LIBRARY OF
WELLESLEY COLLEGE
PRESENTED BY TVof Horafo-rd
HIGHER ALGEBRA A SEQUEL TO
ELEMENTARY ALGEBRA EOR SCHOOLS.
s.
HIGHER ALGEBRA A SEQUEL TO
ELEMENTARY ALGEBRA FOR SCHOOLS
BY
H.
S.
HALL,
M.A.,
FORMERLY SCHOLAR OF CHRIST'S COLLEGE, CAMBRIDGE, MASTER OF THE MILITARY AND ENGINEERING SIDE, CLIFTON COLLEGE AND S.
K
KNIGHT,
B.A.,
FORMERLY SCHOLAR OF TRINITY COLLEGE, CAMBRIDGE, LATE ASSISTANT-MASTER AT MARLBOROUGH COLLEGE.
FOURTH EDITION.
Honfcon:
MACMILLAN AND
CO.
AND NEW YOKE. 1891 [The Right of Translation
is
reserved.}
;
2-
TO
422
12/
First Printed 1887.
Second Edition with corrections 1888.
Third Edition revised and enlarged 1889. Reprinted 1890. Fourth Edition 1891.
}
PREFACE. The
present work
is
intended as a sequel to our Elementary
Algebra for Schools. The first few chapters are devoted to a fuller discussion of Ratio, Proportion, Variation, and the Progressions, which in the former work were treated in an elementary manner and we have here introduced theorems ;
and examples which are unsuitable
a
for
first
course of
reading.
From part new
work covers ground for the most the student, and enters upon subjects of special
this point the
to
importance
:
these
we have endeavoured
to treat
minutely
and thoroughly, discussing both bookwork and examples witli that fulness which we have always found necessary in our experience as teachers. It has
as
been our aim to discuss
completely as possible
within
all
the
the essential parts limits
volume, but in a few of the later chapters possible to find
room
in all such cases first
it
of
a
single
has been im-
more than an introductory sketch our object has been to map out a suitable for
;
course of reading, referring the student to special treatises
for fuller information.
In the chapter on Permutations and Combinations we
much indebted to the Rev. W. A. Whitworth for permission to make use of some of the proofs given in his Choice and Chance. For many years we have used these proofs in our own teaching, and we are convinced that this
are
;
PREFACE.
vi
part of Algebra
made common is
far
more
intelligible to the
beginner
sense reasoning from first principles by a system of than by the proofs usually found in algebraical text-books. The discussion of Convergency and Divergency of Series
always presents great difficulty to the student on his first The inherent difficulties of the subject are no reading.
doubt considerable, and these are increased by the place it has ordinarily occupied, and by the somewhat inadequate treatment it has hitherto received. Accordingly we have placed this section somewhat later than is usual; much thought has been bestowed on its general arrangement, and
on the selection of suitable examples to illustrate the text and we have endeavoured to make it more interesting and intelligible by previously introducing a short chapter on Limiting Values and Vanishing Fractions. In the chapter on Summation of Series we have laid
much
stress
on the " Method of Differences" and
important applications.
known formula
The
basis of this
its
method
in the Calculus of Finite Differences,
wide and is
a well-
which in
the absence of a purely algebraical proof can hardly be considered admissible in a treatise on Algebra. The proof of the Finite Difference formula which 396,
we
believe to be
of the Difference
new and
Method from
we have given
original,
this
in Arts. 395,
and the development
formula has enabled us to
many interesting types of series which have hitherto been relegated to a much later stage in the student's reading. introduce
We
have received able and material assistance in the
chapter on Probability from the Rev. T. C.
Simmons
of
and our warmest thanks are due to him, both for his aid in criticising and improving the text, and for placing at our disposal several interesting and Christ's College, Brecon,
original problems. It
is
Analytical
hardly possible to read any modern treatise on
Conies or Solid Geometry without some know-
PKEFACE. ledge of Determinants and
their
yii
therefore given a brief elementary discussion
nants in Chapter
xxxm.,
in the
We
applications.
hope that
have
of Determi-
may
it
provide
the student with a useful introductory course, and prepare him for a more complete study of the subject.
The
last
chapter contains
all
the most useful propositions
Theory of Equations suitable for a first reading. The Theory of Equations follows so naturally on the study of Algebra that no apology is needed for here introducing pro-
in the
which usually find place in a separate treatise. In fact, a considerable part of Chapter xxxv. may be read with advantage at a much earlier stage, and may conveniently be studied before some of the harder sections of previous positions
chapters. It will
be found that each chapter
complete in
itself,
is
as nearly as possible
so that the order of their succession can
be varied at the discretion of the teacher *
mended
that
all
sections
marked with an
;
but
it is
recom-
asterisk should be
reserved for a second reading.
In enumerating the sources from which we have derived assistance in the preparation of this work, there to
which
it is
difficult
to
say
how
far
we
is
are
one book indebted.
Todhunter's Algebra for Schools and Colleges has been the recognised English text-book for so long that it is hardly
any one writing a text-book on Algebra at the present day should not be largely influenced by it. At the same time, though for many years Todhunter's Algebra has been in constant use among our pupils, we have rarely adopted the order and arrangement there laid down; in many chapters we have found it expedient to make frequent use of alternative proofs; and we have always largely supThese notes, plemented the text by manuscript notes. which now appear scattered throughout the present work, have been collected at different times during the last twenty possible that
H. H. A.
b
;
PREFACE.
Viii
years, so that
ment
impossible to
it is
make
definite acknowledge-
where assistance has been obtained from But speaking generally, our acknowledge-
in every case
other writers.
ments are chiefly due to the treatises of Schlomilch, Serret, and Laurent; and among English writers, besides Todhunter's Algebra, we have occasionally consulted the works of De Morgan, Colenso, Gross, and Chrystal. To the Rev. J. Wolsienholme, D.Sc, Professor of Mathematics at the Royal Indian Engineering College, our thanks due for his kindness in allowing us to select questions from his unique collection of problems and the consequent
are
;
we
gain to our later chapters It
remains
for
us to express our thanks to our colleagues
and friends who have
so largely assisted us in
correcting the proof sheets
C
the Rev. H.
Watson
**'
1887
;
in particular
we
reading and
are indebted to
of Clifton College for his kindness in
revising the whole work, in every part of
gratefully acknowledge.
and
for
many
valuable suggestions;
it.
"
H.
S.
S. R.
HALL, KNIGHT.
PREFACE TO THE THIRD EDITION. and examples are substantially the same as in previous editions, but a few articles havej been recast, and all the examples have been verified again. We have also added a collection of three hundred Miscellaneous Examples which will be found useful for advanced These examples have been selected mainly but students. not exclusively from Scholarship or Senate House papers much care has been taken to illustrate every part of the subject, and to fairly represent the principal University and In
this edition the text
Civil Service Examinations. March, 1889.
j <
;
CONTENTS. CHAPTER
ratio.
I.
PAGE
Commensurable and incommensurable quantities Ratio of greater and less inequality
2
.
3
i
a
_c _e
/pan + qc n +re n +
_
...\
n
b~d~f~"'~\pb n + qd n + rf n +...J + a 2 + a 3 +... + an b l + b 2 + b.i + ... + b n
a1
lies
'
between greatest and least of fractions
flh
V
an
Cross multiplication
8
Eliminant
9
Examples
of three linear equations
10
I
CHAPTER
proportion.
II.
and Propositions Comparison between algebraical and geometrical Case of incommensurable quantities Examples II. Definitions
CHAPTER If
Ace B, then
III.
13
1G
definitions
17 19
VARIATION.
A = mB
21
Inverse variation
22
Joint variation
23
li
Ace
B when G
is
constant,
and A
& C when B
is
constant, then
A=mBG Illustrations.
Examples
23
Examples on
joint variation
.
.
.
.
.21
.
20
III
b-1
.
CONTENTS.
CHAPTER
IV.
ARITHMETICAL PROGRESSION. PAGE
Sum
n terms of an arithmetical Fundamental formulae Insertion of arithmetic means Examples IV. a Discussion of roots of dn~ + (2a
28
series
of
29 31
31
-d)n-2s =
33 35
Examples IV. b
f'
CHAPTER
Insertion of geometric
Sum Sum
V.
GEOMETRICAL PROGRESSION.
....
means
n terms of a geometrical series of an infinite geometrical series Examples V. a. of
38 39
.
40 41
Proof of rule for the reduction of a recurring decimal
43
Sum
44
n terms Examples V. b of
of
an arithmetico-geometric
series
45
CHAPTER VI.
HARMONICAL PROGRESSION. THEOREMS CONNECTED WITH THE PROGRESSIONS.
......
Reciprocals of quantities in H. P. are in A. P.
Harmonic mean
Formulae connecting A. M., G.M., H.M. Hints for solution of questions in Progressions
Sum of squares of the natural numbers Sum of cubes of the natural numbers 2 notation Examples
Number
VI.
a.
of shot in
.
..... pyramid on a square base
Pyramid on a triangular base Pyramid on a rectangular base Incomplete pyramid Examples VI. b
.....
CHAPTER
VII.
5J
scales of notation.
Explanation of systems of notation
Examples VII.
57
a.
Expression of an integral number in a proposed scale Expression of a radix fraction in a proposed scale
59 .
59 01
CONTENTS.
XI PAGE
The
difference between a
number and
tho
sum
of its digits
is
divisible
by r - 1
62
Proof of rule for " casting out the nines " Test of divisibility by r
C3
+1
64
Examples VII. b
65
CHAPTER
SURDS AND IMAGINARY QUANTITIES.
VIII.
a
nationalising the denominator of sjb
07
+ jc + s/d
Rationalising factor of fJa±Z/b Square root of a + Jb + *Jc + Jd
68
69
Cube root of a + *Jb Examples VIII. a. Imaginary quantities
70 72 74
J -ax J -b= - sjab
75
.
If
a + ib = 0, then a = Q, b =
If
a
+ ib = c + id, then a = c,
Modulus
of product
Square root of a + Powers of i
is
75
.
b
=d
75
equal to product of moduli
77
ib
77
79
.
Cube roots of unity Powers of u Examples VIII. b.
;
1 + w -f or =
79
80
.
81
CHAPTER A
IX.
THE THEORY OF QUADRATIC EQUATIONS.
...
quadratic equation cannot have more than two roots
Conditions for
Sum
of roots
=
real, equal,
— b
a
imaginary roots
84
c
,
product of roots = a
Formation of equations when the roots are given
.....
Conditions that the roots of a quadratic should be tude and opposite in sign,
(2)
;
(1)
88 in general the
same 90
Examples IX. b
...
Definitions of function, variable, rativnnl integral function
Condition that ax 2 + 2hxy+ by 2 + 2gx
+ 2fy + c may
92 93
be resolved into two 9
linear factors 2
+ bx + c = and
a'x- + b'x
+ c' = may have
a
i
common 96
root
Examples IX.
86
88
exceptions
Condition that ax
85
equal in magni-
reciprocals
Examples IX. a For real values of x the expression ax 2 + bx + c has sign as a
83
c.
96
CONTENTS.
Xll
CHAPTER
MISCELLANEOUS EQUATIONS.
X.
page
Equations involving one unknown quantity
97 100
.
Reciprocal equations
Examples X. a
101
Equations involving two unknown quantities
103
Homogeneous equations Examples X. b
104
....
106
Equations involving several unknown quantities
107
Examples X.
109
c.
Indeterminate equations
;
111
easy numerical examples
113
Examples X. d
CHAPTER XL
permutations and combinations.
Preliminary proposition
•
•
.115
115 Number of permutations of n things r at a time 117 Number of combinations of n things r at a time The number of combinations of n things r at a time is equal to the .119 number of combinations of n things ?i-rata time Number of ways in which m + n +p + ... things can be divided into .
.
classes containing m, n, p,
...
things severally
Examples XI. a Signification of the terms 'like'
Number
of arrangements of
and 'unlike'
n things taken
all at
....
..... a time,
when p
of
permutations of n things r at a time, when each
may
.
124
125
be
repeated
126
The total number of combinations of n things To find for what value of r the expression n Gr is Ab initio proof of the formula for the number
127 greatest
.
.
.
127
of combinations of n
things r at a time
128
Total number of selections of
p + q+r+
...
things, whereof
p
are alike
of one kind, q alike of a second kind, &c
Examples XI. b
129 131
CHAPTER Illustrations of the
122
things
are alike of one kind, q things are alike of a second kind, &c.
Number
120
XII.
mathematical induction.
method of proof
133
Product of n binomial factors of the form x + a
134
Examples XII
135
CONTENTS.
CHAPTER XIII.
Xlii
BINOMIAL THEOREM.
POSITIVE INTEGRAL INDEX.
....
Expansion of (x + a) when n is a positive integer General term of the expansion The expansion may be made to depend upon the case in which the first term is unity Second proof of the binomial theorem Examples XLII. a The coefficients of terms equidistant from the beginning and end 11
,
are equal
PAGE 137 139
140 141
142 143
Determination of the greatest term
143
Sum Sum
146
of the coefficients of coefficients of
odd terms
equal to
is
sum
of coefficients of even
terms
146
Expansion of multinomials Examples XIII. b.
CHAPTER
XIV.
146 147
ANY INDEX.
BINOMIAL THEOREM.
.....
Euler's proof of the binomial theorem for any index
General term of the expansion of
(1
+ x)'
1
150 153
Examples XIV. a Expansion of (l-rx) n is only arithmetically intelligible when x
(1
-
Particular cases of the expansions of
_n .r)
(1
....
157 158
Approximations obtained by the binomial theorem Examples XIV. b.
Number
of
homogeneous products of
r
155
157
- x)~ n
Numerically greatest term in the expansion of
155
(l
159 161
+ x)
n
162
.
dimensions formed out of n 164
letters
Number of terms in the expansion of a multinomial Number of combinations of n things r at a time, repetitions being allowed
105
Examples XIV.
107
c
CHAPTER XV. General term in the expansion of
MULTINOMIAL THEOREM. (a
+ bx + ex 2 + dx 3 +
...) p ,
when ^
is a
170
positive integer
General term in the expansion of (a + bx + cx- + is
a rational quantity
Examples
XV
166
+
...)
n ,
when
//
171
173
CONTENTS.
XIV
CHAPTER
XVI.
LOGARITHMS. PAGE
Definition.
N=a)og a N
175 176
Elementary propositions Examples XVI. a Common Logarithms
178
Determination of the characteristic by inspection
.179
..... •
•
....
Advantages of logarithms to base 10 Advantages of always keeping the mantissa positive Given the logarithms of all numbers to base a, to find the logarithms to base b
.
.
.
180 181
182 183
log a &xlog 6 a = l
183
Examples XVI. b
185
CHAPTER Expansion e is
XVII.
of ax .
the limit of
( (
Series for l\ n 1
+-
V
Expansion of
EXPONENTIAL AND LOGARITHMIC SERIES.
log,, (1
V )
,
187
e
when n
188
is infinite
+ x)
191
Construction of Tables of Logarithms
Rapidly converging series for
log,,
(n
192
.
+ 1) - log n
194
e
The quantity e is incommensurable Examples XVII
CHAPTER
XVIII.
195 195
INTEREST AND ANNUITIES.
and Amount of a given sum at simple interest Present Value and Discount of a given sum at simple interest Interest and Amount of a given sum at compound interest Nominal and true annual rates of interest Interest
.
.
.
.
.198
.
.
198
.
.
199
....
Case of compound interest payable every moment Present Value and Discount of a given sum at compound interest Examples XVIII. a Annuities.
Amount Amount
.
.
200 200 201 202
Definitions
202
of unpaid annuity, simple interest
203
of unpaid annuity,
203
compound
interest
Present value of an annuity, compound interest Number of years' purchase Present value of a deferred annuity, compound interest
Fine for the renewal of a lease Examples XVIII. b
204 204 .
.
.
.205 206 206
CONTENTS.
CHAPTER
XIX.
XV
INEQUALITIES. PAGE 208
Elementary Propositions
mean of two positive quantities
Arithmetic
mean The sum of two
is
greater than the geometric
209 quantities being given, their product
they are equal
:
sum
product being given, the
is
is least
greatest
when
when they
are
equal
210
The arithmetic mean of a number of positive quantities is greater than the geometric mean Given sum of a, &, c, ...; to find the greatest value of am b n c p Easy cases of maxima and minima Examples XIX. a The arithmetic mean of the ?/i th powers of a number of positive quantities
except
is
when
greater than
m th
power of and 1
m lies between
their arithmetic
211
212 212
213
mean, 214
b
If a
and
b are positive integers,
and a>b,
(
1
+-
)
>
1
(
+^
216
)
? 1> * > » >0'Vrr| > vrrfy 'a
217
+ b\ a+b
217
Examples XIX. b
218
CHAPTER XX.
LIMITING VALUES AND VANISHING FRACTIONS.
....
Definition of Limit
+ a x x + a 2 x" + a 3
+ ...
a when x is zero By taking x small enough, any term of the series a + a rr + a^x- + ... may be made as large as we please compared with the sum of all that follow it; and by taking x large enough, any term may be made as large as we please compared with the sum of all that Limit of a
precede
Method
it
x3
is
some
peculiarities
in the
solution
equations Peculiarities in the solution of quadratic equations
Examples
of
.
.
.
221
simultaneous
....
XX
CHAPTER
222
222
.
of determining the limits of vanishing fractions
Discussion of
220
226
227 228
XXI.
CONVERGENCE AND DIVERGENCY OF
Case of terms alternately positive and negative Series is convergent
if
u Lim ~ n «u
-i
is less
than
1
SERIES.
.....
230 232
.
CONTENTS.
XVI
PAGE
Comparison
The
of 2rtn with
auxiliary series
^p
an auxiliary
.
234
.
235
.
237
.
238
.
238
.
241
.
243
.
244
]
.
245
{n)
.
247
.
248
2v n
series
.
+ 2P + 3~p +
Application to Binomial, Exponential, Logarithmic Series log
Limits of
Product of
71
n an
and nx n when n
infinite
Examples XXI.
a.
number
of factors
.
w-series is convergent
when
Series is convergent
Lim
if
u
convergent
is
Series
20
The
(n)
if
]n
(
2
Series is convergent if
two infinite Examples XXI. b.
n
I
—— - 1
v
if
u ji-i
compared with
auxiliary series
Product
Lim
convergent,
v- series is
<
Series
is infinite
v n-l
)
\un+i
J
n log
—— >
series
2a
)
,l
(log n) p
Lim
\n
of
series
CHAPTER
XXII.
(
—~
1
J
-
l| log
n
248
249 252
UNDETERMINED COEFFICIENTS.
f(x)-=0 has more than n roots, it is an identity Proof of principle of undetermined coefficients for finite series
If the equation
.
.
254
.
.
254
Examples XXII. a
256
Proof of principle of undetermined coefficients for infinite series
Examples XX1T. b
CHAPTER
XXIII.
.
.
257 2C0
PARTIAL FRACTIONS.
Decomposition into partial fractions
261
Use of partial fractions Examples XXIII
265
in expansions
CHAPTER XXIV.
265
recurring series.
Scale of relation
267
Sum
269
of a recurring series
Generating function
269
Examples XXIV
272
.
CONTENTS.
CHAPTER XXV.
XV ii
continued fractions.
....
Conversion of a fraction into a continued fraction Convergents are alternately less and greater than the continued fraction Law of formation of the successive convergents 1)n
Pn&w-l-Pn-l4n=(~ Examples XXV. a. The convergents gradually approximate to the continued fraction Lhnits of the error in taking any convergent for the continued fraction Each convergent is nearer to the continued fraction than a fraction .
with smaller denominator
—
Pp' ,:>
P as->
or
qq
.
PAGE 273 275
275 27G 277
278 279
280
.
p'
<—
or
281
.
q
q
Examples XXV. b
281
CHAPTER XXVI.
indeterminate equations of the first DEGREE.
Solution of ax-bi/ = c
284
Given one solution, to find the general solution Solution of ax + by = c
286
Given one solution,
287
Number
to find the general solution
of solutions of
ax + by = c
287
.
Solution of ax + by + cz = d, a'x + b'y +
Examples XXVI.
286
...
c'z
= d'
289
290
.
CHAPTER XXVII.
recurring continued fractions.
Numerical example
A
292
periodic continued fraction
is
equal to a quadratic surd
.
.
.
Examples XXVII. a
21)4
Conversion of a quadratic surd into a continued fraction
The quotients recur The period ends with a partial quotient 2a x The partial quotients equidistant from first and The penultimate convergents of the periods Examples XXVII. b. .
CHAPTER
293
XXVIII.
.
.
.
.
295
296 297 last are equal
.
.
298 299 301
indeterminate equations of the second DEGREE.
Solution of ax 2 + 2hxy
+ by* + 2gx + 2fy + c =
The equation # 2 - Ny2 =l can always
be solved
303
304
xviii
CONTENTS.
CONTENTS.
XIX PAGE
If
a
prime to b, then different remainders is
......
a, 2a, 3a,
(6-
...
a when divided by
1)
6 leave
350
(p(abcd...)=
352
PO-»'(i-i)(i-J)(i-l)
352
Wilson's Theorem
A
1
:
+
\p
-
1
=M
(p)
where p
property peculiar to prime numbers
a prime
354
.... is
354
.......
Wilson's Theorem (second proof)
355
Proofs by induction
35G
Examples XXX.
357
b.
CHAPTER XXXI.
the
general
theory
of
continued
FRACTIONS.
Law
of formation of successive convergents
—— —-
a,+
a.2
+
...
has a definite value
if
Lim
n4' 1
'" y
The convergents
to
h
\
. .
a l ~ a 2~
ing order of magnitude,
if
.
359
.
>0
362
.
n+l
are positive proper fractions in ascend-
a n
+ bn
363
General value of convergent when a n and b n are constant Cases where general value of convergent can be found
364 365
.
a x + a2 +
is
incommensurable,
if
— <1
366
cl
Examples XXXI. a
367
Series expressed as continued fractions
369
.
'Conversion of one continued fraction into another
371
Examples XXXI. b
372
CHAPTER XXXII. and illustrations. (Examples XXXII. a /Compound Events Definitions
probability.
Simple Events
373
.
376 377
I
Probability that two independent events will both happen
[
The formula holds
also for dependent events
is
pp'
.
Chance of an event which can haj^pen in mutually exclusive ways Examples XXXII. b Chance of an event happening exactly r times in n trials Expectation and probable value "Problem of points" .
.
....... .......
378 379 381
383 385
386
388
CONTENTS.
XX
PAOE
Examples XXXII.
389
c.
391
Inverse probability
Statement of Bernoulli's Theorem Proof of formula
.
P P Q r = ^rj-jn
Concurrent testimony Traditionary testimony
392
.... ....
396
Examples XXXII. d
899
Local Probability. Geometrical methods Miscellaneous examples
401
402
Examples XXXII.
405
e
CHAPTER XXXIII.
dktkrminants.
.....
Eliminant of two homogeneous linear equations Eliminant of three homogeneous linear equations Determinant is not altered by interchanging rows and columns Development of determinant of third order .
Sign of a determinant
is
altered by interchanging
.
fcw<
.
.
.
.
.
.
411
412
.
.
Keduction of determinants by simplification of rows or columns Product of two determinants
.
.
112
.
.
412
.
.
413
.
.111 417
Examples XXXIII. a Application to solution of simultaneous equations
Determinant of fourth order Determinant of any order Notation
Sia^-^
Examples XXXIII.
b.
... .
.
CHAPTER XXXIV.
.
.
.
.
.
.
List of useful formula?
worked out .
419 422 423 42jl
425 \-r,
miscellaneous theorems and examples.
Horner's Method of Synthetic Division Symmetrical and Alternating Functions of identities
... ... .
Keview of the fundamental laws of Algebra f(x) when divided by x - a leaves remainder Quotient of / (x) when divided by x - a Method of Detached Coefficients
Examples
410
adjacent rows or
columns If two rows or columns are identical, the determinant vanishes A factor common to any row or column may be placed outside Cases where constituents are made up of a number of terms .
409
.410
.
429
/»
432 433
434 434 435
437
438
CONTENTS.
XXI PAGE
Examples XXXIV. a Identities proved
438
by properties of cube roots of unity
440
Linear factors of a 3 + 6 3 + c 3 - Sabc
441
Value of an + bn + cn when a + b + c = Q
442
Examples XXXIV.
442
b.
.
Elimination
444
Elimination by symmetrical functions
444
Euler's
method
of elimination
Method
445
....
Sylvester's Dialytic
Bezout's method
446
.
446
Miscellaneous examples of elimination
447
Examples XXXIV.
449
c.
.
.
.
CHAPTER XXXV.
theory of equations.
Every equation of the n th degree has n roots and no more Kelations between the roots and the coefficients These relations are not sufficient for the solution Cases of solution under given conditions Easy cases of symmetrical functions of the roots
Examples XXXV. a. Imaginary and surd roots occur in pairs Formation and solution of equations with surd
.
.
.
452 452 454 454 455
456
....
roots
457
458
Descartes' Kule of Signs
459
Examples XXXV. b
460 462
Value of /(.r + Calculation of
/
(
x)
changes
If f(a)
An An
f(x+h) by Horner's
its
463
process
464
value gradually
and/ (b)
a and
Derived Functions
//).
are of contrary signs, f(x)
=
has a root between 464
6
465
equation of an odd degree has one real root equation of an even degree with
its last
term negative has two real 465
roots
/ (x) =
has r roots equal to Determination of equal roots If
/'(*)_
I
(X)
x-a
J
Sum
1
f (x) =
has r - 1 roots equal to a
466 467 468
,
x-c
an assigned power of the roots Examples XXXV. c Transformation of equations Equation with roots of sign opposite to those of f(x) = Equation with roots multiples of those of f{x) =0 of
.
1 ,
x-b
a,
468
.
.
470 471 .
.
.
471
•
•
•
472
CONTENTS.
XX11
Discussion of reciprocal equations
=
....
Equation with roots reciprocals of those of /
(x)
.
475
.
475
.
476
.
477
.
Equation with roots squares of those of f(x) = Equation with roots exceeding by h those of f (x) = Bemoval of an assigned term Equation with roots given functions of those of f{x)-. Examples XXXV. d Cubic equations. Cardan's Solution .
.
478 480 481 482 483 484 486 486
.
487
.
.
Discussion of the solution
.
Solution by Trigonometry in the irreducible case
Biquadratic Equations.
.
PAGE 472 473
Ferrari's Solution
.
.
.
.
Descartes' Solution
.
Undetermined multipliers
.
Discriminating cubic
;
roots all real
x
Solution of three simultaneous equations
Examples XXXV. e. Miscellaneous Examples Answers
y =1, a+\ + b + \ + c + \
&c.
488 490 525
HIGHER ALGEBRA. CHAPTER
I.
RATIO.
Ratio is the relation which one quantity Definition. bears to another of the same kind, the comparison being made by considering what multiple, part, or parts, one quantity is of the 1.
other.
•
The and
A
A
B
is usually written A to The quantities B. are called the terms of the ratio. The first term is
ratio of
B
:
called the antecedent, the second
To
2.
by -^
,
B
;
find
what multiple or part A
hence the ratio
and we
term the consequent.
A
B may
:
is
of B,
we
divide
A
be measured by the fraction
shall usually find it convenient to
adopt this notation.
In order to compare two quantities they must be expressed in terms of the same unit. Thus the ratio of £2 to 15s. is measured
2x20 .... by the traction —^—
or
8 -
.
A
ratio expresses the number of times that one quantity contains another, and therefore every ratio is an abstract quantity.
Note.
3.
Since by the laws of fractions,
a b
=
ma mJ'
follows that the ratio a b is equal to the ratio ma mb that is, the value of a ratio remains unaltered if the antecedent and the consequent are multiplied or divided by the same quantity. it
:
H. H. A.
:
1
;
HIGHER ALGEBRA.
2
or more ratios may be compared by reducing their Thus suppose equivalent fractions to a common denominator.
Two
4.
bx x and — = =— hence h by by y the ratio a b is greater than, equal to, or less than the ratio x y according as ay is greater than, equal to, or less than bx. _
a
_
and x
o
:
y are two J
:
a a JNow - = ~V xt
x-
ratios.
,
i
:
3
:
:
The
5.
two fractions can be expressed as a
ratio of
C
Ch
of
two
Thus the
integers. °
—
ratio
ratio
— a
:
b
is
measured by the
a
—
fraction
,
or =—
:
and
therefore equivalent
is
to
the
ratio
be
c
d ad
be.
:
or both, of the terms of a ratio be a surd quantity, then no two integers can be found which will exactly measure their ratio. Thus the ratio J'2 1 cannot be exactly expressed by any two integers. If either,
6.
:
If the ratio of any two quantities can be Definition. expressed exactly by the ratio of two integers, the quantities otherwise, they are said to be are said to be commensurable incommensurable. 7.
;
Although we cannot find two integers which will exactly measure the ratio of two incommensurable quantities, we can always find two integers whose ratio differs from that required by as small a quantity as we please.
J5 = V J5 > — mm(>
2-236068...
Thus
-.
4
,
,
.
and therefore
™
A1I , -559017...
=
4
559017
so that the difference
559018
,
and <
-jooOOOO
between the
;
559017
ratios
:
1000000 and
than -000001. By carrying the decimals further, a closer approximation may be arrived at.
J5
:
4
is less
Ratios are compounded by multiplying toDefinition. gether the fractions which denote them ; or by multiplying together the antecedents for a new antecedent, and the consequents for a new consequent. 8.
Example.
Find the
ratio
2a
compounded :
Sb, Q>ab
:
of the three ratios
5c 2 c ,
:
a
KATIO.
m The .
,
,.
required ratio = .
2a
Gab -x x 6b a be c
--„ 1
_4a ~ DC
'
Definition. When the ratio a b is compounded with 2 2 and is called the duplicate ratio itself the resulting ratio is a b 3 3 Similarly a b is called the triplicate ratio of a of a b. b. 9.
:
:
,
:
:
Also a
2
b
:
2"
is
Examples.
:
called the subduplicate ratio of
a
:
b.
The duplicate ratio of 2a 3b is 4a 2 96-. The subduplicate ratio of 49 25 is 7 5. The triplicate ratio of 2x 1 is 8a; 3 1.
(1)
:
(2)
:
:
(3)
:
A
Definition.
:
:
said to he a ratio of greater inequality, of less inequality, or of equality, according as the antecedent is greater than, less than, or equal to the consequent. 10.
A
ratio
is
of greater inequality is diminished, and a ratio of inequality is increased, by adding the same quantity to both
11. less
ratio
terms.
its
a , ,, Let T be the
..
,
ratio,
and
,
,
let
6
adding x to both
its
Now
a+x = be the new ratio formed by J b + x
terms. erms.
a
a+x
ax — bx
b
b
+x
b(b+x) x(a — b) ~b(b + x)
}
and a - b is positive or negative according as a less than b. a +x a y ence it a > b, T > ^ o b + x
H.
is
greater or
/,
;
«
j
d.it
a
a 7-
<
b
x
a
~t~
b
+ x
;
which proves the proposition. can be proved that a ratio of greater inequality increased, and a ratio of less inequality is diminished, by taking Similarly
is
the
it
same quantity from both
its terms.
two or more ratios are equal many useful propositions may be proved by introducing a single symbol to 12.
When
denote each of the equal ratios.
1—2
;
HIGHER ALGEBRA.
4
The proof of the following important theorem the method of procedure. a b
//
c
e
d
f
will illustrate
'
1
.
,
.
.
each of J these ratios
where
r,
p, q,
/pan r-
n
+ qc + re + j\pb n + qd n + rtsn + 11
= —
-
—,
b
d
7
a—
bk,
^
pan =pbn
~>—
.
. .
,
)
/
= dk,
c
qc"
—& j
•
• •
')
J
= qd"k
e n
=fk, ...; re
,
n
= rf"k",... n
n
pa" + gc" + re + '''
.
quantities ivhatever.
ace = —
Ijet
whence
.
(
n are any
then
.\ n
n
_pb"k" + qd"k + rf"k +
...
H
pb + qd"+rf+...
n fb +
'
qcl
n
+r/i +...
= k"; i
'pa"
e
2)b"
a c + re" + .\ n _ = k = ^ = -,= d b + qd" + ?'/" + .../
+
qc"
.
,
.
giving different values to p, q, r, n many particular cases of this general proposition may be deduced ; or they may be proved independently by using the same method. For instance,
By
a
_c
e
b-~d'f-'" each of these ratios
b+d +f+
a result of such frequent utility that the following verbal equivalent should be noticed When a series of fractions are equal, each of them is equal to the sum of all the numerators divided by the sum of all the denominators. :
Example
(I
1.
If b
= C- =€>
d
,
shew that
J
+ 2c 2 e - Sae 2/ _ ace ~b 4 + 2^/-36/3 ~bdf az b
Let
_ _^._A,, «-£-£-X;. 6
then
a = bk,
rf
c
= dk, e =fk
;
RATIO.
*
W
aa6+2c»g-3qgy _ "" k 4 + 2tl-f - Bbf3
'*
5
fc
+2d?fk* - 3bf 3 k 3 4 + 2r/-/ - 36/8 a
...
c
e
ace
=
bdf'
Example
=f=
a
If
2.
#+a it
z
y
c
=r=-=
A;
,
prove that
,
y2 + &2 y+b
*2 + a2 x Let a
c
b
2
+ c2 _ z+c
3
so that x = «£,
s a + a3
„
then
(
.c
+ y +2 2 + (a + & + c) 2 a; + ?/ + 2 + a + &+c )
= 6/c, = ch
?/
2;
—ah + a
= aW+a* =
(k*
+a
x*+a* ar
+a
y
a
?/
+y +
ga +e»_
•
&
z
+
c
(ife
a
'
+ l)a —
L__ ' Jc+1
-
a:
;
+l)o +1
(&
;
+ l)&
2
(fc
£+1
/c
2
&
+ l)c +1
Jfc 2 + l)(a + 6 + c) fc+1 8
Jfc
(a+6+c) 8 +(a+ 6+c) a
&(a + & + c) + a +
+c
+ he) % + (a + b + c) 2 (ka + kb + kc)+a + b + c (x+y+z)*+(a+ b + cf
_ (lea
_
6
+
kb
x+y+z+a+b+c 13.
an equation
If
quantities,
we may
is
homogeneous with respect
for these quantities substitute in the equation
any others proportional to them. lx is
homogeneous in
portional to
x, y, %
x
Put h = — = a
75
£
3
instance, the equation
Let
a, j3,
y be three quantities pro-
respectively.
z
= -
,
so that
x-
ak,
y=
/3k, z
= yk
;
y Ia f3k
4
+ ma(3 2 yk* + n^y'k4 = 3
is,
For
2 2 y + mxifz + ny z —
x, y, z.
3
then that
11
to certain
7a /?
0,
2
2 + ma/3 2 y + nj3 y =
;
an equation of the same form as the original a, /?, y in the places of x, y, z respectively.
one, but with
—
.
.
HIGHER ALGEBRA.
6
The following theorem
14. .
If
.
y*
.
y~
,
nominators are
n
unequal fractions, of which the de-
be
r-
n
3
2
1
Q
i
.— ,....
,
important.
is
of the same sign, then
all
+ a8 + a 3 + b +b 2 +b 3 + a,
...
+ an
•'•
+bn
l
lies
in magnitude between the greatest
Suppose that
all
and
least
of them.
the denominators are positive.
and denote
least fraction,
the fraction
a
— /c
a 1
>k
—
ko r
a> kb
.-.
:
=*
be the
then
;
.'.a r
i
'
b y-
by k
it
Let
b
«
>
'
i
l
a2
> k;
a 2 > kb 2
.-.
ba
and .*.
by
so on;
addition,
a ,+«2
+ an >
+ «3 +
(
+b + K +
b.+b 29 + b.+ 3
+b n
we may prove
^
is tlie
'
br
that
+ an
+*.+*
+K
l
.
+
<
at
V
greatest of the given fractions.
In like manner the theorem may be proved when denominators are negative. 15.
'>
ar
.
a + a2 + a 3 + 6
+K) k
,
+ au
1
Similarly
bl
a + a 2 + a3 + l
where
;
The ready application
all
the
of the general principle involved
Art. 12 is of such great value in all branches of mathematics, that the student should be able to use it with some freedom in any particular case that may arise, without necessarily introducing an auxiliary symbol. in
Example
If
-
b
prove that
—=
X 1.
+ c-a
x+y+z a+b
+
c
V c
+ a-b
z
=
,
a
+ b-c
= *&+*)+? (*+*)+ * (*+V) 2(ax + by + cz)
=
;
;
RATIO.
t Each i
,i i= ofe the given efi actions •
_
sum °f numerators — __ sum of denominators x+y+z
'
a
"
+b+c
(
''
Again, if we multiply both numerator and denominator of the three given fractions by y + z, z + x, x + y respectively,
+ z) = j \ (y + z)(b + c-a)
?(« + *>
{l
each fractions
-
(z
+ x)
(c
'(* + *
-
__.
+ a-b)
(x + y)
)
(a+b-e)
sum of numerators sum of denominators x
=
.'.
from
(1)
and
a
prove that
If
2.
_x
(y
+ z)+y
+ b + c~
(2).
+ x)+z (x + y) (ax + by + cz)
2
(z
m(nc + la-mb)
l(mb + nc-la)
m
I
x(by + cz-ax)
We have
+ y (z + x ) +z {x + y) 2ax + 2by + 2cz
+z)
(2),
x+y+z
Example
(y
y
(cz
y
I
m
mb + nc —
nc
la
(la
+ mb - nc)
'
n
+ ax-by)
x
n
z(ax + by -cz) z
+ la-mb
n la + mb —
nc
v z -+-
=m
n
'"2/a"
= two
similar expressions
ny + mz _lz + nx b a Multiply the y,
of these fractions above
first
and the third by
z
_ mx + ly c
and below by
.r,
the second by
then
;
nxy + mxz ax
_
+ nxy _ mxz + lyz
Jyz
by
cz
= _2lyz by
+ cz- ax
= two
m
I
x
(by
+ cz -ax)
similar expressions
y
(cz
+ ax-by)
z
n (ax + by-cz)'
:
HIGHER ALGEBRA.
8
we have two
equations containing three quantities in the first degree, such as 16.
If
a x+b y+ l
a2 x +
we cannot
c
z=Q
l
l
b 2y
+ c2 z =
solve these completely
;
unknown (1), (2),
but by writing them in the
form
we
can,
II X by regarding - and - as the unknowns, solve in the
z
z
ordinary
way and x %
or,
obtain
- b2c " afi 2 - a 2 b b c2 l
y
i '
z
__
"
l
cxa
2
afi 2
- c2 a
-a
b 2
l
>
' l
more symmetrically,
b c2
x - b2c
x
y cla
x
2
-
c a, 2
afi 2
-
a_p x
'
,(3).
when we have two equations of the type and (2) we may always by the above formula
It thus appears that
represented by (1) write down the ratios x y z in terms of the coefficients of the equations by the following rule :
:
Write down the coefficients those of y; and repeat these as
of x, y, z in order, beginning with in the diagram.
Multiply the coefficients across in the way indicated by the arrows, remembering that in forming the products any one obtained by descending is positive, and any one obtained by ascending is negative. The three results h i cz- h fv
are proportional to
This
is
c x
a2-
c
2
an a
A- a b 2
>
x, y, z respectively.
called the
Rule of Cross Multiplication,
RATIO. Example
Find the
1.
x
ratios of
7x=4y + Qz
By
down
from the equations 3z = 12x + Uy.
y
:
t
:
z
we have 7x - Ay - 8-2 = 0, 12x + lly-Sz = 0.
transposition
"Write
9
the coeilicients, thus
-4
-8 -3
11
-4
7
12
11,
whence we obtain the products
(-8)xl2-(-3)x7,
(-4)x(-3)-llx(-8), or
-75,
100,
x
=
-
is,
4
Example
2.
Eliminate
y
-*-
-3
(2)
and
+ ^ 1 + c 12 = a^ + ^y + c^^O
(2),
Ogaj+fegy+c^^O
(3).
Example
3.
(1),
?/
*__ " Vs
_ C 2«i
denoting each of these ratios by and dividing out by A-, we obtain
(Va -
This relation
.
5
by cross multiplication,
(3),
k> C 3
Oj
= z?
from the equations
x, y, z
a 1 a;
From
z
y
x
,.
that
125;
100 ~ ^75~"125'
•'*
x,
7 x 11 - 12 x (-4),
63ca)
+ &i
k,
(^'"3
y ~ C 3«2
„
j*
«2 6 3
.
~ ll ih
'
by multiplying up, substituting in
- c 3 a a) + ('i
(«
A-
" A-)
= °-
called the eliminant of the given equations.
is
Solve the equations
ax + by + cz =
(1),
= hex + cay + abz = (b - c) x+ y+
From
(1)
and
(2),
.-.
Substituting in
z
(2),
(c-a) (a-b)
(3).
by cross multiplication, x z —y
b-c
= ^— = c-a
x = k (b-
c),
— k, suppose a-bT y — k (c - a), z — k(a- b). :
(3),
k {bc(b-c) + ca
(c
-
a)
+ ab (a -
k{-{b-c)(c-
b)}
a) {a - &) .-.
^ln'nce
(1),
\
={b- c)
-
a) {a
= (b-e) [e - a)
fcss-lj
x = c -b, y — a-r,
(c
z
= b - a.
{a
-
b),
-
b)
;
HIGHER ALGEBRA.
10
z
=
1,
equations (1) and (2) become
ax+
bxy
+
c
= 0,
v+
h 2y
+
c2
we put
If in Art. 16
17.
x
and
(3)
t
=°
>
becomes
x -
b x c2
y cx a
b 2c l
a l
b2
-a b
a
- c2a *
-a b
afi 2
l
- a2 b
aj> %
i
2
' ]
l
Hence any two simultaneous equations involving two unknowns in the first degree may be solved by the rule of cross multiplication.
Example.
By
5x-3y -1 = 0, x + 2y = 12.
Solve
= 0,
5x - 3y - 1
transposition,
x + 2y -12 = 0; x 36 + 2
*'•
=
x=
whence
1
y
- 1 + 60 ~ 38
10 + 3
;
59
= is' y lS'
EXAMPLES.
I.
Find the ratio compounded of
1.
and the duplicate
(1)
the ratio 2a
(2)
the subduplicate ratio of 64
(3)
the duplicate ratio of
#+7
36,
:
2a -j-
:
9,
and the
/6a? :
-
M
-,
and the
If
3.
Find two numbers in the ratio of 7
:
ratio
:
8,
ab.
:
ratio 27
2 (# + 14) in the duplicate ratio of 5
2.
:
ratio of 9b 2
:
Sax
56.
:
2by.
find x.
12 so that the greater
exceeds the less by 275. 4.
to
make
What number must it
equal to
y=3
5.
If x
6.
If 15 (2a-2
:
1
:
:
4,
3
be added to each term of the ratio 5
\
find the ratio of
- y 2 ) = *7xy,
7x-4y
find the ratio of
x
:
:
3x+y. y.
:
37
—
"
RATIO.
7
?=£ = «
If
+ 3a -5eV = _ _^__^__
2rt 4 &
prove that
= = 6ca
-7
If v
8.
11
2 2
2
e
prove that
,
"
is
-j
a
equal
l
t<»
y a 9.
If
If
r+p-q
q + r-p
shew that 10.
y
_
- r) x + (r - p) y + (p - q) z = 0.
(q
——
x-z
—
==-
p + q-r
=-
-
find the ratios of
,
y
z
Khew
#
if
2 (*+?+*)
tliat
12.
a+o + c
tliat
-
+a
—„-
If
prove 1 16.
If
3
„ t 2
y
3
+6
3
t- ;,, y* + b 2 -
c
z3 + c 3 _(.y + + *) 3 + (q + &+c) + 2 + c- — (>c+y + \« /_ _\s 5) + (a + o + c)3
?/
.,
2
,
.,
i
i
26 +
2c-a
= 2c
+ 2a"-
.
_ — 2A-+2y-g =
6
j.
•
,
J
2a + 26-c
+y 2 + ^2 = («.v+^ + ^) 2 x a=y b = z c. I (my + rut - Ix) = m (nz + Ix - my) = n (Ix + my - nz\ y+z-x = z+x-y x+y-z = n (a 2 +6 2 + c2 )
(.i-
2
)
:
:
,
:
—— -
-j
m
I
Shew
-
•
that the eliminant of
ax + cy +
bz
= Q, a3
is
17.
o
2y + 2g-.v _ — 2g + 2.-c-y
II
shew that 15.
6c +
a
BheW that 14.
z.
:
i'=^ = 2-,
.r-fa 1
'y
_ (6+o).r+(C +«)y+(« ± i)i
If .tfS
shew
:
y+ z = z+ v == r+ ^ pb + qc pc + qa pa + qb' '
ii
x
Eliminate ctx
x, y, z
= 0, bx +
from the equations
+ hy + (/z = 0, hx + by-\-fz = 0, gjc+fy+C2=0.
:
HIGHER ALGEBRA.
12 18.
x = cy + bz, y = az+cx z=bx + ay,
If
}
X z = = j—* y 1 - a \-b \—c
19.
2
II
shew that
n
i •> l
1
9 L
Given that a(y + z)=x, b(z + x)=y, c(x+y)=z,
prove that
bc
+ ca + ab + 2abc = l.
Solve the following equations
3x-4y + 7z =
20.
:
z,
3x-2y+17z =
0,
x* + 3f + 2z
tyz + 3sa?=4an/, 2tys - Sac = 4ry,
22.
x+y=
21.
0,
2x-y-2z = 0, to?-f+£=l8. 23.
I
Ja+Jb
5.0-4^ + 73 = 6.
m
+
Jb+Jo
—— =
shew that
= l67.
- 3y 2 -I5z 2 = 0,
7a*
+-^L^
.--*
If
s
3x 2 - 2y 2 + oz 2 = 0,
ff+2y+32=19. 24.
•
+
*„<>, n
=
?==-
(a-b)(c-\/ab)
(b
- c)
(a -
'
V be)
=(c
- a)
(b
-
\J ac)
Solve the equations
ax + by + cz = 0, bcx + cay + abz = 0, xyz + abc (a3x + b 3y + &z) = 0.
25.
a.-£+&y
26.
+ C2=a 2# + & 2y +
6' 2
2==0,
x + y + z + (b-c)(c-a) (a-b) = 0. 27.
a(y+x)=x,
If
b(z + x)=y, c(x+y)=z,
X — 7— = — = — a -be) b(l-ca) c(l-ab) 2
prove that 1 28.
If
?/
-
,
2
*
s2
{I
ax + ky+gz = 0, kx + by + fz^0, gx+fy + cz = 0,
prove that
^
x2 bc-f 2
(2)
(be
-f
y
2
z2
ca-g 2 2 )
{ea -
g
ab-h 2
2 )
(ab
- h 2 ) = (fg - eh)
(gk - af) (A/- bg).
;
CHAPTER
.
II.
PROPORTION,
When
Definition.
18.
ft
if
-
c
= - then
saying that a
d are
a, b, c,
,
is
to b as c
or
The terms a and d are
is
ratios
proportionals.
to d,
a
:
b
a
:
b
:
:
This
extremes
Let
is
equal
a, b, c,
Then by J
—c
d be the
written
:
d.
and
c
the
the means.
product of
the
proportionals.
—
=.
any three terms
Thus
found.
—
•
d
b
ad =
may be
is
product of the means.
to the
definition
if
expressed by
:
called the extremes, b
whence
Hence
is
and the proportion c d
If four quantities are in proportion,
19.
fourth
are equal, the four are said to be proportionals. Thus
them
quantities composing
two
be.
of
if a, c,
a proportion are given,
d are
given, then b
=
the
—
Conversely, if there are any four quantities, a, b, c, d, such that ad = be, then a, b, c, d are proportionals ; a and d being the extremes, b and c the means ; or vice versa.
Quantities are said to be in continued Definition. proportion when the first is to the second, as the second is and so on. Thus to the third, as the third to the fourth are in continued proportion when a, b, c, d, 20.
;
a
b
c
bed
HIGHER ALGEBRA.
14
If three quantities a,
are in continued proportion, then
c
b,
a
:
and
c
:
\
[Art. 18.]
.
be a mean proportional between a and said to be a third proportional to a and b.
In this case \
b
= b2
ac
.-.
c
—
b
c is
b is said to
three quantities are proportionals the first is third in the duplicate ratio of the first to tJie second.
If
21.
be Let the three quantities 1
a
a
then T = -.
be
a. b, c:
b
cue
Now
=r
x -
a
a
=
X
b
that
a
is,
a
=
F,;
6
= a2
c
:
2
:
b
2 .
It will be seen that this proposition is the of duplicate ratio given in Euclid,
If
22.
a
:
b - c
„ .b
:
a or
F b
d and
=
c -
d
,
e
g
j
h
ae
Cor.
=
v.
then will ae
:
bf= eg dh. :
bf=
eg
:
c
d,
:
and
b
:
then
a
:
known
y
dh
:
If
the theorem
definition
eg
bf
a
is
h,
:
same as the
and >=!-: ae
or
Book
'f=g
e
•'*
This
the
to
=
b
x=d
x=
c
:
dh.
:
v/,
:
y-
as ex cequali in Geometry.
d form a proportion, many other proportions may be deduced by the properties of fractions. The results of these operations are very useful, and some of them are often quoted by the annexed names borrowed from 23.
If four quantities a,
Geometry.
b,
c,
:;;
PROPORTION. If a
(1)
For - = -
then b
d,
:
therefore
;
1
a=d
:
=1 =
a
If
(2)
.
For
-
acZ
b=c
:
be
d,
:
then a
a
a If «
(3)
lor
=
7-
o
-,
=
6
:
:
d
c
:
s-
=
b
a+
+1 =
o
is
7
—=
o
a+ b
or
a
If
(4)
For
= -,
=-
,
"
:
d
b
6
:
=
c
:
d,
b
-,
d
=
b
:
+
c
+d
d.
:
[Componeudo.']
1
—+d=—d c
c
:
+d :
:
d.
= c- d
b
—1 =- - 1 d a— —7— = — -=—
:
d.
[Divideudo.]
7
bc-d
a-b
«
If
For by /
1
and by .
by J
d.
:
b
is,
(5)
.'.
=
therefore -
or
1
:
a-b
then
.
that
6
:
:
b
a+b that
b
a
c
:
d, tlien
therefore
= ;
-,
c
or
[Alternando.]
c.
:
c
:
=
-
is,
d.
:
—j = —
therefore
;
[Invertendo.]
c
a=d
:
.
,
that
.
-
a b
c.
d'
-
is
:
-_
-r-
b
or
•!
=-
-f-
d'
b
that
=c
b
:
15
,
:
6
-
c
:
df,
.
a—
-j
c
-d
:
d.
+6 a — b=c+d:c— d. :
bc—d
^-j
« +&
.
division,
=
a+
or
-
r = -j-
(3)
(4) .
b
then a
\
*
:
3
This proposition
is
b
:
=
c
+d •
a-b c-d' a-b = c + d c-d. :
usually quoted as
Componeiuh
a) id JJivi-
dendo.
Several other proportions
may
be proved in a similar way.
'
:
HIGHER ALGEBRA.
16
results of the preceding article are the algebraical equivalents of some of the propositions in the fifth book of Euclid, and the student is advised to make himself familiar with them 24.
The
For example, dividendo may be quoted as
in their verbal form. follows
When
there are
the second is
fourth 25.
of
the first above
second, as the excess of the third above the
the
to
the excess
four proportionals,
is to the fourth.
We
shall
now compare
the algebraical definition of pro-
portion with that given in Euclid. Euclid's definition
is
Four quantities are
as follows
:
said to be proportionals
when
if
any
equi-
multiples whatever be taken of the first and third, and also any equimultiples wJiatever of the second and fourth, the multiple of the third is greater than, equal to, or less than the multiple of the fourth, according as the multiple of the first is greater than, equal to, or less than the multiple of the second.
In algebraical symbols the definition may be thus stated
Four quantities
=
according as p>a I.
qb,
a,
b,
p and
d
c,
are in proportion
q being
any positive
To deduce the geometrical
when
:
p>c
= qd
integers tcJudever.
definition of proportion
from
the algebraical definition. Since
a -z
c
- -
,
by multiplying both
pa
2)C
qb
qd
sides
u by -
,
we
obtain
hence, from the properties of fractions,
pc = qd according as pa =
qb,
which proves the proposition. II.
To deduce
the algebraical definition of proportion from
the geometrical definition.
Given that pc = qd according as pa = a b
=
c ~d'
qb, to
prove
PROPORTION. If
is
-j-
Suppose
which
g
lies
not equal to -
>
-^
;
then
it
,
between them, q and
and
from
one of them must be the greater.
will be possible to find
Hence
From
17
some fraction 2
P
being positive integers.
b
>
-
< ?
-
0).
p
(2>
pa>qb;
(1)
2)c
(2)
and these contradict the hypothesis. Therefore y and - are not unequal; that
is
- =
-•
which proves
the proposition. It should be noticed that the geometrical definition of proportion deals with concrete magnitudes, such as lines or areas, 26.
represented geometrically but not referred to any common unit of measurement. So that Euclid's definition is applicable to incommensurable as well as to commensurable quantities ; whereas the algebraical definition, strictly speaking, applies only to commensurable quantities, since it tacitly assumes that a is the same determinate multiple, part, or parts, of b that c is of d. But the proofs which have been given for commensurable quantities will still be true for incommensurables, since the ratio of two incommensurables can always be made to differ from the ratio of two integers by less than any assignable quantity. This lias been shewn in Art. 7 ; it may also be proved more generally as in the
next
article.
Suppose that a and b are incommensurable; divide b into m equal parts each equal to /?, so that b = m/3, where m is a positive integer. Also suppose f3 is contained in a more than n times and less than n+ 1 times; 27.
nB a (n+1) B - > -^ and < * /^ .
i,
then
o
that
-=
is,
lies
mp
between
o
so that
-j-
differs
H. H. A.
from
— by
imp
—
m
and
,
m
;
a quantity less than
—
.
And
since
2
we
1
,
'
HIGHER ALGEBRA.
18
can choose B (our unit of measurement) as small as we please,
Hence
be made as great as we please. as
can be made as small
and two integers n and m can be found whose ratio express that of a and b to any required degree of accuracy.
we
will
—1
m can
please,
28.
in Art. 23 are often useful in particular, the solution of certain equa-
The propositions proved
In solving problems. tions is greatly facilitated by a skilful use of the operations componendo and dividendo. Example If
1.
(2ma + 6mb
d are proportionals. 2ma + Gmb + Snc + 9nd _ 2ma + Qmb - Snc - 9nd 2ma - bmb + Snc - \)nd 2 ma - 6mb - Snc + 9nd componendo and dividendo, 2 (2ma + Snc) _ 2 {2ma - Sue) 2~{Gmb + 9nd) ~ 2 (Smb - 9m/)
prove that a,b,
.*.
+ Snc + 9wtZ) (2ma - Gmb - Snc + 9nd) = (2ma - 6mb + Snc - 9/uZ) (2mm + Gmi - Snc - dnd),
Alternando,
c,
—
Gmb + (.)nd
2ma + Snc -
=— = -— -— ,. n 2ma-Snc bmb-vna :
Again, componendo and dividendo,
Ama _ \2mb
~
One a
whence
c
a
or
Example
2.
b
:
lQnd
;
b
=
-,
a
—c
:
d.
Solve the equation
Jx+l + Jx^l _ 4a; -1 2 Jx + l- Jx-1 We have,
componendo and dividendo,
Jx+l _ 4a; + + l_ x-1 "
16a; 2
.r
*'•
16a;
2
+
8a;+l
-24a; + 9
*
Again, componendo and dividendo, 2
_ 32a; 2 - 16a; + 10 ~ ~ 32a; - 8
X
~
2x
" whence
16a;
16a; 2
-8a; + 5
16a;- 4 2
-
4a;
•
'
= 16a; - 8a; + 5 2
•
5 x = -.
;
'
c
:
PROPORTION.
19
EXAMPLES.
II.
1.
Find the fourth proportional to
2.
Find the mean proportional between
3.
and
6
(1)
24,
3, 5, 27.
36'0a 4
(2)
X -f
-
y If
a
:
b
=c
4.
a2 c + ac2
5.
pa 2 +
6.
a-c
7.
\/a 2 ~+"c2
If a, 6,
+ bd2 = (a + c) 3
:
*Jb
2
pc 2 — qd2
:
+ d2 <
\/^+d^=jS /ac + ^
:
^Jbd+j.
:
are in continued proportion, prove that
11.
If b is a
:
.
.
10.
9.
a
.
y
+ ^=03
8.
--
x
+ df.
(b
:
— qb 2 =pc 2 + qd2
£>a 2
:
b-d=*Ja 2 + c 2
:
c, o?
b 2d
:
- and
prove that
d,
:
.
x
II
'
Find the third proportional to
and 250a 26 2
6
:
:
:
.
2
.
)
)
mean
proportional between a and •
c,
prove that
a 2_fr2 + c2
a- 2 -6- 2 + c- 2 12.
If
a
:
6=c
:
and
d,
ae + bf
e
:
/=#
:
h t prove that
ae-bf=cg + dh
:
:
cg-dh.
Solve the equations
3^-^ + 507-13
13.
2afi-3afi+a;+l 2073-3072-07-1
14.
Zx*+x2 - 2o7 - 3 _ 5o74 + 2o72 -7o7 + 3 ~~ 3ot* - x 2 + 2o; + 3 5o? 4 - 2o7 2 + 7o- - 3
307 3
(m-\-n)x — (a- b) 15.
(m-n)x — (a + 6)
16.
If a,
&, c, o?
-072
-507+13*
(m + n)x + a + c (wi — n)x + a-c'
are proportionals, prove that 7
.
+ d=b + c +
a-
17.
If a, b,
c,
'
d, e are in
(a K
— b)(a — c)
^
-.
a
continued proportion, prove that
(ab + be + cd + e&) 2 = (a 2 + 6 2 + c 2 + rf 2 ) (6 2 +
2
+ d + e2 2
).
2—2
HIGHER ALGEBRA.
20 18.
by x +
If the work done by x — 1 men in x 2 men in x - 1 days in the ratio of 9
+ 1 days is :
to the
work done
10, find x.
Find four proportionals such that the sum of the extremes is 21, the sum of the means 19, and the sum of the squares of all four numbers is 442. 19.
B
Two casks A and were filled with two kinds of sherry, mixed 20. in the ratio of in the cask A in the ratio of 2 7, and in the cask What quantity must be taken from each to form a mixture 5. 1 consist of 2 gallons of one kind and 9 gallons of the other \ shall which
B
:
:
Nine gallons are drawn from a cask
wine; it is then filled with water, then nine gallons of the mixture are drawn, and the cask is again filled with water. If the quantity of wine now in the cask be to the quantity of water in it as 16 to 9, how much does the cask hold? 21.
full of
If four positive quantities are in continued proportion, shew that the difference between the first and last is at least three times as great as the difference between the other two. 22.
In England the population increased 15*9 per cent, between 1871 and 1881; if the town population increased 18 per cent, and the country population 4 per cent., compare the town and country popula23.
tions in 1871.
In a certain country the consumption of tea is five times the consumption of coffee. If a per cent, more tea and b per cent, more coffee were consumed, the aggregate amount consumed would be 1c per cent, more but if b per cent, more tea and a per cent, more coffee were consumed, the aggregate amount consumed would be 3c per cent, 24.
;
more
:
compare a and
b.
Brass is an alloy of copper and zinc bronze is an alloy 25. containing 80 per cent, of copper, 4 of zinc, and 16 of tin. fused mass of brass and bronze is found to contain 74 per cent, of copper, 16 of zinc, and 10 of tin find the ratio of copper to zinc in the composition ;
A
:
of brass.
A
crew can row a certain course up stream in 84 minutes; 26. they can row the same course down stream in 9 minutes less than they could row it in still water how long would they take to row down with the stream ? :
CHAPTER
III.
VARIATION. Definition. One quantity A is said to vary directly as another B, when the two quantities depend upon each other in such a manner that if B is changed, A is changed in the same 29.
ratio.
The word
Note.
and
directly is often omitted,
A
said to vary
is
as B.
a train moving at a uniform rate travels 40 miles in 60 minutes, it will travel 20 miles in 30 minutes, 80 miles in 120 minutes, and so on; the distance in each case being increased or diminished in the same ratio as the time. This is expressed by saying that when the velocity is uniform
For instance
if
:
the distance is ptroportional to the time, or the distance varies as the time.
The symbol
30.
A
on
B
is
read
so
\
that
A is equal to B multiplied by some
varies as B, tlien
If A.
31.
"A
used to denote variation varies as B." is
oc
constant quantity.
For suppose that values of A and B. mi
i
/»
i
a,
a
lt
V
«,
6,
TT
Hence that
is,
-=
= =* = y^= 62
a3 ...,
a b — = =-
•,•
Inen, by deimition,
/. s-i
a„,
,.j-
63
any value r .
:
7u,
b
b 2 , b3
, x
a b — =— ^ K
a —
;
...
=
»3
are corresponding
b -r
,
;
and
so on,
K
each being equal to T
.
b
of
A
.
=—=
_
the corresponding value ot
— —
b,
where .'.
m
always the same
B
is
is
constant.
A=mB.
:
;
HIGHER ALGEBRA.
22 any pair
If
the constant
of corresponding values of
m can be
For instance,
determined.
^=12, we have
3
=m x
A and B if
are known, A = 3 when
12;
A = \B.
and
One quantity A is said to vary inversely Definition. as another Z?, when A varies directly as the reciprocal of B. 32.
Thus
if
A
A = -^
varies inversely as B,
The following is an do a certain work in 8
,
where
m is
illustration of inverse variation
constant. :
If 6
men
men would do
the same work in and so on. Thus it appears that 4 hours, 2 men in 24 hours when the number of men is increased, the time is proportionately decreased; and vice-versa. hours, 12 ;
Example 1. The cube root of x x=8 when y = 3, find x when y = l^.
By
supposition £/x=
—
,
where
varies inversely as the square of y
;
if
m is constant.
if
Tit
Putting x = 8, y = 3, we have
2
.*.
=n>
?;t
= 18,
v *jx
and
hence, by putting y = ^, we obtain
a;
18
=—
r ,
= 512.
Example 2. The square of the time of a planet's revolution varies as the cube of its distance from the Sun; find the time of Venus' revolution, assuming the distances of the Earth and Venus from the Sun to be 91-J and 66 millions of miles respectively. Let
P
of miles
;
be the periodic time measured in days,
P aD 2
we have
where k
is
the distance in millions
,
P*=kD 3
or
D
3
,
some constant.
For the Earth,
365 x 365 = k x 91± x 91| x 91£, k=
whence
4x4x4 365
_4 x 4 x 4 " rp2 ~ 365 " .
'
.
VARIATION.
23
pa^i^ili x 66 x 66 x 6G
For Venus,
3 b.)
P = 4x66
whence
;
/264
V
::
365
= 264 x a/*7233, = 264 x -85
approximately,
= 224-4. Hence the time
of revolution is nearly 224£ days.
One quantity is said to vary jointly as a number of others, when it varies directly as their product. Thus A varies jointly as B and C, when A = mBC. For instance, the interest on a sum of money varies jointly as the Definition.
33.
and the rate per
principal, the time,
3 -i.
versely as C,
B
when A
B
then
is constant,
said to vary directly as
is
varies as -^
7/*A varies as
35.
when
A
Definition.
cent.
A
and
in-
.
when C
tvill
B
is constant,
BC
vary as
and
A varies as C
ivhen both
B and C
vary.
B and partly on these latter Suppose variations that of C. to take place sepaturn producing its its own effect on A rately, each in also let a, b, c be certain simultaneous values of A, B, C. The variation
of
A
depends partly on that of
;
Let C be constant while B changes to b ; then A must undergo a partial change and will assume some intermediate value a\ where 1
"=2.
Let
B
be constant, that
changes to c ; then intermediate value
From
(1)
and
A must a' to its
(2) x '
that
—
(1)
value b, while C change and pass from its
let it retain its
is,
complete its final value a, where
x
a
- = a
—
A = =-
is,
x
-
b
:
c
.
BC,
be
or
A
varies as
BC.
;
;
illustrations of the
theorem proved in
;
HIGHER ALGEBRA.
24
The following are
36.
the last
article.
work done by a given number of men varies directly as the number of days they work, and the amount of work done in a given time varies directly as the number of men therefore when the number of days and the number of men are both variable, the amount of work will vary as the product of the number of men and the number of days. The amount
of
Again, in Geometry the area of a triangle varies directly as its base when the height is constant, and directly as the height when the base is constant and when both the height and base are variable, the area varies as the product of the numbers ;
representing the height and the base.
The volume of a right circular cone varies as the square of the when the height is constant, and as the height when the base the radius of base is constant. If the radius of the base is 7 feet and the height 15 feet, the volume is 770 cubic feet ; find the height of a cone whose volume is 132 cubic feet and which stands on a base whose radius is 3 feet. Example.
Let h and r denote respectively the height ani radius of the base measured in feet also let V be the volume in cubic feet. ;
Then
By
V=mr
2
where
h,
m is
constant.
770 — m x
supposition,
2 7 x 15
;
22
m=—
whence
.*.
by substituting
V=
132, r = S,
we
;
get
22
132=- xOxft; —X
whence and therefore the height 37.
case in
7i= 14 is
14
feet.
The proposition of Art. 35 can easily be extended to the which the variation of A depends upon that of more than
Further, the variations may be either direct or The principle is interesting because of its frequent ocinverse. For example, in the theory of currence in Physical Science. gases it is found by experiment that the pressure (p) of a gas varies as the "absolute temperature" (t) when its volume (v) is constant, and that the pressure varies inversely as the volume when the temperature is constant ; that is
two
variables.
2? oc
t,
when v
is
constant
;
, ,
VARIATION.
and
From
p
-
we should have
p
cc
-
when
,
t
constant.
is
v
we should expect
these results
variable,
cc
,
25
or
that,
when both
t
and v are
the formula
pv = kt, where k
and by actual experiment
is
constant
found to be
this is
the- case.
Example. The duration of a railway journey varies directly as the distance and inversely as the velocity; the velocity varies directly as the square root of the quantity of coal used per mile, and inversely as the number of carriages in the train. In a journey of 25 miles in half an hour with 18 carriages 10 cwt. of coal is required; how much coal will be consumed in a journey of 21 miles in 28 minutes with 16 carriages? Let
t be the time expressed in hours, d the distance in miles,
v the velocity in miles per hour, q the quantity of coal in cwt., c the number of carriages.
We have
v
t oc
and
v
*!l
oc
c
whence or
—
t oc
t
— —7- where
k is constant.
,
Substituting the values given,
we have
_ k x 18 x 25 ~ 2 jm 1
that
k
is,
Hence
t
=
=
;
25x36"
^—cdT — 2o 36 Jq v/lO
^
Substituting now the values of question, we have
.
.
x
t,
c,
d given in the second part
710x16x21 60" 25x36^2 28
a
••
that
is,
n/10x 16x21
s/q=—15x28 /
whence
Hence the quantity
q of coal
is
,-
=5^10,
— -=- = 6|.
6|cwt.
^
.
'
of the
HIGHER ALGEBRA.
26
EXAMPLES. If
x
2.
If
P
3.
If the square of
varies as y,
varies inversely as Q,
find the value of y
A
4.
x
varies as
B
and i?
If .4 varies as C, each vary as C. 5.
6.
If J. varies as
7.
P
8.
If
a'
:
(^
and
find
P when
x—3 when y = 4,
if
A=2
3
when # = - and
10 C=—
,
= 3. and
i? varies
BC, then
find
when # = 3,
.
C jointly;
as C, then
±Z? and
J.
\/
AB will
C
Z>
varies inversely as -7
Q and
varies directly as
R =—
^ = - and
P=7
and
x when y = 10.
find
varies as the cube of y,
when #=-y-
when A = 54 and
find
#=8 when y = 15,
and
1.
III.
R\
inversely as
.
2
also
P = o~
when
when P=a/48 and jR=\/<5.
varies as y, prove that
x2 +y 2
varies as
x2 -y\
y varies as the sum of two quantities, of which one varies x and the other inversely as x and if y = 6 when x=4, and y = 31 when x = 3 find the equation between x and y. If directly as 9.
;
;
If 3/ is equal to the sum of two quantities one of which varies 10. as x directly, and the other as x2 inversely; and if y 19 when x=2, or
=
3
;
find 11.
y
in terms of x.
If
A
and
C=g
12.
B
varies directly as the square root of and inversely as when 24 if 3 when .£=256 and C=2, find
the cube of C, and
4=
B
A=
.
Given that x + y varies as
z
+-
,
and that x — y varies as
z
z
find the relation
between x and
z,
—
,
z
provided that
z
=2 when x =3
and
y = \.
B
and C jointly, while If J. varies as varies as D. varies inversely as A, shew that 13.
B
varies as Z> 2 ,
and
C
A
varies as the sum of three quantities of which the first is constant, the second varies as .r, and the third as x 2 and if y = when x=l, y l when x=2, and y 4 when x = 3; find y when x=7. 14.
If
y
;
=
=
When a body falls from rest its distance from the starting 15. point varies as the square of the time it has been falling if a body falls through 402^ feet in 5 seconds, how far does it fall in 10 seconds ? Also how far does it fall in the 10th second? :
VARIATION.
27
Given that the volume of a sphere varies as the cul>c of its 16. radius, and that when the radius is 3-&- feet the volume is 179rj cubic feet, find the volume when the radius is 1 foot 9 inches.
The weight
of a circular disc varies as the square of the radius when the thickness remains the same; it also varies as the thickness when the radius remains the same. Two discs have their thicknesses in the ratio of 9 8 find the ratio of their radii if the weight of the first is twice that of the second. 17.
:
18.
At
19.
The
;
a certain regatta the number of races on each day varied jointly as the number of days from the beginning and end of the regatta up to and including the day in question. On three successive days there were respectively 6, 5 and 3 races. Which days were these, and how long did the regatta last? price of a
diamond
varies as the square of its weight.
Three rings of equal weight, each composed of a diamond set in gold, have values «£«., £b, £c> the diamonds in them weighing 3, 4, 5 carats respectively. Shew that the value of a diamond of one carat is
the cost of workmanship being the same for each ring.
Two
persons are awarded pensions in proportion to the square One has served 9 years root of the number of years they have served. receives pensio?i greater by ,£50. If the a longer than the other and length of service of the first had exceeded that of the second by 4| years How long 8. their pensions would have been in the proportion of 9 had they served and what were their respective pensions ? 20.
:
The
attraction of a planet on its satellites varies directly as the mass (M)of the planet, and inversely as the square of the distance (D) also the square of a satellite's time of revolution varies directly as the distance and inversely as the force of attraction. If m v d v t v d2 £ 2 are simultaneous values of J/, D, T respectively, prove and 2 21.
;
m
,
,
,
that
of revolution of that moon of Jupiter whose distance is to the distance of our Moon as 35 31, having given and that the Earth, the that the mass of Jupiter is 343 times that of Moon's period is 27*32 days.
Hence
find the time
:
The consumption of coal by a locomotive varies as the square of the velocity; when the speed is 10 miles an hour the consumption of 22.
coal per hour is 2 tons if the price of coal be 10s. per ton, and the other expenses of the engine be lis. 3c/. an hour, find the least cost of a journey of 100 miles. :
.
CHAPTER
IV.
ARITHMETICAL PROGRESSION. Quantities are said to be in Arithmetical Definition. Progression when they increase or decrease by a common dif38.
ference.
Thus each Progression
15,
3,
7,
11,
8,
2,
-4, -10,
a,
a+
a + 2d, a + 3d,
d,
by subtracting any term of from that which follows it. In the first of the above difference is found
examples the common difference the third
we examine a,
we
is
notice that in
a+
d,
any term
Thus the
and, generally, the
?i
in the second it
a+
2d,
the coefficient
is
—6
;
in
we have
a + 2d;
6 th
term term
is
a + 5d ;
is
a+
I9d',
is
a+
(
40.
To find
of terms,
I
the
=a+
is
always
less
by one
seiies.
is
th p term
. .
of d
term
n be the number
term,
a + 3d,
3 rd
20 th
If
;
the series
than the number of the term hi the
th
4
it is d.
If
39.
forms an Arithmetical
series
:
The common the series
the following
of
p—
and
(n — 1)
\)d. if
I
denote the
last,
or
d.
sum of a number of
terms in Arithmetical
Progression.
Let a denote the the
number
first
of terms.
term,
Also
d the common
let I
and n term, and s
difference,
denote the last
ARITHMETICAL PROGRESSION. sum
the required 8
;
= a+(a +
29
then d)
+
+ 2d) +
(a
+
...
(I
-
2d)
+ (l-d) +
l;
and, by writing the series in the reverse order, s
= I+
(I
-
d)
+
(I
- 2d) +
Adding together these two 2s
•'•
s
= (a +
l)
...
+ (a + 2d)+
(a + d)
+
a.
series,
+
(a
= n (a +
I),
= ^(a +
l)
+
l)
+ (a + l)+
...
to
n terms
(1);
a
l~a + (n-l)d
and .-.
In
s
(2),
= -^{2a + (n-l)d\
(3).
we have
three useful formula; (1), in each of these any one of the letters may denote (2), (3) the unknown quantity when the three others are known. For instance, in (1) if we substitute given values for s, n, I, we obtain an equation for finding a ; and similarly in the other formulae. But it is necessary to guard against a too mechanical use of these general formulae, and it will often be found better to solve simple questions by a mental rather than by an actual reference to the requisite formula. 41.
tlie
last
article
;
Example 1. Find the sura of the series 5^, Here the common difference is 1^; hence from the
=
sum
\
2 x
GJ, 8,
to 17 terms.
(3),
— + 16 *li|
= y (11+20) 17x31 ~2
= 263£. Example 400
:
If
The first term of a series is 5, the last 45, and number of terms, and the common difference.
2.
find the
n be the number of terms, then from 400 = "
whence
(5
(1)
+ 4r>);
n = 10.
the
sum
;
.
;
HIGHER ALGEBRA.
30 If
d be the
common
difference
45= the
16
term = 5 + 15d;
th
d = 2f
whence 42.
any two terms
If
an Arithmetical Progression be
of
given, the series can be completely determined; for the data furnish two simultaneous equations, the solution of which will give the first term and the common difference.
The 54 th and 4 th terms
Example. 23
rd
of
an A. P. are - 61 and 64
;
find the
term. If
a be the
first
common
term, and d the
difference,
- 61 = the 54 th term = a + 53d 64 = the
and
d=
whence we obtain
4
term = a + 3d
th
5 -jr, a = Hh.
;
;
;
and the 23 rd term = a + 22d = 16£.
Definition. When three quantities are in Arithmetical Progression the middle one is said to be the arithmetic mean of the other two. 43.
Thus a 44.
is
the arithmetic
To find
Let a and
Then
b
mean between a — d and a + d.
the arithmetic
mean
betiveen
be the two quantities
since a, A, b are in
each being equal to the
A
the arithmetic mean.
A. P. we must have b - A = A — a,
common a A—
whence
;
two given quantities.
difference
+
b
2
Between two given quantities it is always possible to insert any number of terms such that the whole series thus formed shall be in A. P. and by an extension of the definition in 45.
;
Art. 43, the terms thus inserted are called the arithmetic means. Example.
Insert 20 arithmetic
means between 4 and
67.
Including the extremes, the number of terms will be 22 so that we have which 4 is the first and 67 the last. ;
to find a series of 22 terms in A.P., of
Let d be the then
common
difference
67 = the 22 nd term = 4 + 21d
whence d = S, and the series and the required means are
is 4, 7, 10, 7, 10, 13,
;
61, 64, 67 58, 71, 64.
;
;
;
;
ARITHMETICAL PROGRESSION. To
46.
two
31
a given number of arithmetic means
insert
betiveen
given quantities.
Let a and
b
be the given quantities, n the number of means.
Including the extremes the number of terms will be u + 2 so that we have to find a series of n + 2 terms in A. P., of which a is the first, and b is the last.
Let d be the common difference
= the
then
b
whence
d=
(n
r
71+
1
+
2)
th
term
:
'
and the required means are b
a+ Example their squares
-
n+l
a
,
— — n+l
293
;
a)
2 (b
a+
=-'
*
H
The sum
1. is
—a
of three find them.
,
numbers
Let a be the middle number, d the d, a, a + d.
nib — a) — — _-
*
'
n+l
and the sum of
in A.P. is 27,
common
difference
.
;
then the three
numbers are a -
a-d + a + a + d = 27
Hence
whence a = 9, and the three numbers are 9 (9-rf) 2 + 81
.-.
d, 9,
$
+ d.
+ (9 + d) 2 = 293;
d=±5;
whence and the numbers are
w"'
;
4, 9, 14.
Example 2. Find the sum is 3n - 1.
of the first
p terms
of the series whose
term
By
putting
n=l, and n=p first
.-.
respectively,
term = 2,
last
term =3p — 1
sum=|(2 + 3i>-l)=|(3p + l).
EXAMPLES. 1.
Sum
2, 3|, 4J,...
2.
Sum
49, 44, 39,... to 17 terms.
3.
Sum-,
3
4
2
-, o
—
to 20 terms.
7
I
—
we obtain
,...
to 19 terms.
IV.
a.
.
HIGHER ALGEBRA.
32
Sum
4.
7
-, If,... to n terms. o
3,
3'75,
7.
Sum Sum Sum
8.
Sum
-.-
9.
Sum
-j=
Sum Sum
a - 36, 2a - 56, 3a - 76,
5.
6.
10. 11.
35,
-Tl, -7, -6J,... to 24 terms. -3-1, -7-5,... to 10 terms.
1-3,
12 3 x/3, -75 »... to 50 terms.
6
,
3
4 -tt
,
2a -
6,
+6 Sum -£n
12.
16 terms.
3-25,... to
,
V5
4a -
tt
,
a,
to 25 terms.
;-..
6a -
36,
3a-6 ^
——
56,.
. .
,,
,
,...
to 40 terms.
. . .
n terms.
to ,
to 21 terms.
13.
Insert 19 arithmetic
means between - and — 9|.
14.
Insert 17 arithmetic
means between 3^ and — 41§.
15.
Insert 18 arithmetic
means between -
16.
Insert
17.
Find the sum of the
first
18.
In an A. P. the
term
as
arithmetic
first
36.17
means between x2 and
and
S.v.
1.
n odd numbers. is 2,
the last term 29, the
sum
155;
find the difference.
The sum
19.
ence
is
5
20. find the
;
an A. P.
is
600,
and the common
differ-
find the first term.
The
sum
third term of an A. P. of 17 terms.
The sum
21.
504
;
of 15 terms of
is 18,
and the seventh term
of three
numbers
in A. P.
of three
numbers
in A. P. is 12,
is 27,
is
30
and their product
;
is
find them.
The sum
22.
cubes
is
408
;
and the sum of
their
find them.
23.
Find the sum of 15 terms of the
24.
Find the sum of 35 terms of the
25.
Find the sum of p terms of the
26.
Find the sum of n terms of the series 2 a2 - 1 6a2 -5 3 4a a a a ,
,
whose nth term
series
series
series
,
whose p ih term
whose n th term
.
.
is 4?i4- 1.
is
is
-
^ + 2. + b.
;
;
ARITHMETICAL PROGRESSION.
33
In an Arithmetical Progression when s, a, d are given, to determine the values of n we have the quadratic equation 47.
= ^ <2a + (n- l)d\
s
when both
;
and integral there
roots are positive
in interpreting the result corresponding to each.
no difficulty In some cases
is
a suitable interpretation can be given for a negative value of Example. taken that the
How many
-6, -3,...
must be
n--ln-U = Q,
is,
(n-ll)(n+4)=0;
or
.'.
If
we take 11 terms
sum
of which
?i=ll or -
of the series,
- 9, the
-9,
? {-18 + (»-l) 3}=66;
Here that
terms of the series G6 ?
sum may be
n.
6,
-
4.
we have
3, 0, 3, 6, 9, 12, 15, 18,
21
is 66.
we begin
at the last of these terms and count backwards four terms, the and thus, although the negative solution does not directly answer the question proposed, we are enabled to give it an intelligible meaning, and we see that it answers a question closely connected with that to which the positive solution applies. If
sum
is
also 66;
We
can justify this interpretation in the general case in the following way. 48.
The equation
to determine
n
is
dn 2 + (2a-d)n-2s =
(1).
Since in the case under discussion the roots of this equation have The last opposite signs, let us denote them by n and - n term of the series corresponding to n is .
l
a + (n l
1
)
d
term and count backwards, the common terms is difference must be denoted by - d, and the sum of if
we
beirin
at
this
yi.,
|{2 and we
shall
H. H. A.
(«
shew that
+ »,-!
(»,
this is equal to
-!)(-
3
—
'
HIGHER ALGEBRA.
34
=-
For the expression
?
\
= ^ 1
2a + (2n
— n — l)dl 2
{
2an2 + 2n n 2 d - n 2 (n2 + x
1)
d
1
= ^ I 2n n 2 d - (da* - 2a - d .n2 )\ x
= l(4s-2s) = s, since
— n2
satisfies
dn 2 + (2a — d)
n—
2s
=
0,
and — n n 2
is
the
}
product of the roots of this equation.
When
the value of n is fractional there is no exact number of terms which corresponds to such a solution. 49.
How many
Example.
amount
to 71
n ~ {52 +
Here that
terms of the series 26,
5)i2
is,
be taken to
(n-l)(-5)} = 74; - 57u + 148 = 0,
(n-4)(5n-37) = 0;
or
.*.
Thus the number while the
is greater,
50.
We
Example ratio of
a„,
21, 16, ...must
?
7?t
terms
is 4.
or 1%.
It will
be found that the
sum
of 7
terms
of 8 terms is less than 74.
:
The sums of n terms of two arithmetic 4« + 27; rind the ratio of their 11 th terms.
Let the first term and d 2 respectively.
common
difference of the
|M^* +
"
We have Now we
=4
add some Miscellaneous Examples.
1.
+l
of
sum
?i
have
to find the value of
series be
a v d x and
£+1
{n-l)d 2
2a 2
two
series are in the
—
4?i
+ 27
tttt',' hence,
a 2 + l0d 2
by putting n—21, we x
obtain
2^ + 20^ _ 148 _ 4 "~ 3 2a 2 + 20d 2 ~ 111 thus the required ratio
Example series whose 1, 3, 5, 7,...
;
2.
If
is
4
Su S2
,
:
_
3.
sums of n terms of arithmetic and whose common differences are
S&...S,, are the
terms are 1, 2, 3, 4,... find the value of
first
#L +
+£3 +. + £„. ..
ARITHMETICAL PROGRESSION.
S^
We have
{2
+
(n - 1)}
= nAH+D
,
?i
.S>^{2i>+(»-l)(2i ;-l)}= - {(2p-l)n+l}; •.
sum=
the required
=
w
- {(n m
+
l)
(3n
+ l) +
(2/>- 1
.
n+
1)}
~
?l
-
+
{
(n
+ 3n + 5n +
.
.
.2p
-1
.
;/)
+ p)
1
= ±{n(l + 3 + 5+...21>-l)+p} = r2
2
(»l>
+P)
EXAMPLES. a=
-2, c?=4 and
1.
Given
2.
How many
make 208
.5
terms of the
= 100,
IV.
b.
find n.
series 12, 16, 20,...
must be taken
to
?
In an A. P. the third term sixth term is 1 7 find the series. 3.
is
four times the
first
term, and the
;
4.
respectively 5.
2n
The ;
31 st
*1
,
,
and
find the first
The 4 th 42nd and ,
,
respectively
;
last
terms of an A. P. are 7j, 5 and
-6j
term and the number of terms. terms of an A. P. are 0, - 95 and - 1 25 term and the number of terms.
last
find the first
6.
A man
7.
Between two numbers whose sum
arranges to pay off a debt of £3600 by 40 annual instalments which form an arithmetic series. When 30 of the instalments are paid he dies leaving a third of the debt unpaid: find the value of the first instalment.
2£ an even number of of these means exceeds their is
means is inserted; the sum number by unity how many means are there The sum of n terms of the series 2, 5, 8. arithmetic
:
2
8,... is !>">0
:
find
n.
3—2
HIGHER ALGEBRA.
36
Sum
9.
the series -
If the sum of 7 find the sum of n terms. 10.
11.
If the
p
th
that 12.
p
find
;
13.
945
;
,
r
-_
,
terms
...
to
n
terms.
and the sum of 17 terms
is 49,
is
289,
th r ih terms of an A. P. are a, b, c respectively, shew q (q-i')a + (r-p)b+(p-q)c = 0. ,
The sum of p terms of an A. the sum ofp + q terms.
The sum
.— ,
,
,
P. is q,
of four integers in A. P.
and the sum of q terms
is 24,
and
is
their product is
find them.
Divide 20 into four parts which are in A. P., and such that the product of the first and fourth is to the product of the second and third in the ratio of 2 to 3. 14.
15.
m
tb
The p th term
of an A. P.
is q,
and the q th term
is
p
find the
;
term. 16.
How many
terms of the
must be taken
series 9, 12, 15,...
to
make 306? 17.
'in
2
If the
sum
of
n terms
If the sum of 18. 2 ?i to , shew that the
of an A. P.
2n + 3n 2
is
m terms of an A. P. m term is to the n th
is to th
the
,
find the
sum
term as 2m —
of
tth ?
n terms as
1 is to
Prove that the sum of an odd number of terms in A. P. to the middle term multiplied by the number of terms. 19.
20.
If 5 = n (pn
-
3) for all values of
n
t
find the
term.
2n —
is
1.
equal
p th term.
of terms in an A. P. is even the sum of the odd the even terms 30, and the last term exceeds the first by terms is 24, of 10 1 find the number of terms. 21.
The number
;
:
22.
There are two sets of numbers each consisting of 3 terms in A.
P.
and the sum of each set is 15. The common difference of the first set is greater by 1 than the common difference of the second set, and the product of the the numbers.
first set is
to the product of the second set as 7 to 8
:
find
Find the relation between x and y in order that the ,th mean between x and 2y may be the same as the th mean between 2x and y, 23.
?
?-
n means being 24.
that
its
If the
sum
inserted in each case.
sum of an A. P. is the same p + q terms is zero.
for
for
p
as for q terms, shew
CHAPTER
V.
GEOMETRICAL PROGRESSION. Quantities are said to be in Geometrical Definition. Progression when they increase or decrease by a constant factor. 51.
Thus each gression
of the following series forms a Geometrical Pro-
:
3,
G,
-
1
12,
24,
1
I
3'
9'
ar 2 ar
a, ar,
,
-I 27' 3 ,
The constant factor is also called the common ratio, and it is found by dividing any term by that which immediately iwecedes it.
In the
the second
first of it is
—-
the above examples the ;
in the third it is
common
ratio is 2
;
in
r.
o 52.
If
we examine
the series 3 ar2 ar
a, ar,
we
notice that in
tlian the
,
any term
the
number of the term in
Thus
index of
the 6
term
is is
the 20 th term and, generally, If
term,
the
n be the number we have
ih p term
ar
2
ar
s
is
is
of terms, l
r is
always
less
by one
the series.
the 3 rd term th
ai A ,
,
;
;
ar
a?^
and
19 ;
-1 .
if I
denote the
last,
or
n ,h
= ar"~\
When three quantities are in Geometrical Definition. Progression the middle one is called the geometric mean between the other two. 53.
;
;
HIGHER ALGEBRA.
38 To find
the geometric
Let a and
Then
b
mean
between two given quantities.
be the two quantities
;
G
the geometric mean.
since a, G, b are in G. P.,
_G G~ a' b
each being equal to the
common
G 2 = ab; G = Jab.
.-.
whence To
54.
insert
ratio
a given number of geometric means between
two given quantities.
Let a and
In
all
series of
n
b
be the given quantities, n the number of means.
we have
there will be n + 2 terms ; so that + 2 terms in G. P., of which a is the
Let r be the common ratio then
b
= the «r"
"
first
and
to find a b
the
last.
;
(n
+
2)
th
term
+1 ;
~a' i
••"©" Hence the required means value found in Example.
We
are
<» of,
2
a?- ,...
ar
,
where r has the
(1).
Insert 4 geometric
means between 100 and
have to find 6 terms in G. P. of which 160
sixth.
Let r be tbe
n
common
ratio
tben 5 = the sixth term
= 160?'5 . *
whence
and the means are
1
~32'
'
r= o' 80, 40, 20, 10.
;
is
5.
the
first,
and 5 the
GEOMETRICAL PROGRESSION. 55.
To find
the
sum of a number
o!)
of terms in Geometrical
Progression.
Let a be the first term, r the common Then terms, and s the sum required. 8
= a + car + ar +
multiplying every term by
~2
n the number
+ ar"~
of
l
;
we have
r,
+ ar"~ 2 + ar"" +
= ar + ar 2 +
rs
Hence by
n + ar
2
ratio,
1
ar*,
subtraction,
— s = ar n — a
rs .-.
;
(r-l)s = a(r"-l);
,..-5fe^a r
-
(i).
1
Changing the signs in numerator and denominator,
.-?S=*3 1 -r Note. for
s.
using
be found convenient to remember both forms given above in all cases except when? isj^ositive and greater than 1.
It will
1
(2)
Since ar'^ 1
^
1,
the formula
(1)
may S=
a form which
Example.
is
sum
be written
rl-a 7--T
:
sometimes useful.
Sum
The common
the
(2).
ratio
3
2
the series 3 = --
;
,
-1, -,
to 7 terms.
hence by formula
(-23
=—
2187]
II 2
~3
X
403
128
2315
2
128 * 5
I
(2)
HIGHER ALGEBRA.
40
111
n
Consider the series
56.
r,
1,
^-2
~3
,
,
2 The sum
to
=
n terms
>-
-H
1
27 2
.22
«-i
•
appears that however many terms be taken the sum of the above series is always less than 2. Also we see that, by making n sufficiently large, we can make the fraction
From
njr-^i
this
result
sma U a s we
as
it
please.
Thus by taking a
terms the sum can be made to from 2.
In the next
From
57.
more general case
article a
we have
Art. 55
by as
differ
of
s
=
is
number
sufficient little as
we
please
discussed.
\ -r 1
a 1
Suppose r
n the
smaller
therefore by
n terms
we
ar"
—t
1
—r
'
is
a proper fraction; then the greater the value of
is
the value of
making n
?•",
and consequently
from
1
of
we can make
sufficiently large,
of the series differ
ar'
;
sum
the
by as small a quantity as
^
is
usually stated thus
:
the
sum of an
number of terms of a decreasing Geometrical Progression
infinite
is
^
1
more
briefly, the
Example product
is
1.
sum
:
—
r
a to infinity is
1-r'
Find three numbers in G. P. whose sum
is
19,
and whose
216.
Denote the numbers by -, r
the
of
please.
This result
or
and
numbers are r
,
6, 6r.
a,
ar; then - x a x ar = 216 r
;
hence a = 6, and
'
.
.
GEOMETRICAL PROGRESSION. 6 -
r
+ 6 + 6r=19;
6-13r + 6r2 = 0;
.-.
3
whence
r
Thus the cumbers Example the
sum
The sum
2.
of their squares
Let a denote the is
1
45
r
and the sum of
;
-r
common
the
their squares is
of terms in G. P. is 15,
ratio
1
(1)
by
(2)
and
(1)
l z
(3)
+r
(1),
(
9
= 5;
2
,
... — 20 10 — o y ,
,
EXAMPLES.
112
1.
Sum -,-,-,...
2.
Sum
3.
Sum ^t,
l£,
3,...
to 8 terms.
-4,
8,...
to 10 terms.
6.
Sum Sum Sum
7.
4. 5.
A
O
V.
to 7 terms.
9
-2, 2^, -3i,... to 6 terms.
2,
16'2, 5-4, 1-8,... to 7 terms. 5,
Sum
3,
-4,
8.
Sum
1,
N/3,
9.
Sum
—
,...
p
terms.
to 2n terms.
o 3,...
1
-j/2 v
to
25,...
1,
,
-2,
to 12 terms.
8 -jr s '2
,...
to 7 terms.
2)-
(3),
whence r=x and therefore a = 5. an, Thus +1 the series is o,
then the sum of the
-r*
a2 45 _ 72 =
~=
;
„
-z
,—=15 1 - r
Dividing
and
ci^
1
Hence
and from
number
find the series.
;
term,
first
-
6, 9.
of an infinite
is
2
= - or
'
(l
terms
are 4,
41
a.
:
HIGHER ALGEBRA.
42
11
3 3, -j,.- to ^ terms.
10.
Sum -~,
11.
Insert 3 geometric
means between 2^ and -
12.
Insert 5 geometric
means between 3f and
13.
Insert 6 geometric
means between 14 and
Sum
the following series to infinity
14.
|, -1, ?,...
15.
-45,
16.
1-665, -1-11, -74,...
17.
3" 1 3~ 2
18.
3,
19.
7,
20.
The sum
4
the
first
v/3,
1,...
3 terms
The
term of a G. P.
40|. -
I —
.
64
-015, -0005,...
,
N/42,
of the first 6 terms of a G. P. find the common ratio.
;
.
3-',...
,
6,...
is
9 times the
and the second term
sum
of
is
24; find
The sum of a G. P. whose common ratio is 3 is 728, 22. last term is 486 ; find the first term.
and the
21.
fifth
is 81,
the series.
term 448, and the sum
In a G. P. the first term is 7, the last 23. 889 find the common ratio. ;
The sum
24.
of three
numbers
in G. P.
is 38,
and their product
is
1728; find them.
The continued product of three numbers in G. P. is 216, and the sum of the product of them in pairs is 156 find the numbers. 25.
;
If
26.
p the
s
sum
denote the sum of the series l+rp + r2p +... of the series 1 — rp + r2p - ... ad inf., prove that
Sp
/Op
27.
If the
p
th
th ,
q
that
,
ad
inf.,
and
+ Sp == ^*ij'2p'
r th terms of a G. P. be a, a«- r 6 r -*c*-«=l.
b, c
respectively, prove
The sum of an infinite number of terms of a G. P. of their cubes is 192 ; find the series.
is 4,
and the
Recurring decimals furnish a good illustration of Geometrical Progressions.
infinite
28.
sum
58.
Example.
Find the value of
"423.
•423 =-4232323
4
~ io +
23 iooo
+
~io + IP +
23
iooooo
10
5+
+ ;
'
GEOMETRICAL PROGRESSION. •
•
-«3-
that,*,
4
,
10
_
23 /
+
_4
io V
23
+
10
1
+
1033
10
io
+
2
43
\
1
+ io-*
)
1
:: '
_ _1_ 10"
100
23
4_
"io" "!^3 1
4
~~
"
99
_23
+ 10 990
_419
"
990
which agrees with the value found by the usual arithmetical
The general
rule.
any recurring decimal to a vulgar fraction may be proved by the method employed in the but it is easier to proceed as follows. last example 59.
rule for reducing
;
Tojind
the value
of a recurring decimal.
P
denote the figures which do not recur, and suppose them j> in number; let Q denote the recurring period consisting of q figures let D denote the value of the recurring decimal ; then
Let
;
D = 'PQQQ 10>xD = P'QQQ
.-.
and
10T+'
therefore,
that
*D = PQ-QQQ
by subtraction, (10 p+y - lC) 10" (10'
is,
;
-
1)
. ' '
D = PQ - P
>
D = PQ-P; ;
PQ-P D _ (10''- 1)10'''
Now
10"- 1 is a number consisting of q nines; therefore the denominator consists of q nines followed by p ciphers. Hence we have the following rule for reducing a recurring decimal to a vulgar fraction
For the
:
numerator subtract nonrecurring fgures from the
the integral the integral
number number
consisting of consisting of
recurring figures ; for the denominator take a number consisting of as many nines as there are recurring jig n n 8 followed by as many ciphers as there are non-recurring figures. the
non-recurring
and
r;
r
HIGHER ALGEBRA.
44 To find
60.
the
+
(a
a,
sum ofn
d)
(a
r,
terms of the series
+ 2d)
2
r
(a
,
+ 3d)
r
3 ,
in which each term is the product of corresponding terms in an arithmetic and geometric series.
Denote the sum by
S
;
then
S=a+(a + d)r+(a + 2d)r .-.
+
—
n dr(\-r ^ =a+
1
.
~
.
.
.
(a
.
+ n~^ld)r"-'
+(a+ n-2d)r n + (a + n -ld)r n .
S in
a
) =
—
l
l-r
-
r
involve r n
can
—
-z
1-r
+
be 7^
(1
In this
made
(a
+ n-lct)
1
)
+ n - Id)N r „ '
;
(a
+
;
(l-r)*
case,
n~ld)r\ T^r
small as
we
assuming that
so small that
they
r„ for the sum to
please by taking all the
may
infinity. J
If
x <1, sum the
series
+ 2ar + 3x 2 + 4x 3 +
to infinity.
S = l + 2a; + 3a: s + 4a s +
Let .-.
•
we
shall refer
infinity series of this class it is usually best to
l
•
terms which
be neglected,
We
n
XXI.
proceed as in the following example. 1.
r" '
T^r
r" as
to this point again in Chap.
Example
n
'
_ ( a + n~^\d )
-r)
In summing to
-
,
(a v
dr"
(l-ry~
sufficiently great.
)
the form
dr
+
1
l
dr(l-r"+ l-r (1-r) 2 Write
Cor.
obtain
.
+ dr""
a
•'•
if
+
subtraction,
2 S(l - r) = a + (dr + dr +
then
...
~
ar + (a + d)r2 +
rS=
By
2
xS=
x + 2x* + 3x*+
S(l-x) = l + x + x 2 + x*+ 1
~l-x •
a-
I
]
;
;
J
GE0METK1CAL PROGRESSION. Example
Sum
2.
+
the series 1
+
-
S== ii
Let
+i+
1
+
5
4
3
1
.
to n terms.
Sn-2 _..
7
3n - 5
52
+ 53+
+ -57^-+
/3
,
+
. .
4
1
5*-
•"•
.
o-
_+ _7 + 10
^
»
T
+ —+ 0"*
-,
o
3
3
A
1
3
1 f
45
5
3n-_2
\
+5«-»j
5
,—
3n-2
\
1 "
7-"1
3n-2
"
"~5«~
" ,
=
+
1
*
3
DN
'>-i '
=
3
+
7
12w + 7 ~ 4 5* .
~4 6
•'*
~
—1\
/
" I (1 5
1
3«-2
- -5.-
J
_
;
12/t+7
35 16
16
5"- 1
.
'
EXAMPLES. 2a 4- 3a2 + 4a 3 4-
V.
1.
Sum
2.
Sum 1 + - 4-
3.
Sum 1 + 3.r + 5d'2 + 7o? + 9.z4 + ...
4.
Sum 1 + - 4-
5.
Sum 1 + + + Q + 2 4 o
6.
Sum l + 3^ + 6 f
1 4-
3
4
2
-,
3
lb
3 -
+ ^, + zoo c^rr. + 64 +
2
to
5
4 -3
n terms.
31
15
7 77.
. . .
b.
+
. . .
to
. . •
to infinity.
to infinity.
n terms.
7
-7
l
2
...
to infinity.
4-10ji>3 4-... to infinity.
Prove that the (n + l) th term of a G. P., of which the first term th term of a G. P. of is a and the third term b, is equal to the (2»+l) which the first term is a and the fifth term 6. 7.
whose first term is a and common ratio r is equal to the sum of n of a G. P. whose first term is b and common ratio r 1 Prove that b is equal to the sum of the first two terms of the first series. 8.
The sum
.
of
2n terms of a G.
P.
HIGHER ALGEBRA.
46
Find the sum of the
9.
l
r
;
and
+ (l + b)r + (l + b + b
2
)r 2 +
{l
+ b + b 2 + b 3 )r3 +...,
b being proper fractions.
The sum
10.
The
11.
4,
;
two terms of an
first
and every term
numbers in G. P. is 70 if the two extremes and the mean by 5, the products are in A. P.
of three
be multiplied each by find the numbers.
the
infinite series
is
3 times
infinite G. P. are together equal to 5, the sum of all the terms that follow it; find
series.
Sum
the following series
12.
.r+a,
13.
x
14.
«+o
15.
2 3
16.
454545
17.
If a,
,v
2
+ if)
(x
..
)
2
3
+ ^2 +
7
+ 2
3« - -
j
~
72 +
:
33
73
b, c,
3
+ 34 + -
n terms.
to
— +... to 2p terms.
ha +
,
. . .
2 35
3
+ p + --
74 + 75
„
.
+
76
to mfinity-
.,
to llifinity-
d be in G. P., prove that (b - cf + (c - a) 2 + (d - b) 2 = {a- d) 2
.
If the arithmetic mean between a and b is twice as great as the 18. geometric mean, shew that a 6 = 2 + ^/3 2-^3. :
19.
:
Find the sum of n terms of the
series the r th
term of which
is
(2r-f-l)2'\
Find the sum of 2n terms of a series of which every even term is a times the term before it, and every odd term c times the term before it, the first term being unity. 20.
21. a,
If
Sn
and common 22.
If
whose
first
Sv
r,
terms are
2' 3' 4
23.
If r
<
1
and
Hence shew that nr n
*
'
'
'
and
^Ti
+ S2 + S3 +
positive,
(2»i
+
sums
are the
1, 2, 3,..,j2,
&\
.
,
S3 ,...SP
prove that
is
r
ratio JS2 ,
sum
of n terms of a G. P. whose first term find the sum of Slf S3 /8'5 ,.../8 2B _ 1
denote the
l)r
. . .
and wl
of infinite geometric series, whose common ratios are respectively,
+ Sp =f (p + 3).
m is a positive
(l-r)
is indefinitely
small
2wi
integer,
shew that
+1 .
when n
is indefinitely great.
CHAPTER
VI.
THEOREMS CONNECTED WITH
HARMON ICAL PROGRESSION.
THE PROGRESSIONS. DEFINITION.
61.
Three quantities
Harmonical Progression when - =
Any number
——
are said to be in
.
7
o
c
a, b, c
c
quantities are said to be in Harmonical every three consecutive terms are in Har-
of
Progression when monical Progression.
The reciprocals of quantities in Harmonical Progression
62.
are in
A rithmetical
By
Progression.
definition, if «,
b, c
are in Harmonical Progression,
a—
a
b }
~c^~b^~c' .'.
a(b —
c)
=
c (a
—
b),
dividing every term by abc,
1111 c
b
b
a'
which proves the proposition.
Harmonical properties are chiefly interesting because of their importance in Geometry and in the Theory of Sound in Algebra the proposition just proved is the only one of any importance. There is no general formula for the sum of any number of quantities in Harmonical Progression. Questions in H. P. are generally solved by inverting the terms, and making use 63.
:
of the properties of the corresponding A. P.
;
.
HIGHER ALGEBRA.
48 64.
To find
Let
a,
then -
a
11
,
between two given quantities.
7tiean
H their
be the two quantities,
b
-~
,
harmonic
the
harmonic mean;
A. P.
T are in b
1
I
I
I
''11
a~b
IV
11 +
2
H~ a
&'
2ab
,,
a+b Example.
Here 6
common
is
Insert 40 harmonic
means between
na term of an A. P. whose the 42
difference
;
and ^
first
term
is -
=^ +
Thus the arithmetic means
41d ; whence d 2
are -
3
-
,
41 ---
,
;
and therefore the har-
7 *
b,
a+
b (!)•
~Y~
G = Jab
(2).
H=^ a+b a+ Tr All = —-
is,
b — 2
.
„
Therefore
G
From
is
the geometric
these results .
A-
and harmonic
we have proved
A=
that
d be the
3£, 2\,...~.
means between a and
_,
let
= -.
If A, G> II be the arithmetic, geometric,
65.
;
then 6
monic means are
7
we
2ab .
T
a+b
(3). v '
~
= ab = G 2 7
:
mean between A and
//.
see that
b- 2 Jab ~ a+b = a + g-^_ G=-~-Jab ,-z-
:
;
HARMONICA! PROGRESSION.
49
a and b are positive; therefore the arithmetic mean of any two positive quantities is greater than their geometric mean.
which
is
positive
if
Also from the equation G*--A1I, we see that G is intermediate in value between A and 11; and it lias been proved that
A>
G, therefore
G>
II ; that is, the arithmetic, geometric, and between any tioo positive quantities are in descending
harmonic means order of magnitude.
Miscellaneous questions in the Progressions afford scope for skill and ingenuity, the solution being often neatly effected by some special artifice. The student will find the following 66.
hints useful.
same quantity be added to, or subtracted from, all an A P., the resulting terms will form an A. P. with
If the
1.
the terms of the same common difference as before.
[Art. 38.]
terms of an A.P. be multiplied or divided by the same quantity, the resulting terms will form an A. P., but with a new common difference. [Art. 38.] If all the
2.
If all the terms of a G.P. be multiplied or divided by the quantity, the resulting terms will form a G.P. with the
3.
same same common If a,
4.
ratio as before.
portion^ since,
by
are in G.P., they are also in continued pro-
d...
b, c,
[Art. 51.]
definition,
a
b
c
1
bed
'
r
Conversely, a series of quantities in continued proportion be represented by x, aw, xr'2
may
,
in
Example H. P.
1.
By adding a*
that .-.,
is
ab
If a
2 ,
+ ac + bc
+ ab + ac + bc, {a
b
2
+ b)
(a
dividing each term by (a
b
that
is,
H. H. A.
+c b
+ c,
.
c
+a
c
+ a,
a a
+b
+b
b
+
c,
c
+ a, a +
see that 2
c'
+ ca + cb + ab
+ c)(b + a), (c + a) + b)(b + c) (c + a),
,
-.
we
+ ba + bc + ac,
{b
shew that
are in A. P.,
to each term,
b2
+ c),
c
,
2
are
(c
+ b)
are in A.P.
are in A. P.
m A. P.
are in H.
P 4
b
are
'
HIGHER ALGEBRA.
50
of
Example 2. If n terms of an A.
the last term, d tlie common difference, and s the sum 2 P. be connected by the equation Sds={d + 2l) prove that I
,
d = 2a. Since the given relation
true for any
is
a=
Hence by
then
,
(d-2ay- = 0;
d — 2a.
.-.
Example q, q
-
If the
3.
p
th
th ,
q
r
,
th
s
,
th
terms of an A. P. are in G.
shew that
P.,
r - s are in G. P.
r,
we have
"With the usual notation
a+
(p-l)d_ a + (g-l )d_a+ (r-l)d
LAlt bb -
^V(q^lJd-^T¥^l)~d-aT(^l)d .*.
;
= s.
l
8ad = {d + 2a) 2
substitution,
or
p-
number of terms, put n= 1
-
(4)J
'
each of these ratios
+ (p-l)d}-{a+(q-l)d\ _ {a+ (q- 1) ~ {a + (q - 1) d\ - {a+ (r- 1) d) " {a+(r - 1) {a
= Hence p -
p-q^q-r q-r
q, q
-
natural numbers the
first
r
r,
r
-
—8 s
are in G.P.
The numbers
67.
n terms
68.
- \a + (r- 1) d} d] - {a + {s - 1) d\ d]
the
;
is
To find
1,
n
th
are often referred to as the term of the series is n, and the sum of 2,
3,
- (n +1).
sum of
the
the squares
of
the first
numbers.
Let the sum be denoted by
£=l
2
+
n3 - (n -
2
We have
2
S
+ a
l)
;
3'
then
+n
+
= 3n - 3n+ 2
2
1
.
;
and by changing n into n—l,
n _ \y _ 3 (w - 2) -
(
similarly
_ (71 (n
3 2 1
3
3 3
= 3(™ - l) 2 - 3(w 2 3 3) = 3(w - 2) - 3(n 3
2)
-2 =3.3 -3.3+l; -l 3 =3.2 -3.2 + l; -0 =3.1 -3.1 + 1. 3
2
2
3
2
1)
+
1
2)
+
1
;
;
n natural
;;
THE NATURAL NUMBERS.
51
Hence, by addition,
^3 =
3(l
— 6b
2
2 + 3 + ...+»')-3(l + 2 + 3 + ...+n) + * 3n(n + l) + it.
+
2
3
—
a .
=n -n+
3aS
•.
-±-
—
a
= n(n + „ S=
.
••
C9.
To fiml
—
1 4- ;;);
n(n+ l)(2n +
l)
6
sum of
the
1) (n
cubes
the
of the frst
n natural
numbers.
Let the sum be denoted by
£=l +2 3
We
have
-
4 (n - 1 ) - (?* w _ 2) 4 - ( w ( n*
(n
4
3 2 1
4
4
3
S
then
',
+ 33 +
+n\
- Gn 2 + 4n - 1 4 3 8 2) = 4 (n - 1 ) - 6 (n - 1) + 4 (n - 1) 3 3)« = 4(ra- 2) - 6 (n- 2) + 4 (n - 2) 4
l)
=
4?i
3
;
1
2
1
-2 = 4.3 -6.3 + 4.3-l; -l = 4.2 -G.2 + 4.2-l; -0 = 4.1 -6.1 + 4.1-1. 4
3
2
4
3
2
4
3
2
Hence, by addition,
.\
w4 = 4#-6(l 2 + 2 2 +...+?i 2) + 4(l +2 + .. + 7i)-n; 4S = n 4 + n + 6(l 2 + 2 2 +...+7t 2 )-±(\+2 + ...+n) = n* + 7i + 7i (n+ 1) (2n+ 1) -2n(7i + 1) = 7i (71 + 1) (?r -7i+\ + 2n+\-2) = 7i (n + 1 ) (?r + n) ;
w'(n + l) a g _ *
,
'
Tims
•
"
4
_f W (n+l) )' ~
(
2
*
J
Me
cz^es of the f7'st n natural 7iumhers is equal to the squa7'e of the siwi of these 7iumbers. ^Ae s?
and the two preceding articles may be the sum of the squares, and the sum of the cubes
The formulae
of this
applied to find of the terms of the series «,
a+
d,
a+
2d,
4—2
;;
}
:
HIGHER ALGEBRA.
52
In referring to the results we have just proved it will be convenient to introduce a notation which the student will freWe shall denote the quently meet with in Higher Mathematics. 70.
series
+
1
1* l
3
2
+
+
3
+ 2* + 3* +
+ n by
. .
.
...
+ 2 3 + 3 3 +...
2,n
+na by %n* +«8 by 2n
3 ;
where 2 placed before a term signifies the sum which that term is the general type. Example
1.
Sum
of all terms of
the series 1
.
2
+2
+3
3
.
.
4+
.
.
.to
n terms.
The w th term=ra(n+l)=n2 +«; and by writing down each term in a similar form we shall have two columns, one consisting of the first n natural numbers, and the other of their squares. .•.
sum = 2m2 + 2?i
the
_w(m+1)
(2m + 1)
6
T
n(n +
l)
2
'
n(n+l) j2n+l
)
n(n + l)(n+2) 3
Example
2.
Sum
to
n terms the
series
whose M th term
1_1 is 2'
Let the sum be denoted by S then S = 2 2»- 1 + 82>i3 -62n 2 ;
1 _ 2" " 2- 1
+
8m 2
+ 1) 2 _ 4~
(m ~~
6m ( m+1)(2m +
1)
6
= 2» - 1 +« (m + 1) {2m (m + 1) - (2m + 1) = 2' -l + n(n + l)(2n 2 -l). l
EXAMPLES. 1.
Find the fourth term (1)
2,
(2)
2,
2J, 21,
(3)
2,
2f,
VI.
a.
in each of the following series
3i,... 3,...
3i,...
2.
Insert two harmonic
means between
5
3.
Insert four harmonic
means between
2 - and o
and
11.
—2 1«3
.
+ 8m 3 - 6m 2
.
EXAMPLES OX THE ively,
53
and 9 l are the geometric and harmonic means, respectbetween two numbers, find them. If 12
4.
riiOGltESSlOXS.
:
If the harmonic mean between two quantities is to their geo5. metric means as 12 to 13, prove that the quantities are in the ratio of 4 to 9. If a,
6.
be in H.
b, c
P.,
a lh
If the
7.
:
shew that a—b=a+ c
term of a H.
iii
:
P. be equal to n,
equal to m, prove that the (m + n) th term If the
8.
p
th
that If b
9.
th <7
,
(
-
,
is
equal to
mean between a and
111
1
+ — =-+o —a b a c c ,
Find the sum of n terms of the
i.
mn
m+n
=
j-
(
and the u lh term be
rth terms of a H. P. be a, b, c respectively, prove r) be + (r — p) ca + (p-q) ub 0.
the harmonic
is
a — c.
series
c,
prove that
.
whose n th term
is
10.
3n*-n.
11.
ns +^n.
12.
»(»+2).
13.
»2 (2»+3).
14.
3" -2".
15.
3
16.
If the
and
P.,
on,
(4' l
+ 2; 2 )-4/i i
:J .
(?^+l) th and (r+ l) th terms of an A. P. are in n r are in H. P., shew that the ratio of the common 2
(m+iy\
,
y
term
difference to the first
in the A. P. is
—n
.
If I, m, n are three numbers in G. P., prove that the first term 17. th th of an A. P. whose £th terms are in H. P. is to the common , and ?i difference as m-\-\ to 1. ,
sum
of n terms of a series be the nature of the series.
If the
18.
m
term and
a + bu + cri2
,
find the
19.
Find the sum of n terms of the series whose n th term 2 4?i(?i 2 +l)-(6>i -fl).
20.
If
nth
is
between any two quantities there be inserted two arithmetic
two geometric means G ly G2 and two harmonic means means A u A H1 7/2 shew that 6^0',, II1 H.2 = A l + A 2 I^ + IL,. ;
;
21.
If
and q the
p
n arithmetic means between two numbers, n harmonic means between the same two numbers,
be the
first
of
first
of
prove that the value of q cannot 22.
that
it
:
:
;
,
lie
between p and
f
—
p.
Find the sum of the cubes of the terms of an A. is exactly divisible by the sum of the term-.
J
P.,
and shew
;
;
.
HIGHER ALGEBRA.
54)
Piles of Shot and Shells.
To find
71.
number of
the
pyramid on a square
arranged in a complete
shot
base.
Suppose that each side of the base contains n shot then the 2 2 number of shot in the lowest layer is n in the next it is (n—l) 2 and so on, up to a single shot at the in the next (n-2) ;
\
;
;
top. .-.
S^n =
2 + (n-l) 2 + (n-2) +... +
2
l
n(n+l)(2n + l) 6
To find
72.
pyramid
the base
number of shot arranged in a complete of which is an equilateral triangle. the
Suppose that each side of the base contains n shot
number
(n
-
1)
n(n + —V«
•
that
is,
+ (n 1)
this result write
number
the
2) 1
+
+
1
j
,
2 or - [n
—
In
then the
of shot in the lowest layer is
n+ xi
;
+ n)
—
n — 1, » — 2,
for n,
of shot in the 2nd, 3rd,
and we thus obtain
layers.
S=i($n* + 2,n) >(n + l)(» + 2) M
.-.
To find
73.
pyramid Let
the base
m and
the
number of
of which
is
a
shot
[Art7a] arranged in a complete
rectangle.
n be the number
of shot in the long
and short
side
respectively of the base.
The top layer
m—
and
n+1
so
on
consists of
a single row of
m — (n — l),
shot
— n + 2);
in the next layer the
number
is
2
in the next layer the
number
is
3 (in
number
is
n (m — n + n).
(in
—n+
3)
;
in the lowest layer the
or
AND
PILES OF SHOT
SHELLS.
S= (m -01 + 1) + *2(m-n + 2) + 3(w-w + 3) +
.-.
= (m - n) (1 + 2 + 3 + (wi
- n) n (n +
1 )
...
+
2
n) + (l + 2
w (n +
+
l)
+
3
s
+
...
+n(rn-n + n)
...
+ n2)
6
n (n + 1 ){3(m-n) + 2n +
_n(n + =
2
55
(2n +1)
1)
2
=
)
;
;
(3m- n+
l}
1) '
6
To find
74.
2>yramid
the base
Let a and top layer,
n
the
number of
of which
is
b denote the
the
number
a
an incomplete
shot arranged in
rectangle.
number
two
of shot in the
sides of the
of layers.
In the top layer the number
of shot is ab
;
in the next layer the
number
is
(a
+
1) (6
+
1)
in the next layer the
number
is
(a
+
2) (b
+
2)
\
and so on in the lowest layer the
number
is
(«
+ n --
1) (b
ab + (a + b)(n-l) + ()i-l)
or .-.
+n—
1)
2 .
S = abn + (a + 6) % (n- 1) + % (n- Vf ("-l)w(a + 6) (n-l)n(2 .n-l + + = +
abn
2
=
| {6ab +
3 (a
+
6 b) (n
In numerical examples following method. 75.
-
1)
it is
+
(w
-
1) (2m
-
1)}.
generally easier to use the
Example. Find the number of shot in an incomplete square courses, having 12 shot in each side of the top.
we
If
the base,
.*.
pile of 16
place on the given pile a square pile having 11 shot in each side of we obtain a complete square pile of 27 courses;
and number of shot also
1
number
in the complete pile
=
^
11 x 12 x 23
of shot in the added pile=
number of shot
in the incomplete pile =6424.
„
= <)'.)30 = 506;
;
[Art. 7
1
.]
HIGHER ALGEBRA.
56
EXAMPLES.
VI.
b.
Find the number of shot in 1.
2. 3.
A square pile, having 15 shot in each side of the base. A triangular pile, having 18 shot in each side of the base. A rectangular pile, the length and the breadth of the base con-
taining 50 and 28 shot respectively.
An incomplete triangular pile, a side of the base having 25 shot, side a of the top 14. 4.
and
5.
An
incomplete square pile of 27 courses, having 40 shot in each
side of the base.
The number of shot in a complete rectangular pile is 24395 6. there are 34 shot in the breadth of the base, how many are there in length ?
if
its
The number
of shot in the top layer of a square pile is 169, in the lowest layer is 1089 ; how many shot does the pile contain ? 7.
and
;
Find the number of shot in a complete rectangular pile of 8. 15 courses, having 20 shot in the longer side of its base. Find the number of shot in an incomplete rectangular pile, 9. number of shot in the sides of its upper course being 11 and 18, and the number in the shorter side of its lowest course being 30. What is the number of shot required to complete a rectangular 10. the
having 15 and 6 shot in the longer and shorter upper course?
pile its
side, respectively, of
The number of shot in a triangular pile is greater by 150 than number of shot in a square pile, the number of layers in each being the same; find the number of shot in the lowest layer of the tri11.
half the
angular
pile.
Find the number of shot in an incomplete square pile of 16 12. courses when the number of shot in the upper course is 1005 less than in the lowest course. 13.
number
Shew that the number of shot in a square pile is one-fourth the of shot in a triangular pile of double the number of courses. number
of shot in a triangular pile is to the number of shot in a square pile of double the number of courses as 13 to 175 ; find the number of shot in each pile. 14.
If the
The value of a triangular pile of 16 lb. shot is ,£51 if the 15. value of iron be 10s. 6d. per cwt., find the number of shot in the lowest layer. ;
If from a complete square pile of n courses a triangular pile of 16. the same number of courses be formed ; shew that the remaining shot will be just sufficient to form another- triangular pile, and find the number of shot in its side.
.
\
CHAPTER
VII.
SCALES OF NOTATION.
The ordinary numbers with which we are acquainted in Arithmetic are expressed by means of multiples of powers of 10; 76.
for instance
25-2 4705 = 4
+ 5; 3 10 + 7 x 10 2 +
x 10 x
10 +
x
5.
This method of representing numbers is called the common or denary scale of notation, and ten is said to be the radix of the scale. The symbols employed in this system of notation are the nine digits and zero.
manner any number other than ten may be taken as the radix of a scale of notation thus if 7 is the radix, a number 3 expressed by 2453 represents 2x7 + 4x7" + 5x7 + 3; and in In
like
;
this scale no. digit higher
than 6 can occur.
Again in a scale whose radix is denoted by r the above number 2453 stands for 2r 3 + 4? ,2 + hr + 3. More generally, if in the scale whose radix is r we denote the digits, beginning with that in the units' place, by a a,, a2 ,...aj then the number so formed will be represented by tt
n
a r +a
n
,r
~
1
+a
,
y~~ +
.
.
.
+
a/
2
+
a,r
+ a,
where the coefficients a a ,,...«,. are integers, all which any one or more after the first may be zero. ,
less
than
r,
of
Hence
in this scale the digits are r in number, their values ranging from to r — 1
The names Binary, Ternary, Quaternary, Quinary, Senary, Septenary, Octenary, Nonary, Denary, Undenarv, and Duodenary 77.
are used to denote the scales corresponding to the values three,... twelve of the radix.
fae»,
HIGHER ALGEBRA.
58
In the undenary, duodenary, scales we shall require symbols to represent the digits which are greater than nine. It is unusual to consider any scale higher than that with radix twelve ; when necessary we shall employ the symbols t, e, T as digits to denote ten eleven and twelve .
'
'
It
is
'
'
',
'
ten
.
'.
worthy
especially
symbol not for
.
',
of notice that in every scale 10 is the
but for the radix
itself.
The ordinary operations of Arithmetic may be performed in any scale but, bearing in mind that the successive powers of the radix are no longer powers of ten, in determining the carrying figures we must not divide by ten, but by the radix of the scale 78.
;
in question.
Example
In the scale of eight subtract 371532 from 530225, and multiply the difference by 27. 1.
530225 371532
136473
136473
1226235 275166
27
4200115 After the first figure of the subtraction, since we cannot Explanation. take 3 from 2 we add 8 thus we have to take 3 from ten, which leaves 7 then 6 from ten, which leaves 4 then 2 from eight which leaves 6 and so on. ;
;
;
;
Again, in multiplying by
7,
we have
3x7 = twenty one = 2x8 + 5; we
therefore put
down
and carry
5
2.
7x7 + 2 = fifty
Next put down 3 and carry 6
one = 6x8 + 3;
and so on, until the multiplication
;
In the addition, 3
we
therefore put
down
and carry
1
1.
+ 6 + l = nine=l x 8 + 1; 6 + l + l = eight = lx8 + 0;
Similarly
2
and and so on.
Example
+ 6 = nine = lx8 + l;
Divide 15et20 by 9 in the scale of twelve.
2.
9)15
lee96...G.
Explanation.
we put down Also 8 x
we
1
Since 15 = 1 x
and carry
T + 5 = seventeen = 1 x9 + 8,
8.
T + e = one hundred and
therefore put
down
e
and carry
8;
seven = e x 9 + 8
and so on.
;
is
completed.
;
SCALES OF NOTATION. Example
3.
Find the
.square root of
59
442641 in the scale of seven.
442641(646 34
134 1026 G02
1416112441 12441
EXAMPLES. Vila. 1.
Add
2.
Find the sum of the nonary numbers 303478, 150732, 2G4305.
3.
Subtract 1732765 from 3673124 in the scale of eight.
4.
From 3^756
5.
Divide the difference between 1131315 and 235143 by 4 in the
together 23241, 4032, 300421 in the scale of
five.
take 2e46t2 in the duodenary scale.
scale of six. 6.
Multiply 6431 by 35 in the scale of seven.
7.
Find the product of the nonary numbers 4685, 3483.
8.
Divide 102432 by 36 in the scale of seven.
9.
In the ternary scale subtract 121012 from 11022201, and divide
the result by 1201. 10.
Find the square root of 300114 in the quinary
11.
Find the square of
12.
Find the G. C. M. of 2541 and 3102 in the
13.
Divide 14332216 by 6541 in the septenary scale.
tttt
scale.
in the scale of eleven. scale of seven.
Subtract 20404020 from 103050301 and find the square root of the result in the octenary scale. 14.
15.
Find the square root of ce^OOl in the scale of twelve.
16.
The
nary
following
numbers
are in the scale of six, find
without transforming to the denary scale
rules,
ordi-
:
(1)
the G. C. M. of 31141 and 3102
(2)
the L. C. M. of 23, 24, 30, 32, 40, 41, 43, 50.
79.
by the
To express a given integral number in any proposed
scale.
Let iV be the given number, and r the radix of the proposed scale.
Let a u an a 2 ,...a be the required digits by which iV expressed, beginning with that in the units' place; then _1 a n r" + a n — ,r" + ... + ar~ + a.r + a 2 ,
t
N=
We
10
.
lt
1
have now to find the values of «
,
a,, "_...",,
is
to be
HIGHER ALGEBRA.
60 Divide
N by
then the remainder
r,
+ a71—1 r
l
a nr"
n
+
by
If this quotient is divided if
2
+a
...
and the quotient r+a
is
a
,
l
.
the remainder
r,
and
is
is
a a2 i
the next quotient so on, until there
is
;
;
no further quotient.
the required digits a a ag1 ...an are determined by successive divisions by the radix of the proposed scale.
Thus
all
,
,
x
Example
Express the denary number 5213 in the scale of seven.
1.
7)5213 7)7447
5
7)106.
2
7 )15.
1
2
1
5213 = 2x7 4 +lx7 3 + lx7- + 2x7 + 5; Thus and the number required is 21125.
Example
Transform 21125 from
2.
scale seven to scale eleven.
e)21125 e)1244T
t
~e)Gl3. .-.
the required
number
is 3t0t.
In the
Explanation.
t
first line
of
work
21 = 2x7+l=fifteen = lx
Next
by
e
4x7 + 1 = twenty
therefore
we put down nine = 2 x
we put down 2 and carry
7
e ;
1
+7
and carry so on.
Example 3. Reduce 7215 from scale twelve and verify the result by working in the
In scale of ten
-
4.
;
and
scale ten,
r
+ 4;
7215
to scale ten by scale twelve.
f)7215
working in
1
JL2
«)874.
80
t)t^.
12
t)10.
.4
1033 12
1.
.2
1
In scale of twelve
J
1 12401
Thus the
result is 12401 in each case.
7215 in scale twelve means 7 x 123 + 2 x 12 2 +
1 x 12 + 5 in calculation is most readily effected by writing this expression in the form [{(7 x 12 + 2) } x 12 + 1] x 12 + 5 thus we multiply 7 by 12, and add 2 to the product; then we multiply 86 by 12 and add 1 to the product; then 1033 by 12 and add 5 to the product.
Explanation.
scale ten.
The
;
;
SCALES OF NOTATION.
f>l
Hitherto we have only discussed whole numbers; but fractions may also be expressed in any scale of notation thus 80.
;
•25 in scale ten denotes
5 2 — + — 10*' 10 ,
2
•25 in scale six denotes
+
-=
— 2
•25 in scale r denotes
h
r
5
^ G
—5 r
;
.
Fractions thus expressed in a form analogous to that of ordinary decimal fractions are called radix-fractions, and the point The general type of such fractions in is called the radix-point. scale r is
~ r
r
where 6 b 2 6 a ... are or more may be zero. 81.
Let
,
,
,
To
F
express
3
~
r
3
~
integers, all less
>
than
of
r,
which any one
a given radix fraction in any proposed
scale.
be the given fraction, and r the radix of the proposed
scale.
Let b then left
,
b
,
be the required digits beginning from the
b 3 ,...
;
b b ^ F J-Xr + A+ r r3+
We
have now to find the values of
6 p b2
,
Multiply both sides of the equation by r
rF=b+-
2
Hence
the fractional part by
F
x
,
then
h
+ -l+
b is equal to the integral part of l
;
63 ,
;
rF
;
and,
if
we denote
we have
H-i. + J + the integral part of rF ; and similarly by successive multiplications by r, each of the digits may be found, and the fraction expressed in the pro posed scale.
Multiply again by r\ then, as before, b x
is
4
HIGHER ALGEBRA.
62
the successive multiplications by r any one of the products is an integer the process terminates at this stage, and fraction can be expressed by a finite number of digits. the given & But if none of the products is an integer the process will never terminate, and in this case the digits recur, forming a radixin
If
fraction analogous to a recurring decimal. 13
Example
Express
1.
-^ as
13
a radix fraction in scale
six.
13x3
7. x6= -8- = 4 + 8' ft
16 7
a
7x3
Kj.
1
a
lx3
I..
1
1
^x6 = 3. 4
.-.
the required fraction
13
5
=g + ^+ p + = •4513.
Qi
Transform 16064-24 from scale eight
Example
2.
We must
treat the integral
and the
to scale five.
fractional parts separately, '24
5 )16064 5 )2644 ...
5
5)440...
1*44
5)71. ..3
J>_ 2-64
5)13... 2
2...1
£_
4-04 5_ 0-24
After this the digits in the fractional part recur; hence the required
number
is
212340-1240.
In any scale of notation of which the radix is r, the sum of the digits of any whole number divided by r - 1 will leave the same remainder as the whole number divided by r — 1. 82.
N
a n the digits begindenote the number, a a lt a 2 Let ning with that in the units' place, and S the sum of the digits; then ~ = a Q + a r + a2r 2 + + a„_/ 1 + arn ,
N
S=a r .tf-S=a
1
,
,
;
x
+ a n _ + an + a +a 2 + (r-l) + a 2 (r°--l)+ + •„_, (i*- - 1) + «, (f - 1). x
l
SCALES OF NOTATION.
Now
every term on the right hand side
iV-S =-
* .
that
.
r-
is,
r
when; /
is
some integer
j
1
-
03
is
by r —
divisible
1
•
— an integer y 6 ;
-=/ + r
1
-
which proves
Hence a number in scale r will be sum of its digits is divisible by r — 1.
'
1
tlie
proposition.
divisible
by
?•
—
1
when
the
By
taking ?-=10 we learn from the above proposition that a number divided by 9 will leave the same remainder as the sum of its digits divided by 9. The rule known as " casting out the nines " for testing the accuracy of multiplication is founded 83.
on
this property.
The
rule
may
be thus explained
:
Let two numbers be represented by da + their product by P; then
b
and 9c
-f
d,
and
P^Slac + %c + 9ad + bd. Hence — has the same remainder
as -^
;
and therefore the
of the digits of /*, when divided by 9, gives the same remainder as the sum of the digits of bd, when divided by 9. If on trial this should not be the case, the multiplication must have been incorrectly performed. In practice b and d are readily found from the sums of the digits of the two numbers to be
s?nn
multiplied together. Example.
Can the product
of 31256
and 8127 be 263395312
?
The sums of the digits of the multiplicand, multiplier, and product are 17, 21, and 31 respectively; again, the sums of the digits of these three numbers = 8x3 = 24, which has 6 for the sum of the are 8, 3, and 7, whence digits; thus we have two different remainders, 6 and 7, and the multiplication
M
is incorrect.
N
D
denote the scale of'r, and the difference, supposed positive, between the sums of the digit* in the odd and the even places; then — or + is a multiple, of 84.
If
denote
any number in
N D
r+
1.
N D
HIGHER ALGEBRA.
64 Let a
,
«!,
a
a n denote the
,
digits beginning
with that
in the units' place; then
JV= a +
a,r
+ a r 2 + ar* +
Ar -a + a -a + a3 -
-1
+ a
,r"
-
+a
r".
+ a 3 (r3 + 1) + ...; n and the last term on the right will be aw (r"+l) or a n (r — 1) Thus every term on the right is according as n is odd or even. hence divisible by r + I .-.
A
...=<*,
2
(r+1) + « 2
(r
2
1)
;
^
!
r
NOW
+
i
%
a0~ a +fl 2~
CC
}
=— is
.'.
= an
'-
1
3
mteo-er. °
—^D)
+
an integer;
which proves the proposition. Cor.
sum
the
by r +
sum
of the digits in the even places is equal to = 0, and is divisible of the digits in the odd places, If the
N
D
1.
Example whose radix
1. is
Prove that 4 •41 is a square greater than 4.
Let r be the radix
=4+- + r
thus the given number 2.
in
any
scale of notation
then
;
4-41
Example
number
is
i=(2 + -Y;
rz
rj
\
the square of 2*1.
In what scale
is
the denary
number 2-4375 represented by
213? Let r be the scale
;
then
13 + 2=2-4375=2-^; 7
2+ -
r
whence that
"
16
r-
7r 2 -
16r-48 =
;
(7r+12)(/--4) = 0.
is,
Hence the radix
Sometimes Example by 101215 ?
3.
is 4.
it is
best to use the following method.
In what scale will the nonary number 25607 be expressed
The required scale must be less than 9, since the new number appears the greater also it must be greater than 5 therefore the required scale must be 6, 7, or 8; and by trial we find that it is 7. ;
;
SCALES OF NOTATION.
65
Example 4. By working in the duodenary scale, find the height of a rectangular solid whose volume is 364 cub. ft. 1048 cub. in., and the area of whose base is 46 sq. ft. 8 sq. in.
The volume 264-734 cub.
The
We
area
is
364-^i? cub.
ft.,
which expressed in the
scale of twelve is
ft.
is
46^ 4
sq.
have therefore to
which expressed in the scale of twelve is divide 264*734 by St-OS in the scale of twelve. ft.,
3<-08.
3*08)26473-4(7-e
22*48 36274 36274
Thus the height
is 7ft. lliu.
EXAMPLES.
VII.
b.
1.
Express 4954 in the scale of seven.
2.
Express 624 in the scale of
3.
Express 206 in the binary
4.
Express 1458 in the scale of three.
5.
Express 5381 in powers of nine.
6.
Transform 212231 from soale four to scale
7.
Express the duodenary number 398e in powers of
8.
Transform 6£12 from
9.
Transform 213014 from the senary to the nonary
five.
scale.
five.
10.
scale twelve to scale eleven. scale.
10.
Transform 23861 from
11.
Transform 400803 from the nonary to the quinary
12.
Express the septenary number 20665152 in powers of
13.
Transform
14.
Express
15. 16.
Transform 17 "15625 from scale ten to scale twelve. Transform 200 "2 11 from the ternary to the nonary
17.
Transform 71*03 from the duodenary to the octenary
18.
Express the septenary fraction
ttteee
scale nine to scale eight.
from scale twelve to the common
scale. 12.
scale.
—3 as a radix fraction in the septenary scale. scale. scale.
1552 — — as a denary vulgar fraction
in its lowest terms. 19.
Find the value of
20.
In what scale
is
21.
In what scale
is
H. H. A.
and of '42 in the scale of seven. the denary number 182 denoted by 222? *4
the denary fraction
25 -—
denoted by -0302? 5
HIGHER ALGEBRA.
66
Find the radix of the scale in which 554 represents the square
22.
of 24.
In what scale is 511197 denoted by 1746335 ? Find the radix of the scale in which the numbers denoted by 24. 479, 698, 907 are in arithmetical progression. In what scale are the radix-fractions *16, "20, '28 in geometric 25. 23.
progression ?
The number 212542
26.
is
in the scale of six; in
what
scale will
it
be denoted by 17486?
Shew that
27.
radix
is
is
a perfect square in every scale in which the
greater than eight.
a perfect square in any scale whose radix and that the square root is always expressed by the
Shew that 1234321
28. is
148'84
greater than 4
same four
;
is
digits.
Prove that 1-331 is a perfect cube in any scale whose radix is greater than three. Find which of the weights 1, 2, 4, 8, 16,... lbs. must be used to 30. weigh one ton. Find which of the weights 1, 3, 9, 27, 81,... lbs. must be used 31. to weigh ten thousand lbs., not more than one of each kind being used but in either scale that is necessary. 29.
32.
radix
is
Shew that 1367631 greater than seven.
is
a perfect cube in every scale in which the
Prove that in the ordinary scale a number will be divisible by 8 if the number formed by its last three digits is divisible by eight. Prove that the square of rrrr in the scale of s is rm^OOOl, where 34. 33.
any three consecutive integers. and a new number N' be taken in the scale If any number 35. digits way, shew that the altering the order of its in any by formed be and N' is divisible by r — 1. difference between
q, r, s
are
N
?*,
N
If a number has an even number of digits, shew that it is 36. divisible by r+1 if the digits equidistant from each end are the same. 37.
If in the ordinary scale
St
be the
sum
of the digits of a
number
and 3#2 be the sum of the digits of the number 3iV, prove that the is a multiple of 3. difference between aS^ and 2 Shew that in the ordinary scale any number formed by 38. writing down three digits and then repeating them in the same order is a multiple of 7, 11, and 13. JV,
>S'
In a scale whose radix is odd, 39. digits of any number will be odd if the
shew that the sum of the number be odd, and even if
number be even. If n be odd, and a number in the denary scale be formed 40. by writing down n digits and then repeating them in the same order, shew that it will be divisible by the number formed by the n digits, and also by 9090... 9091 containing n- \ digits.
the
+
CHAPTER
VIII.
SURDS AND IMAGINARY QUANT1T1KS. 85.
In the Elementary Algebra, Art. 272,
it is
the denominator of any expression of the form
proved that
-rr
r-
Jb + Jc rationalised
can be
by multiplying the numerator and the denominator
by Jb — Jc, the surd conjugate
to the denominator.
Similarly, in the case of a fraction of the form
Jb + Jc + Jd surds, we may by '
where the denominator involves three quadratic two operations render that denominator rational.
multiply both numerator and denominator by Jb + Jc — Jd; the denominator becomes (Jb + Jc) 2 — (Jd) 2 or Then multiply both numerator and denominator b + c- d + 2 be. For,
first
J
by (b + c - d) — 2 J be; the denominator becomes which is a rational quantity. Example.
(b
+
c
- d) 2 — Abe,
12
Simplify
3+^/5-2^/2
The me expression
12 _
(
3
+ y/5 + V*)
(3+ ^ 5)S _
(V2 )»
^ 12(3 + ^/5 + 2^/2 6 2 (3
)
+ 6^5
v/5+ 2^/2)^5-1)
U/5+l)U/5-l)
2+V5+V10-V2 = 1+^5+^/10-^2.
5—2
HIGHER ALGEBRA.
68
To find
86.
which will rationalise any given bino-
the factor
mial surd.
Case
Suppose the given surd
I.
xn and y
n
x,
- $b.
£]a
n be the
?Jb = y, and let are both rational.
Let ZJa =
is
l.c.m. of
p and
q
n xn — y is divisible by x - y for all values of n, and ~ n ~' + y"" ). +x n 2y + xn Sf + ar- _ y" = (x-y) {x
Now
1
Thus the
rationalising factor
is
~2
n y + x ~y + n and the rational product is x — y'\
+ xn
1
.X'""
Case Let (1)
Suppose the given surd
II.
+ y"~
n have the same meanings
x, y,
p
Ja +
is
l ;
fjb.
as before; then
n is even, xn — y" is divisible by x + y, and ~ n~ n n x - y = (x + y) (xn - x 2y + + xf~* - y"" If
l
Thus the
rationalising factor l
is
x"
n
y"
is
(2)
If
is
odd, x"
n n x +y =
Thus the
(x
+
+
y) (x
n~1
i
—y
divisible
;
by x + y, and
- xy n ~ 2 + y n ~').
- xn Sj +
x"
1
.
is
-x y+ is
r
n
n~2
and the rational product 1.
x
+ ay"-'- 3
rationalising factor
x
Example
n~
-xy n +y
Find the factor which
-2
+ y"-
1
;
n .
will rationalise ^/S
+ ^/5.
i
Let x = 3 2 y = 5 5 ,
then x b and y 6 are both rational, and
;
x e - y 6 — (x + y) (x 5 - x*y + xhj 2 - xhj 3 + xy i - y 5 ) thus, substituting for x
and
y,
the required factor
541 3223 53+
32
32
32-9.
55
+ 6
and the rational product
.
53
13
5
or
32
.
is
32
32
2
32~.
.
is
14
53+ 32
.
14
53
5
-
6
53,
5
53-15 + 32. 53-5 5
- 5 s = 3 3 - 5 2 = 2. "
).
is
ur -ary + and the rational product
1
;
;
;
then
V
i
SURDS AND IMAGINARY QUANTITIES. Example
(&+&) *
Express
2.
69
\5 5 -9s )
as an equivalent fraction with a rational denominator. i
To 34
=y
rationalise the denominator, ;
the required factor
-y A = [x -
x4
then since
is
3
2
5-
+ 52
denominator
rational
/ .•.
=
•
+Vl
the expression
i\
i
5 V5
.
3*
3
+ 3~4
2 .
3 .
3~4 .
'
'-
52 + 2
;
12
1
2 3^ + 5+ 3 V—\& + 5 —
4 _
+ 5]
- 3-* = 5 2 - 3 = 22.
3
/
i
4
52
is
12
1 3-*
.
i
equal to 5" -3*, put 5 2 = x,
is
+ xhj + xy 2 + if')
y) (x*
4
and the
which
2
1
5 2 .3 j
+ 2.
3\
—
+3
'
13
2
5 .3 4"+2.5 5 .3 j ,J
4
+3
t
22
3
113
1
_l l + o 2
.
3j +
5.3 2 +5 2 .3 j
11
We
87.
have shewn in the Elementary Algebra, Art. 277,
how to find the square root of a binomial quadratic surd. We may sometimes extract the square root of an expression containing more than two quadratic surds, such as a + Jb + Jc 4- Jd. Assume .'.
a+
Ja + Jb + Jc + Jd = Jx + Jy + Jz
Jb + Jc + Jd = x + y + z +
If then
2
Jxy =
Jxz =
Jb, 2
2
Jxy +
Jc, 2
j
Jxz +
2
2
Jyz.
Jyz = Jd,
and if, at the same time, the values of x, y, z thus found x + y + z = a, we shall have obtained the required root. Example.
Assume .'.
21 -
Find the square root
- 4^/5 + 8^/3 - 4^/15.
V 21 - V 5 + V 3 - V 15 = slx + Jy - slz
\
4^5 + 8^/3 - 4J15 = x + y + z + 2 Jxy- 2 Jxz - 2 Jyz.
2jxy = 8JS, 2jxl = 4J15, 2jyz = ±Jo\
Put by multiplication,
whence
of 21
satisfy
it
follows that
xyz = 240
;
,Jx = 2j3,
And since these values root is 2^3 + 2-^/5.
Jxyz=4 s/lo Jy = 2, „Jz = s/5.
that
is
satisfy the equation
;
x+y+z
= 21,
the required
;;
,
HIGHER ALGEBRA.
70
Jh = x + Jy,
88.
If J a, +
For,
by cubing, we obtain ci
+
3
Jb=x
then
ivill
J
a,
— Jh = x — Jy,
+ 3x 2 ^/y + 3xy + y Jy.
Equating rational and irrational parts, we have
= 3x 2 Jy + y Jy a- Jb = x3 - 3x2 Jy + 3xy -y Jy;
a = x3 + 3xy, Jb .'.
that
J a - Jb = x— Jy.
is,
Similarly, it
may
\
by the help
be proved that
of the
if
Ja + Jb = x + Jy, where n
is
any
Binomial Theorem, Chap. XIII.
Ja - Jb = x- Jy,
then
positive integer.
By
the following method the cube root of an expression of the form a ± Jb may sometimes be found. 89.
Ja + Jb = x + Jy
Suppose
Ja- Jb = x- Jy.
then
Jtf^b=x -y 2
.-.
Again, as in the last
(1).
article,
3 a = x + 3xy
The values In (2)
of
x and y have to be determined from
suppose that obtain
(1)
we
(2).
Ja
2
— b=c; then by
a = x3 + 3x (x 2 — that
If trial,
kx 3 — 3cx —
is,
from
(1)
and
(2).
substituting for y in
c)
a.
x can be determined by obtained from y = x 2 — c.
this equation the value of
the value of y
is
We
do not here assume sjx + sly for the cube root, as in the extraction of the square root; for with this assumption, on cubing we should have
Note.
a + Jb = xjx
+ Sxjy + Syjx + yjy
and since every term on the right rational and irrational parts.
hand
y
side is irrational
we cannot equate
1
,
;
SUKDS AND IMAGINARY QUANTITIES. Find the cube root of 72 - 32 x/5.
Example.
Assume
;
^72 + 32 s/5 = x + s/y.
By
multiplication
(1).
72 - 32^/5 = .c 3 - fkcPJy
(1)
and
(2)
72 = x
3
72 = x
:i
is,
By
;
± = x'--y
whence
From
x 5 = a; 2 - y
^5184 - 1024
,
is,
Again
that
-'62^5 = x - ^/y
sf 72
then
that
71
ar
trial,
we
+ Sxy - y^'y
;
+ 3.t//
(2).
+ Sx (x- - 4)
-3x = 18.
}
find that x = S; hence y = o,
and the cube root
is
When
3-^/5.
the binomial whose cube root we are seekhi" consists of two quadratic surds, we proceed as follows. 90.
Find the cube root of 9 N/3 + ll v/2.
Example.
By
.-.
proceeding as in the last
the required cube root
article,
=*J3
(
1
+
we
find that
/-
*
J
= v/3+v/2. 91.
We
'Example
1.
add a few harder examples in
surds.
Express with rational denominator
The expression
4
=—
^
3 - 33 fj
+
(J + l) l3 3 + l) (3 3
-3 +
iM±i]-3 ~d 3
+1
ri
3
+l +1
"
l)
4
NV9-^3
+
r
:
HIGHER ALGEBRA.
72 Example
Find the square root of
2.
l(x-l) + j2x -7x-i.
f
The expression = \
{3x - 3
+ 2 J(2x + l)(x-4)
}
a
= ±{(2x + l) + {x-±) + 2jc2x + l){x-£)}; hence, by inspection, the square root
Example
Given
3.
^5 = 2-23607,
is
find the value of (5
J2 + J7- 3J5' Multiplying numerator and denominator by >J2,
^6-2^/5
the expression 2
+ ^14-6^/5
2
+ 3-^/5
n/5-1
EXAMPLES. Villa. Express as equivalent fractions with rational denominator 1 i L -
1
2 A
+ V2-V3'
=
-
1
3.
4
.
sfa + s/b + s/a + b'
J2+J3-J5'
^^
*Ja-l-\/2a + *Ja + l (j3 + x/5)(j5 + ^/2)
^10 + ^5-^/3 Find a factor which
^
fi
will rationalise i
7.
10.
#3 -a/2. 3
N/3-l.
8.
11.
^/5
2
+ ^/2.
+ 4/7.
9.
12.
i
06+6*.
4/5-^3.
=
,
SUKDS AND IMAGINARY QUANTITIES.
73
Express with rational denominator:
*/3
16
v 8 + ^4
17
Find the square root
W
i«
of
19.
16-2 N /20-2
21.
G
23.
a+36+4+4^/a-4^6-2V3oS
24.
21+3 N /8 - 6 N /3 - Jl - v'24
v/28 + 2 N /l3.->.
24+4^15-4^21-2^35. 5- x /10- N /15 + N /G.
20.
+ ,/12-^24-,/8.
22.
(5
- N /56 + 2 N /21.
Find the eube root of 25.
28.
10+6 JZ. 38^14-100^2.
26.
38 + 17^5.
27.
99-70^/2.
29.
54^3 + 41^5.
30.
135^3-87^6.
Find the square root of 31.
a + x + \J%ax + x2
33.
l
35.
If
a
36.
If
x
.
32.
2a - \/3a 2 - 2ab - b'2
34.
l+(l-« 2 )" 2
i
+ « 2 + (l+a 2 + a 4 ) 2
= ——j-
,
6
i .
= —-i—
find the value of
,
S
jl'jl
,
.
y = /|z7^
.
7«2 +
1 1
find the value of 3
»
ab -
2
lb'
.
~ 5j y + 3^-
Find the value of
V26-15J3 5V2-V38T573'
/
V33-19 N
1
39.
41.
(28
6 + 2V3 /3-
"
- 10 N/3) - (7 + 4 v/.3) 2
2
1
2 .
40.
(26 + 15 s fzf
_>
- (26 +
Given s/b = 2-23607, find the value of lOx/2 N /10 + N/18 '
N/18
- a/3+V5
N /8
x*+ 1 + 3# #2 by * - 1 +
42.
Divide
43.
Find the cube root of 9a6 2 +
44.
Evaluate
V^'
2
-1 ,
x-s/x*-\
when
(b 2
2.r;
+ V3 - V5 */2.
+ 24a 2 ) ^6* -3a8
.
= Ja+ -i\'"
15 N/3)
3
HIGHER ALGEBRA.
74
Imaginary Quantities. Although from the rule of signs it is evident that a negative quantity cannot have a real square rootlet imaginary quantities represented by symbols of the form J- a, J- 1 are of frequent occurrence in mathematical investigations, and their 92.
We
therefore proceed to explain use leads to valuable results. in what sense such roots are to be regarded.
When the quantity under the radical sign is negative, we can no as indicating a possible arithmetical longer consider the symbol a may be defined as a symbol which obeys operation ; but just as a x Ja = a, so we shall define J— a to be such that the relation
J
J
J
J— a x J— a = - a, and we shall accept the meaning to which this assumption leads us. It will be found that this definition will enable us to bring imaginary quantities under the dominion of ordinary algebraical rules, and that through their use results may be obtained which can be relied on with as much certainty as others which depend solely
on the use
93.
By
J- I
definition, .-.
that
of real quantities.
Ja.J-l
x
J-I = -
x
1.
J-l^a^l);
Ja.
2
is,
(
Thus the product
J a J- 1) = - a. .
J a J— 1 may
the imaginary quantity
.
be regarded as equivalent to
J— a.
It will generally be found convenient to indicate the imaginary character of an expression by the presence of the 94.
symbol
J- 1
;
thus
JZjtf = Jja
We
2
x
(
-
1)
= a J7 J-T.
always consider that, in the absence of any statement to the contrary, of the signs which may be prefixed before a radical the positive sign is to be taken. But in the use of imaginary quantities there is one point of importance which 95.
shall
deserves notice.
SURDS AND IMAUiNAltY QUANTITIES. (- a) x (-
Since
by taking the square
—
b)
75
ab,
we have
root,
J- a x J- b = ± Jab. Thus
J
in funning the product of
a and
J— b
would appear
it
— might be placed before Jab.
that either of the signs + or This is not the case, for
J -a x J- b = J a J-
I
.
x \/b
.
J-l
= - Jab. 96.
It
is
usual to apply the term
imaginary
c
to all expres
'
a+bj—l may
sions which are not wholly real. Thus as the general type of all imaginary expressions. are real quantities, but not necessarily rational.
be taken Here a and b
In dealing with imaginary quantities we apply the laws of combination which have been proved in the case of other surd 97.
quantities.
+b
J - 1±
Example
1.
a
Example
2.
The product
+d
(c
J-
1)
= a ± c + (b ± d)
J - f and
of a 4 b
c
J-
1.
+ dj-l
= (a + bj^l)(e + dj^l) = ac - bd 4- (be + ad) 98.
For,
If a + b
J — 1 = 0,
2/tew
0,
-
1.
em(£ b
=
0.
J^\ = 0, bJ—\=-a;
a+
if
then
b
—6" = a";
.'.
.'.
Now
a=
*/
2
a +
b
2
= 0.
are both positive, therefore their sum cannot be zero unless each of them is separately zero that is, a — 0,
« 2 and
b
2
;
and
6
= 0.
99.
7f& + bJ^T = c + d J~l
For, by transposition, a tlierefore,
that
is
by the
—c+
last article,
a=
c,
1,
then a
= c, andh
— d) J- 1 = a — c = 0, and 6 — ^ = 0;
and
(b
6
;
HIGHER ALGEBRA.
76
Thus in order that two imaginary expressions may necessary and sufficient that the real parts should be
is
the
imaginary parts should
be equal it
equal,
and
be equal.
When two imaginary expressions differ Definition. 100. only in the sign of the imaginary part they are said to be conjugate. Thus a — b J — Similarly
1 is
J — 1.
conjugate to a + b
^2 + 3^-1
conjugate to
is
J - 3 J- 1. '2
The sum and the product of two conjugate imayinary 101. expressions are both real.
J -\ +a-b J-\ = 2a. (a + b J- 1) (a - b J- 1) = a - (- b a +
For (
Again
b
2
= a + b2
2
)
2
102. 2
a +
b
2
is
Definition. The positive value of the square root of called the modulus of each of the conjugate expressions
a +b 103.
J— 1
and a —
b
J—
the
Let the two expressions be denoted by their
1.
product of two imaginary expresthe product of their moduli.
The modulus of
sions is equal to
Then
.
product = ac
— bd +
(ad
a+bj— 1
+ bc)
and c+dJ—\.
J — 1,
which
is
an
imaginary expression whose modulus
= J(ac -
bd)
2
+ (ad +
be)
2
= Jasc* + b*
J (a
=
Ja
2
2
+
+
b
b
2
2
)
x
(c
2
+
dr)
2 2 Jc + d
;
which proves the proposition. form a + bj— 1, may be rationalised by multiplying the numerator and the 104.
it
If the
denominator of a fraction
is
of the
denominator by the conjugate expression a —
b
J—
1.
)
;
SURDS AND IMAGINARY QUANTITIES.
77
For instance
dj - 1 a + b J=l c
+
+
(c
dJ-\)(a-bJ -\
~(a + b J~i)(a-bJ-T) — be) J —
ac + bd + (ad
a +
+ bd a + b~
1
72
2
b
ad — be v a' + b
ac
-i
•
_•
Thus by reference to Art. 97, we see that the sum, difference, product, and quotient of two imaginary expressions is in each case an imaginary expression of the same form. 105.
To find
of a + h
the square root
J—
1.
J a + b V— 1 =x + y s/ — 1,
Assume
where x and y are real quantities.
By
therefore,
bj—\=x -y 2
a+
squaring,
by equating
real
x2
(x
.-.
2
and imaginary
-y =
a
^!/ =
b
2
+y
2 2 )
=
(x
From
(1)
and .
x
(3),
x s + if =
we
-*Ja
2
parts,
(2);
- ff + (2xyY
2
2
.
1
(1),
=a + .•
+ 2xy J —
2
,
Ja
2
2.2
b
+
;
6"
(3).
obtain
+2b 2 + a
—
Thus the required root
is
Ja +i b -a 2
.
.»"=*—
2
;
obtained.
Since x and y are real quantities, x 2 + y- is positive, and therefore in the positive sign must be prefixed before the quantity *Ja 2 + b-.
Also from (2) we see that the product xy must have the same sign as hence x and y must have like signs if b is positive, and unlike signs if b negative.
(3)
b
;
is
;
;;
HIGHER ALGEBRA.
78 Example
1.
Find the square root
-
of
7
- 24
J-
1.
J- 7- 21*/- l = x + y J- 1
Assume
-7-24 N/T l =
then
•••
* 2 -?/ 2
and
2
-^
2a;?/
-2/
2
+ 2^
N/
T l;
=-7 =
(1),
-24.
= 49 + 576 = 625; .\ar + 2 = 25
(2).
2/
From
(1)
and
(2),
x- = 9
and .-.
Since the product xy
is
x-
Thus the that
negative,
±4.
we must take
y = - 4 j or # = -
3 - 4 „/ - 1
2.
To
and -3 + 4
find the value of ^/
remains to find the value of
Assume
?/==
3,
*J
?/
= 4.
-
1
7 - 7 - 247"TT= ± (3 - 4 J ~i).
is,
Example
It
roots are
3,
= 16 x= ±3, 2
?y
- 64a 4
\/ ±*J -
1.
.
;
SURDS AND IMAGINARY QUANTITIES.
70
— 1 is often represented by the letter i; but 106. The symbol until the student has had a little practice in the use of imaginary
J
quantities he will find
useful to notice the successive powers of
(7-1)^1,
J—
symbol
easier to retain the
it
J—
1
or
i
;
1.
It is
thus
i*=ij
and since each power is obtained by multiplying the one before by J — 1, or we see that the results must now recur.
it
?',
We
now
investigate the properties of certain imaginary quantities which are of very frequent occurrence. 107.
shall
x - ^1
Suppose that .
(x
is,
\
x—
either
whence
1
then x 3 =
;
=
-
1
)
(x
or
0,
2
or
1,
x+
+
1
=
1
;
;
-W-3
x=
=
1)^0.
x2 + x +
35=1, or
x3 —
.
may
be shewn by actual involution that each of these Thus unity has three cube values when cubed is equal to unity. It
roots,
-1-733~'
-l+JZTs "'
2
two
of
2
which are imaginary expressions.
Let us denote these by a and
then since they are the roots
;
ft
of the equation
x2 + x +
=0.
l
their product is equal to unity
that
aft=
is, .
that
is,
other, it
;
2 aft = a = a2 ft
;
8
since a
,
.
Similarly 108.
\
1
we may shew
that a =
= 1.
2
ft
.
Since each of the imaginary roots is thr, square of thr is usual to denote the three cube roots of unity by 1,
~.
r
,
;
1
HIGHER ALGEBRA.
80 Also
2
the equation x + x +
satisfies
a)
.
that
is,
the
sum of the
+
1
•.
to
+ w2 =
=
1
;
;
three cube roots of unity is zero.
—w =
2
Again,
to
o>
.
1
therefore (1) the product of the two imaginary roots is unity 3 (2) every integral power of w is unity.
it
must be If
of the
n be not a
3m
form
2
and w for, and to" = w3m —
to, ;
;
multiple of
KO
that the
notice
It is useful to 109. integral powers of a> are 1,
3, it
if
;
successive positive n he a multiple of 3,
1.
must be
of the
form
3m +
1
or
3m + 2.
n n = dm + 1 O w = om +-,
-rt>
It
3m +
Wn —
1
o>
=to
3m
.to
=
to.
w=w 3m +2 =w3m .w=to. "
.
2
2
We now
see that every quantity has three cube roots, two of which are imaginary. For the cube roots of a 3 are those 2 7 Similarly the cube roots of a x 1, and therefore are a, ao>, aw 2 7 of 9 are ^9, o> ^9, a> ^9, where ^ 9 is the cube root found by the In future, unless otherwise stated, ordinary arithmetical rule. 110. "
.
the symbol %ja will always be taken to denote the arithmetical cube root of a. (9
Example
1.
Reduce
4.
*
3
/
_
1 \2
,
to the
.
form A +
2 + N/-l
The expression
^
Bj - 1.
4-9 + 12^-1 2
+ v/ :=~l
(-5 + 12 N/Jl)(2-V^l) (2+
J-l)(2-J-T)
-10 + 12 + 29 4+1
=2+ 5 which
is
/—
W-
1;
of the required form.
Example
2.
Resolve x 3 + y 3 into three factors of the first degree.
x 3 + if =
Since .'.
for
29
J~l
(x
+ y)
-xy + y 2) (x + toy) (x + ury) (x 2
+ y 9 = (x + y) w + w2 = -1, and w 3 = l.
x*
;
;
: ;
;
SURDS AND IMAGINARY QUANTITIES. Example
3.
Shew
(a
+ wb + arc)
that
Zr
and
+ w2 6 + toe) - a 2 + b2 + c 2 -
(a
In the product of the coefficients of
81
c
a
+ wb + arc and
2
are or, or 1
a
60 - ca
- ab.
+ orb + wc,
= w2 + o>4 = or + o> = - 1 coefficients of ca and a& = 2 + o> = - 1 (a + cob + arc) {a + urb + wc) = a 2 + b 2 + c 2 - be - ea -
the coefficient of be the
o>
.*.
Example
Shew
4.
that
+ «-«>)'-(l-M+U?)*=0.
(l
Since
1 (1
ab.
+ u- w 2
3 )
+ w + o> 2 = 0, we have
-
(1
- w + wa)3=( - 2m2) 3 -
- 2o>) :J
(
= -8o> 6 + 8a/ = -8 + 8
{
= 0.
EXAMPLES. 1.
Multiply 2 \/~~3 + 3
2.
Multiply 3
3.
Multiply
A
AT
4.
V 3^
V ^7 - 5 V^
VIII.
b.
by 4 *J^3- 5 a/^2. by 3
V^+ 5 V^.
e^ -1 +e'^~ 1 by e^ _1 -e -V-*.
l+V 3^
IX' Multiply I
a;
5
l-V^ =
.
by #
.
Express with rational denominator _ 0.
— 3-V-2 ,
3+2
r
l
+ V-l)a .v-V-1
(.f
o.
•
V~l
2-5\/ g
3 a/~2~ + 2 *J~-h
1
2
+
5V :r l'
(W- l)^ .r
12.
Find the square of
\/
-
l) 4n
+
a+rV^l
a-x»f-i
a-WisT-i
a+a?V^-l"
(a +
V-lja-Cft-V-l)^ (a + \/-l) 2 -(«- V-l) a#
ia
+ V-1
Find the value of ( -
.
,
3V-2-2V-5
3-2 V^l
11.
H. H. A.
,
3 ,
when w
is
a positive integer.
Jd + 40 V"-T+ V9 - 40 V -?. (j
a
HIGHER ALGEBRA.
82
Find the square root of 13.
-S + ISV^L
14.
-ll-COV 17
16.
-SV^l.
17.
a 2 -l+2a^^l.
18.
±ab-2(a 2 -b 2 )*/^T.
Express in the form iy *
2-3r
09
a + O8
If
2 1, co, g>
15.
-47 + 8V-3.
A + iB ZU>
zu
2V3-i\/2' (^ + ^)
9<*
2
1-T
(«-^) 2
a — ib
—i
3
!".
a + io
are the three cube roots of unity, prove
24.
(l+co 2 ) 4 = co.
26.
(1
- co)
27.
(2
+ 5co + 2co 2 6 = (2 + 2co + 5a> 2)6 = 729.
28.
(l-co
29.
Prove that
(1
-
25.
co-
)
(1
- CO 4 )
(1
(l-co + co 2 )(l+co-or)
= 4.
- co 5 ) - 9.
)
+ co 2 )(l-co 2 +
co
4
)(l-co 4 + co 8 )... to
2>i
factors = 2 2 ».
A3 +yZ + £> _ 2#gZ = (x+y+z) (# + i/a> + Za> 2 ) (x +y
x=a+b
If
t
y — aw + Z>co 2
,
s=«co 2 +
6co,
shew that (1)
(2) (3)
31.
If
shew that
(
xyz=a3 +b3
.
+ 2 + 5 2 = 6a6. a3 +y3 +s3 =3(a3 +&3) ax + cy + bz = X, ex + by + az = I", Zu- + ay + gs = if, 2 + 6 2 + c2 - be - ca - ab) (x2 +y2 + z2 -yz- zx - xy) = X 2 +Y2 + Z - YZ- XZ- XY. ^- 2
?/
k
2
2
CHAPTER
IX.
THE THEORY OF QUADRATIC EQUATIONS. After
111.
suitable reduction every quadratic equation m;iy
be written in the form (Lif Jrbx
and the solution
C
—
(1),
of the equation is
x=
We
+
-
± Jtf^iac
b
\2a
(2). v '
now prove some important
shall
with the roots and coefficients of
propositions connected
equations of which (1)
all
is
the type.
A
112.
quadratic equation cannot have more than
ax 2 + bx +
tiro roots.
=
have three Then since each of these values must different roots a, f3, y. satisfy the equation, we have For,
possible, let the equation
if
c~0
(1),
=
(2),
ay 2 + by + c =
(3).
aa 2 2
afi
From
(1)
and
(2),
-t-
ba +
+ bp +
c
by subtraction,
a(a 2 divide out
c
-p
2 )
+ b(a-P) = 0;
by a — fi which, by hypothesis,
is
not zero; then
a (a + ft) + b = 0. Similarly from (2) and (3)
a .•.
{fi
is
y)
+
b
=
a (a - y) —
by subtraction
which
+
;
;
by hypothesis, a is not zero, and u Hence there cannot be three different roots.
impossible, since,
not equal to
y.
G—
is
.
HIGHER ALGEBRA.
84 In Art. Ill
113. /3,
the two roots in (2) be denoted by a and
let
so that
a
=
P
, '
2a
a and
(2)
2
reducing in this case to (3) (4)
2
If b
If b
2
- Aac
—
is
Aac
is
positive,
equal, each
— 77-
2a
negative, a
and
ft
are imaginary and unequal.
a perfect square, a and
is
_
'
:
- Aac (the quantity under the radical) are real and unequal. 2 If b — Aac is zero, a and ft are real and If b
(3
2a
'
then we have the following results (1)
—b- Jb — Aac = 2
2 Jb - Aac
-b +
/3
are rational and
unequal.
By
applying these tests the nature of the roots of any quadratic may be determined without solving the equation. Example
Shew that the equation 2x 2 -6a; + 7 =
1.
by any real values of
cannot be
satisfied
x.
a=
Here
2, b
= - 6, c — 7
so that
;
&2_ 4ac= (_6)2-4.2.7=-20. Therefore the roots are imaginary.
Example
2.
2
If the equation
The condition
a?
+2
(k
+ 2) x + 91c = has equal roots, find
l\
for equal roots gives (fc
+ 2) 2 = 9£,
fc2_5ft + 4=0,
(fc-4)(fc-l)=0j .-.
Example
Shew
3.
k = A, or
1.
that the roots of the equation
x 2 - 2p3 +p 2
-q 2 + 2qr- r2 =
are rational. 2 2 2 roots will be rational provided (-- 2p) - 4 (p - q + 2qr-r2) is a 2 But this expression reduces to 4 (q -2qr + r2), or 4:(q-r) 2 perfect square. Hence the roots are rational.
The
.
- - ,
114.
D Since
we have by
.
a=
-
b
+ Jb 2 - Aac
^
?=
'
-b- Jb
2
- Aac '
2a
addition -
*
+
b
+ Jb 2 - Aac
^ __M_b 2a
a
-b- Jb - Aac 2
2a
0);
THE THEORY OF QUADRATIC EQUATIONS.
85
and by multiplication we have n = (- b
4ac
_
~4a
=
2
J¥~r^
+
* qUadratiC
4**5d?
:
in the
°
c
a
a
e(
l
uation
form '
«**" **
the roots
is
the product of the roots
(ii)
coefficient
is
__ =a+ £
Since
the equation ar+
- a; + - =
of
the first
equal t0 the coefficient of
may
term
with
and £
not contain the 8nta0
be written
(1).
Hence any quadratic may also be expressed in the form 2 x - (sum of roots) x + product of roots =
(2).
(1)
we have
(x-a)(x-p) = Q
We
may now
Example
1.
The equation or
easily
Form is
™
*
x 2 -(a + f])x + ap =
Again, from
(3).
form an equation with given
the equation whose roots are 3 (* - 3)
and -
roots.
2.
(*+ 2)=0,
««-*- 6=0.
When metnou.
the roots are irrational
is
equal to the third term.
*£%* £»"& ££&£**" 115.
(2).
also be expressed as follows.
unity,
its
_ j,/r^-c)
«
2
may
5
c
By writing the equation
these results
(_.
)
it is
easier to use the following ~>
.
;
HIGHER ALGEBRA.
86 Example
Form
2.
the equation whose roots are 2
We have
sum
of roots
+ ^3 and
2 -
^3.
= 4,
product of roots = 1 .
•
.
the equation
x-
is
by using formula
(2) of
- Ax + 1 = 0,
the present article.
By
a method analogous to that used in Example 1 of the last article we can form an equation with three or more given 116.
roots.
Example
1.
Form
the equation whose roots are
-
7
3,
and ^ o
The required equation must be positions
2,
satisfied
by each of the following sup-
:
#-2=0, therefore the equation
# + 3 = 0,
£--7 = ();
must be
(*-2)(*+3)(*-|)=0j that
[x
is,
or
- 2) (x +3) (5a;- 7) =0,
5a;
Example
2.
Form
The equation has
3
-2a; 2 -37a; + 42 = 0.
the equation whose roots are
.
c
x=-a, x=}
b
;
it is
x that
±«, j
by
to be satisfied
x = 0, x = a. therefore
0,
(x -J- a) (x
- a)
(
x--
j
=
;
- a 2) (bx - c) = 0, bx 4 - ex 3 - a~bx- + a-cx = 0. x
is,
or
The
117.
(x 2
114 are most important, and they to solve problems connected with the
results of Art.
generally sufficient In such questions the roots should never be roots of quadratics. considered singly, but use should be made of the relations obtained by writing down the sum of the roots, and their product, in terms of the coefficients of the equation. are
Example (l)a 2
1.
If
a and
+ /3 2 (2)a s + /3 3
We
,
/3
are the roots of
x--px + q = 0,
.
a + (2=p,
have
a
a(3 .-.
a2
= q.
+ /3 2 =(a + /3) 2 -2a/3 =p*-2q.
find the value of
m
.
;
:
THE THEORY OF QUADRATIC EQUATIONS. Again,
a
;{
+ ft = (a + /3)
(a 2
+ p" 2 -
87
a/3)
a =i>{(a + /3) -3a/3]
=*(?»- 89). Example
If a,
2.
are the roots of the equation
p"
equation whose roots are
—
sum
mx + 7i = 0,
find the
-
,
p "We have
/.r'-'-|
a of roots
=
+
product of roots =
-
=
-
p
^
.
,
ap
a.
-=1
p a .-.
by Art. 115 the required equation
apx2 -
or
(a
is
+ p~2 £ + 0/3 = 0.
2
)
O
O
As
in the last
7
=—
example o2 +j8B =
and
,
a/3
=
L
n
,.
.,
.*.
the equation
?ji
2
V
Example
When
3.
and shew that
it
will
.r
—
x+
v
n /x- 2 -
or
=
2
(
?i
y
_
= 0,
it
- 2nZ) x + nl = 0,
^
be unaltered
.
t
-2wZ -=
x-
is
--
,
find the value of
if
2x 3 + 2x 2 -7x+l'2
be substituted for
£-
;
x.
a
Form the
sum
hence the equation
2x 2 -6x + 17
is
is
;
=3
of the roots
17 =—
the product of the roots
.*.
^
the quadratic equation whose roots are
2.r 2
-
6.r
;
+ 17 =
;
a quadratic expression which vanishes for either of the
3*5^/-"!
^
, values
•
m
Now
2a* +
2.t
2
- Ix + 72 = x
(2.r 2
-
C>.c
+
17)
+4
(2.r
2
- Cx + 17) + 4
=xx0+4x0+4 = 4; which
is
the numerical value of the expression in each of the supposed cases.
HIGHER ALGEBRA.
88 To find
118. 2
ax + bx + in sign,
c
=
Vie condition
that
roots
the
of the equation
and
should be (1) equal in magnitude
opposite
(2) reciprocals.
The roots will be equal in magnitude and opposite hence the required condition is their sum is zero
in sign
if
;
_-= a
or b
0,
= 0.
Again, the roots will he reciprocals when their product unity ; hence we must have c
i — = 1. or
a
The
c
is
= a.
these results is of frequent occurrence in Analytical Geometry, and the second is a particular case of a more general condition applicable to equations of any degree. first of
Example. Find the condition that the roots of ax2 + bx + c = may be both positive, (2) opposite in sign, but the greater of them negative.
We (1)
have
have
a
If the roots are
— b
+ B=
(1)
c
a8=-
,
a
.
a
both positive,
a/J is positive,
and therefore
and a
c
like signs.
Also, since a
+ fi
is positive,
—
is
negative; therefore b and a have unlike
signs.
Hence the required condition and opposite to the sign of b. (2)
is
that the signs of
and
ft
If the roots are of opposite signs, a/3 is negative, signs.
c
should be
and therefore
c
like,
and
a have unlike
Also since a +/3 has the sign of the greater root fore - is positive; therefore b
and a have
it is
negative,
and
there-
like signs.
(X
Hence the required condition and opposite to the sign of c.
is
that the signs of a and b should be
EXAMPLES. Form
the equations whose roots are
.
4
3
m
n
o
/
n
m
4.
7±2
N /5.
5.
±2«/3~5.
IX.
a.
p-q p+q 6.
p+q
p—y
-p±2s/Tq.
like,
=
)
.
:
THE THEORY OF QUADRATIC EQUATIONS. 7.
-3±5l
10.
-3, |, i.
13.
Prove that (
1
(2)
14.
-a±ib.
8.
11.
tlio
|,
-b).
12.
2±«/3,
4.
roots of the following equations are real
x2 - 2ax +a a - 6a -
c2
- b + c) r- +4 (a -
(a
±i(a
9.
-|.
0,
89
If the equation
- 0, 6)
.v
+ (a - b - c) = 0.
x2 - 15 -m(2x-8) =
equal roots, find the
lias
values of m. 15.
For what values of
m will the
equation
x2 - 2x (1 + 3//0 + 7 have equal roots 16.
(3
+ 2m) =
?
For what value of
m
will the equation
x*
m- 1 m+ 1
— bx
ax-c
have roots equal in magnitude but opposite in sign 17.
Prove that the roots of the following equations are rational: (1) (2)
If
?
a, /3
+ c-b)x 2 + 2cx + (b + c-a) = 0, abc 2 x2 + 3a 2 cx + b 2 ex - 6a 2 -ab + 2b' = 0. (a
1
are the roots of the equation
ax2 + bx + c = 0,
find the values of 2
18.
»,+£.
19.
aW + aV.
20.
.
(|-f)
Find the value of
= 1 + 2/.
21.
a3 + s2 - X + 22 when
22.
x3 - Zx2 - 8x +15 when x
23.
.t-
3
.r
- «.r2 + 2a 2.r + 4« 3 when
3+ ?
-= 1 - J - 3. a
If o and /3 are the roots of 24. whose roots are (a- ft) 2 and (a + /3) 2
x*+px+q=O
t
form the equation
.
25.
Prove that the roots of (x — a) (.»; -b) = h 2 are always red.
26.
I f
.'-, ,
x%
are the roots of ctx*+bx
+ b)- 2 + («x + b)- 2
(1)
(ax l
(2)
(ax^byt+iaxt+b)-*.
i
,
+
c
= 0,
find tho value
i
>f
HIGHER ALGEBRA.
90 27.
Find the condition that one root of ax2 + bx-\-c =
shall
be
n times the other. 28.
If
a, (3
roots are a
2
are the roots of
+ /3 and o~ 2
2
+/3
-2
ax2 + bx + c = 0, form the equation whose
,
Form
the equation whose roots are the squares of the of the difference of the roots of 29.
2x* + 2 30.
sum and
(m + n) x + m2 + n 2 =0.
Discuss the signs of the roots of the equation
px2 + qx + r = 0. The following example illustrates a useful application 119. of the results proved in Art. 113. Example. can have
If
x
is
a real quantity, prove that the expression
numerical values except such as
all
lie
Let the given expression be represented by
between 2 and
x- + 2x
— 11 -.
6.
y, so that
a 2 + 2:r- ll_
2(s-3)
~ y;
then multiplying up and transposing, we have rr
2
+2.r(l-?/)
+ 6f/-ll = 0.
a quadratic equation, and in order that x may have real values 4(1 -i/) -4(Gy — 11) must be positive; or dividing by 4 and simplifying, 2 Hence 8*/ + 12 must be positive ; that is, (y - 6) (y - 2) must be positive. ?/ the factors of this product must be both positive, or both negative. In the former case y is greater than 6; in the latter y is less than 2. Therefore y cannot lie between 2 and 6, but may have any other value.
This
is
2
In this example it will be noticed that the quadratic expression lie between the roots y — 8y + 12 is positive so long as y does not 2 of the corresponding quadratic equation y — Sy + 12 = 0. 2
This is a particular case of the general proposition investigated in the next article.
For
120. the
all real values
Case
I.
a,
except
+ bx+c has 2 equation ax +bx + c =0
expression ax 2
;
them.
Suppose that the roots of the equation
ax 2 + bx + are real
tlie
when the roots of the and unequal, and x has a value lying between
same sign as
are real
of x
denote them by a and
ft,
c
=
and
let a
be the greater.
THK THEORY OF QUADRATIC EQUATIONS. 2
Then
((.r
4
bx +
c
—a
(
x* + -
a
V
= a {x2 = a (x -
Now
(tt
X+ +
91
)
aj
ft)
X+
aft
- ft).
a) (x
the factors x — a, x — ft are both if x is less than ft, the factors x — a, x — ft are both positive negative; therefore in each case the expression (x — a)(x — ft) is 2 But if x has a positive, and ax + bx + c lias the same sign as a. value lying between a and ft, the expression (./• - a) (x - ft) is negative, and the sign of ax" + bx + c is opposite to that of a.
x and
is
II.
If a
;
Case
greater than
if
and
ft
a,
are equal, then
ax 2 + bx + (x - a)
and
2
is
c
= a(x—
a)
2 ,
positive fur all real values of
has the same sign as
x
;
hence ax 2 + bx +
a.
Suppose that the equation ax 2 + bx + imaginary roots ; then
Case
III.
ax 2 + bx +
c
— alx* {-•
a
4a
2
b
2
—
4«c
.
is
.
is positive,
x + ^-
&)
Aac
+
lias
a\
iae-b')
—
b
;
hence
2
\a 2 2
therefore ax + bx + This establishes the proposition.
positive for all real values of a.
=Q
and the expression
(
same sign as
c
negative since the roots are imaginary
2
is
x+
b\-
(/
But iac-b*
c
x
;
c
has the
From
the preceding article it follows that the expression ax + bx + c will always have the same sign whatever real value x may have, provided that b 2 - Aac is negative or zero; and if this condition is satisfied the expression is positive or negative accord121.
2
ing as a
is
positive or negative.
Conversely, in order that the expression ax + bx + c may be always positive, b 2 — Aac must be negative or zero, and a must be 2 positive; and in order that ax + bx + c may be always negative 1 I - Aac must be negative or zero, and a must be negative. 2
——
—
;
;
HIGHER ALGEBRA.
92
Find the limits between which a must
Example.
lie in
order that
ax 2 - Ix + 5 5x 2 - Ix + a
may
be capable of
x being any
all values,
real quantity.
ax -lx + 5
_ Put
1
o
i-
(a-5?/)a:2 -7.r(l-?/)
then
=v;
rr
+ (5-a?/):=0.
In order that the values of x found from this quadratic
may
be real, the
expression
49
(1
- y)'2 - 4 (a - 5y)
(5
- ay) must be
positive,
- 20a) y + 2 (2a + l)y + (49 - 20a) must be positive hence (2a 2 + 1) 2 - (49 - 20a) 2 must be negative or zero, and 49 - 20a must be that
is,
2
(49
2
positive.
Now
(2a2 +
2
1)
2 (a 2
that
is,
-
- 20a) 2
(49
is
- 10a + 25) x 2 4 (a -
according as
negative or zero, according as
(a 2
o) 2 (a
+ 10a - 24)
is
negative or zero
- 2)
is
negative or zero.
+ 12)
(a
This expression is negative as long as a lies between 2 and - 12, and for such values 49 - 20a is positive; the expression is zero when a = 5, - 12, or 2, but 49 -20a is negative when a = 5. Hence the limiting values are 2 and - 12, and a may have any intermediate value.
EXAMPLES.
IX.
b.
Determine the limits between which n must
1.
lie
in order that
the equation
2ax (ax + nc) + (n 2 - 2) c2 =
may have
real roots.
If
2.
x be real, prove that x ^•2
5
3.
Shew
4.
If
and 5.
that -=
x be
.77
4-
x x — bx + 9 l
1
'-
x- + x+\
lies
prove that x
must
between
1
and - r^
.
11
between 3 and - for
•
all real
values of x.
71 = =- can have no value between —x++34a?— 2# — 7 3
^ 1
9.
Find the equation whose roots are s]a ± sja
6.
1
.
lie
3
sb
real,
'-
-5
If
a, /3
are roots of the equation
x2 — px+q=0,
(1)
atitfp-i-fl + ptfPa-i-a),
(2)
(
a
-p)-* + (P-p)-\
-b find the value of
THE THEORY OF QUADRATIC EQUATIONS. 7.
If the roots of ht?+
8.
If
x be
except such as 9.
real,
lie
nx+n=0 be in
the expression
between
2)i
those of the equation Ax'
-
:
q prove that t
admits of
r n 2 (x - n) .
all
values
and 2m.
ax2 + 2hx + c=() be
Tf the roots of the equation y
the ratio of p
!>:{
+ 2Ux+C=0
be
a-ffi
and
fi
+ d,
a
and
(3,
and
prove that
b*-ae_ B*-AC ~~a 2
10.
values
~
A2
Shew that the expression — when x
is real,
provided
'
—
—5
will be capable of all L
p + 3x - 4x* that p has any value
between
1
and
7.
.#4-2
11.
Find the &greatest value of n 2 n 2x + 3x + 6
12.
Shew
*
that
if
x
is real,
(x 2
the expression
-bc)(2x-b-c)~
has no real values between b and
(a
will
+ c) (ax 2 + 2bx + c) = 2
be impossible, and
14.
Shew that
i
c.
ax2 + 2bx + c =
If the roots of the roots of 13.
for real values of x.
be possible and different, then (ac
-
b2 )
(.r
2
+ 1)
vice versa.
the expression
- -
{-)-
will be capable of all
,! — a) (ex — a) and c 2 — d 2 have the same
fl
(ox
values
when x
is real, if
a2 - b 2
sign.
We
*122. shall conclude this chapter with some miscellaneous It will be convenient here to introduce theorems and examples. a phraseology and notation which the student will frequently meet with in his mathematical reading.
Definition. Any expression which involves x, and whose value is dependent on that of x, is called a function of X. Functions of x are usually denoted by symbols of the form f(x), F(x),(x).
Thus the equation y =f(x) may be considered as equivalent to a statement that any change made in the value of as will produce a consequent change in and vice versd. The quantities x and y are called variables, and are further distinguished as the independent variable and the dependent variable. ;//,
HIGHER ALGEBRA.
94
An
independent variable is a quantity which may have any value we choose to assign to it, and the corresponding dependent variable has its value determined as soon as the value of the inde-
pendent variable %
123.
An
is
known.
expression of the form
pjs"
n +p x
]
x
+ pjf
2
+
.
.
.
+ pn _ x + p n t
where n is a positive integer, and the coefficients p p lt p a ,...p n do not involve x, is called a rational and integral algebraical function of x. In the present chapter we shall confine our attention to lt
,
functions of this kind.
A
*124. function is said to be linear when it contains no higher power of the variable than the first ; thus ax + b is a linear function is said to be quadratic when it function of x. contains no higher power of the variable than the second ; thus ax2 + bx + c is a quadratic function of x. Functions of the third, fourth,... degrees are those in which the highest power of the Thus in the last variable is respectively the third, fourth, th degree. article the expression is a function of x of the n
A
The symbol fix, y) is used to denote a function of two 2 2 variables x and y thus ax + by + c, and ax + bxy + cy + dx + ey +f are respectively linear and quadratic functions of x, y. *125.
;
The equations fix) = ratic,
...
are said to be linear, quadfix, y) — according as the functions f(x), f(x, y) are linear, quad0,
ratic,....
We
have proved in Art. 120 that the expression ax + bx + c admits of being put in the form a (x — a) (x — fi), where a and j3 are the roots of the equation ax2 + bx + c — 0. *126.
2
Thus a quadratic expression ax 2 + bx-\- c is capable of being resolved into two rational factors of the first degree, whenever 2 has rational roots ; that is, when the equation ax + bx + c = 2 b - iac is a perfect square. *127.
may
To find
the condition that
a quadratic function ofx,y
be resolved into two linear factors.
Denote the function hy f(x,
y)
where
2 z f{x, y) = ax +'2hxy + by + 2gx+ 2fy +
c.
THE THEORY OF QUADRATIC EQUATIONS. Write
powers of
this in descending
and equate
x,
95 it
to zero;
thus ax*
+ 2x (hy +
+ by 2 + 2fy +
y)
Solving this quadratic in x
c
-
.
we have
- (h + (j) ± J{hy + y)* - a (by 2 + _ x—
2fy
+
c) ,
a
<
ax + hy + g = ±
>v
Jy
2
(h*
-
+ 2y
ab)
(hy
—
a/)
+
2
(g
-
ac).
Now
in order that J\.r, y) may be the product of two linear factors of the form px + qy + r, the quantity under the radical
must be a perfect square
hence
;
- a/) 2 =
(kg
Transposing and dividing by abc + 2fyh
which
is
-
(h-
we obtain
a,
— af — 2
by
—
2
ch
2
=
;
the condition required.
This proposition
To
*128.
is
of great
may have
a
importance in Analytical Geometry.
find the condition that the equations
ax 2 + bx +
common
c
— 0, ax 2
f b'x
+
-
c
root.
Suppose these equations are both aa.
2
+ ba +
2
c
a'a +b'a + c' .*.
- ac).
ab) {
=
satisfied
by x
a
;
then
0,
= 0;
by cross multiplication a
a"
be
To eliminate equate
it
—
ca — c'a
b'c
— ab
'
square the second of these equal ratios and to the product of the other two ; thus
(ca .'.
is
ab'
a,
a
which
1
(ca
a
— c'a) — ca) 2 = 2
1
(be
—
b'c)
(be
—
b'c) (ab'
'
(ab'
— ab)
'
— ab),
the condition required.
easy to prove that this is the condition that the two 2 quadratic functions ax 2 + bxy + cy 2 and a'x + b'xy + c'y' may have It
is
a common linear
factor.
1
HIGHER ALGEBRA.
06
^EXAMPLES. 1.
For what values of
m will
IX.
c.
the expression
2 y + 2xy + 2x + my - 3
be capable of resolution into two rational factors
m
which will make equivalent to the product of two linear factors. 2.
Find the values of
3.
Shew that the expression
always admits of two real linear 4.
If the equations
common
root,
shew that
x2 + p'x + q' = must be
it
p'l'-p'q nr
q-q 5.
either
9-q
p-p
Find the condition that the expressions Lv2 + mxy + ny 2 ,
may have 6.
a
common
l'x2
+ m'xy
1
-f-
If the expression
2Pxy + 2y 2 + 2ax - 4y +
P
can be resolved into linear factors, prove that roots of the equation P'2 + 4aP + 2d1 + 6 = 0.
must be one of the
Find the condition that the expressions
ax2 + 2hxy + by 2
may
n'y'
linear factor.
%a? +
7.
+ mxy + 3y 2 - 5y - 2
2
2.v
factors.
x2 + px + q = 0, have a
?
,
a'x2
+ 2k'xy + b'y 2
be respectively divisible by factors of the form y 8.
Shew
-mx, my + x.
that in the equation
x 2 - Zxy + 2y 2 - 2x - 3y - 35 = 0, for every real value of x there is a real value of y, value of y there is a real value of x. 9.
If
x and y are two
real quantities connected
9x2 + 2xy +y 2 then
will
lie
between 3 and
6,
for every real
by the equation
- 20y + 244 = 0,
and y between
1
and
10.
If (ax2 + bx + c)y-\-a'x 2 + b'x + e' = 0, find the condition that be a rational function of y.
10.
may
x
92.r
and
x
CHAPTER
X.
MISCELLANEOUS EQUATIONS. In this chapter we propose to 129. consider some miscellaneous equations ; it will be seen that many of solved by the ordinary rules for quadratic equJtions, but others require some special artifice for their solution
the^l
_
Example
3
3_
Solve
I.
Multiply by
.r
2n
8x 2n
-8x~^=63.
and transpose; thus i. 8x n - 63x 2 '*-8 = 0;
—
L
(a?"-8)(8x^+l) = 0;
-
«2n = 8,
1
or--; 8' 2n
=(*)* «(-p)*; .-.*=«», or Example
2.
Solve
2
/-+
V
3
«
A. ^
6a 6
«
I
/- =
V«
.•.%+! = * +
«
2<% 2 -6a 2?/-& 2 + 3a& = 0;
(2ay~&)(ty-3a)=0; 3a
6
*
&2
4a
H. H. A.
9a 2
"l '
&a
*
Z
HIGHER ALGEBRA.
98 Examples.
(*-5)(a:- 7)(«
Solve
We have
(x
2
+ 6)
(*
- x - 20) (x 2 - x - 42)
which, being arranged as a quadratic in x 2 (a2
- x) 2 - 62
.-.
(x2
(x2
130.
Any
- x) + 336 =
-56 =
-7.
8,
equation which can be thrown into the form 2 ax 2 + bx + c + p J ax + bx +
may
Putting y =
be solved as follows.
Let a and
;
x, gives
or a2 -a;
f
x = S, -2,
whence
= 504
-a:-6)(x2 -x-56) =
X *- X -Q =
.-.
+ 4) = 504.
ft
c
J ax
—q
2
+ bx +
c,
we
obtain
be the roots of this equation, so that
J ax
2
+ bx +
from these equations we
c
= a,
Jax
2
+bx + c = ft
;
shall obtain four values of x.
prefixed to a radical it is usually understood that it is to be taken as positive; hence, if a and ft are both positive, all the four values of x satisfy the original equation. If however a or ft is negative, the roots found from the resulting
"When no sign
is
quadratic will satisfy the equation
ax2 + bx +
— p J ax + 2
c
bx + c =
q,
but not the original equation. Example.
Add
Solve x2 - ox + 2
3 to each side
;
rc
Jx 2 - 5z + 3 = 12.
then 2
-5a; + 3
+ 2 N/^-5a; + 3 = 15.
Jx2 -5x+3 = y, we obtain y 2 + 2y - 15 = Thus *Jx 2 - 5x + 3 = + 3, or Jx 2 -6x + S = -5.
Putting
;
whence y = 3 or -
ing quadratics, we obtain from the Squaring, and solving the resulting
ic=6 or -1; and from the second satisfies the given equation,
= x^ :
—^
but the second pair
x2 - 5x -2
.
The
satisfies
Jx2 -5x + 3 = 12.
first
5.
first
pair of values
the equation
;
.
;
MISCELLANEOUS EQUATIONS.
99
Before clearing an equation of radicals it is advisable to examine whether any common factor can be removed by 131.
division.
We
+ 10a'2 - Jx- + ax- 6a- — x- 2a.
Solve *J x'2 - lax
Example. have
*J(x-2a)(x-5a) - J{x-2a) (x+Sa) = x-2a.
The
now
factor *J x - 2a can
-5a- Jx + 3a = „Jx - 2a
sjx
.'.
be removed from every term
x - 5a + x + 3a - 2 *J(x - 5a)
(x
+ 3a) = x - 2a
x = 2 Jx'2 - 2ax - 15a'2
;
;
3ar-8aa;-60a 2 = 0; {x
-6a) (3a; + 10a) =0; x — cba, or
trial it will be
the roots are
10a
Jx - 2a, we
Alsoxby equating to zero the factor
On
——
found that x = 6a does not
—— and
obtain x = 2a.
satisfy the equation
:
thus
2a.
D
The student may compare a similar question discussed in the Elementary Algebra, Art. 281. 132.
The following Solve J'3x- -
Example.
We
artifice is
4.x
sometimes useful.
+ 34 + JSx'2 - 4x -
11
=9
(1).
have identically (3x--4a; + 34)-(3a; 2
Divide each
member
of (2)
J'dx- -
-4x-ll) = 45
by the corresponding member of
4.r
+ 34 - JSx2 - 4x -
11
(2).
(1);
thus
=5
(3).
Now (2) is an identical equation true for all values of x, whereas (1) is an equation which is true only for certain values of x hence also equation (3) is only true for these values of x. ;
From
(1)
and
(3)
by addition 2 v/3x -4a;
whence
as
+ 34 = 7;
= 3,
or
--.
7—7
.
.
HIGHER ALGEBRA.
100 133.
The
solution of an equation of the form
ax4 ± bx 3 ± ex 2 ± bx + a =
0,
which the coefficients of terms equidistant from the beginning and end are equal, can be made to depend on the solution of a in
Equations of this type are known as reciprocal equaand are so named because they are not altered when x is
quadratic. tions,
changed into
its
reciprocal -
For a more complete discussion of reciprocal equations the student is referred to Arts. 568 570.
—
Example.
12a; 4
Solve
- 56x 3 +
-
89a; 2
+ 12 = 0.
56.x
Dividing by x2 and rearranging,
12/W- ) -56^+^+89 = 2
x + -=z: then x
Put
.-.
12
whence we obtain
By
(z 2
z
-
=-
2)
2
solving these equations
we
6a; 4
Solve
We have
6(a;
.-.
x-
-=-
.
1
5
13
a;
2
6
13-
find that x = 2, -
25a; 3
,
2
,
-
.
though not reciprocal may be
+ 12a; 2 + 25a; + 6 = 0.
2 (^ +^i) - 25 fx - -\ + 12 = 0;
6
whence
-
+ — = z 2 -2;
u
The following equation 134. solved in a similar manner. Example.
2
-56^ + 89 = 0;
or
,
a;
0.
2
—
j
-25
(a;
(^--^-3 = 0,
whence we obtain
a;
= 2, - -
,
— 1+24 =
or 3
3,
0;
fx- -]-8-0;
- -
When one root of a quadratic equation is obvious by 135. inspection, the other root may often be readily obtained by making use of the properties of the roots of quadratic equations proved in Art. 114.
MISCELLANEOUS EQUATIONS. Example. This
is
Solve
(
1
-
+ a) - 2a
a-) {x
a quadratic, one of whose roots
Also, since the equation
may
is
-
1
-
3.
2jx + 2x
5.
3"+6=5#».
(1
+ a 2 ) = 0,
X.
9
2.
-
6a?*~7**-8a7
6.
3.f 2n
JL
6 x/a=5a
-13.
11.
32*+ 9« 10.
13.
2 2* + 8 + 1
15.
,/*+£-*
3*.
= 32.2'.
1
-.rri -2=0.
1+8.^ +
12.
5 (5*
14.
2 2* + 3 -57 = 65(2*-l).
16.
^.-#=5A-
19.
20.
(2a-7)(a 2 -9)(2a + 5) = 91.
21.
A'
22.
3a2 - 4a + s/'3x i -4x-6 = l8.
23.
3a2
24.
8+9
25
^- +s
18.
(x
- 7)
(.v
)
)
2
+ 2 >/a + 6a = 24 - 6x. 2
-7 + 3
(N
/3sa -16a? + 21
J(&v -1) 2
.
- 3) {x + 5)
(a-
(x
-2)
= 16a;.
= 3.c 2 -
/,,,_- 5 , +3=
9^ =
10.
+ 1 = 1 680. (x + 9) (x - 3) (x - 7) {x + 5) = 385. x (2x + 1 (.v - 2) (2a - 3) = 63.
17.
*.
\/S + \/?'
a
»>/;+Vj-"*
_J
1
i
9.
—
+ a- 4 = 10a--'.
4.
1
2
—
a.
3
1
7
is
:
§ =5-
2
= 0.
and therefore the other root
;
Solve the following equations
a- 2 -2x~ 1 = 8.
ar)
be written
EXAMPLES.
1.
-
clearly a.
is
2ax- + (1 - a 2 ) x - a
the product of the roots
(
101
7a.
^:.
0.
+ 5-*) = 26.
2*.
.
—
.
HIGHER ALGEBRA.
102
26.
7.-^hs±i_c» x x \*j
W
.Y. j
Jx2 - 9.
27.
J4x2 -7x-lb -
>Jx2 -3x=
28.
>/2^2 -9^ + 4 + 3
J%v-l = J2x2 + 2\x-l\.
29.
V 2^ + 5^-7 + V3(a;
30.
s/a
31.
J2x2 + bx - 2 - V2#2 + 5a; - 9 = 1.
32.
2 2 x/3^ - 2x + 9 + x/3.r -
33.
V2^ 2 -7a;+l -
34.
>/ 3^
35.
o^ + a?-4o;2
36.
x*
38.
10(o7t +l)-63a?(a;2
2
2
J7x2 -6x-l = 0.
-7a; + 6) -
2
+ 2ax-3x2 - Ja 2 + ax-6x 2 = ^/Sa 2 + 3ao; - 9a;2
2
./2a;2
- 7^ - 30 -
2o;
-4 = 13.
-9a; + 4 =
*/2o; 2
- 7x -
5
1.
=
a?
-5
+ a;+l = 0.
8
+ ^x2 + l = 3x3 +3x.
3*+l-3(«s +#)=2tf*.
37.
9
-l) + 52a;2 = 0.
x+J\2a-x _*Ja+\
.
1
a;
+ sjx 2 - 1
-
1
x + a/^ 2 -
1
^/.r 2
\
1
42.
>/^+#I = Jtf-x
44.
2*
2
46
2 2* = 8
:
^/a?-5
|. 2
2
48. 49.
>/a;2
50.
^B+-»-^El. ##+
a2 *(a 2 +
= (a?* + a*)a.
I)
2a;+l
_ 250 V^+T "
3n/7^3
1
(a
2
- a;2 )"3
.
+ aa;-l - Jx2 + bx-l = J a - K/b.
V^'
4
45.
2 a;
18 (7a; - 3)
- *) = 5
.v
./». £".+ -l V#
'
2
51.
43.
'
(a + a;) 3 + 4 (a "
2
1
- 2.v3 + a; = 380.
\/a'2
'
-
,—
= V3a?-7 x-b
3a;-7
a
= 8x jx 2 - 3x + 2.
1.
:
J'a2 - 4x2 _ bx
a+2x- J a 2 - 4a;2 ~
-
a;
,
x - sjx 2 -
a + 2a; +
sJct-V
x -
.
8a
1
52.
27^ + 2U- + 8 =
.
MISCELLANEOUS EQUATIONS.
We
136.
two unknown Example
now
shall
108
some simultaneous equations
discuss
quantities. Solve
1.
(z
x+2+y+S+ J(x + 2) (y + 3) = 39. + 2) 2 + (y + 3) 2 + (:r + 2)( / + 3)=741. 2
Put x + 2 = m, and y + 3 = v
;
then
u+v + Juv = Sd
(1),
w + v + wv = 741 2
hence, from (1) and
(2),
2
we obtain by
(2),
division,
u + v - Juv = 19
From
(1)
and
(3)
u+t?=29;
(3),
Juv = 10,
and
wv = 100;
or
whence
w = 25,
thus
x = 23, or 2; y=l, or 22.
Example
2.
or 4; v
4
Solve
.r
= 4,
or 25
;
+ y*= 82
(1),
ar-y=2
(2).
# = w + t>, and y = u- v;
Put then from
(2)
we obtain
Substituting in
v (w
(1),
.-.
whence
= l.
+ l) 4 + (u- 1) 4 = 82; 2(m4 + 6m2 + 1) = 82; u 4 + 6u 2 -40 = 0; w2 = 4, or — 10 ;
u= ±2,
and
or
± >/~ 10*
x=s, -l, i± V^iO;
Thus
ysal, -3,
JEa;ampZe3. e
-li^-10.
- —^ = 2A f^ Sx-yx +
Solve
(1),
10
y
7x + 5y = 29
From
(1),
15
(2a;
2
+ Sxy + y* - 3z 2 + Axy - y-) = 38 (3.x 2 + 2xy .-.
.-.
Hence or
(2).
129o;
2
-29xy-38?/
2
-
y*)
;
= 0;
{Sx-2y)(iBx + 19y)=0. Sx = 2y
43# = -19y
(3),
(1).
of
;
HIGHER ALGEBRA.
104
= y __7x = + 5y
x
From
(3),
2
29
3
= 1, .-.
Again, from
by equation
x = 2, y = 3.
x
y
19 ~ ^43 ~
(4),
(2).
7x + 5y - 82 29
= -gg, by •*'
~ 82 82 ,V
x = 2, y = 3; or
Hence Example
4.
Solve
(2),
_ 1247
551
X~
equation
x-
-
—
"
1247
551
= -^-
11
,
•
4# 3 + 3a; 2 f/ + ?/ 3 =8, 2z 3 -2a; 2?/ + £?/ 2 = l.
Put y = mx, and substitute in both equations. z 3 (4 + 3m + m 3 ) = 8
(1).
z 3 (2-2m + m2) = l
(2).
•*'
4 + 3m + m z _
2-2m + m2 ~
(/;i-l)(?/i-3)
is,
.*.
(i)
Take
From
m = l,
and substitute # 3 = 1;
(2),
(m-4) = 0;
or
or
3,
4.
in either (1) or .*.
x = l;
y=mx=x=l.
and (ii)
m=l,
'
-12 = 0;
m3 -8»i9 +19m that
Take
m = 3,
and substitute
in (2) 3
/l
thus
5:r
x=
\/
and
y = vix = 3x = 3
*/
-.
3
1
3
/I
(iii)
Take7?& = 4;
3
= l;
.*.
k'->
3/1
we obtain 10.r 3 =l;
and
Thus
.-.
/ x=^-;
y = mx = 4x=4. /r^.
(2).
MISCELLANEOUS EQUATIONS. Hence the complete solution
is
* =1,
•"
v
V5'
=1
s '
The ahove method
Note.
L05
4
\/l>
To*
\/^*
may
of solution
always be used when the
equations are of the same degree and homogeneous.
Example
5.
3 lx 2 y 2
Solve
-7y 4 - 112^ + 64 =
(1),
x 2 -7xy + 4y 2 + 8 =
From
we have -8 = x2 - Ixy + 4//'-
(2)
3\x2 y 2 - 7# 4 + .-.
that
{x 2
and, substituting in
;
- Ixy + Ay 2 +
{x 2
)
(1),
- Ixy + 4 j/")- =
+ (x 2 - Ixy f Ay 2 ) (Uxy + x 2 - Ixy + Ay 2 =
31x-y 2 - 7 j/ 4 .-.
Uxy
('2).
)
;
;
Slx 2 y 2 -7y* + (x2 + 4y 2 ) 2 -(7xy) 2 = 0;
s*-10sy+9y4 =0
is,
.'.
(x2
-y 2 )(x 2 -9y 2 ) = 0;
x=±y,
hence
(3).
ov
x=
±3y.
Taking these cases in succession and substituting in
(2),
we obtain
x = y=±2;
x=-y=± ^J
-
x=±3, y=±l\ 3
>/-17'^ =T \/
-
yj
Note. It should be observed that equation (3) is homogeneous. The method here employed by which one equation is made homogeneous by a suitable combination with the other is a valuable artifice. It is especially useful in Analytical Geometry.
Example
6.
Solve
(x+yft+2
{x
- ?/)* = 3
{x 2
-
(1).
y*fi
3x-2y=13
(2). i
i
Divide each term of
(1)
by (x 2 - y 2
)
,
or {x
i
i
.
(
x +y\s
\x-yj
+y)*
+2 (-~ \.v
y
Y=3
+ gj
i
(x
-
r
-
y)
;
HIGHER ALGEBRA.
106
i
This equation
is
+ v\ a
[x
a quadratic in
(
1
from which we easily
,
i
x
±y-Y = 2oTl; ( \x-yj
whence
^=8 x-y
Combining these equations with
(2),
we obtain 13
x=9, y = 7;
or x = -^,y=0.
EXAMPLES. Solve the following equations
3x-2y = 7,
2.
4
a,
X.
:
bx -y = 3,
+ #y+2/ 4 = 931,
7.
+y 2 =2275.
3#2 -5y 2 = 7,
8.
9.
Zxy - 4y 2 = 2.
16.
.r
+y 4 =706, x+y = 8.
4
,r+i = l, y 4
14.
17.
11 ^3
+y = ie. 3
2
= 84,
20.
+ 165 = 16.ry, 7^y + 3y 2 = 1 32.
3.r
2
#2 + y 2 -3 = 3.zy,
12.
2x2 - 6 + y2 = 0. xA +y* = 272,
£+£-£, x 2 y 3
„
x +y = 1072,
10.
15.
33
^- 2
18.
+y
2
=65.
5
= 5. |+t 2 5 -2 + -5 = 7;5
= 1.
11 xy^+yx^=20,
^-y = 992, x-y = 2.
x-y = 2.
e ?/+-=25.
19.
+3/
bxy - 6x2 = 6.
Zlxy - 3x2 -bf = 45. 13.
ocy
x +y = 7 + \A?y, x2 +y 2 = l33-xy.
5y 2 -7^ = l7,
3x2 +xy+y 2 = l5,
11.
12^ + 13y 2 = 25.
x - *Jxy+y =6.
x + Jxy +y = 65, #2 + #y
4^-3^ = 1,
3.
x2 +
5.
x2 — xy +y2 = 19. 6.
b.
2 2 y - 6# = 25.
xy = %). 4.
;
7x = 9y, or y = 0.
.'.
1.
or 1
find,
11 # +y2 = 5, 2
21.
6(.i?
11 = 2
+y
2
)
5.
.
.
MISCELLANEOUS EQUATIONS. 22.
Jx+y+J7-y = 4,
23.
y+
24
JZ+JZ^,
25.
f~f
.
27.
+ 4y 2 - 15* = 10 (3y - 8), xy = 6. 2 2 2 2 .r y + 400 = 41ay, y = 5.ry - 4.r
28.
4i- 2
29.
9. c
30.
(.-v
31
2.v 2
32.
y—-% + +
26.
107
Jx 2 - 1 = 2, +
v^
= i_7
,
tf*
.
+ 33.r -12 = 1 2xy - 4,y 2 + 22y
2
2
7#-lly = 17.
+ 5y=6+2Qay-25ya + 2.v,
- y2 )
(.r
- y) = 1 6a^,
- xy +y 2 =2y,
(.r
y)
—% =
,
2
(#-y) 2
(a4
- y4
)
-o8
5.v
,
-4xy + 2x 2 ) + 8 = 0.
34.
3a-3
35.
2 2 y (4v - 108) = x (x3 - 9y 3 ),
36.
6xi + x 2y 2 + l6 = 2x(\2x+y 3 ),
37.
x (a + x)=y(b+y),
38.
xy + «Z> = 2ax,
39.
fir_
40.
6.v3
41.
2a(--'A+4a2 = 4:X2 +^-t
- 8ay2 + if + 2 1 = 0,
\y
6.r
a2
(y
- x) = 1
2x 2 + 9xy + y 2 =
L
y—a
108.
x2 + xy-y 2 = 4.
xhf + a2 b 2 = 26 2y2
>
2
x(j/ 2
ax + by = (x + y) 2
+ .lzi = _J b2 x—b
= 10a
- y 2 ) = 64Ga?y.
- 7y = 4.
2 y(y -3.r#-.r2 ) + 24 = 0,
a2
(a3
- ovy = 18.
2x 2 + 4xy = 5^.
33.
a
at?
,
.
.
L
.
a—b
=0
.
ay 3 = 10ab 2y + 3b 3x.
+ 3a3y,
xj
2a
i
a*
Equations involving three or more unknown quantities can only be solved in special cases. We shall here consider some of the most useful methods of solution. 137.
Example
1.
Solve
From
(2)
and
Put u
for
x+y
(3), ;
+ y +z =13 .7^ + 2/2 + 22 = 65 xy = 10
x
(x
+ yf + * 2 = 85.
then this equation becomes
u*+z*=85.
(1), (2),
(3).
HIGHER ALGEBRA.
108 Also from
u +z =13;
(1),
whence we obtain u = l or 6;
z
=6
x+y=
Thus we have
or 7,1
#?/ = 10
Hence the
7.
£ + ?/ =
and
6,
acy = 10
\
solutions are
x=5,
or
2,'|
y = 2, or
5,1
=6
*
.r
= 3db\/-l.,
y^W^T,
or
z=l.
5
J
Example
Solve
2.
(a;
{y
+ y) {x + z) = 30, + z)(y + x) = 15,
[z+x)(z+y) =18. Write
m,
1;,
w
for
?/
+ 2, viv
a
+
as,
— 30,
a;
+y
respectively
tvu = 15,
;
thus
mv = 18
(1).
Multiplying these equations together, we have
wVu>2 _ 30 x 15 x 18 = 15* x 6 2 .*. uvw = ±90. Combining
;
this result with each of the equations in (1),
u = 3, v = 6, w = 5\ or w = -3, v = -6,
Example
ce=4, y = l, 2 3.
Solve
y
= 2;
2
or
x=-i, y=-l, «=-2.
+ ys + 2 2 = 49
+ z:r + a; = 19 x* + xy + y 2 =39 2
Subtracting
(2)
from
2
2
(1)
and
(4)
and
(5),
by division
y-*- 3 «-« whence
:
(3).
(4).
(3)
[z-x){x+y+z)*=10 Hence from
(2),
-x2 + z{y -«)=30;
(y-x){x + y + z) = 30
is,
Similarly from
(1),
2
(1)
y that
w=-5;
y+z=-S,\ y + z=3,\ z + x = $, > or z+x = -d>,\ x + y = -5,i x + y = 5))
.-.
whence
we have
y
.
= 3z-2x.
(5).
;
;
MISCELLANEOUS EQUATIONS. Substituting in equation
(3),
10f)
we obtain
z*-8xa+8zs =13.
From
x 2 + xz
(2),
+
z~
= 19.
Solving tbese homogeneous equations as in Example
a;=±2, jc= ±-ts,
or
Example
Solve
4.
z
= ± 3 and ;
2= ± -t^ .t
2
Multiply the equations by c
2
y, 2,
.r
Multiply the equations by
(1)
and
(2),
a;
-zx = 6 2
z2
,
— xy = c 2
and add
respectively
.
;
then (1).
//
z, x,
y respectively and add
;
then
+ c-y + a*z =
(2).
by cross multiplication,
~^¥c = V^W = 2
^W = 2
k su PP° se
Substitute in any one of the given equations k 2 (a 6
x
a 4_^2c 2
+ b 6 + c 6 - 3a2
^4_ c
Z>
2
c 2)
;
2
ft2
C 4_
'
then
=1
z
11
1 i
a 7/j
EXAMPLES. Joh
= ±5
+ « 2 + & 2z =
b2x
From
y^
,
we obtain
y= T —-,
and therefore
;
-yz = a 2
therefore y
4, Art. 130,
*Ja*+b*+c*-3a?tP
X.
c.
HIGHER ALGEBRA.
110 9.
10. 11.
#VW=8,
x*y*zhi=\% aPy*z=12
t
^-
3
x*yz2 u2 =
l,
=54, .*% 32 2 = 72.
ay+#+y=23, xz+x + z = 4l, yz + ij + z = 27.
2^-4?+2/ = 17, 3yz+y-6z = 52, §xz + 3s + 2#= 29.
12.
13.
xz+y^lz, yz + x=8z, x + y + z = l2.
14.
.r
15. 16.
3xy 2z2 u 2 = 4.
+y 3 + ^3 =a3 ^2 +y 2 + 22 = a2 # + # + s = a. ^2 +y 2 +22 =3/^ + 2^ + .«y = « 2 3.r-# + 2 = a*/3. #2 +y2 -M2 = 21a2 ys + ^-.ry = 6a 2 3x+y-2z = 3a. 3
,
,
,
,
,
Indeterminate Equations. Suppose the following problem were proposed for
138.
tion
solu-
:
A
person spends .£461 in buying horses and cows; if each horse costs £23 and each cow £16, how many of each does he buy 1
?
Let
x,
y be the number
of horses
23a;
+ 16^ =
and cows respectively
;
then
461.
Here we have one equation involving two unknown quantities, and it is clear that by ascribing any value we please to x, we can obtain a corresponding value for y thus it would appear at first sight that the problem admits of an infinite number of solutions. But it is clear from the nature of the question that x and y must and with this restriction, as we shall see be positive integers later, the number of solutions is limited. ;
;
unknown
quantities is greater than the number of independent equations, there will be an unlimited number of solutions, and the equations are said to be indeterminate. In the present section we shall only discuss the simplest kinds of indeterminate equations, confining our attention to positive integral values of the unknown quantities ; it will be seen that this restriction enables us to express the solutions in a very If the
number
of
simple form.
The general theory in Chap. xxvi.
of indeterminate equations will be found
;
;
INDETERMINATE EQUATIONS. Example
1.
Solve 7# +
Divide throughout by
12j/
= 220
x + y+^-~
5y-S
=31
= integer
...
(1)
;
7
that
thus
we must have
Since x anil y are to be integers,
l%-9 =
and therefore
;
y+^ =31 + -;
x+
.-.
Ill
in positive integers.
the smaller coefficient
7,
;
:
integer
*-=-= integer; %-l+ w-2
is,
1/-2
and therefore
:
integer
=p
suppose.
y-2 = 7p, y = lp + 2
.-.
or
Substituting this value of y in
(2).
(1),
+ 7p + 2 + 5> + l = 31; x = 2§-l2p
.r
that
is,
.(3).
If in these results we give to p any integral value, we obtain corresponding integral values of x and y; but if p > 2, we see from (3) that x is negative and if p is a negative integer, y is negative. Thus the only positive integral values of x and y are obtained by putting p = 0, 1, 2.
The complete
solution
may
be exhibited as follows
p=
0,
= 28, y= 2, a:
Note. to
make
artifice
When we
obtained
1,
2,
16,
4,
9,
16.
5y-S integer,
we multiplied by
3 in order
the coefficient of y differ by unity from a multiple of 7. A similar should always be employed before introducing a symbol to denote
the integer.
Example
2.
Solve in positive integers,
14x -
Divide by 11, the smaller coefficient; thus
x+
3x-7 11
Sx 11
i/-2 + ir
= 2 - x + y = integer
;
11//
= 29.
(1).
HIGHER ALGEBRA.
112
12£ - 28
—^
hence
= m teg er .
——^ = integer
*>
g*
that
x - 2+
is,
—=
'— ^ Qfc
.-.
integer =_p suppose;
X = \\p + §
.*.
and, from
;
y
(1),
— 14p + 5 !
called the general solution of the equation, and by giving to p positive integral value or zero, we obtain positive integral values of x
This
any and y
;
is
thus we have
p = 0,
1,
= 6, y = 5,
number Example
3,
17, 28, 39,
.t
the
2,
33, 47,
19,
of solutions being infinite. 3.
In
how many ways can £5
be paid in half-crowns and florins?
Let x be the number of half-crowns, y the number of florins
;
then
5^ + 4y = 200; •••
x +y+\=
5 °;
x 2 — integer = 2^ suppose .
.'
.
x=4p,
.*.
and
;
y
= 50-5p.
Solutions are obtained by ascribing to p the values 1, 2, 3, ...9; and If, however, the sum may be paid either therefore the number of ways is 9. in half-crowns or florins, p may also have the values and 10. If ^ = 0, then x = 0, and the sum is paid entirely in florins if p = 10, then y = 0, and the sum is paid entirely in half-crowns. Thus if zero values of x and y are admissible the number of ways is 11. ;
Example
4.
man
paid there of each?
each
The expenses 5s.,
each
of a party numbering 43 were £5. 14s. Qd. if 2s. 6d., and each child Is., how many were ;
woman
Let x, y, z denote the number of men, women, and children, respectively; then we have x + y + z= 43 (1), 10.r
Eliminating
The general
z,
we obtain
+ 5?/ + 2z = 229. 8x + By = 143.
solution of this equation
is
x=Sp + l, y = 45-8p;
:
:
INDETERMINATE EQUATIONS. Hence by substituting
in
(1),
Here p cannot be negative or from 1 to 5. Thus
p= x-
113
we obtain z = 5p-3.
zero, but
may have
1,
2,
4,
5;
4,
7, 10, 13,
16;
y = 37, 29, 21, 13,
5;
2=2,
3,
positive integral values
*
7, 12, 17, 22.
EXAMPLES.
X.
d.
Solve in positive integers
of
1.
3.i
+ 8y = 103.
2.
5#+2y=53.
3.
7.>;+
4.
l&P+lly=414
5.
23a?+25y=915.
6.
4L>;
+ 47y = 2191.
Find the general solution in positive integers, and the
least values
x and y which
satisfy the equations
5.v-7y = 3.
7.
8.
I7y-13#=0.
10.
12y=152.
11.
6a?-13y=l.
9.
19y-23a?=7.
13.
A farmer spends £752 in
14.
In
8#-2ty=33. 77y-3Qa?=295.
12.
buying horses and cows if each horse costs £37 and each cow £23, how many of each does he buy ?
how many ways can £5 be paid
including zero solutions
What
16.
pay
in shillings
and sixpences,
?
Divide 81 into two parts so that one
15.
and the other of
to
;
may
be a multiple of 8
5.
is
105. 6d. to
the simplest
another
way
who has
for a person
who has only guineas
only half-crowns
?
Find a number which being divided by 39 gives a remainder 17. and by 56 a remainder 27. How many such numbers are there ?
What
16,
the smallest number of florins that must be given to discharge a debt of £1. (5s. 6d., if the change is to be paid in half-crowns 18.
is
only? Divide 136 into two parts one of which when divided by 5 leaves remainder 2, and the other divided by 8 leaves remainder 3. 19.
I
20.
at
£17
:
buy 40 animals consisting
if I
spend £301, how
my
many
of rams at £4, pigs at £2, of each do I buy ?
and oxen
have 27 coins, which are sovereigns, half-crowns and the amount I have is £5. 05. 6d. how many coins of each sort have I ? In 21. or shillings,
H. H. A.
pocket
I
;
8
;
;
CHAPTER XL Permutations and Combinations. Each of the arrangements which can be made by taking 139. some or all of a number of things is called a permutation. Each of the groups or selections which can be made by taking some or all of a number of things is called a combination. Thus the •permutations which can be made by taking the letters a, b, c, d two at a time are twelve in number, namely, ab,
ac,
ad,
be,
bd,
cd,
ba,
ca,
da,
cb,
db,
dc
each of these presenting a different arrangement of two
a,
letters.
The combinations which can be made by taking the namely, b, c, d two at a time are six in number
letters
:
ab,
ac,
ad,
be,
bd,
cd;
each of these presenting a different selection of two
letters.
forming combinations we are only concerned with the number of things each selection contains whereas in forming permutations we have also to consider the order of the things which make up each arrangement; for instance, if from four letters a, b, c, d we make a selection of three, such as abc, this single combination admits of being arranged in the
From
following
this it appears that in
ways
:
abc,
and
acb,
bca,
bac,
cab,
cba,
so gives rise to six different permutations.
PERMUTATIONS AND COMBINATIONS. Before
140.
discussing
the
general
115
propositions
chapter there is an important principle which we explain and illustrate by a few numerical examples.
of
this
proceed to
If one operation can be performed in m ivays, and (when it has been performed in any one of these ways) a second operation can then be performed in n tvays ; the number of ways of performing the two operations ivill be m x n. performed in any one way, we can associate with this any of the n ways of performing the second operation and thus we shall have n ways of performing the two operations without considering more than one way of performing If the first operation be
:
the
and
first;
corresponding to each of the
so,
m
ways
of per-
forming the first operation, we shall have n ways of performing the two; hence altogether the number of ways in which the two operations can be performed is represented by the product
m x n. There are 10 steamers plying between Liverpool and Dublin; in how many ways can a man go from Liverpool to Dublin and return by a different steamer?
Example
1.
There are ten ways of making the first passage and with each of these there is a choice of nine ways of returning (since the man is not to come back by the same steamer) hence the number of ways of making the two journeys is 10 x 9, or 90. ;
;
may
extended to the case in which there are more than two operations each of which can be performed in a given number of ways. This principle
Example
2.
in how different hotel?
hotels;
easily be
Three travellers arrive at a town where there are four many ways can they take up their quarters, each at a
The first traveller has choice of four hotels, and when he has made his selection in any one way, the second traveller has a choice of three ; therefore the first two can make their choice in 4 x 3 ways and with any one such choice the third traveller can select his hotel in 2 ways hence the required number of ways is 4 x 3 x 2, or 24. ;
;
To find
141.
the
number of permutations of
\\
dissimilar things
taken r at a time.
we
This is the same thing as finding the number of ways in which can fill up r places when we have n different things at our
disposal.
The things
first
may
may be tilled up in n ways, for any one of the n taken when it has been filled up in any one of
place
be
;
8—2
HIGHER ALGEBRA.
116
these ways, the second place can then be filled up in n - 1 ways ; and since each way of filling up the first place can be associated with each way of filling up the second, the number of ways in which the first two places can be filled up is given by the product n (n - 1). And when the first two places have been filled up in any way, the third place can be filled up in h — 2 ways. And reasoning as before, the number of ways in which three places can be filled up is n (n - 1) (n - 2).
Proceeding thus, and noticing that a with each new place filled up, and that of factors is the same as the number of have the number of ways in which r equal to
n (n- l)(n— and the r th factor
new
factor is introduced at any stage the number places filled up, we shall
places can be filled
to r factors
2)
or
1),
n — r+1.
Therefore the number of permutations of a time is
n{na time
The number
(n-
2)
(n-r +
of permutations of
n
things taken r at
1).
n things taken
all
at
is
or is
142.
to
n
factors,
3.2.1.
usual to denote this product by the symbol
read "factorial n."
of
1)
n (n - 1) (?i - 2) n(n — Y)(n—2) It
;
is
n — (r—
Cor.
up
We
Also n\
is
sometimes used for
shall in future denote the
n things taken
r at a time
by the symbol
"Pr = w(w-l)(w-2) "P =
also
number n
P
r
,
\n,
which
is
\n.
of permutations
so that
(n-r + 1);
\n.
In working numerical examples it is useful to notice that the n suffix in the symbol Pr always denotes the number of factors in the formula we are using. 143.
a time
The number
may
also be
of permutations of
n things taken
r at
found in the following manner.
Let "Pr represent the number of permutations of n things taken r at a time.
PERMUTATIONS AND COMBINATIONS.
117
Suppose we form all the permutations of n things t;iken r — at a time ' the number of these will be "Pr—l ;
1
.
each of these put one of the remaining n — r + 1 tilings. Each time we do this we shall get one permutation of u things r at a time; and therefore the whole number of the permutations n of n things r at a time is x (n - r + 1) that is, r_
With
P
;
]
By
r—l
writing
for r in this formula,
"P_ = '^ 1
x(n-r-f2),
_2
= 'Pr _a
'P
similarly,
r
we obtain
x (n
-r+
3),
"P^P.x (71 -I), "P
x
=7l.
Multiply together the vertical columns and cancel like factors from each side, and we obtain n
Example seats
;
in
The
P
= n(n-l)(n-2)
r
(n-r+l).
Four persons enter a railway carriage how many ways can they take their places ? 1.
may
in
which there are
six
and then the second person seat himself in 6 ways in 5 ; the third in 4 and the fourth in 3 and since each of these ways may be associated with each of the others, the required answer is 6x5x4x3, or 360. first
person
;
;
Example
;
How many different numbers
2.
of the nine digits
can be formed by using six out
1, 2, 3, ...9?
Here we have 9 different things and we have to find the number of permutations of them taken 6 at a time ;
.
*
.
the required result = 9 P6
=9x8x7x6x5x4 = 60480. To find taken r at a
144. tilings
the
number of combinations of n
dissimilar
time.
Let "Cr denote the required number of combinations.
Then each
of these combinations
dissimilar things which |r
ways.
[Art. 142.]
consists of
can be arranged among
a group of r themselves in
HIGHER ALGEBRA.
118
Hence "C r x things taken
\r
rata
number
equal to the
is
time
that
;
of arrangements of
n
is,
*C x\r = "Pr r
|
=n
(n — 1) (n
—
2)
.
.
.
(n - r + 1)
;
tt(w-l)(w-2)...(w-r+l)
_
V
|r
'"
n This formula for C r may also be written in a different form ; for if we multiply the numerator and the denominator by \n — r we obtain
Cor.
n (n -
1) (n
-
2) \r
The numerator now numbers from n to 1
...
{n
-r+
1) x
n—r \
n—r
consists of the product of all the natural
;
\n .'.
"C r =
~
.
(2).
It will be convenient to remember both these expressions for C r using (1) in all cases where a numerical result is required, and (2) when it is sufficient to leave it in an algebraical shaj)e. n
,
Note.
If in
formula
(2)
we put r = n, we have \n n
but n Cn =l, so that if the formula be considered as equivalent to 1.
i
~jn|_0" |0' is to
be true for r = n, the symbol 10 must
Example. From 12 books in how many ways can a selection of 5 be made, (1) when one specified book is always included, (2) when one specified book is always excluded ? Since the specified book is to be included in every (1) have only to choose 4 out of the remaining 11.
Hence the number
of
ways = n C4 1 1x10 x9x8 ~ 1x2x3x4
= 330.
selection,
we
;
PERMUTATIONS AND COMBINATIONS.
119
Since the specified book is always to be excluded, we have to select the 5 books out of the remaining 11. (2)
Hence the number
of
ways = n C6 _ 11x10x9x8x7
1x2x3x4x5 = 462. 145.
equal
to
The number of combinations of n things r at a time the number of combinations of\\ things n — r at a time.
is
In making all the possible combinations of n things, to each group of r things we select, there is left a corresponding group of
n-r
things
;
that is, the number of combinations of n things the same as the number of combinations of n things
r at a time is n — r at a time
.-.
The proposition may "0
"C r =
_r
C
n—
.
r
proved as follows
also be
=
n
\n =—
n—r n—
(n
- r)
:
[Art. 144.1
n n—
Such combinations are Note.
The
Put r=w, then result
tt
we have
r
r
called complementary.
C = n Cn =l. just proved
is
useful in enabling us to
abridge arithmetical work. Example.
Out
of 14
men
in
how many ways can an
eleven be chosen?
The required number = 14 C U
14 x 13 x 12
1x2x3 = 364. If
we had made
use of the formula
uC
n we should have had ,
expression whose numerator and denominator
to reduce each contained 11 factors.
au
—
.
HIGHER ALGEBRA.
120
number of ways in which m + n things can be divided into two groups containing in and n things respectively. This is clearly equivalent to finding the number of combinations of ra + n things ra at a time, for every time we select one group of ra things we leave a group of n things behind. Ira + n = number h= Thus the required 1 146.
Tojind
the
ra
Note. different
If
ways
\7b
n = m, the groups are equal, and in ~ —\2m —
of subdivision is -
[9
for in
;
any one way
to interchange the two groups without obtaining a
147.
this case the
new
To jind the number of ways in which
be divided into three
groups containing m,
of
possible
is
distribution.
m + n + p things can
p things
n,
it
number
severally.
m
First divide ra + n + p things into two groups containing and n + p things respectively the number of ways in which this :
\m +
can be done
is
-r=
\m
n+p n+p
Then the number of ways in which the group of n+p things can be divided into two groups containing n and p things respec-
\n+p tively
is
Hence the number groups containing m,
of
n,
n p ways in which the subdivision into three
p
things can be
m + n+p n+p
in
Note.
If
n+p x
,
,
made is \m + n + ]>
or
, 5
\n \p
we put ?i=p = m. we obtain
\n \p
Ira
J3wi
:
—r=-|—
;
hut this formula regards
as different all the possible orders in which ~th.e three groups can occur in any one mode of subdivision. And since there are 13 such orders cor-
responding to each mode of subdivision, the number of different ways in |3ot
which subdivision into three equal groups can be made
is
771
Example.
The number
of
ways in which 15
—
r^fm 771
r^ |3
•
recruits can be divided into
115
three equal groups is
,
-
--;
and the number
of
ways in which they
—
15
can be drafted into three different regiments,
I
five into each, is -_—Hr [6 5 [6 J
PERMUTATIONS AND COMBINATIONS.
121
In the examples which follow it is important to notice that the formula for 'permutations should not be used until the suitable selections required by the question have been made. 148.
Example
1.
be formed; in
From 7 Englishmen and 4 Americans a committee of is to how many ways can this be done, (1) when the committee con-
tains exactly 2 Americans, (2) at least 2 (1)
"We have to
Americans ? choose 2 Americans and 4 Englishmen.
The number of ways in which the Americans can be chosen is 4 C, and the number of ways in which the Englishmen can be chosen is 7 C4 Each of ;
.
the
first
groups can be associated with each of the second
the required
number
of
;
hence
ways = 4 C 2 x 7 C4 li
=
\1
X
|~2"[2
TTJ3
17
= 210.
'J^
|2|2|3 (2)
The committee may contain
2, 3,
or 4 Americans.
"We shall exhaust all the suitable combinations by forming all the groups containing 2 Americans and 4 Englishmen then 3 Americans and 3 Englishmen; and lastly 4 Americans and 2 Englishmen. ;
The sum of the three results will = *C2 x 7 C4 + 4 C3 x number of ways 17 [2
1
2
17
|4
+
X TTT^ [4
give the answer. 4 7 (7 C4 x 7 C, 3 +
j_3
Hence the required 17
TK X rl ^ + 1 X j_3 |3|4 [2)5
= 210 + 140 + 21 = 371. In this Example we have only to make use of the suitable formulae for combinations, for we are not concerned with the possible arrangements of the members of the committee among themselves.
Example
made each
2. Out of 7 consonants and 4 vowels, containing 3 consonants and 2 vowels?
The number number of ways
how many words can
be
of ways of choosing the three consonants is 7 C3 and the of choosing the 2 vowels is *Ca and since each of the first groups can be associated with each of the second, the number of combined groups, each containing 3 consonants and 2 vowels, is 7 C 3 x 4 C2 ,
;
.
Further, each of these groups contains 5 letters, which among themselves in [5 ways. Hence the required
number
of words = 7 C 3 x
4
~|3|4
C2 x
Jo
X
= 5x|7 r
= 25200.
X
[2]2
"
may
be arranged
HIGHER ALGEBRA.
122
How many
words can be formed out of the letters article, so places? the even that the vowels occupy
Example
3.
Here we have
nants in the 4 remaining places
and the second in the required
and the 4 consooperation can be done in 3 ways,
to put the 3 vowels in 3 specified places,
1
4
number
.
;
the
first
1
Hence
=|3x[4
of words
= 144. In this Example the formula for permutations is immediately applicable, because by the statement of the question there is but one way of choosing the vowels, and one way of choosing the consonants.
EXAMPLES
XI.
a.
In how many ways can a consonant and a vowel be chosen out of 1. the letters of the word courage?
There are 8 candidates for a Classical, 7 for a Mathematical, and In how many ways can the 4 for a Natural Science Scholarship. Scholarships be awarded? 2.
3.
Find the value of
4.
How many
8
P
25 7,
different
of the letters of the
P
24 5
,
<74
19 ,
CU
.
arrangements can be made by taking 5
word equation ?
If four times the number of permutations of n things 3 together is equal to five times the number of permutations of n — 1 things 3 together, find n. 5.
How many
permutations can be made out of the letters of the word triangle? How many of these will begin with t and end with e ? 6.
How many
different selections can be made by taking four of many different numbers can be formed the digits 3, 4, 7, 5, 8, 1 ? with four of these digits ? 7.
How
8.
9.
10.
If 2n C3
:
n Oj
= 44
:
3,
find n.
How many changes can be rung with a peal of 5 bells ? How many changes can be rung with a peal of 7 bells, the tenor
always being last
?
On how many nights may a watch of 4 men be drafted from a 11. crew of 24, so that no two watches are identical ? On how many of these would any one man be taken?
How many
arrangements can be made out of the w ord draught, the vowels never being separated ? 12.
r
letters of the
PERMUTATIONS AND COMBINATIONS.
1
In a town council there are 25 councillors and 10 aldermen be formed each consisting of 5 councillors
13.
;
how many committees can and 3 aldermen
?
Out
14.
of the letters A, B, C, p, q, r how many arrangements can (1) beginning with a capital, (2) beginning and ending with a
be made capital
23
]
15.
Find the number of combinations of 50 things 4G at a time.
16.
If n
C12 = n Cs
17.
In
how many ways can
arranged,
find n C17 ,
22
<7n .
the letters of the word vowels be the letters oe can only occupy odd places ]
From
18.
chosen
if
,
4 officers and 8 privates, in how many ways can 6 be to include exactly one officer, (2) to include at least one
(1)
officer?
In how 10 persons 19.
many ways can
a party of 4 or more be selected from
?
™Cr = ls Cr + 2
If
20.
find'<75
,
.
Out of 25 consonants and 5 vowels how many words can be 21. formed each consisting of 2 consonants and 3 vowels ? In a library there are 20 Latin and 6 Greek books; in how can a group of 5 consisting of 3 Latin and 2 Greek books be placed on a shelf ? 22.
many ways
In how
23.
persons
many ways
can 12 things be divided equally among 4
?
From 3 capitals, 5 consonants, and 4 vowels, how many words 24. can be made, each containing 3 consonants and 2 vowels, and beginning with a capital ? At an
election three districts are to be canvassed by 10, 15, and 20 men respectively. If 45 men volunteer, in how many ways can they be allotted to the different districts ? 25.
26.
In how
27.
A
many ways can
4 Latin and 1 English book be placed on a shelf so that the English book is always in the middle, the selection being made from 7 Latin and 3 English books?
boat
is
eight men, of whom 2 can only rowon stroke side; in how many ways can
manned by
to be
on bow side and 1 can only row the crew be arranged ?
There are two works each of 3 volumes, and two works each of 2 volumes in how many ways can the 10 books be placed on a shelf so that volumes of the same work are not separated ? 28.
;
29.
that the
In
how many w ays can 10 examination papers
befit
r
and worst papers never come together?
be arranged so
HIGHER ALGEBRA.
124
An eight-oared boat is to be manned by a crew chosen from 11 of whom 3 can steer but cannot row, and the rest can row but can-
30.
men, not
In how many ways can the crew be arranged, can only row on bow side?
steer.
men
if
two of the
Prove that the number of ways in which p positive and n negative signs may be placed in a row so that no two negative signs shall be together is p + 1 Cn 31.
.
32.
If
56
Pr +
54 6
:
Pr + = 30800 3
1,
:
find
r.
different signals can be made by hoisting 6 differently coloured flags one above the other, when any number of them may be hoisted at once ? 33.
How many
34.
U^C
2r
24 :
C2r _ 4 = 225
:
11, find
r.
Hitherto, in the formulae we have proved, the things have been regarded as unlike. Before considering cases in which some one or more sets of things may be like, it is necessary to point out exactly in what sense the words like and unlike are When we speak of things being dissimilar, different, unused. like, we imply that the things are visibly unlike, so as to be On the other hand we easily distinguishable from each other. shall always use the term like things to denote such as are alike For to the eye and cannot be distinguished from each other. consonants and the vowels may instance, in Ex. 2, Art. 1-48, the things united by a common be said each to consist of a group of characteristic, and thus in a certain sense to be of the same kind; but they cannot be regarded as like things, because there is an individuality existing among the things of each group which makes them easily distinguishable from each other. Hence, in the final stage of the example we considered each group to consist of five dissimilar things and therefore capable of [5 149.
arrangements among themselves.
[Art. 141 Cor.]
Suppose we have to find all the possible ways of arranging 12 books on a shelf, 5 of them being Latin, 4 English, and the remainder in different languages. 150.
The books
may
be regarded as belonging to one class, united by a common characteristic ; but if they were distinguishable from each other, the number of permutations would be )12, since for the purpose of arrangement among themin each language
selves they are essentially different.
'
PERMUTATIONS AND COMBINATIONS.
125
however, the books in the same language are not distinguishable from each other, we should have to find the number of ways in which 12 things can be arranged among themselves, when 5 of them are exactly alike of one kind, and 4 exactly alike, of a second kind a problem which is not directly included in any of the cases we have previously considered. If,
:
To find the number of ways in which n arranged among themselves, taking them all at a 151.
things time,
may
be
when p
of one kind, q of them exactly alike of another kind, r of them exactly alike of a third kind, and
of
the things are exactly
alike
the rest all different.
Let there be n letters suppose p of them to be to be b, r of them to be c, and the rest to be unlike. ;
a,
q of them
Let x be the required number of permutations ; then if in any one of these permutations the_p letters a were replaced by p unlike letters different from any of the rest, from this single permutation, without altering the position of any of the remaining letters, we could form p new permutations. Hence if this change were made in each of the x permutations we should obtain x x \p I
permutations. Similarly,
number
the
if
the q letters b were replaced by q unlike letters,
x In
we
like
would be
of permutations
x \p x
|<7.
manner, by replacing the r
should finally obtain x x \p x
\q x
letters c \r
by
r unlike letters,
permutations.
But the things are now all different, and therefore admit permutations among themselves. Hence x x
that
\p
x
x—
is,
\q
~.
x
—r
\p \g
which
is
Any
the required
number
r-
\n;
•
p
of permutations.
case in which the things are not
treated similarly.
of \n
all
different
may
be
HIGHER ALGEBRA.
126
permutations can be made out of the of the word assassination taken all together ?
Example letters
We
1.
How many
have here 13
different
letters of
which 4 are
s,
3 are a, 2 are
i,
and 2 are
n.
Hence the number of permutations
~|^[3|2j£
= 13.11.10.9.8.7.3.5 = 1001 x 10800 = 10810800. Example
How many
2.
1, 2, 3, 4, 3, 2, 1,
The odd
so
numbers can be formed with the that the odd digits always occupy the odd places?
digits 1, 3, 3, 1
digits
can be arranged in their four places in
way s
(1) -
l^2 The even
digits 2, 4, 2
can be arranged in their three places in 13
y^
Each
of the
ways
in
(1)
ways
(2).
can be associated with each of the ways in 14
Hence the required number = y^=x- x
-j^
= 6 x 3 = 18.
To find the number of permutations of n time, when each thing may be repeated once, twice, times in any arrangement. 152.
Here we have
to consider the
places can be filled disposal, each of the
(2).
13
number
of
things r at a
up
to
r
ways in which r
up when we have n different things at our n things being used as often as we please in
any arrangement. The first place may be filled up in n ways, and, when it has been filled up in any one way, the second place may also be filled up in n ways, since we are not precluded from using the same thing again. Therefore the number of ways in which the first The third place can two places can be filled up iswxn or n2 also be filled up in n ways, and therefore the first three places in .
n3 ways. Proceeding in this manner, and noticing that at any stage the index of n is always the same as the number of places filled up, we shall have the number of ways in which the r places can be filled
up equal to n r
.
PERMUTATIONS AND COMBINATIONS.
127
Example. In how many ways can 5 prizes be given away to 4 boys, when each boy is eligible for all the prizes?
Any one of the prizes can be given in 4 ways; and then any one of the; remaining prizes can also be given in 4 ways, since it may be obtained by the boy who has already received a prize. Thus two prizes can be given away in 4a ways, three prizes in 4 ways, and so on. Hence the 5 prizes can be given away in 4 5 or 1024 ways. :!
,
To find
153. to
make a Each
number of ways in which
the total
selection by taking
some or
all
of
it is
possible
things.
\\
may
be dealt with in two ways, for it may either be taken or left; and since either way of dealing with any one thing may be associated with either way of dealing with eacli one of the others, the number of selections is tiling
2x2x2x2 But
to
this includes the case in
n
which
therefore, rejecting this case, the total
This of
n
all
the things are
number
often spoken of as "the total
is
of
number
ways
is
left,
2"-l.
of combinations"
things.
A man
has 6 friends more of them to dinner? Example.
He ways
factors.
has to select some or
is
2s
-
1,
;
in
how many ways may he
all of his 6 friends
;
invite one or
and therefore the number of
or 63.
This result can be verified in the following manner.
The guests may be invited singly, in twos, threes, number of selections = 6 C1 + 6 C2 + 6 C3 + 6 C4 + 6 C5 + <>C6
;
therefore the
= 6 + 15 + 20 + 15 + 6 + 1 = 63. 154.
of
To find for what value of
n things Since
r at
"C
,
a time =
_
^( ?l -
1
- 2) 1.2.3
)( n
n(n-l)(n-2)
= nC
.
it
x
(w-r +
2.3
n—
r+
(w-r + 2)(n-r + (r-l)r
l)
2)
(r-1) 1
r
The multiplying which shews that
number of combinations
is greatest.
1.
"C
r the
factor
may be
decreases as r increases.
written
Hence
—-
-
1,
as r receives
—
;
HIGHER ALGEBRA.
128 the values
in succession,
3
1, 2,
n
G
r
continually increased
is
71 4- 1
until
1
becomes equal to
1
or less than
1.
r
Now
1^1, r
71+1
i
^
>z
so long as
r
that
——
is,
We
>
-
;
r.
have to choose the greatest value of r consistent with
this inequality. (1)
Let n be even, and equal to 2m; then
n+
-2
and
for all values of
Hence by putting combinations
is
—
r
"Cn
2m +1
1
2—
up to
?•
m= —
;
s
?n inclusive this is greater
we
,
1
+
find that the greatest
than
number
r.
of
.
2
(2)
Let n be odd, and equal to
— =-5— n+
and
2m +
up to
m
2
m+ 1 *C.= n Cm mi+I
-
» + li
: '
that
inclusive this
is,'
"C n++ 1
2
and therefore the number things are taken
same
in the
two
then
;
greater than r the multiplying factor becomes equal to 1, and
for all values of r
but when r -
2m +
1
1
—
—
,
or
n
is
C7i—l
•
J
2
of combinations is greatest
—^—
when the
at a time; the result being the
cases.
The formula for the number of combinations of n things r at a time may be found without assuming the formula for the numbes of permutations. 155.
Let "C r denote the number of combinations of n things taken r at a time; and let the n things be denoted by the letters a, b, c, d,
PERMUTATIONS AND COMBINATIONS.
remaining letters we cm form combinations of n— 1 letters taken r - 1 at a time. With of these write a; thus we see that of the combinations tilings r at a time, the number of those which contain
Take away a; then with "~
X
C
eaeli
of
a
n
w~ l
is n~
b is
120
x
C
C
,
x :
tin
1
the number of those and so for each of the n letters.
Tlierefore
which
similarly
\
n
number
x "~*Cr _ l is equal to the
contain
of combinations
r at a time which contain a, together with those that contain those that contain c, and so on.
b,
But by forming the combinations
in this manner, each parFor instance, if r=3, the ticular one will be repeated r times. combination abc will be found anions; those containing a, amonir those containing b, and among those containing c. Hence
*c= n - cr— x-. x
r
r
By
writing
u—
1
1 i
.,
and r — 1 instead
of
n and
ni
r respectively,
1
r-l°
-V^ = ^G _ x »-2
Similarly,
r
n— r + 2/~1 2
and
n - r+1
finally,
z
_n-r + \ri ^i
U—
T
+
o
>.
2
;
C = »-r + 1. 1
Multiply together the vertical columns and cancel like factors from each side thus ;
"C
n (rc-l)(n-2)
To find
156. to
.
make a
selection
r(r-l)(r-2)
(n-r+
l)
1
number of ways in which by taking some or all out qfip +
the total
it is
c
x
-possible
+r +
are alike of one kind, q alike of a second kind, r alike of a third kind; and so on. tilings, ivJierenf-p
The p things may be disposed of in p + 1 ways for wo may take 0, 1, 2, 3, p °f thorn. Similarly the q things may be disposed of in q + \ ways; the r things in r+1 ways; and ;
so on. H.
II.
A.
9
:
HIGHER ALGEBRA.
130
But taken
ways
;
may be
Avays in which all the tilings
Hence the number of disposed of is (^ + 1) (q +
1) (r
+
1)
this includes the case in therefore, rejecting this
which none case,
the
of the things are
total
number
of
is
(jp
+ l)fe+l)(r +
-1.
.l)
A
general formula expressing the number of permuta157. tions, or combinations, of n things taken r at a time, when the
things are not all different, may be somewhat complicated particular case may be solved in the following manner.
;
but a
Example. Find the number of ways in which (1) a selection, (2) an arrangement, of four letters can be made from the letters of the word proportion.
There are 10
letters of six different sorts,
may
In finding groups of four these
Three
(1)
p,
one
(3)
(4)
All four different.
alike, alike,
r, t, is
two others alike. the other two different.
selection can be made in 5 ways for each of the five letters, can be taken with the single group of the three like letters o. ;
The
selection can be of the three pairs o, o; p, p; (2)
made r, r.
in 3 C2 ways for we have to choose This" gives 3 selections. ;
This selection can be made in 3 x 10 ways 3 pairs, and then two from the remaining 5 letters. (3)
(1)
letters
;
one of the
select
This gives 30 selections.
total
number
of selections
is
5
+ 3 + 30 + 15
In finding the different arrangements of 4 letters possible ways each of the foregoing groups.
Thus
we
for
two out
This selection can be made in 6 C4 ways, as we have to take 4 different This gives 15 selections. to choose from the six o, p, r, t, i, n.
Thus the
all
r; t; i; n.
be classified as follows
The n,
r,
different.
Two Two
(2)
(1)
alike,
namely o, o,o; p,p;
=
gives rise to 5 x
(2)
gives rise to 3 x -^=^
,
(3)
gives rise to 30 x
or 360 arrangements.
(4)
gives rise to 15 x j4
number
is,
53.
we have
to
permute in
or 20 arrangements.
(1)
the total
that
;
,
-=,
,
or 18 arrangements.
or 3G0 arrangements.
of arrangements is 20
+ 18 + 360 + 360;
that
is,
758.
PERMUTATIONS AND COMBINATIONS.
EXAMPLES.
XI.
131
b.
Find the number of arrangements that can he made out of the of the words
1.
letters
independence,
(1)
superstitious,
institutions.
(3)
In how
(2)
many ways
can 17 billiard balls be arranged, them are black, 6 red, and 4 white % 2.
if
7 of
A
room is to be decorated with fourteen flags ; if 2 of them are 3. blue, 3 red, 2 white, 3 green, 2 yellow, and 2 purple, in how many ways can they be hung?
How many numbers
4.
the digits
2, 3, 0, 3, 4, 2,
greater than a million can be formed with
3?
Find the number of arrangements which can be made out of the letters of the word algebra, without altering the relative positions of vowels and consonants. 5.
On three different days a man has to drive to a railway station, 6. and he can choose from 5 conveyances in how many ways can he make ;
the three journeys
?
have counters of n different colours, red, white, blue, in I make an arrangement consisting of r counters, supposing that there are at least r of each different colour ? I
7.
;
how many ways can
In a steamer there are stalls for 12 animals, and there are 8. cows, horses, and calves (not less than 12 of each) ready to be shipped; in how many ways can the shipload be made?
how many ways can n things be given to p persons, when no restriction as to the number of things each may receive ?
In
9.
there
is
10.
persons 11.
many ways
In how
can
five
things be divided between two
?
How many different arrangements can be made out of tl ie letters
in the expression
a z b 2 c* when written at
full
length?
A
letter lock consists of three rings each marked with fifteen 12. different letters find in how many ways it is possible to make an unsuccessful attempt to open the lock. ;
13.
Find the number of triangles which can be formed by joining
three angular points of a quindecagon.
A
library has a copies of one book, b copies of each of two 14. books, c copies of each of three books, and single copies of d books. In how many ways can these books be distributed, if all are out at once I 15.
How many
eight digits
numbers
1, 2, 3, 0, 4, 5, 6,
7
less
than 10000 can be made with the
?
In how many ways can the following prizes be given away to a class of 20 boys: first and second Classical, first and second Mathematical, first Science, and first French ? 16.
9—2
HIGHER ALGEBRA.
132 17.
A telegraph has 5
arms and each arm is capable of 4 distinct what is the total number of
positions, including the position of rest signals that can be made ?
;
In how many ways can 7 persons form a ring? In how many ways can 7 Englishmen and 7 Americans sit down at a round table, no two Americans being together? 18.
possible to draw a sum of money from a bag containing a sovereign, a half-sovereign, a crown, a florin, a shilling, a penny, and a farthing? 19.
In how
many ways
is it
3 cocoa nuts, 4 apples, and 2 oranges, how tions of fruit can be made, taking at least one of each kind 20.
21.
n equal
many
From
Find the number of different ways of dividing
selec-
?
mn
things into
groups.
How many signals can be made by hoisting 4 flags of different 22. colours one above the other, when any number of them may be hoisted at once ? How many with 5 flags ? Find the number of permutations which 23. the letters of the word series taken three together ?
can be formed out of
There are p points in a plane, no three of which are in the same 24. straight line with the exception of q, which are all in the same straight line; find the number (1) of straight lines, (2) of triangles which result from joining them. points in space, no four of which are in the same plane with the exception of q, which are all in the same plane; find how many planes there are each containing three of the points. 25.
There are
p
There are n different books, and p copies of each; find the number of ways in which a selection can be made from them. 26.
Find the number of selections and of arrangements that can be made by taking 4 letters from the word expression. 27.
How many
28. letters of the 29.
permutations of 4 letters can be made out of the
word examination ?
Find the sum of
using the digits
1, 3, 5, 7, 9,
Find the sum of
numbers greater than 10000 formed by no digit being repeated in any number.
all
numbers greater than 10000 formed by using the digits 0, 2, 4, 6, 8, no digit being repeated in any number. If of p + q + r things p be alike, and q be alike, and the rest 31. different, shew that the total number of combinations is r (p + l)(q+l)2 -l. Shew that the number of permutations which can be formed 32. from 2n letters which are either a's or 6's is greatest when the number of a's is equal to the number of Z>'s. If the n -f 1 numbers a, b, c, d, 33. be all different, and each of them a prime number, prove that the number of different factors of the expression ambcd is (m + 1) 2 W — 1. 30.
all
CHAPTER
XIT.
Mathematical Induction.
Many
important mathematical formula? are not easily demonstrated by a direct mode of proof; in such cases we frequently find it convenient to employ a method of proof known as mathematical induction, which we shall now illustrate. 158.
Example of the first
1.
Suppose
it
is
n natural numbers
required to prove that the is
equal to
<
—^——
sum
of the cubes
.
'J-
We can easily see by trial that the statement is true in simple cases, such as when re=l, or 2, or 3 and from this we might be led to conjecture that the formula was true in all cases. Assume that it is true when n terms are taken that is, suppose ;
;
13
Add 13
the («+
+ 23 + 33 +
+ 2 3 + 33 +
l) th
to
term, that
to
itteims=|
(n+
is,
n + 1 terms =j
n
1)
3
^ +1
)j
to each side ^ |
2
3
H ( ;t+1
;
.
then
+(n+iy\-
= {n + iy-('j+n + l\ (n+l) 8 (na +4n+4) 4 \
-\
(n
+ l)(K + 2)
)\ '
2
!
which is of the same form as the result we assumed to be true for n terms, n + 1 taking the place of n in other words, if the result is true when we take a certain number of terms, whatever that number may be, it is true when we increase that number by one; but we see that it is true when 3 terms are taken therefore it is true when 4 terms are taken it is therefore true when Thus the result is true universally. 5 terms are taken; and so on. ;
;
;
x
:
HIGHER ALGEBRA.
134 Example
To determine
2.
the product of n binomial factors of the form
x + a.
By
actual multiplication (x
+ a)
(x+a)
(x
(x
+ b)
+ b)
(x
we have
+ c) = x 3 + (a + b + c) x 2 + (ab + bc + ca) x + abc
(x
+ c)
(x + d)
;
= x*+(a + b + c + d)x 3 + (ab + ac+ ad + bc+ bd + cd) x~ + (abc + abd + acd + bed) x + abed.
In these results we observe that the following laws hold
The number
1.
of terms
binomial factors on the
on the right
one more than the number of
is
left.
2. The index of x in the first term is the same as the number of binomial factors and in each of the other terms the index is one less than that of the preceding term. ;
The
the coefficient of the second the coefficient of the third term is the sum of the letters a, b, c, ; term is the sum of the products of these letters taken two at a time; the coefficient of the fourth term is the sum of their products taken three at a time and so on the last term is the product of all the letters. 3.
coefficient of the first
;
term
is
unity
;
;
these laws hold in the case of n - 1 factors that is, suppose ' ~ ~ (x + a) (x+b)... (x + h) = x71 1 +p 1 x n 2 +p.2 x n 3 +p.i x n i + ... +p> n
Assume that
;
^
p 1 = a + b + c+
where
...h;
= ab + ac + + ah + bc + bd+ p 3 = abc + abd+ p.2
,
...
;
;
p n_x = abc...h. Multiply both sides by another factor x + k (x
= x n + (p + k) x
Now
^i
p.2
x
n~l
+ a)
...
+ (p.2 +p x k)
thus
(x + h) (x + k)
n~* +
n (p 3 + pJc) x
+ A;:=(a + & + c + ..+/*) + & = sum of all the n letters
~3
+... +l^ n- x k.
.
a, b, c,...k;
+p k=p.2 + k (a + b + ... + h) 1
= sum n p.A
(x + b)
;
of the products taken two at a time of all the
letters a, b, c,
...
k;
+ ah + bc + .) +p.2 k =p 3 + k (ab + ac + = sum of the products taken three at a time of . . .
the n letters 2? n _ 1 A*
= product
a, b, c,
.
...
.
k;
of all the n letters a,
b,
c,
...
k.
all
MATHEMATICAL INDUCTION.
135
If therefore the laws hold when ?t-l factors are multiplied together they hold in the case of n factors. But we havo seen that they hold in the case of 4 factors; therefore they hold for 5 factors; therefore also for 6 factors and so on thus they hold universally. Therefore ;
;
[x + a) (x + b) {x
where
S^the sum
+c)
+ k) = x + ,V U_1 + Stfp-* + S.A x n ~* + 11
(x
...
of all the n letters a,
.
.
+ 8n
Js;
...
of the products taken two at a time of these n letters.
S n =the product
of all the n letters.
Theorems relating
159.
b, c
.
may
to divisibility
often be esta-
blished by induction.
Shew
Example. values of
By J if
u
-l
n
-l
.-c
x division
But x' fore x 1
by
divisible
is
x-1
for all positive integral
n.
therefore x n 4,
that
1 is
~l -
x-1
divisible by x -
-
z' l
by x -
^
l-1
-l
—
x-1
*;
also divisible by x - 1. therefore x - 1 is divisible by x - 1 ; there1 1, and so on ; hence the proposition is established.
1 is divisible
1 is divisible by^r
= xn ~
1,
then x* - 1
is
3
;
Other examples of the same kind Theory of Numbers.
will be
found in the chapter on the
From
the foregoing examples it will be seen that the only theorems to which induction can be applied are those which admit of successive cases corresponding to the order of the natural numbers 1, 2, 3, n. 1G0.
EXAMPLES. Prove by Induction
:
+ (2n-l) = n 2
1.
1+3 + 5+
2.
l2
3.
2
4.
+ o~q + q-~T + T~o 1.22.33.4
5.
Prove by Induction that
even.
+ 2 2 + 32 +
+ 22 + 23 +
XII.
.
+ n 2 =i?i(n+l)(2tt+l). + 2» = 2(2'
l
-l).
ton terms = .r
n
— yn
is
—
-^ n+1
.
divisible
by
x+y when
n
is
CHAPTER Binomial Theorem. It
161.
may be shewn by
XIII.
Positive Integral Index.
actual multiplication that
+ a) (x + b) (x + c) {x + d) = x4 + (a + b + c + d) x3 + (ab + ac + ad + + (abc + abd + acd + bed) x + abed (x
bc
+ bd + cd) x* (1).
We
may, however, write down this result by inspection ; for the complete product consists of the sum of a number "of partial products each of which is formed by multiplying together four If we letters, one being taken from each of the four factors. examine the way in which the various partial products are formed, we see that (1)
the term x
4
is
formed by taking the
letter
x out
of each
of the factors.
out
x3 are formed by taking the letter x every way possible, and one of the d out of the remaining factor.
(2) the terms involving of any three factors, in
letters a,
6, c,
the terms involving x 2 are formed by taking the letter x out of any two factors, in every way possible, and two of the letters a, b, c, d out of the remaining factors. (3)
the terms involving x are formed by taking the letter x out of any one factor, and three of the letters a, b, c, d out of the remaining factors. (4)
(5) «, b,
c,
the term independent of x d.
is
the product of
all
the letters
.
+ 3) (x - 5) (x + 9) = x 4 + (- 2 + 3 - 5 + 9) z 3 + (- 6 + 10 -18 -15 + 27 -45) a 2 + (30 - 54 + 90 - 135) x + 270 = x4 + 5a; 3 - 47.
Example
1.
(x
-
2) (x
BINOMIAL THEOREM. Example
POSITIVE INTEGRAL INDEX.
137
Find the coefficient of x* in the product (x - 3) (* + 5) [x - 1) (x + 2) (x - 8).
2.
The terms involving x* are formed by multiplying together the x in any three of the factors, and two of the numerical quantities out of the two remaining factors
;
of the quantities
hence the coefficient is equal to the sum of the products - 3, 5, -1,2, - 8 taken two at a time.
Thus the required
coefficient
= -15 + 3- G + 2-1- 5 + 10-40- 2 + 8- 10 = -39. 1G2.
If in equation (1) of the preceding article
b=c=d=a, we
we suppose
obtain
(x
+
a)
4
= x4 + iax* + 6a
V + 4a
3
as
+ a4
.
The method here exemplified of deducing a particular case from a more general result is one of frequent occurrence in Mathematics for it often happens that it is more easy to prove ;
a general proposition than
We
it is
to prove a particular case of
it.
next article employ the same method to prove a formula known as the Binomial Theorem, by which any binomial of the form x + a can be raised to any assigned positive integral power. 163.
shall in the
To find
the
n expansion of (x + a) ivhen n
is
a positive
integer.
Consider the expression (x
the
number
+ a)
(x
+ b)
(x
+
c)
(x
+
k),
of factors being n.
The expansion of this expression is the continued product of the n factors, x + a, x + b, x + c, x + k, and every term in the expansion is, of n dimensions, being a product formed by multiplying together n letters, one taken from each of these n factors. The highest power of x is x n and is formed by taking the letter x from each of the n factors. ~ The terms involving xn are formed by taking the letter x from any n—\ of the factors, and one of the letters a, b, c, ... k ,
l
~
n in the from the remaining factor thus the coefficient of x final product *is the sum of the letters a, b, c, k; denote it 1
;
by^. ~
The terms involving x n 2 are formed by taking the letter x from any n — 2 of. the factors, and two of the letters a, b, c, ... k n from the two remaining factors thus the coefficient of x ~ in the final product is the sum of the products of the letters a, b, c, ... k taken two at a time; denote it by S2 ;
.
\
;
1
HIGHER ALGEBRA.
138
n~r
are formed by taking generally, the terms involving x — the letter x from any n r of the factors, and r of the letters a, b, c, ... k from the r remaining factors ; thus the coefficient of
And,
r x"~ in the final product is the sum of the products of the letters a, b, c, ...k taken r at a time; denote it by Sr .
The
last
term in the product
Hence
is
abc
...
k; denote it
by
(x + k) + a)(x + b)(x + c) + SjxT* + ••• + £rx"~ + ...+Sn— ,x + Sn
12 (x
n~l
= x n + Sx
Sn
.
r
In $j the number of terms
n
is
in
;
S2
.
number of terms
the
is
the same as the number of combinations of n things 2 at a time n n that is, C2 ; in S3 the number of terms is C3 ; and so on.
Now "Ca:
suppose
b,
l
S, becomes "C\a
(x
+
a)
n
each equal to a; then S becomes becomes "Cjf: and so on: thus
k,
...
c, 2
S
:
= xn + n C ax n
-
+
1
n
n
x»
«-il
»
n n (x+a) = x"+nax
we
...
2,
n(n—l)
n+
~2
+ "C^aV" 3 +
.
.
.
+ "Ca"
;
obtain
- —r-^oV
+
the series containing
„_„J
„
+
—
— \)(n—2) n(n /v v l 1
„
n
a 3x n
„
3
+...
+ an
,
terms.
1
the Binomial Theorem, and the expression on the right said to be the expansion of (x + a)*.
This
is
C
2
2
l
substituting for *Clt
C a xn
is
The Binomial Theorem may
164.
By
we can
also be proved as follows
:
the product of the n factors x + a, x + b, x + c, ...x + k as explained in Art. 158, Ex. 2; we n can then deduce the expansion of (x + a) as in Art. 163. induction
The
165.
find
in the expansion of (x
coefficients
+
a)" are n
very n
C3 ... C n conveniently expressed by the symbols "C,, "C 2 shall, however, sometimes further abbreviate them by omitting With this notation we have n, and writing (7,, C 2 C 3 ... C n ,
,
.
We
,
(x If
(x
+ a) n =
we
x"
+ C ax
—a
write
1
l
x
+ C 2 a2 x n
~2
~
l
-l
+
C
2
2 n x + C„a 2
(-a) 2x
-2
+ C3 a 3x n
we
in the place of a,
n -a) n = x" + C\(- a) x
= xn - C,axn
.
,
n~
n-2
~3
+
+ Ca\
...
obtain
+C
3
3 n ~3 - C,a x + 3
3 (-a) 3 xn - +... + Cn (-a)
...
n
+ (- IYG na\ \
/
n Thus the terms in the expansion of (x + a) and (x — a) are numerically the same, but in (x - a)' they are alternately positive and negative, and the last term is positive or negative according as n is even or odd. n
1
BINOMIAL THEOREM. Example
By
1.
POSITIVE INTEGRAL INDEX.
Find the expansion of
.
+ yf = x« + 8tfi xhj + 6 C^xY + C3 afy8 + B C 4 .r 2 4 + 8 <7e x,f + »Ce ,/' = z? + 6.1-V + loo; 4 2 + 20a; 3 3 + loxhf + Gxi/> + if', fi
*/
?/
on calculating the values of 6 C1
Example -
2.
2x) 7
?/
G ,
C2
6 ,
C3
Find the expansion of
= a7 - 7C
X
a c (2x) '
,
-
(a
2.r) 7 .
+ 7 C2 a 5 (2a;) 2 - 7 C3 a 4 (2a;) 3 +
Now remembering that n Cr = n Cn _ r after calculating C 3 the rest may be written down at once; for 7 GX = 7 C^ ,
7
39
the formula, {x
[a
+ y) 6
{x
1
,
to 8 terms.
the coefficients up to Cr — 7 C
7
Hence (a
- 2x) 7 = a7 - 7a B
{2x)
+ jp| a 5 (2xf -
\^-\ «4 (2a;)
3
+
= a7 - la 6 (2x) + 21a 5 (2a;) 2 - 35a 4 {2xf + 35a 3 (2a-) 4 - 21a 2
= a 7 - Ua 6x + 84a5 Example
3.
a;
2
(2a;)
5
+ la
(2s) 6 - (2.r)~
- 280a 4 .r 3 + 560a 3a; 4 - 672aV + USaafi -
128a; 7 .
Find the value of (a
+ Jtf^ly + (a- Ja- -
1)".
We have here the sum of two expansions whose terms are numerically the same ; but in the second expansion the second, fourth, sixth, and eighth terms are negative, and therefore destroy the corresponding terms of the first expansion. Hence the value = 2 {a7 + 21a 5 (a2 - 1) + 35a 3 (a 2 - l) 2 + la (a 2 = 2a (64a 6 - 112a 4 + 56a 2 - 7
l) 3 }
).
In the expansion
+
n
the coefficient of the second term is C of the third term is G 2 ; of the fourth term is "C3 ; and so on ; the suffix in each term being one less than the number of the term to which it applies hence "C r is the coth This is called the general term, terin. efficient of the (r + l) because by giving to r different numerical values any of the n coefficients may be found from C r ; and by giving to x and a their appropriate indices any assigned term may be obtained. Thus the (r + l) th term may be written 166.
of (x
a)
,
n
n
;
l
;
Cjrw,
or
»(»-l)("-2)-(»
—
+»,,-,„,.
t In applying this formula to any particular case, it should 1><> observed that the index of a is the same as the svffix of C, and that the sum of the indices ofx and a is n.
—
;
.
.
HIGHER ALGEBRA.
140 Example
1.
Find the
fifth
term of
+ 2a; 3 ) 17
(a
= 17 C4 a13 (2a; 3
The required term
4 )
17.16.15.14
1.2.3.4
= 38080a 13 x 12 Example
2.
.
xl6ft13 .T 12
.
Find the fourteenth term of
(3
-
a) 15 .
= 15 C13 (3) 2 - a) 13 = 15 C2 x(-9a13 = - 945a 13
The required term
(
[Art. 145.]
)
.
The simplest form of the binomial theorem is the expansion of (l+x) n This is obtained from the general formula of Art. 163, by writing 1 in the place of x, and x in the place of a. Thus 167.
.
+
(1
x)
n
= l+ H C x + "C2 x 2 + n(n-\) 1 ^ 1
— 1.2
+ nx +
+ "C x + ..+ "Cx n r
.
.
.
i
2
r
4- r" +
+
zr—i ar
~
'
}
the general term being
(n-r+
n(n—l)(n—2)
1)
,
tb
.
The expansion of a binomial may always be made upon the case in which the first term is unity thus
to
depend
;
{x
+y
yJ(X (i + l)J = xn (l + z) n where ,
Example
1.
Find the
coefficient of
We have
(a;
2
-
2a;)
a;
10
16
=
z
= Vx
in the expansion of
a;
20
(1 - -
a;
20
-
multiplies every term in the expansion of
(
1 -
Hence the required
coefficient
= 10 C4 10
.
(
-
,
9
.
term which contains
2)
8
.
4
7
1.2.3.4
xl6
= 3360. In some cases the following method
is
10 .
2\ 10 we have in J
this expansion to seek the coefficient of the
2a;)
V /
and, since
2
(as
simpler.
—
BINOMIAL THEOEEM. Example
Find the
2.
POSITIVE INTEGRAL INDEX.
coefficient of
x r in the expansion of
Suppose that x r occurs in
tlio
The
= *CP (x-) n -i> (iY
(p
+
term
1)°'
(p
+ l) ,h
./•-
(
141 -
i
j
.
term.
= n Cp x"' -*". 1
But
this
term contains x r and therefore ,
2n-5p = r,
2*1
or
p=-
—
r
5
Thus
the required eoellicient = n CiJ = n Co, l _,. 5
n
g(2n-r)
Unless
2n — — — -
= (3n
+
?•)
i'
is
a positive integer there will be no term containing x r in
the expansion.
In Art. 163 we deduced the expansion of (x + «)" from the product of n factors (x + a) (x + b) ... (x + k), and the method of proof there given is valuable in consequence of the wide gene1G8.
But the following shorter proof
rality of the results obtained.
of
the Binomial Theorem should be noticed. It will be seen in Chap. xv. that a similar to obtain the general
term
To prove
161).
the
is
used
of the expansion of
+
(a
method
+ c+
b
)".
Binomial Theorem.
The expansion of (x + a)' is the product of n factors, each equal to x + a, and every term in the expansion is of n dimensions, being a product formed by multiplying together n letters, Thus each term involving one taken from each of the n factors. r x"~ a is obtained by taking a out of any r of the factors, and x Therefore the number of out of the remaining n — r factors. terms which involve x"~ a must be equal to the number of ways that is, the coellicient in which r things can be selected out of n n~ "6' and by giving to r the values 0, 1, 2, 3, ... n in cC is of x Hence succession we obtain the coefficients of all the terms. 1
r
r
r
;
r
r
(x
since
+
a)
n
=x
,
l
4-
m
C X*- a + 1
n
n
C,,x -°-a
J
*C and "C n are each equal
2
+
.
.
.
to unity.
+
n
C x"-a + r
r
...+
a",
:
HIGHER ALGEBRA.
142
EXAMPLES. Expand the (#-3) 5
4.
(l-3a 2 ) 6
.
.
2.
(3^ + 2y) 4
5.
{a?+x)\
n
g*-js)'-
13.
The 4th term
15.
The 12 th term
The
of
18.
The
5
19.
The
V th
-
13 .
- 1) 13
(a \ U + 96 J /
term of (2a -
term of (^' -
.
The
5 th term of
(
{Zx-yf.
6.
(1-^j/) 7
12 -
.
NT
14.
The 10 th term
of (1 - 2x) 12
16.
The 28 th term
of (5x + 8y) 30
.
.
10 .
b\ 8 .
-J
^-Y 5.
20.
3.
:
5)
of (2#
4 th term of
th
(a?
.
(H'-
-
Write clown and simplify
17.
a.
following binomials
1.
">•
XIII.
.
8
— - V-
%
x
Find the value of 22.
(V^^+^-CV^3^-^)
24.
(2-Vr^) 6 + (2 + v I^^) 6
21.
(x + s/2y +
23.
(
25.
Find the middle term of
f
26.
Find the middle term of
(
27.
Find the
coefficient of
a.
28.
Find the
coefficient of
x 18
29.
Find the
coefficients of
30.
Find the two middle terms of 3a -
v/2
(x-j2)\
+ l) 6 -( N/2-l) 6
.
/
cV\ 10
a
-+\x a
^»
1
- "—
18 in L'V2
.
j
+—
in (axA
-
/
.
bx) 9
x 32 and #~ 17
(
]
.
in
1\ 15 ( A x - -g
(
J
a3\ 9
—
)
.
.
5 -
BINOMIAL THEOREM.
POSITIVE INTEGRAL INDEX.
31.
Find the term independent of x in
32.
Find the 13th term of
33.
If x* occurs in the expansion of
34.
Find the term independent uf
35.
If
Ux -
\-\
lx+-\
....
,
eihcient
in
b3
.
)
.
f
find its coefficient.
,
1 \
/ ;/;
—
-x2 —
(
1
x— -.,
1\'-' / xp occurs in the expansion of ( xr+-
3»i .
j
1
I
prove that
,
its co-
\2n -
.
is
.
1
(4
"-^ \@n+p)
j3
170. In the expansion of (1 4- x) u the coefficients of terms equidistant from the beginning and end are equal.
The
coefficient of the
(r
+
l)
th
term from the beginning
is
"C..
Tlie
(r+l) th term from the end has n + 1— (r+1), or
n-r
therefore counting from the beginning it is the (n — r + l) term, and its coefficient is "Cn _ r which has been Hence the proposition shewn to be equal to "Cr [Art. 145.]
terms before
it;
th
,
.
follows.
171. (l + x)»
The
To find
the
greatest
coefficient of the general
term of
have only to find for what value of r
By
Art. 154,
when n
in
coefficient
(1
the
+x)"
expansion of
is
m
C
r
j
this is greatest.
even, the greatest coefficient
is
and we is
"C n
;
i
and when n
is
odd,
it
is
"C
,,
or "C
2
,
,
;
these
two
coefficients
2
being equal. 172.
We
To find have
therefore, since
the greatest
(x
+
a)"
term in the expansion of (x +
= x" (l +
-Y
n x multiplies every term
sufficient to find the greatest
a)".
;
in
( 1
+ -j
,
it
term in this latter expansion.
will
be
;
HIGHER ALGEBRA.
144 Let
the r th
The (r+l) th .
—
:
x
r
The
and (r+l) th be any two consecutive terms. term is obtained by multiplying the r th term by
that
is,
by
1
(
r
\ Vh
+
factor
]
—
-
)
decreases
1
[Art. 166.1J L
.
x
J
_
increases
r
as
hence the
;
r
(r+l) th term until
is 1
(
\
r
not always greater than the rth term, but only )
J
- becomes equal x
to
—r
-
/n +
Now
\
n+
,
so long as
1 1
a
J x 1
.,
1
>
n+
1
is,
r
—
—
or
x a
>-+
1.
1•j
x
>-
a
..
that
\
1
or less than
1,
;
1,
>r
( 1 ).
a If
—— -+ a
be an integer, denote
it
by
j>
j
then
——
71+1
- +
r
—
])
the
1
multiplying factor becomes 1, and the (p + l) th term th and these are greater than any other term. /> ;
If
if
be not an integer, denote
its
is
equal to the
integral part
by q
1
a
then the greatest value of r consistent with (1) th term is the greatest. (q + 1 )
is
q\ hence the
Since we are only concerned with the numerically greatest term, the investigation will be the same for (x-a)"; therefore in any numerical example it is unnecessary to consider the sign Also it will be found best of the second term of the binomial. to work each example independently of the general formula.
-
BINOMIAL THEOREM. Example
If
1.
Denote the
?•"'
POSITIVE INTEGRAL INDEX.
x = -, find the greatest term in the expansion
and
(/-
+ l) tu terms by Tr and Tr± 1
9-r
l
9-r
.
so long ° as
(1+
to
I
respectively; then
,
4
->1;
x
r
3
36 - 4r > 3r,
is
3G>7r.
or
The is
of
1
4
T7^. > Tr
hence
that
;
greatest value of r consistent with this
the sixth, and
its
is
5
;
hence the greatest term
value
3i4 243~
Example
2.
Find the greatest term in the expansion of (3- 2a:)9 when
(3-
2
2^ = 3^1
|J;
(2rV -— 1
.
J
„ Here
„ -*r+i
= 9-r+l
—
10 - r
^r,.
3
10 ~ r so ilong as
Tr
...
,
numerically,
;
,
2
x ->
i
1
6
r
20>5r.
is,
Hence
x
2
X
Tr+1 > Tr
iience
that
2* -~o~
for all values of r
up to
T r+x = T r> and
3,
we have Tr+l >Tr
;
but
if
r=4, then
these are the greatest terms. Thus the 4"' and 5 th terms are numerically equal and greater than any other term, and their value
=3"x»C,x f| H. H. A.
J
=36 x 84x8 =489888. 10
HIGHER ALGEBRA.
146 173. To find of (I +x)".
In the identity put x =
1
sum of
the
(1
+ a?)n =
+ C x + G 2 x 2 + C3 x 3 +
1
expansion
in the
the coefficients
. . .
x
+C
af, t
thus
;
2*=l + Cx + C2 + C3 +... + Cn = sum of the coefficients. +C +C + 12
Cor.
C\
that is "the total [Art. 153.]
174.
...
q 3
number
To prove
the coefficients
g
Cn =T-l;
+
'
of combinations of
n things"
+
that in the expansion of (1
of the odd terms
equal
is
to the
x)
n
is 2"
sum of the
1.
sum of
the
,
—
coefficients
of the even terms.
In the identity put x = - 1 thus
(
1
+
x)
n
=
1
+ C x + C2x2 + C 3x3 +
...
x
+ C x\
;
= l-C + a8 -(7a + (74 -C6 +
;
1
...
i
+ 1+ c4 +
;.....
-0
1
1
=
+'Ca + C.+
(sum of
-
all
the coefficients)
—
=
2
n-l
175. The Binomial Theorem may also be applied to expressions which contain more than two terms.
expand
:
Example.
Find the expansion of
Regarding 2x -
+ 2x- l) 3
(x z
1 as a single term, the
expansion (2x - l) 2 + (2x -
= (x-) + 3 (a (2x - 1) + 3a; = x 6 + 6a; 3 + 9a; 4 - 4c 3 - 9a; 2 + 6x — l, 3
2 2
2
)
176.
The following example
Example.
If
find the value of
and
The
series (l)
is
.
3
l)
on reduction.
instructive.
+ x) n = c + c x x + c#? + +c n xn c + 2c 2 + 3c 2 + 4c 3 + + n +l)cn 2 2 c 1 + 2c 2 + 3c 32 + +nc n2
(1
,
(1),
(
= {c + c + c 2 +
+ c n + (c x + 2c 2 + 3c3 + )
1
=2 w +
?i
Jl
+ (/t-l) +
= 2 n + n(l + l) n ~ -2n +w.2»-1 .
1
v
J_L
/
+
(?).
+nc n
+ il
)
4
— BINOMIAL THEOREM. To
(2),
+ 3c 3 x* +
+ ncn x n 8>
*» +
+ *»-}
;
x
,
we have 1
&+£+!*+ X X' X s
cQ
IT
1
hence, hv chauging x into -
Also
1
we proceed thus
=»«{l-Hn-l),+ <"-^= nx (1 + x)"-
:
POSITIVE INTEGRAL INDEX.
find the value of the series c x x + 2c2 x 2
;
+ ^=!(i + i)*n
(s) .
+ en z% =(l+z) n
(4).
X \
X
+ c 1 x + c.2 x 2 +
W
xj
we multiply together the two
series on the left-hand sides of (3) and (4), see that in the product the term independent of x is the series (2) hence n l\ u_1 the series (2) = term independent of x in - (1 + x) n (f 1 + If
we
;
•
I
— )l
term independent of x in
= coefficient
= ?ix 2n -
of x n in
n
(l
+ x)-'
1
'1
+ .r) 2>l-l
(1
1 <
i2n-l
n-1
In-
1
EXAMPLES.
XIII.
In the following expansions find which
3.
when #=11, y = - 3y) 28 when x = 9, y = 4. ( 2x (2a + b) u when a =4, 6 = 5.
4.
(3
1.
2.
(x
y)
is
b.
the greatest term
30
5
+ 2x) lb when x — -
.
ss
In the following expansions find the value of the greatest term 2
5.
6.
(1
+ x) n when x = -
(«
+ #)* when a= s
o
,
,
n = 6.
.r
= -,
??=9.
10—2
:
HIGHER ALGEBRA.
148 7.
middle term of (l + x) 2n is of the coefficients of the two middle terms of coefficient of the
Shew that the
sum
equal to the "- 1 (1+tf) 2 .
be the sum of the odd terms and n even terms in the expansion of (x + a) prove that
A
If
8.
,
10.
Find the expansion of
11.
Find the expansion of
12.
Find the
+ 2x - x )\ (Zx 2 -2ax + 3a
sum
)
.
.
2n +
l
In the expansion of (1 + #) 43 the coefficients of the (2r + th the (r + 2) terms are equal; find r.
Find the relation between r and n in order that the 2n may th th be equal. of the 3r and (r + 2) terms of (l+x) 15.
Shew that the middle term
in the expansion of (1
.3.5...(2n-l)
sn
hi
If c
,
Cj, c 2 , ...
c,
18.
io iy.
+ +i- 4-
c
c.2c
—
i
o
M 21.
a 2c
(
,
H +
(
2 2c,
ncn
cn _
3
,
)
(c n
+
1
4
+ c« =
cn
c,c, 1 2
= --
+
?
)
O A
. 1 + cn )
—+
.
n{n+\
x
2 4c,
+ CjC r + + c 2cr + 2 + j
—
.
N
2 3c2
f+c +c2 + 1
n+
.
= 2 -l — n+l n+\ cn
+~
r
Cl + c 2
+ —2 i+
c cr
.
C2
cx
(co+ej
c
3c8 1
20.
.
3
2
v
23.
c, -2 4-
2
c
22.
+ncn =n.2n - 1
^ + 2^ + 303 +
1
(n+l)'-H
^
—n +-p^
2n +
cn l
3' 1 +
1
-1
-=-..
n+l
|2w i
7i
]^
th
and
+x) 2n
is
^
prove that 17.
l)
coefficients
denote the coefficients in the expansion of
n
.
240, 720,
14.
1
n )
2 3
(]\ x--j
16.
of the
2
(1
term from the end in (x + a) n
r th
the
(x+y) n are
th 2 nd, 3 rd , 4 terms in the expansion of 1080 respectively ; find x, y, n.
The
9.
B
A 2 -B2 = (x2 -a2
•
+cn _ r cn = --
\2n =~~
.
(1
+x) n
,
CHAPTER
XIV.
Binomial Theorem.
Any
Index.
chapter we investigated the Binomial Theorem when the index was any positive integer; we shall now consider whether the formula? there obtained hold in the case of negative and fractional values of the index. 177.
In the
Since,
by Art. 167, every binomial may be reduced to one
common
type,
last
will be sufficient to confine our
it
binomials of the form
By
(1
actual evolution,
(1
.
1
+ ^ X - - X2 +
yr.
x3 -
;
division,
- x)~ 2 =
-
7^
-
xa
=
1
+ 2x + 3x* + ix 3 +
:
[Compare Ex. and
to
we have
+ xf = V 1 + X =
and by actual (1
+x)
attention
n
number
in each of these series the
of terms
is
1,
Art. CO.]
unlimited.
In these cases we have by independent processes obtained an i
expansion for each of the expressions (1 + x) and (1 + x)~~. We shall presently prove that they are only particular cases of the n general formula for the expansion of (1 + x) where it is any 2
,
rational quantity.
This formula was discovered by Newton.
Suppose we have two expressions arranged powers of x, such as 178.
,
I
and
+
mx+
l+n.v +
m (m v
-
g l
1 )
„
'x-+
m x(m -
/x(m
1 )
K J -
>x-+-± ]
-
2)
, 'a?+
a?
+
in ascending
(I
).
(2).
—
xx
;
HIGHER ALGEBRA.
150
The product
two expressions cending powers of x\ denote it by 1+ Ax +Bx2 + Cx3 + Dx4 + of these
will be a series in as-
;
m
are functions of and n, in any particular and therefore the actual values of A, B, C, and n in that case. But case will depend upon the values of the way in which the coefficients of the powers of a; in (1) and (2) combine to give A, B, C, is quite independent of and n ; in other words, whatever values in and n may have, A, B, C, If therefore we can determine preserve the same invariable form. the form of A, B, C, for any value of and n, we conclude that A, B, C, will have the same form for all values of
then
it
is
clear that A, B, C,
m
m
m
m
and
n.
The
principle here explained
We
shall
often referred to as an example ; of "the permanence of equivalent forms " in the present case we have only to recognise the fact that in any algebraical product the form of the result will be the same whether the quantities involved are whole numbers, or fractions ; positive, or negative. is
make
use of this principle in the general proof of The proof which Ave the Binomial Theorem for any index. give is due to Euler.
To prove
179.
the
Binomial Theorem ivhen
the
index
is
a
positive fraction.
Wliatever be the value of m, positive or negative, integral or fractional, let the symbol f(m) stand for the series ,
+ mx +
1
m (m ^- 1)-x- — m (m-Y) (m — 2)
—y—
„
+
v
' v
s 3
'
+
...
then.y(n) will stand for the series -
1
+ nx +
n(n — l)„2 \
'
x +
— n(n — l)(n ' v v
2) '-
„ 3
+
....
we
multiply these two series together the product will be another series in ascending powers of x, whose coefficients loill be and n may be. unaltered inform whatever If
m
To determine this invariable form of the product we may give to m and n any values that are most convenient for this purpose suppose that m and n are positive integers. In this casey(m) m andy*(?i) is the expanded form of is the expanded form of (1 + x) n and therefore (1 +x) ;
,
;
BINOMIAL THEOREM. f(m) xf(n) but when
m
and n are
+
x)
m
x (1
+ a?)" =
(m +
v
n)
+ x) m+ \
(1
(m + n I
This then
1.51
positive integers the expansion of (1
/
-,
(1
ANY INDEX.
.
1)
+ x)"
,
+
"
.
—
the form of the product of f(m) x/(><) in o# cases, whatever tlie values of and n may be; and in agreement with our previous notation it may be denoted hyf(m + n) ; therefore for all values ofm and n is
m
/(m) xf(n)=f(m + n). /(w) x/(n) x/(^) =/(w +
Also
x/( p)
»)
=f(m + n +p), Proceeding in
f(m) xf(n)
tliis
way we may shew
x/(j;)...to k factors
that
=/(»» + n +p
Let each of these quantities m,
?i,
similarly.
+...to k terms).
be equal to =
j),
,
rC
where h and k are positive integers
but since h
but
y* (
y
)
is
a positive integer,
;
f (h) = (1 + x)
;
stands for the series
,, .*.
(
1
+
k\k
h
,
vi a;)
=
,
1
+
h T
«
x+
J
2
k \k x
J /
,
x-
+
,
1.2
which proves the Binomial Theorem index.
h
for
any positive fractional
;
HIGHER ALGEBRA.
152 180.
To prove
the
Binomial Theorem when
the
index
any
is
negative quantity. It has
been proved that
f(m) x/(w) =/(w* + n) for all values of positive),
m
and
Replacing
n.
in
by — n
(wliere
n
is
we have f(-n) •xf(n)=f(-n + 7i) =/(0)
=% since all terms of the series except the first vanish
/( - n)
•'•
but/(w) = (l +
x)'\ for
or
1
(1
+
for the series
+ (- n) x +
«.)-
=
1
of n;
+ *)"" =/(-*)•
^
+ (_ W )
= L
'-f—, 1
...
'
any positive value
(1
But f(—n) stands
/hr
a.
.
ar
+
<" " " + (rg )
;
*>
g»
+
;
which proves the Binomial Theorem for any negative index. Hence the theorem is completely established. 181.
The proof contained
in the
two preceding
articles
may
not appear wholly satisfactory, and will probably present some ficulties to the student. There is only one point to which shall
now
dif-
we
refer.
In the expression iov f(in) the number of terms
is finite
when
a positive integer, and unlimited in all other cases. See Art. 182. It is therefore necessary to enquire in what sense we vi is
BINOMIAL T11EOUEM.
ANY INDEX.
153
are to regard the statement thaty(m) x/(n) =f(m + n). It a\ ill be seen in Chapter xxi., that when x< 1, each of the series/^/), /( n )i/( + n ) * s convergent, and/(m + «) is the true arithmetical equivalent of f(m) *f(n). But when sol, all these series are divergent, and we can only assert that if we multiply the series denoted by/(m) by the series denoted by f(u), the first r terms of the product will agree with the first r terms of f(m + n), whatever finite value r may have. [8ee Art. 308.]
m
3
Example
Expand
1.
2.
- xf2
to four terms.
,.J(H(S-)
Id- 1 ),
3
Example
(1
Expand
(2
+ 3a;) -4
(2
+ 3z)- 4 = 2-<(l + ^)~
to four terms. 4
In finding the general term we must now. use the
182.
formula
m(w-1)(w-2)
written in full
when n
is
;
for the
(n-r +
l)
xr
symbol "C can no longer be employed r
fractional or negative.
Also the
coefficient of the general
one of the factors of
numerator
term can never vanish unless
zero; the series will thereterm, when n — r + 1 is zero ; that is, when fore stop at the r r=oi+ l but since r is a positive integer this equality can never Thus the hold except when the index n is positive and integral. expansion by the Binomial Theorem extends to w+1 terms when n is a positive integer, and to an infinite number of terms in all other cases. th
;
its
is
.
HIGHER ALGEBRA.
154
1
Example
1.
Find the general term in the expansion
The (r+l) th term-
—-
of (1 +x)'\
L± r
(-2r + 3)
-5)
af.
2 r lr
of factors in the numerator is r, and r - 1 of these are negatherefore, by taking -- 1 out of each of these negative factors, we may
The number tive
;
write the above expression
(-i)~
1
-
8
-
-<»-V
6
i
Example
The
(r
2.
Find the general term in the expansion of
+ 1)'- term = »
A
V"
"
M
/
E w
_
Example
-
1) (2«
(_1)»-
3.
i)r-i
(1-r-l.n)
(_
-
1)
l)r-i
3'(- 4 >'-_5
_
~ by removing
.
[
= (_ i)2r-i = _ 1#
1.2.3
(1
- x)~ 3
-,)r
)
(
r
}
(r
+ 2)
(1)ffa,
X r
(r+l)(r+2) . *» 1.2
like factors
^
....(^l.n-l)
)^-(- 3 -'-+ 1
= (1)r 3.4.5 ~
^
(r-l.n-1)
(n-l)(2n-l)
Find the general term in the expansion of
The(r + irterm=<-
^
_
(
(n
since
_
i)r
- «»)r
r
I
l(l-n)(l-2n)
(
(
/
(1-F^Un)
= !(!-«) (l-ar.)r
=
£
(l-nx) n .
from the numerator and denominator.
.
BINOMIAL THEOREM.
ANY INDEX.
EXAMPLES. XIV. Expand
to 4 terms the following expressions: 1
1.
s
a.
(l+xf.
155
HIGHER ALGEBRA.
156 and,
under what conditions the expansion of
not,
if
be used as
its
-x)~ = r
(1
=
in this equation put x
(-l)~
1
2
;
1
n——
+ x + x2 + x 3 +
n
may
=l+2
+
2
2
+
2
3
+
2
is sufficient
n(n-l) —~— l+nx+—
as the true arithmetical equivalent of (1
=
(1) v '
1 -z
I
(1);
+
shew that we cannot
+ x) n
in all cases.
a geometrical proof the first r terms of the
sum
that the
4
to
from the formula for the sum
we know
+
x2 +
'
\
x*
obtain
,
series
x)
we have
l; then
we then
This contradictory result take
gression,
+
true equivalent.
Suppose, for instance, that
Now
(1
of
- xr
—x r
x 1 - x
1
1
and, large
a
when x
numerically less than
we can make number
sufficient
little
is
we
as
-X
x
1,
y
by taking r
sufficiently
r
we
as small as
^
^
;
that
sum can be made
of terms the
please from
please
But when x
.
is,
by taking
to differ as
numerically
is
r
greater than
1,
the value of
x ^ r increases with 1 - x
no such approximation to the value
of
is JL
any number
r.
and therefore
obtained by taking
vC
of terms of the series 1
+ X + Xs + X3
4-
It will be seen in the chapter on Convergency and Divergency of Series that the expansion by the Binomial Theorem
(1+x)" in ascending powers of telligible when x is less than 1.
of
But
if
x
is
greater than
a?
is
1
+ nx
n(n-\) H
„
x"
.j
I
.
in-
then since the general term of
1,
the series ,
always arithmetically
-
+
BINOMIAL THEOREM.
ANY INDEX.
157
contains x it can be made greater than any Unite quantity by taking r sufficiently large in which case there is no limit to the value of the above series; and therefore the expansion of (1 + x) n as an infinite series in ascending powers of x has no meaning arithmetically intelligible when x is greater than 1. r
,
;
We
184.
may remark
by the Binomial Theorem either of the
;
we can always expand (x + y)" we may write the expression in
that for
two following forms x"
('*!)'
:
'(•ff.
and we obtain the expansion from the according as x is greater or less than y. To find in its expansion of (1 — x) -u 185.
simplest
form
first
or second of these
general term in the
the
.
The
(r
+
l)
th
(-
term
n)(-n-
I)
(-71-
2)...
(-n-r+1) (-*y
+1 Hw + 2) -(** + »•= (- iy »(* = (_ I)*
n
From (1
-x)~*
(n
ttv*+l)(tt+2)...ytt +
+
1)
(n + 2)
this it appears
...
(n + r
-
1)
1)
r-l )
xr
r
that every term in the expansion of
is positive.
Although the general term in the expansion of any binomial may always be found as explained in Art. 182, it will be found more expeditious in practice to use the above form of the general term in all cases where the index is negative, retaining the form n(n- l)(n-2) ... ( n - r + 1 ) , x i
t only in the case of positive indices.
—
;
HIGHER ALGEBRA.
158
Find the general term in the expansion
Example.
—-
1
The
(r
= (l-3x)
of
_
-
.
.
- 13
.
+ l) th term
1.4.7
(3r-2) 3rrr
1.4.7
^
(Sr-2)
^H
w
:
•
r _i
formula for
+ Sx)
we should have used the same the general term, replacing Sx by - 3x.
If the given
expression had been
(1
The following expansions should be remembered
186.
(1
- x)' = 1 + x + x 2 + x 3 + - x)~ 2 = 1 + 2x + 3x2 + ±x 3 +
(I
- x)~ 3 =1 + 3x + 6x* + 10x 3 +
:
+ xr +
1
(1
3
+
(r
+
+
(
1)
r+
xr +
J%—K +
l
r
n when n is unrestricted in value, will be expansion of (1 + x) found in Art. 189 ; but the student will have no difficulty in applying to any numerical example the method explained in ,
Art. 172. Example. 2 x=3
,
Find the greatest term
when
and n — 20.
We have
fi j_
^
^V+i— -
19+r
•"•
'r+l >
,xxTr r
?
,
numerically,
•
-'r»
—>
2 (19 + r) £
so long as
that
in the expansion of (l+a;)~ n
1
38 >r.
is,
Hence
for all values of r
I^k = T,. and
up
to 37,
we have
jrr+1
>Tr
;
but
if
r=38, then
these are the greatest terms. Thus the 38 th and 39 th terms are equal numerically and greater than any other term. ,
ANY INDEX.
BINOMIAL THEOREM.
159
Some
useful applications of the Binomial explained in the following examples. 188.
Example
Find the
1.
first
Theorem are
three terms in the expansion of _i
i r
(l
+ 3*) -(l-2x)
Expanding the two binomials
/3
,
1
=1 +
13 -Q
X+
as far as the term containing x'\
/8
2\
55 72
3.
3
we have
2
.
X "'
so that ar = -000004, we see that the third term is a decimal fraction beginning with 5 ciphers. If therefore we were required to find the numerical value of the given expression correct to 5 places If in this
of decimals
Example ^='002,
it
would be
sufficient to substitute *002 for
ing the term involving x 2
1
+
-
x, neglect-
o
.
When x is so small that its square find the value of be neglected, Example
x in
2.
and higher powers may
J(± + xJ* Since x- and the higher powers may be neglected, it will be sufficient to retain the first two terms in the expansion of each binomial. Therefore i
the expression
_tl±±±l b(i+|.)
-K-S-). the term involving x- being neglected.
:
HIGHER ALGEBRA.
160 Example
3.
Find the value of -rj=
to four places of decimals.
x/47 _i --
1
-^ = (47)
--
2
1 /
\
2
*=(7*-2)*=-(l-n)
1/
^
3
1
5
:L_
-7^ + 72 + 3 -74 + 2 -7G+--7 + 73 + 2 To obtain
+
75
*
7+ "" 2 *7
the values of the several terms
we proceed
1)1 7
)
-142857
7
)
-020408
7
)
-002915
7
)
-000416
!
=t, '
= 7-3,
= ^;
•000059
and we can
5
term -
see that the
as follows
=1
.
is
a decimal fraction beginning with
5 ciphers.
-i-
.-.
= -142857 + -002915 + -000088
\/47
= •14586, and
this result is correct to at least four places of decimals.
Example
4.
Find the cube root !
(126)3
of 126 to 5 places of decimals.
= (5 3 + l) a 1
5
~5
/
1
t
3"5
V 1 ~
3
+
~ _
= 5-1^
~ 5 2
9
J. '
55
3*"l0 -04
9
=5-f -013333
= 5 '01329,
1
9'10 + 81 h
3
81 *5 7
5
-00032
...
\ '")
1
_1
+
1^1^ 2
1
5
9'5« + 81*5 9
:J
1
1_ '
M
1
•"
W_ "•• *10 7
-0000128 — 81
- -000035
...+...
to five places of decimals.
~
.
.
BINOMIAL THEOREM.
ANY INDEX.
EXAMPLES. XIV. Find the (r+1)" term !•
4.
(l+#)
7.
10
-
-I 2 .
J3
2.
(l-.t-)
5.
8.
(a+fo?)" 1
.
11.
(
1
+ .v)
(
1
+ a?) 2 when
(l+3.e)
(l+.r2 )-3.
6
(i-2.v)~*.
(2-.r)~ 2
9.
.
-
.
.
:] .
tt{rf-x*)\
f/ 3/
*
12.
(l-3.^
V&Z^ ,
each of the following expansions
in
4 x=— lo
when
7
— 74?)
(1
3.
N
—
15.
i
-5
*
7=A=.
Find the greatest term
14.
:
3 .
13.
b.
in each of the following expansions
1
(l+#)
101
:
.
2
a?= 5
11
4
1
wheu# = -. o
16.
(2a?
17.
(5
+ 5J/) 12 when
-
4.v)
~7
a?
=8
and y = 3.
when v=t
25
18.
(3-r2
Find to 19.
+ 4/) - n when x = 9, y = 2,
five places of
v98.
20.
«=
1 5.
decimals the value of 4/
3/
4/998.
21.
\ 1003.
22.
\ 2400.
26.
tfilla
3
1 1
23.
^=.
24.
(1^)3.
If x be so small that its square find the value of 1
27.
(l-7tf) s (l
3
+ 2a?)"*.
25.
(630)
*.
and higher powers may be neglected,
HIGHER ALGEBRA.
162
V^+C+jj
31.
^T^-^1^
'
32
.
(1+5*)*+
Prove that the
33. is
coefficient of sf in the
(4+|Y
expansion of (l-4r)
*
'
v2
31
Prove that
35.
Find the
(1
first
+*)*=2-
|l
-—
+
^
-^ (f^
)
f
three terms in the expansion of 1 '
(1
36.
Find the
first
+ x) 2 Vl + 4x
three terms in the expansion of 3
(! + #)* +
37.
Shew that
the
double of the (n-l) 189.
of (1 + x)
n
nth
*Jl
+ bx
coefficient in the
expansion of
(1
- x)~ n
is
th .
To find the numerically greatest term in for any rational value of\\.
the expansion
,
Since we are only concerned with the numerical value of the greatest term, we shall consider x throughout as positive.
Case
Let n be a positive integer.
I.
The (r+l) th term by
.
x
;
that
is
obtained by multiplying the r th term
by
is,
1
f
J
terms continue to increase so long as
'n+ 11
-1
Or
(n+ l)x
.,
that
)- 1
— > 1 + x,
is.
,
r
or
—x >r.
*— + 1 1+02
(n
)
'
x
;
and therefore the
BINOMIAL THEOREM.
—X
(ll 4- 1
—
If
ANY INDEX.
103
^
be an integer, denote
it
by p; then
multiplying factor is 1, and the (;>+l) th term th and these are greater than any other term. ^>
is
if
r=p,
the
equal to the
,
—
( 71 4,
If
1
)
—X
be not an integer, denote
then the greatest value of r
is
and the
7,
integral part by q
its
+
(q
l) th
term
is
;
the
greatest.
Case
As rm
Let n be a positive
II.
,
_.
—
(n +
term by
(
th
(r+
before, the 1
t
I
fraction.
term
l)
is
obtained by multiplying the
\
)x.
x be greater than unity, by increasing r the above multiplier can be made as near as we please to - x so that after a certain term each term is nearly x times the preceding term numerically, and thus the terms increase continually, and there is no greatest term. If
(1)
;
than unity we see that the multiplying factor continues positive, and decreases until r > n + 1 and from this point it becomes negative but always remains less than 1 numerically therefore there will be a greatest term.
x be
If
(2)
less
,
;
As
multiplying factor
the
before,
will
be greater than
1
(n + l)x
-^
so Ions: as
1 ( Jl 4-
If
^
1
\ X -— be an
the (p + l) th term any other term. ( 7t 4"
If
the (q
^p 4-
Case
l)
th
1
)
X •
Let n - —
is
r.
by p
integer, denote it th
equal to the
£>
be not an integer,
term
III.
is
—>
+x
;
then, as in Case
I.,
and these are greater than
,
let q
be
its integral
part; then
the greatest.
Let n be negative. in,
so
that
m
is
positive nil _L
value of the multiplying factor
m-l (
r
is
+
f
;
then the numerical J
—
.
x
;
that
is
\ 1
)
x.
J
11—2
HIGHER ALGEBRA.
164
x be greater than unity we may shew, as in Case that there is no greatest term. (1)
If
(2)
If
x be
greater than
than unity, the multiplying factor will be
less
so long as
1,
(m that
II.,
—x >
1
)
is,
1
- x,
r
(m—\)x — — > r. I -x
or
-.
lyn.
If
^—
x
— be — x 1
)
CC
a positive integer, denote
th (p + l) term is equal to the any other term. (fjr
If
1
gral part
If
1
— be -x £C
)
i
;
then the '-
—
th
by p
tlien the
\
term, and these are greater than
positive but not an integer, let q be its inte-
'
*
p
it
(q
+
l)
th
term
be negative,
the greatest.
is
tlien
m
is less
than unity
writing the multiplying factor in the form (1
;
and by
—
x,
we
J
hence each term is less than see that it is always less than 1 the preceding, and consequently the first term is the greatest. :
190. sions that
To find the number of homogeneous products of v dimenand their can be formed out of the n letters a, b, c,
powers.
By
division, or
= 1
by the Binomial Theorem, we have
— ax 1
1
— bx 1
1
—
ex
2 2 3 3 + ax + a x + a x +
=
1
=
1
+ bx +
b
=
1
+ ex +
c
2
2
x2 +
b
x2 +
c
3
,
x3 +
,
x3 +
,
3
;
ANY INDEX.
BINOMIAL THEOREM.
KJ5
Hence, by multiplication, 1
_ ax
1
aV +
=
(1
+ ax +
=
1
+ x (a +
b
+
c
1
...)
+
— bx
1
+ bx +
(1
+x
...)
1
2
(a
2
—
1
+
b*x*
ex
+ ex +
...) (1
+ ab + ac +
2
b'
c°x
+ bc±c 2
2
+
4-
.
...) ... )
. .
=
+ S x + Sjfx? + Saxa +
1
Slt
SaJ
>S'.,,
duets of one, two, a,
b,
...
suppose
t
where
+
homogeneous prodimensions that can be formed of
sums
are the three,
of the
and their powers.
c,
To obtain
the
number
of these products, put a,
each ; 1, and the l9 S S so obtained give the number of the Sl9 2 homogeneous products of one, two, three, dimensions.
equal to 1 values of
each term in ,
:i
S2 S
JS
,
,
1
Also 1
(1
-
r
n(n+
— bx
1
—
1
—
or (1
— a;)
ex ".
x)
S = coefficient ~
1
1
— ax
becomes
Hence
,
:i
b, c,
now becomes
of
r x in the expansion
(n+r-
l)(n + 2)
of (1
— x)~
1)
jr
n + r—1 \r
\n—
1
To find the number of terms in the expansion of any multinomial when the index is a positive integer. 191.
In the expansion of (at
+a )",
+ aB + aB +
r
every term is of n dimensions; therefore the number of terms is the same as the number of homogeneous products of n dimensions and their ... a that can be formed out of the r quantities a,, a powers ; and therefore by the preceding article is equal to ,
I?'
+ n—
n
r
1
—1
r
,
HIGHER ALGEBRA.
166
the result of Art. 190 we relating to the number of combinations of
From
192.
may deduce n
a theorem
things.
then if we were to write Consider n letters a, b, c, d, ; down all the homogeneous products of r dimensions which can be formed of these letters and their powers, every such product would represent one of the combinations, r at a time, of the n letters, when any one of the letters might occur once, twice, thrice,
up
...
to r times.
Therefore the number of combinations of n things r at a time when repetitions are allowed is equal to the number of homogeneous products of r dimensions which can be formed out of n \n
+r—
and therefore equal to
letters,
,
n—\
\r
That
number
1 -
,
or
of combinations of
n+r
*CT
.
things r at a time when repetitions are allowed is equal to the number of combinations of n + r— 1 things r at a time when repetitions are excluded. is,
the
We
193.
shall
it
conclude this chapter with a few miscel-
laneous examples.
Example
Find the
1.
coefficient of
The expression = (1 - Ax + 4.x 2)
The
-4,4
(1
x r in the expansion of
+PyC +p^xr +
...
+p rxr + ...)
xr will be obtained by multiplying p r and adding the results hence
coefficient of
respectively,
But
p r =(- iy fe±afc±9
Hence the required
coefficient
. (
1)
r
(r+lHr + 2) _ 4( _
1)r . 1
+ 4p r _ 2
.
~
.
suppose.
p r - x p r -» by ,
,
1,
.
[Ex 3 Art. 182.] .
,
rJ^ + 4( _ ira I ^r
= ^-[(r + l)(r + 2) + 4r(r + l)+4r(r-l)] f-l) r
- 2a;) 2
;
the required coefficient =p r - 4p r _ x
=
(1
(
3
;
ANY INDEX.
BINOMIAL TIIEOltEM. Example
Find the value
2.
„ 2 4-
3
= 2 + 3— —
.
.
5 2
2
~
.
3
3
3
z 2
2' 2 2
z 2
+
*3
:J
•••
:
3
5 '
;!
2
2
3
2 2J '3 :i+
5
3
.
34
14
7 '
~|3~
1
'3?*"
|2
+
-4-
4
1
3
Ll?
1
5.7.!)
1
2
-Q
|3.3
^
2
.
[2
3
5.7
3.5.7 —1 3.5.7.9 1 — —- _1 + + + 3-
5
.
expression v
T
.
|_2.
m U The
of the scries
5
,
Hi?
5 '
2
7 '
2
1)
'
24
2
*3 5+
]i~
••'
7
5
2*2*2 2 2 2 /2V
a
/2\/2\
"~J2~
1
o-.r-ffl" = 35 =V3 Example 3. If ?t is any n (3 + Jl) is an odd number. Suppose I
I+f=3 +C
Now 3- N/7
is
+
a2 S
and
less
i
1
S"- s/7
positive
shew that the integral part
positive integer,
to denote the integral
n
Then
-
and/
n~2 .
the fractional part of
7+(78
than
1,
3*-«^7) 8
therefore
+
(3
+ a/7)'
Add
together
-C
1
(1)
and
,
3' l
-V7 + C'
3' l
-2
2
(S-^)'
.7+C3 3' -3( v/7) 3 +
1
is
.
= an But since/ and/' are proper :\
.
1.
xm
fractions their
I=an odd
'
- xf 4 + 2a - a 2 (1
2.
an
3.
«* in the expansion of
in the expansion of
.
sum must
integer.
coeflicient of
in the expansion of
.
and we have
even integer.
EXAMPLES. XIV. Find the
(2).
;
X "T" X
c.
be
.
a proper
l
the irrational terms disappear, " I+f+f = 2 (3» + C2 3' 1 2 7 + ) (2)
1
(1).
fraction; denote it by/';
.•./' = 3 n
of
1
HIGHER ALGEBRA.
168 4.
Find the
xn
coefficient of
in the expansion of
2 4- x + X2 (
5.
2
1.3 + 271
1 *
2
'
1
1.3.5
2*
2.4.6'2 3
2_
'
~
V
2n(2n + 2) (2n + 4)
3.6.9
+ 3 + _ 3T6~ +
+
+
3.6.9
J
Prove that 7'
(^-1) n{n-\)(n-2) h + i7 ++ ^ 7.14 + 7.14.21 + ?
1
.
J
±n Ji
%
(2 .
j.
J
»(*+!) , n(» + l)(n + 2) + + 2.4 2.4.6
\
"
Prove that approximately, when x
"7!
1+
2
( 10.
,
Prove that
~2
9.
V 3'
3.5.7 3^5 = 1 + 3- + H t^~; H + \ \n+ 4 4.8.12 4. 8
2n 2n(2n + 2) + ~3 + ~ 3.6 " +
8.
1-3.5.7 1^ 2.4.6.8'2*
+
Prove that N/8
7.
+ ^J
Prove that 1
6.
1
Shew that the
9
very small,
is
256'
\2
•
J
'
r6 'V
integral part of (5
+2
>JQ)
n is odd, if
n be a
positive integer. 11.
Shew
that the integral part of (8 + 3 V/7) H
is
odd,
if
n be a
positive integer. 12.
Find the
coefficient of
xn
in the expansion of
(l-2.v + 3.v2 -4.v 3 + 13.
of
xn
Shew
in the expansion of (1
14.
-Ax)
^"
- xf n + 3nx (1
(
1\ 4 1 '
x+-
1
is
equal to the coefficient
2 .
— x^) n may be put into - xf n ~ 2 + 3n @ n 3) x i (i _ xyn - 4 +
Prove that the expansion of (1
/
that the middle term of
)-*.
(1
the form
3
\
_
BINOMIAL THEOREM. Prove that the coefficient of
15.
1,0,
-
1
according as n
ANY INDEX.
at*
3m -
In the expansion of (a + b + c) s find the sum of the coefficients of the terms. 16.
(2)
l\
+.f)
n-l
If c
18. (1
111
Prove that
17.
u ,
when n
n be an even
n-
\'S
\b\
or 3//<.+
1,
is
1.
number
the
(1)
,
of terms,
integer,
2"" 1
1
n-5
u-\
|rc
,1
'
\
the coefficients in the expansion of a positive integer, prove that fn are
C2 ,
(',,
,
if
—
the expansion
ill
of the form 3m,
is
L69
is
I//-1
-c l+ c 2 -c3+
c
a)
+(-mv-(-i)'-
,
1/
(2)
^-2^ + 3^-4^+
+ (_i)n (/i4
(3)
c*- c *+c£-c*+
+ (-l)»cn2 =0,
according as n 19.
„
20.
If
sum
= 0>
1)t M .
(-1)^,
or
(l-;r)- 3 = ^ + %^ + ^.^+
(2)
2 (*! *, B + 82*2, _j +
1.3.5.7 fr-
2 .4. 6
(!)
?2n +
(2)
2 { ?2n
l
+V»~
1
+...
j2^ + 4
+ 8n8n +
l)
= -—
—^
.
(2)i-l) ,
.
2n
8
P*>™
+ 2n- \
+
- ?1 y,
(l
_
j
+ g^a.
2
that
+ - 1)" -
+
1
(
tj
n_
Find the sum of the products, two at a time, of the the expansion of (1 +x)n when n is a positive integer. 21.
in
+4
n If (7 /3, where n v/3) proper fraction, shew that (1 -f3)(p
23. (1
#„ + J
coefficients
,
22.
a
.
of the first n natural numbers, prove that
(1)
T.
;^r_ 1
odd or even.
is
denote the
If *„
.
|
If c
=p +
+#)*, where
is
?i
c2
2
c .
integers,
and
|9
+ p) = l.
are the coefficients in the expansion of a positive integer, shew that
c^,
,
and p are positive
\
3
rn
(-I) n_1 fn
,11
n
2
3
1
n
—
,
CHAPTER
XV.
Multinomial Theorem.
We
have already seen in Art. 175, how we may apply the Binomial Theorem to obtain the expansion of a multiIn the present chapter our object is not nomial expression. so much to obtain the complete expansion of a multinomial as to find the coefficient of any assigned term. 194.
Example.
Find the
coefficient of (a
a4 b"c 3 d 5 in the expansion of
+ b + c + d) u
.
the product of 14 factors each equal to a+b + c + d, and every term in the expansion is of 14 dimensions, being a product formed by 5 4 2 taking one letter out of each of these factors. Thus to form the term a b c*d , we take a out of any four of the fourteen factors, b out of any hco of the remaining ten, c out of any three of the remaining eight. But the number of ways in which this can be done is clearly equal to the number of ways of arranging 14 letters when four of them must be a, two 6, three c, and five d that is, equal to
The expansion
is
;
114
TTralg 3 5
A
[Art. 151.]J L
.
412
therefore the number of times in which the term a 4 b 2c*d 5 appears in the final product, and consequently the coefficient required is 2522520.
This
is
To find pansion of (a + b + 195.
the coefficient of any assigned term in the exp where is a positive integer. c + cl + .
..)
p
,
The expansion is the product of p factors each equal to a + + c + cZ + ..., and every term in the expansion is formed by taking one letter out of each of these p factors ; and therefore the number of ways in which any term a a b^cyd8 ... will appear in the final product is equal to the number of ways of arranging p letters when a of them must be a, (3 must be b, y must be c; and so on. That is, ft
the coefficient of
aabPcyds
— o ~f~^ p \y 6 \p
...
is
=
...
where
a+
j3
+ y+
S
+
...
=p.
MULTINOMIAL THEOREM.
171
Jn the expansion of
Cok.
(a
+ bx +
the term involving a"b&cyd 6
+ da? +
cx~
...
)'',
is
...
^L-.a-^^v^). or
aWcyd5
i—T5-Ht^t-
where a + This
/3
+y+
may be
$
+
The general term
..
= p.
eallecl tlte
Find the
Example.
...
xfi+2y+M +
...
general term of the expansion.
coefficient of «* in the
expansion of
(a
+ i.c + ex 2
)-'.
of the expansion is
-ii-«a
^V
+2?
(i),
where a + p + y = \).
We which
have to obtain by trial all the positive integral values of /3 and 7 satisfy the equation fi + 2y = 5; the values of a can then be found from a + /3 + 7 = 9.
the equation
= 1, and a = G; putting 7 = 1, we have /3 = 3, and a = 5; putting 7 = 0, we have /3 = 5, and a = 4.
Putting 7 = 2, we have
The required expression
coefficient will be the
/3
sum
of the corresponding values of the
(1).
Therefore the coefficient required 19
|9
19
= 252a 6 6c 2 + 5Q4a*&c + 12Ga 4 b\ To find
19G.
the general (a
vjhere
n
By
is
any
term in the expansion of
+ bx + ex 2 +
clx
3
+
n .
.
.)
,
rational quantity.
the Binomial Theorem, the general term
n(n-l)(n-2)...(n-p + l)
,,_
(
where jp
is
a positive integer
v
is
+ rf +
^+
HIGHER ALGEBRA.
172
And, by Art.
195, the general term of the expansion of (6a;
+
+ dx 3 +
ex'
...)''
\P
\pjy_\o_—
where
ft,
y, 8
.
.
.
are positive integers whose
Hence the general term pression
is p.
in the expansion of the given ex-
is
where
/?
197.
sum
+y+
+
S
...
+ ex 2 + dx 3 +
Since (a + bx
= /?.
may
..)"
be written in the
form 6 „A +-x ail
a
\
+
c
a*
-x+-ar+ 2
3
...
a
a
y ,
J
be sufficient to consider the case in which the hrst term of the multinomial is unity. it will
Thus the general term (1
3 + bx + ex 2 + dx +
(n-p +
(n-l)(n-2). n -
is
\p \v
where Find the
.
.
.)"
^ +9f+u+
l)
bpcyd8 8 \
fi
Example.
of
+ y + &-\-...=p.
coefficient of
x 3 in the expansion of
(l-3z-2.r 2 + 6x' 3 )3.
The
general term
is
+o S(S-0(t-»)...(|-* V ,.,
iO-sA-^e) /-^^
,,
We
8
(i).
have to obtain by trial all the positive integral values of /3, 7, 5 which satisfy the equation j3 + Zy + 35 3 ; and then p is found from the equation 5. The required coefficient will be the sum of the corresponding 2>=/3 + 7 +
=
values of the expression
(1).
MULTINOMIAL THEOREM.
173
In finding /3, 7, 5, ... it will be best to commence by giving to 5 successive integral values beginning with the greatest admissible. In the present case the values are found to be
= 0, p=l; 7 =1, 0=1, p=2; 7 = 0, 0=3, p = 3. 7 = 0,
8=1,
= 0, 5 = 0, 5
Substituting these values in
(1)
18
the required coefficient
^)<^)(->)<- 3"- 2>+^#^ (-3)
s
4_4_4 3
3~3
Sometimes
198.
more expeditious
it is
to use the Binomial
Theorem. Find the
Example.
The required
coefficient of
x 4 in the expansion
of (1
- 2x + 3.r 2 ) -3
.
found by picking out the coefficient of x x from the first few terms of the expansion of (1 - 2x - Sx 2 ) * by the Binomial Theorem that is, from coefficient is
;
1
we stop than
+ 3 {2x - Sx 2 + 6 )
at this
term
(2.r
- 3x2 ) 2 + 10 {2x -
3.r-) :J
+ 15
(2.r
required coefficient = 6
=
.
9
+ 10
.
3
2
(2)
(
-
+ 15
3)
(2)
-66.
EXAMPLES. XV. Find the
coefficient of
(a-b — c+d) w
1.
a 2 Pc4 d in the expansion of
2.
a 2 b d in the expansion of (a + b — c — d) s
3.
a?b s c in the expansion of (2a + 6-f 3c) r
4.
x~yhA in the expansion of
5.
x3
6.
xA in the expansion of
.
ry
7.
.'•"
8.
A"
8
in the expansion of
in tlie
expansion of
in the expansion
(if
{cub
.
.
- by + cz)9
(l+3# — 2a2 )3
.
.
+ 2.r + 3.r 2 10 2 (1 + 2.v - x 2 (1 - 2.r + 3# - 4.r'
(l
3.r
2 4 )
;
terms involve powers of x higher
for all the other
x*.
The
-
)
.
)'\
4
!
)
.
4
.
HIGHER ALGEBRA.
174 Find the 9.
.r
coefficient of
2x + 3x2 - x4 - .i/') 5
in the expansion of (1 -
23
.
i
5
-2x + 3x
10.
x
11.
x
12.
x8
13.
x* in the expansion of (2 -
in the expansion of (1
2
2
.
)
i
3
- 2x + 3x 2 - 4a3 ) 2
in the expansion of (1
2
(
in the expansion of
(
1
X*\ -X — + )
.
~2
'-»
.
4x + 3x2 ) ~ 2
.
3
s
14.
X in the expansion of
15.
x 12
(
1
in the expansion of (3
+ Ax + 1 Ox + 20^G ) " * 4
2
- 15x* + 18^') - l
.
i
Expand
16.
(1
- 2x - 2x
2
)*
as far as
x2
.
2
17.
Expand
(1
+ 3x - 6x*)
18.
Expand
(8
- 9^ + 1 8a4 ) 3
19.
If
2
3
as far as
x5
as far as
x8
.
4 "
+xP) n = a
(l+x + x2 +
.
+ a x + a.^v +
a xn
2
l
r>,
llf>
prove that
a
(2)
a1 +2a2 +3a8 +
If
20.
of
(1
+x+x a
21.
+a^=(p+l)n
+a1 +aa +
(1)
2
a
,
n
2 )
,
a 15 a2
>
ft 3 •••
.
+«p.a«p=5»i>(p+l)*.
are the coefficients in order of the expansion
prove that
+ (-l)n - 1 aU 1 =^an {l-(-l)^an}.
-a 2 + a 2 -a 2 + If the
+x + x2 n + a x+a 2x2 + ... +araf +
expansion of
a
be
(1
)
r
...
l
+a 2n x2n
,
shew that «o + a 3 + a6
+
...
=a + a + a-+ l
4
...
=« + a6 +a + 2
8
...
= 3 n_1
.
;
CHAPTER
XVI.
Logarithms. The logarithm of any number to a given Definition. base is the index of the power to which the base must be raised Thus if a x = JV, x is called in order to equal the given number. the logarithm of to the base a. 199.
N
Examples.
(1)
Since 3 4 = 81, the logarithm of 81 to base 3
(2)
Since
the natural numbers to base 10. 1000,
10 2 = 100, 103 = 1000,
lO^lO,
1, 2,
is 4.
are respectively the logarithms of 10, 100,
3,...
The logarithm of iV to base a is usually written log a jy, 200. so that the same meaning is expressed by the two equations x a =
From
these equations
an identity which Example.
is
N; x = \og
a
N.
we deduce
sometimes
useful.
Find the logarithm of 32 £/!
to base 2 N/2.
Let x be the required logarithm; then,
by
definition,
(2
x/2)« = 32 4/4 2
1 .-.
(2. 2*)*
=2
s
2
3 .-.
hence, by equating the indices,
2*
.
2^ = 2 5 ^; 3 27 - x = -r
.'.
x=
;
—o =
3-6.
)
.
HIGHER ALGEBRA.
176
When
understood that a particular system of logarithms is in use, the suffix denoting the base is omitted. Thus in arithmetical calculations in which 10 is the base, we instead of log 10 2, log l0 3, usually write log 2, log 3, 201.
is
it
Any number
might be taken as the base of logarithms, and corresponding to any such base a system of logarithms of all numbers could be found. But before discussing the logarithmic systems commonly used, we shall prove some general propositions which are true for all logarithms independently of any particular base.
202.
logarithm of
TJie
For
= 1 for may be.
a°
the base
values of a
For a = a 1
;
therefore loga a
To find
201.
=
itself is 1.
1
let
a be the base of the system, and
suppose a:
so that
a*
=
y = \oga J\T;
log. J/,
= M,
a*
= N.
MN==ax x
Thus the product
ay
= ax+y ; whence, by definition, log a
MN = x + y = 100^1/"+
Similarly, \og a 3INP
and so on
for
= \oga M+
any number
low
N.
loga iV+ loga P;
of factors.
log 42 = log (2 x 3 x 7
Example.
= log2 + log3 + log7. To find
205.
Let
so that
M be the -zz
the logarithm
fraction,
x= ax =
\oga
M
t
of a fraction.
and suppose
M i
1-0, whatever
of a product.
the logarithm
MN be the product;
therefore log
;
The logarithm of the base
203.
Let
all
1 is 0.
2/
= loga iT;
ay = N.
LOGARITHMS. Thus the
whence',
by
fraction
definition,
loga
*—-=x — y
'
HIGHER ALGEBRA.
178
3
r
Example log
Express the logarithm of
1.
a —^ m
terms of log
a, log b
and
c.
a2
Ja*
.
3
= log a 2 - log (c^ 2
)
= 3-log«-(logc + logfc 5
2 )
3
= = log a - 5 log c - 2 log &. Example
Find
2.
from the equation a x c~2*=&3a!+1
a;
.
Taking logarithms of both
sides,
.-.
x
we have
= (Sx + 1) log 6 (log a - 2 log c - 3 log b) = log b
x log a - 2x
.
log c
;
;
lo 2 6
_
log a - 2 log
c
- 3 log
EXAMPLES. XVI.
b
a.
Find the logarithms of 1.
16 to base J2, and 1728 to base 2 v '3.
2.
125 to base 5 v/5, and *25 to base
3.
base 2 x/2, and
stt. to
256
'0625 to base
5.
-0001 to base '001, 4
6.
/~*r
kI gp
,
3
i
—
j
,
to base
and 1000
4.
2,
'3
and i
4.
9.
to base -01. to base 9^/3.
r~-^
*/
a
2
to base a.
a? 7.
Find the value of l0g8 128,
l0g6
^,
log-frgj,
log3 43 49
'
Express the following seven logarithms in terms of log a, logb, and logo. 8.
log(N^)
fi
.
9.
log{Va 2
xyb s
).
10.
logflcFW).
=
"
LOGARITHMS.
179
11.
log^o^x^oJR).
12.
log(^a
13.
log
14.
logj^J +
15.
Shew
that log |f
'
;
£^ = 1
VW7J2
4
logV
V
16.
Simplify
17.
Prove that log
72!)
logo - | log2
V63 -j-VP7a).
-
.
5 75 — - 2 log - +
ffi
3
2 ,
log 3.
3
5
9" 1 27 3
f
.
—= 39
>g
1<
h .g
2.
Solve the following equations: 18.
o«=c&*
19.
a2».63*=cs
90 U
°^ - "& &*-*
21
a 2* m6 3*.62 a »=m10J
'
c
*
=a
and log- = 6,
22.
If \og(x'2y 3 )
23.
If
24.
Solve the equation
a3 " *
.
V> x
(a*
}
= ax + \
b 3x ,
-
•
.
^
find log*-
l
and
'
log//.
shew that x log (-") = log a.
- 2a*b* + b*) x - 1 = (a -
ft)
2*
(a + 6) -
-'.
Common Logarithms. Logarithms to the base 10 are called Common Logarithms; this system was first introduced, in 1615, by Briggs, a contemporary of Napier the inventor of logarithms. 208.
x
the equation 10 - JV, it is evident that common logarithms "will not in general be integral, and that they will not
From
always be positive.
3154 > 10 and < 10 4
For instance
'
.*.
log
3151«=3 + a
;
fraction. ion.
12—2
.
HIGHER ALGEBRA.
180
-06
Again, .*.
> 10~ 2 and < 10 _l
log *06
=-
2
+a
;
fraction.
The integral part of a logarithm is called Definition. 209. the characteristic, and the decimal part is called the mantissa. characteristic of the logarithm of any number to the base 10 can be written down by inspection, as we shall now shew.
The
To determine
210.
number
the characteristic
of the logarithm of any
greater than unity.
10 - 10, 1
Since
2
10 =100, 3
10 -1000,
number with two digits in its integral part lies 2 between 10' and 10 a number with three digits in its integral 3 2 Hence a number part lies between 10 and 10 ; and so on. _I 10" and 10". with n digits in its integral part lies between it
follows that a
;
Let
N be
a
number whose
integral part contains
n
digits;
then J\T— in(tt-l)+ a fraction
.*.
log
iV=
(n — 1) + a fraction.
characteristic is n — 1 \ that is, the characteristic of the logarithm of a number greater than unity is less by one than the number of digits in its integral part, and is positive.
Hence the
To determine the characteristic of the logarithm of 211. decimal fraction. Since
10°=
1(rs
10-8
1,
=iJcr
01 '
=i=' 001
>
a
LOGARITHMS.
181
follows that a decimal with one cipher immediately after the decimal point, such as -0324, being greater than -01 and less -1 than -1, lies between 10~ 2 and 10 ; a number with two ciphers after the decimal point lies between 10 _:i and 10""; and so on. Hence a decimal fraction with n ciphers immediately after the decimal point lies between 10~ " + 1) and 10~". it
(
D
Let
be a decimal beginning with n ciphers
_
/) .-.
Hence the
f)~(w + l) +
1
= — (n +
log J)
characteristic
is
l)
thou
fraction.
+n
- (n+
;
fraction. 1)
that
;
is,
the characteristic
of a decimal fraction is greater by unity titan the number of ciphers immediately after the decimal point, and is of
the logarithm
negative.
The logarithms
212.
to base 10 of all integers from
200000 have been found and tabulated
most Tables they are the system in practical
in
j
given to seven places of decimals. This use, and it has two great advantages
to
1
is
:
From
the results already proved it is evident that the characteristics can be written down by inspection, so that only the mantissse have to be registered in the Tables. (1)
The mantissse are the same for the logarithms of all numbers which have the same significant digits; so that it is (2)
sufficient to tabulate the mantissse of the logarithms of integers.
This proposition we proceed to prove.
N
Let be any number, then since multiplying or dividing by a power of 10 merely alters the position of the decimal point without changing the sequence of figures, it follows -~ 10 7 that x 10''. and where p and q are any integers, are numbers whose significant digits are the same as those of N. 213.
N
N
Now
log
(N x
,
10 p )
= log N+p log 10
= log J\r +p Again,
log (AT -
9
1
)
- log N - q log
(1 1
= logiV-7 In
an integer subtracted from log (1)
is
N
;
added to that
is,
).
logiV^,
(2).
and
in (2)
an integer
is
the mantissa or decimal portion
of the logarithm remains unaltered.
;
;
HIGHER ALGEBRA.
182
In this and the three preceding articles the mantissse have been supposed positive. In order to secure the advantages of
we arrange our work so as always to keep the mantissa positive, so that when the mantissa of any logarithm Briggs' system,
has been taken from the Tables the characteristic is prefixed with its appropriate sign according to the rules already given.
In the case of a negative logarithm the minus sign is written over the characteristic, and not before it, to indicate that the characteristic alone is negative, and not the whole expression. 214.
Thus 4-30103, the logarithm of -0002, is equivalent to -4 + -30103, and must be distinguished from — 4*30103, an expression in which both the integer and the decimal are negative. In working with negative logarithms an arithmetical artifice will sometimes be For instance, necessary in order to make the mantissa positive. a result such as - 3*69897, in which the whole expression is negative, may be transformed by subtracting 1 from the Thus characteristic and adding 1 to the mantissa.
- 3-69897 - - 4 +
(1
- -69897) = 4-30103.
Other cases will be noticed in the Examples. Example
1.
Required the logarithm of -0002432.
In the Tables we find that 3859636 is the mantissa of log 2432 (the decimal point as well as the characteristic being omitted) and, by Art. 211, the characteristic of the logarithm of the given number is - 4 ;
.-.
Example
2.
log -0002432 = 4-3859636.
Find the value of ^-00000165, given log 165 = 2-2174839, log 697424=5-8434968.
Let x denote the value required
;
then I
log
a-
= l©g (-00000165) 5 =
l
= log (-00000165) o
= i (6-2174839) the mantissa of log -00000165 being the characteristic being prefixed by the rule.
Now
- (6-2174839)
=-
(10
same as that of log
+ 4-2174839)
= 2-8434968
165,
and the
;
J
'
LOGARITHMS.
183
and -8434908 is the mantissa of log 007424; hence x is a number consisting of these same digits but with one cipher after the decimal point. [Art. 211.
Thus
= -0097424.
a:
The method of calculating logarithms will be explained in the next chapter, and it will there be seen that they are first found to another base, and then transformed into common loga215.
rithms to base 10. It will therefore be necessary to investigate a method for transforming a system of logarithms having a given base to a new system with a different base.
Suppose that the logarithms of
216.
are known and tabulated, to base b.
Let
N
it is
all numbers to base a required to find the logarithms
be any number whose logarithm
to
base
b
re-
is
quired.
Let y = log6 iV, so that
y
=N
log. (&")
••
that
b
= logJT;
ylog£ = log,JV;
is,
J — log r X log 0u N. a6 1/
'
k&^wK*
or
1U Oa°
^
10
C
N
and b are given, los: N and from the Tables, and thus log^V may be found.
Now
since
Hence to base b
it
known
appears that to transform logarithms from base a
we have only J
constant quantity and modulus. 217.
log b are
1 )-
Tn equation
to multiply r J is
them
all
given by the Tables;
;
i
i
,
On
log/t x log 8 /j
Oa
=
1
.
this
is
a
log b
it is
(1) of the preceding article
thus .
by J
known
as the
put a for
N\
:
HIGHER ALGEBRA.
184 This result
may
also be
proved directly as follows
x = log/?, so that
Let
then by taking logarithms to base
a*
=b
;
we have
b,
x \og b a = \ogb b
.-.
218.
loga 6xlog 4 a
The following examples
=
l.
will illustrate
the utility of
logarithms in facilitating arithmetical calculation ; but for information as to the use of Logarithmic Tables the reader is referred to works on Trigonometry. 4
Example
1.
Given
log 3 = -4771213,
27
The
required value
5
find log {(2-7) 3 x (-81)»-H90)*}.
4 81 5 - log - log 90 1Q
= 3 log j= +
-=
= 3(l<^3»-l)+|oog3*-2)-|(log3*+l)
KM)"*a-K +
t)
= ^log3-5H
= 4-6280766-5-85 = 2-7780766.
The student should
notice that the logarithm of 5 powers can always be obtained from log 2 ; thus log 5
Example
2.
— = log 10 - log 2 =
= log
Find the number of log 2
log (875 16 )
= -3010300,
digits in
log 7
,
= -8450980.
=47-072128; is 48.
-
[Art. 210.]
log
875 1C given
= 16 log (7x125) = 16 (log 7+ 3 log 5) = 16(log7 + 3-31og2) = 16x2-9420080
hence the number of digits
1
2.
and
its
;
.
;
;
LOGARITHMS. Example 3. Given log 2 and log value of x from the equation
Taking logarithms of both (3 .-.
.-.
.r (
(3
-
4.r)
sides,
- 4a) log G +
(log 2
+ log 3) +
- 4 log 2 - 4 log 3 + 2 log
find to
two places of decimals the
we have (x
+ 5)
log 4 = log 8
= 3 log 2
(x
+ 5)
2)
= 3 log 2 - 3 log 2 - 3 log 3
= 10
.r
3,
185
2 log 2
- 10 log 2
+ 3 log 3 2 + 4 log 3
log 2
2 log
_ 4-4416639 ~2-al054a2
= 1-77...
EXAMPLES. XVI. 1.
Find, by inspection, the characteristics of the logarithms of
21735, 23-8, 350, '035,
%
-87, -875.
The mantissa of log 7623 is '8821259 2. of 7-623, 762-3, -U07623, 762300, '000007623. 3.
b.
How many
;
write
down
the logarithms
digits are there in the integral part of the
numbers
whose logarithms are respectively 4-30103, 4.
1-4771213,
3-69897,
-56515
1
(Jive the position of the first significant figure in the
numbers
whose logarithms are 2-7781513,
5-4871384.
-6910815,
Given log 2 = -3010300, log 3 ='4771213, log
7
= -8450980,
find
the
value of 5.
log 64.
6.
log 84.
8.
log -0125.
9.
log 14-4.
10.
log 4^.
logW —
13.
logN/ 0l05.
11.
14.
log^l2.
12.
\
7.
log -128.
4
:
-i
Find the seventh root of -00324, having given that log 44092388 = 7-6443036.
15.
Given log 194*8445 = 2'2896883,
find the eleventh root of (39-2) 2
.
HIGHER ALGEBRA.
186 16.
Find the product of 37-203, 3-7203, -0037203, 372030, having
given that log 37-203
= 1-5705780,
and log!915631 = 6:28231 20. 3
17.
Given log 2 and log
find log
3,
//3 2 5**\ /( —y )
.
X
1
18.
Given log 2 and log 3, find log (#48 x 108
19.
Calculate to six decimal places the value of
V
given log 20.
2,
log
-f ^6).
/294 x 125 \ 2 '
V 42 x 32 J log 7; also log 9076-226 = 3-9570053.
3,
Calculate to six places of decimals the value of
(330^-49) 4 ^\/ 22x70;
given log
2,
log 3, log 7
log 11
also
;
= 1-0413927,
21.
Find the number of
22.
Shew
that
and logl7814-1516 = 4-2507651. digits in 3 12
/21\ 100 (
—
is
x 28
.
greater than 100.
J
23.
Determine how many ciphers there are between the decimal
point and the
/1\ 1000 first significant digit
in
(
j
Solve the following equations, having given log 24.
3*~ 2 = 5.
27.
2F = 2 2 * +
29.
2x + y
= 6»
3*
=3
25. 1
.5 3
'
22,
5*
= 10l 28.
2*. 6*- 2
"I
30.
3l
Given log 10 2 = -30103, find log25 200.
32.
Given log 10 2 =
~l
= -84509,
55
and
log
~ 3 *=2* + 2
=5 2 *. 7
~ x - y =4-y
2 2x
31.
-.30103, log 10 7
log 3,
26.
-.
+ ij-
2,
1
7.
.
"*.
\
=3 3j/_a;J
find log 7N/2
and logV27.
CHAPTER
XVII.
EXPONENTIAL AND LOGARITHMIC
SERIES.
In Chap. XVI. it was stated that the logarithms in common use were not found directly, but that logarithms .are first found to another base, and then transformed to base 10. 219.
In the present chapter we shall prove certain formulae known as the Exponential and Logarithmic Series, and give a brief explanation of the way in which they are used in constructing a table of logarithms.
To expand a 1 in ascending powers of x.
220.
By
the Binomial Theorem,
n>l,
if
K)" =
——
nx(nx—\) v
1
1
+ nx - + n
1
r
.
.
2
x (x
-» n"
1) (nx- 2) J ~± -
1 .
3
x (x
)
— nx (nx *
+
)
-s6
n
+
(x — -\ (i).
3 I?
By
putting sb=1,
I
we
.
obtain
(-')•
HIGHER ALGEBRA.
188 hence the seizes
1
+ x+
the xih power of the series (2); that
(1) is
+
,„
is,
rz
3
and
however great n may increased we have
this is true
indefinitely
1+ * +
x2
+
+
]3
|2
The is
x3
xA
/-'ill 1+
=
+
1
1
usually denoted by
+ e
+
1
1
—
l
let
y
1
1
hence
;
+«+
x-
+
X3
+
X4
+
for x, then 6** =
Now
n be
— +—+
+
5
,= Write ex
+ U ^ ^
+
+
+
(
|_4
series
If therefore
be.
e
e
=
«,
-
CX 4
so that c
cV tjj-
cV +
+
-ry
= log/*
by substituting
;
for c
we
obtain a'
= l+x\og a + e
—Vo
+
If
This
is
lr
the Exponential Theorem.
When n
Cor.
+
,»
is infinite,
the limit of
(
1
+-
)
= e.
[See Art. 266.]
Also as in the preceding investigation,
when n
is
it
indefinitely increased, (,1+
x\ n
n)
-
=1+X+
X2 Y2
+
x3 ]3
+
x4 \i
+
may
be shewn that
.
EXPONENTIAL AND LOGARITHM!*! SERIES. tli.tt is,
when n
x
Now m
is
1
—=
l>y putting
,
+- — nj V
the limit of
is infinite,
1
(
c
)
180
T .
we have
H)--K)~={K)7 when n (x\ — — is infinite;
infinite
n
1
Hence the 221.
= e~\ (1 (-.')'
limit of
)
In the preceding
the value of x; also since -
no restriction
article
deserves
placed upon
[Art. 183.]
intelligible.
another point in the foregoing proof which
is
We
notice.
have assumed that when n
1\7
/
nj the limit or
/ :)•••(• nj
\
is
infinite
r-V
2\
n
\
J
xr
.
is -r-
,
\r
for
is
than unity, the expansions we
is less
have used give results arithmetically
But there
*.
e
J
all values
\r
of r.
Let us denote the value of iB
a!
(
Then Since n
"3( ~3 {
—u
™,
1 / = -lx
r— 1\ =
H(*^) z-
u
x-
a,
-^r) by u 1
as
)
n
J
that
is,
r \
,
r
n
r
.
r
—1
+
nr
we have
is infinite,
U X — —= u ,
r
.
;
x
.
u ~ — ur r
.
l
a
It is clear that the limit of ft>
,-x-;
that of
uA
u
is r^-;
hence the limit of u 3
x is .—r-
;
and generally that
of
ut
x is .—
.
is
HIGHER ALGEBRA.
190
The
222.
series
111
ii
+
E
H
+
+
'
~@
which we have denoted by e, is very important as it is the base Logarithms to this to which logarithms are first calculated. base are known as the Napierian system, so named after Napier their inventor. They are also called natural logarithms from the fact that they are the first logarithms which naturally come into consideration in algebraical investigations.
When
logarithms are used in theoretical work it is to be remembered that the base e is always understood, just as in arithmetical work the base 10 is invariably employed.
From
the series the approximate value of e can be determined to any required degree of accuracy ; to 1 places of decimals it is
found to be 2-7182818284. Example
Find the sum of the
1.
1
-,
We
have
e
and by putting x = -
1 in
e
hence the
sum
= l + l+
" 1=1 -
of the series is
1+
1
2.
- ax — x o
Find the
--
+—+
- (e
)
-rg
+
;
ex ,
i2-i3
+ e~
coefficient of
= (1 - ax - x 2
1
1
the series for
a
Example
infinite series
+
n-
x ).
x r in the expansion of
ex
e~x
« «* n n fi = (l-a,-^)|l-, + -^-- +
(-l) rx r
...
+ L_L_ +
1
j.
-
-
EXPONENTIAL AND LOGARITHMIC The
coefficient required
r-1
r
-l) r
223.
7V>
From
expand
log, (1
+
r-2
{l + ar-r(r-l)},
ascending powers of
tt) ira
\.
Art. 220, a"
=-•
1
+ y 1< >g
e
r6
/r (loge a)
8
^ y ^
3
+
•
+
-
(log. c v
|
2
L +x
this series write 1
lii
(-1)'--
(-l)'-ia
(-!)>•
—
101
SERIES.
4+ '
'
for a; thus
(1 +x)'J
=
1
+ y log,
(1
f2 {log, (1
+ *) +
£ {log
+ *)} +
e
Also by the Binomial Theorem, when x <
(i+«y-i+» + Now
g^*+ y fr-
in (2) the coefficient of
,+ 1.3* + 1.2J .
that
r**^
rp&
%K/
\K/
is,
Equate
This
is
Example.
—
known If
+Ct')
1 >fr- 8)
rf+
(1).
(2).
+
'
-
+
±
t
y in
(1)
;
thus
we have
-- -+
as the Logarithmic Series. {log,, (1
+ x)} 9
in ascending powers of
equating the coefficients of y 2 in the series (1) and required expansion is double the coefficient of y'2 in ?/(!/-!) y (y ____. r2 x +
1) (y
2)
.
+
(2),
y(y-i)(y-2)(y-3)
,
1.2.8
is,
...
we have
By
that
+
•>••*
= 7J--+
x < 1, expand
.
3
«C»
this to the coefficient of
l0g t,(l
a;)}
1.2.3.4
o
2i
+
is
+
£ — -- + —
m
y
1
(1
1.2.3.4
we
.t.
see that the
^+
'
double the coefficient of y in
(y-l)(y-2) y-1 + 1.2* 1.2.3
Thn8
{log.(l +
* +
(y
-
1) (y - 2) (y
"1.2.3.4
-
3)
+
*)P=2{^-i(l+l)^(l + l + l)«*-
}.
.
HIGHER ALGEBRA.
192
Except when x is very small the series for log e (l + x) 224. can, however, is of little use for numerical calculations. deduce from it other series by the aid of which Tables of Logar-
We
ithms
may
By
be constructed.
writing
—
obtain log.
1
writing
for
hence
:
1
1
n+1 )- lo& w =
lo S.(
By
we
for x
1
+
S"2? 3^"
x we obtain log e
n -
1
(1)
-
hence, by changing
;
signs on both sides of the equation, log 8 n
From
(1)
log.("
and
-
log e (n
-
1)
= - + s—a +
#7
—3 +
(2).
by addition,
(2)
+ l)-log,(n-l) =
2(-
+_+__+
(3).
...J
obtain log e 4 — log e 2, that is log e 2 ; and by effecting the calculation we find that the value of log 6 2- -69314718...; whence log e S is known.
From
by putting n =
this formula
Again by putting n =
9
we
we
3
obtain log e 10
— log e 8; whence we
findloge 10 = 2-30258509....
To convert Napierian logarithms
we
multiply by
common
,
which
its
value
=-j=
.
system, and
is
into logarithms to base 10
the modulus [Art. 216] of the 1
is
—
^' oOJjOoDk) J .
we
shall denote this
modulus by
,
or '43429448...;
.
/x.
In the Proceedings of the Royal Society of London, Vol. xxvn. page 88, Professor J. 0. Adams has given the values of e, /x, log e 2, log e 3, loge 5 to more than 260 places of decimals. 225. If we multiply the above series throughout by /x, we obtain formulae adapted to the calculation of common logarithms.
Thus from
(1), /x
loge (ra
+
1)
- /* log
e
?i
=£ _
^ + JL
-. ...
•
;
EXPONENTIAL AND LOGARITHMIC SERIES. that
193
is,
log I0 ( M + 1)
Similarly from l
From
-
log,
n=£-
-t-
+
Jt.
m
_
(2),
^-lo >-l)^ + ^ + + ^ g]
either of the above results
we
ot one of
see that
if
(2)
.
the logarithm
two consecutive numbers be known, the logarithm of the other may be found, and thus a table of logarithms can be ° constructed. " ld
Sl
«*"«*«*
e
that the above formula are
only needed ; +^ to J2 calculate the logarithms of prime numbers, for the logarithm ot a compose number may be
logarithms of
component
its
obtained by adding together the
factors.
In order to calculate the logarithm of any one of the smaller prime numbers, we do not usually substitute the number in either of the formula or (1) (2), but we endeavour to find some value ot n by which division may be easily performed, and such that
either 7^+ then find
1
or
n-l
contains the given number as a factor. or log(w -l) and deduce the logarithm of °
We
log(n+l)
the given number. Example.
By
Calculate log 2 and log 3, given ^=-43429448.
putting n = 10 in 1
-
2 log 3
(2),
we have the value
of log
10-
log 9; thus
= -043429448 + -002171472 + -000144765 + -000010857 + -000000868 + -000000072 + -000000006 1-2
;
log 3 =-045757488,
log 3 = -477121256.
Putting M = 80 in 4 log 3 - 3 log 2 - 1
(1),
we obtain
log 81 -log 80; thus
= -005428681 - -000033929 + -000000283 - -000000003 = -908485024 - -005395032, log 2 = -301029997.
3 log 2
In
the
next
article
we
shall
give
another
series
for
+ l)-\ ge n which is often useful in the construction of Logarithmic Tables. For further information on the subject the
iog9
{7i
reader is referred to Mr Glaisher's article on Logarithms in the hncyclopcvdia Britannica. H. H. A.
•> I
x
x
HIGHER ALGEBRA.
194
In Art. 223 we have proved that
226.
3
2
log e (l
changing x into -
cc,
+
x)
=
x~2
+
3"-~-"'
we have
log.(l -«)=-*—2 ~
By
—
3
2
J"
subtraction,
Put
—
=
=
-
l-x +
los* oeV(n
Note.
log. &e
{
x3
a /
x—
so that
,
n
—
\);
+x
1
.
n = 2< 5
(2w +
-x
Zn +
+
7
x5
\
= 1
777^
=
we thus
va
+ l) 3
3(2?i
l
;
+
obtain
j-t=
^-r.5
5(2w +
This series converges very rapidly, but in practice
+
...}-.
l) is
J
not always
so convenient as the series in Art. 224.
The following examples
227. chapter.
Example that
8 are the roots of the equation ax 2 + bx + c = 0, shew
If a,
1.
a 2 ,02
3
-J—x2 + a—^- x 3 -...
\og(a-bx + cx'1 ) = loga+(a. + p)x
+ 8=
—
,
a/3
a
= - we ,
a
=a log (a
(1
+
{a
+ B)x + aBx 2 )
+ cur)
)
3
+ px). (1 + ax) + log (1 + Bx)
(1
- bx + ex 2 = log a + log 2
ffi
have
a - bx + cx 2 =a {l
.-.
.
o
-
Since a
subject of the
illustrate the
B2x 2
3
B3x 3
x a _ + a_-... = loga + ax- a-x' + Bx-'^-+^--...
= \oga + {a + B)xExample log (1
log (1
+ x + x2
2.
is
2
a
+ *+^
Prove that the coefficient
—
2
)
^^
a
1
or - according as
—— =log
1-x 3
+ x + x 2 = log-)
Q
X6
X9
(1
n
is
or
- x'3) - log
(1
n
n
X 3r
of is
(
3
-
.
...
xn in the expansion
not a multiple of
3.
x)
X2
X3
xr
\
of
'
3
'
'
EXPONENTIAL AND LOGARITHMIC SERIES. If
n
a multiple of
is
from the
first series,
denote
3,
it
by
I
together with
g coefficient is
If
.
is
- n
+
n
then the eoeffieient of *»
3,-;
fr
om
1
93
j,
_I
^ J
the second 3eries;
or _ ?
n
not a multiple „f
the required coefficient
is -
3,
*» does not ocour in the
first series,
therefore
.
n
To prove
228.
For
that e is incommensurable.
not, let e
if
= ™ where m
^ui..
then
1
1
and n are positive integers;
.
7i
If
1
1
1
(w
|n+l
"
multiply both sides by \n;
•
'•
m ^irJ But
is
___J___
= integer + -i- +
w+1
—L + _____J_____
n+
(n
1
+
it is
(n+])(jw.2)(n+S)
+ -"
1
l)(n+ 3)
a proper fraction, for
1
*
(n+l)(n+2)
(n
+ 1 )~(^T2)^T3) + *
greater than
and
*
less
'
than the
geometrical progression _
n+\ that
is,
less
+
_i (n+1)
i__ 2
'
(n+iy\3 +
than I; hence an integer
a fraction, which
is
absurd; therefore
•••
'j
equal to an integer plus
is
e is
incommensurable.
EXAMPLES. XVIL 1.
Find the value of
-02^3 + 5-7
4
2.
+ 5~6 ? -.-+...
Find the value of
2
2 . 22
^3
.
23
i72 4
+
5
.
25
"
* '
15
9
o
+
1
;
.
HIGHER ALGEBRA.
19G 3.
Shew
that
a? .+_+ hge (n+a)-logt (n-a)=2l-a + ^
a?
*\*&
/>»o
»
4.
O"**
2
Shew a
6.
4 +•••»
3
^+
x =y +^- +
shew that 5.
*t/
%A/
y=*'-2 +-3 -
if
\
. . .
that
+ ~2
:
(-
+ )+^P 3 V
+...
' )
a
V~^" /
Find the Napierian logarithm of
= log a-log e
——
e
Z>.
correct to sixteen places
of decimals. 7.
Prove that
8.
Prove that
e" 1
= 2 (/ .-1 +
iog,d + xr« ( i -»•)'-'=« 9.
3
2 .-=
\
+ nr +....)
+
(£ +
.
o
+••••)
•
Find the value of
H
-f
''- 2
j2
(-'
4
-#
4 )
+
|i
^-
//,m -
+
• •
Find the numerical values of the common logarithms of and 13; given ^ = -43429448, log 2 = '30103000. 10.
11.
Shew that
12.
Prove that logc (l
and
— 2
are each less unity
)
term of the
—+—
—+
...
series.
Prove that ,
1
find the general
14.
ax2 and
+ 3a-+2^2 = 3.r-
find the general
13.
and
if
Expand
+ 3.?
„
term of the
5x2
S5X3
65.iT4
series.
—^— in a series of ascending powers of
x.
7, 1
;
EXPONENTIAL AND LOGARITHMIC SERIES. Express -
15.
(e ix
+ e~ ix
in ascending
)
powers of
.r,
-where i'=
197
— 1.
25
16.
Shew that
17.
If a
and
/3
x 2 -jtxr + tf = 0, shew that
be the roots of
a 2 4- ft 2
1
of the series
2
„
i
log,
i\ n A =1 -) (1 +
nj
the form
2(» + l)
2.3(>4-l) 2
#,
Shew
3 -
if
De expanded in
3 3
coefficient of
n be
""
a series of ascending
o:
—
1
n is
n be odd,
if
form 4m.
of the
that
+
23
+ ,
]2
33
43
+
(4
J3
+ - =5e
-
Prove that 2 log,
n - log,
Shew
that
(«
— ft
+
+1
1)
If log,
= Yq
-
2 = 7a - 26 +
and calculate
-
log, (»
—
_ 1)=-
—+
1
2(?i+l) 2 1
1_
n
2
2?i
'
«,
3c,
loge
=~ 25
log, 3
=1
1
—+ _+ "
1
3n 3 81 1()
?; '
S«
go
a - 36 + 5c,
log, 2, log, 3, k>g e 5 to
+
3(7i+I) 3
24
<)
log,
+ l) 3
.t-
shew that the
4m + 2. and
3.4(>t
,
x
1
24.
.
i
^,^^,2^ l+.r + .^ + .,
powers of
23.
4
3
,
i
-,
If log,
20.
22.
ft 3
Shew that
19.
21.
Bum
If .r
18.
n 3 4-
= C sheW '
log, 5
that
= 1 6a - Ah + Vc
8 places of decimals.
or of
CHAPTER
XVIII.
INTEREST AND ANNUITIES.
In this chapter we shall explain how the solution of 229. questions connected with Interest and Discount may be simplified by the use of algebraical formulae.
"We
terms
Value in their ordinary arithmetical sense ; but instead of taking as the rate of interest the interest on ,£100 for one year, we shall find it more convenient to take the interest on £1 for one year. shall use the
Interest, Discount, Present
230. To find the interest given time at simple interest.
Let
is
P be
n
year,
and amount of a given sum in a
the principal in pounds, r the interest of £1 for one the amount. the number of years, I the interest, and
M
The interest Pnr that is,
of
P
for one year is Pr,
and therefore
for
n years
;
/
M=P(l+nr)
is,
From or P,
?i,
231.
(2).
and (2) we see that if of the quantities P, n, M, any three be given the fourth may be found.
(1)
r,
To find
the present value
due in a given time, allowing simple
Let
(1).
M = P + I;
Also that
=Pnr
P be
£1
discount of a given
sum
interest.
D
the discount, the present value, for one year, n the number of years.
the given sum,
r the interest of
V
and
r, 7,
I
'
INTEREST AND ANNUITIES.
199
Since V is the sum which put out to interest at the present time will in u years amount to P, we have
P=
V(\+nr);
1
+ nr
P
D=P-
And
1
+ nr
'
Pnr 1 + nr The value of D given by this equation is called the true discount. But in practice when a sum of money is paid before it is due, it is customary to deduct the interest on the debt instead of the true discount, and the money so deducted is called the banker's discount; so that Note.
Banker's Discount = Pnr.
Pnr
True Discount =
+ nr'
1
Example. The difference between the true discount and the banker's discount on £1900 paid 4 months before it is due is 6s. 8d.; find the rate per cent., simple interest being allowed. Let r denote the interest on £1 for one year; then the banker's discount 1900r
^~
——
1900
•
is
~~
''
and «. the *true adiscount* i
-
,
is
i* 1900r
~3~
1900r
""•
~3~
1
7~T~3 l r
;
+i
1900r 2 =3
whence *'•
t,
_ ~ 1 ±Jl + 22800 ~ 3800
.
•
•
r
Rejecting the negative value, .-.
+ >-;
1±151 3800
'
152
1 we nave f—aSui = o? ,
rate per cent.
«
= 100r = 4.
To find the interest and amount of a given sum in a given time at compound interest. 232.
Let
n
the
P denote
number
the principal, 7? the amount of £1 in one year, the amount. of years, I the interest, and
M
.
HIGHER ALGEBRA.
200
P at
the end of the first year is PR ; and, since this is the principal for the second year, the amount at the end of 2 x Similarly the amount at the or PR the second year is 3 end of the third year is PR and so on ; hence the amount in n years is PR" \ that is,
The amount
of
PR R
.
,
M=PR"; I=P(R -l). n
.'.
If r denote the interest
Note.
on £1
for
one year, we have
R = l+r. In business transactions when the time contains a 233. fraction of a year it is usual to allow simjyle interest for the Thus the amount of ,£1 in ^ year is fraction of the year. v reckoned 1 + - ; and the amount of in 4f years at compound
P
interest
n+
PR*
is
(1 + ^ r
Similarly
.
the
amount
of
J
P
in
—m/
—
m years is PR" (In \
)
more than once a year there is a distinction between the nominal annual rate of interest and that actually received, which may be called the true annual rate ; thus if the interest is payable twice a year, and if r is the nominal If the interest is payable
annual rate of
interest, the
amount
of
£1
in half a year
and therefore in the whole year the amount or
1
+
r
+
r
is 1
of <£1 is (1
r +-^
,
+ -J,
2
—
;
so
that
the
true
annual
rate
of
interest
is
4 r
2
If the interest is payable q times a year, and if r is the nominal annual rate, the interest on .£1 for each interval is
234.
r -
,
and therefore the amount
In (f
of
P
in
n
years, or
qn
intervals, is
this case the interest is said to be "converted into principal"
times a year.
INTEREST AND ANNUITIES. If the interest
201
convertible into principal every moment, then q becomes infinitely great. To find the value of the amount, r 1 put - = — so that q - rx thus x q is
:
,
the
amount =
P
(l
+-Y =P(l + -Y" = P {(l + i)T' = Pe nr
since
x
is infinite
when q
To find
235.
is infinite.
and discount of a given allowing comjwund interest.
the present value
due in a given time,
stun
D
P
be the given sum, V the present value, the discount, the amount of £1 for one year, n the number of years.
Let
R
[Art. 220, Cor.,]
,
V
is
time, will in
n
Since
the sum which, put out to interest at the present years amount to P, we have
P=VR
n ',
it
D = P(l-R-).
and
The present value of £672 due in a certain time is £126; interest at 4£ per cent, be allowed, find the time; having given
Example.
compound
log 2
= -30103,
log 3 = -47712.
Hol) = iI' *»**=!'
Here
Let n be the number of years
;
then
672=126 25
.
•'•
?ll0g
=
24 .
or
100
?ilog
96 .-.
n (log 100 - log 96)
thus the time
is
°g
.
i26-'
16
=logy
;
= log 16 - log 3,
n=
"
;
672
.
1
y
=
4 log 2 - log 3 2 - 5 log 2
•72700 -01773
very nearly 41 years.
- log 3
=
Veiy nea y '
'
if
HIGHER ALGEBRA.
202
EXAMPLES. When
XVIII.
required the following logarithms log 2 = -3010300, log 7
= -8450980,
a.
may
be used,
log 3 = '4771 213, log 11
= 1-0413927.
Find the amount of £100 in 50 years, at 5 per
1.
cent,
compound
interest; given log 114-674 = 2-0594650.
interest the interest on a certain sum of money is ,£90, and the discount on the same sum for the same time and at the same rate is £80 ; find the sum. 2.
At simple
3.
In how
compound
cent,
many
years will a
interest
sum
of
money double
itself at 5
per
?
Find, correct to a farthing, the present value of £10000 due 4. at 5 per cent, compound interest given hence years 8 log 67683-94 = 4-8304856. ;
In
5.
compound 6.
mean
how many interest
years will £1000 become £2500 at 10 per cent,
?
Shew that
at simple interest the discount between the sum due and the interest on it.
is
half the harmonic
Shew that money will increase more than a hundredfold a century at 5 per cent, compound interest. 7.
What sum
8.
amount
to
£1000
of money at 6 per cent, Given in 12 years ?
log 106
= 2-0253059,
log 49697
compound
in
interest will
= 4-6963292.
A man
borrows £600 from a money-lender, and the bill is renewed every half-year at an increase of 1 8 per cent. what time will elapse before it reaches £6000 1 Given log 1 18 = 2-071882. 9.
:
amount of a farthing in 200 years at 6 per cent, log 106 = 2-0253059, log 11 5-0270 = 2-0611800. Given compound interest? 10.
"What
is
the
Annuities. *
An annuity is a fixed sum paid periodically under 236. certain stated conditions ; the payment may be made either once Unless it is otherwise a year or at more frequent intervals. stated
we
shall suppose the
payments annual.
An
annuity certain is an annuity payable for a fixed term of a life annuity is an years independent of any contingency annuity which is payable during the lifetime of a person, or of the survivor of a number of persons. ;
1
;
;
INTEREST AND ANNUITIES.
203
A
deferred annuity, or reversion, is an annuity which does not begin until after the lapse of a certain number of years ; and when the annuity is deferred for n years, it is said to commence after n years, and the first payment is made at the end of n + 1 years. If the annuity is to continue for ever it is called a perpetuity if it
does not commence at once
it is
An
annuity left unpaid for a certain number of years to be forborne for that number of years.
To find
237.
amount of an annuity
the
number of years, allowing simple
A
;
called a deferred perpetuity.
left
is
said
unpaid for a given
interest.
be the annuity, r the interest of £1 for one year, n the number of years, the amount.
Let
M
the end of the first year A is due, and the amount of this sum in the remaining n - 1 years is A + (n — 1) rA ; at the end of the second year another A is due, and the amount of this sum in the remaining (a— 2) years is A + (ii — 2) rA and so on. Now is the sum of all these amounts
At
•
M .-.
M={A + (n-l)rA} + {A + (n-2)rA} + n terms
the series consisting of .-.
+ (A + rA) + A,
J/=Wil+(l + = nA +
2
+ 3+
—— — K
'-
+n-l)rA
rA.
238. To find the amount of an annuity left given number of years, allowing compound interest.
Let the
A
be the annuity,
number
R
the
amount
M the amount.
of years,
unpaid for a
of <£1 for
one year, n
the end of the first year A is due, and the amount of this n~ sum in the remaining n— 1 years is at the end of the ; second year another A is due, and the amount of this sum in the n~ 2 remaining n - 2 years is and so on.
At
AR
AR
.-.
M = AR
n
= A(l
~
x
m R =A R^l.
;
2
+ AR"- +
+R + R
x
2
+
+AR to
2
+
AR + A
n terms)
;
HIGHER ALGEBRA.
204
of annuities it is always
In finding the present value
239.
customary to reckon compound interest; the results obtained when simple interest is reckoned being contradictory and untrustworthy. On this point and for further information on the subject of annuities the reader may consult Jones on the Value of Annuities and Reversionary Payments, and the article Annuities in the Encyclopaedia Britannica. 240.
a
(jiven
To find t/ie present value of an annuity to continue for number of years, allowing compound interest.
be the annuity, R the amount of £>\ in one year, n the number of years, V the required present value.
A
Let
The present value
is
AR~
in 2 years
is
AR~'J
due in 3 years
is
AR~
A due
of
the present value of
A due
the present value of
A
and so
in
year
1
l
;
;
3 ;
[Art. 235.]
on.
Now V
is
the
sum
of the present values of the different
payments
V=AR- +AR1
.-.
= AR~
=A Note.
1
+ AK- 3 +
tow terms
- R~"
l-R-
1
1-R-"
R-l may
This result
given in Art. 238, by
l
3
Rn
also be obtained [Art. 232.]
.
Cor. If we make n infinite of a perpetuity
we
R-V
by dividing the value of M,
obtain for the present value
'
r
mA
is the present value of 241. If years' purchase. is said to be worth
an annuity A, the annuity
m
In the case
of a perpetual annuity 1
m=r
=
mA — —
100 rate per cent.
;
hence
— INTEREST AND ANNUITIES.
205
that is, the number of years' purchase of a perpetual annuity obtained by dividing 100 by the rate per cent.
is
As
instances of perpetual annuities we may mention the income arising from investments in irredeemable Stocks such as
many Government
A
Securities, Corporation Stocks, and Railway good test of the credit of a Government is fur-
Debentures. nished by the number of years' purchase of its Stocks ; thus the 2| p. c. Consols at 96} are worth 35 years' purchase Egyptian 4 p. c. Stock at 96 is worth 24 years' purchase while Austrian 5 p. c. Stock at 80 is only worth 16 years' purchase. ;
;
To find
242.
commence
at the
value
present
the
end of p years and
of a deferred annuity to continue for n years, allow-
to
ing compound interest.
Let
A
be the annuity,
R
the
amount
£1
of
in one year,
V the
present value.
The
first
payment
made
is
at
the
end
of
(;>
+
l) years.
[Art. 236.]
Hence the present values
of the
first,
second, third... pay-
ments are respectively
AR- * +l \ AR- (p+2 \ AR- (p+3 \ {
.'.
V=AR- (p+l) + AR- {p+ »+AR1
= AR~ (p+1)
AR~ V
-
(1,+3
>+
...
ton terms
7?~"
—
A R- p -"
Cor. The present value of a deferred perpetuity to commence after p years is given by the formula V
243.
A freehold
estate
~R-V is
an estate which yields a perpetual
annuity called the rent ; and thus the value of the estate to the present value of a perpetuity equal to the rent.
is
equal
It follows from Art. 241 that if we know the number of years' purchase that a tenant pays in order to buy his farm, we obtain the rate per cent, at which interest is reckoned by dividing 100 by the number of years' purchase.
~
HIGHER ALGEBRA.
206
Example. The reversion after 6 years of a freehold estate is bought for £20000; what rent ought the purchaser to receive, reckoning compound Given log 1-05 = -0211893, log 1-340096 = -1271358. interest at 5 per cent. ?
The years,
rent is equal to the annual value of the perpetuity, deferred which may be purchased for £20000.
Let
for 6
'
£A
be the value of the annuity; then since
20000^* ^° 5
.R
= l-05, we
have
'- 6 ;
•0o
A x (1-05) 6 = 1000; log A -6 log 1-05 = 3, .-.
log .-.
A = 3-1271358 = log
A = 1310-096, and
1340-09G.
the rent
is
£1340.
Is. lid.
Suppose that a tenant by paying down a certain sum lias obtained a lease of an estate for p + q years, and that when q years have elapsed he wishes to renew the lease for a term p + n years the sum that he must pay is called the fine for renewing n years of the lease. 244.
;
Let A be the annual value of the estate then since the tenant has paid for p of the p + n years, the fine must be equal to the present value of a deferred annuity A, to commence after ;
p years and
to continue for .
n
the fine
n years
—
AR-* = -= A—1
EXAMPLES. The
interest is supposed
compound
;
that
is,
~
AR- p n A—1
XVIII.
r
.
4
[Art. 242.1
b.
unless the contrary
is stated.
A
person borrows ,£672 to be repaid in 5 years by annual instalments of ,£120; find the rate of interest, reckoning simple interest. 1.
Find the amount of an annuity of ,£100 in 20 years, allowing compound interest at 4| per cent. Given 2.
log 1-045 = -0191163,
log24-117 = 1-3823260.
A
be
freehold estate is bought for £2750 3. let so that the owner receive 4 per cent,
may
A
freehold estate worth 4. rate of interest.
£120 a year
is
at what rent should it on the purchase money ? ;
sold for £4000; find the
INTEREST AND ANNUITIES.
How many years' purchase should be being calculated at 3i per cent.? interest estate,
207
5.
given for a freehold
6.
If a perpetual annuity is worth 25 years' purchase, find the of an annuity of £625 to continue for 2 years.
amount
perpetual annuity is worth 20 years' purchase, find the annuity to continue for 3 years which can be purchased for £2522. 7.
If a
8.
When
9.
Find what
the rate of interest is 4 per cent., find what sum must be paid now to receive a freehold estate of £400 a year 10 years hence; having given log 104 = 2-0170333, log 675565- -8296670.
sum
will
interest being payable every
amount to £500 in 50 years moment; given e _1 = '3678.
at 2 per cent.,
If 25 years' purchase must be paid for an annuity to continue and 30 years' purchase for an annuity to continue 2?i years, find the rate per cent. 10.
n
years,
A
man borrows £5000 at 4 per cent, compound interest if the 11. principal and interest are to be repaid by 10 equal annual instalments, find the amount of each instalment having given ;
;
log 1-04 =-01 70333
and log 675565 = 5-829667.
A man
has a capital of £20000 for which he receives interest at 5 per cent. if he spends £1800 every year, shew that he will be ruined before the end of the 17 th year; having given 12.
;
log 2
= '3010300,
log 3
= '4771213,
log 7 = '8450980.
The annual rent
of an estate is £500 if it is let on a lease of 20 years, calculate the fine to be paid to renew the lease when 7 years have elapsed allowing interest at 6 per cent. having given 13.
;
;
logl06 = 2-0253059, 14.
tinue n,
If a,
log4-688385 = '6710233,
log3'118042 = '4938820.
years' purchase must be paid for an annuity to conyears respectively; shew that
b, c
2/i, 3?i
a2 — ab + b 2 = ac.
What is the present worth of a perpetual annuity of £10 15. payable at the end of the first year, £20 at the end of the second, £30 at the end of the third, and so on, increasing £10 each year; interest being taken at 5 per cent, per annum ?
c
; ;
CHAPTER
;
XIX.
INEQUALITIES.
Any
245.
quantity a
said to be greater than another positive; thus 2 is greater than -3, is
quantity b when a -b is because 2 - (- 3), or 5 is positive. Also b is said to be less than a when b- a is negative; thus -5 is less than -2, because — 5— (— 2), or - 3 is negative.
In accordance with this definition, zero must be regarded as greater than any negative quantity. In the present chapter we
shall suppose (unless the contrary is directly stated) that the letters always denote real and positive quantities.
246.
If
a >
b,
then
evident that
it is
>
b
c
>
b
ac
>
be
a+ a—
a
that
is,
+c —c
;
b
an inequality
will still hold after each side has been increased, diminished, multiplied, or divided by the same positive quantity.
247.
a-ob,
If
by adding
c to
each
side,
a>b+ c; which shews that in an inequality any term
from one If
may
be transposed
side to the other if its sign be changed.
a >
b,
then evidently
b
< a
that is, if the sides of an inequality be transposed, the sign of inequality must be reversed.
;
INEQUALITIES.
is,
If a
>
—a—
(—
then a - b
b,
b) is
a >
b,
—a
then
is
negative; that
and therefore
negative,
;
an inequality
hence, if the signs of all the terms of the sign of inequality must be reversed. if
b-a
and
is positive,
—a < —b
Again,
209
< — b, and
— ac < —
be changed,
therefore
be
that is, if the sides of an inequality be multiplied by the same negative quantity, the sign of inequality must be reversed. 248.
a.>b,,
If
a.
>
>b
a
a^>b.,,
boi
it
,
clear
is
that
a + a 2 + a 3 +...+ am > l
and
:
11
If
a>b, and
V
=-
a'
that
;
is
b'
;
a
.
n a~ <
.
than
is,
the arithmetic
their geometric
The
—
,>
mean of
b
2
>
is, a'
>
b'\
where n
is
any
b~".
2ab +
-—^
Similarly
that
2
a2 +
•
.
;
b r
1
that
The square of every real 250. Thus (a therefore greater than zero. .
+ bm
V
>
1
1
—<
...
p, q are positive integers, then ^/a>^Jb,
if
or a > b 9 ; and therefore positive quantity.
Further,
+ b^+ba +
>h AK-'-
a a 2 a,--' a ,n 249.
6,
b
quantity
2
b)
>
2
is
is
positive,
positive
and
;
;
2ab.
> Jxy
;
tivo positive quantities is greater
mean.
inequality becomes an equality
when
the quantities are
equal.
251. The results of the preceding article will be found very useful, especially in the case of inequalities in which the letters
are involved symmetrically. H. H. A.
li
>
x
HIGHER ALGEBRA.
210 Example
If a,
1.
denote positive quantities, prove that
b, c
a 2 + b 2 +c 2 >bc + ca + ab; 2 (a 3 + b 3 + c 3 )>bc (b
and
+ c) + ca(c + a) + ab (a + b).
& 2 + c 2 >2bc
For
(1);
+ a 2 >2c«; a 2 + b 2 >2al); a 2 + b 2 + c 2 > be + ca + a&. c2
whence by addition It
may
be noticed that this result
Again, from
b2
(1)
down
writing
2 (a 3 It
and that
Example .r
3
+l
+ b 3 + c3
2.
If
if
+ c 3 >bc(b+c)
(3).
+ c) + ca
(3) is
+ a) + ab{a+b).
obtained from
(2)
by introducing the
this factor be negative the inequality (3) will
may have any
x
[c
we obtain
real value find
which
is
no
the greater,
or x 2 + x.
x3
Now
[x
-
+l- (x 2 + x) =x 3 - 2 - (x - 1) = (x 2 -l)(x-l) = (.r-l) 2 (* + l).
l) 2 is positive,
hence
x3 + 1
+1
according as x
x—
If
b3
be (b
)
b, c.
(2);
the two similar inequalities and adding,
should be observed that
factor b + c, longer hold.
true for any real values of a,
-bc + c 2 >bc
.-.
By
is
-
1,
is
>
or
<
x2 + x
positive or negative; that
is,
according as x
>
or
<
-
1.
the inequality becomes an equality.
252. Let a and b be two positive quantities, and their product ; then from the identity
$
their
sum
P
4a6 = (a + bf - (a - b)\
we have
iP = S - (a - b) 2
Hence, given,
S is
if
S
least
is
given,
2
P
,
and S 2 = is
greatest
±P+(a- b)
2 .
when a — b\ and
if
P
is
when
a= b; if the sum of two positive quantities is given, their product is greatest when they are equal ; and if the product of two positive quantities is given, their sum is least when they are equal.
that
is,
:
INEQUALITIES.
To find
253.
211
of a product the
the greatest value
sum of whose
factors is constant.
Let there be n factors a, b, sum is constant and equal to s.
c,
...
and suppose that their
k,
and suppose that a and b are we replace the two unequal factors a+b a+b ..
Consider the product abc any two unequal factors. If
...
k,
Tii by the two equal factors —— ii>
a, b
-—
,
—
,
,
,
the product
is
increased
while the sum remains unaltered ; hence so long as the product contains two unequal factors it can be increased tvithout altering the sum of the factors ; therefore the product is greatest when all In this case the value of each of the n the factors are equal. s /s\" factors is - and the greatest value of the product is ( - ) or
n
,
\nj \n/
/a a+b+ \
Cor.
If «,
b, c,
...
+k\ +k\"
...
)
k are uneqiud,
/a + b +
+ n
c
(
\
that
c+ c + n
, '
+k\ n
...
)
7
7
> abc
...
k
;
J
is,
a + b + c+
...
+k
>
n
By an
(abc
...
k)".
extension of the meaning of the terms arithmetic this result is usually quoted as follows
mean
and geometric mean the arithmetic
mean of any number of positive
quantities is greater
than the geometric mean. Example. where
r is
.
that (l r + 2 r + S r +
.
.
.
+ nr n > nn ( \nY )
;
real quantity. lr
c Since
.'.
any
Shew
+ 2 r+Zr +
+nr
n (
)
whence wo obtain the
1
>(l r .2 r .3 r
>l r .2 r .3 r
« r )'
n r that ,
1
is,
;
>(|»)
r ;
result required.
14—2
HIGHER ALGEBRA.
212
m
254. To find the greatest value q/'a b"cp is constant; m, n, p, ... being positive integers.
. .
.
when a + b +
c
+
...
Since m, n, p,... are constants, the expression am b n cp ... will
— (-) be & greatest when \mj \nj expression is the product of (
m — )+ n (—) +p \nj * (
\mj
stant.
are
—) +
(
that
is,
of
o
c
ni
n
p
b
c
m
n
p
n
np
The given expression
sum
a+
Thus the
+
.
.,
.
is
and therefore con-
.
.
.
is
b
b
+ c+
)
greatest
63
M4*4*+"
...\
(
(a - x)* is greatest
c
is
/a +
p
+
+ m+n+p +
when
(
of the factors of this expression
hence {a + x) 3
c
whose sum
when
Example. Find the greatest value of x numerically less than a.
the
+
b
a
Thus the greatest value
m
+
factors
...
this last
be greatest when the factors
will
a
m
m + n +p + or a
.. .,
But
greatest. &
is
...
)
\pj
Hence am b n c p ...
all equal,
\pj
(
)
(a
—— is
when —^— = . *
84
greatest value is
a7
+ x) s
3
(a -a:) 4 for
( J
(
—^—
—j—
—^- J + ,
or
x=
j
4
any
is
I
--
real value
greatest
—^—
)
,
;
but
or 2a;
.
.
The determination of maximum and minimum values 255. may often be more simply effected by the solution of a quadratic equation than by the foregoing methods. Instances of this have already occurred in Chap. ix. ; we add a further illustration.
Example. is
a
Divide an odd integer into two integral parts whose product
maximum.
Denote the integer by 2/i + 1 ; the two parts by x and 2n + 1 - x the product by y then (2n + 1) x - x* = y ; whence ;
2x = (2n +
1)
±
V^h + I) -^ 2
;
;
and
11
.
;
INEQUALITIES.
213
but the quantity under the radical must be positive, and therefore y cannot be greater than
-
value must be n-
(2/t
+ l) 2
+ n\
in
+n+-
2
or
,
n'
and since y
;
which case x = n+
1,
or n
is
integral its greatest
thus the two parts are n
;
and n+1.
Sometimes we may use the following method.
256.
Find the minimum value
Example.
of
—
'
'
c
Put
c+x=y
..
.
the expression
;
—
'
+x
then
=
(a-c + y){b-c + y) ^-^ y
_ (a -
a~
(
is
- c)
c) (b
b
C
+y+a-c+b-c ~ C)
Jy
-^yy + a-c + b-c + 2j (a-c)(b-c)
Hence the expression is a minimum when the square term when y=J(a -c)(b- c). Thus the minimum value
and the corresponding value of x
*J(a -
-
is */(«
c) (b
c) {b
-
-
c)
EXAMPLES. XIX. 1.
Prove that (ab + xy) (ax + by) > 4abxy.
2.
Prove that
3.
Shew
reciprocal
is
sum
never less than
If
5.
If « 2 + 6 2 + c 2 =l,
of
any
real
a>
b,
-c.
a.
positive quantity
2.
l,
and x2 +y 2 + z2 = l, shew that
ax + by + cz < If
c)
and x2 +y 2 = l, shew that ax + by<\.
4.
6.
1.
shew that a a b b > a b b a and loe - < ,
8.
^ + y 2 z + z2x) (xy 2 +yz 2 + zx2 ) > Find which is the greater 3«6 2 or aP+263
9.
Prove that a3 6 + ab3 < « 4 +
7.
Shew that
(.r
that
+ c) (c + a) (a + b) > 8abc.
that the
a 2 + b2 =
;
is
a-c + b-c + 2
(b
zero
is
.
2
losr =
D.'-'-V - 2.
.
64
.
10.
Prove that 6abc < be (b + c) + ca(c + a) + ab (a + b).
11.
Shew that
b 2 c°-
+ c2 a2 + a 2 b 2 > abc (a + b + c).
.
and
its
HIGHER ALGEBRA.
214 12.
Which
13.
Shew
the greater x3 or x2 +x + 2 for positive values of x%
is
that x3 + lSa 2x
>
hax* + 9a3
Find the greatest value of x greater than x^ + Hx.
if
,
x>
a.
in order that
14.
may
7x2 + 11
Find the minimum value of x2 - 12#+40, and the value of 24? - 8 - 9x2 15.
be
maximum
.
16.
Shew that ( \nf > »• and
17.
Shew that
18.
Shew that n* >
19.
If
(x +y + ,s) 3 1
.
3
.
Shew
(2)
7
4
.
> 27^^. 5 .(2n - 1 . .
6.
. .
2?i< (w +
l) n .
).
>l+?iV2 ,7_1
2,
shew that
.
that
(x+y +z) 3 > 27 (y + z- x) (z + x - y) (x+y - z). xyz>(y+z-x)(z + x-y)(x+y-z).
(1)
Find the
22.
.
be a positive integer greater than
?i
2 ft
21.
2
maximum value of
(7
- x) A
(2
+ #) 5 when # lies
between
and - 2. no
—
We
To prove m —+ b\
/a
>
•
v
*257.
am +bm
since
(5 off
Find xu the minimum value T7-
23.
(
—
have
—~—
is
,
)
•
1
m a" + 6 = 1
less
and b are
a that if " except
than
L
f
m
when
-y- +
—
—
,
.
xs
m
•'•"~2~ =
.
and
positive
unequal,
a positive proper jr action.
+
-^-J
(
-^
g-J
we may expand each
of
;
and these
-
expressions in ascending powers of
a" + 6
x)(2 + x) + =-*-* 1+*
m /a + b\
v~2-J
+
——
.
2
[Art. 184.]
m (m - 1) (a + b\"—- fa - b\* 1.2
\~r)
\~r)
m (m - \)(m - 2)(m-3) fa + bV"+ 1.2.3.4
"A
2
J
4
4 fa - b\ + '" j
12
— INEQUALITIES.
a positive integer, or any negative quantity, the terms on the right are positive, and therefore If
(1)
all
m
215
is
n
a" +
>
=
m
If
(2)
and
positive
is
the right after the
first
m
——
>
If on
1
and
\2~)
'
than
less
1,
all
the
terms on
are negative, and therefore
a + --
(3)
fa + b6 V" s
b'
b
m
fr fa + b\
<
m=-
put
positive,
m
where n <
1
;
then
76
i
m
fa +
{-2
m m
b \
=
) 1
'a
m
i 7l
fd" + b n \ nH fa
(-2-)
1
+ bm \ m
— o—
i
n
(a*)
;
1
m
+
>
(b»)
H
/ox
,
o
i
]j
y
(
2)
i >
.*.
Hence the
+
b'"\"
1
>
a+b
—~
— — > fi-7 =
proposition
established.
is
If
m = 0,
or
1,
the
inequality becomes an equality.
If there are n 'positive quantities a, b, c, ...k, then a m +b m + c m + ... + k m /a + b + c+...+k > n n \
*25&.
)
unless
m&
positive proper fraction.
rt
Suppose
on to
and
have any value not lying between
1.
m m Consider the expression a + b + c" + ... + k"\ and suppose that a and b are unequal ; if we replace a and b by the two equal 1
....
quantities
——
a+ _
i
altered,
b
,
a —+ -
b
..
,
,
„
,
the value or a +
+ c+...+fc remains un7
-j
m but the value of a" + b m + c + 1
...
+
k'" is
diminished, since
HIGHER ALGEBRA.
216
Hence
any two of the quantities a, b, c,...&are unequal m m m the expression am + b + c + ... + k can be diminished without altering the value of a + b + c + ...+k; and therefore the value m n of a" + b' + c + + km will be least when all the quantities a, b, c,...k are equal. In this case each of the quantities is equal so long as
1
.
. .
a+
to
b
+
c
+
+k
...
;
n
and the value
of
m a + b m + cm +
n
fa +
...
b
+
c
+
.
then becomes .
.
+ k\ m •
»
{
Hence when
+km
a, b, c,...k
)
are unequal,
am + b m + cn +...+le m
m c + ...+fc ... + k\ /a + b + c+ > fa
n If ?n lies
between
and
1
we may
in a similar
manner prove
that the sign of inequality in the above result must be reversed.
The proposition may be stated verbally
is
as follows
:
The arithmetic mean of the m th powers of n positive quantities greater than the m th power of their arithmetic mean in all cases
except
when
*259.
m
lies
between
If a and h
and
1.
;
'
INEQUALITIES. X
*2G0.
To prove
and y
if'x
/I + 'I±^-
>
that
i-x'
217
/ lh V
v
i-y
5
and ])ositive, and x >
are proper fractions
y.
x
/l+x .71±f >flr<
For
1
,.
according as
+a
1
.
-lo
l-x
But
//g*
|
1 1 +y > or < - log y °l-2/'
Ilog^2(l + !\ & 3 1-2/ 5 \
and
7
2/
1
and thus the proposition
To prove
•261.
<
1
>-
l-x
is
.
log
y
—
'
\+y
-
°
t
1
-
y
,
proved.
+
that (1
777 i'
1 4- a;
.
- log x *
d
AAM ^
S^lzS" 2 ^ +? + ?+•••)»
x)
,+x
(1
-x) 1_x >l,
if
x
/a + b\ a+b k aa bD > ( _ ,
luce that
J
Denote
(1 +jb)
1+ *
(1
-a)
1- *
logP = (l+a)log(l + x) + = x {log (1 +x)=
2x^ r\
--.
Hence that
is,
I
/
log
P is
3
+
\
+
JC
+*)
log
_ / *C
...)-2^
+
+
i//
+
and therefore 1
(1
1
r
(l
P> 1
-*)'-'>!.
+ log (l—x)
x)
+
\
Jit
positive, (1
- x)) +
log (1
^
-aj)log(l-a;)
(l
JO
**s
+
•*/
by P; then
\
SI/
-
6
+
...)
to
1
HIGHER ALGEBRA.
218
f9
put x = — where u > z u
In this result
,
Z
1
l
;
y-^Y>r>rl u
+ z\
'u
u
put u + z =
\
J
+
(u
.'.
Now
z\ ~u
^-i) %1
u)
{
\
Z
sY + w/,
+
then
;
z)
u+
J
*(u-z) u
u — z = b,
a,
.
-z
>u 2u
.
w = ———
so that
;
(TJ * EXAMPLES. 1.
Shew
that 27 (a 4 + & 4 + c4 )
2.
Shew
that
3.
bers
is 4.
n (n + l) 3 < 8
(l
> 3
XIX.
b.
(a + b + c)\
+ 2 3 + 3 3 + ... + n 3 ).
Shew that the sum of the m ih powers greater than n (n + l) m if m > 1.
of the first
n even num-
,
and
If a
are positive quantities, and a
/3
if
n>
that
-SM-jr-
(
Hence shew that
> /3, shew
the value of
1
1
(
+-
lies
)
between 2 and
2-718... 5.
If a,
b, c
are in descending order of magnitude, shew that
/a + c\ a
fb + c\
^.
\a-c) 'a a
6.
Shew
7.
Prove that
8.
If
ii
that (
is
b
\b-cj
+ vb + vc+...+k\ a + b + ~
~
c
+
--
+ ,i
'
< a a b b<*.
'
)
- log (1 + am <
- log
lib
Ih
)
a positive integer and 1 _ #n + J 1
?&
+
<
(1
+ a n ),
x < 1, shew _ A.n n
if
.
m > n.
that
.>&*.
)
:
INEQUALITIES. If a,
9.
b,
c are in
Find the
10.
H.
P.
maximum
»> 1,
and
219
shew that n n -\-c n > 2b n
x3 (4a - .r) 5
value of
i
than 4a; and the
maximum
if
x
is
positive
and
less
i
x*(\—xf when x
value of
.
a proper
i.s
fraction.
If
11.
x
is positive,
shew that
< x and
log (1 +.r)
"
1+.?;
x + y + z=l, shew that the
If
12.
and that
(1
- x) (1 - y)
(1
least value of -
y
h-
is
i)
z
- z) > 8xyz.
(a+b+c+d) (a 3 + 6 3 + c 3 + a73 ) >
13.
Shew
that
14.
Shew
that the expressions
(a 2 + 6* + c2 + cl2 f.
a(a-b)(a-c) + b (b- c) (b-a) + c (c-a) and
-\
x
(c
- b)
+ b 2 (b-c){b-a) + c2 (c-a){c-b)
cfi{a-b)(a-c)
arc both positive.
that (xm + y m ) n
15.
Shew
16.
Shew that abfr <
17.
If
at,
6, c
(.t'
fa 4-h\
(^p)
n
a
+y n
+
,
if
m > n.
.
denote the sides of a triangle, shew that
cannot be negative; p, (2)
MI )
b
a 2 (p-q)(p- r) + b 2 (q - r)
(1
18.
<
q, r
(q
-p) + c2
(r
being any real quantities;
ah/z + b 2zx + c2 xy cannot be positive, if
Shew that
[1
\2n-l >
j3 15
p
L
\a\bJ±\ c is
the quotient and r
»
P "r (|g)
x + y + z = 0.
(\n)n
are positive integers, If a,b,c, d, to n, shew that the least value of 19.
where q
-p) (r - q)
whose sum r
is
equal
(|g+ 1 ) > the remainder when n is divided by £>.
;
CHAPTER XX. LIMITING VALUES
AND VANISHING FRACTIONS. a
262.
If a be a constant
be made as small as is,
we can make
we
when x
the fraction
by
sufficiently increasing
a - approximate to zero as nearly as
is infinite
;
--
can
x
please
by taking x large enough "
finite quantity,
this is usually abbreviated
the limit of
-
is
x
we
;
that
please
by saying,
zero."
x
Again, the fraction - increases as x decreases, and by making
x
x as small as we please we can make thus
when x
is
as large as
x
zero - has no finite limit;
we
please
this is usually ex-
JO
pressed by saying, "
when x
is
zero the limit of -
is infinite."
When we
say that a quantity increases without limit or is infinite, we mean that we can suppose the quantity to become greater than any quantity we can name. 263.
when we say that a quantity decreases without limit, we mean that we can suppose the quantity to become smaller than any quantity we can name. Similarly
used to denote the value of any quantity is used to which is indefinitely increased, and the symbol denote the value of any quantity which is indefinitely dimi-
The symbol
nished.
go
is
LIMITING VALUES.
The two statements
204.
symbolically as follows if
if
221
may now
of Art. 2G2
be written
:
x
is co
x
is
,
then -
is
,
then
is
;
x co
.
x
But in making use of such concise modes of expression, it must be remembered that they are only convenient abbreviations of fuller verbal statements.
The student will have had no difficulty in understanding the use of the word limit, wherever we have already employed it; but as a clear conception of the ideas conveyed by the words limit and limiting value is necessary in the higher branches of Mathematics we proceed to explain more precisely their use and 26~>.
meaning. If y =f(x), and if when x approaches a Definition. value a, the function f(x) can be made to differ by as little as we please from a fixed quantity b, then b is called the limit of
266.
y when x — a. For instance,
1+2 +
2
2
+
Here S as
n
we
2~J
is
please
+
S
if
'"
;
denote the sum of n terms of the series
then
'
S'
=
2
~2^-
a function of n, and
by increasing n
;
that
^— is,
t
,
Art 56 *1
can be made as small
the limit of
S
is
2
when
is infinite.
We
shall often have occasion to deal with expressions 267. consisting of a series of terms arranged according to powers of some common letter, such as
a + a x + a ax" + a 3 x3 + x
where the
coefficients
independent of
x,
o
,
a,,
a2
,
and the number
a3
,
...
are
of terms
quantities be limited or
finite
may
unlimited. It will therefore be convenient to discuss some propositions connected with the limiting values of such expressions under
certain conditions.
,
HIGHER ALGEBRA.
222 The
268.
limit of the series
a + a x + a 2 x 2 + a 3x 3 + x
when x
Suppose that the
is
a
.
an
series consists of
infinite
number
be the greatest of the coefficients alf asi o3 , us denote the given series by a + S ; then
Let
let
diminished
is indefinitely
b
S
and
if
x<
1
Thus when x small as
we
please
is ;
+
...
...
;
and
;
bx
S< — 1 x
we have
,
3
of terms.
-=
.
S
can be made as hence the limit of the given series is a indefinitely diminished,
.
number
If the series consists of a finite
of terms,
S
is less
than in the case we have considered, hence a fortiori the proposition
is true.
In
269.
the series
a
+ a,x + a2 x + a 3 x,3 +
.
.
.
by taking x small enough ive may make any term as large as we please compared with the sum of all that follow it ; and by taking x large enough we may make any term as large as we please compared with the sum of all that precede it.
The
ratio
of
the term an x
n
to the
sum
of all that follow
it is
ax an+1 x
When
x
n+l
n
a
+a n+2x"
+2 '
+
or '
...
a n+1 x + a u+2 x*+..
denominator can be made the fraction can be made as large
indefinitely small the
is
as small as we please as we please.
that
;
is,
Again, the ratio of the term a nx precede it is
a xn a n—l,cc
where u y=x
.
n
l
n
+an — »x 2
2
+...'
or
n
to the
sum
of all that
a 2 +...' a n — ,y + a n—2
—
.
LIMITING VALUES.
When
223
indefinitely largo, y is indefinitely small hence, fraction can be made as large as as in the previous case, the
we
x
is
;
please.
270. position
The following particular form is
foregoing pro-
of the
very useful.
In the expression a x + a H — ,x ii
H-i
+
+ a.x + a
, »
1
1
consisting of a finite number of terms in descending powers of x, by taking x small enough the last term a can be made as large as we please compared with the sum of all the terms that precede it, and by taking x large enough the first term ax* can be made as large as we please compared with the sum of all that follow it.
Example 1. By taking n large enough we can make the first term of n 4 - 5/i3 -7/i + 9 as large as we please compared with the sum of all the other terms that is, we may take the first term ?i 4 as the equivalent of the whole expression, with an error as small as we please provided n be taken large ;
enough.
Example
3.t
—
3
Find the limit of =-=
2.
2x'2 :
—4
— when
(1)
x
is infinite
:
(2)
x
is
zero.
In the numerator and denominator we
(1)
3a; 3
the
first
;
hence the limit
^-s
is
,
3 or ^
OXr
(2)
When
-4
x
is
P
indefinitely small the limit
Find the limit of
is —--
—
/ 1- + x - when x V 1—x
^
/
denote the value of the given expression
log
P=i
X
{log (1+ x)
Hence the is
2 e' .
limit of log
P
terms but
is 2,
;
1
,
or - 2
is
zero.
by taking logarithms we
-log (1-x)}
^(l + ^' + '^+.-.V required
all
.
*
3.
disregard
O
8
Example Let have
may
[Art. 226.]
and therefore the value of the
limit
.
HIGHER ALGEBRA.
224
VANISHING FRACTIONS. Suppose
271.
required to find the limit of
it is
2 x 2 + ax — 2a
x 2 - a2
when x = a. If
we put x = a + h, then h
approaches the value
2
x
and when h
is
approach the value zero as x
a.
Substituting a + h for
x +
will
x,
ax— 2a 2
-a
3ah + h 2
2
=
2
2ah + h
2
=
3a + h
2a~+h'
indefinitely small the limit of
this
expression
3
.
-.
is
a
There
is
however another way x 2 + ax - 2a2 x 2 — a2
and ^r
,
if
of regarding the question; for
- a) (x + 2a) (x — a)(x + a)
x + 2a x+a
(x
we now put x = a
value
the
of
the
'
expression
is
as before.
-j
2
If in the given expression
simplification
it
fi
.
x^ — a
found that
will be
2
~— we
pat x = a before
assumes the form -
it
,
the
value of which is indeterminate ; also Ave see that it has this form in consequence of the factor x -a appearing in both numerator and denominator. Now we cannot divide by a zero factor, but as long as x is not absolutely equal to a the factor x - a may be removed, and we then find that the nearer x approaches to the value «, the nearer does the value of the 3 fraction approximate to ^
,
or in accordance with the definition of
Art. 266,
when x = a, i
,
!
i
.
.
,
n
the limit of
x t ax — ^
— Jia
.
x~ -^a"
.
is
o
2
^
.
.
225
VANISHING FRACTIONS. If
272.
f(x) and
two functions of x, each of which some particular value a of x, the
(x) are
becomes equal to zero for fraction
^~
(a)
and
takes the form Ki v
called
is
a
Vanishing
Fraction.
Example
x = S, find the limit of
If
1.
T 3 -5:r 2 + 73-3 3 x- — ox— '6 .T
When x = 3,
the expression reduces to the indeterminate form ^; but by
removing the factor ~ 2x+1 becomes 2 x + 2x + 1 ,
x-3 from numerator and When x = S this reduces to
denominator, the fraction ,
which
is
therefore the
4
required limit.
Example
To
The
2.
fraction
J'^-a-Jx + a x-a
becomeg
when % _ a
numerator and denominator by the surd conJ'dx-a- Jx + a; the fraction then becomes
find its limit, multiply
jugate to
(Sx-a)-(x + a)
2 ^
,
or
[x- a )(Jdx-a + Jx + a)' whence by putting x = a we
find that the limit is
1
Example
3.
The
J'6x-a+
fraction
-
—j=
21 x
becomes ^ when x=l. 1 _% x
To find its limit, put x = l + h and expand by Thus the fraction 1
-
(1
+
fe)*
l-(l + /0i
_
3
V
the Binomial Theorem.
9
l-(l+J*-^»F+-.) 1
3 1
+
-5 +
Now h = when
>Jx +
a
1, -;<
9 2
7
/l
-
25 5
*ael; hence the required limit
is
-
Sometimes the roots of an equation assume an indeterminate form in consequence of some relation subsisting between the coefficients of the equation. lo H. H. A. 273.
HIGHER ALGEBRA.
226 For example,
ax + b = ex +
if
(a
— c)x = d — b, x=
But
= a, then x becomes
c
if
a simple equation
d,
d-b a—
—— j-
c
,
or go
indefinitely great
is
that
;
is,
the root of
the coefficient of x
if
is
indefinitely small.
The
274.
solution of the equations
ax + by + c = be'
ab'
If ab'
— a'b =
— =—= m
0,
c—
m
suppose
ax + b J
ii
c'
=
H
c,
are
—
now
— c'a '
ab'-a'b
equal to ^
c.
infinite.
for
a',
In this case
b\
second
the
the two equations ax + by +
c
=
and
only in their absolute terms, and being J
differ
Ml
m is
0,
c — = 0.
inconsistent cannot be satisfied
If
ca' 1
by substituting
;
not equal to
is
- b'e — ab
+c =
then x and y are both
equation becomes ax + by +
If
a'x + b'y
0,
we have
by any
finite values of
-=-=-, a
b
x and
y.
and the two equations
c
identical.
Here, since be —
b'e
=
Q each assume the form -
,
and
ca'
— c'a —
and the solution
the values of x and y is
indeterminate.
In
the present case we have really only one equation involving two unknowns, and such an equation may be satisfied by an unlimited number of values. [Art. 138.] fact,
in
The reader who
acquainted with Analytical Geometry will have no difficulty in interpreting these results in connection with the geometry of the straight line. is
;
:
VANISHING FHACTIONS.
227
We
shall now discuss some peculiarities 275. arise in the solution of a quadratic equation.
which may
Let the equation be
ax 2 + bx + If c
c
-
0.
= 0, then 2 ax + bx = 0;
x=
whence that
one of the roots
is,
If 6 sign.
=
0,
is
0,
or
—
:
a
zero and the other
is finite.
the roots are equal in magnitude and opposite in
[Art. 118.]
If
that
a= in
namely —
the equation reduces to bx + c = this case the quadratic furnishes 0,
=- .
for
we
appears one root,
it
only
But every quadratic equation has two
order to discuss the value of the other root
Write —
and
;
roots,
and in
proceed as follows.
x in the original equation and clear
of fractions
*J
thus 2
cy + by + a
Now
put a =
0,
=
0.
and we have cy
the solution of which
is
y J
—
2
+ by = 0;
—c> b
0,
or
;
that
is,
x=
oo,
or
—
ofx
2
T
.
b
'
Hence, in any quadratic equation one root will become if the coefficient
c
infinite
becomes zero.
This is the form in which the result will be most frequently met with in other branches of higher Mathematics, but the student should notice that it is merely a convenient abbreviation of the following fuller statement
In the equation ax 2 + bx + c = 0, if a is very small one root is very large, and as a is indefinitely diminished this root becomes indefinitely great. In this case the finite root approximates to
-y
as its limit.
o
The
may
cases in which more than one of the coefficients vanish be discussed in a similar manner.
15—2
HIGHER ALGEBRA.
228
EXAMPLES. XX. Find the limits of the following expressions, (1) -
when #=oo,
(2s
-3) 7#2 -
(3
-5*)
(2)
when x = 0. (3r?-
I)*
)
CHAPTER
XXI.
NVERGENCY AND DIVERGENCY OF
SERIES.
terms are formed expression in which the successive at if the series terminate is called a series ^ ! lr law w some reguJ« by number q{ lb ca'iea a lt:
Ax
076
some
assigiMgl
terms
^e
^m
t^
unfimitfd, it
is
is
^
;
called
an
In the preset chapter we an expression c- ° ue form u + n2 + 3 +
^J.
B infinite series.
shall usually denote a series
i/-
by
+u +
x
/
consisting of w terms. Suppose that we have a series function of n; if n increases The sum of the series will be a equal to a certain
tends to become indetinitely, the sum either becomes infinitely great. finite fcmi*, or else it
An
infinite series
is
said to be convergent
numerically the first n terms cannot quantity however great n may be.
when
the
exceed some
sum finite
of
be divergent when the sum of numerically greater than any finite the first n terms can be made great. quantity by taking n sufficiently infinite series is said to
An
of a given we can find the sum of the first n terms divergent or we may ascertain whether it is convergent becomes in-
978 series
If
the W examining whether made finite,
when n
series
remains
finite,
or
indefinitely great.
is
For example, the sum
of the first
n terms of 1-*" .
1
2 + x + x + x* +
...
is
.
_a
•
.
the series
T 230
HIGHER ALGEBRA.
If
x
numerically
is
j—-
finite limit
x
If
—y
is
d e greater cUve?ge n t.
serie's
If
,
and by taking n
""*
™Y
&n * e
d
1,
sum
series is therefore
r i^entUm * ^ ** x= -
the
1,
numerically greater than
is
n terms
and the
,
than
less
sum
the
1,
;
of
sufficiently great,
^
*
-,
the
converge,^!
l""*^
"
appro.
thus
"»
*" d
«»
t„ first
tk can *>•»
***»
is
^
the series becomes
1-1+1-1+1 -1+ The sum of an even number of terms is of an odd number of terms is 1 and thiw +\L between the values and 1. Thi t,fe Ac which may be called o-iBo^ or
,;L the sum •„
I
X A^c^ •
T
^™ ^
^
5
are many 0ases in which we'haic sum of, the first n terms of a series. therefore to investigate rules by which we off
here /^' ?i finding the.
S: 280.
4n
Let the
°f a
tra/mfe series
series
^
«*•
,
6 giVe
"
h od
can test the cTt
-*Hout
u,
effecting
its
+ u -M +
SerfeS
may
5
....
be Written in each of the following
K-«,)+(«,-«0 +(».-«,) +
».-K-«J-(«4 -«,)-K-«r)From
p"
w* taAicA *Ae ferw are alternately
w1 >^>^ M a o >w,> * 4
for,™
We
be denoted by
M -%+%where
,„.
1
^ (
2 ).
(1) we see that the sum of any number of terms is a positive quantity; and from (2) that the sum of any nnmber of terms is less than «, ; hence the series is convergent.
CONVEltGENCY AND DIVERGENCE OF SERIES. For example, the
281.
,
1
is
convergent.
sum
log e
is
By
series
11111 f-
h
1
2
231
4
3
putting x
6
5
in Art. 223,
1
-•
we
see that its
2.
Again, in the series
23 +
T~2
6
5
_7 + ~6
'
numerically less than the preceding term, and the therefore convergent. But the given series is the sum of
each term series is
is
i
1
and
Now as the
_5 3"4 +
4
(1) is
number
11111 -2 3-4 5"6
+
1-1+1-1
+
+
equal to log e of terms is
2,
+
and
+
1-1
(2) is equal to
even or odd.
m '
,
(1) ' (2).
or 1 according'
Hence the given
series
convergent, and its sum continually approximates towards log,, 2 if an even number of terms is taken, and towards 1 + log 8 2 if an odd number is taken. is
282. An infinite seizes in which all the terms are of the same sign is divergent \f each term is greater than some finite quantity
however small. each term is greater than some finite quantity a, the sum of the first n terms is greater than na and this, by taking n sufficiently great, can be made to exceed any finite
For
if
;
quantity.
Before proceeding to investigate further tests of convergency and divergency, we shall lay down two important principles, which may almost be regarded as axioms. 283.
remain convergent, and if divergent it will remain divergent, when we add or remove any finite number of its terms for the sum of these terms is I.
If a series is convergent it will
;
a finite quantity.
which
the terms are positive is convergent, then the series is convergent when some or all of the terms are negative ; for the sum is clearly greatest when all the terms have the same sign. If a series in
II.
We
all
shall suppose that all the
contrary
is
stated.
terms are positive, unless the
;
HIGHER ALGEBRA.
232
infinite series is convergent if from and after some the ratio of each term to the preceding term is numerically
An
284.
fixed term less than some quantity zuhich
Let the
series
+u + 12 c>
U
—<
let
U
r,
:
where
r
<
< u. ~1
< 1
-r
,
+ u^4 +
U1 U — < r, -*A < Us U2
r,
it,
since r
Hence the given 285.
u^3
+u + u+u 12
u,
1
is,
than unity.
r
1.
Then
tliat
less
beginning from the fixed term be denoted by u,
and
numerically
is itself
3
/_.
u9
u.
V
^
(1
<
+
r
A 4
+
ua u x
+ r2 + r 3 +
n u u u 3 u 2 Ul
1
)
1.
series is convergent.
In the enunciation
of the preceding article the student
should notice the significance of the words " from and after a fixed term."
Consider the 1
— CONVERGENCY AND DIVERGENCY OF
SERIES.
'2Xi
An
infinite series in which all the terms are of the same 286. sign is diverge) it if from and after some fixed term the ratio of ea<-li
term
to the 'preceding
term
is
greater than unify, or equal to unify.
Let the fixed term be denoted by t* If the ratio is equal to of the succeeding terms unity, each is equal to u and the sum is equal to of n terms nu hence the series is divergent. .
,
l
;
than unity, each of the terms after the and the sum of n terms is greater fixed term is greater than u than nu hence the series is divergent. If the ratio is greater
x
,
;
}
In the practical application of these tests, to avoid having to ascertain the particular term after which each term is greater or less than the preceding term, it is convenient to find 287.
the limit of
—
be denoted by
n—\ A.
If If
X< \>
when n
-
indefinitely increased; let this limit
is
1,
the series
is
convergent.
1,
the series
is
divergent.
[Art. 284.] [Art. 286.]
the series may be either convergent or divergent, and a further test will be required ; for it may happen that If
—— < n—
X=l,
1
but continually approaching to
1
as
limit ivhen
its
n
is
m
indefinitely increased.
quantity r which
Hence the
is
In this case we cannot name any finite itself less than 1 and yet greater than X.
test of Art.
tinually approaching to Art. 286.
We
1
shall
"the limit
of
284
If,
however,
—u — >
uH —
as its limit, the series
1
use " Liin
fails.
—— "
un —
—
U — when n un —
is
an abbreviation
as
1
but con-
I
divergent by
of the
words
,
1
is infinite."
.
1
Example 1
1.
Find whether the
series
whose n lh term
—
-
.,
di-
vergent or divergent.
„
is
?/
n
('„_!
(n
+ l)a: n n
2
ru^1 {h
-
1)
him
(n
+ l)(n-l)-
2
n*
— "n
—x\ I
—
is
con-
;
;;
HIGHER ALGEBRA.
234 hence
If
x < 1 the
series is convergent
if
x > 1 the
series is divergent.
—
—u =1, and
x = l, then Lim
Example
if
a further test
required.
is
Is the series
2.
l2
+ 2 2x + 3 2 x 2 + 4?xs +
convergent or divergent?
_ Here
T Lim .
Hence
If
x = 1 the
Example
series
~ n2 x n l —. (n-l)-x n„
—un
=Lim-.
7jL
un-i
9 2
=x.
if
x < 1 the series
is
convergent
if
x> 1
is
divergent.
the series
becomes
l2
+ 2 2 + 32 + 4 2 +
. .
.
,
and
is
obviously divergent.
In the series
3.
~ a+(a + d)r+{a + 2d)r 2 +... + (a + n-1 d)r n 1 + .
d (n —wn—=Lim a— -^--.r = a (n-2)d
, Lim
T
.
—.
+
»*-i
thus
if
r< 1
the series
is
1)
-t-
.
...,
convergent, and the
sum
r;
is finite.
[See Art. 60, Cor.]
two infinite series in each of which all the terms are ])Ositive, and if the ratio of the corresponding terms in the two series is always finite, the two series are both convergent, 288.
If
there are
or both divergent.
Let the two
infinite series
be denoted by
u + ua + ua + w4 +
,
x
+ + 12
and
v,
The value
v,
v,3
+
v.
4
+
of the fraction
+n
u + u , + ua^ i
lies
between the greatest and
\ and
is
therefore
Hence
if
a,
one
series is infinite
proposition.
n
least of the fractions
-»,
-*,
finite quantity,
L
[Art. 14.1
say
series is finite in value, so is the other; if
one
which proves
the
in value, so
is
the other;
CONVERGENCY AND DIVERGENCY OF
235
SERIES.
The application
289.
of this principle is very important, for can compare a given series with an auxiliary
by means of it we series whose convergency or divergency has been already established. The series discussed in the next article will frequently be found useful as an auxiliary series. The
290.
infinite series
1111 y T
2P
3P
always divergent except when p
is
4.1'
is positive
and
greater than
Case I. Let;? > 1. The first term is 1 the next two terms together are ;
2 j—; Z
1.
than
less
4 the following four terms together are less than-—; .
the
fol-
4
lowing eight terms together are
—
than
less
;
and so
Hence
on.
o
the series
4
2 is less
than
I
8
+ t^+th+ttt, P 4' 2
that
is,
less
is less
Case The
than II.
series
1,
since
p>
1
hence the series
;
is
common
ratio
convergent.
Let_p=l.
now becomes
1
+^ + 2
The
+•••;
1
than a geometrical progression whose
2
~j
o
-^
+ - + =;+ 4
3
...
5 2
1
t
^ —
third and fourth terms together are greater than - or
;
4 1 the following four terms together are greater thau ^ or the o 2 8 1 following eight terms together are greater than and so or ;
—
Hence the
on.
1111 +
and
is
than
series is greater
2
+
2
;
+
2
+
'"'
2
[Art. 2^6.]
therefore divergent.
Case III. Let p<\, or negative. Each term is now greater than the corresponding term Case
II.,
therefore the series
is
divergent.
always divergent except in the case when positive and greater than unity.
Hence the
p
is
in
series is
— HIGHER ALGEBRA.
236
Prove that the series
Example.
4
n+1
with 1 + «
+ «+v
3
2
is
divergent.
Compare the given
Thus
series
•"••••
"^
th terms of the given series and the n and v n denote the n
if
«*
we have
auxiliary series respectively,
un i'
— =1,
_n + l n
n
_ w+ 1
1
.
2
re
'
?i
7/
hence
Zi-m,
and therefore the two
series are
both convergent or both
divergent. But the auxiliary series is divergent, therefore also the given series is divergent.
This completes the solution of Example
In the application
291.
—
limit of
should be finite
the
,
nth term
highest powers of n.
u
- is finite
288
of Art.
necessary that the
it is
this will be the case if
;
way
auxiliary series in the following
Take u
Art. 287.
1.
find our
:
of the given series
Denote the
we
and retain only the
by v n
result
then the limit of
)
by Art. 270, and v may be taken as the
7i
th
term of
the auxiliary series.
Example r
1.
Shew that
the series
whose n th term
3/2n2 is
,,
Z/S?vi +
-
= 1
is
2n+5
divergent.
As n
increases,
un approximates
l/w 1
Hence,
if
v„=-r ,we ~-
have
to the value
or '
Lim
n1
therefore the series
series.
But
divergent.
whose nth term
4/3
i
*
n 12 3
— = ^r, 3 v u
/2
n
— 1
is
which
is
a finite quantity;
v/
may
this series is divergent [Art. 290]
;
be taken as the auxiliary
therefore the given series
is
CONVERGENCY AND DIVERGENCY OF Example
Find whether the
2.
vn = is
series in
^/;<
:}
which
+l -n
convergent or divergent.
"»=«
Here
\\/ * + tf ~ 1
//
+
(
~3n 2 If
we take v n =
=
,
5
1
)
+
we have
9n'J
3
— =x. v„
Luti
3
n
But the auxiliary
*J
»-» + -;"
9><
v MN
series
Jl JL P + 22 + 3 2+
JL
is
237
SERIES.
'"
convergent, therefore the given series
is
l1+ "
-
n
convergent.
To shew that the expansion of (1 292. T/teorem is convergent when x < 1. ?* th
Let u r u r+l represent the pansion ; then ,
u wr .
When r>?6+l,
+ x) n by
the
and (?-+l) th terms
Binomial
of the ex-
n-r+1
,
r
negative; that is, from this point the terms are alternately positive and negative when x is positive, and always of the same sign when x is negative.
Now when x <
this
ratio
is
7/
r
is
infinite,
— =x
Lim —
numerically
;
therefore
the series is convergent if all the terms are of the same sign; and therefore a fortiori it is convergent when some of [Art. 283.] the terms are positive and some negative. since
1
To shew that the expansio?i of a x in ascending 293. of x is convergent for every value of x.
Here
—W
*-
=
#
— —^ 1°#«
;
•
7-
and therefore Lim 1
-
1
n-1 «„_, the value of x; hence tlie series
•
—U
=-
«*__, is
convergent.
<
p owers
11 whatever 1
1
be
.
HIGHER ALGEBRA.
238
To shew that the expansion of log (1 + x) in ascending powers of x is convergent when x is numerically less than 1. 294.
— —= u
-i
ni
Here the numerical value of
n't
I
is
equal to x If
a5
the series becomes
is less
1— k + 77-t+--> 4 3 2
an ^
than is
1.
con "
[Art. 280.]
vergent. If
which in the limit
hence the series is convergent when x
\
= l,
x.
n
,
x~ — 1,
the
becomes —
series
— -—
1
_
t"
•••>
an(^
*s
q a 4: O This shews that the logarithm of zero is divergent. [Art. 290.] infinite and negative, as is otherwise evident from the equation
e-°°=0. 295.
and
The
results of the
two following examples are important,
will be required in the course of the present chapter.
—2--x when x
\q
Example
Find the limit
1.
of
is infinite.
Put x = ev; then logs
y
y
~ eV
X
i
-
when x
Example Let
x=-
is infinite
Shew
2.
so that
,
y
is infinite
that
when n
yi
y
y'
\2
y also
y*
;
2
^
3
+"
hence the value of the fraction is infinite
the limit of nx n = 0,
is
zero.
when x
y>l;
if
also let y n =z, so that
n\ogy = logz; then
—
^^ = logy' fu^=—n = iz'logy y
Now when
n
is infinite z is infinite,
and
logz .
z
—s_ = 0;
also
logy
is finite;
z
Lim nx n = 0.
therefore
296.
It
is
sometimes necessary to determine whether the
product of an infinite number of factors
is finite
or not.
Suppose the product to consist of n factors and to be denoted by io uMAia then if as n increases indefinitely u <<1, the product will ultimately be zero, and if u n > 1 the product will be infinite ; hence in order that the product may be finite, u- must tend to the limit 1 ;
2
.
CONVERGENCY AND DIVERGENCY OF Writing
+ vn
1
239
SERIES.
u n the product becomes
for
,
(l+* )(l+*,)(l+*8)
(l+O-
1
P and
Denote the product by
take logarithms
j
then
logP = log(l+v ) + log(l+v 8 ) +...+ log(l + vJ
(1),
1
and
in order that tlie product
may
be
finite this series
must be
convergent.
Choose as an auxiliary
series
+v n
v,+v 2 + v3 +
_1
/
log(l +
Now
product
(2) is
if
_.
r-
the limit of
convergent, (1)
un
"
/==1,
vn
\
when
is
.
2** +
^ = Lim\
vn
since the limit of v n
Hence
t;)
r Lim-2-l .
(2).
I
is 1
convergent, and the given
is
finite.
Example.
Shew
that the limit,
13
3
5
when n
'4*1*6' 6
2'
of
2n-l 2n + l ~JT~'~2ir
7
5
is infinite,
is finite.
Uj,
The product consists of 2n factors; denoting the m 2 Ug,... and the product by P, we have
successive pairs by
,
P=u
v 2 u3
u n>
2n-l 2«+l
,
•
1
—
s n = — 2m
where
x
**
—=
-5
2n
1
- t-?; 4«-
logP = logM 1 + log« 2 + logM 3 + ...+logM n
but
and we have
Now
to
shew that
this series is finite.
log« n = log (l
therefore as in Ex.
2,
(1),
-^)=-
~^
'•'
32/i-1
Art. 291 the series is convergent,
and the given product
is finite.
In mathematical investigations infinite series occur so 297. frequently that the necessity of determining their convergency or divergency is very important ; and unless we take care that the series we use are convergent, we may be led to absurd conclusions. [See Art. 183.]
HIGHER ALGEBRA.
240 For example,
we
we expand (l—x)~ 2 by
if
the Binomial Theorem,
find (
1
- a;)- 2 =
1
3 + 2x + 3ar + 4a +
obtain the sum of n terms of this series as explained in Art. 60, it appears that
But
we
if
in + 1
+
2.*;
O
9
+
3ar
n-l
...
+ nx
^
*
=
t=
(1
^^ 1 - x
—=
ra
- x)
:
whence +'2x+ 3x~ + -'—'—
I
(l-x) 2
+ nx -
...
4-
^+
7z
1-*
(1-a?)9 i
ByJ making° n
infinite,
we
(l-x) 2
garded as the true equivalent of the
nx x vanishes. ri + =— l-x -x)~
-rz
(1
If
n
is
or aj>l,
that
infinite series
+ 2x + 3x2 + ix3 +
1
when
-a can only be re-
see that -z
it is
infinite,
this quantity
becomes
infinite
when x=l,
and diminishes indefinitely when a,*
\
+ 2x + 3x2 + 4#3 +
Ta =* 1
and we should be
to
inf.
295], so
j
if we were to use 2 x)~ by the Binomial Theorem as if it were the expansion of (1 In other words, we can introduce the true for all values of x. 2 infinite series 1 + 2x + 3x + ... into our reasoning without error if
the series
is
divergent.
The
is
led to erroneous conclusions
convergent, but
we cannot do
so
when the
series
divergent series have compelled a distinction For to be made between a series and its algebraical equivalent. 2 example, if we divide 1 by (1 - x) we can always obtain as many terms as we please of the series difficulties of
,
l
whatever x
may
be,
2 + 2a;+3£ +4a;3 +
and
so in a certain sense
called its algebraical equivalent
;
yet, as
valence does not really exist except
-p.
^
niay be
we have seen, the when the series is
equi-
con-
—
1
CONVERGENCY AND DIVERGENCY OF vergent.
It
therefore
is
more appropriate
SERIES.
to speak of
241
— (l
—
X)
as the generating function of the series +2a,-
1
+ 3a 2 +
being that function which Avhen developed by ordinary braical rules will give the series in question.
The use
alge-
term generating function will be more fully explained in the chapter on Recurring Series. of the
EXAMPLES. XXI.
Ill
Find whether the following .
1
#
x and a
series are 1
x+a
x-^-2a
convergent or divergent.:
_ 4.
1
x
a.
+ 3a
.v
being positive quantities. 1
1
1
1
+ + + + 1.2 273 371 475 -
x and y
1
4
(*+i)(y+i) >+a)(y+*)
xy
(*+3)(y+3)
+
'
being positive quantities.
x
4
1
1
_1_
6
x3
x2
x* h
1.2^2.3^3.4^4.5 1
1
o»o
/)»2
/>»
tf
1
o***
\Mj
\Mj
.
%A/
+ + + + T72 3T4 576 778 n
,
6
-
7
-
8.
1+
2
__
11.
I+ I + I+
+ toe + bx + la? + 9af* +
1
2
ia
42
32
\/l + \/i + \/f + \/1 +
y *
22
"i^
1
1 + i. + A + + 2/' 4p 3p
+
x+
+ 2
+ 5
Ib
+ - + ,^TT +
—
3 2 8 , 15 A - x + -x* + x + 5 10 17
H. H. A.
.,
.
.
.
+
n 2 -\ -.,—-..
nl +
xn + 16
HIGHER ALGEBRA.
242 -.n
-,
12.
l
2.r+— + -- + 8
/2 2
1C 15 '
14
,
,
—
-2*p-i+
2' l
— — —
1 14.
6
2
+ g * +5 *» +I ^P + „. +
2\-!
(p-l)
+
--
-,—
..+
?t
/3 3 (2-3
3\- 2
-2)
3
2
+
.
27
+
3
4\- 3 _4\
4
/4 /4_
+«= \3 4
=" 3^
4
4 2 3 +- + - +-+
16.
1
17.
Test the series whose general terms are
Jn*+l-n.
(1)
18.
jtF+l- Jnt-^i.
(2)
Test the series 1
1
1
1
.r
A+l
a+2
x+3
1
1
/1N
11
/on (2)
1
—-+ -+ # -1r-+— T5+ a 2 A+l ^ -25+— +
A-
x
being a positive fraction. 19.
Shew that the
series
1+ is
convergent for 20.
all
2"
+
3^
+
4"
I I E
+
values of p.
Shew that the
infinite series
ux + 2i 2 + u 3 + u± + is
convergent or divergent according as Lim^fu n 21.
Shew
is finite
22. infinite,
1
when n
Shew
<1, or >1.
that the product 2
•
iti
2 *
4
4
3' 3' 5
'
6 5
2ti-2 2tt-2 2n 271-3' 2»-l"S^Ti
is infinite.
that when x=\, no term in the expansion of (1 +#)" is except when n is negative and numerically greater than unity.
CONVERGENCY AND DIVERGENCY OF
SERIES.
The
*298.
tests of convergency and divenrencV & cxltv Art* 9x7 ogi 11 «» . proved in the next article enables of sriven
m
6
"
venter
when
<
;
divergent if
**
ddit£0U,a
onrf
—
5-
U n-i
Let
I.
3?
+
,7
^
wUch
Me „*»*»
+
---
S01 » eti »' es
> _-n V.,
"»
fa-d
con-
some particular term
AWjori
«,<« J,
b
1
+
•••
the v-servs is convergent if after
Let us suppose that
Case
1
2»
l»
— ^
wp nave i.™ we
n/^T ^ ST^
I +.1 +
tergent
243
«,/t(!re the v-sertes is
'
Wj
and
are the particular terms.
»,
* < Ei &
.
then
2
w,
+
w + u3 + fl
=
2t
w
V
that
<
is,
Hence, vergent.
Case
II.
if
—
(v. -f
i
**
2
w,
+v +
-y
y
)
the ^-series is^convergent the w-series
Let
- > 3» 2
v,
**,
^^ tt,
a
i
M + ^9 + i
V
V,
;
«
is
also con-
then
2
U,
Va
+
27,
J
16—2
:
HIGHER ALGEBRA.
244 that
—
>
is,
Hence,
if
(v
+ v2 + v 3 +
l
the ^-series
...).
divergent the
is
it-series
is
also di-
vergent.
We
*300. have seen in Art. 287 that a series is convergent th or divergent according as the limit of the ratio of the ?i term In the to the 'preceding term is less than 1, or greater than 1. remainder of the chapter we shall find it more convenient to use this test in the equivalent form :
A series
convergent or divergent according as the limit of th the ratio of the n term to the succeeding term is greater than 1, or less than
is
1
Similarly
;
that
is,
theorem
the
1,
preceding
the
of
—— >
Lim
according as
or
<
article
1.
may be
enunciated
The
when the
w-series will be convergent
Lim
provided that
—u— > Lim —v—
vergent when the v-series
Lim *301.
The
"When p >
is
—— -
(
—
thatis,if
.
that
is, if
1,
convergent or di-
< 1.
convergent, and in this
is
+ iy.
or (l
,
/
is
or
n?
u n+i
n
un
> l + g + -J CP-lV+ „ >
B
di-
.
(n+iy
JS
be
with the auxiliary series
series
the auxiliary series case the given series is convergent if
l
is
>>
1 \
1
un Un +
it-series will
^^ Lim ^.
Let us compare the given
whose general term v
convergent
divergent provided that
is
Lim \ n
vergent according as
and the
;
whose general term
series
v-series is
n
u
2n~
,\ l
KCr
r
Lim \n I
(
p+
—
-
Wh
^
p (p-l)
1
)
1
J)
+
>)>.
'
.
'
CONVERGENCY AND DIVERGENCY OF But the auxiliary
series is
245
SERIES.
convergent if y; is greater than 1 ; hence the first part of the
by a Unite quantity however small proposition is established.
When p<
the auxiliary series is divergent, and by proceeding as before we may prove the second part of the proposition. 1
Find whether the
Example.
a;
l is
series
^ + L3
1 2* 3
+
x=
2.4*
1.3 .5
+
5
x[
2~i.d'T + '"
convergent or divergent.
Here Lim the series If
is
1 —— =-; it
u n+l
hence
x"
x
convergent, and
the series
is
5
(2w-
3)
1
6
(2n -
2) ' 2~/T=T
if
if
x>l
divergent.
x= 1, Lim
u — —= u
In this case
1.
n+l
1 M" _
~2
*302.
a;
=1
.
......
(2n - 1) (2n
'•
hence when
4
.
-
2n(2n+l)
wn un+1
and
3
-
"Urn
the series
1)
(2n-l)2
J"
'
convergent.
is
whose general term
T/ie series
vergent, according as
-
Lim ( n
log
—-
j
>
un
is
or
1,
is
<
convergent or di-
1.
Let us compare the given series with the series whose general
term
is
— n
-
l
When
p>
1
the auxiliary series
case the given series
is
convergent
u-sn+
that
or
is, if
log
!
if
if
lv + ij
/„
>
1
[Art. 300.]
;
I
—— > p log (1
log
convergent, and in this
is
—
**
'u
.,
71+
>P « '
" 1
h
)
:
7J
^-52 +
2n
"3
.
HIGHER ALGEBRA.
246 that
:
——
Lim In log
is, if
>p.
)
Hence the first part of the proposition is established. When p < 1 we proceed in a similar manner in this ;
auxiliary series Example.
is
divergent.
Find whether the
series
5 5 x5
4*r 4
3 s x3
2 2z 2
is
case the
convergent or divergent. (n + l) n+l /pW+1
n n
„ Here
—un*- = n— x
wn+l
|n_
.-.
Hence
if
a?<- the
.
'
Lim
[n
H)'
3l = -1
wn+1
[Art.
#>-
the series
is
divergent.
u n+l
•.log
-n— = loge-wlog( 1
3n8+ "
~2n un
.
.
hence when x = - the series
If
1
)
'
1
Lim [ n log —— is
+-
J_
_1
1
=-
divergent.
Lim -^- = 1, and wn+1
also
Liminf-^- - l)) =
\ u n+ ) tests given in Arts. 300, 301 are not applicable.
series
220 Cor.l.
^St—
e
To
A lyV
(n+l)**
+l
series is convergent, if
If:r=-,then
*303.
7^
'- v
discover a further test
whose general term
is
we
—n
shall
make
r-
(log n) p
.
i
1,
the
J)
use of the auxiliary
In order to establish
the convergency or divergency of this series proved in the next article.
we need
the theorem
,
CONVERGENCY AND DIVERGENCY OF
247
SERIES.
*304. If $ (n) is positive for all positive integral values of n and continually diminishes as n increases, and if a be any posit ive integer, then the two infinite series (l)
and
a<£ (a)
+
<£(2)
+
+
(3)
3 2 + a <£(a 2 ) + a
(a
3
)
+
...
+
.
. .
+
(n)
...,
+ a n <£ (a n ) +
.
. .
are both convergent, or both divergent.
In the
us consider the terms
first series let
(«*+ 1), (a
k
+
k
2),
(a
+
i+1
S),
beginning with the term which follows
0)
)
(«*).
The number of these terms is ak+l - ak or ak (a- 1), and each them is greater than <£(a* +1 ); hence their sum is greater than ,
of
1
ak (a-
k+1
1) (a
By
);
that
is,
greater than
x
giving to k in succession the values 4>(2)
ak+l
(a
cf>
0, 1, 2, 3,...
k+1 ).
we have
++W>^x«*W;
+ 4>(3)-f
Co
<]>(a
+
+
1)
<£(«
+
+
2)
(a+
by addition,
therefore,
—
$!
+ <£(«*)>
3)+
<£(1)
>
S
2
x
a 2(a 2 )
;
,
ct
where
S
denote the sums of the first and second series respectively; therefore if the second series is divergent so also is the £,
,
2
first.
Again, each term of (1) is less than <£(«*), and therefore the sum of the series is less than (a— 1) x ak (ak ).
By
giving to k in succession the values
{2) (a
+
+
I)
therefore,
<£(3)
+
(a
+
4>(4)
+
2)
+
+
+
(a
+
3)
<£(«)
+
< (a-
0, 1, 2, 3...
we have
1) x <£(1); 2
+(a
)<(a-
1) x a(a);
by addition
4-+(l)<(«-l){4 + *(l)}; hence
if
the second series
is
convergent so also »'
is
the
first.
Note. To obtain the general term of the second series we take (») the general term of the first series, write a n instead of n and multiply by a n .
HIGHER ALGEBRA.
248 The
*305.
>
if p
and
1,
whose general term
series
divergent if p
=
or p
1,
<
^ — r— n(logn)
is
convergent u
is
p
1.
By
the preceding article the series will be convergent or divergent for the same values of p as the series whose general
term
11
is 1
1 c\y
ft" \l
a"(loga") p
The constant
factor
X
(\V
(n\oga) p
'
common
is
r_
7=
'
np
(log a)'
'
to every term
there-
;
fore the given series will be convergent or divergent for the
values of p as the series whose general term
required result follows.
The
*306.
Lim
vergent according as
\\\
Let us compare the given
term
is
— j)
>
—
-
(
—
series
1
Hence the
]
is
un
—
1
is
convergent or di-
log
>
n
>
1,
or
<
1.
with the series whose general
Now when n log (n
+
l)
convergent, and in this convergent by Art. 299, if
the auxiliary series
1
is
un M,+i is
is
(w + l){log(n+l)}'
n
(!)•
{log n) v
very large,
= log n + log
Hence the condition
(1)
un + .
is,
.
-.
case the given series
that
-
n (log ny
When
thatis,
—
[Art. 290.]
whose general term
series
is
same
,
l
(
+
1
J
=
log
n+-
becomes
nj
VN
'
n
VN
log O
n
, '
^>(l + l)(l+P nj \ nlog n log n) \
u H+l
u — - > u .,
n+l
i
1
1 +- + ?i
P
wlogw o
'
,
nearly;
CONVERGENCY AND DIVERGENCY OF 1
)
>
P
+
1
240
SERIES.
;
.
l0g?4
or
Hence the second part Example.
CSr
1
)- 1 } 10*"^
part of the proposition is established. The be proved in the manner indicated in Art. 301.
first
may
Is the series
22
2 2 .4 2 .6 2
2 2 .4 2
^3 2 ^3 2 .5 2 ^3 2
.5 2 .7 2
convergent or divergent?
A. = *£* . 1 + I +
Here
—*- =1, and we proceed to the next
* (1).
ti
Lt«i
.-.
Fromfl),
.-.
Lim In
(
»fe-l)=l+5 -1 - -1)1=1, and we
*-»
ffe-
1
—^— =
[Art. 295];
n
<
pass to the next
)- 1 } 108
•••^"[ffesince Lt/u
test.
1
2>-
test.
^'^
)-
1
108
}
"] 30
hence the given series
is
'
divergent.
We have shewn in
Art. 183 that the use of divergent series in mathematical reasoning may lead to erroneous results. But even when the infinite series are convergent it is necessary to exercise caution in using them.
*307.
For instance, the
series JC
-
Ou
%)C
+
+
4/2~J/3 convergent
is
series 1
by
474~^5
+
'"
when x=l.
itself, 1
Jb
[Art. 280.] But if we multiply the 2n the coefficient of x in the product is 1
+
1
+
1
»
'
.
HIGHER ALGEBRA.
250 Denote
this
by a2n
;
then since 1
1
a„2
> "
—Jn— ;
, '
and
J^ when n
therefore infinite
is
is infinite.
x=l, the product becomes %-a +a3 - aB f ... + a - a 2n+1 + a2>I+a -
If
gJ1
x
._ and since the terms ol, a2/i+.,.1' a2;i+2 2h' no arithmetical meaning. ffl
...,
are infinite, the series has
...
'
This leads us to enquire under what conditions the product of two infinite convergent series is also convergent. *308.
Let us denote the two
infinite series
a + a x 4- a 2 x + a 3x +
.
+ a2 x
. .
x
bQ
by
A
and
If
B
+
b^x
+
b 2x
2
+
b3 x
3
+
.
.
+
.
b 2n x
2n
+
+
.
.
.
.,
.
respectively.
we multiply
these series together
we
obtain a result of
the form
a
A + a(A) x +
(
a J>o + a fii +
afiz)
x2 +
...
Suppose this series to be continued to infinity and let us denote it by G ; then we have to examine under what conditions C may be regarded as the true arithmetical equivalent of the product AB.
A and B
First suppose that all the terms in
are positive.
JO
Let A„ B„2« C„2« denote the series formed by taking the 2w + 1 terms of A, B, C respectively. 2/1
,
,
first
multiply together the two series A 2ai B2ni the coefficient of each power of x in their product is equal to the coefficient of the like power of x in C as far as the term x 2 " ; but in A 2n B, n 2n there are terms containing powers of x higher than x whilst 2n x is the highest power of x in C0n ; hence If
we
,
C
^o B* > 2/i
If
we form
C 2n includes all besides ; hence
2/1
2 2/1
.
the product A B the last term is a b x 2n but the terms in the product and some other terms ;
C.%n
>A B ii
.
ii
CONVERGENCY AND DIVERGENCY OF
C
Thus
intermediate in value between 1 whatever be the value or n. T
Let
is
2"
•
B
and
-4
/.
be convergent series
A = A-X, B
;
SERIES.
251
A B B B and A B 2/»
,
2/i'
put
=B-
Y.
X
where and Y are the remainders after n terms of the series have been taken; then when n is infinite and Y are both
X
indefinitely small. .-.
A n B H = (A-X)(B-Y) = AB-BX-AY+XY' the limit of A B is AB. since A and B are botli finite. }
therefore
Similarly, the limit of
it lies
is
AB.
which is the limit of C 2n must be equal to between the limits of A B B B and A 2« n B„
Therefore since
A 2n B„ a
C
AB
.
2;«
Next suppose the terms
in
A and B
are not
all of
the same
sign.
A 2n the inequalities > C„2b > A b B b are n Bn I 2b necessarily true, and we cannot reason as in the former case. In
this
case
not
Let us denote the aggregates of the positive terms in the two series by P P' respectively, and the aggregates of the negative terms by iV, N'; so that t
A = P-N, B^F-N'. Then
each of the expressions P, vergent series, the equation if
P\
JV,
N' represents a con-
AB = PF- NF- PN' + NN\ has a meaning perfectly
each of the expressions PP\ NF, PN\ NN' is a convergent series, by the former part and thus the product of the two series A and of the proposition B is a convergent series. intelligible, for
;
Hence that the
the
sum
product of two series will be convergent provided of all the terms of the same sign in each is a con-
vergent series.
But
N
each of the expressions P, P', N' represents a divergent series (as in the preceding article, where also = = and N' N), then all the expressions PF, NF, PN\ NN' are divergent series. When this is the case, a careful investigation is necessary in each particular example in order to ascertain whether the product is convergent or not. if
y
F P
HIGHER ALGEBRA.
252
^EXAMPLES. XXI. Find whether the following 1
1.
.r
1.3.5.7.9
#*
£«
1
12
+
3.6.9 ^ 3.6.9.12 2 A + + ^ 10 +7.10.13* 7.10.13.16
3.6
+ ^+7. 22
o
series are convergent or divergent
1+ 2*4 + 2.4.6'8 + 2.4.6.8.10' 3
2.
1.3.5
2
b.
2 2 .4 2
a
2
2 2 42 6 2 .
G
o
.
3.
^+374^+3.4.5.6^+3.4.5.6.7.8*°+
4 *'
n 2# ^ 3
2
.?
2
1
|2
12
^
|3
^
|4
12
^h
,5
14
13
.3 2
54 1
12
1
l2
43 1
.3 2 .5 2
+ 2 2 .4 2 r + 2 2 .4 2 .6 2 ^ 2 + 22 '
*
7
g(l-a)
i
'•
,
,
"*"
X "T
(
*
+ a)«(l-g)(2- g)
l
~
12
l2
(2
.
22
+ q)(l+a)q(l-a)(2-,a)(3-q) I2
a being a proper 8
9
a+x *
—
.
22 32 .
fraction.
(a + 2#) 2 12~
IT*—
(a + 3ai) 3 + "13"" +
^ ^MM,
1+ 1
.
+
.
1
y
.
Z
.
y (y+1) a(a + l)(a + 2)/30+l)(/3 + 2)
1.2.3.y(y+l)(y + 2) 10.
x1 (log 2)* + a?3 (log 3)i + a?4
11.
i
12
If -
integer,
^_a_i
+a+
^;
(log 4)*
__^+—_-^ =
+ +
'^r^w^S^' + + +
shew that the series w j s positive, and divergent if 1
"*" '
?^
2
«3
A -a- 1
is
where *
is
a positive \
convergent negative or zero is
if
CHAPTER
XXII.
Undetermined Coefficients. 309. In Art. 230 of the Elementary Algebra, it Avas proved that if any rational integral function of x vanishes when x = a, it is divisible by x — a. Cor.] [See also Art. 514. n n p x +p x
Let
" '
x
+ pjf
"2
+pn
+
be a rational integral function of x of n dimensions, which vanishes when x is equal to each of the unequal quantities «!>
«*,
%i
«„•
Denote the function hy f(x); by x - a we have
tlien
since
f(x)
is
divisible
,
l
i
f(x)=:(x-a )(p x"- + quotient being of n — 1 dimensions.
),
l
the
Similarly, since f(x) is divisible
2W
n~X
by x -a,
7 ,
we have
= (x-aj-(pjf- +
+
the quotient being of n
)«
— 2 dimensions; and
Proceeding in this way,
we
shall finally obtain after
n
di-
visions
f(x)
=p
(x
- a) (x-a}(x-aa)
(x- aH).
If a rational integral function of\\ dimensions vanishes for more than n values of the variable, the coefficient of each power of the variable must be zero. 310.
Let the function be denoted hyf(x), where f(x)
!>x"
+p x"~ )
x
+p,c'-' +
+p
n
;
~
:
HIGHER ALGEBRA.
254
and suppose tha,tf(x) vanishes when x a n ; then unequal values a lt a2i a3
equal to each of the
is
f( x) =Po (x - a i) ( x ~ a 2 ) ( x ~ °0
(
x ~ a ,)-
be another value of x which makes f(x) vanish since f(c) = 0, we have
Let
c
Po
(
c
~ a i)
and therefore p = 0, factors
is
a*)
(°
(
G
~ a s)
Hence f (x) reduces to "~ n-2 +p2x + 2) 3X 3+ +Pn
of
the other
equal to zero.
2\x
n-
x
-
By
hypothesis this expression vanishes for more than and therefore p = 0.
of x,
2>o,
then
(c-«J = 0;
by hypothesis, none
since,
;
n values
x
In a similar manner we may shew that each of the P3 Vn mus t be equal to zero.
coefficients
,
may
This result
be enunciated as follows
also
If a rational integral function of n dimensions vanishes for more than n values of the variable, it must vanish for every value of the variable. If the function f(x) vanishes for more than Cor. has more than n roots. (x) — of x, the equation
n
values
f
Hence roots
it is
also,
an
if
an
equation of n dimensions has more than
n
identity.
Prove that
Example.
(x
-
-
c)
(x
- c)
(x
(a
-b) (a-
c)
{b
-c)
(6
b) (x
- a) - a)
(x
-
a) (x
— b) _
1
(c-a) (c-b)~
This equation is of tivo dimensions, and it is evidently satisfied by each of the three values a, 6, c ; hence it is an identity.
If two rational integral functions of n dimensions are equal for more than n values of the variable, they are equal for 311.
every value of the variable.
Suppose that the two functions 2)
xn
+p x
n-1
1
2
+pH
+2> 2 x"- +
n 2 qox + q^"- + q 2 x- + 1
are equal for
more than n values '1
U>»
- %) x +
(Pi
-
?i)
x
"~ l
+
of
+ qmt
x ; then the expression
"~ - ad x + 2
(p»
,
+ (p* -
?.)
UNDETERMINED COEFFICIENTS. n values
vanishes for more than preceding article,
that
255
and therefore, by the
of x;
is,
2\
= %>
Pi=9li>
Pi^Vv
Hence the two expressions are
>
and therefore are
identical,
equal for every value of the variable.
*
=
l> n
Thus
if two rational integral functions are identically equal, we equate the coefficients of the like powers of the variable.
we assumed
This is the principle Art. 227.
in the
may
Elementary Algebra,
This proposition still holds if one of the functions Cor. For instance, if of lower dimensions than the other. x"
p
+ pff~ + pjf~ 2 + pjf~* + = q 2x n ~ 2 + q 3 x n ~ 3 + +q n l
is
+pn ,
only to suppose that in the above investigation q o = and then Ave obtain
we have q =
0,
^o=°> Pi=°> P2=vs Ps=q 3 >
The theorem
312.
p,,
>
=
0,
q»-
of the preceding article is usually referred
The application to as the Principle of Undetermined Coefficients. of this principle is illustrated in the following examples. Example,
Assume
1.
Find the sum of the series 1.2 + 2.3 + 3.4+
+n(n+l).
that
1.2 + 2. 3 + 3. 4 + ... + n(n + l)=A + Bn+Cn 2 + Dn3 + Eni +..., where A, B, C, D, E,... are quantities independent of n, whose values have to be determined.
Change n 1. 2
into
n+
1
;
then
+ 2.3+...+?i(;i + l) +
+ l) (n + 2) = A+B(n + l) + C(n+l)* + D(n + l)3 + E(n + iy+....
By subtraction, (n + 1) [n+2) = B+C
(?t
{2n + l)
+D
(3}v>
+ 3}i + l) + E
{±-n*
+ 6ri- + ±n + l)+
..
.
This equation being true for all integral values of n, the coefficients of the respective powers of n on each side must be equal thus E and all succeeding ;
coefficients
must be equal
to zero,
3D = 1;
and
3D + 2C = 3; 1
whence
1)
=-
, •
o
(7=1,
D + C + B = 2; B=
2 -
o
.
HIGHER ALGEBRA.
256
=A + — + n2 + - n 3 o
Hence the sum
.
o
To
find A, put
n = l; the
series
2=A
Hence
1 .2
Note.
+2
.
3
+ 3.
4
then reduces to
+ 2,
or
its first
term, and
A = 0.
+ ... + n(;i + l) = - n
(n
+ 1)
(n + 2).
be seen from this example that when the n lh term is a rational integral function of n, it is sufficient to assume for the sum a function of n which is of one dimension higher than the w th term of the It will
series.
Example
Find the conditions that x3 +px2 + qx + r may be x 2 + ax + b.
2.
Assume
x3 +px2 + qx + r=(x + k)
Equating the
(x 2
by
+ ax + 6).
powers of
coefficients of the like
divisible
x,
we have
k + a=p, ak + b = q, kb = r. the last equation k =
From
-
;
b r
r that
is,
r
=b
hence by substitution we obtain
+ a=p, andn
ar — +b = q; ,
(p-a), and ar = b (q-b);
which are the conditions required.
EXAMPLES.
XXII.
Find by the method of Undetermined 1.
l 2 +3*
2.
1.2. 3 + 2. 3. 4 + 3. 4. 5 +
3.
1. 2 2
+ 5*+7*+...to n
4.
2
Coefficients the ,sum of
terms. ..
.ton terms.
+ 2.3 + 3.4 + 4.5 +... to n I 3 + 33 + 5 3 + V 3 + .to n terms. l 4 + 2* + 3 4 + 4 4 + ...to?i terms. 2
a.
2
terms.
. .
5.
Find the condition that x3 -3px + 2q may be factor of the form a?+%ax + a2 6.
divisible
by a
.
7.
Find the conditions that ax3 + hv2 -\-cx + d may be a perfect cube.
8.
Find the conditions that a2 AA + bx3 +cx2 + dx+f 2 may be a
perfect square. 9. 1
if b'
Prove that ax2 + 2bxy + cif- + 2tlv + 2ey +/
= ac,
d- = a/, e2
= cf.
is
a perfect square,
UNDETERMINED COEFFICIENTS. + bx2 + cx + d is
:i
10.
If
a.<
11.
If
3tP
12.
Trove the identities
(
}
w /
n
— f>qx+4r
by x2 + h 2 prove that
divisible
divisible
by
,
(x
— c) 2 shew ,
= bc.
that g*=r*,
:
a2 (x-b)(x— c) b 2 (x-c)(x — a) + ~(b-c){b-a)~ (a-6)(«-c)
c2 (x
- a) (x - b) _ "
+ ~Jc^aJ(c-b)
2
(ff-c)(#- eg) (.?-«)
(y-^>)(^-c)<.y-cQ t"
(rt-6)(a-c)(a-J)" (b-c)(b-d)(b-a)
+ 13.
is
257
(x - d) (x - a) (x
~b)
(x
- a)
(x
- b) (x - c)
+ \d-a){d-b)\d-c)** {c-d){c-a)(c-b)
'
Find the condition that
ax2 + 2/ixy + by 2 -f 2gx + 2fy + c
may
be the product of two factors of the form jfctf+gy+r,
14.
If £
= lx + my + nz,
r)
jt/.t'
+ ^'y + r'.
= nx + ly + mz, £=mx + n// +
same equations are true for all values of x, changed with x y, 2 respectively, shew that
y, z
when
£,
l~, 77,
and
if
the
£ are inter-
t
l
15.
2
+2mn = l,
m 2 + 2ln = 0,
Shew that the sum
quantities a, a
3
2
(
,
a
,
+
1
«
y
(a
n 2 + 2lm=0.
of the products
-//
-
/•
together of the n
,..a n is
-l)(tt*- + a
-1)
(a 2
-
-l)...(a»-l)
1).. .(a*-'-
a
i(„-r)(»-r+l).
1)
3
2
313. is equal If the infinite series a + a x + a.,x + a 3 x + to zero for every finite value of x for which the series is convergent, tit en each, coefficient must be equal to zero identically. 2
Let the series be denoted by S, and let S\ stand for the ex2 pression a + a 2 x + a x + then S = a + xS and therefore, = l>y hypothesis, a + xS for all finite values of x. But since S is convergent, #, cannot exceed some finite limit; tlierefore by taking x small enough xS may be made as small as we please. In this case the limit of & is a but S is always zero, therefore a Q must be equal to zero identically. ;
t
:i
l
,
t
x
;
Removing the term a we have xS = for all finite values 2 that is, a + a 2x + ajc + vanishes for all finite values of ,
x;
x
x
Similarly, coefficients a n
H. ir.A.
we may prove a.,,
a
is
succession that each equal to zero identically. in
of
17
of x.
the
—
. .
;
HIGHER ALGEBRA.
258
314. If tivo infinite series are equal to one another for every finite value of the variable for which both series are convergent, the coefficients of like powers of the variable in the two series are equal.
Suppose that the two
series are
denoted by
3 2 a + ax x + a x + a 3 x +
A +A
and
x
x+
A x + Aj? + 2
;
2
then the expression «o
" Ao +
(«i
"A
x + (a2 -
)
i
A
+ (« - A a)
2 2)
°°
°f
+
vanishes for all values of x within the assigned limits by the last article
a that
aB -A a = 0, a3
-A = O a.-A^Q, t
« =
is,
^
°i-^n
>
a2 =
A
a.
2,
d
= A3
;
therefore
-A a = 0, ;
,
which proves the proposition. Example
Expand
1.
5 as the term involving x
where a
,
fl
a
,
a.2
a
,
2
:i
,...
„
in a series of ascending powers of x as far
.
+ X2
— =a
2
r—
Let
+ x2
2 -=
'-
2
+ a x x + a 2 x'2 + a.jx? +
are constants
+ x 2 — (1 + x - x 2
)
whose values
(a Q
...,
are to be determined; then
+ Oj a; + a 2 ar + o a
re
3
+
)
. . .
In this equation we may equate the coefficients of like powers of x on n each side. On the right-hand side the coefficient of x is a n + a u _ 1 - a n _ 2 2 and therefore, since x is the highest power of x on the left, for all values of ?t>2 we have ,
this will suffice to find the successive coefficients after the first three
To determine
been obtained.
a
= 2, a
whence Also
a3
a1
we have the equations
+ a = 0,
a.2
+ a1 -a = l;
= 2, ^=-2,
a 2 =5.
+ a 2 -a 1 = 0, whence
a4 + a 3
~~
a 3 = -7;
a 2 = 0, whence a 4 = 12
;
a 5 + a±-a 3 = 0, whence a 5 = - 19
and 2
thus
these
,
+ X~ = 2 2x + 5x 2 „ 2
l + ic-a;
7.t 3
+ 12x 4 - 19a 5 +
. .
have
—
.
.
UNDETERMINED COEFFICIENTS. Example
Prove that
2.
if
n and
r are positive integers
*-.fr-y+«fez3 .-y- -fr-?)fr-«> fr (
3
£
is
equal to
We
if r
250
~+
.„
I
be less than n, and to
|w if r
=n
have
= x n + terms containing
higher powers of x.
.
.(1).
Again, by the Binomial Theorem,
(g*-l)n =c»w -ne(»-l)ai + ^_(±Ll) e(n-2)*_
By expanding each of
xr in
of the terms
e
nx ,
e (n
~ l)X y
we
...
(
j
find that the coefficient
(2) is
nr
(»-l) r
|r
[r
n(n-l) (n-2) r
and by equating the
w(m-1)(w-2) (n-3)*
1
|r
j2
coefficients of
r
|3
x r in
and
(1)
the result follows.
(2)
y = ax + bx 2 + ex3 + express x in ascending powers of y as far as the term involving y 3
Example
3.
If
,
x=py + qy 2 + ry 3 +
Assume and substitute y = a{py
in the given series
;
an = 1
ar
Thus
Cor.
is
;
whence q =
+ 2bpq + cp 3 —
;
whence
the series for
?/
.
r
—
5
a6
= —5= a
+
(2&
-
-
-
.
,
a
1
ac) y
?—
Series.
be given in the form
y= k
+ ax + bx 2 + ex? + ...
y-k = z;
put then
from which x
+....
a
an example of Reversion of If
t
we have
whence p = -
;
V &'V" # = •'---4-
m,
y,
+ bp- =
aq
This
thus
powers of
coefficients of like
.
,
+ qy* + ry 3 +...) + b(py + qy 2 +...y2 + c{2>y + qif+...y
Equating
z
may
— ax + bx- + ex3 +
.
.
2 ).
;
be expanded in ascending powers of
z,
that
is
of y - k.
17—2
—
I
HIGHER ALGEBRA.
260
EXAMPLES.
"'
3+* '
-x—x
1
be (Sn-2)xn
Find
a + bx + cx
~
+ ax — ax2 — .r3
l+x 2+.r + .r2'
.
so that the coefficient of
2
,
3
n- + ,
If cxz
% =#
(3/
xn
--,
(l-.r)-
in the expansion of
+ 1),
+ ax -y = 0,
shew that one value
of
y
is
+ s .r-js S A +
shew that one value of x e?/
3
a4
3c2;/5
12c3;/ 7
a7
«'°
i
x= -00999999 is an approximate x + 100.? -1 = 0. To how many places of
Hence shew that 3
result correct
bv
l.
y a equation
-4-
,
|.r 9.
a 7-
.
a, b, c
If y 2 +
far
1
—(l-.r)^— mayJ be 8.
l -
x as
th Find « and b so that the n term in the expansion of
6
7.
*
i_#_6#2-
5
2
2
b.
l-&g
1+2^; 1_^_^.24
XXII.
following expressions in ascending powers of
Expand the
may
«'
solution of the
decimals
is
the
?
the In the expansion of ( 1 + x) ( 1 + ax) ( 1 + a\c) ( 1 + a\v) < shew and a that the coefficient infinite, of being 1, factors number of 10.
Xr
,
.
I
1S
11.
(1
12.
\
(l-a)(l-a )(l-« ) When a < 1, find the coefficient 2
If
n
is
- ax)
(1
3
(l-O of
xn
—a 2x) (1 — dAx)
n
hr(r-l)
in the expansion of
to inf.
a positive integer, shew that
(1)
nn+1 -n(n-l)n+1 +
(2)
n n -(n+l)(n-l) n +
n ^~ 1 ' K
——
^-
(n-2)* +1 -
(n-2)«-
the series in each case being extended to 7t
^~ 1
3a -
(3)
l"-»2»+
(4)
(n+p) n -n(n+p-l) n
<
n terms
=(-l)w
+ —^—
'
=jn\ n+.l =1;
;
and
\n;
(n+p-2) n -
l±
the series in the last two cases being extended to
= '— \n;
n + 1 terms.
;
CHAPTER
XXIII.
Partial Fractions. In elementary Algebra, a group of fractions connected by the signs of addition and subtraction is reduced to a more simple form by being collected into one single fraction whose denominator is the lowest common denominator of the given 315.
But the converse
process of separating a fraction into a group of simpler, or jwtial, fractions is often required. For — 3 5a; ^--„ in a series of ascendexample, if we wish to expand 1 — iX -r OXT ing powers of x, we might use the method of Art. 314, Ex. 1, and so obtain as many terms as we please. But if we wish to find the general term of the series this method is inapplicable, and it is simpler to express the given fraction in the equivalent form 1 2 -1 and (1 — 3aj) -1 Each of the expressions (1 —a;) fractions.
—
1-
.
— x l — ox can now be expanded by I
the Binomial Theorem, and the general
term obtained. 316. In the present chapter we shall give some examples illustrating the decomposition of a rational fraction into partial fractions. For a fuller discussion of the subject the reader is referred to Serret's Cours d'Algebre Superieure, or to treatises on In these works it is proved that any the Integral Calculus. rational fraction may be resolved into a series of partial fractions;
and that to any
linear factor
x—a
denominator there
in the
to any linear X cc the denominator there correspond
responds a partial fraction of the form factor
x-
b occurring twice in
— x— 7?
two partial fractions, times, there is
——
—— 7?
l
-j
b
cor-
and
-.
(x
*__
.
by
an additional fraction
If
;
x — b occurs
hnl au d (x-b)" .
so
on
three
-
To
s
;
HIGHER ALGEBRA.
262 any
quadratic
+px + q —Px + Q if
fraction of the form twice, there
is
'
Here the shall
+q
Av B
quantities
P x + QL
lt
a
partial
2 the factor x + vx + q occurs
:
x' +])x
corresponds
there
fraction y—~ a second partial r (x-+2)x
independent of
We
x2
factor
P B 2
,
+
—
P, Q,
3,
;
and so
on.
q)
Pv Q
are
all
x
x.
make
the examples that
use of these results in
follow.
Example
5x — 11 Separate =-^ ^ into partial fractions.
1.
+ x - 6 = (x + 2) (2x - 3), we assume 5.r-ll B A + x+2 2.c-3' 2x 2 + x-§
Since the denominator
B
where A and
2.r
2
x whose values have
are quantities independent of
to be
determined. Clearing of fractions,
5x-ll = A (2x-S) + B(x + 2). Since this equation powers of x thus
we may equate
is identically true,
coefficients of like
;
~SA + 2B=-U;
2A+B = 5,
B= -1.
A = 3,
whence
5.r-ll '"'
Example
2.
2x
Resolve
;
might now equate
r,
(x
+ b)
2x-B'
into partial fractions.
z-.
Since
A and B
In
put
coefficients
are independent of x,
x-a = 0,
(1).
.
and find the values proceed in the following manner.
simpler to
(1)
- a)
-.
.'
is
1
A B mx + n =-r = h x-a x + b' (x-a)(x + b) mx + n = A {x + b) +B (x-a)
Assume
We
3
+ x-6~ x + 2 r~.
(x
.
2
of
t
n = 0,
or
x—
ov x = a; then
-
,
b,
— B— nib-n -r,
.
CI "T*
mx + n (x - a) (x + b)
1
B, but
it
we may give to x any value we please.
ma + n A= a + brputting x + b
A and
/ma + n
~ a b + \
x-a
mb-ii\ x+
b
J
—
r
;
.
PARTIAL FRACTIONS. Example
23
- 11
v
2 r'
-.-rrx- —^ mto 7^7— y£x — I) (J — x
Resolve
3.
.
• .
By
23x-ll.r 2
=
5 rm r= n 2.c-lr+ 3 + x (3 + x)(3-.r)
^-^
(2.c-l)
23x - lLc2 = .1
+ x)
(3
partial fractions.
ABC
j
Assume
-x)+B
(3
+ x— 0,
3
Example
.•
4
1_
3+x
3 -a:'
3s 2
Let let
a;
1
—— ^— + s-2 —- — —-^
%x-
2
&)"
s-7 2x)
+ x-2 = A
1) (3
1)
5
+ x).
find that
'
,
[x —
i
+x—2
-
(x
1
2
2)
into partial fractions.
B + x-2«
k~ H 2x
~k
"
:
&x\
(J.
+£
(1
-
2x) {x
G (x
-
-
2)
2)
2
'
+ C (1 - 2x).
A= --
- 2x = 0, then
-2 = 0,
To
—
'
-.
z-^-n (x - 2)- (1 .
= 0, we
- lis 2 1 + 2 )~2x-l (2.c-l)(9-x 3.t
Assume
3 -#
-1.
Resolve
4.
3-x
4 = 1, B = i, C= 23.c
w
H 5
+ G (2x -
(2x - 1) (3 - x)
2^-1 = 0,
putting in succession
•'•
203
o
C=-4.
then
find B, equate the coefficients of x 2
3
;
thus
= A - 2B whence B = - ^ ;
o
3.r'
Example 1
'
(x-
+x-
2)
a
2
- 2x)
(1
3(1- 2x)
- 2)
(x
-
2)
2
'
42 —— into partial fractions. +l)(x-4] 19a;
5.
Resolve
[x
2
-r
-r-,
42-19.C
Assume
-7-3
tt,
.-.
42 -
r;
+ 1)^-4)
(.^
19.r
Ax + B —3 + x-+l
—
(x-
C
-
4)
.i--4'
+ C (x* + l).
C=-2;
equating coefficients of
2
x'
,
equating the absolute terms,
= A + C, and .4=2; 42 = - 4Z? + C, and B = - 11,
42 - 19a •'"
The
=
L
= (Ax +B)
Let x = 4, then
317.
3 {x
2s -11
2
p+l)(x-4)"^TT *-4*
artifice
employed in the following example
sometimes be found useful.
will
;
,
HIGHER ALGEBRA.
2G4 -;
=rr-
7^t~.
(a:-2) 4 (a:+l)
—
into partial fractions.
A
9z 3 -24;r 2 +48;c —. ^tx~? l) (x-2)*(x +rrv
Assume where A
9x* — 24rc2 + 48#
Resolve
Example. r
f(x)
— —n + x+1
{x-2) 4
some constant, and / (x) a function
is
of x
'
whose value remains
to
he determined.
9x3 -24£ 2 + 48x = ,l (x-2)*+(x + l)f{x).
.-.
Let
x= - 1,
Substituting for (x
A=
then
-
1.
and transposing,
^4
+ 1) / (a) = [x - 2) 4 +
9a3 - 24s3 + 48x = x4 + x* + 16* + 16
.'./(*)
= £ 3 + 16.
—+ —
•r
To determine .r
then
the partial fractions corresponding to 3
+16
(2
+ 2) 3 + 16
(x-2) 4
23
~z
+
6 z^
3
16
\x—2)
,
put
x-2 = 2;
+ 6^ 2 +122 + 24 z*
Z*
1
-
12
+
+
~z^
24
~
4
~z
24
12
6
1
+ /. nva2 + /_ n\n + ~ x-2^(x-2)' (x-2f^ (x-2f '
9x-
"
3
-24j; 2 + 48*
(x-2) 4
(x
+ l)
1,1,+
= "
-\
x+1 -i
«
x-2
6
Z
12
,
+ (x-2)" /
Svi
TZ
^-J
(x-2f
+
24
(x-2) 4
*
the preceding examples the numerator has been of lower dimensions than the denominator ; if this is not the case, we divide the numerator by the denominator until a remainder is obtained which is of lower dimensions than the denominator. 318.
In
Example. *
By
all
2 -7 6r 3 + 5# =- into partial fractions. ox- - 2x - 1
—
Resolve
-zr-=
division,
-
2
3a:
2.x
-
v-
1
= 2x + 3 +
-4 = — 3x -2x-l
5
8a;
and
^-= 2
'*
319. fractions
We
pr
^
s
3.c
+
= l
Sx 2 - 2x -
+
1
1
x-1'
1 5 6^ + 5^-7 = 2.r + 3 + - - + 2 *-l' 3.T+1 3x -2x-l
shall
may be
now
explain
how
resolution
into
partial
used to facilitate the expansion of a rational fraction in ascending powers of x.
PARTIAL FRACTIONS.
—-——— _ 2— ;;
Example
Find the general term of
1.
series of ascending
By Ex.
4,
powers of
Art. 316,
3(1-2*)
(*-2) (l-2*)
r+6
+
7
Expand
2.
(1
when expanded l
in a
+
(*-2) 2 4 (2-a?) a
is
r+l\
2r
•
4
3(2-*)
1
6
3
V
Example
.
(*- 2)-(l - 2x)
of the expansion
/
;•
3(*-2)
3(1-2*)
Hence the general term
-|-
15 15
we have
2
,'-
x.
+ .r-2
3.r 2
265
sr
+*
r-^r— *) (I + *~) .
y
in ascending powers of *
and
find
the general term.
—+ 1+x
+* -— 7
.
Assume
H
(1
.\
=J
;
)
A = 3;
then
equating the absolute terms, equating the coefficients of *
+* _ 2 (1 + *)(1+* )
7 2 ,
= A + C, = A + B,
whence C = i whence
;
B— - 3.
4-3*
3
+ 1 + *^ 1+* 2
= 3(1 + .r)- + (4 = 3{l-* + * 2 1
+ (4-3*) To
+C 1 + *2
JB*
+ *) (1+* ) 7 + * = J(l + * 2 + (E*+C)(l + *).
Lctl + *=:0,
7
.4
find the coefficient of xr
3*) (1
+ x 2 )~l
+ (_l)P;C P +
{l-.r 2 + *-»-
...j
+ (-1)p*'^+...}.
:
r
(1)
If
/•
is
even, the coefficient of * r in the second series is
4(-l) 2
r
therefore in the expansion the coefficient of x r is 3
+4 (-
1)
2 .
r-l (2)
If r is odd, the coefficient of *r in the
second series
is
-3(-
1)
'-
r+l
and the required
coefficient is 3
(
-
1)
2
-
EXAMPLES. Resolve into partial fractions ,
lx-\ l-bj; + 6jf-
3.
XXIII.
:
46+13.r '
12.t 2
-lU--15'
l+3. r + 2 .r 2 (1 -2.r) (1
-.//-')'
;
}
HIGHER ALGEBRA.
266 -
2x*+x2 -x-3 x(x-l)(2x + 3)'
+13 2 (x-l)(x -5x+6)' 2
.y
'
10a;
9 6
(a;-l)(^ + 2) 2
*
26^2 + 208o;
(#+l) 2 (#-3) 2^2 -lLr + 5
Q
+ l)(^ + 5)' 3^-8x2 +10 (a;
x*- 3x* -3a;2 + 10
„ 7*
'
(07-1)*
+ 2^-
(^--3) (^ 2
2
5)
5^ + 6.r 2 + 5.r
,,
(^
*
2
-l)(^+l) 3
'
Find the general term of the following expressions when expanded in ascending powers of x.
+ 3# l + llo; + 28^* 2#-4
5a;
l
12#
15.
t^
(1
2 - x5tt^ ) (1
3
19
.
21
.
7,
[
24.
I
(i+^)(i+^2 )
22
*
'
(1
+x)
When
(l
+
(l-x)
Sum
to
find the
+ (l-x3 )
)
'
»-«• (2
- 3.r + a2) 2
'
series
— + + (i+^2 )(n-^) (i+^)(i+^4 )
+ ax) (1 + a%)
< 1,
a?
+ x - 2x 2
1
(1-tf) 3
. '
x (1 - ax)
.
1
25.
(1
- co;)
Find the sum of n terms of the
(l)
^
1 (1
(
+ 3a;)(l+.r) 2 1 -* +i *
20.
- te)
+ 7a- + uy
(2
*"* „cw?) (1
2
4 + 7x
no 18.
• *
(^-1)(^2 +1)"
23.
.
—
,
.
- x)
(1
2.r)
w ttx* (l+a?)(l-4a?) 2
17.
16.
+ 2x-x2
tf
4 + 3^+2a'2
.« • *
#2 + 7;f + 3
u
(2+a?)(l-#)'
'
—- ^
+6
ax
+ (1
sum
+ ax)
n terms the
-a?)
(1
series
- a2x)
+a%)
(1
+ a 3.r)
+
of the infinite series
x2 (1
(1
(1
-.r5 )
+
xA (1-tf5 )
whose p th term
(1
-^) +
is
xp(1+xp + 1 )
(l-^)(l-.^ + 1 )(l-^ + 2 )' Prove that the sum of the homogeneous products of n dimen26. sions which can be formed of the letters a, b, c and their powers is
an + 2 (b -c) + bn + 2 (c- a) + cn + 2 (a-b) a2 (b-c) + b 2 (c-a) + c2 (a-b)
*
CHAPTER XXIV. Recurring Series. 320.
A series
u + u + u 2 + u3 + l
which from and after a certain term each term is equal to the sum of a fixed number of the preceding terms multiplied respectively by certain constants is called a recurring series. in
321.
In the
series 1
+ 2x +
3ar
+ 4a? +
5a;
4
+
,
each term after the second is equal to the sum of the two preceding terms multiplied respectively by the constants 2x, and - x 2 j these quantities being called constants because they are Thus the same for all values of n.
5x4 that
= 2x
.
4a;
3
+ (- x2 )
3a;
.
2 ;
is,
u4 = 2xn3 — x2u 2 and generally when n is greater than with the two that immediately precede
uh —
2xiin— 1 ,
it .
,
1
each term is connected by the equation
1,
— x 2 u n—2*
u H — 2xu n — + x 2 u
or
;
ii
— „2
=
,
0.
In this equation the coefficients of u n «*,_,, and l*,_ a taken with their proper signs, form what is called the scale of relation. ,
,
Thus the
series 1
is
a recurring series
+ 2x + 3a; 2 + 4a; 3 + 5x 4 + in which the scale of relation 2 1 - 2x + x
is
.
322.
If the scale of relation of a recurring series is given,
any term can be found when a
sufficient
number
of the preceding
HIGHER ALGEBRA.
268
As the method of procedure is the same terms are known. however many terms the scale of relation may consist of, the following illustration will be sufficient. 1
If is
- px - qx2 - rx 3
the scale of relation of the series
a + a
we have "~ 1
~
~
3 2 + rx3 a n - 3x + x " a n - 2x am =pan_, + &».- + «- 3 or thus any coefficient can be found when the coefficients
anx*=px a n -i x
'l
-
'l
•
-
m
i
5
of the
three preceding terms are known.
Conversely, if a sufficient number of the terms of a series be given, the scale of relation may be found. 323.
Find the scale of relation of the recurring series 2 + 5x + 13x2 + 35x 3 + Let the scale of relation be 1 -px - qx*-, then to obtain p and q we have 13 - 5p - 2q = 0, and 35 - 13p - 5q = the equations Example.
;
whence p = 5, and q= -
6,
thus the scale of relation 1
- 5x + 6a;2
is
.
If the scale of relation consists of 3 terms it involves 324. 2 constants, p and q ; and we must have 2 equations to deTo obtain the first of these we must know termine p and q. at least 3 terms of the series, and to obtain the second we Thus to obtain a scale of must have one more term given. two constants we must have at least 4 terms relation involving 'O
given. 2 3 to find the the scale of relation be 1 — px — qx - rx To obtain the first of 3 constants we must have 3 equations. these we must know at least 4 terms of the series, and to obtain the other two we must have two more terms given hence to find a scale of relation involving 3 constants, at least G terms of the
If
,
;
series
we
must be
given.
Generally, to find a scale of relation involving must know at least 2m consecutive terms.
Conversely,
if
2m consecutive
terms are given, we
for the scale of relation 1
m
~ l\ x ~ l\ x * ~ lhx* ~
-PJ**
constants,
may assume
— RECURRING To find
325.
the
sum ofn
2G9
SERIES.
terms of a recurring
series.
of finding the sum is the same whatever be the scale of relation ; for simplicity we shall suppose it to contain only two constants.
The method
Let the
be
series
a u + axx + a 2 x 2 + aj£ +
(1)
the sum be S ; let the scale of relation be so that for every value of n greater than 1, we have
and
let
Now S—a + a.x + a,x + ...+ a — px S= — pa x — pa x* — ... — - qa? S= - qajt? - ... -qa 2
it
2
1
2^>ci
x
...
(i
- px _
qtf)
S-
a + {a
x
— px — qx*
'
1
H_2
x
n~ l
_ 3x*-
— pa
x*t
-qa H _ x n -qa H _ x
1
i
u
+
l
-pa ) x - {pan _ + qan _ a ) x n - qa^x** x
for the coefficient of every other
power
;
,x"~\
/»—
H
1
of
x
is
\ 1
,
zero in consequence
of the relation
a n-P a n-l-<2 a «-2=
s_
.
% + («, -P<-Q x 1
-px— qx
Thus the sum nominator
is
(P a
,t
-,
2
'
H+l + qan - 3 ) x" + qa n _ x 2 1 - px — qx }
of a recurring series is a fraction
whose de-
the scale of relation.
32G. If the second fraction in the result of the last article decreases indefinitely as n increases indefinitely, the sum of an
number
infinite
—\—
of terms reduces to
1
we develop
—
^—
— px — — qx" !
-
.
ascending powers of x as many terms of the series as we please; original for this reason the expression If
this fraction in
we
explained in Art. 314,
1
is
shall obtain as
—px — qx
2
called the generating function of the series.
From
327.
a X —P—%— — =a -px —
an + (v a °,
1
the result of Art. 325,
,
'
£
.)
qx'
we
o
lt
°
+ a.x + ax+ 2 '
1
...
obtain
+a
+ xn.xi
l
" -1
- px— qx
HIGHER ALGEBRA.
270
from which we see that although the generating function
— px —
1
qx 2
be used to obtain as many terms of the series as we please, can be regarded as the true equivalent of the infinite series
may it
a + a x + a 2x 2 +
,
l
only
if
the remainder
+qan-
( I**,.-,
+ (2 a n-^" +l
— poj — qx 2
1
vanishes when when the series
xn 2)
n
is
is
convergent. o
indefinitely increased
;
in other
words only
v
When the generating function can be expressed as a 328. group of partial fractions the general term of a recurring series may be easily found. Thus, suppose the generating function can be decomposed into the partial fractions
ABC h
1— ax Then the &general term
1
h
+ bx
(I— ex) 2
'
is
r
{Aa + (-
l)
r
M+ r
+
(r
r
1)
Cc } x\
may be found without
In this case the sum of n terms
using
the method of Art. 325. Example. Find the generating function, the general term, and the sum to n terms of the recurring series 1
- Ix - x 2 -
-
3
43.-C
-px - - + 7
Let the scale of relation be
whence p = l,
5 = 6;
and the
1
;
scale of relation is 1
Let S denote the sum of the
-x-
series
;
6.r 2 .
then
S = l-lx- x 2 -4Sx s -
-xS= -Qx S= 2
.-.
which
is
- x + 7x 2 +
-6x 2 + 42.r 3 +
(l-x-6x 2)S = l-8x, J" 8 * s-
the generating function.
x*+
—
'
RECURRING SERIES. If
1
we separate
whence the (r+
-
8.r
fractions, ^-„ into partial ±
1-x-U.rtU
l)
or general term
271
—+ 2
we obtain
1
-
-
1
2a;
;
1-305'
is
{(-lyw^-v ].<>. Putting the
sum
=
{
2
to
r ?i
= 0,
-1,
1, 2,...n
terms
- 2 x + 2%2 -... 2
~_ 2 + - I)'
1"1
(
2 n+1
+ (-
xn
I)"" 1 2" a;'
1
"1 }
-
(1
+ 3a + 3%* +
+ 3"- xn~ 1
.
.
.
l
)
_ 1_- 3* xn
l+lte
1
'
- 3x~
term and sum of n terms of the we have only to find the recurring series a + a + a_,+ general term and sum of the series a + a l x + a2x2 + and put x — 1 in the results. 329.
To
find the general
,
i
,
Example.
Find the general term and sum of n terms of the 1
The
+ 6 + 24 + 84+
scale of relation of the series 1 + 6.r 1 +x *—
and the generating function
—
is
is
—
-
— OX + OX"
1
This expression
1
.
.
.
is 1
- ox + Ooj2
,
.
3
- Sx
1
- 2a;
be expanded in ascending powers of x the general
If these expressions
of
+ 24x 2 + 84x3 +
equivalent to the partial fractions
4
term
series
is
(4
Hence the general term n terms is
.
3r
-
3
.
2 r ) xr.
of the given series is 4 2 (3' 1 - 1) - 3 (2' 1 - 1).
.
3r
-3.
2r ;
and the sum
We
may remind the student that in the preceding 330. article the generating function cannot be taken as the sum of the series
+6x + 24:x +8±x 3 + 2
1
when x has such a value as to make the series convergent. Hence when x = 1 (in which case the series is obviously divergent)
except
the generating function But the general term of 1
+
is
6
not a true equivalent of the
+ 24 + 84 +
independent qfx, and whatever value x be the coefficient of x" in
is
1
We
series.
may
have
it will
always
+ Gx + 24* 2 + 84a 3 +
therefore treat this as a convergent series and find general term in the usual way, and then put x = 1.
its
a
:
HIGHER ALGEBRA.
272
EXAMPLES. XXIV. Find the generating function and the general term of the following series 1.
l
+ 5.r + 9.r 2 +13.r 3 +
2.
2-.v + 5.r2 -7.r3 +
3.
2
+ 3x + 5x2 + 9x3 +
4.
7
5.
3+
6a?
-6x + 9x2 + 27x4 +
+ Ux2 + 36.r* + 98.^ + 276.1-5 +
Find the n th term and the sum to n terms of the following 6.
2
+ 5 + 13 + 35+
8.
2
+ 7^ + 25^ + 91^+
9.
1
+ 2.v + 6x2 + 20# 3 + 66x* + 212^ +
7.
10.
-^ + 2 + + 8+
11.
Shew that the
series
:
-l+6.v2 + 30.v 3 +
series 2
+ 2 2 + 3 2 + 42 +
+ n2
,
13
+ 23 + 3 + 4 +
+n
,
1
3
3
3
are recurring series, and find their scales of relation. 12.
Shew how
to
deduce the
sum
of the first
n terms
of the re-
curring series
a from the sum to 13.
+ a x + a2 x2 + a^v3 + x
infinity.
Find the sum of 2n +
1
terms of the series
3-1 + 13-9 + 41-53+ 14.
The
scales of the recurring series
+ vv+ a^x2 + a3.r3 + b + b 1x+b^c2 -{-b 3 3 + 2 l + rx + sx2 respectively; shew 1 +px+qx a
.v
are general term
is l
15.
,
,
+
(
(p
+6 n )^"
is
,
,
that the series whose
a recurring series whose scale
is
+ r)x + (q + s +pr) x2 + (qr +ps) x3 + qsx*.
If a series be
formed having
for its
nih term the sum
of
n terms
of a given recurring series, shew that it will also form a recurring series whose scale of relation will consist of one more term than that of the given series.
CHAPTER XXV. CONTINUED
All expression of
331.
FllACTIONS.
the form a +
is
a
c
+ e
continued fraction here the letters a, b, quantities whatever, but for the present the simpler form a +
integers.
a3 +
we
shall only consider
...
This will be usually written in the more compact form 1
a,
When
332.
...
where a n a 2i « 3 ,... are positive
,
x
2
a
may denote any
c,
;
+
called
the
+
1
a2 + a3 +
number
of quotients a
a
,
« 3 ,...
is finite
the
continued fraction is said to be terminating ; if the number of quotients is unlimited the fraction is called an infinite contirmed fraction. It is possible to reduce every terminating continued fraction to an ordinary fraction by simplifying the fractions in succession
beginning from the lowest.
To convert a given fraction into a continued fraction.
333. tn
Let
—
be the tnven fraction
quotient and
j>
the remainder
;
;
be the
thus
m — — a. +-p n
divide in by n, let a
1
=a, + — n n
:
P si.
H.
A
18
HIGHER ALGEBRA.
274
n by ^»,
divide
^
«„
be the quotient and q the remainder
;
thus
;
and
n
1 - = as + - ; - = a. + q
V
V
'
'
P 9.
p by
divide on.
q, let a.6
be the quotient and r the remainder
.so
Tims
— = a.
1
rn,
n
+
m is less
a.
+
i
1
+
«o
If
=
than
?t,
1
1
a2 + a3 +
a3 +.
the
first
quotient
7)1
1
n
ti
is zero,
and we put
m and proceed as
before.
It will be observed that the above process is the same as that and n ; hence if of finding the greatest common measure of and n are commensurable we shall at length arrive at a stage
m
m
where the division is exact and the process terminates. Thus every fraction whose numerator and denominator are positive integers can be converted into a terminating continued fraction.
Example.
Reduce
251
^^
Finding the greatest we have
to a continued fraction.
common measure
process,
5
of 251
and 802 by the usual
CONTINUED FRACTIONS.
275
335. To shew that the convergents ewe alternately greater than the continued fraction. 1
Let the continued fraction be a + l
The
first
convergent
is
aa +a3 +
is «,,
and
a 2 + a3 + too small because the part
is
convergent ;
is
a
-i l
—a
,
and
is
k
too great, because the denominator aa
great
1
The second convergent °
omitted.
and
less
and
is a,
-\
,
a 2+
CC
and
is
is
The
too small.
too small because a
3
-\
—%
third is
too
so on.
When
the given fraction is a proper fraction a = ; if in this case we agree to consider zero as the first convergent, we may enunciate the above results as follows t
:
The convergents of an odd order are of an even order are
To
336.
the convergents
than the continued fraction.
all greater,
establish the
and
all less,
law of formation of
the successive con-
vergents.
Let the continued fraction be denoted by 1
a + x
then the
first
1
1
a 2 + a3 + a 4 +
three convergents are a. 1
a x a3 + a2
1
o, (a, a,
a3
.
+
!)
a2 +
+ «, 1
and we see that the numerator of the third convergent may be formed by multiplying the numerator of the second convergent by the third quotient, and adding the numerator of the first convergent also that the denominator may be formed in a similar manner. ;
Suppose that the successive convergents are formed, in a similar way; let the numerators be denoted by^,^.,, p 3 ,..., and the denominators by q lt q q 3 ,... ,
Assume that
is,
that the law of formation holds for the
»tt
convergent;
suppose 1\
= »J».-i +P»-i
In
=
<*
n
?.-,
+
Q„- 2
-
18—2
HIGHER ALGEBRA.
27g
The (*+
l)
th
convergent
±
the quotient a n +
differs
from the
in the place of
ft*
aj hence
only in having
the (» + 1)-
«*
vergent
+ ^»-i by ?„_!' + ?„
^n-n ^»
supposition.
?
« n+1
If therefore
we put
^^ ^ tttVs
th
conand denominator of the (» + l) numerator co. th«t the we to hold in the case of which was supposed P f
^ ^*
conhold in the case of the third so on; therefore * holds for the fourth, and
vergent, hence it holds universally. 337.
to call It will be convenient
aH the n* partial quotient;
at this stage being a n + a„ 4 the complete quotient +1 + «« +2 quotient at'any stage by usually denote the complete
ft.
We shall We have seen that
denoted by m ; then x differs from be fraction continued let the quotient ft instead of the partial only in taking the complete
&
quotient a„
;
thus
X_
ft
j^i-l
~kq
338
// Eb
6e
tfl6
n_x
+ ff»-2 + qn -2
n th convergent
to
'
a continued fraction, then
Q
denoted by Let the continued fraction be a,1
+
111
a Q + a 3 + a4 +
CONTINUED FRACTIONS.
277
then
= ("
But
p 2 q -]\ x
hence
q, />„
1
=
2
(P.-
)
(
2
% + 1) - «x
g^, -#,_,
When the hold
still
if
-iV,
9«-a
continued fraction we suppose that a x =
=
g,
•
(-
a,
^-2)1 similarly,
= 1 = (-
l)
2 J
1)".
than unity, this result will and that the first convergent
is less
0,
is zero.
When we
Note.
are calculating the numerical value of the successive convergents, the above theorem furnishes an easy test of the accuracy of the
work.
Cor. q n had a
Each convergent is in its lowest terms for iipn and common divisor it would divide pn q nl — pn_ l q ni or unity
which
impossible.
1.
;
;
is
Cor.
2.
The
between two successive convergents
difference
a fraction whose numerator
q»
is
unity
;
q n qn ^
?«_i
is
for
q,,q n -i'
EXAMPLES. XXV.
a.
Calculate the successive convergents to 1.
2.
3.
2
+
l
l *
l
'
6+ 1+ 1+ 11+
1111111 111111
2+ 2+ 3+ 1+ 3+
44-
2
2+
3+ 1+ 2+ 2+ 1+
6
9"
Express the following quantities as continued fractions and find the fourth convergent to each.
729 4.
HIGHER ALGEBRA.
278
A metre
39*37079 inches, shew by the theory of continued fractions that 32 metres is nearly equal to 35 yards. 12.
series of fractions converging to "24226, the excess in of the true tropical year over 365 days.
13.
days
14. .
A,
is
Find a
A kilometre 18
5
..
is
^
the fractions -,
very nearly equal to "62138 miles; shew that 23 64 the == , ^z are successive approximations to «.
•*•*
.
,
ratio of a kilometre to a mile. 15.
Two
16.
If
scales of equal length are divided into 162 and 209 equal parts respectively; if their zero points be coincident shew that the th 31 st division of one nearly coincides with the 40 division of the other.
—
that the quotients are cessive convergents. 17.
converted into a continued fraction, shew
is
s
n3 + n u + n + l
n — 1 and n+l
find the suc-
Shew that Pn + \~Pn - 1
_ Pn
-1
9.n
(!)
2n + 1
9.n
2
(2)
(^-O^-fH-vrPn Pn U / \
\
18.
and
alternately,
If
—
-1
\ cJn
+
g«-l
X ,
the n th convergent to a continued fraction, and a n the
is
corresponding quotient, shew that
Each convergent
339.
any of the
is
nearer
to the
continued fraction than
'preceding convergents.
Let x denote the continued
fraction,
and
*—
"-
,
9*
^-*±J
—"-±2
?« + !
^+2
three consecutive convergents; then x differs from *-a±l only in
taking the complete (n + 2) th quotient in the place of a this
and
by
k:
x=
thus
^^ ~ Pn +
&+
1
1
a;
? n+l +Pn
;
= Pn^l^n-Pn^l ?„ +
,
(%» +l +
7.)
;
1 "
y. + ,
(%„ + + ,
?„)
denote
—
«
CONTINUED FRACTIONS.
Now
k
greater than unity, and qm
is
is less
accounts the difference between
botli
—" and
between
difference
x: that
is,
--"
270 than q
and x
'
'
;
is less
every convergent
lience
on
than the is
nearer
to tlie continued fraction than the next preceding convergent, and therefore a fortiori than any preceding convergent.
Combining the
result of this article with that of Art.
.°>3.>,
it
follows that
convergent of an odd order continually increase, hat are always less than the continued fraction ; tli^
covrergents of an even order continually decrease, hut are always greater than the continued fraction. tin'
To find
340.
for
the
made in taking any convergent
limits to the error
continued fraction.
—
p r_n±2
p-^
p
Let
Y ,
1
]
)0
three consecutive convorgents, and let
k denote the complete (n + 2) th quotient;
x=
then
^^
t
k
p
1
>+ n J
'.(*•«
Now p.. — is
k
greater than
is
.
i
•
less
—
therefore the difference between x and
1,
,
and greater than
than
+
l
><7„,
the error in taking -" instead of x
taking
and greater than 77-0—
v. +
From the p —
last
.
,
article it
appears that the error in 1
instead of the continued fraction
is
qm or (I
a i+l
is,
II
less
than
,
?.?.+,
;
a/ (a V
is
1 5
?.
341.
-
p
<7, 1
1
less
i
..
than
Again, since
+9 i)
7
., ill +1
+q2
;
;
that
,) Ft— 1/
is,
less
than
« 11+ .7ili
3 "
:
hence the larger
.
I
the nearer does £2 approximate to the continued fraction; /
.
'
HIGHER ALGEBRA.
280
any convergent which immediately precedes a a near approximation to the continued fraction.
therefore,
quotient is
Again, since the error
than
is less
—
g
,
it
large
follows that in order
to find a convergent which will differ from the continued fraction
by
than a given quantity a
less
successive convergents
—
up to
,
we have only
where q n 2
,
to calculate the
greater than
is
a.
The
properties of continued fractions enable us to find two small integers whose ratio closely approximates to that of two incommensurable quantities, or to that of two quantities whose exact ratio can only be expressed by large integers. 342.
Example. Find a series of fractions approximating to 3* 14159. In the process of finding the greatest common measure of 14159 and 100000, the successive quotients are
3-14159 =
The
3+1
Thus
7, 15, 1, 25, 1, 7, 4.
1
1
111
1
7+ 15+ 1+ 25+ 1+ 7+
4
successive convergents are
1
this last convergent
333
22
3 '
7
'
106
'
355 113
which precedes the large quotient 25
approximation, the error being less than
25TP5)-
»
•° 00004
^-
is
a very near
and therefore
,
than
less
-
343. Any convergent is nearer to the continued fraction than any other fraction whose denominator is less than that of the convergent.
Let x be the continued fraction,
V
—
,
°n
P '-*=* two consecutive ?.-,
r
- a fraction whose denominator convergents, °
s is less
than q
8
If possible,
let
v r - be nearer to x than — «
7)
nearer to x than
-Ji
r
,
then - must be s
?»
^1 [Art. 339]
;
and since x
.
lies
between
9.-X
it
P -I"
In - J £-5=?
. "
follows that S
must
lie
between
— and —
'
%
?»-!
and
+
.
.
;
.
CONTINUED FRACTIONS.
281
Hence
P»-*P.
r
rqn _ x ~ spn _ x
.'.
that
is,
an integer p —
Therefore
&
<£
l
;
than a fraction
which
impossible. r be nearer to the continued fraction than -
must
P If -
344.
Pn -i fWi<5 ^
less
;
is
*
P' —
,
fraction x, then
be
—
two consecutive conver gents
x 2 according as
greater or less than
is
,
a continued
to
,
- is
—
greater or less than
,
q
Let k be the complete quotient corresponding to the convergent immediately ° J succeeding °
'
''
5
"*=
=
(tfp'q
,
factor ky'q'
—>
;
q" ,
WW^YY w
then x — -f—. lcq '
{hq
'
+ qY " "'
—+
,
q
w
pY]
-pq)(pq'-2>'q)
qq'(kq'
The
—
- pq
is
+ q) 2
positive, since p'
>p,
q'
>q, and
k>
I
pp'
lience
that
is,
,
or
< x 2 according
as
<
—
,
according as - > or
]iq'
—p'q
is
positive or negative
;
,
from the above investigation that the ex2 2 2 2 2 2 q' x —p' have the same pressions ]iq'—2 q VP ~ cL cL^-> p - q M Cor.
It follows )/
,
i
sign.
EXAMPLES. XXV.
b.
—
222 1.
Find limits to the error
a metre, given that a metre
is
in taking
yards as equivalent to
equal to 1-0936 yards.
)
HIGHER ALGEBRA.
282
Find an approximation to
2.
JL J_ J- JL JL + 3+ 5+ 7+ 9+ 11+ which
differs
from the true value by
less
than -0001. 99
3.
of continued fractions that =- differs from
Shew by the theory
1*41421 by a quantity less than
4.
„ Express 1
.
,. ,. , n a 3 + 6a 2 + 13a+10 as a continued fraction, and rs , A « \ 1K 3 4 +6a a + 14a--+15a + 7 ,
.
,
,
find the third convergent. 5. is
Shew
that the difference between the
and n th convergent
first
numerically ecpial to 1
1
(-l)n
1
+
Mi Ms Wh 6.
Shew
that
an
if
a ^
Pn-1~
'
«n-l+ «u- 2 + 0»-3+
"
8.
In the continued fraction
If
1
(
2)
—
tt
?i
th
...
«
p
s-5 ,
a2+
J_
1.
«3+
«2
1111 — «+ a+ + +
+P\ + 1 =Pn - lPn + 1 + £>„£>„ + 2 Pn = q n -lthe
«3+
n-3+
Pn
is
'"
L_
an-l+ «»-2+
qn-1
(
9n-l2n
the quotient corresponding to
_i- =an+ _L_ ^2
(2 )
7.
is
...+
,
'
«1
shew that
«
>
convergent to the continued fraction
111111 a+ a+ «+ b+
shew that 9.
b+
q 2n =p 2n + u
6
+
a
#».. -n q2n - 1 = r br
•
In the continued fraction
1111 a+ 6+ «+ 6+
'
shew that Pn + 2~ ( ah + 2 ) P n +Pn-2 = °i
9n + 2
~ ( ab + 2 )
?u + •?•*- 2 = °-
,
''
CONTINUED FRACTIONS. 10.
111
Shew that / a[a\ x+
a.v.j,+
\
ar
3
+
= «.v,H :r.,+
11.
If
-r;
-
,
iV
oa?3
2+ a S+
respectively,
If
/
.v,+
,
l)
th .
*+ a B +
'
111
W 4+
tt
(
'
':5+
"4+ a 5+
'
shew that T
—
iV
= (a^ + 1) P + aJL
the n th convergent to
is
i_
i
j.
"
a+ «+ a+ shew that
(?i-2) th convergent* to the
111
J/= OjP + 5, 12.
\
to 2/i quotients.
+
,
111 ((
+
oa;4
L
A3
continued fractions
a i+
x.
to 2/i quotients
- are the n tU (n —
,
(^
283
'
and q n are respectively the
pn
coefficients of
xn
in the
expansions of
# 1
Hence shew that equation
13.
t
2
If
-
at
—
-
is
1
«.#
.
— ax — x 2
a pn —
and
— «-r — x 2
1
n
+ x2 '
_ Qn i
>
where
a, /3
are the roots of the
= 0.
the n th convergent to
9n
_l
a+
shew that
1
1
b-\-
a+
1_
"
and q n are respectively the
pn
'
b-t-
coefficients of
expansions of
x + bx2 — ^ ax+(ab + l)x2 — xA and - (ab + 2) x2 + x* 1 - (aft + 2) x 2 + xA .
1
Hence shew that op, n = bq 2n _ x = ab
where
a, /3
are the values of
a
,
x2 found from the equation
l-(ab + 2)x2 + xA = 0.
'
xn
in the
CHAPTER XXVI. INDETERMINATE EQUATIONS OF THE FIRST DEGREE. 345. In Chap. X. we have shewn how to obtain the positive integral solutions of indeterminate equations with numerical coefficients; we shall now apply the properties of continued fractions
any indeterminate equation
to obtain the general solution of the first degree.
of
Any
equation of the first degree involving two unknowns x and y can be reduced to the form ax±by = ± c, where a, 6, c are positive integers. This equation admits of an unlimited number of solutions but if the conditions of the problem require x and y to be positive integers, the number of solutions may be 346.
;
limited.
It is clear that the equation ax + by = — c has no positive integral solution ; and that the equation ax — by = — c is equivalent
— ax — ax ±by — c.
c) hence it will be sufficient to consider the equations
to by
have a factor m which does not divide c, neither of the equations ax±by = c can be satisfied by integral values of x and y for ax ± by is divisible by m, whereas c is not.
and
If a
b
;
If a,
so that
have a common factor
b, c
we
that a and 347. equation
shall suppose a, b
b,
can be removed by division; to have no common factor, and
c
it
are prime to each other.
To find
the general solution
ax — by —
in positive integers of the
c.
Let - be converted into a continued fraction, and 6
the convergent just preceding j
let
— denote q
;
then
aq—bp = ±l.
[Art. 338.]
;
.
INDETERMINATE EQUATIONS OF THE FIRST DEGREE. If aq
I.
— bj) —
ax — by — c (aq —
a(x —
.-.
Now divisible
that
b
;
cq)
-b
(y
—
written
;
c/>).
and
b
x=
is,
x- cq
y — cP.
b
a
+
bt
y—
cq,
at
+
cj)
\
may
be obtained by giving any positive integral value, or any negative integral value
from which positive integral solutions to
b]j)
l»e
have no common factor, x — cq must be hence x — cq = bt, where t is an integer,
since a
by
may
the given equation
1,
285
t
numerically smaller than the less of the two quantities also
t
may
II.
be zero; thus the number of solutions
If
aq — bp — —
1,
.
=
(aq
—
bji)
x + cq =—
+ cp — = t, an = y
x=
—
\
unlimited.
;
cq)
-
b (y
+
cj))
',
integer
a
o
lience
c
a(x +
—
• .
,
we have
ax — by — — .'.
is
cd
en -j-
bt
— cq, y — at —
cp;
from which positive integral solutions may be obtained by giving to t any positive integral value which exceeds the greater of the CO
CD
o
a
two quantities -=-,—; thus the number
III.
of solutions
If either a or b is unity, the fraction
j-
unlimited.
is
cannot be con-
verted into a continued fraction with unit numerators, and the investigation fails. In these cases, however, the solutions may be written down by inspection; thus if 6 = 1, the equation becomes ax — y = c; whence y = ax—c, and the solutions may be found by ascribing to x any positive integral value greater than -
a
Note. It should be observed that the two arithmetical progressions in which the respectively.
and y form are b and a
series of values for x
common
differences
;
HIGHER ALGEBRA.
286 Example.
Find the general solution in positive integers of
In converting 13 is -jr
;
we have
— into a continued fraction
29.r
-
42*/
= 5.
the convergent just before
—
therefore
.-.
29xl3-42x9 = -l; 29x65-42x45 = - 5;
combining this with the given equation, we obtain 29 (* + 65)
=42(# + 45);
x + 65 •*•
hence the general solution
-£j- =
u
+ 45 29"
= *»
an mte 8 er
5
is
= 42«-65,
a:
ij
= 20t~4o.
Given one solution in positive integers of ax — by = c, to jind the general solution. 348.
Let
h,
ax-by = c;
k be a solution of
then ah — bk =
ax — by = ah - bk
.'.
the equation
c.
;
a (x — h) — b(y — k);
.'.
x—h — — = y—k = a
.'.
z
t.
an integer
;
b
.'.
which
is
bt,
y—k+
at
the general solution.
To Jind
349.
x=h+
the general solution in positive integers
ax + by =
equation
fraction,
and
b
I.
If aq
D
let
— be the q
;
then aq — bp = ±
1.
— bp=l, we have ax +by =
(aq
— bp);
a(cq — x) = b(y +
.'.
cq .'
c
.
c2));
—x
— = y—+—acp— = L an integer °
-=-=
b . ' .
the
c.
a Let t be converted into a continued convergent just preceding j
of
x = cq — bt, y — at - cp '
;
;
'
; ;
;;
INDETERMINATE EQUATIONS OF THE FIRST DEGREE.
may be
from which positive integral solutions
obtained by giving
—
CI)
to
t
positive integral values greater than
Thus the number
If aq
— bp = -
1,
x + co
—— -
•
—
c
(aq
—
•
.
en =— « y=
a
JL
o
x=bt — cq,
.
y
bp)
no
solution.
number
If either
III.
no integer
;
;
at
may be
co positive integral values greater than -~ before, the
is
an integer 7
t.'
= cj) —
from which positive integral solutions
As
there
.
a(x + cq) = b(cp-y);
.-.
t
CO
than j-
less
we have
ax + by =
to
if
and
no solution.
fulfilling these conditions there is
II.
and
of solutions is limited,
287
of solutions
a or
is
limited,
obtained by giving
and
than
less
cP
and there may be
b is equal to unity, the solution
may
be
found by inspection as in Art. 317. Given one solution in positive integers of Ike equation ax + by = c, to find the general solution. 350.
Let
A,
k be a solution of ax
.'.
by —
c
;
then ah + bk =
.
a (x — h) — b (k -
y)
7— =- —-— ——hk—y o a
x .'.
.'.
which
is
equation
x=h +
bt,
y-k—
t,
an integer
;
at
the general solution.
To find
351.
c.
ax + by — ah + bk
'
.
-f
the
number of solutions in positive ax + by = c.
integers
Let T be converted into a continued fraction, and b
convergent just preceding j
let
- be the q
;
then aq — bp
= at 1.
of the
.
HIGHER ALGEBRA.
288 Let aq -bp
I.
=
l
then the general solution
;
x = cq-bt, y = at-
is
ep.
[Art. 349.]
Positive integral solutions will be obtained by giving to positive
not
values
integral
than °f
greater
and
,
not
t
less
o
—a
than
c
c
a
b
Suppose that - and T are not integers. °
(i)
— = m+f. J
Let
=n+
-±
a
b
a. y
'
where m, n are positive integers and J\ g proper fractions then the least value t can have is m+ 1, and the greatest value is n; ;
therefore the
number
of solutions is
cq
cp
b
a
n-vi = -±-
Now
this is
an
c — +fg=-—+f-g. J J J J .
j.
ab
and may be written
integer,
— + a fraction, ab
—r- a
fraction, according as /is greater or less
number
of solutions
as/ or g
according
In
this case
the
this,
is
Suppose that
(ii)
g-
number
0,
the integer nearest to
is
than
—
,
g.
the greater. z-
is
an
integer.
and one value
of
x
is zero.
If
C
^7+1
or -j
,
according as
include
-r+f, which must be an
Suppose that -
of solutions
we
c
(iii)
we
c
of solutions is
Hence the number
C
Thus the
greater or less
ao teger.
or
is
is
in-
the greatest integer in
include or exclude the zero solution.
an
integer.
cc
case/=0, and one value of y is zero. If we include the least value of t is m and the greatest is n; hence
In this,
the
this
number
of solutions is
71
—m+
l.
or —r
ab
-
q+
1.
Thus the
INDETERMINATE EQUATIONS OF number
c
of solutions is the greatest integer in -7
ab the zero include or exclude solution.
we
cording as
(iv)
Suppose that
c -
and
+1
289 c
or —=. ae-
ab
c
7 are
both integers.
and y = 0, and both x and y have a zero If we include these, the least value t can have is m, and
In this case value.
the greatest
-y + ab
DEGREE.
TJIE FIRST
1.
is
n
f— ;
hence the number of solutions
If
we exclude
If
aq
the zero values the
is
number
11-111+
1,
or
of solutions is
4-i.
ab
II.
-bp= -
1,
x=
and similar
the general solution bt
—
cq
is
y— cp — at,
}
results will be obtained.
352. To find the solutions in positive integers of the equation ax + by + cz — d, we may proceed as follows.
transposition ax + by = d — cz ; from which by giving to z we obtain equations of in succession the values 0, 1, 2, 3, the form ax + by = c, which may be solved as already explained.
By
we have two simultaneous equations ax + by + cz=d, ax + b'y + cz = d\ by eliminating one of the unknowns, z say, we obtain an equation of the form Ax + By = C. Suppose that x —f, y — g is a solution, 353.
If
then the general solution can be written
x=f+Bs, y = g-As, where
s is
an
integer.
Substituting these values of x and y in either of the given equations, we obtain an equation of the form Fs + Gz = II, of which the general solution is 8
Substituting for
s,
=h + we
Gt,
= k - Ft
say.
obtain
x=f+Bh + BGt, and the values
z
y = g-Ah-AGt;
of x, y, z are obtained
by giving to
t
suitable
integral values. H. H. A.
19
:
HIGHER ALGEBRA.
290
If one solution in positive integers of the equations
354.
ax + by +
=
cz
ax + b'y + c'z = d',
d,
can be found, the general solution Lety,
g,
be obtained as follows.
h be the particular solution
af+ bg +
By
may
ch
=
d,
a'f+
then
;
+ ch = d'.
b'g
subtraction,
a(x-f) + b(y-g)+c(z- h) = 0, a'(x-/) + b'(y-g) + c'(z-h) = 0;
1
j
whence
x-f be
where be
t
— b'c
y-g _ z-h _
=
ca
— c'a
ab'
— a'b
t
k
'
an integer and k is the H.C.F. of the denominators Thus the general solution is ca — c'a, ab' — a'b. is
—
b'c,
x
=f+
(be'
—
b'c)
j
,
y—g+
(ca'
—
c'a)
=• ,
z
=h+
(ab'
- a'b) ?. fc
fc
/c
EXAMPLES. XXVI. j
Find the general solution and the 1.
775.r-711y =
4.
In how
crowns 5.
l.
many ways
least positive integral solution, of
2.
455#-519y=l.
can
,£1. 19s. 6d.
3.
436#-393y = 5.
be paid in florins and half-
?
Find the number of solutions in positive integers of
lLe+15y=1031. 6.
Find two fractions having 7 and 9
such that their 7.
and 8 8.
of
sum
is 1
for their denominators,
Find two proper fractions in their lowest terms having 12 for their
A
denominators and such that their difference
certain sum consists of x shillings ; find the sum.
pounds y
shillings,
is
and
— 24
Solve in positive integers 21 6#+ty + 4s=122\ lhr + 8y- 6^=145 5J
10.
-
12.r-lly 1 2x 1 \y
.
it is
y pounds x
9.
and
£-§-.
= 221 4^=22 + 4z
'
-4.v+ 5y+ z=ll)
half
INDETERMINATE EQUATIONS OF THE FIRST DEGREE. 11.
by
20^-21^=381 3y+ 4s =34/
12.
13^ + 1 Is =103)
'
7z
+ 4y + 19^ = 84.
"
- by=
23.r+17.y +
4J
lU = 130.
13.
7.r
15.
Find the general form of
5, 7,
8 leave remainders
16.
Find the two smallest integers which divided by
remainders
A
1,
14.
3, 2,
291
all positive
integers which divided
5 respectively. 3, 7, 11
leave
6, 5 respectively.
number
of three digits in the septenary scale is represented in the nonary scale by the same three digits in reverse order if the middle digit in each case is zero, find the value of the number in the 17.
;
denary
scale.
If the integers 6, «, b are in 18. possible values of a and b.
harmonic progression, find
all
the
Two
rods of equal length are divided into 250 and 243 equal parts respectively if their ends be coincident, find the divisions which are the nearest together. 19.
;
commenced
to toll at the same time, and tolled at intervals of 23, 29, 34 seconds respectively. The second and third bells tolled 39 and 40 seconds respectively longer than the first ; how many times did each bell toll if they all ceased in less than 20 minutes?
7.r
20.
Three
21.
Find the greatest value of c in order that the equation have exactly six solutions in positive integers.
bells
+ 9y = c may
Find the greatest value of c in order that the equation 14r + lly=c may have exactly five solutions in positive integers. 22.
Find the limits within which c must lie in order that the equation 19x + 14y = c may have six solutions, zero solutions being 23.
excluded.
Shew that the
greatest value of c in order that the equation ax + by = c may have exactly n solutions in positive integers is (n + l)ab-a-b, and that the least value of c is (n-l)ab + a + b } zero solutions being excluded. 24.
in
o
CHAPTER
XXVII.
RECURRING CONTINUED FRACTIONS.
We
have seen in Chap. XXV. that a terminating continued fraction with rational quotients can be reduced to an ordinary fraction with integral numerator and denominator, and but we shall prove that a therefore cannot be equal to a surd quadratic surd can be expressed as an infinite continued fraction whose quotients recur. We shall first consider a numerical 355.
;
example. Example. Express ^19 as a continued fractions approximating to its value. x/19 = 4 v/19
5
+ 4_ 2
V19 + 2
+ 2_ 1j ^19-3 = , 1 +
=1+^—
.
_
+ 3_ 1
1
5
2
\/19 + 3'
1
1
9-2_ 1
~ + N/ 1
+
3 (
~
N/19-4)
1
,
.
3
1
0i
\/19 + 4'
=8+
after this the quotients 2, 1, 3, 1, 2, 8 recur; 1
2
^£9-8 +V
+4=8+
'
\/19 + 3'
n iN/19-4 2
,
5
5
N /19
5
+,x/19 z _2_ 3
/L9 + 2
series of
3
,/19 + S 2
v/19
and find a
+ ( v/19-4) = 4+ Tl9 - -; v +
3 N /19
fraction,
Jl_£
hence 1
2. •••
V 19 - 4 + 2+ 1+ 3+ 1+ 2+8+
be noticed that the quotients recur as soon as we come to a quotient which is double of the first. In Art. 361 we shall prove that this is always the case. It will
»
,
RECURRING CONTINUED FRACTIONS,
293
[Explanation. In each of the lines above we perform the same series of operations. For example, consider the second line we first find the :
greatest integer in -
—-—
;
this is 2,
and the remainder
—
-
is
is
^— ^
—
We
.
o
new
that
then multiply numerator and denominator by the surd
^19-2,
conjugate to
2,
6
o
5
so that after inverting the result
.
,
we begin a
line with a rational denominator.]
The
first
seven convergents formed as explained in Art. 336 are 4
48
13
9
170
61
1421 '
1
The less
'
2
3
'
11
'
'
14
eiTor in taking the last of these
than
.
—
-
,
or
and
,
'
39
is less
a fortiori
326
'
than
less
,
'
_
,
and
than -00001.
is
therefore
Thus the
seventh convergent gives the value to at least four places of decimals.
Every periodic continued fraction is equal to one of the of a quadratic equation of which the coefficients are rational.
356. roots
Let x denote the continued and suppose that
x=
a+
1
1
z
b
fraction,
,
+
c
+
and y the periodic
part,
:
—
:
HIGHER ALGEBRA.
294
+ {s — r) y — r = 0, which gives the value of if the positive value of y, has its roots real and of opposite signs v'y + p — on rationalising the denominator be substituted in x = -,
The equation
s'y
2
;
—
ii
qy+q
the value of x
B
is of
,
—
~G
the form
being positive since the value of y Example.
Express
where A, B,
,
is real.
,1111 — — ~— +s
l
^-—
5
C are integers,
as a surd.
...
Let x be the value of the continued fraction
—+ O— +
then x - 1 = =
;
1
£>
1
\X
—
—
7 = 0. continued fraction is equal to the positive root of this equation,
;
1)
whence 2x + 2x 2
The is
therefore equal to
—
^—-
and
.
EXAMPLES. XXVII.
a.
Express the following surds as continued fractions, and hnd the sixth convergent to each 1.
v/3.
2.
^5.
3.
y/6.
4.
s/8.
5.
v/11.
6.
x/13.
7.
x/14.
8.
V22.
9.
2^3.
10.
j&-
14
13
-
4 v/2.
3^5.
11.
15 -
V33-
\/s-
-
4 N/10.
16 -
\/n-
268
—— is
17.
Find limits of the error when
18.
916 Find limits of the error w hen '—-
19.
Find the
taken for N /17.
65
T
first
12.
is
taken for
convergent to N /101 that
is
v/23.
correct to five places
of decimals. 20.
Find the
first
convergent to
VI 5
that
is
correct to five places
of decimals.
Express as a continued fraction the positive root of each of the following equations
2x-l = 0.
a8 -4*?- 3=0.
21.
x* +
24.
Express each root of x2 - 5^ + 3 =
25.
Find the value of 3 + 5— x— x--
26.
Find the value of
22.
Ill
6+ 6+
—
,
6+
-
1+ 3+ 1+
3+
23.
la?- 8x- 3=0.
as a continued fraction.
RECURRING CONTINUED FRACTIONS.
3+
Find the value of
27.
28.
Find the value of 5 +
29.
Shew that *+
2
,
+
111111 3+ 3+ 5+
1+
let
10 +
;
continued fractions
infinite
1+ 5+
1+
""
To convert a quadratic surd into a continued fraction.
N be
a positive integer which is not an exact square, a be the greatest integer contained in then
JN
x
N/iV = «,
Let
+
111111 3+ 3+ 5+
•"'
5+ 1+
*357.
Let
3
2
i+6+ i+ e+""~*\ 1+ a+ 2+ 3+ 2+
1+
and
111111 1+ 3+ 1+ + 1111 1+ 1+
Find the difference between the
30.
295
+ (Jff- a,) =
-j£—
+
«,
,
if r,
j
W- »,\
=
be the greatest integer contained in —
b
'
;
then
r i
JM+a
l
=
b
JN-b r x
x
+a
^
x
|
«2 = b r
where
i
— a and
1
x
JN-a
h
,
r r2 x
^h +
2
= N — a„ 2
.
Similarly 2
r2
so
on
;
—
,
.
b M-l ,
+
JN-a v —
H— 1
an =
Hence and thus "We
N— a
2
3
;
and generally
JN+a-s=i = where
JN + a./
-
2
«3 = bf2 — as and r 2 r 3 —
where
and
»*
'
&„_,/•„_,
*JN= a, +
-
= b "-' + ,
(1-1
and
a„_
"
jy + a ,
>
?•„_,*•„
=
1111 1
r
.
"
' it
N - a/.
JN can be expressed as an infinite continued fraction.
shall presently prove that
this fraction consists of
re-
curring periods ; it is evident that the period will begin whenever any complete quotient is first repeated.
;
HIGHER ALGEBRA.
296
We
shall call the series of quotients T
JA the
first,
*358.
JAr + a,
JN + a
r
r2
.
x
JJST+ a 3
2
)
r3
complete quotients.
second, third, fourth
From
the preceding article it appears that the quantities a v rv b v b b are positive integers; we shall now prove 3 that the quantities a2 a 3 a4 are also positive inr8 r 3 r 4 ,
,
,
,
,
,
, .
,
.
.
tegers.
p p Let — , —.. q
q
p —r.
q
be three consecutive convergents to JN. and ° x
— be the convergent
P" let
The complete quotient
corresponding to the partial quotient b n
at this stage
— r
v^=
p +p
parts,
=P
,t
Clearing of fractions and
is
—
;
.
hence
JW+a„P+r p v
equating rational and irrational
we have
«y + rnP = ^Y>
ck
+ rn q =p
;
whence a n ( pq - pq) =pp* ~
;
Since two convergents precede the complete quotient
-
r*
this investigation holds for all values of
*359.
To prove
that the complete
n greater than
and partial
1.
quotients recur.
In Art. 357 we have proved that rn r n _ = N—a 2 Also r n and r n _ are positive integers ; hence a n must be less than ^/JV, thus an cannot be greater than a v and therefore it cannot have any .
l
l
values except 1, 2, an cannot exceed a x
'
3, ...a x }
that
is,
the
number of different values of
.
Again, a n+1 =rv b u -a that is rn b n = a n + a n+v and therefore r n b n cannot be greater than 2a also b n is a positive integer hence rn cannot be greater than 2a v Thus rn cannot have any values except 1, 2, 3,...2a ; that is, the number of different values ofi\ cannot exceed 2a r h
,
l
1
;
1
1
1
' ;
.
RECURRING CONTINUED FRACTIONS. Thus the complete quotient 2a*
different values
;
—
is,
—
b n is the greatest integer in
must
quotients
and
also recur,
cannot have more than
rn
some one complete quotient, and must recur.
that
therefore all subsequent ones,
Also
—
297
—
rn
hence
;
number of partial
the
the partial
quotients in
2
each cycle cannot be greater than 2a
']
To prove
*3G0.
We
that a,
have
< au +
1
t
5
;
N"-a;=rnrn _
But
;
l
+ a «= or >?, ,-i
a posit ive integer
is
.
+ an = bH _ rn _
«,_,
«»_i
since 6„_ l
rn
a ~ a n < rn i
-
l i
,
which proves the proposition. *361. To shew that the period begins loith the second partial quotient and terminates ivith a partial quotient double of the first.
we have
seen in Art. 359, a recurrence must take place, let us suppose that the (n+ l) th complete quotient recurs at the (*+ l) th ; then Since, as
a.
we
shall
r,
,
=r
,
=
and
b.
— 1',
b, *—
b
;
prove that a,4 — 1
=a n —
r* =
N -a,
.
We
=a
,
rm»—
.
1'
,
=r
,
ii
=
b —
,
ii
have r.* —
, i
2 '
»
—
iV
— a = r H — ,r —r 2
it
v
i
ii
H
—
. l
r,*
=r
Again,
a .-.
,
"
_a
'
n-1
*
~
1
7
-b
— om = zero, 7
,
.
or an integer.
1
HIGHER ALGEBRA.
298
But, by Art. 360, a a.
-a.
,
-a
therefore a
' ,
l
:
n_
,
l
<.r _
- aa
and a
,
ii
x
,
.
l
-a _
hence
;
s
l
that
;
i
—
-^
is
is less
n-
than unity, and therefore must be
Thus
«,_!
= «„_!, and
also 6#_1
zero.
=6
fl
_i.
th the (n + l) complete quotient recurs, the ?^ comth plete quotient must also recur; therefore the (n- l) complete quotient must also recur; and so on.
Hence
th
if
This proof holds as long as n is not less than 2 [Art. 358], hence the complete quotients recur, beginning with the second
—
It follows therefore that the recurrence rx begins with the second partial quotient b ; we shall now shew that it terminates with a partial quotient 2a
quotient 1
-
.
x
x
Let
.
"be the complete quotient which
-
second complete quotient -
when
it
just precedes the
recurs
;
then
—— H
1
ancj
v
l
a
ar e
two consecutive complete quotients
r,
hence rn =
;
therefore
»",
but
N- a* = Again, a
y
;
— aH <
?'„,
that
1.
is
««
Also an + a
= rn bn — bn
;
<
1
hence a - an -
;
0,
x
that
is
= «,•
hence
bn
= 2a
i
;
which establishes the
proposition.
*362. distant
any period the partial quotients equibeginning and end are equal, the last partial
To shew
from
the
that in
quotient being excluded.
Let the
last
—
complete quotient be denoted by *
;
r.
rn =l,
ci n
=a x
,
bn
=2a r
We shall prove that ^- 2 =^ 2)
«h-2=«3
>
^.-2= & 2^
then
+
.
;
RECURRING CONTINUED FRACTIONS.
We
299
have r«-x
=
N - a; -
=-N- a,;
rH_t
r*
r,
Also
+a =
»„_,
and
+ a H = rm_,
«„_,
x
«,+«., =
.
-=-
.
-
&„_,
=
r, &„_,
?•,&,;
= 0, —
= zero,
o M-1
or an integer,
i
"-' I
But unity
^
<
~ a "~
CT i
Similarly
that
,
thus a 2 - an_x =
;
l
= r2
rn _ 2
hence
•
a,,.!
= «3
«„_ 2
,
is
<
a '~ a '-
1
which
,
= « 2 and ,
6„_ 2
,
-
=
o,,^
b
is less
than
.
l
b2
and
;
so on.
*363. From the results of Arts. 3G1, 362, it appears that when a quadratic surd v/iVr is converted into a continued fraction, it must take the following form
1
J_ J_ J_ + & +6 + &i
°3
3
3
To obtain
*364.
J_ J_ J +
+
&2
1_
+2a,+
°i
the penultimate convergents
of
the recurring
periods.
Let n be the number of partial quotients in the recurring period then the penultimate convergents of the recurring periods are the ?iih 2n th 3n th convergents ; let these be denoted by ;
,
,
,
V-\ ^=,
Now
xt
/v = a + JiV v l
^,
respectively.
—11 — + +
—111 —+ —+ + j
i
b
b2
i
b _
b3
7l
-
2a
l
—
7)
so that the partial quotient corresponding to
l
+1
is
2a
;
hence
t
Pn+X
SWl
= ^Pn+Pn-l " 2 «1 9n + ?n-l
Tlie complete quotient at the
2«,+T bi
K
same stage :
-r
+
+
'
6 «-i
consists of the period
,
HIGHER ALGEBRA,
300 and
is
therefore equal to a + x
J'N
hence
;
Clearing of fractions and equating rational and irrational parts, we obtain
*iP.+JV-i = -jfy«i a i9 H + 9 n-i=^« ,
Again
r
— can be obtained from — and
-^
(!)•
by taking
for the
quotient
2* 1
which
is
equal to
rtj
+
—
+
1
C 1
1
V?V^ Thus
.
in
&=1 **
=
2^
U+%)qn + q^ .
& P« + %.qn
,
from
l(A + *&)
(1);
(2)>
?2 „
In
like
manner we may prove that
if
—
-
is
the penultimate
icn
convergent in the
c th
«i ^c«
recurring period,
+Fcn-i
= Nqmi
a, q cn
+ qm_l =#*,
and by using these equations, we may obtain £—
,
—
,
suc-
cessively.
It should be noticed that equation (2) holds for all multiples of
n
;
thus
Ol
the proof being similar to that already given.
In Art. 356, we have seen that a periodic continued fraction can be expressed as the root of a quadratic equation with rational coefficients. *365.
y
RECURRING CONTINUED FRACTIONS. we might prove by
Conversely,
the method of Art. 357 that
—
an expression of the form
301
tt~~
>
where
B,
-<4>
C
are positive
B
and
not a perfect square, can be converted into a recurring continued fraction. In this case the periodic part will not usually begin with the second partial quotient, nor will the last partial quotient be double the first. integers,
For further information on the subject
we
of recurring continued
refer the student to Serret's
Cours cVAlgebre Superieure, and to a pamphlet on The Expression of a Quadratic Surd as a Continued Fraction, by Thomas Muir, M.A., F.R.S.E.
fractions
^EXAMPLES. XXVII.
b.
Express the following surds as continued fractious, and find the fourth convergent to each :
2
+ l.
1.
N/a
4.
V/T7T.
7.
Prove that
Ja* -a.
2.
5
y«"^f
.
J9a* + 3 = 3a+ and
N /«--l.
3.
6
.
^l
.
—— -i
2a + 6a + 2a + 6a +
find the fifth convergent.
Shew
8.
that 2
p 9.
V
p+
If
Cl
2
+
Ja'2 +
tt
\
3 + P9 4 + Ct
=P a + aCL \
/
Tf If
+
PaZ +
OCl \
+
be expressed as a continued fraction, shew that
1
.%'=
?/ •
,
1111
—
2p n = q n _ l + q n + l ...,
1111 1111
a x + « 2 + a i+ a 2 +
~2a + 1
2« 2 + 2a x + 2a 2 +
3^ + 3« 2 + x {f- - z-) + 2y (z 2 - .r2 + 3z {a? ~3tf 1
shew that
—
l
2(a*+l)q n =p n _ 1 +p n + 1 11 11.
r-
111—
that
a \-\
P9
—
i—»
p
/111 Shew
\ 10.
i+
1111 i+
+~3a 2 +
)
""'
'"' 1 )
= 0.
.
HIGHER ALGEBRA.
302 Prove that
12.
V————
JL J_ Jl J_ ^ =+ a+ b + a+ '")\b+ a+ b+ a+ '")"b
(
b
X ~ a+
U
13.
1
J_ J_ J_ b+ b+ a+ a+
••'
J_ J_ J_ J_ y- b + a+ a+ i+ b+ + a + b)x- (a b + a + b)y = a -b2 •••'
shew that
(ab 2
— be the n
If
14.
2
P2
2
2
th
Ja 2 + l.
convergent to
shew that
+P32 +'"+P 2n + l = Pn + lPn + 2-PlP2
Shew that
15.
— (— \a+ a+ + b
v
If
16.
c+
c
'
"/ \
— denote the r
ih
+
1
1
^+
a+
c
_i_j_2_ a+ b+ c+ =
1
18.
number
l
+
/
l+ctb'
—
? 3 + ?5 + -
Prove that the difference of the
equal to
\
,
shew that
2
Pi+Pi>+ >~+P°.n-l=P2n-p
+ bc
1
convergent to ^—— &
qr
is
.
+ ao
i •'
1
continued fractions
infinite i
+ &» - = ?8» ~ ft.
i
b+ a+ c+
'••'
.
If s/JV is converted into a continued fraction, of quotients in the period, shew that
and
if
n
is
the
De converted into a continued fraction, and if the penultimate convergents in the first, second, ...kth recurring periods be denoted by n lt n2i ...nk respectively, shew that 19.
If
\/^
*CHAPTER
XXVIII.
INDETERMINATE EQUATIONS OF THE SECOND DEGREE. *366. The solution in positive integers of indeterminate equations of a degree higher than the first, though not of much practical importance, is interesting because of its connection with In the present chapter we shall confine the Theory of Numbers. our attention to equations of the second degree involving two variables.
To shew Iww
*367.
obtain the positive integral values of x arid y which satisfy the equation to
ax 2 + 2hxy + by 2 + 2gx + 2fy + a, b, c, f, g,
h being
c
=
0,
integers.
Solving this equation as a quadratic in
x,
as in Art. 127,
we
have
ax +
Now
hy+g = ±J(h
2
-ab) y 2 +
2 (hg
- af)y+(g 2 -ac)...(l).
x and y may be positive integers, the expression under the radical, which we may denote kv py 2 + 2gy + r, must be a perfect square that is in order that the values of
;
2 py + 2qy + r =
z
2 ,
suppose.
Solving this equation as a quadratic in y, 2 2 py + q = ± Jq -jjr+pz
we have
;
and, as before, the expression under the radical square ; suppose that it is equal to t 2 ; then 2
t
where
t
and
must be a perfect
- pz 2 = q 2 - pr^
z are variables,
and
j\ q, r are constants.
HIGHER ALGEBRA.
304
Unless this equation can be solved in positive integers, the original equation does not admit of a positive integral solution.
We
shall return to this point in Art. 374. If
a,
b,
h are
positive, it is clear that the
all
number
of
limited, because for large values of x and y the sign 2 2 of the expression on the left depends upon that of ax + 2hxy + by [Art. 2G9], and thus cannot be zero for large positive integral
solutions
is
values of x and
Again,
if
and
negative, solutions
is
Example.
y.
—
2
negative, the coefficient of y in (1) is by similar reasoning we see that the number of
h*
ab
is
limited. Solve in positive integers the equation a;
2
- 4xy + &y* - 2x -
20*/
= 29.
Solving as a quadratic in x, we have
x = 2y + 1 ± ^30 + 24// - 2y\
But 30 + 24?/ - 2j/ 2 = 102 - 2 (y - G) 2 hence (y - 6) 2 cannot be greater than By trial we find that the expression under the radical becomes a 51. 8 perfect square when (y-6) =l or 49; thus the positive integral values of y ;
are
5, 7, 13.
When
?/
= 5,
x = 21
or
1;
when y = 7, x = 25
or
5;
when y = 13,
x = 29 or 25.
We
have seen that the solution in positive integers *3G8. of the equation 2 ax 2 + 2hxy + by + 2gx + 2fy +
c
=
can be made to depend upon the solution of an equation of the form
x2 ± where
iV*
and a are positive
Ny = ± a, 2
integers.
The equation x 2 + Ny* = — a has no real roots, whilst the 2 2 equation x + Ny = a has a limited number of solutions, which
we shall therefore confine our attention be found by trial 2 2 to equations of the form x - Ny = ± a.
may
;
*369. To sJiew that the equation x solved in positive integers.
Let
JN
2
-Ny =l 2
can always
be converted hito a continued fraction, and
be
let
2_
l-
'— be any three consecutive convergents; suppose that
q
q
9
y
U
1
INDETERMINATE EQUATIONS OF THE SECOND DEGREE. 305 t
17"
,n
a,/
the complete quotient corresponding to
is
*. (/"/
But
=
r„
1
~
— JVq
]>
"
] } (1
2
—]
then
;
[Art. 358].
-P"
any period [Art.
at the end of ..
W
V'q)
„
.'3(51]
)(
2
)
/
,
being the penultimate convergent of any recurring period. If the
number
convergent, and
pq —pq = 1.
thus
;
and therefore x=]>\ y = q p —
Since
period, the If the
is
number
is
than
Jn
v/iV,
i.s
an even
and therefore
this case p'*
— N"q' a = J,
a solution of the equation xr — Ny*
=
1.
penultimate convergent of any recurring
the
number
greater
therefore
is
P
greater than
of quotients in the period is even, -,
of solutions
is
unlimited.
of quotients in the period
is
odd, the penultima
1
convergent in the tirst period is an odd convergent, but the penultimate convergent in the second period is an even convergent. Thus integral solutions will be obtained by putting x=p', y — q\
—
where
the penultimate convergent in the second, fourth,
is
q
Hence
recurring periods. of solutions is unlimited. sixth,
*370.
As
To obtain a
also in this case the
solution, in positive inte
in the preceding article,
number
of the equation
we have
v -Jq =pq-pq> f£
the
If
number
~KT
'2
'
I
of quotients in the period is odd,
and
if <1 >
is
an odd penultimate convergent in any recurring period, -,-<-,
pq —pq — — 1. case p' — Nq -
and therefore
In this equation x 2 —
where
'
2
X
1
=—
2
1
will
-\, and integral solutions of the be obtained by putting x =p\ y — q ',
—
is the penultimate convergent in the first, third, q recurring periods. nappeu i.H.10 we can discover ,, „ 11.
n. A.
2fc
—
fifth...
,
irv
1:0
HIGHER ALGEBRA.
306
Solve in positive integers x 2 - 13y 2 = ±1.
Example.
We
can shew that
11111 1+ITl+
^13-3 + Here the number
of quotients in the period is
18 vergent in the
first
6+
1+
period
is -=-
hence
;
a;
= 18,
odd
y=o
;
the penultimate con-
a solution of
is
x 2 -13y 2 =-l.
By
Art. 364, the penultimate convergent in the second recurring period is 1
/ 18
+
2U hence # = 649, y = 180
By forming periods
is
5
649 ,u * thatlS
io\ Xl3
•
'180' J' a solution of x 2 - 13y 2 =l. 18
the successive penultimate convei'gents of the recurring of solutions of the equations
we can obtain any number x2 -
13?/
2
=
-
and x 2 - lSy 2 = + 1.
1,
one solution in positive integers of x 2 — Nif = 1 lias been found, we may obtain as many as w e please by th following method.
When
*371.
r
Suppose that x = h, y = k is a solution, h and k being positive n 2 integers; then (A — Nk 2 ) = 1 where n is any positive integer. ,
x2
Thus .
•.
(x
Put x +
+ yjN)
-Ni/= (h* - m?y - yJJST) =
(x
(h
+
.
kJN)
yJN = (h + kJX)' x-y JN 1
,
=
n
(h (h
- k s !X)\ - kJN)"
;
n
2x = (h +
kJN) + (h-kJJYy; 2 Us in = (h + kjiry - (h - kjNy. .-.
Tlie values of
x and y so found are positive integers, and by
ascribing to n the values can be obtained.
Similarly
Xy = —
x —
2
2
1,
x=h and if n
if
i
x*
Thus the values
n
is
of
as
1, 2, 3,...,
many
solutions as
y = k is a solution of the is any odd positive integer,
- Ntf =
(h
2
we
please
equation
- Nk 2 )\
x and y are the same as already found, but
restricted to the values
1, 3,
5,
By putting x = ax', y = ay the equations x — Ny — ± a 3 become of 2 — IFyf* = d= 1, which we have already shewn how to *372.
solve.
2
2
,
INDETERMINATE EQUATIONS OF THE SECOND DEGREE
We
*373.
Hence
if
307
have seen in Art. 3G9 that n P ~ N(f~ = ~ rn ( V<1 ~ V
a
is
a denominator of any complete quotient which
JX
occurs in converting
and
into a continued fraction,
—
if
•
i
is
q
convergent obtained by stopping short of this complete 2 quotient, one of the equations x~ — Xy — ±a is satisfied by the the
values
x =p\ y = q'-
Again, the odd convergents are
even convergents are
greater than
all
than
all less
JX
;
hence
JN, and —
if
,
the
an even
is
<2 t
convergent, is
x=p, y = q
an odd convergent, x *374.
a solution of x*
is
=p
,
y—q
method explained
is
Tlie
-Xif — a: and
if —.
'
q
— Xy = — a.
a solution of Xs
2
in the preceding article enables
±a
us to find a solution of one of the equations x 2 — Xy 2 — only when a is one of the denominators which occurs in the process of into a continued fraction. converting For example, if Ave convert J7 into a continued fraction, we shall find that
JX
and that the denominators
The
and
if
of the complete quotients are 3, 2,
3, 1.
successive convergents are 2
3
5
8
37
45
1'
1'
2'
3'
IT
17'
we take the
x 2 - ty 2 = — I"
127
18
'
;
cycle of equations
O 3,
2
ar
— Iff = 2, *"
2
CI
o
ar
we
shall find that they are satisfied
for
x the values
2, 3, 5, 8,
and iovy the values *375.
82 31'
1,
1,
f o — 7^ = — 3, o
o
cc
- y2 = h'
1
/
1
by taking
37, 45, 82, 127, 2, 3,
14, 17, 31, 48,
number of cases in which solu2 2 the equations x — Xy = ± a can be obtained
It thus appears that the
tions in integers of with certainty is very limited.
In a numerical example however, sometimes happen that we can discover by
it
may,
trial
20—2
a
HIGHER ALGEBRA.
308
positive integral solution of the equations x — Ny = =*= a, when a is not one of the above mentioned denominators ; thus we easily 2
2
# 2 -7?/ 2 = 53
by y=2, # = 9. one solution in integers has been found, any number of
find that the equation
When
solutions
may be
satisfied
is
obtained as explained in the next
article.
Suppose that x =f, y = g is a solution of the equation x _ Ny = a and let x = h, y - k be any solution of the equation x2 - JSfy 2 = 1 then *376. 2
2
;
;
~ Ny 2 = (f 2 - Kg ) (h 2 - Nk ) = (fh±Ngk) 2 -N{fk±gh)\
By
2
2
x*
x
putting
-fh ± Kgk, y -fk ± gh,
to h, k their values found as explained in Art. 371, obtain any number of solutions.
and ascribing
we may
2
form x - n y = 2
2
N
is not a perfect has been supposed that is a perfect square the equation takes the which may be readily solved as follows.
*377. Hitherto square ; if, however, a,
it
N
Suppose that a = be, where b and then of which b is the greater
c
two
are
positive integers,
;
(x
Put x + ny =
+ ny) (x — ny) =
x - ny =
be.
the values of x and y found from these equations are integers we have obtained one solution of the equation ; the remaining solutions may be obtained by ascribing to b and c all their possible values. Example.
b,
c
;
if
Find two positive integers the difference
of
whose squares
is
equal to 60.
Let
x,
Now
y be the two integers
60
is
;
then
,xr
- y 2 = 60
;
that
is,
the product of any of the pair of factors
1,60;
2,30; 3,20;
4,15; 5,12; 6,10;
and the values required are obtained from the equations ic
+ y = 30,
x-y=
# + y = 10,
x-y=
2;
the other equations giving fractional values of x and y.
Thus the numbers
are 16, 14; or
8, 2.
6;
(as
+ y)
(x
-
y)
= 60.
;.
INDETERMINATE EQUATIONS OF THE SECOND DEGREE.
may
In like manner we
Cor.
309
obtain the solution in positive
integers of ax' if
+ 2hxy + by 1 + %jx +
member can be
the left-hand
*2/'y
+
= k,
c
two
resolved into
rational linear
factors.
*378. If in the general equation a, or b, or both, are zero, instead of employing the method explained in Art. 3G7 it is simpler to proceed as in the following example. Example.
Solve in positive integers 2.ry
Expressing y in terms of
- 4a- 2 + V2x - 5y = 11
x,
we have
+ V= —^r-r— 4a; 2
- 12* 11 „ 6 , =2tf-l+; 2x - 5 2x -
5
n
may be an
In order that y
integer
=
2iX
—O
must be an integer
± 2, or ± 3, or ± G. The cases ±2, ±6 may clearly be rejected; hence of x are obtained from 2x - 5 = ± 1, 2x - 5 = ± 3
must be equal
to
± 1,
whence the values
of
;
hence
2.r
-5
or
.x-
Taking these values
are
the admissible values
3, 2, 4, 1.
in succession
we obtain the
x = S, y = ll; s=2, y = -3; # = 4,
?/
= 9;
solutions
ar=l,
y=
-1;
and therefore the admissible solutions are a; = 3, y = 11; x = 4, y = 9.
The
principles already explained enable us to discover for what values of the variables given linear or quadratic Problems of this functions of x and y become perfect squares. kind are sometimes called DiopJiantine Problems because they
*379.
were first investigated by the Greek mathematician Diophantus about the middle of the fourth century. Example 1. Find the general expressions for two positive integers which are such that if their product is taken from the sum of their squares the difference is a perfect square. Denote the integers by x and y x.-.
This equation
is satisfied
;
-xy + y'2 = z'2 suppose x(x-y) = z
m
and n are
2
;
-y-.
by the suppositions
mx= n {z + y), where
then
positive integers.
n (x -y)
= m (z
-
y),
,
HIGHER ALGEBRA.
310
From
nx + (m -n)y- mz - 0.
mx - ny - nz = 0,
Hence
these equations
we
obtain by cross multiplication
_
x
m
2mn - ri
2
and since the given equation
2
V - n2
2
.
m
- mn + ri1
2
'
homogeneous we may take
is
for the general
solution
x = 2mn-n 2
y=m
,
2
-n 2
z
,
= m 2 -vin + n 2
.
Here m and n are any two positive integers, »i being the greater; thus ?n = 7, n = 4, we have x = ±0, y = SS, 3 = 37.
if
Find the general expression for three positive integers in arithmetic progression, and such that the sum of every two is a perfect
Example
2.
square.
Denote the integers by x-y,
x + y; and
x,
2x-y=p 2
let
2x = q 2 2x + y = r 2
,
,
then
2 2 2 p + r = 2q
or
i*-q*=q*-p>. This equation
is satisfied
m
From
and n are
,
by the suppositions,
m (r - q) = n (q - j>), where
;
n
(r
+ q) = m (q +p),
positive integers.
these equations
we obtain by
V w2 + 2mn-m Hence we may take
2
=
cross multiplication
_
m +n 2
2
r
m + 2mn- n 2
'
2
for the general solution
p=n* + 2mn-m*,
q
= m2 + n 2
r
,
= )u 2 + 2mn-u 2
;
x = = {m 2 + w 2 ) 2 y = inin (m 2 - w2)
whence
,
and the three integers can be found.
From odd
;
the value of also their values
x it is clear that m and n are either both even or both must be such that x is greater than y, that is, (m 2 + n 2 ) 2 >8mn{m 2
mz (m - Sn) + 2inn
or
which condition
m
is satisfied if
2
-n 2 ),
+ 8m n* + n 4 >
;
m>Sn.
= 9, w=l, then a- = 3362, y =2880, and the numbers are 482, 33G2, If 6242. The sums of these taken in pairs are 3844, 6724, 9604, which are the squares of 62, 82, 98 respectively.
:
INDETERMINATE EQUATIONS OF THE SECOND DEGREE. 311
*EXAMPLES.
XXVIII.
Solve in positive integers 1. 3.
5.
= 77. y 2 -4.ry + 5.r2 -10.i- = 4. 3.y + 3.ry-4j/ = 14. 5a- 2 -10.iv/
+
7?/
2
2.
7^-2^+3y2 =27.
4.
xy - 2.v - y = S.
6.
4^ 2 -y 2 =315.
Find the smallest solution in positive integers of 7.
10.
-14y 2 =l. x2 - 61/ + 5 = 0. .r
2
8.
11.
Find the general solution 12.
.r
2
-3/=l.
^-19^=1.
13.
9.
2
= 4iy
2
- 17y 2 = -
.t-
2
-l.
x 2 -7y 2 -9 = 0.
in positive integers of
x 2 -5y 2 =l.
14.
.v
1.
Find the general values of x and y which make each of the following expressions a perfect square 15.
x2 -3xy + 3y 2
.
:
16.
afi+2xy + 2f.
17.
5^+y2
.
Find two positive integers such that the square of one exceeds the square of the other by 105. 18.
19.
Find a general formula
for three integers
which may be taken
to represent the lengths of the sides of a right-angled triangle.
Find a general formula to express two positive integers which are such that the result obtained by adding their product to the sum 20.
of their squares
is
a perfect square.
my
" There came three Dutchmen of acquaintance to see me, 21. being lately married they brought their wives with them. The men's ;
Claas, and Cornelius; the women's Geertruij, Catriin, and Anna but I forgot the name of each man's wife. They told me they had been at market to buy hogs each person bought as many hogs as they gave shillings for one hog; Hendriek bought 23 hogs more than Catriin; and Claas bought 11 more than Geertruij likewise, each man laid out 3 guineas more than his wife. I desire to know the name of each man's wife." (Miscellany of Mathematical Problems, 1743.)
names were Hendriek, :
;
;
22.
Shew
that the sum of the first n natural numbers is a perfect equal to k 2 or k' 2 - 1, where k is the numerator of an odd,
square, if n is and k' the numerator of an even convergent to N ^2.
CHAPTER XXIX. SUMMATION OF Examples
380.
of
summation
SERIES.
of certain series
have occurred
previous chapters ; it will be convenient here to give a synopsis of the methods of summation which have already been explained.
in
(ii)
Arithmetical Progression, Chap. IV. Geometrical Progression, Chap. Y.
(iii)
Series
(i)
which are partly arithmetical and partly geo-
metrical, Art. 60. (iv)
Sums
Series, Arts. (v)
of the
powers of the Natural Numbers and
allied
68 to 75.
by
means
Series,
Chap.
Summation
of
Undetermined
Coefficients,
Art. 312. (vi)
We
Recurring
XXIY.
proceed to discuss methods of greater generality ; but in the course of the present chapter it will be seen that some of the foregoing methods may still be usefully employed.
now
th If the r term of a series can be expressed as the dif381. ference of two quantities one of which is the same function of r that the other is of r - 1 the sum of the series may be readily ,
found.
For
and
its
let
the series be denoted by
sum by S
the form v r -v r _
1 ;
,
and suppose that any term u r can be put in
then
^.=(«i-0+(w.-«i)+(*.-«f) + »- + (w.-i-0 + (*.- ,,--i) !
=v -v
n
.
SUMMATION OF Example.
Sum
313
SERIES.
terms the series
to n
1
1
1
+ (l T + „ + U »„v„ (l + s)(l+2s) + 2u-)(l + n3.r) (l + 3j-)(l + 4ar) .
If
we denote the
series
.
.
,
.
by
*-±(—-—\ + + 3*/' 2x
a;\l
-
_!/
Ws
1
1
a;\l
1
x\l + nx b}'
SL=-
addition,
I
\
1
+ 3#
i+ n+ i.x/
—
^
ar\l
'
+ 4*,/
+ a;
I
+w + l.a?/
l
n (1
+ x)
+n+ l
(1
..r)
Sometimes a suitable transformation may be obtained by separating u into partial fractions by the methods explained 382.
XXIII.
in Chap.
Example.
Find the sum
+ x)(l + ax)
nu, The
ft*term= n th
+
n,
(l
+ .'.
putting 1
+
+
x~n z~r + ax)(l+a*x)
t-,
(1
an
«" _1 .r, 1
~x
+ a u x equal
.
-a'
B= —1 -a'
1/1
1
a
- a \1 + a: 1
a
/ ,
1
/
1
+ ax J
a — -5—
2
*~l-o\l + *
.
aw
a"- 1
~l-a Vl+a*"
\ 5
1
*
l
l
+
an
+ a n x)'
.
to n terms.
suppose:
**
1
nn
,
t**=;
'"•
.
to zero in succession,
=—
....
Wn
+
B
7
1
similarly,
v
1
a n-l
u,1
5—
+ a 3 x)
A
A=Hence
(1
= —.t- + — w + a— n n x) l + a"- ^ l+a x -A (1 + a**) + B (1 + a '- *).
n, a n 1 x)(l .
,,
>
.,
+ a-x)
« n_1
— {l
By
a2
a
1 (l
of
\
j/
we obtain
;
HIGHER ALGEBRA.
314 To jind
sum
of\\ terms of a series each term of which is composed of r factors in arithmetical progression, the first factors of the several terms being in the same arithmetical progression.
383.
Let the where u„
-
(a
be denoted by u x + u2 + us +
series
+ nb)
the
(a
+n+
n—
Replacing n by k «„ »-i !
=
(a
.'.
+n—
(«
+
?i
1
—
.
1
1
b)
Replacing n by n +
b) (a
+ nb) (a + n +
un = 1
b)
.
(a
...
(a
+n+
/•
1
—
.
1
6) .
.
b)
..
+ r.
6)tf» =
«Il+1 j
by subtraction,
(r+l)b un = vn+ i-vn .
Similarly,
+n+r—
(7+1)6. wB_, =
r
/(
-
(a
?.«.„_!
we have
(a + w therefore,
+n+2
,
1
.
b).
we have
1,
&) (a .
.
+ un
.
/<„_,,
+n+r— 2
=
vn
,
say.
.
b)
— SUMMATION OF
Find the sum of n terms of the
Example.
315
SERIES. series
1.3.5+3.5.7+5.7.9+ The n th term
{2n - 1) (2«
is
+
+ 3)
1) (2n
hence by the rule
;
^
- _ (2w-l)(2n + l)(2n + 3)(2n+5) n >\+C7.
To determine
C, put n
we have
=1
then the series reduces to
;
= ——
15
—
:
-
-j
8
S *~
•'•
,
f-
C
whence C
;
The sum
8
(2n-l)(2n+l)(2n+3)(2n+5) 8
= n (2n3 + 8«2 + In - 2), 384.
=—
its first
term, and
;
+
15 8"
after reduction.
of the series in the preceding article
may
found either by the method of Undetermined Coefficients [Art. 312] or in the following manner.
also be
We
have u n = (2w .
\
1)
(2w +
\S
3
3)
= $n 3 +
2
12>i
2
- 2m -
3;
3m,
;
2
= 2m 2
U
(2w +
Sm = 82?i + 122^ - 22m -
using the notation of Art. 70 .
1)
(m + l) + 2m (n
+
1)
(2n + 1) - n (n +
1)
- 3m
= w(2m 3 + 8m + 7?i-2). 2
It should be noticed that the rule given in Art. 383 is 385. only applicable to cases in which the factors of each term form an arithmetical progression, and the first factors of the several terms are in the same arithmetical progression.
Thus the sum 1.3.
of the series
+ 2.4.
5
6
+ 3.5.
7
+
n terms,
to
may
be found by either of the methods suggested in the preceding article, but not directly by the rule of Art. 383. Here
+
4)
-n
= n{n+ l)(™ +
2)
+ 2n(n+
\)
+ u(a +
= n (m +
2)
+ 3m (n +
1)
+ 3m.
un = n
The
(m
rule can
Sn = \n (m+ \
)>
(r
h
+
2) (m
1)
(m
now be
1)
+
(m
+1 +
1)
(m
applied to each term
(m + 2)(m + 3)
l)(/r»-l) (» +
+n (n+
5),
1) (m
+2+
;
+
2)
'2)
+
2/4
thus 2)
+ |« (»+
the constant being zero.
1)
+C
;
HIGHER ALGEBRA.
316
sum
of\\ terms of a series each term of which is composed of the reciprocal of the product of r factors in arithmetical progression, the first factors of the several terms being in tlie same arithmetical progression.
To find
386.
Let the where
series
the
be denoted by u +
— = (a + nb) (a + n + 1
b) (a
.
Un
+ un
+u +
it,
x
:i
+n+2
(a
+ n + r-l
b) ...(a
+ n + r—2
b)
.
,
.b).
4i
Replacing n by n -
—=
(a
+n-
1
.
1,
+ nb)
b) (a
+n+
(a
1
.
.
b)
M„_i n-l l
.'.
(a
+n+
r-l
.
Replacing n by n +
un =
b) 1,
(a
+n-
1
by subtraction,
Similarly
(r
-
1) b
(r
— l)b
(r—
, ,
,
addition,
.
(r
un _ x =
.
.
1) b
.
v _l ll
—
u.2
=
Wj
=v —
v.2
x
.
is
vn
,
vSt v.2
.
-
x
?W _ rU
_
(
*"-(r-l)6~
C
-
,
— 1) b Sn = v — v n+1
^i
Q
tnatis
where
say.
;
(r-l)b. un = vn -vn+1
By
un _ x = vni
we have
(a+nb)un = vn+1 therefore,
b)
.
;
a + nb) un
(r-l)6
a quantity independent of
'
which may be found by
n,
ascribing to n some particular value.
Thus
Sn = C- ,
=
* .
(r-l)6
(a
.
Hence the sum may be found by the following th Write doivn the n term, strike off a factor divide by the number offactors so diminished difference, change the sign and add a constant.
The value
of
C= (r — Vv~7o -,
1)
each case to determine
C by
r-l.
+ n+L.b)... (a + n +
=
t
(r
—
tti
u
i
'>
rule
,
b)
:
from the beginning, and by the common
but
**
is
advisable in
1) 6
ascribing to
n some
particular value.
:
.
SUMMATION OF Example
Find the sum of
1.
1.2.3.4 The
term
1
re'
'
hence, hy the rule,
terms of the series
re
+ 2.3.4.5 + 3.4.5.6 + 1
-
is
317
SERIES.
»(n + l)(n + 2)(n + 3)'
we have
3(n+l)(w + 2)(» + 3) rut »=1, then
^=0-3^; X
5
•
By making n Example
"
*
18
3(re
Find the sum
+ l)(re + 2)(re + 3)
we obtain
indefinitely great,
2.
whence (7=1;
=—
fi^
to n terms of the series 5
4
3
—
r—5 + + a—T7—- + 3.4.6 r 1.2.42.3.5 t
Here the rule is not directly applicable, because although 1,2,3, , the first factors of the several denominators, are in arithmetical progression, In this example we may the factors of any one denominator are not. proceed as follows (n+2) 2
n+2 "
re(re+l)(re re (re
+ 3)
4-1)
n{n+l)
(n
+ 2)
(re
+ 3)
+ 3re + 4
:
re(re
+ l)(re + 2)(re + 3)" 3
1 (re
Each to
which
+ 2)(w + 3)
(re
+
l)(re
of these expressions may the rule is applicable.
•
-c
S
+ 2)(re + 3)
now
w(re+l)(n + 2)(re + 3)'
be taken as the
n+S
2(re
+
th
term of a
4
3
l
?i
2)(re
+ 3)
3 (re+ 1)
(re
+ 2)
(re
+ 3)
'
put re=l, then
3-13
17271=
C
_29 n
36
4
29
"4 " 27174 " 372.3 .4' Whence C = 36 3
1 re
+3
2
(re
+ 2) (re + 3)
;
4 3
(re
+ 1)
(re
+ 2)
(re
+ 3)'
series
;
HIGHER ALGEBRA.
318
386 are directly applicable, instead of quoting the rules we may always effect the summation in the following way, which is sometimes called the 387.
In cases where the methods
of Arts. 383,
'
Method
of Subtraction.'
Example.
Find the sum of n terms
of the series
2.5 + 5.8 + 8.11 + 11.14+ The
arithmetical progression in this case
is
5,8, 11, 14,
2,
In each term of the given series introduce as a new factor the next term of the arithmetical progression denote this series by &", and the given series by S; then ;
S' .-.
= 2.
5. 8
+ 5. 8.
£'-2.5.8 = 5.8.11 +
By
+ 8.
11
8. 11.14
11.
14+
+(3w-l)(3n+2)(3»+5);
+ 11. 14.17+...
to
(u-1) terms.
subtraction,
_2.5.8=9[5.8 + 8.11 + 11.14+...to(»-l)terms]-(3n-l)(3n+2)(3n+5), -2
.
5
.
8 = 9 [S - 2
9S =
(3/i
-
.
5]
1) (3/i
-
-
(3/i
+ 2)
(3n+2) (3n+5),
1)
(3/i
+ 5) -2. 5.
8
+ 2, 5.0,
fif=n(3n3 +6n+l).
When the nth term of a series is a rational integral 388. function of n it can be expressed in a form which will enable us readily to apply the method given in Art. 383. (n) is a rational integral function For suppose dimensions, and assume
cf)(n)
where A, number.
=A JB,
+Bn+ C,
Cti(n +
D,
of
n
+B)i(u+ l)(n + 2)+
1)
are undetermined constants
of
p
,
p+
l
in
values of n, we may equate the coefficients of like powers of n; we thus obtain ^> + 1 simple equations to determine the p + 1 constants.
This identity being true for
Example.
all
Find the sum of n terms
of the series
n*+6n3 + 5w2
whose general term
is
.
Assume 7i
4
+ 6/i3 + 5/t 2 = A + Bn+ Gn
[n + 1)
+ Dn [n +
at once obvious that ,4=0, 2? = 0, 0. successively, we obtain C = - 6, J)
it is
=
« 4 + 6» 3 + 5/< 2 =//(n +
l)
1) (n
E=1
;
+ 2) + En (n + 1) (n + 2) (w + 3)
and by putting n = -
Thus
(n+2)
(?i
+ 3)-6/t(/i + l).
2,
n=- 3
)
SUMMATION OF Sn = s n
Hence
(/t
o
+
l)(»
310
SERIES.
+ 2)(n + 3)(?i + 4) - 2n(n + l)(n + 2)
= \n(n+l)(n+2){n' + 7n + 2). i
o
Polygonal and Figurate Numbers. •
n+ \n(n— l)b, which is the sum arithmetical progression whose first term is 1 of n terms of an we give to b the values 0, 1, 2, 3, and common difference b, •> we get u, \n (n + 1), n* \n (Bn — 1)n If in the expression
389.
j
s
which are the u ih terms
of the Polygonal
Numbers
of the second,
orders; the first order being that in which unity. The polygonal numbers of the second, third, orders are sometimes called linear, triangular
third, fourth, fifth,
each term
is
fourth, fifth,
square, pentagonal
)
390. To find the j>olygonal numbers.
The n ih term .-.
sum of the
first
of the rtb order is
n terms of
n + \n (n -
th the r order
1) (r
$ =$n + l(r-2)%(n-l)u = \n (n + 1) + 1 (r -2)(n-l) n (n +
—
of
2);
i
= in(n + l){(r-2)(n-l) + 391.
If the
sum
of
n terms 1,
be taken as the
?*
th
1,
1,
1) [Art.
383]
n terms
of the
$}.
of the series
1,1,
term of a new
,
series,
we
obtain
1,2,3,4,5, If again
we take
n in + —
1 ,
which
is
the
sum
of
-j
last series, as the
?t
th
term of a new 1, 3, 6,
By
series,
we
obtain
10, 15,
proceeding in this way, we obtain a succession of series such that in any one, the nih term is the sum of n terms of the preceding series. The successive series thus formed are known as Figurate Numbers of the first, second, third, ... orders.
HIGHER ALGEBRA.
320
To find the n th term 392. order offigurate numbers.
and
the
sum of n
terms of
ilie
rth
th term of the of the first order is 1; the n th n; the n term of the third order is Hn, that is
The nih term second order
\n
(n
+
is
n
1); the
tYl
term
2 "V
of the fourth order is
1
n(n+l)(n+2) L± 1.2.3
-—
..
.
xl
that
is
the
•
'-
u th term tIl
,
—
^-n4
2
it is
.
~
.
M
3) ;
and
A«rain, the
which
is
the
—^5
2,
sum of n terms n (n + 1) (n +
w th term
of tlie (r
.
2
.
o
,
of the r th order is
|rc+r-2
w(?*+l )(w + 2)...(n + r-2) .
r-1
is
so on.
n th term
easy to see that the
tnat
^ n(n+l) —*-= (n+2)
.
,
of the fifth order is
1
w(w+l)(n + 2)(M — + —
Tims
,
>
2
.
01
i
n—
1
I
r
—
1
of the r th order is 2)
+
.
. .
(w
+
r
-
1)
l) th order.
In applying the rule of Art. 383 to find the sum of n terms of figurate numbers, it will be found that the constant is always of order any
Note.
zero.
The properties of figurate numbers are historically 393. interesting on account of the use made of them by Pascal in his Traite du triangle arithmetique, published in 1665. The following
table exhibits the Arithmetical Triangle in its
simplest form 1
1
1
...
;
SUMMATION OF
SERIES.
321
Pascal constructed the numbers in the triangle by the following rule
:
Each number immediately thus
is the
sum of
that immediately above
and
it
that
to the left
of it; 5 + 10, 28 = 7 + 21, 126 = 56 + 70.
=
15
From
the mode of construction, it follows that the numbers in the successive horizontal rows, or vertical columns, are the hgurate numbers of the first, second, third, orders. .
A
.
.
drawn so as to cut off an equal number of units from the top row and the left-hand column is called a base, and the bases are numbered beginning from the top left-hand corner. Thus the 6th base is a line drawn through the numbers 1, 5, 10, and it will be observed that there are six of these num10, 5, 1 bers, and that they are the coefficients of the terms in the expansion of (1 + x) 5 line
;
.
The properties
of these
numbers were discussed by Pascal
with great skill in particular he used his Arithmetical Trianyle to develop the theory of Combinations, and to establish some :
interesting propositions in Probability. The subject treated in Todhunter's History of Probability, Chapter n.
is
fully
"Where no ambiguity exists as to the number of terms in a series, we have used the symbol % to indicate summation but in some cases the following modified notation, which indicates the limits between which the summation is to be effected, will be found more convenient. 304.
Let
cf>
(x)
be any function of
x,
then 2
(x)
denotes the
sum
x=l
terms obtained from (x) by giving to x inclusive. values from I to tive integral 'a
of the series of
all posi-
m
1
For instance, suppose
required to find the terms of the series obtained from the expression it is
sum
of all the
(p-l)(p-2)...(p-r)
by giving to p H.H.
A.
all
integral values from r
+
1
to
j>
inclusive.
21
7
1
HIGHER ALGEBRA.
322
Writing the factors of the numerator in ascending order, *=* - r)—(p - r + — sum = 2 (p
.
the required
1)
...
(p —
-
1)
= i{1.2.3.. ..r+2.SA....(r+l)+...+(p-r)(p-r+l)...(p-l)}
(p-r)(p-r + l) „(p-l)p
=l =
+
r
\r
[Art. 383.]J L
1
y~(-l)(y-2)...(^-r) jr+1 i
Since the given expression is zero for all values of r inclusive, we may write the result in the form p
%
•(p- r) _ p(p-l)
(p-l) ( p-2)
r |
EXAMPLES. XXIX. the following series to n terms
1.2.3 + 2.3.4 + 3.4.5 +
2.
1.2.3.4 + 2.3.4.5 + 3.4.5.6 +
3.
1.4.7 + 4.7.10 + 7.10.13 +
4.
1.4.7 + 2.5.8 + 3.6.9 +
5.
1.5.9 + 2.6.10 + 3.7.11 the following series to I
1
7 '
II +
174
4.
1
n terms and
3.4^ 1
+ 77l0 + 1
1_ + + + 1.3.5 3.5.7 5.7.9 1
1.4.7
1_ 1 + 4.7.10 + 7.10.13 +
4
10
1.2.3^2.3.4
11 *
io \9,
6
5 1
1-
4-
-
.3.4.5^
J_ + _1_ + _JL + 4.5.6 5.6.7
3.4.5
— 1.2.3
1 ——
3
5 -l
2.3.4
—
3.4.5
a.
+
1
1.2^2.3
+
:
1.
Sum
ff-2) ...(p-r)
\r
v-\
Sum
(
*""' 7
i
4.5.6
u
p from
to infinity
:
1
to
SUMMATION OF Find the sum of n terms of the 1
14.
(?i
)
sum
»*(»*?i
2
2
n terms of the
-l)
(?i'
n* + .,
Shew
equal to the r
+ DR + 4)(n 2 + 5n + H). + 2/> 3 + h -1 2
*&•
_ 2)i
•
v,
u-
+
/i
7i*+n 2 + l zu.
.
,
iv th
term of the term of the n order.
that the th
4
?* •
is
(n*
16.
1).
whose na term
series
n*+3n?+2n+2
21.
22. the (n
of
A 9 2 -li 4w
1Q iy.
:
-l 2 + 2<> 2 -2 2 ) + 3(> 2 -3 2 ) +
2
Find the
17.
series
,3.22 +2.4.3a +3.5.4*+
13.
15.
323
SERIES.
r
?i
+n
th
order of figurate numbers
n th term of the r th order of figurate numbers term of the (>-2) th order, shew that r=n+%
If the
+ 2)
23.
th
is
tXx
is
equal to
Shew that the sum
numbers from the
of the first n of all the sets of polygonal linear to that of the ? ,th order inclusive is
{r-\)n(n + \),
„
oN
Summation by the Method of Differences. Let un denote some rational integral function of », and w 3 tt4 ,... denote the values of u n when for n the values
395. let Mj, u.2 1
,
,
2, 3, 4,
,
.
.
.
are written successively.
We
proceed to investigate a method of finding un certain number of the terms u u.2 w3 u4 ,... are given. ,
x ,
From by
when
a
,
the series u x u 2 u 3 u A u 5 ,... obtain a second series subtracting each term from the term which immediately
follows
The
,
,
,
,
it.
series u.2
— w,, u s -u.
2
,
u4 — u3
,
u 5 -u 4 ,...
thus found is called the series of the first order of differences, and may be conveniently denoted by
Aw,,
&u~
&u A
,
At*4 ,...
By
subtracting each term of this series from the term that immediately follows it, wr e have Attg — Awg,
Am.,- A?*,,
which may be called the and denoted by
series
A../',,
of
A.,?'.,,
Aw4 — Awj,...
the second order
AjWg,...
of
differences,
)
.
HIGHER ALGEBRA.
324
From
we may proceed
this series
third, fourth, fifth,... orders these series being Aur 3ur ,
A
From
A
of
A
,
5
to form the series
of the the general terms of
differences,
respectively.
?t r) ...
the law of formation of the series Uj
u3
u.2 ,
t
Au.2
Attj,
A.y^ .
A
A2w
,
3
Au3
,
u5
u±,
,
Aw4
,
A.m3
2,
Au
,
Au
,
A u,,
Wj,
2
A u3
3
u6
,
3
i
5
,
,
,
,
appears that any term in any series is equal to the term immediately preceding it added to the term below it on the left. it
Thus
By
u.2
= «j
addition, since u.2
Au ly and
4-
Ait.2
= Au + i
A.m^
.
+ Au.2 = u 3 we have ii.j
= t^ + 2Au +
A.2 u x
±
.
In an exactly similar manner by using the second, third, and fourth series in place of the first, second, and third, we obtain
Au3 = Au + x
By
+
A^.
u3 + Au3 = u4i we have
addition, since ?f 4
2A.2u 1
- u + ZAu + x
x
SA^ + A^
.
So far as we have proceeded, the numerical coefficients follow the same law as those of the Binomial theorem. We shall now prove by induction that this will always be the case. For suppose that
un+i = «i + mAmj +
v
A,u +
9
X
+ "CVA^j +
...
l
+
AnWj
i
-J
then by using the second to the (n + 2) th series in the place of the th first to the (n + l) series, we have
——
1
it (11
Au n+1 = A%! + nA. u + -A. 2
By M»+a
addition, since
=
Mj
+ fa +
jr-f-
}
1)
A 3Wj +
+ B C r_ 1 A^w1 + f
.
.
.
.
.
.
+
A n ^u
u n+l + Au n+1 = un+2i we obtain
Awj +
.
.
.
+
n (
Cr + *Gr _j) A ru
x
+ ...+ A„ +1«,
Y
.
)
)
SUMMATION OF
^
(—
*Cr + HJr -i -
But
hut
+ l)
w4
it is
=
"„
Ui
.
1X 1)
, + (n
To
39G.
A?^ +
sum
find the
»C r _ x = ?i±i
x
therefore
,
(w-l)(w-2) £*_ *
—
of
in terms of the differences of
Suppose the
32.",
law of formation holds for u n+l
if tlie
true in the case of therefore universally. Hence f£ n+8 ,
SERIES.
x "C,,.,
l)w(w-l) ...(w+l-r+1) _ 1.2. 3... (r-l)r
(n +
Hence
;
.
series u^, u.2
,
w terms
ul
also holds for
it
holds for
it
u rn and
.
A
2
+
?^
+ An.iWi.
...
of the series
.
u 3 ,...
the
is
order of differences
first
of the series
then v n+1 = (vn+1 •
Hence
'•
- vn) + ^»+l
v3
v.2)
Vl,
- vn _ t) +
(v n
= u a + u n-l + v3i
va1
••
3
Aw1? Aw 2 the law of formation •'
Wj
=
+ w3 + u z +
- nu + x
«»+i
is
v4
2
1
is,
...
+
(v2
-v) +
v x identically
x
+ u 2 + u\ + v
l
•
in the series 0,
that
v4 ,...,
,
,
4
J
Aw
J
3
the same as in the preceding article;
+ «Wi + -4 ...
v5
,
—s— Aw
x
+
.
.
.
+ A„?^
+ un
n (n—\) n(n—l)(n-2) — -^ A M! + —r — AWj + -— y
The formula)
t
>
2
of
;
this
and the preceding ,
,
.
article
.
+
A„?f
,
may be
ex-
a is the first the first terms of the sucterm of the given series is
pressed in a slightly different form, as follows
term of a given series, (I x d2 d3 ,... cessive orders of differences, the n th obtained from the formula
.
:
if
'
HIGHER ALGEBRA,
326 and the sum of n
^i"^"2
1
terras is
»(»-l )("- 2
)^
f/
4
j3
Find the general term and the sum
Example.
12,
The
n(»-l)(tt-a)(n-3)
)
,
of
?*
terms of the series
40,90, 168, 280, 432,
successive orders of difference are 28, 50, 78, 112, 152,
22, 28, 34, 40, 6,
6,
6, 0,...
0,
Hence the n th term
™, ,x = ,c 12 + 28 (re - 1) +
—
22(re-l)(re-2) K -
'
~P
+
-
l±
= 2re
?i
M
M
6 (re1) (re- 2) v
(re
-3) -
II
+ 5re2 + 6>t.
3
The sum of n terms may now be found by writing down the value of + 52re2 + 62re. Or we may use the formula of the present article and
3
obtain
S^ia^ 28"'"- 1
+
'
22 "'"- 1)( "- 2 »
+
«M»-D (»-2) (-8)
= ^(3re2 + 26re + 69re + 46),
= in(re+l)(3n 2 + 23re + 46). It will be seen that this
397.
method
of
summation will only
when the series is such that in forming the orders of differences we eventually come to a series in which all the terms This will always be the case if the n th term of the are equal. succeed
function of n. series is a rational integral ^» j
For simplicity we will consider a function of three dimensions; the method of proof, however, is perfectly general. Let the
series
+ ua2 + ua3 +
u.1
+ u +u n +.,+u n+2„ + u u + „3 + .
,
\
ii
u " = An 3 + Bn 2 + Cn + D.
where
and
be
let v n' ,
w
, ii*
denote the
% ii
orders of differences;
?i
th
term of the
first,*
second, third *
SUMMATION OF then that
vh
SERIES.
m m+1 — un = A(3n* + 3n+
-
l)
327
+ 2?(2» +
1)
3Au 2 + (3A + 2B) n + A + 11 + C = v — v = 3A (2n + I) + 3A + 211
vn =
is,
w
Similarly
+C:
;
,
.
and
=w »ti.— iv
zH
=6-4. ii
Thus the terms and
generally,
in the third order of differences are equal; the n ih term of the given series is of p dimensions,
if
the terms in the
p
th
order of differences will be equal.
Conversely, if the terms in the ]j th order of differences are equal, the u tu term of the series is a rational integral function of ii of p dimensions. Find the « th term of the
Example.
The
series
-1, -
successive orders of differences are
-2,
6, 20, 40, GO,
20,20,
8, 14,
6,
6,
Thus the terms assume where A, B, G,
0,
in the third order of differeLces are equal it
D have
= A+Bn+Cn + Dn 2
H
1,
2,
If a
series is 3
S
Let
S (1
-
- 3n - 2n 2
12
ax + ajx
+
series, ivhose scale
2
+
...
--
+ n9
;
.
a o + (a - a )x
+
{a,
+ a xn »i
of relation
denote the sum of the series
x)
four simultaneous
D—\
a rational integral function of p dimensions
is ri
n, the series
a recurring
may
,
3,
a, is
hence we
to be determined.
hence the general term of the
398.
;
3
4 for 7i in succession, we have equations, from which we obtain A =3, B = -3, C = - 2,
Putting
in
23, G3, 129,
3, 3,
;
- ajx* +
is (I
x) p+1
—
.
then .
..
+
(a,
- a ,_>" - ax" +
l
x
=a +
x + bjc2 +
b
...
+ bx" - ax" + \ say;
t
here b n
=a —a h
it
—
,
1
, '
so that 6 n is of xp
-
1
-
x,
"•
Multiplying this last series by
1
dimensions in
n.
we have
S(i-xy
=s+(^-a„)*+(^-^K+..-+(6n-6„-iK-(« ,+6>"
+l
J
= c^+{b-a )x+c^ + c i x^...+cX-{a +b 2
i
here cn -b n
-b u u
so that cn
is
of
p-
y^+ ,
i
+«X
,+
a
:c i
2 dimensions in n.
+a
% say;
HIGHER ALGEBRA.
328
follows that after the successive multiplications by n products 1 — x, the coefficients of x in the first, second, third, orders of differences are general terms in the first, second, third,
Hence
it
.
.
.
.
.
.
of the coefficients.
hypothesis a n is a rational integral function of n of p dimensions ; therefore after p multiplications by 1 - x we shall arrive at a series the terms of which, with the exception of p terms at the beginning, and p terms at the end of the series, form a geometrical progression, each of whose coefficients is the same.
By
[Art. 397.]
S (1 - xf = k(xp + x>' +1 +
Thus
...+ x") +/(a?),
f
(x) stands for the a constant, and the beginning and p terms at the end of the product.
where k
is
r.Sil-xy that
a=
is,
thus the series
(l-x) p+1
.
is
p terms
at
J^l^K/ix);
kx»(l-x"-^) + (l-x)f(x) (1
-x)
p+l
^
'
a recurring series whose scale of relation
is
[Art. 325.]
not given, the dimensions of an are readily found by the method explained in Art. 397. If the general
Example.
term
is
Find the generating function of the 3
series
+ 5a; + 9a; 2 +15a; 3 + 23a; 4 + 33a; 5 +
Forming the successive orders
of differences of the coefficients,
we have
the series 2,
4,
2,
G,
2,
8,
10,
2,
2,
;
thus the terms in the second order of differences are equal hence a n is a rational integral function of n of two dimensions ; and therefore the scale We have of relation is (1 - a;) 3 ;
.
S = 3 + 5x +
+ 15.r 3 + 23a; 4 + 33a; 5 + - SxS = - 9.r - 15a; 2 - 27.x- 3 - 45a; 4 - 69^ 5 4 2 3 9a; + 15a; + 27.r + 45a; + Sx 2 S = - 3^- 5a;4 - 9a; 5 -xs S= By
addition,
(
9a;
1
2
- a;) 3 S = 3 -
•*•
b
4a;
3-4.r +
~
+ 3a; 2
3a;
(1-a;) 3
2 *
;
SUMMATION OF
329
SERIES.
We
have seen in Chap, xxiv. that the generating function of a recurring series is a rational fraction whose denomiSuppose that this denominator can nator is the scale of relation. then the be resolved into the factors (1 — ax) (1 — bx) (1 — ex) generating function can be separated into partial fractions of the 399.
-
to rm
,
1
ABC 1-
- ax
1
—
bx
1
;
- ex
can be expanded by the Binomial Theorem in the form of a geometrical series; hence in this case the recurring series can be expressed as the sum of a number of geometrical series.
Each
of these fractions
however the scale of relation contains any factor more than once, corresponding to this repeated factor there If
A—
A
—
1
- ax
will
be
which r=, - ax) when expanded by the Binomial Theorem do not form geometrical series; hence in this case the recurring series cannot be expressed partial
fractions
of
the form
-^
(1
as the
sum
of a
number
-7,
-axy
...
:
(1
of geometrical series.
The successive orders of differences of the geometrical 400. progression a,
ar,
ar
2 ,
ar
3 ,
ar\ ar
n ,
«(r-l), a(r—l)r, a(r-l)r 2 a(r—\)r?
are
'
,
a(r-l) 2 a(r-l) 2r, a(r-\fr 2 ,
,
which are themselves geometrical progressions having the same
common 401.
ratio r as the original series.
Let us consider the
series in
which
where (rc) is a rational integral function of n of p dimensions, and from this series let us form the successive orders of differences. Each term in any of these orders is the sum of two parts, one arising from terms of the form ar n ~\ and the other from terms of the form <£(?i) in the original series. Now since (n) is of ;; dimensions, the part arising from (n) will be zero in the (p + l) th and succeeding orders of differences, and therefore these series will be geometrical progressions whose common ratio is r. [Art. 400.]
HIGHER ALGEBRA.
330 Hence
if
the
few terms
first
of a series are given,
and
if
the
order of differences of these terms form a geometrical progression whose common ratio is r, then we may assume that the general term of the given series is ar"" +f(n), where f(n) is a rational integral function of n of p - 1 dimensions.
p
th
1
Example.
Find the n th term 23,
10,
The
of the series 60, 169, 494,
successive orders of differences are 13, 37, 109, 335,
24, 72,
Thus the second order the
common
ratio is 3
;
a geometrical progression in which assume for the general term
of differences
hence we
may
un —a
To determine
216,
the constants a,
.
S n -^ b, c,
is
+ bn + c. make n equal
to
1, 2,
3 successively;
a + b + c=10, 3a + 2b+c = 23, 9a + 3b + c = 60;
then
a = 6, 6=1,
whence un = 6
Thus
.
3' 1
"1
c
+n+3=2
= S. .
3» + n +
3.
In each of the examples on recurring series that we 402. have just given, on forming the successive orders of differences we have obtained a series the law of which is obvious on inspec-
and we have thus been enabled to the ?4 th term of the original series.
tion,
for
find a general expression
however, the recurring series is equal to the sum of a are number of geometrical progressions whose common ratios n n~ + Cc ~\ «, b, c, ..., its general term is of the form Aa"' + Bb and therefore the general term in the successive orders of differences is of the same form ; that is, all the orders of differIn this case to ences follow the same law as the original series. find the general term of the series we must have recourse to the more general method explained in Chap. xxiv. But when the coefficients are large the scale of relation is not found without considerable arithmetical labour ; hence it is generally worth while to write down a few of the orders of differences to see whether we shall arrive at a series the law of whose terms is If,
1
l
evident.
403.
We
add some examples in further
preceding principles.
illustration of the
SUMMATION OF Example
Find the sum
1.
SERIES.
of n terras of the series
1.2'3 + 2.3'33+ 3.4'35 + 4.5
„
2« + 3 "
we
1l(ll
—2n + 3
tt Hence
t/,.
"
/3 \n
=
\
1
n + 1)
+ _3_ + 3. 7
3
term
_5 3. 7. 11
.,„,., 3.7 11
is
11—
,
.
n +
-
.
n+1
(4n _
7
+
.
series
7
3. 7. 11. 15
5) (4 n
_i)
~3
A .
(n
.
3"'
1
3' 1
•,
2n-l
ssume 3
-
3"- 1
r, kt: (An5) (4/i-l)
.
.
— 1
.
Find the sum of n terms of the
2.
1
ri h
=1
,
1
-1.
n
#,, n
=
n+
B=
1 — = 13"
)
(
and therefore
The
7?
.
A =3,
find
Example
3"
+ l) .4
=
+
*3^
1
=-+ n(u+l) n
Assuming
33]
+
•
+ 1) + B
An + B
7 ......4»-l
"
3.7
(4„ - 5)
'
2rc-l = ,4n + (J+I> )-(.-t» + .B)(4?i-l). >
.-.
On equating coefficients we have three equations involving the two unknowns A and B, and our assumption will be correct if values of A and B can be found to satisfy
Equating
all three.
coefficients of n 2
,
we obtain ^1=0.
Equating the absolute terms, -1 = 2B; that is B = -%; and found that these values of A and B satisfy the third equation. 1
1 V,l
""'
~2 *3.7
hence
S„
"
Example
3.
Sum G. 9
By
the
method
of the series
and the
;«
th
=
to
2
21
+ 20.
37
(4»-l)
series
+ 30.
57 + 42. 81
of Art. 396, or that of Art. 397, 12, 20, 30, 42,
6,
term of the 9,
;
—
3.7.11
n terms the
+ 12.
(4»-5)(4»-l)
.
2
be
1
1
2'3.7
(4»-5)
will
it
is
?r
+
we
find that the
+ 3» + 2,
series
21, 37, 57, 81,
is2n*+6n+l.
;t
th
terra
;
HIGHER ALGEBRA.
3.32
+ 2) {2m (m+3) + 1} = 2m [n + 1) (?i + 2) (»+ 3) + (n + 1) (m + 2)
»„=(« + 1)
Hence
•'•
(a
S«=ln(»+l)(»+2)(n+3)(n+4)+|(n+i)(n+2)(n+8)-2.
Example
Find the sum
4.
of
??.
terms of the series
2.2 + 6.4 + 12.8 + 20.16 + 30.32+ In the series
the
2, 6, 12, 20, 30,
un = {n + 2
hence
Assume
(rc
2
dividing out by
and equating
A=2,
whence
wn = (2?i2 - 2n + 4)
2n
S n = (2m2 - 2m + 4)
and
B=
n2 +
is
coefficients of like
2 = 2A+B,
t
term
{A (n-l) 2 + B
1
2' 1_1
th
2.
8,
26, 54, 92, 140,
3.
2,
12, 36, 80,
4.
8,
16, 0,
5.
30,
7.
8.
9.
10.
{
l) 2
- 2 (n -
XXIX.
1)
198,
-64, -200, -432,
144, 420, 960, 1890, 3360,
+ 3x + 7x2 +13.^ + 21a4 + 31a6 + 2 3 1 + 2a + 9a + 20a + 35a4 + 54a 3 + 2 + 5a + 10a2 + 1 7a-3 + 26a4 + 37a-5 + - 3a + 5a2 - 7 Xs + 9a4 - 11a6 + I 4 + 2% + 3 4 a 2 + 4 4^ + 5 4 a 4 + 1
1
11.
3
+ 2
12 1Z>
series
i2 *
_?5
32
+
33
infinite series
+
g4 2
+
62
+ ??_iV + 5« _ 5* + 53 52
"
:
}
b.
150, 252,
Find the sum of the
1)
+ C\
:
series
2"- 1 ;
we have
n,
+ 4 2"2 n - 4 = (na - n + 2) 2*« - 4. - 2 (n -
Find the generating functions of the 6.
-
0=4.
-2,
114,
14, 30, 52, 80,
(n
powers of
Find the n th term and the sum of n terms of the 4,
;
0=C-A + B;
EXAMPLES.
1.
n
n) 2 n .
+ m) 2' = (An2 + Bn+ C)2 n 2=A
.-.
?i
1 j
SUMMATION OF
333
SERIES.
Find the general term and the sum of n terms of the series 13.
16, 29, 54,
9,
103,
14.
-3, -1,
15.
2,
5,
12,
16.
1,
0,
1,
17.
4,
13, 35, 94, 262,
167,
11, 39, 89,
31,
8(i,
29, 80,
8,
193, Tr»5
Find the sum of n terms of the
+ 8* + 3.>/- +
+
4./,-' ;
1
19.
1+ 3.i- + 6x2 + lO.f' + 1 5.r* + 5
onJLi 1.2*2 + 21
'
:
+
1
18.
5.t-
series
4
2.3
4+
2T3-
i£i-
6 3.4'2 + 4.5'2 4 + 5
1
"2
f
' :2
4S+
1
1
!
0- 44+
5- 4 ' +
4^
22.
3.4 + 8.
23.
1.3 + 4.7 + 9.13 + 16.21+25.31
24.
1.5 + 2.15 + 3.31+4.53 + 5.81
o C 25
'
nn 26
'
+ 15.20 + 24.31+35.44+
11
1
1.3^1.3.5 1.2
A
— 1.3.5.7.9
k
1.3.5.7
^- + 14- + -T5- + -T6- +
27.
2.2 + 4.4 + 7.8 + 11.16+16.32 +
28.
1
.
3
+3
.
.
'
2.
32
+ 5 3 3 + 7 34 +9. .
.
+
3>
...
1.3.5.7
1.3.5
1.3
1
rtr
30 ^
4-
4.2 4
3.23
2.2-'
+ 4
3
2
1
+
42. 4. 62.4. 6. 82. 4. 6. 8. 10
+—
l -± + 2+i5L 2 3+ 2 + l 2+ 3.4' 92, + 2.3' 4.5" 1.2
_4_
1 1 _5_ + + 2 5" 2.3.4' 3 3.4. 33
1
(J
1.2.3*3
±+A +H+^+ (3^
32
|5
|4
19
33 '
1
.
2
|6
28
I .
3
'
+2 4
.
3
1 .
4
"
8
+
_39_ 3
.
4
.
5
J_ *
16
+
52 4
.
5
1 .
6
'
32
+
:
— HIGHER ALGEBRA.
334
There are many series the summation of which can be brought under no general rule. In some cases a skilful modification of the foregoing methods may be necessary ; in others it properties of Avill be found that the summation depends on the 404.
certain
as those obtained
known expansions, such
by the Binomial,
Logarithmic, and Exponential Theorems. Example
Find the sum of the
1.
2 [I
term of the
12
+
28
+
|2
infinite series
+
|3_
50
78
+
|I
+
|5
series 2, 12, 28, 50, 78.
..
v
is 3n-
.
3h 2 +j«-2
3h(h-1)+4»-2
]n
|n
a »"
+ n -2; hence
2
;i-2
Put n equal
+
n-1
to 1, 2, 3, 4,... in succession
2
",
= 4--;
„2 = 3
4 + ri i
and so
Example
,Sf„
= Se + 4e - 2 {e -
2.
If (1
4
3
2
2
+ ^ - -gj i
«3 =j
;
r2
in Art. 398 l2
= 5e + 2.
1)
+ a;) n = c + c rr + c 2 .r 2 + l-c 1
we may
.
.
+ c nx n
.
. .
.c. 2
easily
£n
-2
+ c^
the coefficient of x 11
x
in
+
,.,
(1
11
'1
+c
n+1 .r) .„
xn =
.r)- 3
the given series =
-
(n
+
jc
. ,
- x) A
The only terms containing 2"+! (1 -
find the value of
shew that
Multiply together these two results; (l
,
+ 2 2 c 2 + 3 2 c 3 +... + n\v
+ 2 2 .r + 3 2 .r 2 + &x 3 +...+ n-x n ~ l +
Also cn + c n _ x x +
.-.
then we have
;
on.
Whence
As
In"
that
(1
. .
=
.
+ .r) n
.
then the given series (2 - 1 - x) n+1
is,
.
in
-
J=
-
7
(1
-
x) 3
—
is
.
n-1 in this expansion arise from
l 1) 2> (1
l - .t)" 2 + \!l±Jl %*-i
L^+3 2»+i _ „ („ + 1) gn + ?ii"±:
f
-n(w+l)2«- !
l
.
)
(i
_ ^-l.
2 h-i
equal to
—
—
MISCELLANEOUS METHODS OF SUMMATION. Example IP _ (n _
= a + l, and »-»>(»-«>
If b
3.
-+
1) „,,.
_ C-3)(»-4)(»-5)
.
rfj.
|2
By
\6
we
the Binomial Theorem,
(n-5)(n-4)(»-3)
.r'
.r
,
n_4
1
-*
in the expansions of (1
5
.r'
,
,
Hence the respectively. (1-x)-*, (l-.r) -4 to the coefficient of x* in the expansion of the scries
(1-.t)- 2 e
1-2 ,
sum
,
,
1
ax*
1-bx
{1-bx)
(1-fcc) 8
'
may
required
(1
=
series
, ;
—
•-
+
( 1
1-bx
number
finite
\
=
)
,
1-bx J
z
Hence the given
series
-
+ l)x + ax-
(a
z
of x n in
= coefficient
of
_
since b
,
= coefficient
a H+l
— a+1.
(l-x)(l-ax)
xn in
a-
-
=
(
1
\1
-ax
-
~ 1—x)
1
a-1 ,
1
+
If the series
4.
x3
xe
+
are denoted by a,
w
is
+
'
b, c
X*
c3
Now
'
+
>/
+
lob
a
+
.t
(a
+ wb + w'-c
2
+
~\9
+ OJ-C-1+ C0X+
•.
,
+ wc = c
io'-b ,
.,
,
,
o
xz
.t
+
4
\3
Tl
—+
ur\r-
-r^\
+
)
(a
Xs ~\5
+
w-b
w 4 .c 4
+ —T- +
w'.r'
-r=~
\
0)=X
bc
=
X 1,
uX since l
co
2
X
+ ojc)
+
I
,
+
|8_
of unity,
\
similarly
+
|5
X8
shew that a8 + 63 + c8 -3o6c=l.
+ c = 1+x +
=e
X5
|2_
w-.r-
and
+
'
- Sabc = {a + b + c)
lA ~ h
2
x'
|7
respectively,
an imaginary cube root
a3 + b 3 +
+
X7
]5
JG
J3
If
+
r
•
is
of terms,
1-bx + ax"
i
1
',
'
- bx)*
be considered to extend to infinity.
But the sum of the
Example
x)
a*x 6
a-x 4
+ 3
and although the given expression consists only of a this series
+
see that
(n-8) (n-2) are the coefficients of x n
^
a positive integer, find the value of
is
)i
335
(l+u> + w ! )x
+ w + ur = 0.
\
.
-
}
)
.
'
HIGHER ALGEBRA.
336
To find
405.
sum of
the
the r th
powers of
n natural
the first
numbers.
Sn
Let the sum be denoted by
SH =V+2 + r
then
;
r
3
+
...
+n
r .
Assume that Sn =A nr+i +An + A n where A A^ A A 3 ... r
r
n
1
2
1
,
2
,
+An ~
2
r
+
3
+An + A
...
r
r
(1), / \
+1 + 1
are quantities whose values have to be
,
determined.
Write n + (n + l)
=A
r
n and
in the place of
1
{(n
+
l)
r+1
- n r+1 ] + A }
+
A
(n
+
l)
+ A 2 {(n+ l)- -n'1
Expand (?i+l) r+ \ efficients of like
1
powers of
3
+
{{n x
(n+l) r_1
r ,
- nr ]
1)'
r + iy~ 2 -n
{(n
By
n.
subtract; thus
,
-2
+
}
...
+A
r ...(2).
and equate the
...
co-
equating the coefficients of n r
,
we have
l=A.
By
+
(r
n
equating the coefficients of
=
r
Equate the
A (r+
— +A —l)r
°
n
coefficients of
r
1
A =
so that
1),
T
a
.
we have
!
,
1
r x
r
whence A = ^
;
p ,
x
substitute for
A and A Jf and
multiply both sides of the equation by \P
r(r-l)(r-2)
...
{r- 2)+
;
1)
we thus obtain i
~p In
n =A r
+
+A
+ 2
l
(1) write
{n
r+l
'4
p+
1
w—
+A
1
2
^ r(r -
in the place of r
{?i l
coefficients of
+
+
'r(r-l)
-(n-iy +i }+A
Equate the
o
'r
n and
-(n-l) } + A 2
n
r
r
~p ,
(r-2)
1)
+
subtract; thus
{n
r
'
1
and substitute
-(n-iy- } + 1
for
A A ,
^-^gzi) +i /^;);^) 2
r
3
r
4
(?•
1
v
"^
(?•
1) (r
2)
1
;
...
thus
-....
w
MISCELLANEOUS METHODS OF SUMMATION. From
p+
2
and
(3)
"r
1
by .addition
(4),
337
.and subtraction,
r(r-l)(r-2)
*
0'-i)(p-g(^-3) o^/_^)^/ + 3
r (r
r
1)
1) (r
(?•
2) (r
w
(6).
3)
By
ascribing to }> i n succession the values 2, 4, 6, from (G) that each of the coefficients A.^ A 5> A.,... to zero; and from (5) we obtain .
r
1
r
___1_
6 " 1^
,
. .
we
see
equal
is
(r-l)(r-2) ,
'
J
30 .
li
r(r-l)(r-2)(r-3)(r-4) _J_ 8 ~42"
;
|6
By
equating the absolute terms in
\=A^A
(2),
we
obtain
+A
+ A% + AZ +
X
-
r
n= 1 in equation (1), we have +A + A r+l 1 = A + A + A + A9 + a
and by putting
r
l
A r+1 =
thus
0.
The result of the preceding expressed by the formula, 406.
n r+x
„
1
r+1
"
„ r
,
l
2
r
_
3
2
most conveniently
article is
_ r(r-l)(r-2)
x
;
r
_3
4
r(r-l)(r-2)(r-3)(r-4) '6
w lprp }
7?
-
i
7?-
1
i
7?
B B
i
7?
7?
—
^
+
5
quantities 2? ... are known as Bernoulli's Numbers; , 3 5 for examples of their application to the summation of other series the advanced student may consult Boole's Finite Differences.
The
,
,
x
Example. ttt
We
Find the value of n6
S„ = ^r.
,
have
_?t6
l5
+ 25 +
^,5
n5
35
~6 +
n5 "2
+
5?i 4
.
_5 4j—3 n* + C,_ —
+ ^ + ^ -^ n* - Ba .
+ n5
-f
.
.
„
n2
l2~r2'
the constant being zero. II. ii.
A.
22
1
HIGHER ALGEBRA.
EXAMPLES. XXIX. series:
the following Find the sum of
l
5.
+ ^+-\T'T
A ^
^
JL + ^
2
|1
p 6.
£± £ + f^l.2- +
q
p
3
ii
-3
r-i
r
c.
r/3
to r
+
1
terms.
*rz
TX^" 1+tm;
7-
+2 ^ "•(1+^)"
»^"2) + .r) _ "'
(1
12
X
_
1+3a?
?i(?i-l)(™- 2 )
2n + l
o,2
9.
2
W
,
2
( 7l
2/i + K /
_
Y+
x
2 (ft2
12)
7i
-
12
i-j[i+-ii7?
...
2
2
)^
is.
terms.
^
2 )
to
+
w + 1 terms.
.2*.3
+ 23
+ 3^T5 + 5T677 + r2T3 2
12.
n
2
1
1L
to
to n terms.
-
ji
+
3^6 +
+
11
+
|4
]|
[3
2a 8
^+W
1
18
+
|5
23s 5
"[7-'|r
121s 6 _ 16
+-J2-|3 14
Without
W
fuming
!«+*+*+
the formula, find the +»«•
«
17
sum
of the series:
+ 2; + 3? +
+ "-
SUMMATION OF Find the sum of
15.
Shew
16.
fl
Y
8 .-
("
1
2
-l)(« 2 -4)
if" If
17.
n
is
n
53
B
I*
I*
— + _+
c+ (n*-lKn*-4)(n*-9) * + ,
11
1
-
18.
a multiple of
is
(
- 8)
2
^- (»- 8 )(»- 4 )»- 6
3,
)
g
\o
?i
is
a positive integer greater than
rf+ «flga (.-
^+
shew that
»- 1)+ (»-»H»-3) _(»-8)(»-4)(»-6) +
If
}'
a positive integer, find the value of
»-i)^ + e»- g 2
if
3,
=( _ 1)n
shew that
y+ "(»-i)<«-«)(— *) ,l 4y + (
...
SP*" 4.
=»»*(» + 3) 19.
is
)
1
\
and
.,
(l-X) 2 -r.r
[7
)(' t
(
43
+ 23 + - +
/-
1
33
339
that the coefficient of x n in the expansion of
"'- 1
+
l3
SERIES.
Find the sum of 1
??.
terms of the series 2
:
3
W
i
(2)
_5__J_+JL__L+i3 _JLL+
+ i2 +
+
+2
l
i"4
2.3
2
+ 2 4+ l+3 2 + 3 4 +
3.4
5.6
4.5
6.7
17
7.8 (-l) n + xn ?i(n+l)(n + 2) 1
20.
Sum
to infinity the series
— +
If (1 21. x)n Cq c vv integer, find the value of
+
(n 22.
^
is
+ cn#n n
+ c^v 2 + CyV3 +
,
being a positive
- \)\ + (n - 3) 2 c3 + (?i - 5) 2 c5 +
Find the sum of n terms of the
„N
23.
whose n th term
series
:
2
4
8
16
32
1.5
5.7
7.17
17.31
31.65
17
7
1.2.3
Prove that,
49
31
+ 2.3.4 3.4.5 if
a < 1,
ax
(
1
+ or)
(
—
4.5.6
_71_ 5.6.7
+ A) ( 1 + a?x) ....
1
aPx3
a*x2 =5wi 2
= 1 + 54 + 2 (l-« 2 )(l-a 4 )(l-a' (l-« )(l-a=K 1-a-,+T: '
'
)
!
"
)
22—2
HIGHER ALGEBRA.
340 If
24.
A
the coefficient of xr in the expansion of
is
r
2/ (i
*A22
2
+ ,f(i +
(i
prove that
If
25.
n
^l r
n
=
2s
2^
|)
,,
-^~\i
—
-
|)
6,
+ ^r-
2)
+ 2~ J
w(w-l)(w-2)
>
"3^5
of the series
shew that each
3+
-
3
"
1
,
*3 +
n(n-l)(n-2)(n-S)(n-4)
1 ""'32
•-
|5
equal to zero. If
26.
n
is
pti
a positive integer, shew that
+1
_ qn + 1
equal to
is
27.
.
(n-r+^-1),
P =(w-r)(»-r+l)(n-r+2)
If
r
(r+^-1),
&=r(r+l)(r+2) shew that
P& 28. 1
1072
j and ^4 =
[3
,
'
3
;
[5
11
is
^.\2
/
(i
,
v
,
(^4 r -i
a multiple of
is
+
»-3
"^"
+ P »-i^-i =
+ P2Q2 + P3Q3+
If
?i
is
a multiple of
3,
(w-5)(w-6)(w-7)
|3
H
3 to - or equal u
^
n
29.
If
x 1_^2
x
is
— 1
n
,
k \n-l+p + q
|>
+ g +l|n-2
shew that
(m-4)(w-5)
+ "~
+ (-!) is
ho
(n-r-l)(w-r-2)...(tt-2r + l)
according as n
is
odd or even.
a proper fraction, shew that
xz T l_#6
x
x5 l_a?io
1
Xs
x3
+.v2^1+^
,
"'"•••'
u.
'
l+.r10
;
CHAPTER XXX. Theory of Numbers. In this chapter we shall use the word number as equivalent in meaning to positive integer. 407.
A
number which
not exactly divisible by any number except itself and unity is called a prime number, or a prime; a number which is divisible by other numbers besides itself and unity is called a composite number \ thus 53 is a prime number, and 35 is a composite number. Two numbers which have no common factor except unity are said to be prime to each other thus 24 is prime to 77.
We shall
408. propositions,
is
make frequent use
of the following elementary
which arise so naturally out of the definition of a prime that they may be regarded as axioms. of
number a divides a product must divide the other factor c.
If a
(i)
factor
some
b, it
be
and
is
prime to one
For since a divides be, every factor of a is found in be; but since a is prime to b, no factor of a is found in b; therefore all the factors of a are found in c that is, a divides c. ;
prime number a divides a product bed..., it must divide one of the factors of that product ; and therefore if a prime number a divides b", where n is any positive integer, it must divide b. If a
(ii)
a is prime to each of the numbers b and c, it is prime to the product be. For no factor of a can divide b or c therefore the product be is not divisible by any factor of a, that is, a is prime to be. Conversely if a is prime to be, it is prime to eacli (iii)
If
;
numbers
and c. Also if a is prime to each of the numbers b, c, d, ..., it is prime to the product bed... and conversely if a is prime to any number, it is prime to every factor of that number.
of the
b
;
HIGHER ALGEBRA.
342 a and integral power of a
prime to each other, every positive is prime to every positive integral power of b. This follows at once from (iii). (iv)
If
a
If
(v)
prime to
is
n and
lowest terms,
- are any two equal c
b are
b,
the fractions
=-
and j- are in their ft
m being any positive integers. and
fractions,
and d must be equimultiples
of a
The number of primes
409.
bo
j
if
j and
in its lowest terms, then
is
and
Also
b respectively.
is infinite.
p be the greatest prime number; then the product 2 3 5 7 11 .p, in which each factor is a prime numand therefore ber, is divisible by each of the factors 2, 3, 5, .p the number formed by adding unity to their product is not hence it is either a prime divisible by any of these factors number itself or is divisible by some prime number greater than p in either case p is not the greatest prime number, and thereFor
not,
if
.
.
let
.
.
. .
.
;
.
;
:
fore the
410.
numbers
number
No
of primes is not limited.
rational
algebraical
formula can
represent
prime
only. 2
3
the formula a + bx + ex + dx + ... represent the value of prime numbers only, and suppose that when x = the expression is ]), so that If possible,
let
m
3 2 p — a + bm + cm + dm +
when x = m + np that
is,
;
the expression becomes
+ np) +
3 {m + np) 2 + d (m + np) +
a +
b (m
a+
bm + cm 2 + dm3 +
p+
or
thus the expression
is
c
.
+ a multiple
. .
...,
of p,
a multiple of p,
divisible
by
£>,
and
is
therefore not a prime
number. 411.
A number
can
be resolved into
prime factors in only one ,
way.
Let
N
a, b, c, d, ...
where
a,
/3,
denote
the
number; suppose
N = abed...,
Suppose also that are prime numbers. Then y, 8, ... are other prime numbers. abed...
=
a/3yS...
;
where
JV = a/3yS...,
.
THEORY OF NUMHEHS.
343
hence a must divide; abed... but eacli of the factors of this product is a prime, therefore a must divide one of them, a suppose. But a and a are both prime therefore a must be equal to a. Hence bed. =/3yS. and as before, /? must be equal to one of the ;
;
.
.
;
.
.
factors of bed... J and so on. Hence the factors in a/3y<$... are equal to those in abed..., and therefore iV can only be resolved into prime factors in one way. the
N
Let a, b,
number of divisors of a composite number. denote the number, and suppose N"=apbg
412.
c, ...
integers.
(l+a + a' +
...+a'')(l+b +
b
2
+
+ V)
...
(I
+
c
+
c
2
+ ...+c r )...
a divisor of the given number, and that no other number is a divisor hence the number of divisors is the number of terms in the product, that is, is
;
(f>+l)fe+l)(r + l) divisors, both unity and the number
This includes as
To find
413.
the
itself.
number of ways in which a composite number
can be resolved into two factors.
N=
Let N" denote the number, and suppose a, b, c... are different prime numbers and ]), integers. Then each term of the product (I
+ a + a2 +
...
+
of) (1
+
b
+
b
2
+
+ b' ) 1
. . .
(1
a'tyc'
c
where
,
. .
q, r...
+c+
.
2
+
are positive
.
.
.
+
r
c
)
.
.
a divisor of iV; but there are two divisors corresponding to each way in which iV can be resolved into two factors hence the required number is
is
;
}(!>+l)& + l)(r + l)
N not
This supposes quantities^, q,
N
r, ...
a perfect square, so that one at least of the is
an odd number.
a perfect square, one way of resolution into factors is x/iVx JNj and to this way there corresponds only one divisor JX. If we exclude this, the number of ways of resolution is If
is
!{(p+l)(? + l)(r + l)...-l}, and
we must add required number
to this
for the
the one
way
\{(P + !)(
JN x
l)- +
N
/iV; thus
lj
we obtain
—
-
;
HIGHER ALGEBRA.
344 To find
414.
number can
number of ways in which a composite
the
two factors which are prime
be resolved into
to
each
other.
N
r
= av b qc .... Of the two factors before, let the number one must contain ap for otherwise there would be some power of a in one factor and some power of a in the other factor, and thus
As
,
Similarly b q the two factors would not be prime to each other. must occur in one of the factors only ; and so on. Hence the required number is equal to the number of ways in which the product abc... can be resolved into two factors; that is, the
number the
ways
of
number
of different
To find
415.
-(1 + 1)(1 + 1)(1 +
is
the
sum of the
2
a divisor, and therefore the product ) that is, is
the
sum
required
=
a
_
.
a
Example
1.
r
sum
is
1
b'
)
(1
+
+
c
c
2
Then each
+ ...+c
r
)...
of the divisors is equal to this
c
b-l
*
as before.
...,
+
...
&»+'_!
i
-1
21600 = 6 3
.
102 = 2 3
the number of divisors = ..
where n
r+1
-l c-1
"
Consider the number 21600.
Since
the
,
divisors of a number.
2 a + ...+ar )(l+b + b +
+a +
1
prime factors in N.
Let the number be denoted by ap b q c term of the product (1
or 2""
1)...
sum
...
,.
(5
.
33
22
.
+ 1)
(3
26-1
of the divisors = — ? 2 — 1 .
.
52 = 2 3
+ 1)
(2
33
.
5 2,
+ 1) = 72
—— — ——-l
3*-l .
.
5
53
.
o
1
-
5
1
= 63x40x31 = 78120. Also 21600 can be resolved into two factors prime to each other in 2 3_1 or 4 ways.
Example
2.
If
n
is
odd shew that n
We have Since n
one of them
n(n2 is
l)
(n2
= {n7i
-
1)
1) is divisible
and n+1 are two consecutive even numbers by 2 and the other by 4.
;
hence
Again n - 1, n, n + 1 are three consecutive numbers hence one of them by 3. Thus the given expression is divisible by the product of 2, and 4, that is, by 24. ;
is divisible
3,
24.
(n+1).
odd, n - 1
is divisible
by
,
THEORY OF NUMBERS. Example
Find the highest power of 3 which
3.
34".
is
contained in
100. J
Of the first 100 integers, as many are divisible by 3 as the number of times that 3 is contained in 100, that is, 33 and the integers are 3, G, 9,... 99. ;
Of these, some contain the factor 3 again, namely 9, 18, 27,... 99, and their number is the quotient of 100 divided by 9. Some again of these last integers contain the factor 3 a third time, namely 27, 54, 81, the number of
them being the quotient
One number
of 100 by 27.
only, 81, contains the
factor 3 four times.
Hence the highest power required = 33 + 11 + 3 + 1 = 48. This example
is
a particular case of the theorem investigated in the next
article.
To find
416.
contained in
of a prime number a which
the highest 'power
In.
n — 2 iii
Let the greatest integer contained in -, Cv
be denoted by / 1,2, 3,
(
-]
,
the numbers
a,
/ ( - which
2a, 3a, 4a,
and so
tJ
...
(
— g
respectively
CL
Similarly there are
...
)
I[-A
which contain « 3 at
Hence the highest power
on.
—n
contain a at least once, namely
j
contain a 2 at least once, and I
Ct
,
Then among the numbers
/(-,], /(-§),...
n. there are
...
is
which
least once;
of a contained in \n
is
'©'©)*'6) + ~ In the remainder of this chapter we shall find venient to express a multiple of n by the symbol Jl(n). 417.
To prove
418. divisible
Let least of
by
P
n
that the prodicct
integers is
stand for the product of r consecutive integers, the is n ; then
which
and •
1>
con-
|r.
Pn = n(n+l)(n + 2)
.-.
r consecutive
of
it
\
...
(u + r-l),
Pn+l = (n+l)(n + 2)(n+3) nPm+i = (n + r) P = nP + rP n
H
...(n+r); ;
p -P =lsxr = r times the product
of r
—
1
consecul ive integer-.
)
.
HIGHER ALGEBRA.
346
Hence \r
;
;
:
—
1,
the product of r
if
—1
consecutive integers
is
by
divisible
we have
Pm+1 -P
= rM(\r-l)
m
= M(\r).
Now also P.
P, =
P
,
,
.
.
are multiples of
.
the product of
P
and therefore
|?',
r— 1
is
2
a multiple of
We
(r.
\r
therefore
\
have thus proved that
consecutive integers
is
divisible
by
\r
—
1,
if
the
product of r consecutive integers is divisible by \r but the product of every two consecutive integers is divisible by 2 ;
1
therefore the product of every three consecutive integers by 3 ; and so on generally.
is
divisible
1
This proposition
By means
is
may
also be proved thus
we can shew
that every prime factor contained in \n + r as often at least as it is contained in \n \r.
we
This
of Art. 416,
leave as an exercise to the student.
If p
419.
is
a prime number,
the expansion q/*(a
+
b) p
,
except the first
"With the exception of the efficient of the form
of every term in is divisible by p.
the coefficient
and
and
first
last,
last,
p(p-l)(p-2)...(p-r +
every term has a co-
l)
'-
where r may have any integral value not exceeding p — 1. Now this expression is an integer; also since p is prime no factor of r is a divisor of it, and since p is greater than r it cannot divide any factor of \r that -is, (p — 1) (p — 2)... (p - r + 1) must be Hence every coefficient except the first and divisible by |r. the last is divisible by p. \
;
420.
If p (a
Write
ft
is
a prime number,
+b+c+ d+
for b
+c+
. .
...)p ;
= a5 +
to
prove that
b + 1'
cp
+ dp +
.
.
.
+ M(p).
then by the preceding article
py = a* + p' + M(p). p = (b + c + d+ = (b + y) p suppose (a +
Again
p
J3
.
.
.
= bp + y + M{p).
By
proceeding in this
way we may
establish the required result.
=
;
THEORY OF NUMBERS. 421 prime to
We
[Fermat'a Theorem.]
If p is a prime number and N"" -lis a multiple of p.
is
have proved that
+
b
+
d+
+
c
...y^a' + V+c* + d"+
each of the quantities «, 6, Cj 4 pose they are in number ; then let
...
N
,V
N
1
p, then
(a
But
347
T
M
(p);
be equal to unity, and sun J P '
prime top, and therefore iV'- -
is
+
...
1 is
a multiple of p.
Si °°\ ^ is P ™°> P-li* an even number except r when />=J. lherefore '
}
Hence
either
that
.V
is
2^ + = 7^ ±
-•
tins result is
Example Since 7
Shew
1.
K
«
1
a multiple of ft
some positive
is
integer.
is
1)
it
a prime,
n (n +
is
divisible
M
n7 - n
n:=n
(»
1) is divisible
G
-l) =
by
|3
;
by
42.
(7)
>i(n
+ l)(n-l)(n* + nS + l).
hence n? - n
is divisible
by 6 x 7, or
42.
iS pHme DU ber shew that the difference of the p" of^J; ^ thG dlfference of 7 tSL* mimbeiS GXCeedS the numbers
Dowfra
\
mXpleof^ Let
where
that n 7 - n
n? -
-
1,
S^ -
or
sometimes more useful than Fermat's theorem.
a 1S ° (n
1
It should be noticed that in the course of Art. 421
422.
T Now
l
.r,
'
b/a
y be the numbers then * p -x=M(p) and y»-y=M (p), *p -yp -(*-y)=^(p); ;
thatls
>
whence we obtain the required
Example Tf
v
V -
3.
UOt
iS -
result.
Prove that every square number
m t0 ? L lG r
5
x'J
e
have
AT2
= 5;i
~l S£r^PTi?5n 1 is a multiple i n»- 1 or N*+l of 5 eitner
l
•
where »
n ultipIe 0f 5
iS
i*
;
is
that
is,
^
of the form Sn or on
w some
± 1.
positive integer
Fermat '« theorem tf»=5w ± 1.
;
thus
HIGHER ALGEBRA.
348
EXAMPLES.
i
respectively,
which
Find the
2.
will
make the products
3. is
which
will
the products perfect cubes.
If x and y are positive integers, divisible by 4.
Shew that the
4.
539539
109350,
make
perfect squares.
numbers
least multipliers of the
7623, respectively,
74088
18375,
4374,
3675,
is
a.
Find the least multipliers of the numbers
1.
a?—y 2
XXX.
aud
if
x—y
is
even,
between any number and
difference
shew that square
its
even. If
5.
by
-ix-y
is
a multiple of
shew that 4x 2 + 7xy - 2y 2
3,
is divisible
9. 6.
Find the number of divisors of 8064.
7.
In how
two factors
many ways can
number 7056 be
the
resolved into
?
8.
Prove that 2 4 -
9.
Prove that n
'1
(?i
+ 1) (n + 5)
Shew that every 10. the same remainder. n
even,
12.
Shew that n (?i 2 - 1)
13.
If
is
15.
a multiple of its
(Sn + 2)
greater than
2,
6.
cube when divided by 6 leave
shew that n (;i 2 + 20)
If
n
is
number and
11.
is
by
1 is divisible
is divisible
is divisible
by
by
48.
24.
shew that n 5 — 5n 3 -f 4?i
is
divisible
by
120. 14.
Prove that 3 2n + 7
15.
If
n
a multiple of
if
17.
n
is
a multiple of
number
a prime
8.
greater than
3,
shew that
2
?i
-
1
is
24.
Shew that n5 — n
16.
240
is
is
is divisible
by 30
for all values of n,
and by
odd.
Shew that the
numbers greater than
6
difference of the squares of is divisible
by
any two prime
24.
18.
Shew that no square number
19.
Shew that every cube number
is
of the form
is
3?i
of the form
— 1.
9?i
or
9n±L
THEORY OF NUMBERS. Shew that
20. is 0, 1
or
21.
form 7n 22.
a cube
if
number
is
349
divided by
If a number is both square or 7?t+l.
Shew
24. may be. 25.
the remainder
that no triangular
and cube, shew that
number can be
If 2»
4- 1
Shew
that ax
l
is
+ a and
2 ,
of the
it is
of the form 3u
a prime number, shew that divided by 2>i+l leave different remainders. 23.
7,
6.
-
1.
when
2 2 , 3 2 ,...n 2
a* - a are always even, whatever a and
Prove that every even power of every odd number
is
x
of the
form 8r + l. 26.
Prove that the 12 th power of any number
is
of the form I3)i
is
of the form I7n
or 13ft+l. 27.
or
Prove that the 8 th power of any number
I7n±l.
If n is a 28. divisible by 240
prime number greater than
If n is any prime 1 is divisible by 168.
29.
n — G
30.
and
Show
that
?i
36
-
number
greater than
1 is divisible
by 33744
shew that n 4 —
5,
if
3,
n
except
is
7,
1
is
shew that
prime to
2, 3,
19
37.
When p + l and 2p + l are both 31. «** — 1 is divisible by 8(p l)(2/) + l), if
+
prime numbers, shew that x is prime to 2, £> + l,and
2p+h p
If 32. divisible by 33.
If
p
m
is r .
is
pV p1 a prime, and X prime to p, shew that x ~
a prime number, and
prove that
am is
-2
a, b
+ am ~ 3 b + am - 4 b'i +
two numbers
less
-
1
is
than m,
+ bm ~ 2
a multiple of m.
N
any number, then any other number may be expressed in the form = aq + r, where q is the integral quotient when is divided by a, and r is a remainder less than a. The number a, to which the other is referred, is sometimes called the modulus ; and to any given modulus a there are a different 423.
If
a
is
N
N
;
;
:
HIGHER ALGEBRA.
350
number iV, each form corresponding to a different value of r. Thus to modulus 3, we have numbers of the form 3q + 2; or, more simply, 3q, 3q±l, since 3q + 2 is 3<7, 3q + l, In like manner to modulus 5 any numbe^ equal to 3 (q+ 1) - 1.
forms
of a
will be
one of the
five
forms
5q,
±
5q
1,
±
5q
2.
424. If 6, c are two integers, which when divided by a leave the same remainder, they are said to be congruent with respect to the modulus a. In this case b — c is a multiple of a, and following the notation of Gauss we shall sometimes express this as follows b = c (mod. a), or b - c=0 (mod. a).
Either of these formulae 425.
pb and
called a congruence.
is
are congruent with respect pc are congruent, p being any integer.
If
b, c
to
modulus
For, by supposition, b - c = ?za, where w is some therefore ])b — pc — pna ; which proves the proposition.
If a
426.
is
prime
to
and
b,
then
integer
the quantities
(b
2a, 3a,
a,
a,
—
1
)
a
are divided by b, the remainders are all different.
For
ma when
two of the quantities ma and leave the same remainder r, so that
possible, suppose that
if
divided by b
ma = qb + r,
m'a = q'b + r
(m - 7/1') a = (q-q')b
then
;
m
— m', since therefore b divides (m — m') a ; hence it must divide it is prime to a ; but this is impossible since and m' are each
m
less
than
b.
Thus the remainders are quantities
is
exactly divisible by
terms of the series
and since none of the the remainders must be the
all different,
b
1, 2, 3,
b,
—
1,
but not necessarily in this
order.
Cor. of the a.
If
a
is
prime to
b,
and
c is
any number, the
p. c,
c
+
a,
c
+
2a,
c
+
(b
—
1) a,
b
terms
.
THEORY OP NUMBERS. when
351
divided by b will leave the same remainders as the terms
of the series
c+
c,
c+
1,
though not necessarily in mainders will be 0, 1, 2, b3
c
2,
+ (b-
order
this
b-
1),
and therefore the
;
re-
1.
are respectively congruent to c n c„, c wn7A regard to modulus a, £/te?i //te products b,baba ..., ^c.c.^ (o-tf also congruent. 427.
-(/"b., b.j
...
,
,
...
}
...
For by supposition, b
where n lt n 2 .
•.
-c = n
1
l
n3
,
bx baba
x
-c = u
b2
a,
2
...
are integers;
...
=(
=
+
c,c c 2
2
»,») (ca
b
+
rc «) 2
M
+
...
3
-c
a,
3
:i
(ca
u./i,
+w
a)
...
.
.
:j
(a),
which proves the proposition.
We
428.
If p
be
can
now
give another proof of Fermat's Theorem.
a prime number and
N
prime
to p,
then
N
1
'
-1
—
1
is
a multiple of p. Since JV and
p
are prime to each other, the numbers 2tf,
if,
when divided by p
(p-l)iV
3.V,
(1),
leave the remainders 2,
1,
(p-1)
3,
(2),
though not necessarily in this order. Therefore the product of all the terms in (1) is congruent to the product of all the terms in (2),
p being the modulus.
That is, |^—1 N''~ and divided by p ; hence i
but i^—l
is
prime to p
;
\p
-
leave the
therefore
JP" 1
We
1
1
=
it
M
same remainder when
follows that
(j>).
denote the number of integers less than a number a and prime to it by the symbol (a) ; thus (2) = 1 ; <£(13) = 12; (18) = G; the integers less than 18 and prime to It will be seen that we here it being 1, 5, 7, 11, 13, 17. consider unity as prime to all numbers. 429.
shall
,;
HIGHER ALGEBRA.
352
To shew that if the numbers 430. each other, (f>
(abccl
.
.)
.
= <£ (a)
.
b,
a,
(b)
d,
c,
<£ (c)
.
.
.
.
...
are prime to
.
Consider the product ab then the first ab numbers can be written in b lines, each line containing a numbers ; thus ;
(&_
a,
k,
2,
1,
a+l,
a+
2,
a+
k,
a+
a,
2a +1,
2a +
2,
2a +
k,
2a +
a,
a+
2,
a+
a.
\)
a+
(6-
1,
1)
(b-l)a +
...
k,
...
-
(b
1)
Let us consider the vertical column which begins with h ; if k is prime to a all the terms of this column will be prime to a but if k and a have a common divisor, no number in the column Now the first row contains <£ (a) numbers will be prime to a. prime to a \ therefore there are <£ (a) vertical columns in each let us suppose that the of which every term is prime to a vertical column which begins with k is one of these. This column is an A. p., the terms of which when divided by b leave remainders 0, 1, 2, 3, ... 6 — 1 [Art. 426 Cor.]; hence the column contains to b. <£ (b) integers prime ;
Similarly, each of the term is prime to a contain
<£
table there are (a) cj> (b) also to by and therefore to ab .
<£
Therefore
cf>
(abed
(ab)
-
...)
=
columns in which every hence in the (b) integers prime to b integers which are prime to a and (a) vertical
;
;
<£
that (a)
.
(a)
.
(a)
.
is
<£ (6).
(bed
. .
.)
=
cj>
=
(a).(b).(c).<}>(d)....
431. To find the number given number, and prime to it.
(f)(b)
.
(cd
...)
of positive integers
less
than a
Let JV denote the number, and suppose that JV = ap bq c ... where a, b, c, ... are different prime numbers, and p, q, r ... Consider the factor a ; of the natural numpositive integers. p p — the only ones not prime to a are bers 1, 2, 3, ... a 1, a r
1
'
,
a,
2a,
3a,
...
(a*-
1
-
I) a,
"
(a1
1
)
a,
;
.
THEORY OF NUMBERS. and the number
of these
4>
Now .
= a" -
(av )
the factors ap
all
\
i
is a''~
(a)Vc
r .
.
.)
=
,
(a
1
hence
;
a'-' c\
b'\
')
.
353
=a?(l-
(b
.
are prime to each other
...
-^
1
)
(c
.
r
)
.
.
-H)-H)-H)^••K)H)H)that
^
is,
to
= iir(i-i)(i-J)(i-I)....
Shew that the sum ^N
Example.
and prime
W
which are
of all the integers
less
than
N
it is
N
If x is any integer less than integer less than and prime to it.
N
Denote the integers by
1,
p, q,
and prime
r, ...
,
and
to
their
it,
then
sum by
N-x
is also
an
number
of
S; then
S = l+p + q + r+... + (N-r) + {N-q) + (N-p) + {N-l), (N) terms.
the series consisting of
Writing the series in the reverse order,
S = {N-l) + (N-p) + (N-q) + (N-r)+...+r + q+p + l; .-.
2S = N + N + N+
by addition,
.-.
...
to
(N) terms;
S = $N
From the last article it follows that the 432. integers which are less than J¥ and not prime to it is :
'-'(.4>(>-i)(» 3("i)-' tli at is,
N N N N N N _++_+..._ a ao ac be b
. .
c
Here the term
N —
gives the
a,
la, da,
number ...
.
+
N+
....
abc
of the integers
— .a (t
which contain
N
a as a factor; the term —= gives the number of ao
H.
II.
A.
23
HIGHER ALGEBRA.
354
N
.
.
-j ab, which contain ab as a factor, ao and so on. Further, every integer is reckoned once, and once only ; thus, each multiple of ab will appear once among the multiples of a, once among the multiples of b, and once negatively among the multiples of ab, and is thus reckoned once only. iV Again, each multiple of abc will appear among the j- ,
the integers ab, 2ab, Sab,
...
N N — — a o c ,
terms which are multiples of a, b, c respectively; among the iV JV =- terms which are multiples of ab, ac, be respectively r r J
& — — ,
,
ab'
ac'
and among the -j- multiples
of abc; that
is,
[Wilson's Theorem.]
»
is divisible
by
By
2,
Ijp-1
Ex.
= (P~
3-3+1
=
1,
Similarly, other
a prime number,
be
If -p
'
since
each multiple of abc occurs once, and once only. cases may be discussed.
433.
;
'
be
1
+ \p -
1
p.
Art. 314 i)""
we have
- (P -
1
i)
Jp-l)(p
(P -
2r
+
'
2)(p-3) {p
_
and by Fermat's Theorem each of (p-2) p ~\ (p-2>y~\ ... is of the form
p-l = M(p) +fl-(p-l) + (P~
l
^z^ipD (p - $y-> irl+ tlie 1
top _ lterms
expressions
(j)
-
l) p
.
~
l
,
+M(p)-, thus
)(P~
2)
-...top-
I
terms!
=M(p) + {(i-iy->-(-iy->} = M(p) — Therefore
1
1,
since
+ pI
1
p—
1 is
M
(p).
=
even.
This theorem is only true when p is prime. For suppose p has a factor q; then q is less than p and must divide \p — 1 hence ;
1
+
\p
—
1 is
not a multiple of
q,
and therefore not a multiple
of p.
Wilson's Theorem may also be proved without using the result quoted from Art. 314, as in the following article.
1
.
.
THEORY OF NUMBERS. [Wilson's Theorem.]
434.
by
is divisible
If p
be
,'}:>5
a prime number,
+ lp—
1
p.
Let a denote any one of the numbers 2,
1,
then a
prime to p, and
is
4
3, if
...
5
(p-1)
(1),
the products
\.a, 2. a,
3. a,
(^;
—
1 )
a
divided by p, one and only one of them mainder 1. [Art. 426.]
are
leaves
the re-
ma; then we can shew that the numbers m and a are different unless a=j)~ 1 or 1- For if a were to give remainder 1 on division by^>, we should have (mod. p), a~ - 1 = Let
this be the product
2
and since p is prime, this can only be the case when a + or a — 1 — 0; that is, when a=p— 1 or 1.
1
- p,
Hence one and only one of the products 2a, 3a, ... (p — 2) a gives remainder 1 when divided by p that is, for any one of the series of numbers in (1), excluding the first and last, it is ;
possible to find one other, such that the product of the pair
the form
M
+
(p)
is
of
1
Therefore the integers 2, 3, 4, ... (p-2), the number of which is even, can be associated in pairs such that the product of each pair is of the form (j?) + 1
M
Therefore by multiplying
2.3.4
...
thatis,
1.2.3.4
...
whence
\p
or
1
+
1^
—
Cor.
by 2p +
1
is
If 2p
-
1
all
these pairs together,
we have
(p-2) = M(p) + l;
(p-l) = (p-l){M(p) + = (p) +p - 1
M
l} ;
j
a multiple of p.
+
l
is
number
a prime
/jp\
2
+
(-
iy
is
divisible
l.
For by Wilson's Theorem 1 + \2p n = 2p + 1, so that p+ 1 = n —p then
is
divisible
by 2p +
.
Put
1,
and
1
;
= 1.2.3.4 p(p+l)(p + 2) (n-1) = 1 (w-1) 2(n-2) 3(»-3) ... p (n-p) = a multiple of n + (- iy (\p) 2
\2p
.
Therefore
+
1
therefore (|^)
2
(
-
l) p (\p)
+ (— l/
is
2
is
divisible
divisible
by n or 2p +
by 2;;+l.
23—2
;
:
HIGHER ALGEBRA.
356
Many
435.
theorems relating to the properties of numbers
can be proved by induction. Example
1
.
p
If
is
a prime number, x p - x
Let x p - x be denoted by f(x)
by p.
is divisible
then
;
/ (x + 1) -/ (x) = {x + 1)p - (x + 1) - {xP - x) 1) =px p ~ l + P
J.
=a therefore/^)
and this
is
if
p
by^i, so also is/(.r
is divisible
/(2) =
a multiple of p
by p, therefore /(4)
multiple of p,
is
prime
[Art. 419.]
f(x + 1) =f(x) + a multiple of p.
.-.
If
.
a
•
is
but
~ 2 = (1 + 1)^-2,
2*5
when p is prime
divisible
+ l);
[Art. 419]
;
therefore / (3)
is divisible
by^, and so on; thus the proposition
is
true
universally.
This furnishes another proof of Fermat's theorem, for
it
~ follows that x p J
Example
24?i
if
x
is
prime to
p,
1 is a multiple of p.
Prove that 5 2,l+ 2 -
2.
Let 5 2n+2 -
-
- 25
24/i
is
by 576.
divisible
- 25 be denoted by f(n) /(?i+l) = 5 2n+4 -24(w + l)-25
then
= 5 2 .5 2w+2 -24n-49; .-.
f(n+l) - 25/ (n) =25
(24n + 25) - 24u - 49
= 576 (n + 1). Therefore if f(n) is divisible by 576, so also is /(u + 1); but by trial we see that the theorem is true when n = l, therefore it is true when n=2, therefore it is true when ?i = 3, and so on; thus it is true universally.
The above
result
52n+2
may
also be proved as follows
_ 2in - 25 = 25 M+ 1 -
- 25 = 25 (1 + 24)" -24rc-25 24;i
= 25 + 25 n 24 + M (24 2 = 576n + iW(576) .
.
)
- 24n - 25
= i)/(576).
EXAMPLES.
XXX.
4" +
divisible
1.
Shew that 10 n + 3
2.
Shew
that 2
3.
Shew
that 4 6 n
4.
Shew that
.
7n
.
8
.
7n
.
+3
.
2
5H
+5
-5
is is
b.
by
9.
a multiple of 24.
+ 5 n + x when divided by 20 leaves remainder
+ 4" + 2
is
of the form 24 (2r -
1).
9.
b
THEORY OF NUMBERS. p
5.
If
6.
Shew
7.
Shew
prime, shew that 2
is
that a v, +
-a
l
is
\p-3 + l
is
357
a multiple of p.
divisible 1>y 30.
that the highest power of
2
contained in
2r
1
is
2''-;--l.
that 3 4 + - + 5 2 +* is a multiple of 14.
8.
Shew
9.
Shew that
'1
'1
3** +6 +160»a
- 56n- 243
is divisible l»y
512.
Prove that the sum of the coefficients of the odd powers of x 10. 2 in the expansion of (l+ # 3 + .r 4 ) n "" 1 , when n is a prime number other than 5, is divisible by n. <
If n is a 11. divisible by 504. If
12.
n
is
r+# +
prime number greater than
an odd number, prove that
shew that n°-l
7,
5
?i*
+ 3>i 4 + 7>i 2 - 11
is
is
a
multiple of 128.
p
a prime number, shew that the coefficients of the terms of (H-a?)*-* are alternately greater and less by unity than some multiple of p. If
13.
is
a prime, shew that the sum of the (p-l) th powers of any p numbers in arithmetical progression, wherein the common difference is not divisible by p, is less by 1 than a multiple of p.
p
14.
If
15.
Shew
is
that a 12 -
12
is divisible
by
91, if
a and
b are
both prime
to 91. If
16.
p
is
a prime, shew that \p
n—
If 17. 1, n + 1 are 2 that n(?i -4) is divisible
-2r
2r
-
1
-1
1
is divisible
by p.
both prime numbers greater than 5, shew by 120, and ?i 2 (>i 2 + 16) by 720. Also shew
that n must be of the form 30£ or 30^ + 12.
Shew
18. ,
is
that the highest power of n which
n r — nr + r-
.
equal to
19.
number
If c2
n- 1
is
20.
1 is
1
1 .
p
is a prime number, and a prime to ]), and if a square can be found such that c 2 — a is divisible by pt shew that
l(p-D
a2
contained in \nr ~
divisible
by p.
Find the general solution of the congruence 98a;-
1=0
(mod. 139).
— HIGHER ALGEBRA.
358
Shew that the sum of the squares and prime to it is a given number 21.
N
of all the
numbers
less
than
?(i--30-J)0-9- + ?ci-.»a-.)a-*-. and the sum of the cubes
is
?(i-3(x-])(t-9- + ?a-^ci-«a-*., a,b,c... being the different prime factors of
and q are any two positive by (|£>)«. |# and by (\q) p \p.
If
22.
divisible
jt?
iV.
integers,
shew that \pq
is
.
Shew that the square numbers which
23.
are also triangular are the squares of the coefficients of the powers of x in the ex-
given by
an<^ ^hat the square numbers which are also r— \)X -f" X2> pentagonal by the coefficients of the powers of x in the expansion of
pansion of
L
1_
Shew that the sum of the fourth 24. and prime to it is less than
N
the numbers
all
a
\
5
powers of
-gg(l-<*)(l-i»)(l -«*)..., being the different prime factors of
a, 6, c,...
If
25.
prime to
(iV) is
(f>
it,
and
if
the
x
number prime to
is
^^26.
If
dv d2 ds ,
,
...
(f>
Shew
of integers JV, 1
=
A".
which are
less
shew that (mod.
JV).
denote the divisors of a number
(dj
+
(c? ) 2
than JV and
JV,
then
+ <£ (d3 + ... =iV. )
also that
(!)
r-
—9~0( 3 r — + 0( 5 )T^ 1
;
)
;
fi
™-
---
~—kJ.
odinf. = -?
l
.;
*
CHAPTER XXXI.
The General Theory of Continued Fractions. In Chap. xxv. we have investigated the properties of
*-436.
Continued Fractions of the form
a,
+
—
-
-
%+
«,+
.
.
.
,
where a2 a 3 ,
. . , '
'
are positive integers, and a^ is either a positive integer or zero. shall now consider continued fractions of a more general
We
type.
The most general form
*-i37.
~' a
=*=
1
„ "i rZl * a « 2 3
'
=*=
where a i> a2> a3>
a continued fraction
of
••> *,»
KK
•••
is
represent
any quantities whatever.
The
fractions
— a,
,
—
a2
—
,
a3
We
.
,
.
.
are called components of the
shall confine our attention to two cases; continued fraction. (i) that in which the sign before each component is positive (ii) that in which the sign is negative.
To
*438.
investigate the
law of formation of
the successive
convergents to the continued fraction
b i
The
first 6,
b2
b3
three convergents ergents are
a2 b
x
a 3 .a 2 b
i
+b 3 .b
l
AVe see that the numerator of the third convergent may be formed by multiplying the numerator of the second convergent by a 3 and the numerator of the first by b 3 and adding the results also that the denominator may be formed in like together manner. ,
;
n
;
HIGHER ALGEBRA.
360
Suppose that the successive convergents are formed in a similar way; let the numerators be denoted by p it p 2 p 3 ..., and the denominators by q lt q 2 q3 ... ,
,
,
that the law that is, suppose
Assume vergent
;
p In
-anlp
+
n—\ ,
b n-lp n — 2> „,
The (n+l) th convergent a
—a—
h
n+
formation holds for the nth con-
of
n-Ln —
J-
,
1
;
hence
,,p
,,
+
bq h-Ih—2_.
wth
from the
differs
in the place of a
=a q
q
only in having
\
the (n+ l) tb convergent
we put
If therefore ?>
^,=a
nP +b
q
+
,
=a
+,q
+b
,,q
,,
numerator and denominator of the (u + l) th convergent follow the law which was supposed to hold in case of the th But the law does hold in the case of the third convergent ?t hence it holds for the fourth and so on therefore it holds
we
see that the
.
;
;
universally. *-439.
In the case of the continued fraction b2
b,
«1
we may prove Vn
-
b3
a2 ~
CC
~ 3
that
= anPn-l ~ kPn-*
,
Qn
=
»«?«-!
~
k
',
a result which may be deduced from that of the preceding article by changing the sign of b n .
*440.
In the continued fraction
h
K
K
a +
a2 +
«3 +
1
we have
seen that
p n =a nlp
J-
ii
,+bp
—\
n-L
„, n — 2'
q J-n
^an-ln q — ,+bq n-J-n—2 1
a.
GENERAL THEOHY OF CONTINUED FRACTIONS.
Qn-J'
\9n
?„ + ,
Ml
but
and
is
therefore a proper fraction: hence
"
+
1
—
—
&.+1 less
than
— — —— In
,
and
numerically
is
ft
of opposite sign.
is
In-}
By
reasoning as in Art. 335, we may shew that every convergent of an odd order is greater than the continued fraction, and every convergent of an even order is less than the continued fraction ; hence every convergent of an odd order is greater than every convergent of an even order.
Thus 2--'^ 32/1+]
^
is
positive
1
2--is 1
^
than £ss=l -
2
2ft*-J
2
22/1 —
less
9an 2
Also ?*=i -
and
2/1
+
1
2/1
+
1
positive
1
2/1
—
1
and
less
than
^=* - &=s 22/1 — 1
2
2
2/1
2
2/1
Hence the convergents
hence
2*2/1
2 2/1-1
22/1
2
" ;
2/1
—2
2/1
—2
an odd order are
hence
22/1-2
greater than the continued fraction but continually decrease, and the convergents of an even order are all less than the continued fraction but continually increase. of
all
Suppose now that the number of components is infinite, then the convergents of an odd order must tend to some finite limit, and the convergents of an even order must also tend to some these limits are equal the continued fraction tends to one definite limit ; if they are not equal, the odd convergents tend to one limit, and the even convergents tend to a different limit, and the continued fraction may be said to be oscillating; in this case the continued fraction is the symbolical representation of two quantities, one of which is the limit of the odd, and the other that of the even convergents. finite limit
;
if
a
.
HIGHER ALGEBRA.
362 *441.
To shew
that the continued fraction
a +
a 2 + a3
x
has a definite value if the limit of
rf""
n+I
when n
+
is infinite
is
when n
is
greater than zero.
fraction will have a definite value
The continued
infinite if the difference of the limits of
-^
— is equal to zero.
and
?«
9n+l
fPn
n+lffn-l
Now
_ Pn-1
W«
?«+!
£«-!>
whence we obtain Pjt+X
_ £» =
(_
)-l 6 "+^»-l & ng»-3
1
KV* KVl (P* _ Pl\
<2n+l
k.-i?
But
k.-M? n+lin-1 li+1 !7u
+
an+l Qn
^B+lS'n-l
k^g an+l q n
an(
a.
K?-i + &«£«-*) _ «*»«+!
i
^n+lSn-l
"n+lSn-l
also neither of these "
n+1
is
^n+1
-J**^
is
less
is
x
1
:
if
K
the limit of
"+
the limit of
than
J
^n+A^-2 b^ q
terms can be negative; hence
greater than zero so also
case the limit of
+
+
;
in
which
and therefore ^±i_ -^
Qn+i
q„+i
m
t
£a
qn
the limit of the product of an infinite
number
of proper fractions,
and must therefore be equal to zero
that
-^
:
is,
?«+:
the same limit .For
;
and
-- tend to
qn
which proves the proposition.
example, in the continued fraction
V
T-
3-
n~
3+5 a.
Lim -f-^- = Lim*«+i
2n+l +
1
7
\, (n+iy .
2
•
-'
•
»
=4
and therefore the continued fraction tends to a
:
definite limit.
;
GENERAL THEORY OF CONTINUED FRACTIONS. In
*442.
3G3
continued fraction
the
H;„ A
cl
c\i,
2
1
if the denominator of every component exceeds the numerator by unity at least, the convergents are positive fractions in ascending order of magnitude.
By
in
supposition
—
'
—
3
-*
,
,
.
,
.
are positive proper fractions
.
a2 a ? each of which the denominator exceeds the numerator by
a
unity at
i
The second convergent
least.
is
1
a exceeds
by unity at
ft
and
least,
-* is
,
and since a
i
--
2
a proper fraction,
it
follows
t
2
tliat «,
greater than
is
2
a positive proper fraction
is
,
the third convergent
proper fraction
—
is
-.
!
«1
p
i
'
ii
denote
may be shewn by f, then
it
therefore a positive proper
—— —— —
that
— anlp
—
ii
,
K
K
%~
a,
K -
CV
a positive proper fraction
Again, O
is
;
it
is
is
a positive
hence also the fourth convergent
;
ft,
is
and
,
we may shew
Similarly
fraction.
the second convergent
is,
In like manner
a positive proper fraction. that
that
ft,;
—
1
ft
;
2
~
and
so on.
q—aa
p
2' »/ n — .,
2
iuii-
ii
,
—
I
ft
q „2 niii-
;
'
^ 2 hence ^s±J -
But P °~
^
- - ^=*
;
n
g
^=
92
hence ^ 2
and
«\
>^ ^ > ^-
9jt
proposition.
1
(
l,
7,
-^
r 'i
,
^>^ <74
%
tl ie
h
?3
2
,
7,
- - =
have
;
,
same
and
is
sign.
therefore positive
2
and so on; which proves the
—
;
,
;
.
HIGHER ALGEBRA.
364
If the number of the components is infinite, the conCor. vergents form an infinite series of proper fractions in ascending order of magnitude ; and in this case the continued fraction must tend to a definite limit which cannot exceed unity.
From
*443.
Pn =
the formula
+ hPn-t*
<*nPn-i
= a&n-, +
9n
&.&_«>
we may always determine in succession as many of the convergent^ as we please. In certain cases, however, a general expression can be found for the
11
th
convergent. c
To
Example.
find the
We have p n = Sp,^ - 6p n _ any three consecutive terms
2
;
of
c
c
to -
w th convergent
5- 5- 5-
hence the numerators form a recurring which are connected by the relation
series
Pn " 5p«-! + §Pn-2'
S =p1 +p.2x +Prf? +
Let
But the
first
~ +p nx n 1 +
S= Pl1+-(P2~ 5x
we have
then, as in Art. 325,
• • •
two convergents are -
18
Similarly
we
S'
if
whence
gn
3"- 1 - 12
.
2' 1
"1 =
1
- 2x
x
___ = ___ _
=9
.
3*- 1 - 4
.
'
6 (3" - 2").
= q + q& + q 3x* +...+ q nxn ~ x +
^=
find
.
12
l-3x
- hx + Qx-
p n = 18
whence
2
=-^
,
6 1
.
X
5Pl)
+ 6a;
.
.
;J
i
r
2*- 1
= S'^ 1 - 2' + l
. .
] .
w y w _ 6(3"-2 ) """
ffn
~ 3n+1 - 2?l+1
'
This method will only succeed when a„ and b n are constant Thus in the case of the continued fraction for all values of n.
a+
a+ a+
...
,
we may shew
that the
numerators of the
successive convergents are the coefficients of the powers of
the expansion of
^
7-2
,
x in
and the denominators are the ft
coefficients of the powers of x in the expansion of
^
-I-
1 ~~ OjX
ti /Y*
7-2 — ox
.
1
GENERAL THEORY OF CONTINUED FRACTIONS.
305
*444. For the investigation of the general values of pn and qn the student is referred to works on Finite Differences ; it is only The in special cases that these values can be found by Algebra.
following method will sometimes be found useful. Find the value
Example.
12
3 -— =— 5— o 1 z
of
+
+
+
The same law of formation holds for p n and u n = nun _ x + nu n _ 2 either of them then ;
qn
us
let
;
tal
n to denote
i*
,
un -
or
(n
+ 1) «„_! = -
(uB_, -
?w n _ 2 ).
i^j - RU _s = - (« u _ 2 -ra- 1 « n_ 3 ).
Similarly,
tt
whence by multiplication, we obtain u n -(n + l)u n _ 1 =
The
2
1
first
two convergents are 1
(-iy^(u 2 -3u
T -A
,
Tims
^n 7i
Pn-1
(" !)
m
+l
?l-l
l)"- 8
|n-l
In
0n
gn-1
iw+l
to
In
9»-2 n -
1
2
[3'
ffn-1
In
Ps_Pi
?3
|2'
13
-(n + l)q n -i = (- I)""2
qn
lra+1
At! _ Art = (~
_
3
2
).
hence
;
n p n -(n + l) Pn ^=(-l) -\
i
1
3i
2'
'2
2
_
2 (~ I)""
j»
|2'
Lit n+1
1
|n
+l
12
+ |4
|3
. '
j
7a
n+1 By making n
infinite,
,
is
1
|2
|3
~a
1
,
(- l) n
n+1
|1
we obtain
2n
which
1
= 1-7^+1i- "77 +
e
e)
e-V
therefore the value of the given expression
+l
In
whence, by addition
+
-
'
—
*
;
HIGHER ALGEBRA.
366
——+ — — —— a + a +
If every component of
*445.
3j
o
j
For
-7
—
=
^-
*—m
...
C=
-= Similarly J
B
f
*-=
—*- —*-
.
a3 + a +
.
.
D
before, it follows that .
.
Again,
—
than less
,
ax than
B -7
C =
,
which
—
;
is
and
*—-
is
D ,
denotes the infinite continued fraction
hence
;
jy
...
,
com-
are positive integers
—
=^
-
/
C
therefore
=
—S^
*
=
,
:
=
B
7* ;
continued
suppose
and so
less
;
and as
;
on.
5
,
..
is
,
infinite
proper tractions
a proper fraction
A, B, «x 6 X
a positive integer.
is
a positive integer are
Now
suppose.
where fa denotes the
,
+f
a2
B
where A and
,
is positive,
— b
f,
f=
hence
;
-^ XL
where
,
are integers and
fraction
the continued
possible, suppose that the continued fraction is
if
mensurable and equal to then
3
and denominator,
fraction with integral numerator fraction is incommensurable.
a proper
is
...
,
.
tor -j is less
than -* as
;
-^
is
O
so on.
form an infinite series of positive integers which is absurd. Hence the in descending order of magnitude given fraction cannot be commensurable.
Thus A, B, C, D,
...
;
holds if some of the components are not proper fractions, provided that from and after a fixed component all the others are proper fractions.
The above
result
For suppose that
still
—
and
all
the succeeding components are
n
proper fractions
;
thus, as
we have
tinued fraction beginning with -s
an
just proved, the infinite conis
incommensurable
;
denote
k
7)
it
by
k,
then the complete quotient corresponding to
—n
is
in
and therefore the value
of the continued fraction
is
^-^ 9n-l
/ —+ hn-2
^
"~ 2 .
GENERAL THEORY OF CONTINUED FRACTIONS. This cannot be commensurable unless
V —n_1
P = l-^^
tfn-l
?«-«
condition cannot hold unless ?2=a = *-=2 , Pn= B ?n-a ?n-3 Qn-z finally
P P — H=
%
that
;
is
£>6
=
0,
which
this
;ultl
t
9n-4
impossible
is
and
•
= P^*
3G7
;
hence the
?,
.
given fraction must be incommensurable.
—- — —K a a a 1
*446.
//* eirary
component of
1
«-
i
""
a
*-
~
...
?'s
a proper
""
3
fraction with integral numerator and denominator, and if the value of the infinite continued fraction beginning with any component is less than unity, the fraction is incommensurable.
The demonstration
* 1.
Shew that
is
similar to that of the preceding article.
EXAMPLES. XXXI.
in the continued fraction
\ ax
_h
h_ - a2 - a 3 -
Pn = a nPn - 1 ~ ^nPn - 2 2.
Convert
"
rators.
3.
Shew
m '^-
|
-
)
J
V^=«-
2 |_
In the continued fraction
"?i?n
-2
nume-
^ ±r.
.......
—— —— ——
«1~ a%~ a3~
.... if
component exceed the numerator by unity at
p n and qn 5.
Qn = a n9.n -\~
that
(2 )
of every
'
into a continued fraction with unit
« V*+6=«+^
4.
a.
the denominator least,
shew that
increase with n.
If a lf a,, rtg,..^,, are in
an
1
^"2^
harmonical progression, shew that 1
1
1
2^
2~="
2^
«2
SJ'
—
1
.
HIGHER ALGEBRA.
368 Shew that
6.
cc+ V
and
(
7.
a
—- -—-
-
+
...
2a+ 2a +
J
— ... \/ a + s2«+ 2a+"7\ (
)
\
1
2 ,
J
o^--)= "7 a 2a-
7T2a-
= 2a
...
2a- 2a-
V 1
-
:
[a -
, ~
2a2 -
2a'2
In the continued fraction b
b
b
a+ a+ a+
pn + = 6an
shew that 8.
.v
x
Shew that
— a+ b
number
being the
ax
b
b
=
bq n + 1 - apn +
,
%
n_
x
— 8P
—p
= 6.-^7 x + — ^n x+
-
a+ a+
1
a
of components,
and
a, /3
.
i
1
the roots of the equation
k2 — ak — b = 0. 9.
Prove that the product of the continued fractions
J_ L _L A_ 6+ c+ d+ a+ is
equal to
—
J_
_1_ -c +
,7
'"'
J_
-6+ -a+
_x _
-0?+
'"'
1
Shew that 1
10.
11.
12.
13
.
15.
16.
64
9
(?i
i !> «* 1- 3- 4- 5-
ji£2- 3- 4-
1+
2+ v,
=
'
-2=1 w+l-
=.-1. 2(e2 -l) e2
3(n + 2)
f
,
Uo
=
+l _'6(2 e3+l)
n+
3+ Ucy
6
!
2/i
3.4 3.5
+ l)(w + 2)(2/i + 3)
?i
+2 n+
8
3.3
=
(n
£±L 5±|- 1+ i+i« + |. + ...+|«. I— L_ + l— + 2
?i
1+ 2+ 3+
If u,1
-l) 2
^2 ~1 _ ?*fo+ 3 ) 2» + l~ 2
-L JL _§_ 1- 5- 7-
6
2
»2 +(»+l) 2
1- 5- 13- 25-
4 14.
4
"
—f
5e 3
,
-2
each successive fraction
a + 26 bein» formed by taking the denominator and the sum of the numerator and denominator of the preceding fraction for its numerator and denomia
a+
e>
,
•*
—
nator respectively, shew that u „=**—=
.
CONVERSION OF SERIES INTO CONTINUED FRACTIONS. 309 Prove that the n ih convergent to the continued fraction
17.
f
J'
y'H
)'
is
r+l- r+l- r+l-
—
Find the value of
18.
cij
a19 a2 a 3v being >
19.
1 ;)
lh
convergent to 6
»* +1 -l
to
1
—
--
hence shew that
equal to
,
to
3-d e-2
3
2
2+ 3+ 4+
e lies
is
-
— -— — -2—+ 1+2+1+ n
1111111 1
'
and greater than unity.
Shew that the 3n th convergent
Shew that
'
a 3 +l-
n is
5- 2- 1- 5- 2- 1- 521.
y.
%
%l-
a.,+
Shew that the n lh convergent
— the (2m v
20.
positive
\
+ l-
+1
between 2§ and 2 r8T
3>i
+l
'
.
Conversion of Series into Continued Fractions. *417.
It will be convenient here to write the series in the 1 —1 +—
u2
^6 1
1
Put
+ 1
•
—1 +..
u3
—
form
HIGHER ALGEBRA.
370
— —=— —11111 —
Similarly,
u
+
x
u2
+
u3
12 +
=
1
u — — u +u +x 2
y
us + x2 u u UQ2 u - u + u2 — u2 + u3
u
x
x
x
2
x
x
and
so
on 1
;
hence generally 1
1
x
'
a
CONVERSION OF SERIES INTO CONTINUED FRACTIONS.
,
—=
x
1
By
putting
we obtain
1 ,
"n x
•
.
V,l=
.',71
a n+l - a nx
;
hence we have
X
.T rt
l
3
.T (l
2
3
XA (J
3
X (
i
a 2~ a l X +
+
Vx
x
+ *) =
log (!
.-.
'l
2
x a3~ U,2 X +
(IfX
2
a ,~ a i x
+
3 2 .r
2°-x
1+ 2-x+ 3-2x+ ~i-Sx +
In certain cases we may simplify the components of the continued fraction by the help of the following proposition *44S.
:
The continued
is
fraction &.
K
h
K
ax +
a2 +
a3 +
a4 +
equal to the continued fraction C
cxa x
where
cn
c2 ,
c
3,
c
4
+
c 2a
Let
denote
——
-
«2
Similarly, c is
f
=
...
•
+
;
-
C 3 rt 3
...
G
;
+/2
'^-~
+
ci«i
+/i
£ Q Q -
tlien
=
3
denote
c/c4
7
«i
/„
a + 3 3
—— ——
the continued fraction
Let
c
C 3C
are any quantities whatever.
,
7
/"
+
2
A
^3
5 CA
A
J
;
+
c i/i
then
C 2<\.
and
+
C X2
so on;
'
whence the proposition
3
established.
24—2
-a
HIGHER ALGEBBA..
372
^EXAMPLES.
XXXI.
b.
Shew that 1.
1111 +
UQ
m3
u2
Wj
+ (-l) u— n
+ 1
2.
&n
—+ X
X2
ClfiCCt
Cl^Qz-tCto
1 -+-
Mj
+
r—\ ?•
-2
r
5.
6.
7.
8.
9.
l
+ s2 + 3« +
+
11 i»+oa+
l2
e
x
+
22
'
x
• • •
^n
1
*^
+2 r + 2...
n quotients.
,
to
-.
1114
111 1-
(n + 1) 2
x
n2
9
1- 3- 5- 7-
?i+l
l
2x
2
2n+l
4
?i
+ 22 -
4
n*+(n+Yf'
3x
1111 r
a
ab
,
1
+
la
— + ...= a+ abed b-l+ c— 1+ d—l + 5—,
-i
abc
1
1
rn — + 1— r
1
1
...=i+ l+- + -i + -a + -iB+ r lb r— r
a x + a2 + if
&JI
= l + 1x + 2- x + S- x+4-
?**
u.
lv
?'
,
11
,
Un ~ u n-l
+
?<
iX-xJu
1111 .2^ — — — — — ^—^ n+l 1-4-1—4-.
u 2 n-1 n_
a?
Ctry.0
r+1 »~ r + 1-
4.
u t2 -
U.2
CIqCC -iCt-v
r
~~
- UQ +
+
J.
3.
1
Jl
2
ic
+
«
,.
,
?
r3
,J
a3 +
p=4 a+
'
1+
an
,4- 4b+ c+
'
«x +
qc
r,55 + ,
«3 +
« 2 -f
r5
7-3
,
1
— ryr-r +1—
' '
b+ c+ d+
•••
<(„_ H-l
'
P (a+ 1 + Q) = + Q.
shew that
o
12.
Shew that
M2 Ms Mt
9i
—— 9*
tinued fraction
\-
1
/>»
...
is
equal to the
con-
/>»
....
where
denominators of the successive convergents.
#.,
q„,
a*.
...
are
the
,
'
CHAPTER
XXXII.
PROBABILITY. Definition. If an event can happen in a ways and fail in b ways, and each of these ways is equally likely, the probability, 449.
or the chance, of
a+
its
happening rr °
is
z
a+
,
and that
b
of its failing G
is
b
For instance,
if
in a lottery there are 7 prizes
and 25 blanks,
.
the chance that a person holding
and
his chance of not
winning
ticket will
1
win a
prize
is
—7
25 is
—
.
Oa 450. bability
The reason may be made
for the
clear
mathematical definition of pro-
by the following considerations
:
an event can happen in a ways and fail to happen in b ways, and all these ways are equally likely, we can assert that the chance of its happening is to the chance of its failing as a to b. Thus if the chance of its happening is represented by ka, where k is an undetermined constant, then the chance of its failing will be represented by kb. .-. chance of happening + chance of failing = k (a + b) Now the event is certain to happen or to fail therefore the sum of the chances of happening and failing must represent certainty. If therefore we agree to take certainty as our unit, we have If
;
1
.-.
= k (a + v
b), '
or
k
— a+
T b
:
the chance that the event will happen
and the chance that the event
will not
is
happen
a+
b b
is
a+b
the probability of the happening of an event, the probability of its not happening is 1 — p.
Cor.
If
p
is
;
HIGHER ALGEBRA.
374
Instead of saying that the chance of the happening of
451.
it is sometimes stated that the odds are a T a+o in favour of the event or b to a against the event.
an event
is
,
to
b
,
The definition of probability in Art. 449 may be given 452. in a slightly different form which is sometimes Useful. If c is the total number of cases, each being equally likely to occur, and of these a are favourable to the event, then the probability that the event will happen
is
-
,
and the probability that
it
will not
c
happen
is
1
.
Example 1. What is the chance of throwing a number greater than 4 with an ordinary die whose faces are numbered from 1 to 6? There are 6 possible ways in which the die can fall, and of these two are favourable to the event required
=-= -
therefore the required chance
.
Example 2. From a bag containing 4 white and 5 black balls a man draws 3 at random what are the odds against these being all black ? ;
The total number of ways in which 3 balls can be drawn is 9 <73 and the number of ways of drawing 3 black balls is 5 C 3 therefore the chance ,
;
of drawing 3 black balls
~*C% ~ 9.8.7 Thus the odds against the event Example
3.
throw with two
The
possible
=
are 37 to
'
42 5.
Find the chance of throwing
at least
one ace in a single
dice.
number
of cases is 6 x 6, or 36.
on one die may be associated with any of the 6 numbers on the other die, and the remaining 5 numbers on the first die may each be associated with the ace on the second die thus the number of favourable cases
An
ace
;
is 11.
Therefore the required chance
Or we may reason
is
— 3b
as follows
.
:
There are 5 ways in which each die can be thrown so as not to give an That is, the chance ace ; hence 25 throws of the two dice will exclude aces. 25 of not throwing one or more aces is so that the chance of throwing one ; 36
^
ace at least
is 1
-
^ do
,
or
^, oo
5
;
;
;
PROBABILITY. Example
Find the chance
4.
375
throwing more than 15 in one throw with
of
3 dice.
A
throw amounting to 18 must be made up of 6, G, G, and this can occur in 1 way; 17 can be made up of G, G, 5 which can occur in 3 ways; 16 may be made up of G, G, 4 and 6, 5, 5, each of which arrangements can occur in 3 ways.
Thereforo the number of favourable cases 1
And
the total
number
+ 3 + 3 + 3,
is
or
10.
of cases is 6 3 or 21G; ,
therefore the required chance
= =^ 108 21G
A
has 3 shares in a lottery in which there are 3 prizes and share 6 blanks B has 1 in a lottery in which there is 1 prize and 2 blanks shew that A's chance of success is to ZJ's as 1G to 7.
Example
5.
:
;
A may draw he
way
3 prizes in 1
may draw
2 prizes
3
and
2
blank in —^— x 6 ways • m
1
:
JL
he
may draw
1 prize
6 2 blanks in 3 x -r-^r JL m .
and
ways
•
the
sum
win a
of these
prize.
numbers
is 64,
which
-4's
Z»"s
chance
:
B's
chance of success C :
L
= 16
the chance of which
:
20 is —-.
84
therefore A's chance of success = 1
,
is
,
ways
in
which A can
or 84 ways
—r = —-
84
.
21
clearly o
;
O 7.
Or we might have reasoned thus: A ; '
'
of
1
chancer— —
20 ways J
'
chance of success =
1
A 's
number 9.8.7
the
Also he can draw 3 tickets in therefore
therefore
is
or
6.5.4 will get all blanks in
5 — 21
*
'
,
or
;
-— = -—. —1
ZL
Suppose that there are a number of events A, B, C,..., of which one must, and only one can, occur ; also suppose that a, b, c, ... are the numbers of ways respectively in which these events can happen, and that each of these ways is equally likely to occur it is required to find the chance of eacli event. 453.
;
The and
total
of these
number of equally possible ways is a + b + c+ ..., the number favourable to A is a; hence the chance
;
HIGHER ALGEBRA.
376
A
that
will
happen ri
will
happen rr is
is
=
a+
b
+ c+ :
-.
a+
+
b
c
+
.
Similarly the chance that J
B
...
and
so on.
...
From the examples we have given it will be seen that 454. the solution of the easier kinds of questions in Probability requires nothing more than a knowledge of the definition of Probability, and the application
of the laws of
Permutations and Combina-
tions.
EXAMPLES. 1.
XXXII.
a.
In a single throw with two dice find the chances of throwing
(1) five, (2) six.
Prom
a pack of 52 cards two are drawn at random chance that one is a knave and the other a queen. 2.
;
find the
3.
A
4.
If four coins are tossed, find the chance that there should be
bag contains 5 white, 7 black, and 4 red chance that three balls drawn at random are all white.
two heads and two 5.
find the
balls:
tails.
One
of two events must happen given that the chance of the two-thirds that of the other, find the odds in favour of the other. :
one
is
will
from a pack four cards are drawn, find the chance that they be the four honours of the same suit. 6.
If
Thirteen persons take their places at a round table, shew that five to one against two particular persons sitting together.
7. it is
There are three events A, B, C, one of which must, and only 8. one can, happen; the odds are 8 to 3 against A, 5 to 2 against B: find the odds against C.
Compare the chances and 12 with three dice.
9.
dice,
of throwing 4 with one die, 8 with two
In shuffling a pack of cards, four are accidentally dropped the chance that the missing cards should be one from each suit. 10.
;
find
A
has 3 shares in a lottery containing 3 prizes and 9 blanks B has 2 shares in a lottery containing 2 prizes and 6 blanks compare their chances of success. 11.
:
12. Shew that the chances of throwing six with 4, respectively are as 1 6 18, ;
;
3,
or 2 dice
'
PROBABILITY.
377
There are three works, one consisting of 3 volumes, one of 4, and the other of 1 volume. They are placed on a shelf at random prove that the chance that volumes of the same works are all together 13.
;
3 18
140 14.
-1
and
B
throw with two dice
;
if
A
throws
9,
find
i>'s
chance
of throwing a higher number.
The what
15.
a row
:
forming the word Clifton are placed at random in the chance that the two vowels come together ?
letters is
In a hand at whist what 16. held by a specified player ]
a
17. line
:
crowns
is
the chance that the 4 kings are
There are 4 shillings and 3 half-crowns placed at random in shew that the chance of the extreme coins being both halfis -
.
Generalize this result in the case of
m
shillings
and
n half-crowns.
We
455. have hitherto considered only those occurrences which in the language of Probability are called Single events. When two or more of these occur in connection with each other, the joint occurrence is called a Confound event.
For example, suppose we have a bag containing 5 white and 8 black balls, and two drawings, each of three balls, are made from it successively. If we wish to estimate the chance of chawing first 3 white and then 3 black balls, w^e should be dealing with a
compound
event.
In such a case the result of the second drawing might or might not be dependent on the result of the first. If the balls are not replaced after being drawn, then if the first drawing gives 3 white balls, the ratio of the black to the white balls remaining is greater than if the first drawing had not given three white; thus the chance of drawing 3 black balls at the second trial
by the
the
But
the balls are replaced after being drawn, it is clear that the result of the second drawing is not in any way affected by the result of the first. is
affected
We are thus
result of
first.
if
led to the following definition
:
Events are said to be dependent or independent according as the occurrence of one does or does not affect the occurrence of the others. Dependent events are sometimes said to be contingent.
;
;
;
HIGHER ALGEBRA.
378
If there are two independent
456. babilities
of which are known,
to
find
events the respective prothe probability that both will
happen.
event may happen in a ways and fail in b ways, all these cases being equally likely ; and suppose that the second event may happen in a' ways and fail in b' ways, Each of the a + b cases may all these ways being equally likely. be associated with each of the a + b' cases, to form (a + b) (a! + b')
Suppose that the
compound
first
cases all equally likely to occur.
In aa' of these both events happen, in bb' of them both fail, in ab' of them the first happens and the second fails, and in a'b Thus of them the first fails and the second happens. aa (a
is
the chance that both events happen
is
the chance that both events
is
the chance that the
+ b){a'+b') bb'
(a
+ b)(a+b') ab'
(a
first
fail
happens and the second
+ b)(a'+b') fails
a'b is
(a
+ b)(a'+b')
the chance that the
first
and the second
fails
happens.
two independent events are will happen is pp'. Similar the case of any number of independent
Thus if the respective chances p and p\ the chance that both
of
reasoning will apply in Hence it is easy to see that if p lf p 2 p 3 ... are the events. respective chances that a number of independent events will separately happen, the chance that they will all happen is p p 2 p 3 ... ; the chance that the two first will happen and the rest fail is 2\Po (1 — P ) (1 —pj'- > an d similarly for any other par3 ,
,
x
ticular case.
the chance that an event will happen in one trial, the chance that it will happen in any assigned succession of r trials is p ; this follows from the preceding article by supposing 457.
If
p
is
r
'
P =P 2 =P 3 = 1
To
find the chance that
happen we proceed thus
=P-
some one at
least of the events will
the chance that all the events fail is (1 -p (1 -]).,) (1 -p 3 ) -.-j and except in this case some one x) of the events must happen ; hence the required chance is :
=
'
PROBABILITY.
379
Example 1. Two drawings, each of 3 balls, arc made from a bag containing 5 wbitc and 8 black balls, the balls being replaced before tbe second find the chance that the first drawing will give 3 white, and the second trial 3 black balls. :
The number
ways
of
which 3
in
balls
C3
;
5C.,;
3black
*C Z
first trial
compound event =--
.
= •—^-f1.2" 1.2.3
8.7.6 1.2.3
3 black at the second trial
therefore the chance of the
is 13
be drawn
3white
Therefore the chance of 3 white at the
and the chance of
may
13. 12. 11 _ :
=
143
<
throw must give either head or
first
second gives the opposite to the is
the same as the
first is
^
first is
-
,
tail
28 143* •j
>
.
20449
Example 2. In tossing a coin, find the chance of throwing alternately in 3 successive trials. Here the
=
1.2.3
"-
x
14o
143
head and
tail
the chance that the
;
and the chance that the third throw
.
Therefore the chance of the
compound
event =- x = 2 2
=
-j
.
4
Example 3. Supposing that it is 9 to 7 against a person A who is now 35 years of age living till he is 65, and 3 to 2 against a person B now 45 find the chance that one at least of these persons will be living till he is 75 alive 30 years hence. ;
The chance the chance that
that
B
A
will die within
30 years
is
—9
;
3 will die within 30 years is ^
therefore the chance that both will die is
3 9 x ^ lb o
;
',
or
27 —
-
8U
therefore the chance that both will not be dead, that .
..
be alive,
.
,
27
>sl-8o>
is
;
that one at least will
53 or-.
By a slight modification of the meaning of the symbols 458. in Art. 45G, we are enabled to estimate the probability of the concurrence of two dependent events. For suppose that when the first event has happened^ a denotes the number of ways in which the second event can follow, and b' the number of ways in which it will not follow then the number of ways in which the two ;
.
:
;
'
HIGHER ALGEBRA.
380
events can happen together
aa concurrence
is
-.
(«
+
7
,
,
o) (a
aa\ and the probability of their
is
—+ tr
•
,
o
)
the probability of the first event, and p' the contingent probability that the second will follow, the probability of the concurrence of the two events is pp
Thus
if
p
is
Example 1. In a hand at whist find the chance that a specified player holds both the king and queen of trumps. Denote the player by 13
^;
A
The chance
then
12 — ol
:
for the
m * « Therefore the
that,
A
then the chance that
has the king
is clearly
can be dealt in 52 different ways, 13 of which
for this particular card
to A.
;
when he has the
king, he can also hold the queen is
queen can be dealt in 51 ways, 12 of which .
13
12 — x ^ = ol 52 17
= chance required u ,
Or we might reason
n
fall
fall
to A.
1
---
as follows
.
:
The number of ways in which the king and the queen can be dealt to A is equal to the number of permutations of 13 things 2 at a time, or 13 12. And similarly the total number of ways in which the king and queen can be .
dealt is 52
.
51.
Therefore the chance =
13 „
.
*
12 „„
52.51
=— 1
as before.
,
17
Example 2. Two drawings, each of 3 balls, are made from a bag containing 5 white and 8 black balls, the balls not being replaced before the second trial: find the chance that the first drawing will give 3 white and the second 3 black balls. At the first trial, 3 balls may be drawn in and 3 white balls may be drawn in 5 C3 ways;
When
balls
balls
may
.
12
.
11
5
1.2.3
143
;
may be drawn 8 C ways 3
therefore at the second trial 3 balls
and 3 black
13
drawn and removed, the bag contains
3 white balls have been
and 8 black
C3 ways
5.4 1.2"
therefore the chance of 3 white at first trial
2 white
13
be drawn in
in
10
C 3 ways
;
;
therefore the chance of 3 black at the second trial
8.7.6
"1.2.3 therefore the chance of the
.
'
10.9.8 1.2.3
_
~
1_
m
15
compound event 5
143
The student should compare
x
7 =-=
15
.
=
7
429
this solution with that of
Ex.
1,
Art. 457.
;
.. .
PROBABILITY.
381
459. If an event can happen in ttvo or more different ways which are mutually exclusive, the chance that it wilt happen is the sum of the chances of its happening in these different ways.
This is sometimes regarded as a self-evident proposition arising It may, howimmediately out of the definition of probability. ever, be proved as follows :
Suppose the event can happen in two Avays which cannot concur
and
;
let
^
,
be the chances of the happening of the
=*
event in these two ways respectively. Then out of bfi 2 cases there are a b 2 in which the event may happen in the first way, and a b ways in which the event may happen in the second; and tliese ivays cannot concur. Therefore in all, out of b l b 2 cases there are a,b„ + a ,b, cases favourable to the event: hence the chance that the event will happen in one or other of the two x
J
k
ways
is
a b 2 + a2 b x
a,
a_ x
bh 12
x
bf
6,
Similar reasoning will apply whatever be the clusive ways in which the event can happen.
number
of ex-
an event can happen in n ways which are mutually exclusive, and if plt p p^ ---Pn are the probabilities that the a event will happen in these different ways respectively, the probability that it will happen in some one of these ways is
Hence
if
,
Pi+Pl+Pa* Example with two
+Pn-
Find the chance of throwing 9
1.
throw
at least in a single
dice.
9 can be
made up
in 4 ways,
and thus the chance
of throwing 9 is
4 .
,
3
^
10 can be
made up
in 3 ways,
and thus the chance
11 can be
made up
in 2 ways,
and thus the chance of throwing 11 is--
12 can be
made up
in 1 way,
2
and thus the chance of throwing 12
Now the
chance of throwing a number not separate chances .*.
of throwing 10 is
the required chance =
less
ou
than 9
=
n
lb
is
.
the
sum
is
-^
of these
;
:
HIGHER ALGEBRA.
382
and 3 shillings, a second purse contains 2 sovereigns and 4 shillings, and a third contains 3 sovereigns and 1 shilling. If a coin is taken out of one of the purses selected at random, find the chance that it is a sovereign.
Example
One purse contains
2.
Since each purse the
first is
-
;
is
1 sovereign
equally likely to be taken, the chance of selecting
and the chance of then drawing a sovereign
chance of drawing a sovereign so far as - x j , or =^ 12 3 4
.
is
depends upon the
it
-
;
hence the
first
purse
Similarly the chance of drawing a sovereign so far as
depends on the second purse
is
chance of drawing a sovereign
is
12 13 - x -
- x 3 6
1
or 9
,
it
and from the third purse the
;
1
or -
,
the required chance =
.-.
is
— + -y + x*5
-.
=-
tc
o
.
In the preceding article we have seen that the pro460. bability of an event may sometimes be considered as the sum of the probabilities of two or more separate events ; but it is very important to notice that the probability of one or other of a series of events is the sum of the probabilities of the separate events only when the events are mutually exclusive, that is, when the occurrence of one is incompatible with the occurrence of any of the others.
Example.
drawn
at
From
random
:
20 tickets marked with the first 20 numerals, one find the chance that it is a multiple of 3 or of 7.
The chance that the number it is
a multiple of 7 .
,
is
—2
;
and
a multiple of 3
is
is
—
,
and the chance that
these events are mutually exclusive,
—6 + -^2
,
required chance
is
,
2 or -
hence the
.
But if the question had been: find the chance that the number multiple of 3 or of 5, it would have been incorrect to reason as follows Because the chance that the number chance that the number
is
a multiple of 5 6
it is
a multiple of 3 or 5
is
is
4
^+^
1
,
might be a multiple both of 3 and of are not mutually exclusive.
or 5,
.
is
a multiple of 3 is
—4
,
is
—
,
is
a
and the
therefore the chance that
For the number on the
so that the
ticket
two events considered
should be observed that the distinction between simple and compound events is in many cases a purely artificial 461.
It
:
PROBABILITY.
383
one in fact it often amounts to nothing more than a distinction between two different modes of viewing the same occurrence. ;
Example. A bag contains 5 white and 7 black balls; if two drawn what is the chance that one is white and the other black?
balls arc
Regarding the occurrence as a simple event, the chance
(i)
= (5*7H.=C 2 = -
.
66
(ii)
The occurrence may be regarded compound events
as the
happening of one or other
of the two following
drawing a white and then a black
(1)
ball,
°r 132 12 * 11 (2)
drawing a black and then a white 7
i2
And
X
5
ir
0r
the chance of which
is
;
ball,
the chance of which
is
35 132'
since these events are mutually exclusive, the required chance
_35 - j$5 + ^5_ 132~66' 132 be noticed that we have here assumed that the chance of drawing two specified balls successively is the same as if they were drawn simultaneously. A little consideration will shew that this must be the case. It will
EXAMPLES. XXXII. What
b.
the chance of throwing an ace in the successive throws with an ordinary die ? 1.
is
first
only of two
Three cards are drawn at random from an ordinary pack 2. the chance that they will consist of a knave, a queen, and a king.
:
find
The odds against a certain event are 5 to 2, and the odds in 3. favour of another event independent of the former are 6 to 5 find the chance that one at least of the events will happen. ;
The odds against A solving a certain problem are 4 to 3, and 4. the odds in favour of B solving the same problem are 7 to 5 what is the chance that the problem will be solved if they both try 1 :
What
the chance of drawing a sovereign from a purse one compartment of which contains 3 shillings and 2 sovereigns, and the other 2 sovereigns and 1 shilling ? 5.
A
is
bag contains 17 counters marked with the numbers 1 to 17. 6. counter is drawn and replaced; a second drawing is then made: what is the chance that the first number drawn is even and the second
A
odd?
HIGHER ALGEBRA.
384
Four persons draw each a card from an ordinary pack: find the chance (1) that a card is of each suit, (2) that no two cards are of 7.
equal value. 8.
Find the chance of throwing
six
with a single die at least once
in five trials.
be favourably reviewed by three independent critics are 5 to 2, 4 to 3, and 3 to 4 respectively what is the probability that of the three reviews a majority will be favourable ? 9.
The odds that a book
will
;
A
bag contains 5 white and 3 black balls, and 4 are successively drawn out and not replaced what is the chance that they are alternately 10.
;
of different colours 11.
%
In three throws with a pair of dice, find the chance of throwing
doublets at least once. 12.
If 4 whole
numbers taken at random are multiplied together
shew that the chance that the .
1S
last digit in the
product
is 1, 3, 7,
or 9
16
625'
In a purse are 10 coins, all shillings except one which is a 13. Nine coins are taken sovereign in another are ten coins all shillings. from the former purse and put into the latter, and then nine coins are taken from the latter and put into the former find the chance that the sovereign is still in the first purse. ;
:
two coins are tossed 5 times, what be 5 heads and 5 tails
14.
will
If
is
the chance that there
\
If 8 coins are tossed, 15. one will turn up head?
what
is
the chance that one and only
C in
order cut a pack of cards, replacing them after each cut, on condition that the first who cuts a spade shall win a prize find their respective chances. 16.
A, B,
:
A and B draw from a purse containing 3 sovereigns and 17. 4 shillings find their respective chances of first drawing a sovereign, the coins when drawn not being replaced. :
A party
of n ^persons sit at a round table, find the odds against two specified individuals sitting next to each other. 18.
A
one of 6 horses entered for a race, and is to be ridden by one of two jockeys B and C. It is 2 to 1 that B rides A, in which case all the horses are equally likely to win if C rides A, his chance what are the odds against his winning ? is trebled 19.
is
;
:
If on an average 1 vessel in every 10 is wrecked, find the chance that out of 5 vessels expected 4 at least will arrive safely. 20.
.
PROBABILITY.
385
The probability of the happening of an event in one being known, required the probability of its happening once,
462. trial
twice, three times,
exactly in
...
n
trials.
Let p be the probability of the happening of the event in a single trial, and let q = 1 -p\ then the probability that the event will happen exactly r times in n trials is the (r + l) th term in the expansion of (q + p)*.
For
we
any particular
set of r trials out of the total number n, the chance that the event will happen in every one of r q"~ [Art. 456], and as these r trials and fail in all the rest is if
a set of r
select
can be selected in
n
p
Cr
ways, all of which are equally applicable to the case in point, the required chance is trials
C rp q
we expand
If
2f
+ "C 2) n
-
1
(/;
1
q
+
q)"
.
by the Binomial Theorem, we have
+ n C jS'- 2 q 2 +
...
+"C n _ p
r
q"- r
r
+
...
+ qn
;
thus the terms of this series will represent respectively the probabilities of the happening of the event exactly n times, n — 1 times,
n—
463.
2 times,
If
inn
...
trials.
the event happens
n
or
times,
(n — r) times, it happens r times or more ; chance that it happens at least r times in n trials is
twice,
...
n n P + "Cy-*q + "C aP -Y+
or
the
sum
of
the
first
n—r+
1
...
tv^r.
terms of the expansion of
Example 1. In four throws with a pair of throwing doublets twice at least ?
dice,
c
In a single throw the chance of doublets failing to
throw doublets
is
5 ^
.
Now
the
sum
-^ do
is
H. H. A.
is
the chanco of
i ,
or ^ o
;
;
and
the chance of if
doublets
therefore the required chanco
of the first three terms of the expansion of
Thus the chance =
what
the required event follows
are thrown four times, three times, or twice is
only once, therefore the
fails
/l 5\ h+d
4 .
19 1 — (1 + 4.5 + 6.5-)= -^
25
:
;
HIGHER ALGEBRA.
386
A
bag contains a certain number of balls, some of which are replaced, another is then drawn and replaced; chance of drawing a white ball in a single trial, find is the on if and so p the number of white balls that is most likely to have been drawn in n trials.
Example
2.
white; a ball
is
drawn and
:
of drawing exactly r white balls is n C rp r q n value of r this expression is greatest.
The chance what
find for
nC
Now
p
r
r
q
n- r
> n Cr - l p r - l q n - (r ~
(n +
or
n
and we have
to
\
l)p>(p + q)r.
But p + 5 = 1; hence the required value of p[n + l). If
,
(n-r + l)p>rq,
so long as
is
successes
l
-r
such that pn
and qn
is
an
integer,
r is
the greatest integer in
the most likely case
is
that of
pn
failures.
Suppose that there are n tickets in a lottery for a prize of £x; then since each ticket is equally likely to win the prize, and a person who possessed all the tickets must win, the money value of 464.
pay
x
£-
words
this
for each ticket; hence a person
who
each ticket
is
n
:
in other
£— TX
reasonably expect
would be a
fair
sum
possessed r tickets might
as the price to be paid for his tickets
any one who wished to buy them; that
is,
£- x
convenient then to
n
as the worth of his chance.
troduce the following definition
It
to
is
by
he would estimate in-
:
represents a person's chance of success in any venture and the sum of money which he will receive in case of success, the sum of money denoted by is called his expectation. If
p
M
pM
In the same way that expectation is used in reference 465. to a person, we may conveniently use the phrase probable value applied to things. Example 1. One purse contains 5 shillings and 1 sovereign a second purse contains 6 shillings. Two coins are taken from the first and placed in the second then 2 are taken from the second and placed in the first find the probable value of the contents of each purse. :
;
The chance that the sovereign is in the first purse is equal to the the chances that it has moved twice and that it has not moved at all
sum
of
8 PROBABILITY. that
is,
the chance
=
112 - + 5 .1=-t. 8
-
.
6 .*.
the chance that the sovereign
Hence the probahle value
is
of 25*.
4
.*.
4
4
3
.
in the second purse =-r.
of the first purse
3
=T
,387
+
1 .
4
of 6*.=£1.
O.s-.
3r/.
the probable value of the second purse
=31*.-2Q£«.=10*.
Or the problem may be The probable value
<>
solved as follows
of the coins
=s
:
removed
of 25s.
= 8^s.;
the probable value of the coins brought back
=^of
W,
(Gs.+S V>\)=3 r :
.\ the probable value of the first purse
= (25-81 + 3^)
shillings
= £1.
Example 2. A and B throw with one be won by the player who first throws 6.
Ck
3d., as before.
die for a stake of If
A has
the
first
£11 which is to throw, what are
their respective expectations?
In his
first
throw A' a chance
1
is
b
;
in his second
5
it is
5
1
x - x o 6
-^
,
because
each player must have failed once before A can have a second throw in his 1 /5\ 4 third throw his chance is x ^ because each player must have failed ;
(
J
twice; and so on.
Thus A's chance
is
the
sum
of the infinite series
5MKi© Similarly #'s chance
is
the
sum
A' a chance
--
and =y, and
is
}•
of the infinite series
WM» .-.
4+
4
,+
G)
* J-
to 7>"s as G is to 5; their respective chances are therefore their expectations are
£6 and £5
respectively.
26—
HIGHER ALGEBRA.
388
We
shall now give 466. and interesting results.
two problems which lead
to useful
Example 1. Two players A and B want respectively m and n points of winning a set of games their chances of winning a single game are p and q respectively, where the sum of p and q is unity the stake is to belong to the player who first makes up his set determine the probabilities in favour ;
;
:
of each player.
Suppose that A wins in exactly m + r games; to do this he must win the The chance of last game and m-1 out of the preceding m + r-1 games. ^~ ™+^1 r 1 m 1 m r p™Cm-1 p q . this is q 2h or m_ 1 the set will necessarily be decided in m + n - 1 games, and A may 1 games, ... or + n - 1 games; win his m games in exactly m games, or therefore we shall obtain the chance that A wins the set by giving to r the Thus A s chance is values 0, 1, 2, ... n - 1 in the expression m+r 1 Cm _ 1 p m q r
Now
m+
,
m
J
.
similarly B's chance is
1.2
+ n-2 jm-1 \m
n(nA-l\ *
)
ii
This question is known as the " Problem of Points," and has engaged the attention of many of the most eminent mathematicians since the time of Pascal. It was originally proposed to Pascal by the Chevalier de Mere in 1654, and was discussed by Pascal and Fermat, but they confined themselves to the case in which the players were supposed to be of equal skill their results were also The formulae we have given are exhibited in a different form. assigned to Montmort, as they appear for the first time in a work The same result was afterwards obof his published in 1714. tained in different ways by Lagrange and Laplace, and by the latter the problem was treated very fully under various modi:
fications.
Example 2. There are n dice with / faces marked from 1 to /; if these are thrown at random, what is the chance that the sum of the numbers exhibited shall be equal to p? the
number
may
be exposed on any one of the n dice, of ways in which the dice may fall is / n
Since any one of the
/
faces
.
Also the number of ways in which the numbers thrown will have sum is equal to the coefficient of x p in the expansion of
p
for
their
{x l
+ x* + x 3 + ... + xf
n )
\
for this coefficient arises out of the different ways in 1, 2, 3, .../can be taken so as to form p by addition.
which n of the indices
PROBABILITY.
Now
=x
the above expression
11
(l
+ x + x2 +
3-S!)
...
+ xf
')"
-(£?)" We
have therefore to find the (I
and
<1
-
v
.r)
« , -" =1
x') n (I
n(n-l)
t
.
-
*
- x)~ n
„
' a;
2
.
n(n-l)(n-2)
.,.
»(n+l)
+ nx +
x p ~ n in the expansion of
coefficient of
+
.,,
—
?t(»+l)(w + 2) _3 x +... 1 ^ 3 .
~ Multiply these series together and pick out the coefficient of x p n in the product we thus obtain ;
n(n+l)...{p-l) \
P -n
it
n(n+l)...(p-f-l) p - n -f \
+
n
(n
-
M(;t
1)
+ l)...(j>-2/-l)
•"
1.2
\
p-n-2f
where the series is to continue so long as no negative factors appear. required probability is obtained by dividing this series by/ n
The
.
This problem in 1730
j
is
due to
it illustrates
De Moivre and was
a method of frequent
published by him
utility.
Laplace afterwards obtained the same formula, but in a much more laborious manner he applied it in an attempt to demonstrate the existence of a primitive cause which has made the planets to move in orbits close to the ecliptic, and in the same direction as the earth round the sun. On this point the reader may consult Todhunter's History of Probability, Art. 987. ;
EXAMPLES. XXXII. In a certain game A'a skill is to winning 3 games at least out of 5.
1.
2>'s
of
.1
is
coin whose faces are marked 2. the chance of obtaining a total of 12 ?
A
2,
c.
as 3 to 2
3
is
:
find the chance
thrown 5 times
:
what
In each of a set of games it is 2 to 1 in favour of the winner 3. the of previous game what is the chance that the player who wins the first game shall win three at least of the next four ? :
There are 9 coins in a bag, 5 of which are sovereigns and 4. the rest are unknown coins of equal value find what they must be if the probable value of a draw is 12 shillings. ;
HIGHER ALGEBRA.
390
A coin
5.
present itself
is
n times, what an odd number of times ? tossed
is
is
the chance that the head will
From a bag containing 2 sovereigns and 3 shillings a person 6. allowed to draw 2 coins indiscriminately; find the value of his ex-
pectation.
Six persons throw for a stake, which is to be won by the one who first throws head with a penny ; if they throw in succession, find the chance of the fourth person. 7.
8.
Counters marked
9.
A coin whose faces are marked 3 and
3 are placed in a bag, and one is withdrawn and replaced. The operation being repeated three times, what is the chance of obtaining a total of 6 ?
are the odds against the
1, 2,
sum
5
is
tossed 4 times
numbers thrown being
of the
less
:
what
than 15?
Find the chance of throwing 10 exactly in one throw with
10.
3 dice. players of equal skill, A and B, are playing a set of games they leave off playing when A wants 3 points and wants 2. If the stake is £16, what share ought each to take \
Two
11.
B
;
12.
A
and
B throw with
of throwing a higher
number
3 dice
:
if
A
throws
8,
what
is Z?'s
chance
?
A had in his pocket a sovereign and four shillings taking out 13. two coins at random he promises to give them to B and C. What is the worth of (7's expectation ? ;
14. (1)
In
five
throws with a single die what
is
the chance of throwing
three aces exactly, (2) three aces at least.
A
B of 5s. to 2s. that in a single throw with two dice he .will throw seven before B throws four. Each has a pair of dice and they throw simultaneously until one of them wins find B's 15.
makes a bet with
:
expectation.
A person
one the common cube, and the other a regular tetrahedron, the number on the lowest face being taken in the case of the tetrahedron; what is the chance that the sum of the numbers thrown is not less than 5 ? 16.
throws two
dice,
A
bag contains a coin of value J/, and a number of other coins whose aggregate value is m. A person draws one at a time till he draws the coin 31 find the value of his expectation. 17.
:
6n tickets numbered 0, 1, 2, 6n- 1 are placed in a bag, and three are drawn out, shew that the chance that the sum of the numbers on them is equal to 6?i is 18.
If
3?&
(6n-l)(6n-2)'
;
PROBABILITY.
3Dl
*Inverse Probability. all the cases we have hitherto considered it lias been supposed that our knowledge of the causes which may produce a certain event is sucli as to enable us to determine the chance of the happening of the event. have now to consider problems of a different character. For example, if it is known that an event has happened in consequence of some one of a certain number of causes, it may be required to estimate the probability of each cause being the true one, and thence to deduce the probability of future events occurring under the operation of the
*467.
In
We
same
causes.
*468.
we
Before discussing the general case
shall
give a
numerical illustration.
Suppose there are two purses, one containing 5 sovereigns and 3 shillings, the other containing 3 sovereigns and 1 shilling, and suppose that a sovereign lias been drawn it is required to find the chance that it came from the first or second purse. :
Consider a very large number iV of trials ; then, since before the event eacli of the purses is equally likely to be taken, we may
assume that the
and
first
purse would be chosen in ^ iV of the
5 in - of these a sovereign
would be drawn
;
8
would be drawn
5 -
o
1
x ~iV, or 1
5 times from the —N lb
The second purse would be chosen 3
j
of these a sovereign
in -
would be drawn
;
N
trials,
thus a sovereign ° first
purse.
of the trials,
and
in
thus a sovereign would
3
be drawn -JV times from the second purse.
Now
very large but is otherwise an arbitrary number let us put iV-16n; thus a sovereign would be drawn 5 n times from the first purse, and Qn times from the second purse; that is, out of the lln times in which a sovereign is drawn it comes from the first purse bn times, and from the second purse 0?i JV
is
:
HIGHER ALGEBRA.
392 times.
Hence the
— 5
first
purse
is
,
probability that the sovereign
and the probability that
it
came from the
came from the
6
A'is rr. second *469. It is important that the student's attention should be directed to the nature of the assumption that has been made in Thus, to take a particular instance, the preceding article. although in 60 throws with a perfectly symmetrical die it may not happen that ace is thrown exactly 10 times, yet it will doubtless be at once admitted that if the number of throws is continually increased the ratio of the number of aces to the number of throws will tend more and more nearly to the limit
—
.
There
no reason
is
why one
face should appear oftener than
6
hence in the long run the number of times that each of the six faces will have appeared will be approximately equal. another
;
The above instance is a particular case which is due to James Bernoulli, and was
theorem given in the Ars
of a general first
Conjectandi, published in 1713, eight years after the author's Bernoulli's theorem may be enunciated as follows death. event happens in a single trial, then if the number of trials is indefinitely increased, it becomes a certainty that the limit of the ratio of the number of successes to the number of trials is equal to p ; in other words, if the number of trials is
an
is the
probability that
N,
number of successes may
If p
the
be taken to be
pN.
A
proof See Todhunter's History of Probability, Chapter vn. of Bernoulli's theorem is given in the article Probability in the Encyclopaedia Britannica.
An
observed event has Jiappened through, some one of a number of mutually exclusive causes : required to find the probability of any assigned cause being the true one.
*470.
.
Let there be n causes, and before
suppose that the probability of the existence of these causes was estimated Let p r denote the probability that when the at P P2 P3 ... Pn r* cause exists the event will follow after the event has occurred th cause was the it is required to find the probability that the r true one. x ,
1*
,
,
the event took place
.
:
'
393
PROBABILITY.
Consider a \ cry great number JV of trials then the first cause of these, and out of this number the event follows exists in P similarly there are p^^N trials in which the event in p Px j follows from the second cause; and so on for each of the other causes. Hence the number of trials in which the event follows is ;
N
x
X
N
;md the number in which the event was due to the r th cause 'P,.I\N
',
lience after the event the probability that the r
was the true one
th
is
cause
is
pJPjr+NUpP); tli at
cause
is,
the probability that the event was produced by the
r"'
PrK
is
Mvn
*471.
necessary to distinguish clearly between the probability of the existence of the several causes estimated before the event, and the probability after the event has happened of any The former are usually called assigned cause being the true one. P a priori probabilities and are represented by ... n \ probabilities, and if we denote the latter are called a posteriori them by Q t1 Q„, Q 3 ... Q Hf we have proved that It
is
P
,
x
,
P
P
.
,
Qr
2 ( P P)
'
where p r denotes the probability of the event on the hypothesis th of the existence of the r cause.
From
=
which is otherwise evident as the event has happened from one and only one this result it appears that
S
(Q)
lj
of the causes.
We
now
give another proof of the theorem of the preceding article which does not depend on the principle enunciated in Art. 469. shall
An
observed event has happened through some one of a member of mutually exclusive causes : required to find the probability of any assigned cause being the true one.
*472.
Let there be n causes, and before
suppose that the probability of the existence of these causes was estimated at P P2 Pz , ... P n Let p r denote the probability that when the ?- th cause exists the event will follow ; then the antecedent probability that the event would follow from the r th cause is p r P r t
,
,
the event took place
.
.
;
HIGHER ALGEBRA.
394
th
be the a posteriori probability that the r cause was the th true one; then the probability that the r cause was the true one
Let
Q
r
proportional to the probability that, would produce the event
is
A._JL =
.
"
'"
pA pA
_
.
if
in existence, this cause
s(
Hp p
p. p ,
i
_
.
s(P P)'
)
Pr pr
2 (pP)
appears that in the present class of problems the product Prp r will have to be correctly estimated as a first step; in many cases, however, it will be found that P lt P 2 P3 ... are all equal, and the work is thereby much simplified.
Hence
it
,
,
,
Example. There are 3 bags each containing 5 white balls and 2 black and 2 bags each containing 1 white ball and 4 black balls a black ball having been drawn, find the chance that it came from the first group. balls,
:
Of the
If
five bags, 3
a bag 2
ball is 7
;
if
is
belong to the
selected
from the
first
first
group and 2 to the second
group the chance of drawing a black
from the second group the chance
" Hence the chance
lh
p 1_
6
3o'
hence
;
lh
that the black ball
2
is
4 -
2
;
thus
o
p x = -/
,
p.2
=
4 o
;
^
~~25'
came from one
of the first
group
is
8\15 JL^/A \35"h 35
25/
'
43*
When
an event has been observed, we are able by the method of Art. 472 to estimate the probability of any particular cause being the true one ; we may then estimate the probability of the event happening in a second trial, or we may find the probability of the occurrence of some other *473.
event.
For example, pr is the chance that the event will happen from the rth cause if in existence, and the chance that the r th cause is the true one is Qr hence on a second trial the chance Therefore that the event will happen from the rth cause is p r Qr the chance that the event will happen from some one of the causes on a second trial is 2 (2} Q)' ;
.
PROBABILITY.
395
A
purse contains 4 coins which arc either sovereigns or drawn and found to be shillings: if these are replaced the chance that another drawing will give a sovereign?
Example.
shillings; 2 coins are
what
is
This question
may
be interpreted in two ways, which we shall discuss
separately. If we consider that all numbers of shillings are a priori equally likely, shall have three hypotheses; for (i) all the coins may be shillings, (ii) three of them may be shillings, (iii) only two of them may be shillings. I.
we
P^P.^P.^;
Here
^=
also
Hence probability of
iirst
1,
J>a
=g, P»=q-
hypothesis
= 1-5- (1 + o
probability of second hypothesis =
o
probability of third hypothesis — -
+ r) —
^ (*+2 -f-
(
1
+
+~ +
f )
.-
)
tTv^ Qi>
=
To
~ ^'
J'
=T7\=Qy
Therefore the probability that another drawing will give a sovereign
II.
If
each coin
is
1
3_
~4
*10
/l -
lb
-
- 18*
,
we
see that
4 ,
of
two shillings
10
also, as before,
iJ i
P
2
= l>
6
'
12
'
6
chance of four
-± ~1G'
Qi +
is
—6
-.
10
Pi— a*
Qi_Q-2_Q*_
the
J
of three shillings is 77:
P_l il_ 16'
Hence
IV
+
. ,
'To~40
4
I
1 r-^
5^
1_
equally likely to be a shilling or a sovereign, by taking
the terms in the expansion of
shillings is
2
+
;
thus
_A. P ^»-16' Ps—r'
Q2+Q* 24
_
1 24'
Therefore the probability that another drawing will give a sovereign
= (<2ix0)+(q,x^ + ~
8
+
16
4
((?
:j
x|)
HIGHER ALGEBRA.
396
We
shall now shew how the theory of probability may *474. be applied to estimate the truth of statements attested by wit-
We
shall nesses whose credibility is assumed to be known. be truth, the suppose that each witness states what he believes to observation, deduction, or whether his statement is the result of or experiment; so that any mistake or falsehood must be attributed to errors of judgment and not to wilful deceit.
problems we shall discuss furnishes a useful intellectual exercise, and although the results cannot be regarded as of any practical importance, it will be found that they confirm
The
class
the verdict of
of
common
*475. When speaks the truth
sense.
asserted that the probability that a person p, it is meant that a large number of statehas been examined, and that p is the ratio
it is is
ments made by him of those which are true to the whole number.
independent witnesses, A and B, whose probabilities of speaking the truth are p and p' respectively, agree in making a certain statement what is the probability that the statement is true % *476.
Two
:
Here the observed event is the fact that A and B make the same statement. Before the event there are four hypotheses for A and B may both speak truly or A may speak truly, B falsely; or A may speak falsely, B truly or A and B may both speak The probabilities of these four hypotheses are falsely. ;
;
;
PP\
Hence
p( l ~P\
P'Q-P)*
(
1
-P)( 1 ~P')
after the observed event, in
which
respectively.
A and B make
the
same statement, the probability that the statement is true is to that the probability that it is false as pp to (1 - p) (1 -p') ;
is,
the probability that the joint statement
is
true
is
pp'
pp' + (l-p)(l-p')' Similarly if a third person, whose probability of speaking the truth is p", makes the same statement, the probability that the statement is true is ./„//
ppp
ppY + {1-p){i-p')(i-p") and so on
for
any number
of persons.
}
PROBABILITY.
397
In the preceding article it lias been supposed that we have no knowledge of the event except the statement made by A and B if we have information from other sources as to the probability of the truth or falsity of the statement, this must be taken into account in estimating the probability of the various *477. ;
hypotheses.
For instance, if A and B agree in stating a fact, of which the a priori probability is P, then we should estimate the probability of the truth and falsity of the statement by Ppp* and (1 - P) (1 — p>) (1 — p') respectively. Example. There is a raffle with 12 tickets and two prizes of £9 and £3. A, B, C, whose probabilities of speaking the truth are ^, §, f respectively, A and B assert that he has report the result to D, who holds one ticket. won the £9 prize, and C asserts that he has won the £3 prize; what is D's expectation? Three cases are possible; D may have won C may all have spoken falsely. Now with the notation of Art. 472, we have
£9, £3, or nothing, for A, B, the a priori probabilities
P-i P-A *i- 12 ^ ~12' *a-l2' P-™. 3
«
also
12 24 X X
Pi~2
3
5-30>
1133 ~
**~~2 X 3 X 5
" hence D's expectation
4
3
1
30
20
=—4 of £9 + —3 of £3
'
1
A_ 2 * 3
X
2_ 2 5~3()
;
27'
=£1.
13s. id.
*478. With respect to the results proved in Art. 47G, it should be noticed that it was assumed that the statement can be made in two ways only, so that if all the witnesses tell falsehoods they agree in telling the same falsehood.
not the case, let us suppose that c is the chance that the two witnesses A and B will agree in telling the same falsehood then the probability that the statement is true is to the probability that it is false as pp' to c (1 —p) (1 — p'). If this
is
;
As
general rule, it is extremely improbable that two independent witnesses will tell the same falsehood, so that c is usually very small; also it is obvious that the quantity c becomes These consmaller as the number of witnesses becomes greater. siderations increase the probability that a statement asserted by two or more independent witnesses is true, even though the credibility of each witness is small.
a
.
HIGHER ALGEBRA.
398
Example. A speaks truth 3 times out of 4, and B 7 times out of 10; they both assert that a white ball has been drawn from a bag containing 6 balls all of different
colours
:
find the probability of the truth of the assertion.
There are two hypotheses
their coincident testimony is true,
(i)
;
(ii) it
is
false.
PP 2 ~6' 5
Px =•16' ~
Here
,
^ 1- 4 X
P2 ~25 X
10'
•
4
X
10
;
B
will
P po,
and
for in estimating p.2 we must take into account the chance that A and both select the white ball when it has not been drawn ; this chance is
11 X 5
Now
5
1
°r
*
25
the probabilities of the two hypotheses are as
P^
to
therefore as 35 to 1; thus the probability that the statement is true
2
—
35
is
*479. The cases we have considered relate to the probability of the truth of concurrent testimony; the following is a case of traditionary testimony.
A
states that a certain event took place, having received an account of its occurrence or non-occurrence from B, what is the If
probability that the event did take place
The event happened
1
they both spoke the truth, (2) if and the event did not happen if only
(1) if
they both spoke falsely ; one of them spoke the truth.
Let p, p denote the probabilities that A and B speak the truth ; then the probability that the event did take place is pp' +
and the probability that
it
(l-p)(l-p)
}
did not take place
is
p(l- 2))+p'(l-p). *480. The solution of the preceding article is that which has usually been given in text-books; but it is" open to serious objections, for the assertion that the given event happened if both A and B spoke falsely is not correct except on the supposition that Moreover, the statement can be made only in two ways. although it is expressly stated that A receives his account from B, this cannot generally be taken for granted as it rests on A'& testimony.
:;
PROBABILITY.
A
399
the different ways of interpreting the question, and of the different solutions to which they lead, will be found in the Educational Times Reprint, Yols. XXVII. and XXXII. full discussion of
^EXAMPLES. XXXII.
d.
There are four balls in a bag, but it is not known of what 1. colours they are one ball is drawn and found to be white find the chance that all the balls are white. ;
:
In a bag there are six balls of unknown colours; three balls are drawn and found to be black; find the chance that no black ball 2.
is left in
the bag.
A letter is known to have come either from London or Clifton 3. on the postmark only the two consecutive letters ON are legible what is the chance that it came from London ? ;
Before a race the chances of three runners, A, B, C, were estimated to be proportional to 5, 3, 2 but during the race A meets with an accident which reduces his chance to one-third. What are now the respective chances of B and C ? 4.
;
A
purse contains n coins of unknown value a coin drawn at random is found to be a sovereign; what is the chance that it is the only sovereign in the bag ? 5.
;
A man
has 10 shillings and one of them is fcnown to have two heads. He takes one at random and tosses it 5 times and it always what is the chance that it is the shilling with two heads ? falls head 6.
:
A
bag contains 5 balls of unknown colour; a ball is drawn and replaced twice, and in each case is found to be red if two balls are now drawn simultaneously find the chance that both are red. 7.
:
A
purse contains five coins, each of which may be a shilling 8. or a sixpence two are drawn and found to be shillings find the probable value of the remaining coins. ;
9.
thrown 10.
A die is
A
15
:
:
thrown three times, and the sum of the three numbers find the chance that the first throw was a four.
is
speaks the truth 3 out of 4 times, and B 5 out of 6 times will contradict each other in .stating
what is the probability that they the same fact ?
;:
HIGHER ALGEBRA.
400
speaks the truth 2 out of 3 times, and B 4 times out of 5 they agree in the assertion that from a bag containing 6 balls of different colours a red ball has been drawn find the probability that the state11.
A
:
ment
is true.
One of a pack of 52 cards has been lost ; from the remainder 12. of the pack two cards are drawn and are found to be spades ; find the chance that the missing card is a spade. with 10 tickets and two prizes of value £5 holds one ticket and is informed by that he has won the £b prize, while C asserts that he has won the ,£1 prize is denoted by §, and what is A's expectation, if the credibility of that of C by f ? 13.
and £1
There
is
a
respectively.
raffle
B
A
B
A
purse contains four coins two coins having been drawn are found to be sovereigns find the chance (1) that all the coins are sovereigns, (2) that if the coins are replaced another drawing will give a sovereign. 14.
;
:
P
makes a bet with Q of ,£8 to £120 that three races will be the three horses A, B, C, against which the betting is 3 to 2, 4 to 1, and 2 to 1 respectively. The first race having been won by A, and it being known that the second race was won either by B, or by a horse against which the betting was 2 to 1, find the value of P's 15.
won by
D
expectation.
From
a bag containing n balls, all either white or black, all numbers of each being equally likely, a ball is drawn which turns out to be white; this is replaced, and another ball is drawn, which also turns out to be white. If this ball is replaced, prove that the chance 16.
of the next
draw giving a black
ball is - (n
— 1)
(2n + l)~ l
.
m
mn
purses, n into each, coins have been distributed into will found in the same coins be find (1) the chance that two specified 17.
If
(2) what the chance becomes when r purses have been and found not to contain either of the specified coins. examined
purse; and
A, B are two inaccurate arithmeticians whose chance of solving a given question correctly are -£ and y1^ respectively if they obtain the same result, and if it is 1000 to 1 against their making the same mistake, find the chance that the result is correct. 18.
;
witnesses, each of whom makes but one false statement in six, agree in asserting that a certain event took place ; shew that the odds are five to one in favour of the truth of their statement, even 19.
Ten
although the a
'priori probability of the
event
is
—r
as small as ^9
•
.
1 :
.
PRoUAl'.ILITY.
The application
1
Geometrical Methods.
Local Probability. *481.
41
Geometry
to
questions of Probability requires, in general, the aid of the Integral Calculus; there are, however, many easy questions which can be solved by of
Elementary Geometry. Example 1. From each of two equal lines of length I a portion is cut random, and removed what is the chance that the sum of the
off at
:
remainders
is less
than
I?
Place the lines parallel to one another, and suppose that after cutting, the right-hand portions are removed. Then the question is equivalent to asking what is the chance that the sum of the right-hand portions is greater than the sum of the left-hand portions. It is clear that the first sum is equally likely to be greater or less than the second; thus the required probability
is
a
Cor.
Each
of two lines
chance that their sum
is
is
known
to be of length not exceeding
not greater than
Z
is
I:
the
a
Example 2. If three lines are chosen at random, prove that they are just as likely as not to denote the sides of a possible triangle. Of three
one must be equal to or greater than each of the other two denote its length by I. Then all we know of the other two lines is that the length of each lies between and /. But if each of two lines is known to be of random length between and 1, it is an even chance that their sum is greater than /. [Ex. 1, Cor.] lines
;
Thus the required
result follows.
Example 3. Three tangents are drawn at random to a given circle shew that the odds are 3 to 1 against the circle being inscribed in the triangle formed by them.
P O
Draw draw
three random lines P, (), 11, in the same plane as the circle, to the circle the six tangents parallel to these lines.
H. H.A.
2G
and
.
HIGHER ALGEBRA.
402
of the 8 triangles so formed it is evident that the circle will be escribed to 6 and inscribed in 2 and as this is true whatever be the original directions of P, Q, R, the required result follows.
Then
;
Questions in Probability may sometimes veniently solved by the aid of co-ordinate Geometry.
be con-
*4:82.
a rod of length a + b+c, lengths a, b are measured at find the probability that no point of the measured lines will
On
Example.
random: coincide.
Let AB be the line, and suppose AP = x and PQ = a; also let a be measured from P towards B, so that x must be less than b + c. Again let AP' = y, P'Q' = b, and suppose P'Q' measured from P' towards B, then y must be less than a + c. Now in favourable cases we must have AP'>AQ, or else AP>AQ\
y>a + x,
hence Again for
all
the cases possible,
or
x>b + y
we must have
x>0, and 2/>0, and
<& + c)
Take a pair of rectangular axes and equal to a + c.
Draw
the line y
= a + x,
(1).
represented by
TML
equal to
b
in the figure;
+ c, and
OY
and the
line
x = b + y represented by KB.
n
Q
W
P'
•A
Then YM,
B
f
EX are each equal to c,
031,
-0.—Q
OT are
The
b
-
each equal to
K
X
a.
MYL
conditions (1) are only satisfied by points in the triangles and ItXR, while the conditions (2) are satisfied by any points within the rectangle OX, OY; .*.
=
the required chance
c2
{a + c)(b
*483.
We
—
+ c)
shall close this chapter
with some Miscellaneous
Examples. Example 1. thrown
balls are
A
box
at
random
is
m
divided into equal compartments into which n find the probability that there will be p com;
partments each containing a balls, q compartments each containing r compartments each containing c balls, and so on, where
Z>a+qb + rc +
=n.
b balls,
PROBABILITY.
403
Since each of the n halls can fall into any one of the m compartments n and these are the total number of cases which can occur is all equally , likely. To determine the number of favourable cases we must find the number of ways in which the n balls can be divided into p, <1, r, ... parcels containing a, b, c, ... balls respectively.
m
First choose
the
number
of
any
g
of the compartments,
ways in which
where
done
this can be
is
s
stands for \m
p + q + r + ...
—\m-s ——
(1).
-.
-v
\s
Next subdivide the s compartments into groups containing p q, severally; by Art. 147, the number of ways in which this can be done is t
r
\»\1
;
'
r, ...
(2). ..
Lastly, distribute the n balls into the compartments, putting a into each of the group of p, then b into each of the group of q, c into each of the group of r, and so on. The number of ways in which this can be done is In
-
(\a)*(\b)«(\c_)
(3).
Hence the number
of ways in which the balls can be arranged to satisfy the required conditions is given by the product of the expressions (1), (2), (3). Therefore the required probability is
\m
m"
(\a)>>
(\b)i ([£)-•
t
m-p-q-r-
|p|j|r. \
Example 2. A bag contains n balls and the ball on each occasion is found
;
made in succession, find the chance that the the balls are replaced after
k drawings are
to be white
next drawing will give a white ball'; (i) when each drawing (ii) when they are not replaced.
:
;
Before the observed event there are n + 1 hypotheses, equally likely; bag may contain 0, 1, 2, 3, ... n white balls. Hence following the notation of Art. 471, (i)
for the
= -Pj — P-2 = P 3 =
1
Hence
.
.
.
= Pn
;
after the observed event, 7-*
Qr = l»+2*+ 3*+. ..+n*
Now
the chance that the next xt drawing will give giv< a white ball
thus the required chance
= n
=2 - Qr
p + 2* + 3* + ...+n*
and the value of numerator and denominator may be found by
Art. 405.
26—2
\
—
1
'
HIGHER ALGEBRA.
404
In the particular case when k = 2,
—
=- <—*-=
the required chance
^
-4-
S
3 (n + 1)
~2(2n+l)' n
If
indefinitely large, the chance
is
V^ '
n
k
+2 fe
and thus the chance
is
r
'
fc
+ 1.'
+1
r
-
r- 2
1
n n— 1 n- 2 '
"
r "
(r-k + l)(r-k + 2)
p,
Q r =i it
is in-
not replaced,
r
and
when n
&+2*
If the halls are
(ii)
equal to the limit,
«*+!
2
1
finite, of
is
'
-
A;
+
»— k+1
(r-l)r
r=K
y (>--£+l)(r-ifc + 2)
r
(r-l)r
r=0
(u-ifc
The chance
+ l)(n-Jfe + 2)
(n-l)n (n+1)
that the next drawing will give a white ball=
2 r =0
(;i
_
-
A) (u
-
/c
+ 1) fc
(n + 1) r=0
+1
(n-k)(n-k + l)
(/i-A-)(n-/v
n(n +
*
l)
Qr
Ii
(r-l)r
s"(r-fc)(r-fc+l)
?i
.
U—
+ l) 2~ k~+
n(»i
+ l)
Jfc+1
~k + 2' which
independent of the number of balls in the bag at
is
first.
A
Example 3. person writes n letters and addresses n envelopes ; if the letters are placed in the envelopes at random, what is the probability that every letter goes wrong ? Let un denote the number of ways in which all the letters go wrong, and represent that arrangement in which all the letters are in their let abed own envelopes. Now if a in any other arrangement occupies the place of an assigned letter b, this letter must either occupy a's place or some other. .
. .
Suppose b occupies a's place. Then the number of ways in which all the remaining n - 2 letters can be displaced is u n _ 2 and therefore the numbers of ways in which a may be displaced by interchange with some one of the other n- 1 letters, and the rest be all displaced is (n - 1) «„_ 2 (i)
,
.
PROBABILITY.
405
(ii) Suppose a occupies i>'s place, and b does not occupy a's. Then in arrangements satisfying the required conditions, since a is fixed in &'s place, the letters b, c, d, ... must be all displaced, which can be done in h__j ways; therefore the number of ways in which a occupies the place of another letter but not by interchange with that letter is (n - 1) u n - l ; .-.
v n = (n-l)
from which, by the method of Art. Also n 1
= 0,
tig
=1
;
(M n _!
4-44,
we
+ «„_„);
find u n - nu n _ 1
=
(
-
l) n (ttj
-
Uj).
thus we finally obtain ,
f
1
(- 1 )'
i
i
1
!
Now
the total number of ways in which the n things can be put in n therefore the required chance is places is In ;
11 + [2
|£
1
_
|4
'••
+
(- 1)" in
'
and in lias maintained a permanent place some of its many It was first discussed in works on the Theory of Probability. by Montmort, and it was generalised by De Moivre, Euler, and problem
Tlie
involved modifications
liere
is
of considerable interest,
Laplace.
*484. The subject of Probability is so extensive that it is impossible here to give more than a sketch of the principal algebraical methods. An admirable collection of problems, illustrating every algebraical process, will be found in "NVliitworth's Choice and Chance; and the reader who is acquainted with the Integral Calculus may consult Professor Crofton's article Probability in the Encyclopcedia JJritannica. complete account of the origin and development of the subject is given in Todhunter's History of the Theory of Probability from the time of Pascal to that of Laplace.
A
The
practical applications of
the theory of Probability to commercial transactions are beyond the scope of an elementary treatise ; for these we may refer to the articles Annuities and Insurance in the JEncyclopcedia Britannica.
^EXAMPLES. XXXII. L
What
e.
are the odds in favour of throwing at throw with two dice ?
lea.st 7 in
a single
In a purse there are 5 sovereigns and 4 shillings. If they are drawn out one by one, what is the chance that they come out sovereigns und shillings alternately, beginning with ;t sovereign? 2.
:
HIGHER ALGEBRA.
406
If on an average 9 ships out of 10 return safe to port, the chance that out of 5 ships expected at least 3 will arrive 1 3.
is
what
the tickets are blanks but one; each person draws a ticket, and retains it shew that each person has an equal chance of drawing the prize. 4.
In a lottery
all
:
One bag contains 5 white and 3 red balls, and a second bag 5. contains 4 white and 5 red balls. From one of them, chosen at random, two balls are drawn find the chance that they are of different colours. :
E
throw a die in the order named Five persons A, B, C, B, until one of them throws an ace find their relative chances of winning, supposing the throws to continue till an ace appears. 6.
:
Three squares of a chess board being chosen at random, what the chance that two are of one colour and one of another 1 7.
is
A person throws
one the common cube, and the other a regular tetrahedron, the number on the lowest face being taken in the case of the tetrahedron find the average value of the throw, and compare the chances of throwing 5, 6, 7. 8.
two
dice,
;
A's skill is to 2?'s as 1 3 9. 3 ; to Cs as 3 2 ; and to Z)'s as 4 find the chance that in three trials, one with each person, will succeed twice at least. :
:
:
A
10.
A
certain stake is to be
an ace with an octahedral die chance of the last ? 11.
Two
if
the first person there are 4 persons
who throws what
is
the
B
of equal skill are playing a set of games ; A to complete the set, and wants 3 games: compare
players A,
wants 2 games
won by
:
B
their chances of winning.
A
purse contains 3 sovereigns and two shillings a person draws one coin in each hand and looks at one of them, which proves to be a sovereign shew that the other is equally likely to be a sovereign or a shilling. 12.
:
;
A and B play for a prize A is to throw a die first, and is to he throws 6. If he fails B is to throw, and to win if he throws If he fails, A is to throw again and to win with 6 or 5 or 4, 6 or 5. and so on find the chance of each player. 13.
win
;
if
:
Seven persons draw lots for the occupancy of the six seats in 14. a first class railway compartment find the chance (1) that two specified persons obtain opposite seats, (2) that they obtain adjacent seats on :
the
same 15.
A
chance of 16.
side.
number
its
consists of 7 digits
being divisible by 11
whose sum
.4 —
is
is
59
;
prove that the
.
Find the chance of throwing 12 in a single throw with 3
dice.
PROBABILITY.
A
17.
bag contains
407
marked with the numbers
7 tickets
A
0, 1, 2, ...G
ticket is drawn and replaced find the chance that respectively. after 4 drawings the sum of the numbers drawn is 8. ;
There are 10 tickets, 5 of w hich are blanks, and the others are marked with the numbers 1, 2, 3, 4, 5 what is the probability of drawing 10 in three trials, (1) when the tickets are replaced at every T
18.
:
trial, (2) if
If
19.
the tickets are not replaced
n
integers taken at
?
random
are multiplied together, shew
that the chance that the last digit of the product An _
the chance of
and of
its
being
its beinc:
2, 4, 6,
or 8 is
is 1, 3, V,
or 9
Kn ;
—
;
o
9>i
—=——
is
.pi
of its being 5 is
10 H -8'l -5 n + 4 n 10*
is
A
purse contains two sovereigns, two shillings and a metal dummy of the same form and size ; a person is allowed to draw out one at a time till he draws the dummy find the value of his expectation. 20.
:
A
sum of money persons A, B, C who first throws to throw in the order named until 21.
certain
be given to the one of three 10 with three dice; supposing them the event happens, prove that their is to
chances are respectively
Two
22.
2 -
5
and -
/8\ 2
56
(ja)'
W>
/7\ 2 [&)'
persons, whose probabilities of speaking the truth are
respectively, assert that a specified ticket has been
of a bag containing 15 tickets: the assertion ?
23.
.
and
A
bag contains
what
—
-
is
drawn out
the probability of the truth of
counters, of which one
is
marked
1,
two are marked 4, three are marked 9, and so on a person puts in his hand and draws out a counter at random, and is to receive as many ;
shillings as the
number marked upon
it
:
find the value of his ex-
pectation. 24.
If 10 things are distributed
among
3 persons, the chance of
a particular person having more than 5 of them
is
_
....
.
If a rod is marked at random in n points and divided at those points, the chance that none of the parts shall be greater than 25.
— th n
of the rod
is
—
an
.
;;
.
HIGHER ALGEBRA.
408
There are two purses, one containing three sovereigns and a and the other containing three shillings and a sovereign. A coin is taken from one (it is not known which) and dropped into the other and then on drawing a coin from each purse, they are found to be two shillings. What are the odds against this happening again if two more are drawn, one from each purse 1 26. shilling,
If a triangle is formed by joining three points taken at random in the circumference of a circle, prove that the odds are 3 to 1 against its being acute-angled. 27.
Three points are taken at random on the circumference of a what is the chance that the sum of any two of the arcs so determined is greater than the third ? 28. circle:
A
random
into three parts, that they form the sides of a possible triangle ? 29.
line is divided at
what
is
the chance
Of two purses one originally contained 25 sovereigns, and the other 10 sovereigns and 15 shillings. One purse is taken by chance and 4 coins drawn out, which prove to be all sovereigns what is the chance that this purse contains only sovereigns, and what is the probable value of the next draw from it? 30.
:
31. find the 32.
a straight line of length a two points are taken at random chance that the distance between them is greater than b.
On
A straight
random
points taken at
a is divided into three parts by two find the chance that no part is greater than b.
line of length ;
on a straight line of length a + b two lengths a, b are measured at random, the chance that the common part of these lengths 33.
If
2
—c r
shall not exceed c is
where
,
ab
that the smaller length b
c is less
lies entirely
than a or b
;
also the chance
within the larger a
is
—
.
(Jj
on a straight line of length a + b + c two lengths a, b are measured at random, the chance of their having a common part which 34.
If
d
shall not exceed
is
T (c
—
.
.
.
7
,
+ a)(c+6)'
where d
is less
than either a or
b.
Four passengers, A, B, C, D, entire strangers to each other, are 35. travelling in a railway train which contains I first-class, second-class, and n third-class compartments. and are gentlemen whose respective a priori chances of travelling first, second, or third class are represented in each instance by X, fi, v, C and are ladies whose similar a priori chances are each represented by I, m, n. Prove that, for all values of X, fi, v (except in the particular case when
m
A
B
D
X
:
p
:
v=l.
company
:
m
of the
:
oi),
A
and
B
same lady than
more
likely to be found both in the each with a different one.
are
;
CHAPTER
XXXIII.
Determinants.
485.
The
48G.
Consider the two homogeneous linear equations
devoted to a brief discussion of determinants and their more elementary properties. The slight introductory sketch here given will enable a student to avail himself of the advantages of determinant notation in Analytical Geometry, and in some other parts of Higher Mathematics fuller information on this branch of Analysis may be obtained from Dr Salmon's Lessons Introductory to the Modern Higher Algebra, and Muir's Theory of Determinants. present chapter
a x+
is
b l
]
y=
0,
a 2 x + b 2 y = 0; multiplying the first equation by bsi the second tracting and dividing by x, we obtain
This result
is
by 6
,
sub-
sometimes written a
-0,
b x
a„
x
b„
and the expression on the left is called a determinant. It consists of two rows and two columns, and in its expanded form each term is the product of two quantities; it is therefore said to be of the second order.
The
« determinant, and letters
,
b
tile
aa b 2 are called the constituents of the terms «,/>,,, ab. are called the elements, ,
HIGHER ALGEBRA.
410 Since
487.
a.
a
A - «A =
a.
a„
a.
follows that the value of the determinant is not altered by changing the rows into columns, and the columns into rows. it
488.
«
Again,
it is
easily seen that
411
DETERMINANTS. hence *i
a that
is,
the value
of
^3
the
rows into column*, and 491.
From
a.,
6.
«.,
ft..
*,
K
\
<\
C
C3
2
not altered by changing the the columns into rov)S.
determinant
is
the preceding article,
=«, c2
C3
(I.
HIGHER ALGEBRA.
412 492.
The determinant
= «, (K C3 ~ K C 2) + ="
b l(a2CB
~
«« C »)
a
x
bl
c
<**
h2
C2
%
b3
C3
x
~ Cl
l
A
- C /h) + C (« " «3 6 2 ) C C C b (K a 3 ~ 6A) ( 2 3 - J>>) -
h l ( C 2«3
,
l
>
hence «,
K
a2
b2
ax
bx
a (i
c.
3
C
3
appears that if two adjacent columns, or rows, of the determinant are interchanged, the sign of the determinant is changed, but its value remains unaltered.
Thus
it
we denote
If for the sake of brevity
a (l
b
x
h K
«3
by
(a b 2 c 3 ), then the result x
cx
t
2
the determinant
C
2
°3
we have
just obtained
may be
written
(VsO = - (»Ac
s)-
Similarly
we may shew (
c
that
i«A) = - ( a
Ab
3)
= + («A C 3 )-
493. If two rows or two columns identical the determinant vanishes.
of
tlie
determinant are
For let D be the value of the determinant, then by interchanging two rows or two columns we obtain a determinant whose value is — D; but the determinant is unaltered; hence J) = — D, that is D = 0. Thus we have the following equations, a A, — 1
1
aA 2
n
2
+ aJL m = D. 3 3
494. If each constituent in any row, or in any column, is multiplied by the same factor, then the determinant is multiplied
by that factor.
DETERMINANTS. For
tna
ma ma
x
2
a
by
413
HIGHER ALGEBRA.
414
may
be generalised; thus if the constituents of the three columns consist of m, n, p terms respectively, the determinant can be expressed as the sum of mnp
These results
easily
determinants.
Example
1.
Shew
that
The given determinant b
a
+a
a- b b-c
+b
c
—a
c
b
+c
c
a
b
— Babe -a 3 - b 3 — c 3
.
:
415
DETERMINANTS.
and the last two of these determinants vanish [Art. 494 Cor.]. Tims we see that the given determinant is equal to a new one whose constituents of first column is obtained by subtracting from the the first column of the original determinant equimultiples of the corresponding constituents of the other columns, while the second
and third columns remain unaltered. Conversely,
a + j)b + qc x
{
a + PK +
C2
a
.,
b }
<7 C 2
{
^2 c„
c.
and what has been here proved with reference to the first column hence it appears is equally true for any of the columns or rows that in reducing a determinant we may replace any one of the rows or columns by a new row or column formed in the following ;
way of the row or column to be replaced, and increase or diminish them by any equimidtij)les of the corresponding constituents of one or more of the other rows or columns.
Take
the constituents
After a little practice it will be found that determinants may often be quickly simplified by replacing two or more rows or columns simultaneously for example, it is easy to see :
that
a +2 i
}
b
% + Ph
b
}
}
- qc
c,
x
2
K ~ C1 C
a3 +2 jb 3
K-Q c
€l
c
s
2
b2
a..
b„
Ct
°2
2
l
3
c.,
but in any modification of the rule as above enunciated, care must be taken to leave one row or column unaltered.
on the left-hand side of the last identity the constituents of the third column were replaced by c l +rali c 2 + ra^ respectively, we should have the former value inc, + ra creased by Thus,
if
a + 2>b x
t
b
x
— qc
ra x
x
ra„
« a + i'K
K - vc
ra.. i
HIGHER ALGEBRA.
416 and
which this may be resolved one which does not vanish, namely
of the four determinants into
there
is
ra,
pbs
Example
1,
Find the value of
-
I
!
26-4l = -3x4x
-6
31
9
54
-4 -8
.
29
26
22
25
31
27
63
54
46
The given determinant 3
ra „9
qc 2
1
DETERMINANTS.
417
[Explanation. In the first new determinant the first row is the sum of the constituents of the three rows of the original determinant, the second and third rows being unaltered. In the third of the new determinants the first column remains unaltered, while the second and third columns are obtained by subtracting the constituents of the first column from those of the second and third respectively. The remaining transformations are sufficiently obvious.]
Before shewing how to express the product of two determinants as a determinant, we shall investigate the value of 497.
A + ^7,
»A + h A + c i?2 Vi + KPt + c*y, %% + h A + c*y a a + h A + Wi %% + b A + ^ y «i«i
3
+
&
3
i
a a3 + b i
A+
A+
r i7-s
2
« 2 <* 3 +
h
2
«3 a 3
&A + c y
From
+
c
a 3
ya 3
Art. 495, we know that the above determinant can be expressed as the sum of 27 determinants, of which it will be sufficient to give the following specimens :
*1
« 2 ft 3
a3a 2
«3 a 3
a a x
{
HIGHER ALGEBRA.
418
X = a^ + a x \ ^=A*i+/W we have Substituting for X, and X in
where
2
l
(at a
+ x
b2 P
}
)
+ (a^+bfij x 2 = 0\ x + (a 2 a, + bfi 2 ) x2 = Oj
x,
a^ But equations
(3).
x
may simultaneously zero, we must have = + bfr a a + bfi
In order that equations values of x and x2 other than x
.(2).
(1),
2
(a^ + 6^)
2
(3) will
(3)
2
x
hold
2
hold for
(4).
equations (1) hold, and this
if
will be the case either if (5),
aa or
X
if
which
=
b2
X
and
l
2
= 0;
last condition requires that a.
ft
a.
ft
=o
.(6).
Hence if equations (5) and (6) hold, equation (4) must also hold and therefore the determinant in (4) must contain as ;
factors the determinants in (5) and (6) ; and a consideration of the dimensions of the determinants shews that the remaining
factor of (4)
al
must be numerical
;
hence «i
a i
+
&
A
»i a,
+
h
A
the numerical factor, by comparing the coefficients of afyafl, on the two sides of the equations, being seen to be unity.
DETERMINANTS.
4
419
HIGHER ALGEBRA.
420 12.
Without expanding the determinants, prove that x y z b a b c 1
x
y
P 13.
z
x
a
P
P
r
z
c
r
a
r b
c
Solve the equations 1)
14.
15.
16.
1
a
a"
1
b
b2
1
c
18.
1
1
1
a
b
c
19.
y
x2
y
yz
zx
2
-la b+a +a
=
(b
b).
- c)
(c
- a)
(a
- b)
(a
+ b + c).
= (y-z)(z-x) (x -y)(yz + zx + xy).
6 *2
xy a+b
a+c
-26 c+b
b
^2
4(b + c)(c + a)(« + b).
+c
-2c
(b+cY 62
20.
-
c3
x
c
(a
i
a? 17.
= (b-c)(c- a)
2dbc{a+b+cf.
a{c
+ af
62
(a+bf
Express as a determinant
f c
b
c
b
DETERMINANTS. 22.
Fi
421
HIGHER ALGEBRA.
422
The properties of determinants may be 499. ployed in solving simultaneous linear equations.
usefully em-
Let the equations be
+ c z + d = 0, a 2X + b 2 y + c 2 z + d2 = 0, aja + bg + cji + d^O; ax+
h xy
x
x
x
multiply them by A lt -A s A.A respectively and add the results, A , A A 3 being minors of alt a2i aa in the determinant ,
j
D=
a,
*,
a„
«3
The
coefficients of
in Art. 493,
^
y and z vanish in virtue of the relations proved
and we obtain
(Mi Similarly
^3
-
M* + M
we may shew
3)
(6,5,
-6
A+
6
*+
(M
-
M
Mb) =
°-
+ (<*A - <*A + <*A) =
0,
i
2
+
that
A)
2/
and fcff,
Now
- e,C, + c 3 C3
+
(dfi,
- dfia + d3 C,) = 0.
«A - a A, + «A - - (6,5, - 6,5, + 6A) t
hence the solution
x <*,
)
may be
—y
written z
423
DETERMINANTS. a x + b y + c,s + d u — a2x + b 2 y + c 2 z + d2u = a 3 x + b 3 y + c 3 z + d3u = = ax + b Wy + c 4 z + du 4 4 x
A
From the x
K
last three of these,
—y
x
x
A
we have
0, 0,
0, 0.
as in the preceding article z
—u
HIGHER ALGEBRA.
424
The left-hand member of this equation consists of n rows and n columns, and is th II
the
-
a determinant which called a determinant of
is
order.
discussion of this more general form of determinant is beyond the scope of the present work ; it will be sufficient here to remark that the properties which have been established in the case of determinants of the second and third orders are quite general, and are capable of being extended to determinants of
The
any
order.
For example, the above determinant
of
the
n th
order
is
equal to
a or
1
ct l
A -b B + c C -d D 1+ A -a A + a A -a4A + 1
1
1
2
l
2
1
B
1
1
4
3
K
...
+ (-l)"- k
...
+ (-1)"- a n A n
1
1
l,
1
,
according as we develop it from the first row or the first column. Here the capital letters stand for the minors of the constituents denoted by the corresponding small letters, and are themselves Each of these may be exdeterminants of the (n-l) th order. th pressed as the sum of a number of determinants of the (n — 2) order ; and so on ; and thus the expanded form of the determinant may be obtained.
Although we may always develop a determinant by means of the process described above, especially when our object the whole determinant, as elements. 502.
not always the simplest method, is not so much to find the value of to find the signs of its several it is
The expanded form a
l
of the
determinant
DETERMINANTS.
425
can be deduced from the leading element by an even or odd number of permutations of two suffixes ; for instance, the element a 3 b 2 c is obtained by interchanging the suffixes 1 and 3, therefore its sign is negative ; the element ajb c 2 is obtained by first interchanging the suffixes 1 and 3, and then the suffixes 1 and 2, hence its sign is positive. cording as
it
1
1
The determinant whose leading element may thus be expressed by the notation 503.
%^aJ>aeBdA the
a
is
x
b 2 c 3 dA
...
,
2 * placed before the leading element indicating the aggregate which can be obtained from suffixes and adjustment of signs.
of all the elements
interchanges of
by suitable
it
Sometimes the determinant is still more simply expressed by enclosing the leading element within brackets; thus (a^crf ...) is used as an abbreviation of 5 ± a,b„c„d .... A
Example. the element
In the determinant
(a^c^e^ what
sign is to be prefixed to
a^c^e.,1
From
the leading element by permuting the suffixes of a and d we get from this by permuting the suffixes of b and c we have a 4 b 3 c 2 d 1 e 5 ; by permuting the suffixes of c and d we have a i b.i c 1 d2 e 5 finally by permuting the suffixes of d and e we obtain the required element rt 4 & 3 c 1 rf 5 ? 2 and since we have made four permutations the sign of the element is positive. a 4 b2c 3 d x e 5
;
;
;
504.
If in Art. 501, each of the constituents 6
k is in other words ,
c
,
...
equal to zero the determinant reduces to a A it is equal to the product of a and a determinant of the (n — l) th order, and we easily infer the following general theorem. ;
y
the first row or column of a determinant is zero except the first, and if this constituent is equal to m, the determinant is equal to times that determinant of lower
If each of
the constituents
of
m
order ivhich is obtained by omitting the first column row,
and
first
Also since by suitable interchange of rows and columns any constituent can be brought into the first place, it follows that if any row or column has all its constituents except one equal to zero, the determinant can immediately be expressed as a determinant of lower order. This is sometimes useful in the reduction and simplification of determinants.
426 Example.
HIGHER ALGEBRA. Find the value
of
DETERMINANTS.
EXAMPLES.
XXXIII.
Calculate the values of the determinants 1.
3.
1
1
I
2
1
3
1
4
a 1 1
1
427
b.
b
:
k
HIGHER ALGEBRA.
428 Solve the equations 12.
x+ y+ 0=1, ax + by + cz=k, a 2x + b y + c z = l 2
2
13.
ax + by + cz=k, a2x + 2y + c2z = 2 a?x + b 3y + c?z= P.
,
2 .
x + y+ z+ u=l, ax + by + cz+ du=k, a2x + b 2y + c2z + d2 u = k 2 3 s 3 cfix + b y + c z + d u = P.
14.
,
15.
Prove that
b+c—a—d c+a—b—d a+b-c-d
+ d)-ad(b-t-c) ca(b-\-d) — bd(e + a) ab (c + d) -cd{a + b)
— ad ca — bd ab — cd
be (a
be
= -2 (b- e) (c-a) (a-b) (a-d) (b - d) (c-d). 16.
Prove that
a2
a 2 -(b — c) 2
be
b2
b2
-(c-a) 2
ca
2
c2
- (a -
ab
c
b) 2
= (b-c)(c-a){a-b)(a + b + c)(a 2 + b 2 + c2 17.
where
\
Shew
that
a
b
c
d
e
f
f
«
b
c
d
e
ABC CAB
e
f
a
b
c
d
B C A
d
e
f
cc
b
c
c
d
e
f
a
b
b
c
d
e
f a A=a 2 -d + 2ce 2
B=e -b 2
2
).
-2bf,
+2ac-2df,
C=e -f + 2ae-2bd. 2
2
If a determinant is of the ?i th order, and if the constituents of its first, second, third, ...nth rows are the first n figurate numbers of the first, second, third, ...nth orders, shew that its value is unity,. 18.
CHAPTER XXXIV. MISCELLANEOUS THEOREMS AND EXAMPLES.
We
with some remarks on the permanence of algebraical form, briefly reviewing the fundamental laws which have been established in the course of the work. 506.
shall begin this chapter
In the exposition of algebraical principles 507. at the outset we do not lay down new analytically :
new
but
we begin from our knowledge
we
proceed
names and
abstract Arithmetic ; we prove certain laws of operation which are capable of verification in every particular case, and the general theory of these operations constitutes the science of Algebra. ideas,
Hence
of
usual to speak of Arithmetical Algebra and SymIn the bolical Algebra., and to make a distinction between them. former we define our symbols in a sense arithmetically intelligible, and thence deduce fundamental laws of operation ; in the latter we assume the laws of Arithmetical Algebra to be true in all cases, whatever the nature of the symbols may be, and so find out what meaning must be attached to the symbols in order that they may obey these laws. Thus gradually, as we transcend the limits of ordinary Arithmetic, new results spring up, new language has to be employed, and interpretations given to symbols which were not contemplated in the original definitions. At the same time, from the way in which the general laws of Algebra are established, we are assured of their permanence and universality, even when they are applied to quantities not arithmetiit is
cally intelligible.
Confining our attention to positive integral values of the symbols, the following laws are easily established from a priori 508.
arithmetical definitions.
:
HIGHER ALGEBRA.
430 I. (i)
The Law of Commutation, which we enunciate as follows Additions and subtractions may be made in any order. a+
Thus (ii)
Thus
b-c = a-c + b = b-c + a.
Multiplications
and
divisions
may
axb=bxa; axbxc = bxcxa = axcxb'
}
ab-±- c
II.
x b
-=-
c
=
(a
-f-
x b
c)
made in any
and so
and
divisions
may
order.
on.
= (b + c)
xa.
The Law of Distribution, which we enunciate
Multiplications
and
=a
be
as follows :
be distributed over additions
subtractions.
Thus
{a (a
-b+
—
b)(c
c)
m = am —bm + cm,
— d) =
ac
— ad —
bc
+
bd.
[See Elementary Algebra, Arts. 33, 35.]
And
since division is the reverse of multiplication, the distributive law for division requires no separate discussion. III. (i)
The Laws of Indices. n m+n am xa = a
m
H
a +a =a (n)
[a
)
=a
3
m -".
.
[See Elementary Algebra, Art. 233 to 235.]
These laws are laid down as fundamental to our subject, having been proved on the supposition that the symbols employed are positive and integral, and that they are restricted in such a way that the operations above indicated are arithmetically intelligible. If these conditions do not hold, by the principles of Symbolical Algebra we assume the laws of Arithmetical Algebra to be true in every case and accept the interpretation to which this assumpBy this course we are assured that the laws of tion leads us. Algebraical operation are self-consistent, and that they include in their generality the particular cases of ordinary Arithmetic.
From the law of commutation we deduce the rules 509. for the removal and insertion of brackets [Elementary Algebra, Arts. 21, 22] ; and by the aid of these rules we establish the law
|
; ;
MISCELLANEOUS THEOREMS AND EXAMPLES. For example,
of distribution as in Art. 35.
(a
proved that
it is
-b)(c — d)~ac — ad—bc +
431
bd,
with the restriction that a, b, c, d are positive integers, and a Now it is the province of greater than b, and c greater than d. Symbolical Algebra to interpret results like this when all restricHence by putting a = and c = 0, we obtain tions are removed. (— b) x (— d) = bd, or the product of two negative quantities is and c= 0, we obtain a x (—d) =—a
"We are thus led to the Rule of Signs as a direct consequence of the law of distribution, and henceforth the rule of signs is included in our fundamental laws of operation. 510. For the way in which the fundamental laws are applied to establish the properties of algebraical fractions, the reader is referred to Chapters xix., xxi., and xxn. of the Elementary Algebra
there be seen that symbols and operations to which we cannot give any a priori definition are always interpreted so as to make them conform to the laws of Arithmetical Algebra. it will
The laws
Chapter xxx. Elementary Algebra. When m and n are positive integers and m > n, we prove directly from the definition of an index that 511.
of indices are fully discussed in
of the
a
m
xa n = am+n
am
;
1
-r a'
= am
~n j
(a
m
n )
= am ".
We
then assume the first of these to be true when the indices are free from all restriction, and in this way we determine meanings for symbols to which our original definition does not apply. p
The interpretations
a~" thus derived from the first law are found to be in strict conformity with the other two laws and henceforth the laws of indices can be applied consistently and with perfect generality. 512.
for a\ a
In Chapter
,
we =— 1
defined the symbol
vill. 2
i
or
J— 1
as
obeying the relation i From this definition, and by making i subject to the general laws of Algebra we are enabled to discuss the properties of expressions of the form a + ib, in which real and imaginary quantities are combined. Such forms are sometimes called complex numbers, and it will be seen by reference to Articles 92 to 105 that if we perform on a complex
number the operations of and division, the result is
.
addition, subtraction, multiplication, in general itself a complex number,
HIGHER ALGEBRA.
432
Also since every rational function involves no operations but those above mentioned, it follows that a rational function of a complex number is in general a complex number. x+ly \og(x±iy) cannot be fullyExpressions of the form a treated without Trigonometry; but by the aid of De Moivre's theorem, it is easy to shew that such functions can be reduced to complex numbers of the form A + iB. ,
x+iy is of course included in the more general The expression e x+i form a \ but another mode of treating it is worthy of attention.
We
have seen in Art. 220 that
(x\ —nj 1 H
n
)
,
when n
is infinite,
x being any real quantity the quantity defined by means of the equation ;
*+iy
e
= Lini
x and y being any
Handbuch
H
n
)
,
when n
similarly
is infinite,
J
real quantities.
complex numbers will be discussed in Chapters x. and XI. of Schlomilch's
The development found fully
(1 \
may be
x+i!/
e
of the theory of
der algebraischen Analysis.
We
513. shall now give some theorems and examples illustrating methods which will often be found useful in proving identities, and in the Theory of Equations.
To find the remainder ivhen any rational integralfunction divided by x - a.
514.
of x
is
Let fix) denote any rational integral function of x ; divide f(x) hyx-a until a remainder is obtained which does not involve x ; let Q be the quotient, and R the remainder then ;
f(x) Since value
R we
= Q(x-a) + R.
does not involve x it will remain unaltered whatever give to x ; put x = a, then
f(a)
now Q
is finite
= QxO + R;
for finite values of x, hence
:
.MISCELLANEOUS THEOREMS AND EXAMPLES.
R
Cor.
f(a) =
x—
a,
433
;
it
== 0, that is If f{x) is exactly divisible by x - a, then hence if a rational integral function of x vanishes when
is divisible
by x -
a.
The proposition contained
515.
in the preceding article
we
give another proof of of exhibiting the form of the quotient.
useful that
is
Suppose that the function is of n dimensions, and denoted by "~ }+ 2 n 3+ -+P»> P^"~ +P-^"" p x +2\ x
let it
then the quotient will be of n -
by
c"-
q{f
let
R
x
+ qi xn
-2
dimensions
1
+q x
n -3
2
+
...
be the remainder not containing x
pjf +p x1
1
+2> 2 x'-
2
3
+p.ax"~ +
= (x-a)
X^
(q
;
denote
;
+q
H_
l
it
be
;
then
+P„ + qi x"~ 2 + q 2x"~ 3 +
•••
...
+ qa _
x
)
+
Multiplying out and equating the coefficients of like powers of we have
- a Qi=P 2 q3 2 = ihi
or
& = «
R - oqn - =Pn
or
R = aq
9. 2
or qs
>
n
,
%
= a4i+Pa 2
n
+ ih _l
Ct(
«?0
% Thus
q
P3
P2
Pi
lx
%
x
Cl(
l2
In-2
+p n
;
by the the
be
Pa (l(
?.-,
v,
x,
;
Pa-X
Cl(
R
'>
thus each successive coefficient in the quotient is formed multiplying by a the coefficient last formed, and adding The process of finding next coefficient in the dividend. successive terms of the quotient and the remainder may arranged thus Po
so
which has the advantage
it
2n-l
x
R = aq^ +p - « («
1
=P
n
~2
+
•••
+P,r
x + a the same method can be used, only
this case the multiplier is - a.
H. H. A.
28
in
HIGHER ALGEBRA.
434
is
Example. Find the quotient and remainder when divided by x + 2.
Here the multiplier
is
-
3
Thus the quotient mainder 516.
00
21
14 -28 -6
12
-24
5 6
-6
12
- 3
11
31
14
3
-
3.r 6
is
- a; 6 +
31a: 4
+ 21a; + 5
and we have
2,
3-10 -6 -7
3a;7
7a;
5
+ 14a; 4 + 3a; 3 - 6a; 2 +12a;-3, and
the
re-
is 11.
In the preceding example the work has been abridged
by writing down only the
coefficients of the several terms, zero
being used to represent terms corresponding to powers This method of Detached Coefficients may of x which are absent. frequently be used to save labour in elementary algebraical processes, particularly when the functions we are dealing with The following is another illustration. are rational and integral. coefficients
Example.
Divide 1
3a; 5
-
8a; 4
-
5a; 3
+ 26a;2 - 33a; + 26 by
a;
3
-
+ 2 + 4-8)3-8- 5 + 26-33 + 26(3-2 + 3 3 + 6 + 12-24 -2 +
-2-
2a;
2
- 4a; + 8.
MISCELLANEOUS THEOREMS AND EXAMPLES.
435
we form the next horizontal
line, and add the terms in the third column; which is the coefficient of the third term of the quotient. By adding up the other columns we get the coefficients of the terms in
this gives 3,
the remainder. ]
Example.
Divide 6a 5 + ba*b - 8a?b 2 - 6a 2 b 3 - 6a ¥ by 2a 3 + 3a 2 6 -
to four terms in the quotient.
2
bz
HIGHER ALGEBRA.
436
A=
z = 0, then 3 pansion of (x + y)
Put
3,
2 being the coefficient of x y in the ex-
.
Put x = y = z =
l,
and we get 27 = 3 +
(3 x 6)
+B
;
whence
B = 6. Thus (x + y +
z)
3
= x3 + y 3 + z 3 + 3x 2y + 3xy 2 + 3y 2z + 3yz 2 +
3
3z x
2 + 3zx + 6xyz.
A
function is said to be alternating with respect to its variables, when its sign but not its value is altered by the interchange of any pair of them. Thus x — y and 520.
2 a 2 (b-c) + b (c-a) +
2
c (a
-
b)
are alternating functions.
no linear alternating function involving more than two variables, and also that the product of a symmetrical function and an alternating function must be an It is evident that there can be
alternating function.
Symmetrical and alternating functions may be concisely denoted by writing down one of the terms and prefixing the symbol % ; thus %a stands for the sum of all the terms of which a is the type, %ab stands for the sum of all the terms of which ab is the type; and so on. For instance, if the function involves 521.
four letters a,
b, c,
d
}
^a-a + b + c + d; %ab = ab + ac + ad +bc + bd+ cd;
and so
on.
Similarly
the function involves three letters
if
$a
2
(b
2
-c) = a (b-c)±
%a 2 bc = a 2 bc + and
%a
2
2
b ca
2
b (c
+
2
c
- a) +
2
c
(a
a, b,
- b)
c,
•
ab;
so on.
It should be noticed that when there are three letters involved b does not consist of three terms, but of six thus :
2
2
= a 2 b + a c + b 2c + b2 a + 2
2
c
a+
2
c b.
The symbol 2 may also be used to imply summation with regard to two or more sets of letters; thus %yz (b-c) = yz (b~c) + zx (c-a) + xy (a - b).
MISCELLANEOUS THEOREMS AND EXAMPLES.
437
The above notation enables us to express in an abridged form the products and powers of symmetrical expressions thus 522.
:
(a+b + (a
+
b
+
(a
+
c
3
c)
= %a 3 + 32a 2 b +
j
+ df = 2« 3 + 3$a 2 b + Gtabc;
b
+
4
c)
= %a A + i%cfb + 6Sa"6 + 1 2%a 2 bc; fl
%a x 2« 2 = 2a 3 + %a 2 b Example
Gabc
.
Prove that
1.
(a + b) 5
- a5
-
b*
= 5ab (a + b) (a 2 + ab + b 2
).
Denote the expression on the left by E then E is a function of a which hence a is a factor of E similarly 6 is a factor of E. when a = Again E vanishes when a— - b, that is a + b is a factor of E; and therefore E contains ab(a + b) as a factor. The remaining factor must be of two dimensions, and, since it is symmetrical with respect to a and b, it must be of the form Act? + Bab + Ab'z thus ;
vanishes
;
;
;
+ b) 5 - a 5 -
(a
where
A and B
b5
= ab (a + 6) and
are independent of a
(Aa*
+ Bab + A b~),
b.
= 1, we have 15 = 2A + B a = 2, b = - 1, we have 15 = 5A - 2B putting whence A = o, J5 = 5; and thus the required result Putting a =
1, b
;
;
Example
Find the
2.
(&3
+ c 3)
at once follows.
factors of
(b-c)
+ (c 3 + a 3 (c-a) + (a 3 + b 3
Denote the expression by
)
E
)
(a
- b).
E
then is a function of a which vanishes therefore contains a - b as a factor [Art. 514]. Similarly it contains the factors b-c and c-a; thus contains (b c) (c a) (a - b) as a ;
when a = b, and
E
factor.
E is of the fourth degree the remaining factor must be of the degree; and since it is a sj^mmetrical function of a, b, c, it must be of the form M{a + b + c). [Art. 518]; Also since
first
.-.
E = M (b-c)
(c-a) (a-b)(a + b + c).
M
we may give to a, b, c any values that we find most conTo obtain venient; thus by putting a = 0, 6 = 1, c = 2, we find M=l, and we have the required result. Example (x
3.
Shew
that
+ y + zjt-x5 -y b - z? = 5
(y
+ z)
(z
+ x)
(x
+ y)
(x2
+ y 2 + z~ + yz + zx+ xy).
Denote the expression on the left by E then E vanishes when and therefore y + z is a factor of E; similarly z + x and x + y are ;
therefore
E
contains
(y
+ z)
(z
+ x)
[x
+ (/)asa
factor.
Also since
E
y=-z, factors;
is
of the
HIGHER ALGEBRA.
438
degree the remaining factor is of the second degree, and, since symmetrical in x, y, z, it must be of the form fifth
A Put»=2/=z=l; x=2, y=l,
put
(x
thus 2
2
+ y2 + z 2 + B )
+ zx + xy)
.
10=^1+5;
= 5A + IB A=B = 5,
= 0;
thus 35
whence
and we have the required
(yz
it is
;
result.
We
collect here for reference a list of identities which 523. are useful in the transformation of algebraical expressions; many of these have occurred in Chap. xxix. of the Elementary Algebra.
^bc (b
— c) = —(b-c)(c- a)
(a
-
b).
$a2 (b-c) = -(b-c)(c-a)(a-b). $a(b 2 -c 2 ) = (b-c)(c-a)(a-b).
2a 3 (b-c) = -(b-c) (c-a) (a-b) (a + b + c). s 3 3 a + b + c - 3abc = (a + b + c)(a2 +b 2 + c 2 - bc-ca- ab). This identity 3
a +
b
3
+
3
c
may be
given in another form,
-3abc = l(a +
b
+ c){(b-c) 2 + (c-a) 2 + (a-b) 2 }.
(b-c) 3 + (c-a) 3 + (a-b) 3 = 3(b-c)(c-a)(a-b). (a
+
b
+
c)
+
b
+
2
2
+
2c
2b c
-a 3 -b -c 3 = 3(b + c)(c + a)(a + b). 3
+
+ 2abc =
+ c)(c + a)(a +
b).
%a 2 {b + c) + 2abc =(b + c)(c + a) (a +
b).
Hbc
(a
3
c) (be
(b
c)
(b
+ ca + ab) - abc =(b +
V + 2a
2 2
b
c)(c
b
+
c)(b
+ c-a)(c + a-b)(a+b-c).
EXAMPLES. XXXIV. Find the remainder when
by x + 5. 2.
a.
3^ + 1 1^ + 90#2 - 19# + 53
Find the equation connecting a and b in order that
2xi -7x3 +ax + b
may
b).
-tf-fr-c*
= (a +
1.
+ a) (a +
be divisible by x - 3.
is
divided
MISCELLANEOUS THEOREMS AND EXAMPLES. 3.
Find the quotient and remainder when jfi
_ 5#4 + 9 iV3 _ q xi _ iq v + 13 tf _ 2xA
5.
Expand
x2 - 3v + 2.
divided by
j >s
may
Find a in order that x3 -7x + 5
4.
439
be a factor of
- 4^ + 19.V2 - Six + 12 + a.
^.^^.g ^
descending powers of x to four
terms, and find the remainder.
Find the factors of 6.
a(6-c) 3 + 6(c-a) 3 + c(a-6)3.
7.
a4 (6 2 - c2) + 6 4 (c2 - a 2 ) + c4 (a2 -
8.
(a
9.
a (6 - cf + &
6 2 ).
+ 6 + c) 3 -(6 + c-a) 3 -(c+a-6) 3 -(a + 6-c) 3 4
- c4 ) + b
10.
a
11.
(6c
12.
(a + 6 + c) 4
13.
(a + 6
14.
(tf
(6
(c
.
- af + c(a- 6) 2 + 8a6c.
(c 4
- a4 + c(a i - 6 4 )
+ ca + a6) 3 - J 3
3 **
- c%3 - a 3 6 3
).
.
-(6 + c) 4 -(c + a) 4 -(a + 6) 4 + a 4 + 6 4 + c4
+ c) 5 -(6 + c-a) 5 -(c + a-6) 5 -(a + 6-c) 5
- a) 3 (6 - cf + (x - b) s (c - af + (x - c) 3 (a - 6) 3
Prove the following identities
.
.
.
:
15.
2 (6 + c - 2a) 3 = 3(6 + c- 2a) (c + a- 26) (a + 6- 2c).
i0,
c{a-bf _ a(b-cf He-*)* fl|M (c-a)(a-6r (a-6)(6-cr (6-c)(c-a)
17'"
J^6
18.
2 a2(& + c)-2a3 -2a&c=^& + c-a)(c + a-6)(a + &-c)-
iy
*
a+
_?L 6
+c
c
2c
(6-c)(c-a)(a-6)_
+a
(6
.
g
3
+ c)(c + a)(a+6)
(a-6)(a-c)^(6-c)(6-a)^(c-a)(c-6)
20.
42(6-c)(6 + c-2a) 2 = 92(6-c)(6 + c-a) 2
21.
ty+z)*(e+x)*(x+y)*=tx*(y+zY+2(^z)3 -2^
22.
^ («6 - c2 (ac- 6 2 = (26c) (26c - 2a2 ).
23.
«6c (2a) 3
24.
5(6- c) 3 (6 + c - 2a) =
.
2z*'
)
)
- (26c) 3 = abc 2a 3 - 26 3 c3 = (a 2 - 6c) (6 2 - ca) (c2 - a6). ;
hence deduce
2
- y) (£ + 7 - 2a ) 3 = °-
"
HIGHER ALGEBRA.
440 25.
(b
+ cf+(c + af+(a + bf-Z(b + c)(c + a)(a + b)
= 2(a3 -\-b 3 + c3 -3abc). 26.
If
x=b+c-a, y = c + a-b, z = a + b-c, shew that #3 + ^3 + # _ g^g = 4 (a 3 + 6 3 + c 3 - 3a6c).
Prove that the value of a 3 + b 3 + c3 - 3a6c is unaltered substitute s-«. s- b, s-c for a, 6, c respectively, where
if
27.
3s
we
= 2(a + 6 + c).
Find the value of 28
.
« w ,w i (a-b)(a-c)(x-a) ,
a 2 — b 2 — c2
29.
30.
31.
33. 'jp
,y«
»_ w _ » + + /1 (b-c)(b-a)(x-b)
— cz — a 2 ;+77 7 + ft7 wt (b -c)(b — a) (a -b)(a- c)
c2
b2
x
(
a +P)( a +
^
3 (a -b)w(a- w(a — d)' ^ c) If
2
Z>
-a) (c-b)
(b+p)(b + q) (b-c)(b-a) (b + x)
,
(a-b)(a-c)(a+x)
__
(c
— a2 -
s (a-
32.
.
(c-a) (c-b) (x-c)'
+
'
(
c
+p)(c + (J )
(c-a) (c-b)(c + x)
b) (a
— c) (a-
'
'
d)
x + y + z = s, and #yz =£< 2 shew that ,
A /£. _ #\ + /£ _ A /> _ y\ + fp__ p) \zs pj\xs p) \xs pj\ys p)
_y\(p__z\ p)\zs
4 '
8
Miscellaneous Identities. 524. Many identities can be readily established by making use of the properties of the cube roots of unity; as usual these will be denoted
Example.
by
The it
2 .
Shew that (x
hence
1, w, o>
+ yf -x7 -y7 = Ixy
expression, E, on
must contain xy
(x
the
left
+ y)
(x
(x 2
+ y)
vanishes
+ xy + y 2
2 )
.
when x = 0, y = 0, x + y = 0;
as a factor.
Putting x = coy, we have
E = {(1 +
ta)7
- W7 -
1}
y7=
{(_
w 2)7 _ w 7 _
!} y
i
= (_ w 2 - w -l)y7 = 0; hence
E
contains x - wy as a factor and similarly tains x - ury as a factor; that is, is divisible by ;
we may shew that
E
(x-
ury)
(x
-
2 to ?/),
or x^
+ xy + y 2
.
it
con-
=
;
;
MISCELLANEOUS IDENTITIES.
441
E being of
seven, and xy(x + y) (x2 + xy + y2) of five dimensions, the remaining factor must be of the form A (x 2 + y'2 ) + Bxy thus
Further,
;
+ y)
(x
Putting
- x - y7 = xy {x + y) (x
2,
y=-
1,
.-.
aa +
b
3
+
we have a*
+
b
2
c
we have 21 = 5^1 -2B;
3
hence a + b 3 + thus
a3 +
b
3
Example.
3
c
2
c
3,
b
+
c) (a
2
+
+
c
2
-be- ca —
ab)
;
— 3abc can be
resolved into three linear factors;
+ c 3 - 3abc = Shew
(a
+
b
+ ub +
x
+c) (a + mb +
(x
a;
3
+
arc) (a
2
+
+
b
2
b
wc)
+
wc).
+ y 3 + z3 - Sxyz
A 3 + B 3 + C3 - SABC.
+ y + z)
+ wb + ore) (a + w 2 & + wc) (x + uy + urz) (x + w 2 y + uz).
(a
taking these six factors in the pairs (a
+ u>b + w2 c)
ore) (a
that the product of
= [a + b + c)
we obtain the three
2
(a
can be put into the form
(a
b
Art. 110, that
a3 + b 3 + c 3 - dabc and
By
.
— be — ca — ab =
3
The product
2 )
elementary Algebra that
- 3abc = (a+
seen in Ex.
+
)
have 21 = 2^+5;
We know from
525.
+ xy + y (Ax + Bxy + Ay2 ). 2
2
A = 7, B = 7 2 2 (x + y) 7 - x 7 - y7 = Ixy (x + y)(x + xy + y
whence
also
2
7
= l, y = l, we
a;
x
putting
7
(x
+ cry + uz)
;
and
(a
+ b + c)
(x
+ urb + uc)
+ y + z);
(x
+ wy + urz),
partial products
A + B + C, A + wB + u-C,
A+u B + u)C, 2
A = ax + by + cz, B — bx + cy + az, C = cx + ay + bz.
where
Thus the product = (A + B +
+ wB + u 2 C) = A 3 + B 3 +C3 -SABC. C) [A
(A
+ orB + «C)
In order to find the values of expressions involving c when these quantities are connected by the equation + c = 0, we might employ the substitution
526. a,
b,
a+
b
a—h+
k,
b
=
ioh
+
2 (x>
k,
c
=
ufh
+
u>k.
however the expressions involve a, b, c symmetrically the method exhibited in the following example is preferable. If
1
1
1
.
HIGHER ALGEBRA.
442
+ b + c = 0, shew that 3 3 3 5 6 (a 5 + b 5 + c ) = 5 (a + 6 + c )
If a
Example.
We have
(a 2
+ 6 2 + c 2 ).
identically
+ bx)
(1
p — a + b + c,
q
+ ax)
(1
where
(1
+ cz) = 1 +px + qx 2 +rx3
,
r — abc.
= bc + ca + ab,
Hence, using the condition given, (1
(a n
'
n
of
)
of
putting
xn in
= 2,
rc
3,
)
6
we have
~
T- = -
'
«
•
= 7 -a,
= a-/3,
c
.
.
identically for all values of a,
/3,
is
satisfied;
hence
y
5
}
{{§-
y?+ (y- a) 2 + (a-/S) 2
}
is,
(/3-7)
5
+ (7-a) 5 + (a-^) 5 =5( 8-7)( 7 -a)(aJ
compare Ex.
i
3)(a 2 +
Shew
2.
(a +
,
c) 3
b.
is
a positive
+ a>c) 3 = (2a - b - c) (2b - c - a) (2c - a - b). (x+y) n -x -y n is divisible by xy(x2 + xy+y2
+ (a +
a>
2
b
1l
an odd positive integer not a multiple of
),
if
3.
Shew that
4.
(bz
-/37-7a-a^;
that 2
Shew that
3.
2
+ b + cf = a3 + bz + c3 shew that when n + + cf n + = a2n + + b 2n + + c2n + K
If (a integer (a b 1.
^+7
3, Art. 522.
EXAMPLES. XXXIV.
a3
we have
'
the given condition
)
is
,
a 5 + b 5 +c 5
= 5{(/3- 7 3 + (Y-a) 3 + (a-/3) 3 }
n
xn
result at once follows.
5 6{(iS-7) 5 + (7-«) +("-/3)
that
of
coefficients
+ rx 3 ) 2 + ^ (qx2 + rx3) 3 -
3— =r
—=
a=fi-y,
{qx2
a3 + b 3 + c 3
»-
and the required
.
we obtain
5
j— =*•
whence
= l + qx 2 + rx3
xn in the expansion of log(l + qx 2 + rx3)
+ rx3 - ^
[qx2
a2 + b2 + c 2
If
bx)(l + cx)
+ b n + c n = coefficient
= coefficient By
(1 -t
and equating the
logarithms
Taking (~
+ ax)
-
cy) 3
+ b 3 (ex - azf + c3 (ay - bx) = Sabc (bz - cy) (ex - az) (ay - bx). 3
6
c
MISCELLANEOUS IDENTITIES. 5.
(6
Find the value of
-c)(c — a) (a-b) + (b6.
may
a>c) (c
- a>a)
(a
- cob) + (6 - eo 2 c)
— ca - ab)
(c
- arc<)
(a
-
2 a> 6).
2 2 y + z — yz - zx - xy) the form A 2 + B 2 + C 2 -BC-CA- AB.
Shew that
be put into 7.
443
(a2
+ b 2 + c2 -
be
(x2
-f
(a2 + ab + b 2 ) (x2 + xy + y 2 ) can be put into the form find the values of A and B.
Shew that
A + AB + B and 2
2
,
Shew that 8.
2 («a + 26c) 3 - 3 (a 2 + 26c) (6 2 + 2ca) (c2 + 2ab) = (a3 + 6 3 + c3 - 3a6c) 2
9.
2 (a 2 - fc) 3 - 3 (a2 - be) (b 2 - ca) (c 2 - ab) = (a 3 + 6 3 + c3 - 3a6c) 2
.
+ 6 2 + c 2 3 +2(6c + ca + a6) 3 -3(a 2 + 6 2 + c2 )(6c + ca + a6) 2 = (a?+b3 + c3 -3abc) 2 If a + 6 + c = 0, prove the identities in questions 11 — 17. 10.
(«
2
)
11.
2(a 4 + 6 4 + c 4 ) = (a2 + 6 2 + c2 ) 2
12.
a5 + 6 5 + c5 =
13.
a6 + 6 6 + c6 = 3a
14.
3(a 2 + 6 2 + c2 )(a 5 + 6 6 + c5 ) = 5(a 3 + 6 3 + c3 )(« 4 + 6 4 + c4 ).
,_
a7 + b 7 + c7
15.
m
/6-c 16.
17.
18.
2 (6 c
5a6c (6c + ca + ab).
W
- 2 (6c + ca + a6) 3
5 5 5 = a + 6! + c
c--a
a
-
.
a-b\
b
c
a0
.
+ 62 + c2
•
•
-R
( a
b
J\b — c
c—a
+-—c a
\
-J
=9-
+ c2 a + a2 6 - 3a6c) (6c 2 + ca2 + ab 2 - 3abc) = (be + ca + ab) 3 + 27a2 62 c2
.
25 {Q, - zf + (z - x) 7 + (x - y) 7 } {{y - zf + (z- xf + (x- yf)
= 21 19.
{(y
- zf + (z - xf + (x- yf] 2
{(y-z) 2 + (z-xf + (x-y) 2 } 3 -54:(i/-z) 2 (z-x) 2
.
(x-yf
= 2(y + z-2x) (z+x- 2y) 2 (x +y- 2z) 2 2
20.
(6
- cf + (e - a) 6 + (a - 6) 6 - 3
(6
(6-c)
7
+ (c-a) + (a-6) = 7(6-c)(c-a)(a-6)(a2 + 6 2 + c2 -6c-ca-a6) 2 7
If
a + 6 + c = 0, and
.
7
x+y + z = 0, 4 (ax + by + czf -3(ax + by + cz) (a2 + 2 + 22.
.
- c) 2 (c - a) 2 (a - bf
= 2 (a2 + b2 + c2 -be- ca- ab) 3 21.
.
.
shew that 2 )
(x2 +y 2 + z 2 )
-2(b-c)(c-a)(a-b)(y-z)(z- x) (x-y) = 54abcxyz.
.
+ HIGHER ALGEBRA.
444 If
a+
b
+ c + d=0, shew
g? + b 3 + (? + d3 -•
a5 + &5 +c5 + c#>
24.
25.
3—
g
23.
(a3
+
If 2s
3 Z>
that
+ c3 + d3 2 = 9 (M
a2 + b 2 + c2 + d2 2
cda + da& + abc) 2 = 9 (be- ad) (ca - bd) (ab -
)
= a2 + 6 2 + c2 prove that 2 2 2 (4s2 + 5 (s - b) (s - c) (a - a + 5a6cs = {s -
=a+b+c
and
2o-
2
,
2
cr )
)
26.
Shew that
(a?
cd).
2 a- ).
+ 6.2% + 3.vy 2 - y3 3 + (v 3 + 6xy 2 + 3x2y ~ x 3 = Zlxy (x+y) (x 2 + xy +y 2 3
)
)
3 )
.
5
27.
Shew
a that
2 (a — b)(a — c) (a - d)
= a + b + c + d2 + ab + ac + ad+bc + bd+cd. 2
28.
2
2
Resolve into factors
2« 2 6 2 c2 + (a3 + b 3 + c3 ) abc + Z> 3 c3 + c-%3 + a 3 Z> 3
.
Elimination.
In Chapter xxxiii. we have seen that the eliminant of a system of linear equations may at once be written down in the form of a determinant. General methods of elimination applicable to equations of any degree will be found discussed in treatises on the Theory of Equations in particular we may refer the student to Chapters iv. and VI. of Dr Salmon's Lessons Introductory to the Modern Higher Algebra, and to Chap. xm. of Burnside and Panton's Theory of Equations. These methods, though theoretically complete, are not always the most convenient in practice. We shall therefore only give a brief explanation of the general theory, and shall then illustrate by examples some methods of elimination that are more practi527.
;
cally useful.
Let us first consider the elimination of one unknown quantity between two equations. 528.
Denote the equations by f(x) = Q and <£ (x) = 0, and suppose that, if necessary, the equations have been reduced to a form in which f(x) and <£ (x) represent rational integral functions of x. Since these two functions vanish simultaneously there must be some value of x which satisfies both the given equations ; hence
ELIMINATION.
445
the eliminant expresses the condition that must hold between the coefficients in order that the equations may have a common root.
Suppose that x = a, x = J3, x = y,... are the roots of f(x) = 0, then one at least of the quantities (a), (/?), (y), must be equal to zero hence the eliminant is ;
4>
(a)
The expression on the
tf)
=0.
(y)
symmetrical function of the roots of the equation fix) = 0, and its value can be found by the methods explained in treatises on the Theory of Equations.
We
left is a
now
explain three general methods of elimination it will be sufficient for our purpose to take a simple example, but it will be seen that in each case the process is applicable to equations of any degree. 529.
shall
:
The
principle illustrated in the following
example
is
due to
Euler. Eliminate x between the equations
Example.
ax* + bx 2
+ cx + d = 0, fx 2 + gx + h = 0.
Let x + k be the factor corresponding to the root and suppose that
common
to both equa-
tions,
ax 3 + bx 2 + ex + d =
fx 2 + gx + h =
and k,
I,
m, n being
From
unknown
these equations,
(axs + bx 2
Equating
(x
(x
+ k)
+ k)
(ax 2
+ lx + m),
(fx + n)
,
quantities.
we have
identically
+ cx + d)(fx + n) = (ax 2 + Ix + m)
coefficients of like
powers of
x,
(fx
2
+ gx + h).
we obtain
-an + ag-bf=0, gl +fm -bn + ah- cf= 0,
fl
Jd
+ gm- en
- df= 0,
hm-dn
=0.
From n,
we
these linear equations by eliminating the obtain the determinant
a
/ g
f
b
h
g
c
h
d
ag-bf = 0. ah-cf -df
unknown
quantities
I,
in,
HIGHER ALGEBRA.
446
can of the equations f(x) = 0, (x) = be very easily expressed as a determinant by Sylvester's Dialytic shall take the same example as Method of Elimination.
The eliminant
530.
We
before.
Eliminate x between the equations
Example.
axs + bx2 + cx + d = 0, fx 2 +gx + h = 0. 2 Multiply the first equation by x, and the second equation by x and x in succession we thus have 5 equations between which we can eliminate the 4 quantities x 4 x z x 2 x regarded as distinct variables. The equations are ;
,
,
,
ax* + bx 2 +cx +
d=0,
axi + bx3 +cx 2 + dx
=0,
fx 2 + gx + h = 0,
= 0,
fxs + gx 2 + hx fx 4 + gx 3 +
Hence the eliminant
is
a
/
The
=0.
ltx 2
a
b
c
b
c
d
f
9
/
9
h
9
h
= 0.
d
h
method
due to Bezout; it has the advantage of expressing the result as a determinant of lower order than either of the determinants obtained by the preceding methods. We shall choose the same example as before, and give Cauchy's mode of conducting the elimination. 531.
Example.
principle of the following
Eliminate x between the equations
ax 3 + bx 2 + cx + d=0, fx 2 + gx + h = 0.
From
we have
these equations,
a
_ bx 2 + ex + d gx2 +hx
f ax + b
cx
fx +g~
whence and
(ag
-
bf)
'
+d
Jix
'
x 1 + {ah ~cf)x- df= 0,
(ah - cf) x2 +(bh - eg - df) x - dg = 0.
Combining these two equations with fx 2 +gx + h = 0,
is
ELIMINATION. and regarding
x'z
and x as
532.
If
-
bh - eg
cf
for the eliminant
=o.
h
g ah -cf
ag - bf ah
we obtain
distinct variables,
f
447
- df - dg
-df
we have two equations
of
the form
<£,
(x,
y)
— 0,
2/)=0, then y may be eliminated by any of the methods already explained; in this case the eliminant will be a function of x. <£.,(#,
we have
If
three equations of the form
0, (*» y> z)
=
°>
2
(^
=
y> z )
°>
03
(
a; >
y» *)
=
°>
eliminating z between the first and second equations, and then between the first and third, we obtain two equations of the form l>y
we
If
=
^
°>
(
x
y)
>
= °-
eliminate y from these equations
we have
a result of
= 0.
the form/* (a:)
By
V)
(»>
•A,
reasoning in this manner
n variables between n +
1
follows that
it
we can
eliminate
equations.
533. The general methods of elimination already explained may occasionally be employed with advantage, but the eliminants so obtained are rarely in a simple form, and it will often happen that the equations themselves suggest some special mode of elimination. This will be illustrated in the following examples.
Example
Eliminate
1.
lx
By
squaring the
first
Z
2
= cos0,
Z
2
+ m2 =l.
+ m-x 2 + »»V + *V = a2 + & 2 (Z
hence the eliminant
the equations
two equations and adding,
is,
If
m between
+ my = a, vix-ly = b,
7-.c
that
Z,
2
+
/»
2 )
is
m= sin $,
(.t
2
.t
+ y*) = a 2 + 2
+
2 ?/
the third equation
>
2 Z,
;
= a 2 + ZA
is
satisfied identically; that
the eliminant of
x cos 6 + y sin 6 = a is
,
x sin 6 - y cos
x 2 + y* = a° + b*.
=
Z>
is,
;
;
HIGHER ALGEBRA.
448 Example
2.
Eliminate
z2
y*+z*=zayz, v
We have
* z
between the equations
x, y, z
+ x 2 =bzx, x 2 + y* = cxy. x y =c = h -+" x
x ~
z
z
+ -=a, -x + y
.
,
5
y
z
by multiplying together these three equations we obtain,
2
hence
Example
3.
w2
z2
z2
t/
2
z2
z2 a;
z2
a
2
2
s
2/
y
2 .
X'
+ (a 2 - 2) + (6 2 - 2) + (c 2 -2) = abc 2 2 2 .-. a + & +c -4 = a&c.
Eliminate
#,
?/
between the equations
x 2 -y 2 =px-qy, ±xy = qx+py, x 2 + y 2 =l. Multiplying the
first
equation by
and the second by
x,
y,
we obtain
x s + Sxy~=p {x 2 + y 2 )\ hence, by the third equation,
p = x 3 + Sxy 2 2 q = Bx y
Similarly
p + q={x+y)
Thus .:
Example
4.
(p + q)*
Eliminate
We
have
y
a+o+c=
>
+ (p-
x, y, z
v z ?--= a, z
3
.
+ ys p-q={x-y) 3
q)*
.
\
= {x + y)* + {x - yf = 2(x2 + y 2 );
between the equations
z
x
x
z
x = b, --^y = c. T
y
x(y 2 -z2)+y(z2
x
-x2 )+z(x 2 -y 2
)
xyz
_{ y-z){z-x) (x-y) xyz If
we change the
sign of x, the signs of b
and
c
sign of a remains unaltered
hence
Similarly,
and
a-b-c—
(y-z){z + x)(x + y) xyz
b-c-a = (y
+ z){z-x)(x + y)
c-a-b = (y
+ z)(z + x){x^y)
xyz
xyz
are changed, while the
.
ELIMINATION. .-.
{a^b+c)(b + c-a){c + a-h){a + b-c) = -
441) {ul
~
"
Z )2
= -a?b- c 2
2 .
26V + 2c 2 a 2 + 2a2 6 2 - a 4 - i 4 - c 4 + a 26 2 c 2 = 0.
EXAMPLES.
m
Eliminate
1.
z) \y
y) \x
\z
.• .
^~fJ^ztl
XXXIV.
c.
from the equations m2x — ?ny + a=0 my + x=Q. }
2.
Eliminate m, n from the equations
m\v — my + a = 0, 3.
Eliminate p,
q, r
),
nx + my = 2amu,
Eliminate
(1
+
??i),
wy - x— a (1 - m).
from the equations zx=b 2 xy = c 2 x2 +y 2 + z 2 = d 2
y, z
a:,
yz = a
2 ,
,
,
p-q = k(l+pq),
xpq = a.
y from the equations x — y = a, x2 — y 2 = b 2 x3 —y 3 = c3
Eliminate
x,
,
10.
.
Eliminate p, q from the equations
x(p + q)=y, 9.
a 2 + x2 - 3ax = 0.
m from the equations
Eliminate
y + mx=a
8.
1
Eliminate x from the equations
ax2 - 2a 2 x + 1=0,
7.
2
2
a(qr+rp+pq) = 2a-x, qr= — 1.
apqr=y,
6.
m +n =
from the equations
p + q + r — Of 5.
mn + 1 = 0.
Eliminate m, n between the equations
mx — ny — a (m 2 — n 2 4.
n2x — ny + a = 0,
.
y from the equations x+y = a, x2 +y 2 = b 2 #*+#*=c*.
Eliminate
x,
,
11.
Eliminate
u from the equations x = by + cz + rfw y=cz + cfo + a#, 2 = cfti + a# + fry, w = cu; + by + cs. x, y,
z,
}
12.
Eliminate
x, y, z
from the equations
x+y + z = 0, x 2 +y 2 + z2 = a 2 aP+ff+sP^fc, ^ +y 5 + 25 = c5
,
,5
.
n. h. a.
29
x)
+
y
HIGHER ALGEBRA.
450 Eliminate
13.
y 14.
s
#
z^x^y
Eliminate 2
(y+z)
=y
2
+ x) = g2 fo+ff) a
(z
c3
b3
^
a!l(
x,
Eliminate #, y, 2
x, y, z
xy =
2 .
from the equations
z
= 4a 2yz,
Eliminate
y)
abc
y from the equations 4 (.r2 + 2 ) = ax + Z>y, 2(x 2 -y 2 ) = ax - by,
Eliminate
(y -f- z) 17.
xj\x
zj\z
\y
'
from the equations
#, y, z
a3
16.
from the equations
#, y, z
'
ff
15.
c
(2;
+ #) 2 = 46^,
(a;
y)
2
= 4c2#y.
from the equations
(x+y - z) (x-y + z) = ayz, (y + z - x) (y - z + x) = 6s#, (z+x—y) (z — x + y) = cxy.
is
y from the equations x2y=a, x(x-{-y) = b, 2x-\-y = c.
18.
Eliminate
19.
Shew that (a+6 + c) 3 -4
a?,
+ c) (c + a) (a + 6) + 5a&c=0
(b
the eliminant of cm;2
20.
+ fry 2 + cz2 = ax + by + cz =yz + zx + xy = 0.
Eliminate
#,
y from the equations
ax2 -t-by 2 =ax+by = 21. is
& 3c3 + c% 3 + a 3
Shew that
——
=c.
P= 5a b c2 2
2
the eliminant of
ax+yz = bc, 22.
Eliminate
by + zx=cai
x, y, z
cz
+ xy = ab,
xyz=abc.
from
x2 +y 2 +z2 =x + y + z=l,
^(x-p)=-(y-q)= 23.
Employ
Bezout's
method
ax3 + bx2y + cxy 2 + dy 3 = 0,
C
-{z-r).
to eliminate x,
y
from
a'x 3 + b'x2y + c'xy 2 + d'y3
= 0.
2
CHAPTER XXXV. THEORY OF EQUATIONS.
Ix Chap. ix. we have established certain relations between the roots and the coefficients of quadratic equations. We shall now investigate similar relations which hold in the case of equations of the nth degree, and we shall then discuss some of the more elementary properties in the general theory of equations. 534.
Let 2? x" + 2) x"~ 535. integral function of x of 1
y
=
1
+
)
2
X
"~ 2
+ Pn-ix+ Pn ^ e a
+
n dimensions, and
ra tional
us denote
let
it
by
the general type of a rational integral equan degree. Dividing throughout by^> we see that without any loss of generality we may take
f(x); then tion of the
(a?)
is
th
,
xn +2) x"~ +2, o^"~ 2 + 1
i
+2
:>
n-i
x
'
]
coefficients
536. Any value of x which root of the equation f(x) = 0.
£>,
Q
any degree.
as the type of a rational integral equation of
Unless otherwise stated the be supposed rational.
=
-2) n
,
^>
,
.
.
.
pn will
makes f(x) vanish
is
always
called a
In Art. 514 it was proved that when f(x) is divided by x-a, the remainder is f(a) hence if f (x) is divisible by x — a without remainder, a is a root of the equation f{x) = 0. ;
537.
We shall assume that every equation of the form f(x) =
has a root, real or imaginary. The proof of this proposition will be found in treatises on the Theory of Equations ; it is beyond the range of the present work.
29—2
HIGHER ALGEBHA.
452
nth
the
Every equation of
538.
degree has
n roots, and no more.
where Denote the given equation by/(a;) = 0, n ~°' + P*' + x +2> «? +PJXT =p 2 f(x) Q 1
let this be has a root, real or imaginary; so that is divisible by x-a
The equation f(x) = denoted by
then/(a)
a,;
,
}
f(x) = (x-a
i
) l
(x),
n-1 dimensions a rational integral function of = has a root real or Igain, the equation that is divisible by x-a 2 so this be denoted by a2 then <£»
where
(x) is
"0^^**
*»
,
;
fa^^ix-aj^x), where
a
a rational integral function of
(x) is
=
/(a>)
Thus
(«
~
«,) (*
/(«) = PoO* ~
°i)
dimensions.
O *b(*>
obtain, as in Art. 309,
we
Proceeding in this way,
-
n-2
(* -
a " ^)
(
a-)*
since f(x) vanishes Hence the equation f(x)= has n roots, when sc has any of the values a a 2 a 3 ,...a n roots; for if x has Also the equation cannot have more than n .
x
,
,
from any of the quantities a xi a2 a ...«„, all from zero, and therefore the factors on the right are different of x. f(x) cannot vanish for that value ,a ,...a n In the above investigation some of the quantities a ,a 2 3
any value
different
,
l
that the may be equal; in this case, however, we shall suppose all different. equation has still n roots, although these are not investigate the relations between the
To
539. coefficients
roots
and
j
the J
in any equation.
Let us denote the equation by
xn
+p
x
n-
1
l
+2> 2 x
n ~2
+Pn-lX + P» =
+
> j
and the
roots
x"+p x n
by
-l
1
a, b,
+2)
k; then
c,
xn ~'+
we have
identically
+Pn -ix+ P» = (x-a) (x-b)(x-c)
hence, with the notation of Art. 163, n -* H~l n
x
+p x l
+pax
- wT - S x x
n~l
+
(x-k)'}
we have
+Pn.^+Pm
+ Ssx*~* -
+ (-
iy-%-^ + (" !)"£,
;
THE011Y OF EQUATIONS. Equating
powers of x in this identity,
coefficients of like
- %> — S ~ sum x
453
of the roots
l
pa = S„ = sum
taken two at a
of the products of the roots
time;
—pB — Sa -sum
taken three at a
of the products of the roots
time
;
(— \)*p **SU = product of the roots. If the coefficient of x" is
then on dividing each term by
poi
the equation becomes
]? ui
+ ;- xn 1
aj"
-
+ P»=1 X + P» =
+ ]-^x"- 2 +
l
Po
Po
Pu
we have
and, with the notation of Art. 521,
& = -&,
2aft=&, %abe = -%>.
Po
Example
Po
,
abc...k
=
(-
1)"^
P
V,
.
Po
Solve the equations
1.
x + ay + a 2z = a 3 x + by + b 2z = b*, x + cy + c 2 z = c*. 1
From
we
these equations
see that a,
are the values of
e
&,
t
which
satisfy the cubic equation t
hence
z
Example
= a + b + c,
form the equation whose
,
The required equation (x 2
or that
is,
But hence
(x
-
a) (x
(x {x
Thus
(a;
b) (x
- a)
(x
(y
-
2
-
(x
b) (x
+ a)(x + b)
- a 2)
6 2 ) (x 2
- c)
the required equation
(x
x3 +p1x3 +-PzC+p a =0,
,
(y
-
b 2 ) (y
- c 2 ) = 0,
+ a)
(x
+ b)
- c 2 )=0, if
(x
y = x2
;
+ c) = 0.
- c) = x 3 +p 1 x 2 +p 2 x +pt
+ c) = x3 - p
x
;
x 2 +p.& - p v
is
-p xx2 +p*x -p 3 = 0, (x3 +p&) 2 - (p xx2 +p 3 2 = 0, x6 + (2ft - j^ 2 x 4 + (p. 2 - 2p 1 p.i x 2 - p. 2 =
(x3
or
+p 1 x' +p 2 x +p 3 z
)
(x3
)
)
or
and
is
- a2 ) -
x = abc.
are the roots of the equation roots are a2 6 2 c 2 .
If a, b, c
2.
-yt-x = 0; y= -(bc + ca + ab),
-zt 2
3
)
)
if
we replace x 2 by
y,
;
we obtain
f + (2p,-p
2 )
2 y + (p.? - 2p dh ) y
-p 2 =0. :
.
HIGHER ALGEBRA.
454
540. The student might suppose that the relations established in the preceding article would enable him to solve any proposed equation; for the number of the relations is equal to the number little reflection will shew that is this not the of the roots. for suppose we eliminate any n - 1 of the quantities case a, b,c,...k and so obtain an equation to determine the remaining one; then since these quantities are involved symmetrically in each of the equations, it is clear that we shall always obtain an equation having the same coefficients; this equation is therefore the original equation with some one of the roots a, b, c,...k sub-
A
•
stituted for x.
Let us take for example the equation 3 x + p x2 + p2 x + p 3 =
;
x
and
let a, b, c
be the roots; then
a+ ab
b
+
c
= —]
+ ac + bc=
:> x
>
ps)
abc= — p 3 2 Multiply these equations by a - a, I respectively and add thus .
;
,
rf
that
= -l\rf-l\u>-p 3i
+ p 2 a + p 3 = 0, equation with a in the place 3
a +p
is,
which
is
the original
a?
x
The above
process of elimination applicable to equations of any degree. 541.
If
two or more
is
of x.
quite general,
and
is
an equation are conthe properties proved in Art. 539
of the roots of
nected by an assigned relation, will sometimes enable us to obtain the complete solution. Solve the equation 4.r 3 - 24a; 2 + 23x that the roots are in arithmetical progression.
Example
1.
+ 18 = 0, having
given
then the sum of the roots is 3a ; the b, a, a + b of the products of the roots two at a time is 3a 2 - 6 2 and the product of the roots is a (a 2 - 6 2) hence we have the equations
Denote the roots by a -
;
sum
;
;
3a = 6,
from the
first
3a 2
-Z>2
= ^,
a( a2 -6 2 )=-|;
equation we find a = 2, and from the second
since these values satisfy 1 Thus the roots are - - , 2, 2
the third, the three 9 -
£
5
6=±-,
and
a equations are consistent.
p
THEORY OF EQUATIONS. Example
Solve the equation
2.
-
24a; 3
14.r 2
-
(ftx
455 + 45 = 0, one
root being
double another.
Denote the roots by
then we have
a, 2a, b\
Sa + b = ^, 2a 2 + 3a&=--^,
From
the
first
two equations, we obtain 8a 2 a=
.-.
It will
2a-6=-^.
be found on
-2a-3 = 0;
1 5 25 -or--and&=--or-.
3
,
,
25
1
that the values
trial
the third equation 2a 2 6 = ~
15 — -
a= --,
6
=^
do not satisfy
hence we are restricted to the values
;
o
3
a=-
v
Thus the
roots are 7
~
,
-o
>
5
b=--.
•
Although we may not be able to find the roots of an 542. equation, we can make use of the relations proved in Art. 539 to determine the values of symmetrical functions of the roots. Example
Find the sum of the squares and of the cubes
1.
x
of the equation
Denote the roots by
3
a, b, c
;
then
a + b + c=p, bc
+ ca + ab = q.
a 2 + b 2 + c 2 = (a + b + c) 2 - 2
Now
of the roots
-px + qx-r — 0. 2
(bc
+ ca + ab)
—p 2 - 2q. Again, substitute a3 +
b3
«, b, c for
x in the given equation and add; thus
+
=p 3 - Spq 4- dr. Example
2.
d are the roots of
If a, 6, c,
x A +px* + qx 2 + rx + s = 0, find the value of
Ha 2 b. a+b+c+d=-
(1),
+ ac + ad + bc + bd + cd = q
(2),
We have ab
abc
+ abd + acd + bcd — -r
(3).
x
:
HIGHER ALGEBRA.
456 From
we have
these equations
-pq = Sa 2 6 + 3
= Sa .-.
2
(dbc +
aid + acd + bed)
6-3r;
2a-b = 3r-pq.
EXAMPLES. XXXV. Form
a.
the equation whose roots are
1.
|, |,
3.
2, 2,
±V& -2, -2,
0,
5.
0,
2.
0,
4.
a + b, a-b,
2,
2,
-3, -3.
-a + b, -a-b.
Solve the equations
-16x3+ 86x2 - 1 76# + 105 = 0, two
roots being
and
5.
a?
6.
4r3 + 16.r2 - 9x - 36 = 0, the sum of two of the roots being
4^ + 2(Xr2 - 23.r + 6 = 0, two of the roots being Sx3 — 26x2 + 52.27 — 24 = 0, the roots being in
7.
8.
1
7.
zero.
equal.
geometrical pro-
gression.
2a?
9.
3:
—
2
— 22#- 24 = 0, two
of the roots being in the ratio of
4.
24x* + 46.2?2 + 9# — 9 = 0, one root being double another of the
10. roots.
&r 4 - 2^ -27^,2 + 6# + 9 = 0, two of the roots being equal but
11.
opposite in sign.
54^ -39^ -26^7+ 16 = 0, 2
12.
the roots being in geometrical pro-
gression. 13.
32^3-48^2 +22^-3 = 0, the
roots being in arithmetical pro-
gression.
6#* - 29^ + 40a 3
14.
being
- 1x -12 = 0, the product
of two of the roots
2.
#* - 2x* -21.r2 + 22.27 progression. 15.
16.
27.27
4
-195.27 3
+ 40 = 0, the
roots being in arithmetical
+ 494.r2 - 520.27 + 192 = 0,
the roots being in geo-
metrical progression. 17.
18a 3 +
other two,
8U + 121.37 + 60 = 0, 2
one root being half the
sum
of the
THEORY OF EQUATIONS. If a, the value of 18.
If a,
19.
'
£>,
c
2
c2
are the roots of
.r
3
+
{h-cy + (c-a)* + (a-b)\
(1)
r
(2)
'
2
i + i,+t W
w
+ 1^+1 6 a
(1)
Xs -paP+qx - r = 0,
are the roots of the equation
b, c
g'.r
(2)
457
c 2 rt 2
find
aaja
+ r=0, find the value of )-i + c «)-i (& + c + + a + ?,)-i. (
(
Find the sum of the squares and of the cubes of the roots of
20.
#*
+ qx2 + rx + s = 0.
Find the sum of the fourth powers of the roots of
21.
x3 +qx+r=Q. 543.
an equation with
7?i
imaginary
real coefficients
roots
occur in pairs. is an equation with real coefficients, Suppose that f(x) = and suppose that it has an imaginary root a + ib we shall shew that a — ib is also a root. ;
The
(x
— a — ib)
(x
— a + ib),
or (x
Let f(x) be divided by (x — a) 2 + b 2 Q, and the remainder, if any, by Rx +
;
E
f(x) = Q{(xIn (x
this identity
- a) + 2
b
2
=
;
a)
2
+
by hypothesis
is
real
is
= a-ib
is
+
b
is
2 .
denote the quotient by then
+ Rx + E. by hypothesis
and imaginary
;
also
parts,
Rb = 0;
R=
and
exactly divisible by (x (x
hence x
2
roots
not zero, .-.
Hence f (x)
}
',
a)
E
Ra + E=Q, b
b
2
—
put x = a + ib, then f(x) — = 0. hence R (a + ib) +
Equating to zero the
and
two
factor of f(x) corresponding to these
—a-
ib)
(x
#=
0.
— a) 2 +
— a + ib)
b
2 ,
that
is,
by
\
also a root.
In the preceding article we have seen that if the equa544. 2 2 tion f{x) = has a pair of imaginary roots a ± ib, then (x — a) + b is a factor of the expression f(x).
;
HIGHER ALGEBRA.
458
Suppose that a±ib, c±id, e±ig,... are the imaginary roots (x) is the product of the of the equation f(x) - 0, and that quadratic factors corresponding to these imaginary roots; then cf>
(x)
= {(x-a) 2 + b 2 }{(x-c) 2 + d 2 }{(x-e) 2 + /}....
Now
each of these factors is positive for every real value of x; hence (x) is always positive for real values of x.
As in Art. 543 we may shew that in an equation with 545. rational coefficients, surd roots enter in pairs; that is, if a + Jb is a root then a- Jb is also a root. Example
Solve the equation 6z 4 - 13x 3
1.
that one root
is
- 35z 2 - x + 3 = 0, having given
- ^3.
2
Since 2-^/3 is a root, we know that 2+^/3 is also a root, and corre2 sponding to this pair of roots we have the quadratic factor x - 4# + 1.
6z 4 - 13z 3 -
Also
- x + 3 = {x 2 - 4x + 1)
35a: 2
(6a; 2
+ 11* + 3)
hence the other roots are obtained from 6a;
2
+ 11a; + 3 = 0, 3
1
--
thus the roots are
Example coefficients,
2.
Form
or
--
,
2
,
(3a;
is ,J2
(2z + 3)=0;
+ ^/3, 2-^3.
the equation of
one of whose roots
+ 1)
the fourth degree with
rational
+ sj - 3.
Here we must have /J2 + /J-3, J2-J- 3 as one pair of - >/2 + ^/ - 3, - J2 - 3 as another pair.
roots,
and
J
Corresponding to the first pair we have the quadratic factor x 2 - 2 /v/2x + 5, and corresponding to the second pair we have the quadratic factor
x 2 + 2 J2x + 5. f
Thus the required equation (x 2
or
or
Example
3.
Shew
is
+ 2,J2x + 5) (x 2 -2 J2x + 5) = 0, (x2 + 5) 2 -8a2 = 0, a^ + 2x 2 + 25 = 0. f
that the equation
A2 + B2 C2 7 + x-a x-b x-c +
H ,=&, — x-h 2
...
+
'
,
has no imaginary roots. If possible let these values for x
p + iq
be a root and subtract the
B2
A2 2 i)
{(p-a + q which
is
2
(p-b) + q 2
impossible unless q =
2
;
then
p - iq
first result
is
also a root.
from the second
C2 ^(p-c) 2 + q 2
;
Substitute thus
H* '
""
'
(p-h) 2 + q
THEORY OF EQUATIONS.
459
To determine the nature
of some of the roots of an equation it is not always necessary to solve it ; for instance, the truth of the following statements will be readily admitted.
546.
If the coefficients are all positive, the equation has no 5 positive root ; thus the equation # + x 3 + 2x + 1 = cannot have a (i)
positive root. (ii)
sign,
If the coefficients of the even
and the
coefficients
powers of x are all of one of the odd powers are all of the contrary
equation has no negative root; thus the equation
sign, the
x7 + x 5 - 2x4 + x3 - 3x 3 + 7x-
5
=0
cannot have a negative root. If the equation contains only
even powers of x and the coefficients are all of the same sign, the equation has no real root thus the equation 2x8 + 3x* + x 2 + 7 = cannot have a real (iii)
;
root.
odd powers of x, and the coefficients are all of the same sign, the equation has no real root except x = ; thus the equation x9 + 2x 5 + 3x3 + x = has no real root except x = 0. If the equation contains only
(iv)
All the foregoing results are included in the theorem of the next article, which is known as Descartes' Rule of Signs.
An
equation f(x) = cannot have more positive roots than there are changes of sign in f (x), and cannot have more negative roots than there are changes of sign in f (-x). 547.
Suppose that the signs of the terms in a polynomial are we shall shew that if this polynomial + H is multiplied by a binomial whose signs are A there will be at least one more change of sign in the product than in the original 1
—
1
1
;
—
,
polynomial.
Writing down only the signs of the terms in the multiplication, we have
+
:
.
HIGHER ALGEBRA.
460
Hence we (i)
see that in the product
an ambiguity replaces each continuation
of sign in the
original polynomial;
the signs before and after an ambiguity or set of ambiguities are unlike; (ii)
(iii)
a change of sign
is
introduced at the end.
Let us take the most unfavourable case and suppose that all the ambiguities are replaced by continuations; from (ii) we see that the number of changes of sign will be the same whether we take the upper or the lower signs; let us take the upper; thus the number of changes of sign cannot be less than in
+ - + -+,
+
+ +
this series of signs is the same as in the original polynomial with an additional change of sign at the end.
and
the factors corresponding to the negative and imaginary roots to be already multiplied together, each factor x — a corresponding to a positive root introduces at least one change of sign; therefore no equation can have more positive roots than it has changes of sign. If then
we suppose
Again, the roots of the equation f(—x) = are equal to those but opposite to them in sign; therefore the negative of /(^) = roots of f(x)-0 are the positive roots of /*(-#) = 0; but the number of these positive roots cannot exceed the number of changes of sign in f{— x) that is, the number of negative roots cannot exceed the number of changes of sign in of f(x) = ;
/(-
>
Example.
Consider the equation
Here there are two changes of
a;
9
+ 5x 8 - x* + Ix + 2 = 0.
sign, therefore
there are at
most two
positive roots.
— x 9 + 5x 8 + x 3 -7x + 2, and
Again /(- x)=
here there are three changes of sign, therefore the given equation has at most three negative roots, and therefore it must have at least four imaginary roots.
EXAMPLES.
XXXV.
b.
Solve the equations 1.
3xA —
2.
6s4 - l&e 3 - 35#2 - x + 3 = 0, one root being 2 - N/3.
3.
xA + 4-r3 + 5x2 + %x -2 = 0, one
3
lO.'o"
+ 4x 2 - a — 6 = 0,
one root being
root being
^
- 1 + ,J ^1
.
;
THEORY OF EQUATIONS. +
+ G.f2 + 4x + 5 = 0,
4.
X*
5.
Solve the equation x5
^3 and
4./,-"
another
Form
1
one root being «/-l.
-xA + 8x2 -9x — 15 = 0,
one root being
the equation of lowest dimensions with rational coefficients, is
s iZ+J^2.
-J2-J^2.
8.
L
—2J- 1.
one of whose roots 6.
4G
7.
-J^l+J5.
9.
N /5
+ 2 x/6.
10.
Form
the equation whose roots are
± 4 a/3,
11.
Form
the equation whose roots are
1± >/- 2, 2±
12.
Fomi
is »J2
2 */
-
1.
J -3.
+ J3 + x/ — 1.
13.
Find the nature of the roots of the equation 3xi + l2x2 + bx-4 = 0.
14.
Shew
that the equation
=l
with rational co-
the equation of the eighth degree
one of whose roots
efficients
5
2.v 7
- xA + 4.V3 - 5 =
has at least four
imaginary roots.
What may
15.
be inferred respecting the roots of the equation 10
a-
Find the
16.
least possible
number
of imaginary roots of the
x° — o?> + xA + x 2 + 1 = 0.
equation 17.
-4a 6 + xA -2.y-3=0?
Find the condition that x3 -px 2 + qx - r = two roots equal but of opposite sign (1) (2)
may have
the roots in geometrical progression.
the roots of the equation xl +p.v3 -\-qx 2 + rx + arithmetical progression, shew that p 3 — 4pq + 8r=0; and in geometrical progression, shew that p 2s = r 2 18.
If
s
=
if
are in
they arc
.
19.
If the roots of the equation
xn -
1
=
b, c
are the roots of the equation x3
20.
Za 2 b 2
22.
S (* + !)•
If a,
b, c,
24.
%a*b&
.
d
1, a, /3, y,
.,
shew that
-px2 + qx -r = 0,
find the
+ c)(c + a)(a + b).
21.
(b
23.
$a 2 b.
are the roots of xA +px3 + qx- + rx + s 25.
. .
=n.
(l-a)(l-/3)(l- 7 ) If a, value of
are
$a\
= 0,
find the value of
HIGHER ALGEBRA.
462
To find the value of f (x 548. integral Junction ofx. ~
"~ 2
n n Let f(x) =p Q x +p x +P 2 X n fix + h) = p (x + h) +2\ (x + l
l
h)
+
h),
when
+2> 2 (x
+
a rationed
(x) is
+Pn-i x + Pn
+ H~ }
f
h)"-*
then
i
+
+2\-A x + h ) + Pn Expanding each term and arranging the powers of h, we have n
result in ascending
~
~l
+2^xn 2 + ... +p n . x+p n n ~3 1 + ...+ + h {np^- + (n- ljjyrf- + (n-2) p 2 x
Pox -\-2\x
n
l
1
+ ^{n(n-l)p + +
x"-
2
p^}
3 + (n-l)(n-2) Pl x'>- +... + 2p n _ 2 }
.
^{n(n-l)(n-2)...2.1 2> \n
}-
l
This result
is
usually written in the form
/(» + *)=/(*) + hf{x) + *J/» + *j/-» + and the functions
f
...
+
*i/»,
are called
f"(x), f"(x),... derived functions oifix). second, third,... (x),
the
first,
The student who knows the elements
of the Differential Calculus will see that the above expansion of f(x + h) is only a (x), f" (x), particular case of Taylor's Theorem; the functions f'"{x) may therefore be written down at once by the ordinary
f
rules for differentiation: thus to obtain f'(x) £romf(x) we multiply each term in f(x) by the index of x in that term and then
diminish the index by unity. Similarly
J
by successive
we
differentiations
obtain fix),
\X), ....
By
writing
—h
in the place of
h,
we have
h
f(x-h)=f(x)-h/'(x) + 'f" (x )Jff'"( x)+ ... + { -
to
The function f(x + h) x and h; hence fix + h) =/(h) +
is
xf
I)-
%-f{x).
evidently symmetrical with respect
+ (h)
i
*r £
,n
(h)
+
...
4
(h). f/* \n
I
; .
;
THEORY OF EQUATIONS.
463
,
Here the expressions f'(h) f"{Ji),f "{1b) ... denote the results obtained by writing h in the place of x in the successive derived i
i
functions f'(x), /"(#), f"(x),.... Example.
If
/ {x) = 2x* - xs - 2xz + 5x -
/ (x) = 2x* - x* - 2x* + 5x -
Here
/' (x)
= 8z3 - Sx°- - 4x + 5,
1
find the value of
,
so that
1,
ana /'
(3)
/ (x + 3).
/ (3) = 131
= 182
^- =12a»-3aj-2, and -^ = )
97;
3) and /-^ = 23;
QS-te-1,
ii 4 / (x + 3) = 2s + 23r* + 97x2 + 182* + 131
Thus
The
calculation may, however, be effected process, as explained in the next article.
549.
Let
put x — y +
Now
h,
=p xn +p
f{x)
1
x
n~
1
more systematically by Horner's
+ p 2x"~ 2 +
...
+pn _ x + pn l
;
and suppose that f [x) then becomes
y = x — h; hence we have the identity ~
~ +p xn + p xn 2 +... +pn _ x + p (x- h) +q (x- h)- + + q _ (x - h) + qn
p
= qo
x"
l
2
1
n
1
n
1
.
.
.
n
x
x
therefore q n is the remainder found by dividing f(x) also the quotient arising from the division is l q (x-h)*-
Similarly q n _ expression by x-
is
l
h,
9o( x
and
l
i
by x-h;
.
the remainder found by dividing the last and the quotient arising from the division is
-h
so on. Thus q n plained in Art. 515. equal to j) -
+q {x-hy-'+...+qH _
;
,
2 3 T' + QA x - h T~ +
qn _
1 ,
The
qn _ a
,
•••
-
+ Qn- 2
'
}
may be found by
last quotient is q
,
the rule exand is obviously
:
HIGHER ALGEBRA.
464 Find the
Example.
result of changing
x into x + 3 in the expression
2a;4_^_2x +5x-l. 2
Here we
divide successively
2-1-2
5
-1
15
39
132
6
by x -
3.
Or more
briefly thus
THEORY OF EQUATIONS.
4G5
intermediate values; but since f(a) and f(b) have contrary signs for some the value zero must lie between them; that is, f(x) = value of x between a and b. It does not follow that f(x) = and b; neither does it follow that
signf(x) =
has only one root between a if f(a) and /(b) have the same
has no root between a and
b.
Every equation of an odd degree has
553.
at least one real
root whose sign is opposite to that of its last term.
In the function f(x) substitute for x the values + successively, then
/( + oo) = + If
— oo and and + co
if
,
pn
f(0)=pnJ has a f(x) =
then is negative f(x) =
is positive,
pn
co,
co
,
0,
-co
/(-oo) = -oo.
root lying between and has a root lying between
.
last
of an even degree and has its two real roots, one positive and one
Every equation which
554.
term negative has at
least
is
negative.
For in
this case
/(+co) = +co, but
pn
and +
is
co
,
f(0)=p n
,
f(-co) = + co;
negative; hence f(x) = has a root lying between and a root lying between and - co .
and
have contrary signs, an odd number of roots of f (x) = will lie between a and b; and «/*f(a) andi(h) have the same sign, either no root or an even number of roots will lie between a and b. 555.
If
the expressions f (a)
Suppose that a
f (b)
than b, and that a, /3, y, k represent all the roots of f(x) = which lie between a and b. Let (x) be the quotient when f(x) is divided by the product (x — a) (x — /3) (x — y) ... (x — k ) then is
greater
.
.
.
;
Hence
f(x) — (x — a)(x—/3)(x- y)
...
(x
— k ) <£ (x).
— y)
...
(a
-
f (a) = (a — a) (a —
/3)
(a
k)
(«)•
/(8)=(5-a)(6- 3)(6- r)...(ft-K)*(5). J
Now
must be of the same sign, for otherwise a = 0, and therefore of f (x) = 0, would root of the equation lie between a and b [Art. 552], which is contrary to the hypo<{>(a)
and
(b)
(.x')
H. H. A.
30
HIGHER ALGEBRA.
466
Hence
thesis.
/(a) and /(b) have contrary
if
signs, the
ex
pressions
- a) (a -fi)(a-y) ... (a - k), (b-a)(b-P)(b-y)...(b- K )
(a
must have contrary signs. Also the factors in the first expression are all positive, and the factors in the second are all negative; hence the number of factors must be odd, that is the number of roots a, /?, y, ... k must be odd. /(a) and /(b) have the same sign the number of factors must be even. In this case the given condition is satisfied k are all greater than a, or less than b thus it does if a, /?, y, has a root between a and b. not necessarily follow that y* (as) = Similarly
if
;
. . .
556.
If a,
b, c,
...k are
/ (x) = j? If
(x
the roots of the equation /(x)
— a)(x-b)(x — c)
Here the quantities a, r of them are equal to a,
b,
s
c, ...
to
b,
— k).
k are not necessarily unequal. then t to c, .
.
.
/(x) = p (x — a) r (x -
In
(x
...
= 0, then
—
s
b) (x
,
c)'
speak of the equation as having n roots, each of the equal roots being considered
this case it is convenient still to
/{x) — a distinct root.
557.
equation
1/
the equation f(x)
=
—
1
f (x) =
will have r
has r roots equal roots equal to a.
Let <£(#) be the quotient when /(x) r then /(x) = (x — a) $>{x).
Write x + h
j
in the place of x; thus r
+ ¥'(x) + %/"(x)+..
= Ux-a) + r(x-a) r
In
then the
divided by (x — a) r ;
is
/(x + h) = (x-a + h) 4>(x+ .-../(»)
to a,
this identity,
r
-
x
h)
;
.
h + ...\U(x) + hcf>'(x)+~
by equating the
<}>"(x)+ ...]
coefficients of A,
/'(x)=r(x - ay-'^x) +
(x
- a)
r
.
we have
$ (x).
Thus/'(aj) contains the factor x-a repeated r-\ times; that is, the equation /' (x) = has ?• - 1 roots equal to a.
;
;
'
THEORY OF EQUATIONS.
4G7
Similarly we may shew that if the equation (x) = has s (x) = has s — 1 roots equal to b roots equal to b, the equation
f
f
and so on.
From the foregoing proof we see that if f{x) contains 558. -1 a factor (x — a)\ then (x) contains a factor (x — a)* and thus ; r f{x) and f'(x) have a common factor (x — a) ~\ Therefore if f(x) and fix) have no common factor, no factor in f(x) will be repeated ; hence the equation f (x) = has or has not equal roots, according as f (x) and f (x) have or have not a common factor involving x.
f
From the preceding article it follows that in order to 559. obtain the equal roots of the equation f(x) = 0, we must first find the highest common factor of f(x) and /*'(#). Example
Solve the equation
1.
ar
1
- llx3 + 44ar -76x+ 48 = 0, which has
equal roots. 4 3 / {x) = x - lis + 44a? - 76a + 48,
Here
= 4x3 - 33a2 + 88x - 76
/' {x)
and by the ordinary hence (a;) is x - 2
/'
;
rule (x
-
find that the highest common faetor of f(x) a factor of f(x) and
we 2)
2
is
/(a;)
thus the roots are
Example
2.
may have two
= (a:-2) 2 (a:2 -7x+12) = (a;-2) 2 (a;-3)(a:-4);
2, 2, 3, 4.
Find the condition that the equation
aar3
+
36a;
2
+ 3ca; + d —
roots equal.
In this case the equations f(x) = 0, and /'
axs + Sbx 2
+
ax 2
-f
3cx
(x)
= 0,
that
By combining
(1)
and
is
+d=
2bx +
c
(1),
=Q
(2)
must have a common root, and the condition required eliminating x between these two equations.
will be obtained
we have
(2),
bx2 + 2cx + d =
From
(2)
and
and
;
(3),
(3).
we obtain a;
2
2 (bd
thus the required condition (6c
-
_
-
c
~
2 )
x
_
1
be-all ~ 2{ac -
2)
ft
is
adf =4 (ac - b-)
(bd -
<-).
30—2
by
HIGHER ALGEBRA.
4fi8
has r roots the equation/^) = But has r - 1 roots equal to a. equal to a, the equation /(a) = (x); hence the equation r(x) is the first derived function of to a; similarly the equation f'\x) = must have r-2 roots equal IheseL f>t x \ = o must have r - 3 roots equal to a; and so on. discover the equal considerations will sometimes enable us to = with less trouble than the method ot Art. 559.
We
560
have seen that
if
f
roots of
f(x)
561.
If
prove that
= f(x) v '
We
are the roots of the equation f (x)
a, b, c, ...k
M MM +
x-a
have fix) = (x-
x-b
f(x +
hence f(x)
is
member
h)
=/(«)
,
...
+ lW
x-kk
.
(x-k);
.+
4-
h)(x-
c
+ h)
(x-k + h)
...
hf (x) + r^ /" (»)+...
...
(1).
',
right-hand
to the coefficient of h in the
equal
of (1); therefore, as in Art. 163,
f(x)= (x-b) (x-c) that
...
0, to
x,
f (x + h) = (x - a + h)(x - b But
x-c
+
-b) (x-c)
a) (x
writing x + h in the place of
+
=
(x-k) +
...
x A w
*,,
•
n
f'(x)
is,
The
562.
to find the
Example. equation
find the value of
Let
-a)(x- c)
Ax + A ^-x +^-{ x-b x — c
) = J^^x—a
) J
)
+
...
(x-k)+
x) / ^/ + f(
...;
-
...
x—k
.
result of the preceding article enables us very easily
sum If
(x
of
an assigned power
of the roots of
Sk denote the sum of the fcth powers x 5 +px* + qx 2 + t = 0, S 4 S 6 and S_ 4 ,
an equation.
of the roots of the
.
f(x)=x 5 +px + qx* + li
then
/'
(
t',
x ) = 5x* + 4ps 3 + 2qx.
fix) 3 Now Z^=rf+(a+P)x + (a2 + ap)x2 +(a* + a2p + q)x + a4 + a?p + aq-, a so
and similar expressions hold
for
fw />) /w x-b' x-c' x-d' x-e'
/(*)
'"
p
THEUltY OF EQUATIONS. Hence by
409
addition,
+ 4px* + 2qx =5x* + (St + 5p)x? + (S 2 +pSJ x 2 + {Si +pS2 + 5q) x + {S4 +p83 + qSJ. By equating coefficients,
5ar*
S 1 + 5p = ±p, whence 8X .— - p S 2 +pS l = 0, whence S 2 =p z
;
;
S3 +pS 2 + oq = 2q, whence S 3 = S 4 +pS 3 + qS x = 0, To
find the value of
Sk
Multiplying the given equation by #*~ 5 x*
+px
-Sq;
whence S 4 =p i + 4j)q.
for other values of k,
k~ 1
%
we proceed as
follows.
,
+ qx*~ 3 + to* -5 = 0.
Substituting for x in succession the values a, b, we obtain S k +pS k _ x + qS k _ 3 + tS k _ 5 = 0.
c, d, e
and adding the
results,
Put k = 5
;
S 5 +pS4 + qS2 + 5t = 0,
thus
S5 = -p 5 - op 2 q -
whence
Put k = 6
;
S 6 +pS 5 + qS3 + tS x = 0,
thus
S6 =p 6 + 6p'*q + Sq 2 + bpt.
whence
To
find
bt.
5_4 put k = 4,
succession; then
3, 2, 1 in
,
S i +pS3 + qS 1 + *#_! = (), whence S_ x = 0; S 3 +pS 2 + 5q + tS_ 2 = 0, whence S.2
S_.2 = -
— 2,/
;
+pS 1 + qS- 1 + tS_ 3 = 0, whence S_ 3 = 0;
S 1 + 5p + qS_ 2 + tS_ 4 = 0, whence £_ 4 =
%-
When the coefficients are numerical 563. ceed as in the following example.
Ap .
we may
-
Example.
Find the sum of the fourth powers of the roots of
x*-2x 2 + x-l = 0.
= x*-2x2 + x-l, 2 f'(x) = Sx -±x + l.
Here
Also
f(x)
*
/
-.-'
f(x)
=
x- a
+
x—b
+
x-c
2
= 2 /l-
a a a3 +_++ —+... \x x- X x*
O
= -+— X
s
Oo
i>,
X~
H
X
3
+
o« "
X*
+
\
J
also pro-
:
:
HIGHER ALGEBRA.
470 hence #4
which
is
is
equal to the coefficient of
in the quotient of f'{x)
^
very conveniently obtained by the
method
by f(x),
of synthetic division as
follows
3-4 + 1 6-3 + 3 -1 4-2 + 41 1
2
2 2
+2 10-5 + 5 3 + 2 + 2 + 5 + 10 +
3
Hence the quotient
is
- +
2 + 2
2 -3
10 5 + -4 + -g +
S.
thus
= 10.
XXXV.
EXAMPLES.
= xA + 10^ + 39# 2 + 76o; + 65,
1.
If f{x)
2.
If
3.
If /(#)
= 2# 4 - 13#2 + 10a; - 19,
4.
If f(x)
=x* + 16^ + 12x2 + 64a; -
5.
If f(x)
= ax9 + bx5 + ex +
6.
Shew
f(x)=xi - I2x3 + 11x2 -9x + 1,
find the value of f(x - 4).
of/(# + 3).
find the value
find the value of f(x +
1).
129, find the value of f(x
find the value of /(#
that the equation 10a*3 - 17#2
and -
between
c£,
c.
- 4).
+ /i) —f(co - It).
+#+6=0
has a root
1.
the equation x* - 5x3 + Sx2 + 35# - 70 = between 2 and 3 and one between - 2 and - 3.
has a root
=0
has a root
7.
Shew that
Shew that the equation x* - l<2x2 + I2x - 3 between - 3 and - 4 and another between 2 and 3. 8.
Shew that x5 + 5x* - 20x2 - 19a; - 2 = and a root between - 4 and - 5. 9.
3,
has a root between 2 and
Solve the following equations which have equal roots 4
-9a; 2 + 4a;+12 = 0.
5
- 13#*+67#3 -
^-6^ + 12^_ 1007 + 3 = 0.
10.
a;
12.
a,-
13.
x5 -x3 + 4x2 -3x + 2 = 0.
15.
xG -3x5 + 6x3 -3x 2 -3x + 2 = 0.
16.
x6 - 2x* - 4xA +
17.
xi -(a + b)x*-a(a-b)x 2 + ai(a + b)x-a3b = Q.
17la;2
3
12a,*
11.
+ 216^- 108=0. 14.
8^ + 4^3-18^+11^-2=0.
- Sx2 - 18a; + 18 = 0.
1
:
TRANSFORMATION OF EQUATIONS.
471
Find the solutions of the following equations which have common roots
21.
+ x2 + 3x -6 = 0, 4#* - 2s 3 + 3x -9 = 0. 4#* + 2x* - #2 -15.*; = 0, 6^ 4 + 1 3a; - 4a; 2 - 1 bx = 0. Find the condition that x -px 2 + r=0 may have equal roots. Shew that x + qx2 + s = cannot have three equal roots.
22.
Find the
2s* -
18.
2a- 3
3
19.
1l
20.
i
may have
(1)
Shew
23.
ratio of b to
a
in order that the equations
ax2 + bx + a = and xs -2x2 + 2x-l one, (2) two roots in common.
=0
that the equation
xn + nxn
~1
+ n (n - 1) x n ~ 2 + ... + \n =
cannot have equal roots. If the equation x5 that ab* - 9a 5 + c5 0.
24.
shew
- l0a 3x 2 + b ix + c5 = Q has three equal
roots,
=
xA + ax3 + bx2 + cx + d=0 has three equal
If the equation
25.
shew that each of them
is
equal to
— —^r
roots,
^
.
+ =
prove that one of
has two equal roots, t If x5 -hqx3 + rx 2 26. them will be a root of the quadratic 15rx2 - 6q 2x + 25* - 4qr = 0. 27.
In the equation x3 - x -
28.
In the equation xi
and
S6
1
=0,
find the value of
- x3 -1x2 + x + 6 = 0,
S6
.
find the values of
£4
.
Transformation of Equations. discussion of an equation is sometimes simplified by transforming it into another equation whose roots bear some assigned relation to those of the one proposed. Such transformations are especially useful in the solution of cubic equations. 564.
The
565.
To transform an equation
those
into another ivhose roots are
of the proposed equation with contrary
Let f(x) =
Put
-y
signs.
be the proposed equation.
for x; then the equation
every root of f(x) equation is f(—y) =
with 0.
its
f(—y) -
sign changed
;
is
satisfied
by
thus the required
HIGHER ALGEBRA.
472
If the proposed equation is
then
it is
py
n
evident that the required equation will be
-py~
l
2 + p 2 f~ -
+ (-
ir A-,y + (- W. = o, x
obtained from the original equation by changing the sign of every alternate term beginning with the second.
which
is
To transform an equation into another whose roots are 566. equal to those of the proposed equation multiplied by a given quantity. Let f{x) =
be the proposed equation, and
Put y — qx,
given quantity. equation
is
so that
x— -
let q
denote the
then the required
,
f ( - = 0. J
The
chief use of this transformation is to clear
an equation
of
fractional coefficients.
Example.
Remove
from the equation
fractional coefficients
*»-•*-* .+1-0. Put x = - and multiply each term by q 3
By
thus
13
3
;
putting q = 4 all the terms become integral, and on dividing by
2,
we obtain y
s
-Sy 2 -y + Q = 0.
567. To transform an equation into another whose roots are the reciprocals of the roots of the proposed equation.
Let f(x) = X —
-
;
be the proposed equation
then the required equation isy( -
)
=
;
put y = -
,
so that
0.
y
One
of the chief uses of this transformation is to obtain the
values of expressions which involve negative powers of the roots.
symmetrical functions of
TRANSFORMATION OF EQUATIONS. Example
473
If a, b, c are the roots of the equation
1.
Xs -px2 + qx — r = 0,
-+r + b-
find the value of
Write - for y
multiply by
x,
has for
.
c-
and change
y'\
all
the signs; then the re-
-1 = 0,
ry 9 - qy2 +py
suiting equation
-s
.,
a2
111
its roots
''
a' b'
c
2- = ^, S-= =-:
hence
a
ab
r
1
V
r
2
q
a2
Example
- 2pr r-
If a, b, c are the roots of
2.
-3x-l = 0, a -3 + b~ 3 + c~ 3
«3 + 2x 2 find the value of
Writing - for
.
x, the
transformed equation
-2y-l = 0;
y* + Sy 2
and the given expression
is
is
equal to the value of
Ss
in this equation.
S 1= -3;
Here
£2 =(-3) 2 -2(-2) = 13; and
S3 + 3S.2 -2S1 -3 = 0;
whence we obtain
S,= -42.
568. is
1
an equation
If
is
unaltered by changing x into —
,
it
called a reciprocal equation. If the given equation is
x + Pl xn n
-l
2
+p x"- +
+1
2
x +Pn _ iX + ^=0, i
->
n_2
the equation obtained by writing - for
x,
and clearing
of fractions
is n V£? + l\-p
If these
Fl
from the
~
x
2
+ p n -X~ +
•
•
•
+l\n
2
+PF +1 = 0.
two equations are the same, we must have
p
'
?' 2
last result
*
'••' '
V »-*~
we have p
classes of reciprocal equations.
p
=*fc 1,
'
*-»-«
'
P "-p
>
and thus we have two
HIGHER ALGEBRA.
474
pn =l
If
(i)
then
t
P*=P*-»
Px=Pn-^
Pb=P*-B>
'>
the coefficients of terms equidistant from the beginning and end are equal.
that
is,
If
(ii)
2> n
=~
1)
then
p,=-ps _
1
,
p 2 =-p n Ps=-p n - 3 is of 2m dimensions p = — pmi m 2
,
,
;
or £> TO =0. hence if the equation In this case tlie coefficients of terms equidistant from the beginning and end are equal in magnitude and opposite in sign, and if the equation is of an even degree the middle term is wanting.
Suppose that
569.
f (x) =
If
is
f (x) =
f
f(x) = 1
and of an odd degree If (x) divisible by aj + 1.
is is
a reciprocal equation of the
of the second class
and
of
of the second class
and
of
has a is the
it
first class
an odd degree, it in this case f(x) is divisible by as— 1, and as is a reciprocal equation of the first class and of
is
has a root + before (x) =
a reciprocal equation.
of the first class
(x) root —1; so that quotient, then (x) =0 and of an even degree.
If
is
;
an even degree. If f(x) has a root 2
x —
=
is
+
and a root -
1
and as before (x) = Q and of an even degree.
1,
class
it
in this case f{x) is divisible by a reciprocal equation of the first
1
is
an even degree,
;
Hence any
reciprocal equation is of an even degree with its last term positive, or can be reduced to this form; which may therefore be considered as the standard form of reciprocal
equations.
570.
duced
to
A
reciprocal equation of the standard form can be re-
an equation of half
its
dimensions.
Let the equation be
ax2m + bx2m m
dividing by x
a xm +
-
]
+ cx 2m
~2
+
...
+ kx m +
...
+ ex 2 + bx + a = 0;
and rearranging the terms, we have
i) + 6(^' +5 L) + .(.r-' +; l,)+... + *=a.
x
TRANSFORMATION OF EQUATIONS.
475
Now
* +,+ hence writing
^^4)K)-(*-
% for
x+ -
,
and giving
p
to
,+
^);
in succession the values
we obtain
1, 2, 3,...
- = s*
x2 +
2
3 a + -j = z x
-
(z
2
2,
-2) -z =
- = z (z -
-
3
x4 +
4
2>z)
(z
z
2
m and so on; and generally x +
—
therefore the equation in z
m
To find
571.
is
of
the equation
3
- 3z
-
2)
is
;
= z* -
of
m
iz
2
+
2;
dimensions in
z,
and
dimensions.
whose roots are the squares of those
of a proposed equation. 2 Let f(x) = be the given equation ; putting y = x we have Jy\ hence the required equation is f(Jy) — 0. ,
x—
Find the equation whose roots are the squares of those of the
Example.
vP+p^+ptfc+p-^Q.
equation Putting
x=Jy, and
transposing,
we have
(y+P2)Jy= -(PiU+2h)> 2
whence
{y 3
or
2/
Compare the
+ 2p 2 y +pf) y =p x hj 2 + 2p x p z y + p 3 2
,
+ (3pa - Pl*) tf- + (i> 22 - 2Pnh y-p.A 2 = 0. )
solution given in Ex.
2,
Art. 539.
To transform an equation
another whose roots exceed those of the proposed equation by a given quantity. 572.
Let
into
f (x) =
quantity equation
;
be the proposed equation, and let h be the given put y = x + h, so that x = y — h; then the required
is
f(y — h) — 0.
Similarly f(y + h) = h than those oif(x) = 0.
is
an equation whose roots are
less
by
HIGHER ALGEBRA.
476 Example. equation
Find the equation whose roots exceed by 2 the roots of the 4x* + 32a; 3
+ 83a; 2 + 76a; + 21 = 0.
The required equation will be obtained by substituting x - 2 for a; in the proposed equation ; hence in Horner's process we employ x + 2 as divisor, and the calculation is performed as follows :
4
TRANSFORMATION OF EQUATIONS Sometimes
477
be more convenient to proceed as in the
it will
following example.
Remove
Example.
the second term from the equation
px? + qx 2 + rx + s = 0. Let
a,
7 be the
/9,
each of the roots by £dp will be
equal to - -
+
p
roots, so that a
+ p + y= --.
Then
in the transformed equation the
,
-
that
;
is,
if
sum
we
increase
of the roots
the coefficient of the second term will
p
be zero.
Hence the required transformation
will be effected
by substituting
x--~ 6p
for
x in the given equation.
574. From the equation f(x) = we may form an equation whose roots are connected with those of the given equation by some assigned relation.
Let y be a root of the required equation and let cf>(x, y) = denote the assigned relation; then the transformed equation can be obtained either by expressing a; as a function of y by means and substituting this value of x in of the equation <£ (x, y) = f(x) = §; or by eliminating x between the equations f(x) = Q
and
(x,
y)
Example
= 0. If a, b, c are the roots of the equation
1.
11
x 3 +p>x 2 + qx + r=0,
form the equation whose roots are a -
—
,
b
When x = a ,
in the given equation,
r
.
ab
ca
be
—
1
c
,
y = a-
in the transformed equation
=-
j
1 a a = a —— =a + -; a-— abc be r
,
but
and therefore the transformed equation x y v =x+
will be obtained
or
,
x=
r
thus the required equation 2
r y
3
ry ~1
+r
by the substitution
;
is
+Pr{l + r)y 2 + q{l + r) 2 y + (l + r)* = 0.
Form
the equation whose roots are the squares of the differences of the roots of the cubic
Example
2.
x 3 + qx + r = Q. Let
a,
b,
equation are
c
be the roots of the cubic (b
-
c)
2 ,
(c
- a) 2
,
;
then the roots of the required (a - b) 2 .
;
:
HIGHER ALGEBRA.
478
(b-c) 2 = fc 2 + c 2 -2fcc
Now
= a2 + & 2 + c 2 -a 2 -
= (a + b + c) 2 -
2 (be
+ ca + ah) -
2a6c
a
a2
2abc
a
= -2 5 -a 2 + ^; a
also
when x = a
given
the
in
equation,
y
= (b-c) 2
the transformed
in
equation .*.
Thus we have
?/
= - 2o-
a;
2r
. J
H
.
x
to eliminate x between the equations
xs + qx + r=0,
and
.r
By
3
+
(2#
+ y) x - 2r = 0.
(#+?/)# = 3r
subtraction
Substituting and reducing,
y Cor. If a, &, c are 27r + 4g 3 is negative.
s
+6
^2 (2
real, (&
or
;
a:
=
.
we obtain
+ 9^ + 27,-2 + 4^3 _ - c) 2
,
(c
-
a) 2, (a
-
6)
2
.
are all positive
;
therefore
2
Hence in order that the equation x z + qx + r = real 27r2
+ 4<73 must be
If 27r 2
negative, that
is
(-)
may have
+(f) must
all its roots
be negative.
+ 4# 3 =0
the transformed equation has one root zero, therefore equation has two equal roots. the original If 27r2 + 4g 3 is positive, the transformed equation has a negative root [Art. 553], therefore the original equation must have two imaginary roots,
since it is only such a pair of roots which can produce a negative root in the transformed equation.
EXAMPLES. 1.
XXXV.
d.
Transform the equation x3 - 4#2 + - x — - =
integral coefficients,
and unity
for the coefficient of the first term.
Transform the equation 3xA - 5x 3 + x2 - x + the coefficient of whose first term is unity. 2.
Solve the equations 3.
2x4 + x3 -6x2 + x + 2 = 0.
4.
^-10^ + 26^-10^+1 = 0.
5.
6.
into another with
x*-5xi + 9x3 -9x2 + 5.^-1 = 0. 2 A 3 4#fi - 24^ + 57x - Idx + 57 x - Mx + 4 = 0.
1
=
into another
:
TRANSFORMATION OF EQUATIONS. 7.
Solve the equation
479
- 22.r2 + 48.r - 32 = 0, the
3./,-3
roots of which
are in harmonica! progression. 8.
gression
The
roots of find them.
;
x3 - lLr2 + 36#-36 =
2 If the roots of the equation x3 - ax progression, shew that the mean root is 3b.
9.
10.
Solve the equation 4(Xr4
are in harmonica!
pro-
+x—b=0 are in harmonica!
-22^-2Lr2 + 2.t*+l =0,
the roots of
which are in harmonica! progression.
Remove the second term from the equations 8 a? - &c*+ 10a?- 3=0. 11.
13.
x*+4o?+ 2a2 -4# -2=0. 2 A 3 - 1 = 0. afi + 5x + 3X + x + x
14.
afi
15.
x Transform the equation a^-j
12.
- 12s5 + 3.v2 - 1 7.v + 300 = 0. 3
— 7=0 mto
one whose roots
3 exceed by - the corresponding roots of the given equation. 22
16.
Diminish by 3 the roots of the equation
Find the equation each of whose roots of the equation x3 - bx2 + 6x - 3 = 0. root than a
greater
is
17.
by unity
18.
Find the equation whose roots are the squares of the roots of x* + x3 + 2x2 + x+ 1 = 0.
19.
Form
the equation whose roots are the cubes of the roots of
x3 + 3x2 + 2 = 0. If a, b, c are the roots of roots are 20.
ka~\ hb-\ hr\
+c
b
24. ~n 26.
x3 + qx + r-0, form the equation whose
c
21.
a+b
+a
,
«(6 + c), &(c+a), c(a + b c
+
c
r b
c
,
a
a
b 2 c\ c 2 a 2
b).
,
a2 b 2
.
1
1
1
b
25.
«3
,
6 3 , c3
.
b
- + -, T + -a c a b
Shew that the cubes of the roots of x3 + ax 2 +bx + ab=0 are given by the equation x3 + a 3x 2 + b 3x + a3 b 3 = 0. Solve the equation x* - bx* - bx3 + 2bx 2 + 4a - 20 = 0, whose 28. roots are of the form «, —a,b, — b, c. 27.
29.
gression,
x3 + 3px2 + 3qx + r = shew that 2gs =r(3pg — r).
If the roots of
r
are in harmonica] pro-
HIGHER ALGEBRA.
480
Cubic Equations.
The general type
575.
x3
of a cubic equation is
+Px
2
+ Qx + fi=0,
but as explained in Art. 573 this equation can be reduced to the simpler form x3 + qx + r = 0,
which we
shall take as the standard
To
576.
solve the equation x
Let x = y +
z
x3
;
form of a cubic equation.
+ qx + r =
3
0.
then
=y + z + 3yz (y + z) = y3 + z3 + 3yzx, 3
3
and the given equation becomes 3 3 y + z + (3yz +
q)x+r =
0.
At
present y, z are any two quantities subject to the condition that their sum is equal to one of the roots of the given equation if we further suppose that they satisfy the equation thus obtain 3yz + q = 0, they are completely determinate. ;
We
3
y + hence y
3 ,
z
3
z
3
=-r,
y
3
z
3
=
o
3
-^;
are the roots of the quadratic
Solving this equation, and putting
y
=-\
sr
~
z
we
2
+
J V
r
i
+
» ™
it
27
4
= y + z; thus
obtain the value of x from the relation x 1
r
-2 +
fr
2
V4
+
q
3
Y
27}
+
f
r
11r
H-V" '
2
q
3 ^
+ 97 4
The above solution is generally known as Cardan's Solution, as it was first published by him in the Ars Magna, in 1545. Cardan obtained the solution from Tartaglia; but the solution of the cubic seems to have been due originally to Scipio Ferreo, about
;
CU13IC EQUATIONS.
An
1505.
481
interesting historical note on this subject
will
be
and Panton's Theory of Equations.
found at the end of Burnside
By
Art. 110, each of the quantities on the right-hand side of equations (1) and (2) of the preceding article lias three cube roots, hence it would appear that x has nine values ; this, 577.
however,
For since yz = — ^, the cube roots are
not the case.
is
to be taken in pairs so that the product of each pair is rational. Hence if y, z denote the values of any pair of cube roots which fulfil this condition, the only other admissible pairs will be wy, ta*z unity.
and
2 ii>
where
y, a)Z,
Hence the
y+
wz.
= 126.
15.r
for x, then y"*
put
+ z* + {3yz-15)x = 126; 3f/2-15 = 0,
y^z3 = 126; y*zs = 125
then also
hence
2
wy + w 2z,
z,
Solve the equation x 3 -
Example.
imaginary cube roots of
or are the
roots of the equation are
y+
Put y + z
co,
y's , z :i
are the roots of the equation i
2
= 125, sfc=l; y = 5, 2 = 1.
.-.
Thus
j/
-126£ + 125 = 0; 2/3
+ 2 = 5 + l = 6;
u,y
+ u~z =
^—-
5
+-
2
= -3 + 2^/^3; w-y
and the roots are
+ wz = - 3-2^/^3; 6,
-3 + 2*7-3,
-3-2 J -3.
To explain the reason why we apparently obtain nine 578. values for x in Art. 576, we observe that y and z are to be found from the equations y 3 + z3 + r =
0,
yz = — \ o
j
but in the process of 3
solution the second of these
H.H.
A.
was changed into y
J
z
3
= - q^
,
which
31
;
;;
HIGHER ALGEBRA.
482 would also hold
yz =
if
—~
2 ,
or yz
=
^
;
hence the other six
values of x are solutions of the cubics 3
x3 + wqx + r —
2
x +
0,
+r=
ou
0.
We
proceed to consider more fully the roots of the equation x + qx + r = 0. 579.
3
v If
(i)
y and
3
2
q + ~ h .
-r
cube
# represent their arithmetical
y+
are
The
then y 3 and z3 are both real;
positive,
first of
these
z,
is real,
wy +
2 oy z,
2
(o
y+
roots,
let
then the roots
wz.
and by substituting
for
a>
and w 2 the
other two become
r (ii)
2
If -j
3 <7
+
.
is
^=-
zero,
the roots become 2y, 2/(w + r
(iii)
If
2
co
2/(00
),
+
2
or 2y, —3/,
to ),
?/
—
= z, and ?/.
3
2
—
then y 3 — z3 \ in this case
~ <7
+
is
negative, then
pressions of the form a + ib roots of these quantities are the cubic become
3 ?/
and a — ib. m + in and
m + in + m — in, (m + m) o> + (m — in)
,
2
3
are imaginary ex-
Suppose that the cube m — in; then the roots of
or
2m
or
—m—w
(m + m) co + (m — in)
and
^/3
—m + n ^3
As however there is no general arithmetical or algebraical method of finding the exact value of the cube root of imaginary quantities [Compare Art. 89], the solution obtained in Art. 576 is of little practical use when the roots of the cubic are all real and unequal. which are
all real quantities.
This case
is
sometimes called the Irreducible Case of Cardan's
solution.
In the irreducible case just mentioned the solution may be completed by Trigonometry as follows. Let the solution be 580.
1
x = (a +
ib)
3
1
+
(a
- ib)
3 ;
BIQUADHATIC EQUATIONS. a = r cos
put
b
6,
then
= r sin 0,
(«
Now
3 + ^) =
so that r 2
+
{r (cos
= a2 + i
483
=
b\ tan
-
sin 6)f.
by De Moivre's theorem the three values of
this ex-
pression are r3 (^cos - + ,
;.,:
*
sin
+ 2tt J/f cos l^fl 0+27A H + i sin
*\ -\
.
,
<9
and
r3
(cos
+
4tt
.
+
1
.
sin
+
.
4
1
where r 3 denotes the arithmetical cube root of smallest angle found from the equation tan
=
and
r,
-
the
.
a' 1
The three values ot
of (a - ibf are obtained by changing the sign in the above results ; hence the roots are
1
24cos^T,
2r!co8'
I
2,icos^ti^
o
3
Biquadratic Equations. 581
We
shall
now
give a brief discussion of some of the methods which are employed to obtain the general solution of a biquadratic equation. It will be found that in each of the methods we have first to solve an auxiliary cubic equation ; and
thus
be seen that as in the case of the cubic, the general solution is not adapted for writing clown the solution of a given numerical equation. it will
58 2
he solution of a biquadratic equation was ; J tained by Ferrari, a pupil of Cardan, as follows. •
first
i
ob-
Denote the equation by x4 + 2px 3 + qx2 + 2rx + s=0; add to each side (ax + so as to
make
2
the quantities a and b being determined the left side a perfect square; then b)
,
x 4 + 2px3 +(q + a 2)x 2 + 2(r + ab)x +
s
+
b
2
=
(ax +
b)
2 .
Suppose that the left side of the equation is equal to (rf+px+k)*then by comparing the coefficients, we have p* + 2k
- q + a2
,
pk =
r
+
ab,
If
=s+
2
b'
;
31—2
;
;
HIGHER ALGEBRA.
484 by eliminating a and
- r) 2 =
(pk
2k3
or
From
from these equations, we obtain
b
-qk + 2
(21c
+p* - q)
-
k +p
2 ( pr
s)
2
(k
- s),
2
- qs - r 2 = 0.
s
one real value of k can always be found [Art. 553]; thus a and b are known. Also this cubic equation
(x .'.
and the values
of
2
+px +
W=
ax + W'> x 2 +px + k = ±(ax + b); (
x are to be obtained from the two quadratics
x2 + (p — x2 +
and
(k
— b) =
+ a)x+
(k
+ b) = 0.
(2)
x+
a)
t
Solve the equation
Example.
xi -
Add a 2 x 2 + 2abx + b 2 x* - 2x* + (a2
2.x 3
- 5x- + Hh; - 3 = 0.
to each side of the equation,
- 5) x 2 + 2
{ab
and assume
+ 5) x + 6 2 - 3 = (x 2 - x + k) 2
•
then by equating coefficients, we have
a2 =2fc
= -k-5, 2 (2fc + 6)(fc + 3) =
+ 6,
.-.
2k* + 5k 2
.-.
By
trial,
we
find that
k= - 1
But from the assumption,
it
(x2
;
-M-
1
z 2 -3.r + l = 0,
—~—
b,
= 4, ab= - 4.
.
= ± (2x - 2) and x 2 + x-3 = Q;
^—
,
.
-
The following solution was given by Descartes
Suppose that the biquadratic equation
x + qx 2 + rx + 1
assume
;
we have the two equations
a
583.
&2
-x + k) 2 =(ax + b) 2
x2 - x -
whence the roots are
+ 5) 2 7 = 0.
follows that
}
is,
=k 2 + 3;
(7c
hence a 2 = 4,
Substituting the values of k a and
that
b2
ab
x4 + qx2 + rx + s =
(x
2
s
=
+ kx + 1)
is
reduced to the form
;
(x
2
in 1637.
- kx + m)
BIQUADRATIC EQUATIONS. then by equating
+
I
From
vi
the
we have
coefficients,
— k 2 — q,
— l) =
k (m
two
first
Im = s.
r,
we
of these equations,
2m = AT +
485
v
obtain v
q+ t
2l=k 2 + q -j;
,
hence substituting in the third equation, (k
3
+ qk + r)
¥ + 2qk
or
4
+
(k
3
2
(q
+ qk -
r)
- 4s) k 2
= 4sk 2 r
2
,
= 0.
a cubic in k 2 wliich always has one real positive solu2 tion [Art. 553]; thus when k is known the values of I and are determined, and the solution of the biquadratic is obtained by solving the two quadratics This
is
m
x2 + kx + I = Example.
and x 2 — kx +
0,
m=0.
Solve the equation
z 4 -2a; 2 + 8j;-3 = 0. x4 -
Assume
2a; 2
+ 8# - 3 = (x 2 + kx + 1) (x 2 - kx + m)
;
then by equating coefficients, we have l
+ m-k 2 = -2,
whence we obtain
{k 3
= - 12& 2 6 -4fc 4 + 16fc 2 -64 = 0.
- 2k + 8)
or
fc
(A;
3
- 2k -
clearly satisfied when sufficient to consider one of the values of k
This equation
Thus hence
and
as*
7n-l = 4]
- 2.r 2 + 8.r a;
2
3=
+ 2.r-l = 0,
therefore the roots are
584.
8)
is
m+l = 2,
The general
lm = -Z;
k(m-l) = 8,
-
1
that
-4 = 0, or k— ±2. It will be putting k = 2, we have
A;
;
is,
,
2
l
= -l, m = 3.
+ 2x - 1 {x 2 - 2x + 3) and x--2x + 3 =
(.r
2
)
;
;
± J2, 1± J^2.
equations of a degree higher than the fourth has not been obtained, and Abel's demonstration of the impossibility of such a solution is generally accepted by Mathematicians. If, however, the coefficients of an equation are numerical, the value of any real root may be found to any required degree of accuracy by Horner's Method of approximation, a full account of which will be found in treatises on the Theory of Equations. algebraical
solution
of
HIGHER ALGEBRA.
486
We
shall conclude with the discussion of 585. neous equations.
Example
Solve the equations
1.
some
miscella-
:
x + y + z + u = 0,
+ by +cz + du = 0, a2x + b 2 y + c 2z + d2 u = 0, a?x + b 3 y + c 3 z + d3 u = k. ax
Multiply these equations, beginning from the lowest, by 1, p, q, r respectively ; p, q, r being quantities which are at present undetermined. Assume that they are such that the coefficients of y, z, u vanish ; then
x whilst
(a 3
+pa2 + qa + r) = k,
d are the roots of the equation
b, c,
t
Hence
a3
and therefore
Thus the value x
3
+pt2 + qt + r = 0.
+pa2 + qa + r = (a-b){a-c){a- d) (a -b)(a- c) {a -d)x = h. is
found, and the values of y,
z,
;
u can be written down
by symmetry. Cor.
If the
equations are
x + y + z + u = l,
ax + by + cz + du = I;
+ b 2 y + c*-z + d 2 u = k 2 a 3 x + b s y + c 3z + dhi = 3 before, we have x (a 3 +pa 2 + qa + r) = k 3 +pk 2 + qk + r; 2
a x
,
A;
by proceeding as
,
(a-b)(a-c)(a-d) x = (k- b)(k-c)(k-d).
.'.
Thus the value
of x is found,
and the values
of y,
z,
u can be written
down by symmetry. The
solution of the above equations has been facilitated by the use of Undetermined Multipliers.
Example {x
are
2.
Shew that the
-a){x-
b) (x
- c)
roots of the equation
-f2 (x -a)-g (x-b)- h 2 {x -c) + 2fgh = 2
all real.
From
the given equation,
we have
{x-a){(x-b)(x-c)-f*}-{g*{x-b) + h*(x-c)-2fgh}=0. Let p, q be the roots of the quadratic
{x-b)(x-e)-f*=0,
;
BIQUADRATIC EQUATIONS. and suppose ^
than
to be not less
By
q.
487
solving the quadratic,
we have
2x = b + c±J(b-c)* + tf
:i
now
the value of the surd
or
and
c,
q
than
is less
is
~
greater than b
(1);
c,
p
so that
is
greater than h
b or c.
In the given equation substitute for x successively the values
+ °°»
-
q>
v,
30
;
the results are respectively
+ °°
-fajp^b-h Jp ~ c
,
since
{p
-b)(p-
2
+{
>
)
Jb-q- h Jc - q)~,
=f* = (b - q)
c)
Thus the given equation has three _p and q, and one less than q.
(c
-
real roots,
-
cc
,
q).
one greater than
2',
one
between
=
2 If p q, then from (1) we have (6-c) In this case the given equation becomes
(x
thus the roots are
-b){{x-
a) (x
+ 4/ 2 =
-b)-g*-
lr}
and therefore
b
= c,f=0.
=0
all real.
If p is a root of the given equation, the above investigation fails for it only shews that there is one root between q and + oo namely p. But as before, there is a second real root less than q bence the third root must also be real. Similarly if q is a root of the given equation we can shew that all ;
,
;
the roots are real.
The equation here discussed is frequently in Solid Geometry, and
it occurs of considerable importance is there known as the Discriminating ;
Cubic.
586.
The following system
of
equations occurs in
branches of Applied Mathematics. Example.
Solve the equations
x
z
y
a+\
b
x a
:
+\
c
b
+
+\ z
y
-\-fji.
/j,
c
+ fx
— + —+ c+v = a+f b+ x
,
y J
z
-,
1.
v
Consider the following equation in
x a x, y, z
+
+
y b
+d
+ c
6,
z
(0-X)(g-ft)(g-y)
+
(a
being for the present regarded as
.
+ e)(b + 6){c + 0y known quantities.
many
::
HIGHER ALGEBRA.
488
This equation when cleared of fractions is of the second degree in 6, and by the three values 8 = \ 6 = p., d = v, in virtue of the given equations hence it must be an identity. [Art. 310.] is
satisfied ;
To
find the value of #, multiply
..
thus
-
up by a+0, and then put a +
(--X)(-^)(-"-'0 (b
that
-
a) (c
-
= 0;
.
a)
. = fe+£Lfe+i4!ttd.
is,
(a-
(a-c)
b)
By symmetry, we have
y=
(b
+ \){b + fx)(b + v) {b-c)(b-a)
{c
and
+ \){c + fi)(c + v) (c
-a)
(c
-
b)
XXXV.
EXAMPLES.
e.
Solve the following equations 1.
a3 -18a = 35.
2.
a?+ 7207- 1720=0.
3.
a3 + 63a- 316 = 0.
4.
ff
6.
&s -15#8 -33ar+ 847=0.
+ 21# + 342 = 0.
7.
28^-9^+1=0. 2a3 + 3a2 + 3a + 1=0.
8.
Prove that the real root of the equation a3 + 12a -12 =
5.
is
3
2^/2-^4. Solve the following equations 9.
a4 -3a2 -42 a -40 = 0.
11.
a4 + 83? + 9a2 -8a -10 = 0.
12.
a-
13. 15.
16. 17.
10.
a4 - 10a2 - 20a- 16 = 0.
+ 2a3 - 7 a2 - 8a + 1 2 = 0. **- 3^-6^-2=0. a*-23?-12afi+10x + 3=0. 14. 4a4 - 20a-3 + 33^ 2 - 20a + 4 = 0. a6 -6a4 -17a3 + 17a2 + 6a-1 = 0. a + 9a3 + 1 2a 2 - 80a - 1 92 = 0, which has equal roots. 4
4
Find the relation between q and r in order that the equation ^A A + + r=0 may be put into the form a 4 = (a2 + «a+&) 2 18.
3
.
Hence
solve the equation
8a3 -36a + 27 = 0.
BIQUADRATIC EQUATIONS.
489
jfi+3pafl+3qx+r and x*+2px+q have a common factor, shew that 4(p 2 -q) (q 2 —pr) - (pq-r) 2 = 0. 19.
If
If they
20.
If
common factors, shew that 2 p -q=0, q 2 -pr=0.
have two
the equation
shew that each of them 21.
axs + 3bx2 + 3cx + d=() has two equal
is
equal to —. rs? 1 2 (etc - b 2 )
.
Shew that the equation x4 +PX3 + qx2 + rx + s =
as a quadratic
if
roots,
may
be solved
r2 =p 2s.
Solve the equation
22.
one of whose roots If a,
23.
/3,
gfl
- 1 SxA +
is
J6 — 2.
y, 5
find the equation
1 6.1*
3
+ 28x2 - S2x + 8 = 0,
are the roots of the equation xA + qx2 + r.t- 4- s = 0,
whose roots are /3+y + d + (/3y§) _1
,
&c.
In the equation x4 — px3 + qx2 - rx + s = 0, prove that if the sum of two of the roots is equal to the sum of the other two p 3 - 4pq + 8r = and that if the product of two of the roots is equal to the product of the other two r2 =p 2 s. 24.
;
25.
unity
:
The equation x° - 209.£ + 56 = determine them.
26.
Find the two roots of
27.
If a,
b, c,...k
has two roots whose product
^ — 409^ + 285 =
whose sum
is
is 5.
are the roots of
Xn +p 1Xn ~ 1 +p2Xn ~ 2 +
+Pn-l$ +Pn = °>
shew that
(l+a 2 )(l+b 2 ) The sum
{
l
+ k 2 = (l-p,+p±- ...) 2 + )
(
Pl
-p,+p,-
...)
2 .
two roots of the equation 4 - 8.r> + 21^2 - 20a- + 5 = is 4 explain why on attempting to solve the equation from the kuowled^e of this fact the method fails. 28.
of
.-r
;
)
:
MISCELLANEOUS EXAMPLES. If s l s 2i *3 are the sums of n, 2n, Sn terms respectively of an arithmetical progression, shew that s 3 = 3 (s 2 — sj. 1.
,
Find two numbers such that their are to one another as 1, 7, 24.
difference,
2.
3.
In what scale of notation
4.
Solve the equations
5.
first
p
is
sum and
product,
25 doubled by reversing the digits?
:
(1)
(#+2)(#+3)(a;-4)(#-5)=44.
(2)
x(y + z) + 2 = 0,
z(2x-y) = b.
y(z-2x) + 2l=Q,
In an A. P., of which a is the first term, if the terms = 0, shew that the sum of the next q terms
sum
of the
a{p + q)q ^
p—l
[R.
M. A. Woolwich.]
Solve the equations
6. (
11
(a + b) (ax +
1
(2)
x*
b)(a- bx) = (a 2 x - b 2 )
(a + bx).
i
+ (2x-Zf={l2(x-l)Y.
[India Civil Service.]
Find an arithmetical progression whose first term is unity 7. such that the second, tenth and thirty-fourth terms form a geometric series. 8.
If a,
fi
are the roots of
a 2 + a/3 + /3 2 9.
10.
If
,
x-+px+q = 0, find a 3 + /3 3 a 4 + a 2/3 2 +
the values of 4
,
.
2x — a + a~ 1 and 2y = b + b~ 1 find the value of xy + *J(x2 -\)(y 2 - 1). ,
Find the value of (4
3
3
2
2
3
3"
+ Vl5)" + (4-Vi5)'
_
2
(6
+ V35)" -(6-\/35)' [R.
11.
If a
and
/3
M. A. Woolwich.]
are the imaginary cube roots of unity, a
4
+ ^ + a4
1
^-
1
= 0.
shew that
:
MISCELLANEOUS EXAMPLES. Shew that in any scale, whose radix is 12. number 12432 is divisible by 111 and also by 112.
401
greater than
B
4,
the
B
A and run a mile race. In the first heat A gives a start 13. in the second heat A gives of 11 yards and beats him by 57 seconds a start of 81 seconds and is beaten by 88 yards in what time could ;
B
:
each run a mile ? 14.
Eliminate
x, y, z
between the equations
x 2 —yz— a 2 y 2 - zx = b 2 ,
- xy = c2
z2
,
:
x -f y + z<= 0.
,
[R. 15.
M. A. Woolwich.]
Solve the equations
ax 2 + bxy + ey 2 = bx 2 + cxy + ay 2 = d. [Math. Tripos.]
A
waterman rows
and back in 14 hours: he finds that he can row 4 miles with the stream in the same time as 3 miles against the stream find the rate of the stream. 16.
to a place 48 miles distant :
17.
Extract the square root of
+ ab + be + ea) {be + ca + ab + b 2
(1)
(a 2
(2)
l-.r+\/22^-15-8^2
)
(be
+ ca + ab + c2
).
.
10
18.
Find the
coefficient of
term independent of x in
(
xG
-x2 -
\^S
19.
Solve the equations
in the expansion of (1 - Sx)
— AX
)
:i
,
and the
.
J
:
3^-8
/1N
2.r-3
(2)
x2 -y 2 = xy — ab,
ff+3
n
(x + y) (ax + by)
= 2ab(a + b).
[Trin. Coll. Camb.] 20.
Shew that
if
a(b-c) x2 + b (c-
square, the quantities a,
b, c
a)
xy + c(a-b)y 2
is
a perfect
are in harmonica! progression.
[St Cath. Coll. Camb.] 21.
If
(y-z) 2 + (z-x) 2 + (x-y) 2 = (y + z-2x) 2 +
and
x, y, z are real,
22.
find in
shew that
x=y =
z.
(z
+ x-2y) 2 + (x+y-2z) 2
,
St Cath. Coll. Camb.]
Extract the square root of 3e582Gl in the scale of twelve, and
what
scale the fraction o
would be represented by
-17.
—
:
:
HIGHER ALGEBRA.
492
Find the sum of the products of the integers 1, 2, 3, ... n taken two at a time, and shew that it is equal to half the excess of the sum of 23.
the cubes of the given integers over the
A man
sum
of their squares.
his family consume 20 loaves of bread in a week. If his wages were raised 5 per cent., and the price of bread were raised 2\ per cent., he would gain 6d. a week. But if his wages were lowered 7^ per cent., and bread fell 10 per cent., then he would lose \\d. a week find his weekly wages and the price of a loaf. 24.
and
:
25.
The sum
of four
numbers
the product of the extremes find the numbers. 26.
Solve the equations (1) fr%.
(2) v '
27.
if
in arithmetical progression is 48 and to the product of the means as 27 to 35 :
:
a{b-c)x2 + b(c-a)x+c(a-b) = 0.
— x-a —
— x-c-d
(x-a)(x-b) 1± l — (x-c)(x-d) \ ^ —£ b
b
If /s/a-x
+ ^/b-x+\/c-x=0
(a + b
and
is
i
[Math. m Tripos.]J r, r L
.
..
shew that
+ c + 3x) (a + b + c-x) = 4(bc + ca + ab)-,
^a + 4/6+4/c = 0, shew
that (a + b + c) 3 = 27abc.
A train,
an hour after starting, meets with an accident which detains it an hour, after which it proceeds at three-fifths of its former rate and arrives 3 hours after time but had the accident happened 50 miles farther on the line, it would have arrived l£ hrs. sooner find the 28.
:
:
length of the journey. 29.
Solve the equations
2x+y = 2z, 9z-7x=6y, x3 +f + z 3 =2l6. [R.
M. A. Woolwich.]
Six papers are set in examination, two of them in mathematics in how many different orders can the papers be given, provided only that the two mathematical papers are not successive ? 30.
31.
In how
many ways can
£5.
consisting of half-crowns, shillings
4s. 2d.
be paid in exactly 60 coins,
and fourpenny-pieces ?
Find a and b so that x3 + ax2 + llx + 6 and x3 + bxi + l4x + 8 may have a common factor of the form x2 -\-px + q. 32.
[London University.]
do
In 33. in six
it
the time 1
what time would A,B,C together do a work if A alone could hours more, B alone in one hour more, and C alone in twice
:
MISCELLANEOUS EXAMPLES. 34. .,
35.
sion of 36.
ax + by = \, ex 2 + dy% = 1 have only >ne solution a b ___ „, and x = - y = -, [Math. Tiuros.]
If the equations a2 b2 , .
prove that
—
+-7
= 1,
i
,
.
,
Find by the Binomial Theorem the
(l-2x + 2x2 )~'2
shew that p - q (3p -
terms
first five
in the
expan-
'
If one of the roots of 3
493
x 2 -f- px + q —
the square of the other,
is
+ q = 0. 2
1)
[Pemb. Coll. Camb.] 37.
Solve the equation
xi -5x^-6x-b = 0. [Queen's Coll. Ox.] 38.
Find the value of a
for
which the fraction
x3 - ax2 + 19.27 - a — 4 x?-(a + l) x2 + 23x-a~7 Eeduce
admits of reduction. 39.
If a, b,
c,
it
x, y, z are real quantities,
(a + b + c) 2
and
= 3 (be + ca + ab- x2 -y
a = b = c, and x = 0, y = 0,
shew that
[Math. Tripos.]
to its lowest terms.
2
2
- z2),
= 0. [Christ's Coll. Camb.] i
40.
What
is
the greatest term in the expansion of
x
is
-
the value of
(
1
-
-
x]
when
[Emm. Coll. Camb.]
?
Find two numbers such that their sum multiplied by the sum of their squares is 5500, and their difference multiplied by the difference of their squares is 352. [Christ's Coll. Camb.] 41.
42.
If
x = \a,
y = (k-l)b,
2 2 1 = (\-3)c, X= —z—b0"2+ 3c ~p _|_
s
'
,
Qj
x2 +y 2 + z 2
in its simplest
form in terms of
a, b,
"T"
C
,
,
express
c.
[Sidney Coll. Camb.] 43.
Solve the equations (1)
xa + 3j*=16x + 60.
(2)
y
2
+ z 2 -x = z 2 + x' -y = x 2 +y -z = \. i
2
[CoRrus Coll. Ox.] 44.
If x, y, z are in harmonical progression,
log (x + z)
+ log
{x
shew that
-2y + z) = 2 log (x - z).
:
HIGHER ALGEBRA.
494 Shew
45.
1
that
1.3.5 /lV
1.3/1\
4
._
/oN
,
,_
[Emm. Coll. Camb.]
3a-26~36-2c~3c-2a' b(x+y + z) (5c + 46 -
then will
3a) = (9x + 83/
+
13^) (a
+ b + c).
[Christ's Coll. Camb.] 5 vowels, how many letters can be formed having 2 different vowels in the consonant (repeated or different) at each end?
With
47.
17 consonants
and
words of four middle and 1
A question was lost
on which 600 persons had voted ; the same persons having voted again on the same question, it was carried by twice as many as it was before lost by, and the new majority was to the former as 8 to 7 how many changed their minds? [St John's Coll. Camb.] 48.
:
Shew
49.
that l-x
(l+x)
2
5x*
9^5
+ 2.3 + 4.5 + l+£?-^
13^7
6.7
+ "'
[Christ's Coll. Camb.]
A body of men were formed into a hollow square, three deep, was observed, that with the addition of 25 to their number a solid square might be formed, of which the number of men in each side would be greater by 22 than the square root of the number of men in each side of the hollow square required the number of men. 50.
when
it
:
51.
52.
Solve the equations (1)
V (a + x) + 2 V(a^0 = 3 \/a
(2)
(x - a)* (x
2
2
- 6)2
- {x
2
^
2 .
- c)i (x - d)% = (a - c)%
(6
- d)K
Prove that 3/,
v/4 N
= H^62 t
2
,
-
5
1
6.12
v
2.5.8 — — + 6.12.18 [Sidney Coll. Camb.]
53.
Solve $6(5a? + 6)-^5(6#-ll)=l.
[Queens' Coll. Camb.]
:
MISCELLANEOUS EXAMPLES.
405
A
vessel contains a gallons of wine, and another vessel contains b gallons of water: c gallons are taken out of eaeh vessel and transferred to the other; this operation is repeated any number of times shew that if c(a + b) ab, the quantity of wine in each vessel will always remain the same after the first operation. 54.
=
:
m
The arithmetic mean between
55.
mean between a and of a and
b are each equal to
m+n
-: find
m and n in terms
b.
If x, y, z are such that their
56.
and n and the geometric
sum
is
constant, and
if
(z+x-2y)(x+y-2z) varies as yz, prove that 2 (y
+ z) - x
varies as yz.
[Emm. Coll. Camb.] Prove that,
57.
1.2. M CV2.3.'
l
if
n
greater than
is
3,
C_ + 3.4.«
+ (-l)'-(r+l)(/-+2)=2.»- 3 C
r
1
.
[Christ's Coll. Camb.] Solve the equations
53.
(1)
*J'2x
- 1 + */&v -
2
= *J~4x - 3 + *Jbx^~i.
3
I
a a 8 (2) 4{(s -16)*+8}=# +16(# -16)*
[St John's Coll. Camb.] 59. .,
another,
Prove that two of the quantities y-
.j.
z
f l+yz
if
must be equal
x, y, z
to one
-x — — + x y =n — l+zx l+xy z
h
2-
-
0.
In a certain community consisting of p persons, a percent, can read and write ; of the males alone b per cent., and of the females alone find the number of males and females in c per cent, can read and write the community. 60.
:
61.
If
!•=?•'-" [Emm. Coll. Camb.]
62.
Shew that the
(1
63.
x4n
coefficient of
—x+ x — x 2
3 )' 1
in the expansion of is
unity.
Solve the equation
x-a + x-b = a
b
x —a
+
a
x-b' [London University.]
Find (1) the arithmetical series, (2) the harmonical series of 64. n terms of which a and b are the first and last terms and shew that tb the product of the r* term of the first series and the {n — r+ l) term of ;
the second scries
is ab.
=
.
:
:
HIGHER ALGEBRA.
496
If the roots of the equation
65.
1
" q+
(1
+q) x+q ( q ~
shew that p 2 = 4q.
are equal,
If a 2 + b 2
66.
^J
* 2+p
= lab, l°g
1)
+ f =0 M. A. Woolwich.]
[R.
shew that
jg (« +
V)
}
=g
(
lo S
a + loS h )[Queen's Coll. Ox.]
If
67.
n
is
a root of the equation x- (1
- ac) - x (a2 + c 2 ) - (1 + ac) = 0,
n harmonic means are inserted between a and c, shew that the difference between the first and last mean is equal to ac {a — c). and
if
[Wadham If n + 2
68.
W" 2 8
:
P = 57 4
:
Coll. Ox.]
16, find n.
A
person invests a certain sum in a 6rr per cent. Government the price had been £3 less he would have received \ per cent, more interest on his money ; at what price was the loan issued ? 69.
loan
:
if
Solve the equation
70.
{(^
2
+ ^ + l) 3 -(^2 + l) 3 -^3}{(^2 -^ + l) 3 -(^ 2 + l) 3 +^3} = 3 {(^4 + x2 + 1) 3 - (#*+ If - a6} [Merton Coll. Ox.]
If
71.
by eliminating x between the equations
x2 + ax + b
an d xy + 1 (x + y) + m = 0,
a quadratic in y is formed whose roots are the same as those of the original quadratic in x, then either a =21, and 6 = m, or b + m=al. [R.
M. A. Woolwich.]
Given log 2 = '30103, and log 3 = -47712, solve the equations
72.
(1)
6*=y-6-«.
(2)
V5M-V5-*=|q.
Find two numbers such that their sum is 9, and the sum of [London University.] their fourth powers 2417. 73.
A
walk at the rate of 4 miles an hour after he had been walking 2| hours, B set out to overtake him and went 4£ miles the first hour,4| miles the second, 5 the third, and so gaining a quarter In how many hours would he overtake A of a mile every hour. 74.
set out to
;
l
a
75. factor.
+ 1 as Prove that the integer next above (^3 + l) 2m contains 2 m
-
;
MISCELLANEOUS EXAMPLES. The
76.
407
numbers is divided into groups 1 2, 3, 4 prove that the sum of the numbers in the
series of natural
and so on 5, 6, 7, 8, 9 th group is (?i- l) 3 + n 3 ?i ;
:
;
.
Shew
77.
2
+
that the
sum
of
[3
\2/
+
|2_W
is
n terms of the
series
+
+ |4
,,
1.3.5.7 =—
,
to equal 1
\2J
1
—(2n-l)
:
2'*
[R.
Shew that the
78.
coefficient of
xn
M. A. Woolwich.] 1
in the expansion of
+ 2x
j—2
is
w-2
n-1
n
.
\n
(-l)S 3(-l)3, 2(-l)3, according as n
is
of the form 3m,
Solve the equations
79.
(1 )
.„.
£ a
x y
3m + 1,
3«i
+ 2.
:
= ^_ 2 _
yyz
b
c
x+y + z
ii
z
v
z
x
z
x
x
y
z
[Univ. Coll. Ox.]
The value
80.
of xyz
is
b is arithmetic or harmonic to be positive integers.
:
7£ or 3f according as the series a, x, y, z, find the values of a and b assuming them [Merton Coll. Ox.]
If ay-bx=c \/(x -a) 2 (y- b) 2 , shew that 2 will satisfy the equation unless c2 < a 2 + b
+
81.
and y
no real values of x
.
If
82.
(#+l) 2
is
greater than 5x
- 1
and
less
than 7#-3, find the
integral value of x.
P
is the number of integers whose logarithms have the If characteristic p, and Q the number of integers the logarithms of whose reciprocals have the characteristic - q, shew that
83.
log 10 P-log 10 #
= p-2 + l.
In how many ways may 20 shillings be given to 5 persons so 84. t lat no person may receive less than 3 shillings ?
A man
wishing his two daughters to receive equal portions came of age bequeathed to the elder the accumulated interest of a certain sum of money invested at the time of his death in 4 per and to the younger he bequeathed the accumulated cent, stock at 88 interest of a sum less than the former by £3500 invested at the same time in the 3 per cents, at 63. Supposing their ages at the time of their father's death to have been 17 and 14, what was the sum invested in each case, and what was each daughter's fortune ? •
85. rilen they
;
11.
11.
A
32
HIGHER ALGEBRA.
498
A
of three digits in scale 7 when expressed in scale 9 has its digits reversed in order find the number. [St John's Coll. Camb.] 86.
number
:
m
If the sum of to the sum of the next 87.
terms
;
terms of an arithmetical progression is equal n terms, and also to the sum of the next p
prove that (m + n)(
i
= (wi +p)(
)
•
[St John's Coll. Camb.]
Prove that
88.
1
1
/
1
—
V
1
1
1
+ + 7—4 + 7-—. = 7— + (y-z? {z-xf (x-y)vi2 \y-z z-x7 x-y) [R.
If
89.
m
M. A. Woolwich.]
and greater than + (2n-l) m >nm + 1
negative, or positive
is
l
m + 3™ + 5 m +
1,
shew that
.
[Emm. Coll. Camb.] If
90.
each pair of the three equations
x2 -p 1 x + q l = 0, aP-ptfC+q^Q, x 2 -p 3x+q3 =0, have a common
root,
2
Pi +P-?
prove that
+ P32 + 4
(?i
+ ft + ft) = 2 (P2P2 +P?,Pi +PiP
and B travelled on the same road and at the same rate from Huntingdon to London. At the 50 th milestone fioin London, A overtook a drove of geese which were proceeding at the rate of 3 miles in 2 hours and two hours afterwards met a waggon, which was moving at the rate of 9 miles in 4 hours. B overtook the same drove of geese at the 45 th milestone, and met the waggon exactly 40 minutes before he came to the 31 milestone. Where was B when ^4 reached London ?
A
91.
;
st
[St John's Coll. Camb.]
Ifa + 5 + c + c?=0, prove
92.
that
abc + bed + cda + dab = *J(bc- ad) (ca
- bd) [R.
An
93.
terms
:
A.
P.,
shew that
a G.
(?i
+ 2)
—
1
P. have a and b for their first terms will be in G. P. if
and an H.
P.,
their
77-0
th
«tn
ba(b 2n -a2n )
94.
Shew that the
— cd). M. A. Woolwich.] {ah
=
coefficient of
[Math. Tripos.]J
.
L
n
xn
x
in the expansion of
,
(x in ascending 01power of in the expansion of r
x
a n — bn is
(l-#)L3
-,,
—^
a-b is
— a
I
.
7-
n bn
;
'
two
and that the
2 2 n_1 hi + 4w + 2l » '
r — a) (x - 0) r-,
coefficient of
v
x2n
__ r r [Emm. Coll. Camb.] ,
,
-.
z
MISCELLANEOUS EXAMPLES. Solve the equations
95.
1
/
;
#--1
/
*/x - y
Find the value of
96.
499
Till 1
1
+
1
sF+y*
,
1
:
ay =34
:
15.
[St John's Coll. Camb.]
1
...
ratio surd.
in the
[R.
form of a quad-
M. A. Woolwich.]
Prove that the cube of an integer may be expressed as the difference of two squares that the cube of every odd integer may be so expressed in two ways and that the difference of the cubes of any two consecutive integers may be expressed as the difference of two squares. [Jesus Coll. Camb.] 97.
;
;
Find the value of the
98.
infinite series
1 + 1 + J3 + i + '" |5
13
"
x—
If
99.
|7
b
a
+ d+ b+ d + a
and
y
d+
b
a
+ d+ b+
bx-dy=a-c.
then 100.
n th term
If a,
6, o
[Christ's Coll. Camb.]
and the
are in H. P., then
a+b
c+b
(1)
2^> + 27^ >4
(2)
b 2 (a-c) 2 =2{c2
by x - a
'
Find the generating function, the sum to n terms, of the recurring series 1 + 5# + 7x 2 + 1 7.V3 + 31. 4 +
101.
102.
[Emm. Coll. Camb.]
|9
-
(b-a) 2 + a 2 (c-b) 2 }.
If a, 6, c are all real quantities, and and also by x - b ; prove that either a
[Pemb. Coll. Camb.]
x3 - 3b 2x + 2c3
= b = c,
or a =
is divisible
— 26 = — 2<\
[Jesus Coll. Ox.]
i
that the sum of the squares of three consecutive odd umbers increased by 1 is divisible by 12, but not by 24. 103.
Shew
104.
Shew that
according as a
is
is
the greatest or least value of ax2 + 2bx + c,
negative or positive.
x*+yA + zi +y 2 z 1 + z2x2 + x2y 2 = Zxyz (x+g + z), and x, y, z arc all [St John's Coll. Camb.] shew that x=y=z.
If real,
32—2
-
n
HIGHER ALGEBRA.
500
"
105.
Shew that the expansion
JL3 + 2?4' 2
106.
If a,
V
of
2~
^ + 1.3.5.7
x 1S
/ l-Vl-a;2 a»
+ 2. 4. 6.8* 10
6
are roots of the equations
/3
x2 +px + q = 0, x2n +p nxn + q n = 0, where n
is
an even integer, shew that ~ — are roots of ,
.r»
107.
a
P
+ l + (#+l) n = 0.
Find the difference between
[Pemb. Coll. Camb.] squares
the
the infinite
of
continued fractions
a+-—b
2a+
b
b
^—,2a +
d
,
^..
2a+
"'
"
j
aD0
-
c
d
+ 2c+
d 2c +
2c +
[Christ's Coll. Camb.] 108.
A
109.
Solve the equations
distributed amongst a certain number of persons. The second receives Is. more than the first, the third 2s. more than the second, the fourth 3s. more than the third, and so on. If the first person gets Is. and the last person £3. 7s., what is the number of persons and the sum distributed 1
sum
K
'
(2)
110.
If
of
a
money
b+c
~
is
:
b
c
+a
c
a+ b
2
&&+"»=*&
+ x*+f= l3 i>
a and b are positive and unequal, prove that
a*-b n >
(«
b) (ab) 2
.
[St Cath. Coll. Camb.] 111.
hence find the least
Express ^r^ as a continued fraction;
values of x and
y which
satisfy the equation 396.t'— 763y
= 12.
a certain work, a workman A alone would take many days as B and C working together B alone would take n times as many days as A and C together C alone would take p times as many days as A and B together shew that the numbers of 112.
m times
To complete
as
;
;
:
days in which each would do Prove also
m+l
H
n+l
it
alone are as
+ -^— = 2. p+l
m + 1 »+l :
m
[R.
:
jp
+ 1.
,_ , M. A. Woolwich.]
,,
.
—
;
MISCELLANEOUS EXAMPLES.
501
The expenses of a hydropathic establishment are partly con113. stant and partly vary with the number of boarders. Each boarder pays £65 a year, and the annual profits are £9 a head when there are 50 boarders, and £10. 13s. 4d. when there are 60: what is the profit on each boarder when there are 80 ? 114.
If
x2y = 2x — y, and x2
is
not greater than
shew that
1,
[Peterhouse, Camb.] 115.
X V — c2 shew that when a and i = Ti and xv h — —sr ° 2 a -y- a -x2 o
—
If -sl
.
c
are unequal, (a 2
116.
If
-c 2 2 -b 2 c2 = 0,
(1
+ x + x 2 f r = 1 + k\x + 2 x2 + l'
. .
(1)
= a*" - c^' ~ + c^s* " 2 = 1, \—k + k2 -
(2)
l-k^ + hc., —
and
(x
prove that
a2 + c 2 -b 2 = Q.
or
)
-
1 ) 3r
.,
1
. .
.
x
!3r
=± \r\2r [R.
117.
Solve the equations (1) {x
118.
:
— y) 2 + 2ab = ax+by,
x2 -y 2 + z 2 = 6,
(2)
M. A. Woolwich.]
xy + ab = bx + ay.
x-y + z = 2.
2yz-zx + 2xy = 13,
n positive quantities alt a 2 ,... a n and if the their products taken two together be found, prove
If there are
square roots of that /
all
—
,
n—\ <— «— («i + « +
/
N
,
Vaia2 + V«i«3+
+ an);
2
hence prove that the arithmetic mean of the square roots of the products two together is less than the arithmetic mean of the given [R. M. A. Woolwich.] quantities. 119.
If
6¥ + «V=a
2 6'i ,
and d2 + V = x 2 +y 2 = \, prove that
Wx6 + a*y G = (b 2xA -f a 2y 4 120.
Find the sum of the (1)
~
r
first
2 )
[India Civil Service.]
.
n terms of the
|~_,
(2)
series
whose
r th
terms
(a+r*6)*-' [St John's Coll. Camb.]
121.
Find the greatest value of
x+ 2 o
.
2iX" t~ *iX
+~a
— HIGHER ALGEBRA.
502
Solve the equations
122.
:
l+^ 4 = 7(l+#) 4
(2)
3#y+20=ff0+6y=2^s+3d?=O.
123.
(1)
.
a2 a 3> ai are an y
If «x,
f° ur consecutive coefficients of
y
an
expanded binomial, prove that
—— I
124.
Separate
find the general
3
I
—=
'
\
£-
/
,
[Queens' Coll. Camb.1
.
=r into
„
3x — 8
term when
is
2
partial fractions
;
and
expanded in ascending powers
of X.
125.
In the recurring series 5 4
:
- lx + 2x* + lx3 + bx 4 + 7x + >
2
the scale of relation is a quadratic expression determine the unknown coefficient of the fourth term and the scale of relation, and give the [R. M. A. Woolwich.] general term of the series. ;
126.
and
If x, y, z are unequal,
if
2a-3z^^^ z
2a-3v 9 = -^2,and y (
(v
2a -3.?=—
then will
-
*)
,
2
and
,
x+y + z = a.
[Math. Tripos.]
Ob
Solve the equations
127.
:
(1)
xy + 6 = 2x-x2
(2)
{ax)^ a = {by)^ h
,
xy-9 = 2y-y 2 b Xo & x
,
.
= a lo%y.
Find the limiting values of
128.
(1)
,
.
(2)
x \fx2 + « 2 -
*JxA
——++ x-—?—— \fa
\/Za
+ a when x = oc 4
,
2x—\/3x
University.]T [London tt rr
,
,
.
when x—a.
2sjx
There are two numbers whose product is 192, and the quotient of the arithmetical by the harmonical mean of their greatest common measure and least common multiple is 3f | find the numbers. 129.
:
[R.
M. A. Woolwich.]
:
MISCELLANEOUS EXAMPLES. 130.
Solve the following equations
yiar + 37- J/l3.r-37=
(1)
:
J/2.
6Vl-2 2 + c\/l-y 2 = «,
(2)
c \/l
- #2 + « Vl - 22 = 6,
a*Jl-y 2 + b*Jl-x 2 =c. 131.
503
Prove that the 1
2^3 "
1.3 24)4
sum
_
to infinity of the series
1.3.5
+ ~Wb
*
* "
.23
2
1S
3*
~ 24
,
n
[Math. Tripos.]
A
number consisting of three digits is doubled by reversing 132. the digits; prove that the same will hold for the number formed by the first and last digits, and also that such a number can be found in only one scale of notation out of every three. [Math. Tripos.] 133.
Find the
x 12 and xr
coefficients of
n_
1+x3 2ui_
\
an( * 1
in the product of
-*+*"
[R-
M. A. Woolwich.]
A
purchaser is to take a plot of land fronting a street ; the 134. plot is to be rectangular, and three times its frontage added to twice its depth is to be 96 yards. What is the greatest number of square yards he 135.
may
[London University.]
take ?
Prove that
+ b + c + dy + (a + b-c-dy + (a-b + c-dy + (a-b-c + d)* - (a + b + c - d)* - (a + b - c + d) A - (a - b+c+df - (- a + b + c + d)* (a
= 192 abed. [Trin. Coll. Camb.] 136.
Find the values of
pressions xt + aaP + square. 137.
bx'Z
which will make each of the exand xA + 2ax3 + 2bx2 + 2cx + 1 a perfect [London University.]
a, b, c
+ cx+l
Solve the equations
(1)
4^S
(2)
\j2x2 +\
f
= 3(
+ \l& -
1
^=65.
= V3 - 2j--
A farmer sold 10 sheep at a certain price and 5 others at 10*. per head; the sum he received for each lot was expressed in pounds by the same two digits find the price per sheep. 138.
less
:
)
:
;
HIGHER ALGEBRA.
504
Sum
139.
n terms
:
(1)
(2»~l)+2(2»-3)+3(2»-5)+....
(2)
The squares
(3)
The odd terms
If a,
140.
that
to
3 (a
2
/3,
+ /3
2
of the terms of the series of the series in
a?(%-5)=
(1)
15
[Trin. Coll. Camb.]
(2).
3 y are the roots of the equation x + qx + r=0 prove 2 5 5 5 = 3 3 3 4 + y ) (a + /3 + y ) 5 (a + /3 + y ) (a + £ 4 + y 4 ). [St John's Coll. Camb.]
Solve the equations
141.
1, 3, 6, 10,
:
41
(2)
y(2A + 7) = 27J'
A3 +y3 + z 3 = 495)
ar+y+*=15V. Ay2=105 [Trin. Coll. Camb.]
142.
If a,
equation whose 143.
are the roots of the equation x3 + qx2 +r = 0, form the roots are a + b-c, b + c — a, c + a-b.
b, c
Sum
the series
(1)
n + (n-l)x + (?i-2)x2 +...+2zn - 2 + x n - 1
(2)
3 - x - 2x -
(3)
6+9
2
1 6a-
3
- 28^ - 676a + ... 4
+ 14 + 23 + 40 +
..
.
5
to
;
to infinity
n terms.
[Oxford Mods.] 144.
Eliminate
a, y, z
from the equations
x-i+y-i + z- 1 = a~ 1 .v
and shew that be equal to d. 145. all
2
+ y 2 + z 2 = c2
if a, y, z
,
are all finite
The
unequal
:
x+y + z=b. A3 +3/ 3 + 3 = c?3
roots of the equation find them.
,
£r
,
and numerically unequal, b cannot [R. M. A. Woolwich.] 3a2 (a2 + 8) +
16(a-3
[R.
A
-
1)
are not
M. A. Woolwich.]
traveller set out from a certain place, and went first day, 3 the second, 5 the next, and so on, going every 146.
= 1
mile the
day 2 miles After he had been gone
more than he had gone the preceding day. three days, a second sets out, and travels 12 miles the first day, 13 the second, and so on. In how many days will the second overtake the first?
147.
Explain the double answer.
Find the value of
11111 3+ 2+ 3+ 2+ 1+
1
1
+
""
)
:
MISCELLANEOUS EXAMPLES.
505
Solve the equation
148.
x3 + 3ax2 + 3
-
(a 2
be)
as
+ a 3 + b 3 + c3 - Zabc = 0. [India Civil Service.]
a prime number which will divide neither «, b, nor ~ ~ ~ ~ that a n 2 b — a n b 2 + a n i b 3 — ...+ab n 2 exceeds by 1 a a+ [St John's Coll. Camb.] multiple of n. If 149. b y prove
n
is
:i
Find the ?t th 150. sum to infinity is (1 -
term and the sum to n terms of the abx 2 ){\ — ax)~ 2 (l — bx)~ 2
series
whose
.
[Oxford Mods.] If a,
151.
b, c
x3 + px + q = 0,
are the roots of the equation b
whose roots are equation 1
2
+c
c
2
2
-,
,
a
—
—+ a
2
a
2
+b
find the
2 .
,
b
c
[Trin. Coll. Camb.]
Prove that
152.
(y
+ z- 2xY + (z + x-2y) i + (x+i/-2z) = 18 (x2 + y 2 + z2 - yz - zx - xy) 2 i
.
[Clare Coll. Camb.] Solve the equations
153. (
1
(2)
x3 - 20x 4-133 = 0, by Cardan's method. x5 - 4t4 - KU-3 + 40.i' 2 + 9x -36 = 0, having roots of the form
+ a,
±b,
c.
It is found that the quantity of work done by a man in an 154. hour varies directly as his pay per hour and inversely as the square root of the number of hours he works per day. He can finish a piece of work in six days when working 9 hours a day at Is. per hour. How many days will he take to finish the same piece of work when working 16 hours a day at Is. 6d. per hour ?
If
155.
sn
denote the
sum
n terms
to
of the series
1.2 + 2.3 + 3.4+...,
and o^-! that to n —
1
terms of the series
1
1.2.3.4 shew that
+
1
1
+ + 2.3.4.5 3.4.5.6
1 8s n cr n _
x
-
"'
- s n + 2 = 0. [Magd. Coll. Ox.]
Solve the equations
156.
:
(1)
(12a?-l)(&p-l)(4a?-l)(&e-l)=5.
(2);
I fo+^ fo-S) 5 (x + 2)(x - 4)
^
+
1
9
x+3)(x-5) (x + 4) (x- 6)
(
_2_
(a?+5)(a?-7)
3
{x + 6)(* - 8)
"" 1
92
~ 585
*
[St John's Coll. Camb.]
:
'
HIGHER ALGEBRA.
506
A cottage at the beginning of a year was worth £250, but it 157. was found that by dilapidations at the end of each year it lost ten per cent, of the value it had at the beginning of each year after what number of years would the value of the cottage be reduced below £25 ? Given log 10 3 = -4771213. [R. M. A. Woolwich.] :
Shew that
158.
+
the infinite series
1.4 4.8
1
4
1.4.7
+ 4.8.12 +
i+?_l + + ll? 6
6
.
+6 12
1.4.7.10 + 4.8.12.16 '"'
2.5.8 .
12
.
+6 18
2.5.8. 12
.
.
18
.
24
+ '••
[Peterhouse, Camb.]
are equal.
159.
11
Prove the identity
H
x(x -
x{x - a) (x - /3)
a)
~aPy~
a/3
[
|
a
\
x(x + a)(x + p)
x(x+ a)
X
r
+ |
\ |
a(5
a/3-y
J
x2 (x2 - a2 ) _ x2 (x2 -a 2 )(x 2 -^) +•••• 2 2 „2/92„2 „2R2 a a 2(3 2 y ^
_x> a
[Trin. Coll. Camb.] 160.
If
n
is
a positive integer greater than
1,
shew that
n*-57i 3 + 60n 2 -56n is
[Wadham
a multiple of 120.
Coll. Ox.]
A number of persons were engaged to do a piece of work 161. which would have occupied them 24 hours if they had commenced at the same time; but instead of doing so, they commenced at equal intervals and then continued to work till the whole was finished, the payment being proportional to the work done by each the first comer :
received eleven times as 162.
much
as the last
;
find the time occupied.
Solve the equations
x (1)
(2)
y
2 y -3
x2 -S
-7 x3
+f
+ z2_ x{]/ + z) = a ^ z 2 + x2 — y (z+x) = b 2
y2
,
2
x' -t-y
2
- z (x +3/) = c
2 .
[Pemb. Coll. Camb.]
:
:
'
.
507
MISCELLANEOUS EXAMPLES. Solve the equation
163.
a 3 (6 - c) {x also
-c) + b 3 (c- a) {x - c) (x - a)
b) (x
shew that
+
3 c-
__+-_ + -j- = 0. —
Sum (1)
(2)
s
165. a b
the series
Shew
b)
=
;
+
+ il
[St John's Coll. Camb.1 L
s/b~ sfc
:
1.2.4 + 2.3.5 + 3. 4.6+...
S
- a) (x -
the two roots are equal
if
s]a 164.
(a - b) (x
5
that, if a,
+-
n terms.
toiuf
-
d be
6, c,
to
four positive unequal quantities and
= + + c + d, then (s
- a) (s — b)(s — c) (s -d)> 8labcd. [Peterhouse, Camb.]
Solve the equations
166.
— \Jy - a = - v/a, \/x—a-\!i/ Jta=-Ja.
(1)
\/x + a
(2)
x + i/ + z = x 2 +f + z* = ^(x3 + i/ + z 5 ) = Z. [Math. Tripos.]
Eliminate
167.
m, n from the equations
I,
lx+ my + nz = rax + ny + lz = nx + ly + mz = Jc 1 {I'1 + m2 + n 2 ) = 168.
1
Simplify
a (b + c - a) 2 + a 2 (b + c-a) +
+ + (b + c - a) (c + a - b) {a + b - c) ... + ... -(6 + c-a)(c + a-6)(a + 6-c) .
. .
. . .
[Math. Tripos.] 169.
Shew that the
(x 2 - yz) 3 + (y 2 is
expression
- zx) 3 + (z 2 - xy) 3 -
a perfect square, and find
its
square
3 (x2 root.
- yz) (y 2 -
zx)
(z
2
— xy)
[London University.]
There are three towns A, B, and C; a person by walking to B, driving from to C, and riding from C to A makes the journey in 15^ hours by driving from A to B, riding from to C, and walking from C to A lie could make the journey in 12 hours. On foot he could make the journey in 22 hours, on horseback in 8|- hours, and driving in 11 hours. To walk a mile, ride a mile, and drive a mile he takes altogether half an hour: find the rates at which he travels, and the distances between the towns. 170.
from
B
A
;
B
:
HIGHER ALGEBRA.
508 Shew that 171. integer not less than 172.
?t
7
-7n5 +14?i3 -8?i
is divisible
by
840, if
n
is
an
3.
Solve the equations six 1 + 1 2y + *Jy
(1)
w ,~x
(2)
2
+ I2x= 33, x+y=2S.
y(u — z) z(y-x) u(y-x) ^ — = = &, — =a, -^ —
x(u-z)J
—-^-y = d.
,
'
2
-
2-W
—W
#
c,
#
7
[Math. Tripos.] 173.
sum
If s be the
+
of
n positive unequal quantities
— +—+ =
>
...
-
a, b,c...,
then
[Math. Tripos.]
.
A merchant bought a quantity of cotton this he exchanged which he sold. He observed that the number of cwt. of cotton, the number of gallons of oil obtained for each cwt., and the number of shillings for which he sold each gallon formed a descending geometrical progression. He calculated that if he had obtained one cwt. more of cotton, one gallon more of oil for each cwt., and Is. more for each gallon, he would have obtained £508. 9s. more whereas if he had obtained one cwt. less of cotton, one gallon less of oil for each cwt., and how Is. less for each gallon, he would have obtained .£483. 13s. less actually did he receive much ? 174.
;
for oil
;
:
175.
Prove that
2 (b + c - a - x)*(b -
c)
(a-x) = 16 (b -c)(c- a) (a -b)(x- a) (x - b) (x - c). [Jesus Coll. Camp,.]
176.
If a,
/3,
y are the roots of the equation st?— paP+r =0, find the
equation whose roots are
——
,
a
^-~ p
—-.
,
TR.
M. A. Woolwich.]
y
If any number of factors of the form a2 + b 2 are multiplied 177. together, shew that the product can be expressed as the sum of two squares.
Given that (a2 + b 2 )(c2 + d 2)(e 2 +f2 )(c/ 2 + h'2 )=p 2 + q 2 find p and q in terms of a, 6, c, d, e,f, g, h. [London University.] ,
178.
Solve the equations
x2 +y 2 =6l, a*-y*=91. 179.
M. A. Woolwich.]
A man
goes in for an Examination in which there are four maximum of marks for each paper; shew that the of ways of getting 2m marks on the whole is
m
papers with a
number
[R.
-
(m + 1 ) (2m2 + Am + 3).
[Math. Tripos.]
1
1
MISCELLANEOUS EXAMPLES.
509
are the roots of «8 +jw?+l=0, and y, S .are the roots of x2 + qx+l=0; shew that (a - y)(/3 - y)(a + 8)(/3 + 8) = J2 - jo 2 If a,
180.
j3
.
M. A. Woolwich.]
[R.
Shew
181. (1
that
a m be the coefficient of xm in the expansion of
if
+#)*, then whatever n be, «
,s -« + «.,-...+(-l)-/
i
1
1
(n~l)(n l). + , A — 2)...(n-m ; 2U(-i)--i. «m _ 1 = ^ >
|w _^
[New Coll.
A
number
Ox.]
the product of three prime factors, the whose squares is 2331. There are 7560 numbers (including unity) which are less than the number and prime to it. The sum of Find the its divisors (including unity and the number itself) is 10560. 182. sum of
certain
is
[Corpus Coll. Camb.]
number.
Form an equation whose roots shall be the products of every 183. two of the roots of the equation x3 - ax2 hx c 0.
+
Solve completely the equation A 2afi + x + x + 2
=
+ =
2x* + 2x2
.
[R. 184.
Prove that
if
n
is
a positive integer,
nn -n(n-2) n + 185.
If
prove that 186.
(6V6 + 14)
NF =20
2>t
2n +
1
\
=:jr,
+1 .
Solve the equations (1) (2)
M. A. Woolwich.]
= 2 B [w.
-'(n-4) n -
and
if
F
be the fractional part of N, [Emm. Coll. Camb.]
:
x+y+z = 2, x2 +y 2 + z 2 = 0, x3 +y 3 + z 3 = - 1. x*-(y-z) 2 = a 2 y 2 -(z-x) 2 = b 2 z 2 -{x-y) 2 =cK ,
,
[Christ's Coll. Camb.]
At a general election the whole number of Liberals returned was 15 more than the number of English Conservatives, the whole number of Conservatives was 5 more than twice the number of English Liberals. The number of Scotch Conservatives was the same as the number of Welsh Liberals, and the Scotch Liberal majority was equal to twice the number of Welsh Conservatives, and was to the Irish The English Conservative majority was 10 Liberal majority as 2 3. The whole number of the whole number of Irish members. than more members was 652, of whom 60 were returned by Scotch constituencies. Find the numbers of each party returned by England, Scotland, Ire187.
:
land,
and Wales,
188.
respectively.
[St John's Coll. Camb.]
Shew that a 5 (c - b) + b5 (a - c) + & (b - a)
= (b- c){c - a)(a - b) (2a 3 + 2a*b + abc).
:
HIGHER ALGEBRA.
510
3c*2
a3
Prove that
189.
a
3a
a
a
2
13
If
190.
— a
1
?
j
c
progression, unless b = a
= (a-l)«
1 1
3
a—b H c—b
|
1
+ 2a 2a+l 2a+l « + 2
2
[Ball. Coll. Ox.]
1
=0, prove that
a, b, c are in
+ c.
harmonical
[Trin. Coll. Camb.]
Solve the equations
191.
(1)
(2)
- 13# 2 +1 5x + 189 = 0, having given that one ceeds another root by 2. 4 2 .r - Ax + 8x -f 35 = 0, having given that one root 3
.r
2 + \/-~3.
[R.
root ex-
is
M. A. Woolwich.]
Two numbers a and b are given two others a v b ± are formed 192. the by relations 3a 1 = 2
,
;
,
.
If
193.
x +y + 2 + w = 0, shew that
mr (w + a;)2
-+-
yz (w
— x) 2 + wy(w+y) 2
+ zx(io - yf + wz(w + z) 2 + xy (w - z) 2 + 4xyzw = 0. [Math. Tripos.] If
194.
a +
-a 2 a2 + + c2 be
be not altered in value by interchanging a
fc2
not equal to each other, it will not be altered by interchanging any other pair; and it will vanish if a + b + c=\. [Math. Tripos.]
pair of the letters a,
b, c
B
a quadruple line of rails between two termini A and y trains start at 6.0 and 6.45, and two up trains at 7.15 and If the four trains (regarded as points) all pass one another 8.30. simultaneously, find the following equations between xlt x2 , x3i x4 their rates in miles per hour,
On
195.
two down
,
Am + 5#o
*53/i)
«VO
where
m
196.
is
the
Jb
number
Am
-+-
1 Ox,
-t
of miles in
AB.
[Trin. Coll. Camb.]
Prove that, rejecting terms of the third and higher orders,
^-4
*
+ (1 ~ y)
i+V(i -#)
(i
2
-y)
= l + ^+y) + ^(3.* + ^ + 3y 8 2 2
2 ).
[Trin. Coll. Camb.]
.
MISCELLANEOUS EXAMPLES. Shew that the sum
197.
a,
a — b, a -2b,
,
a — {n -l)b, of the form
is
3m8 — 1,
(m + 1)6.
even, and a + /3, a-/3 are the middle pair of terms, that the sum of the cubes of an arithmetical progression is If
198.
shew
2)
511
of the products of the series
taken two and two together vanishes when n
and 2a = (3m -
;
n
is
na{a 2 + (w 2 -l)/32 }. If «,
199.
b, c
are real positive cpiantities,
111
a
b
g8
shew that
+ 68 + C8 a3 b 3c 3
c
[Trin. Coll. Camb.]
A, B, and C start at the same time for a town a miles distant A walks at a uniform rate of u miles an hour, and B and C drive at a uniform rate of v miles an hour. After a certain time B dismounts and walks forward at the same pace as A, while C drives back to meet A J A gets into the carriage with C and they drive after B entering the town at the same time that he does shew that the whole time occupied a 3v + u rT. r 200.
:
was -
.
v
.
-
3u+v
.
-.
[Peterhouse, Camb.]J L
hours.
'
The streets of a city are arranged like the lines of a chess201. There are streets running north and south, and n east and board. Find the number of ways in which a man can travel from the west. to the S.E. corner, going the shortest possible distance. N.W.
m
[Oxford Mods.] 202.
Solve the equation
*/ x
+ 27 + v 55 - x— 4. [Ball. Coll. Ox.]
203.
Shew ab +
(a
that in the series
+ x) (b + x) + (a + 2x) (b + 2x) +
to 2 n terms,
the excess of the sum of the last n terms over the sum of the first n terms is to the excess of the last term over the first as ril to 2n — 1 204.
Find the n th convergent to (1)
a
)
:
9
HIGHER ALGEBRA.
512 If
206.
a,
#,
qx+r = 0,
y are the roots of x3 +
ma + n ma — n in terms of m, n, q,
m{3 + n
my+n
m(3 -
my — n
n
find the value of
[Queens' Coll. Camb.]
r.
In England one person out of 46 is said to die every year, 207. and one out of 33 to be born. If there were no emigration, in how many years would the population double itself at this rate ? Given log 2
= '3010300,
+ x + x2 n =
If (1
208.
7i
is
(n
= 3-1849752, x
—1
n a = (n- r ) « '
What
3.
prove that
,
+ (- 1
+ -y72~ «r-2-
a multiple of
log 1518 = 3-1812718.
+ a x + a^c2 +
)
« P -wa r -i
unless r
log 1531
is its
r )
r!
>
\
value in this case 1 [St John's Coll. Camb.]
209. In a mixed company consisting of Poles, Turks, Greeks, Germans and Italians, the Poles are one less than one-third of the number of Germans, and three less than half the number of Italians. The Turks and Germans outnumber the Greeks and Italians by 3; the Greeks and Germans form one less than half the company while the Italians and Greeks form seven-sixteenths of the company determine the number of each nation. ;
:
210.
Find the sum to (n + If
211.
n
n
is
infinity of the series
whose n th term
l)n- 1 (?i+2)- 1 (-x) n+1
is
[Oxford Mods.]
.
a positive integer, prove that
n(n2 -l) [2
+
n(n 2 -l)(n2 -2 2 ) |2J_3
*K
n{n 2 -l){n2 -V) ;
\r \r
(n2
-r2 )
~
+l
k
; '
[Pemb. Coll. Camb.] 212.
Find the sum of the (1)
6,
(2)
4
(
213.
)
to
24, 60, 120, 210, 336,
n terms.
- 9x + 16x2 - 25^3 + 36^ - 49^ +
1.3
3
series
-x
3.
55. 7
7.
Ax 6x + 2 8.r+l
inf.
.
tomf
+ ^r + ^3-+-^r +
Solve the equation
to
-
Qx + 2
8#+l
+3
12# l6x + 2
9.r
12.r
=0.
[King's Coll. Camb.]
.
MISCELLANEOUS EXAMPLES. 214. (1)
Shew
that
+ ^ + ^ 2 (l+c2 + c2 (l+rt 2 )>6«6^ ?2(rtP + « + Z)P + « + ^ + «+...)>(« + 6' + C"+...)(^+^ + C^+...),
a2 (l
2
)
)
,
,>
(2)
the
number 215.
of quantities a,
being
6, c,...
n.
Solve the equations yz = a{y + z)
+ a\
zx=a(z+x) +/3>. xi/ = a(x+y) + y\ 216.
513
n be a prime number, prove that
If
l(2»-^l) + 2^- 1 is divisible
[Trin. Coll. Camb.]
by
+^+3f4«-^!U...+(»-l)^-^^ [Queen's Coll. Ox.]
n.
In a shooting competition a man can score 5, 4, 3, 2, or 217. points for each shot: find the number of different ways in which he [Pemb. Coll. Camb.] can score 30 in 7 shots.
Prove that the expression x> - bx3 + ex2 + dx product of a complete square and a complete cube if 218.
be the
e will
126_9^_5e_^ 5
"
b
~
~ c2
c
*
A
bag contains 6 black balls and an unknown number, not 219. greater than six, of white balls three are drawn successively and not replaced and are all found to be white; prove that the chance that ;
ft*7*7
a black ball
220.
will
Shew
squares of the
be drawn next that the
first
sum
[Jesus Coll. Camb.]
is jr—r
of the
of the products of every pair
n whole numbers
is
—
n(n 2 — l)(4?i2 —
-
l)(5?i
+ G).
[Caius Coll. Camb.]
221.
If
x—a
+
—x-b —
'4.£_i
^
x-c
=o
has equal roots, prove
that a(b-c) ±/3 (c -a)±y (a- b) = 0. 222.
Prove that when n
is
a positive integer,
». 2 .-..^ y -. + "-»X»-4) 8 .., <
(n-4)(»-5)(»-6) «,_,. j3
"
+•,
[Clare Coll. Camb.] H.
H.A.
33
)
+
.
:
'
HIGHER ALGEBRA.
514
Solve the equations
223.
2
.r
(1
(2)
+ 2yz = if- + 2z.r =g»+ %xy + 3 = 7G. .v+y + z = a + b + c S
? + + -=3 c b a f ax + by + cz = bc + ca + ab
-
[Christ's Coll. Camb.]
Prove that if each of m points in one straight line be joined to each of n in another by straight lines terminated by the points, then, 224.
excluding the given points, the lines will intersect
-mn{m-\){n—\) [Math. Tripos.]
times.
Having given y = x + x 1 + r>, expand x 2 3 i s y + ay + by + ey + dy +
225.
and shew that a 2 d- 3abc + 2b 3 = -
in the form ;
[Ball. Coll. Ox.]
1.
A farmer spent three equal
sums of money in buying calves, more than a pig and £2 more calf cost Each £1 sheep. pigs, and than a sheep altogether he bought 47 animals. The number of pigs exceeded that of the calves by as many sheep as he could have bought for £9 find the number of animals of each kind. 226.
;
:
Express log 2 in the form of the
227.
22
1
1
32
1+ 1+ 1+ 1+
infinite
continued fraction
n2 1
[Euler.]
+
In a certain examination six papers are set, and to each are assigned 100 marks as a maximum. Shew that the number of ways in which a candidate may obtain forty per cent, of the whole number 228.
of
marks
is II
!—
(1245 ;
1144
_ 6< r
'
.
[139
[5 {[240
=
~ , f Mods.] [Oxford J L
143) i \ +ll>
-|38j
Test for convergency
229.
x
x*
1.3.5.7
1JJ + 2.4"6" + 2.4.6.8 2
x*_ ,
1.3.5.7.9.11
+ 2.4.G.8.10.12* 10
Find the scale of relation, the n th term, 230. terms of the recurring series 1 + 6 + 40 + 288 +
Shew for its r
-,
r
>
th
also that the
sum
term the sum of
r
of
n terms of the
4 (2* - 1) i (2* - 1) - ^
.
14
+
and the sum of n
series
terms of this series
x*_
formed by taking
is
[Caius Coll. Camb. ]
:
MISCELLANEOUS EXAMPLES.
515
It is known that at noon at a certain place the sun is hidden 231. by clouds on an average two days out of every three find the chance that at noon on at least four out of five specified future days the sun :
will
be shining.
[Queen's Coll. Ox.]
Solve the equations
232.
x 2 + (j/ -z) 2 = a 2 2 2 2 y + (z — x) — b z 2 + {x - i/) 2 = c 2 Eliminate
233.
x 2 — x
x,
z
?/,
S—^— =
a
[Emm. Coll, Camb.]
from the equations
y 2 -yz — yx
=s*
^
z2
-zx — zii k
b
.
,
c
and ax + by J + cz = 0. 7
[Math. Tmros.] Tf
234.
two roots of the equation
of opposite signs, 235.
Sum (
\
236.
1
shew
If (1
+ px2 + qx + r =
be equal and [Queens' Coll. Camb.]
:
+ 2\v + 3 V +
+ ?ih; n ~ \
2
25 12 o.{ o:? is.23.33
)
that pq = r.
the series 1
)
.v5
5?i 2
52 ~i" '
o3 22
Tx* 3 .3 .4 o33
+12/i + 8
"•
*
w2 (w+1)3(tH-2) 3 [Emm. Coll. Camb.]
'
+«V) (1 + a\i*)(l + a°x
lc
+a*x**)
>)(\
= l+A ix4 + A 8x8 + A l2x l2 + prove that A gn + i = (rA Sn} of the expansion.
&ndA 8n = a 2n A in and ;
terms [Corpus Coll. Camb.] find the first ten
is no current from A to B but a rows down stream from A to C in 3 hours, and up stream from C to A in 3^ hours had there been the same current all the way as from B to C, his journey down stream would have find the length of time his return journey would occupied 2 1 hours r ave taken under the same circumstances.
237.
On
current from
a sheet of water there
B
to
C
;
a
man
;
;
238.
Prove that the
?i
3
3
3
th
2+ 2+ 2+
convergent to the continued fraction .
is
3» +1 + 3(-l)" +1 3» +1 -(-l)* +1
*
[Emm. Coll. Camb.] 239.
If all the coefficients in the equation
xn + p xxn ~ 1 +p 2 xn ~ 2 +
+pn =f(x) = 0,
be whole numbers, and if/(0) and/(l) be each odd integers, prove that the equation cannot have a commensurable root. [London University.]
:
HIGHER ALGEBRA.
516 Shew
240.
that the equation is]
ax + a + \]bx + /3 + *J ex + y =
reduces to a simple equation
if
Ja± s/b± Jc = 0.
f
f
Solve the equation \f6x 2 -
1 5.i-
- 7 + V4.r2 - 8x -
11
- \/2x2 - 5#+ 5 = 2# - 3.
A
bag contains 3 red and 3 green balls, and a person draws random. He then drops 3 blue balls into the bag, and again Shew that he may just lay 8 to 3 with draws out 3 at random. advantage to himself against the 3 latter balls being all of different [Pemb. Coll. Camb.] colours. 241. out 3 at
Find the sum of the at - lx + 4x - 3 = 0. 242.
fifth
2
243.
p
th
tYl ,
q
,
A
powers of the roots of the equation [London University.]
Geometrical and Harmonica! Progression have the same shew that b, c respectively
rth terms a,
:
a(b-c)\oga + b (e-a)
log b + c(a-b)\ogc = 0.
[Christ's Coll. Camb.]
Find four numbers such that the sum of the first, third and fourth exceeds that of the second by 8 the sum of the squares of the first and second exceeds the sum of the squares of the third and fourth by 36; the sum of the products of the first and second, and of the third and fourth is 42 the cube of the first is equal to the sum of the cubes of the second, third, and fourth. 244.
;
;
245.
If
Tw Tn + Tn+2 be 3 consecutive terms of a recurring series Tn + = aTn + — bTn prove that l,
connected by the relation
1 {T\ 246.
Eliminate 1
x
+
1
l
fi
,
-aTn Tn + + bTn*} =a 1
x, y, z
from the equations
+-+y z
Xs + y 3 +
constant.
z3
=-,
.r*+y2 +
a
= c3
2
=
Z>
2
xyz = d 3
,
. i
[Emm. Coll. Camb.] 247.
Shew that the x*
are in proportion.
roots of the equation
— px3 + ox2 - rx + —„ =
Hence
solve
.r
4
—
1 2.r
3
+ 47.^2 — 72.r + 36 = 0.
+
J
:
z
MISCELLANEOUS EXAMPLES.
517
can hit a target four times in 5 shots; U three times in 1 shots; and twice in 3 shots. They fire a volley: what is the pro248.
A
And
that two shots at least hit? bility that it is C who has missed? liability
249.
Sum
each of the following series to n terms:
1+0-1+0 + 7 + 28 + 70+
(1)
2-2 I.2.3.4
(2)
3
(3)
two hit what is the pro[St Cath. Coll. Camb.]
if
;
l-»
»»
6- 23
T 2.3.4.5 T 3.4.5.6 T 4.5.6.7 ,
,
,
.
.
'
+ x + 9x* + x3 + 33x* + a* + 29^ + ;
1
[Second Public Exam. Ox.] 250.
Solve the equations 2 2 y +yz + z =ax,\
(1)
x(g + z-x) = a, y(z + x -y) = b,\
(2)
+ zx + x* = ay, I x + xy+y2 = az.) z
2
2
z
+y - z) = c.
(.
[Peterhoisk, Camb.] If
251.
—
a
h t b
+ = c
1
and a
, J
111 + + -= 7-„
bn
aH If
—
a+b+c
c
n
is
an odd integer, shew that °
1 '
an + b n + c n
u 6 - v G + 5 tt¥(«2 - v2 ) + 4md (1 - u*v *) = 0, prove that (
252.
w2-v2 )6=16^V(l-w8)(l—p8
[Pemb. Coll. Cai
).
..
x+y-\-z=3p yz + zx + xy = 3q, xyz = r, prove that 3 (y + z - x) (z + x - y) (x +y - z) = - 27js + 36pg - 8r,
If
and
J
(.'/
2 ~"
Find the
253. {a (b + c)
x ) 3 + (s + # — y) 3 + (#+# factors, linear in x, y,
x2 + b(c + a)y 2 + c(a + b)
2 2 }
z,
*')
3
= 27j9 3 - 24/-.
of
- Aabc (x2 +y 2 + z2 )(ax2 + by 2 + cz 2 ). [Caius Coll. Camb.]
254.
Shew that
\
255.
By means
—x+y+z J
(
I
>.r*yy.s»>( J
^
)
3
\
J [St John's Coll. Camb.]
of the identity \l
-
,
'
,„
-
"
==
—
-
,
prove that
r=n
*
r=1 <
1;
r!(r-l)!(»-r)!
"
[Pemb. Coll. Camb.]
:
HIGHER ALGEBRA.
518 256.
Solve the equations
(1)
ax-\-by+z=zx-\-ay-\-b=yz + bj.
(2)
x -fy +z
~u =
12,\
+ a = 0.
,
MISCELLANEOUS EXAMPLES. 263.
A
264.
Prove that the equation
519
farmer bought a certain number of turkeys, geese, and ducks, giving for each bird as many shillings as there were birds of that kind; altogether he bought 23 birds and spent £10. 11*.; find the number of each kind that he bought. *o*
(y+z-8xfi+(z+x - §y)i+(#+y - 8^ = 0, is
equivalent to the equation
[St John's Coll. Camb.] If the
265.
equation 1
x+a
H
.
x+b
=
are then
c
+ = + abed
or d, or else -
- a, — a,
:
'
'
-
b,
— b,
;
-
-
r
or 0,
0,
'
'
have a pair of L
is
equal to one of
x+d
equal roots, then either one of the quantities a or b the quantities
,
1
x+c
Prove also that the roots ,
a+b
.
[Math. Tripos.] Solve the equations
266.
:
(1)
x + y + z = ab, x- l + y- 1 + z- l = a-
(2)
ay z + by + cz = bzx + cz + ax = cxy + a.
1
b,
>;
xyz=az
.
+ by = a + b + c.
[Second Public Exam. Oxford.] Find the simplest form of the expression
267.
(a-j8)(a-y)(a-*)(a-*)
+ +
„„
>„
*
^— +
(0- a)((3 -y)(/3 - S)(/3 - c)
...
^_ '
(*-«.)(« -/3)(e-
7 )(e-
8)
[London University.] In a company of Clergymen, Doctors, and Lawyers it is fcund that the sum of the ages of all present is 2160; their average a;e is 36; the average age of the Clergymen and Doctors is 39; of the 268.
and Lawyers 32^; of the Clergymen and Lawyers 36f. If each Clergyman had been 1 year, each Lawyer 7 years, and each Doctor 6 years older, their average age would have been greater by 5 years find the number of each profession present and their average 1
octors
:
ages.
269.
Find the condition, among ciyX*
+ Aa^xhf + Ga.s v-y- + -i't...ry 3 + «
should be reducible to the expressions in x and y.
sum
that the expression
its coefficients,
4
y
4<
two linear [London University.]
of the fourth powers
of
HIGHER ALGEBRA.
520 Find the
270.
real roots of the equations
=a 2 2 2 2 2 y -f w + u = b x2 + v 2
z
2
,
vw-hu{y + z)=^bc,
,
wu + v (z+x)=ca,
,
uv + w(x+y)=ab.
-\-w2
+u2 +v2 =c2
[Math. Tripos.] no consonant or group of consonants can stand immediately between a strong and a weak vowel the strong vowels being a, o, u and the weak vowels e and i. Shew that the whole number of Gaelic words of n + 3 letters each, which can be formed It is a rule in Gaelic that
271.
;
;
ft + 3 — ——
2 of
n consonants and the vowels
aeo
is
1
where no
-
+2
ft
peated in the same word.
Shew
272.
that
if
[Caius Coll. Camb.]
x2 +y 2 = 2z 2 where ,
r,
I,
273.
and k are
x, y, z are integers,
2y = r(k2 + 2lk-l 2 ),
2x = r{l 2 + 2lk-k 2 ),
where
letter is re-
2z=r(l 2 + k 2 ) [Caius Coll. Camb.]
integers.
112
Find the value of
then
4
6 "
1+ 1+ 3+ 5+ 7+
to inf.
[Christ's Coll. Camb.] 274.
Sum
the series
»-2
(1)
2.3
2.1-3
+ -— + 3.4
:
—
3.^
B inf. .
+ 4.5 -
to
-
|1
[2 -^ + a+l (a + l)(a +
(2)
275.
[ft
+
+ (a
2)
Solve the equations
+ l)(a + 2)...(a + n)
:
2^ + 3 = (2^-l)(3y + l)(42-l) + 12
(1)
= (2x+l)(3y- l)(4g + l) + 80 = 0. 3ux -2oy = vx + uy = 3u 2 + 2v 2 = 14
(2)
276.
Shew
that
a2 + \ ab
is divisible
by X 3 and
ab b
2
+X
ac
be
ad
bd
c2
;
xy = 10«v.
ac
ad
be
bd
+X
cd
find the other factor.
cd
d +\ 2
[Corpus Coll. Camb.]
MISCELLANEOUS EXAMPLES. If c, b, c,... are the roots of the
277.
equation
of os +6s +c8 +..., and shew that
sum
find the
521
2
a"
b'
a*
2
c2
2 I/
a
c
(t
£>
c
Pn-iOr -2/*.,) J
b
c
2> n
[St John's Coll. Camb.]
Hy the expansion of
278.
l-3»+
wlien n
j
,
or otherwise, prove that
j-g "
-cVo.-(-l),
1.2.3.4
an integer, and the series stops at the
is
279.
+ 2a'
(3/t-2)(3/t-3)(3w-4) lT 273
(3m -1) (3m -2)
+
1
Two sportsmen A and B went birds. The sum of the squares
first
term that vanishes. [Math. Tripos.]
out shooting and brought
number
was 2880, and the product of the numbers of shots fired by each was 48 times the product of the numbers of birds killed by each. If A had fired as often as B and B as often as A, then B would have killed 5 more birds than A find the number of birds killed by each.
home
10
of the
of shots
:
280.
Prove that 8 (« 3 + 6s + c 3 ) 2 > 9 (a 2 + be) (b 2 + ca) (e* + ab). [Pemb. Coll. Cams.]
281.
Shew
that the 2
n lh convergent
4
6
.
3- 4- 5What
is
282.
the limit of this
If
—
is
the
?i
th
... is '"
when n
2» +1
2»2 r (n-r)\
'
[Kino's Coll. Camb.]
is infinite?
convergent to the continued fraction
111111 a+ a+ b+ b+ c+
shew that
_ 2-
to
p 3n + 3 = bp 3n + (bc+l)q 3n
c-\-
[Queens' Coll. Camb.]
.
Out of n straight lines whose lengths are 1, 2, 3, ...n inches 283. respectively, the number of ways in which four may be chosen which will form a quadrilateral in which a circle may be inscribed is -L {2n
(/i
-
2)
(2* - 5)
-3+3 (-
1 )"}
.
[Math. Tripos.]
y
;
:
HIGHER ALGEBRA.
522
If u 2 , u3 are respectively the arithmetic means of the squares cubes of all numbers less than n and prime to it, prove that 6nu2 4m 3 0, unity being counted as a prime.
284.
and ?i
3
—
=
+
[St John's Coll. Camb.] If n 285. is divisible by
shew that
&m - 1 shew that {y - z) n + (z- x) n + (x - y) n x 2 +y 2 + z 2 -yz — zx-xy; and if n is of the form 6m +1,
is
it is
of the form
by
divisible
(x2 +
2
+ z2 -yz — zx — xy
.
P
m
th powers, the sum of the products If S is the sum of the a a a a shew that ... together of the n quantities lt 2 n 3,
286.
m
2 )
,
n- 1 S > .
\
!
n-
,
m
.
\jm
P.
.
[Gaius Coll. Camb.]
Prove that
287.
the equations
if
x3 + qx-r = if
?' 2
=Q
common
have a
and
and rx3 — 2q 2x 2 — 5qrx — 2q 3 —
each
root, the first equation will have a pair of equal roots of these is a, find all the roots of the second equation.
[India Civil Service.]
x V2a2 - Sx2 +y */2a 2 -Sy 2 + z \/2a2 - 3z 2 = 0, where a 2 stands for x2 +y' + z 2 prove that 288.
If
,
(x+y + z)(-x+y + z)(x-y + z)(x+y-z) = Q. [Thin. Coll. Camb.] 289.
Find the values of x { x2 ...x n which ,
,
satisfy the following
system of simultaneous equations til
ax -
/C\)
x\
a1 ~
W
Cl
bx
a x -b n
b2
X.,
+
OC-%
an -
—
Oj
bx
2
+a
~~
n
+...+
^2
-
bo
+ ...+
X a2 -
~
r ,
h
bn
x,
an ~
K [London University.]
290.
yz - xl
Shew that
zx
- y2
xy — z 2 where
r2
= x2 +y 2 + z 2
,
- yL xy -z 2 yz — x2 zx
xy -
z-
yz- x* zx - y 2
and u 2 =yz + zx + xy.
r2
MISCELLANEOUS EXAMPLES.
523
A piece of work was done by A, B, C\ at first A worked alone, 291. lmt after some days was joined by />', and these two after sonic days were joined by C. The whole work could have been done by II and (", if they had each worked twice the number of days that they actually The work could also have been completed without B'h help if A did. had worked two-thirds and ('four times the number of days they actually did; or if A and B had worked together for 40 days without C; or if The all three had worked together for the time that B had worked. number of days that elapsed before B began to work was to the number that elapsed before C began to work as 3 to 5 find the number of days that each man worked. :
292.
Shew that
if
>S' r
is
the
sum
of the products r together of
l
o H _ r = /6
then
.
.
.'-
(
[St John's Coll. Camb.] If a, b, c are positive and the 293. the third, prove that
sum
any two greater than
of
'^'T(»t)'('*"-"i'«' [St John's Coll. Camb.] 294. (a
Resolve into factors
+ b +c) (6+ e - a) (c + a - b) {a + b-c) (a2 + ¥ + c8
)
- 8a 2 b-c 2
.
Prove that 4{a 4 + /y 4 + y
1
+ (a + ^-ry) }=(/3 + y) + (y + a) + (a + i3) 4 + 6(^ + y) 2 (y + a) + 6(y + a) 2 (a + 2 + 6(a + ^(^ + y)-. 4
t
l
2
/i{)
[Jesus Coll. Camb.]
Prove that the sum of the homogeneous products of r dimen295. sions of the numbers 1, 2, 3, ... w, and their powers is
L^L.M_^,
2
,,.-^(^;)(;-^3^M,,,, t0ittcn
J
[Emm. Coll. Camb.] 296.
Prove that,
if
n be a positive integer
i-fc+»?yq-*
,
<» t
r??
t
~ B)
+""»(-»>'[Oxford Mods.]
297.
If
x=y = z = u
x(2a -.»/)=.?/ (2a-z) = z (2a-u)=n (2a-#) = 6*, shew that
unless o2 =2a2 and that equations are not independent. ,
if
this condition
is
satisfied the
[Math. Tkipos.]
524 298.
HIGHER ALGEBRA. Shew
that
if a, b, c
ax+yz + z = 0,
are positive and unequal, the equations
zx+by+z=Q, yz+zx+c=0,
give three distinct triads of real values for x, y, z products of the three values of x and y is b (b - c)
;
:
and the ratio of the a {c - a).
[Oxford Mods.] 299.
A = ax -by-cz,
If
prove that
D=bz + cy, E= ex + az, F= ay + bx,
B=by-cz- ax, C = cz- ax - by, ABC- AD - BE? - CF 2 + 2DEF = (a 2 + b2 + c2 (ax + by + cz) (x2 +y2 + z 2
)
2 ).
[Second Public Exam. Oxford.]
A
student found it necessary to decipher an old manuscript. During previous experiences of the same kind he had observed that the number of words he could read daily varied jointly as the number of miles he walked and the number of hours he worked during the day. He therefore gradually increased the amount of daily exercise and daily work at the rate of 1 mile and 1 hour per day respectively, beginning the first day with his usual quantity. He found that the manuscript contained 232000 words, that he counted 12000 on the first day, and 72000 on the last day and that by the end of half the time he had counted 62000 words find his usual amount of daily exercise and work. 300.
certain
;
:
ANSWERS. Paces 10—12.
I.
1.
4 10.
20. 25. 26.
(1)
546
:
a.
(2)
9
:
7.
*
5
5.
11,
bx
(3)
i
2.
13.
6.
Z
= y = *-,or^-± = Q.
17.
18.
5
:
385, 600.
3.
6 or - 3
:
5.
abc+2fgh-af*-bg*-cti>=0.
i
22. 7,3,2. -3,4,1. 21. 3,4,1. 2 2 ± a (62 - c 2 ), ± b (e 2 - a 2 ), ± c (a - ft ). be(b-c) ca(c-a),ab(a-b).
3,4,1.
23.
t
Pages
II.
'
:
ay.
:
45
2.
5-
«
14.
(1)
12.
19, 20.
(2)
300a3 6.
15.
0,j 3,> 8.
13 li.
u, o,
18
8
21.
45 gallons.
25.
64 per cent, copper and 36 per cent. zinc.
.
19
.
5 parts of bronze.
g, 9, 10, 15.
23.
20.
z3 3.
3 gallons
17:3.
.
,
—
-.
«,.
.
7T—
•
cm- bm -2aii from A 8 gallons from a = ±b. 24. ;
B.
3 parts of brass are taken to 26.
63 or 12 minutes.
HIGHER ALGEBRA.
526
IV. 1.
277£.
153.
2.
a.
Pages
3.
0.
31, 32.
4.
n (v 10 -*>'
30.
5.
.
o 6.
10.
-42.
820a -16806.
-j,
16.
z2 -a:+l,
20.
612.
^
1325^/3.
8.
11.
n
--,..., -9|. a;
2
14.
-2a; + 2,
...,
21 12.
1,
-l£,...,-39.
15.
x.
—(11a -96). a
o 2a
Uff5.
b.
?i
22.
1, 4, 7.
26.
n(»+l)a--.
4.
First term 8,
5.
First
6.
Instalments £51, £53, £55,...
...,a%
3.
19.
5.
23.
495.
24.
160.
x
Pages
-31z,
18.
.
a
35, 36.
8 or -13.
10 or -8.
2.
2
-33a;,
17.
+1
(
1.
3.
2,5,8,...
8.
25.
number of terms 59. term 1\, number of terms 54.
fi
2(l-.r)
(2
13.
3, 5, 7, 9.
14. 20.
2,4,6,8. 10p-8.
22.
3, 5,
+ rc~ 8 •^r)
12.
%2
*
12,
-te + 2)-
numbers a - M, a-d, a + d, a + 3d.] 12 or - 17. 16. 17. 6r-l. p + q-vi.
[Assume 15.
7.
10#
•
for tbe
7; 4,5,6.
V.
1.
75^/5.
+ l)a-n2 b.
{n
4, 9, 14.
21.
IV.
9"
9.
3
1 •13.
25.
-185.
7.
a.
21.
8 terms.
23.
ry = (n + l-r)x.
Pages
41, 42.
Series 1|, 3, 4^,.
1
1
)
ANSWERS.
1-g"
*
9
1-a'
(l-a)2
8 3'
10
10>
4/2 a +
Pages
46, 40.
-
(i
2ft '
'
10
^ ^i'"
-
r=(.r 5 »-l)
x--l
l'\
15.
^r.
16.
li.
21
,
.
r
VI.
Pages
a.
« -
1
*
xy-1
23
1-pJ. 9 (i + .)i«y-d f
«c -
a 7/ (x"//"
.
2
2 /
1H
3
n(n+l) a
*-l
2o
Zl
(l-r)fl-H s(.r»-l)
14.
b.
mi"
' -
V.
527
B.2"+f -2*« + 2.
19.
1
'-"!S'":"" -
1
?•*-
(
1
52, 53, 2
1.
4.
11.
13.
(1)
5.
Gaud
(2)
3*.
(3)
2.
24.
^w(?t +
0^, 79-
l)(?i 2
3.
4:9.
5.
10.
|,
12.
gn(n+l)(2»+7).
-n(n + l){n2 + 3/t + l).
14.
+1 J (3
4" +1
18.
The n th term = + c
+
,l
-
1)
J, I
>r (»
+ ?i + 3).
15.
11
1).
1
2' 1 + 1 .
-4-w(tt + l)(n 2 -«-l). fr
term 19.
3H-
a
is
+ b + c;
v\
22.
1.
12,0.
6.
52.
.
(2n —
n greater than
for all values of
1),
the other terms form the A. P. b
I
(2a
+ n^ld)
la*
VI.
Page
b.
+ (n -
1)
+ 3c,
b
"
1}
ad + -
(
1140.
3.
16646.
4.
2170.
7.
11879.
8.
1840.
9.
11940.
300.
12.
18296.
15.
120.
16.
n-1.
d*
.
[
a.
Tack
59.
2.
728G26.
3.
1710137.
4.
*7074.
6.
334345.
7.
1783212G.
8.
1625.
9.
2012.
10.
13.
1456.
14.
7071.
15.
16.
(1)
121.
12.
231.
(2)
122000.
VII.
b.
190.
10.
333244.
M90001.
21321.
5.
1.
11.
fr
Triangular 364; Square 4900.
14.
VII.
5c,
Tho first + 7c,....
56.
2.
11.
-f
1.
Pages
65,
112022.
5.
842. -
66.
1.
20305.
2.
4444.
3.
11001110.
4.
2000000.
6.
34402.
7.
6587.
8.
8978.
9.
26011.
5.
10.
i
87214.
HIGHER ALGEBRA.
528 11.
30034342.
12.
710^3.
16.
20-73.
17.
125-0125.
20.
Nine.
2714687.
13.
14.
-2046.
15-1*6.
15.
5
25.
31.
18.
Four.
21.
Twelve.
22.
Twelve.
30.
2 n + 27 + 2 6
Ten.
26.
39
Pages
a.
Eleven.
72, 73.
^6 + ^15
3+ z.
.
- \/ a6
a*/ 6 + & \/ a
+ 6)"'
(«
~~2^
~~
a-1
.
4
5
2
33
+ 33 2^+3
8.
56
-
9.
a6
3
+ 33
541322 .
56
.
23+
.
56
19
10
11
2
23-
.
.
22
56
+
2
.
-2 2
12.
10
11
5
2
3
4
+5
3
.
.
22
18.
33
.
10
.
.
21
11
1-33 + 33
+3*.
17
2
22
2
+
1 \
33
5 .
22-
1
33
.
23.
31262116 17#
25
11
+26+2 6 +26+26+2 6 + l
J 5
3*+3»+l.
+ 33.23-33.22.
3
+ 36_ 3 6 +3
2 \ 3
.
10.
2
5
4
33
21
11
15 22- 3 2+ |S&4 33 1\ 6_ i(36_36 22-
23".
23
.
1
17-33.22 + 33.2 2 -3.
32
145 -
56
3 4 +. ..+53.34
4
3
.
4~.
1 .
*
2 2 + 22.
.
13 "-7
.7 4 +2.7 2
53+53
32
:1
110
1
1
23
(a - 1)
5
1
+3
-a 6 6*+a%2"-...+a«6*-6 4
11.
+ N/2a
g
1
7.
y/a 2 - 1
^2 + ^/3 + „/5
6.
•
^
+
a-1
3^/30 + 5^/15 - 12 - 10^/2
.
24.
.
6
5.
16
Eight.
,
.
4
15.
5 o
2
5
+ ^/2 + ^/6
2
23.
5 6 o
19.
-3 8 -3 7 -3 -3 + 3 3 + 3 + l. 6
l.
14.
2 .
o
VIII.
3*
=
.
gj
1
+ 8 6 + 86
19.
.
Q o
aA " 4/^
*
+
25.
1
+ ^/3.
3-2^/2.
28.
JU-2J2.
30.
3^/3-^/6.
31.
a/^^+a/|-
36.
289.
37.
5v/ 3
V 5 -\/7 + 2^3.
21.
1+^/3-^/2.
23.
2
24.
3-
26.
+ Ja-JSb. 2 + ^/5.
27.
29.
2^/3 +
35.
ll
+ 56 x/3.
+ ^7-2. 22.
20.
^/5.
v/5
^7 + ^2 -
-
•
.
ANSWERS.
41.
3^3 + 5. 3 + ^/5 = 5-23007.
43.
Sa+Jlr^^r.
38.
1.
0-2^/6.
4.
.i-
-x
+
_8_
7 -
29* 3a 2 -
10.
1
1
39.
529 40.
3.
42. 44.
a^ +
8^3.
l+^+a
x^/2
s 2.
HIGHER ALGEBRA.
530
IX. ±7.
1.
-2.
7.
(aa'
- bb'f + 4 (ha'
\bb'
-
10. (ac'
- a'c)
2.
2
2ac'
-
2a' c)
= (ab' -
2
5.
+ Jib)
= (b
a'b) (be'
2
-
(Jib'
96.
(ln'-l'nf=(lm'-l'm)(mn'-m'n).
+ Ji'a) = 0.
- 4ac)
(b'
2
- 4a'c')
;
which reduces
b'c).
X.
1.
Page
c.
a.
Pages
101, 102.
to
y
'
.
ANSWERS. X. 8
-p
1.
*=5,
3.
*=1, -
5.
.r
= 8,
—
15 -—
y=4, y
;
PAGES
b.
= l, -^.
2; y = 2,
8.
531 L06,
4.
.i= ±o,
13.
ar=5,
14.
a?=4, -2, ±JS /~15
= 4,
15.
.r
16.
.t'=r>
-2,
4
d=
!/
;
=3
>
8.
~;y = 20,
17.
5.
o
21. 23.
x = l,
25.
26.
*=6, 2,4,3; y = l,
27.
x=±5,
28.
a.^
29.
a:= -
30.
.r
32.
*=5,|,0; y=3, -^' ~f
s*-
*=i.V^;y=2,3^/i.
35.
#=±3, ±^-18; y=±3,
36.
.r=?/=±2.
37.
x=o,
= 0,
3,
>
5L-
^
/a ,
b
~3
y
;
= 2,
2 -.
i25; y = ±9.
4 31.
/rt •
~u
=f
.c
•
= 0,
33>
1,
*=
^ 2,
;
y=0,
•Vl,
2;
2,
^.
T/---2,
v/-18.
„ =0
-^*
--£^
((Sa-
_//-' ;
——
1± 3^/^143 >
3, 9.
a (26 -a) •
~
2.
|,
'>
y=0,
'
5
48
,
3;
9,
(
±-, ±2; y=±5, ±4, ±10, ±8.
107
tr fe
9.
— ;^!,-. —+ ^143 V-
. 1 6,
2.
16.
.r
;
y=l,
x = 729, 343; y = 343, 729. x = 9, 4; ?/=4, J.
24.
= l,
as=2, 1;
19.
# = 5; y =±4.
±4,
±J.
15.
22.
1; y
= ±3, ±5.
+ l; y=2, -4, ±„/~15-l. -4, ± J -11-1.
20.
= 9,
vy
r v -ll + l; y=2,
s=6, 4; y=10, a; =16, 1; y=l,
18.
y
97
4=f>/-97'
>
1
o
;
-
x=±5, ±3; y=±3,
10.
5
±3
.7,
,,
.r
^±^,±^^=0,^6^/1. 4± v/-^7
2,
= 45, 5; y = 5, ]:,. x = ±2, ±3; 7/= ±1, ±2.
6.
12.
3,
8 -_
x=
11.
9.
„
2.
x=9, 4; y=4, 9. .r= ±2, ±3; ?/= ±3, ±4. s=±2, ±1; y=±l, ±3.
7.
107.
'
~T
34—2
2-/4, 6.
.
HIGHER ALGEBRA.
532 40.
41.
±« x/7,
x = 0,
#= ± 1, ±
2a
2
.
1.
x = ±3;
3.
af=5,
5.
.x
6.
x = ±3;
8.
# = 8, -8; y = 5, -5;
= ±5; z—
Pages
c.
2±Vi5i ^
.
3,
?/
*=1;
?/
12.
.t=1,
-2;
13.
# = 4,
—
;
2
?/
*/
4,
= 3,
# = 8, -3; ;
7.
jr=
9.
.c
11.
-3;
= 6, —;
2
?/
= 2,
-
-
= 7. = 3; 2 = 3,
1; 2
?/
±5; y= ±1;
= 3;
?/
2
2
= 4;
2
«=a, -2a,
±1. o
.r
= 5, -7;
y = 3, -5;
.r
= a,
y = 0,
14.
0, 0;
^
a, -
2
y = 2,
Page
d.
a, 0; 2
"
a;
= 2a,
-4a, (l± v/-15)
a?=l,
3, 5, 7, 9;
2/
3.
ar-20, 8; y = l,
8.
4.
a?=9, 20, 31; y = 27, 14,
5.
# = 30, 5; ?/=9,
32.
6.
.t
7.
x=7p-5,
8.
s=l$p-2,
y=5p-4,
5, 8, 11.
19, 14, 9, 4.
1.
3; y
= 3,
13.
11 horses, 15 cows.
16.
To pay
17.
19.
an infinite To pay 17 florins 37,99; 77,59; 117,19.
20.
28 rams, 1 pig, 11 oxen; or 13 rams, 14 pigs, 13 oxen.
21.
3 sovereigns, 11 half-crowns, 13 shillings.
18.
1147
;
?/
3 guineas
14.
10. 4.
12.
1.
44.
11; y=6p-l, 5. = 17/), 17; = 13^, 13. x = llp- 74, 3; y = 30p-25, 5.
11.
.t
y = 8p-5, 3. = 23^-19,
= 50,
= 21^-9, 12; x = 19p-W, 3;
9.
a.
113.
2.
2;
0, a.
' a-
z = 29,
= 24,
= 0,
3
1.
21, 13, 5;
-8.
'
-6.
X.
= 6,
2
3
a; y = ±a,
^
«==
= -; w = -.
^-^3' 16.
8.
11
= 3,
= 2,
-
11 —
a
= 3.
= 7,
4.
1
-3.
=
# = 5;
i5. 2
109, 110.
2=7151
.
y = S,
= t2; 2=
= 2;
10.
_
;
-"2 - 1
16 * 4
2.
±4.
f
.
.
V
-1; y = l — 5; «=2.
= 4,
a
= ± „2a, =
?/
;
,716a4 -a8 -1
X. ?/
±«; y = 0, ^bs/7, ±6^/13, T &, T 36.
ift^/13, ±3a,
-£
101.
?/
15.
56, 25 or 16, 65.
and receive 21 half-crowns. number of the form 1147 + 39 x and receive 3 half-crowns.
56p.
'
ANSWERS. XI.
533
Tacks 122—12-4.
a.
1.
12.
2.
221.
3.
40320, 0375600, 10626, 11628.
4.
6720.
5.
15.
6.
40320; 720.
8.
6.
9.
120.
15, 860.
7.
10.
720.
11.
10626, 1771.
15.
230300.
12.
1440.
13.
6375G00.
14.
360, 144.
16.
1140, 231.
17.
144.
18.
224, 896.
19.
848.
20.
56.
21.
360000.
22.
2052000.
23.
3(19600.
24.
21000.
25.
yJ^-p.
26.
2520.
27.
5700.
28.
9466.
29.
2903040.
30.
25920.
32.
41.
33.
1956.
34.
7.
XL
Pages
b.
3326400.
(2)
129729000.
151351200.
4.
360.
5.
nr
8.
531441.
9.
1.
(1)
3. 7.
1663200.
.
131, 132.
(3)
72. n
V
11.
1260.
12.
3374.
13.
15.
4095.
16.
57760000.
17.
19.
127.
2.
4084080.
6.
125.
10.
.
455.
30.
14.
yunrwr 1023.
720; 3628800.
18.
inn
315.
20.
21.
'
24.
*Ji£l>-i<£zi> + l;
(1)
P(p-l){p-
(2)
113; 2190.
2454.
28.
XIII.
6 26.
2. 3.
4. 5. 6.
7.
6666600.
29.
Pages
a.
+ 90a; 3 - 270.r 2 + 405a; - 243. 81a:4 + 216a; 3 + 216.r-// 2 + 96.r?/ 3 + 16// 4
««
-
(p + l)*-l. 30.
142, 143.
15a; 4
//
32.c
<
6 ,
1.
e
6
6
5
- 80x*y +
80.r
3
y° -
40.r-y
3
+ lO.r//
4
.
- if.
+ 135a - 540« + 1215a 8 - 1458a 10 + 729aM 10 + 5a; 9 + 10a; 8 + 10a; 7 + 5a; 6 + a; 5 a; 2 2 ~ - tlxhf + 7«V - x7 'f' 35a; 3 3 + 35a; 1 - 7.iv/ + 2 la; 1
-
4
18rt 2
G
.
.
V
*/
*/
16-48a; 2 + 54ar -27.c 6 +
81a;8
l
16
8.
729o« - 972a 5 + 540a 4 - 160a 3 +
9
1+
Ix
21x 2
T + ~T~
35a; 3
35a; ~8_ +
4
l
+
^
21x n ~32~
-
+
23.
\ii
3>(g-l) (p-2) _ g(g-l)(g-2) ! hi. 27.
64; 325.
22.
'
{\m) n
^+~ 7xG 64
+
x7 128"
5199960.
42.
.
HIGHER ALGEBRA.
534 64a;
10
6
32a; 4
a
1
7ft
135 + 4a; 2 _
20a-
+
729~~2~r
*
^~ 2
45
10
35ft 5~ + n +— 8
X
X'
a'
13.
- 35750a: 10
16.
J~(5a;) 3 (8?/)- 7
.
64a; 6 ,
x
1
'
_
.
+ 7ft 6 + 4ft 7 +
—
r+
a;
_
729
210
252
210
5-
+
4
_
5
7ft
+— x*
120
1- — + -Z ,
12.
8a;
4
7ft 3
+ ytv + t? ^T^ 2ob lb lb +^T 4
11.
243
120
^
v
14.
- 112640a 9
17.
40ft7 6 3 .
+
_
ft
8 .
45
10
x
iC
:8
130 .
x,.io
-
15.
.
*
9
1120 18.
81
;27|3_
10500 19. a:
20.
3
-
(16a; 4
22.
2x
24.
2 (3 65
27.
110565ft 4
30.
+ 5ft 4 ).
20a; 2 ft 2
28.
.
21
'"I6
8
70x 6 y 10
- 363a; + 63a; 2 -a; 3 ).
189ft17
312a; 2
a 19
252.
25. 84ft 3 & 6
31.
.
MK
21.
2x4 + 24a; 2 + 8.
23.
140^2.
26.
-^a; 14
29.
1365, -1365.
32.
18564.
34.
(
7 18*
.
Bn
In
33.
|i(»-r)!4(H + r)'
9 th
The
5.
The3 rd = 6|.
10. 11.
b.
The 12 th
2.
.
Pages
The 4 th and
6.
The
3.
.
147, 148.
14.
_
5.
5 th =Jrj>
9.
x = 2,
+ 8.r + 20a; 2 + 8a; 3 - 26a; 4 - 8.r 5 + 20a; 6 - 8a;7 + x\ 27a; 6 - 54«a; 5 + 117ft 2a;4 - 116« 3.c 3 + 117ft 4 a; 2 - 54ft 5 a; + 27ft 6 r-1 n-r+1
xr-i a n-r+l.
14.
-2 5* 1
3
2
__8_
25*
- x - a; 2 - - a; 3
x-
a.
13.
(- l) p
15.
2r = n.
Page
155.
?/
3
3
1
„
+ 2 a- + _a;-__a:-.
4.
l-2a; 2 + 3a;4 -4a; 6
6.
l
14
.
7.
l-aj+ga^-g^3
9
1+X+
6-U-
- z*>-*"+\
, „., p + 1 2n -p
i
125
+ a; + 2a; 2 + —
a;
.
3 .
o
-
8.
10.
1
.
?
— a; + ;r
a;
-
—
l-2ft + -ft-
—
ar
«r
11*.
= 3, n=5.
I2n+1
-
2.
o
-
The 10 th and
1
XIV.
3>
4.
6 th .
n 12.
1)» |»| 2/i*
XIII. 1.
-
ANSWERS.
U.
4(l + .-|.. + |..). 2a
18.
1040
-
lr
(r
21-
a
\
-i
2
a-
lb
250
10//
-
19.
2 43rt
+ l)(r+2)(r + 3) a?-. _, lr *J
i(l + * + §*« + |«»).
!4.
._
a«
20.
,.
1
3
.
(-1)
22.
2 3
x
_
.
5
(r
+ 1)**
(2r-3)
...
r
11.8 5.2 .1 .4,..(3r-14) .
23 »'
(
24.
-1848.1 13
I
2
535
*•
3r| r
10719
-i^-z<\
25.
.
XIV.
Pages
b.
1.3.5.7...(2r-l) *•
3
[
(>•
^y
x >
161, 162.
+ !)
(
r
+ 2)(r + 3)(r + 4 ) •
•
•
^1.2.5... (3r-4)^,
p 2.5.8. ..(8r-l)
4> 3'-
r
3.5.7... (2r +
l)
t /_nrfe±llt±2) *~. 5 r -r *•{-*)
fi 6.
-
xr
xr
.
.
^_ br
r+1
^
2 .1.4...(3r-5) 9<
V>--'
3'lr
2.5.8... (3r-l)
17.
The 3**. The 4 th and
20.
9-99333.
13.
24.
14.
5* h .
21.
1.3.5... (2,-1)
U "
(n
18.
5 th
}
\r
r
The
[
15.
.
The
10 00999.
3 r<1
.
22.
+ l)(2n + l)... (r-l.» + l) The 13 th
.
19. G- 99927.
16.
The
7 ,h .
989949. 23.
-19842.
.r
r
J HIGHER ALGEBRA.
53G /
14. 18.
2\-'
v
1V H
/
Deduced from (1 - a; 3 ) - (1 - #)3 = 3:r - 3.r 2 16. n r Equate coefficients of x in (1 + x) (1 + re)- 1 = (1) .
(2)
Equate absolute terms in
20.
Series
21.
2 2 "- 1
on the
[Use
(c
left
+ (
1)» q n 2
{l
+ x) n f 1 + -
= coefficient
)
45.
(1)
(1
(2)
+ x)9^ 1 .
=z 2 (l +
of x- n in (1 -
.t
o;)
n-2.
2 )-*
.
|2w
1
- J
2
'
In
ire
+c1 + c2 +
..
2 .c
-
XV. 1.
|2n
2 (Coc,
+ c lCs + ...) = c 2 + Cl 2 + c 22 +.
Pages
173, 174.
.
.c
2
n
].
6561.
answers. XVI.
Pages
b.
537 185, 186.
1.
4,1,2,2,1,1,1.
2.
-8821259, 2-8821259, 3-8821259, 5-8821259, 6-8821259.
3.
5, 2, 4, 1.
4.
Second decimal place
5.
1-8061800.
9.
1-1583626.
13.
units' place
;
1-9242793.
6.
decimal place.
fifth
;
1-1072100.
7.
2*0969100.
8.
10.
-6690067.
11.
-3597271.
12.
-0503520.
f-5052973.
14.
-44092388.
15.
1-948445.
16.
1915631.
17.
1-1998692.
18.
1-0039238.
19.
9-076226.
20.
178-141516.
21.
9.
25.
4-29.
28.
4-562.
301.
23.
3-46.
24.
.-_£«». log 3 -log 2'
29.
31og3-21og2
32.
££-««, 2
9.
e*~-cy~.
7i
= 50
12
.
(
in
.
10.
= 10
?*
L
(2.r)
2
^.
2 ji +
/)»4
/>»-
Pages 195—197.
log, 3
-log,
2.
in (1);
and
?i
= 1000 13 .
In Art. 225 put
in (1) respectively. "' 1 r 3r + 2r
<-
»
^.
(2x)- r
(2x) 4
+ i_X + ... + i_^ + ...{. )
/y»*-7*
/y»6
}
,*•
1^+iog.a-*).
E a=
-69314718; 1-09861229; 1-60943792;
= - log,
-0020000000666670.
r
1"
b
6.
-8450980; 1-0413927; 1-1139434.
H" 1L~ 1124.
.,«*,
J = 5-614.
lr ..r±i>. r
.
„
14
;
,
log 2
2.
(2)
-log 2'
log 3
l0g
XVII. log, 2.
(1
-
A^ - -040821995 XVIII.
a.
;
c
-log, (l -
= log,
Page
f1
+
M
=-105360516;
iA = -012422520.
202.
1.
£1146.
14s. 10J.
2.
£720.
3.
14-2 years.
4.
£6768.
7s.
10hd.
5.
9-6 years.
8.
£496. 19*.
9.
A
little less
than 7 years.
10.
XVIII. 1.
6 percent.
4.
3 per cent.
5.
8.
£6755. 13s.
9.
13.
14-200.
27.
lo S 2
log3
?,
log 7
1.
fr J
1-206.
26.
2.
£1308. 12s. 4 U.
b.
£3137.
28J years. £183. 18s. 15.
£4200.
£119. 16s. 4^7.
Pack 2s. 6.
10.
4frf.
207.
2U.
3.
£1275. 3} per cent.
£110. 7.
£920.
11.
£616.
2s.
9s.
l£d
538
HIGHER ALGEBRA. XIX.
8.
14.
22.
a 3 + 2& 3
The 4
4 .
5
is
the greater.
greatest value of 5 ;
12.
x
is 1.
when x = 3.
XIX. ,n
3 3 .5 5
a.
.
/3
3
/2
b.
Pages x s > or
213, 214.
< # 2 + x + 2,
15.
4
23.
9,
Pages
;
according as x > or
8.
when #=1.
218, 219.
< 2.
ANSWERS. 5.
x1,
8.
x<-
1
x> -
11.
or x
,
—
e
xl, convergent
10.
x-1,
1
convergent;
y
e 9.
or
y- a - (3
if
;
5:5!)
divergent.
1
Divergent,
7.
,.
divergent.
,
e
divergent. If x = l and if 7~a-/3 is positive, negative, 01 zero, divergent.
is
.rl, or x = l, divergent. values of q, positive or negative.
The
results hold for all
a negative, or zero, convergent; a positive, divergent.
XXII.
Page
a.
256.
jn(n+l)(n+2)(n+3).
l.
-n(4n«-l).
2.
3.
in(n+l)(n+2)(3n+5).
4.
?t
5.
in(w + l)(2/i + l)(3» 2 + 3»i-l).
6.
p3
7.
&3
8.
ad=bf, 4a%-0»=8ay.
13.
= 27a 2
- a/ 2 -
+ 2/^ft
flic
c*=27acP. 3 6jy -
2 c/i
l
+ Zx + 4x2 + 7x s
1
-
+ x-- x - + - X 3.
5.
1
- aX + a
6.
a = l, b
9.
The next term
3.
3
1
Page
b.
.
3 -
+
«
= 1, &=-l,
X3
1)
260.
4.
,
+ 1) x* - (a 3 + 2a 2 -
= 2.
11
5
-
X+s
,
X2
-
:i
.
21
+mX
. .,.
.
7.
is
2
l-7x-x 2 -4Zx
l
(a
^
2.
.
1
„
(2n 2 -l).
= 0.
XXII. 1.
2
c
= 2.
+ -00000000000003.
an 11. (1
-
- a2)
a) (1
(1
- a3 )
(1
4 «r^
i
l-3.c 2
4.
zr
X-l
—
+
7.
+3 X*+l
41.r
5
10.
*•'
•
1-2.T* -
a:
i.
3.r-5
—
^. -3' a:
4
5 }
?;
3.
•
4x+H'
,11 1+--
5.
.r
5 (x
l-2x
+2
(x
+ 2) 2
1)
'
11
17
16(s + l)
4(.c+l)
17 2
1(5
(x- 3)*
15
x + 5' 7
3.r
13 (.r-lf
T I-l
'
.r
,
2
+ 2.r-5
1-x' 8
-
3 "
;,T TTl-7— 7T,+ (.l•-l) (*-!)«
-,
«
265, 266.
4
3
1
- 1
x-2 +
8.
7 n
n x -2
1 .t
5 ^
*
Pages
XXIII. !•
- a")
.r-3"
5(2.r
+ 3)
HIGHER ALGEBRA.
540
12
'
13 • ld
1
X-l
X+l
{x
+ l)
2
{X
BiiWsoW 4
_il +
1
'
v
l
8(I^j-3(I^) + (r^y-
"•
4lI^)
18
Tfx + (TT^-2T3i' (-V
-
19 '
1 4(l
31
2la^I)
+
— 3a; ;? 2(rT^)
2
3 d
(l-sc)
*
j
(a
-
(l-x)
b) (a
2
+ llr
-
+ c)
(6
- c) (6^o)
2
1
*
11)
a:'-
)
(+•-£?)*
1{{
eVen '2
'
-(2^p-2^ + (I^-)2 +
f9\
'
!{9r + 8+(-l)'2~}V.
12
1-sc 2 *" +
1
23
(
2^+V
-2 r +2}^.
1)2
*
!
_JL J_l_ 2 n
l
c^" 2
2
r^; f+»-
5
7.
+x
1
*
(4r+1)
x (l-3
(l-.r) 2
^
2
'
xn+l i_ xn+i
x*
l-^
272.
l
l-4x
'
-
+ x-2x*
{
>
1
1
-
6'
3-1 -3. 2-1)^-1- 2(1-3^") 1-3* ;
'
+ ax)
(
'
v
X
W*
1
<
l=^flE?r®5^ + tr-+ 1)" r (2.
{
—
1)2 }XT
1
3 - 12* + Use2
'
+ -
1
XXIV. Page (TT^25
-h
1
5r + 9)
(
1
25
)
(c-a)(c-b) 1
+ aP+tx
1
.-r(l-a•)(l-.'c 2)•
+
1
(1-a) \l + a x
*'
3
- 3}a;r;r0dd '
-
5
'
;4M
+1
'
2
^r+2
j
22
^p
-1
(- l)' r
{1+
5
>'-V
3V5'-
"
+
1
V~* J
1
3(x + 2)'
-L. - -1_ - ,-4l-# 1 + a; 1-2*
{X+l)*'
flll(+
+ *)' 3\
3(2
3(* + 5)
+ 1)3
±Pn-r-4n*.
•
3(1-*)
1
15.
3
3
1
1-Sx
+(-
1 ) r2r
^
r -
3- + 2n-l;-(3«-l) + 2n-l. 3(1-2^*) 1-2*
*
'
"
ANSWERS.
11.
12.
-3» n _ 1 + 3» u _o-j/ n _3=:0; Sn =5aD -S, where 2 = smn
?/
may
13.
(2n+ 1)3+| (2**+i+l).
L
2
13
1'
6'
'
7
2'
5'
3
10
1
3
'
+
*•
+
9
323 150'
074 313*
232'
1497'
1174
95
85
121
T
11
26
37
1
' '
'
1
1+ 1
1
1
1
+
2
+
2
6.
JL _L JL JL J_ JL *• 3+ 3+ 3+ 3+ 3+ 3+3'
7
_1_JLJ^JL_! 3+~
3T 5+ IT 1+ J__l_ J_ J_
10.
n
16.
"30"
^ 109"
Ill
2+ 1+
_1_1.
11
35"
;
5
_7
;
1
19"
11111111 3
3+ 3+ 3+ 6+ 1+ 2+ 1+ 4
'
'
;
3
2+ 1+ 2+ 2+ 1+
12
'
157
1
D
4+ 3+ 2+ 1+
"359
1117
1
1
+ 2+
'
'
*
'
+
J__l_l 3+ o+
n - 1
+7
(n
3' r-
10'
=
jt
,
7i
- 1
+l
and the
;
n2
?j
3
XXV. j
*
11 a+ (203)
4.
and
(«
1
"'
151
11
"2(1250)**
+ !)+
- n2 + n -
Packs 281—283.
b.
+2) + a +
"
3'
a
a
3
5
?5i 223'
39
161'
47 194*
three convergents are
first
~n*~
1
(a
33'
29'
i'
+ l) + (n-l)+
8
7
1
~T~' n+~l'
2
7+ 5+ 6+ 1+
X
208*
GO*
—
+ J_J^AJl_
G3
259
//
1
277, 278.
JU3
36
2
Pages
43 105'
13
1
+
a.
22'
'
term.
to agree with the result in Art. 325.
f7'
1 2
'
7
.r"
to
XXV. 28 13'
15
12
2
shewn
easily be
l
M»-4«n_1 +6Mw_a -4ti _a+«»_4=0. to infinity beginning with (n + l) th
n
This
l-2'
l-3».r"
l-.r»
,
541
115'
a*+3a+3 + 3a- + la + 2
'
'
HIGHER ALGEBRA.
542
XXVI.
7m + 100,
1.
a=
2.
x = 519t-73,
3.
x = 3934 + 320,
4.
Four.
8.
290, 291.
= 775f + 109; a=100, y = 109. = 455t-64; a=446, y = S91. = 436t + 355; x=320, = 355.
2/ ?/
?/
?/
5
Seven.
5.
^3117 12'
Pages
8*'
12'
8
°
;
\ 8'
r
£6. 13s.
L 12
^ ;
4
-.
1_ 12*
8'
x = 9, y = S,
9.
-,
6.
= l.
= S.
z
10.
x = 5, y = 6,
= 9,
s=7.
11.
x = 4,
13.
14.
aj=l,
15.
280« + 93.
17.
19.
Denary 248, Septenary 503, Nonary 305. a=ll, 10,9, 8, 6,4, 3; 6 = 66, 30, 18, 12, 6,3, 2. v The 107 th and 104 th divisions, reckoning from either end.
20.
50, 41, 35 times, excluding the first time.
21.
425.
18.
= 3,
y=2,
z
7, 2, 6, 3,
2; y
1; y
= 5,
= ll,
3; z = 2, 4,
1,
+
J '
5
d -
7.
1
899.
2T4T'-
11
+ 3+"6+"
3-f
;
d -
4
198
2 -
3970 1197
;
1
,
2889 1292"
1
IT4+ -
+
;
99 35"
1 1 J_ + J. J_ 1+ 1+ 1+ 1+ 6+"-
6 '
119 ;
33
""'
31
197 .
"'•'
*
42
1351 '••
1
4+""
116
1+ 2+ 1+ 6+
+ _1_ J_ 2+ 6+
294, 295.
n
+ _L J_ JL J_ JL J_ 1+ 2+ 4+ 2+ 1+ 8+ 9
Pages
a.
„
1111
1829 and 1363.
23.
485 ;
1, 2, 2, 3.
181,412.
••"15*
1
?/
= 7.
3.
26
1
"
3
= 1,
16.
XXVII.
1+2+
2
4, 8, 1, 5;
22.
1
as=2,
12.
2
J_
JL_
"390"'
111111 11111111 111111
JL
1_
1+ 1+ 1+ 10+
-
198 ;
*
35
161
11.
12.
6
1+ 2+ 2+ 2+ 1+ 12+
12 1
13.
ie ±D
'
21
1+ 1+ 1+ 5+ 1+ 1+ 1+ 24+
4+ 1+ 1+ 2+ 1+ 1+ 8+ "" _1_
'
+
'
Jj_
J_
_1
1_
.
'"•'
253 '
'
20
12 55*
47_
1111111111
5+ 1+ 2+ 1+ 10+
_1
10+ 2+
270'
1+ 3+ 1+ 16+ 1+ 3+ 2+ 3+ 1+ 16+
1_
280 .
'•' 351*
5291 >
'"' 4830*
ANSWERS
4030 401
9 '
„„
'
1
!
1
3+3T-
;
21,
*
483
-
1111 ——
a 4+
24
1(577
20
*
—+ r— 1+ 1+ 1+ 4+
(
4
22.
"»4.*>
2
23.
... '
11111 rT2+8+3+3+-
26.
Positive root of x-
28.
4^/2.
+ 3« -3 = 0.
+
1
+
—111 + 3+ 1+
2
2+2 +
""
"
^
25 '
•
1
I
27.
Positive root of 3x 2 - lOx
30.
-.
-4 = 0.
a
XXVII. 1
1
a+
1.
2
2a
+ 2a+ 2a+
+
*
*
1
4.
+
a+
6.
-
:,
-8a + l 8a-4 2a- -1 8a 2
1
1
1
1
1
301, 302.
l
2T 2(a-l)T 2+ 2(a-l)+
,;
*
1
1+ 2(a-l)+ 1+ 2(a-l)+ "' .
1111 1111
+ 8a + l 8a 2 + 4a 2a2 6 2 + 4a& + l
"'
'
'
1111
b+
2a + b
+ 2a +
2a
2
8a
2a+ 2+ 2a + 2+
a-l-t
6.
8a 4 + 8a2 + ""' 8a + 4a
1_
J.
,
a-l +
3.
1
Pages
b.
'
2a&*+26
'
2a/i-l ""
1+ 2(n-l) + 1+ 2(a-l)+ 432a + 180a 3 + 15a 141a 4 + 36a 2 +1
'
"
2/i
5
7.
Page
XXVIII.
s=7
l.
or
1,
2/
= 4; s=7
or
5,
311.
y=6.
4.
x=3, y=l, 11; *=7, y=9, 19j x = 10, y = 18, 22. 5. x = 3, 2; y=l, x=2, 3, 6, 11; y=12, 7, 4,3.
6.
x = 79,
27, 17, 13, 11, 9;
7.
x = 15,
?/
9.
x=32, y=5.
3.
12.
2.r
y=157,
x = 170, y = 39.
8.
x = lG4, y = 21.
10.
= (2+ x/3) +(2- v/3)»;
2V
'3
.
2^/5
.
4.
61, 29, 19, 13, 3.
= 4. n
# = 2, y = l.
2.
y = {2 + s/3)
11.
n
-
(2
x=4, y = l. - v/3)»;
/t
being any
integer. 13.
2x = (2 + v/5) n +(2-^/5)' even positive integer.
14.
2x =
(4
1
+ v/17) n +(4- v/17) n
any odd positive integer. The form of the answers
mode
;
;
^=
2 V/17.
—
(2
?/
+ ^5)"- (2 - v/5)»;
n being uny
= (4 + x /17)» - (4 - v/17)»;
n being
to 15 17, 19, 20 will vary according to the of factorising the two sides of the equation.
.
HIGHER ALGEBRA.
544 x = 11 fi -
15.
3/t 2 ,
y = m? - 2mn.
x=2mn, y = 5m?-n*. m--n-\ 2mn; m 2 + w 2 Hendriek, Anna Claas,
17. 19.
16.
x=-
18.
53, 52; 19, 16; 13, 8; 11, 4.
20.
m2 -n2
.
21.
;
m 2 + 2mn + n~; 2mn + n'
y = m2 -
?i
2 .
2
j
.
Catriin; Cornelius, Geertruij.
XXIX.
Pages
a.
321, 322.
1
1.
^n(n + l)(n + 2)(ti + S).
3.
—
-
(3>i
2) (3n
+
1) (3/i
2.
+ 4)
(3n
+
+ l)(n +
n
fc
n 3;i+l'
'
l
=^^
t-t^
^r
;
+ 1) (2/7 + 3)' 5 2/1 + 5 + l) (?? + 2) 4*
4(2/1
5
'
« 9.
,~. 12"
'
2
(7i
S-nT2 + a(n + l)(i» + 2)
;
16.
— (n + 1) (n + 2) (3/r + 36n
1?
(n-l)«(n+l)(n+2) 6(2n+l)
2
+
4).
50)
3)
,
3
2
2
?i(7i
+ l);
+ l) 2
5.
rc(»+l)(n+2)(n+4);
(l-.r) 3
+ 3)(h + 4)
»(n+l)(n+2)
n
3
n+1" 20
Z)'
-
Pages
b.
-77 (/7+1) (n 2 - 3u
6*
1
H+
1
-7l+~r
332, 333. 5/t
2
+ 3/7;
-
o
n
(n
+ 1)
(5/7
+ 7).
l-a; + 7
-
2).
^n(n+l) (n+2)(n+3)(4»+21).
2 *
1 ;
— «(n+l)(n+2)(3n+l).
-4raa (n-3);
.r
(u
n j (»-l)(w + l)(n + 2)(2n+l).
2.
4.
+
2
15.
.
7t
l
24*
'
4)
1
3.
(/i
+
j^+l><« + >H« + «>(* + «>'
ig
XXIX. 3» 2 + 7i;
h+3
+
l) (3//
+ 151n+ 240) - 32.
+ + 2~^2~ ~^~ (n+l)(n + n(/t
12.
(/t
^
i'
1
_
3)
1
1 ^T 24 6(3/i +
6
-?i2 (n 2 -l).
1.
+
1 3*
11
5
14.
19,
(«
-(n + l)(n + 8)(n+9).
7.
l
t^ 12
12
5.
1.
n+1'
'4
+ 7).
6)(/i
, ;
1
o
8.
+2)
1) (n
11
-( n
6.
n (n+
+ ^ = ~ {21 n* + 90/i 2 + 45u -
7)
71
4.
-
6.x-
2
-2.r 3
-
8
*
-
(l-a;)3
2-.r + .r 2 (I-*; 3
1-aJ
1 + lLc + lla^+g3
9
(1+.t)-
(l-z)
4
gj.
13.
3.2» +
/7
+ 2; 6(2"-l) +
W
'
^+5
^ .
+
1
ANSWERS.
545
14.
n«-(n+l) s ; ^(Sn»+2n*-15n-26).
16.
2»+>-n s -2tt; 2»+ 2 -4- t»(u+1)(2«+7).
17.
3--1 + 1 n - xn
1 18.
n
'
+ 1)
(n
(3
_
„ +1
8)
+
2n
(3k3
n - xn >i.r " (1-*)" (1 - xf n + n- 1 4 1 2
1
19.
21
'
'
+ 27k 2 + 58n + 2) 23.
)i
"
+
(n
1)
2(1 -a?)
+
60
k(k + 1)(9h + 13k + 8) 2
1
25.
12
1 - -
2
2'1.3.5.7
(2k
+ 1)'
2»t+i
1-
26.
27.
+ 2'
|k
(n-l)3»+1 +3.
28.
n 30.
31.
4
1
n+
1
(e
(e
x
- e~ x ) - x.
x
- c~ x
+ x)e x
(l
5.
1-
33.
XXIX. -
1-3-5 2.4.6
2
2""^T2
1.
-k + 4)2"-4.
+ ie~ ix ).
1+
J
2(k+1)(k + 2)
*3"'
{p
k+4 (K + l)(K+2)
'
1 n 2 +1
'
—-log(l-ar). 1
4.
6
.
(2k + 2)"
Pages 338—340.
c.
2.
ie ix
(2/t+l )
l
J
1 32>
(k 2
*
29
2».
n+
(r-2)|r-T
+ q) r 7.
1.
0.
10.
4.
3(<3-l).
13.
e*-log(l +
II 8.
k(2«-1).
9.
2-£,
12.
lo g<
11.
...
14
>
<
15.
7
+
n
2
5 ?i ?f
3 71*
n
+ 2--6 +
n8
K 42'
17.
(1
2 \
+
-'
)2
1
+ k+
lo-a+.T)-
n2J
v
7
?i
(J)
lot'.
x
20.
6 71°
?l'
X '
(1)
7kb
7k 4
12
24
+
.r).
ir
12
n + 1.
k+
'
^.
3
21.
»(n-fl)2»~».
4.r
22 -
W
1
3
I
H. H. A.
2»+i
xn
7TT2-~^~ 3 n (k + 1) (12k 3 + 33na + 37n + 8)
15 24.
]
»(» + l)(. + 5) _ „
1
n+1
22.
1
.
11
1 20.
+g)
nx 1 - x
- a;) 2
(1
(n
3*+ n~+"
S*-i + n;
15.
+ (-!)«+'}•
<-)
+ (_1)
5l2
(n+l)(n+2)/'
35
'
HIGHER ALGEBRA.
54G
XXX. 1.
3, G, 15, 42.
7.
23.
(jl,
where
t
is
2.
1h
—-
10.
2 to 3.
2197 20825 11
16.
18.
Pages 356—358.
a.
Pages 367—369. 18.
2.
6.
integer.
.
XXXII.
5.
b.
an
XXXI.
348, 349.
8987.
33.
x = 139t +
Pages
1617, 180, 1859.
2.
:.
20.
a.
a.
1
;
it
can be shewn that qn =l +j> n
Pages 376,
377.
-
ANSWERS. 11.
14.
A £5; B MX
(1)
20
£11.
12.
...
;
1
I, ,
shillings.
276
(2)-
w
XXXII. 1.
13.
27
250
7770
.j
15.
4a".
16.
17.
31+
in.
2
d.
Pages
309, 400.
47
HIGHER ALGEBRA.
548 26.
27.
The determinant
is
u
w'
v'
w
v
w
v'
v!
10
equal to
=0.
a?
ANSWERS
XXXIV. 1.
x3 + xy 2 + ay- = 0.
2.
4.
y-
= a(x-Sa).
5.
7.
9.
6 e
4
4
4
4
4
2
+a
+
1
+6
.
-3=1. 1 1+
+c
1
13.
ab = l
15.
(a+6)*-(a-5)$=4c$.
17.
abc = (ic
2
(o
+
c.
+b-
a-b - c) 2 1)-
-
c
b
x
+
-
(a
-b)cr+(a- c)
bq
1) (a-
(b
-
c)
+y»=o* ««+ys =2as
.
?/-- loa;=fc2 (a:+a)".
-2a 2 & 2 -& 4 + 2c 4 = 0.
10.
«4
12.
5a 2 6 3 = 6c 5
14.
a3 + 6 3 a°
+
1 i
3
aB
.
+ c 3 +abc = 0.
+ c 2 ±2abc = l. 18. « 2 -4a&c + ac 3 + 46 3 -&V = 0. - 2a6 + & 2 - a - b) + ab = 0.
1
22
6.
16.
.
{a+
3.
8.
abed
l
20.
= 0.
rt
4
11.
449, 450.
a*-a*=l.
+ c a + a 6 = cf-b-c-J a -4
Pages
c.
z+
549
ap +
(b
-
a) cr
b2
1
+ (c
-
a) bq
+ (c - b) ap 1
bcqr + carp + abpq ab'
23.
-
ac'
a'b
- a'c ad' - a'd ac'
-
ad' - a'd + be' bd'
XXXV.
-
3.
4.
re
6.
4
-2
(a 2
+ b 2 x 2 + {a 2 )
b2 )2
cd'
1
l
2'
2'
_3 2'
4'
113 2' 4*
c'd
+ 2xr - lis*-
1,3,
p ~°-
3
-
= 0.
456, 457. x*
5.
.
V
b'c
b'd
2. 6x 4 -13x 3 -12x 2 + 39.r-18 = 0. 4 3 2 6 ox* - 8x + 40x + 1 G.r - 80r = 0. x
1.
-
Pages
a.
- a'd bd! - b'd
ad'
a'c
1
3"
i
5, 7.
12.r 3
+ 3Gx 2 = 0.
'
.)
HIGHER ALGEBRA.
50
9.
11.
s4 - 10^+ 1=0. 3 2 ar» - 6a; + 18x - 26a; + 21 = 0.
4
10.
re
12.
8 a;
-
10a; 3
-19.r 2 + 480a;
1 6.r 6
+ 88a;
4
+192.t
-1392 = 0. 2
+ 144 = 0.
15.
One One
16.
Six.
one negative, two imaginary. [Compare Art. 554.] one negative, at least four imaginary. [Compare Art. 2 20. 17. (1) pq = r; (2) ph'=q*. q -2pr.
21.
pq-r.
22.
^-3.
24.
pr-ls.
25.
2 2 i p -4p q + 2q + 4:pr-4LS.
13.
positive, positive,
1. 3.
5.
10.
4 a,-
2a;
-6a; 3 + 15a: 2 -12a; + 4
+ 8a;
16aa;7i 2, 2,
(.r
1
13.-2,
3
l.
2.
-a; -8a; -20.
+ 7a;
1,
-
4
/*
2
+ 7a;
1, 1,
1,
17.
a, a,
-a,
19.
0, 1,
-|, -|
22.
(1)
+ h + 2bh )
4
- 37a; 2 -
2.
1
^
16.
± x/3,
2
(2)
0, 1,
-|, -|.
-1.
27.
d.
20.
3, 3, 3, 2, 2.
2
'
i^/3, 1
l±J-7
;
+ 2ch.
4
7i )
1
14.
>
- 110.
-24a; -l.
12.
—
/3 ^Wg'"
18.
6.
123a:
2
+ 10.r 2 /r +
1
XXXV. 1.
a;
4 (5.T
4
1, 1, 1, 3.
2
-1, -1,
15.
/t
6
470, 471. a:
4. 4
1±J~Z ,
2
2
11.
3.
= ^/^3
-2;
Pages
c.
2
6
pq-Sr,
23.
XXXV.
554.]
2
= ^/^1.
/3 ;± \/2'
l±,7r23
n"j*-* = 4p»{n-2)»-*. 28.
5.
Pages 478, 4
99,795.
479.
-5?/ 3
+ Sy'2 -9y +27=0.
2.
?/
4.
3 ±2^/2, 2
6-
2,2l,L(l±Jl3).
= ^/3.
4
ANSWERS.
XXXV. d*±£Lll.
5,
1.
2
3±4 7^~3.
-6,
-ld=7^1.
-2,
4,
13.
1±72, -1±7^1.
15.
2,2,|, |.
17.
-4, -4, -4,
22.
-2±76, ±72,
25.
2±73.
28.
.t
4
-
8.r 3
±72.
+ 21a; 3 -
2 y -
+5
5//
I-
sV + 2s(l-*) + r(l-s) °—^?. 26. ?/
4.
(1)
1±75; 1±275.
(2)
35=1, y = S,
we merely reproduce the
z= -
5
;
or
x=
a + 2b
8.
9.
14
.
15.
a4 +
+ a-^"
&4
(p
1
2
10.
).
:
+ (l-4 4 = 0.
7.
original equation.
Pages 490—524.
Eight.
-
y=
1,
-
3; 2
First term 1
;
= 5.
common
1 -
difference
.
-q)(p°-3q).
^.
A, 7 minutes;
13.
ZJ,
8 minutes.
=
or
,-=£;
c-a a-b 2 be -ca- ab) = d. Jc-a (a- + b- + c a+b+c
;
16.
One mile per hour.
17.
(1)
13.
(1)
{b+e)(e+a){a+b).
^
(2)
^/-g- +
^
"
~.
'
2
18.
^
;
22G8.
105
(2)x = y=±7^; 22.
j/
+ c 4 = 6 2 c 2 + c 2 a 2 + a 2 62<
1 x- = J y =
where
3.
(2)
.
^-3; -pCp8 -^); -(oft
3
;
3.
-,
2
2
23.
respectively, so that
-q
3
+ 5 = (x 2 - 5x + 5) (a; 2 - 3a; + 1) on putting x = 4 - y, 2 and a;- - 3a; + 1 become y 2 - 3// + 1 and a; - 5a; + 5
6, 8.
1,
11,11,7.
20a;
2.
(1)
6.
-^. 3 g»+8r*=0;?, -^-5
MISCELLANEOUS EXAMPLES.
6.
4,
4 ±715, -
1,
18.
3.
2
-2±5j~^S.
3.
12. 1,2,-2,-3. * 1,-4*^/6. 14. 1, -3,2±75.
U.
16.
expressions
the
488, 489.
--.-iVJ?.
5.
10.
Pages
e.
10,-5±7V^8,
-
a 4.
551
1«*5; nine.
a^6= 23.
i {(1
-(J+26) + 2 + 3+
..
=± .
+
\/y+a6-a«-
/<)-- (l 2
+ 2-* + 3 2 +
...
+n-)\.
—
-
HIGHER ALGEBRA.
552 24.
Wages
15s.; loaf 6d.
29.
x = 3fc,
?/
= 4fc,
= 5fc; where
z
6, 10, 14, 18.
25.
F = l,
so that fc=l, w, or or.
32.
Either 33 half-crowns, 19 shillings, 8 fourpenny pieces; 6 shillings, 17 fourpenny pieces. or 37 half-crowns, 33. 40 minutes. a = 6, ft = 7.
35.
1
31.
37
+ x + ^x 2 -- x - — x\ i
-l^y-3
1^/gj-
pr
^
.
[a,.4_.r
1
a = 8;^—-.
38.
480.
30.
_5 (a;2 + a; + l) = 0.] The
40.
term.
first
.r- 5
+ 4ft 2 c 2 + 9c 2 a 2 + a 2 .,,.,, , a 2 o- + c-
l 41.
42.
13,' 9.
43.
ft
2 •
-|-
-
(1)
3,
(2)
x = l,
2,
— ^^—-|, -1,
ssl, -|,
0,
2=1, -\,
0,
[Add
.
0,
0;
-1,
0;
a-
+4
2
to each side.]
0,-1.
5780.
17.
a
150 persons changed their mind; at
48.
majority 350. 51
-
-
'
[Put (a - c)(6- d)
ad -be
.
(2) K
C
53.
a-b-c+d
= {(x - c) -
--577-.
6,
936 men.
50.
„. _ 2 m -l 1 ) 0, 7Ji* '2 m +lrrt.
.
(x
- a)} {(x -
55.
m = -r—
30
^/a
±4 ^[putting
2
(1)1.
60.
—,
64.
Common
difference of the A. P. is
which
the reciprocal of the H.P.
——
b-c
——
males; ^—
is
b-c
a(n-r)— b(r-l) + -^ — '-
-
n-1
68.
...
;
,-
(x
,
-
ft)}
;
»=-, v/a
then square.]
+-^St. ^/6
,
the in - r +
63.
n-1
,... th
l)'
69.
common
;
-=->
is
ab (n -
.
,
term
is
—
a
0,
.
difference of the A.P.
—
.
[The
r th
term
1)
aft(n-l)
— —
a(n-r)e + brr (r -.
— a+b
+ b,
= 0.]
4)
rv -1
1)
£78.
-1*^-3
l±V-3
*
'
[(a
—
-
+ ^/6
females.
19.
U *
a;
d)
- 16 = y 4 we find y*-16 -4y(?/ 2 -
58.
(2)
the minority was 250, the
first
2
+
ft)
3
'
2
-a 3 -ft 3 = 3aft(a +
ft),
and (a-
ft)
3
- a 3 + 63 = -3aft(a -
ft).]
is
.
ANSWERS. 72.
73. -.
79.
*4U
.,*]«»
(1) v '
log b
*
a -=*f=-=sO, !f
a
'
o
or
85.
Sums
86.
503 in scale seven.
95
T 6 5 *-
100.
?/
29
.. , ,. r Generating function is
l) n }
+6-
-
a-
109.
(1)
111.
1+ JL -=-
117.
(1)
c-
-„
1-x- 2i-
a(J
1)
.,
;
= 3.
126.
84.
/5 96
'
! "21'
\/3'
"
——
l-(-l)».r» 2(l-2'\r») = -' \ 1 - 2.c 1+x
suru= -^
12 persons, £14. 18s.
x = 3, or
(2)
d
+ 122.
.
x = 0, y = 0,
(2)
1;
|f=l, or
3.
x = 948, *y = 492.
{*»+"-
113.
£12. 15*.
2
+ **+> - (n + 1) 2 * 2 + (2*« + 2li - 1) x - ny-
(1)
t
lO.r-1
- 3x -
3 (**+*+
= 1;
1)
scale of relation
125.
1
127.
(1)
128.
(1)^".
130.
(1)
x=
(2)
-
.r=-6,
a
2; y
= 9,
2^+1
'
.
±3.
'
- x - 2x- general term is ;
-3.
(2)^.
(2)
129.
*=-;
2' 1 ~ 3 {
+
(- l) n_1
}
.r"-'.
y=y
12, 16; or 48, 4.
±7.
= I = - = ± JL b
c
r-1.
133.
11,
137.
(1)*=±-^, V2' £3.
is 1
1)
1
r+4
t
2 (.r
^-
or
= 0; x=±2, y=±l 2=
- 23
13.c
3
ar
x = a, y = b; x = a, y = 2a\ x = 2b, y = b. x = 3 or 1, y = 2, 2=1 or 3;
W l ~^TW"'
124
82.
x n ~K
-L
-i-
= v.]
2y
+ 4x
-z
= c.
z
b
10^
108.
(2)
138.
5
,— 1; 1+12+1+1+1+9
-
+
'
d.
x = a, y = b,
(2)
121.
and y -
25
3 '
l
.
,
107.
.
and £3500: the fortune of each was £1100. 91. 25 miles from London.
ItthE}.>*>-_ d3
! *»
'
n ih term= {2" + ( -
12 °-
x-a = u
invested were £7700
_
= y = 2 = l. ,
2)
.
[Put
81.
1.
...
-
a6c
a = 3, 6 =
_„
a+6+c
c
80.
'
8 hours.
74.
** * (1) v
l0 2) *=* 2(^? ~*l-189. 1 log S 2
(2)
2.
7,
553
2s. at
134.
the
where & 2 = 26 2c 2 + 2c 2a 2 + 2a 2 6 2 - a 4 -
2abc
384sq.yds.
y=±^. y"
(2)
v/2
first sale
and £2.
136.
a= ±2,
±^; *^/^12*. at the
second
sale.
64
-
c 4.
6 = 3, c
= ±2.
.
HIGHER ALGEBRA.
554 139.
141.
142.
(1)
i«(«+l)(2« + l).
(3)
|n(»+l)(4w-l).
(1)
x = l or
(2)
x, y, z
y
3
?/
=3
may have
y.
or
the permutations of the values 3,
y/
(1)
w
1)
2' l +i
3
x-l'
(x-iy
(3)
(
-2, -2, -2, 3*
146.
A
overtakes
A
5
nth term
is
A= ,
where
(an — b n ) n L -*
— a-b
a;"
n n a(l-na x —— y
1
- ax
a).
12, 13,
14,15,16,17,18,19,20,
-1
miles,
(a
+
2 (o'
b
;
A
+ wc).
Sum = A - B,
;
a2x
) '-
-
11,13,15,17,19,21,23,
-(a + b + c), -(a + wb + u) 2 c), -
148.
*
~
c 2) (b
and passes him on the third day and overtakes him on B's 9 th day.
^6-1 147.
-
2
in 2 days
B
subsequently gains on
(&
1, 3, 5, 7, 9,
B walks
B
'
- d3) = 3
2 (6 3
144.
walks in successive days
so that
+ Ua; -157a: 2
+ 5x -50a; 2 -8a; 3
1
'
+ -u(/t + 7)-2.
145.
150.
5, 7.
+
143.
y;
~n(n + l)(n + 2)(Bn 2 + 6n + l).
(2)
H
(1
- an- 1 x n ~
*_ (1-aa;) 2
i
)
and
B
denotes a corre-
sponding function of B.
- 2p 2y 2 - 5pqy - 2p 3 -
151.
qif
153.
(1)
-7,
156.
(1)
|
„.
,
,
7±8 y~ 8 -
^ ,„„
0± T92
22 years nearly.
162.
(1)
_
(2) [It
(
2)
2
= 0.
^
±8
89
^4
,
157.
x -=r=
-
q
[(12a;
.
11
2
-
>
r*-
;
154.
«
1) (12a;
-
2) (12s
-
3 days.
3)
(12* - 4) = 120.]
"I
161.
-7±V217
4
44 hours.
.s=±i, ±2; y==F2, Tl;ar=-y=±V3
x = h (6 4 + c 4 - a 2 6 2 - a?e 2) &c. where 2/i; 2 (a6 + b G + c 6 - 3a 26 2c 2) = 1 2 is easy to shew that a x + b 2 y + c 2 z = 0, and a 2y + b 2 z + c 2x = a; 3 + y'i + z'i - Sxyz = a 2z + b 2x + c 2 y .] ,
,
+ b + c)x = (be + ca + ab) ± J (be + ca + ab) 2 - 4abc (a + b + c). [Equation reduces to (a + b + c) x 2 - (be + ca + ab) x + abc = 0.]
163.
2 (a
164.
(1)
~n(n + l)(n + 2)(Sn + 13).
(2)
2e-5.
;
555
ANSWERS.
166
.
J + f* 1
l )x {
#,
(2)
/2
z are
?/,
(*+y+*) 9 =3*
170
He
172.
(1)
9.
168.
AB = 37h £C =
= 10
i/
M
b(d-c)
* M fa ± / + ( 6c * ad} t? T '?) - ac * bd if T eh>' q = (be T ad) (0
7>
=
(
ac
/l
*»+*»+** -Says.
10 miles per hour.
a(d-c)
,
''
)
)
(
. = 6, -5;
178.
g
,
rr + 3^- + (3,-r)^r-0.
"6.
£3200.
177.
.,
or 13.
c(.-ft).
d ia-b)
-
rides
drives 7* miles, 30, CA = 15 miles.
or 10,
2,
169.
2.
3| miles,
walks
= 13
[Eliminate..]
the permutations of the quantities
167.
a;
1
a,y= £.
£
if
)
'
>
2
14*^/^74
-18±*/^47
y = 5,
-6;
[Put
- y = M and ay =0, then u + 2y = 61 u(61 +v) = 91.]
.
2
g
2
,
a;
3=^5
1 183.
8987.
182.
?/
3
-
2 ft»/
-aci/ -
c
2
= 0.
-1, -8, -j, I
186.
(1)
x,
?/,
s
quantities are the permutations of the a(Z> 2
+ c2
1,
+ n/^3- 1-n/£? g—
_
)
Irish 35, Welsh 11. Conservatives; English 286, Scotch 19, English 173, Scotch 41, Irish 68, Welsh 19. Liberals
191,
(1)
-3.
2±J-3, -2±J-h
(2)
a-ft
a~ b
-n J. I.-"—-. 2b n =a+b--gr.
2aw =.+6+-yr;
_
201.
07,
.
192.
n
-26, 14±840 N /-1.
54,
202.
3rw 3 + nm2 g-3Ti3 206.
m+ 3
2
nvi q
+
24 7 Poles, 14 Turks, 15 Greeks,
210 J1U
1_? _l+^iog(l + 6
212.
213.
2
(1)
4
v
2.c
jm + n-2 |m _ x
4«+ 1
)n
204.
^,
207.
81 years nearly.
Germans, 20
4 „+! _(_!)»
Italians.
a;). '
jn(n+l)(n+2)(n+3);
W
"Ti+iyr
'
_x
+ 4(-l) w +
n8
209.
*
2
,
187.
7, 9,
2
(3)
»•
1
.
556
(l)^ = l (±15±V 33 or* = 4,
6,
y = 6,
4,
-6; -6, -4;
= 5,
5,
-5,
2
~
a;
(2)
)j
(±15W ^
);
-5.
-rt)~c^rT6)- X where (&- c )( C -a)(«- & )x = a2 62 + + c2 _
of
6( C
'
relation
2*»-i
^n = —s
23«-i
TH
3
Lim
&c
_ ca
_^
f (^-l)} =|;
convergent.
Xtenn = -1 {«« + 8—
5
"21
7
ar =
232.
2T3-
229.
1-12., + 32.-
is
^
11
231
=1
- V~ b _ g-c
fl
a(6-c)
Scale
230.
,
-4,
12calves, I5pigs, 20 sheep.
226.
±
~
P+?*
N /a=*
+ 6«- c a, & c
+ V + cS=aZ(b + c) + bZ(c + a) + c*(a + b). (1) (l-*)^=l+4o;+^-.(n + l)3a.n +(3B8+6|lf _
.
a3
233.
235.
^
4)
I
- (3m + 3n - 3k + 3
(2)
2
(n
fl
3hours51min.
237.
1)
«*M + »%»+«
1
J 8
+ l)2( 7i + 2 )31 + a*x* + a *x* + «x" + ««*!« + a*V» + aux* + „^ 28 + '
236.
240
2 or ^
.
242
.
^
+ a20;r,6<
_ 140
244.
3,4,5,6.
246.
a*(c*-Sd?)*=:(ab*+2&) (ab*-£f.
247.
2,
248.
±
6,
3.
1,
.
.
13 249.
(1)
2*-H-2-jU(n + l)(2,t + l).
(2)
-_^!L_ _ 2 (n
<
+ l)(n + 3)
T^
3 >
when n 250.
is
+
S'
*££P -he- . U even
(l)* = 2/ = * = 0or|.
(2)
,
If*^ + 'Jiz*^)
odd. If
however X > + y > + z * + yz+zx xy = +
* + 2/ + * = -a, and the solution
—,
a(-a + b + c)
V
is
0)
then
indeterminate. z
b(a-b + c)-^(^b^c) 1
±
flM
253.
.
HIGHER ALGEBRA.
223.
-
'
+ + cjja -b + + -&+BE1 C,H-A, + B, + c.)<4s/l£+ ,- B , + + (i
a
c) (a
*>
, )(Alf
A=
y/a
(6
-
c),
&c.
b~^~c)
Xre
;
;
ANSWERS. 256.
(1)
.>—
(2)
z=-(a + b), x = S, or 7 y = 7, or 3
w,
1,
to-
;
+ W), -(i/w'- + /yu). 2 = 0, or -4 m = 4, or -6
-(rtw
| )
257.
262. 266.
To
places.
Tea,
258.
2s. Go".;
Coffee, 1*. 6d.
a
11 turkeys, 9 geese, 3 ducks. 263. 2g -6pr+24». (1) x, y, z have the permutations of the values
#=
as=f/=« = l;
(2)
\ a (b -
la(b-l+sJb*-2b-3),
a,
268.
3r-2
at least
55?
j_
(i
b
=— c
1
-
JJfi^W^
3).
-f-
;
&c.
267.
16 Clergymen of average age 45 years
0.
;
24 Doctors of average age 35 years 20 Lawyers of average age 30 years. 269.
- a 32 = (afy - a 22 ) 2 2 a.2 a i + 2a a. a - a a afaA - a 2 3 = 0. x 2 3 3
a 2 - afl
(a
or a 270.
X= ±
(a.
2
a4
)
—
:
Va + 6 + c 2
274.
275.
2
;
'\fl-?Viog(l-a;)-2. xj *=?,
(1)
?,
4'
4'
z=±4,
(2)
*
=± 5 3 2
279.
^, 6 birds; B, 4 birds.
287.
a,
294. 300.
-^ a-1
jl(
#
(a +
,
c~$.
273.
w '^—
,
j
l)(a + 2)...(a + n))
w=±l.
^ t2 v!' u= 4v-3'
Vi>
a2 +
+ c 2 + d 2 + \.
-5a,
(2)
2/= ±5, t<=±2,
276.
ft
&c
T
-1;
-g, 3
'
—
2;
3 2
±-
V« 2 + ^ 2 + c 2
(1) v
?/=-l,
291.
u=
&C.
,
2
-5a.
277.
-p 8 + 3^0- Bp 3
281.
2.
289.
S1= - ft"?
1
u=± \^.
^'fl'-fr"?? &* ,
worked 45 days i>, 24 days; C, 10 days. 2 (ft + c a 2 ) (a 2 - 2 + c 2 (a2 + ft- - c 2). Walked 3 miles, worked 4 hours a day or walked 4 miles, worked 3 hours a day. .4
;
2
CAMBRIDGE
ft
:
PRINTED BY
)
C.
J.
CLAY, M.A. AND SONS, AT
THE UNIVERSITY TRESS.
£L
tV f
Date (\i
'DJEC
Due
4
JAM 3^
-
1
*-*"
*£&**•
mm 12.M.
MAY
N»«
1
1
6 4
m W •
P 2 7
\JU^s^'
>)
£L _L
_.
L. B. x:at. No. ii 37
WELLESLEY COLLEGE LIBRARY
3 5002
03079 0690
QA Hi 8
AUTHOR
Hall & Knight
4?212
TITLE
Higher al gebra BORROWER'S NAME
DATE DUE
—r~v t
V^
.,
2jt-AW-$-:
H^APp"
Vy
/
422*£
