(

-1)* ~, ?c

the sign being determined by the fact that p n <>JNq n according as n is even or odd. Since is a and is to prime pn prime q n the last fraction is in its lowest terms. Hence p n 2 - Nq n 2 = ( - l) n . 2 or ( - l) n and the >

N

,

,

latter alternative

is

impossible.

MULTIPLE-CYCLES * ^ p n - Nq n = -

Therefore

Vfi^

Hence, NOTE. and noting

(

When N that p c /q c

Thus we

and the

;

in

is

find that

its

>/21 ^

and

2

.

rest follows

p c and

not a prime,

is

n l)

537

q c =p n qn

by

(1), (iii), p.

may

be

qc

c

= 6,

= 4 + -i- -i-

L

*

_L I

1+1+2+1+1+8

9,

713

o

C^Q

(6)

and

.

n o y ^

The Convergent

c is the

number of

quotient

is

j9 6

_r t>5,

g6

= 12.

V Pn/q n the n-th convergent of
is

(i)

quotients in the cycle, then

Pn+tc =PnPtc

For the

-

anc* therefore

fo iw

.

and

2,

#3

,

*

p ~ 81+21.4 = 55 2/

found by using equation (K)

lowest terms.

# so that for \/21,

534.

+ N(lnqtc

(c + l)th quotient a x + N/A therefore

is

and 2a x

q n + tc

=p n q + q np tc

(L)

tc

and the corresponding complete

r

,

and equating rational and

fractions

Removing Again,

11

1_

irrational parts,

^_ %

the last quotient being the

(te

we have

From

+ l)th.

this,

H- _p n

by means

?n

of equations

(I),

Pn+tc

If JC is

(

a l + Pn/
the numerator and

Y

X-N Y= Pn Ptl?-Nq

PnPtc

+ Afyngtc

the denominator of the last fraction,

and

PtcY -q teX = q n ( Pt

-Nq *). Now Ptc 2 -^9tc 2=: il ^7 (5)> (iii), therefore any common factor of -X and y divides p n and y n which are prime to one another. Hence Xj Y is in its lowest terms so also is p n+t c/
qtc

(

t

*)

t

,

:

tc

which are the equations in question.

As a

special case

we have

P(s+t)c=PscPtc

in particular,

+ N qsc q tc 2

P 2 tc-Ptc + Nq tc

and 2

and

q( S + t )c=Psc
q^ ic

+ y8cPtc'>

......... (

N

)

^^p tc q tc ..................... (0)

ABBREVIATED CALCULATION

538 (ii)

JN

Since

is

irrational, equations (L) are equivalent to the single

relation

Putting

t

= 1 and n = c,

In Ex.

Find a 1, p.

t,

)

533,

solution in positive integers of x*

^61

- Sly 2 = 1.

Hence the smallest solution

The mid-elements of the cycle are the 5th and 58. Hence by equations (F),

(p 22 , q 22 ).

;

) (p c + JNq c y ............. (Q)

expressed as a continued fraction, the number of quotients of the equatiort in question is

is

in the cycle being 11.

2? 6

+ V-%c = (Pc + J^Vc) 3 -, .......... (P) p tc + JNq tc (p c + JNqc P3C

p n+tc + JNqn+tc = (p n + JNqn

and therefore

find that

(

generally, for every positive

Ex. L*

,

we

in succession,

...

+ >/#fee = (Pc + V^c) 2

Pte

and

2c, 3c,

6th,

and p 6 ~164, q 6 ~2l,

^453, # e

Pu -Ms +Pe?6 = 164

.

+ 453 58 - 29718,

21

qn

.

Using equations (0) and observing that p n

*

= - 1, q 22 = 2p n qu = 226153980.

Pn ^Pn + 61 ^u = 2^ii + 1 = 1766319049, 2

2

Thus the 7.

where

2

least solution is

(1766319049, 226153980).

The Cycle belonging is

r'

= q* + g c2 - 21 2 + 58 2 = 3805.

- 61g n a

to z

=

(

N/N

+

Suppose that

b)/r.

a positive integer, then the simple continued fraction correno acyclic part (Art. 3), and the cycle can be calculated

sponding to z has as follows.

Rule.

Let

(x> y)

be a solution in positive integers of

x

equations

2

-Ny*=

the latter pair being used only

when

or

1

r,

r'

4,

any one of

the

...........................

(A)

are both even and N,

are odd.

b

x/y for .JN in the given surd. Express the result as a simple continued fraction with an even or an odd number of quotients according as Substitute

on the right of the chosen equation is + or or more of the cycles belonging to (JN + b)/r. one of the sign

Since

Proof.

N =6

2

-f-rr',

the surd rs 2

(1)

We

shall

such that

where y *

may

the positive root of

......................... ....... (B) ,

f

y

pq'-p'q=l

For equation if

is

This fraction consists

prove that this equation can be written in the form

z~(pz f p')/(qz + q'), where p

with (B)

(JN + b)/r

-2fe-r' =

.

(C) is

the same as qz 2

= q ry,

q p', q are positive integers y

and (p

p-q' = 2by,

......

(C)

q'^q .......................... (D) -

q')z

p'

= r'y, p'

= 0, which

is

identical

........................ (E)

be any rational.

Thi8 problem was set by Fermat as a challenge to the English mathematicians of his time.

GENERAL RULE From

we have -Kp + q')

(D), (E)

where x

^Ny

2

I

=,r

2 (A) x, y are positive integers such that x

(B) x

/2,

539 2

........................... (F)

,

All possible cases are included in the following

rational.

is

2

y = 77/2 where

,

77

are

2 ~-Ny =

1.

odd integers such that

2

- A7 2 -= ??

N

'

In case (B), divisible

by

4,

r

r,

and

f

are both even, for T

if

A

is

odd

so also

In both cases, from (E) and

(F),

we

p = x + by, and

p/q

It remains to be

(i)

2 // x

(ii)

Ny

2

iind that

=x-b\j .............................. (G)

q'

This will be so

if

x/y^.b +

r.

:

the least value of y

,

therefore

N**:(b

x/y^\/N -i-l^b + r where

~/2, y^rj/2

//

is

- Afy 2 = - 4, then or xly<^JN
Now N<(b + r) 2

(iii)

b2

*

1

^=I,

Hence

r'y.

= (x/y -f b)/r ................................. (H)

Three cases must be considered

- Ny 2 = ~ // x

p'

ry,

J-/1.

is 6,

shown that q'^q.

2

g=

:

z

-Nr)

and

is 1,

+ r) 2 and

2

--^i,

1.

Hence

(f^q* the least value of

77

is

land

xA/
Hence (2) It

r

and

x/7/
follows that

Up

q'^q.

11

-=a x + -----

..-

an

2 -f

(7

............................... ( J )

,

even or odd according as pq -p'q = the convergent immediately preceding p/q. (Ch. be determined by z = al + ... -7 a n -f z 2 +

where n

f

is

-111

l

-1, then

or

XXIV,

12.)

is

p'/q'

Now

f

let

z

,

'

then

z

= pz + P' ? 2/

Hence belonging

2;'=^

and

+ 2'

and

^jtP'^^tP', '

J2

'

+ ?'

2 21

the fraction (I) consists of

+ ?'

one or more of the cycles

to z.

The rule can be used (as in Ex. 2) to express any quadratic surd as a continued fraction, beginning at the stage where recurrence commences.

2M

B- C

-

A

EXAMPLES OF USE OF RULE

540 Ex.

Express (*JQ2l +21)/18 as a simple continued fraction.

1.

21< s/621<21 +

Since

there

18,

no acyclic part.

is

Also

18r'

= 621 -21 2 = 180,

r' =. 10.

giving

Thus

r,

r'

are even

and

6 is odd, also

25 2 -621.

I

2

^4.

Using the rule, we substitute 25/1 for ^/621 in the given surd, and express the result as a simple continued fraction with an even number of quotients, thus

+ 21

25

23

1

_9 +

1

1

~~T8~~""" r+ r+ s/621 ^ -

and

+21 -

18

2.

1 -

-

v 1+1+4

* Ex.

1

1

= 2n +

4' .

#

Express (^13-+ 7)/4 as a simple continued fraction.

Since 7> N /13, there is an acyclic part. The reckoning on the right shows that the second complete quotient is (/s/13 + l)/3, and since l<*yi3
3

Since

r --3,

-13.1

2

^-4

2

first

+

l)/3,

jy/13

fraction with an odd

number

in (;s/13

18

\1._

5/3

is

I

]

23 j^

_

5

.

and express the

~

, 1

15

-

2

-

7,

1

4,

3

a

2,

1

1.

Taking the second, we

useless.

is

equation

of quotients, thus

___L This

and

an odd number, the

substitute 18/5 for

18 - 13 2

b r

result as a simple continued

1111 ___

___-

-j

1

+

-

1

+

6

-

+

1*

the cycle belonging to the given surd, and s/13

+

7

"

L JL _L

+

4

1

|

1

_L IT IT 6+

I'

EXERCISE LVI 1. Verify some of the following results, where the values of as far as the middle of the first cycle

6, r,

a are given

:

fb. 0.3. 1.2 (i)

s/13

I

,

(Hi)

1

.

4

.

3

.

3

a 3

.

1

.

1

.

1

r

-j

.

.

Cb. 0.4. 2. 3. 3 (ii)

I

0.5. 1.4. 5.

Cb

.

r

.

1

.

6

.

5

.

3

.

2

.

3

I

a

.

5

.

1

.

1

.

3

.

6

.

3

f

6

^31

v/19 \ r

.

(v) s/91 \ r

.

I

.

[ a

.

9

.

^43^

(iv)

[

9.1.8. 7.7 10

.

9

.

3

.

14

.

3

1.1.5. 1.5

8 7.5.9. 7 8 9.5. 12.7.4. 15.3. {ba. 10. 2.3. 1.2.4. 1.6. .

r

.

.

1.

10

.

.

.

.

3

.

5

.

2

.

5

.

2

.

1

.

3

.

1

T6.0. 6. 1.5. 4. 5.

5

j

.

I

.

a 4

.

.

10 3

6

5

r

.

1

.

7

.

6

.

3

.

9

.

2

.

9

a

.

6

.

1

.

1

.

3

.

1

.

5

.

1

GENERAL CASES r

N/^Tl

(vii)

__

VaM- 2

(ix)

J

6

.

r

.

.

o

.

a

.

1

.

1

1

541

O.a-l.a-1

("6. \l~tf~- 1 J

(viii)

r

1

.

2(a -

.

l.a.2rt.2a

[a.a-1.

fb.Q.a.a

f

J I

r

a

.

1

.

2

.

1

.

a

.

a

.

2a

Va 2 - 2

(x)

-|

I

b

.

r

.

a

.

1)

1

.

1

.

2(a-~l)

.a-l.a-2.a-2 I

a-

1

.

2a - 3

.

.

1

.

2 a -2

.

.

2a - 3 1

(a>2) 6

.

r

.

r

Va(a4 1H

(xi)

1

.

a

.

a

.

a

.

1

Va^Ht

(xii)

la.a.2.2a

(xiii)

*Ja(a

4 4)

r 6

.

r

.

.

1

.

la.a +

1.

^

p.O. r

-I

1

.

a

.a -2. 2

4

.

.

a

La.a.i(a-l). (a> 1 and odd)

a41 2a - 1

.

a-2 4

.

1

.

-l(a

-

1)

a

-1

a

.

a

.

.

a

.

.2

.

1

(a> 1 and odd)

4

(-!)

.

In certain cases, a solution of one of the equations x- can be written down at once thus

Ny 2 ~-

2,

1

or

i4

;

if

A = a~

if

A =a

a2 -

T

if if

Apply

3,

this to write

2

,

then

J- 2,

then

1

A7 = a (a 4- 1 A ^ a i 4, 2

down

the

then

),

2

(a

(2a

: j- 1 )

-f 1

2

2

-

)

N -

1

A

.

-

2

T

Aa

2

22

T1 1

;

-= 1

;

;

a --A.I ~ T 4. then -A 1 when integral solutions of x N = 5l, 47, 56. 2

2

T

2

2

?/

Find a solution of x 2 -109?/ 2

-4, and apply the H.C.F. process to show

that

+7 __. ^

v/109

*-

A

JL J_ JL _JL Y^ 2+ T+ 94 14

^-

4,

Obtain the following

* *

2

*

:

I4l4l4l42

;

^

1+

6

84 24 14 34

2

;

*

/17_ = "9 +

V3 5.

111111 24 F+ ff IT 24*4*

Find two positive integral solutions of 2 2 2 ^ 2 -7, (i) 5^ ~13t/ ^7, (ii) 5x -13i/

(iii)

3x 2 -

2

17i/

^7.

Find positive integral solutions of x 1 ~~ 621?/ 2 ~4 and of a; 2 621z/ 2 1. Referring to Art. 7, verify that if 623/25 and 7775/312 are substituted for v/621 in the fraction ( N /62l421)/18, the first gives two cycles and the second three 6.

cycles of the fraction.

SUNDRY THEOREMS

542 If A=:(2ra

7.

+ 3) -4, prove 2

-l

_

~~-

that

JYI

111__

__

-I

___ m-f 1+ 4(w-f-l)+

2+

* If a

8,

'

1

#

odd, find a positive integral solution of

is

x 2 -(a 2 + 4)y 2 = a;

(i)

(ii)

x 2 -(a 2 + 4)*/ 2 -^ -a.

9. Prove that r n ^2a lf and that if r n ~2a l9 then a w term of the reciprocal part of the a cycle. T Show also that if A is an odd prime and rn = 2a l9 then

= 2fl,,

therefore

6n

+ 6n

1

and a n

is

the mid-

A must be equal =a n rn = 2a so r

<], 1 b n ~b n+l Hence a n is the mid-term of the reciprocal part of the a cycle. 1. is an odd prime, we must have a n a.i or a t Hence a x = 1 or 2, and N = 3, 5, 7. On trial it is found that JV 3.] rw

[If

n <(N/-V-f'6J/rw

,

1

,

.

to 3.

that If

N

10. If r n = r nn = a 19 then a n = a w}1 = l, so that a n a n+1 are the mid- terms of the reciprocal part of the a cycle. ~ I(a + b n )la l ~\ or 2, according as b n or b n = a l If 6 n a x l [a n ~ = &n+i 2! -6 n ffj, so that 6 n & w ,i. This is impossible. Hence a n = l and ,

^a

l

.

,

= J(ffi + &nu)/i = l-]

4i

11. If aj

is

an odd prime, or a power of such a number, and

rn

a^ then

=!.

X-bn +l9

[For r n r nn of

a;

2

=JV(mod

= (b n + b n

therefore a n

r n -ir n since each

and

Gfj),

t

}Ir n

l

N-b n

two

and

irrationals z

p', q 9 q'

therefore b n and 6 nfl are solutions a l is an odd prime, b n ~\~b n+1 ~a l9

continued fractions which are equivalent

are identical, prove that

(pz+p')/(qz + q')

z'

where p,

z'

,

^aja^ --!.]

12. If, after a certain stage, the simple

to

2


Show

are integers.

where

pq'~p'q=^:l

also that the converse

is

true.

number (c) of elements in the cycle belonging to \/N is even, prove that Pc/zf Jci2 = (Pc-^^)l^c^ a^ c rding as c is or is not divisible by 4. = 21 illustrate the fact that equations (J) of Art. 6, (5), (iv), (ii) By taking = 19. is a prime, arid verify the equations for are not necessarily true unless 13. If the

(

(i)

N

9

N

14. If the (i)

Also, (ii)

and

N

by taking

A'

If

ra

verify T

[(i)

number

Prove that

A

is

=

J

(c

when

1

iV

N

of elements in the cycle belonging to *JN is odd sum of two squares which are prime to one another. 205, show that the converse of this is not generally true.

(c)

is

)

and n

= 29.

a factor of

:

the

p r 2 + L]

J

(c -f 1),

prove that

MISCELLANEOUS EXERCISES

(A)

L Show

that the product of any two numbers, as 19 and 35, can be found as In the left-hand column, each number is the jg quotient when the preceding number is divided by 2. Any 70 9 remainders are disregarded. In the right-hand column, 4 each number is twice the preceding. 2 1 560 The numbers 140 and 280 which are opposite even numbers ~ 665 on the left are crossed out, and the sum of the other numbers 1 9 x 35 on the right is 19 x 35. [The reckoning shows that

on the

right.*

2 19x35^(1 +2 + 2

2. Show that n 8 is the sum of n consecutive odd numbers, and find or the two middle numbers.

tjie

middle

3. Take any sum of money less than 9 8s. 7d. Reverse the order 12, say of pence and pounds, obtaining 7 8s. 9d. The difference between these sums is 1 19s. lOd. Reversing the order of pence and pounds, we obtain 10 19s. Id. Adding the last two sums, we have 12 18s. lid. Prove that the result will be the same whatever sum less than 12 is chosen.

4.

r t r2 ,

,

a/b is a proper the remainders

If ...

rn

show that

fraction and q l9 q 2 ... q n are b is divided by a, r 19 r 2 ,

quotients and

the

,

when

.

.

?"n~i>

respectively,

<^

ql

Rn =

where

(

1)

.

b

tfn (ii)

For some value of form

n,

Rn = 0,

h

I

p where p, (iii)

5.

and

q

r

9

s

are positive integers, their

q, ...

Express in this

can always be expressed in the

so that a/b

1111

way

Prove that a -f b

-f

2 (a 4

-4-

and

f-f

J-f

number being

finite.

.

+ d and a f 6 - c - d are factors of 6 4 + c 4 -f d*) - (a 2 + 6 2 + c 2 -f d 2 ) 2 4- Sabcd, c

find the remaining factors.

6.

Solve

7.

If x, y, z are real, prove that (x

cannot 8.

(i)

among (ii)

lie

between

1

and

2

-f

y

2

+ z 2 )/ (yz + zx + xy)

2.

Find the simplest equation with integral coefficients which has -Vf + s/f and -Vf-Vf

its roots.

What

Prove that 4(z

2

are the other roots of the equation ?

+ #4 l)*-21x*(x+ l) 2 ^(z-

*Known

as the Russian peasants'

method

2

l) (2a;

+

2

l) (z

+ 2) 2

of multiplication.

.

MISCELLANEOUS

544 9. If #-ft/

+ 2 = o;

10. If a~\~b

+c+d

2

3

-f ?/

4-2

2

and

(A)

-2, show that

a;

f

?/

|

2

t

|

= 0,

prove that the results of rationalis-

ing the equations *Jax

*Jby

+ *Jez 4- V^ ~- 0,

*Jay + \'dz

v/fa: 4-

0,

Find the equation whose roots arc given by a a a %=*<-*, + (ai + s* + S3 s ~ are the roots of x x -f 4 ~ 0. 2

(i=l,

),

where x l9 x 2

equation

being given that 13. If

^

x

15. If

it

has a pair of roots whose

------- f -----/ + x -me x f tno x + nm value of x is m(nc + bd)/(a + b).

-ac')
a< 6 < c <

r/,

and a -

are real for all values of

+m+n

0,

positive

}-d

0,

c>

^

/>

-

r/,

c)

prove that

&' 2

>a'c'.

the roots of

^k(x

-

b)(x-d)

k.

show that the expression b 2 m'z + c'2 /i 2 2brmn ~ 2canl - 2ablm

a 2/ 2 if

zero.

2 prove that 6 >ac and

z

(ca'

l

is

and oH-6-f c

_-f

(x a)(x

16. If

sum

nid

the only finite 14. If

2, 3),

2

x%

,

12. Solve the

is

-f >/cJ

State the two other equations which lead to equivalent results.

are equivalent. 11.

-f

-I-

a&c(a + & + c)

is

positive.

Show

that the expression (ex ~-az) z - (ay -bx)(bz -cy) is the product of two linear functions of x, y, z, and that, when a, b, c are real, the coefficients in the linear functions are real if 6 2 5? 4ac. 17.

abed

from the equation

18. If the cubic derived

-

+ --, -- -- -f ----x+b x+c x+d

-,

x-\-a

has a pair of equal roots, then either one of the numbers the numbers c, d or

a, b is

equal to one of

1111 abed

-+T=-+119.

20.

Express in factors - a 2 ) (ca - b 2 (be )

Show

that

if a, b, c

+ (ca- b 2

)

(ab

- c2)

-f

(ab

are real, the roots of

are real. 21.

Eliminate

x, y, z

from the equations x

y

z

y

z

x

__

x z

:

y x

y

z

y

~

c 2 ) (be

- a 2 ).

MISCELLANEOUS 22,

10 n - (5

Prove that

23. (a) Solve

If a:,

(6)

i/,

(xz are

3

l)

divisible

is

... -f

n+1 by 2

.

(x-n)^-O.

unequal and

show that each product

- 1 and that equal to ~ yz + zx -f xy 3 (a: -f y 4- 2) -f 9 =

is

xyz-2(x + y + z) + 9~0.

and 24.

+ */l7) n - (5 - \/17) n

+ (z-2) 3 + (x -3) 3 +

545

(A)

By the method of partial fractions, prove sum of n fractions of the type

that, if a, 6,

c, ...

are

ft

/&

numbers,

then the

is

(a-b)(a~~c)(a-d) ... (a-k) and is unity if

m is an integer less than n- I

zero if

24

21

25. If

n

is

m

n -

I

.

a positive integer,

+ 2\n-2 \n _ _ ,

26.

Under what conditions

27.

Show

9

2

+4

jn-f3|n-3 -

*

.

-

...

are the following equations consistent

that, if the roots of x*

n terms

to

'

?

- ax 1 -f 6x - care not in A.P., then there are form x y-\-\ such that the transformed

in general three transformations of the cubic in y has its roots in G.P.

28. (a)

Rrove algebraically that,

if a, 6, a,

j8

are real numbers, then

the positive root being indicated by the radical sign. (b) If a and b are positive and a-f-6=-l, show that \ 2

29.

Examine,

1

/

\2

for different positive or negative values of

x and p, the conver-

gence or divergence of "

x r\ r (1)27-;

rN/(^ +

r-\

1

--

)-Vw

(11)^^5

.

30. Show how to find the nth term and the sum of n terms of a recurring series of which the scale of relation is of the form ^n -f jm n _ 1 + n _ a = ^ anc* of which the first four terms are known. Apply the method to the series

^

(i) (ii)

31.

(a)

Show

(b)

Eliminate

1,2,5, 14..., 1,2,5, 12....

that 2Z'6 2 c 2 (6 2

--

x, y, z 2

z

= a3

,

q = 3)

b--2, ^=-1)

- 6 2c 2 ) - a 4 [(6 4 - c 2a 2 ) (a 2 - 6 2

c 2 ) (a 4

- 27 (a 26 2 4- a 2c 2 x*y 4-

(i>=-4,

)

)

4

-f (c

- a 26 2 ) (c 2 - a 2 )].

from

y

2

z 4-

2 2/

a:

= 68

,

z zx

+ z 2y = c 3

,

JM/Z

=

3

rf .

MISCELLANEOUS

546

Show

32. (a)

If #

(b)

Show

a, 6, c, r

is less

.

33.

number n can be expressed

that any root of the

where each of the quantities

than >/A #

1S

(A)

is

...

n, 1 or

n~ l

but greater than

,

nearer *to

Ns

*u than

AT

is

form

.

%JN

-^ either

'

in the

9

show that

-

-

or x.

that

a z n(ai

~

___

-a

y)

2

(a-a 2 )" 2 (a-a 3 )-

Write out the terms of the product in the numerator, and give the resulting expression its correct sign. 34. Discuss the reality of the roots of

x 4 + 4z 3 - 2x* - I2x + a = 0, for all real values of a.

[Denoting the equation

by/(#)0,

35. Solve the equation 46u; side as the sum of two cubes. 36. If

(

1

+ x) n

1

+ c^x

{-

3

c zx z

-f

72#

2

consider the roots of f'(x)~Q.] -f

+ ...+ c nx n show ,

[Denoting the expressions by ^ and 37.

Prove that there

18&- 11=0, by expressing the left-hand that

show that f-==^-, ttic ct^c

v,

only one set of real values of

is

x,

y

t

z

etc.]

which

satisfy the

equation

and

find them.

[Writing

a, 6, c, d, for

o + 6 a + c 2 4-d a = i 38. If

39.

oj is

a

;

fifth

Prove that,

(l-x), (x-y), (y-z)> 2 hence (a-b) = 0.]

Z

root of 2 and

if r 2

p 2^,

a;

= eo + w a

,

z,

we have a+b + c\-d = l and

prove that

the product of a pair of roots of the equation

x 1 ~\-px* 4- qx 2 4- rx + 5 is

equal to the product of the other pair. x* - # 3 - 16# 2 - 2x + 4 = 0. Solve 40. Obtain the roots of

in the

41.

form Prove

= (3 + and (4?i-f l)a = 27r.

that

positive integer

a;

4

"* 1 +

1

77r " 2

.

2

1 (a:

+ 2o:

cos r a -1),

where n

is

a

MISCELLANEOUS numbers

42. If three real

satisfy the equations

x+y+z=5

prove that none of them 43.

by

yz 4- zx

4-

xy = 8,

nor greater than 2^.

1

when divided 3 respectively, and show that 206 is the

for all the positive integers which,

leave remainders

1, 2,

them.

leant of

44.

and

than

is less

Find a general formula

5, 6, 7, will

547

(A)

Show

that a determinant of order n can be expressed as one of "order

n-l

as follows.

a l^l c l

If

^1

,

then Jrt 1 "- 2

[Multiply A by

=

at

-&!

...0 ...0

tfi

a 1 ...0

-c,

-^ 45.

where as r

Show that, if n is a positive integer,

^

and

when n

7? 2

is

is

1,

7 (a-f

n JL

is

~\~

where

number of

the

3 36 2 or

rr^-TT 4 44

solutions of the equation

N

N

or is equal to + 1 according denotes the quotient and r the remainder

N

3 3a 2

^3^3

Prove that

47. If

j

are positive integers or zero,

not equal to divided by 6.

or

is

46.

...

~rr 4 4

r>)

a b being positive,

>

>

a positive integer,

OIL

3.4 ~ n(n-\) ~ ~ *

~\~

1

.2

4.5 w(n-l)(n-2) 3 1.2

r ^r

~"7^

I

48. If a, 6,

d are

c,

real

and b2

- ac

ad -be

show that each

ratio

= (~

)

ad -be r 2 - bd

'

and that a*c + bd* + a*d*

0,

the real values of the

\a/

roots being taken in 49.

all cases.

Prove that the condition that

(ax + by -\-cz) (bx + cy + az)(cx -\-ay + bz) can be expressed in the form P* + Q* where P and Q are linear functions of x y> z is a 3 + 6 3 + c 3 ^3a6c; 9

and

find values of

50. If

Zx m ^n

x i9 x 2 if

,

P and

...

xn

Q.

are

positive

numbers such that

Zx=n,

prove that

MISCELLANEOUS

548 51. If r

to i(r-f

terms

1)

/ } 4,

is

# n

52. If (

equal to

-I

j

+

,

7,

-

prove that

>

}

|.

-l-a^-f a^-f tf 2 # (r

-

-

an odd integer and q r =^ ~>

is

(A)

1

2

-h...

r

+ #r

-f ...

a r+l - 2na r - (r -

)

1

show that

,

)

fl^ - 0.

Also show that

53.

Sum

54.

Examine the convergence of

27*

the series

ft

;

th

series

x n /(x* n for

any value of

55. If

56.

m

is

Prove

a positive integer, show that

that, as the real variable x changes steadily (x '

where a and 6 are real and r/ an interval of length 4(6 a) Prove that,

,/;,

if

a,

58.

ft,

-"V-6

is

f

'

al]

values except those in

locate this range of values precisely.

a

a3

6

1

fr

a4 b*

c3

^*

-

1

|-0,

1

1

'*

The graph of y~---

Prove that,

if

,

----.

l)(x

x^

then x^

100,

61. If f'(x)

is

Prove that

show that

positive,

1)}.

has a turning point at P(2, -1). is

a

maximum

-%((x+ f(x)

is

1 ft

i-

(x

at P.

-

1 ft] -=

increasing.

2x + x cos x - 3 sin x >0

if

0< x<

.

4

Prove that

T75 + 5. 11.31.3.5 ^ 44.64.6.8 ^

-f

-

-

4-

#,

Find

4)

and show that y

6,

i-

Sketch the curve. *x~

the approximation being correct to at least eight decimal places.

and that

-h oo

in the expansion of (1 \-x) n (I in ascending powers of x)' a positive integer, the sum of the first n -t- r coefficients is

the values of a and

62,

to

c are all different, then

(x

60.

oo

1

+ 3)+4r(r 59.

-

a)

Prove that,

where n

from

2

assumes twice over

and

;

-

r

r,

and

is

x.

the function

57.

whose wth term

+x n +1)

-

:

4-

. . .

to infinity =

1

.

(nearly),

MISCELLANEOUS 63. Solve the equations

:

y*

+ z*-x(y + z)^a,-} )~ b

[Show that 64.

6

+ c - a = 2 (x 2 - yz),

By expanding + (n -

2

.

3

ax + by + cz =

65. If

that this

is

>

.

etc.]

2

1

in

)~

two

different ways, prove that

2-' +

.

1

which - H x y

x, y, z, for

Show

1)

-2x-3x

(1

549

(A)

h

and

a, 6, c

-

stationary, are given

is

z

maximum

a true

are

or

show that the values of

positive,

minimum,

by

if

xyz>Q>

tan x~ I + x 66. Show that the equation has an infinite number of real roots, and find graphically the approximate value of the smallest positive root. 67.

Assuming that

+ a;)}~ 1 can be expanded

a;{log(l

in

ascending powers of

a:,

f~\d the first four terms of the expansion. By help of this result prove that a capital

interest at r per cent, per

annum

/23026

\

h

(

sum accumulating at compound be increased ten-fold after

will

1-15

years approximately.

j

68. If

x

y~y

3

2

-\-2y

-y*

and

9

it

is

assumed that y can be expanded

in

ascending powers of x show that 9

69.

Draw graphs

of 2 _2(z ~6z)

70.

in

2(x*-$x)

2

By expanding

two ways, prove that

71.

Show

maximum difference

that

(a

--- #)(4-3o:

value and one

between them

minimum / is

J

(

a

V

for various values of a 72.

Prove that

has four real roots,

where a

),

value.

1\ }

is

a positive constant, has one

Find these values and show that the

3 .

What

is

a/

?

x* if

+

2

- ISx 2 + 4dx + 9 = d*

^ 1728.

the least value of this difference

MISCELLANEOUS

550

r*x^aR + bW,

73. If

and the quantities denoted by

all

is

Prove that,

if

1

approximately

where

the letters are real and positive, prove that jc,

74.

(A)

^ 2^ab.

small, one root of x* + x 2 -(I+2c*)x-I^O

is

f ie 2

and

,

find corresponding approximations to the other

roots.

75. If

,

/?,

y are the roots of x 3 - ax 2 4- bx

c

= 0,

the area of the triangle, of which the lengths of the sides are a,

If the triangle

is

right- angled,

a(4ab 76. If

n

is

show that

-a 3 - 8c)(a

2

- 26) ^

_<3

71

(ii)

71

P.2

2

.3 3

... rc

n

when n

<(^-

/?/

78.

y

y, is

a positive integer, prove that nn 2n n - 1 when __ <

(i\

77. If

/?,

is

positive,

Prove that

all

show that

log

y

between 2

lies

f

-

1\ -

y

and

2 -~

j

the roots of the equation

8< k<

are real if

11.

7

f

79. Find a function (ax H- 2bx + c)/(a x 2 -f 2b x + c') which has tm ning values 2 respectively, and has the value 6 when x = 0. 3 and 4 when x = 2 and 2

80. If a, 6 are

81.

Show

i

unequal positive quantities, show that

that two pencils of straight lines in one plane, containing p and q no two lines being coincident or parallel, divide the plane into

lines respectively,

pq + 2p + 2q-l

parts.

Show

that the number 30 may be divided into three unequal parts in 61 ways, while if equal parts are allowed, there are 75 ways. 82.

83. In

how many ways can a batsman make

more than 4 84.

off

any

Prove that the equation

has no real roots 85.

14 runs in 6 balls, not scoring

ball ?

if r

4


.

Find the minimum values of a m + a.m /a + x\ I

_.

2

_

1

AT/

for positive values of x,

where

,

CITlH

-

3 a, 6,

m are positive and

~~3 m>l.

MISCELLANEOUS 86.

551

(A)

Find the condition that ax + b/x can take any

of x.

real value for real values

- (x - a] (x ~

b)/ (x -c)(x-d) in terms of y where y~(x~ d)f (x-c); and hence, or otherwise, show that f can take all real values if (c-a)(c~b)(d~a)(d-b)

Express f

is

negative. 87.

maximum and minimum

Find the

values of

and draw a rough graph of the curve. 7 2n - 48n - 1 is divisible by 2304. 88. Prove that 89. If

4

then

I

#,_

yl

3

2/3

I

-c).

that an approximate solution of x log x 4- x -

90.

Show

91.

Find the

--y3 ~ j3

limits of

greater than

Prove that

f/^a;-a;

(ii)

,

ar =

1 -f r~ -

+ o ai +
94.

Show

95. If

sum

that a

farthings in (n

+

2

I)

of

e is small,

y-^x-x*.

(iii)

,

2

+ 6 2 + c 2 -6c-ca-o6

is

-- (a-c), where a>b>c. &

+

Li 1

where

v/3

or than

if

then

3

prove that square root of a

|(2a-6~c)

= e,

as x, y tend to zero along the curves

,

2

92. If a, 6, c are real,

1

I

x ~y

y=x-x

(i)

93.

-x

(x l

b""2

Z

-^

-f ... -f

y-

,

l

a 2+--- to

oo

=2^6.

n pence can be made up of

pennies, halfpennies

and

different ways.

x>2, prove that n n(n + l)

^^

" t

"'""

n *~*~

n(n-l)

+"

and deduce that

n+ n(n-l)

(r-l)

^

96.

_

n(n-l)(n-2) (r-l)(r-2)

^+

j

_n(n +..._

+ l)

_

...(n

+ r-1) .

Find approximations to the roots of

where a 97. If

is

small,

3wn -7w n _ 1 + 5w n _ 2

~u n _ 3

and w ~l>

^i

8,

w2

17,

prove that

MISCELLANEOUS

552 98.

Expand

approximation for

Deduce that

+ V(9 + 3z 2 )}/(3 _#)

{2a;

the fifth power of

ei

and show that

x, e

i

(A)

n ascending powers of x, as far as x it leads to a good

for small values of

x .

= 1-2840

....

99 If a, 6 are positive, and p and q are positive rationals such that 1 /p + prove that ab ^ a p !p + b //qt .

I,

1 /q

<

By means

100.

of the expansions of

ex

and

log (1 +x), prove that,

when n

is large,

1 H

Show

that e

is

)

given approximately by

with an error of about 0-46 per cent. 101. If

a time,

p

Show

102.

is

r>p>q, show that the number of combinations of p+q things r at things being of one sort and q of another, is p + q -r + 1. that one root of the equation

approximately 1-0103, and find the other roots correct to two places of decimals.

Prove that,

103.

if

n

is

a positive integer, the

coefficient of

xn

in

(l+x + x z

n )

is

1

104.

The number of ways of selecting

105.

Having given that

r pairs

from n different things

is

prove that 106. If a,

then a 2

,

fc

2

107. If

and x 2 ^ 3.

,

6, c

are real

and

c 2 are all greater

than

1

or

all less

than

x + y + z~xyz and x 2 =yz, then y and

1.

z

may have any

real values

108. Solve 109.

Show

that the result of eliminating x and y from the three equations

1.11

x-a y-a

a

1 9

x-b

11

y-b

9

b

and is

110.

Prove that

log e {log, ( 1

+ *)*} = - \x + ^-a: 2 - |a;3 -

....

Find, without using tables, the value of log e (log e 1-01) correct to five places of decimals, having given log e 10 2-302585.

MISCELLANEOUS 111.

--

~ in a -that the coefficient of a371 * 1 in the expansion of (x + 4)(x +&) # of is ascending powers

Show

series of

Find to

112.

553

(A)

# ~^

113. If

equation

five places of

3

is

a

3

given by

+ x + 1 = 0.

decimals the real root of x3

where ^ is small, show that an approximate root of the x a{l+%p log e a + ^n 2 log e a ( I + log e a)}. -J-

i

114.

Prove that

(

1

+

x^ =

1 -f

x + x* + f* 3 + V* 4 +

115. The difference between the A.M. and the G.M. of a set of nearly equal numbers is to a first approximation equal to the quotient of half the difference between the square of their A.M. and the A.M. of their squares by any of the

numbers. 116.

Find the sum of the terms

after the

nth in the expansion of

(1

+ #)/(! ~x) 2

powers of x. Prove that the ratio of this sum to the sum of the corresponding terms in the 1 expansion of (l-x)" can be made equal to any given number A, which is than suitable choice of x. Explain clearly why the restriction 2n, by greater

in ascending

upon A

is

necessary.

m

marks are 117. An examination consists of three papers, to each of which assigned as a maximum. One candidate obtains a total of 2m marks on the three papers. Show that there are f (m + l)(w + 2) ways in which this may occur. 118.

Show

that the 2n-th convergent of

3434 ------

-

43+4-3+ n n

...

differs

from the

12 + 4( -

value of the continued fraction by 13/{9 l) }. with a like [Prove that (i) p 2 n 1 Ijp2n-z + 12# 2n _ 4 - l) n and ? 2 n-i/4 = 2n 2n + ( (ii) ? 2 (<7 -#an-i)/5 .

,

result

for

the

q'a;

n^

119. If the infinite product '

2/ is

expressed in the

form

a

then 120.

(2

Find the

l+xz

_

1

- z - xz z 1

r +*

. . .

- l)a i r+

~f

2V

V

a rx r +

. . .

,

= 4(ar + ar _j).

expansion in ascending powers of z of

,

show that

f/l + V4a?+l\) H

n+ a

j/ ,

I

+ wo; H

v

/l-Vto-fl\ n+2

-(2;

..-x M ..-2) o: j^r

2

l

/

(n-2)(n-3)(7t-4)^ rx3 +

H

. . .

.

Prove that

where m and ?i are positive integers, where

2V

V

2 c^x + a 2 x -f

coefficient of z n in the

2

121.

-f

p r = m(m + n -r-

1)

is

divisible

(r^m),

by >

r

(x

-

2

I)

,

and that the quotient is

~(r-f-l)n

(r
MISCELLANEOUS

554 122.

among

123.

(A)

Prove that the number of ways in which n prizes may have one at least is

may

Prove that when x

is

sufficiently small

1

Jog (1 +#)~~

Also, if n

>1

1

1

#

2

11 24*

12*

,

"

#

(n

L__ |

+ 2)

"

'

\n 'L

A man

owes a sum of a, and repays b at the end of each year, partly interest at 100?* per cent, and partly to repay capital. Show that the will be repaid in A years, where A is the integer equal to or next greater than

124. to

be distributed

q people so that everybody

pay the

sum

125.

Show

which n are

that the number of combinations taken n axe b, and the rest unlike is (

and that

n together of 3n

this

sum

is

%

4-

26'Jt-

i

w equal to 2

+ 3CJU + (n + 2).

.

. .

~1

of

+ nC? + n + l,

126. If a, b are positive integers, the probability that ^(a 2 f 6 2 )

integer

letters

a,

is

a positive

gV

is

x + (b+c)y + bcz = b~c, x + (c + + caz - c ~ a,

127. If

a)y

x + (^

-

&2

4- ft)y 4-

- a - 6,

where a^b^c, prove that 2;o;-2/ 2 = 3. 128. Ladders 15 ft. and 10 ft. long are placed in a passage, each with its top against a wall and its foot against the opposite wall. If the point where the ladders cross is 5 ft. from the ground, find the width of the passage. (Sunday Times. ) [If the width of the passage b ft. from the ground, then -

a

Hence

if

6

5(t/-|-l),

i/

ft.

+ ! = ?> and b

and the tops of the ladders are a a2

-& 2 -125.

5

show that 0
giving

x

is

4

4-2?/

3

+ 5i/2 --2?/-1^0,

= 0-5761287, 6^7-880644, ^ = ^(100-

6 2 )^ 6- 155929.]

ft.

and

MISCELLANEOUS EXERCISES

(B)

N

A

of party consisting of 7 robbers ami a monke}' had stolon a number In the night a nuts, which were to be divided equally among the robbers. robbor woke and decided to take his share. He found that, if he gave one to the monkey, the remaining nuts could be divided into 7 equal lots. He gave one to the monkey and took his share. The other robbers woke in succession, found that the same thing was possible, and each gave one to the monkey and took one-seventh of the remaining nuts. In the morning they divided what v,ere left, and they found that, if they gave one to the monkey, the nuts could be divided 8 is 7 - 6 = 5764795. into 7 equal lots. Prove that the least value of - 7, - 6 of i\ satisfies the numerical conditions 6 fur value [The ~ - 6. Thus - 6 is unaltered - 7 ~-7 = -7- the set of 1.

N

I

;

1,

by

1)

(

Hence the values of

corresponding to what each robber did.

2,

Let a/b be a proper fraction and

c l9 r 2 , c 3 ,

...

an

N

infinite

operations

are given by

sequence of

r n the If q l9 g 2 positive integers. remainders when ac L9 r L c z nc 3 ... r n _^r n respectively, are divided by 6, show that ,

,

,

--

......

,

c^ qr^

CL

b

,

qca...^

E n = --

r ~

1

where

Ci<- a

(iiiJV-ar^ (iv) If a

is

...c n

-

;

b

...c

(mod6);

a prime to b and a<6, then

a

^? +

_??

CL

c^'o

1

6

-

(h .j..

|

c,c 2 c 3

where the of

n

series terminates if c x c 2 ... c n is divisible by otherwise the series is convergent.

:

Express the following fractions in the form

3.

ft

3

3

8

+ J'- 4--? + 3.5 3.5.7 (i)

In the 4.

two

last

Show

~

6.

Use Ex.

Sum

(ii)

;

i

;

some value

:

JLn

4.

3.5... (2n +

cases, find the value of q n

for

(iii)

'

l)

J.

.

that "

"

32

8 5.

J

4--

6,

2

3"

.

2, (iv), to

show that

e is

irrational.

the series

n 2 + 2(w-l) 2 -f3(rc-2) 2 -f..., where n

is

a positive

Prove that

integer.

$I

+ TS+-"+ 12

where

2N

Sn

is

the

sum

-

n

n

+

-

of the squares of the

to infinity ' first

,

6

n positive

integers. B.C.A.

MISCELLANEOUS

556 7.

Prove

(B)

and x + y + z = l, then

that, if x> y, z are positive

\

\y

From

8.

a cask containing x apples, a

man

half the contents of the cask

sells

and half an apple to one customer, half the remainder and half an apple to a second, and so on, always selling half of what are left and half an apple. In no case is an apple divided. After n sellings, a apples are left. Show that z = 2 n (af 1)-1.

Show

9.

mon

line

m 10.

if,

n

TA-

II

/?

if,

= --<

.,

y~"n

d

Cp

have a com-

y

where ad -

3 d

-f-

d are connected by the equation

a, 6, c,

mx~ly~v

.

COL -f

and

ny-mz = \, h-nxp, ZA + mfji + nv =

that the planes

and only

a2 +

+6 + cy

=-ay

.

prove that

.

,

e/

Show

also that a,

y are the roots of

/?,

x

_

.

ex

|2r-2

where

r

A:

-ri

r

,

^), ^,

13. If

-dx + b _____-ex

-a

-

Ca - a

,

k r =kjc r

that

^+

-

-da + -6

aa +6 ---COL f a

=a

.

+

...

a positive integer greater than unity, and

is

10 ou .u . 12. Show that

where

s show

-

11. If

.

+d

-

a P + 6 p + cP + a+& m---> ~ c

r are positive quantities

a-j-^-t .

m

such that

x r = x(x-l)(x-2) ...(x-r+l), prove that (2n

-

!)_!

.

(2n-3) w _3

.

(2n-5) n _ 5 ...>nn

.

nw _ 3 nn _ B .

.

7i n

_7

each product being continued as long as the suffixes are greater than zero. 14. If

L

TH,

Z',

w', I"

and

X/

7W

are integers,

and

if

/)9

is

not rational, and

show that / = Z and m m'. Also show that no two of the numbers r

,

(2Z

+ 5m)a + /^,

X

-f

(2Z

5m' -f

/X

1

)a

-f

Z^,

-f

(2Z

5m") a -f (Z

/x

-f 1) jS

can be equal. 15. is

One

root of the equation

3 - 33 - 33"

;

x 6 - 9x 5 + 18x*

f

9o;

3

+ 27 x 2 - 54x -36-0

find all the roots.

+ c ^V, x) + c 2 = O

a 1 yz + b l (y + z)

16. If

a&x

4-

6 a (z

4-

are true for an infinite series of values of x, y, c 2 --

26 2 6 3

,

agCj, 4-

1

t

z,

a x c 3 = 26 1 6 3

,

prove that, in general, a 1 c 2 + a 2 c 1 ~26 1 6 2 .

if

MISCELLANEOUS

557

(B)

show that

17. If a, b, c are unequal,

ax

b~ c

4-

+ cxa ~ b ^a + b + c.

~ bxc a

18. If

a (z 4- x + 6)

where y and

z are unequal,

bzx (z

4-

= 0,

x)

+ ?/ 2z 2 + fy/z (y + z) = 0.

Prove that 2

a4-4-y4-5

a + j3 + y

is

4-

prove that

a (y 4- 2 + 6 ) 19.

z*x 2

4-

+8

a(y 4- 8) 4- y8(a

2(a4-]8)(y4-8)

equal to zero. 20.

Prove that rt

~(L-x where

A r ~(

according as n 21.

-

7""' 1

I)

cos n

sin 2

~

4

is

and

-

r&

TIT

,

X"

r

,

4-

tan-

5

= 1,

n ...

2, 3,

^(n

~

or

1)

-^

odd or even.

Determine the condition that

(x

n

shall be divisible

by

1).

22.

Show

that the

(

are identical

four terms in the expansions of

first

1 4- x) n 4-

a

1 4-

f

-

and j

\

if

a

b

c

p

q

r

'

and

find the values of a, 6, c in terms of p, q,

23.

a

b

pz

q

2

c r2

'

r.

Find the limiting values of i

(cos x)

i

.1

x

(cos x)

,

xz

(cos x)

,

x3 ,

as x tends to zero through positive or negative values. 24.

Show

that, if

I

l9

m

l9

n^

;

/2 ,

m

2,

n2

-f

w

a

;

13 ,

m

3,

n 3 are

real quantities satisfying

the six relations, 2

!

-f

m^ 4- ^! = 2

/2

2

a

-h

n 22 = /

=l 3 -f n 2 r? 3

4-

then

4-

and

n 3?3

ll

m

l

4-

^

w^a

0,

MISCELLANEOUS

558 25. If z 3

-j-3/fcz

3

+ <7~

[/x(z-hi/)

-v(z-f

3 /x) ]

A = g 2 + 4th 3

where

Show 26. If

that the cubic 4z 3

u

(i)

If

u

(ii)

If

u

for all values of

z,

show that

.

has two equal roots.

27a 2 (z-f a)

- c) n + (c- a) n + (a- 6) n where n is a positive integer, prove that 2 divisible by 27a Z!bc, then n is of the form 3k 1, where k is an

(b is

(B)

,

integer,

divisible

is

2 2 by (Ha - Hbc) then u ,

is

of the form 3& +

1.

27. If all the roots of the equation

are rational

28.

and negative, show that

Prove that,

if

n

is

a positive integer, and 1

m -2xn

x'

29. Investigate the

r

~(a-f 2r7r)/n, then

^tt-xsina-tfsin (oc~6 r ) r "=0 x 2 ~-2x cos O r + l

_

cosa + l~~nsina

maximum and minimum

5

l) /(x

+ 1), and

real roots if

0
values of (x+

5

trace its graph.

Prove that the equation (x+ I)*~m(x 5 + 1) has three and only one real root ( - 1) if m<0 or m>16. 30. If

divisible 31.

n and n + 2 are both prime numbers, prove that by n (n + 2)

(n -2) in

-1-2

is

.

Prove that,

if e is small,

has a root nearly equal to

1

the equation

-f

4e 2

,

and

find approximations to the other

two

roots.

32. If a, b,

show that

JJL

c, ... k, I

lies

are positive

numbers arranged -Z /xa

between the greatest and a/6, 6/c, c/d,

33. Discuss the

least of the ...

,

in descending order,

and

numbers

k/l.

convergence of the series whose nth terms are (J*)!*-.

(-!)

w+1

|^, .

34.

where If z

Find the seventh roots of unity and show their positions in the z plane z x + iy. is any one of the imaginary roots, find the equation whose roots are z2

and

MISCELLANEOUS 35. If

and

559

(B)

d denotes the determinant ax

a2

a3

a4

a

a6

!

#2

aa

a

a4

a5

ax

aa

a

D denotes the determinant obtained by A = a + 2a 2

x

!

2

a 6 + 2a 3 a 4 , a52

At

4-

^4

= a 4 + 2a 2

2

2ia 4 -f 2a 2 a 3

x

^4 5

,

changing a r into A r where ^ 3 = a 2 2 -f 2a 1 a 3

a 2 + 2a 3 a 5

,

= a 3 2 -f 2^(15 4- 2a

2

a4

,

show that d*~D. 36. If

n, p, g are positive integers

m,

and

if x, y, z all

^(yn^g^ + ytn^n^^^gm^^yn) a;(y-^)+yP(2-a^) + p (a^-y)

wre (

tend to

m,n

prove that

a,

n

)

__

2:

Prove that,

37.

in the expansion of

where - !<#
Show

38.

zero)

l)x

2 .

that the number of distinct sets of three positive integers (none is the odd integer 2n -f 1, is given by the least integer containing

whose sum

39.

Show that

whose sum 40.

is

the

number of distinct

an even integer 2n

is

positive for positive values of x, 41.

which

sets of three positive integers (none zero) the integral part of \ri*.

Prove that

I-x is

m(m

+ ix*-(l+x)e~ x

z

and increases

as x increases.

Find to 6 places of decimals the root of lies

between 5 and

Prove that the in powers of * is 42.

6.

coefficient of

xm

in the

expansion of

1

-

(1

-

v

7

,

*) (1

- *)( 1 *)

43. If the volume of a right circular cone is 20 cubic feet, prove that, when the cone is such that the curved surface is a minimum, the radius of the base is approximately 2-381 feet. 44. Arrange the following numbers in order, .so that as x -> a each number to the preceding may tend to infinity :

ar

2 ,

2X ,

xx

,

ex ,

a;lo8

r ,

(log

2 a;) ,

^

the ratio of

MISCELLANEOUS

560 45.

---

(B)

Prove that

n

/

-r

'

v

|r)2[n-r

and deduce, or otherwise prove, that ln

+1

lrc

1

46.

Prove that,

+2

a, 6, c are

+3

ln-f-4 "

n-3"

|3

|w^4~~'"

[3 [4

if

a

where

|2

[n-2

[2

|n

"

b

c

a

b

not zero and no two of

bm

al

c

-

-

m x y j+ m + n^o. ,

,

I

en

~ are equal, then

n

I

47. If a 1 a 2 ^3 an = the a's is greater than -

An

prove that the

,

48. Solve the equations

sum

of the products r together of

:

x + y + z + w = l,

+ dw~ A, + ^ 2t^ = A 2 a*x + 6 3 y -f c 3 ^ + d 3 w = A 3 c 22

/ 4-

Prove that, if a, 6, c, d, A are all than three of x, y, z, w are positive. 49.

Prove that 2 t/

-f

yz

+

real

,

.

and unequal, at

least

two and not more

if

z2

a2

z2

,

+ 2#-f # 2 ~6 2

,

x z + xy + y z

y2 -f zx 4- ^2/ = 0,

c 2,

a&c-0.

then 172 50.

Prove that 7^=7-, -

51.

Prove that

if

1 is divisible

by

73.

m is prime and p
8f

52. If a 2 -fj8 2 = *a0, j3 2 are all different, then

53.

-1 =

+ y 2 = /c)5y, y 2 + 8 2 -fcy8,

82

+

2

^AcSe,

and

Prove that

1 1

ttfo"

1

)

n(n-l)(n-2)(n-3)

_

2n-f2r

54. If

ar

prove that

and that

1

+3a t 4- 6a 2 +

... -f

a 2n_r

;

(4w + l)a,n = (2n

+ 1)(2 +f

n )

,

if a,

)3,

y,

MISCELLANEOUS

561

(B)

55. If of 3n letters there are n a's, n 6's and n c's, the number of combinations of these r together is equal to the number of combinations 3n ~ r together. Also, if n>r>2n + 1, the number of combinations is

|(n + l)(n + 2) -f (r n)(2n number of combinations is

and the maximum according as 56.

n

is

|{3(n-f even or odd. 1 4- -

Assuming that

+~

-f

2

-f

l)

--

H-

. . .

or

1}

r)

-J(n-f I)

- log n tends

2 ,

a limit v as

to

n -*

oo

,

prove that (i)

lim

^

57.

Expand

lim

n v j' -a:

l/-vf* /"I 1 V

2

obtain an expression for

1

^^~3

\

,

,

g

n~

7

-/

*

# 4 and ,

if

'

>

-2

t^

and prove that

n,

-

r hm

- (2x ~ x 2 )

_ ^03

--(j*

f

'

)*

/yi

v

lim

'""

as far as the term containing f%\ Ulj H 2-
Findi

Find

1

+

)

T-"! 1

59.

^

2

"~

lo g

KQ 58.

h

n ~

(

log(l -a:

^ n ~ "2

^

2

(n

1 (li)

,

where

,

,_>o c^-rf*

a, 6, c,

are positive

c

* 2 + y a +2 2 = f a + i7 2 -f

60. If

and show that

a;

+ y + 2 = f 4-^-f 3a:

2

2

^=(i7-{)

#

and

2

and

= 1,

+
2

4-f

2

^!.

61. If n points are taken in space so that not more than 3 of them lie in any plane and not more than 2 in any straight line, show that the points define \n(n - 1) straight lines and \ \n 1 polygons of n sides.

------

- in which a is x - 1 and is repeated any number of times, must have one of three equal to that if x satisfies the equation 2# 3 + 3# 2 - 3x - 2 ~ 0, the fraction and values, 62.

a

Prove that the continued fraction

a~ a

...

a

satisfies this equation.

63. If (1

+ x) n ~Co -cx+ -\

P

...c n x n ,

where n

a positive integer, prove that

.

'

38

2*

is

(n

+

w+lV

a

l)

2

3

n+1/'

64. If a, 6, c are real and I, ra, n are integers, find the values of l ~ b) m (x - c) n has maxima or minima. (x ~a) (x

I,

x

Determine which of these values give maxima and which minima, m, n are all even, (ii) when I, m, n are all odd. 65. Solve the equations

y + z = a(l -yz),

^

66.

o Sum

i

-

the series

which

(i)

when

:

+ x~b(I

-zx),

2

8

z

,

for

3

13

x + y + z -xyz~c(\ 30

55

+ 7^ + 77; + "l+7T1 + T + T^H~r7 4 |5 3 o |2 1

~xy~ yz-zx).

to

-

MISCELLANEOUS

562

(B)

convergence of the series

67. Discuss the

I2n

E.

,n

Z

!

w) 2 (l for all values of z, real or complex, distinguishing cases of absolute

and conditional

convergence.

The equations of four

68.

(1) (3)

lines are

y = a, z = a'; x = c, y = c';

= 6, * = &';

(2) 3

x=y=z.

(4)

Find the equations of the line drawn from the point (k, k, k) to cut the lines (1) and (2). Prove that two lines can be drawn to cut all the four given lines, and that these k where lines cut the line (4) in two points determined by x = y = z 9

(k-a)(k~b)(k-c)^(k-a')(k-h')(k-c'). 69. Discuss the reality

and equality of the roots of x* - 2A(3* 2 - 4x + 3) + 3A 2 =

for different real positive values of A.

A

70. bag contains 21 balls of G different colours, there being 6 of the first Show that the number of colour, 5 of the second, 4 of the third, and so on. different selections of 6 balls which can be made from the bag is 259.

Af(a)=f(a + w)-f(a)

71. If

f(a + xw) =f(a)

prove that

Use

A 2f(a) = AJ(a + w) -

9

+ xAf(a) + -^r^-

1)

J/(o), etc.

4 /() -f 2

- - .

.

given that

this to find log 7-063,

J 7

= 0-845098

log 7 -2

= 0-857332

log

--

x(x

J2

log

J3

log

log

12234

-334 11900

log 7-4

16

= 0-869232

-318 11582

log 7-6 = 0-880814 72.

A

series is

such that the

sum

of the rth term and the

(r

+ l)th

is

4 always r

.

Prove that (i)

(ii)

73. If

the rth term the

n

is

sum

is

%r(r

l)(r

2

r(r 2

of r terms

is

-r1

->

- -

-

1)

W^r

2

(

r l) c

7^ '-

;).oO

;

r

+ ^{l-(~ 2

r l)

}.

a positive integer,

f

7)

74.

-

Prove that,

if -

%

,

- are fractions s

nator of any fraction whose value

lies

such that qr-ps

between

p q

and

\,

then the denomi-

r

- is

s

not

less

than q -f s.

Prove that there are two and only two fractions with denominators which lie between jf and -J.

less

than 19

MISCELLANEOUS

563

(B)

Determine ranges of x for which the function (log x)jx (i) increases and decreases as x increases. Hence prove the following theorems, wherein n is a given positive number and only positive values of x are considered 75.

(ii)

:

The equation n x

(i)

x has two

roots,

one root or none according to the value

of n.

The inequality n x >x n

(ii)

a consequence of either

is

x>n

x
or

accord-

ing to the value of n. State the critical values of n.

From

a bag containing 9 red and 9 blue balls, 9 are drawn at random, the Show that the probability that 4 balls of each colour will be included is a little less than -J. 76.

balls being replaced.

For

77.

-

continued fraction

the

(I

altered if a l9 a 29

An

78.

...

approximate value of

provided that <^< error

is

a n are permuted

OTT.

approximately

------

prove that

...,

^ -f-

cyclically.

measured

<^,

radians,

in.

when

Establish this result

<^

is

3 sin

(f>

<^)

/180. fc

i

nearly equal to

cos

r>

<

Hence express approximately the acute angles of a right-angled 2 terms of the sides and deduce that, if c 2 ~a'2 then

is

/ (2 -f

small and show that the

is

values of a and

\TT for all positive

triangle in

,

h.

79. Given a l things of one kind, a 2 things of another kind, etc., a r things of an rth kind, show that the number of different groups which can be formed from one or more of the above things is (a 1

and show that n members is |-

80.

and

if

verify

if

[(n

When n^2,

r~4 and -M) (n

-f

+

l)(a t

+

l)...(a r +l)-l,

a l
2) (n

+ 3) - (n - ax

)

tl

(n

,

- Oj +

1)

(w

- a x + 2)].

Zn^Vn-i + ^n-z*

if

xQ ~I,

xl

a^, yQ by induction that xn+l yn -xn yn+l = ( ~

Q, l)

1,

2/i

n+1 ,

and deduce that

2/n_i2/n

2/22/3

81.

number of groups having

the

Find the equation determining the values of x for which

stationary.

Hence, or otherwise, exceeds

m

show

that,

if

m

is

an

integer,

sin mx/sin.

sin 2 war/sin 2

x

is

x never

2 .

82. A quadratic function of x takes the values y l9 y zt y%> corresponding to three equidistant values of #. Prove that if 2/i + ,V 3 >2?/ 2 , the minimum value of the function is 2 _ ,

83.

Show

that

(n

.

+ l) n ~ l (n 4-2) n >3 n ( ]n ) 2

MISCELLANEOUS

564 84. The any order. by 11. 85. If

a number are 1, 2, 3, 4, 5, 6, 7, 8, 9, written at random in that the odds are 115 to 11 against the number being divisible

digits of

Show

w, n are

show that

positive,

\

86.

(B)

The

w-f

1

x 2n in the expansion of

coefficient of

87. If

-f

(

1 -f x*)

n /

(

1

- x) 3

=

cz 2 2

2

is

= d,

&z + by 2 prove that

88.

Z2

2/3

23

Prove that

sin a

sin

cos a

cos

sin 2a

sin 2/J

sin

sin 3 a

sur

sin j

sin a

sin

sin

cos a

cos

89.

2/2

-4

^

sin

p

cos y

t

y

J

()3

y) sui

^(y

~ a)

sin v(a

~

.

j8)

j3).

2y --

y

- sin

(p -y) sin (y a) sin (a

/?)

sin (a

-f-

/?

+ y).

y

Denoting the number of combinations of n

unlike, the

Show from 3n

by

W 6' r

,

show that

number of combinations

letters taken r together, all the of the letters^are alike, the rest being

if r

is

that the number of combinations in groups of n which can be formed letters, of which n are a and n are b and the rest unlike, is

VV

and show that


this

Prove that 1

if

n

is

.

+ Vn-s + - + ~ 2 n (w + 2). 1

a positive integer,

n n(n-l) x+

t(w -!)(*_- 2) 3

1T2

TT2i"

1.

.

91. Investigate the convergence of the stories

whose nth term

is

92. Prove that

^__

"* ,/V

and

93. If

A,

nn ;

_ 2n(2n

n(w-fl)

where

2 sin (a +

COS y

,

letters being unlike,

90.

sin

/Lt,

t

2

+ at~b both have

v are integers

[For instance,

^

2

and

+ 5^-f6 and

jtx

is

prime to

rational factors, then

v.

MISCELLANEOUS For the fraction

94.

p

2

2

3

565

(B)

2 . . .

prove that

,

95. Discuss the reality of the roots of

16* 4 + 24* a +16^+9 = 0, for all real values of k, 96. If ax*

for

all real

and solve the equation when k = ^2.

+ 2hx + b>0

show that

for all real values of x,

values of x, y,

z,

the coefficients being real.

97. 2 n players of equal skill enter for a tournament. They are drawn in pairs, and the winners of each round are drawn again for the next. Find the chance that two given competitors will play against each other in the course of the tournament. Also show that if n -7 the chance that a given competitor either wins or is beaten by the actual winner is tV

98. If thore are four relations,

for i = 1, 2 j= 1, 2 A fy + B& + Cfy + Djdj = (B,C)_(C A)_(A,B)_(A D)_(B,D)_(C,D) ~ ~ ~ ;

show that

(a~d) 99. If sin

""(ft, d)

(c,

d)

(b, c)

x~xy 2 where x 2 and y - I

(ii)

;

9

9

(c,

a)

(a,

are both small,

'

ft)

show that

x*= -12(t/~l)+(y-l) 2 approx. c

of Craps, each player in turn acts as banker.' The banker throws with two dice numbered 1 to 6. If he throws 7 or 11 he wins. If he throws If he throws any other number, he throws again and con2, 3 or 12, he loses. tinues to throw until either the number he threw first or 7 turns up. In the first case he wins, in the second he loses. Show that the odds against the bank are 251 to 244. 100, In the

101

.

where

If

a,

/?

game

un u n _i + aun + bu n

_!

+ c = 0, show

that

are the roots of

unless a = jS, in which case 1

Un - a Hence obtain the general results so obtained agree

102. If

solution of the first equation,

with those in Exercise

rv

cosx

1

COS

XXXVIII,

X

1

COS a

COS

COS a

1

COS

cos 2 2 2 prove that sin y 8in z = cos a

-f

ft

oos 2 /3

cos y

-2

oos a oos

ft

y

1

ft

oos y.

and show that the 15, p. 372.

MISCELLANEOUS

566 103. If o

*-.**-?* =*, x x d +

+y

(B)

!=,<) or z = a-c. *++ o a

then either

y

a 2x* + b*y* + cW = 0,

104. If

aW + b y z

l-

and

2

+c

a z

3

z*^Q,

2

-f

6

3

4 ?/

-f c

4 3

z

a e z 3 + & 6 */ 3 4- cV = a 4 ^ 2

and

1 ,

2

y 4

2

=--& = - -c

a;

prove that

3

= 0,

&y + c z 4

-f-

2 .

Y are two places on a road, 4 miles apart. A man A starting from X at any time between one and four o'clock, walks to Y at 4 miles an hour. A man B, starting from Y at any time between one and four o'clock, walks to X at 105.

and

J

y

4 miles an hour. Show that the odds in favour of the men meeting on the way are as 5 4, all times of starting being equally likely. [Suppose that A and B start at x hrs. and y hrs. past one, respectively. Taking axes Ox, Oy at right angles, set off OL~3 units along Ox and draw a square on OL. The total number of cases is represented by the area of the square, and the number of favourable cases by the part of the square between the lines :

*-y= 106.

1.]

that, in a game of whist, the chance that a hand is void of a suit is Also the chance that in 25 deals a certain player has a hand void of

Show

about -$. a suit at

least twice is

about 0-358.

[Chance that a hand 107. If

is

void of a suit

4

.

u~ax* + 4bx* + 6cx + 4:dx + e and

show that the

z

roots of

w=

are given

by

Gy3J t

-rC'JJ.]

is

any root of

ANSWEES EXERCISE

I

PAGE

10.

7.

28 2 -27 2

82

;

11. 23. 6,28,496.

-3 2

10. 212, 213,

.

27. 120.

28. 360.

...

,

22. 20, 4836.

219, 220.

30. (i)**'"**; -

(ii)

(p-1)

EXERCISE 22. 19.

(?)

5-394, 0-006

(ii)

;

2-594

;

II

0-002

;

EXERCISE 31.

1.

5* 3 -20z 2 + 80a;-314; 1258.

3.

ia-V*' + tfa-tt

5.

3z 2 + a;-3;

8.

2.

(iii)

11. 13.

15.

2-276

0-002.

;

III

z 4 - 3* 3 - 6* 2 - &x - 10

;

-21.

fc^-i^ + fcc'-fta + A; ~^A-3z~7. 6. 3z 3 + o;-3; 5a + l. 7.13. 3 2 3 (i)(x + 5) -13(z + 5) -f56(*4-5); (ii) (x + I) ~(x+ l) + 80 4.

;

;

+ 5^2.9. (ii)jf-75y + 7; (iii) -n + 7n(n-hl)-67i(7i + l)(n4-2)4

9. (1)2/3

10.

T(p-1)

~^;

0,^.

3

-

i/

2

15?/ -f

257.

12.

(x-3y)(2x + y)(x* + xy + y 7z 2 - lla; - 6.

z ).

16.

+ 4x 2 - 6x - 2. 3* - Ixy + 4y 2

23.

x 2 - 3y 2 + 3z*

14. 3a; 3 2

.

18.

20. a( 21.

a=3, 6=-3,

c

= l.

32. 22. p* - 3q z + 3r 2 - s 2 = 0. 24.

26.

a=5, 6- -3, c = l. a = 8, 6=0.

25. a

-w = 0. 2

= l, 6-2.

27.

29. a 2 + 2)(a? + 2y-3). 31. (i) (x-l)*(x + 5); x + 3)*(x-3). A = l,(2a? + y-2)(aj-y + l); A = - 7/8, (* - 7y + 4) (x - y - 4)/8. A=-l,(y-l)(y-2); A- -2, -(

28. (3a;-4y 30. 33. 34.

(

EXERCISE IV .

16. 120.

21. 23.

(i)

17. 60.

#i"("+i)

;

(n)|tt

18. 15.

20.

(i)

2 2 ".

+ l.

(-l)Ca ifn=2m; (-l)(n + l)05, if n=2m + l. f

HIGHER ALGEBRA

568

EXERCISE V PAGE

43.

1.

5.

a

(s-l) 2 (s-1)

8. JC:=:r 3

10.

.

2.

2*4-

1.

3.

.

6.

3z~4.

7.

+ l, 7=3.

x*

+ x-3.

4.

= i(x + 2), y = J(a;-2). Z^l-fz 2 7^1 JC

'

9.

,

A=-2 = l; 4= -*,=-!, 0=1, !>=-*. ft

EXERCISE VI 50.

3.

18.

51.

-2x* + 3Zx

2

p^a+f-g-fi, 25.

24. 6.

-2.

26.

31.

-4(6-c)(c-a)(a-6)(a + 6 + c). l

33.

l

2 ~(6-c)(c-a)(a-6)(Z a + 3Z 6c). 2 2 2 (l-a&c)(l-a ~& -c -f2a&c).

36. (16)(25)(34)

(12)(24)(36)(65)

;

(16)(65)(53)(24).

;

EXERCISE 60.

1. 2, 7T/6;

4. (i) (5

61.

5.

-7T/6; 2,

2,

^O

(x

+ 1/ 2

5 F 2 )

(a:- iy)(x*

(i)1625;

/

cos

/

*>

2

;

(iii)

(iii)

+1

)

\

+ y*)

(ii)

y\ --

tan"" 1

7i

L_

(x -2/ 7.

13/125.

2

7.

3.

i( 1

+ 0/25.

*

(

-

(18

+ cot-

sin n( tan~" x \

y i*

1 ._

l~2t; i(ltV3).

13.

~7r/3.

EXERCISE 67.

VII

-. D

2,

;

-0/13; (ii)(- 2 + 2)/8; 2

(i)

(ii)

6.

I2xyz.

-

30.

+ a + 26)(a + 6 + 2c).

32. 3(6H-c-f-2a)(c

62& 2 y-

-f

27.

-(6-fc-2a)(c + a-26)(a + 6-2c).

35.

3

q 1.

29.

34.

2z

4.

y-12xyz.

VIII

2-i, -4-3i.

EXERCISE IX 78.

3.

KX + 4r = (l+* + iy)/(l-*-iy), then X = and

2 a (l -a: -i/ )/[(l

-

x 2 + t/ 2 = l.

Z^O, y-2?//[(l-a;) + 2/ ]; Geometrically, take any line ^4OJ5, so that BO OA, and draw AL, 9 parallel to Ox to meet a line through 1, parallel to BOA, in L and con2

2

if

BM

M

struct

OMN

similar to

opposite directions.

1+3,

1-zand

hence L 5.

If

OM

is

If

OL1, with L* ION,

A

represents (l+z)/(l-z). If |z| =

a right angle

a=p + iq, px + qy=p + q 2

pendicular to qx =py.

2 ,

;

and

z,

l,

LOM

:

described in

N

then L, M, represent O^lLl, J5O13f arerhombi;

N lies on Oy.

which passes through

(p, q)

and

is

per-

ANSWERS EXERCISE X

569

PAGE

86.

1.

1 = 0; 4z - llz + 9z - 2=0

4x- 3

(i)

4

2x*-3x-

3

(iii)

(i)4z

3.

t(3

-8z 2 + 8;r-3=0;

3

N/6)

-

2, 9/2,

t(

;

-

1

(iv)

6.

4.

-

2, 1/3,

-2,

9.

8. 4, 2, 1, i.

+ 9x 2 +

llx-f 3=0

2z - 9x*+llx

;

-3=0;

x3 -4

(ii)

N/5).

1/2.

2x* 3

;

2.

5.

(ii)

2

3/2, 1(1

\/5).

7.

1/2.

5, 3, 1,

10. 3,

3, 6.

-2,

-

1.

1/2,

-

1/2.

2 4 x - Sx - 15 (ii):r -23x 4-59x-52=0. 3 3 2 14z + lla; -75 = 0; (ii) a; 4 25* 3 + 375z 2 - 1260*- 11700=0. z 3

11.

(i)

12.

(i)

;

EXERCISE XI 94.

3.

4,

7.

V,

-2. -1.

4.

2.

11,

-3. 15, 3,6, -4. 8.

14. 6. 18. 2, 2, 2/3,

-

3-1,

5.

4,

9.

7,

-

2,0.

<>.

6.

-0.

13.

4.

17.

3/2,2/3.

16. 2/3. 19. 3/2, 2/3.

1/2.

EXERCISE XII 99. 17.

iv3a 2 + 2a& + 36 2 ).

-a, -6, -i(a4~&

EXERCISE i

3

'

i

i

2.

^ i; J5___^___? T x

9 ^


1 i

*-3 x-2

^T

1

\2

(x-2)

^_

3

(x-2)

4

'

3

'

J-+-

x-l

i

x

2

* 5.

XIII

ft

5.

x-2

x-3

X4-2

|

-1) Q

(a:

_ _____

j^

_i_

______ 4.

3

x + 2) 3 *

!

10.

"' 13

.

x"2

-fl" (x 2 -fl) 2

_1

x 2 +~2

x

" L(x+l)

2

x 2 + x-fl'(x 2 -fx+l) 2

1,1,1,1

-I)

14.

JLr

1 3

2(x-fl)

3

HIGHER ALGEBRA

570 PAGE

104.

15.

*_

-l + -JL_ + 1 -ax

-ox

I

;

'

'x-a'

(a-b)(a-c) 18.

Put x^din Ex.

19.

x + a + b + c + Z. r. 7-7-7 (a~b)(a-c)

1.

\

2.

j

+ l)(2n- I).

l-n(2n

7.

Tzn(n + l)(?i-f 2)(3n + 13). 10. 2926. ^(47i 2 -f7i-l). J Miw 1). !)(/<- f2)(3/* iV/t^j

14.

pui>x~d.

;

19.

*

2

a

(i)

(n

})

i-/i(6n

+

8.

(ii)

iw(n

23. 715.

(B)

2.^.

1.^IV\UIV

A

I

I

13. 4300.

- l)(9w a -9w -2), where

21. 344.

EXERCISE XIV 117.

12. 50336.

%n(n + l)(n + 2,).

15.

'2).

-3-l);

20. 220.

(A)

11. 286.

t-

-f-

16. -iVw("

'

4.

3.

9.

x-a

1

x-a

EXERCISE XIV 116.

-b)(a-c)

17.

a*

105.

'

(a

3.

_

f

6.

n(29n 2 +

5

138r*4- 157) *

"

36(ft4-l)(rZ'4-2)(tt4-3)

10

*

2.5....(3n + 2j"

2

llf

'

1 _ Li?_i -Jil^^tl) 2 2.4.6 .... (2w + 2)"

~

*

1~,

17.

^

^y/2.

.

18.

-

19.

140

4(2n + 7)(2ra

i

Q|

i __.

r

i

i I

i I

si

I

n(n-f 2)

ANSWERS

571

PAGE '

(l-.r)

n)'

~

26.

-

,

fc

where k^=2 n

(I-*) 2

2(1-.*:)

-

------

27.

.

3

^

,

where & = 2 r

28. 1

EXERCISE XV 129. 8. 131.20.

a3 + & 3

(i)

- (a 3 + 6 3

(ii)

;

2

(0-y)(y-a)(y-|8);

(i)

36.

(iii)

;

)

-

(ii)

(jB

y)(y

- a)(a0)(a +

+

y)

EXERCISE XVI 143. 146.

1.

19.

2

V|7

0,

39

2

+ &2 + c2

).

= 30 2 + 20 + 14 2 + 5 2

2 .

EXERCISE XVII 158.

1.

^ ii

-Tir;

-2, -1,3.

(ii) 1,

-2:

3:

2.

10.

3.

-3/2.

EXERCISE XVIII 165.

3.

4. (1

*N/2,

5. (0, 0),

V,

IT J-N/2),

-f

9. (1,1,

(

(

\/3, q= N/3),

(

7. 2, 1, i,

in

any

(

- iTt'), i>/57~, t,

(

-2

-2

iN/2,

T 0-

order.

6. 2, 3, 4,

in

8. (3, 5),

-f,

(

any

tD

2

x4:

or -4^36

2

;

y, a) y,

(W54^, W^T ii^S), or

a;

= -y

wy

*v2).

-|),

order.

-^).

(f,

2 or -*

(a-iVSdbl^, i^ v

a fourth root of f

.

4 2 2 2 2 2 4 [Use the identity, # + x y + y = (# + xy + y ) (x

12.

T

- i=F iV57).

1),(-|, -f,7).

10. x, MX, 11.

-1

(2, 0), (0, 2),

The only

solutions are (0, a), (a, 0).

xy0 or a

2 ,

and that the

[Remove

-xy + y

fractions

rangement of

= i{l

with ftny ar _

signs.

+ 26 + 2c - 2a)}, etc., with any arrangement of signs. b 2 + a 2c 2 -b 2 c*)labc, etc., the upper or lower sign to be taken

14.

a;

15.

x

16.

&*=a a (& a + c 2 -a 2 ), etc., where

(a

).]

and show that

latter alternative is impossible.] _ t

13. (1, 1),

2

s/(l

z

throughout. 17. (a,6,c)

2o

or

a

a Jfc

2 2 2 a 2 2 2 2 a =(6 + c -a )(c + a -6 )(a + 6 -c ).

a;=(a6-a6 c~l)/(a6

2

c2

-6 2c + 6),

etc.

[Put

s=

HIGHER ALGEBRA

572 PAGE

165.

= (a + 6 + c)/(a-&-c),

or

18. (1,1,1)

etc.

(-3, -3, -3), (>J2, dW2, 1TV2), in any order.

19. (1,1,1),

20. (i,|, -i).

J^V(&

21. If

2

+ c 2 ~a 2 ),

B=^/(c + a*-b 2

(Voc,

),

2z~A+B,

2z=J3-hC', 2y=C+A, 2 2 [Show that 2x(y - 2) = 6 - c .]

then

22.

2

N/(

giving eight solutions.

(i(a-6+c)VQ, J(6 + c-a) Q = Za*~2Ebc.

N/6c),

C=

;

F

VG) where

Zbc^Q, the only solutions are (0, 0, 0), (a, 0, 0), (0, 6, 0), (0, 2bc=Q, the equations are not independent. [Show that

23. If

0, c).

If

x(y-z)/bc =

n>i

,

24.

0, 0,

rt ,

--

x=

or

-

.

..

-

z

= ....]

-

-

a(ca-b )(ub-c

25.

(-!,-!,

26.

The values of

27.

*=y=a: =y =a.

/

etc.

(

-2,

3, |, f).

28. (f,V).

166. 29. (2,1), (-1, -2), (-5,4). 31. a 3 -f-6 3 + c 3 ^a 2 (6 + c)-f6 2 (c + a) (i)

,

)

-1), (i,i,i), (if,0). x, y, u, v are (3, -2, |, |) or

/

32.

-

2

+ c 2 (a + 6).

Subtract the second equation from the first, and divide by x-y; 4 3 5 then, from the fact that prove (x + y*)(l -xy}~ 2a -a ; 2 z are roots of x, y (2a-l)t -\-a(2a-l)t + a (a-l)=Q, obtain (ii)

(2a~l)(o;

from

(i)

2

+ 2/ 2 )=a 2 (2a- l)(z 3 -ft/ 3 )=a 3 (a-2), and z xy = 2a* -a + l =a (a l)/(2a 1) ,

for

;

(iii),

(iv),

(v),

multiply the result

- 2a -

obtain 3a 4

39.

40.

a 4 + 6 4 =c 2 (a 2 + 6 2 ).

167. 42. 16:5: -21.

46.

EXERCISE XIX 1.

i(3dbv/5),

3.

i(-lW3), i(-lW15). 1, i(-3V15), i(-lW15).

5. 7.

-1, -1,

8.

-1,

9.

-1, -1,

t,

2.

i(5iV21).

*,

i(3

(-

4. i( 6.

-1,

+ K/7).

i[3-f N/33

>/(6 N/33-42)J,

K-1 + W3).

16.

i[3

k=-3.

EXERCISE XX 184.

(ii)

1, and the results follow. 2 = ...^...-Za + 2i;bc. 2x(b + c-a) 2 3 3 4 2 2 4 2c 3a& 3a 6 4ac 3 ) + 16d 3 (a 2 - 6 2 ) 2 -0. -f -f 36 (a ) (a

34.

177.

thus

2-68061

0-80125

4-22524.

1.

(i)

2.

2 cos 40 or 1-53209, 0-34730, -1-87939. (ii) 0-69073, 0-11479, -6-30553. (i)

;

(ii)

;

(ii)

by a 4-1, and

ANSWERS

573

EXERCISE XXI PAGE

198.

9.

10.

-lN/2, l2i. i(-l

11. t(l

13. 14.

12. |(1

*s/3).

-2>/2, -4 2*. a 2 3x + 3), i(# -'# + 1)(# 2

\

199.

1(1

N/13)",

{x

2

s/5),

1(3

N/17).

- x(2 -
i>/3) 4-

i

-

{a;

- V5

v2 + 2\^5,

18. s/5

v 2 - 2^5

the other substitutions are given

;

by g + 2# + 5^0. 2

20.

Roots as in Ex. 10

the other substitutions are given by

;

-3=0.

EXERCISE XXII 214.

7.

8.

= t(y -9y); z*-10a; 2 + 1^0; 8

s/2

- i(i/ 3 - lly) -18i/- 110 = 0;

-s/3= 3 7/

N/

;

z9

- 15z 6 - 117z 3 - 125^0.

EXERCISE XXIV 233.

1.

Oscillates

(i)

between

- 2 and + 2

;

(iv)

1

and 3

(ii)

;

converges to 2

diverges

;

;

(iii)

oscillates

between

(v) diverges.

EXERCISE XXVI 254. 255.

1.

(i)

7.

Divergent.

Convergent

Convergent.

15.

Convergent.

Divergent

if

19. Convergent.

23. a?
24.

oscillates finitely.

(ii)

8.

11.

18.

;

12.

Convergent.

16. Divergent.

p-q+

1

^ 0, 25.

10. Divergent.

Convergent.

14. Divergent.

17.

Convergent.

convergent

20. Convergent.

x^l.

Divergent.

13.

9.

Divergent.

if

2>--fl<0. 22. Convergent.

21. Divergent.

x^l.

26.

x
29.

x>a.

30.

x^l.

31.

39.

.

40.

&.

41. -h.

27.

x>I. 38.

28.

x
A-

EXERCISE XXVII 264.

10.

265.

13.

(i)

Between i(5-Vil) and

(ii)if

or between 2

i,

and

|a:|

1-t.

EXERCISE XXVHI 282. 283.

1.

15.

i, i.

2. 0,

-oo.

f, 0.

3.

4.

1.

5.

7/3.

6.

GO.

(

(ii)

-(l-&r(l+aO

(v)

-n

"" cos n 1 x sin

x

2 ;

;

(iii)

(vi)

(a

-2z 2 )/(a 2 + z 2 );

n tann-1 x

sec 2 x.

(iv)

7.

fa;*.

8.

1.

HIGHER ALGEBRA

574 PAGE

283.

-(hx + by+f)l(ax + hy + g);

16.

(i)

21.

y^ax.

-ay)l(ax-y*).

(x*

(ii)

when z^l/3.

24. 4/27,

the graph has a point of inflexion at (1, 0), where the gradient is zero; another at (3/2, -1/16), where the gradient is -J; a minimum at (7/4, -27/256).

25. 1,7/4;

284.

27.

Between x~ -2 and x 0, and for x>3 at ( - 2, - 65) min. at (3, - 190).

;

max. at

(0,

-1); min.

;

29. 18 cubic inches.

EXERCISE XXX a

= 2,

s=

304. 305.

18.

315.

2.

2e

9.

+z+ -[(z~l)e* 3?

3.

l,

Zx^

Use

1,

or a = if, * = 3/5, 3/5, 5/6. substitute the values of 27s 3 , 2J^ 2 , 2k.

5/2;

3 and -272:(a; + 3) ,

EXERCISE XXXI 3.

c-1.

4.

e.

5.

6.

3e/2.

15e.

7.

1-2/e.

8.

-1/c.

l].

EXERCISE XXXII 322.

5. 9.

(ii)

0-84510.

6.

-3/4m; + l/(4w +

1-04139.

(ii)

13

12.2-log,2.

o^o

323.

,r

&1

c--,.-;

17.

1-010299956?

19.

^

20.

-4

(x

t

+ 2) +

6

6

C

15.

1

7.

r

1

-

1-11394.

s

1

1

,

(ii)

21og e 2-l.

11.

10. log, 2.

l).

-^-^.^l.c^-;a^. 0-3010299957.

;

^

r

+ _L_ (5 _ a;2)lo g i_ a;) (

+ 3^ + 2^) +

%x

x2

1

+

-^

(8

.

- x 3 log )

(1

- *).

EXERCISE XXXIII 339.

10.

Convergent for

13.

(i)

If

all

values of

y>a + ]8+l;

(ii) if

(see Art. 11,

y>a +

Ex.

1).

]8.

EXERCISE XXXIV 349.

1.

8th term.

5.

6th and 7th terms.

7.

f

<
23. 1-5860098; 5

.

10~

5th term.

3.

8th term.

6.

5th and 6th terms.

4.

llth term.

8.

.

350. 21. 10-0033322284; 5 25. 1-2431626;

llth term.

2.

10~ 13

.

8

4.10~ 8

.

5

22. 1-414214;

.

24. 1-70998;

.

26. 1-319508;

3

.

3

.

10~ 7

10~ 6 .

.

.

10~ 7

.

ANSWERS

575

EXERCISE XXXV PAGE

356.

10.

(n-f 12.

P-
13.

P-g,

where

_

+ 2+

+

Q--^ 2

^

~

1.

57i

n+ ^9

2

-?i~9 ~~

II* 2 + 26^ + 57a: 4 + I20x 5 2 n 2 ~ - 3. - 1 + 3 2* - 3 W + 2o; + 2x 2 - 4.T3 - 34z 4 - I48,r 6

(i) 1 4- 4o; 4-

1 (ii)

f

7?

;

.

;

EXERCISE XXXVII 365.

1.

366.

4.

1 (iJKl + S"- ),

(20/i

- 27

-f

i(2?i

~ 3 3 n ), J (20w 8

5.

K(23-3n)2^ + (-l)].

370.

1.

2n

371.

8.

n

- 34n -f 27 - 3 3 ~ n ). 6.

-J + l +

EXERCISE XXXVIII

10.

4>

~2 .

=

A .2 n -f-.3 n + i(4n-f7).

W2

2n-l 2n-3 "~'

1

11.

a

2n

EXERCISE XXXIX 384.

1.

s 3 + 2a?-l.

2.

(i)J(4n-l);

(ii)

and

and

EXERCISE XXXVI 359.

*

15

.

+

131 ""'

13n

HIGHER ALGEBRA

576

EXERCISE XL PAGE

394.

1.

+2

H,

|,

(i) J,

-ffs

A

Q

1

,

J.

+3

+4 -3' -"12

'

5 D

T+r+BTsTi* +

1

'

l

A

J.

111

*

6.

(ii) 1,

;

a

1

a-2+

'

^~2^3~4-

-I

1

-^-^-

3

2a 2

a

5*

-a~l *

'a-2'a-l'

2a 2

-

3a

EXERCISE XLI 412.

8.

10.

-f& = 0-242253 12

yards

. . .

^=

,

0-242268

....

3 inches.

;

EXERCISE XLII 419.

127, 55

(i)

3.

7

5.

29 + 41*, 12+17*.

7.

155

+ 8*,

13. 3,

8

;

4

(ii)

2.

+ 27*.

+ 225*,

10. 39, 8

420.

30, 13.

1.

137

+ 199*. 11.

18.

17. 67.

20. 14, 2, 3

22. 2, 1, 3

20 + 29*.

4.

9+

6.

24 + 7*, 11 +

13*,

12*. 9.

1,

20;

8, 9.

A-A; #-

12.

3, 2,

2

;

x = 205 -17*;

13,

12,

x = 201 -

17*.

21. 11, 3, 7. 23.

4, 3, 1.

2.3.7.

1149 + 1540*.

24. 1149;

26.

25.

27.

+ 8*.

f + A.

1, 13, 8.

;

;

1

14. 57, 43,

13, 10.

;

+ 27*,

8. 6, 1.

10, 20.

;

19

One

12 12s. 8d.

;

EXERCISE XLIII 428.

12. 2,

9

;

3,

6

;

4, 13

;

5,

7

;

8,

15

;

10, 12

;

11, 14.

EXERCISE XLIV 434.

1.

(i)9+13*

3.

435.

8.

11.

;

(ii)75

+ 179* (i)78 + 1217*; + f + A-117 + 77 k.

2. (i)

A

149

;

+ 77*;

(iii)

379 + 770*.

(ii)

30+179*.

(ii)

1800+ 1861*

;

9.

(iii)

f+

12. 3

540+ 1009* + H-2.

;

A

+ 23*, -7 + 23*.

(iv)

1683 + 1901*. 10.

ANSWERS

577

PAGE

435.

15. 2,

14. 1735 + 3465*, + 66*, 9 + 66*. 2 x + l=s(inod 7) has no solution. 4

16.

2,

etc.

13

13.

;

-3, -4.

17. 3.

EXERCISE XLV 442.

1.

those of 19 in reverse order

1, 10, 5, 12, etc., i.e.

since 7

number

the 6th

is

2 1 2 5 , 2 7 2 11 2 13 , 2 17 , or ,

2.

1, 3, 9, 10, 13, 5, 15,

number or

444.

12.

,

,

is

divisible

2, 20, 36, 32, 37,

14. Tf

16.

is

2n-fl

since

preceding,

3 2n +*,

;

is

i.e.

every other

prime

to

17-1,

6.

41, hence recurring periods contain 5 digits the remainders in the division of 2 by 41.

by

= 271; w = 123, ^' ^903342366757. p-1 occurs as a remainder.

7i=41, w

Because

and 6

1,

2, 13, 14, 15, 3, 10.

11, 16, 14, 8, 7, 4, 12, 2, 6

10,5, 11, 14,7, 12,

3,

90999

the

in

after

a period of 3, a factor of (19-1);

1, 7, 11,

;

7

EXERCISE XLVI 445.

4. 25.

1.

1, 7, 11.

2.

15.

3.

5.

5.

8, 33.

6.

2, 10, 19.

7.

19fc9, 19&2

8.

(25, 214), (54, 2157), (67, 4120).

18& + 4.

EXERCISE XLVII 447.

1.

10.

2.

15.

3.

5. 5.

4. 4.

4.

EXERCISE XLVIII 456.

2.

(

-

2,

4.

(

6,

6.

(

- 3,

8.

Two

10.

(

- 1), (I, 2), (3, 4). - 7) and two in (0, - 4) and two in (5,

(

1).

5.

(

6).

7.

(2, 3), (3, 4).

9.

(-2, -

in (1,2).

- 6, -

5),

-

1, 0), (4, 5).

(

- 8, - 9), (13, 14), (14, 15). - 9, - 10) and two in (3, 4).

3.

11.

One

1), (0,

root in

l)and two in

(1, 2).

EXERCISE XLIX 466.

2-6306166.

1.

1-3569, 1-6920.

2.

1-7838.

4.

16-0428539.

5.

0-3472964, 1-5320862, -1-8793826.

6.

0-1147994, 0-6907309.

7.

13-8440609, 14-2895592.

5-2100150,5-2973245.

9.

17-7459667. -1-9216606.

8.

10. 1-2783089, 1-5511616.

3.

11. 1-4007219, 1-5823564.

12. 1-0156820, 1-52858586.

13. 3-4334634, 3-6617081.

14. 1-7320508, 1-8666161.

15. 4-4707625, 4-7814825.

16. 2-59861938, 2-76190631.

17. 2-1010205, 2-1084952.

18. 7-3355540.

19.

(i)

3-3548487

;

(ii)

4-4641016

;

(in)

0-6386058.

(1, 2).

;

HIGHER ALGEBRA

578

EXERCISE

LIII

PAGE

c+

504.

1.

(i)6720;

9.

1024.

10.

116280.

505.

13.

(i)

20.

(i)44; (ii)20. n

17.

27.

(i)

506.

480.

(ii)

2.

22.

.

\\n.

57

47

(ii)

;

1771

(i)

(ii)

;

a~ l

.C+*>-i c m

1895040; (ii) 145680. 969 (iv) 165 (iii) 885 ;

;

28. 23,

15.

(iii)

;

7.

\n-lj\r.

29. 42.

(v) 552.

;

30. 15.

EXERCISE LIV 506.

1.

C%-('$.

EXERCISE LV 523.

1.

6.

;

(ii)

5/12

(iii)

5/9.

1/12;

(ii)

125/1296;

(iii)

1/36

(i)

8. (i)

10.

15.

(ii)

23.

12.

18.

(ii)(ft-l)/2;

(n-

(iii)

1/8.

9.

25.

;. (ii)

2/7.

(m + l)(2n-m -2)/n(n-

(iii)

(i)

7.

20/27,

(ii)

21. (n

l)2w,

1).

496/729.

+ l)(3n + 2) pence.

2257/54145.

(ii)

;

1/7

(i)

2(n-7w-l)/n(fl-l);

2162/54145

(i)

4. 5/33.

14/33.

155/648.

l/2w.

19. (i)l//*;

22.

;

236/270725.

13. (i)2//i;

524.

3.

(i)6/55; (ii)4/ll.

(i)

Zp

(H)

24. 73/648.

525.

25. (0-55) 7

OC Ab.

/

;

\

(i)

.

A p-,

O

I

]6

.

13 5i

3-0

2197

13

13 .

.

4U~ 2-0-82.I

/' 1_2 P = 4 tf

/;; \ 11 ) I

i

13

I

I

'

13

49

5"6"

^

13

~4.*=P

.

>

|

(iii)

97 Z/.

Tsp +

J ~ U/ ?71--^

(\\

i$p'-"-p. 2fi

'

2

-^01-30

~ 833

2_^__3_26..

49

/::\ 11 /

>

V

P /_ O. a

24 50

_2_R

51

.

.

26 49

_2 _5 .

__

.

48'-/?,

-

526.

29.

(i)6/w(w~l); (ii)6(-3)/?i(w-l);

33.

(i)

35.

1/2.

36.

(i)

1/6

;

(ii)

pp'p" to

34.

7/9.

(1

-p)(l

-j>')(l

-XO

(i)

;

(iii)

1/4

(n-3)(n-4)/n(w-

;

(ii)

^'(1 -X')

(ii)

to (1 -jp)(l

EXERCISE LVI 541.

542.

5.

(i) (2, 1),

6.

(i)

8.

(i)[|a(a-l)

(50, 31)

;

(ii) (3,

(25, 1), (623, 25),

...

;

2), (29, 18)

(7775, 312),

+ l, i(a-l)];

(ii)

[Ja(a

;

(iii) (5, ...

2), (12, 5).

.

+ 1) +

1,

1).

2/3.

J(a

+ 1)].

-^)X'.

ANSWERS

579

MISCELLANEOUS EXERCISES (A) or

n -! and n 2 + l. 2

4.

2.

ri*

5.

a + c-6-d, a + d-6-c.

i-i + i-A + iis;

(iii)

*=-i

6.

2

-

8.

10.

1L

^. V _,, + 1 o=o.

21. a6 26.

29.

.

>/o

= c + l.

a?=i(n +

23.

or

l)

(a~6) + c a ~c 3 =:2a~6. Convergent if a? <1 or if a;= 1.

a-4-6^0 and (i)

c1

|

the series

is

Convergent

(ii)

30.

12.

1(14-3^);

(i)

p> i,

if

Divergent divergent if

|

p> 1,

if

convergent

divergent

if

a? |

|>

1

.

When

a?=

p^%.

if

(ii)

31. d 6 (

>9, none

34. If 35.

46z= -24 + 15. -.

3.

y

4,

/

- 7 sea <0, four

real,

io

4* A;

+ 10

5, 2

49.

Equate the factors to 4

59.

X

a=

l,

6

or

= l,

A:

43.

i(5N/17).

P 4- Q,

4

- x)

r- log (1

where

or

o>

o>

real. 2 .

4.

x=~2*J2 ic

2* A; 2

- 1 two

!

39.

53.

.

a<

real,

P-fcuQ,

Convergent unless x =

54.

.

X

206 + 2 lOn.

= 0. [y0, ~l,x= 4, are asymptotes; a;

-

66

etc -

-

y is a minimum at (

2

(i)

(ii)

-1+e-l 4

/ 76.

Since

1

\m

+


2 .'.


6 * 05

74.

71. 32/9.

56.

1.

_l- -i

.

.

4

/4\ 3

/3\ 2

r

2 ,

/

n

" X*1 1

...()

<8.

(l (|) (|) L) Consider the arithmetic and geometric means of the factors of P.2 2 .3 3 ..... nn

when

written at full length thus

1.2.2.3.3.3.... 83.31.

85.

Minimum

87.

H

is

a

values

maximum

when

(i)

value

;

x~a

9

(ii)

a;=^(a-

and - 4^f

are

minimum

values.

2,

-i).]

HIGHER ALGEBRA

580 91.0,

2,

96.

QO.

Put p^llm, g = l/7i.

99.

l+l

+

+

?-^

108. 4*, 6*,

-0-68233.

112.

-4-61015.

110.

-1047.

102. 946,

3

2-a,

^>

123.

MISCELLANEOUS EXERCISES 3

ul+^+^i+ir^

-

(iii)

& = 0,

g 4w+a

= 3 4- 6n, 2

(

;

=n

* )q

(B)

'>

g 4w ^3 = 5 + 6n,

+ 2w, for* = 0,

1, 2, ....

6.

15. 3

-3?k-3*k 2 and of the form 6k

21.

n

22.

The values of - a,

23.

31.

is

1,

3^fc,

+

where

has the values

A;

-

1.

6, c

are

pz

~-~

2 .

and similar expressions.

:

(p-q)(p-r)

and

16

29.

e~i, 0.

2+8e-f32e 2

1,

0.

--3-7c.

and

e

33.

(i)

(ii)

x < 4, divergent Convergent Oscillates between finite limits. if

|

35.

D = IJ(a

36.

Put

l

+ a 2 o> + a a w 2

-i-

64.

_

a 4 o> 8 + a 5 a> 4

x=a + h, y=a + k, z=a +

x ^ 4,

l,

2 )

,

where

where

# -= -

oscillates if

4.

-1.


5

=

1

.

h, k and Z->0.

44. (logs) 2 , 2lg*,

41. 5-674619.

58.

if

\

a:

2 ,

X x^e*, 2 9

e

x ,

x*.

59. (Ioga-log6)/(logc-logd;).

Let a, j9, (a<)3) be the roots of Zl(x~b)(x-c)=Q. (i) If I, m, n are odd, x a gives a max. and x=j8 a min. value, (ii) If Z, m, n are even, cc=a and x = j3 give max. values, and x=a, 6, c give min. values, each zero.

c-a Put

c-6

=

67. Converges absolutely if

|l
verges, but not absolutely.

diverges

if

||>J.

If

||=t,

it

con-

ANSWERS

581

'

(a-k)(b-k)

(a'~ k) (b'-k)

(a'

> 2, the roots are all real if A < 0, all are imaginary

69. If A

-k)(b-k)'

< A < 2, two are real and two imaginary if A = 1 or 2.

if

; ;

equal roots

71. 0-8489892. 73.

For the

first part,

use Chap.

show that the second part, r

sum

15, p. 223, putting

of the series <

6 to oo

to e

and

+l

w= -1.

H

=

4n-t-l decreases as

3?t

increases as x increases from

75. (logo;)/a;

from

XIV,

<

(ii)

One root if < n < 1 two roots if nx > xn if x < n < e, or if e < n < x. ;

A rough graph of

(log x)/x should

84. Consider the number of sum is 17 or 28.

85. This

is

so if

n

/ I

-

n

ways

\wM-i

1

rr

(

in

)

P<4,

all

;

< n < ee

x increases 1

;

no root

digits

the roots

* = N/2, -

if

n > ee

.

can be chosen so that their

then use Chap. XIV, 11.

and comparison test (iii), p. 250. are imaginary; if A; 2 ^4 two are real and two

imaginary. If

.

3

be drawn.

which 4

>1

1

91. Convergent; use Euler's constant, p. 311, 95. If

-

.

1 (i)

For the

=-oo.-8in tcos- + Binwhere Bin-j

or

;

INDEX The numbers Abel's inequality, 330 test for convergence, 331 theorem on multiplication of series, 338 Absolute error, 15 least residues, 421

value, 13

residues

of

terms, 8

Approximations, "M, 15-17 to roots of equations, 456-461 to sums of series, 260 Approximate value as a fraction ?'(l +x) (Newton), 347, 348 of log e (l+a;), 323 of log e (a/6) (Napier), 323 Archimedes' axiom (Eudoxus), 13 Associated residues, 425

of

Asymptote, 473 Bernoulli's numbers, 113, 114, 378

theorem, 115 Bessel's interpolation formula, 381 Bicycle -gear as revolution counter, 435 Bilinear substitution, 80

Binomial equations, 170, 171 solution of

a"-l,

series, 263,

reciprocal

form,

Tschirnhauseri's transformation, 105

Bordered determinant, 138 Bounds of a function, 286 of a sequence, 237 of,

Cardan's solution of cubic, 180 Cauchy's condensation test, 325 test for convergence, 253 Change of order of terms, 240, 258 Cofactors of determinant, 123 Combinations and permutations, 403

Commutative law, 13 Comparison tests for convergence, 250 Complex functions of a real variable, 270 Complex numbers, 52 et seq. modulus and amplitude, 56 product and quotient, 60 use as operator, 76

Complex sequences, 243 Complex variable, 201, 334 in binomial series, 328 limits and continuity, 202

Associative law, 13

Binomial

into

258

60, 70

Arithmetic mean, 221 Arithmetical progression,

transformation 104

Brackets, introduction and removal

Absolutely convergent series, 256, 261 Aggregate, 14, 211 Alternating functions, 45 Amplitude of complex number, 56

Argand diagram, 56, 77 product and quotient,

refer to pages.

173 328

for complex variable, 334 Binomial theorem, 34, 204 general statement, 340 elementary proof, 342 Euler's proof, 341 Biquadratic equation, 186-190

condition for four real roots, 102, 453 Euler's solution, 100 Ferrari's and Descartes' solutions, 189-

102 functions /, J, and the discriminant A, 187, 188 reducing cubic, 187-101

roots of equations, 66 Composite numbers, 4 Conjugate numbers, 60 partitions, 402 Congruences, linear, 428 reduction of, 433 x a ~b(modp), 445 Continued fractions, 380 approximations by, 402 equivalent surds, 401, 527 calculation of convergents, 535 Continued product, 33 Continuous functions, 260-271 variable, 266 Continuum, 212 Convergence, general principle of, 235, 240, 244, 257 Convergence of infinite products, 486 Convergence of sequences, 231, 235, 244, 257 Convergence of series, 247 et seq. binomial series, 263, 328, 334

INDEX comparison test, 250 hypergeometric series, 328 (for other tests see under the respective proper names) Cube roots of unity, 65 Cubic as sum of two cubes, 182 Cubic equation, 85 auxiliary quadratic, 182, 184

Cardan's solution, 180 character of roots, 180 functions 0, H, and the discriminant J, 179 irreducible case, 180 roots as power series, 469, 472 Curve tracing, 474 et seq. Cusp, 474 Cyclic expressions, 46

D'Alembert's test for convergence, 252 Dedekind, definition of irrational, 201 theorem on section of system of real numbers, 211 De Gua's rule, 90 Do Moivre's theorem, 59, 63

De Morgan's and

Bertrand's test for con-

vergence, 330 Derivatives of polynomials, 273 x n 309 of a x , log x, Descartes' rule of signs, 88 ,

Determinants, 119 bordered, 138

Expansion off(x + k), 35, 290 Exponential curve, 309 function and series, 19, 306, 312 inequalities and limits, 232, 306, 310 values of sine and cosine, 313

Exponential theorem, 307 Expectation, 514 Eudoxus, theorem of, 13 Euler's constant, 311 solution of biquadratic, 199

theorem on polynomials, 301 Factorial n, powers of primes in, 9 Factors of cos mj> - cos nO t 68 of x n l, 64 of x 2n -2x n coaO + l, 68 Fer mat's theorem, 424, 433 Euler's extension, 425, 437 Ferrari's solution of biquadratic, 189 Finite difference equations, 363, 367 Fourier's theorem on position of roots of

an equation, 454 in connection with Newton's method, 457 Function of a function, 271, 278

rule

Fundamental laws of arithmetic,

13, 53,

of order, 12

et seq.

Gauss's test of convergence, 327 General principle of convergence, 235,

expansion, 132, 137

reciprocal, 136

symmetric, 138-140 use of remainder theorem, 126 Differences, method of, 106 Differentiation, 273, 276 Dirichlet's test for convergence, 331 Discontinuous functions, 269 Discriminant of biquadratic, 188 of cubic, 179 of quadratic in x, y, 30

Displacements and vectors, 70 Distributions, 494, 496

of,

240, 244, 257 Generating function, 362 Geometric series, 248, 262 Gradient of tangent, 274 Greatest term in binomial expansion, 345

Hessian of cubic, 183 Higher derivatives, 280 Highest common factor, 40 ^Homer's method, 461 319 Hyperbolic functions, Hypergeometric series, 328

Imaginary roots of equations, number of, 89, 90

Distributive law, 13 Division (mod n), 432 Division transformation, 24

number and sum

Errors in excess and defect, 15 Exclusive events, 509

54

minors and cofactors, 193 product of two, 134

Divisors,

583

6

Double points on curves, 473 Double roots of equations, 42 Elements of

set, 14 Endless decimals, 16, 203 Equation of a plane, 154

Equations, cubic, biquadratic, 85 equivalent systems, 149 numerical, 447 rational roots, 92 symmetric functions of the roots, 95

65, 82

Implicit functions, 467 Independent events, 511 Indeterminate equations, 415 Indetermination, limits of, 239 Induction, method of, 9 Inequalities, fundamental, 216, 219

A.M.^Q.M., 221 exponential, 239 logarithmic, 306, 310 Infinite continued fractions, Infinite products, 485

390

derangement of factors, 488 expansion as series, 488

INDEX

584

sums of powers of roots of an equation,

Infinite sequences, 15

Inflexion, points of, 282, 473 line of inflexions, 484 Integration, 289 Interdependent events, 512

297 upper limits to roots, 446 Node, 473

Intermediate convergents, 402 Interpolation, Lagrange's formula, 379 Interval, 210 of convergence, 333 Inverse functions in general, 83, 284 circular functions, 285 hyperbolic functions, 321 Inverse probability, 518, 528 Inversions, 48 Irrational numbers, 12, 17 Dedeldnd's definition, 201 Irrational indices, 209

Operators A, E, D, 373, 378, 382 cos 6 + 1 sin 6, or E(id), 75 Order and weight of functions, 298

Rummer's

test for convergence, 306

Lagrange's interpolation formula, 379 theorem on divisibility, 426, 433 Least positive residues, 421 Leibniz'

theorem on differentiation of

product, 383 Limits, exponential and logarithmic, 232, 306, 310 of a function, 228, 229 of indeterrnination, 239 of sequence, 15, 16 to roots of equation, 90 Line at infinity, 152 Linear congruences, 429 Linear difference equations, 369

Logarithmic curve, 309 function and series, 306, 310-316 Logarithms, 210, 317

Maclaurin's theorem, 291 Magnification, 80 Matrices, 135

Maxima and minima, by

five sides,

178

of seventeen sides, 175 Power-series, 332 criterion for identity, 334 Prime and composite functions, 42 Primitive roots, 438, 444

Principal value of amplitude, 56 of inverse circular functions, 285 of inverse hyperbolic functions, 321

Principal nth root, 207 Pringsheim's theorem on convergence, 257 Probability, 508 of causes, 518

geometrical methods, 522, 526 Product of consecutive integers, 7 Lagrange's theorem, 426, 433

Quadratic function of

.r,

y,

29

Raabe's test for convergence, 326 Radius of convergence, 333 Ratio and proportion, 212 Rational fractions, 100, 357 integral functions, 23 roots of equations, 92

differentiation,

inequalities, 223

Mean

value theorem, 288 Minors of determinants, 123, 132 Modulus of complex number, 56-59 Monotone sequences, 230 Multinomial theorem, positive integral index, 35-37 real index,

Pfaffian, 141

Polygon of

Real numbers,

280

by

Partial derivatives, 299 Partial fractions, 100, 302 Partition of numbers, 500 Perfect numbers, 11

354

Multiple roots of equations, 82 Multiplication of series, 335-338 Abel's theorem, 338

E(x)xE(y)=E(x + y), 336 Merten's theorem, 337

Newton's approximation to >/l 4- x, 348 approximation to roots of an equation, 456 method of divisors, 92 parallelogram, 479

18,

201

Reciprocal equations, 168 determinants, 136 Rectangular arrays, 135 Recurring continued fractions, 392, 409 decimals, 439 series,

360

Relative error, 15 Remainder theorem, 26 use in determinants, 126 Residues, associated, 425 of terms of A. P., 8 of terms of G.P., 436 Reversion of series, 467 Rolle's theorem, 287

deductions from, 295 Roots of equations, number position

and character

of,

81

of, 87,

Roots, of complex numbers, 62 of i 1,64 of congruences, 428 Rule of Sarrus, 128

448, 454

INDEX Scalar quantities, 71 Scale of relation, 360 Section of rational system, 200 of real system, 211 Sequence, definition of, 15

bounds, upper and lower limits, 237, 239 general principle of convergence, 235,

244 monotone, 230 Series, approximation to sum, 260 of complex terms, 261

summation

of,

106

Taylor's, Maclaurin's, 290 Series for cos 6, sin 0, 313 for cosw0, sinn#, 61, 360 for cos n 0, sin n 6, 62 for Jog c 2, 312 for tan- 1 .*-, 294 Series summablc by the binomial theorem,

351 exponential series, 3 1 4 logarithmic series, 317 n r in terms of
,

...

T 6(6

+

(a

1). ..(&

+ r-

1)

+ r-l)

,

16),

106

110

Simple continued fractions, 391 Smallness, orders of, 14 Skew- symmetric determinants, 138, 140 n 1, 171 Special roots of x n 515 Stirling's approximation to |

Sturm's theorem, 448, 452 Substitutions, 47 Successive events, 515 Sums of powers of roots, 297 of two squares, 407, 408

585

Surds, 208 expressed as continued fractions, 401, 427, 435 JP, 7, 8 Symbols, I(r/y), ,

[n,

0(x), 14 t as an operator, 75 l /- (*),83

Symmetric continued

fractions,

406

determinants, 138

Symmetric functions, 44 of roots cf equations, 95 order and weight of, 298

Synthetic division, 24, 84 Systems of equations, linear, 149 of any degree, 160

Tangent to a curve, 274 Tangents and asymptotes, 473 Taylor's theorem and series, 290 for polynomials, 300 Testimony, value of, 521 Test-series for convergence S\jn'p 251 Z'l/n(lo g n), 325 Tetrahedron, volume of, 159 Three planes, 156 Transformation, Z = (az + b)f(cz+d), 80 Transformation of equations, 82-84 Transpositions, 44 ,

;

Tschirnhausen's transformation for cubic, 186 for biquadratic, 195

Undetermined

coefficients,

28

Vandermoride's theorem, 340 Vectors, 70-76

Weierstrass' inequalities, 216 Whit worth's theorem, 499 Wilson's theorem, 425, 433