Transcript half range expansions

DEPARTMENT OF
MATHEMATICS
[YEAR OF ESTABLISHMENT – 1997]
DEPARTMENT OF MATHEMATICS, CVRCE
MATHEMATICS-II
TEXT BOOK: ADVANCED
ENGINEERING
MATHEMATICS BY ERWIN
KREYSZIG [8th EDITION]
LECTURE-15
Half Range Fourier Series Expansions
[Chapter – 10.4]
DEPARTMENT OF MATHEMATICS, CVRCE
Definition
Fourier series as half range expansions
Let us consider the Fourier series expansion of a
function f ( x ) which is periodic, defined in an interval
c  x  c  2 L of length 2L. Now we wish to expand a
non-periodic function f ( x ) defined in half of the above
interval say 0 < x < L of length L, such an expansions
are known as half range expansions or half range
Fourier series.
DEPARTMENT OF MATHEMATICS, CVRCE
A
half range expansion containing only cosine
term is known as half range Fourier cosine series of
f ( x ) in the interval 0  x  L .
A
half range expansion containing only sine
term is known as half range Fourier sine series of
f ( x ) in the interval 0  x  L .
DEPARTMENT OF MATHEMATICS, CVRCE
Fourier cosine series as the half range
When we wish to express the function f ( x ) as a
cosine series in the interval 0  x  L
, then we
extend the function f ( x ) reflecting it in the y-axis,
so that f (  x )  f ( x ) which is symmetrical about yaxis. Then the extended function is even in the
interval  L  x  L and its expansion will give the
required Fourier cosine series which is given by the
formula as:
DEPARTMENT OF MATHEMATICS, CVRCE
Fourier cosine series as the half range
 n x 
f ( x )  a 0   a n cos 

L


n 1

(1)
with the Fourier coefficients
a0 
1
L

L
f ( x ) dx
(2)
0
and a n 
2
L

L
0
 n x 
f ( x ) cos 
 dx
 L 
(3)
DEPARTMENT OF MATHEMATICS, CVRCE
Fourier sine series as the half range
To express the function f ( x ) as a sine series in the
interval 0  x  L , then we extend the function
f ( x ) reflecting it in the origin, so that f (  x )  f ( x )
which is symmetrical about the origin. Then the
extended function is odd in the interval  L  x  L
And the expansion will give the required Fourier
sine series which is given by the formula as:
DEPARTMENT OF MATHEMATICS, CVRCE
Fourier sine series as the half range

 n x 
f ( x )   bn sin 

 L 
n 1
(4)
with the Fourier coefficient
bn 
2
L

L
0
 n x 
f ( x ) sin 
 dx
 L 
(5)
Some Solved Problem Based on Half Range Expansion
1. Find the Fourier cosine series as well as the
Fourier sine series of the following function:
f ( x)  x,
Solution :
(0  x  L )
E x p an sio n o f
f ( x )  x as h alf ran g e
co sin e series in 0  x  L .
The graph of f ( x )  x , (0  x  L ) is the line O A.
DEPARTMENT OF MATHEMATICS, CVRCE
Some Solved Problem Based on Half Range Expansion
We know that the half range Fourier cosine series of
the given function f(x) in the prescribed range is given
by
 n x 
f ( x )  a 0   a n cos 

 L 
n 1

(1)
with the Fourier coefficients
a0 
1
L

L
f ( x ) dx
0
 a0 
1
L
x dx 

L
0
1
1
L
x  
L  0 



0
2L
2L
2
2
L
2
DEPARTMENT OF MATHEMATICS, CVRCE
Some Solved Problem Based on Half Range Expansion
an 
2
L

L
0
 n x 
f ( x ) cos 
 dx
 L 

2  L 
n x

 

 x sin 
L    n 
 L


2
L
 n x 
x cos 
 dx
 L 
L

0

 L 

 n x
   
 co s 
0 
 L
  n 
L
2


2
 L 

0  
  cos n  1 
L 
 n 

 0, if n b e even

 an   4 L
.
 2 2 , if n b e o d d
n 
2

2L
n 
2
L




0




 (  1)  1 


n
2
DEPARTMENT OF MATHEMATICS, CVRCE
Some Solved Problem Based on Half Range Expansion
Substituting the values of an’s in the eqn.(1), we get
f ( x) 
L
2

2L

2


n 1
1
n
2
 n x 
 (  1)  1  co s 



L


n

4L 
1
1
x
 3 x 
 5 x 

 2  co s 
  2 co s 
  2 co s 
  ... 
2
 
 L  3
 L  5
 L 

L
which is the required Fourier cosine series of f(x) over
the half range 0<x<L.
DEPARTMENT OF MATHEMATICS, CVRCE
Some Solved Problem Based on Half Range Expansion
Expansion of
f ( x )  x as half range sine series in 0  x  L.
The graph of f(x)=x in 0<x<L is the line OA. Let us
extend the function f(x) in the interval –L<x<0, shown
by the line OA’, so that the new function
is
symmetrical about the origin which represents an odd
function in –L<x<L that contain sine series only and is
given by the formula as:
 n x 
f ( x )   bn sin 
        (2)
 L 
n 1

with the Fourier coefficient
DEPARTMENT OF MATHEMATICS, CVRCE
bn 
2
L
L

0
 n x 
f ( x ) sin 
 dx
 L 
Some Solved Problem Based on Half Range Expansion
 bn 
2
L
L

0
 n x 
x sin 
 dx
 L 
L
L
2

2
 L 
 n x  
 L  
 n x   
  
  x cos 
  
  sin 
 
L   n  
 L   0  n  
 L   0 


2
n
 L cos n
 0 
2 L
n
  1
n
 2 L
 n , if n be even

.
 2 L , if n be odd
 n
DEPARTMENT OF MATHEMATICS, CVRCE
Some Solved Problem Based on Half Range Expansion
Putting the values of bn’s in eqn.(2), we obtain the
desired Fourier sine series over the half range 0 < x < L
as
f ( x)  
2L



n 1
1
n
  1
n
 n x 
sin 

 L 

2L 
 x  1
 2 x  1
 3 x 

  sin 
  sin 
  ...  .
 sin 
 
2
3
 L 
 L 
 L 

DEPARTMENT OF MATHEMATICS, CVRCE
Some Solved Problem Based on Half Range Expansion
2. Find the Fourier cosine series as well as the
Fourier sine series of the following function.
f ( x)    x,
Solution :
(0  x   ).
f ( x )  x as h alf ran g e
E x p an sio n o f
co sin e series in 0  x   .
We know that the Fourier cosine series as a half range
is given by
 n x 
f ( x )  a 0   a n cos 
  a0 
 L 
n 1


a
n
cos nx
( L )
(1)
n 1
DEPARTMENT OF MATHEMATICS, CVRCE
Some Solved Problem Based on Half Range Expansion
with the Fourier coefficients a 0 
1
L

L
f ( x)dx
0

1




1 
x 

 x 

 
2 0
2
f ( x)dx
0
1 


 
2

 


2 
2
2
and
an 
2

f ( x ) cos

L
0
n x
L
2   (  x ) sin nx
 
 
n
dx 
2



f ( x ) cos nxdx 
0
  cos nx

2
n
 



0
2


 (
 x ) cos nxdx
0
2 
1

  0  2  cos n  1  
 
n

DEPARTMENT OF MATHEMATICS, CVRCE
Some Solved Problem Based on Half Range Expansion

2
   1   1
2

n  
n
 0 , if n b e e v e n

  4
.
, if n b e o d d
 2
n 
Substituting the values of an’s in eqn.(1), we get
f ( x) 

2

2



n 1
1
   1   1  co s n x
2 

n
n

4 
1
1



co s x  2 co s 3 x  2 co s 5 x  ... .

2
 
3
5

is the required half range Fourier cosine series of f(x).
DEPARTMENT OF MATHEMATICS, CVRCE
Some Solved Problem Based on Half Range Expansion
Expansion of f ( x )  (  x ) as half range sine series in 0  x   .
We know that the Fourier sine series is given by

f ( x) 
b
n
sin
n 1
n
L

x
b
n
sin nx      (2)
n 1
where
bn 
2

f ( x ) sin

L
0
n x
L
dx 
2



0
f ( x ) sin nx dx 
2


 (  x ) sin nx dx
0



2 
 
 2
2  (  x ) cos nx 
 sin nx  
  
0    0  .
 
  
2


  
n
0  n
 0 
 
n 
 n
DEPARTMENT OF MATHEMATICS, CVRCE
Some Solved Problem Based on Half Range Expansion
Substituting the values of bn’s in eqn.(1), we get

f ( x)  2
n 1
1
s in n x
n
1
1


 2  s in x 
s in 2 x 
s in 3 x  ... 
2
3


is the required half range Fourier sine series of f(x).
DEPARTMENT OF MATHEMATICS, CVRCE
Some Solved Problem Based on Half Range Expansion
3. Find the half-range Fourier sine series of
f ( x )  x (  x ) in 0  x  
Solution:
We know that the Fourier sine series is given by

f ( x) 
b
n
sin
n

x 
L
n 1
b
n
sin n x      ( 2 )
n 1
where
bn 
2

f ( x ) sin

L
0
n x
L
dx 
2


2

 f ( x ) sin nx dx    x (
0
0
 x ) sin nx dx
Some Solved Problem Based on Half Range Expansion

2


  x sin nx dx 
0
2



2
x sin nx dx
0

2


2  x
2
 2  x sin nx dx    
cos nx  
  n
0 n
0




0

x cos nxdx 



1
 x

 2   cos nx  2 sin nx 
n
 n
0


2


2  x
2 x
1

  
cos nx    sin nx  2 cos nx 
  n
n n
n
0

0





Some Solved Problem Based on Half Range Expansion
2



 
 2
 2   cos n     
cos n
 n
   n

2
cos n 
n

4
n 
3
2
 2 1

   2  cos n  1   

 n n
cos n 
n
1  cos n 
4
 cos n
 1
n 
 8
if n is odd
 3
 n 
 0 if n is even

3
Some Solved Problem Based on Half Range Expansion
Substituting the values of bn’s in eqn.(1), we get
f ( x) 
4



n 1
1
n
3
1 
c o s n
 s in
nx
8 
1
1


s in 3 x 
s in 5 x  ... 
 s in x 
3
5
 
3
3

is the required half range Fourier sine series of f(x).
Assignments
(1) Find the Fourier cosine series as well as the
Fourier sine series of the following functions:
(i ) f ( x )  x
(0  x  L )
3
( ii ) f ( x )  e
x
(0  x  L ).
(2) Find the two half-range expansions of the functions
L
 2k
x , if 0  x 

 L
2
f ( x)  
.
 2 k ( L  x ), if L  x  L

2
 L
DEPARTMENT OF MATHEMATICS, CVRCE
Assignments Contd …
(3) Find the half-range Fourier sine and cosine series
for the function as f ( x )  x in 0  x   2 and
f ( x )  (  x ) in  2  x   .
DEPARTMENT OF MATHEMATICS, CVRCE