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Unit VIII
Fourier Series
Unit VIII
Fourier Series
Fourier series are series of cosine and sine terms. Since sine and cosine are periodic functions
we shall study Trigonometric series representation of what we call periodic functions.
7.1 Periodic Functions; Trigonometric Series.
Definition 7.1 A function f is said to be periodic if there is some positive number p
such that:
f (x + p) = f (x) for all x  dom. f.
The number p is called the period of f and the smallest number p is called the primitive
(or fundamental) period of f.
Example 1 Let f (x) = sin x.
Now dom. f = , f (x + p) = f (x) x 
 sin (x + p) = sin x x  
 sin x cos p + sin p cos x = sin x x  
 sin x (1  cos p) = sin p cos x x  
If x =

, then 1  cos p = 0.
2
Now 1  cos p = 0  p = 2n ,where n  N.
Therefore, 2n, where n  N is a period of f and 2 is the primitive period of f.
Example 2 Let g (x) = cos x.
Now dom. g = , g (x + p) = g (x) x 
 cos (x + p) = cos x x  
 cos x cos p  sin p sin x = cos x x  
 cos x (cos p  1) = sin p sin x x  
If x = , then cos p  1= 0.
Now cos p  1= 0  p = 2n, where n  N.
Therefore 2n, where n  N is a period of g and 2 is the primitive period of g.
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Example 3. Let f (x) = tan x.
Now dom. f =x   x  (2n + 1)

, n  Z.
2
f (x + p) = f (x) x dom. f.  tan (x + p) = tan x x  dom. f.

tan x  tan p
= tan x x  dom. f.
1  tan x tan p

tan x + tan p = tan x  tan p tan 2 x x  dom. f.

tan p (1 + tan 2 x) = 0 x  dom. f.

tan p = 0 x  dom. f.
 p = n, where n  N.
Therefore n, where n  N is a period of f and  is the primitive period of f.
Example 4. Let f (x) = sin (x), where   0.
Now dom. f = .
f (x + p) = f (x) x  .
 sin (x + p) = sin x x  .
 sin x cos p + sin p cos x = sin x x  .
 sin x (1  cos p) = sin p cos x x  .
If x =

, then 1  cos p = 0.
2
Now 1  cos p = 0  p = 2n, where n  N.
Therefore
2n 

, where n  N is a period of f and
2

is the primitive period of f.
Graphs of Periodic Functions.
The graph of a periodic function of period p is obtained by periodic representation of its
graph in any interval of length p.
Example 1. Sketch the graph of f (x) = tan x.
Solution. The primitive period of f is .
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Example 2 Sketch the graph of g (x) =  sin x .
Solution Dom. g. = .
Now g (x + p) = g (x) x  .
 sin (x + p) =  sin x  x  .
 sin (x + p) = sin x x   or sin (x + p) =  sin x x  .
 p = 2n, where n  N or sin x (1 + cos p) =  sin p cos x x  .
 p = 2n, where n  N or p = (2n 1), where n  N
 p = n, where n  N.
Therefore  is the primitive period of g.
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Example 3. Sketch the graph of f, where
 1
f (x) = 
 1
if 1  x  0
if 0  x  1
and f (x + 2) = f (x) x  .
Properties of Periodic Functions.
1. If f is a periodic function with period p, then for any n  N np is a period of f.
Proof: (By the principle of mathematical induction)
Since f (x + n p) = f (x) it is true for n = 1.
Assume for n = k, i.e. f (x + k p) = f (x).
We need to show for n = k + 1, i.e. f (x + (k + 1) p) = f (x).
f (x + (k + 1) p) = f (x + k p + p) = f (x + p + k p) = f (x + p) = f (x).
Therefore for any n  N, np is a period of f.
2. If f and g are periodic functions with period p, then for any a, b  
h (x) = a f(x) + b g(x) x  dom. h
is a periodic function of period p.
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Proof: h (x + p) = a f(x + p) + b g(x + p)
= a f(x) + b g(x)
= h (x) x  dom. h.
Therefore h is a periodic function with period p.
3. If f (x) is a periodic function with period p, then for any b   and b  0
a) f (b x) is periodic with period
b) If (
p
.
b
x
) is periodic with period  b  p.
b
Trigonometric Series.
Definition 1.2 The system of functions:
1, cos x, sin x, cos 2x, sin 2x, cos 3x, sin 3x, …
is called a trigonometric system.
The series

a0 +
 (an cos nx  bn sin nx)
n 1
where a0, an’s and bn’s are constants is called Trigonometric series, the
constants are called coefficients.
Definition 1.3 ( Orthogonality of the Trigonometric System)
1. Let f and g be functions defined on some interval a, b. f and g are said to be
orthogonal on a, b if and only if
b
 f ( x) g ( x) dx = 0.
a
2. The set of functions n ( x) : n  N  is said to be orthogonal on a, b if and
b
only if  n ( x)m ( x) dx = 0 for n  m.
a
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Example 1. Show that the set of functions n ( x) : n ( x)  sin nx, where n  N  form an
orthogonal set on  , .
Solution. Let n, m  N and n  m.
n (x ) = sin nx.




 n ( x)m ( x) dx =
 sin nx sin mx dx

1
=  cos (n  m) x  cos (n  m) xdx
2 

 
1  sin (n  m) x
sin (n  m) x
= 


2  nm
nm


= 0.
Therefore n ( x) : n ( x)  sin nx, where n  N  form an orthogonal set on  , .
Example 2. Show that the trigonometric system 1, cos nx, sin nx, … , where n  N,
form an orthogonal set on  , .
Solution. We need to show that:
a)
 sin nx
dx =
 cos nx
dx =
 sin nx cos mx dx = 0 for n, m  N and m = n.



b)









 sin nx cos mx dx =
 sin nx sin mx dx =
 cos nx cos mx dx = 0 where n, m  N
and m  n.
When m = n.




1
1
 cos nx
=  sin nx dx =  cos nx dx = sin nx
= 0,
n
n
  




and
 sin nx cos nx dx =



1
1
=

= 0.
sin
2
nx
dx
cos 2nx
2 
4n

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When m  n.


1
 sin nx cos mx dx = 2  sin (n  m) x  sin (n  m) xdx


 
 cos (n  m) x 
cos (n  m) x


 = 0.
nm
nm



1
=
2


1
 sin nx sin mx dx = 2  cos (n  m) x  cos (n  m) xdx



 
1  sin (n  m) x
sin (n  m) x
= 

 = 0.
2  nm
nm




1
 cos nx cos mx dx = 2  cos (n  m) x  cos (n  m) xdx


=

 
1  sin (n  m) x
sin (n  m) x


 = 0.
2  nm
n

m


Therefore 1, cos nx, sin nx, … for n  N form an orthogonal set on  , .
Definition 1.4
1. A function f (x) is said to be normal or normalized in a, b if and only if
b
  f ( x)
a
2
dx = 1
The norm of a function f (x) denoted by f (x) on a, b is defined by:
b
f (x) =
  f ( x)
a
2
dx
2. An orthogonal set of functions n (x): n  N is said to be an orthonormal
set on a, b if and only if
b
a n ( x)
2
dx = 1  n  N.
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Example 1. The set of functions n (x): n (x) =
1

sin nx, n  Nis an orthonormal
set on  , .
Solution. Let m, n  N and m  n.


1
cos (n  m) x  cos (n  m) xdx
sin nx sin mx dx =

2 
 
1

 
1  sin (n  m) x
sin (n  m) x
= 

 = 0.
2  nm
nm


Therefore n (x): n (x) =
1
sin nx, n  Nis an orthogonal set on  , .

For m = n.

1 
1
sin 2nx  
=
=
(
1

cos
2
nx)
dx
sin
nx
dx
x

1


 
2 
2 
2n 

1

2
Therefore n (x): n (x) =
1

sin nx, n  Nis an orthonormal set on  , .
Exercises. 1.1
1. Show that the following functions are non-periodic functions.
i)
f (x) = x
ii)
g ( x) = 2x2
iii)
h (x) = e x
2. Sketch the graphs of the following functions, which are assumed to be periodic of
period 2 and, for    x  , are given by the formulas:
1. f (x) = e  x 
2. f (x) = x2
3. f (x) =  x 
 x
4. f (x) = 
 x
if    x  0
if 0  x  
if    x  0
 0
5. f (x) = 
 sin x if 0  x  
3. For the trigonometric system find the corresponding orthonormal system on  , .
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1.2 Fourier Series of Functions of Period 2
Let us assume that f (x) is a periodic function of period 2 that can be represented by a
trigonometric series.

f (x) = a0 +
 an cos nx  bn sin nx
(1)
n 1
i.e. the series converges and has f (x) as its sum.
Now we need to determine the coefficients a0, an and bn .
i) To determine a0 integrate (1) from   to .




 
 f ( x) dx =  a0 dx + n1 an cos nx  bn sin nx  dx

 f ( x) dx

= 2 a0

 a0 =
1

2  
f ( x) dx
ii) To determine an, n  N. Multiply (1) by cos mx and integrate from   to  where
m N, fixed.




 
 f ( x) cos mx dx =  a0 cos mx dx + n1 an cos mx cos nx  bn sin nx cos mx  dx
If m  n, then



cos mx dx =

 cos mx cos nx dx =


sin nx cos mx dx = 0.


If m = n, then

 sin nx cos mx dx


and

=

 sin nx cos nx dx = 0.

 cos nx cos mx dx =  cos

2
nx dx
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
=
1
1 cos 2 nx dx
2 
=


1
1
x
+
sin 2nx


2
4n
= .
Therefore an =
1

 
f ( x) cos nx dx  n  N.
iii) To determine bn ,  n  N. Multiply (1) by sin mx where m  N and integrate from
  to .



f ( x) sin mx dx =

 a0 sin mx dx +

 
  an sin mx cos nx  bn sin nx sin mx  dx
n 1  
If m  n, then


 a0 sin mx dx =  sin mx cos nx dx


=
 sin nx sin mx dx = 0.

If m = n, then

 a0 sin mx dx = 0,


 sin nx cos mx dx


and

=
 sin nx sin mx dx =

Therefore bn =
1
 sin nx cos nx dx = 0.


 sin

2
nx dx = .

 
f ( x) sin nx dx  n  N.
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Therefore we get what we call Euler’s Formulas.
a0 =
an =
bn =

1
2  
1

 
1
f ( x) dx
f ( x) cos nx dx

 
f ( x) sin nx dx , where n  N.
Definition 1.5 The numbers a0, an and bn are called the Fourier coefficients of f (x).
The trigonometric series

a0 +
 an cos nx  bn sin nx
n 1
Where a0, an and bn are called the Fourier coefficients is called the Fourier
Series of f (x) (Regardless of convergence).
Remark: In the Euler’s Formulas the interval of integration may be replaced by any
interval of length 2.
Example 1 Find the Fourier coefficients of each of the following periodic functions.
 k
i) f (x) = 
 k
ii) f (x) =
if    x  0
if 0  x  
and f (x + 2) = f (x)
x2
for 0  x  2 and f (x + 2) = f (x).
4
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1

2  
Solutions i) a0 =
f ( x) dx =
an =


 f ( x) cos nx dx
=
=

1
2  0
1
k dx


 k cos nxdx
0

0
k
k
+
sin nx
sin nx
n
n

0
f ( x) sin nx dx =
=
+


 
+

  k cos nx  dx

=
and bn =
2
0
1

1
  k  dx
kx 
kx 0
+
= 0,
2 
2 0
=
1
0
1
0
  k sin nx  dx
1

+

= 0,
1


 k sin nxdx
0

0
k
k
+
cos nx
cos nx
n
n

0
2k
( 1  cos n)
n
 4k

Thus bn =  n
 0
if n is odd
.
if n is even
Therefore the Fourier series of the square wave function f is:
4k  sin 2n  1x
.

 n 1 2n  1
ii) a0 =
an =
2
1
 f ( x) dx =
2
2
2
2
x2
x3 2
=
.
 dx = 2 4
3
0 4
0
2
1
x2
 4 cos nx dx =  2 n 
0
1

and bn =
0
1
1

2
2
 x sin nx dx =
0
1
n2
2
1


x2
 4 sin nx dx =  n + 2 n   x cos nx dx =  n .
0
0
Therefore f (x) =

2
1

+  ( 2 cos nx  sin nx) .
3
n
n 1 n
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Convergence and Sum of Fourier series
Suppose f is any function of period 2 for which the integrals
a0 =
an =

1
2  
1
f ( x) dx

 
f ( x) cos nx dx  n  N.

1
 
and bn =
f ( x) sin nx dx  n  N.
exist. Then the Fourier series of f (x) is given by:

a0 +
 (an cos nx  bn sin nx)
n 1
(1)
i) If the series in (1) converges to f (x), then we write:

f (x) = a0 +
 (an cos nx  bn sin nx) .
n 1
ii) If the series in (1) doesn’t converge to f (x), then we write:
f (x)


a0 +
 (an cos nx  bn sin nx) .
n 1
Question. When is the Fourier series in (1) converges to f (x)?
To answer this question first we need to define piecewise continuity.
Definition 1.6 Piecewise Continuity.
A function f (x) is said to be piecewise continuous on an interval I if
i) the interval can be divided into a finite number of subintervals in each of
which f (x) is continuous.
and ii) the limits of f (x) as x approaches the end points of each subinterval are
finite.
Remark: A piecewise continuous function is one that has only a finite number of jump
discontinuities.
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Unit VIII
Fourier Series
Examples Consider the following functions.
n

i) f (x) =  0
4

for x  n  1, n  where n  N and n  2.
for x  0
for x  2
 x

ii) g (x) =  x  1
  x2  1

if  1  x  0
if 0  x  1
if 1  x  2
f and g are examples of piecewise continuous functions.
Example. h (x) = tan x for x  (
 3
,
).
2
4
lim h ( x)
lim h ( x)

    and
     , h is not
Since h (x) is discontinuous at x =
and
x
x
2
2
2
piecewise continuous on (
 3
,
).
2
4
Theorem 1.1 (Representation by a Fourier series)
If a periodic function with period 2 is piecewise continuous in the interval
   x   and has a left hand derivative and a right hand derivative at each
point of that interval, then the Fourier series (1) of f (x) is convergent. Its sum
is f (x) except at a point x0 at which f (x) is discontinuous and the sum of the series
is the average of the left and the right hand limits at x0.
Example 1. Find the Fourier series of the function f where
if    x  0
 k
f (x) = 
if 0  x  
 k
and f (x + 2 ) = f (x).
Solution. The Fourier coefficients are:
 0

a0 = an = 0 and bn =  4k

 n
if n is even.
if n is odd .
n  N
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Fourier Series
Therefore the Fourier series of f (x) is

 (an cos nx  bn sin nx) =
a0 +
n 1
sin 2n 1 x
.
 n 1
2n 1
4k


Since f is piecewise continuous on ( , ) this series converges to f (x) for each
x  ( , ) \ 0.
For x = 0, the Fourier series is given by
lim f ( x)
1 lim f ( x)
1


 k + k = 0
 =

x 0
x 0
2
2
sin 2n 1 x
= 0.
 n 1
2n 1
4k
and indeed at x = 0,
Therefore f (x) =
If x =

sin 2n 1 x
 x  ( , ) \ 0.
 n 1
2n 1
4k




, then f ( ) = k.
2
2
Hence k =



1
 1 n 2n 1
4k
 n0

1
 1 n 2n 1
n0
=

.
4
Therefore

1
 1 n 2n 1
n0
=

.
4
Example 2. For each of the following, find the Fourier series expansion of the function f (x)
that has period 2.
i) f (x) = x2 ,  x   .
ii) f (x) = x sin x , 0  x  2.
iii) f (x) = x  x2 ,  x   .
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Unit VIII
Fourier Series

1
x 2 dx

2  
Solutions. i) a0 =

x3
6

2
 x cos nx dx



2
3
 x 2
 
2x
2

sin nx  2 cos nx  3 sin nx 
 n
  
n 
n

1
an =


4
 2 cos n .
n
bn =
1

 

2


2x
2
 x

 
cos nx  2 sin nx  3 cos nx 


n 
n
 
 
2
x sin nx dx
= 0.
Therefore f (x) =
For x = ,
2
3

4


n 1 n
2
cos n cos nx for  x   .
lim f ( x)
lim f ( x)
  2 and
2


x 
x 
Hence 2 =
2
3
4


4
n 1 n
2
cos 2n
Therefore

2
=
 2 6.
n 1 n
2
ii) a0 =
an =
1
x sin x dx
2 0
1

1


2
1
 x cos x  sin x   1.
2
0
2
2
1

x sin x  nx   sin x  nx  dx
2 0
 x sin x cos nx dx
0
If n  1, then an =
bn =
1
2
2
2
n 1
 x sin x sin nx dx
0
and if n = 1, then a1 = 

1
.
2
2
1
x cos x  nx   cos x  nx  dx
2 0
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Unit VIII
Fourier Series
If n  1, then bn = 0 and if n = 1, then b1 = 
Therefore f (x) = 1 

1
2
cos x   sin x   2
cos nx .
2
n2 n  1
If x = , then f () = 0.


Thus
n2
1 n
2
n 1
1
.
4
=
Therefore
1 n


n2
1
an =


x  x 2 cos nx dx


2

x3
2
x




.


2
3
3



 
1
2




n 1
1
.
4
1




x cos nx dx 
1


x

2
cos nx dx
1 x
1
4
4
 
 2 cos n   2 cos n .
 sin nx  2 cos nx 
 n
n
n
 
n

bn =



1
x  x 2 dx
2 
iii) a0 =
=
2
1






x  x 2 cos nx dx

2
cos n .
n
Therefore f (x) = 


2
3
+2
1

 1
n 1



x sin nx dx 
1


x

sin nx
n  1  2 cos nx


n
 n2
2
sin nx dx

 x    ,  

If x = 0, then

 1
n 1
n 1
2
=
.
12
n2
1
Therefore

n 1
 1
n 1
2
=
.
12
n2
1
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Unit VIII
Fourier Series
1.3 Functions of any Period p = 2L
Theorem 1.2 If a function f of period p = 2L has a Fourier series representation, then
the series is:

 (an cos
f (x) = a0 +
n 1
n
n
x  bn sin
x)
L
L
where the Fourier coefficients of f (x) are given by the Euler formulas:
L
1
a0 =
2L
an =
and bn =
L
L
1
L
1
L
 f ( x) dx

f ( x) cos
L
L

f ( x) sin
L
n
x dx  n  N,
L
n
x dx  n  N.
L
Proof. (Refer page 578 Kreyszig)
Remark: The interval of integration in the above formulas may be replaced by any
interval of length p = 2L.
Example 1. Find the Fourier series of the function f (x) with period p = 4, where
 0

f (x) =  2
 0

if  2  x  1
if 1  x  1
and f (x + 4) = f (x)
if 1 x  2
Solution. p = 4  L = 2.
1
2
1
1
1
1
Now a0 =  f ( x) dx =  2 dx = x
= 1.
4 2
4 1
2 1
2
an
=
1
1
 n 
 n 
f ( x) cos 
x  dx   cos 
x  dx .

2 2
 2 
 2 
1
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Unit VIII
Fourier Series

2
4
 n  1
 n 
sin 
x

sin 
,  nN ,
n
 2  1 n
 2 

 0
= 
1 n 1 n4


if n is even.
i f n is odd .
2
bn
1
1
 n 
 n 
=
f ( x) sin 
x  dx   sin 
x  dx .

2 2
2
 2 


1

2
 n  1
cos 
x
 0,  nN.
n
 2  1
Therefore f (x) = 1 +
4 
1 n1

cos 2n  1 x .
 n  1 2n  1
2

If x = 0, then f (0) = 2.


1 n1
n  1 2n  1
=

.
4
Example 2. Find the Fourier series representation of the periodic function f (x) with period
p = 1, where
f (x) =  sin x for 0  x  1 and f (x + 1) = f (x).
Solution. p = 1  L =
1
.
2
1
1
0
0
Thus a0 =   sin x dx   cos x
 2.
1
an = 2   sin x cos 2nx dx
0
1
1
0
0
=   sin 2n  1x dx    sin 2n  1x dx
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Unit VIII
Fourier Series
=
1
1
1
1


cos 2n  1x
cos
2
n

1

x
+
2n  1
2n  1
0
0
2
2

2n  1
2n  1
=
4
=
1  4n 2
Thus an =
4
n  N.
1  4n 2
1
bn = 2   sin x sin 2nx dx
0
1
1
0
0
=   cos 2n  1x dx    cos 2n  1x dx
=
1
1
1
1
sin 2n  1x

sin 2n  1x
= 0.
2n  1
2n  1
0
0
Thus bn = 0 n  N.

Therefore f (x) = 2 + 4
cos 2nx
 1  4 n2
n 1
.
If x = 0, then


1
2
n 1 4 n 1
If x =

1
.
2
1
, then
2


 1 n
n 1 1  4 n
2

 2
4
.
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Unit VIII
Fourier Series
1.4 Even and Odd Functions.
Definition 1.7 (Even and Odd Functions)
1. A function f (x) is even if
f ( x) = f (x)  x  dom. f.
2. A function g (x) is odd if
g ( x) =  g (x)  x  dom. g.
Example 1. f (x) = cos nx , x  , where n  N is even.
g (x) = sin nx , x  , where n  N is odd.
2.
Note that:
1.
If f (x) is an even function, then
L

L
f ( x) dx = 2  f ( x) dx
L
2.
0
If g (x) is an odd function, then
L
 g ( x) dx
= 0
L
3.
The product of
i) even and odd functions is odd.
ii) two odd functions is even.
iii) two even functions is even.
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Unit VIII
Fourier Series
Theorem 1.3 (Fourier series of Odd and Even Functions)
a. The Fourier series of an even function f of period 2L is a Fourier cosine series.

f (x) = a0 +
 an cos
n 1
L
n
x
L
with coefficients
L
1
2
n
a0 =
f ( x) dx and an =
f ( x) cos
xdx  n  N


L 0
L 0
L
b. The Fourier series of an odd function f of period 2L is a Fourier sine series.

f (x) =
 bn sin
n 1
n
x
L
with coefficients
L
2
n
bn =
f ( x) sin
xdx  n  N.

L 0
L
Theorem 1.4 (Sum of Functions)
The Fourier coefficients of a sum of two function f1 and f2 are the sum of the
corresponding Fourier coefficients of f1 and f2.
The Fourier coefficients of cf are c times the corresponding Fourier
coefficients of f.
 2k
Example 1. Let f (x) = 
 0
if 0  x  
if    x  0
and f (x +2) = f (x)  x  dom. f.
Find the Fourier series of f (x).
 k
Solution. Let f1 (x) = 
k
if 0  x  
if    x  0
and f1 (x +2 ) = f1 (x)  x  dom. f
and let f2 (x) = k  x  . Then (f1 + f2) (x) = f (x)  x  dom. f.
But f1 (x) =
4k  sin 2n  1x
and f2 (x) = k.

 n  1 2n  1
4k  sin 2n  1x
Therefore f (x) = k +
 x  dom. f.

 n  1 2n  1
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Unit VIII
Fourier Series
Example 2. Find the Fourier series of the function f (x) where f (x) = x +  if    x  
and f (x + 2) = f (x)  x  dom. f.
Solution. Let f1 (x) = x and f2 (x) =  if    x   and be periodic functions with period 2.

Now f1 (x) = 2
  1
n 1
n 1
Therefore f (x) =  + 2
sin nx
and f2 (x) = .
n

  1
n 1
n 1
sin nx
 x  dom. f.
n
Example 3. Find the Fourier series representation of
f (x) = x + x2 on   ,  .
Solution. Let f1 (x) = x and f2 (x) = x2 for x    ,   and f1 and f2 be periodic functions with
period 2.
Now f1 (x) is odd hence it has a Fourier sine series
1 x
1
 
and bn =
x sin nx dx    cos nx  2 sin nx 
 
  n
n
 

1
=  1 n 1
2
n
nN

Thus f1 (x) = 2
  1
n 1
n 1
sin nx
.
n
f2 (x) is even, hence it has a Fourier cosine series representation.


1
1
x 2 dx   x 2 dx
Now a0 =

2  
0
and an =

0

2
3

 2
 
1
2 cos nx dx   x sin nx  2 x cos nx 
x

cos nx dx
2
2 
 n

 
n

n



 
1

=  1 n
Thus f2 (x) =
x3
3

2
Therefore f (x) =
3
4
n2
nN

+4
  1
n 1
2
3

+
  1
n 1
n
cos nx
n2
.

2
n  4
.
cos
nx

sin
nx
 n2

n


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Unit VIII
Fourier Series
Note that:
If g (x) is defined  x  , then the function
P (x) =
1
 g ( x)  g ( x) is even
2
and q (x) =
1
 g ( x)  g ( x) is odd.
2
Hence g (x) = p (x) + q (x) and g ( x) = p (x)  q (x).
Exercises 1.4
Represent the following functions as a sum of even and odd functions.
i)
ii)
iii)
1
1 x
1
1
1  x 2
1  x 2
e
1
1
=
1 x
kx
e
kx
=
1  x2

1  x2
1  x2

2
+
+
x
1  x2
2x
1  x2 2


= cosh kx + sinh kx.
1.5 Half-range Expansion.
Sometimes it is required to extend a function f (x) in the range (0, L) in a Fourier series of
period 2L, and it is immaterial what the function may be outside the range 0  x  L. We could
extend f (x) with period 2L and then represent the extended function by a Fourier series.
i) If we extend the function f (x) by reflecting it in the y-axis, then the Fourier series
expansion contains only the cosine terms.
ii) If we extend the function f (x) by reflecting it in the origin, then the Fourier series
expansion contains only the sine terms.
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Unit VIII
Fourier Series
The cosine halt-range expansion is:

f (x) = a0 +
a
n
n 1
cos
n
x
L
L
L
1
2
n
where a0 =  f ( x) dx and an =  f ( x) cos
x dx  n  N
L0
L
L0
The sine halt-range expansion is:

L
2
n
n
f (x) =  b sin
x dx  n  N
x , where bn =  f ( x) cos
n
L0
L
L
n 1
Example. Find the two half-range expansions of the function
2k

x

L
f (x) = 
 2k L  x 
 L
if 0  x 
L
2
L
if
 x L
2
where k  0.
Solution. f (x) can be extended to (L, L) into two ways.
I. Make f (x) even on (L, L). i.e. f (x) = f ( x)  x  (L, L), to get Fourier cosine
expansion.
II. Make f (x) odd on (L, L). i.e. f ( x) =  f (x)  x  (L, L), to get Fourier sine
expansion.
Fourier cosine expansion
L
Now a0 =
L
2
L
1 2k
1 2k
1
x dx + 
L  x dx
f ( x) dx = 

L0 L
L0
L L
L
2
L
L
k
k
k
k
1  2k 
1 
= 2 x 2 2 +   Lx  x 2   =
+
=
L
2
4
4
L L 
2 
L
0
2
Prepared by Tekleyohannes Negussie
167
Unit VIII
Fourier Series
L
L
1
n
2
n
f ( x) cos
x dx =
f ( x) cos
x dx
an =


L L
L
L
L
0
L
2
L
n
4k
n
x dx + 2  ( L  x) cos
= 2  x cos
x dx
L
L
L 0
L L
4k
2
L
xL
n
L2
n
sin
x
cos
x) 2
= 2 (
2
L
L
L n
(n )
4k
0
L
L2
n
xL
n
L2
n
 2 (
sin
x
sin
x 
cos
x) L
2
L
n
L
L
L n
(n )
4k
2
=
n

4k

 cos n  1 
 2 cos
2

n 2 
Thus an =
8k
n

 cos
 cos
2
n 2 
2 n 
  n  N.
2 

Therefore f (x) =
k
8k 
n
n
2 n 
+ 
cos

cos
cos
x.


2
2
2
2
L




n

n 1
II. Fourier sine expansion.
L
Now an =
L
1
n
2
n
f ( x) sin
x dx =
f ( x) sin
x dx


L L
L
L
L
0
L
2
L
n
4k
n
x dx + 2  ( L  x) sin
= 2  x sin
x dx
L
L
L 0
L L
4k
2
L
xL
n
L2
n
cos
x
sin
x) 2
= 2 (
2
n

L
L
L
(n )
4k
0
L
L2
n
xL
n
L2
n
 2 (
cos
x
cos
x 
sin
x) L
2
n
L
n
L
L
L
(n )
4k
2
Prepared by Tekleyohannes Negussie
168
Unit VIII
=
Fourier Series
8k
n
2
sin
2
n 
Thus bn =
8k
n 
2
Therefore f (x) =
sin
n
 n  N.
2

n
n
sin
sin
x.
2
L
 2 n  1 n2
8k

1
lim f ( x)
Since f (x) is piecewise continuous and
x
L

lim f ( x )
L  = k.
=
x
2

0
2 n
2 n
=
k,
but
sin

sin

2
2
 2 n  1 n2
1
8k
1


2
if n is even
if n is odd
2
Thus 
=
.
2
8
n  1 2n  1
1
Therefore


n0
2
=
.
8
2n  12
1
Exercises 1.5
1. Represent each of the following functions by a Fourier sine and cosine series extensions,
where L  + .
i) f (x) = x,
0  x  L.
0  x  L.
ii) f (x) = x2,
iii) f (x) = sin

x,
L
0  x  L.
2. Obtain a half-range sine and cosine series for
i) ex in 0  x  1.
ii) x2  x
in 0  x  .
Prepared by Tekleyohannes Negussie
169
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