1. CHAPTER 2
FOURIER SERIES
PERIODIC FUNCTIONS
A function f (x ) is said to have a period T if for all x, f ( x + T ) = f ( x) , where T is a
positive constant. The least value of T>0 is called the period of f (x) .
EXAMPLES
We know that f (x ) = sin x = sin (x + 4 π ) = … Therefore the function has period 2 π , 4
π , 6 π , etc. However, 2 is the least value and therefore is the period of f(x).
Similarly cos x is a periodic function with the period 2 π and tan x has period π .
DIRICHLET’S CONDITIONS
A function f (x ) defined in c ≤ x ≤ c+2l can be expanded as an infinite trigonometric
a nπx nπx
series of the form o + ∑ a n cos + ∑ bn sin , provided
2 l l
1. f (x) is single- valued and finite in (c , c+2l)
2. f (x) is continuous or piecewise continuous with finite number of finite
discontinuities in (c , c+2l).
3. f (x) has no or finite number of maxima or minima in (c , c+2l).
EULER’S FORMULAS
If a function f (x) defined in (c , c+2l) can be expanded as the infinite trigonometric
ao ∞
nπx ∞ nπx
series
2
+ ∑ an cos
n =1 l
+ ∑ bn sin
n =1 l
then
c + 2l
1 nπx
an =
l ∫
c
f ( x) cos
l
dx, n ≥ 0
c + 2l
1 nπx
bn =
l ∫c
f ( x) sin
l
dx, n ≥ 1
[ Formulas given above for a n and bn are called Euler’s formulas for Fourier coefficients]

2. DEFINITION OF FOURIER SERIES
ao ∞
nπx ∞ nπx
The infinite trigonometric series
2
+ ∑ an cos
n =1 l
+ ∑ bn sin
n =1 l
is called the
Fourier series of f (x) in the interval c ≤ x ≤ c+2l, provided the coefficients are given by the
Euler’s formulas.
EVEN FUNCTION
If f (x) = φ (x) in (-l , l) such that φ (− x) = φ (x) , then f (x ) is said to be an even
function of x in (-l , l).
φ1 ( x ) in (−l ,0)
If f ( x) = 
φ 2 ( x) in (0, l )
Such that φ1 (− x) = φ 2 ( x) or φ 2 ( − x ) = φ1 ( x ) , then f (x) is said to be an even function of x in
(-l , l).
EXAMPLE
y = cos x , y = x 2 are even functions.
ODD FUNCTION
If f (x) = φ (x) in (-l , l) such that φ (− x) = - φ (x) , then f (x) is said to be an odd
function of x in (-l , l).
φ1 ( x ) in (−l ,0)
If f ( x) = 
φ 2 ( x) in (0, l )
Such that φ1 (− x) = - φ 2 ( x ) or φ 2 ( − x ) = - φ1 ( x ) , then f (x) is said to be an odd function of x in
(-l , l).
EXAMPLE
y = sin x , y = x are odd functions.
FOURIER SERIES OF EVEN AND ODD FUNCTIONS
1. The Fourier series of an even function f (x ) in (-l , l) contains only cosine terms
2

3. (constant term included), i.e. the Fourier series of an even function f (x) in (-l , l) is
given by
ao nπx
f (x) =
2
+ ∑a n cos
l
,
nπx
l
2
where a n = ∫ f ( x ) cos dx.
l 0 l
2. The Fourier series of an odd function f (x) in (-l , l) contains only sine terms, i.e.
the Fourier series of an odd function f (x ) in (-l , l) is given by
nπx
f (x) = ∑b n sin
l
,
nπx
l
2
where bn = ∫ f ( x) sin l dx.
l 0
PROBLEMS
1. Find the Fourier series of period 2l for the function f (x ) = x(2l – x) in (0 , 2l). Deduce
1 1 1
the sum of f (x) = 2
− 2 + 2 −
1 2 3
Solution:
ao ∞
nπx ∞ nπx
Let f (x ) =
2
+ ∑a
n =1
n cos
l
+ ∑ bn sin
n =1 l
in (0 , 2l) …………(1)
nπx
2l
1
an =
l0∫ x(2l − x) cos l dx
2l
  nπx   nπx   nπx 
  sin   − cos   − sin 
1 2  l  − (2l − 2 x ) l  + (−2) l  ,
= (2lx − x )
l  nπ   n 2π 2   n 3π 3 
      
  l   l2   l3  0
using Bernoulli’s formula.
2
1
= [ − 2l cos 2nπ − 2l ] = − 4l 2
n 2π 2 n 2π
2l
1 x3 
2l
1 4
a o = ∫ x(2l − x )dx = lx 2 −  = l 2 .
l0 l 3 0 3
3

4. nπx
2l
1
bn =
l0∫ x(2l − x) sin l dx
=0
Using these values in (1), we have
2 2 4l 2 ∞
1 nπx
x (2l - x) = l − 2
3 π
∑n
n =1
2
cos
l
in (0, 2l) ……………..(2)
1 1 1
The required series 2
− 2 + 2 − … ∞ can be obtained by putting x = l in the Fourier
1 2 3
series in (2).
x = l lies in (0 , 2l) and is a point of continuity of the function f (x) = x(2l – x).
∴ [ Sum the Fourier series in (2) ] x =1 = f(l)
∞
2 2 4l 2 1
i.e. l − 2
3 π
∑n
n =1
2
cos nπ = l(2l - l)
4l 2  1 1 1  l
2
i.e.. -  − 2 + 2 − 2 + ...∞  =
π2  1 2 3  3
∴
1 1 1 π2
− 2 + 2 − …∞ =
12 2 3 12
2. Find the Fourier series of period 2 π for the function f (x) = x cos x in 0 < x < 2 π .
Solution:
∞ ∞
ao
Let f (x ) =
2
+ ∑ an cos nx + ∑ bn sin nx
n =1 n =1
.……..…………(1)
2π
1
an =
π ∫ x cos x cos nxdx
0
2π
1
=
2π ∫ x[ cos(n + 1) x + cos(n − 1) x]dx
0
1  sin( n + 1) x cos(n + 1) x  2π  sin( n − 1) x cos(n − 1) x  2π 
=  x. +  +  x. +  ,
2π  n +1 (n + 1) 2  0  n −1 ( n − 1) 2  0 
 
if n ≠ 1
=0, if n ≠ 1
ao = 0
4

5. 2π 2π
1 1
an = ∫ x cos xdx = 2π ∫ x(1 + cos 2 x)dx
2
π 0 0
2π
1 x sin 2 x cos 2 x 
2
=  +x +  = π.
2π 2 2 4 0
2π
1
bn =
π ∫ x cos x sin nxdx
0
2π
1
=
2π ∫ x[ sin(n + 1) x + sin(n − 1) x]dx
0
1  − cos(n + 1) x sin( n + 1) x  2π  − cos(n − 1) x sin( n − 1) x  2π 
=  x. +  +  x. +  ,
2π  n +1 (n + 1) 2  0  n −1 (n − 1) 2  0 
 
if n ≠ 1
1 1  1 1  2n
=− − = − + =− 2 , if n ≠ 1
n +1 n −1  n + 1 n − 1 n −1
2π 2π
1 1
b1 =
π ∫ x cos x sin xdx =
0
2π ∫ x sin 2 xdx
0
2π
1   − cos 2 x  sin 2 x  1
=  x +  =−2
2π   2  4 0
Using these values in (1), we get
∞
1 n
f(x) = π cos x − sin x − 2 ∑ sin nx
n = 2 , 3,... n − 1
2
2
3. Find the Fourier series expansion of f (x) = sin ax in (-l , l).
Solution:
Since f (x) is defined in a range of length 2l, we can expand f (x ) in Fourier series of
period 2l.
Also f ( − x) = sin[a(-x)] = -sin ax = - f (x)
∴ f (x) is an odd function of x in (-l , l).
Hence Fourier series of f (x ) will not contain cosine terms.
∞
nπx
Let f (x ) = ∑b
n =1
n sin
l
………………….(1)
5

6. 1   nπ  nπ
l
  
= ∫ cos l − a  − cos l + a  xdx
l 0     
l
  nπ   nπ  
1  sin  l − a  x sin  l + a  x 
=    −   
l nπ nπ
−a +a 
 l l 
 0
1  nπ  1  nπ 
= sin  − a l − sin  + a l
nπ − la  l  nπ + la  l 
=
1
nπ − al
{− (−1) n sin al} − nπ 1 al {(−1) n sin al}
+
 1 1 
= (−1) n +1 sin al  + 
 nπ − al nπ + al 
(−1) n +1 2nπ sin al
=
n 2π 2 − a 2 l 2
Using these values in (1), we get
∞
(−1) n +1 n nπx
sin ax = 2π sin al ∑ sin
n =1 n π −a l
2 2 2 2
l
4. Find the Fourier series expansion of f (x) = e − x in (−π , π ) . Hence obtain a series for
cosec π
Solution:
Though the range (−π , π ) is symmetric about the origin, e − x is neither an even function
nor an odd function.
∞ ∞
ao
∴ Let f (x) =
2
+ ∑ an cos nx + ∑ bn sin nx
n =1 n =1
..…..…………(1)
in (−π , π ) [ the length of the range is 2π ]
6

7. π
1
∫π e cos nxdx
−x
an =
π−
π
1  e −x 
=  2 ( − cos nx + n sin nx ) 
π n +1  −π
=−
1
{e −π
(−1) n − e π (−1) n }
π ( n + 1)
2
2( −1) n
= sinh π
π (n 2 + 1)
2 sinh π
ao =
π
π
1
∫π e sin nxdx
−x
bn =
π−
π
1  e −x 
=  2 ( − sin nx − n cos nx ) 
π n +1  −π
=−
n
{e −π
(−1) n − e π (−1) n }
π ( n 2 + 1)
2n(−1) n
= sinh π
π (n 2 + 1)
Using these values in (1), we get
sinh π 2 sinh π ∞
(−1) n 2 sinh π ∞
(−1) n n
e−x =
π
+
π
∑ n 2 + 1 cos nx + π
n =1
∑ n 2 + 1 sin nx
n =1
in (−π , π )
[ Sum of the Fourier series of f ( x )] x =0 = f (0),
[Since x=0 is a point of continuity of f(x)]
sinh π  ∞
(−1) n 
i.e., 1 + 2∑ 2 −0
 = e =1
π  n =1 n + 1
 −1 ∞
(−1) n
i.e., π cos ech π = 1 + 2  + 2∑ 2
 2  n=2 n + 1
2 ∞ (−1) n
i.e., cos ech π = ∑
π n=2 n 2 + 1
7

8. HALF-RANGE FOURIER SERIES AND PARSEVAL’S THEOREM
(i) The half range cosine series in (0 , l) is
ao ∞
nπx
f (x) =
2
+ ∑a
n =1
n cos
l
l
2
l∫
where a o = f ( x )dx.
0
nπx
l
2
an = ∫ f ( x) cos l dx.
l 0
(ii) The half range sine series in (0 , l) is
∞
nπx
f (x) = ∑b
n =1
n sin
l
,
nπx
l
2
where bn = ∫ f ( x) sin l dx.
l 0
(iii) The half range cosine series in (0 , π ) is given by
∞
ao
f (x) =
2
+ ∑a
n =1
n cos nx
π
2
where a o = ∫ f ( x )dx.
π 0
π
2
π∫
an = f ( x ) cos nxdx.
0
(iv) The half range sine series in (0 , π ) is given by
∞
f (x) = ∑b
n =1
n sin nx ,
π
2
where bn = ∫ f ( x) sin nxdx.
π 0
8

9. ROOT-MEAN SQUARE VALUE OF A FUNCTION
Definition
c +2 l
1
∫y
2
If a function y = f (x ) is defined in (c , c+2l), then dx is called the root mean-
2l c
square(R.M.S.) value of y in (c , c+2l) and is denoted by y.
c + 2l
1 2
Thus y = ∫y
2
dx.
2l c
PARSEVAL’S THEOREM
If y = f (x ) can be expanded as a Fourier series of the form
ao ∞
nπx ∞ nπx
2
+ ∑ an cos
n =1 l
+ ∑ bn sin
n =1 l
in (c , c+2l), then the root-mean square value y of y = f (x)
in (c , c+2l) is given by
1 2 1 ∞ 1 ∞
y = a o + ∑ a n + ∑ bn
2 2 2
4 2 n =1 2 n =1
PROOF
ao ∞
nπx ∞ nπx
f (x) =
2
+ ∑ an cos
n =1 l
+ ∑ bn sin
n =1 l
in (c , c+2l) ....……………….(1)
∴ By Euler’s formulas for the Fourier coefficients,
c + 2l
1 nπx
an =
l ∫ c
f ( x) cos
l
dx, n ≥ 0 ..…………………(2)
c + 2l
1 nπx
bn =
l ∫
c
f ( x) sin
l
dx, n ≥ 1 …....……………..(3)
Now, by definition,
c + 2l c + 2l
1 1
∫ [ f ( x)]
2
∫ y dx =
2
y = 2
dx
2l c
2l c
c + 2l
1 a ∞
nπx ∞ nπx 
= ∫ f ( x)  o + ∑ a n cos + ∑ bn sin dx, using (1)
2l c  2 n =1 l n =1 l 
ao1 c + 2l  ∞ a 1 c + 2l nπx  ∞ bn 1
c + 2l
nπx 
=  ∫ f ( x)dx  + ∑ n  ∫ f ( x) cos dx  + ∑  ∫ f ( x) sin dx 
4  l c  n =1 2  l c l  n =1 2  l c l 
∞ ∞
ao an bn
= .a o + ∑ .a n + ∑ .bn , by using (2) and (3)
4 n =1 2 n =1 2
9

10. ∞2 ∞ 2 2
ao an b
= +∑ +∑ n .
4 n =1 2 n =1 2
EXAMPLES
1. Find the half-range (i) cosine series and (ii) sine series for f (x ) = x 2 in (0 , π )
Solution:
(i) To get the half-range cosine series for f (x ) in (0 , π ), we should give an even
extension for f (x) in ( − π , 0).
i.e. put f (x) = ( − x ) 2 = x 2 in ( − π , 0)
Now f (x) is even in ( − π , π ).
∞
ao
∴ f (x) =
2
+ ∑a
n =1
n cos nx ………………….(1)
π
2
an =
π ∫ f ( x) cos nxdx.
0
π
2 2
π∫
= x cos nxdx
0
π
2   sin nx   − cos nx   − sin nx 
= x2   − 2 x  + 2 
π  n   n
2
  n
3
 0
4 4(−1) n
= .π (−1) n = ,n ≠ 0
πn 2 n2
π π
2 2 2 2 2
ao = ∫ f ( x)dx = π ∫ x dx = 3 π
π 0 0
∴ The Fourier half-range cosine series of x 2 is given by
π2 ∞
(−1) n
x2 = + 4∑ 2 cos nx in (0 , π ).
3 n =1 n
(ii) To get the half-range sine series of f (x ) in (0 , π ), we should give an odd extension
for f (x) in (- π , 0).
i.e. Put f (x ) = - ( − x ) 2 in (- π , 0)
= - x 2 in (- π , 0)
Now f (x) is odd in (- π , π ).
10

11. ∞
∴ f (x) = ∑b
n =1
n sin nx ……………….(2)
π π
2 2
bn = ∫ f ( x) sin nxdx = ∫ x 2 sin nxdx
π 0 π 0
π
2   cos nx   sin nx   cos nx 
= x2  −  − 2 x −  + 2 
π  n   n 2   n 3  0 
2 π 2 
 (−1) + 3 {(−1) − 1} 
n +1 2
= n
π n n 
 2 π 2 4 
  − , if n is odd
= π  n n 3 
− 2π , if n is even
 n
Using this value in(2), we get the half-range sine series of x 2 in (0 , π ).
2. Find the half-range sine series of f (x) = sin ax in (0 , l).
Solution:
We give an odd extension for f (x) in (-l , 0).
i.e. we put f (x) = -sin[a(-x)] = sin ax in (-l , 0)
∴ f (x) is odd in (-l , l)
∞
nπx
Let f (x ) = ∑b
n =1
n sin
l
nπx
l
2
bn = ∫ sin ax. sin l dx
l 0
1   nπ  nπ
l
  
= ∫ cos l − a  x − cos l + a  x dx
l 0     
l
  nπ   nπ  
1  sin  l − a  x sin  l + a  x 
=    −   
l   nπ   nπ  
  l − a  + a 
    l  0
1 1
= (−1) n +1 sin ( nπ − al ) − sin ( nπ + al )
nπ − al nπ + al
11

12. 1 1
= (−1) n +1 sin al + ( −1) n +1 sin al
nπ − al nπ + al
2nπ
= (−1) n +1 sin al. 2 2
n π − a 2l 2
Using this values in (1), we get the half-range sine series as
∞
(−1) n +1 .n nπx
sin ax = 2π sin al ∑ 2 2 sin
n =1 n π − a l
2 2
l
3. Find the half-range cosine series of f (x ) = a in (0 , l). Deduce the sum of
1 1 1
2
+ 2 + 2 + ∞ .
1 3 5
Solution:
Giving an odd extension for f (x) in (-l , 0), f (x ) is made an odd function in (-l , l).
nπx
∴ Let f(x) = ∑b n sin
l
..……………(1)
nπx
l
2
bn = ∫ a sin dx
l 0 l
l
 nπx 
 − cos l 
=
2a 
 nπ
l 

 =
2a
nπ
1 − ( − 1)
n
{ }


 l 0

 4a
 , if n is odd
=  nπ
0,
 if n is even
Using this value in (1), we get
4a ∞ 1 nπx
a= ∑5 n sin l in (0 , l )
π n =1,3,
Since the series whose sum is required contains constant multiples of squares of bn , we apply
Parseval’s theorem.
l
1 1
∑ bn = l ∫ [ f ( x)] dx
2 2
2 0
12

13. ∞
1 16a 2 1
i.e. .
2 π2
∑ ( 2n − 1)
n =1, 3, 5
2
= a2
∞
8a 2 1
i.e.
π2
∑ ( 2n − 1)
n =1
2
= a2
∞
1 π2
∴ ∑ ( 2n − 1) 2 8 .
n =1
=
4. Expand f (x) = x - x 2 as a Fourier series in -1 < x < 1 and using this series find the
r.m.s. value of f (x ) in the interval.
Solution:
The Fourier series of f (x ) in (-1 , -1) is given by
∞ ∞
ao
f (x) =
2
+ ∑ an cos nπx + ∑ bn sin nπx
n =1 n =1
.………………(1)
1 1
a o = ∫ f ( x)dx = ∫ ( x − x 2 ) dx
1
1 −1 −1
1
 x2 x3  1 1  1 1
=  −  =  − − + 
 2 
 3  −1  2 3   2 3 
−2
ao = ..........................(2)
3
1 1
∫1 f ( x) cos nπx dx = −∫1( x − x ) cos nπx dx
1
an = 2
1−
1
  sin nπx   − cos nπx   − sin nπx 
= ( x − x 2 )   − (1 − 2 x )  2  + (−2) 3 
  n   n   n  −1
− cos nπ 3 cos nπ
= −
n2 n2
4 cos nπ
an = − ……………….(3)
n2
13

14. 1 1
∫1 f ( x) sin nπx dx = −∫1( x − x ) sin nπx dx
1
bn = 2
1−
1
  − cos nπx   − sin nπx   cos nπx 
= ( x − x 2 )   − (1 − 2 x )   + (−2) 3 3 
 nπ  nπ  n π  −1
2 2
  
− 2 cos nπ 2 cos nπ 2 cos nπ
= − + 3 3
n 3π 3 nπ nπ
n +1
2(−1)
bn = ..........................( 4)
nπ
Substituting (2), (3), (4) in (1) we get
1 ∞ 4(−1) n +1 ∞
2(−1) n +1
f (x) = − + ∑ cos nπx + ∑ sin nπx
3 n =1 n 2 n =1 nπ
We know that r.m.s. value of f(x) in (-l , l) is
1 2 1 ∞ 1 ∞
a o + ∑ a n + ∑ bn
2 2 2
y = ……………….(5)
4 2 n =1 2 n =1
From (2) we get
−2 2 4
ao = ⇒ ao = .………………..(6)
3 9
From (3) we get
4( −1) n +1 2 16
an = 2
⇒ an = 4 ………………..(7)
n n
From (4) we get
2(−1) n +1 2 4
bn = ⇒ bn = 2 2 ..………………(8)
nπ nπ
Substituting (6), (7) and (8) in (5) we get
1 1 ∞  16 4 
+ ∑ 4 + 2 2 
2
y =
9 2 n =1  n nπ 
5. Find the Fourier series for f (x) = x 2 in − π < x < π . Hence show that
1 1 1 π4
+ 4 + 4 + =
14 2 3 90
Solution:
The Fourier series of f (x ) in (-1 , 1) is given by
π2 ∞
4(−1) n
f (x) =
3
+ ∑ n 2 cos nx
n =1
14

15. The co-efficients a o , a n , bn are
2π 2 4(−1) n
ao = , an = , bn = 0
3 n2
Parseval’s theorem is
(a )
π ∞
1 1 2 1
∫ [ f ( x)] dx = ao + ∑
2 2 2
n + bn
2π −π
4 2 n =1
 ao 2 1 ∞ 2 
( )
π
∫ [x ] + ∑ a n + bn 
2 2
dx = 2π 
2
∴
−π  4
 2 n =1 

π
 x5   π 4 1 ∞ 16 
i.e., 
 5  = 2π 
 + ∑ 4
  −π  9 2 n =1 n 
2π 5 2π 5 ∞
16
i.e., − =π∑ 4
5 9 n =1 n
8π 4 ∞
16
=∑ 4
45 n =1 n
∞
1 π4
i.e., ∑ n 4 = 90
n =1
1 1 1 π4
i.e., + 2 + 2 + ∞ =
12 3 5 90
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 ao , a n , and bn of the
function y = f (x) in (0 , 2 π ) are given by
a o = 2[mean value of y in (0 , 2 π )]
a n = 2[mean value of y cos nx in (0 , 2 π )]
bn = 2[mean value of y sin nx in (0 , 2 π )]
(i) Suppose the function f (x) is defined in the interval (0 , 2l), then its Fourier series is,
ao ∞
nπx ∞ nπx
f (x) =
2
+ ∑a
n =1
n cos
l
+ ∑ bn sin
n =1 l
and now, a o = 2[mean value of y in (0 , 2l)]
 nπx 
a n = 2 mean value of y cos in (0 , 2l )
 l 
15

16.  nπx 
bn = 2 mean value of y sin in (0 , 2l )
 l 
(ii) If the half range Fourier sine series of f (x) in (0 , l) is,
∞
nπx
f (x) = ∑b
n =1
n sin
l
, then
 nπx 
bn = 2 mean value of y sin in (0 , l )
 l 
(iii) If the half range Fourier sine series of f (x) in (0 , π ) is,
∞
nπx
f (x) = ∑b
n =1
n sin
l
, then
bn = 2[ mean value of y sin nx in (0 , π )]
(iv) If the half range Fourier cosine series of f (x) in (0 , l) is,
ao ∞
nπx
f (x) = + ∑ a n cos , then
2 n =1 l
a o = 2[mean value of y in (0 , l)]
 nπx 
a n = 2 mean value of y cos in (0 , l )
 l 
(v) If the half range Fourier cosine series of f (x) in (0 , π ) is,
ao ∞
nπx
f (x) =
2
+ ∑a
n =1
n cos
l
, then
a o = 2[mean value of y in (0 , π )]
a n = 2[ mean value of y cos nx in (0 , π )] .
EXAMPLES
1. The following table gives the variations of a periodic function over a period T.
0 T T T 2T 5T T
x 6 3 2 3 6
f (x) 1.98 1.3 1.05 1.3 -0.88 -0.25 1.98
2πx
Show that f (x ) = 0.75 + 0.37 cos θ +1.004 sin θ , where θ =
T
Solution:
Here the last value is a mere repetition of the first therefore we omit that value and
consider the remaining 6 values. ∴ n = 6.
16

17. 2πx
Given θ= ..………………..(1)
T
T T T 2T 5T π 2π
∴ when x takes the values of 0, , , , , θ takes the values 0, , ,
6 3 2 3 6 3 3
4π 5π
π, , . (By using (1))
3 3
Let the Fourier series be of the form
ao
f ( x) = + a1 cos θ + b1 sin θ , ………………(2)
2
∑y
where a o = 2 ,
 n 
 
 ∑ y cos θ 
a1 = 2 ,
 n 
 
 ∑ y sin θ 
b1 = 2 , n=6
 n 
 
θ y cos θ sin θ y cos θ y sin θ
0° 1.98 1.0 0 1.98 0
π
1.30 0.500 0.866 0.65 1.1258
3
2π 3 1.05 -0,500 0.866 -0.525 0.9093
π 1.30 -1 0 -1.3 0
4π 3 -0.88 -0.500 -0.866 0.44 0.762
5π 3 -0.25 0.500 -0.866 -0.125 0.2165
4.6 1.12 3.013
∑y
a o = 2  = 1.5, a1 = 2 ∑ y cos θ = 0.37
 6  6
 
2
b1 = ∑ y sin θ = 1.00456
6
Substituting these values of a o , a1 , and b1 in (2), we get
∴ f (x) = 0.75 + 0.37 cos θ + 1.004 sin θ
2. Find the Fourier series upto the third harmonic for the function y = f (x) defined in
(0 , π ) from the table
x 0 π 2π 3π 4π 5π π
6 6 6 6 6
17

18. f (x) 2.34 2.2 1.6 0.83 0.51 0.88 1.19
Solution:
We can express the given data in a half range Fourier sine series.
f ( x) = b1 sin x + b2 sin 2 x + b3 sin 3 x ..………………...(1)
x y = f(0) sin x sin 2x sin 3x y sin x y sin 2x y sin 3x
0 2.34 0 0 0 0 0 0
30 2.2 0.5 0.87 1 1.1 1.91 2.2
60 1.6 0.87 0.87 0 1.392 1.392 0
90 0.83 1 0 -1 0.83 0 -0.83
120 0.51 0.87 -0.87 0 0.44 -0.44 0
150 0.88 0.5 -0.87 1 0.44 0.76 0.88
180 1.19 0 0 0 0 0 0
4.202 3.622 2.25
 ∑ y sin x  1
Now b1 = 2   = [ 4.202] = 1.40

 6  3

 ∑ y sin 2 x  1
b2 = 2   = [ 3.622] = 1.207

 6  3

 ∑ y sin 3 x  1
b3 = 2   = [ 2.25] = 0.75

 6  3

Substituting these values in (1), we get
f (x) = 1.4 sin x + 1.21 sin 2x + 0.75 sin 3x
3. Compute the first two harmonics of the Fourier series for f(x) from the following data
x 0 30 60 90 120 150 180
f (x ) 0 5224 8097 7850 5499 2626 0
Solution:
Here the length of the interval is π . ∴ we can express the given data in a half range
Fourier sine series
i.e., f ( x) = b1 sin x + b2 sin 2 x ………………………(1)
18

19. x y sin x sin 2x
0 0 0 0
30 5224 .5 0.87
60 8097 0.87 0.87
90 7850 1 0
120 5499 0.87 -0.87
150 2626 0.5 -0.87
 ∑ y sin x 
Now b1 = 2  = 7867.84

 6 

 ∑ y sin 2 x 
b2 = 2   = 1506.84

 6 

∴ f (x) = 7867.84 sin x + 1506.84 sin 2x
4. Find the Fourier series as far as the second harmonic to represent the function given in
the following data.
x 0 1 2 3 4 5
f (x ) 9 18 24 28 26 20
Solution:
Here the length of the interval is 6 (not 2 π )
i.e., 2l = 6 or l = 3
∴ The Fourier series is
ao πx 2πx πx 2πx
f ( x) = + a1 cos + a 2 cos + b1 sin + b2 sin …………………..(1)
2 3 3 3 3
πx 2πx πx πx 2πx 2πx
y cos y sin y cos y sin
x 3 3 y 3 3 3 3
0 0 0 9 9 0 9 0
1 π 3 2π 3 18 9 15.7 -9 15.6
2 2π 3 4π 3 24 -12 20.9 -24 0
3 π 2π 28 -28 0 28 0
4 4π 3 8π 3 26 -13 -22.6 -13 22.6
5 5π 3 10 π 3 20 10 -17.4 -10 -17.4
125 -25 -3.4 -19 20.8
19

20.  ∑ y  2(125)
Now a o = 2 = = 41.66,
 6  6
 
2 πx
a1 = ∑ y cos = −8.33
6 3
2 πx
b1 = ∑ y sin = −1.13
6 3
2 2πx
a2 =
6
∑ y cos 3 = −6.33
2 2πx
b2 =
6
∑ y sin 3 = 6.9
Substituting these values of a o , a1 , b1 , a 2 and b2 in (1), we get
41.66 πx 2πx πx 2πx
f ( x) = − 8.33 cos − 6.33 cos − 1.13 sin + 6.9 sin
2 3 3 3 3
COMPLEX FORM OF FOURIER SERIES
∞
The equation of the form f ( x) = ∑c e
n = −∞
n
inπx l
is called the complex form or exponential form of the Fourier series of f (x) in (c , c+2l). The
coefficient c n is given by
c + 2l
1
∫ f ( x )e
−inπx l
cn = dx
2l c
When l = π , the complex form of Fourier series of f (x) in (c , c+2 π ) takes the form
∞
f ( x) = ∑c e
n = −∞
n
inx
, where
c + 2π
1
∫ f ( x )e
−inx
cn = dx.
2π c
PROBLEMS
1. Find the complex form of the Fourier series of f (x) = e x in (0 , 2).
Solution:
Since 2l = 2 or l = 1, the complex form of the Fourier series is
∞
f ( x) = ∑c e
n = −∞
n
inπx
20

21. 2
1
c n = ∫ f ( x)e −inπx dx
20
2
1
= ∫ e x e −inπx dx
20
2
1  e ( 1−inπ ) x 
=  
2  1 − inπ  0
=
1
2(1 − inπ )
{e 2(1−inπ ) − 1}
=
(1 + inπ ) {e ( cos 2nπ − i sin 2nπ ) − 1}
2
2(1 + n π 2 2
)
=
(e 2
− 1)(1 + inπ )
2(1 + n 2π 2 )
Using this value in (1), we get
 e 2 − 1  ∞ (1 + inπ ) inπx
 2  ∑ (1 + n 2π 2 ) e
ex =  
  n =−∞
2. Find the complex form of the Fourier series of f (x) = sin x in (0 , π ).
Solution:
Here 2l = π or l = π 2 .
∴ The complex form of Fourier series is
∞
f ( x) = ∑c e
n = −∞
n
i 2 nx
…………………..(1)
π
1
c n = ∫ sin xe −i 2 nx dx
π 0
π
1  e −i 2 nx 
=  { − i 2n sin x − cos x} 
π 1 − 4n
2
0
=
1
π ( 4n − 1)
2
[
− e i 2 nx − 1 = − ] 2
π ( 4n 2 − 1)
Using this value in (1), we get
2 ∞ 1
sin x = − ∑ 4n 2 − 1 .e i 2nx
π n =−∞
in (0 , π )
3. Find the complex form of the Fourier series of f (x) = e − ax in (-l , l).
Solution:
21

22. Let the complex form of the Fourier series be
∞
f ( x) = ∑c e
n = −∞
n
inπx l
l
1
c n = ∫ f ( x)e −inπx l dx
2l −l
l
1
= ∫ e − ax e −inπx l dx
2l −l
l
1
= ∫ e −( al +inπ ) x / l dx
2l −l
l
1  e −( al +inπ ) x l 
=  
2l  − ( al + inπ ) l  −l
=
1
2( al + inπ )
[
e −( al +inπ ) − e ( al +inπ ) ]
=
2
1
( al + inπ )
[
e al (−1) n − e − al (−1) n ]
[ e ± inπ
= cos nπ ± i sin nπ = (−1) n ]
sinh al (−1) n
=
al + inπ
sinh al.( al − inπ ) (−1) n
=
a 2 l 2 + n 2π 2
Using this value in (1), we have
∞
(−1) n ( al − inπ ) inπx l
e − ax = sinh al ∑ 22 2 2 e
n = −∞ a l + n π
in (-l , l)
4. Find the complex form of the Fourier series of f (x) = cos ax in (- π , π ), where a is
neither zero nor an integer.
Solution:
Here 2l = 2 π or l = π .
∴ The complex form of Fourier series is
∞
f ( x) = ∑c e
n = −∞
n
inx
………………….(1)
22

23. π
1
∫π cos ax.e
− inx
cn = dx
2π −
π
1  e −inx 
=  2 { − in cos ax + a sin ax} 
2π  a − n 2
 −π
=
1
2π ( a − n 2 )
2
[
e −inπ ( − in cos aπ + a sin aπ ) − e inπ ( − in cos aπ − a sin aπ ) ]
1
= (−1) n 2a sin aπ
2π ( a − n )
2 2
Using this value in (1), we get
a sin aπ ∞
(−1) n inx
cos ax =
π
∑
n = −∞ a2 − n2
e in (- π , π ).
UNIT 2
PART – A
1. Determine the value of a n in the Fourier series expansion of f ( x) = x 3 in − π < x < π .
Ans: f ( x) = x 3 is an odd function.
∴ an = 0
2. Find the root mean square value of f ( x) = x 2 in the interval (0 , π ) .
Ans:
RMS Vale of f ( x ) = x 2 in (0 , π ) is
π π π
1  x5 
∫ [x ]
2 1 2 2 1
y = dx = ∫ x 4 dx =  
π 0
π 0 π 5


0
1 π 5  π 4
= =
π5  5
 
3. Find the coefficient b5 of cos 5 x in the Fourier cosine series of the function f ( x ) = sin 5 x in
the interval (0 , 2π )
Ans: Here f ( x) = sin 5 x
Fourier cosine series is
23

24. ∞
ao
f (x) =
2
+ ∑a
n =1
n cos nx , where
π π
2 2
an = ∫ f ( x) cos nx dx = π ∫ sin 5 x cos nx dx
π 0 0
π
2
=
2π ∫ [ sin(5 + n) x + sin(5 − n) x] dx
0
π
− 1  cos(5 + n) x cos(5 − n) x 
= + =0
π  5+n
 5 − n 0 
cos x, if 0 < x < π
4. If f ( x) =  and f ( x) = f ( x + 2π ) for all x, find the sum of the Fourier
50, if π < x ≤ 2π
series of f (x ) at x = π .
Ans: Here π is a point of discontinuity.
∴ The sum of the Fourier series is equal to the average of right hand and left hand limit of the
given function at x = π .
f (π − 0) + f (π + 0)
i.e., f (π ) =
2
cos π + 50 49
= =
2 2
5. Find bn in the expansion of x 2 as a Fourier series in (−π , π ) .
Ans: bn = 0
Since f ( x) = x 2 is an even function in (−π , π ) .
6. If f (x) is an odd function defined in (-l , l) what are the values of a 0
Ans: a0 = 0
a n = 0 since f (x) is an odd function.
7. Find the Fourier constants bn for x sin x in (−π , π ) .
Ans: bn = 0
Since f ( x) = x sin x is an even function in (−π , π ) .
24

25. 8. State Parseval’s identity for the half-range cosine expansion of f (x ) in (0 , 1).
Ans:
1 2
∞
a0
2 ∫ [ f ( x)] dx = + ∑ an
2 2
0
2 n =1
where
1
a 0 = 2 ∫ f ( x) dx
0
1
a n = 2 ∫ f ( x) cos nx dx
0
9. Find the constant term in the Fourier series expansion of f ( x ) = x in (−π , π ) .
Ans:
a 0 = 0 since f (x ) is an odd function in (−π , π ) .
10. State Dirichlet’s conditions for Fourier series.
Ans:
(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 2 π .
(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
f ( x + 0) + f ( x − 0)
(b) if x is a point of discontinuity.
2
11. What you mean by Harmonic Analysis?
Ans:
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 ao , a n , and bn of the
function y = f (x) in (0 , 2 π ) are given by
a o = 2[mean value of y in (0 , 2 π )]
a n = 2[mean value of y cos nx in (0 , 2 π )]
bn = 2[mean value of y sin nx in (0 , 2 π )]
25

26.  2x
1 + π , − π < x < 0

12. In the Fourier expansion of f ( x) =  in (−π , π ) . Find the value of bn ,
1 − 2 x , 0 < x < π
 π

the coefficient of sin nx.
Ans:
Since f (x) is an even function the value of bn = 0.
 2( − x) 2x 
 In − π ≤ x ≤ 0 i.e., 0 ≤ − x ≤ π , f (− x ) = 1 − π = 1 + π = f ( x )
 
13. What is the constant term and the coefficient of cos nx, a n in the Fourier expansion of
f ( x) = x − x 3 in (-7 , 7)?
Ans:
Given f ( x) = x − x 3
f ( − x ) = − x + x 3 = −( x − x 3 ) = − f ( x )
The given function is an odd function. Hence a 0 and a n are zero.
14. State Parseval’s identity for full range expansion of f (x ) as Fourier series in (0 , 2l).
Ans:
c + 2l 2 2 2
1 ∞ ∞
∫ [ f ( x)] dx. = ao + ∑ a n + ∑ bn .
2
2l c 4 n =1 2 n =1 2
where
c + 2l
1 nπx
an =
l ∫
c
f ( x) cos
l
dx, n ≥ 0
c + 2l
1 nπx
bn =
l ∫
c
f ( x) sin
l
dx, n ≥ 1
15. Find a Fourier sine series for the function f (x ) = 1; 0 < x < π .
Ans:
∞
The Fourier sine series of f ( x) = ∑ bn sin nx …………………….(1)
n =1
26

27. π
2
bn =
π ∫ f ( x) sin nx dx
0
π π
2  − cos nx 
2
= ∫ sin nx dx = 
π 0 π  n 0
 =−
2
nπ
( (−1) n − 1)
bn = 0, when ' n' is even
4
= , when ' n' is odd
nπ
∞
4
∴ f ( x) = ∑ . sin nπ
n =1, 3, 5, nπ
0 0< x<π
16. If the Fourier series for the function f ( x ) =  is
sin x 0 < x < 2π
− 1 2  cos 2 x cos 4 x  1 1 1 1 π −2
f ( x) = +  + +  + sin x Deduce that − + − ∞ = .
π π  1.3 3.5  2 1.3 3.5 5.7 4
Ans:
π
Putting x = we get
2
 π  −1 2  1 1 1  1
f = + − + − +  ∞ +
 2  π π  1.3 3.5 5.7  2
−1 2  1 1 1  1
0= + − + − + ∞ +
π π  1.3 3.5 5.7  2
1 1 1 π −2
− + − ∞ = .
1.3 3.5 5.7 4
17. Define Root mean square value of a function?
Ans:
c +2 l
1
∫y
2
If a function y = f (x ) is defined in (c , c+2l), then dx is called the root mean-
2l c
square(R.M.S.) value of y in (c , c+2l) and is denoted by y.
27

28. c + 2l
2 1
Thus y = ∫y
2
dx.
2l c
18. If f ( x) = x 2 + x is expressed as a Fourier series in the interval (-2 , 2), to which value this
series converges at x = 2.
Ans:
Since x = 2 is a point of continuity, the Fourier series converges to the arithmetic mean of
f (x) at x = -2 and x = 2
f (2) + f (−2) 4 − 2 + 4 + 2
i.e., = =4
2 2
19. If the Fourier series corresponding to f ( x ) = x in the interval (0 , 2π ) is
a0 ∞
+ ∑ (a n cos nx + bn sin nx), without finding the values of a 0, a n , bn find the value of
2 n =1
2 ∞
a0
+ ∑ (a n + bn ).
2 2
2 n =1
Ans:
By using Parseval’s identity,
2 2π 2π
a0 ∞
1 1  x3  8
+ ∑ (a n + bn ) = ∫ x 2 dx =   = π 2 .
2 2
2 π 0  3 
π  0 3
n =1
20. Find the constant term in the Fourier series corresponding to f ( x) = cos 2 x expressed in the
interval (−π , π ) .
Ans:
Given f ( x) = cos 2 x
π π π
1 1  1 + cos 2 x  1 sin 2 x 
Now a 0 = ∫π cos x dx = π −∫π  2 dx = π  x + 2  0 = 1
2
π −    
PART B
1. (i) Express f ( x) = x sin x as a Fourier series in 0 ≤ x ≤ 2π .
28

29. 2l  πx 1 2πx 1 3πx 
(ii) Show that for 0 < x <l, x =  sin − sin + sin  . Using root mean square
p l 2 l 3 l 
1 1 1
value of x, deduce the value of 2
+ 2 + 2 +
1 2 3
2. (i) Find the Fourier series of periodicity 3 for f ( x ) = 2 x − x 2 in 0 < x < 3.
(ii) Find the Fourier series expansion of period 2 π for the function y = f (x) which is defined
in (0 , 2π ) by means of the table of values given below. Find the series upto the third harmonic.
x 0 π 2π π 4π 5π 2π
3 3 3 3
f (x ) 1.0 1.4 1.9 1.7 1.5 1.2 1.0
3.(i) Find the Fourier series of periodicity 2 π for f ( x) = x 2 for 0 < x < 2 π .
l 4l  πx 1 3πx 
(ii) Show that for 0 < x <l, x = −  cos + 2 cos +  . Deduce that
2 π2  l 3 l 
1 1 1 π4
+ 4 + 4 + = .
14 3 5 96
l − x, 0 < x ≤ l
4. (i) Find the Fourier series for f ( x) =  . Hence deduce the sum to infinity of
0, l ≤ x ≤ 2l
∞
1
the series ∑ (2n + 1)
n =0
2
.
(ii) Find the complex form of Fourier series of f ( x ) = e ax (−π < x < π ) in the form
sinh aπ ∞
a + in inx π ∞
(−1) n
e ax =
π
∑ (−1) n
−∞ a2 + n2
e and hence prove that =∑ 2
a sinh aπ −∞ n + a 2
.
5. Obtain the half range cosine series for f ( x) = x in (0 , π ).
6. Find the Fourier series for f ( x) = cos x in the interval (−π , π ) .
1 1 π3
7. (i) Expanding x(π − x) as a sine series in (0 , π ) show that 1 − + 3 + = .
33 5 32
(ii) Find the Fourier series as far as the second harmonic to represent the function given in the
following data.
x 0 1 2 3 4 5
29

30. f (x ) 9 18 24 28 26 20
8. Obtain the Fourier series for f (x ) of period 2l and defined as follows
 L + x in ( − L,0)
f ( x) = 
 L − x in (0, L)
1 1 1 π2
Hence deduce that + 2 + 2 + = .
12 3 5 8
9. Obtain the half range cosine series for f ( x) = x in (0 , π ).
1 in (0, π )
10. (i) Find the Fourier series of f ( x ) = 
2 in (π ,2π )
(ii) Obtain the sine series for the function
 l
 x in 0 ≤ x ≤ 2

f ( x) = 
l − x in l ≤ x ≤ l

 2
11. (i) Find the Fourier series for the function
0 in (−1, 0)
f ( x) =  and f ( x + 2) = f ( x ) for all x.
1 in (0, 1)
(ii) Determine the Fourier series for the function
πx, 0 ≤ x ≤1
f ( x) = 
π (2 − x ), 1 ≤ x ≤ 2
12. Obtain the Fourier series for f ( x ) = 1 + x + x 2 in (−π , π ) . Deduce that
1 1 1 π2
+ 2 + 2 + = .
12 2 3 6
13. Obtain the constant term and the first harmonic in the Fourier series expansion for f (x)
where f (x) is given in the following table.
x 0 1 2 3 4 5 6 7 8 9 10 11
f (x) 18. 18. 17. 15. 11. 8.3 6. 5. 6. 9. 12. 15.7
0 3 4 0 4
0 7 6 0 6
14. (i) Express f ( x) = x sin x as a Fourier series in (−π , π ).
30

31. (ii) Obtain the half range cosine series for f ( x) = ( x − 2) 2 in the interval 0 < x < 2.
15. Find the half range sine series of f ( x) = x cos x in (0 , π ).
16. (i) Find the Fourier series expansion of f (x ) = e − x in (−π , π )
(ii) Find the half-range sine series of f (x ) = sin ax in (0 , l).
17. Expand f (x ) = x - x 2 as a Fourier series in -1 < x < 1 and using this series find the r.m.s.
value of f (x) in the interval.
18. The following table gives the variations of a periodic function over a period T.
0 T T T 2T 5T T
x 6 3 2 3 6
f (x) 1.98 1.3 1.05 1.3 -0.88 -0.25 1.98
2πx
Show that f (x ) = 0.75 + 0.37 cos θ +1.004 sin θ , where θ =
T
19. Find the Fourier series upto the third harmonic for the function y = f (x) defined in (0 , π )
from the table
x 0 π 2π 3π 4π 5π π
6 6 6 6 6
f (x) 2.34 2.2 1.6 0.83 0.51 0.88 1.19
20. (i) Find the half-range (i) cosine series and (ii) sine series for f (x ) = x 2 in (0 , π )
(ii) Find the complex form of the Fourier series of f (x) = cos ax in (- π , π ).
31