1
TRANSFORMS AND PARTIAL DIFFERENTIAL EQUATION
UNIT I – FOURIER SERIES
PART – A
1. State the Dirichlet’s conditions for a given function to expand in Fourier series.
Sol. The Dirichlet’s conditions are
(i) f(x) is periodic with period 2l in (c, c + 2l) and f(x) is bounded.
(ii) The function f(x) must have finite number of maxima and minima.
(iii) The function f(x) must be piecewise continuous and has a finite number of finite
discontinuities.
Then the Fourier series of f(x) converges in (c, c + 2l)
2. Write the formula for finding Fourier coefficients.
c  2l
c  2l
c  2l
n x
n x
1
1
1
dx
dx , bn   f ( x) sin
Sol. a0   f ( x) dx , an   f ( x) cos
l
l
c
l
c
l
c
l
3. Define RMS value of a function.
Sol. The RMS value of a function f(x) in (a,b) is defined by
b
1
y
[ f ( x)] 2 dx
b  a a
b
1
y 
[ f ( x)] 2 dx

ba a
2
4. State the Parseval’s identity for Fourier series.
Sol. The Parseval’s identity for Fourier series in the interval (c, c + 2l) is
c  2l
a
1
[ f ( x)] 2 dx  0 

l c
2
2

 (a
n 1
 bn )
2
2
n
The Parseval’s identity for Fourier series in the interval (c, c + 2π) is
c  2
2
a
[ f ( x)] dx  0 

 c
2
1
2

 (a
n 1
2
n
 bn )
2
 2x
1  ,    x  0
5. In the Fourier series expansion of f ( x)   
in (–,),
2x
1
, 0 x


find the coefficient of sin nx
Sol. Since the interval is (–,), let us verify whether the function is odd or even
f ( x)  1 
and f ( x)  1 
2(  x )

 1
2x

in ( , 0)
 f ( x) in (0,  )
2(  x )
 1

2x

in (0,  )
 f ( x) in ( , 0)
Hence f(x) is an even function.
So the coefficient of sin nx (i.e.) bn = 0.
6. Find the mean square value of the function f(x) = x in the interval (0, l).
Sol. Mean square value is
2
l
y 
1
[ f ( x)] 2 dx

l0
l
l

1 2
1  x3 
1 l 3
  x dx       0
l0
l  3 0 l3


l2
3
2
7. Find the value of an in the cosine series expansion of f(x) = 10 in the interval (0,10).
Sol.
10
10
2
nx
2
nx
an 
10

f ( x) cos
0
10
dx 
10
 (10) cos 10
dx
0
nx 

 sin 10 
20
sin n  0  0
 2


n
 n 
 10  0
10
8. What do you mean by Harmonic Analysis.
Sol. The process of finding the Fourier series for a function given by numerical value is known
as harmonic analysis.
9. What is the constant term a0 and the coefficient an in the Fourier series expansion of
f(x) = x – x3 in (–,).
Sol. Since the interval is (–,), 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.
10. Find the constant term in the Fourier series corresponding to f(x) = cos2 x expanded
in the interval (–,).
Sol. Since the interval is (–,), let us verify whether the function is odd or even.
f(–x) = cos2 (–x)= cos2x = f(x). Hence the function is even.
a0 
2


 f ( x) dx

0

2


 cos
2
x dx
0

1  cos 2 x
dx
 0
2
2


1
sin 2 x 
 x 

2  0
1
 (  0)  (0)

1
Hence the constant term in the Fourier expansion is
a0 1

2 2
11. Find the Fourier constant bn for f(x) = xsinx in (–π, π)
Sol. f(–x) = (–x)sin(–x) = (–x)(–sinx) = xsinx = f(x)
Hence the function is an even function. So bn = 0.
12. Find the constant term in the Fourier expansion of f(x) = x2 – 2 in –2 < x < 2
Sol. f(–x) = (–x)2 – 2 = x2 – 2 = f(x), which is an even function
2
a0 
2
2
2
 f ( x) dx   ( x
0
2
 2) dx
0
2
 x3

   2 x
3
0
 8

4

   4   (0)  
3

 3

Hence the constant term in the Fourier expansion is
a0  4 / 3
2


2
2
3
3
13. Find an in expanding e  a x as a Fourier series in (–,).
Sol.
an 

1
 f ( x) cos nx dx




1
 e

a x
cos nx dx

1  e  a x
  2
  a  b 2



 (a cos nx  n sin nx)

 

  e a

1  e  a 
n
[

a
(

1
)

0
]
[ a (1) n  0] 
 2
 2
2
2
  a  b
 a  b


 a (1) n
( e  a  e a  )
2
2
 (a  b )
a (1) n

( e a  e  a )
2
2
 (a  b )
(i.e.) a n 
a (1) n
2 sinh a
 (a 2  b 2 )
14. If f(x) is an odd function in the interval (–l, l), write the formula to find the Fourier
coefficients.
Sol. a0 = an = 0
l
n x
2
bn 
 f ( x) sin
l
0
dx
l
15. If f(x) is an even function in the interval (–l, l), write the formula to find the Fourier
coefficients.
l
l
n x
2
2
dx , bn = 0
Sol. a0   f ( x) dx , an   f ( x) cos
l
l
0
l
0
16. Find a0 if f ( x)  x , expanded as a Fourier series in (–,).
Sol. f ( x)   x  x  f ( x) , which is an even function.
a0 

2


2
f ( x) dx 

0

2


 | x | dx
0

 x dx
0

2  x2 
  
  2 0


2  2
 0  

 2

17. Find a sine series for f(x) = x in (0, π)
Sol.
bn 
2


 f ( x) sin nx dx

0
2


 x sin nx dx
0

2    cos nx 
  sin nx 
  x
  (1)

2
  n 
 n
 0

 2(1) n 1

2    (1) n

0

{
0

0
}



 
n
n


Sine series is

2(1) n  1
 sin x sin 2 x sin 3x

sin nx  2


 .................
n
2
3
 1

n 1

f ( x)   bn sin nx 
n 1
4
18. To which value the half range sine series corresponding to f(x) = x2 expressed in the
interval (0,2) converges at x = 2?
Sol. At x = 2 (which is point of discontinuity), the
x2
Half range Fourier sine series converges to
–2 – x2
2 – (4 – x )2 4
0
2
2
f ( 2  )  f ( 2  ) 2  [  ( 4  2) ]

2
2
44

0
2
19. If the Fourier series of the function f(x) = x + x2, in the interval (–,) is
2


2
4

cos
nx

sin nx , then find the value of the infinite series
2
n
n

 (1) n 
3 n 1
1
1
1
 2  2  .............
2
1
2
3
Sol. Given f ( x) 
2

2
4

  (1) n  2 cos nx  sin nx
3
n
n

n 1
Put x = π in the above series we get

2
4

f ( ) 
  (1) n  2 (1) n  0 --------------- (1)
3
n
n 1

But x = π is the point of discontinuity. So we have
f ( )  f ( ) (   2 )  (   2 ) 2 2
f ( ) 


2
2
2
2
Hence equation (1) becomes
2

4
  (1) n  2 (1) n 
3
n
n 1

4 (1) 2 n
2
2 

3
n2
n 1
 
2

0

2 2
1
1
1

 4  2  2  2  ...............
3
2
3
1

2
6

1
1
1
 2  2  ...............
2
1
2
3
20. The cosine series for f(x) = x sin x in 0 < x <  is given as
x sin x  1 
Sol.
 ( 1) n
1
1
1
1
 


 .... 
cos x  2  2
cos nx . Deduce that 1  2
2
1.3 3.5 5.7
 2
n2 n  1
As n2 –1 = ( n – 1)( n + 1)
x sin x  1 
Put x 

1
 cos 2 x cos 3x cos 4 x cos 5 x cos 6 x

cos x  2 




 ................ 
2
2.4
3.5
4.6
5.7
 1.3


in the above series we get
2
1
1
1
 1

(0)  2 
0
0
 ................ 
2
2
3.5
5.7
1.3


1
1
 1

 1 2 


 ................ 
2
1.3 3.5 5.7

(1)  1 
21. Find the half range sine series for f(x) = 2 in 0 < x < .
Sol.
bn 
2


 f ( x) sin nx dx
0

2


 2 sin nx dx
0

4   cos nx 
4
4
 

(1) n  1 
1  (1) n

  n  0 n
n




5
Half range sine series is
4 [1  (1) n ]
sin nx
n
n 1


f ( x)   bn sin nx 
n 1

4  2 sin x 2 sin 3x 2 sin 5 x




..........
.......

  1
3
5

8  sin x sin 3x sin 5 x



 .................

 1
3
5

22. 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 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.
23. Does f(x) = sin (1/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) = sin(1/x) has minimum or maximum value at odd
multiple of
1 (2n  1)

. (i.e.) when 
x
2
2
2
x
(2n  1)
As n tend to ∞, x = 0. So the function does not have a Fourier series expansion.
24. Without finding the values of a0, an and bn, the Fourier coefficients of Fourier series,
 a0 2
for the function f(x) = x2 in the interval (0, π) find the value of 
 2
2
Sol.
a0

2

 (an  bn ) 
2
2
n 1


2

2


 (a
n 1
2
n
2 
 bn )


[ f ( x)]

2

2
dx
0

[x
2 2
] dx
0

x
4
dx
0

 2 4
2  x5 
2  5
      0 
  5 0   5
5

a0 
2
2
  (an cos n x  bn sin n x) is the Fourier series of f(x) = x in (–1, 1), find a3  b3
25. If
2 n 1
Sol. Since the given function f(x) = x is an odd function, we have a0 = 0, an = 0 (i.e.) a3 = 0
1
bn 
1
2
f ( x) sin n x dx  2  x sin n x dx
1 0
0
1
   cos n x 
  sin n x 
 2( x)
  (1)

2 2
n

 n
 0
 
 (1) n


 2  
 0  {0  0}

 n

n
2 (1)
bn  
n
3
2 (1)
4 (1) 6
4
2
b3  
 b3 

2
3
9
9 2
4
4
2
2
 a3  b3  0  2  2
9
9
6
PART – B
1. Find the Fourier series for the function f(x) = 1 + x + x2 in (–π, π).
1
1
1
2



..........
....

6
12 2 2 3 2
Deduce
Sol. The given function is neither an even nor an odd function.
f ( x) 
a0 

a0
  (a n cos nx  bn sin nx)
2 n 1

1

 f ( x) dx
1




 (1  x  x
2
) dx


1
x2 x3 
 x 
 

2
3  
an 
1 
2 3 
 2  3 


 
    

 
2
3  
2
3 

1
2 3 
2 2
2 
  2

3 
3

1


 f ( x) cos nx dx



1

(1  x  x
2
) cos nx dx


1
 sin nx 
  cos nx 
  sin nx 
 (1  x  x 2 )
  (1  2 x)
  (2)

2
3

 n 
 n

 n
  
 

(1  2 ) (1) n
(1  2 ) (1) n
1 
0


0

0

 0 
 

2
2
 
n
n
 
 cos n  ( 1) n
(1) n
sin n  0
1  2  1  2 

2
n
cos n(  )  cos n
n
n
(1)
4 (1)
 ( 1) n

(
4

)

 n2
n2

bn 
1


 f ( x) sin nx dx


1


 (1  x  x
2
) sin nx dx


1
  cos nx 
  sin nx 
 cos nx 
 (1  x  x 2 )
  (1  2 x)
  (2)

2
3

n


 n

 n   

n
n
2 (1) n  
2 (1) n 
1 
2 ( 1)
2 ( 1)

(
1




)

0



(
1




)

0

 


 
n
n
n3  
n 3 

(1) n

1    2 1    2
n

 2 (1) n
2 (1) n 1
(1) n

(2 ) 

n
n
n
f ( x) 
a0 
  (a n cos nx  bn sin nx)
2 n 1
1
2 2
 2 
2
3
   4(1) n

2(1) n 1
   
cos
nx

sin nx 
2
n
 n 1  n

2
 cos x cos 2 x cos 3 x

 sin x sin 2 x sin 3 x

 1
 4  2 

 ..........  2 


 ..........
2
2
3
2
3
2
3
 1

 1


(i.e.) f ( x)  1 
2
 cos x cos 2 x cos 3x
  sin x sin 2 x sin 3x

 4 2 
 2  ............  2


 ............
2
3
2
3
2
3
 1
  1

Put x = π in the above series we get
2
1
 1 1

f ( )  1 
 4 2  2  2  ............  2(0) --------------- (1)
3
 1
2
3

7
But x = π is the point of discontinuity. So we have
f ( )  f ( ) (1     2 )  (1     2 ) 2  2 2


 1  2
2
2
2
f ( ) 
Hence equation (1) becomes
2
1
1
 1

 4  2  2  2  ............
3
2
3
 1

2

1
1
1

2 
 4  2  2  2  ............
3
2
3
1

1  2  1
2 2
1
1
1

 4  2  2  2  ............
3
2
3
1

2

1
1
1
 2  2  2  ................
6 1
2
3
1  x,
1  x,
2. Find the Fourier series for the function f ( x)  

Deduce that
1
 (2n  1)
n 1
Sol.
2


2 x 0
0 x2
2
8
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 
nx
 f ( x) 
  an cos
2 n 1
2
a0 
2
2
2

2
f ( x ) dx 
0
 (1  x) dx
0
2

x2 
 x  
2 0

 (2  2)  (0)
0
2
an 
2
2

0
nx
nx
f ( x) cos
dx   (1  x) cos
dx
2
2
0
2
2


nx 
nx 

  cos

 sin


2   (1)
2 
 (1  x)
 n 2 2  
 n 






 2 
4

 0

4 (1) n  
4 
 0  2 2   0  2 2 
n  
n  

4
 2 2 1  (1) n
 n

f ( x) 

0  4 [1  (1) n ]
nx

cos
2 2
2 n 1
2
n
4 2
x
2
3x
2
5x

cos  0  2 cos
 0  2 cos
 ........................
2  2
2
2
2
 1
3
5

8 1
x 1
3x 1
5x

f ( x)  2  2 cos  2 cos
 2 cos
 ........................
2 3
2
2
 1
5


Put x = 0 in the above series we get
f (0) 
8 1
1
1

 2  2 ............ --------------- (1)
2  2
 1 3 5

cos n  (1) n
sin n  0
cos 0  1
sin 0  0
8
But x = 0 is the point of discontinuity. So we have
f (0)  f (0) (1)  (1) 2

 1
2
2
2
f (0) 
f(x) = 1 + x
f(0–) = 1 + 0 = 1
Hence equation (1) becomes
1
2
f(x) = 1 – x
f(0) = 1 – 0 = 1
8 1 1
1

 2  2 ............
2  2
 1 3 5

1 1
1


................
8 12 3 2 5 2

2
1
(i.e.)

2
8
n  1 ( 2n  1)

3. Expand f(x) = cos x, 0 < x < π in a Fourier sine series.
Sol. Fourier sine series is

f ( x)   bn sin nx
n 1
bn 
2



f ( x) sin nx dx 
0



2
 cos x sin nx dx

0

1
 2 sin nx cos x dx

0
2SinACosB = Sin(A+B) + Sin(A–B)

1
 [sin( n  1) x  sin( n  1) x] dx ,

n 1
0
cos( n  1)  (1) n 1
cos( n  1)  (1) n1

1   cos( n  1) x    cos( n  1) x 
 


 
n 1
n 1
 
 0



1  (1) n 1 (1) n 1   1
1 





  n  1
n  1   n  1 n  1 
1   1
1 
1 
 1
(1) n 





 
 n  1 n  1   n  1 n  1 
1   1
1 
1 
 1
(1) n 





 
 n  1 n  1   n  1 n  1 
1 
 2 n   2 n 
(1) n  2
 2


 
 n  1   n  1 
2n
bn 
(1) n  1 , n  1
2
 (n  1)



When n = 1, we have
b1 
2



2
f ( x) sin x dx 

0

1


 cos x sin x dx
0

 sin 2 x dx
0

1   cos 2 x 
1
 

(1  1)  0

  2 0
2


f ( x)   bn sin nx  b1 sin x   bn sin nx
n 1
n2

 0
n2
2n [ (1) n  1]
sin nx
 (n 2  1)

2  4 sin 2 x
8 sin 4 x
12 sin 6 x

0
0
 0  ..................

 3
15
35


8  sin 2 x 2 sin 4 x 3 sin 6 x



 ..................

 3
15
35

9
4. Find the Fourier series expansion of f(x) = x2, 0 < x < 2π. Hence deduce that
1
1
1
2
(i ) 2  2  2  .............. 
6
1
2
3
1
1
1
2
(ii ) 2  2  2  .............. 
12
1
2
3
1
1
1
2
(iii ) 2  2  2  .............. 
8
1
3
5
Sol.
Fourier series is

a0
f ( x) 
  (a n cos nx  bn sin nx)
2 n 1
a0 
2
1


f ( x) dx 
0
1

2
x
2
dx
0
2
1  x3 
  
  3 0

an 
 8 2
1  8 3

0
 
  3
3

2
1
 f ( x) cos nx dx


0
1

2
x
2
cos nx dx
0
2
1
 sin nx 
  cos nx 
  sin nx 
 ( x 2 )
  (2 x)
  (2)

2
3

 n 
 n

 n
 0

(4 ) (1)
1 

 0  0  0  0
0 

2
 
n


4
 2
n

bn 
2
1
 f ( x) sin nx dx

0

1

cos 2n  1
sin 2n  0
2
x
2
sin nx dx
0
2
1
  cos nx 
  sin nx 
 cos nx 
 ( x 2 )
  (2 x)
  (2)

2
3

n


 n

 n  0
1  4 2
2 
2 
  
 0  3   0  0  3 
  n
n  
n 
4

n
f ( x) 
a0 
  (a n cos nx  bn sin nx)
2 n 1
  4
4

    2 cos nx 
sin nx 
n

 n 1  n
4 2
 cos x cos 2 x cos 3 x

 sin x sin 2 x sin 3 x

f ( x) 
 4 2 

 .......... .....  4 


 ............
2
2
3
2
3
2
3
 1

 1


1  8 2

2  3
Put x = 0 in the above series we get
4 2
1
1
1

f (0) 
3
 4  2  2  2  ............  4 (0) --------------- (1)
2
3
1

But x = 0 is the point of discontinuity. So we have
f (0) 
f (0)  f (2 ) (0)  (4 2 )

 2 2
2
2
10
Hence equation (1) becomes
2 2 
4 2
1
1
1

 4  2  2  2  ............
3
2
3
1

4 2
1
1
1

2 
 4  2  2  2  ............
3
2
3
1

2
2 2
1
1
1

 4  2  2  2  ............
3
2
3
1

2

6
1
1
1
 2  2  ................
2
1
2
3
       ( 2)
Now, put x = π (which is point of continuity) in the above series we get
4 2
1
1
 1

2
 
 4 2  2  2  ............  4 (0)
3
2
3
 1

2
4
1
1
1

2 
  4  2  2  2  ............
3
2
3
1

2
1
1
1

  4  2  2  2  ............
3
2
3
1


2
12

1
1
1
 2  2  ................        (3)
2
1
2
3
Adding (2) and (3), we get
2
6
2
1
1
1

 2  2  2  2  ............
12
3
5
1


3 2
2
12
1
1
1

12  3 2  5 2  ............


2

1
1
1
 2  2  2  ................
8 1
3
5
5. Find the Fourier series expansion of (π – x)2 in –π < x < π.
Sol. Fourier series is
f ( x) 
a0 

a0
  (a n cos nx  bn sin nx)
2 n 1
1


 f ( x) dx


1


 (  x)
2
dx


1  (  x) 3 
 
   3   


1
0  8 3
 3
8 2

3


1
1
a n   f ( x) cos nx dx   (  x) 2 cos nx dx






1
 sin nx 
  cos nx 
  sin nx 
 (  x) 2 
  [2(  x)( 1)] 
  (2)

2
3

 n 
 n

 n
  



(4 ) (1) n
1


0

0

0

0

 0 


2

n



4 (1) n
n2
11
bn 

1


f ( x) sin nx dx 

1


(  x)
2
sin nx dx


1
  cos nx 
  sin nx 
 cos nx 
 (  x) 2 
  [2(  x)( 1)] 
  (2)

2
3

n


 n

 n   
f ( x) 

n
2 (1) n  
2 (1) n 
1 
2 ( 1)


(
4

)

0




0  0 
 
n
n3  
n 3 

4 (1) n
n
a0 
  (a n cos nx  bn sin nx)
2 n 1
   4(1) n

4 (1) n
   
cos
nx

sin nx 
2
n
 n 1  n

2
4
 cos x cos 2 x cos 3 x

 sin x sin 2 x sin 3 x

f ( x) 
 4  2 

 ...............  4 


 ............
2
2
3
1
2
3
2
3
 1




1  8 2

2  3
(i.e.) f ( x) 
4 2
 cos x cos 2 x cos 3 x

 sin x sin 2 x sin 3 x

 4 2 

 ...............  4 


 ............
2
2
3
2
3
2
3
 1

 1

6. Expand in Fourier series of f(x) = x sinx for 0 < x < 2π and deduce the result
1
1
1
 2


 .......... 
1.3 3.5 5.7
4
Sol. Fourier series is
f ( x) 
a0 
1


a0
  (a n cos nx  bn sin nx)
2 n 1
2
 f ( x) dx

0


1

1

1

2
 x sin x dx
0
x ( cos x)  (1)( sin x)02
(2  0)  (0  0)
 2
an 
1

2
 f ( x) cos nx dx

0
1

1

2


1
2
1
2
1

2
2
 x sin x cos nx dx
0
2
 x(2 cos nx sin x) dx
0
2
 x sin( n  1) x  sin( n  1) x dx
,
n 1
0
2
 x sin( n  1) x dx 
0
1
2
2
 x sin( n  1) x dx
0
2
   cos( n  1) x 
  sin( n  1) x 

  (1)
( x)
2
n

1
(
n

1
)



 0

2
(1) 2 n 2  1
(1) 2 n2  1
  sin( n  1) x 
1    cos( n  1) x 


  (1)
( x)
2
2  
n 1
(
n

1
)


 0
 1   2 (1) 2 n  2



1   2 (1) 2 n  2

 0  0  0 
 0  0  0


2 
n 1
n 1


 2 


1
1

n 1 n 1
12
an 
 (n  1)  (n  1)
(n  1)( n  1)
an 
2
, n 1
n 1
2
When n = 1, we have
a1 
2
1
 f ( x) cos x dx


0
2
1
 x sin x cos x dx

0
2
1

2
 x sin 2 x dx
0
2
   cos 2 x 
  sin 2 x 
  (1)

 x
2
4  0


 
1

2


bn 
1

2

1
f ( x) sin nx dx 

0
1
2
2
 x sin x sin nx dx
0
2
1
2

 x(2 sin nx sin x) dx
0
2
 x cos( n  1) x  cos( n  1) x dx
1
2

   1  

2    0  (0  0)
  2  

1
2
n 1
,
0
2
1

2
1
0 x cos( n  1) x dx  2
2
 x cos( n  1) x dx
0
2
  cos( n  1) x 
1   sin( n  1) x 


  (1)
( x)
2
2   n  1 
(
n

1
)

 0
1

2

1
2

(1) 2 n  2  
1  1
0

 0 


2 
2 
(
n

1
)
(
n

1
)





 2

1  
1  1
 0 

0 
2 
2 
(
n

1
)
(
n

1
)





 2
bn  0 , n  1

2
  sin( n  1) x 
  cos( n  1) x 

  (1)
( x)
2
 (n  1)
 0
  n 1 
1
2

(1) 2 n  2  
1 
0

 0 


2 
(n  1)  
(n  1) 2 


1  
1 
 0 

0 
2 
(n  1)  
(n  1) 2 

When n = 1, we have
b1 
1

2
 f ( x) sin x dx

0


 x sin x sin x dx

1

1

1

2

2
1
0
2
 x sin
2
x dx
0
2

0
 1  cos 2 x 
x
 dx
2


1  2
1 
 1  
2  0     0  0  
2 
2 
 2  

2
 x 2   sin 2 x 
  cos 2 x 
  (1)
 
   x
4

  0
2   2 
13
a0 
  (a n cos nx  bn sin nx)
2 n 1
f ( x) 



a0
 a1 cos x   a n cos nx  b1 sin x   bn sin nx
2
n2
n2


2 1
2
 cos x  
cos nx   sin x  0
2 2
n  2 ( n  1)( n  1)
1
 cos 2 x cos 3x cos 4 x cos 5 x

x sin x  1  cos x   sin x  2 



 ..................
2
2.4
3.5
4.6
 1.3


Put x =
2
in the above series we get

1
1
 1

(1)  1  0   (1)  2   0 
0
 0  ..................
2
3.5
5.7
1.3


1
1
1

   1  2  

 ..................
2
1.3 3.5 5.7

  2  2
1
1
1

 2  

 ..................
2
1.3 3.5 5.7

  2
1
1
1

 2  

 ..................
2
1.3 3.5 5.7

 2 1
1
1



 .................
4
1.3 3.5 5.7
7. Find the Fourier series of periodicity 3 for f(x) = 2x – x2 in 0 < x < 3.
Sol. Fourier series is
f ( x) 
a0  
2n x
2n x 
   a n cos
 bn sin

2 n 1 
3
3 
3
3
1
2
a0 
f ( x) dx   (2 x  x 2 ) dx

(3 / 2) 0
30
3
2  2x 2 x3 
 
 
3 2
3 0


2 
27 
 9    (0  0)

3 
3 

0
1
an 
(3 / 2)
3

0
2
2nx
f ( x) cos nx dx   (2 x  x 2 ) cos
dx
30
3
3
3



2nx 
2nx 
2nx 

  cos

  sin

sin



2
2
3
3
3




  (2  2 x)
 (2 x  x )
 (2)
2 2
3 3
2
n





3
4n 
8n 









3


9
27



 0
2 
 9   
  9  
0  (4) 2 2   0  0  (2) 2 2   0
3 
 4n    
 4 n   
2   54 
  2 2
3  4n  
9
 2 2
n

14
1
bn 
(3 / 2)
3

0
2
2nx
f ( x) sin nx dx   (2 x  x 2 ) sin
dx
30
3
3
3



2nx 
2nx 
2nx 

  sin

 cos

 cos



2
2
3
3
3




  (2  2 x)
 (2 x  x )
 (2)
2 2
3 3
2
n




3
4n 
8n  









3


9
27



 0
f ( x) 

2 
 3 
 27  
 27 
  0  2  3 3   0  0  2  3 3 
(3)
3 
 2 n 
 8n   
 8n  

3
n
a0  
2n x
2n x 
   a n cos
 bn sin

2 n 1 
3
3 
2n x 3
2n x 
 9
 0    2 2 cos

sin

3
n
3 
n 1  n 

(i.e.) f ( x)  
2 x 1
4 x 1
6 x
9 1

cos
 2 cos
 2 cos
 .....................
2  2
3
3
3
 1
2
3


2 x 1
4 x 1
6 x
3 1

sin

sin

sin
 .....................

 1
3
2
3
3
3

8. Expand f(x) = x – x2 as a Fourier series in –l < x < l and using this series find the root
square mean value of f(x) in the interval.
Sol. Fourier series is
a0  
n x
n x 
f ( x) 
   a n cos
 bn sin

2 n 1 
l
l 
l
l
1
1
a 0   f ( x) dx   ( x  x 2 ) dx
l l
l l
l
1  x2 x3 
   
l2
3  l
1   l 2 l 3   l 2 l 3 
        
l   2 3   2 3 
1  2l 3   2l 2

 
l  3 
3
1
nx
1
nx
a n   f ( x) cos
dx   ( x  x 2 ) cos
dx
l l
l
l l
l
l
l
l



nx 
nx 
nx 

  cos

  sin

sin



1
2
l
l
l






 ( x  x )
 (1  2 x)
 (2)
 n 2 2 
 n 3 3 
l
 n 







l


l2
l3



  l
 (1) n l 2
1 
 0  (1  2l ) 2 2
l 
 n
(1) n l 2
1  2l  1  2l 

l n 2 2
  
 (1) n l 2
  0  0  (1  2l ) 2 2
  
 n
4 l 2 (1) n  1
(1) n l
 2 2  4l  
n
n 2 2
 
  0
 
15
1
nx
1
nx
f ( x) sin
dx   ( x  x 2 ) sin
dx

l l
l
l l
l
l
bn 
l
l



nx 
nx 
nx 

  sin

 cos

 cos



1
2
l
l
l




  (1  2 x)
 ( x  x )
 (2)
2 2
3 3



n

l
n
n  









l


l2


 l 3   l
n
 (1) n l 
1 
2(1) n l 3  
2(1) n l 3 
2  ( 1) l 
  0 


  (l  l 2 )


(

l

l
)

0

 

 n 
l 
n 3 3  
n 3 3 
 n 



 (1) n l

l l2  l  l2
l n

n 1
(1) n 1
2l   2 l (1)
n
n

a
n x
n x 

f ( x)  0    a n cos
 bn sin

2 n 1 
l
l 

1   2l 2

2  3
   4 l 2 (1) n  1
n x 2 l (1) n  1
n x 
   

cos

sin
2 2
l
n
l 
 n 1  n 
l 2 4l 2  1
x 1
2 x 1
3 x 1
4 x

(i.e.) f ( x) 
 2  2 cos
 2 cos
 2 cos
 2 cos
 ....................
3
l
l
l
l
 1
2
3
4



2 l 1  x 1
2 x 1
3 x 1
4 x

sin
 sin
 sin
 sin
 ....................

 1
l
2
l
3
l
4
l

RMS value of f(x) in (–l, l ) is
2
a
1
y  0 
4 2
2

 (a
n 1
2
2
n
1   2l 2  1
 
 
4  3  2
2
(i.e.) y 
l4

9
 bn )
2
16 l 4 (1) 2 n  2 4 l 2 (1) 2 n  2 




n 4 4
n 2 2
n 1 


 8l 4
2l 2 


 4 4

n 2 2 
n 1  n 

l

in 0  x 
 x
2
9. Obtain the sine series for f ( x)  
l  x in l  x  l

2
Sol. Fourier sine series is

f ( x)   bn sin
n 1
n x
l
2
nx
f ( x) sin
dx

l 0
l
l
bn 
2

l
l/2

0
nx
2
nx
x sin
dx   (l  x) sin
dx
l
l l/2
l
l
l/2
l
 



nx 
nx 
nx 
nx 

  sin

  sin

 cos
 cos






2
2
l
l
l
l




  (1)
  (1)
 ( x)
  (l  x)
2 2
2 2
n
n





n

n

l 
l













2
2
 

l
l



l
l

 0

 l / 2
16

n

 l . cos
2 
l
  
2
   
l   2  n




n
 2
2l . sin
2
2
 
2 2
l  n

4l
n
bn  2 2 sin
2
n

f ( x)   bn sin
n 1


n 1
4l
n
2
2
n
  2
  l . sin
2

2 2
  n
 
 



n



 l . cos



2
l
  
2

  0  0  {0  0}    

 l
  2  n









n
  2
  l . sin
2

2 2
  n
 
 





n x
l
sin
n
n x
sin
2
l

4l  1

x
3
3 x
5
5 x
1
1

sin
sin

0

sin
sin

0

sin
sin
 0  .......................
2  2
2
2
2
l
2
l
2
l
 1
3
5


4l  1
x 1
3 x 1
5 x

sin
 2 sin
 2 sin
 .......................
2  2
l
l
l
 1
3
5

10. Find the half range cosine series for the function f(x) = x (π – x) in 0 < x < π.
1
1
1
4
Deduce that 4  4  4  ............ 
90
1
2
3
Sol. Half range fourier cosine series is
a0 
  an cos nx
2 n 1
f ( x) 
11.
a0 
2


 f ( x) dx

0
2


 x(  x) dx
0

x3 
2  x 2
 


 2
3 0


2   3  3 
  (0  0)


  2
3 


2  3 
  6 

an 
2

2
3

 f ( x) cos nx dx
0

2


 x(  x) cos nx dx
0

2
 sin nx 
  cos nx 
  sin nx 
 ( x  x 2 )
  (  2 x)
  (2)

2
3

 n 
 n

 n
 0
 
2 
( )( 1) n
( )(1)

 0 

0

0


0



 
n2
n2

 
2

 (1) n  1
2
n





2
(1) n  1
2
n
 
 


 
 
 
17
f ( x) 
a0 
  a n cos nx
2 n 1
1  2
 
2 3


 
2
    2 (1) n  1 cos nx
 n 1 n
2
2 cos 2 x
2 cos 4 x
2 cos 6 x



 2 0 
0
0
 0  .......... .....
2
2
2
6
2
4
6



2
 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
2


2
 [ f ( x)] dx 
0

a0
2
  an
2
n 1
2

2

1  2 
4


[

x

x
]
dx


[( 1) n  1]2

4



0
2 3 
n 1 n
2
2
2 2

(

2
x  x  2 x )dx 
2
4
3
0
4
4
4
4


 4 0  4  0  4  0  4  0  ..............
18
4
6
 2


2   2 x 3 x 5 2 x 4 
 4 16  1 1 1

 
 4  4  4  4  ..............

 
 3
5
4  0 18 2 1 2
3

2   5  5  5    4  1 1
1

  0 


  4  4  4  ..............

  3
5
2   18 1 2
3

2  5   4 1 1
1
 4  4  4  ...................
 
  30  18 1 2 3
4
4
1 1
1
 4  4  .............
4
15 18
1 2
3
4

1 1
1
(i.e.)
 4  4  4  .............
90 1 2
3


11. Find the complex form of the Fourier series of f ( x)  e  x in –1 < x < 1.
Sol. The complex form of Fourier series of f(x) is given by

f ( x) 
C
n 
n
ei n x
2l = 2
l=1
1
1
Cn 
f ( x) e  i n  x dx

2(1) 1
1
1
  e  x e  i n  x dx
2 1
1
1
  e  (1i n  ) x dx
2 1
1
1  e  (1i n  ) x 
 

2   (1  i n )  1


1
e  (1i n  )  e (1i n  )
2(1  i n )

 (1  i n ) 1  i n 
e e
 e 1e i n 
2 2
2(1  n  )
 (1  i n ) 1

e (cos n  i sin n )  e 1 (cos n  i sin n )
2 2
2(1  n  )





18
Cn 
 (1  i n ) 1
e (1) n  e 1 (1) n
2 2
2(1  n  )


(1  i n ) (1) n 1
e  e 1
2 2
2(1  n  )

(1  i n ) (1) n
2 sinh 1
2(1  n 2 2 )



(1) n sinh 1(1  i n )
Cn 
1  n 2 2

(1) n sinh 1(1  i n ) i n  x
 f ( x)  
e
1  n 2 2
n
 1,
 2,
12. Find the Fourier series of f ( x)  
0 x  
  x  2
1 1
1
Hence evaluate the value of the series 2  2  2  ................
1 3
5
Sol. Fourier series is

a0
  (a n cos nx  bn sin nx)
2 n 1
f ( x) 
a0 
1

2
 f ( x) dx

0


1

1

1


1
2
 (1) dx    (2) dx
0
x 0  2 x 2

(  0)  2 (2   )

 1 2  3
an 
1

2

f ( x) cos nx dx 
0
1


 (1) cos nx dx 
0
1

2
 (2) cos nx dx

2
1  sin nx 
2  sin nx 
 
 

  n  0   n  
1
2
 (0  0)  (0  0)


0
bn 
1

2

f ( x) sin nx dx 
0
1


 (1) sin nx dx 
0
1

2
 (2) sin nx dx

2
1   cos nx 
2   cos nx 
 
 

  n  0   n  
1
2

[( 1) n  1] 
[1  (1) n ]
n
n
1

 (1) n  1  2  2(1) n
n
(1) n  1

n

f ( x) 

a0 
  (a n cos nx  bn sin nx)
2 n 1


3  
(1) n  1
  0. cos nx 
sin nx 
2 n 1 
n


3 2  sin x sin 2 x sin 3x

 


 .........................
2  1
2
3

19
When we put x = 0,  , π, 2π we will not get the given series.
2
So, using Parseval’s identity for Fourier series we have
2
1


1

 (1)
2


dx 
x0
2
2
1



0
1
1
0


a
2
2
[ f ( x)] dx  0   (a n  bn )
2
n 1
2

4

x2

4
4
9 1 4

 2  2  0  2  0  2  0  ..................
2  1
5
3

  0  4 2     9 

2
5


2
(1) n  1 
(3) 2  
  0 
(2) dx 

2
n 2 2 
n 1 

2
1
1
4 1

 2  2  ..................
2  2
 1 3 5

1
1
4 1
9

 2  2  2  2  .......... ........
2  1
5
3

1
1
4 1
1

 2  2  2  2  .......... ........
2  1
5
3

2
8

1
1
1
 2  2  .......... .......
2
5
3
1
l

0 x
 x,
2
13. Find the Fourier series expansion of f ( x)  
l
l  x,
 xl
2


1
Hence deduce the value of 
4
n  1 ( 2n  1)
l
2
0xL
 x,
2 L  x, L  x  2 L
Sol. Let 2 L  l  L  , then the given function becomes f ( x)  
a0  
n x
n x 
 bn sin
Fourier series is f ( x)     a n cos

2 n 1 
L
L 
1
a0 
L
2L

0
L
1
1
f ( x) dx   ( x) dx 
L 0
L
2L
 (2L  x) dx
L
L
2L
1  x2 
1  (2 L  x) 2 
    

L  2 0 L   2 L
 1
1  L2
L2 

0

0





L 2
 2
 L
L L
  L
2 2
n x
f ( x) cos
dx
L

1
an 
L
1

L
2L

0
n x
1
0 x cos L dx  L
L
2L

(2 L  x) cos
L
n x
dx
L
L
2L
 



n x 
n x  
n x 
n x  

  cos

  cos

sin
sin






1
1
L
L
L
L








  ( 2 L  x )
 ( x )
 (1)
 (1)
 n 2 2 
 n 2 2 
L   n 
L
 n 







 

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   
n   L  
n   
n  
1 L2
2L

(1) n  1  1  (1) n  2 2 (1) n  1
2 2
Ln 
n 





20
bn 
2L
1
L

f ( x) sin
0
n x
dx
L
n x
1
0 x sin L dx  L
L
1

L
2L

(2 L  x) sin
L
n x
dx
L
L
2L
 



n x 
n x 
n x 
n x 

  sin

  sin

 cos
 cos






1
1
L
L
L
L




  (1)
  (1)
  (2 L  x)
 ( x)
2 2
2 2




n

n

L 
L
n
n












 

L
L



L2
L2

 0

 L
 1

 (1) n L2

1  (1) n L2






0

0

0

0

0

 0 





L 
n
n



 L
0

a
n x
n x 

f ( x)  0    a n cos
 bn sin

2 n 1 
L
L 



n x
L   2 L [( 1) n  1]
  
cos
 0 
2 2
2 n 1 
L
n


x
3 x
5 x
L 2L  2
2
2

 2  2 cos
 0  2 cos
 0  2 cos
 0  .................
2   1
L
L
L
3
5


x 1
3 x 1
5 x
L 4L  1

 2  2 cos
 2 cos
 2 cos
 .................
2  1
L 3
L
L
5

(i.e.) f ( x) 
2 x 1
6 x 1
10 x
l 2l  1

 2  2 cos
 2 cos
 2 cos
 .......... .......
4  1
l
l
l
3
5

Using Parseval’s identity for Fourier series we have
1
L
L
2L


a0
2
2
  (a n  bn )
2 n 1
2
[ f ( x)] 2 dx 
0
1
1
( x) 2 dx 

L0
L
2L

L


2

L2   4 L2 (1) n  1
(2 L  x) dx 
 

0

2 n  1 
n 4 4

2
L
2L
1  x3 
1  (2 L  x) 3 
L2 4 L2  4
4
4



 4  4  0  4  0  4  0  ..................
 


L  3 0 L   3 L
2  1
3
5

 1
1  L3
L3  L2 16 L2  1
1
1


0

0

 4  4  4  4  ..................




L 3
 3 2
 1 3 5

 L
2
2
2
2L
L
16 L  1
1
1


 4  4  4  4  ..................
3
2
 1 3 5

L2 16 L2  1
1
1

 4  4  4  4  ..................
6
 1 3 5

4 1 1 1
 

 .................
96 14 3 4 5 4

4
1
(i.e.)

96 n  1 (2n  1) 4
14. Find the half range cosine series for the function f(x) = x in 0 < x < l.

Hence deduce the value of the series
1
 (2n  1)
n 1
Sol. Half range Fourier cosine series is f ( x) 
l
4
a0 
n x
  a n cos
2 n 1
l

2
2
2  x2 
2 l 2
a0   f ( x) dx   x dx       0  l
l 0
l 0
l  2 0 l  2

l
l
21
n x
n x
2
2
f ( x) cos
dx   x cos
dx

l 0
l
l 0
l
l
an 
l
l
 

n x 
n x  
  cos

 sin


2
l
l





 ( x)
 (1)
 n 2 2 
l   n 



 
l

l2

 0
2 
(1) n l 2  
l 2 
 0  2 2   0  2 2 
l 
n   
n  
2l
 2 2 (1) n  1
n 
a0 
n x
f ( x) 
  a n cos
2 n 1
l



n x
l  2 l [( 1) n  1]

cos
2 2
2 n 1
l
n
x
3 x
5 x
2
2
 2

 12 cos l  0  3 2 cos l  0  5 2 cos l  0  ...................


x 1
3 x 1
5 x
l 4l  1

(i.e.) f ( x)   2  2 cos
 2 cos
 2 cos
 ...................
2  1
l
l
l
3
5


l 2l

2 2
Using Parseval’s identity for half range Fourier cosine series we have

a0
2
2
2
[
f
(
x
)]
dx


an


l 0
2 n 1
l
2


2
l
2
l 2   4l 2 (1) n  1 
2
( x) dx    

l 0
2 n  1 
n 4 4

l
2  x3 
l 2 4l 2  4
4
4


 4  4  0  4  0  4  0  ..................
 
l  3  0 2  1
3
5

 l 2 16l 2  1
2 l 3
1
1


0

   4  4  4  4  ..................
l 3
 1 3 5

 2
2
2
2
2l
l
16l  1
1
1

  4  4  4  4  ..................
3
2
 1 3 5

l 2 16 l 2  1
1
1

 4  4  4  4  ..................
6
 1 3 5

4 1 1 1
 

 .................
96 14 3 4 5 4

4
1
(i.e.)

96 n 1 (2n  1) 4
15. Find the half range sine series of f(x) = x cos x in (0, π).

Sol. Fourier sine series is f ( x)   bn sin nx
n 1
bn 
2


 f ( x) sin nx dx

0



2

1

1

1


 x cos x sin nx dx
0

 x (2 sin nx cos x) dx
0

 x [sin( n  1) x  sin( n  1) x] dx ,
n 1
0

 x sin( n  1) x dx 
0
1


 x sin( n  1) x dx ,
0
n 1
22


  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
 
n 1
n 1


 (n  1)
 0   
 (n  1)
 0
 1

1    (1) n 1
 0  0  0 

  n  1

 
(1) n  2 (1) n


n 1
n 1
1 
 1
 (1) n 


 n  1 n  1

   (1) n 1


 0  0  0


 n  1



2n
 (1) n 

 (n  1)( n  1) 
2n (1) n
(i.e.)bn  2
, n 1
n 1
When n = 1, we have
2
b1 


 f ( x) sin x dx
0


2
 x cos x sin x dx


0
1


 x sin 2 x dx
0

1    cos 2 x 
  sin 2 x 
  x
  (1)

  2 
4  0



1    1  
1
    0  {0  0}  
   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
n2 1
n2
1
  sin x  2
2
16. Prove that 1 
 2 sin 2 x 3 sin 3 x 4 sin 4 x



 ..................
 3
8
15


4 
x 1
3x 1
5x

sin
 sin
 sin
 .............. in the interval 0 < x < l

 
l
3
l
5
l

Sol. Since RHS contains sine series and given 0 < x < l, we have to find half range Fourier sine
series for f(x) = 1

n x
Fourier sine series is f ( x)   bn sin
l
n 1
n x
n x
2
2
f ( x) sin
dx   (1) sin
dx

l 0
l
l 0
l
l
bn 
l
n

 cos
2
l
 
n
l

l
l
x

2  (1) n l   l 


  

 
l 
n   n 

 0
2

(1) n 1  1
n
n 1


n x
2[( 1)  1]
n x
f ( x)   bn sin

sin
l
n
l
n 1
n 1


2 2
x
2
3 x
2
5 x

sin
 0  sin
 0  sin
 0  ...........

 1
l
3
l
5
l

4 
x 1
3x 1
5x

 sin
 sin
 .............
(i.e.) 1  sin
 
l
3
l
5
l


23
 l  x, 0  x  l
l  x  2l
 0,
1 1 1
1 1
1
Hence deduce the value of the series (i) 1     .......... (ii) 2  2  2  ..............
3 5 7
1 3
5

a
n x
n x 

 bn sin
Sol. Fourier series is f ( x)  0    a n cos

2 n 1 
l
l 
17. Find the Fourier series expansion of f ( x)  
1
a0 
l
2l

0
l
2l
1
1
f ( x) dx   (l  x) dx   (0) dx
l 0
l l
l
1  (l  x) 2 
 

l   2 0


1
0 l2
 2l

l
2
l
n x
n x
1
f ( x) cos
dx   (l  x) cos
dx  0
l
l 0
l

1
an 
l
2l

0
l


n x 
n x  

  cos

 sin


1
l
l




 (l  x)
 (1)
 n 2 2 
l
 n 





l


l2

 0
1 
(1) n l 2  
l 2 
 0  2 2   0  2 2 
l 
n   
n  
1 l2

(1) n 1  1
l n 2 2
l
 2 2 (1) n 1  1
n 
l
n x
n x
1
f ( x) sin
dx   (l  x) sin
dx  0
l
l 0
l


1
bn 
l
2l

0


l


n x 
n x 

  sin

  cos


1
l   (1)
l 
 (l  x)
 n 2 2 
n
l







l


l2

 0
 l2

1
 {0  0}  
 0 
l
 n

l

n

a
n x
n x 

f ( x)  0    a n cos
 bn sin

2 n 1 
l
l 

 l [( 1) n 1  1]
n x
n x 
l
l

  
cos

sin
2 2
4 n 1 
l
n
l 
n
x
3 x
5 x
l
l 2
2
2

  2  2 cos
 0  2 cos
 0  2 cos
 0  .................
4  1
l
l
l
3
5



(i.e.) f ( x) 
l 2l

4 2
2 x 1
3 x
l 1  x 1

sin
 sin
 sin
 .................

 1
l
2
l
3
l

x 1
3 x 1
5 x
1

12 cos l  3 2 cos l  5 2 cos l  .................


2 x 1
3 x
l 1  x 1

  sin
 sin
 sin
 .................        (1)
 1
l
2
l
3
l

24
l
2
Put x  (which is point of continuity) in equation (1), we get
l
l
l 2l
l
  2 (0) 
2 4 

3 1
5
1
1
1  1

sin

sin


sin

sin
4


sin
 .................
1
2 2
3
2
4
5
2


1
1
1


1  0  3  0  5  0  7  .................


l l
l  1 1 1

  1     .................
2 4  3 5 7

l
l l
 
2 4 
l
l  1 1 1

 1     .................
4  3 5 7


1 1 1
 1     .................
4
3 5 7
Put x = l in equation (1) we get
l 2l

4 2
f (l ) 
1
 1 1

 12  3 2  5 2  ................ --------------- (2)
But x = l is the point of discontinuity. So we have
f (l )  f (l ) (0)  (0)
f (l ) 

0
2
2
Hence equation (2) becomes
1
1
1

12  3 2  5 2 ............
2l  1
l
1
1

   2  2  2  2 ............
4
 1 3 5

l 2l

4 2
0
2

8
f(x) = l – x
f(l–) = l – l =0
f(x) = 0
f(l) = 0
1
1
1
 2  2 ................
2
1
3
5
18. Obtain the half range cosine series for f(x) = (x – 2)2 in the interval 0 < x < 2.

Deduce that

n 1
1
2

8
(2n  1) 2
Sol. Half range cosine series is
a0 
nx
f ( x) 
  a n cos
2 n 1
2
a0 
2
2
2
2

f ( x) dx   ( x  2) 2 dx
0
0
2
 ( x  2) 3 


 3 0
   8  8
  0    
  3  3
2
2
2
nx
nx
a n   f ( x) cos
dx   ( x  2) 2 cos
dx
2 0
2
2
0
2



nx 
nx 
nx 

  sin

  cos

 sin


2
2
2
2






 ( x  2)
 [2( x  2)]
 (2)
 n 3 3  
 n 2 2 
 n 








 2 
4
8



 0

16


 0  0  0  0  2 2  0
n 



16
 2 2
 n
25
f ( x) 
8  16
nx
  2 2 cos
6 n 1 n 
2
(i.e.) f ( x) 
4 16  1
x 1
2x 1
3x

 2  2 cos  2 cos
 2 cos
 ........................ --------- (1)
3  1
2 2
2
2
3

Put x = 0 in equation (1) we get
4 16  1
1
1

 2  2  2  2  ........................ ------------- (2)
3  1
2
3

f (0) 
But x = 0 is the point of discontinuity. So we have
( x  2) 2  ( x  2) 2
2
2
(0  2)  (0  2) 2 (4)  (4)
f (0) 

4
2
2
f ( x) 
(x + 2)2
–2
Hence equation (2) becomes
4 16  1
1
1

 2  2  2  2  ........................
3  1
2
3

4
4
4 16  1
1
1

 2  2  2  2  ........................
3  1
2
3

8 16  1
1
1

 2  2  2  2  ........................
3  1
2
3

2
6

1
1
1
 2  2  ......................           (3)
2
1
2
3
Put x = 2 in equation (1) we get
4 16

3 2
f (2) 
1
1
 1

 12  2 2  3 2  ........................ ------------- (4)
But x = 2 is the point of discontinuity. So we have
( x  2) 2  (2  x) 2
2
2
(2  2)  (2  2) 2
f (2) 
0
2
f ( x) 
Hence equation (4) becomes
1
1
 1

 12  2 2  3 2  ........................
4
16  1
1
1

   2  2  2  2  ........................
3
 1 2
3

0
2
12
4 16

3 2

1
1
1
 2  2  ......................           (5)
2
1
2
3
Adding (3) and (5), we get
2
2
1
1
1

 2  2  2  2  ............
6 12
3
5
1

2
3
1
1
1

 2  2  2  2  ............
12
3
5
1


2
(i.e.)
8
2
8

1
1
1
 2  2  ................
2
1
3
5


n 1
1
(2n  1) 2
(x – 2)2
0
(2 – x )2
2
4
26
19. 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
(i.e.) 2l = 6  l = 3
Fourier series is
a0  
n x
n x 
   an cos
 bn sin

2 n 1 
3
3 
a
x
2 x
3 x
x
2 x
3 x
f ( x)  0  a1 cos
 a 2 cos
 a3 cos
 b1 sin
 b2 sin
 b3 sin
2
3
3
3
3
3
3
a
x
(i.e.) f ( x)  0  a1 cos   a 2 cos 2  a3 cos 3  b1 sin   b2 sin 2  b3 sin 3 where  
2
3
f ( x) 
θ=πx/3 cosθ
x
y
0
1
2
3
4
5
9
18
24
28
26
20
Total
125
0
π/3
2π/3
π
4π/3
5π/3
1
0.5
–0.5
–1
–0.5
0.5
cos2θ
cos3θ
1
–0.5
–0.5
1
–0.5
–0.5
1
–1
1
–1
1
–1
sinθ
sin2θ
0
0
0.866 0.866
0.866 –0.866
0
0
–0.866 0.866
–0.866 –0.866
sin3θ
ycosθ
ycos2θ
ycos3θ
0
0
0
0
0
0
9
9
–12
–28
–13
10
–25
9
–9
–12
28
–13
–10
–7
9
–18
24
–28
26
–20
–7
ysinθ
ysin2θ
ysin3θ
0
15.588
20.784
0
–22.516
–17.32
–3.464
0
15.588
–20.784
0
22.516
–17.32
0
0
0
0
0
0
0
0
Here n = 6
 y 
125 
a 0  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 
 7
a 2  2 [mean value of y cos 2 ]  2 
  2     2.333
n
 6 


  y cos 3 
 7
a3  2 [mean value of y cos 3 ]  2 
  2     2.333
n
 6 


  y sin  
  3.464 
b1  2 [mean value of y sin  ]  2 
2
   1.155
n
 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
 f ( x) 
 8.333 cos   2.333 cos 2  2.333 cos 3  1.155 sin   0 sin 2  0 sin 3
2
x
(i.e.) f ( x)  20.833  8.333 cos   2.333 cos 2  2.333 cos 3  1.155 sin  where  
3
27
20. Find the Fourier series expansion up to second harmonic from the following data:
x: 0

 /3
2 / 3
4 / 3
5 / 3
2
f(x) : 10
12
15
20
17
11
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
f ( x) 
a0 
  (a n cos n x  bn sin n x)
2 n 1
(i.e.) f ( x) 
a0
 a1 cos x  a2 cos 2 x  b1 sin x  b2 sin 2 x
2
x
y
cosx
cos2x
0
π/3
2π/3
π
4π/3
5π/3
10
12
15
20
17
11
1
0.5
–0.5
–1
–0.5
0.5
1
–0.5
–0.5
1
–0.5
–0.5
Total
85
sinx
sin2x
0
0
0.866 0.866
0.866 –0.866
0
0
–0.866 0.866
–0.866 –0.866
ycosx
ycos2x
ysinx
ysin2x
10
6
–7.5
–20
–8.5
5.5
–14.5
10
–6
–7.5
20
–8.5
–5.5
2.5
0
10.392
12.99
0
–14.722
–9.526
–0.866
0
10.392
–12.99
0
14.722
–9.526
2.598
Here n = 6
 y 
 85 
a 0  2 [mean value of y ]  2 
  2    28.333
6
 n 
  y cos x 
  14.5 
a1  2 [mean value of y cos x]  2 
2
   4.833
n
 6 


  y cos 2 x 
 2.5 
a 2  2 [mean value of y cos 2 x]  2 
  2    0.833
n
 6 


  y sin x 
  0.866 
b1  2 [mean value of y sin x]  2 
2
   0.289
n
 6 


  y sin 2 x 
 2.598 
b2  2 [mean value of y sin 2 x]  2 
2
  0.866
n
 6 


28.333
 f ( x) 
 4.833 cos x  0.833 cos 2 x  0.289 sin x  0.866 sin 2 x
2
(i.e.) f ( x)  14.1665  4.833 cos x  0.833 cos 2 x  0.289 sin x  0.866 sin 2 x
21. Find the Fourier series expansion up to first harmonic from the following data:
x: 0
1
2
3
4
5
6 7
8 9 10
11
f(x) : 18 18.7 17.6 15 11.6 8.3 6 5.3 6.4 9 12.4 15.7
Sol. Here the length of the interval is 12
(i.e.) 2l = 12  l = 6
Fourier series is
a0  
n x
n x 
   an cos
 bn sin

2 n 1 
6
6 
a
x
x
f ( x)  0  a1 cos
 b1 sin
2
6
6
a
x
(i.e.) f ( x)  0  a1 cos   b1 sin  where  
2
6
f ( x) 
28
x
y
θ=πx/6
0
1
2
3
4
5
6
7
8
9
10
11
18
18.7
17.6
15
11.6
8.3
6
5.3
6.4
9
12.4
15.7
0
π/6
2π/6
3π/6
4π/6
5π/6
π
7π/6
8π/6
9π/6
10π/6
11π/6
Total
144
cosθ
sinθ
ycosθ
1
0
18
0.866
0.5
16.1942
0.5
0.866
8.8
0
1
0
–0.5
0.866
–5.8
–0.866
0.5
–7.1878
–1
0
–6
–0.866 –0.5 –4.5898
–0.5 –0.866
–3.2
0
–1
0
0.5
–0.866
6.2
0.866
–0.5 13.5962
36.0128
ysinθ
0
9.35
15.2416
15
10.0456
4.15
0
–2.65
–5.5424
–9
–10.7384
–7.85
18.0064
Here n = 12
 y 
a 0  2 [mean value of y ]  2 
2
 n 
144 
 12   24


  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
x
(i.e.) f ( x)  12  6.002 cos   3.001sin  where  
6
29
Part – A
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.

n x
f ( x)   bn sin
Sol.
l
n 1
2
n 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 < x < π
Sol. f(x) = x2 is an even function.
Fourier series is
f ( x) 
a0 
a0 
  an cos nx
2 n 1
2


 f ( x) dx

0
2


x
2
dx
0

2  x3 
2
   
  3 0 
 3
 2 2

0

 
3
 3

Hence the first term of the Fourier series =
a0  2

2
3
4. If f(x) = 2x in the interval (0, 4) then find the value of a2 in the Fourier series
expansion.
4
1
2 x
a

2
x
cos
dx
Sol.
2

2
2
0
4
  x cos  x dx
0
4
  sin  x 
  cos  x 
  x
  (1)

2
 
 0
   
1  
1 

 0  2   0  2 
     
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
y 
[ f ( x)] 2 dx

ba a
2
a
ba
6. Find the root mean square value of f(x) = x2 in (0, l)
l
1
Sol. R M S  y   [ f ( x)]2 dx
l0
2
l
l
l
 l4
1
1
1  x5 
1 l5
  [ x 2 ]2 dx   x 4 dx       0 
l0
l0
l  5 0 l 5
 5
y
l2
5
30
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 
2
2

0
1
[ f ( x)] dx  y 
2
2
2
2

[ f ( x)] 2 dx
0
8. Find the root mean square value of f(x) = 1 – x in 0 < x < 1
2
1
Sol. R M S  y 
1
[ f ( x)]2 dx

10
1
 (1  x) 3 
  1  1
  (1  x) dx  
 0  
 

  3  0    3  3
0
1
y
3
1
2
9. Find the root mean square value of f(x) = π – x in 0 < x < 2π
1
Sol. R M S  y 
2
2
1

2
2
 [ f ( x)] dx
2
0
2
2
1  (  x) 3 
0 (  x) dx  2   3 
0
2
1

2

y
  3    3 

   
 3    3 
1  2 3   2

2  3  3

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
a) f(x) if x is a point of continuity
b)
f ( x  0)  f ( x  0)
if x is a point of discontinuity.
2
11. What do you mean by Harmonic Analysis.
Sol. 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)]
x , 0  x  1
at x = 1
2 , 1  x  2
12. Find the sum of the Fourier series for f ( x)  
Sol. Here x = 1 is a point of discontinuity
f (1) 
f (1  0)  f (1  0) 1  2 3


2
2
2
31
cos x , 0  x  
and f(x) = f(x + 2π) for all x, find the sum of the Fourier
50 ,   x  2
13. If f ( x)  
series of f(x) at x = π.
Sol. Here x = π is a point of discontinuity
f (  0)  f (  0) cos   50  1  50 49
f ( ) 



2
2
2
2
14. Find the coefficient b5 of cos5x in the Fourier cosine series of the function f(x) = sin5x
in the interval (0, 2π).
Sol. Fourier cosine series is
f ( x) 
b0 
  bn cos nx
2 n 1
bn 
 b5 
1
2
 f ( x) sin nx dx

1
0
2

2
1
 f ( x) sin 5 x dx    sin 5x sin 5 x dx
0
0


1
2
 sin

2
5 x dx
0
2
1  cos 10 x
dx
 0
2
1

1

2
2
sin 10 x 

 x  10 
0
1
(2  0)  (0  0)
2
b5  1


sin nx
. Find the RMS value
n
n 1
15. The Fourier series expansion of f(x) in (0, 2π) is f ( x)  
of f(x) in the interval (0, 2π).
Sol. RMS value of f(x) in (0, 2π) is
2
2
y 
a0 1

4 2

 (a
n 1
2
n
 bn )
2
But here we have a0 and an are zero and bn = 1/n
2
 y 
1
2

1
n
n 1
2
16. If f(x) is discontinuous at x = a what value does its Fourier series represent at that point.
17. Find the root mean square value of the function f(x) = x in (o, l)
x, 0  x  1

at x  1.
18. Find the sum of the Fourier series for f(x) = 
2, 1  x  2
 0, 0  x  
19. If the Fourier series for the function f(x) = 
is
sin x,   x  2
2  cos 2 x cos 4 x cos 6 x
 1


 .........  sin x

   1.3
3.5
5.7
 2
1
1
1
 2
Deduce that


 ......... 
1.3 3.5 5.7
4
f ( x)  
1

20. Find a Fourier sine series for the function f(x) = 1; 0 < x < π
21. Find the constant term in the Fourier series corresponding to f(x) = | cosx | expressed in
the interval (–π, π).
32
Part – B
x, 0  x  
of period 2π.
2  x,   x  2

1. Obtain the Fourier expansion of the function f ( x)  
2. Find the half-range cosine series for the function f(x) = x; 0 < x < π and hence deduce the

sum of the series
1
 (2n  1)
n0
4
3. Expand the function f(x) = sinx, 0 < x < π in Fourier cosine series.
4. Determine the first two harmonics of the Fourier series for the following values
x: 0

 /3
2 / 3
4 / 3
5 / 3
f(x) : 1.98 1.30 1.05 1.30 –0.88 –0.25
5. Find the Fourier series of f(x) = (π – x)2 in (0, 2π) of periodicity 2π.
6. Obtain the Fourier series to represent the function f(x) = |x|, – π < x < π and deduce
1
2


2
8
n  1 ( 2n  1)

7. Find the half-range Fourier cosine series of f(x) = (π – x)2 in the interval (0, π). Hence find
the sum of the series
1
1
1
 4  4  ..........
4
1
2
3
8. Determine the Fourier expansion of f(x) = x in the interval – π < x < π
9. Find the half range cosine series for x sinx in (0, π) and hence find the value of
1
2
2
2
2



 ...........
1.3 3.5 5.7 7.9
10. Obtain the Fourier series for the function
  x, 0  x  1
f ( x)  
 (2  x), 1  x  2
11. Calculate the first 3 harmonics of the Fourier of f(x) from the following data:
xo : 0 30 60 90 120 150 180 210 240 270 300 320
f(x): 1.8 1.1 0.3 0.16 0.5 1.3 2.16 1.25 1.3 1.52 1.76 2.0
12. Find the Fourier series of the function
1
1
1
 0,    x  0


 .........
f ( x)  
and hence evaluate
1.3 3.5 5.7
 sin x, 0  x  