1st Year Maths Notes

48
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 ..................................... 3 1.2 Group ...................................... 3 1.3 Combinatorics .................................. 3 2 Functions of a real variable 5 2.1 Continuity .................................... 5 2.2 Differentiation .................................. 5 2.3 Polynomial functions .............................. 8 2.4 Rational functions ............................... 8 2.5 Trigonometric functions ............................ 10 3 Integration 12 3.1 Interpretation of the integral .......................... 13 3.2 Integration by part ............................... 14 3.3 Change of variable ............................... 14 3.4 Improper integrals ............................... 15 4 Logarithm and exponential functions 18 4.1 Logarithm .................................... 18 4.2 Exponential ................................... 20 4.3 Hyperbolic functions .............................. 21 5 Taylor expansions and series 22 5.1 Approximation of a function around a value of the argument ........ 22 5.2 Radius of convergence and series ........................ 22 5.3 Examples .................................... 23 5.4 Expansion for a composition of functions ................... 24 1

Transcript of 1st Year Maths Notes

Page 1: 1st Year Maths Notes

Mathematics I(Part of 4CCP 1350)

Department of Physics, King’s College LondonDr J. Alexandre, 2009/2010

Contents

1 Introduction 3

1.1 Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3 Combinatorics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2 Functions of a real variable 5

2.1 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.2 Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.3 Polynomial functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.4 Rational functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.5 Trigonometric functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

3 Integration 12

3.1 Interpretation of the integral . . . . . . . . . . . . . . . . . . . . . . . . . . 133.2 Integration by part . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143.3 Change of variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143.4 Improper integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

4 Logarithm and exponential functions 18

4.1 Logarithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184.2 Exponential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204.3 Hyperbolic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

5 Taylor expansions and series 22

5.1 Approximation of a function around a value of the argument . . . . . . . . 225.2 Radius of convergence and series . . . . . . . . . . . . . . . . . . . . . . . . 225.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235.4 Expansion for a composition of functions . . . . . . . . . . . . . . . . . . . 24

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6 Vector calculus 25

6.1 Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256.2 Rotations in two dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . 256.3 Scalar product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276.4 Cross product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 286.5 Scalar triple product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 286.6 Polar coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

7 Complex numbers 30

7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307.2 Complex exponential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307.3 Trigonometric formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 317.4 Roots of complex numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . 327.5 Relation to hyperbolic functions . . . . . . . . . . . . . . . . . . . . . . . . 32

8 Linear differential equations 33

8.1 First order, homogeneous . . . . . . . . . . . . . . . . . . . . . . . . . . . . 338.2 Variation of parameters method . . . . . . . . . . . . . . . . . . . . . . . . 348.3 Second order, homogeneous . . . . . . . . . . . . . . . . . . . . . . . . . . 348.4 Second order, non-homogeneous . . . . . . . . . . . . . . . . . . . . . . . . 378.5 General properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 378.6 Separation of variables method . . . . . . . . . . . . . . . . . . . . . . . . 37

9 Linear algebra 38

9.1 Linear function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 389.2 Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 389.3 Determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 399.4 Composition of linear functions . . . . . . . . . . . . . . . . . . . . . . . . 409.5 Eigenvectors and eigenvalues . . . . . . . . . . . . . . . . . . . . . . . . . . 41

10 Functions of several variables 43

10.1 Partial differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4310.2 Differential of a function of several variables . . . . . . . . . . . . . . . . . 4410.3 Implicit functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4510.4 Double integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4510.5 Triple integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

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1 Introduction

1.1 Numbers

• 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, where pand q 6= 0 are elements of Z. These numbers have either a finite number of decimalsor 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 andrandom decimals. Examples of real numbers, which are not in Q are:

√2, π, e, ....

Density property: Between any two real numbers can be found a rational number, and viceverse.

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 of groups: {Z,+}, or {Q⋆,×}, where Q⋆ is Q without 0.

1.3 Combinatorics

• 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, independentlyof the order, is given by the binomial coefficients

(

nk

)

=n!

k!(n− k)!.

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Indeed, the number of possible ways to order n points is n!, and has to be dividedby the number of ways to order the k chosen elements, which is k!, and also by thenumber of ways to order the remaining n− k elements, which is (n− k)!Some simple properties are:

(

nn− k

)

=

(

nk

)

,

(

n1

)

= n,

(

n0

)

= 1.

• Binomial formula. We show here that

(a+ b)n =k=n∑

k=0

(

nk

)

an−kbk, (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+ b2

=

(

20

)

a2b0 +

(

21

)

a1b1 +

(

22

)

a0b2

=k=2∑

k=0

(

2k

)

a2−kbk.

Second step: suppose that eq.(1) is valid for n, and show that it is then valid forn+ 1:

(a+ b)n+1 = (a + b)k=n∑

k=0

(

nk

)

an−kbk

=k=n∑

k=0

(

nk

)

an−k+1bk +k=n∑

k=0

(

nk

)

an−kbk+1

=

k=n∑

k=0

(

nk

)

an+1−kbk +

k=n+1∑

k=1

(

nk − 1

)

an+1−kbk

= an+1 + bn+1 +k=n∑

k=1

[(

nk

)

+

(

nk − 1

)]

an+1−kbk

= an+1 + bn+1 +k=n∑

k=1

(

n+ 1k

)

an+1−kbk

=

k=n+1∑

k=0

(

n + 1k

)

an+1−kbk.

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2 Functions of a real variable

A function of a real variable f is an operation, which to a real variable x associates thequantity f(x).

2.1 Continuity

Intuitively, a function f of the variable x is continuous is a small change in x leads to asmall change in f(x). More rigorously, f is continuous in x0 if for any ε > 0, one canalways find a δ > 0 such that

|x− x0| < δ ⇒ |f(x) − f(x0)| < ε.

2.2 Differentiation

The derivative of a function f at the point x is the slope of the tangent of the curve y = f(x)at x. In order to calculate it, let’s consider the points M and M ′ with coordinates (x, f(x))and (x+∆x, f(x+∆x)) respectively, where dx > 0 is an increment (see fig(1)). The slopeof the straight line (MM ′) is

slope =f(x+ ∆x) − f(x)

(x+ ∆x) − x=f(x+ ∆x) − f(x)

∆x.

The slope of the tangent of the curve at M is obtained when ∆x → 0. The derivative off at the point x is then

f ′(x) = lim∆x→0

f(x+ ∆x) − f(x)

∆x=df

dx, (2)

where dx denotes the infinitesimal increment in x and df the corresponding infinitesimalincrement in f(x).

Example Let us calculate the derivative of f(x) = axn, where a is a constant and n is aninteger. By definition

f ′(x) = lim∆x→0

a(x+ ∆x)n − axn

∆x

= lim∆x→0

a[xn + nxn−1∆x+ n(n− 1)xn−2(∆x)2/2 + · · ·] − axn

∆x= a lim

∆x→0

(

nxn−1 + n(n− 1)xn−2∆x+ · · ·)

= anxn−1,

where the dots represent higher orders in ∆x.

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x

f(x)

x+ x

f(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

lim∆x→0

f(x) − f(x− ∆x)

∆x,

where ∆x > 0. A function f is said to be differentiable at x if these two definitions leadto the same result. If these two derivatives are different, the function is singular at thepoint x and its derivative is not defined. An example of such a singularity is the functionf(x) = |x| at x = 0. Indeed, for x = 0, the left derivative is -1 and the right derivative is1.Note that a function can be continuous but not differentiable for a given value of x, asshows the previous example.

Extrema of a function Since the derivative f ′(a) of a function at the point a correspondsto the slope of the tangent of the curve of equation y = f(x), we have the followingclassification:

• 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;

• if f ′(a) = 0 and f(x) changes sign at x = a, then f(a) is an extremum of f ;

• if f ′(a) = 0 and f(x) does not change sign at x = a, then the point of coordinates(a, f(a)) is called an inflexion point. At such a point, the second derivative changessign.

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Derivative of a product If f, g are two functions of x, the derivative (fg)′ is given by

(fg)′(x) = lim∆x→0

f(x+ ∆x)g(x+ ∆x) − f(x)g(x)

∆x

= lim∆x→0

f(x+ ∆x)g(x) − f(x)g(x) + f(x+ ∆x)g(x+ ∆x) − f(x+ ∆x)g(x)

∆x

= lim∆x→0

{

g(x)f(x+ ∆x) − f(x)

∆x+ f(x+ ∆x)

g(x+ ∆x) − g(x)

∆x

}

= f ′(x)g(x) + f(x)g′(x). (3)

Chain rule Consider two functions f and g, and the function F defined as F (x) = f(g(x)).The derivative of F is

F ′(x) = lim∆x→0

F (x+ ∆x) − F (x)

∆x

= lim∆x→0

f(g(x+ ∆x)) − f(g(x))

dx

= lim∆x→0

f(g(x+ ∆x)) − f(g(x))

g(x+ ∆x) − g(x)× g(x+ ∆x) − g(x)

∆x

= lim∆x→0

f(g + ∆g) − f(g)

∆g× g(x+ ∆x) − g(x)

∆x

= f ′(g(x)) × g′(x),

where ∆g = g(x+ ∆x) − g(x) is the increment in g(x) corresponding to x→ x+ ∆x.

Derivative of a ratio The derivative of 1/x is −1/x2, such that the derivative of thefunction 1/f is

(

1

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(x)

g(x)

)′

= f ′(x) × 1

g(x)+ f(x) ×

(

− g′(x)

g2(x)

)

=f ′(x)g(x) − f(x)g′(x)

g2(x).

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 todefine the inverse of a function, one needs a one-to-one mapping between x and y. This isusually the case on a given interval for x at least.The derivative of the inverse is then

(

f−1(y))′

= lim∆y→0

f−1(y + ∆y) − f−1(y)

∆y

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= lim∆x→0

x+ ∆x− x

f(x+ ∆x) − f(x)

=1

f ′(x),

where ∆x is defined such that y + ∆y = f(x+ ∆x).

2.3 Polynomial functions

A polynomial function of x is of the form

P (x) =

n=N∑

n=0

anxn,

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

limx→−∞

P (x) = −∞ and limx→+∞

P (x) = +∞,

such that the line representing y = P (x) cuts at least once the axis y = 0, since thepolynomial is a continuous function.A polynomial of degree 2 can have two poles, but might not have any (real) pole:

• P (x) = a(x− z1)(x− z2) has two poles z1, z2. The pole is double if z1 = z2;

• Q(x) = ax2 + bx+ c has no pole if b2 − 4ac < 0.

A polynomial of degree 3 has either one pole or three poles, and can be written, for all x,

• P (x) = a(x− z1)(x− z2)(x− z3) if P has three poles;

• Q(x) = (x− z)(ax2 + bx+ c), with b2 − 4ac < 0, if Q has one pole z.

In general, any polynomial function can be written

P (x) = (x− z1) · · · (x− zn) × (a1x2 + b1x+ c) · · · (amx

2 + bmx+ cm),

where zi, i = 1, ..., n are the poles of the polynomial, b2j − 4ajcj < 0 for all j = 1, ..., m,and n+ 2m is the degree of the polynomial.

2.4 Rational functions

A rational function is the ratio of two polynomial functions P and Q, and has the form,for each x,

R(x) =P (x)

Q(x).

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x

y

z z z1 2 3

Figure 2: A polynomial function of degree 5, with three poles z1, z2, z3.

x

y

z z z z1 2 43

Figure 3: A polynomial function of degree 6, with four poles z1, z2, z3, z4.

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1

1

x

cos x

sin xM

Figure 4: The coordinates of M on the trigonometric circle are (cosx, sin x).

If the degree of P is less than the degree of Q, It is always possible to reduce R as a sumof irreducible rational functions of the form a/(x− z) or (ax+ b)/(x2 + cx+ d).

Example The fraction (x+ 2)/(x2 + 5x+ 4) can be written

x+ 2

x2 + 5x+ 4=

x+ 2

(x+ 1)(x+ 4)=

a

x+ 1+

b

x+ 4,

where a(x+ 4) + b(x+ 1) = x+ 2, such that a+ b = 1 and 4a+ b = 2, which gives a = 1/3and b = 2/3. Finally,

x+ 2

x2 + 5x+ 4=

1/3

x+ 1+

2/3

x+ 4.

2.5 Trigonometric functions

For a given angle 0 ≤ x ≤ 2π, sin x and cos x are defined as the coordinates of the pointM at the intersection of the straight line (OM) with the trigonometric circle (see fig(4)).

Property Using Pythagoras’ theorem, we have sin2 x+ cos2 x = 1.

Trigonometric formula It will be shown in the chapter on vector calculus (subsection

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6.2) that the sine and cosine of the sum of two angles are given by

sin(a+ b) = sin a cos b+ sin b cos a

cos(a+ b) = cos a cos b− sin a sin b. (4)

Important limit We will now show, geometrically, that

limx→0

sin x

x= 1,

and this limit will be very useful in deriving fundamental properties of the trigonometricfunctions.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 ≤ sin2 x+ (1 − cosx)2 ≤ x2,

such that

0 ≤ 1 − cosx ≤ x2

2.

Using this in the inequalities (5), we obtain

sin x

x≤ 1 ≤ sin x

x+x

2,

and the only possibility for this to be valid in the limit x→ 0 is to have sinxx

→ 1.

Derivative of trigonometric functions The first important consequence of the previouslimit is the calculation of the derivative of the sine. From eq.(4) we have

sin′ x = lim∆x→0

sin(x+ ∆x) − sin x

∆x

= lim∆x→0

sin x cos(∆x) + sin(∆x) cosx− sin x

∆x

= lim∆x→0

{

cosxsin(∆x)

∆x+ sin x

cos(∆x) − 1

∆x

}

.

We have seen that 1− cos(∆x) is of order ∆x2, and therefore the second term vanishes inthe limit ∆x→ 0, whereas the first term leads to

(sin x)′ = cosx.

In the same way, one can easily show that (cosx)′ = − sin x. As a consequence, we alsohave (tanx)′ = 1 + tan2 x.

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a

b

c

x

Figure 5: On the figure: a = sin x, b = 1 − cosx and c2 = a2 + b2

3 Integration

The integration corresponds to the inverse operation of the differentiation: F is a primitiveof f if

F (x) =

f(x)dx ⇒ F ′(x) = f(x).

We have∫ b

a

f(x)dx = F (b) − F (a),

and therefore∫ x

0

f(u)du = F (x) − F (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

bf(x)dx = −

∫ b

af(x)dx

•∫ c

af(x)dx =

∫ b

af(x)dx+

∫ c

bf(x)dx

•∫ b

a[c1f1(x) + c2f2(x)]dx = c1

∫ b

af1(x)dx+ c2

∫ b

af2(x)dx, where c1, c2 are constants.

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a bdx

x

f(x )

k

k

Figure 6: Riemann definition of the integral

3.1 Interpretation of the integral

As explained on fig.(6), the Riemann definition of∫ b

af(x)dx, b > a, corresponds to the

surface area between the line y = 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(x), and we have

∫ b

a

f(x)dx = limn→∞

{

b− a

n

n−1∑

k=0

f(xk)

}

where

xk = a+ kb− a

n, k = 0, ..., n− 1,

Equivalence with the definition based on the derivative. We show here that theRiemann definition of the integral, as a surface area, is equivalent to the definition givenpreviously. From the Riemann interpretation, the quantity

F (x) =

∫ x

a

du f(u)

corresponds to the surface area of between the lines y = f(x) and y = 0, from a to x. Theintegral from a to x+ ∆x is then

F (x+ ∆x) =

∫ x+∆x

a

du f(u),

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and the derivative of the function F is

F ′(x) = lim∆x→0

1

∆x

{∫ x+∆x

a

du f(u) −∫ x

a

du f(u)

}

= lim∆x→0

1

∆x

∫ x+∆x

x

du f(u).

The latter expression corresponds to the surface area between the lines y = f(x) and y = 0from x to x+ ∆x, which is equal to ∆x× f(x)+ higher powers in ∆x. As a consequence,we obtain the expected result:

F ′(x) = lim∆x→0

1

∆x

(

∆xf(x) + (∆x)2 · · ·)

= f(x).

As a consequence of this interpretation of the integral, if two functions f, g satisfy

f(x) ≤ g(x) for a ≤ x ≤ b,

then∫ b

a

f(x)dx ≤∫ b

a

g(x)dx.

3.2 Integration by part

The derivative of the product of two functions f, g is (fg)′ = f ′g+fg′, such that we obtain,after integration

f(x)g(x) =

f ′(x)g(x)dx+

f(x)g′(x)dx,

which can be helpful to calculate one of the integrals on the right hand side, if we knowthe other:

∫ b

a

f ′(x)g(x)dx = [f(x)g(x)]ba −∫ b

a

f(x)g′(x)dx

Example Integration by part is very useful for the integration of trigonometric functionsmultiplied by power law functions, as

dx x cosx =

dx x(sin x)′ = x sin x−∫

dx sin x = x sin x+ cosx+ c,

where c is a constant.

3.3 Change of variable

Suppose that one can write x = g(u), where u represents another variable with which theintegral can be calculated. We have then dx = g′(u)du and

∫ b

a

f(x)dx =

∫ g−1(b)

g−1(a)

f(g(u))g′(u)du,

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where g−1 represents the inverse function of g: x = g(u) ⇔ u = g−1(x). For the change ofvariable to be consistent, one must make sure that there is a one-to-one relation betweenx and u in the interval [a, b].

Example In the following integral, one makes the change of variable u = sinφ,for 0 ≤ φ ≤ π/2:

∫ 1

0

du√1 − u2

=

∫ π/2

0

cosφ dφ√

1 − sin2 φ=

∫ π/2

0

dφ =π

2.

Note that, in the interval [0, π/2], we have

cos2 φ = | cosφ| = cosφ,

since cosφ > 0.

3.4 Improper integrals

The domain of integration of an integral might either contain a singular point, wherethe function to integrate is not defined, or might not be bounded. In both cases, thecorresponding integral is said to be convergent if the result of the integration is finite, anddivergent if the result of the integration is infinite. We describe here this situation for theintegration of power law functions.

Case of a non-compact domain of integration We first show that the integral

I1 =

∫ ∞

1

dx

x

diverges. For this, one can see on a graph that

I1 >

∞∑

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 2N , can be written:

2N

n=1

1

n= 1 +

(

1

2

)

+

(

1

3+

1

4

)

+

(

1

5+

1

6+

1

7+

1

8

)

+

(

1

9+ · · · + 1

16

)

+ · · ·

and satisfies

2N

n=1

1

n> 1 +

(

1

2

)

+

(

1

4+

1

4

)

+

(

1

8+

1

8+

1

8+

1

8

)

+

(

1

16+ · · ·+ 1

16

)

+ · · ·

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Page 16: 1st Year Maths Notes

The sum in each bracket is equal to 1/2, and there are N bracket, such that

2N

n=1

1

n> 1 +

N

2,

which shows that the sum goes to infinity when N goes to infinity.

As a consequence,∫ ∞

1

dx

xis divergent.

Consider now the integral, for a 6= 1,

Ia =

∫ ∞

1

dx

xa= lim

x→∞

x1−a − 1

1 − a

As can be seen, the result depends on a:

• if a > 1 then Ia = 1/(a− 1) is finite;

• if a < 1 then Ia = +∞.

Since the integral Ia also diverges for a = 1, it converges only for a > 1.

Case of a singular point Consider the integral, for b 6= 1,

Jb =

∫ 1

0

dx

xb= lim

x→0

1 − x1−b

1 − b

As can be seen, the result depends on the power b:

• if b < 1 then Jb = 1/(1 − b) is finite;

• if b > 1 then Jb = +∞.

The integral Jb also diverges for b = 1 (the surface area is the same as the previous case,with a non-compact domain of integration), it therefore converges only for b < 1. Ingeneral, we have:

∫ 1

z

dx

(x− z)bis

{

convergent if b < 1divergent if b ≥ 1

Example Consider the integral

∫ ∞

1

dx

(x− 1)b(2x+ 3)a

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Page 17: 1st Year Maths Notes

• at x = 1: the integrand is equivalent to 5−a/(x− 1)b, such that there is convergenceif b < 1;

• at x = ∞: the integrand is equivalent to 2−a/xa+b, such that there is convergence ifa+ b > 1;

As a consequence, the integral is convergent only if b < 1 and a+ b > 1 simultaneously.

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4 Logarithm and exponential functions

4.1 Logarithm

We have seen that, for a 6= −1,

xadx =xa+1

a+ 1a 6= −1,

and we still have to define this integral for a = −1. For this, we introduce the logarithmas

lnx =

∫ x

1

du

u,

so that the logarithm gives the surface area between the function 1/u and the horizontalaxis, from 1 to x > 0. The real logarithm is not defined for x < 0, since the correspondingsurface area would be infinite. The number e is defined by ln e = 1 and e ≃ 2.718281828.

Properties:

• We have seen that the integrals∫∞

1dx/x and

∫ 1

0dx/x both diverge, such that

limx→∞

ln x = +∞ and limx→0

ln x = −∞

• From the definition of the logarithm, one can see that

ln(ab) =

∫ ab

1

du

u=

∫ a

1

du

u+

∫ ab

a

du

u= ln a +

∫ b

1

dv

v= ln a+ ln b, (6)

where we make the change of variable u = av.

• One can also see that

ln(xa) =

∫ xa

1

du

u=

∫ x

1

adv

v= a lnx, (7)

where we make the change of variable u = va.

• We have in general, for any differentiable function f ,

dxf ′(x)

f(x)= ln |f(x)| + c,

where c is a constant

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Logarithm in base a The logarithm in base a is defined as

loga(x) =ln x

ln a,

and is equal to 1 when x = a. Note that ln x = loge(x).

Integral of the logarithm To calculate∫

ln x dx, one uses an integration by parts:∫

lnx dx =

(x)′ lnx dx = x ln x−∫

x(ln x)′ dx = x ln x− x+ c,

where c is a constant.

Limits

• When x→ +∞: We show here the important limit

limx→+∞

ln x

xa= 0, a > 0 (8)

which means that any (positive-) power law goes quicker to infinity than the loga-rithm, when x→ +∞.Proof For any u ≥ 1 and for any a > 0, we have

1

u≤ 1

u1−a/2.

Integrating this inequality from 1 to x leads to

0 < ln x ≤ 2

a(xa/2 − 1) <

2

axa/2.

Dividing by xa gives the expected result:

0 <ln x

xa≤ 2

ax−a/2 → 0 when x→ +∞.

• When x→ 0: Another important limit to know is

limx→0

xa ln x = 0, a > 0, (9)

which means that any (positive-) power law kills the divergence of the logarithm atx = 0.Proof For any u satisfying 0 < u ≤ 1 and any a > 0, we have

1

u≤ 1

u1+a/2.

Integrating this inequality from x to 1, we obtain

0 < − ln x ≤ 2

a(x−a/2 − 1) <

2

ax−a/2.

Multiplying by xa gives the expected result:

0 ≤ xa| lnx| ≤ 2

axa/2 → 0 when x→ 0.

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4.2 Exponential

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+ v) = (exp u) × (exp v),

and from property (7), if we note y = ln x, we have

(

exp y)a

= exp(ay).

Derivative of the exponential one can differentiate the definition exp(lnx) = x, which,using the chain rule, leads to exp′(ln x) = x. We therefore conclude that the derivative ofthe exponential is the exponential itself:

exp′ x = exp x.

Exponential of base a This function is defined as

ax = exp(x ln a),

which is consistent with the properties of the logarithm and the exponential. It’s derivativeis then

(ax)′ = (ln a)ax.

One can also define the function xx, with derivative

(xx)′ =d

dx

(

exp(x ln x))

= (1 + lnx)xx.

Limits

• From the limit (8), if we note y = ln x and b = 1/a > 0, we have

limy→+∞

exp y

yb= +∞,

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

limy→+∞

yb exp(−y) = 0,

and the decreasing exponential kills the divergence of any power law.

20

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4.3 Hyperbolic functions

The hyperbolic functions are defined as

• hyperbolic cosine: cosh x = (ex + e−x)/2;

• hyperbolic sine: sinh x = (ex − e−x)/2;

• hyperbolic tangent: tanhx = sinh x/ cosh x;

• hyperbolic cotangent: cothx = cosh x/ sinh x,

and their derivatives are given by

cosh′ x = sinh x

sinh′ x = cosh x

tanh′ x = 1 − tanh2 x

coth′ x = 1 − coth2 x

It can easily be seen that, from their definition, the functions cosh and sinh satisfy, for allx,

cosh2 x− sinh2 x = 1.

Also, it can be easily checked that

cosh(2x) = cosh2(x) + sinh2(x)

sinh(2x) = 2 sinh(x) cosh(x)

21

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5 Taylor expansions and series

5.1 Approximation of a function around a value of the argument

It is sometimes useful to approximate the value f(x) of a function f around f(x0). Thefirst approximation consists in replacing f(x) by a linear function p1 (polynomial of firstorder, representing a straight line) in a small interval around x0:

f(x) ≃ p1(x) = a0 + a1(x− x0).

In order to find the coefficients a0, a1, one imposes the constraints p1(x0) = f(x0) andp′1(x0) = f ′(x0), such that a0 = 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 of secondorder, representing an arc of parabola), which is better than a straight line. One writesthen

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 onewishes to push further the precision of the approximation, one can take the third orderpolynomial

f(x) ≃ p3(x) = a0 + a1(x− x0) + a2(x− x0)2 + a3(x− x0)

3,

and impose the additional constraint p′′′3 (x0) = f ′′′(x0), leading to f ′′′(x0) = 2×3×a3, andso on...Going on like this finally leads to the Taylor expansion of the function f :

f(x) ≃ f(x0) + (x− x0)f′(x0) +

1

2!(x− x0)

2f ′′(x0) +1

3!(x− x0)

3f ′′′(x0) + · · ·,

where the dots represent higher powers in the difference x − x0, which are smaller andsmaller as the order of the Taylor expansion increases. Obviously, such an expansion isvalid only if the function is differentiable a number of times large enough to reach thedesirable order.Note that a polynomial function of order N is exactly equal to its Taylor expansion oforder N .The power n of the first neglected term in the expansion of a function around x0 is denotedO(x− x0)

n, and means “terms which are at least of the power n”.

5.2 Radius of convergence and series

For many functions, the Taylor expansion around x0 can be pushed to an infinite order, atleast in a vicinity of f(x0): if |x− x0| < R, where R is the radius of convergence, then theseries is convergent and one can write

f(x) =

∞∑

n=0

1

n!f (n)(x0)(x− x0)

n,

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Page 23: 1st Year Maths Notes

where f (n)(x0) denotes the n-th derivative of f at x0.

Ratio convergence test Consider the geometric series S =∑N−1

n=0 qn. An expression for

this sum can be obtained by noting that qS = S − 1 + qN , and hence

S = 1 + q + q2 + · · ·+ qN−1 =1 − qN

1 − q

From this expression, we see that, if q < 1, then limN→∞ S = 1/(1 − q) is finite, and ifq ≥ 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 convergencetest:

• if limn→∞ |an+1/an| < 1, the series is (absolutely) convergent;

• if limn→∞ |an+1/an| > 1, the series is divergent;

• if limn→∞ |an+1/an| = 1, one cannot conclude, and each case has to be looked atindividually.

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

limn→∞

f (n+1)(x0)

f (n)(x0)

R

n + 1

= 1. (10)

5.3 Examples

By calculating the different derivatives of the following functions at x = 0, one can easilysee that

cosx =∞∑

n=0

(−1)n x2n

(2n)!, R = ∞;

sin x =

∞∑

n=0

(−1)n x2n+1

(2n+ 1)!, R = ∞;

exp x =

∞∑

n=0

xn

n!, R = ∞;

1

1 + x=

∞∑

n=0

(−1)nxn, R = 1;

ln(1 + x) =

∞∑

n=0

(−1)n xn+1

n+ 1, R = 1.

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Page 24: 1st Year Maths Notes

Counter example. Consider the function f(x) = exp(−1/x). If we note y = 1/x > 0,we have

f(0) = limy→+∞

exp(−y) = 0

f ′(0) = limx→0

1

x2exp(−1/x) = lim

y→+∞y2 exp(−y) = 0

f ′′(0) = limx→0

(

1

x4− 2

x3

)

exp(−1/x) = limy→+∞

(

y4 − 2y3)

exp(−y) = 0

etc...

As a consequence,

f(0) + xf ′(0) +x2

2!f ′′(0) +

x3

3!f ′′′(0) + · · · = 0,

and no Taylor expansion of f can be defined around 0, whereas f(0) = 0 is defined.

5.4 Expansion for a composition of functions

Suppose that two functions f1, f2 have the following expansions around x = 0, to the orderx3:

f1(x) = a1 + b1x+ c1x2 + d1x

3 + O(x4)

f2(x) = a2 + b2x+ c2x2 + d2x

3 + O(x4)

The expansion of the product f1f2 can then be obtained up to the order 3 maximum, andis

f1(x)f2(x) = a1a2+(a1b2+b1a2)x+(a1c2+b1b2+c1a2)x2+(a1d2+b1c2+c1b2+d1a2)x

3+O(x4)

Example To calculate the expansion of tanx up to the order x5, we first expand theinverse of cosx to the order x5:

1

cosx=

(

1 − x2

2+x4

24

)−1

+ O(x6)

= 1 +x2

2+

5

24x4 + O(x6),

and then multiply by the expansion of sinx to the order x5:

tan x =

(

x− x3

6+

x5

120

)(

1 +x2

2+

5

24x4

)

+ O(x7)

= x+x3

3+

2

15x5 + O(x7)

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6 Vector calculus

6.1 Vectors

A vector u has a direction, given by the unit vector u, and a modulus |u|, and can bewritten

u = |u|u.n vectors u1, · · ·,un are said to be linearly independent if

a1u1 + · · · + anun = 0 ⇒ a1 = · · · = an = 0,

which means that these vectors point in different directions, and none of them can beobtained by a linear combination of the others.

A vector space V of dimension d is set of vectors spanned by d independent vectors,and is group for the addition. A set of basis vectors in V is made of d linearly independentvectors i1, · · ·, id, and any other vector can be decomposed onto this basis:

u = a1i1 + · · · + adid,

where (a, · ··, ad) are the coordinates of u in this basis. A change of basis leads to a changeof coordinates.

Addition of vectors Vectors can be added according to the rule (for example in threedimensions)

u1 + u2 = (x1i + y1j + z1k) + (x2i + y2j + z2k)

= (x1 + x2)i + (y1 + y2)j + (z1 + z2)k.

Example The set of polynomials of order N is an (N + 1)-dimensional vector space.Proof Consider the polynomials pn(x) = xn, n = 0, · · · , N , and a set of constants cn suchthat, for any x, we have c0p0(x) + c1p1(x) + · · · + cNpN(x) = 0. Then we necessarily havecn = 0, for all n, since a polynomial of degree N has at most N zeros. As a consequence,the polynomials pn are linearly independent, and span an N -dimensional vector space,where each vector can be written

P =n=N∑

n=0

anpn

and an are the coordinates of P on the basis {pn, n = 0, · · · , N}.

6.2 Rotations in two dimensions

(i, j) form an orthonormal basis in a plane. After a rotation of angle α, the basis haschanged to (i′, j′) where (see fig.7)

i′ = cosα i + sinα j

j′ = − sinα i + cosα j. (11)

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Page 26: 1st Year Maths Notes

i

j

j’

αi’

α

Figure 7: Rotation of the unit vectors

From these relations, one can easily express the vectors i, j in the basis (i′, j′) by makingthe inverse rotation (α→ −α), which leads to

i = cosα i′ − sinα j′

j = sinα i′ + cosα j′. (12)

The vector u = (a, b) is then transformed into the vector u′ = (a′, b′) such that

u′ = a i′ + b j′ = (a cosα− b sinα)i + (a sinα + b cosα)j = a′ i + b′ j,

and therefore

a′ = a cosα− b sinα

b′ = a sinα + b cosα.

Equivalently, we also have

a = a′ cosα + b′ sinα

b = −a′ sinα + b′ cosα.

Trigonometric formulas One way to find the expression for sin(α + β) and cos(α + β)in terms of sinα, cosα, sin β, cosβ is to perform two consecutive rotation, of angle α andβ respectively, and identify the result with a rotation of angle α+ β. We have seen that arotation of angle α of the basis vectors (i, j) gives

i′ = cosα i + sinα j

j′ = − sinα i + cosα j

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Page 27: 1st Year Maths Notes

A second rotation, of angle β, leads to

i′′ = cosβ i′ + sin β j′

= (cosβ cosα− sin β sinα)i + (cosβ sinα + sin β cosα)j

j′′ = − sin β i′ + cosβ j′

= (− sin β cosα− cosβ sinα)i + (− sin β sinα + cosβ cosα)j.

This must be equivalent to

i′′ = cos(α + β) i + sin(α + β) j

j′′ = − sin(α + β) i + cos(α + β) j,

such that

cos(α+ β) = cosα cosβ − sinα sin β

sin(α+ β) = sinα cosβ + cosα sin β.

Don’t learn this by heart, but rather remember how to get the result.

6.3 Scalar product

Let u and v be two vectors in a plane, with coordinates (a, b) and (c, d) respectively. Fromthese 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 caneasily express the scalar product in terms of the coordinates of the vectors, by doing thefollowing. Let’s denote by (a′, b′) and (c′, 0) the coordinates of u and v respectively, inthe orthonormal basis (i′, j′) where i′ is along v. In the basis (i′, j′), the scalar product isobviously given by u · v = a′c′, with

a′ = a cosα− b sinα

b′ = a sinα + b cosα

c′ = c cosα− d sinα

0 = c sinα+ d cosα.

Together with

cosα =c√

c2 + d2

sinα =−d√c2 + d2

,

27

Page 28: 1st Year Maths Notes

one easily obtains a′c′ = ac+ 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

u · v =

d∑

i=1

xiyi.

Example Find the equation of the plane perpendicular to the vector u = (1, 2, 1), andcontaining the point A of coordinates (3, 4, 2).Any point M of coordinates (x, y, z) of this plane is such that AM · u = 0, which reads

(x− 3) + 2(y − 4) + (z − 2) = 0 or x+ 2y + z = 13.

6.4 Cross product

One often needs to define, from two vectors u,v, a third vector which is perpendicular tou and v. The cross product u × v is

u× v = |u||v| sin(u,v)n,

where n is the unit vector perpendicular to the plane spanned by u,v, which defines theanticlockwise direction. If (i, j,k) form the usual orthonormal basis, we have

i × j = k

j × k = i

k × i = j.

From this, it is easy to find the coordinates of the vector product ofu = (a1, a2, a3) times v = (b1, b2, b3), which are

u× v =

a2b3 − a3b2a3b1 − a1b3a1b2 − a2b1

.

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 × v = −v × u

6.5 Scalar triple product

If (u,v,w) are three vectors, one defines the scalar triple product by u · (v ×w), and onecan check that a cyclic permutation does not change the result:

u · (v ×w) = w · (u × v) = v · (w × u)

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θ

i

j

M

O

er

θe

r

Figure 8: Polar coordinates (r, θ) of the point M . The orientation of the basis vectors er

and eθ depend on the position of M , such that er is always along OM and eθ is the imageof er in a rotation of angle π/2.

6.6 Polar coordinates

We denote O the origin of space and (i, j,k) the orthogonal and unit basis vectors ofEuclidean coordinates (x, y, z). Points in the plane (O, i, j) can also be labeled by thepolar coordinates (r, θ) (see fig.8), such that

r =√

x2 + y2 with 0 ≤ r <∞tan θ =

y

xwith 0 ≤ θ < 2π.

The orthogonal and unit basis vectors (er, eθ) in polar coordinates are defined by

er = cos θi + sin θj

eθ = − sin θi + cos θj.

Note that

der

dθ= eθ

deθ

dθ= −er.

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7 Complex numbers

7.1 Introduction

Complex numbers can be seen as two-dimensional vectors in the complex plane, spannedby the basis (1, i), where i2 = −1. In Cartesian coordinates, a complex number z can bewritten

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 thendefined as

z⋆ = a− ib.

Complex numbers can be added, or multiplied, to give a new complex number:

z1 + z2 = (a1 + ib1) + (a2 + ib2) = a1 + a2 + i(b1 + b2)

z1z2 = (a1 + ib1)(a2 + ib2) = a1a2 − b1b2 + i(a1b2 + a2b1).

This is because the set of complex numbers C is a group for both the addition and multi-plication. Finally, the modulus of z is defined as

|z| = |z⋆| =√a2 + b2 =

√zz⋆.

7.2 Complex exponential

Complex numbers, seen as two-dimensional vectors, can be expressed using polar coordi-nates (r, θ):

z = r(cos θ + i sin θ).

Using the series expansion for cosine and sine, we find

z = r

∞∑

n=0

(

(−1)n θ2n

(2n)!+ i(−1)n θ2n+1

(2n+ 1)!

)

= r

∞∑

n=0

(

(iθ)2n

(2n)!+

(iθ)2n+1

(2n+ 1)!

)

= r

∞∑

n=0

(iθ)n

n!

= r exp(iθ).

r is the modulus of the complex z, and θ is its argument, and the last result leads to theEuler’s formula:

cos θ + i sin θ = exp(iθ). (13)

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Page 31: 1st Year Maths Notes

From this, it is easy to find the de Moivre’s formula: noting that [exp(iθ)]m = exp(imθ),where m is any integer, we have

(cos θ + i sin θ)m = cos(mθ) + i sin(mθ).

Example The number −1 has modulus 1 and argument π (in the interval [0; 2π[), and cantherefore be written

−1 = eiπ

This equation relates three fundamental numbers, which are 1, e, π.One also has i = eiπ/2, such that

ii = exp(i ln i) = exp(i× iπ/2) = e−π/2 ≃ 0.208. (14)

Note that the logarithm of a complex number z is a multi-valued function: its definitiondepends on the range of angles in which the argument θ of z is considered. Indeed, ifθ → θ + 2kπ, where k is an integer, z is invariant, but its logarithm changes as:

ln z → ln z + 2ikπ.

As a result, ii as given in eq.(14) is the value when the argument of complex numbers aredefined in [0; 2π[.

7.3 Trigonometric formula

From the Euler’s formula (13), one can express cosine and sine with complex exponentials:

cos θ =eiθ + e−iθ

2

sin θ =eiθ − e−iθ

2i,

and therefore one can also express the nth power of cosine and sine, in terms of cosine andsine of n times the argument. For example:

(cos θ)2 =1

4(e2iθ + e−2iθ + 2) =

1

2+

1

2cos(2θ)

(sin θ)3 =i

8(e3iθ − e3iθ − 3eiθ + 3e−iθ)

=3

4sin θ − 1

4sin(3θ). (15)

These formulas are useful when one needs to integrate expressions involving powers ofcosine or sine. Do not learn these expressions by heart, but derive them whenever you need

them.

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7.4 Roots of complex numbers

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

A = ρ exp(iφ)

z = r exp(iθ).

The equation to solve is then rn exp(inθ) = ρ exp(iφ), which leads to, after identificationof the modulus and the argument of both sides of the equation zn = A,

r = ρ1/n = n√ρ

θ =φ

n+

2kπ

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

(

2iπk

n

)

k = 0, 1, ..., n− 1.

7.5 Relation to hyperbolic functions

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 expansionfor a complex variable. It is therefore possible to understand a function of a complexvariable in terms of a series expansion. We give here two examples.From eqs.(15), we have for any real x

sin(ix) = i sinh x

cos(ix) = cosh x,

which gives a formal way to define trigonometric functions with complex arguments.

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8 Linear differential equations

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 , which implies that fmust have a specific form.

8.1 First order, homogeneous

Let us consider the homogeneous equation

f ′(x) = a(x)f(x), (16)

valid for any value of the argument x of f , and where a is a given function of x. Supposethat f1 is a solution of eq.(16), and suppose that f2 is another solution. We have then

f ′1(x)

f1(x)=f ′

2(x)

f2(x),

such that, after integration,ln |f2(x)| = ln |f1(x)| + k

where k is a constant. Taking the exponential of this, one finds

f2(x) = cf1(x),

where c = ± exp(k), and therefore f2 and f1 are proportional: the set of solutions for theequation (16) is a one-dimensional vector space.For the equation (16), the solution can be derived by using the separation of variables

method, which consists in writing the equation in the form

df

f(x)= a(x)dx,

which, after integration, leads to

ln

(

f(x)

f0

)

=

∫ x

x0

a(u)du,

where f0 = f(x0), such that

f(x) = f0 exp

(∫ x

x0

a(u)du

)

.

Example Consider the equation

f ′(x) = af(x) + b,

where a, b are constants. If one defines g(x) = f(x) + b/a, one sees that g satisfies g′(x) =ag(x) and one can use the previous result to find

f(x) = g0 exp(ax) − b

a,

where g0 = f(0) + b/a is a constant of integration.

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8.2 Variation of parameters method

We consider now the non-homogeneous equation

f ′(x) = a(x)f(x) + h(x), (17)

where h is a given function of x. If we suppose that f1 is a specific solution of eq.(17), wehave

[f(x) − f1(x)]′ = a(x)[f(x) − f1(x)],

such that the general solution of eq.(17) can be written

f(x) = c exp

(∫ x

x0

a(u)du

)

+ f1(x),

where c = f(x0) − f1(x0). In order to find a specific solution f1, one can try

f1 = φ(x) exp

(∫ x

x0

a(u)du

)

,

where φ(x) is a function to be found. Plugging this ansatz into eq.(17), one finds

φ′(x) = h(x) exp

(

−∫ x

x0

a(u)du

)

,

which, after an integration, gives the function φ.

Example Consider the equation

f ′(x) = af(x) + 2xeax,

where a is a constant. The general solution of the homogeneous equation is A exp(ax),and the variation of parameters method consists in finding a specific solution of the formφ(x) exp(ax), which leads to

φ′(x) = 2x.

The general solution is therefore

f(x) = (A + x2) exp(ax).

8.3 Second order, homogeneous

We consider the following differential equation

f ′′(x) + a(x)f ′(x) + b(x)f(x) = 0, (18)

where a, b are functions of x. We will see with several examples that it is possible to finda least two linearly independent solutions f1, f2 of eq.(18). Suppose that f3 is a third

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solution: we show now that, necessarily, f3 is a linear combination of f1 and f2.Proof From eq.(18), we find easily

fi(x)f′′3 (x) − f3(x)f

′′i (x) + a(x)

(

fi(x)f′3(x) − f3f

′i(x)

)

= 0 i = 1, 2,

which, in terms of the Wronskians Wi(x) = fi(x)f′3(x) − f3(x)f

′i(x), read

W ′i (x) + a(x)Wi(x) = 0 i = 1, 2.

These equations can be integrated to give

Wi(x) = Ai exp

(

−∫

a(x)dx

)

i = 1, 2,

and we conclude that

A1

(

f2(x)f′3(x) − f3(x)f

′2(x)

)

= A2

(

f1(x)f′3(x) − f3(x)f

′1(x)

)

.

This equation can be written

f ′3(x)

f3(x)=A1f

′2(x) − A2f

′1(x)

A1f2(x) − A2f1(x),

and leads, after integrating and taking the exponential, to

f3(x) = C1f1(x) + C2f2(x),

where Ci are constants. This shows that f3 is necessarily in the vector space spanned byf1 and f2.

Example Consider the following differential equation

f ′′(x) + 2af ′(x) + bf(x) = 0, (19)

where a, b are constants. In order to find two independent solutions of this equation, weassume the following x-dependence

f(x) = exp(zx),

where z is a constant, which can be complex. This assumption leads to

z2 + 2az + b = 0,

which has the following solutions:

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Page 36: 1st Year Maths Notes

• if a2 > b:z± = −a± k,

where k =√a2 − b. The general solution of the differential equation (19) is then

f(x) = exp(−ax)(

A exp(kx) +B exp(−kx))

= exp(−ax)(

C cosh(kx) +D sinh(kx))

,

where C = A+B and D = A−B are constants.

• if a2 < bz± = −a± ik,

and the general solution is

f(x) = exp(−ax)Re{

A exp(ikx) + B exp(−ikx)}

= exp(−ax)(

A cos(kx) +B sin(kx))

,

where A, B are complex constants, and A =Re{A+ B}, B =Im{B − A}. The latterexpression can also be written

f(x) = f0e−ax cos(kx+ φ0),

where f0 =√A2 +B2 and tanφ0 = −A/B.

• if a2 = b. In this case, z+ = z− and the assumption f(x) = exp(zx) gives one solutiononly, which is exp(−ax). In order to find a second linearly independent solutions ofthe differential equation (19), we assume the form

f(x) = x exp(wx),

where w is a constant, which leads to

2(w + a) + (w2 + 2aw + b)x = 0.

This equation must be valid for any x, such that necessarily

w + a = 0 and w2 + 2aw + b = 0,

for which the only solution is w = −a. Finally, the general solution of the differentialequation (19) is

f(x) = (A+Bx) exp(−ax),where A,B are constants.

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8.4 Second order, non-homogeneous

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 fs is a specific solution of the equation.We have then

(f − fs)′′ + a(x)(f − fs)

′ + b(x)(f − fs) = 0,

and the results derived for a homogeneous differential equation hold for the differencef − fs, such that the general solution of the equation (20) is

f(x) = Af1(x) +Bf2(x) + fs(x),

where A,B are constants, and f1, f2 are linearly independent.

8.5 General properties

In general, the solution of a homogeneous linear differential equation of order n is an n-dimensional vector space, spanned by n linearly independent specific solutions. The nconstants of integration can then be seen as the coordinates of the solutions in the basisof the n linearly independent specific solutions, and their values are given by n boundaryconditions.

8.6 Separation of variables method

We finally give an example of non-linear differential equation, solved by the separation ofvariables method. Consider the following equation,

f ′(x) = x3f 2(x).

This can also be written, when f(x) 6= 0,

df

f 2= x3dx,

such that the left hand side has the variable f only and the right-hand side has the variablex only. Both sides can then be integrated separately, which leads to

−1

f+

1

f0=x4

4,

where f0 = f(0), and the solution is finally

f(x) =f0

1 − f0x4/4

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9 Linear algebra

9.1 Linear function

A linear function l of a variable x satisfies, by definition,

l(ax+ by) = al(x) + 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 Matrices

We have seen in section 6 that the rotation of angle α of the vector of coordinatesu = (u1, u2) in the plane leads to the vector u′ = (u′1, u

′2) with

u′1 = u1 cosα− u2 sinα

u′2 = u1 sinα + u2 cosα.

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 componentsRij , with i, j = 1, 2 such that

R11 = cosα R12 = − sinα R21 = sinα R22 = cosα,

where i represents the line and j represents the row. We have then

u′1 = R11u1 +R12u2

u′2 = R21u1 +R22u2,

which can be written(

u′1u′2

)

=

(

R11 R12

R21 R22

)(

u1

u2

)

,

where the multiplication rule is

u′i =

j=2∑

j=1

Rijuj.

More generally, any linear transformation of a n-dimensional vector u = (u1, ..., un) can bewritten in the form

u′i =

j=n∑

j=1

Mijuj for i = 1, ..., n,

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Page 39: 1st Year Maths Notes

where Mi,j are the components of a matrix M which represents the linear transformation.Besides rotations, other linear transformations can be: projections, scalings, ... as well ascompositions of these.

A matrix S is said symmetric if Sij = Sji, and a matrix A is said antisymmetric ifAij = −Aji. The product of a symmetric matrix with an antisymmetric matrix is zero.

9.3 Determinants

Suppose one has the following system of equations

x′ = ax+ by

y′ = cx+ dy (22)

which can be written(

x′

y′

)

= M

(

xy

)

, with M =

(

a bc d

)

.

One wishes to find (x, y) in terms of (x′, y′), if possible, and therefore the inverse M−1 ofthe linear transformation represented by M:

(

xy

)

= M−1

(

x′

y′

)

.

The system of equations (22) is equivalent to

(ad− bc)x = dx′ − by′

(ad− bc)y = ay′ − cx′, (23)

and leads to the following two cases

• if ad−bc = 0, the previous set of equations is equivalent to dx′ = by′, or ay′ = cx′, suchthat the two equations of the system (22) are equivalent. There is thus an infinityof solutions (x, y), corresponding to the straight line of equation ax + by = x′, orequivalently cx+dy = y′. In this case, the matrix M has no inverse, since there is noone-to-one relation between (x, y) and (x′, y′). A typical example of such a situationis a projection on a given straight line, since all the points on a perpendicular straightline are projected on the same point.

• if ad− bc 6= 0, there is one solution only to the system (23), which is

x =dx′ − by′

ad− bcy =

ay′ − cx′

ad− bc. (24)

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Page 40: 1st Year Maths Notes

Therefore it is essential, in order to find a unique solution to the system of equations (22),and therefore to find an inverse of the matrix M, that the determinant ad− bc of M is notzero.

det M = ad− bc 6= 0,

or in other words: a linear function represented by the matrix M has an inverse, representedby the matrix M−1, if and only if det M 6= 0. From the solution (24), one can see that theinverse of the matrix M is then

M−1 =1

det M

(

d −b−c 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 Composition of linear functions

Given the two linear functions f1 and f2, represented by the matrices M1 and M2, with

M1 =

(

a1 b1c1 d1

)

M2 =

(

a2 b2c2 d2

)

,

we wish to represent the composition of functions

w = f2(v) = f2(f1(u)).

We have, with u = (x, y),

v = M1 · u =

(

v1

v2

)

=

(

a1x+ b1yc1x+ d1y

)

,

and therefore

w = M2 · v =

(

w1

w2

)

=

(

(a1a2 + c1b2)x+ (b1a2 + d1b2)y(a1c2 + c1d2)x+ (b1c2 + d1d2)y

)

.

This can also be writtenw = M2 · M1 · u,

where the product of matrices M = M2 · M1 is defined by

Mi,j =∑

k=1,2

M2 ikM1 kj, i, j = 1, 2

such that

M =

(

a1a2 + c1b2 b1a2 + d1b2a1c2 + c1d2 b1c2 + d1d2

)

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Page 41: 1st Year Maths Notes

Remark In general, the two operations do not commute: f2(f1(u)) 6= f1(f2(u)), and thus

M2M1 6= M1M2.

Determinant of a product The determinant of M = M2M1 is

det M = (a1a2 + c1b2)(b1c2 + d1d2) − (a1c2 + c1d2)(b1a2 + d1b2)

= (a1d1 − b1c1)(a2d2 − b2c2),

such thatdet (M2M1) = det M2 × det M1 = det (M1M2)

The previous properties are also valid for n× n matrices, and the determinant of a matrixis also noted

det

a11 · · · a1n

. . .

an1 · · · ann

=

a11 · · · a1n

. . .

an1 · · · ann

9.5 Eigenvectors and eigenvalues

Given a matrix M, an eigenvector e of M satisfies, by definition,

M · e = λe, with e 6= 0 (25)

where the real number λ is the eigenvalue of M corresponding to e. Therefore the effectof the matrix M on its eigenvector e is simply a rescaling, without change of direction.A n × n matrix, operating on a n-dimensional vector space, can have at most n linearlyindependent eigenvectors. In this case, these vectors can constitute a basis (e1, ..., en), andthe corresponding matrix, in this basis, is diagonal, with the eigenvalues being its diagonalelements:

∆ =

λ1 0 · · · 0 00 λ2 0 · · · 0

. . .

0 · · · 0 λn−1 00 · · · · · · 0 λn

In this case, the determinant is simply the product of the eigenvalues

det ∆ = λ1λ2 · · ·λn

In order to find the eigenvalues of a matrix, the first step is to write the system ofequations (25) in the following way:

[M− λ1] e = 0,

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Page 42: 1st Year Maths Notes

where 1 is the unit matrix. If the corresponding matrix M − λ1 had an inverse, the onlysolution to this system of equations would be e = 0. But if the initial matrix M haseigenvectors, these are not zero, and as a consequence M − λ1 has no inverse. Thereforeits determinant vanishes:

det [M− λ1] = 0.

This determinant is polynomial in λ, and the solutions to this equation give the eigenvalueswhich are expected.

Example For a 2 × 2 matrix, we have

a− λ bc d− λ

= (a− λ)(d− λ) − bc = 0,

such that the eigenvalues λ, if there are, satisfy a quadratic equation.

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10 Functions of several variables

10.1 Partial differentiation

If f is a function of two variables, and associates the value z = f(x, y) to the pair (x, y),one can define the partial derivative of f with respect to x, for a fixed value y, and thepartial derivative of f with respect to y, for a fixed value x. These partial derivatives aredenoted

∂f

∂x= lim

∆x→0

f(x+ ∆x, y) − f(x, y)

∆x∂f

∂y= lim

∆y→0

f(x, y + ∆y) − f(x, y)

∆y.

An important property of partial derivatives concerns their commutativity:

∂2f

∂x∂y=

∂2f

∂y∂x.

Proof From their definition, the partial derivatives satisfy

∂2f

∂y∂x= lim

∆y→0

1

∆y

[

∂f

∂x(x, y + ∆y) − ∂f

∂x(x, y)

]

= lim∆y→0

lim∆x→0

1

∆y∆x[f(x+ ∆x, y + ∆y) − f(x, y + ∆y) − f(x+ ∆x, y) + f(x, y)]

= lim∆x→0

1

∆x

[

∂f

∂y(x+ ∆x, y) − ∂f

∂y(x, y)

]

=∂2f

∂x∂y.

Example For the function f(x, y) = xn cos(ay), where n and a are constants, we have

∂f

∂x= nxn−1 cos(ay)

∂f

∂y= −axn sin(ay),

and of course∂2f

∂y∂x= −anxn−1 sin(ay) =

∂2f

∂x∂y.

Nabla operator One defines the differential operator ∇ as the symbolic vector of com-ponents

∇ =

(

∂x,∂

∂y,∂

∂z

)

,

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Page 44: 1st Year Maths Notes

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:

∇φ(x, y, z) =

(

∂φ

∂x,∂φ

∂y,∂φ

∂z

)

∇ · E(x, y, z) =∂Ex

∂x+∂Ey

∂y+∂Ez

∂z

∇× E(x, y, z) =

(

∂Ez

∂y− ∂Ey

∂z,∂Ez

∂x− ∂Ez

∂x,∂Ey

∂x− ∂Ey

∂x

)

10.2 Differential of a function of several variables

We consider here the example of two variables x, y, and formally use the notations dx anddy for the infinitesimal limits of the increments ∆x and ∆y.If f depends on two variables x, y, the change in f(x, y) has two contributions: one fromthe cange ∆x and one from the change ∆y. A Taylor expansion on both variabls leads to

f(x+ ∆x, y + ∆y) = f(x, y + ∆y) + ∆x∂f

∂x(x, y + ∆y) + O(∆x)2

= f(x, y) + ∆y∂f

∂y(x, y) + ∆x

∂f

∂x(x, y) + O

(

(∆x)2, (∆y)2,∆x∆y)

such that, if we note ∆f = f(x+ ∆x, y + ∆y) − f(x, y), we have

∆f = ∆x∂f

∂x+ ∆y

∂f

∂y+ · · · , (26)

where dots represent higher orders in ∆x and ∆y. In the limit where ∆x → 0 and ∆y → 0,we obtain the definition of the differential

df =∂f

∂xdx+

∂f

∂ydy, (27)

which is an exact identiy, and can be interpreted as a vector in a two dimensional vectorspace spanned by dx and dy, with coordinates ∂f/∂x and ∂f/∂y.

Remark It is important to distinguish the symbols for partial and total derivatives. In-deed, in eq.(27), if y is a function of x, one can consider the function F (x) = f(x, y(x)),which, using the chain rule, has the following derivative

F ′(x) =df

dx=

∂f

∂x+dy

dx

∂f

∂y

=∂f

∂x+ y′(x)

∂f

∂y.

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Page 45: 1st Year Maths Notes

As a consequence,∂f

∂x6= df

dx

Finally, if a function depends on N variables, one can define the partial derivatives withrespect to any of these variables, and these partial derivatives will commute among eachother.

10.3 Implicit functions

If the variables x, y, z are related by an equation of the form g(x, y, z) = 0, where g is adifferentiable function, on can define each variable as a function of the other two (besidespossible singular points). We can show then that

(

∂x

∂y

)

z

=

(

∂y

∂x

)−1

z

,

where the variable in subscript represents the one kept constant in the differentiation.Proof Since g(x, y, z) = 0, we have

dg =∂g

∂xdx+

∂g

∂ydy +

∂g

∂zdz = 0,

and if we consider the case z= constant, we have dz = 0, such that

(

∂x

∂y

)

z

=dx

dy

dz=0= −∂g

∂y

/∂g

∂x=

(

dy

dx

dz=0

)−1

=

(

∂y

∂x

)−1

z

Another important property is(

∂x

∂y

)

z

(

∂y

∂z

)

x

(

∂z

∂x

)

y

= −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 Double integration

If f is a function depending on two variables x, y, one can define the function F (x) as

F (x) =

∫ d

c

f(x, y)dy,

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Page 46: 1st Year Maths Notes

and then the integral of F over an interval [a, b]

I1 =

∫ b

a

F (x)dx =

∫ b

a

(∫ d

c

f(x, y)dy

)

dx =

∫ b

a

dx

∫ d

c

dy f(x, y).

The product dxdy represents an infinitesimal surface are in the plane (0, x, y), and theintegral is thus the volume between the rectangular area of surface |b − a||d − c| and thesurface defined by z = f(x, y).In the simple case where f is a product f(x, y) = φ(x)ψ(y), the latter integral is just aproduct of integrals

I1 =

∫ b

a

dx

∫ d

c

dy φ(x)ψ(y) =

(∫ b

a

φ(x)dx

)

×(∫ d

c

ψ(y)dy

)

.

More generally, one can define a double integral over any area D which is not rectangularby

I2 =

∫ ∫

Df(x, y)dxdy

In this case, one can perform the integrals in whichever order: first over x and then overy or the opposite:

I2 =

∫ x2

x1

(

∫ y2(x)

y1(x)

f(x, y)dy

)

dx,

where the values y1(x), y2(x) are the boundaries of the domain D for a given value of x, or

I2 =

∫ y2

y1

(

∫ x2(y)

x1(y)

f(x, y)dx

)

dy,

where the values x1(y), x2(y) are the boundaries of the domain D for a given value of y.

Example Calculate the volume of a pyramid whose base is an equilateral triangle of sidesa , 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, suchthat the axises (Ox), (Oy), (Oz) are along the edges which meet orthogonally. The base isthen perpendicular to the vector (1, 1, 1), and intersects the previous edges at the distancea/

√2 from the top. Its equation is thus x+ y + z = a/

√2. The volume is then

V1 =

∫ a/√

2

0

dy

∫ a/√

2−y

0

(

a√2− x− y

)

=

∫ a/√

2

0

dy1

2

(

a√2− y

)2

=a3

12√

2. (28)

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Page 47: 1st Year Maths Notes

dr

rdΘ

Figure 9: The infinitesimal surface area in polar coordinates is rdrdθ

Double integral in polar coordinates The infinitesimal surface area in polar coordi-nates is dr × rdθ (see fig(9)), and an integral over a domain D is thus

J =

∫ ∫

Df(r, θ)rdrdθ.

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 + y2 + z2 = R2.This volume of half the ball is then

V2 =

∫ ∫

Cz(x, y)dxdy =

∫ ∫

C

R2 − x2 − y2dxdy,

where C is the disc of radius R, centered in the origin. A change of variables to polarcoordinates gives

V2 =

∫ R

0

rdr

∫ 2π

0

dθ√R2 − r2

= 2π

∫ R

0

rdr√R2 − r2

= 2π

[

−1

3(R2 − r2)3/2

]R

0

=2π

3R3.

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Page 48: 1st Year Maths Notes

dr rd Θ

φ

Θ

r sin dΘ φ

x

y

z

Figure 10: The infinitesimal volume in spherical coordinates is r2dr sin θdθdφ

10.5 Triple integration

Triple integration is a straightforward generalization of double integration, and it cansometimes be useful to use spherical coordinates (r, θ, φ), if the function to integrate isexpressed in terms of spherically symmetric quantities. Using these coordinates, the in-finitesimal volume is rdθ× r sin θdφ×dr = r2dr sin θdθdφ (see fig.10), and an integral overa three-dimensional domain is then

∫ ∫ ∫

Dr2dr sin θdθdφ f(r, θ, φ).

Example The volume of half a solid ball of radius R, which was calculated before, is easierto calculate using spherical coordinates, and is

V2 =

∫ R

0

r2dr

∫ π/2

0

dθ sin θ

∫ 2π

0

=R3

3× [− cos θ]

π/20 × 2π

=2π

3R3

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