MBEYA UNIVERSITY OF SCIENCE AND TECHNOLOGY

COLLEGE OF SCIENCE AND TECHNICAL EDUCATION

DEPARTMENT OF MATHEMATICS AND STATISTICS

SEMESTER II ACADEMIC YEAR 2023/2024
MS 8102/MS 8122: ADVANCED CALCULUS
UQF 8 FIRST YEAR: CIVIL AND ELECTRICAL ENGINEERING
LECTURE 1
PREPARED BY: MOHAMED H. MOHAMED

INTRODUCTION
CONCEPTS:

1) This module covers theree topics which are:

(i) Fourier Series

(ii) Fourier Integrals and Transforms
(iii) Line and Multiple Integrals

2) We shall begin with the topic of Fourier Series which discusses the following concepts:

(a) Periodic Functions

(b) Trigonometric Series
(c) Fourier Coefficients and Fourier Series Representation
(d) Even and Odd Fourier Series Representations
(e) Harmonic Fourier Series
(f) Complex Exponential Form of Fourier Series

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 3) The topic of Fourier Integrals and Transforms covers the following:
(a) Derivation of Fourier Integrals
(b) Cosine and Sine Fourier Integrals
(c) Derivation of Fourier Transforms
(d) Cosine and Sine Fourier Transforms
(e) Laplace Transforms and its Applications
4) And the topic of Line and Multiple Integrals covers the following:
(a) Evaluation of Line Integrals in Cartesian and Polar Coordinates
(b) Evaluation of Multiple Integrals Using Curvilinear Coordinates
5) Now, here we go:

1.0: FOURIER SERIES

CONCEPTS:
1) Fourier series are infinite series that represent periodic functions in terms of cosines and
sines.
2) Fourier series are of greatest importance to the engineers, physicists and applied
mathematicians because they allow the solutions of Ordinary Differential Equa-
tions (ODEs) and the approximation of periodic functions.
3) Morever, Fourier series are more universal than the familiar Taylor series in Calculus since,
many dis continuous periodic functions that come up in applications can be developed in
Fourier series but do not have Taylor series expansions.
4) To define Fourier series, we first need some background material. Now, here we go:

1.1: Periodic Functions

1) Definition 1: (Periodic Function)

⇒ A function f (x) is said to be periodic if for all values of x and some positive number
T called the Period, we have;
f (x + T ) = f (x)

2) T is the interval between two successive repetitions and is called the Period of f (x).
That is, the following is a graph of periodic function with period T .

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 ⇒ Other examples of graphs representing periodic functions are shown below:

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 3) The graph of a periodic function has the characteristic that it can be obtained by periodic
repetition of its graph in any interval of length T .

4) The smallest positive period is often called the Fundamental Period.

5) (i) The following are examples of periodic functions each with period 2π:
(a) f (x) = 1
(b) f (x) = cos x
(c) f (x) = sin x
(ii) The following are examples of functions that are not periodic (nonperiodic functions):
(a) f (x) = x
(b) f (x) = x2
(c) f (x) = ex
(d) f (x) = cosh x
(e) f (x) = ln x

6) A constant function f (x) = c is periodic with any period, where c, is a constant number.

7) If f (x) has period T , it also has the period 2T because the equation f (x + T ) = f (x)
implies:

f (x + 3T ) = f (x + T + T + T ) = f ([x + T + T ] + T )
= f (x + T + T ) = f ([x + T ] + T )
= f (x + T )
= f (x)
⇒ f (x + 3T ) = f (x)

Therefore, for any integer n = 1, 2, 3, ... we have;

f (x + nT ) = f (x),

for all x.

8) If f (x) and g(x) have period T , then af (x) + bg(x) with any constants a and b also has
the period T .

9) The two periodic functions that most of us are familiar are sine and cosine and in fact
we will be using these two functions regularly in the remaining sections.

EXAMPLE

1. Show that each of the following functions is periodic with the period 2π.

(a) f (x) = cos x

(b) f (x) = sin x

SOLUTION

⇒ A function f (x) is periodic with period T if f (x + T ) = f (x) for all x.

⇒ Since, in each case T = 2π, thus we have;

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 (a) f (x) = cos x ⇒ f (x + T ) = f (x + 2π) = cos(x + 2π) = cos x cos 2π − sin x sin 2π. But,
cos 2π = 1 and sin 2π = 0, thus; f (x+T ) = f (x+2π) = cos(x+2π) = cos x(1)−sin x(0) =
cos x = f (x), hence shown.

(b) f (x) = sin x ⇒ f (x + T ) = f (x + 2π) = sin(x + 2π) = sin x cos 2π + cos x sin 2π. But,
cos 2π = 1 and sin 2π = 0, thus; f (x+T ) = f (x+2π) = sin(x+2π) = sin x(1)+cos x(0) =
sin x = f (x), hence shown.

1.2: Trigonometric Series

CONCEPTS:

1) Our problem in the first few sections of this topic will be the representation of various
functions f (x) of period 2π in terms of the simple functions

1, cos x, sin x, cos 2x, sin 2x, ..., cos nx, sin nx. (1)

2) All of these functions have the period 2π. They form a system called Trigonometric
System.

3) Definition 2: (Trigonometric Series)

⇒ A series of the form
∞
X
a0 + a1 cos x + b1 sin x + a2 cos 2x + b2 sin 2x + ... = a0 + (an cos nx + bn sin nx) (2)
n=1

is called the Trigonometric Series, where a0 , a1 , b1 , a2 , b2 , ... are constants, called the
Coeffients of the series.

4) Each term of the series (2) has the period 2π. If the coefficients are such that the series
converges, then its sum will be a function of period 2π.

5) Defintion 3: (Fourier Series)

⇒ Suppose that f (x) is a given function of period 2π and is such that it can be represented
by a series (2), that is, (1) converges, has the sum f (x), then the series
∞
X
f (x) = a0 + (an cos nx + bn sin nx)
n=1

is called the Fourier Series of f (x).

1.3: Fourier Coefficients and Fourier Series Representa-

tion
CONCEPTS:

1) Theorem 1: (Orthogonality of the Trigonometric System)

⇒ The trigonometric system (1) is orthogonal on the interval −π ≤ x ≤ π (hence also
on 0 ≤ x ≤ 2π or any other interval of length 2π because of periodicity);

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 ⇒ That is, the integral of the product of any two functions in (1) over that interval is 0,
so that for any integers n and m, we have;
Z π
cos nx cos mxdx = 0 (n 6= m)
−π
Z π
sin nx sin mxdx = 0 (n 6= m)
−π
Z π
sin nx cos mxdx = 0 (n 6= m or n = m)
−π

Proof:
⇒ The proof follows simply by transforming the integrands trigonometrically from prod-
ucts into sums. That is,
Z π
1 π 1 π
Z Z
cos nx cos mxdx = cos(n + m)xdx + cos(n − m)xdx and
−π 2 −π 2 −π
Z π
1 π 1 π
Z Z
sin nx sin mxdx = cos(n − m)xdx − cos(n + m)xdx
−π 2 −π 2 −π

⇒ Since, m 6= n (integer!), the integrals on the right are all 0. And for all integers m and
n (without exception), we have;
Z π
1 π 1 π
Z Z
sin nx cos mxdx = sin(n + m)xdx + sin(n − m)xdx
−π 2 −π 2 −π
=0+0
=0 Hence proved.

2) Definition 4: (Fourier Coefficients/Euler Formulas)

⇒ The Fourier coefficients a0 , an and bn of the series

∞
X
f (x) = a0 + (an cos nx + bn sin nx),
n=1

are given by the Euler Formulas:

Z π
1
a0 = f (x)dx
2π −π
1 π
Z
an = f (x) cos nxdx; for n = 1, 2, 3, ...
π −π
1 π
Z
bn = f (x) sin nxdx; for n = 1, 2, 3, ...
π −π

3) That is, from the Fourier Series representation of the function f (x)
∞
X
f (x) = a0 + (an cos nx + bn sin nx) (3)
n=1

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 ⇒ Integrating on both sides of (3), from −π to π, we get;
Z π Z π" ∞
X
#
f (x)dx = a0 + (an cos nx + bn sin nx) dx
−π −π n=1

⇒ Integrating term by term, we obtain;

Z π Z π ∞ 
X Z π Z π 
f (x)dx = a0 dx + an cos nxdx + bn sin nxdx
−π −π n=1 −π −π

⇒ Whereby, for n = 1, 2, ... we have;

Z π Z π
cos nxdx = 0 and sin nxdx = 0
−π −π
∞ 
X Z π Z π 
Thus an cos nxdx + bn sin nxdx = 0 and
n=1 −π −π
Z π Z π
f (x)dx = a0 dx = 2πa0
−π −π

Z π
1
Therefore a0 = f (x)dx. (4)
2π −π

⇒ Multiplying on both sides of (3) by cos mx with any fixed positive integer m and
integrating from −π to π, we have;
Z π Z π" ∞
X
#
f (x) cos mxdx = a0 + (an cos nx + bn sin nx) cos mxdx
−π −π n=1

⇒ Integrating term by term, we have;

Z π
a0 cos mxdx = 0
−π
Z π (
am π; for n = m
an cos nx cos mxdx =
−π 0; for n 6= m.
Z π
and bn sin nx cos mxdx = 0; for all n and m.
−π

Z π
1
Therefore am = an = f (x) cos nxdx (5)
π −π

⇒ Multiplying on both sides of (3) by sin mx with any fixed positive integer m and
integrating from −π to π, we have;
Z π Z π ∞
X
f (x) sin mxdx = [a0 + (an cos nx + bn sin nx)] sin mxdx
−π −π n=1

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 ⇒ Integrating term by term, we have;
Z π
a0 sin mxdx = 0
−π
Z π
an cos nx sin mxdx = 0
−π
Z π (
bm π for n = m
and bn sin nx sin mxdx =
−π 0 for n 6= m.

Z π
1
Therefore bm = bn = f (x) sin nxdx (6)
π −π

1.3.1: Transition From Period 2π to any period 2L

CONCEPTS:

1) The transition from period 2π to be period T = 2L is effected by a suitable change of

scale as follows:

⇒ Let f (x) have period T = 2L, then we can introduce a new variable v such that f (x),
as a function of v, has period 2π. If we set,
T 2π π
x= v, so that v = x= x
2π T L
then v = ±π corresponds to x = ±L.
⇒ This means that, f as a function of v, has period 2π, and therefore, a Fourier series of
the form
∞
L X
f (x) = f ( v) = a0 + (an cos nv + bn sin nv)
π n=1

with coefficients given by;

Z π
1 L
a0 = f ( v)dv
2π −π π
1 π L
Z
an = f ( v) cos nvdv
π −π π
1 π L
Z
bn = f ( v) sin nvdv
π −π π

2) We could use these formulas directly, but the change to x simplifies calculations. Since,
π π
v= x, we have dv = dx
L L
and integrate over x from −L to L, consequently we obtain for a function f (x) of period
2L the Fourier Series,
∞
X nπ nπ
f (x) = a0 + (an cos x + bn sin x) (7)
n=1
L L

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 with the Fourier coefficients given by;
Z L
1
a0 = f (x)dx
2L −L
1 L
Z
nπ
an = f (x) cos xdx; for n = 1, 2, 3, ...
L −L L
1 L
Z
nπ
bn = f (x) sin xdx; for n = 1, 2, 3, ...
L −L L

3) We continue to call (7) with any coefficients a Fourier Series, and we can integrate over
any other interval of length T = 2L.

EXAMPLES

1. Given that
(
−k; if − π < x < 0
f (x) = and f (x + 2π) = f (x).
k; if 0 < x < π,

(a) Find the Fourier coefficients of f (x).

(b) Write the Fourier series of f (x).
(c) Write the first two Partial sums of f (x).

SOLUTION

(a) Since f (x) has a period T = 2π, then the Fourier coefficients are given by the Euler
formulas:
Z π
1
a0 = f (x)dx
2π −π
1 π
Z
an = f (x) cos nxdx; for n = 1, 2, 3, ...
π −π
1 π
Z
bn = f (x) sin nxdx; for n = 1, 2, 3, ...
π −π

⇒ Thus, we have;
Z π Z 0 Z π
1 1
a0 = f (x)dx = ( (−k)dx + kdx)
2π −π 2π −π 0
k k
= [−(0 − −π) + (π − 0)] = (−π + π)
2π 2π
a0 = 0.

1 π 1 0
Z Z Z π
an = f (x) cos nxdx = [ (−k) cos nxdx + k cos nxdx]
π −π π −π 0
1 −k sin nx 0 k sin nx π
= [ |−π + |0 ]
π n n
an = 0.

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 1 π 1 0
Z Z Z π
bn = f (x) sin nxdx = [ (−k) sin nxdx + k sin nxdx]
π −π π −π 0
1 k cos nx 0 k cos nx π
= [ |−π − |0 ]
π n n
⇒ Since, cos(−x) = cos x and cos 0 = 1, this gives;

k 2k
bn = [cos 0 − cos(−nπ) − cos nπ + cos 0] = (1 − cos nπ)
nπ nπ
⇒ Since, cos π = −1, cos 2π = 1, cos 3π = −1, e.t.c, in general, we have;

cos nπ = (−1)n

⇒ And thus we have;

2k
bn = (1 − (−1)n )
nπ

(b) The Fourier series of f (x) is given by;

∞
X
f (x) = a0 + (an cos nx + bn sin nx)
n=1

2k
⇒ Where a0 = 0, an = 0 and bn = nπ
(1 − (−1)n ), so the Fourier series of f (x)
becomes;
∞
X 2k
f (x) = 0 + (0 + [ (1 − (−1)n )] sin nx)
n=1
nπ
∞
(1 − (−1)n )
  
2k X
= sin nx
π n=1 n
 
2k 2 2
= 2 sin x + 0 + sin 3x + 0 + sin 5x + · · ·
π 3 5
 
4k 1 1
f (x) = sin x + sin 3x + sin 5x + · · ·
π 3 5

(c) The first two partial sums of f (x) are:

4k
S1 = sin x
π
4k 1
S2 = (sin x + sin 3x).
π 3

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 2. Find the Fourier series of the function f (x) given by:

0; if − 2 < x < −1

f (x) = k; if − 1 < x < 1 T = 2L = 4, L = 2.

0; if 1 < x < 2,


SOLUTION

⇒ Since f (x) has period T = 2L = 4 → L = 2, thus its Fourier Series is given by:
∞
X nπ nπ
f (x) = a0 + (an cos x + bn sin x)
n=1
L L

with the Fourier coefficients given by;

Z L
1
a0 = f (x)dx
2L −L
1 L
Z
nπ
an = f (x) cos xdx; for n = 1, 2, 3, ...
L −L L
1 L
Z
nπ
bn = f (x) sin xdx; for n = 1, 2, 3, ...
L −L L

⇒ Thus, we have;
Z L
1
a0 = f (x)dx, L = 2,
2L −L
1 2 1 −1
Z Z Z 1 Z 2
⇒ a0 = f (x)dx = [ 0dx + kdx + 0dx]
4 −2 4 −2 −1 1
1
= [0 + 2k + 0]
4
k
⇒ a0 =
2Z
1 2 1 1
Z
nπ nπ
an = f (x) cos xdx = k cos xdx
2 −2 2 2 −1 2
2k nπ
= sin
nπ
 2

nπ 0; if n is even

But, sin = 1; if n = 1, 5, 9, ...
2 
−1; if n = 3, 7, 11, ...,


0; if n is even

2k
⇒ an = nπ ; if n = 1, 5, 9, ...

 2k
− nπ ; if n = 3, 7, 11, ...,

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 1 1
Z
nπ
bn = k sin xdx
2 −1 2
1 −2k nπ 1
= [( cos x)|−1 ]
2 nπ 2
−k nπ 1 −k nπ 1
=( cos x)|−1 = (cos x| )
nπ 2 nπ 2 −1
−k nπ nπ
= (cos − cos )
nπ 2 2
= 0 for n = 1, 2, ...

⇒ Hence from the formula

∞
X nπ nπ
f (x) = a0 + (an cos x + bn sin x)
n=1
L L

then the Fourier series of f (x) is:

∞
k X 2k nπ nπ
f (x) = + ( sin cos x + 0)
2 n=1 nπ 2 2
∞  
k 2k X 1 nπ nπ
= + sin cos x
2 π n=1 n 2 2
k 2k π 1 3π 1 5π
f (x) = + (cos x − cos x + cos x − + ...)
2 π 2 3 2 5 2

3. Given that f (x) = x + x2 for −π < x < π:

(a) Find the Fourier series of f (x).
(b) By putting x = π and x = −π in the Fourier series of f (x), deduce that:
∞
π2 X 1
=
6 n=1
n2

SOLUTION
∞
(a) ⇒ Let f (x) = x + x2 = a0 +
P
(an cos nx + bn sin nx). And since f (x) is 2π periodic,
n=1
then we have;
Z π
1
a0 = f (x)dx
2π −π
Z π
1 1 x2 x3 π
= (x + x2 )dx = ( + )|−π
2π −π 2π 2 3
 2 3 2 3

1 π π π π
= ( + )−( − )
2π 2 3 2 3
 2 3 2 3
1 2π 3
  
1 π π π π
= + − + =
2π 2 3 2 3 2π 3
π2
⇒ a0 =
3

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 1 π
Z
an = f (x) cos nxdx
π −π
1 π
Z
⇒ an = (x + x2 ) cos nxdx
π −π
Rπ
⇒ Integrating the integral −π (x + x2 ) cos nxdx by parts, we have from
R R
udv = uv − vdu, that Rlet u = (x + x2 ) and dv = cos nxdx, so that
du = (1 + 2x)dx and v = cos nxdx = n1 sin nx. Thus we have;

1 π
Z
an = (x + x2 ) cos nxdx
π −π
 Z 
1 1 2 1
= (x + x ) sin nx − (1 + 2x) sin nxdx
π n n
R
⇒Again,
R integrating
R the integral (1 + 2x) sin nxdx by parts, we have from
udv = uv − vdu, R that let u = (1 + 2x) and dv = sin nxdx, so that
du = 2dx and v = sin nxdx = − n1 cos nx. Thus we have;
Z Z
1 1
(1 + 2x) sin nxdx = − (1 + 2x) cos nx − − cos nx(2)dx
n n
Z
1 2
= − (1 + 2x) cos nx + cos nxdx
n n
Z
1 2
⇒ (1 + 2x) sin nxdx = − (1 + 2x) cos nx + 2 sin nx
n n
 Z 
1 1 2 1
an = (x + x ) sin nx − (1 + 2x) sin nxdx
π n n
 π
1 1 2 1 1 2
= (x + x ) sin nx − [− (1 + 2x) cos nx + 2 sin nx]
π n n n n −π
 π
1 1 2 1
= [ (x + x2 ) − 3 ] sin nx + 2 [(1 + 2x)] cos nx]
π n n n −π
   
1 1 1
= 0 + 2 (1 + 2π) cos nπ − 0 + 2 (1 − 2π) cos nπ
π n n
cos nπ
= [(1 + 2π) − (1 − 2π)]
πn2
cos nπ 4π cos nπ 4π(−1)n
= [1 + 2π − 1 + 2π)] = =
πn2 πn2 πn2
n
4(−1)
⇒ an = 2
Zn π
1 π
Z
1
bn = f (x) sin nxdx = (x + x2 ) sin nxdx
π −π π −π

⇒ Integration by parts, gives;

−2(−1)n 2(−1)n+1
bn = =
n n

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 ∞
⇒ Put the values of a0 , an and bn into f (x) = x+x2 = a0 +
P
(an cos nx+bn sin nx),
n=1
we get;
∞ 
π 2 X 4(−1)n 2(−1)2

2
x+x = + 2
cos nx − sin nx
3 n=1
n n
2 ∞  ∞ 
(−1)n (−1)2
 
π X X
= +4 cos nx − 2 sin nx
3 n=1
n2 n=1
n
π2
 
2 1 1
⇒x+x = + 4 − cos x + 2 cos 2x − 2 cos 3x + · · ·
3 2 3
 
1 1
− 2 − sin x + sin 2x − sin 3x + · · · (∗)
2 3

π2 P∞ 1
(b) ⇒ To deduce, = 2
, put x = π into (∗), we get;
6 n=1 n

π2
 
2 1 1 1
π+π = + 4 1 + 2 + 2 + 2 + ··· (∗∗)
3 2 3 4

⇒ Now, put x = −π into (∗), we get;

π2
 
2 1 1 1
−π + π = + 4 1 + 2 + 2 + 2 + ··· (∗ ∗ ∗)
3 2 3 4

⇒ Thus, adding the equations (∗∗) and (∗ ∗ ∗), we get;

π2 π2
   
2 1 1 1 1 1 1
2π = + 4 1 + 2 + 2 + 2 + ··· + + 4 1 + 2 + 2 + 2 + ···
3 2 3 4 3 2 3 4
2
 
2π 1 1 1
= + 8 1 + 2 + 2 + 2 + ···
3 2 3 4
2π 2
2π 2 −
3 = 1 + 1 + 1 + 1 + ···
8 22 32 42
∞
π2 X 1
=
6 n=1
n2

0; if − 2 < x < −1,

4. (a) Obtain a Fourier series for f (x) = 1 if − 1 < x < 1,

0; if 1 < x < 2.


(b) By using the Fourier series obtained in (a) above, deduce a series
π
for at the point x = 0.
4

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 SOLUTION

(a) ⇒ Since f (x) has period of 2L = 4 → L = 2, thus we have;

∞
X nπ nπ
f (x) = a0 + (an cos x + bn sin x)
n=1
L L

with the Fourier coefficients given by;

Z L
1
a0 = f (x)dx
2L −L
1 L
Z
nπ
an = f (x) cos xdx
L −L L
1 L
Z
nπ
bn = f (x) sin xdx
L −L L

⇒ Thus, we have;

1 1
Z
1 1
a0 = 1dx = x|1−1 = (1 − −1)
4 −1 4 4
1
⇒ a0 =
2Z
1 1 nπ 2 nπ 1
an = 1 cos xdx = sin x|
2 −1 2 2nπ 2 −1
1 nπ nπ 1 nπ nπ
= [sin − sin(− )] = [sin + sin ]
nπ 2 2 nπ 2 2
2 nπ
⇒ an = sin
nπZ 2
1 1 nπ 2 nπ 1
bn = 1 sin xdx = − cos x|
2 −1 2 2nπ 2 −1
1 nπ nπ 1 nπ nπ
= − [cos − cos(− )] = [cos − cos ]
nπ 2 2 nπ 2 2
⇒ bn =0
∞
(an cos nπ x + bn sin nπ
P
⇒ Put the values of a0 , an and bn into f (x) = a0 + L L
x), we
n=1
get;
∞  
1 X 2 nπ nπ
f (x) = + sin cos x+0
2 n=1 nπ 2 2
∞  
1 2X 1 nπ nπ
= + sin cos x
2 π n=1 n 2 2
 
1 2 π 1 3π 1 5π 1 7π
⇒ f (x) = + cos x − cos x + cos x − cos x + · · ·
2 π 2 3 2 5 2 7 2

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1 2 π 1 3π 1 5π 1 7π
(b) ⇒ Put x = 0 into f (x) = + cos 2 x − cos 2 x + cos 2 x − cos 2 x + · · · ,
2 π 3 5 7
we get;
 
1 2 π 1 3π 1 5π 1 7π
f (0) = + cos (0) − cos (0) + cos (0) − cos (0) + · · ·
2 π 2 3 2 5 2 7 2
 
1 2 1 1 1
1= + 1 − + − + ···
2 π 3 5 7
 
4 1 1 1
2=1+ 1 − + − + ···
π 3 5 7
 
4 1 1 1
1= 1 − + − + ···
π 3 5 7
∞
π 1 1 1 X 1
= 1 − + − + ··· = (−1)n+1
4 3 5 7 n=1
n
∞
π X (−1)n+1
=
4 n=1
n
(
0; − 5 < x < 0,
5. (a) Given that f (x) =
3; 0 < x < 5.
(i) Find the Fourier coefficients of f (x).
(ii) Write the corresponding Fourier series.
(b) Deduce a series for π4 at the point x = π2 from the Fourier series expansion of
f (x) = −1 if −π < x < 0 and f (x) = 1 if 0 < x < π.

EXERCISE

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