### Part 1

```FOURIER SERIES
PERIODIC FUNCTIONS
A function f(x) is said to be
periodic
with
period
T
if
f(x+T)=f(x)  x , where T is a
positive constant . The least
value of T>0 is called the period
of f(x).
f(x+2T) =f ((x+T)+T)
=f (x+T)=f(x)
f(x+nT)=f(x) for all x
Ex.1 f(x)=sin x has periods 2,
4, 6, …. and 2 is the period
of f(x).
Ex.2 The period of sin nx and
cos nx is 2/n.
FOURIER SERIES
Let f (x) be defined in the
interval (l , l ) and outside the
interval by f ( x  2l )  f ( x) i.e
assume that
f (x) has the
period 2l .The Fourierseries
corresponding to
f (x ) is
given by
a0
nx
nx 

   an cos
 bn sin

2 n1 
l
l 

where the Fourier coeffecients are
l
a0
1

f ( x)d x

l l
an
1
nx

f ( x ) co s
dx

l l
l
bn
1
nx

f ( x ) sin
dx

l l
l
l
l
n  1,2,3,....
If f (x) is defined in the interval
(c,c+2l), the coefficients can be
determined equivalently from
a0
1

l
an 
bn
1
l
1

l
c  2l

f ( x)d x
c
c  2l

c
c  2l

c
f ( x ) co s
nx
dx
l
nx
f ( x ) sin
dx
l
DIRICHLET CONDITIONS
Suppose that
1. f(x) is defined and single valued
except possibly at finite number of
points in (-l,+l)
2. f(x) is periodic outside (-l,+l) with
period 2l
3. f(x) and f ’(x) are piecewise
continuous in( -l,+l)
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)]/2 if x is a
point of discontinuity
METHOD OF OBTAINING FOURIER
SERIES OF f (x )

a0
nx
nx 

1. f (x)  2    an cos l  bn sin l 

n 1 
l
1
2. a0 
f ( x)d x

l l
3. an
1
nx

f ( x ) co s
dx

l l
l
4. bn
1
nx

f ( x ) sin
dx

l l
l
l
l
n  1,2,3,....
SOLVED PROBLEMS
1. Expand f(x)=x2,0<x<2 in Fourier series
if the period is 2 . Prove that 1  1  1  ...   2
12
22
32
6
SOLUTION
Period = 2 l= 2  thus l=  and choosing
c=0
1
an 
l


c  2
1

1


c
nx
f ( x) cos
dx
l
2
 f ( x) cos nx dx
0
2
2
x
 cos nx dx
0
2
1  2 sin nx
  cos nx 
  sin nx 
 x
 2 x
  2

2
3
 
n
n
n



 0
4
 2
n0
n
2
2
8

2
a0 
x
dx 

 0
3
1
1
bn 
l


1

1

c  2

c
nx
f ( x ) sin
dx
l
2

f ( x ) sin nxdx
0
2
2
x
 sin nxdx
0
2
1  2  cos nx 
cos nx 
  sin nx 

x 
  2 x
2

2
  
n 
n
n 3  0


 4

n
a0 
f ( x)    an cosnx  bn sin nx
2 n1

4

4
4

2
f ( x)  x 
   2 cos nx  sin nx
3 n1  n
n

2
At x=0 the above Fourier series

reduces to 4 2
4
3

n
n 1
2
X=0 is the point of discontinuity
By Dirichlet conditions, the series converges
at x=0 to (0+4 2)/2 = 2 2
 2



n 1
2

4 2
4

 2
3
n 1 n
1
2

2
n
6
2. Find the Fourier series expansion for the
following periodic function of period 4.
Solution
2  x  2  x  0
f ( x)  
2  x 0  x  2
f ( x  4)  f ( x )
l
a0
1

f ( x)d x

l l
0
2

1 

  2  x d x   2  x d x
2 2
0

0
2

1 
x2 
x2  





2x 

2x 




2 
2 2
2 0 



 2
1
nx
  f ( x ) co s
dx
l l
l
l
an
0
1
n

xd x 
  2  x  co s
2 2
2
n


2

x
co
s


xd x
2

2
0

nx
nx
sin
 co s

1 
2
2  x  n2  (1)

2
( n )
2 


2
4

0





 2

nx
nx

sin
 co s
2  ( 1)
2
  2  x 
2
n

( n )

2
4


4
1  (1) 
n
n 2 2
0 fo r n even

 8
fo r n o d d

2
2
n 
2
 
 
 
 
 
0 
1
nx
bn   f ( x ) sin
dx
l l
l
l
0
2

1
nx
nx
   2  x  sin
dx   2  x  sin
dx
2 2
l
l
0


nx


cos


1 
2

2  x 
n
2 



2



nx


cos



2
2  x 
n




2


f ( x)  1 
8
2


n 1
0
nx 


sin



2
  (1)

2
2
n






4
2
nx  

 sin




2
  ( 1)
  0
2
2
n 

 



4
0 

1 
x 


cos
2
n

1
n2 
2 


2
EVEN AND ODD FUNCTIONS
A function f(x) is called odd if
f(-x)=-f(x)
E x : x 3, s i n x , t a n x , x 5+ 2 x + 3
A function f(x) is called even if
f(-x)=f(x)
E x : x 4, c o s x , e x+ e -x, 2 x 6+ x 2+ 2
EXPANSIONS OF EVEN AND ODD
PERIODIC FUNCTIONS
If f (x) is a periodic function defined in the
interval (l , l ) , it can be represented by the
Fourier series
Case1. If f (x) is an even function
l
a0
1

f ( x )d x

l l
2

l
l

0
f ( x )d x
2
nx
an   f ( x) cos
dx
l 0
l
l
nx


is also even function
 f ( x) cos
l


1
nx
bn   f ( x) sin
dx
l l
l
l
nx


 0  f ( x) sin
is odd function
l


If a periodic function f (x) is even in
(l , l ) , its Fourier series expansion
contains only cosine terms

a0
nx
f ( x) 
  an co s
2
l
n 1
a0
an
l
2

l

2

l
l
f ( x)d x
0

0
nx
f ( x ) co s
dx
l
Case 2. When f (x) is an odd function
l
1
a0   f ( x ) dx  0
l l
1
nx
nx


an   f ( x) cos
dx  0  f ( x) cos
is odd 
l l
l
l


l
2
nx 
nx

bn   f ( x) sin
dx  f ( x) sin
is even
l0
l
l


l
If a periodic function f (x) is odd in
(l , l ) ,its Fourier expansion
contains only sine terms

nx
f ( x) 
b
n 1
2
bn 
l
l

0
n
sin
l
nx
f ( x ) sin
dx
l
SOLVED PROBLEMS
1.For a function defined by f ( x)  x ,    x  
obtain a Fourier series. Deduce that
1
1
1
2
 2  2  .... 
2
1
3
5
8
Solution
f ( x)  x is an even function

a0
nx
f ( x) 
  an cos
2
l
n 1
SOLUTION
a0 

an 
2

2

2



x dx
0


0
2x

xd x 
 
 2
2


x co s n xd x 
0



 
0
2

x co s n xd x


0

2   sin n x    co s n x  

x



2
   n  
n
 0
2
n




1
1
2
n


f ( x) 

2

2


 1


n 1
n
n
cosnx
1
2
At x=0 the above series reduces to
n

2   1  1

2


n 1
n2
x=0 is a point of continuity, by
Dirichlet condition the Fourier
series converges to f(0) and
f(0)=0
0
0

2

2


2

2

  1n  1



2
n
n 1 

2
2
 2

 2  2  2  ....
3
5
1


1
1
1
 2  2  .... 
2
1
3
5
k
PROBLEM 2 f ( x)  

k

2
8
when  3  x  0
when 0  x  3
Is the function even or odd. Find
the Fourier series of f(x)
SOLUTION
f (x) is odd function
a0  0
bn
an  0
l
2

l

2

3
3
0

0
nx
f ( x ) sin
dx
l
nx
k sin
dx
3
nx


k
co
s
2 
3


n
3 
3

2k

[1  ( 1) n ]
n
3




0
[1  (1) ]
nx
f ( x) 
sin

 n 1
n
3
2k

n
nx


k
co
s
2 
3


n
3 
3

2k

[1  ( 1) n ]
n
3




0
[1  (1) ]
nx
f ( x) 
sin

 n 1
n
3
2k

n
nx


k
co
s
2 
3


n
3 
3

2k

[1  ( 1) n ]
n
3




0
[1  (1) ]
nx
f ( x) 
sin

 n 1
n
3
2k

n
HALF RANGE SERIES
COSINE SERIES
A function f (x) defined in (0, l ) can be expanded
as a Fourier series of period 2l containing only
cosine terms by extending f (x) suitably in (l ,0) .
(As an even function)

a0
nx
f ( x) 
  an cos
2
l
n 1
where
2
an 
l
l

0
nx
f ( x ) cos
d x, n  0
l
SINE SERIES
A function f (x) defined in (0, l ) can be expanded
as a Fourier series of period 2l containing
only sine terms by extending f (x) suitably in (l ,0).
[As an odd function]
nx
f ( x )   bn sin
l
u 1
wh ere

bn
2

l
l

0
nx
f ( x ) sin
d x, n  1
l
SOLVED PROBLEMS
Obtain the Fourier expansion of (x sinx )as a
cosine series in (0,  )
.Hence find the value of
1
2
2
2



1.3
3.5
5.7
SOLUTION
Given function represents an even function in ( ,  )

a0
f ( x) 
  an cos nx
2 n1
2
an 
 f ( x) cos nxdx
 0
2
a 
 f ( x)dx
0
 0
2
2
a   x sin xdx  x( cos x) 1( sin x)
0
0 

0
 2
an 
2


0
x sin x cos nxdx

1


x  sin(1  n) x  sin(1  n) x  dx

 0 2

2

  cos(1  n) x


   sin(1  n) x  
 x

  1.
2
1 n
  (1  n)
0 
1  
 


 
  cos(1  n) x    sin(1  n) x  
 
   x
  1.
2
1 n
  (1  n)
  
0 
n 1
 1(1)

n 1
1
1 n

(1)
n 1
1  2 1
 1
 (1) 


2

 n  1 n  1 n  1
n 1
n 1
n 1
if
n 1


2
1
a1   x sin x cos xdx   x sin 2 xdx
0
0

1    cos 2 x    sin 2 x 
  x
  1.

2
  2   2
 0
1     1
  
 2  2

1
(1) n1
x sin x  1  cos x  2 2
cos nx
2
n2 n  1
in
(0,  )
x
At

2

1  2
n2
x


2
the above series reduces to
 1
n
cos
1
2
n 1
n2
is a point of continuity
The given series converges to
 
f( )
2
2

2

 1  2
n2
 1
n
co s
1
2
n 1
n
2
1
1
1
 2



 ........... 
1.3 3.5 5.7
4
2) Expand f ( x)  x, 0  x  2 in half range
(a) sine Series (b) Cosine series.
SOLUTION
(a)
Extend the definition of given function to that of an
odd function of period 4
i.e
 x;  2  x  0
f ( x)  

 x; 0  x  2
an  0
Here
2
nx
bn   f ( x ) sin
dx
l 0
l
l
2
nx
bn   f ( x ) sin
dx
20
2
2
2

nx
nx 
 sin
n
  cos 2

 4( 1)
2
  x(
)  1(
) 
2 2
n
n
n



 0
2
22
 (1) n
nx
f ( x) 
sin

 n 1
n
2
4

(b)
Extend the definition of given function to that of
an even function of period 4
 x;2  x  0
f ( x)  

x
;
0

x

2


bn  0
2
nx
an   f ( x) cos
dx
l 0
l
nx
l
2
2
an   f ( x) cos
dx
20
2
2

nx
nx 
 cos
n
 sin 2

4
(

1
)
1
2
  x(
)  1(
) 
;
2 2
2 2
n

n
n



 0
2
22
n0


2
a0 
 xdx  2
0
f ( x)  1 
4

2


n 1
(1)
n
n
2
 cos nx
1
2
Exercise problems
1.
0;  x  o 
f ( x)  

sin x;0  x   
f (x )
Find Fourier series of
2.
f (x ) e
x
in
(0,2 )
Find Fourier series of
f (x )
3.Find the Fourier series of
in
( ,  )
4.Find the Fourier series of
in
f ( x)  x
3
(-2 ,2)
f ( x)  4  x
2
5.Represent function
x
f ( x)  sin
L
In (0,L) by a Fourier cosine series
6.Determine the half range sine series for
x
f ( x)  
8  x
0 x4
4 x8
PARSEVAL’S IDENTITY
• To prove that
2
1 2 
2
2 


f
(
x
)
dx

l
a

(
a

b
n )
2 0  n
l
n 1


l
Provided the Fourier series for f(x) converges uniformly in (-l, I).
The Fourier Series for f(x) in (-l,l) is


a0
nx
nx
f ( x) 
  an cos
  bn sin
................(1)
2
l
l
n 1
n 1
Multiplying both sides of (1) by f(x)and integrating term from – l to l
( which is justified because f(x) is uniformly convergent)






a0
nx
nx 
2
l f ( x) dx  2  n1 an n1 f ( x) cos l dx  n1 bn n1 f ( x) sin l dx


a0
  la0   an (lan )   bn (lbn )
2
n 1
n 1

l
2


a
2
2 
2
0
  f ( x) dx  l    (an  bn ).....................(2)
l
 2 n1

l
CASE-I
If f(x) is defined in (0,2l) then Parseval’s
Identity is given by
2
1 2 
2
2 
(an  bn )...............(3)
0  f ( x) dx  l  2 a0  
n 1

2l
CASE-II
If half range cosine series in (o,l) for f(x) is

a0
nx
f ( x) 
2
  an cos
n 1
l
.
Then Parseval’s Identity is given by
2

l 1 2
2
an ..............(4)
0  f ( x) dx  2  2 a0  
n 1

.
l
CASE-III
If the half range Sine sereies in (0,l) for f(x) is

nx
f ( x)   bn sin
n 1
l
Then Parseval,s Identity is given by
2
l 
2


f
(
x
)
dx

b
..........
....(
5
)

n


0
2  n 1

l

RMS VALUE OF FUNCTION
If a function y=f(x) is defined in ( c , c+2l ),then
1
2l
c  2l

y 2 dx
c
is called the root mean square value (RMS value) of y in
( c , c+2l ).It is denoted by
y
1
y 
2l
2
c  2l

c
2
f ( x ) dx
.
Equation(2)
becomes
1 2
2
2 
y  2l  l  a0   (an  bn ).
n 1
2


2
y
2
1 2 1  2
2 
  a0   (an  bn ).
2 n1
4

Equation(3)
becomes
1 2
2
2 
y  2l  l  a0   (an  bn ).
n 1
2


2
y
2
1 2 1  2
2 
  a0   (an  bn ).
2 n1
4

Equation(4) becomes
l 1 2  2 
  a0   an .
2 2
n 1

2
y l
y
1 2 1  2 
  a0   an .
2 n1
4

2
Equation(5)becomes
y
2
1  2 
   bn .
 2 n1 
SOLVED PROBLEMS
1) Find the Fourier series of periodic function
f ( x)  x  x
2
in
( ,  )
Hence deduce the sum of series
1 1
1
1
 4  4  4  .......
4
1 2
3
4
1 2


2
6
n 1 n

Assuming that
SOLUTION
a0 
nx 
nx
f ( x)    an cos
  bn sin
.
2 n1
l
l
n 1
( ,  )
in


1
1 2
2
a0   f ( x)dx   x dx 
 
 
3
2



1
 sin nx
 cos nx 
an   f ( x) cos nxdx   ( x 2  x) cos nxdx   x 2 (
)  2x
 
 

n
n 2  
1
1
4
 2 (1) n
n
bn 
1

if
n0
xcosnx is odd function



f ( x) sin nxdx 
1


2
(
x
  x) sin nxdx  0 

1


 x sin nxdx


1   cos nx
 sin nx 
2
n
   x(
)  1(
)

(

1
)

 
n
n2
n

2
 x sin nx
is odd function

(1)
(1)
f ( x) 
 4 2 cos nx  2
sin nx.
3
n
n 1 n
n 1
in ( ,  )
2


n
n
Using the Parseval’s Identity

1
1

2
2
2 
2
  a0   (an  bn ).
y
2 n1
4

y
2
1

2

2
2
(
x
 x) d x 


4
5

2
3
2
4
2
n
n
a0 
; an  2 (1) ; bn  (1)
3
n
n
4
2
4




1 4
1
16 1
4



  4  2
5
3
4 9
2 n 1 n
2 n 1 n
2



n 1
1
4

4
n
90


n 1
1
2

2
n
6
i
2)By using sine series for
Show that
2
8
 1
f ( x)  1
in
0 x 
1
1
1


 .......
2
2
2
3
5
7
SOLUTION

nx
f ( x)   bn sin
 f ( x)   f ( x) sin nx
l
n 1
n 1

bn 
2




0


2   cos nx 
f ( x) sin nxdx   sin nxdx  (
)
 0

n
0
2

1  (1) n
n
2

for
n0
1  (1) n
f ( x)  1  
sin nx
 n 1
n
2

By Parseval’s Identity
y
2
1
2
   bn .
 2 n1 




1
4
2
n
y 
1
dx

1

1

(

1
)


 0
2 n 1 n 2 2
1
2
2
1
1
1

 1  2  2  2  .......
8
3
5
7

2
0 xl
3)Prove that in
l 4l
x 1
3x 1
5x
x   2 (cos  2 cos
 2 cos
 ..........)
2 n
l 3
l
5
l
1
1
1
1
4
and deduce that 14  34  54  7 4  .......  96
SOLUTION
In Half range cosine series
l
l
2
2
a0   f ( x)dx   xdx  l
l 0
l 0
2
nx
2
nx
an   f ( x) cos
dx   x cos
dx
l 0
l
l 0
l
l
l
l

nx
nx 
sin
 cos

2
2l
n
l
l
  x(
)  1.(
)

(

1
)
1

2 2
2 2
n
n
l 
n


 0
l
l2


a0
nx
f ( x) 
  an cos
2
l
n 1
l
1
x 
 2
2

4
nx
cos

2
n
l
n 1, 3, 5. . .


By Parseval’s Identity
y
2
1 2 1  2 
  a0   an .
2 n1
4

2
l
1 l
1
l
y   f ( x) 2 dx   x 2 dx 
l 0
l 0
3
2
l2
l2
1  4l 2
n


  4 4 ( 1)  1
3
4
2 n 1 n 
2


1
1
1
1
4
 4  4  4  4  ....... 
1
3
5
7
96
COMPLEX FORM OF FOURIER
SERIES
The Fourier series of
function of period 2l is
a
periodic
a0  
nx
nx 
f ( x) 
   an cos
 bn sin

2 n 1 
l
l 
e i  e  i 
e i  e  i
cos 
sin  
2
2i
nx
nx
nx
nx
i
i
i
i

a0   e l  e l
e l e l
f ( x) 
   an
 bn
2 n 1 
2
2i






nx
i
 i nl x

f ( x)  c0    cn e  c n e l 
n 1 

1
1
1
c0  a0 , cn  an  ibn , c n  an  ibn 
2
2
2
l
inx
1
cn   f ( x)e l dx for n  0,1,2,....
2l l

The Fourier series can be represented in the
following way
f ( x) 
n 
c
n  
n
e
1
where cn 
2l
inx
l
dx
l

f ( x )e

inx
l
dx
l
SOLVED PROBLEM
1.Find the complex form of the Fourier series
of the periodic function
0 when 0  x  l
f ( x)  
 when l  x  2l
SOLUTION
inx
f ( x )   cn e
l
n  

cn
1

2l
2l
 f ( x )e

inx
l
dx
0
l
inx



1
cn 
  0 e l
2l  0 
 
l


e
2l  in
2l
inx



dx   e l dx


l

inx

l
2l



l



e  2in  e  ni 
2ni



n

1   1
2ni
i
n

1   1
2n

i
inx
n
f ( x)  
(1  ( 1) )e
l
n   2 n


2.Find the complex form of Fourier seriesof
f(x)=sinx in (0,)
SOLUTION
inx
c
f ( x) 


n  
n
e
2
 

 l 

2



2 inx
c
e
 n
n  
cn 

1

1



f ( x )e  2 inx d x
0

 2 inx
sin
xe
dx

0


1  e
 2in sin x  co s x 


2
 1  4n
0
 2 inx
1
 2 in
cn 
e
1
2
 4n  1
2

2
 4n  1




f ( x)  
2


 4n
n  

1
2

1
e
2 inx
in (0,  )
HARMONIC ANALYSIS

a0
f ( x) 
  an cos nx  bn sin nx
2
n 1
a0 
an 
bn 
1

1

1

2
 f ( x)dx
0
2
 f ( x) cos nxdx
0
2
 f ( x) sin nxdx
0
2
 1






a0  2
f
(
x
)
dx

2
m
ean
of
f
(
x
)
in
0
,
2



2

0


2
 1


an  2
f
(
x
)
cos
nxdx


2

0


 2m eanof f ( x) cos nx in 0,2 
 1 2


bn  2
f
(
x
)
sin
nxdx


2

0


 2m eanof f ( x) sin nx in 0,2 
T h e t e r m a 1c o s x + b 1s i n x i s c a l l e d t h e
fundamental or first harmonic,
t h e t e r m a 2c o s x + b 2s i n x i s c a l l e d t h e
second harmonic and so on.
Solved Problem
1.Find first two harmonics of Fourier
Series from the following table
```