Mathematics I (Part of 4CCP 1350) Department of Physics, King’s College London Dr J. Alexandre, 2009/2010
Contents 1 Introduction
3
1.1 Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Combinatorics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Functions of a real variable
2.1 2.2 2.3 2.4 2.5
Continuity . . . . . . . . Diﬀerentiation . . . . . . Polynomial functions . . Rational functions . . . Trigonometric functions
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3 Integration
3.1 3.2 3.3 3.4
Interpretation of the integral Integration by part . . . . . Change of variable . . . . . Improper integrals . . . . .
5 5 8 8 10 12
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4 Logarithm and exponential functions
13 14 14 15 18
4.1 Logarithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Exponential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Hyperbolic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Taylor expansions and series
5.1 5.1 5.2 5.3 5.4 5.4
3 3 3
18 20 21 22
Approoxima Appr ximati tion on of a fun funct ctio ion n aro aroun und d a value alue of the the arg argum umen entt Radius of convergence and series . . . . . . . . . . . . . . . . Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . Expans ansion for a compos position of functions ons . . . . . . . . . . .
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22 22 23 24
6 Vector calculus
6.1 6.2 6.3 6.4 6.5 6.6
Vectors . . . . . . . . . . . . Rotations in two dimensions Scalar pro duct . . . . . . . . Cross pro duct . . . . . . . . Scalar triple pro duct . . . . Polar co ordinates . . . . . .
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7 Complex numbers
7.1 7.2 7.3 7.4 7.5
Intro duction . . . . . . . . . . . Complex exponential . . . . . . Trigonometric formula . . . . . Ro ots of complex numbers . . . Relation to hyper perbo bollic functions
30
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8 Linear diﬀerential equations
8.1 8.2 8.3 8.4 8.5 8.6
First order, homogeneous . . . . Variation of parameters method Second order, homogeneous . . Second order, nonhomogeneous General properties . . . . . . . Separation of variables method Linear function . . . . . . . . . Matrices . . . . . . . . . . . . . Determinants . . . . . . . . . . Compo possition of linear functions Eigenvectors and eigenvalues . .
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10 Functions of several variables
10.1 10.2 10.2 10.3 10.4 10.5
30 30 31 32 32 33
9 Linear algebra
9.1 9.2 9.3 9.4 9.5
25 25 27 28 28 29
38 38 39 40 41 43
Partial diﬀerentiation . . . . . . . . . . . . . Diﬀe Diﬀeren rential ial of a fun function tion of sev several ral var varia iabl blees I mp mplicit functions . . . . . . . . . . . . . . . D ou ouble integration . . . . . . . . . . . . . . T Trriple integration . . . . . . . . . . . . . . .
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43 44 45 45 48
1
Intr Introd oduc ucti tion on
1.1
Num Numbers bers
• Natural numbers N These are all positive integers, including 0. • Integers Z These are the elements of N, plus the negative integers. • Rational numbers Q These are all the numbers which can be written p/q, p/q, where p and q = 0 are elements of Z. These numbers have either a ﬁnite number of decimals or a periodic inﬁnite number of decimals, for example 1 . 3795 3795 3795 3795
······
Q contains Z, which is obvious if one takes q = 1.
• Real numbers R These are the elements of Q plus all the numbers with√ inﬁnite and random decimals. Examples of real numbers, which are not in Q are:
2,π,e,.... ,π,e,....
Density property: Between any two real numbers can be found a rational number, and vice
verse.
1.2
Group
A group G is a set of elements a , together with an operation ⋆, such that:
{}
• if a, b are two elements of G, then a ⋆ b is element of G; • G contains a unit element u such that, for any element a of G, a ⋆ u = a; • for any element a of G, there is an element ˜a such that a ⋆ a˜ = u. Examples Examples of groups: groups: {Z, +}, or {Q , ×}, where Q is Q without 0. ⋆
1.3 1.3
⋆
Com Combina binato tori rics cs
• Permutations The number of ways to choose an order for n elements is the factorial n! = n × (n − 1) × (n − 2) · · · ×3 × 2 × 1. Indeed, there are n possibilities for the ﬁrst element, and for each of these possibilities, there are n 1 for the second element, etc...
−
• Combinations The number of ways to choose k elements out of n, independently of the order, is given by the binomial coeﬃcients
n k
= 3
n! . !(n k)! k !(n
−
Indeed, the number of possible ways to order n points is n!, and has to be divided by the number of ways to order the k chosen elements, which is k!, and also by the number of ways to order the remaining n k elements, which is (n ( n k )! Some simple properties are:
−
n n
−k
−
n k
=
n 1
,
n 0
= n,
= 1.
• Binomial formula. We show here that k =n
n
(a + b) =
n k
k =0
an−k bk ,
(1)
where n is an integer and a, b are real numbers, using a proof by induction : First step: check that eq.(1) is valid for a given value of n, for example n = 2: (a + b)2 = a2 + 2ab 2ab + b2 2 = a2 b0 + 0
k=2
=
2 k
k=0
2 1
2 2
a1 b1 +
a0 b2
a2−k bk .
Second step: suppose that eq.(1) is valid for n, and show that it is then valid for
n + 1: k =n
(a + b)
n+1
− −
= (a + b)
k =0
k =n
=
k=0 k =n
=
k=0
n k
an−k bk
k =n
n k
a −
n k
an+1−k bk +
n k+1 k
b +
k=0 k=n+1
= an+1 + bn+1 +
k=1 k =n
= an+1 + bn+1 +
k=1
k=n+1
=
k=0
n+1 k
4
n k
n+1 k
an+1−k bk .
an−k bk+1
n
k
k=1
k =n
n k
+
1
n
k
1
an+1−k bk
an+1−k bk an+1−k bk
2
Funct unctio ions ns of of a real real var varia iabl ble e
A function of a real variable f is an operation, which to a real variable x associates the quantity f ( f (x).
2.1 2.1
Con Contin tinuit uity
Intuitively, a function f of the variable x is continuous is a small change in x leads to a small change in f ( rigorously ly,, f is continuous in x0 if for any ε > 0, one can f (x). More rigorous always ﬁnd a δ > 0 such that
x − x  < δ ⇒ f ( f (x) − f ( f (x ) < ε. 0
2.2
0
Diﬀer Diﬀeren entia tiatio tion n
The derivative derivative of a function f at the point x is the slope of the tangent of the curve y = f ( f (x) ′ at x. In order to calculate calculate it, let’s conside considerr the points points M and M with coordinates (x, (x, f (x)) and (x (x + ∆x, f (x + ∆x)) respectively, where dx > 0 is an increment (see ﬁg(1)). The slope of the straight line (M (M M ′ ) is slope slope =
∆x) f ( ∆x) f ( f (x + ∆x f (x) f ( f (x + ∆x = (x + ∆x ∆x) x ∆x
− −
− f ( f (x) .
The slope of the tangent of the curve at M is obtained when ∆x ∆x f at the point x is then ∆x) f ( f (x + ∆x ∆x→0 ∆x
f ′ (x) = lim
derivativee of → 0. The derivativ
− f ( f (x) df = ,
(2)
dx
where dx denotes the inﬁnitesimal increment in x and df the corresponding inﬁnitesimal increment in f ( f (x). Example Let us calculate the derivative of f ( f (x) = axn , where a is a constant and n is an
integer. By deﬁnition ∆x)n axn a(x + ∆x lim ∆x→0 ∆x n a[x + nxn−1 ∆x + n(n = lim
−
f ′ (x) =
∆x
→0
2
∆x
= a lim nx − + n(n − 1)x 1)xn−2 ∆x + · · · ∆x→0 = anxn−1 ,
n 1
n 2
where the dots represent higher orders in ∆ x.
5
n
− 1)x 1)x − (∆x (∆x) /2 + · · ·] − ax
f(x+∆ x)
f(x)
x
x+∆x
Figure 1: The derivative is the slope of the tangent. Eq, (2) deﬁnes the “right derivative”, for ∆ x > 0. One can also deﬁne the “left derivative” by f ( f (x) f ( f (x ∆x) lim , ∆x→0 ∆x where ∆x ∆x > 0. A functio function n f is said to be diﬀerentiable at x if these two deﬁnitions lead to the same result. result. If these two two deriv derivatives atives are diﬀeren diﬀerent, t, the function function is singular singular at the point x and its derivative is not deﬁned. An example of such a singularity is the function f ( f (x) = x at x = 0. Indeed, for x = 0, the left derivative is 1 and the right derivative is 1. Note that a function can be continuous but not diﬀerentiable for a given value of x, as shows the previous example.
−
−

Extrema of a function Since the derivative derivative f ′ (a) of a function at the point a corresponds
to the slope of the tangent of the curve of equation y = f ( f (x), we have the following classiﬁcation:
• if f ′(a) > 0, then f is increasing in the vicinity of a; • if f ′(a) < 0, then f is decreasing in the vicinity of a; and f ( • if f ′(a) = 0 and f (x) changes sign at x = a, then f ( f (a) is an extremum of f ; f ; • if f ′(a) = 0 and f ( f (x) does not change sign at x = a, then the point of coordinates (a, f ( f (a)) is called an inﬂexion point. At such a point, the second derivative changes sign. 6
( f g )′ is given by Derivative of a product If f, g are two functions of x, the derivative (f ∆x)g (x + ∆x ∆x) f ( f ( f (x + ∆x f (x)g (x) ∆x→0 ∆x ∆x)g (x) f ( ∆x)g (x + ∆x ∆x) f ( ∆x)g (x) f ( f (x + ∆x f (x)g (x) + f ( f (x + ∆x f (x + ∆x = lim ∆x→0 ∆x ∆x) f ( ∆x) g (x) f ( f (x + ∆x f (x) g (x + ∆x = lim g (x) + f ( ∆x) f (x + ∆x ∆x→0 ∆x ∆x = f ′ (x)g (x) + f ( (3) f (x)g ′ (x).
(f g )′ (x) =
−
lim
− −
−
−
Chain rule Consider two functions f and g , and the function F deﬁned as F ( F (x) = f ( f (g (x)).
The derivative of F is F ′ (x) = = = = =
∆x) F ( F ( F (x + ∆x F (x) ∆x→0 ∆x ∆x)) f ( f ( f (g (x + ∆x f (g (x)) lim ∆x→0 dx ∆x)) f ( ∆x) g (x) f ( f (g (x + ∆x f (g (x)) g (x + ∆x lim ∆x→0 ∆x) g (x) ∆x g (x + ∆x ∆g ) f ( ∆x) g (x) f ( f (g + ∆g f (g ) g (x + ∆x lim ∆x→0 ∆g ∆x f ′ (g (x)) g ′(x),
−
lim
−
− − −
−
×
−
×
×
where ∆g ∆g = g (x + ∆x ∆x)
∆x. − g(x) is the increment in g(x) corresponding to x → x + ∆x Deriv Derivative ative of a ratio The derivative of 1/x 1 /x is −1/x , such that the derivative of the 2
function function 1/f is
′
1 f ( f (x)
=
−
1 f 2 (x)
f ′ (x) ′ × f (x) = − . f 2(x)
As a consequence, the derivative of the ratio of the functions f and g is
f ( f (x) g (x)
′
= f ′ (x)
×
1 + f ( f (x) g (x)
g ′(x) g 2(x)
× −
=
f ′ (x)g (x) f ( f (x)g ′ (x) . g 2 (x)
−
If y = f ( Derivative of an inverse function If y f (x), the inverse function f −1 , when it exists, is deﬁned by x = f −1(y ). Do not confuse the inverse function f −1 with 1/f ! . In order to deﬁne the inverse of a function, one needs a onetoone mapping between x and y . This is usually the case on a given interval for x at least. The derivative of the inverse is then
′ f −1 (y ) =
∆y ) f −1 (y + ∆y lim ∆y →0 ∆y 7
1
− f − (y)
∆x x + ∆x ∆x→0 f ( ∆x) f (x + ∆x 1 = , f ′ (x)
=
lim
−x − f ( f (x)
where ∆x ∆x is deﬁned such that y + ∆y ∆y = f ( ∆x). f (x + ∆x
2.3 2.3
Polyn olynom omia iall funct functio ions ns
A polynomial function of x is of the form n=N
P ( P (x) =
an xn ,
n=0
where an are the coeﬃcients and N is the degree of the polynomial. If N is odd, the polynomial has at least one zero. Indeed, we have then x
lim P ( P (x) =
→−∞
−∞
and
x
lim P ( P (x) = + ,
∞
→+∞
such that the line representing y = P ( P (x) cuts at least once the axis y = 0, since the polynomial is a continuous function. A polynomial of degree 2 can have two poles, but might not have any (real) pole: )(x − z ) has two poles z , z . The pole is double if z • P ( P (x) = a(x − z )(x • Q(x) = ax + bx + c has no pole if b − 4ac < 0. 1
2
1
2
2
1
= z2 ;
2
A polynomial of degree 3 has either one pole or three poles, and can be written, for all x, )(x − z )(x )(x − z ) if P has three poles; • P ( P (x) = a(x − z )(x )(ax + bx + c), with b − 4ac < 0, if Q if Q has one pole z. • Q(x) = (x − z)(ax 1
2
3
2
2
In general, any polynomial function can be written P ( P (x) = (x
2
− z ) · · · (x − z ) × (a x n
1
+ b1 x + c)
2
· · · (a x + b x + c ), where z , i = 1,...,n are the poles of the polynomial, b − 4a c < 0 for all j = 1,...,m, ,...,m, 1
2 j
i
m
m
m
j j
and n + 2m 2m is the degree of the polynomial.
2.4 2.4
Rati Ration onal al funct functio ions ns
A rational function is the ratio of two polynomial functions P and Q, and has the form, for each x, P ( P (x) R(x) = . Q(x) 8
y
z
z 1
z
2
x
3
Figure 2: A polynomial function of degree 5, with three poles z1 , z2 , z3 .
y
z
1
z
z 2
3
z
4
x
Figure 3: A polynomial function of degree 6, with four poles z1 , z2 , z3 , z4.
9
1 M
sin x
x cos x
1
Figure 4: The coordinates of M on the trigonometric circle are (cos x, sin x). If the degree of P is less than the degree of Q, It is always possible to reduce R as a sum of irreducible rational functions of the form a/( (ax + b)/(x2 + cx + d). a/(x z ) or (ax
−
( x + 2)/ 2)/(x2 + 5x 5x + 4) can be written Example The fraction (x x+2 x+2 a b = = + , 5x + 4 (x + 1)(x 1)(x + 4) x2 + 5x x+1 x+4 where a(x + 4 ) + b(x + 1) = x + 2, such such that a + b = 1 and 4a 4a + b = 2, which gives a = 1/3 and b = 2/3. Finally, 1/3 2/3 x+2 = + . 5x + 4 x2 + 5x x+1 x+4
2.5
Trigono rigonome metri tric c functio functions ns
For a given angle 0 x 2π, sin x and cos x are deﬁned as the coordinates of the point ( OM )) with the trigonometric circle (see ﬁg(4)). M at the intersection of the straight line (OM
≤ ≤
Property Using Pythagoras’ theorem, we have sin 2 x + cos2 x = 1. Trigonometric rigonometric formula It will be shown in the chapter on vector calculus (subsection
10
6.2) that the sine and cosine of the sum of two angles are given by sin(a sin(a + b) = sin a cos b + sin b cos a cos(a cos(a + b) = cos a cos b sin a sin b.
(4)
−
Important limit We will now show, geometrically, that
sin x = 1, x→0 x lim
and this limit will be very useful in deriving fundamental properties of the trigonometric functions. Proof: From the deﬁnition of sine and cos, one can see on ﬁg.(5) that sin x
≤ x ≤ sin x + 1 − cos x.
(5)
But one can also see that 0
2
≤ sin
x + (1
2
such that 0
2
− cos x) ≤ x ,
≤ 1 − cos x ≤
x2 . 2
Using this in the inequalities (5), we obtain sin x x
≤ 1 ≤ sinx x + x2 ,
and the only possibility for this to be valid in the limit x
→ 0 is to have
sin x x
→ 1.
Derivative of trigonometric functions The ﬁrst important consequence of the previous
limit is the calculation of the derivative of the sine. From eq.(4) we have sin(x sin(x + ∆x ∆x) sin x ∆x→0 ∆x sin x cos(∆x cos(∆x) + sin(∆x sin(∆x)cos x sin x = lim ∆x→0 ∆x sin(∆x sin(∆x) cos(∆x cos(∆x) 1 = lim cos x + sin x . ∆x→0 ∆x ∆x
sin′ x =
−
lim
−
−
We have seen that 1 cos(∆x cos(∆x) is of order ∆x ∆ x2 , and therefore the second term vanishes in the limit ∆x ∆x 0, whereas the ﬁrst term leads to
→
−
(sin x)′ = cos x. In the same way, one can easily show that (cos x)′ = have (tan x)′ = 1 + tan2 x. 11
− sin x.
As a conseq consequen uence, ce, we also
x a
c
b
Figure 5: On the ﬁgure: a = sin x, b = 1
3
2
− cos x and c
= a2 + b2
Inte Integr grat atio ion n
The integration corresponds to the inverse operation of the diﬀerentiation: F is a primitive of f if
F ( F (x) = We have
f ( f (x)dx
F ′ (x) = f ( f (x).
⇒
b
f ( f (x)dx = F ( F (b)
− F ( F (a),
f ( f (u)du = F ( F (x)
− F (0) F (0)..
a
and therefore
x
0
Make sure never to use the same name for the variable of integration and the limit of the integral.
From the linearity of the diﬀerentiation, integrals have the following properties:
• • •
a b
f ( f (x)dx =
c a
f ( f (x)dx =
b a
−
b a
f ( f (x)dx
f ( f (x)dx +
b [c f (x) + c2 f 2 (x)]dx )]dx a 1 1
c b
= c1
f ( f (x)dx b a
f 1 (x)dx + c2 12
b a
f 2 (x)dx, dx, where c1, c2 are constants.
f(x ) k
a
b
x
dx
k
Figure 6: Riemann deﬁnition of the integral
3.1
Inter Interpre pretat tation ion of the integ integral ral b
As explained on ﬁg.(6), the Riemann deﬁnition of a f ( f (x)dx, dx, b > a, corresponds to the surface area between the line y = f ( f (x) and the straight line y = 0, from x = a to x = b. Indeed, this area can be seen as the sum of the inﬁnitesimal areas dx f ( f (x), and we have b
n
→∞
where xk = a + k
b
− b
f ( f (x)dx = lim
a
×
a
n
− a,
n 1
−
f ( f (xk )
k =0
k = 0,...,n
n
− 1,
Equiv Equivalence alence with the deﬁniti deﬁnition on based based on the deriv derivativ ative e . We show here here that the
Riemann deﬁnition of the integral, as a surface area, is equivalent to the deﬁnition given previously. From the Riemann interpretation, the quantity x
F ( F (x) =
du f ( f (u)
a
corresponds to the surface area of between the lines y = f ( f (x) and y = 0, from a to x. The integral from a to x + ∆x ∆x is then x+∆x +∆x
∆x) = F ( F (x + ∆x
a
13
du f ( f (u),
and the derivative of the function F is 1 F ′ (x) = lim ∆x→0 ∆x 1 = lim ∆x→0 ∆x
x+∆x +∆x
x
du f ( f (u)
a x+∆x +∆x
−
du f ( f (u)
a
du f ( f (u).
x
The latter expression corresponds to the surface area between the lines y = f ( f (x) and y = 0 from x to x + ∆x ∆x, which is equal to ∆x ∆ x f ( ∆x. As a consequence, f (x)+ higher powers in ∆x we obtain the expected result:
×
1 ∆xf (x) + (∆x (∆x)2 = f ( f (x). ∆x→0 ∆x As a consequence of this interpretation of the integral, if two functions f, g satisfy F ′ (x) = lim
f ( f (x) then
···
≤ g(x)
for a
b
b
f ( f (x)dx
a
3.2 3.2
≤ x ≤ b,
≤
g (x)dx.
a
Inte Integr grat atio ion n by by part part
The derivative of the product of two functions f, g is (f ( f g )′ = f ′ g + f g ′, such that we obtain, after integration f ( f (x)g (x) =
f ′ (x)g (x)dx +
f ( f (x)g ′(x)dx,
which can be helpful to calculate one of the integrals on the right hand side, if we know the other: b
f ′ (x)g (x)dx = [f ( f (x)g (x)]b
a
a
b
−
f ( f (x)g ′ (x)dx
a
Example Integration by part is very useful for the integration of trigonometric functions
multiplied by power law functions, as
dx x cos x =
where c is a constant.
3.3 3.3
dx x(sin x)′ = x sin x
−
dx sin x = x sin x + cos x + c,
Chan Change ge of varia ariabl ble e
Suppose that one can write x = g (u), where u represents another variable with which the integral can be calculated. We have then dx = g ′(u)du and g −1 (b)
b
a
f ( f (x)dx =
))g ′(u)du, f ( f (g (u))g
g−1 (a)
14
where g −1 represents the inverse function of g : x = g (u) u = g −1(x). For the change of variable to be consistent, one must make sure that there is a onetoone relation between [a, b]. x and u in the interval [a,
⇔
Example In the following integral, one makes the change of variable u = sin φ,
for 0
≤ φ ≤ π/2: π/2: 1
du = 1 u2
√ 0
−
π/2 π/2
cos φ dφ
π/2 π/2
−   sin2 φ
1
0
=
π . 2
dφ =
0
Note that, in the interval [0, [0 , π/2], π/2], we have
cos2 φ = cos φ = cos φ,
since cos φ > 0.
3.4 3.4
Impr Im prope oper r inte integr gral alss
The domain of integration of an integral might either contain a singular point, where the functi function on to integ integrat ratee is not deﬁned deﬁned,, or might might not be bounded. bounded. In both cases, cases, the corresponding integral is said to be convergent if the result of the integration is ﬁnite, and divergent if the result of the integration is inﬁnite. We describe here this situation for the integration of power law functions. Case of a noncompact domain of integration We ﬁrst show that the integral
I 1 =
∞ dx
1
x
diverges. For this, one can see on a graph that I 1 >
∞
n=2
1 , n
and we show, that the sum of the inverse of the integers is divergent. Proof  from the 14th century! The sum of the inverses of integers, up to 2 N , can be written: 2N
n=1
1 = 1+ n
1 2
+
1 1 + 3 4
+
1 1 1 1 + + + 5 6 7 8
+
1 + 9
1 + 16
···
1 + 16
+
···
and satisﬁes 2N
n=1
1 > 1+ n
1 2
+
1 1 + 4 4
+
1 1 1 1 + + + 8 8 8 8 15
+
···
1 + 16
+
···
The sum in each bracket is equal to 1/2, and there are N bracket, such that 2N
n=1
1 N > 1+ , 2 n
which shows that the sum goes to inﬁnity when N goes to inﬁnity. As a consequence,
∞ dx
is divergent.
x
1
Consider now the integral, for a = 1,
I a =
∞ dx
x1−a 1 = lim xa x→∞ 1 a
1
− −
As can be seen, the result depends on a:
• if a > 1 then I = 1/(a − 1) is ﬁnite; • if a < 1 then I = +∞. a a
Since the integral I a also diverges for a = 1, it converges only for a > 1. Case of a singular point Consider the integral, for b = 1,
1
J b =
0
1 x1−b dx = lim x→0 1 xb b
− −
As can be seen, the result depends on the power b:
• if b < 1 then J = 1/(1 − b) is ﬁnite; • if b > 1 then J = +∞. b b
The integral J b also diverges for b = 1 (the surface area is the same as the previous case, with a noncompact domain of integration), it therefore converges only for b < 1. In general, we have: 1 dx convergent if b < 1 is divergent if b 1 z )b z (x
−
≥
Example Consider the integral
∞
1
(x
−
dx 1)b (2x (2x + 3)a 16
a
b
• at x = 1: the integrand is equivalent to 5− /(x − 1) , such that there is convergence if b < 1;
a
a+b
• at x = ∞: the integrand is equivalent to 2 − /x
, such that there is convergence if
a + b > 1;
As a consequence, the integral is convergent only if b < 1 and a + b > 1 simultaneously.
17
4
Logari Logarith thm m and and expon exponen enti tial al func functi tion onss
4.1 4.1
Loga Lo gari rith thm m
We have seen that, for a =
−1,
xa+1 x dx = a+1 a
a=
−1,
and we still have to deﬁne this integral for a = 1. For this, we introduce introduce the logarithm logarithm as x du ln x = , u 1
−
so that the logarithm gives the surface area between the function 1 /u and the horizontal axis, from 1 to x > 0. The real logarithm is not deﬁned for x < 0, since since the corresponding corresponding surface area would be inﬁnite. The number e is deﬁned by ln e = 1 and e 2.718281828.
≃
Properties:
• We have seen that the integrals →∞
1
lim ln x = +
x
1 dx/x 0
∞ dx/x and
and
∞
both diverge, such that
lim ln x =
x
→0
−∞
• From the deﬁnition of the logarithm, one can see that ab
ln(ab ln(ab)) =
1
du = u
a
du + u
ab
1
a
du = ln a + u
b
1
dv = ln a + ln b, v
(6)
where we make the change of variable u = av. av.
• One can also see that a
xa
ln(x ln(x ) =
1
du = u
x
1
a
dv = a ln x, v
where we make the change of variable u = v a .
• We have in general, for any diﬀerentiable function f , f ,
f ′ (x) = ln f ( dx f (x) + c, f ( f (x)

where c is a constant
18

(7)
Logarithm in base a The logarithm in base a is deﬁned as
ln x , ln a and is equal to 1 when x = a. Note that ln x = loge (x). loga (x) =
Integral of the logarithm To calculate
ln x dx =
where c is a constant.
ln x dx, dx, one uses an integration by parts:
(x)′ ln x dx = x ln x −
x(ln x)′ dx = x ln x
− x + c,
Limits
• When x → +∞: We show here the important limit
ln x = 0, a > 0 (8) x→+∞ xa which means that any (positive) power law goes quicker to inﬁnity than the logarithm, when x + . Proof For any u 1 and for any a > 0, we have 1 1 . a/2 u u1−a/2 Integrating this inequality from 1 to x leads to 2 a/2 2 a/2 0 < ln x (xa/2 1) < xa/2 . a a Dividing by xa gives the expected result: ln x 2 −a/2 0< a 0 when x + . x a/2 x a lim
→ ∞ ≥
≤
≤
≤
−
→
→ ∞
• When x → 0: Another important limit to know is lim xa ln x = 0,
a > 0,
x
→0
(9)
which means that any (positive) power law kills the divergence of the logarithm at x = 0. Proof For any u satisfying 0 < u 1 and any a > 0, we have 1 1 . 1+a/2 2 u u1+a/ Integrating this inequality from x to 1, we obtain 2 −a/2 2 a/2 0 < ln x (x a/2 1) < x−a/2 . a a Multiplying by xa gives the expected result: 2 a/2 0 xa ln x 0 when x 0. xa/2 a
≤
≤
−
≤
−
≤  ≤
→
19
→
4.2 4.2
Expon Exponen enti tial al
The exponential is deﬁned as the inverse function of the logarithm: y = ln x
x = exp y = ey
⇐⇒
From property (6), if we note u = ln a and v = ln b, we have exp(u exp(u + v) = (exp u)
× (exp v),
and from property (7), if we note y = ln x, we have a
exp y
= exp(ay exp(ay)).
Derivative of the exponential one can diﬀerentiate the deﬁnition exp(ln x) = x, which,
using the chain rule, leads to exp ′ (ln x) = x. We therefore conclude that the derivative of the exponential is the exponential itself: exp′ x = exp x. Exponential of base a This function is deﬁned as
exp(x ln a), ax = exp(x which is consistent with the properties of the logarithm and the exponential. It’s derivative is then (ax )′ = (ln a)ax . One can also deﬁne the function xx , with derivative d (x )′ = x
dx
Limits
exp(x exp(x ln x) = (1 + ln x)xx .
• From the limit (8), if we note y = ln x and b = 1/a > 0, we have exp y lim = +∞, → ∞ y y
b
+
and the exponential goes to inﬁnity quicker than any power law.
• From the limit (9), if we note y =  ln x and b = 1/a > 0, we have lim y exp(−y ) = 0, → ∞ b
y
+
and the decreasing exponential kills the divergence of any power law. 20
4.3 4.3
Hyper Hyperbol bolic ic funct functio ions ns
The hyperbolic functions are deﬁned as x
x
• hyperbolic cosine: cosh x = (e + e− )/2; • hyperbolic sine: sinh x = (e − e− )/2; • hyperbolic tangent: tanh x = sinh x/ cosh x; • hyperbolic cotangent: coth x = cosh x/ sinh x, x
x
and their derivatives are given by cosh′ x sinh′ x tanh′ x coth′ x
= = = =
sinh x cosh x 1 tanh2 x 1 coth2 x
− −
It can easily be seen that, from their deﬁnition, the functions cosh and sinh satisfy, for all x, cosh2 x sinh2 x = 1.
−
Also, it can be easily checked that cosh(2x cosh(2x) = cosh2(x) + sinh2 (x) sinh(2x sinh(2x) = 2 sinh(x) cosh cosh((x)
21
5
Taylor ylor expan expansi sions ons and and ser series ies
5.1
Appro Approximati ximation on of a function function around around a valu value e of the argume argumen nt
It is sometimes useful to approximate the value f ( Thee f (x) of a function f around f ( f (x0 ). Th ﬁrst approximation consists in replacing f ( f (x) by a linear function p1 (polynomial of ﬁrst order, representing a straight line) in a small interval around x0 : f ( f (x)
≃ p (x) = a 1
0
+ a1 (x
− x ). 0
In order to ﬁnd the coeﬃcients a0 , a1 , one imposes the constraints p1 (x0 ) = f ( f (x0 ) and ′ ′ ′ p1 (x0 ) = f (x0 ), such that a0 = f ( f (x0 ) and a1 = f (x0 ). If one wants a better approximation, one can choose to approximate f locally by a quadratic function p2 (polynomial (polynomial of second second order, order, represe representi nting ng an arc of parabola), parabola), which is bette b etterr than a straight straight line. One writes writes then f ( f (x) p2 (x) = a0 + a1 (x x0 ) + a2(x x0 )2 ,
≃
−
−
and imposes the additional constraint p′′2 (x0 ) = f ′′ (x0 ), such that f ′′ (x0 ) = 2a2. If on one wishes to push further the precision of the approximation, one can take the third order polynomial f ( f (x)
≃ p (x) = a 3
0
+ a1(x
2
− x ) + a (x − x ) 0
2
0
+ a3 (x
3
−x ) , 0
′′′ ′′′ and impose the additional constraint p′′′ 3 (x0 ) = f (x0 ), leading to f (x0 ) = 2 so on... Going on like this ﬁnally leads to the Taylor expansion of the function f : f :
× 3 × a , and 3
1 1 ≃ f ( ( x − x )f ′ (x ) + (x − x ) f ′′ (x ) + (x − x ) f ′′′ (x ) + · · ·, f (x ) + (x 2! 3! where the dots represent higher powers in the diﬀerence x − x , which are smaller and f ( f (x)
0
0
0
0
2
0
0
3
0
0
smaller smaller as the order of the Tayl Taylor or expansion increases. increases. Obviousl Obviously y, such such an expansion is valid only if the function is diﬀerentiable a number of times large enough to reach the desirable order. Note that a polynomial function of order N is exactly equal to its Taylor expansion of order N . The power n of the ﬁrst neglected term in the expansion of a function around x0 is denoted (x x0 )n , and means “terms which are at least of the power n”.
O − 5.2 5.2
Radi Radius us of con converge ergence nce and and seri series es
For many functions, the Taylor expansion around x0 can be pushed to an inﬁnite order, at least in a vicinity of f ( f (x0 ): if x x0 < R, R , where R is the radius of convergence, then the series is convergent and one can write
 − 
f ( f (x) =
∞
n=0
1 (n) )(x f (x0 )(x n! 22
n
−x ) , 0
where f (n)(x0 ) denotes the nth derivative of f at x0. Ratio convergence test Consider the geometric series S =
this sum can be obtained by noting that qS = S
−1+q
N
N 1 n n=0 q .
−
An expression for
, and hence
1 − qN − = S = 1 + q + q + · · · + q 2
N 1
1
−q
From this expression, we see that, if q < 1, then limN →∞ S = 1/(1 q ) is ﬁnite, and if . q 1, then S diverges when N More generally, for any series n an , one can compare the ratio of two consecutive terms, and conclude, from the behaviour of the geometric series, the following ratio convergence test:
≥
−
→ ∞
• if lim →∞ a • if lim →∞ a • if lim →∞ a
 < 1, the series is (absolutely) convergent; /a  > 1, the series is divergent; /a  = 1, one cannot conclude, and each case has to be looked at
n
n+1 /an
n
n+1
n
n+1
individually.
n
n
The convergence of the Taylor series of a function f about x0 therefore depends on x x0 , and the radius of convergence of the series is deﬁned by
− 
f (n+1) (x0 ) R lim = 1. n→∞ f (n) (x0 ) n + 1
5.3
Exampl ample es
(10)
By calculating the diﬀerent derivatives of the following functions at x = 0, one can easily see that
∞
x2n cos x = ( 1) , (2n (2 )! n n=0
− − − − ∞
n
x2n+1 sin x = ( 1) , (2n (2 + 1)! n n=0 exp x =
∞
n=0
1 = 1+x
∞
n
R=
∞;
R=
∞;
xn , n!
R=
( 1)n xn ,
R = 1;
∞;
n=0
∞
xn+1 ln(1 + x) = ( 1) , + 1 n n=0 n
23
R = 1.
Consider the functi function on f ( we note note y = 1/x > 0, Counter Counter example. Consider f (x) = exp( 1/x). /x). If we
−
we have f ( f (0)
=
f ′ (0)
=
f ′′ (0)
y
lim
→ +∞
lim
x
→0
=
lim
x
etc...
→0
exp( y ) = 0
− 1 exp(−1/x) /x) = x 2
1 x4
2 x3
−
y
lim
→+∞
y 2 exp( y ) = 0
−
y4
exp( 1/x) /x) = lim
−
y
→+∞
− 2y
3
exp( y ) = 0
−
As a consequence, 2
3
2!
3!
x x f (0) f (0) + xf ′ (0) + f ′′ (0) + f ′′′ (0) + · · · = 0, and no Taylor expansion of f can be deﬁned around 0, whereas f (0) f (0) = 0 is deﬁned.
5.4 5.4
Expa Expans nsio ion n for a compos composit itio ion n of functi function onss
Suppose that two functions f 1 , f 2 have the following expansions around x = 0, to the order x3 : f 1(x) = a1 + b1 x + c1x2 + d1 x3 + f 2(x) = a2 + b2 x + c2x2 + d2 x3 +
4
O (x ) O (x ) 4
The expansion of the product f 1 f 2 can then be obtained up to the order 3 maximum , and is +(a1 b2 +b1 a2 )x+(a +( a1c2 +b1 b2 +c1 a2)x2 +(a +(a1 d2 +b1 c2 +c1 b2 +d1a2 )x3 + (x4) f 1 (x)f 2 (x) = a1 a2 +(a
O
Example To calculate the expansion of tan x up to the order x5, we ﬁrst expand the
inverse of cos x to the order x5 :
−1
x2 x4 1 + + (x6 ) 2 24 5 x2 = 1+ + x4 + (x6 ), 2 24
1 = cos x
−
O
O
and then multiply by the expansion of sin x to the order x5 : 5 x3 x5 x2 tan x = + 1+ + x4 + x 6 120 2 24 3 2 x = x+ + x5 + (x7 ) 3 15
−
O
24
7
O(x )
6 6.1 6.1
Vecto ector r calc calcul ulus us Vecto ectors rs
A vector u has a direction, given by the unit vector u ˆ, and a modulus u , and can be written u= uu ˆ.

n vectors u1 ,
· · ·, u

n
are said to be linearly independent if a1 u1 +
···+a u
n n
=0
⇒
a1 =
···=a
n
= 0,
which means that these vectors point in diﬀerent directions, and none of them can be obtained by a linear combination of the others. A vector space of dimension d is set of vectors spanned by d independent vectors, and is group for the addition. A set of basis vectors in is made of d linearly independent vectors i1 , , id , and any other vector can be decomposed onto this basis:
V
V
···
u = a1 i1 +
···+a i , d d
where (a (a, , ad) are the coordinates of u in this basis. A change of basis leads to a change of coordinates.
· ··
Addition Addition of vectors vectors Vectors can be added according to the rule (for example in three
dimensions) ( x2 i + y2 j + z2 k) u1 + u2 = (x1 i + y1 j + z1 k) + (x = (x1 + x2 )i + (y (y1 + y2 ) j + (z (z1 + z2 )k. (N + 1)dimensional vector space. Example The set of polynomials of order N is an (N n Consider the polynomials polynomials pn (x) = x , n = 0, , N , and a set of constants cn such Proof Consider that, for any x, we have c0 p0 (x) + c1 p1 (x) + + cN pN (x) = 0. Then we necessarily have
···
···
cn = 0, for all n, since a polynomial of degree N has at most N zeros. As a consequence, the polynomials pn are linear linearly ly indepen independe dent nt,, and span span an N dimensional dimensional vector space, where each vector can be written n=N
P =
an pn
n=0
and an are the coordinates of P on the basis pn , n = 0,
{
6.2 6.2
· · · , N }.
Rota Rotati tion onss in tw two dime dimens nsio ions ns
(i, j) form form an orth orthon onor orma mall basi basiss in a plan plane. e. Afte Afterr a rota rotati tion on of angl anglee α, the basis has ′ ′ changed to (i , j ) where (see ﬁg.7) i′ = cos α i + sin α j j′ =
− sin α i + cos α j. 25
(11)
α
j j’
α
i’
i
Figure 7: Rotation of the unit vectors From these relations, one can easily express the vectors i, j in the basis (i′ , j′ ) by making the inverse rotation (α (α α), which leads to
→−
i = cos α i′
− sin α j′
j = sin α i′ + cos α j′ .
(12)
The vector u = (a, b) is then transformed into the vector u′ = (a′ , b′ ) such that u′ = a i′ + b j′ = (a cos α
− b sin α)i + (a (a sin α + b cos α) j = a′ i + b′ j,
and therefore a′ = a cos α b sin α b′ = a sin α + b cos α.
−
Equivalently, we also have a = a′ cos α + b′ sin α b = a′ sin α + b′ cos α.
−
sin( α + β ) and cos(α cos(α + β ) Trigonometric formulas One way to ﬁnd the expression for sin(α in terms of sin α, cos α, sin β, cos β is to perform two consecutive rotation, of angle α and β respectively, and identify the result with a rotation of angle α + β . We have seen that a rotation of angle α of the basis vectors ( i, j) gives i′ = cos α i + sin α j j′ =
− sin α i + cos α j 26
A second rotation, of angle β , leads to i′′ = cos β i′ + sin β j′
= (cos β cos α sin β sin α)i + (cos β sin α + sin β cos α) j sin β i′ + cos β j′ j′′ = = ( sin β cos α cos β sin α)i + ( sin β sin β sin α + cos β cos α) j.
−
− −
−
−
This must be equivalent to i′′ = cos(α + β ) i + sin(α sin(α + β ) j j′′ =
− sin(α sin(α + β ) i + cos(α cos(α + β ) j,
such that cos(α cos(α + β ) = cos α cos β sin α sin β sin(α sin(α + β ) = sin α cos β + β + cos α sin β.
−
Don’t learn this by heart, but rather remember how to get the result.
6.3 6.3
Scal Scalar ar produ product ct
Let u and v be two vectors in a plane, with coordinates (a, ( a, b) and (c, (c, d) respectively. respectively. From these two vectors, one wishes to construct a quantity which is unchanged after a rotation (= a scalar ). ). The scalar product of u and v is deﬁned as u v = u v cos(u, v),
·
  
and is indeed unchanged after a simultaneous rotation of both vectors u and v. One can easily express the scalar product in terms of the coordinates of the vectors, by doing the follo followin wing. g. Let’s Let’s denote denote by (a′ , b′ ) and (c (c′ , 0) the coordinates of u and v respectively respectively,, in ′ ′ ′ ′ ′ the orthonormal basis (i , j ) where i is along v. In the basis ( i , j ), the scalar product is obviously given by u v = a′ c′ , with
·
a′ b′ c′ 0
= = = =
a cos α b sin α a sin α + b cos α c cos α d sin α c sin α + d cos α.
−
−
Together with cos α = sin α =
√c c+ d √ −d , 2
2
c2 + d2
27
one easily obtains a′ c′ = ac + bd. bd. The scalar product is then given by the expression u v = ac + bd.
·
More generally, in d dimensions, the scalar product of u = (x1 ,...,xd ) and v = (y1 ,...,yd ) is d
u v=
·
xi yi .
i=1
(1, 2, 1), and Example Find the equation of the plane perpendicular to the vector u = (1, containing the point A of coordinates (3, (3, 4, 2). Any point M of coordinates (x (x , y , z) z) of this plane is such that AM u = 0, which reads
·
(x
6.4 6.4
− 3) + 2(y 2(y − 4) + (z (z − 2 ) = 0
or
2y + z = 13. 13. x + 2y
Cros Crosss produ product ct
One often needs to deﬁne, from two vectors u, v, a third vector which is perpendicular to u and v. The cross product u v is
×
u
× v = uv sin(u, v)n,
where n is the unit vector perpendicular to the plane spanned by u, v, which deﬁnes the anticlockwise direction. If ( i, j, k) form the usual orthonormal basis, we have
× j j × k k×i i
= k = i = j.
From this, it is easy to ﬁnd the coordinates of the vector product of u = (a1 , a2 , a3 ) times v = (b1 , b2 , b3 ), which are u
×v=
a2 b3 a3 b1 a1 b2
−a b −a b −a b
3 2 1 3 2 1
.
Note that the cross product is a vector, unlike the scalar product which is a number . Finally, the cross product is not commutative, since u
6.5 6.5
× v = −v × u
Scal Scalar ar trip triple le produc productt
If (u, v, w) are three vectors, one deﬁnes the scalar triple product by u (v can check that a cyclic permutation does not change the result:
· × w), and one
u (v
· × w) = w · (u × v) = v · (w × u) 28
e
θ
e
r
r
M j O
θ
i
Figure 8: Polar coordinates (r, (r, θ) of the point M . The orientation orientation of the basis vectors vectors er and eθ depend on the position of M , such that er is always along OM and eθ is the image of er in a rotation of angle π/2. π/2.
6.6 6.6
Polar olar coord coordin inat ates es
We denote O the origin of space and (i, j, k) the orthogonal and unit basis vectors of Euclidean coordinates (x,y,z (x,y,z). ). Poin oints in the plane plane (O, (O, i, j) can also be labeled by the polar coordinates (r, (r, θ) (see ﬁg.8), such that
x2 + y 2 with y tan θ = with 0 x r=
0
≤r<∞
≤ θ < 2π.
The orthogonal and unit basis vectors ( er , eθ ) in polar coordinates are deﬁned by er = cos θi + sin θ j eθ =
− sin θi + cos θ j.
Note that der = eθ dθ deθ er . = dθ
−
29
7 7.1 7.1
Comp Compllex numbers bers Intr Introduc oducti tion on
Complex numbers can be seen as twodimensional vectors in the complex plane, spanned by the basis (1, (1, i), where i2 = 1. In Cartesian Cartesian coordinates, coordinates, a complex complex number number z can be written z = a 1 + b i = a + ib,
−
×
×
where a is the real part of z and b the imaginary part. The complex conjugate z ⋆ is then deﬁned as z ⋆ = a ib.
−
Complex numbers can be added, or multiplied, to give a new complex number: ( a2 + ib2 ) = a1 + a2 + i(b1 + b2 ) z1 + z2 = (a1 + ib1 ) + (a )(a2 + ib2 ) = a1 a2 b1 b2 + i(a1 b2 + a2 b1 ). z1 z2 = (a1 + ib1 )(a
−
This is because the set of complex numbers C is a group for both the addition and multiplication. Finally, the modulus of z is deﬁned as ⋆
z  = z  = 7.2 7.2
√
a2
+
b2
√ = zz . ⋆
Comp Comple lex x expone exponen ntial tial
Complex numbers, seen as twodimensional vectors, can be expressed using polar coordinates (r, (r, θ): z = r(cos θ + i sin θ). Using the series expansion for cosine and sine, we ﬁnd z = r
∞
n=0
= r
∞
n=0
= r
2n+1 θ2n n θ ( 1) + i( 1) (2n (2n)! (2n (2n + 1)!
− ∞
n=0
n
−
(iθ) (iθ) iθ)2n iθ)2n+1 + (2n (2n)! (2n (2n + 1)!
(iθ) iθ)n n!
= r exp(iθ exp(iθ)).
r is the modulus of the complex z, and θ is its argument , and the last result leads to the Euler’s formula : cos θ + i sin θ = exp(iθ exp(iθ)). (13)
30
From this, it is easy to ﬁnd the de Moivre’s formula : noting noting that [exp( [exp(iθ )]m = exp(imθ exp(imθ), ), iθ)] where m is any integer, we have (cos θ + i sin θ)m = cos(mθ cos(mθ)) + i sin(mθ sin(mθ)). Example The number
therefore be written
interval al [0; 2π[), and can −1 has modulus 1 and argument π (in the interv −1 = e iπ
This equation relates three fundamental numbers, which are 1 , e , π. π. iπ/2 iπ/2 One also has i = e , such that exp(i ln i) = exp(i exp(i ii = exp(i
π/2 π/2
208. × iπ/2) iπ/2) = e− ≃ 0.208.
(14)
Note that the logarithm of a complex number z is a multivalued function : its deﬁnit deﬁnition ion depends on the range of angles in which the argument θ of z is conside considered red.. Indee Indeed, d, if 2kπ,, where k is an integer, z is invariant, but its logarithm changes as: θ θ + 2kπ
→
ln z
→ ln z + 2ikπ. 2ikπ.
As a result, ii as given in eq.(14) is the value when the argument of complex numbers are deﬁne deﬁned d in [0; 2π [.
7.3
Trigono rigonome metri tric c form formula
From the Euler’s formula (13), one can express cosine and sine with complex exponentials: eiθ + e−iθ cos θ = 2 iθ e e−iθ sin θ = , 2i
−
and therefore one can also express the nth power of cosine and sine, in terms of cosine and sine of n times the argument. For example: 1 2iθ 1 1 (e + e−2iθ + 2) = + cos(2θ cos(2θ) 4 2 2 i 3iθ = (e 3e−iθ ) e3iθ 3eiθ + 3e 8 3 1 = sin θ sin(3θ sin(3θ). 4 4
(cos θ)2 = (sin θ)3
− − −
(15)
These formulas are useful when one needs to integrate expressions involving powers of cosine or sine. Do not learn these expressions by heart, but derive them whenever you need them.
31
7.4 7.4
Roots Roots of of com compl plex ex number umberss
Consider the equation zn = A, where A is a given complex number and z is the unknown. In order to solve this equation, one writes exp(iφ)) A = ρ exp(iφ exp(iθ)). z = r exp(iθ The equation to solve is then r n exp(inθ exp(inθ)) = ρ exp(iφ exp(iφ), ), which leads to, after identiﬁcation of the modulus and the argument of both sides of the equation z n = A,
√
r = ρ1/n = n ρ φ 2kπ + θ = , n n where k = 0, 1,...,n 1. Therefore a complex number has n roots of order n. For example, the nth roots of the unity are
−
zk = exp 2iπ
7.5 7.5
k n
k = 0, 1,...,n
− 1.
Rela Relati tion on to hype hyperbo rboli lic c funct functio ions ns
We have seen that a function can usually be expanded as a series of powers of the argument. Since complex numbers can be multiplied and added, one can express a Taylor expansion for a comple complex x variabl ariable. e. It is therefor thereforee possibl possiblee to understa understand nd a funct function ion of a comple complex x variable in terms of a series expansion. We give here two examples. From eqs.(15), we have for any real x sin(ix sin(ix)) = i sinh x cos(ix cos(ix)) = cosh x, which gives a formal way to deﬁne trigonometric functions with complex arguments.
32
8
Line Linear ar diﬀer diﬀeren enti tial al equat equation ionss
A diﬀerential equation gives a relation between a function f and its derivatives f ′ , f ′′ ,.... ,.... This relation must be valid for any value of the argument x of f , f , which implies that f must have a speciﬁc form.
8.1 8.1
Firs Firstt ord order er,, homo homoge gene neou ouss
Let us consider the homogeneous equation f ′ (x) = a(x)f ( f (x),
(16)
valid for any value of the argument x of f , f , and where a is a given function of x. Suppose that f 1 is a solution of eq.(16), and suppose that f 2 is another solution. We have then f 1′ (x) f 2′ (x) = , f 1(x) f 2 (x) such that, after integration, ln f 2 (x) = ln f 1 (x) + k




where k is a constant. Taking the exponential of this, one ﬁnds f 2 (x) = cf 1(x), where c = exp(k exp(k), and therefore f 2 and f 1 are proportional: the set of solutions for the equation (16) is a onedimensional vector space. For the equation (16), the solution can be derived by using the separ separation ation of variables variables method, which consists in writing the equation in the form df = a(x)dx, f ( f (x)
±
which, after integration, leads to x
f ( f (x) ln f 0 where f 0 = f ( f (x0), such that
=
a(u)du,
x0
x
f ( f (x) = f 0 exp
a(u)du .
x0
Example Consider the equation
f ′ (x) = af (x) + b, where a, b are constants. If one deﬁnes g (x) = f ( f (x) + b/a, b/a, one sees that g satisﬁes g ′ (x) = ag( ag(x) and one can use the previous result to ﬁnd exp(ax)) f ( f (x) = g0 exp(ax where g0 = f (0) f (0) + b/a is a constant of integration. 33
− ab ,
8.2
Variati ariation on of of param paramete eters rs meth method od
We consider now the nonhomogeneous equation f ′ (x) = a(x)f ( f (x) + h(x),
(17)
where h is a given function of x. If we suppose that f 1 is a speciﬁc solution of eq.(17), we have [f ( )[f ((x) f 1 (x)], )], f (x) f 1(x)]′ = a(x)[f
−
−
such that the general solution of eq.(17) can be written x
f ( f (x) = c exp
a(u)du + f 1 (x),
x0
where c = f ( f (x0 )
− f (x ). In order to ﬁnd a speciﬁc solution f , one can try 1
0
1
x
f 1 = φ(x)exp
a(u)du ,
x0
where φ(x) is a function to be found. Plugging this ansatz into eq.(17), one ﬁnds x
φ′ (x) = h(x)exp
−
a(u)du ,
x0
which, after an integration, gives the function φ. Example Consider the equation
2 xeax, f ′ (x) = af (x) + 2xe where a is a consta constant nt.. Th Thee gener general al soluti solution on of the homoge homogene neous ous equation equation is A exp(ax exp(ax), ), and the variation of parameters method consists in ﬁnding a speciﬁc solution of the form )exp(ax), ), which leads to φ(x)exp(ax φ′ (x) = 2x. The general solution is therefore exp(ax f ( f (x) = (A + x2 ) exp( ax)).
8.3
Second Second order, order, homog homogene eneous ous
We consider the following diﬀerential equation f ′′ (x) + a(x)f ′ (x) + b(x)f ( f (x) = 0,
(18)
where a, b are functions of x. We will see with several several examples examples that it is possible to ﬁnd a least two linearly independent solutions f 1 , f 2 of eq.(1 eq.(18). 8). Sup Suppose pose that that f 3 is a third 34
solution: we show now that, necessarily, f 3 is a linear combination of f 1 and f 2 . Proof From eq.(18), we ﬁnd easily
f i(x)f 3′′ (x) − f 3(x)f i′′ (x) + a(x) f i (x)f 3′ (x) − f 3f i′ (x) = 0 which, in terms of the Wronskians W i (x) = f i(x)f 3′ (x) W i′ (x) + a(x)W i (x) = 0
i = 1, 2,
− f (x)f ′(x), read 3
i
i = 1, 2.
These equations can be integrated to give W i (x) = Ai exp
− a(x)dx
i = 1, 2,
and we conclude that
A1 f 2(x)f 3′ (x) − f 3 (x)f 2′ (x) = A2 f 1 (x)f 3′ (x) − f 3 (x)f 1′ (x) . This equation can be written
f 3′ (x) A1f 2′ (x) = f 3(x) A1f 2 (x)
− A f ′ (x) , − A f (x) 2 1 2 1
and leads, after integrating and taking the exponential, to f 3(x) = C 1f 1 (x) + C 2 f 2 (x), where C i are constants constants.. This shows shows that f 3 is necessarily in the vector space spanned by f 1 and f 2. Example Consider the following diﬀerential equation
2 af ′ (x) + bf (x) = 0, f ′′ (x) + 2af
(19)
where a, b are constants. constants. In order to ﬁnd two two independen independentt solutions solutions of this equation, equation, we assume the following xdependence exp(zx)), f ( f (x) = exp(zx where z is a constant, which can be complex. This assumption leads to 2az + b = 0, z2 + 2az which has the following solutions:
35
2
• if a
> b: b: z± =
−a ± k,
√a − b. The general solution of the diﬀerential equation (19) is then exp(kx)) + B exp(−kx) f ( f (x) = exp(−ax) ax) A exp(kx kx) = exp(−ax) sinh(kx)) , ax) C cosh( C cosh(kx kx)) + D sinh(kx where C = A + B and D = A − B are constants. where k =
2
2
• if a
−a ± ik,
and the general solution is
˜ exp( ikx) exp(ikx)) + B f ( f (x) = exp( ax)Re ax)Re A˜ exp(ikx ikx)
− exp(−ax) ax)
=
−
cos(kx)) + B sin(kx sin(kx)) , A cos(kx
˜ B ˜ are complex constants, and A =Re A˜ + B ˜ , B =Im B ˜ where A, expression can also be written
{
}
{ − A˜}. The latter
cos(kx + φ0 ), f ( f (x) = f 0 e−ax cos(kx where f 0 =
√A
2
+ B 2 and tan φ0 =
−A/B. A/B .
2
if a • if a
= b. In this case, z+ = z− and the assumption f ( exp(zx)) gives one solution f (x) = exp(zx only, which is exp( ax). ax). In order to ﬁnd a second linearly independent solutions of the diﬀerential equation (19), we assume the form
−
exp(wx)), f ( f (x) = x exp(wx where w is a constant, which leads to 2(w 2(w + a) + (w ( w 2 + 2aw 2aw + b)x = 0. This equation must be valid for any x, such that necessarily w+a=0
and
2aw + b = 0, w2 + 2aw
for which the only solution is w = a. Finally, the general solution of the diﬀerential equation (19) is exp( ax) f ( f (x) = (A + Bx) Bx) exp( ax),
−
−
where A, B are constants. 36
8.4
Second Second order, order, nonho nonhomo moge geneo neous us
We consider now the equation f ′′ + a(x)f ′ + b(x)f = g (x),
(20)
where g is a given function of x, and suppose that f s is a speciﬁc solution of the equation. We have then (f f s )′′ + a(x)(f )(f f s )′ + b(x)(f )(f f s ) = 0,
−
−
−
and the results derived for a homogeneous diﬀerential equation hold for the diﬀerence f f s , such that the general solution of the equation (20) is
−
f ( f (x) = Af 1 (x) + Bf 2 (x) + f s (x), where A, B are constants, and f 1, f 2 are linearly independent.
8.5 8.5
Gene Genera rall prope propert rtie iess
In general, the solution of a homogeneous linear diﬀerential equation of order n is an ndimensiona dimensionall vector vector space, space, spanned spanned by n linearly linearly independen independentt speciﬁc solutions. solutions. The n constants of integration can then be seen as the coordinates of the solutions in the basis of the n linearly independent speciﬁc solutions, and their values are given by n boundary conditions.
8.6
Separa Separatio tion n of of vari variabl ables es method method
We ﬁnally give an example of nonlinear diﬀerential equation, solved by the separation of variables method. Consider the following equation, f ′ (x) = x3 f 2 (x). This can also be written, when f ( f (x) = 0,
df = x3 dx, 2 f such that the left hand side has the variable f only and the righthand side has the variable x only. Both sides can then be integrated separately, which leads to
−
1 1 x4 + = , 4 f f 0
where f 0 = f (0), f (0), and the solution is ﬁnally f ( f (x) =
1
−
37
f 0 f 0 x4 /4
9 9.1 9.1
Linear ear algebra ebra Line Linear ar func functi tion on
A linear function l of a variable x satisﬁes, by deﬁnition, l(ax + by) by ) = al( al(x) + bl( bl(y ), for any constants a, b and any variables x, y . If x is a number, the only possibility is l(x) = kx,
(21)
where k is a constant. We will now generalize this to linear functions applied to vectors.
9.2
Matr atrice ices
We have seen in section 6 that the rotation of angle α of the vector of coordinates u = (u1 , u2 ) in the plane leads to the vector u′ = (u′1 , u′2 ) with u′1 = u1 cos α u2 sin α u′2 = u1 sin α + u2 cos α.
−
A rotation is linear, and in order to generalize eq.(21), we would like to write it in the form u′ = R u,
·
where R represents the rotation. This can be satisﬁed if R is a 2 2 array with components Rij , with i, j = 1, 2 such that
×
R11 = cos α
R12 =
− sin α
R21 = sin α
R22 = cos α,
where i represents the line and j represents the row. We have then u′1 = R11 u1 + R12 u2 u′2 = R21 u1 + R22 u2 , which can be written
u′1 u′2
=
R11 R12 R21 R22
where the multiplication rule is
u1 u2
,
j=2 j =2
u′i =
Rij u j .
j=1 j =1
More generally, any linear transformation of a ndimensional vector u = (u1 ,...,un ) can be written in the form j= j =n
u′i =
for i = 1,...,n,
M ij ij u j
j=1 j =1
38
where M i,j i,j are the components of a matrix M which represents the linear transformation. Besides rotations, other linear transformations can be: projections, scalings, ... as well as compositions of these. A matrix S is said symmetric if S ij ij = S ji , and a matrix A is said antisymmetric if Aij = A ji . The product of a symmetric matrix with an antisymmetric matrix is zero.
−
9.3 9.3
Dete Determ rmina inan nts
Suppose one has the following system of equations x′ = ax + by y ′ = cx + dy
(22)
which can be written
x′ y′
=M
x y
,
with
M=
a b c d
.
One wishes to ﬁnd (x, ( x, y) in terms of (x ( x′ , y ′ ), if possible, and therefore the inverse M−1 of the linear transformation represented by M:
x y
= M−1
x′ y′
.
The system of equations (22) is equivalent to (ad (ad
− bc) bc)x − bc) bc)y
= dx′ = ay ′
− by′ − cx′,
(23)
and leads to the following two cases
• if ad if ad−bc = 0, the previous set of equations is equivalent to dx′ = by′ , or ay ′ = cx′ , such
that the two two equations equations of the system system (22) are equivalen equivalent. t. There There is thus thus an inﬁnity inﬁnity of solutions (x, (x, y ), corresponding to the straight line of equation ax + by = x′ , or equivalently cx + dy = y ′. In this case, the matrix M has no inverse, since there is no onetoone relation between (x, ( x, y ) and (x (x′ , y ′ ). A typical example of such a situation is a projection on a given straight line, since all the points on a perpendicular straight line are projected on the same point.
• if ad − bc = 0, there is one solution only to the system (23), which is dx′ − by′ ay ′ − cx′ x= y= . ad − bc ad − bc 39
(24)
Therefore it is essential, in order to ﬁnd a unique solution to the system of equations (22), and therefore therefore to ﬁnd an inverse inverse of the matrix M, that the determinant ad bc of M is not zero. det M = ad bc = 0,
−
−
or in other words: a linear function represented represented by the matrix M has an inverse, represented 1 − by the matrix M , if and only if det M = 0. From the solution (24), one can see that the inverse of the matrix M is then
1 M−1 = det M
d c
−
−b a
More generally, a n n matrix has an inverse if and only if its determinant is not zero. The expression for the determinant involves sums of products of n elements of the matrix.
×
9.4 9.4
Compo Composi siti tion on of of linea linear r funct functio ions ns
Given the two linear functions f 1 and f 2 , represented by the matrices M1 and M2 , with M1 =
a1 b1 c1 d1
M2 =
a2 b2 c2 d2
,
,
we wish to represent the composition of functions w = f 2 (v) = f 2 (f 1 (u)). )).
We have, with u = (x, y ), v = M1 u =
·
v1 v2
=
a1 x + b1 y c1 x + d1y
and therefore w = M2 v =
·
w1 w2
=
(a1a2 + c1 b2 )x + (b (b1 a2 + d1 b2 )y (a1c2 + c1 d2)x + (b (b1 c2 + d1d2 )y
This can also be written w = M2 M1 u,
·
·
where the product of matrices M = M2 M1 is deﬁned by
·
M i,j i,j =
M 2 ik M 1
kj ,
i, j = 1, 2
k =1, =1,2
such that M=
a1a2 + c1 b2 b1 a2 + d1 b2 a1c2 + c1d2 b1 c2 + d1d2 40
.
Remark In general, the two operations do not commute: f 2 (f 1 (u)) = f 1 (f 2 (u)), and thus
M2 M1 = M1 M2 .
Determinant of a product The determinant of M = M2 M1 is
det M = (a1 a2 + c1 b2 )(b )(b1 c2 + d1 d2) (a1 c2 + c1d2 )(b )(b1 a2 + d1 b2 ) = (a1 d1 b1 c1 )(a )(a2 d2 b2 c2),
−
−
−
such that det (M2 M1) = det M2 The previous properties are also valid for n is also noted a11 a1n ... det an1 ann
9.5
··· ···
× det M
1
= det (M1M2 )
× n matrices, and the determinant of a matrix ··· a a
=
Eigen Eigenve vecto ctors rs and and eige eigenv nvalu alues es
11
an1
1n
...
···
ann
Given a matrix M, an eigenvector e of M satisﬁes, by deﬁnition, M e = λe,
with e = 0
·
(25)
where the real number λ is the eigenvalue of M corresponding to e. Therefore Therefore the eﬀect eﬀect of the matrix M on its eigenvector e is simply a rescaling, without change of direction. A n n matrix, operating on a ndimensional vector space, can have at most n linearly independent eigenvectors. In this case, these vectors can constitute a basis ( e1 , ..., ..., en ), and the corresponding matrix, in this basis , is diagonal, with the eigenv eigenvalues being its diagonal elements: 0 0 λ1 0 0 λ2 0 0 ... ∆=
×
···
0 0
··· 0 ··· ···
···
λn−1 0 0 λn
In this case, the determinant is simply the product of the eigenvalues det ∆ = λ1 λ2
···λ
n
In order to ﬁnd the eigenvalues of a matrix , the ﬁrst step is to write the system of
equations (25) in the following way: [M
− λ1] e = 0, 41
where 1 is the unit matrix. If the corresponding matrix M λ1 had an inverse, the only solution to this system of equations would be e = 0. Bu Butt if the the initia initiall matrix matrix M has eigenvectors, these are not zero, and as a consequence M λ1 has no inve inverse. rse. Therefore Therefore its determinant vanishes: det [M λ1] = 0.
− −
−
This determinant is polynomial in λ, and the solutions to this equation give the eigenvalues which are expected. Example For a 2
× 2 matrix, we have a−λ b c d−λ
= (a
)(d − λ) − bc = 0, − λ)(d
such that the eigenvalues λ, if there are, satisfy a quadratic equation.
42
10
Funct unction ionss of sev several eral vari variab able less
10.1 10.1
Partial artial diﬀer diﬀeren entia tiatio tion n
If f is a function of two variables, and associates the value z = f ( ( x, y ), f (x, y ) to the pair (x, one can deﬁne the partial derivative of f with respect to x, for a ﬁxed value y , and the partial derivative of f with respect to y , for a ﬁxed value x. These partial derivatives are denoted ∂f = ∂x ∂f = ∂y
∆x, y ) f ( f (x + ∆x, ∆x→0 ∆x ∆y ) f ( f (x, y + ∆y lim ∆y →0 ∆y lim
− f ( f (x, y ) − f ( f (x, y ) .
An important property of partial derivatives concerns their commutativity: ∂ 2 f ∂ 2 f = . ∂x∂y ∂y∂x Proof From their deﬁnition, the partial derivatives satisfy
∂ 2 f = ∂y∂x
1 ∂f ∂f lim (x, y + ∆y ∆y ) (x, y ) ∆y →0 ∆y ∂x ∂x 1 = lim lim [f ( ∆x, y + ∆y ∆y ) f (x + ∆x, ∆y →0 ∆x→0 ∆y ∆x 1 ∂f ∂f = lim (x + ∆x, ∆x, y ) (x, y ) ∆x→0 ∆x ∂y ∂y ∂ 2 f = . ∂x∂y
−
∆y ) − f ( ∆x, y ) + f ( − f ( f (x, y + ∆y f (x + ∆x, f (x, y )]
−
Example For the function f ( cos(ay), ), where n and a are constants, we have f (x, y ) = xn cos(ay
∂f = nxn−1 cos(ay cos(ay)) ∂x ∂f = sin(ay)), axn sin(ay ∂y
−
and of course
∂ 2 f = ∂y∂x
n 1
−anx −
sin(ay sin(ay)) =
Nabla operator One deﬁnes the diﬀerential operator
ponents
∇=
∂ ∂ ∂ , , ∂x ∂y ∂z 43
,
∂ 2 f . ∂x∂y
∇ as the symbolic vector of com
which has to be understood as an operator applied to a scalar quantity φ or a vector E depending on the coordinates x,y,z: x,y,z :
∇× 10.2 10.2
∂φ ∂φ ∂φ φ(x , y , z) z) = , , ∂x ∂y ∂z ∂E x ∂E y ∂E z + + E(x , y , z) z) = ∂x ∂y ∂z ∂E z ∂E y ∂E z E(x , y , z) z) = , ∂y ∂z ∂x
∇ ∇·
−
−
∂E z ∂E y , ∂x ∂x
−
∂E y ∂x
Diﬀer Diﬀeren entia tiall of a function function of sev several eral vari variabl ables es
We consider here the example of two variables x, y , and formally use the notations dx and ∆ x and ∆y ∆y . dy for the inﬁnitesimal limits of the increments ∆x If f depends on two variables x, y, the change in f ( two contrib contributions utions:: one from f (x, y ) has two the cange ∆x ∆x and one from the change ∆y ∆ y . A Taylor expansion on both variabls leads to ∂f (x, y + ∆y ∆y ) + (∆x (∆x)2 ∂x ∂f ∂f = f ( ∆y (x, y ) + ∆x ∆x (x, y ) + (∆x (∆x)2 , (∆y (∆y )2, ∆x∆y f (x, y ) + ∆y ∂y ∂x
∆x, y + ∆y ∆y ) = f ( ∆y ) + ∆x ∆x f ( f (x + ∆x, f (x, y + ∆y
O
O
such that, if we note ∆f ∆ f = f ( ∆x, y + ∆y ∆y ) f (x + ∆x, ∆f = ∆x
− f ( f (x, y), we have
∂f ∂f + ∆y ∆y + ∂x ∂y
···,
where dots represent higher orders in ∆ x and ∆y ∆y . In the limit where ∆x ∆x we obtain the deﬁnition of the diﬀerential df =
∂f ∂f dx + dy, ∂x ∂y
(26)
→ 0 and ∆y ∆y → 0, (27)
which is an exact identiy, and can be interpreted as a vector in a two dimensional vector space spanned by dx and dy, dy , with coordinates ∂f/∂x and ∂f/∂y. ∂f/∂y . important to distinguish distinguish the symbols symbols for partial and total derivativ derivatives. es. InRemark It is important deed, in eq.(27), if y is a function of x, one can consider the function F ( F (x) = f ( f (x, y (x)), which, using the chain rule, has the following derivative F ′ (x) =
df ∂f dy ∂f = + dx ∂x dx ∂y ∂f ∂f = + y ′ (x) . ∂x ∂y
44
As a consequence,
∂f df = ∂x dx
Finally, if a function depends on N variables, one can deﬁne the partial derivatives with respect to any of these variables, and these partial derivatives will commute among each other.
10.3 10 .3
Impl Im plic icit it funct functio ions ns
If the variables x,y,z are related by an equation of the form g (x,y,z) x,y,z ) = 0, where g is a diﬀerentiable function, on can deﬁne each variable as a function of the other two (besides possible singular points). We can show then that
∂x ∂y
∂y ∂x
=
z
−1
,
z
where the variable in subscript represents the one kept constant in the diﬀerentiation. Proof Since g (x,y,z) x,y,z) = 0, we have dg =
∂g ∂g ∂g dx + dy + dz = 0, ∂x ∂y ∂z
and if we consider the case z= constant, we have dz = 0, such that
∂x ∂y
z
dx = dy
dz=0 dz=0
=
−
∂g ∂y
Another important property is
∂g = ∂x
dy dx
∂x ∂y
z
−1
∂y ∂z
∂z ∂x
x
dz=0 dz=0
= y
=
∂y ∂x
−1 z
−1
Proof We have
∂x ∂y
= z
−
∂g ∂y
∂g ∂x
∂y ∂z
=
−
x
∂g ∂z
∂g ∂y
∂z ∂x
= y
−
∂g ∂x
∂g , ∂z
such that the product of these three derivatives is 1.
10.4 10.4
Double Double integ integrat ration ion
If f is a function depending on two variables x, y , one can deﬁne the function F ( F (x) as d
F ( F (x) =
f ( f (x, y )dy,
c
45
and then the integral of F over an interval [a, [a, b] b
I 1 =
b
F ( F (x)dx =
a
d
a
b
d
f ( f (x, y )dy dx =
c
dx
a
dy f ( f (x, y ).
c
The product dxdy represents an inﬁnitesimal surface are in the plane (0 , x , y), y), and the integral is thus the volume between the rectangular area of surface b a d c and the surface deﬁned by z = f ( f (x, y ). In the simple case where f is a product f ( f (x, y ) = φ(x)ψ(y ), the latter integral is just a product of integrals
 −  − 
b
I 1 =
d
dx
a
b
dy φ(x)ψ(y ) =
c
d
×
φ(x)dx
a
c
More generally, one can deﬁne a double integral over any area by I 2 =
ψ (y )dy .
D which is not rectangular
f ( f (x, y )dxdy
D
In this case, one can perform the integrals in whichever order: ﬁrst over x and then over y or the opposite:
x2
I 2 =
y2 (x)
x1
f ( f (x, y )dy dx,
y1 (x)
where the values y1(x), y2 (x) are the boundaries of the domain
y2
I 2 =
x2 (y )
y1
D for a given value of x, or
f ( f (x, y )dx dy,
x1 (y )
where the values x1 (y ), x2 (y ) are the boundaries of the domain
D for a given value of y.
Example Calculate Calculate the volume volume of a pyramid pyramid whose base is an equilater equilateral al triangle triangle of sides sides
a , and the three other faces, of equal surface area, have edges which meet orthogonally. For this problem, let’s consider the top of the pyramid at the centre of coordinates, such that the axises (Ox (Ox)), (Oy) Oy ), (Oz Oz)) are along the edges which meet orthogonally. The base is then perpendicular to the vector (1 , 1, 1), and intersects the previous edges at the distance a/ 2 from the top. Its equation is thus x + y + z = a/ 2. The volume is then
√
√
√
√−
a/ 2
V 1 =
dy
0
√
a/ 2
=
0
=
a/ 2 y
√ − − √ − a 2
0
1 dy 2
a 2
a3 . 12 2
√
x
y
2
y
(28) 46
rdΘ rdΘ
dr
Figure 9: The inﬁnitesimal surface area in polar coordinates is rdrdθ Double Double integra integrall in polar coordinate coordinatess The inﬁnitesimal surface area in polar coordi
nates is dr
× rdθ (see ﬁg(9)), and an integral over a domain D is thus J =
f ( f (r, θ)rdrdθ.
D
Example Calculate the volume of half a solid ball of radius R.
A sphere of radius R, centered on the origin, is given by the equation x2 + y 2 + z 2 = R2 . This volume of half the ball is then V 2 =
z(x, y )dxdy =
C
− R2
x2
C
2
− y dxdy,
where is the disc of radius R, cent center ered ed in the origin origin.. A change change of variabl ariables es to polar coordinates gives
C
√ − √ − − − R
V 2 =
2π
dθ R2
rdr
0
0
R
rdr R2
= 2π
r2
0
= 2π
1 2 (R 3
2π 3 = R. 3
47
R
2 3/2
r )
0
r2
z
rd Θ
dr Θ
r sinΘ dφ
y
φ
x
Figure 10: The inﬁnitesimal volume in spherical coordinates is r2 dr sin θdθdφ
10.5 10.5
Triple riple integ integrat ration ion
Triple integration integration is a straight straightforw forward ard generaliz generalization ation of double integration integration,, and it can sometimes be useful to use spherical coordinates (r,θ,φ ( r,θ,φ), ), if the function to integrate is expresse expressed d in terms of spherical spherically ly symmetri symmetricc quantitie quantities. s. Using these coordinates, coordinates, the in2 ﬁnitesimal volume is rdθ r sin θdφ dr = r dr sin θdθdφ (see ﬁg.10), and an integral over a threedimensional domain is then
×
×
r2 dr sin θdθdφ f ( f (r,θ,φ) r,θ,φ).
D
Example The volume of half a solid ball of radius R, which was calculated before, is easier
to calculate using spherical coordinates, and is R
V 2 =
π/2 π/2
2
r dr
0 3
2π
dθ sin θ
0
R π/2 π/2 [ cos θ]0 3 2π 3 = R 3 =
×−
48
0
× 2π
dφ