# Unit II Fourier Series

```TRANSFORMS AND PARTIAL
DIFFERENTIAL
UNIT II: FOURIER SERIES
Periodic Functions
f
0
T

a0 
f      an cosn  bn sin n
2 n1
n1
Definition of a Fourier series
A Fourier series may be defined as an expansion of a function in
a series of sines and cosines such as

a0 
f  x     an cos nx   bn sin nx
2 n 1
n 1
The coefficients are related to the periodic function f(x)
Henceforth we assume f satisfies the following (Dirichlet)
conditions:
(1) f(x) is a periodic function;
(2) f(x) has only a finite number of finite discontinuities;
(3) f(x) has only a finite number of extrem values,
maxima and minima in the
interval [0,2p].
A Fourier series is a convenient representation of a periodic function.
A Fourier series consists of a sum of sines and cosine terms.
Sines and cosines are the most fundamental periodic functions.
Fourier series
 p ,p 
0 ,2p 


a0
f  x     an cos nx   bn sin nx
2 n 1
n 1
FOURIER COEFFICIENTS FOR (-π,π)
1
a0 
an 
bn 
1
p
1
p
p
p
 p f x  dx

p
 p f x cos nx dx

p
 p f x sin nx dx

n  1, 2, 
n  1, 2 , 
FOURIER COEFFICIENTS FOR
a0 
an 
bn 
1
p 
2p
1
2p
0
p 
0
1
p 
2p
0
(0,2π)
f  x  dx
f  x  cos nx dx
f  x sin nx dx
n  1, 2, 
n  1, 2, 
Fourier series
( l , l )
(0,2l )
a0
 npx 
 npx 
f  x     an cos
   bn sin

2 n1
 l  n1
 l 


FOURIER COEFFICIENTS FOR
1
a0 
l

l
l
( l , l )
f  x  dx
1
 npx
a n   f ( x ) cos 
l l
 l
l

 dx

1
 npx 
b n   f ( x ) sin 
 dx
l l
 l 
l
(0,2l )
FOURIER COEFFICIENTS FOR
1
a0 
l
1
an 
l
1
bn 
l
2l

0
2l

0

2l
0
f  x  dx
 npx
f ( x ) cos 
 l

 dx

 npx 
f ( x ) sin 
 dx
 l 
Interval
( l , l )
(0,2l )
 p , p 
0 ,2 p 
Parsevals Identity
1
l
2
l
a 02
 l  f ( x )  dx  2 
2
2l
a 02
0  f ( x )  dx  2 
1
l
 a

n 1
 a

n 1
1
2
n
2
n
 b n2
 b n2 
2
p

a 02
2
2



f
(
x
)
dx


a

b

n
n 

p p
2
n 1
1
p
2

a 02
2
2


f
(
x
)
dx


a

b


n
n 

p p
2
n 1
1

Even and Odd Functions
Even Functions
The value of the function would be
the same when we walk equal
distances along the X-axis in
opposite directions.
Mathematically speaking -
Odd Functions
The value of the function would
change its sign but with the same
magnitude when we walk equal
distances along the X-axis in
opposite directions.
Mathematically speaking -
Even functions can solely be represented by cosine
waves because, cosine waves are even functions. A
sum of even functions is another even function.
Odd functions can solely be represented by sine waves
because, sine waves are odd functions. A sum of odd
functions is another odd function.
The Fourier series of an even function
f x 
is expressed in terms of a cosine series.
a0
f x  
2



n 1
a n cos nx
The Fourier series of an odd function
is expressed in terms of a sine series.
f x  

b
n 1
n
sin nx
f x 
1. Find the Fourier series expansion of f(x) = x2, 0 &lt; x &lt; 2π. Hence
deduce that
1
1
1
p2
(i )
 2  2  .......... .... 
2
6
1
2
3
1
1
1
p2
( ii ) 2  2  2  .......... .... 
12
1
2
3
1
1
1
p2
( iii ) 2  2  2  .......... .... 
8
1
3
5
.
Fourier series is
a0
f (x) 

2
a0 

1
p


f ( x ) dx 
0
1
p
 x

 3
1  8p

p  3
8p

3
2
3
n
n  1
2p

(a
1
p
3

 0

cos
2p
x
2
dx
0



2 p
0
nx  b
n
sin nx )
2p
1
an 
 f ( x ) cos nx dx
p

0
1
p
2p

x 2 cos nx dx
0
2p
1  2  sin nx 
  cos nx 
  sin nx 
 ( x )
  (2 x)
  (2)

2
3
p
 n 
 n

 n
 0



( 4 p ) (1)
1 



0


0

0

0

0



p  
n2


4
n
2
bn 

1
p
1
p

2p
 f ( x ) sin nx dx
0
2p
2
x
 sin nx dx
0
1  2   cos nx
( x )

p 
n


  sin nx

(
2
x
)


2

 n

 cos nx

(
2
)


3

 n
1   4p 2
2  
2 
 
 0  3   0  0  3  
p  n
n  
n 

4p
n
2p


 0

a0
f ( x) 
  ( a n cos nx  bn sin nx )
2 n 1
1  8p 2    4
4p

    2 cos nx 
 
sin nx 
2  3  n 1  n
n

4p 2  cosx cos2x cos3x

sin x sin2x sin3x

f ( x) 
 4 2  2  2  ...............  4p 


 ............
3
2
3
2
3
 1

 1

Put x = 0 in the above series we get
4p 2
1
1 1

f (0) 
 4  2  2  2  ............  4p (0) --------------- (1)
3
2
3
1

But x = 0 is the point of discontinuity. So we have
f ( 0 )  f ( 2p ) ( 0 )  ( 4p 2 )
f (0) 

 2p 2
2
2
Hence equation (1) becomes
2p 2
2p
2
4p 2
1
1
1


 4  2  2  2  .......... ..
3
2
3
1

4p 2
1
1
1

 4  2  2  2  ..........
3
2
3
1

.. 

2p 2
1
1
1

 4  2  2  2  .......... ..
3
2
3
1

p2
1 1 1
 2  2  2  ................
6 1 2 3
       (2)
Now, put x = π (which is point of continuity) in the above series we get
p2
4p 2
1
1
 1


 4   2  2  2  .......... ..  4p ( 0 )
3
2
3
 1

4p 2
1
1
1

p 
  4  2  2  2  .......... ..
3
2
3
1

2
p
2
1
1
1


  4  2  2  2  .......... .. 
3
2
3
1

p2
1 1
1
 2  2  2  ................        (3)
12 1 2
3
Adding (2) and (3), we get
p
2
6

p
2
1
1
1

 2  2  2  2  .......... .. 
12
3
5
1

3p 2
1
1
1

 2  2  2  2  .......... ..
12
3
5
1

p2
1 1
1
 2  2  2  ................
8 1 3
5
2. Expand in Fourier series of f(x) = x sinx for 0 &lt; x &lt; 2π and deduce
the result
1
1
1
p 2


 .......... 
1.3 3.5 5.7
4
Sol. Fourier series is
a0 
f ( x) 
  (a n cos nx  bn sin nx)
2 n 1
a
0

1
p
2p

0
f ( x ) dx


2p
1
 x sin
p
1
p

x dx
0
x ( cos x )  (1)(  sin x)02p
1
p
(  2 p
 2
 0 )  (0  0 )
an 
1
p

2p

f ( x ) cos nx dx
0
1
p
2p
 x sin x cos nx dx
0
2p
1

2p
1

2p
 x ( 2 cos nx sin x ) dx
0
2p
 x sin(n  1) x  sin(n  1) x dx ,
0
n 1
2p
1

2p
1

2p
1
0 x sin( n  1) x dx  2p
 x sin( n  1) x dx
0
2p
   cos( n  1) x 
  sin( n  1) x  
 
  (1)
 ( x )
2
n 1

 ( n  1)
 0
 
1

2p
Note:
2p
2p
   cos(n  1) x 
  sin(n  1) x 

  (1)
( x)
2
n 1

 (n  1)
 0
 
(  1)
2n2
1
(  1)
2n2
1
 1  2p (1)2n2 

1  2p (1)2n2 
 
 0  0  0  
 0 0  0
2p  n 1


 2p  n 1

1
1


n 1 n 1
 (n  1)  (n  1)
an 
(n  1)(n  1)
2
an  2
, n 1
n 1
When n = 1, we have
a1 
1
p
1

2p
1

2p
2p
 f ( x ) cos x dx
0

1
p
2p
 x sin x cos x dx
0
2p
 x sin
2 x dx
0
2p
   cos 2 x 
  sin 2 x  
  (1)

 x
2
4


 0
 


1 
 1

  0   (0  0)
  2p 
2p  
 2 


1
 
2
bn 
1
p
2p
 f ( x) sin nx dx
0
1

2p
1

2p
1

2p
2p

1
p
2p
 x sin x sin nx dx
0
2p
 x(2 sin nx sin x) dx
0
2p
 x cos(n  1) x  cos(n  1) x dx
,
n 1
0
1
0 x cos(n 1)x dx  2p
2p
 x cos(n  1)x dx
0
1

2p
2p
  sin( n  1) x 
  cos( n  1) x  
 
  (1)
 ( x )
2
 ( n  1)
 0
  n 1 
1

2p
2p
  sin( n  1) x 
  cos( n  1) x  
 
  (1)
 ( x )
2
 ( n  1)
 0
  n 1 
1  1  (1) 2 n2  
1 
1  (1) 2 n2  
 0 
 0
 0 
 0

2  
2 
2  
2p  (n 1)   (n 1)  2p 
(n 1)   (n  1) 2 
1  
1  1 
1  
1 
1 
 0 
 0
 0 
 0

2 
2 
2 
2p  (n 1)   (n 1)  2p  (n  1)   (n  1) 2 
bn  0 , n  1
When n = 1, we have
b1 
1
p

2p
 f ( x ) sin x dx

0
1
p
2p
 x sin
0
2
x dx 
1
p
1
p
2p

0
2p
 x sin x sin x dx
0
 1  cos 2 x 
x
 dx
2


2p
  sin 2 x 
1 x
  cos 2 x   

  x
  (1) 


2p  2
4

 0
  2 
2
1 

  2p
2p  
p
2
1 
 1  
 0     0  0  
2 
 2  

a0
f ( x) 
  ( a n cos nx  bn sin nx )
2 n 1
a
 0  a 1 cos x 
2

a
n2
n
cos nx  b1 sin x 

b
n2
n
sin nx

2 1
2

 cos x  
cos nx  p sin x  0
2 2
n  2 ( n  1)( n  1)
1
cos2x cos3x cos4x cos5x

xsinx  1 cosx p sinx  2



..........
........
2
2.4
3.5
4.6
 1.3

Put x = in the above series we get
p
1
1
 1

(1)  1  0  p (1)  2   0 
0
 0  ..................
2
3 .5
5 .7
 1 .3

p
1
1 1

 p 1  2   
 ..................
2
1.3 3.5 5.7

p  2p  2
2
1
1
 1

 2  

 .......... ........
1 .3 3 .5 5 .7

p  2
1
1
1

 2  

 ..................
2
1.3 3.5 5.7

p 2
4
1
1
1



 .......... .......
1 .3 3 .5 5 .7
3. Find the Fourier series of
 1,
f ( x)  
 2,
0 x  p
p  x  2p
Hence evaluate the value of the series
1
1
1
 2  2  .......... ......
2
1
3
5
Sol. Fourier series is

a0
f ( x) 
  ( a n cos nx  bn sin nx )
2 n 1
a0 
2p
1

f ( x ) dx 
1
x 0
p

0
p

p
1
p
(p

2
p
1
p
p
1
2p
 (1) dx  p p ( 2 ) dx
0
x p2 p
 0)
2
p
1 2  3
( 2p
 p )
an 
1
p
2p
1
p
1
2p
 f (x) cosnxdx  p  (1) cosnxdx  p p (2) cosnxdx
0
0
p
2p
1  sin nx 
2  sin nx 
 
 

p  n  0 p  n  p

1
p
2
(0  0)  (0  0)
 0
p
bn 
1
p
2p
 f ( x) sin nx dx

0
1
p
1
(1) sin nx dx   (2) sin nx dx

p
p p
0
p
2p
1   cos nx 
2   cos nx 
 
 

p  n  0 p  n  p

1
2
[(  1) n  1] 
[1  (  1) n ]
np
np

1

 (  1 ) n  1  2  2 (  1) n
np
(  1) n  1

np
2p

a
f ( x)  0 
2

 (a
n 1
n
cos nx  b n sin nx )

3  
(1) n  1
   0. cos nx 
sin nx 
2 n 1 
np

3 2  sin x sin 2x sin 3x

  


 .........................
2 p 1
2
3

When we put x = 0, , π, 2π we will not get the given series.
So, using Parseval’s identity for Fourier series we have
2p
1

p
0
2

a0
2
2
2
[ f ( x )] dx 
  ( a n  bn )
2
n 1
2
n


1
1
(3)
(1) 1 
2
2
(1) dx   (2) dx 
 0 

2 2

p0
p p
2 n 1 
np

p
1
p
1
2p
4
x0  xp
p
2p
p
4
2

9 1 4
4
4

  2  2  0  2  0  2  0  ..................
2 p 1
3
5

9 4
p  0  2p  p    2
p
p
2 p
1 1 1

12  32  52  ..................


9
4 1 1
1

5   2  2  2  2  ..................
2 p 1
3
5

1 4 1 1 1

 2  2  2  2  ..................
2 p 1 3 5

p
2
1
1
1
 2  2  2  .......... .......
8 1
3
5
4.Find the Fourier series expansion for the function
f ( x)  p  x , in 0  x  2p with period 2p
Hence deduce the sum of 1 
1 1
  ...
3 5
Solution:
a0 
f ( x) 
  (a n cos nx  bn sin nx)
2 n 1
a0 
1
p
2p
 f ( x ) dx
0

1
p
2p
 (p
 x ) dx
0
2p
1  (p  x ) 
 

p 
2
0
2


1
p 2 p 2
2p
 0

an 

1
p
1
p
2p

f ( x ) cos nxdx
0
2p
 (p
 x ) cos nxdx
0
1  
 sin nx
   p  x 
p  
 n
1
 2 1  1  0
np
2p


 cos nx
     1
2
n
 0


2p


 0



bn 

1
p
1
p
2p

f ( x ) sin nxdx
0
2p
 (p
 x ) sin nxdx
0
1  
 cos nx
   p  x  
p  
n

1 1

  p p   0
p n

2

n
2p


 sin nx
     1 
2
n
 0


2p


 0




f ( x)  p  x  2
Hence
n 1
sin nx
n
1 1
p
The required series 1    ... can be obtained by putting x 
3 5
2
p 
p
2

 2
n 1
sin nx
n
p
2p 1
3p 1
4p
 p 1

 2 sin  sin
 sin
 sin
 ...
2
2 2
2 3
2 4
2


p
1 1


 2 1    ... 
2
3 5


 1 1
 p
1 3  5  ...   4

2p
2
p  x

 dx

p 0 2 
1
2p
 (p  x) 


3

0
1

p 3 p 3
12p
p2

6
3
1

4p


p2
6

4.Find the Fourier series expansion for the function
p  x 
2
f ( x)  
 , in 0  x  2p with period 2p . Hence Pr ove that
 2 

1 p2


2
n
6
n 1
Solution:
a0 
f ( x) 
  (a n cos nx  bn sin nx)
2 n 1
a0 
1
p
2p
 f ( x ) dx
0
an 

1

4p
1
p
2p

f ( x ) cos nxdx
0
2p
2
p  x 

 cos nxdx

p 0 2 
1
2p
2p
2p






sin
nx
cos
nx

sin
nx




 
2
   2p  x   2    2 1
 
p  x  
3
n
n
n

 0 

 0 

 0 

2p


1 
 2


 0   2 p  x  cos nx   0 
4 p 

n
0
 2
 p  p

2
4p n


1
n2
bn 

1

4p
2p
1

p
1
p
f ( x ) sin nxdx
0
2p
p  x 
 
2
0
2
 sin nxdx

2p
2p
2p
 






cos
nx
sin
nx
cos
nx




 
2
     2p  x  
    2  1
 
 p  x  
2
3
n
n
n

 0

 0

  0 
 



=0
1 1

p

4p  n
2
 p
2

0 
2



1

1

n3
2
Hence
p2
p  x 


 
12
 2 


n 1
cos nx
n2
putting x  0
p
2
4

p
12


n 1
2



n 1
1
n2
1
p 2
p 2


2
n
4
12

p
6
2
1  cos x in the int erval 0  x  2p
5 . Obtain the fourier exp ansion of
and hence deduce that

 4n
n 1
1
2
1

1
2
Solution:
a0 
f ( x) 
  (a n cos nx  bn sin nx)
2 n 1
1
a0 
1
p
2p
 f ( x ) dx
0
1  cos x in the int erval 0  x  2p
5 . Obtain the fourier exp ansion of
and hence deduce that

 4n
n 1
1
2
1

1
2
Solution:
a0 
f ( x) 
  (a n cos nx  bn sin nx)
2 n 1
1
a0 
1
p
2p
 f ( x ) dx
0
a0 
1
p

2p

1  cos x dx
0
1
p
2p
x
2 sin dx
2

0
2p
x 

cos
2
2


1 
p 
2 0



2
2
p
4
2
p
 1  1 
an 
1
p

2p

f ( x ) cos nxdx
0
1
p

2p

1  cos x cos nx dx
0
1
p
2p

0
x
2 sin cos nxdx
2

2
p
2p

0
1
2
 1

1
 
 sin   n  x  sin   n  x dx

2
 
 2
2p

1

1
 
cos

n
x
cos

n



x 

2
2
 
2
 


1
1
2p 

n
n


2
2
0
Since

cos 


cos 

1

 n 2p   1
2

1

 n 2p   1
2


2   1
1



 1
2p    n 1  n

  2
2
 

  1
1 



  1  n 1  n 
 

  2
2
 


2 2 1
1 


1

2p   n 1  n 
2
2

2 4 

p 1  4 n 2 
1
1

 n   n
2 2
2



1
p 
2

n
4


bn 


p
1
p

2p
1
f ( x ) sin nxdx
0
2p

1  cos x sin nx dx
0
1
p
2p
2p

0
x
2 sin sin nxdx
2
1 1

1
 

cos   n  x  cos   n  x dx

p 0 2 2

2
 
2
2p

1

1
 
sin   n  x sin   n  x 

2
2
 
2
  0


1
1
2p 

n
n


2
2
0
Hence f ( x ) 
2 2
p

4 2

cos nx
2
n 1 p 1  4 n 
putting
f (0) 
x0
2 2
p

 f (0)  0


n 1
1
1

2
4n  1 2


n 1
4 2
2
p 4 n  1 
f ( x )  e ax in 0 , 2p 
6.Find the Fourier series for
Solution:


a0
f ( x) 
  a n cos nx   bn sin nx
2 n 1
n 1
1
a0 
p


2p

f ( x ) dx
0
1
2p
e
p
ax
dx
0
2p
1 e 
p  a  0
ax


1 2ap

e 1
ap
1  e ap  e  ap 



a p  e  ap 
eap
2 sinhap 

ap
e x  e x
sinh x 
2
an 
1
p
2p
 f ( x ) cos nxdx
0

1
p
2p
ax
e
 cos nxdx
0
2p

1  e
a cos nx  n sin nx 
  2
2
p a  n
0
ax
note :
ax
e
ax
e
 cos bxdx  a 2  b 2 a cos bx  b sin bx 
 cos 2 np  1
sin 2 np  0

a
2

1
ae
2
 n p
ap
2 ap
 a
e  e
a


2
2
 ap
a  n p  e
ap
 ap
2 ae
 2
sinh a p
2
a  n p




bn 
1
2p
p

 f ( x ) sin nxdx
0
1
p
2p
ax
e
 sin nxdx
0
2p

1 e
a sin nx  n cos nx 
  2
2
p a  n
0
ax
note :
ax
e

e ax
a sin bx  b cos bx
sin bxdx  2
2
a b



1
2ap
 2 2  ne  n
a  n p

n
2 ap
 2
e
1
2
a  n p

 2 ne a p

sinh a p
2
2
a  n p
7.Find the fourier series expansion for the function f(x) = 1 + x + x2 in
(–π, π). Deduce
1 1 1
p2
 2  2  .............. 
2
6
1 2 3
Sol. The given function is neither an even nor an odd function.
a0 
f ( x)    (an cos nx  bn sin nx)
2 n 1
a0 
1
p
 f (x) dx 
p p
1
p
2
(
1

x

x
) dx

p p
p
1
x
x 
 x 
 
p
2
3  p
2
3
1  p 2 p 3  
p 2 p 3 
 p 
    p 
 
p 
2
3 
2
3 
1 
2p 3 
2p 2
  2p 
  2
p 
3 
3
an 
1
p
p
p

f ( x ) cos nx dx 
1
p
p
2
(
1

x

x
) cos nx dx

p
p
1
  cos nx 
  sin nx 
2  sin nx 
 (1  x  x )
  (1  2 x)
  (2)

2
3
p
 n 
 n

 n
  p
  (1  2p ) (1) n

1  (1  2p ) (1) n
 0 
 0  0 
 0
2
2
p 
n
n
 

(1) n
1  2p  1  2p 

2
pn
(  1) n
4 (  1) n

( 4p ) 
2
pn
n2
cos n p  (  1 )
sin n p  0
cos n (  p )  cos n p
n
 (  1)
n
bn 
1
p
p
p

f ( x ) sin nx dx 
1
p
p
2
(
1

x

x
) sin nx dx

p
1 
  sin nx 
 cos nx
2   cos nx 
  (1  x  x ) 
  (1  2 x ) 
  ( 2 )
2
3
p 
n


 n

 n
p


  p
n
n
n
n



2
(

1
)
2
(

1
)
1 
(

1
)
(

1
)
2
2
  (1 p  p )
 0  3    (1p  p )
 0  3 
p 
n
n
n  
n 

(1) n

1  p  p 2  1  p  p 2
np

 2(1)n 2(1)n1
(1)n

(2p ) 

np
n
n
a0 
f ( x) 
  (an cos nx  bn sin nx)
2 n 1

1
2p 2    4(1) n
2(1) n1
    2 cos nx 
  2 
sin nx
2
3  n 1  n
n

 1
p2
 cos x cos 2x cos3x
  sin x sin 2x sin 3x

 4 2  2  2  ......  2


 .......
3
2
3
2
3
 1
  1

(i.e.)
p2
 cos x cos2x cos3x
  sin x sin 2x sin 3x

f ( x)  1   4 2  2  2  ............  2


 ............
3
2
3
2
3
 1
  1

Put x = π in the above series we get
p2
 1 1 1

f (p )  1  4 2  2  2  ............  2(0)...........(1)
3
 1 2 3

But x = π is the point of discontinuity. So we have
f (p )  f (p ) (1  p  p 2 )  (1  p  p 2 ) 2  2p 2
f (p ) 


 1 p 2
2
2
2
Hence equation (1) becomes
1 p 2  1
p

2
2p
3
p
2
2
p
p2
1
1
 1

 4   2  2  2  .......... ..
3
2
3
 1

2
1
1
 1

 4  2  2  2  .......... .. 
3
2
3
1

1
1
 1

 4  2  2  2  .......... .. 
2
3
1

1
1
1
 2  2  2  .......... ......
6
1
2
3
Find the Fourier series expansion of (π – x)2 in –π &lt; x &lt; π.
. 8.
Sol.
Fourier series is

a0
f ( x) 
  ( a n cos nx  b n sin nx )
2 n 1
a0 
1
p
p
p
f ( x ) dx 

1  (p  x ) 3 


p 
 3



1
0  8p
 3p
3

1
p
p
2
(
p

x
)
dx

p
p
 p
8p

3
2
an 
1
p
p
p
f ( x ) cos nx dx 

1
p
p
2
(
p

x
)
cos nx dx

p
p
1
  cos nx 
  sin nx 
2  sin nx 
 (p  x) 
  [2(p  x)(1)] 
  (2)

2
3
p
 n 
 n

 n
  p
 (4p ) (1) n

1
 0  0  0  0 
 0 
2
p
n


4 (  1) n

n2
bn 
1
p
p

f ( x) sin nx dx 
p
1
p
p
2
(
p

x
)
sin nx dx

p
p
1
  sin nx 
 cosnx 
2   cosnx 
 (p  x) 
  [2(p  x)(1)]
  (2) 3 
2
p
 n 
 n 
 n  p
n
2 ( 1) n  
2 ( 1) n 
1 
2 ( 1)
  0  0 
0
   ( 4p )

3
3
p 
n
n
n
 

4 p (  1) n

n
a
f (x)  0 
2

 (a
n 1
1  8p 2
 
2 3
n

 

cos nx  b n sin nx )


n 1
 4 (  1) n

4 p (  1) n
cos
nx

sin
nx


2
n
n


9.Obtain the fourier series expansion for
2x

1  p
f (x)  
2x
 1
p

,p  x  0
,0  x  p
1
1
p2
and hence deduce that 1  2  2  ....... 
3
5
8
Solution:
f ( x)  1 
2(  x )
p
 1
2x
p
in ( p , 0)
 f ( x ) in (0, p )
2(  x )
2x
and f (  x )  1 
 1
in (0, p )
p
p
 f ( x ) in ( p , 0)
Hence the fourier coefficien t bn  0

a0
f x  
  a n cos nx
2 n 1
a0 

2
p
2
p
p

0
p

f ( x ) dx
0
2x 

1 
 dx
p


p
2 
x2 

x 
p 
p  0
 0
an 
2
p
p
 f ( x ) cos nxdx
0
p
2 
2x 
  1 
 cos nxdx
p 0
p 
p
2 
2 x  sin nx    2   cos nx  
  1 




2
p 
p  n   p  n
 0
p
2   2 cos n p 
2 




2
2 
p 
pn
 pn  0

4
 2 2 1  (  1) n
p n

 0 , when n is even

an   8
 p 2 n 2 , when n is odd
The required Fourier Series
 f x 

8
 np
n1,3,5
2
2
cosnx
8  cos x cos3x cos5x 
i.e., f  x  2  2  2  2  ...
p  1
3
5

Putting x=0 , we get,
8 1 1 1

f 0  2  2  2  2  ...
p 1 3 5

8 1 1 1

1  2  2  2  2  ...
p 1 3 5

p2 1 1 1

  2  2  2  ...
8 1 3 5

10.If a is neither zero nor an int eger , find the fourier series exp ansion of
period 2p for the function f ( x )  sin ax, in  p  x  p
Solution :
f (  x )   sin ax
 f ( x ) is an odd function
Hence a 0  0 , a n  0
f ( x )

 b n sin nx
n 1
bn 
2
p
f ( x) sin nxdx

p
0

2
p
sin ax sin nxdx

p
0
p

1
   cosn  a xdx  cosn  a xdx 
p 0

1  sin n  a x sin n  a x 




p 
na
na
0
p

1  sin n  a p sin n  a p 



p  na
na
1  sin np cos ap  cos np sin ap sin np cos ap  cos np sin ap 
 


p
na
na
1  (  1) n  sin a p  (  1) n sin a p  
 


p 
na
na

1  (  1 ) n  1 sin a p

p 
na

(  1 ) n  1 sin a p  

na

note :
sin a p  , a is not an int eger
(  1) n 1 sin a p   1
1 


 n  a n  a 
p
(  1) n 1 2 n sin ap 
bn 
p n 2  a 2 
11 .Obtain the fourier series to represent the function
1
1
p2
f ( x )  x ,  p  x  p and deduce 1  2  2  ..... 
3
5
8
Solution
:
f (x)  x
f ( x)   x
 x
 f ( x ) is an even

a0
f x  
  a n cos nx
2 n 1
a0 
2
p
p

0
f ( x ) dx
function
2
a0 
p
a0 
2
p
p

x dx
0
p
 xdx
0
2 p
2 x 
   p
p  2 0
an 
2
p
f ( x ) cos nxdx

p
0

2
p
p


x cos nxdx
0
2
p
p
 x cos nxdx
0
p
2   sin nx   cosnx 
  x
 1  2 
p   n   n  0
2  cos np 1 
  2  2
p n
n 


2
n

(

1
)
1
2
pn
 0 , if n is even

an    4
, if n is odd
 p n 2
 f ( x) 
p
2



n  odd
4
cos nx
2
pn
p
4
cos 3x cos 5 x

f ( x)    cos x  2  2  ......
2 p
3
5

put x  0
0
p
2
p2

4
1
1

1



......


2
2
p 
3
5

1
1
 1  2  2  ......
8
3
5
11 .Obtain the fourier series to represent the function
f ( x )  cos x ,  p  x  p
Solution
:
f ( x )  cos x
f (  x )  cos(  x )
 cos x
 f ( x ) is an even

a0
f x  
  a n cos nx
2 n 1
a0 
2
p
p

0
f ( x ) dx
function
a0 
2
p
cos x dx

p

0
 sin x 
p
2
p
0

2
p
1  1
2
2
p
p
2
2
 cos xdx  p p  cos xdx
0
2
  sin x 
p
p

p
4
p
2

an 

2
p
2
2
p
2
p
p

f ( x ) cos nxdx
2
cos x cos nxdx

p

0
0
p
2
p
p
 cos x cos nxdx  p  cos x cos nxdx
0
2
p
1
1
  cos n  1x  cos n  1xdx   cos n  1x  cos n  1xdx
p 02
p 2
2
p
p 

2








1  sin n  1 x sin n 1 x 
 sin n  1 x sin n 1 x  
 






p  n  1
n 1  0  n  1
n 1 p 
2

p
p
p
p









sin
n

1
sin
n

1
sin
n

1
sin
n

1
1
2 
2 
2 
2
 
p 
n 1
n 1
n 1
n 1

p
p

2 sin n  1
2 sin n  1
1 
2 
2
 
p 
n 1
n 1






note :
p

p

sin      cos   sin    
2

2






p
p

2 sin n  1
2 sin 1  n 
1 
2 
2
 
p 
n 1
n 1

np
np

cos
cos
2
2 
2
 
p  n 1
n 1







np
2  1  1 
 n  1 n  1
p


2 cos






np
2  n  1  n  1
 n 2  1 
p
2 cos
np
2 cos
 2 
2

 n 2  1 
p
np
4 cos
2 Pr ovided n  1

p n 2  1
When n  1
a1 
2
p
p

cos x cos xdx
0
p
p


2 2

cos x cos xdx    cos x cos xdx

p 0

p
2


p
p


2 2 2
2

cos
xdx

cos
dx


p 0

p
2


p
p


2  2 1  cos 2 x 
1

cos
2
x

 

dx




 dx


p  0
2
2

 
p 
2


p
p 

2
2  x sin 2 x 
 x sin 2 x 


 
  
 
p  2
4 0
4 p 
2
2

2 p
p
p 




p 4
2
4 
a1  0
np
4 cos
2 
2 cos nx
 f ( x)    
p n  2 p n 2  1
•11.Expand f(x) = x – x2 as a Fourier series in –l &lt; x &lt; l and using this series
•find the root square mean value of f(x) in the interval.
Sol. Fourier series is
a0  
np x
np
f ( x) 
   a n cos
 bn sin
2 n 1 
l
l
l
x


l
1
1
a 0   f ( x ) dx   ( x  x 2 ) dx
l l
l l
l
1x
x 
 
 
l 2
3 l
2
3
1 l 2 l 3   l 2 l 3 
       
l  2 3   2 3 
1  2l 3   2l 2
 
 
l 3 
3
l
l
1
npx
1
n px
2
a n   f ( x ) cos
dx   ( x  x ) cos
dx
l l
l
l l
l

n px

 sin
1
2
l
  ( x  x )
l
 np


l



  (1  2 x )




npx
  cos
l

 n 2p 2

l2



npx

  sin
l
  (2) 

 n 3p 3


l3


 (1) n l 2   
 (1) n l 2  
1 
 0  (1  2l ) 2 2   0  0  (1  2l ) 2 2   0
l 
 np   
 n p  
(1) n l 2
1  2l  1  2l 

2 2
lnp
(  1) n l
 4 l 

2
2
n p
4 l 2 (1) n 1

n 2p 2
l





  l
l
l
1
n px
1
n px
2
bn   f ( x ) sin
dx   ( x  x ) sin
dx
l l
l
l l
l
l


 npx 
npx 
npx 

  sin

 cos

 cos



1
l   (1  2x) 
l   (2) 
l 
 (x  x 2 )
 n2p 2 
 n3p 3 
l
 np 




 3 
2

l


 l

 l
 l
n
n 3
n
n 3 






1
(

1
)
l
2
(

1
)
l
(

1
)
l
2
(

1
)
l
2
2
0 3 3 (l l )
0 3 3 
 (l l )
l 
np  
n p 
 np 
 np 

 (  1) n l

l  l2  l  l2
l np
(  1) n 1

np

2 l 
2 l (  1) n  1

np

a0
np x
np x 

f ( x) 
   a n cos
 b n sin

2 n 1 
l
l 
1   2l2
 
2 3
   4 l 2 (  1) n  1
n p x 2 l (  1) n  1
np x 
   

cos

sin
2 2
l
np
l 
 n 1  n p
RMS value of f(x) in (–l, l ) is
y
2
2
a0
1


4
2
1   2l
 
4 3
2
l4
(i.e.) y 

9
2


n 1
2
2
2
(a n  bn )
 1
 
 2
16 l 4 (1) 2 n  2 4 l 2 (1) 2 n  2 




4 4
2 2
n
p
n
p
n 1 


 8l 4
2l2 

 4 4  2 2
n p 
n 1  n p

l

x
,
0

x


2
f ( x)  
l
l  x ,
 xl
2

12.Find the Fourier series expansion of
Hence deduce the value of
Sol.
Let
2L  l  L 

1

4
n 1 (2n  1)
l
2
then the given function becomes
0xL
 x,
f (x)  
2L  x, L  x  2L
Fourier series is

a0
np x
np x 

f ( x) 
   a n cos
 bn sin

2 n 1 
L
L 
1
a0 
L
2L
L
2L
1
1
0 f ( x) dx  L 0 (x) dx  L L (2L  x) dx
1

L
L
2L
x 
1  (2 L  x) 
   

2
L

2
 0

L
2
2
 1 
1  L2
L2 

 0 

0 

L  2
L

2




L
L

 L
2
2
1
an 
L
2L

0
np x
f ( x ) cos
dx
L
1

L
L
np x
1
x
cos
dx

0
L
L
2L

( 2 L  x ) cos
L
np x
dx
L
L
2L
 



np x 
np x  
np x 
np x  





sin
 cos
 cos

 sin



1  
1
L   (1)
L   ( 2 L  x )
L   ( 1)
L 
 ( x )
 n 2p 2 
 n 2p 2  
L   np
L

 np








 

L
L



L2
L2

 0

 L
1 
( 1) n L2  
L2  1 
L2  
( 1) n L2 
 0  2 2   0  2 2   0  2 2   0  2 2 
L 
n p  
n p  L  
n p  
n p 




1 L2
2L
n
n
n

(

1
)

1

1

(

1
)

(

1
)
1
2 2
2 2
Lnp
np
1
bn 
L
2L

0
1
np x

f ( x ) sin
dx
L
L
L
np x
1
x
sin
dx

0
L
L
2L

( 2 L  x ) sin
L
L
np x
dx
L
2L
 



np x 
np x 
np x 
np x 

  sin

  sin

 cos

  cos


1  
1
L   (1)
L   (2L  x)
L   (1)
L 
 ( x)
2 2
2 2
 np

 np

L   np
L

 np








 

L
L



L2
L2

 0

 L
 1
 (1) n L2 
1  (1) n L2 
 
 0  0  0  0  0  
 0
L 
np
np



 L
0
a0  
np x
np x 
f ( x) 
   an cos
 bn sin

2 n 1 
L
L 

 2 L [(  1) n  1]

L
np x
   
cos
 0 
2 2
2 n 1 
n p
L


 l [(  1) n  1]
l
2 np x 

   
cos
2 2
4 n 1  n p
l 
Using Parseval’s identity for Fourier series we have
1
L
2L

L
0
1
1
2
( x) dx 

L0
L
3 L
2

a0
2
2
2
[ f ( x )] dx 
  ( a n  bn )
2
n 1
2L

L
2
2
n


L
4L (1)  1
2
(2L  x) dx 
 
 0
4 4
2 n 1 
np

2

3 2L
1  x  1 (2L  x) 
L2 4L2  4
4
4

........
   
   4  4  0  4  0  4  0 ..........
L  3 0 L  3 L 2 p 1
3
5

1  L3  1 
L3  L2 16L2
  0  0     4
L 3
p
 L   3 2
1 1 1




..........
........
14 34 54

2 L2
L2
16 L 2  1
1
1






..........
........

3
2
p 4  1 4
34
54
L2 16 L2
 4
6
p
p4
96

1
1 1




..........
........
14 3 4 5 4

1
1
1


 .......... .......
4
4
4
1
3
5
( i .e .)
p
4
96



n 1
1
( 2 n  1) 4
. 13. Find the Fourier series expansion of
 l  x, 0  x  l
f ( x)  
l  x  2l
 0,
Hence deduce the value of the series
1 1 1
1 1 1
1   .......... and 2  2  2  .............
3 5 7
1 3 5
Sol.
Fourier series is

a0
np x
np

f ( x) 
   a n cos
 bn sin
2 n 1 
l
l
1
a0 
l
2l

0
l
2l
x


1
1
f ( x ) dx   (l  x ) dx   ( 0 ) dx
l 0
l l
l
1  (l  x ) 2 
 

l   2 0
1
an 
l
2l

0

1

0l2
 2l


l
2
l
np x
np x
1
f ( x) cos
dx   (l  x) cos
dx  0
l
l0
l
l


np x 
np x 

  cos

sin



1
l   (  1) 
l 
  (l  x )


np
l 


n 2p 2




2

l


l

 0
1  (1) n l 2  
l 2 
 0  2 2   0  2 2 
l 
n p   n p 


1 l2
n 1

(

1
)
1
2 2
ln p
1
bn 
l
2l

0

l
n p
2
2
(1)
n 1

1
l
np x
np x
1
f ( x ) sin
dx   (l  x ) sin
dx  0
l
l 0
l
l


np x 
np x 

  sin

  cos


1
l
l

  (  1) 
  (l  x )
2
2


np
l 


n p




2

l


l

 0
 l2

1
 {0  0}  
 0
l
 np 

l
np
a
f (x)  0 
2
np x
np x 

a
cos

b
sin



n
n
l
l 
n 1 

np x
np x 
l   l [(1) n 1  1]
l


  
cos

sin
2 2
4 n 1 
l
np
l 
np
px 1
3p x 1
5p x
l 2l  1

(i.e.) f (x)   2  2 cos  2 cos
 2 cos
 .................
4 p 1
l 3
l
l
5

l 1 p x 1 2p x 1 3p x

  sin  sin
 sin
 .................        (1
p 1
l 2
l
3
l

Put
l
x 
2
(which is point of continuity) in equation (1), we get
l l 2l
l 1 p 1
1 3p 1
1 5p

l    2 (0)   sin  sinp  sin  sin4p  sin  .................
2 4 p
p 1 2 2
3
2 4
5
2

l l l
 
2 4 p
1
1
1


1

0


0


0


..........
.......


3
5
7


l
l
l  1 1 1

  1     .......... ....... 
2 4 p  3 5 7

l
l 
1 1 1

 1     .......... ....... 
4 p 
3 5 7

p
1 1 1
1   ..........
.......
4
3 5 7
Put x = l in equation (1) we get
l 2l
f (l )   2
4 p
 1 1 1





..........
......
 12 32 52

But x = l is the point of discontinuity. So we have
f (l) 
f (l )  f (l ) (0)  (0)

0
2
2
l 2l  1 1 1

0   2  2  2  2 ............
4 p 1 3 5


l
2l  1
1
1

  2  2  2  2 .......... .. 
4
p 1
3
5

p2
8

1 1 1
 2  2 ................
2
1 3 5
f(x) = l – x
f(l–) = l – l =0
f(x) = 0
f(l) = 0
14.Find the Fourier series for the function
Deduce that


n 1
1
p2

2
8
(2n  1)
1 x,  2  x  0
f ( x)  
0 x2
1 x,
Solution:
f(– x) = 1 – x in (–2, 0)
= f(x) in (0, 2)
and f(– x) = 1 + x in (0, 2)
= f(x) in (–2, 0)
Hence f(x) is an even function.
a0
 f ( x) 

2
npx
a n cos

2
n 1

2
a0 
2
2
2
 f ( x ) dx

0
 (1  x ) dx
0
2

x 
 x 

2 0

2
(22) (0)
0
2
an 
2
2

0
n px
f ( x ) cos
dx 
2
2
 (1  x ) cos
0
n px
dx
2

npx

 sin

2
  (1  x ) 
 np



2


npx

  cos

2
  (  1) 
 n 2p 2




4

 4 (1) n  
4 
 0  2 2   0  2 2 
n p   n p 


4
n
 2 2 1  (  1)
p n

 0 , when n is even
 8
an  
, when n is odd
2
2
 p n
2





 0
0
f (x)  
2


n 1
8
n 2p
2
npx
cos
2
8  1 px 1 3px 1 5px

f (x)  2  2 cos  2 cos  2 cos ..........
..........
....
2 3
2 5
2
p 1

Put x = 0 in the above series we get
8 1 1
1

f (0)  2  2  2  2 ............
p 1 3 5

But x = 0 is the point of discontinuity. So we have
f (0)  f (0) (1)  (1) 2
f (0) 

 1
2
2
2
Hence equation (1) becomes
8 1
1
1

1  2  2  2  2 .......... ..
p 1
3
5

p2
1 1
1
 2  2  2 ................
8 1 3
5
(i.e.)
p2
8


n 1
1
( 2 n  1) 2
•15.Find the Fourier series of periodicity 3 for f(x) = 2x – x2 in 0 &lt; x &lt; 3.
Sol. Fourier series is

a0
2 np x
2 np x 

f ( x) 
   a n cos
 bn sin

2 n 1 
3
3 
3
3
1
2
2
a0 
f
(
x
)
dx

(
2
x

x
) dx


(3 / 2 ) 0
30
2 2x
 
3 2

2
x 


3 
3
3
0

2 
27 
9


(
0

0
)



3  
3 

 0
1
an 
(3 / 2 )
3

0
3
2
2 n px
2
f ( x ) cos nx dx   ( 2 x  x ) cos
dx
3 0
3
3



2npx 
2npx 
 2npx 
  cos

  sin

sin



2
3   (2  2x)
3   (2)
3 
 (2x  x 2 )
 4n2p 2 
 8n3p 3 
3
 2np 







 3 
9
27  0



2 
 9   
  9  
 0  (4) 2 2   0  0  (2) 2 2   0
3 
 4n p   
 4 n p  
2

3
  54 
 4 n 2 p 2 

9
n 2p
2
1
bn 
(3 / 2 )
3

0
3
2
2 n px
2
f ( x ) sin nx dx   ( 2 x  x ) sin
dx
3 0
3

2 n px

 cos


2
3
  ( 2 x  x 2 )
2 np
3



3


2 n px

  sin

3
  ( 2  2 x )
 4 n 2p 2




9



2 n px

 cos
3
  (  2 )

 8 n 3p 3


27


2 
 3 
 27  
 27 
 (3)
  0  2  3 3   0  0  2  3 3 
3 
 2np 
 8n p  
 8n p 

3
np
3





 0
a
f (x)  0 
2
2np x
2np x 

a
cos

b
sin
 n


n
3
3

n 1 

2 np x
3
2 np x 
 9
   2 2 cos

sin

3
np
3 
n 1  n p

HALF RANGE FOURIER SERIES
FOR THE INTERVAL
0 , p 
COSINE SERIES
SINE SERIES
a0 
f  x     an cos nx
2 n 1
1
a0 
an 
2
p
2
p
f  x    bn sin nx
n 1
p

f ( x ) dx
0
f ( x ) cos nxdx

p
0

bn 
2
p
f ( x ) sin nxdx

p
0
HALF RANGE FOURIER SERIES
FOR THE INTERVAL
COSINE SERIES
a0 
 npx 
f  x     an cos

2 n 1
 l 
2
a0 
l
SINE SERIES
 n px 
f  x    bn sin 

 l 
n 1

l

f ( x ) dx
0
2
 npx 
an   f ( x ) cos
dx
l 0
 l 
l
0 , l 
2
 npx 
bn   f ( x ) sin 
dx
l 0
 l 
l
•16. Find the half range sine series for f(x) = 2 in 0 &lt; x &lt; p.
Solution:
bn 
2
p
p
 f ( x ) sin nx dx

0
4   cos nx 
 
p  n 
p
0
2
p
p
 2 sin nx dx
0

 0 , when n is even

bn   8
,
when
n
is
odd
 n p
Half range sine series is
f ( x) 

4

(  1) n  1
np


n 1
n 1
 bn sin nx  
8
sin nx
np
. 17. Expand f(x) = cos x, 0 &lt; x &lt; π in a Fourier sine series.
Sol. Fourier sine series is

b
f ( x) 
n 1
bn 
sin nx
p
2
 f ( x) sin nx dx
p

n
0
1
p

2
p
p
 cos x sin nx dx
0
p
 2 sin nx cos x dx
0
2SinACosB = Sin(A+B) + Sin(A–B)

1
p
p
 [sin( n  1) x  sin( n  1) x] dx ,
n 1
0
p
1   cos(n  1) x    cos(n  1) x 
 


p 
n 1
n 1
 
 0
cos( n  1)p  (  1) n 1
cos( n  1)p  (  1) n 1
1  (1) n1 (1) n1   1
1 
  




p  n  1
n  1   n  1 n  1
1   1
1
1 
n  1
  (1) 




p 
 n  1 n  1  n  1 n  1
1   1
1
1 
n 1
 (1) 




p 
 n  1 n  1  n  1 n  1
1
n  2n   2n 
 (1)  2    2 
p 
 n  1  n  1


2n
n
bn 
(

1
)
1 , n  1
2
p ( n  1)
When n = 1, we have
b1 
2
p
p
 f ( x) sin x dx

0
1

p
2
p
p
 cos x sin x dx
0
p
 sin 2 x dx
0
p
1   cos 2 x 
 

p 
2
0
f ( x) 

b
n 1

n
sin nx  b1 sin x 


n2
1
  (1 1)  0
2p

b
n2
n
sin nx
2 n [ (  1) n  1]
sin nx
2
p ( n  1)
18.Find the half range cosine series for the function f(x) = x (π – x) in 0 &lt; x &lt; π.
Deduce that
1
1
1
p4
 4  4  .......... .. 
4
90
1
2
3
Sol. Half range fourier cosine series is
a0 
f ( x) 
  a n cos nx
2 n 1
a0 
2
p
p
 f ( x) dx
0

2
p
p
 x(p  x) dx
0
p
x 
2 p x
 


p 2
3 0
2
3

2  p 3 p 3 
     (0  0)
p  2 3 

2 p 3 
  
p 6 
an 
2
p

p2
3
p
 f ( x ) cos nx dx
0

2
p
p
 x (p  x ) cos nx dx
0
p
2
  cosnx
  sinnx
2  sinnx 
 (p x  x )
  (p  2x)
  (2) 3 
2
p
 n 
 n 
 n 0
  (p )(1) 
2  (p )(1) n
 0 
 0  0  2  0
2
p 
n
n

 

2p
n

(

1
)
1
2
p n




2
n
 
(

1
)
1
2
n
4
an  
, when n is even
2
n
0
, when n is odd

a0
f ( x) 
  a n cos nx
2 n 1
1 p 2
 
2 3


4
    2 cos nx
 n  even n
p2
 cos 2 x cos 4 x cos 6 x


4 


 .......... ..... 
2
2
2
6
4
6
 2

Parseval’s identity for half range fourier cosine series is
p
a0
2
[ f ( x )] dx 

p 0
2
2
2



n 1
an
2
p
1 p
[p x  x ] dx  

p 0
2 3
2
2
2 2
p
2

 



n  even
16
n4
p4
1 1 1

(p x  x  2p x )dx 
 16 4  4  4  ..............

p0
18
2 4 6

2
2 2
4
3
p
p 4 16 1 1 1
2 p x x 2p x 

 
 4  4  4  4  ..............

 
p 3
5
4  0 18 2 1 2 3

2 3
5
4
2  p 5 p 5 p 5   p 4  1 1 1






0





..........
....




p  3
5
2   18 14 2 4 34
2 p 5  p 4
1
1
1
 4  4  4  .......... .........
 
p  30  18 1
2
3
p4 p4
1 1 1

 4  4  4  .............
15 18 1 2 3
p4 1 1
1
(i.e.)
 4  4  4 ..........
...
90 1 2 3
20.Find the half range sine series of f(x) = x cos x in (0, π).
Sol. Fourier sine series is
f ( x) 

b
n 1
bn 


2
p
1
p
1
p
n
sin nx
p
 f ( x) sin nx dx
0

2
p
p
 x cos x sin nx dx
0
p
 x ( 2 sin nx cos x ) dx
0
p
 x [sin( n  1) x  sin( n  1) x] dx ,
0
n 1

1
p
p
1
p
 x sin( n  1) x dx  p  x sin( n  1) x dx ,
0
n 1
0
p
p
  sin(n  1) x 
  sin(n  1) x 
1    cos(n  1) x 
1    cos(n  1) x 
   x 

bn   x 
  (1) 
  (1) 
2
2
p 
n 1
n 1


 (n  1)
 0 p  
 (n  1)
 0
 1    p (  1) n 1



1    p (  1) n 1
 
 0   0  0   
 0   0  0
p  n  1


 p  n  1

(  1) n  2
(  1) n


n 1
n 1
1 
 1
 (  1) n 


 n  1 n  1


2n
 (  1) 

(
n

1
)(
n

1
)


n
( i .e .) b n
2 n (  1) n

, n  1
2
n 1
When n = 1, we have
b1 
2
p

p
 f ( x) sin x dx
0
1
p

2
p
p
 x sin
0
2 x dx
p
 x cos x sin x dx
0
p

1    cos 2 x 
  sin 2 x  
x

(
1
)





p 
2
4  0




1    1 
1
  p 
  0   {0  0}  
p   2 
2




n 1
n2
f ( x)   bn sin nx  b1 sin x   bn sin nx

2 n ( 1) n
1
  sin x  
sin nx
2
2
n 1
n2
21.Obtain the sine series for

 x
f (x)  
l  x

l
2
in
0  x 
in
l
 x  l
2
Sol. Fourier sine series is
np x
f ( x )   b n sin
l
n 1

l
2
npx
bn   f ( x) sin
dx
l 0
l
2

l
l/2

0
l
npx
2
npx
x sin
dx   (l  x) sin
dx
l
l l/2
l
 
npx 

cos


2 
l
  (1)
 ( x )
np
l  



 
l


np

l
.
cos

2   l 
2
   
l   2  np



np
  2
  l .sin
2

  n 2p 2
 
 
np
 2
2
l
.
sin
2 
2


2
2
l  n p

bn 
4l
n 2p
2
sin
np
2
np
np x
  2 2 sin
sin
2
l
n 1 n p

4l
l

npx  
  sin

l


2 2
 n p



l2

 l / 2



np


l
.
cos


 2
  l 

2
  0  0  {0  0}    

 l
  2  np











np x
f (x)  bn sin
l
n 1

l/2


npx  
npx 

  sin


cos



2
l
l

    (l  x )
  ( 1)
2 2
 n p

np
l 







l


l2

  0
np
  2
  l .sin
2

  n 2p 2
 
 






22.Find the half range cosine series for the function f(x) = x in 0 &lt; x &lt; l.
Hence deduce the value of the series

1

4
n  1 ( 2 n  1)
Sol. Half range Fourier cosine series is
a0 
np x
f ( x) 
  a n cos
2 n 1
l
2
a0 
l
2

l
l

f ( x ) dx
0
l
l
2  x2 
0 x dx  l  2 
0

2 l 2
   0  l
l 2

2
an 
l
l

0
l
np x
np x
2
f ( x ) cos
dx   x cos
dx
l
l 0
l
l


np x 
np x 

  cos

 sin


2
l   (1) 
l 
  ( x )


np
l 


n 2p 2




2

l


l

 0
2 
(  1) n l 2  
l 2 
  0  2 2   0  2 2  
l 
n p  
n p 

2l
n 2p 2
(1)  1
n
  4l
 2 2 , when n is odd
an   n p
 0 , when n is even

a0
np x
f ( x) 
  a n cos
2 n 1
l
l   4l
np x
   2 2 cos
2 n 1 n p
l
Using Parseval’s identity for half range Fourier cosine series we have
l
2
a0
2
2
[ f ( x )] dx 


l 0
2


n 1
an
2
l
2
l 2   16l 2 
2
( x ) dx     4 4 

l 0
2 n 1  n p 
l
2 x 
l 2 16 l 2
    4
l  3 0 2
p
3
2l 2 l 2
16 l 2


3
2
p4
1
1 1




..........
........
14 3 4 5 4

1
1
1




..........
........
 1 4 3 4 5 4

2
16
l
l
1 1 1

 4  4  4  4  ..................
6 p 1 3 5

2
p
4
1
1
1
 4  4  4  .......... .......
96 1
3
5
( i .e .)
p
4
96



n 1
1
( 2 n  1) 4
23.Prove that
4 
px 1
3p x
1
5p x

1
sin

sin

sin

..........
....

p 
l
3
l
5
l
in the interval 0 &lt; x &lt; l
Sol. Since RHS contains sine series and given 0 &lt; x &lt; l, we have to find
half range Fourie sine series for f(x) = 1

f ( x)   bn sin
n 1
2
bn 
l
l

0
np x
l
np x
2
f ( x ) sin
dx 
l
l
l
 (1) sin
0
np x
dx
l
l
np x 

 cos
n



2
2
(

1
)
l   l 
l
 
   
  

n
p
l 
l
n
p
n
p

 



 0
l


2
(  1) n  1  1
np

np x  2 [(1) n 1  1]
np x
f ( x)   bn sin

sin
l
np
l
n 1
n 1

2 2 p x
2 3p x
2 5p x

  sin  0  sin
 0  sin
 0  ...........
p 1
l
3
l
5
l

4 
px 1
3p x
1
5p x

1
sin
 sin
 sin
 .......... ... 

p 
l
3
l
5
l

24.Obtain the half range cosine series for f(x) = (x – 2)2 in the interval 0 &lt; x &lt; 2.
Sol. Half range cosine series is
a0 
npx
f ( x)    an cos
2 n 1
2
2
2
2
a0   f ( x) dx   ( x  2) 2 dx
20
0
2
 ( x  2) 


3

0
3
   8  8
 0    
  3  3
2
an 
2
2

0
n px
f ( x ) cos
dx 
2

n px

sin


2
2
 ( x  2) 
 np




2
2
2
(
x

2
)
cos

0
n px
dx
2
2


npx 
n px  

  sin

  cos


2   ( 2 )
2 
  [ 2 ( x  2 )]
2
2
3
3
 n p

 n p








4
8



 0

16


 0  0  0  0  2 2  0  
n p




16
p 2 n2
8  16
n px
f ( x )    2 2 cos
6 n 1 n p
2
( i.e.) f ( x ) 
4 16  1
px 1
2p x 1
3p x

 2  2 cos
 2 cos
 2 cos
 .... 
3 p 1
2 2
2
3
2

COMPLEX OR EXPONENTIAL FORM OF FOURIER SERIES
In  p  x  p

1
f ( x )   cn e , cn 
2p
n  
inx
p
p
f ( x)e inx dx

In 0  x  2p

1
f ( x )   cn e , cn 
2p
n  
inx
2p
 f ( x)e
0
inx
dx
In  l  x  l
f ( x) 

c e
n  
n
in p x
l
1
, cn 
2l
l
 f ( x )e
 in p x
l
dx
l
In 0  x  2 l
f ( x) 

c e
n  
n
in p x
l
1
, cn 
2l
2l
 f ( x )e
0
 in p x
l
dx
25.Find the complex form of the Fourier series of
in –1 &lt; x &lt; 1
Sol. The complex form of Fourier series of f(x) is given by

f (x) 
i np x
C
e
n
n  
1
1
i np x
Cn 
f
(
x
)
e
dx

2 (1) 1
1
1  x  i np x
 e e
dx
2 1
2l = 2
l=1
f (x)  e x
1
1  (1 i n p ) x
 e
dx
2 1
 (1i np ) x
1

1 e
 
2   (1 i np )  1

1
 (1 i n p )
(1 i n p )

e
e
2(1  i np )
 (1  i np ) 1  i np
1 i np

e
e

e
e
2 2
2(1  n p )



 (1  i n p )  1
1

e
(cos
n
p

i
sin
n
p
)

e
(cos n p  i sin n p )
2
2
2 (1  n p )

 (1  i np ) 1
n
1
n
Cn 
e
(

1
)

e
(

1
)
2(1  n 2p 2 )

(1  i n p ) (  1) n 1
1

e

e
2 (1  n 2 p 2 )



(1  i n p ) (  1 ) n

2 sinh 1
2
2
2 (1  n p )
(  1) n sinh 1 (1  i n p )
Cn 
1  n 2p 2
 f ( x) 


n
(1) n sinh1(1  i np ) i n p x
e
2 2
1 n p

26.Find the complex form of the Fourier series of
f ( x)  cos ax in  p  x  p
Solution:
f ( x) 

i np x
C
e
 n
n 
cn
1

2p
1

2p
p
p
f ( x ) e  inx dx

p
 inx
cos
ax
e
dx

p
p

1  e

 2 2 2  in cosax  a sin ax
2p  i n  a
 p
inx
note :
ax
e
ax
e
 cos bx  a 2  b 2 a cos bx  b sin bx 
1

2p
 e inp

e inp





in
cos
a
p

a
sin
a
p


in
cos
a
p

a
sin
a
p
 2

2
2
2
a

n
a

n




1
inp
inp
inp
inp


in
cos
a
p
e

e

a
sin
a
p
e

e




2
2
2p a  n 


1
2p a  n
2
2

1
2p a  n
2
2
2in cos ap sin np  2a sin ap cos np 

2a sin ap cos np 
n

 1 a sin ap

p a 2  n 2 
 f ( x) 


n  
 1n a sin a p
p a 2  n 2 
e inx
 sin np  0
HARMONIC ANALYSIS
The process of finding the Fourier series for a function given by
numerical value is known as harmonic analysis. In harmonic
analysis the Fourier coefficients a0, an and bn of the function
y = f(x) in (0, 2π) are given by
a0 = 2 [mean value of y in (0, 2π)]
an = 2 [mean value of y cosnx in (0, 2π)]
bn = 2 [mean value of y sinnx in (0, 2π)]
27. Find the Fourier series expansion up to third harmonic from the following
data:
x: 0
1
2
3
4
5
f(x) : 9
18
24
28
26
20
Sol.
Here the length of the interval is 6
l=3
(i.e.) 2l = 6

Fourier series is
a0  
np x
np x 
f (x)    an cos
 bn sin

2 n 1 
3
3 
a0
px
2p x
3p x
px
2p x
3p x
 a1 cos
 a2 cos
 a3 cos
 b1 sin
 b2 sin
 b3 sin
2
3
3
3
3
3
3
a
(i.e.) f ( x)  0  a1 cos   a2 cos 2  a3 cos 3  b1 sin   b2 sin 2  b3 sin 3
2
px
where  
3
f ( x) 
x
y
θ=πx/3
ycosθ
ycos2θ
ycos3θ
ysinθ
ysin2θ
ysin3θ
0
9
0
9
9
9
0
0
0
1
18
π/3
9
–9
–18
15.588
15.588
0
2
24
2π/3
–12
–12
24
20.784
–20.784
0
3
28
π
–28
28
–28
0
0
0
4
26
4π/3
–13
–13
26
–22.516
22.516
0
5
20
5π/3
10
–10
–20
–17.32
–17.32
0
–25
–7
–7
–3.464
0
0
Total 125
Here n = 6
 y 
125
a0  2 [mean value of y]  2 

2
 41.667



 6 
 n 
  y cos 
  25
a1  2 [mean value of y cos ]  2 

2
  8.333



n
 6 


  y cos 2 
a 2  2 [mean value of y cos 2 ]  2 
2
n


 7
 6    2.333
 
  y cos 3 
a 3  2 [ mean value of y cos 3 ]  2 
2
n


 7
 6    2.333


  y sin 
b1  2 [ mean value of y sin  ]  2 
n


  3 . 464 

2
  1 .155



6



  y sin 2 
0
b2  2 [ mean value of y sin 2 ]  2 

2
0



n
6


  y sin 3 
0
b3  2 [mean value of y sin 3 ]  2 
2  0
n
6


41 .667
 8 .333 cos   2.333 cos 2  2.333 cos 3  1 .155 sin   0 sin 2  0 sin 3
2
px
(i.e.) f ( x )  20 .833  8.333 cos   2.333 cos 2  2.333 cos 3  1 .155 sin  where  
3
 f ( x) 
28.Find the Fourier series expansion up to second harmonic from the
following data:
x:
0
f ( x) :
10
p
3
12
2p
3
15
p
20
4p
3
17
5p
3
11
2p
10
Sol. Since the last value of y is a repetition of the first, only the first six values
will be used.
Fourier series is
a0 
f ( x)   (an cos n x  bn sin n x)
2 n 1
a0
(i.e.) f ( x)   a1 cos x  a2 cos 2x  b1 sin x  b2 sin 2x
2
x
y
ycosx
ycos2x
ysinx
ysin2x
0
10
10
10
0
0
π/3
12
6
–6
10.392
10.392
2π/3
15
–7.5
–7.5
12.99
–12.99
π
20
–20
20
0
0
4π/3
17
–8.5
–8.5
–14.722
14.722
5π/3
11
5.5
–5.5
–9.526
–9.526
Total
85
–14.5
2.5
–0.866
2.598
Here n = 6
 y 
 85
a0  2 [mean value of y]  2 
  2    28.333
6
 n 
  y cosx 
 14.5
a1  2 [mean value of y cosx]  2 

2
  4.833



n
 6 


  y cos 2x 
 2.5 
a2  2 [mean value of y cos 2x]  2 
  2    0.833
n
 6 


  y sin x 
  0.866
b1  2 [mean value of y sin x]  2 
  0.289
2

n
 6 


  y sin 2 x 
 2.598
b2  2 [mean value of y sin 2 x]  2 
 0.866
2

n
 6 


28.333
 f ( x) 
 4.833cos x  0.833cos 2 x  0.289sin x  0.866sin 2x
2
(i.e.) f ( x)  14.1665  4.833cos x  0.833cos 2 x  0.289sin x  0.866sin 2 x
29.Find the Fourier series expansion up to first harmonic from the
following data:
x: 0 1
2
f (x) : 18 18.7 17.6
3
15
4
11.6
5
8.3
6
6
7
5.3
8
6.4
Sol. Here the length of the interval is 12
l=6
(i.e.) 2l = 12

Fourier series is
a0  
np x
np x 
f ( x) 
   a n cos
 bn sin

2 n 1 
6
6 
a0
px
p x
f ( x) 
 a 1 cos
 b1 sin
2
6
6
a
p x
( i.e.) f ( x )  0  a 1 cos   b1 sin  where  
2
6
9
9
10
12.4
11
15.7
x
0
y
18
θ=πx/6
0
cosθ
1
ycosθ
18
ysinθ
0
1
18.7
π/6
0.866
16.1942
9.35
2
17.6
2π/6
0.5
8.8
15.2416
3
15
3π/6
0
0
15
4
11.6
4π/6
–0.5
–5.8
10.0456
5
8.3
5π/6
–0.866
–7.1878
4.15
6
6
π
–1
–6
0
7
5.3
7π/6
–0.866
–4.5898
–2.65
8
6.4
8π/6
–0.5
–3.2
–5.5424
9
9
9π/6
0
0
–9
10
12.4
10π/6
0.5
6.2
–10.7384
11
15.7
11π/6
0.866
13.5962
–7.85
Total
144
36.0128
18.0064
Here n = 12
 y 
144 
a 0  2 [mean value of y ]  2 

2
 24



 12 
 n 
  y cos 
36.0128
a1  2 [mean value of y cos ]  2 

2
 6.002



n
 12 


  y sin  
18.0064 
b1  2 [mean value of y sin  ]  2 

2
 3.001



n
 12 


24
 f ( x) 
 6.002 cos   3.001sin 
2
(i.e.) f ( x)  12  6.002 cos   3.001sin 
where  
px
6
1. Write down the form of the Fourier series of an odd function in (– l, l)
and the associated Euler’s formula for the Fourier coefficients.
Sol.
np x
f ( x )   bn sin
l
n 1

2
np x
bn   f ( x ) sin
dx
l 0
l
l

2. If f(x) = 3x – 4x3 defined in the interval (– 2, 2) then find the value
of a1 in the Fourier series expansion.
Sol. Since f(x) is an odd function, an = 0.
a1  0
3.
Obtain the first term of the Fourier series for the function
f(x) = x2, – π &lt; x &lt; π
Sol. f(x) = x2 is an even function.
Fourier series is
a0 
2
p
p

0
f ( x ) dx 
2
p
a
f (x)  0 
2

a
n 1
n
p
2
x
 dx
0
p
 2p 2
2  x3 
2 p 3
    
 0 
p  3 0 p  3
3

Hence the first term of the Fourier series is
a0
p 2

2
3
cos nx
4.If f(x) = 2x in the interval (0, 4) then find the value of a2 in the Fourier series
expansion.
Sol.
1
2p x
a2   2x cos
dx
20
2
4
4


x cos p x dx
0
  sin p x 
  cos p x  
  x
  (1 ) 

2
p
p



 
1  
1 

 0 

0


2 
2 
p
p

 

 0
4
0
5. Define root mean square value of a function
Sol. The root mean square value of f(x) over the interval (a, b) is defined as
b
RM S  y 

[ f (x)]2 dx
b
1
2
y 
[
f
(
x
)]
dx

ba a
2
a
ba
6.Find the root mean square value of f(x) = x2 in (0, l)
Sol.
RM S y
2
l
1
  [ f ( x )] 2 dx
l 0
l
1 22
1 4
1  x  1 l 5  l 4
 [x ] dx   x dx       0 
l0
l0
l  5 0 l  5  5
l
l
5
l2
y
5
7.Find the root mean square value of a function f(x) in (0, 2π)
Sol. The root mean square value of f(x) over the interval (0, 2π) is defined as
1
RM S  y 
2p
2p

0
1
2
[ f (x)] dx  y 
2p
2
2p

[ f (x)]2 dx
0
8.Find the root mean square value of f(x) = 1 – x in 0 &lt; x &lt; 1
Sol.
RM S y
2
1
1
  [ f ( x )] 2 dx
10
3 1
(1 x)    1  1
 (1 x) dx 
 0   

 3  0   3 3
0
1
2
1
 y 
3
9.Find the root mean square value of f(x) = π – x in 0 &lt; x &lt; 2π
Sol.
1
RM S y 
2p
2
1

2p
2p
2p
2
[
f
(
x
)]
dx

0
1  (p  x ) 
0 (p  x ) dx  2 p   3 
2
3
2p
0
1  p 3   p 3  
  
 

 
2p   3    3  
1  2p 3  p 2




2p  3 
3
y
p
3
10.Write the sufficient conditions for a function f(x) to satisfy for the existence
of a Fourier series.
Sol. i) f(x) is defined and single valued except possibly at a finite number of points
in (–π, π)
ii) f(x) is periodic with period 2π
iii) f(x) and
f (x) are piecewise continuous in (–π, π)
Then the Fourier series of f(x) converges to f(x)
a) if x is a point of continuity
b)
f (x  0)  f (x  0)
2
if x is a point of discontinuity.
x , 0  x  1
f ( x)  
2 , 1  x  2
11.Find the sum of the Fourier series for
at x = 1
Sol. Here x = 1 is a point of discontinuity
f (1) 
f (1 0)  f (1 0) 1 2 3


2
2
2
12.State the Parseval’s identity for Fourier series.
Sol. The Parseval’s identity for Fourier series in the interval (c, c + 2l) is
1
l
c  2l
2
a0
2
[
f
(
x
)]
dx


c
2

 ( a n  bn )
2
2
n 1
The Parseval’s identity for Fourier series in the interval (c, c + 2π) is
c  2p
2
a0
2
[ f ( x )] dx 


p c
2
1

 ( a n  bn )
n 1
2
2
12.Write the formula for finding Fourier coefficients.
Sol.
1
a0 
l
c  2l
1
an 
l
 f ( x) dx
c
1
bn 
l
c  2l

f ( x ) sin
c
c 2l

f ( x ) cos
c
np x
dx
l
np x
dx
l
13.Define RMS value of a function.
Sol. The RMS value of a function f(x) in (a,b) is defined by
b
y 
y
2
1
2
[
f
(
x
)]
dx

ba a
b
1
2

[
f
(
x
)]
dx

ba a
14.State the Parseval’s identity for Fourier series.
Sol. The Parseval’s identity for Fourier series in the interval (c, c + 2l) is
1
l
c  2l
2
a0
c [ f ( x)] dx  2 
2

 ( a n  bn )
2
2
n 1
The Parseval’s identity for Fourier series in the interval (c, c + 2π) is
1
p
c  2p
2
a0
2
[
f
(
x
)]
dx


c
2


n 1
2
2
(a n  bn )
15.Find the mean square value of the function f(x) = x in the interval (0, l).
Sol. Mean square value is
l
y
2

1
2
[
f
(
x
)]
dx

l 0
1

l
l

0
1x 
x dx  

l  3 
2
3
l
0

1 l3
 
 0
l  3

l2

3
16.Find the value of an in the cosine series expansion of f(x) = 10 in
the interval (0,10).
Sol.
10
10
2
npx
2
npx
an 
f
(
x
)
cos
dx

(
10
)
cos
dx


10 0
10
10 0
10
10
npx 

sin


20
10
sin np  0  0
 2


n
p
np


 10  0
. 17.
What is the constant term a0 and the coefficient a&shy;n in the Fourier series
expansion of
f(x) = x – x3 in (–p,p).
Sol. Since the interval is (–p,p), let us verify whether the function is odd or even
f(– x) = (– x) – (–x)3
= – x + x3 = – (x – x3) = – f(x)
The given function is an odd function.
Hence a0 = 0 and an = 0.
Find the constant term in the Fourier series corresponding to f(x) = cos2 x
expanded in the interval (–p,p).
Sol. Since the interval is (–p,p), let us verify whether the function is odd or even.
f(–x) = cos2 (–x)= cos2x = f(x). Hence the function is even.
. 18.
a0 
2
p
p

2
f ( x ) dx 
p
0
2

p
p

cos
p
1  cos 2 x
dx
2
2
x dx
0

0
p
1 
sin 2 x 

x

p 
2  0
1
( p  0 )  ( 0 ) 

p
1
Hence the constant term in the Fourier expansion is
a0
1

2
2
19. Find the constant term in the Fourier expansion of f(x) = x2 – 2 in
–2 &lt; x &lt; 2
Sol. f(–x) = (–x)2 – 2 = x2 – 2 = f(x), which is an even function
2
a0 
2
2

0
2
f ( x) dx   ( x 2  2) dx
0
2
x

   2 x
3
0
3
 8

4

   4   (0)   
3

 3

Hence the constant term in the Fourier expansion is
a0  4 / 3
2


2
2
3
20. If the Fourier series of the function f(x) = x + x2, in the interval (–p,p) is
--------------- (1)
p2
3
, then


2

n 4
(

1
)
cos
nx

sin
nx

 n 2

n
n 1
find the value of the infinite series
Sol. Given
f ( x) 
p2
1
1
1


 .......... ...
2
2
2
1
2
3

2
4

  (1) n  2 cos nx  sin nx
3
n
n

n 1
Put x = π in the above series we get
p2

4
f (p ) 
  (1)  2 ( 1) n 
3
n
n 1
n

0

But x = π is the point of discontinuity. So we have
f (p)  f (p) (p  p 2 )  (p  p 2 ) 2p 2
f (p) 


p2
2
2
2
Hence equation (1) becomes
p
2

 4

n
p 
  (  1)  2 (  1)  0 
3
n

n 1
2n
2

4
(

1
)
p
p2

2
3
n
n 1
2
n
2p 2
1
1
1

 4  2  2  2  .......... ..... 
3
2
3
1

p2
1
1
1
 2  2  2  .......... .....
6
1
2
3
21. Does f(x) = tan x posses a Fourier series? Justify your answer.
Sol. For a function f(x) to have Fourier series expansion it must satisfy all the
three criteria in Dirichlet’s conditions. But f(x) = tan x has value ∞ at
p
x
2
and so it is a discontinuous point and moreover it is an infinite discontinuity. So
it does not have a Fourier series expansion.
```