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=
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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=
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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=
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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=
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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=
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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= ? 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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fine
fine
Free coaching of B.Sc (h) maths & JAM
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=
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
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= ? 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
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=
a
? cosnx
=
a
a
cosx a
cos2x a
cos3x
.
/ 0
cos4x
=
0
? cos
1
n
n
)
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27 docs|150 tests
|
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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