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fine 
     
 
 
 
 
 
 
 
 
 
 
 
 
 
   
 
 
fine 
  
 
 
 
 
 
 
 
  
 
 
 
   
  
 
  
 
   
 
 
 
 
 
 
  
  
 
   
 
 
 
  
 
  
 
 
 
  
 
 
 
 
 
 
 
   
  
  
   
 
 
   
 
 
 
 
 
 
   
     
 
 
 
 
      
 
 
 
 
 
 
 
  
     
 
 
      
 
 
 
 
 
 
 
 
  
 
 
  
 
 
 
   
 
 
   
 
 
 
 
   
 
  
     
 
 
     
 
 
     
 
   
   
 
 
 
 
 
 
 
 
 
   
 
 
 
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fine 
     
 
 
 
 
 
 
 
 
 
 
 
 
 
   
 
 
fine 
  
 
 
 
 
 
 
 
  
 
 
 
   
  
 
  
 
   
 
 
 
 
 
 
  
  
 
   
 
 
 
  
 
  
 
 
 
  
 
 
 
 
 
 
 
   
  
  
   
 
 
   
 
 
 
 
 
 
   
     
 
 
 
 
      
 
 
 
 
 
 
 
  
     
 
 
      
 
 
 
 
 
 
 
 
  
 
 
  
 
 
 
   
 
 
   
 
 
 
 
   
 
  
     
 
 
     
 
 
     
 
   
   
 
 
 
 
 
 
 
 
 
   
 
 
 
Free coaching of B.Sc (h) maths & JAM 
For more  8130648819 
 
 =
 
 
 0|(  x)
     
 
|
 
 
 ?
(  )     
 
dx
 
 
1 
 = 
 
 
0|(  x)
     
 
|
 
 
 0
     
 
 
1
 
 
1 
 = 
 
 
[(0 0 ) 0
(  )
 
 
 
 
 
 
 
1] 
 = 
 
  
 
,1 ( 1 )
 
- 
  a 0 = 
 
 
? f(x)dx
 
 
 = 
 
 
?
(  x)dx 
 
 
0 x 
 
 
 
1
 
 
 
 
 
 = 
 
 
0 
 
 
 
 
 
 0 1 
 
 
.
 
 
 
   
  the fourier cosine series of the function f(x) which is continuous on (0, ?? ) is 
 
 
 
 ? a
 
cosnx
 
   
 
 = 
 
 
 ?
 
  
 
,1 ( 1 )
 
-cosnx 
 
 
 
 
 
 .
     
 
 
 
      
 
 
  /
 
   
 
f(x)  =
 
 
 
 
 
 .
    
 
 
 
     
 
 
  / 
now, cosine series at 0 = f(0
+
) 
   f(0
+
) = 
 
 
 
 
 
 .
 
 
 
 
 
 
 
  / 
    ??  = 
 
 
 
 
 
.1 
 
 
 
  / 
      
 
 
  =  
 
 
.1 
 
 
 
  / 
     
 
 
 
 = 1 
 
 
 
 
 
 
 
   
GENERAL FORMON [0,  l] (HALF RANGE SERIES) 
[0, ?? ]  a n = 
 
 
? f(x)cos
   
 
dx
 
 
  
[0, ?? ]  a n = 
 
 
? f(x)cosnxdx
 
 
  
Similarly, for b n 
Que. Obtain the fourier series of the foll
n
 
(a)f(x)= x
2
   0   x   ??  
(b) f(x) ={
 
 
0
 
 
 
          
0 x 
 
 
 
 
 x 
  
 
  
 
 x  
 
Check sv. Notes pg 13     
(c) f(x) = 2
x
2 x
      
0 x 1
1 x 2
     
Check s.v. notes page 17   
b n = 
 
 
? f(x)sin.
   
 
/dx
 
 
 
 = 
 
 
? x
 
sin.
   
 
/dx
 
 
 
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fine 
     
 
 
 
 
 
 
 
 
 
 
 
 
 
   
 
 
fine 
  
 
 
 
 
 
 
 
  
 
 
 
   
  
 
  
 
   
 
 
 
 
 
 
  
  
 
   
 
 
 
  
 
  
 
 
 
  
 
 
 
 
 
 
 
   
  
  
   
 
 
   
 
 
 
 
 
 
   
     
 
 
 
 
      
 
 
 
 
 
 
 
  
     
 
 
      
 
 
 
 
 
 
 
 
  
 
 
  
 
 
 
   
 
 
   
 
 
 
 
   
 
  
     
 
 
     
 
 
     
 
   
   
 
 
 
 
 
 
 
 
 
   
 
 
 
Free coaching of B.Sc (h) maths & JAM 
For more  8130648819 
 
 =
 
 
 0|(  x)
     
 
|
 
 
 ?
(  )     
 
dx
 
 
1 
 = 
 
 
0|(  x)
     
 
|
 
 
 0
     
 
 
1
 
 
1 
 = 
 
 
[(0 0 ) 0
(  )
 
 
 
 
 
 
 
1] 
 = 
 
  
 
,1 ( 1 )
 
- 
  a 0 = 
 
 
? f(x)dx
 
 
 = 
 
 
?
(  x)dx 
 
 
0 x 
 
 
 
1
 
 
 
 
 
 = 
 
 
0 
 
 
 
 
 
 0 1 
 
 
.
 
 
 
   
  the fourier cosine series of the function f(x) which is continuous on (0, ?? ) is 
 
 
 
 ? a
 
cosnx
 
   
 
 = 
 
 
 ?
 
  
 
,1 ( 1 )
 
-cosnx 
 
 
 
 
 
 .
     
 
 
 
      
 
 
  /
 
   
 
f(x)  =
 
 
 
 
 
 .
    
 
 
 
     
 
 
  / 
now, cosine series at 0 = f(0
+
) 
   f(0
+
) = 
 
 
 
 
 
 .
 
 
 
 
 
 
 
  / 
    ??  = 
 
 
 
 
 
.1 
 
 
 
  / 
      
 
 
  =  
 
 
.1 
 
 
 
  / 
     
 
 
 
 = 1 
 
 
 
 
 
 
 
   
GENERAL FORMON [0,  l] (HALF RANGE SERIES) 
[0, ?? ]  a n = 
 
 
? f(x)cos
   
 
dx
 
 
  
[0, ?? ]  a n = 
 
 
? f(x)cosnxdx
 
 
  
Similarly, for b n 
Que. Obtain the fourier series of the foll
n
 
(a)f(x)= x
2
   0   x   ??  
(b) f(x) ={
 
 
0
 
 
 
          
0 x 
 
 
 
 
 x 
  
 
  
 
 x  
 
Check sv. Notes pg 13     
(c) f(x) = 2
x
2 x
      
0 x 1
1 x 2
     
Check s.v. notes page 17   
b n = 
 
 
? f(x)sin.
   
 
/dx
 
 
 
 = 
 
 
? x
 
sin.
   
 
/dx
 
 
 
Free coaching of B.Sc (h) maths & JAM 
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 = 
 
 
 0 x
 
cos.
   
 
/.
 
  
/  ?2x cos.
   
 
/.
 
  
/dx1
 
 
 
 =
 
 
 [0 x
 
cos.
   
 
/.
 
  
/  
  
  
2xsin.
   
 
/
 
  
  ?sin.
   
 
/.
 
  
/dx  31
 
 
] 
 = 
 
 
[ x
 
cos.
   
 
/.
 
  
/  
  
  
{x.
 
  
/sin.
   
 
/ cos
   
 
 .
 
  
/
 
}]
 
 
 
 = 2l( 1 )
 
.
 
  
 
 
  
0
 
 
  
.0 
 
 
(  )
 
( 1 )
 
 
 
 
(  )
 
1 
 = 
  
 
  
( 1 )
   
 
  
 
 
 
 
 
,( 1 )
 
 1 - 
(b)b n= 
 
 
? f(x)sinnxdx
 
 
 
 = 
 
 
0?
 
 
sinnx dx ? 0 .sinnxdx ? 
 
 
sinnxdx
 
    
    
   
   
 
1 
 = 
 
 
0?
 
 
sinnx dx ?
 
 
sinnxdx
 
    
   
 
1 
 = 
 
 
, 
 
  
 
- 
Where I 1 = ? sinnxdx    and   
 
 ? sinnx dx
 
    
   
 
 
Consider, I 2        let y = ?? x i.e. x = ?? y 
?  2 = ?
  /
 
sinn(  y)( dy) 
 = ?
 
  /
sin(n  ny)dy   
 =? sinnx dx
  /
 
   if n is odd 
     ? sinnx dx
  /
 
  if n is even. 
?  2 = {
 
 
  
 
     
if n is even
 if n is odd
 
   b n = {
0
 
 
 
 
  
if n is odd 
 if n is even
 
 =
 
 
? sinnxdx
   
 
 
 =
 
 
  . 
     
 
/
 
   
 
 =  
 
  
0cos
  
 
 cos0 1 
 =  
 
  
 0cos
  
 
 1 1 
 
  
01 cos
  
 
1 
 
  
. 2 sin
 
  
 
/ 
b n = 
 
  
sin
 
  
 
 
put  
       n = 2r 
b 2r =
 
  
sin
 
  
 
 
now,  sin
  
 
  {
±
v 
 
0
  
if r 3k 
 if r 3k
     
  b 2r = {
 
  
.
 
 
0
    
if r 3k 
 if r 3k
 
? b r ={
 
 
0
          
if r 3k 
 if r 3k
 
  the reqd. sine series is  
 = ? b
 
sinnx ? b
  
sin2rx ?
 
 
     
 
 
   
(    
 
   
 
   
 
     
 
 
     
 
 
      
 
   
(c)b n =
 
 
? f(x)sin.
   
 
/dx
 
 
 
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fine 
     
 
 
 
 
 
 
 
 
 
 
 
 
 
   
 
 
fine 
  
 
 
 
 
 
 
 
  
 
 
 
   
  
 
  
 
   
 
 
 
 
 
 
  
  
 
   
 
 
 
  
 
  
 
 
 
  
 
 
 
 
 
 
 
   
  
  
   
 
 
   
 
 
 
 
 
 
   
     
 
 
 
 
      
 
 
 
 
 
 
 
  
     
 
 
      
 
 
 
 
 
 
 
 
  
 
 
  
 
 
 
   
 
 
   
 
 
 
 
   
 
  
     
 
 
     
 
 
     
 
   
   
 
 
 
 
 
 
 
 
 
   
 
 
 
Free coaching of B.Sc (h) maths & JAM 
For more  8130648819 
 
 =
 
 
 0|(  x)
     
 
|
 
 
 ?
(  )     
 
dx
 
 
1 
 = 
 
 
0|(  x)
     
 
|
 
 
 0
     
 
 
1
 
 
1 
 = 
 
 
[(0 0 ) 0
(  )
 
 
 
 
 
 
 
1] 
 = 
 
  
 
,1 ( 1 )
 
- 
  a 0 = 
 
 
? f(x)dx
 
 
 = 
 
 
?
(  x)dx 
 
 
0 x 
 
 
 
1
 
 
 
 
 
 = 
 
 
0 
 
 
 
 
 
 0 1 
 
 
.
 
 
 
   
  the fourier cosine series of the function f(x) which is continuous on (0, ?? ) is 
 
 
 
 ? a
 
cosnx
 
   
 
 = 
 
 
 ?
 
  
 
,1 ( 1 )
 
-cosnx 
 
 
 
 
 
 .
     
 
 
 
      
 
 
  /
 
   
 
f(x)  =
 
 
 
 
 
 .
    
 
 
 
     
 
 
  / 
now, cosine series at 0 = f(0
+
) 
   f(0
+
) = 
 
 
 
 
 
 .
 
 
 
 
 
 
 
  / 
    ??  = 
 
 
 
 
 
.1 
 
 
 
  / 
      
 
 
  =  
 
 
.1 
 
 
 
  / 
     
 
 
 
 = 1 
 
 
 
 
 
 
 
   
GENERAL FORMON [0,  l] (HALF RANGE SERIES) 
[0, ?? ]  a n = 
 
 
? f(x)cos
   
 
dx
 
 
  
[0, ?? ]  a n = 
 
 
? f(x)cosnxdx
 
 
  
Similarly, for b n 
Que. Obtain the fourier series of the foll
n
 
(a)f(x)= x
2
   0   x   ??  
(b) f(x) ={
 
 
0
 
 
 
          
0 x 
 
 
 
 
 x 
  
 
  
 
 x  
 
Check sv. Notes pg 13     
(c) f(x) = 2
x
2 x
      
0 x 1
1 x 2
     
Check s.v. notes page 17   
b n = 
 
 
? f(x)sin.
   
 
/dx
 
 
 
 = 
 
 
? x
 
sin.
   
 
/dx
 
 
 
Free coaching of B.Sc (h) maths & JAM 
For more  8130648819 
 
 = 
 
 
 0 x
 
cos.
   
 
/.
 
  
/  ?2x cos.
   
 
/.
 
  
/dx1
 
 
 
 =
 
 
 [0 x
 
cos.
   
 
/.
 
  
/  
  
  
2xsin.
   
 
/
 
  
  ?sin.
   
 
/.
 
  
/dx  31
 
 
] 
 = 
 
 
[ x
 
cos.
   
 
/.
 
  
/  
  
  
{x.
 
  
/sin.
   
 
/ cos
   
 
 .
 
  
/
 
}]
 
 
 
 = 2l( 1 )
 
.
 
  
 
 
  
0
 
 
  
.0 
 
 
(  )
 
( 1 )
 
 
 
 
(  )
 
1 
 = 
  
 
  
( 1 )
   
 
  
 
 
 
 
 
,( 1 )
 
 1 - 
(b)b n= 
 
 
? f(x)sinnxdx
 
 
 
 = 
 
 
0?
 
 
sinnx dx ? 0 .sinnxdx ? 
 
 
sinnxdx
 
    
    
   
   
 
1 
 = 
 
 
0?
 
 
sinnx dx ?
 
 
sinnxdx
 
    
   
 
1 
 = 
 
 
, 
 
  
 
- 
Where I 1 = ? sinnxdx    and   
 
 ? sinnx dx
 
    
   
 
 
Consider, I 2        let y = ?? x i.e. x = ?? y 
?  2 = ?
  /
 
sinn(  y)( dy) 
 = ?
 
  /
sin(n  ny)dy   
 =? sinnx dx
  /
 
   if n is odd 
     ? sinnx dx
  /
 
  if n is even. 
?  2 = {
 
 
  
 
     
if n is even
 if n is odd
 
   b n = {
0
 
 
 
 
  
if n is odd 
 if n is even
 
 =
 
 
? sinnxdx
   
 
 
 =
 
 
  . 
     
 
/
 
   
 
 =  
 
  
0cos
  
 
 cos0 1 
 =  
 
  
 0cos
  
 
 1 1 
 
  
01 cos
  
 
1 
 
  
. 2 sin
 
  
 
/ 
b n = 
 
  
sin
 
  
 
 
put  
       n = 2r 
b 2r =
 
  
sin
 
  
 
 
now,  sin
  
 
  {
±
v 
 
0
  
if r 3k 
 if r 3k
     
  b 2r = {
 
  
.
 
 
0
    
if r 3k 
 if r 3k
 
? b r ={
 
 
0
          
if r 3k 
 if r 3k
 
  the reqd. sine series is  
 = ? b
 
sinnx ? b
  
sin2rx ?
 
 
     
 
 
   
(    
 
   
 
   
 
     
 
 
     
 
 
      
 
   
(c)b n =
 
 
? f(x)sin.
   
 
/dx
 
 
 
Free coaching of B.Sc (h) maths & JAM 
For more  8130648819 
 
 = ? f(x)sin.
   
 
/dx
 
 
 
 =? xsin.
   
 
/dx
 
 
 ?
(2 x)sin.
   
 
/dx
 
 
 
 =[ xcos.
   
 
/.
 
  
/|
 
 
 ? cos.
   
 
/.
 
  
/dx
 
 
]+[(2 x) cos.
   
 
/..
 
  
/|
 
 
 
?
( 1 )( 1 )cos.
   
 
/.
 
  
/
 
 
] 
 =0 cos.
  
 
/.
 
  
/1 [sin.
   
 
/.
 
  
/
 
]
 
 
 [20 cos.
  
 
/.
 
  
/3 {sin.
   
 
/.
 
  
/
 
}
 
 
] 
 =[ cos.
  
 
/.
 
  
/ {sin.
  
 
/.
 
  
/
 
 0 }] [cos.
  
 
/.
 
  
/ { sin(n ).
 
  
/
 
 sin
  
 
.
 
  
/
 
}]  
 =[sin.
  
 
/ .
 
  
/
 
 0 ] [ sin( n).
 
  
/
 
 sin.
  
 
/ .
 
  
/
 
] 
 = sin.
  
 
/.
 
  
/
 
 [ 0 sin.
  
 
/.
 
  
/
 
] = 2 sin.
  
 
/.
 
  
/
 
= 
 
 
 
 
 
sin.
  
 
/ 
   b n = 0  for n even. 
b n   ( 1)
2n 1
  for n odd.  
Que. Find the fourier series for the function |sin x| in the interval , ?? , ?? ]  
Sol
n
  f(x) = |sin x|  which is an even function 
  fourier series of    f(x) will have cosines terms only. 
We have that |sin x| = 2
 sinx
sinx
 
for   x 0
for 0 x  
    
Now  
 a 0 = 
 
 
? f(x)dx
 
  
 = 
 
 
.2 ? f(x)dx
 
 
  
 
 
? sinx dx  
 
 
 (cosx)
 
 
 
 
 
 
 
 
? a n = 
 
 
? f(x)cosnx dx 
 
 
? sinxcosnxdx 
 
 
  
 
 
 
 
 
 
Where I n =? sinxcosnx dx .sinx
     
 
/
 
 
 
 
 ?
     
 
cosxdx 0 
 
 
[? sinnxcosxdx
 
 
]
 
 
 
 =  
 
 
0. 
     
 
cosx/
 
 
 ?
     
 
( sinx)
 
 
1 
   n  = 
 
 
 
, cosn  1 - 
 
 
 
 ? cosnx
 
 
sinx dx 
   n =  
 
 
 
,1 ( 1 )
 
- 
 
 
 
  
 
 
  .1 
 
 
 
/ 
 
 =  
 
 
 
 (1 ( 1 ))   
   n =  
 
 
 
  
 ,1 ( 1 )
 
- 
  a n =  
 
 ( 
 
  )
,1 ( 1 )
 
-  
Hence, the fourier series of |sin x| is  
Page 5


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For more  8130648819 
 
 
  
 
 
 
 
 
 
 
        
         
 
 
 
 
 
 
 
 
    
 
 
 
   
   
 
 
 
 
        
         
 
 
 
 
 
 
 
 
    
 
 
 
 
  
 
   
 
 
 
        
         
 
 
 
 
 
 
 
 
    
 
 
 
  
 
  
    
 
 
 
     
 
 
 
     
 
 
 
     
 
 
   
 
 
 
 
 
      
   
   
 
 
 
 
 
      
   
   
 
        
         
 
 
 
 
 
 
 
 
    
 
 
 
 
  
 
 
  
fine 
     
 
 
 
 
 
 
 
 
 
 
 
 
 
   
 
 
fine 
  
 
 
 
 
 
 
 
  
 
 
 
   
  
 
  
 
   
 
 
 
 
 
 
  
  
 
   
 
 
 
  
 
  
 
 
 
  
 
 
 
 
 
 
 
   
  
  
   
 
 
   
 
 
 
 
 
 
   
     
 
 
 
 
      
 
 
 
 
 
 
 
  
     
 
 
      
 
 
 
 
 
 
 
 
  
 
 
  
 
 
 
   
 
 
   
 
 
 
 
   
 
  
     
 
 
     
 
 
     
 
   
   
 
 
 
 
 
 
 
 
 
   
 
 
 
Free coaching of B.Sc (h) maths & JAM 
For more  8130648819 
 
 =
 
 
 0|(  x)
     
 
|
 
 
 ?
(  )     
 
dx
 
 
1 
 = 
 
 
0|(  x)
     
 
|
 
 
 0
     
 
 
1
 
 
1 
 = 
 
 
[(0 0 ) 0
(  )
 
 
 
 
 
 
 
1] 
 = 
 
  
 
,1 ( 1 )
 
- 
  a 0 = 
 
 
? f(x)dx
 
 
 = 
 
 
?
(  x)dx 
 
 
0 x 
 
 
 
1
 
 
 
 
 
 = 
 
 
0 
 
 
 
 
 
 0 1 
 
 
.
 
 
 
   
  the fourier cosine series of the function f(x) which is continuous on (0, ?? ) is 
 
 
 
 ? a
 
cosnx
 
   
 
 = 
 
 
 ?
 
  
 
,1 ( 1 )
 
-cosnx 
 
 
 
 
 
 .
     
 
 
 
      
 
 
  /
 
   
 
f(x)  =
 
 
 
 
 
 .
    
 
 
 
     
 
 
  / 
now, cosine series at 0 = f(0
+
) 
   f(0
+
) = 
 
 
 
 
 
 .
 
 
 
 
 
 
 
  / 
    ??  = 
 
 
 
 
 
.1 
 
 
 
  / 
      
 
 
  =  
 
 
.1 
 
 
 
  / 
     
 
 
 
 = 1 
 
 
 
 
 
 
 
   
GENERAL FORMON [0,  l] (HALF RANGE SERIES) 
[0, ?? ]  a n = 
 
 
? f(x)cos
   
 
dx
 
 
  
[0, ?? ]  a n = 
 
 
? f(x)cosnxdx
 
 
  
Similarly, for b n 
Que. Obtain the fourier series of the foll
n
 
(a)f(x)= x
2
   0   x   ??  
(b) f(x) ={
 
 
0
 
 
 
          
0 x 
 
 
 
 
 x 
  
 
  
 
 x  
 
Check sv. Notes pg 13     
(c) f(x) = 2
x
2 x
      
0 x 1
1 x 2
     
Check s.v. notes page 17   
b n = 
 
 
? f(x)sin.
   
 
/dx
 
 
 
 = 
 
 
? x
 
sin.
   
 
/dx
 
 
 
Free coaching of B.Sc (h) maths & JAM 
For more  8130648819 
 
 = 
 
 
 0 x
 
cos.
   
 
/.
 
  
/  ?2x cos.
   
 
/.
 
  
/dx1
 
 
 
 =
 
 
 [0 x
 
cos.
   
 
/.
 
  
/  
  
  
2xsin.
   
 
/
 
  
  ?sin.
   
 
/.
 
  
/dx  31
 
 
] 
 = 
 
 
[ x
 
cos.
   
 
/.
 
  
/  
  
  
{x.
 
  
/sin.
   
 
/ cos
   
 
 .
 
  
/
 
}]
 
 
 
 = 2l( 1 )
 
.
 
  
 
 
  
0
 
 
  
.0 
 
 
(  )
 
( 1 )
 
 
 
 
(  )
 
1 
 = 
  
 
  
( 1 )
   
 
  
 
 
 
 
 
,( 1 )
 
 1 - 
(b)b n= 
 
 
? f(x)sinnxdx
 
 
 
 = 
 
 
0?
 
 
sinnx dx ? 0 .sinnxdx ? 
 
 
sinnxdx
 
    
    
   
   
 
1 
 = 
 
 
0?
 
 
sinnx dx ?
 
 
sinnxdx
 
    
   
 
1 
 = 
 
 
, 
 
  
 
- 
Where I 1 = ? sinnxdx    and   
 
 ? sinnx dx
 
    
   
 
 
Consider, I 2        let y = ?? x i.e. x = ?? y 
?  2 = ?
  /
 
sinn(  y)( dy) 
 = ?
 
  /
sin(n  ny)dy   
 =? sinnx dx
  /
 
   if n is odd 
     ? sinnx dx
  /
 
  if n is even. 
?  2 = {
 
 
  
 
     
if n is even
 if n is odd
 
   b n = {
0
 
 
 
 
  
if n is odd 
 if n is even
 
 =
 
 
? sinnxdx
   
 
 
 =
 
 
  . 
     
 
/
 
   
 
 =  
 
  
0cos
  
 
 cos0 1 
 =  
 
  
 0cos
  
 
 1 1 
 
  
01 cos
  
 
1 
 
  
. 2 sin
 
  
 
/ 
b n = 
 
  
sin
 
  
 
 
put  
       n = 2r 
b 2r =
 
  
sin
 
  
 
 
now,  sin
  
 
  {
±
v 
 
0
  
if r 3k 
 if r 3k
     
  b 2r = {
 
  
.
 
 
0
    
if r 3k 
 if r 3k
 
? b r ={
 
 
0
          
if r 3k 
 if r 3k
 
  the reqd. sine series is  
 = ? b
 
sinnx ? b
  
sin2rx ?
 
 
     
 
 
   
(    
 
   
 
   
 
     
 
 
     
 
 
      
 
   
(c)b n =
 
 
? f(x)sin.
   
 
/dx
 
 
 
Free coaching of B.Sc (h) maths & JAM 
For more  8130648819 
 
 = ? f(x)sin.
   
 
/dx
 
 
 
 =? xsin.
   
 
/dx
 
 
 ?
(2 x)sin.
   
 
/dx
 
 
 
 =[ xcos.
   
 
/.
 
  
/|
 
 
 ? cos.
   
 
/.
 
  
/dx
 
 
]+[(2 x) cos.
   
 
/..
 
  
/|
 
 
 
?
( 1 )( 1 )cos.
   
 
/.
 
  
/
 
 
] 
 =0 cos.
  
 
/.
 
  
/1 [sin.
   
 
/.
 
  
/
 
]
 
 
 [20 cos.
  
 
/.
 
  
/3 {sin.
   
 
/.
 
  
/
 
}
 
 
] 
 =[ cos.
  
 
/.
 
  
/ {sin.
  
 
/.
 
  
/
 
 0 }] [cos.
  
 
/.
 
  
/ { sin(n ).
 
  
/
 
 sin
  
 
.
 
  
/
 
}]  
 =[sin.
  
 
/ .
 
  
/
 
 0 ] [ sin( n).
 
  
/
 
 sin.
  
 
/ .
 
  
/
 
] 
 = sin.
  
 
/.
 
  
/
 
 [ 0 sin.
  
 
/.
 
  
/
 
] = 2 sin.
  
 
/.
 
  
/
 
= 
 
 
 
 
 
sin.
  
 
/ 
   b n = 0  for n even. 
b n   ( 1)
2n 1
  for n odd.  
Que. Find the fourier series for the function |sin x| in the interval , ?? , ?? ]  
Sol
n
  f(x) = |sin x|  which is an even function 
  fourier series of    f(x) will have cosines terms only. 
We have that |sin x| = 2
 sinx
sinx
 
for   x 0
for 0 x  
    
Now  
 a 0 = 
 
 
? f(x)dx
 
  
 = 
 
 
.2 ? f(x)dx
 
 
  
 
 
? sinx dx  
 
 
 (cosx)
 
 
 
 
 
 
 
 
? a n = 
 
 
? f(x)cosnx dx 
 
 
? sinxcosnxdx 
 
 
  
 
 
 
 
 
 
Where I n =? sinxcosnx dx .sinx
     
 
/
 
 
 
 
 ?
     
 
cosxdx 0 
 
 
[? sinnxcosxdx
 
 
]
 
 
 
 =  
 
 
0. 
     
 
cosx/
 
 
 ?
     
 
( sinx)
 
 
1 
   n  = 
 
 
 
, cosn  1 - 
 
 
 
 ? cosnx
 
 
sinx dx 
   n =  
 
 
 
,1 ( 1 )
 
- 
 
 
 
  
 
 
  .1 
 
 
 
/ 
 
 =  
 
 
 
 (1 ( 1 ))   
   n =  
 
 
 
  
 ,1 ( 1 )
 
- 
  a n =  
 
 ( 
 
  )
,1 ( 1 )
 
-  
Hence, the fourier series of |sin x| is  
Free coaching of B.Sc (h) maths & JAM 
For more  8130648819 
 
 =
 
 
a
 
 ? cosnx
 
   
 
 =
 
 
a
 
 a
 
cosx a
 
cos2x a
 
cos3x   
 
 
 .
 
 
/ 0 
 
   
cos4x   
 = 
 
 
0
 
 
 ? cos
   
  
 
  
 
   
1 
 
 
 
      
 
  
 
 
 
   
 
 
 
   
 
 
   
  
 
 
  
 
  
 
  
  
 
 
  
 
 
 
 
  
   
  
 
 
 
  
 
 
 
  
 
  
   
 
  
 
 
  
 
     n
n
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FAQs on Half Range Series - Fourier Series - Topic-wise Tests & Solved Examples for Mathematics

1. What is a half-range Fourier series?
Ans. A half-range Fourier series is a mathematical representation of a periodic function that is defined only over half of its period. It is obtained by taking the Fourier series of the given function over a symmetric interval and then modifying the coefficients to represent the function over the desired half-range.
2. How is a half-range Fourier series different from a full-range Fourier series?
Ans. A half-range Fourier series is only defined over a specific half of the period of a periodic function, while a full-range Fourier series represents the entire period. In a half-range Fourier series, the coefficients are modified to only consider the even or odd terms, depending on the desired half-range, resulting in a more compact representation.
3. What are the advantages of using a half-range Fourier series?
Ans. Using a half-range Fourier series can often simplify calculations and reduce computational effort compared to a full-range Fourier series. Since the series only includes even or odd terms, depending on the half-range, the number of terms needed to approximate the function is reduced, leading to a more efficient representation.
4. Can any periodic function be represented using a half-range Fourier series?
Ans. No, not all periodic functions can be represented using a half-range Fourier series. The function must satisfy certain conditions, such as being symmetric about the y-axis or having odd symmetry, for a half-range Fourier series to accurately represent it. Functions that do not meet these conditions may require a full-range Fourier series.
5. How can a half-range Fourier series be used in practical applications?
Ans. Half-range Fourier series have applications in various fields, such as signal processing, electrical engineering, and physics. They can be used to analyze and approximate periodic signals or functions in a more efficient manner. By representing a function over a half-range, computational resources can be optimized, making it useful in real-time systems or applications with limited resources.
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