Find the Fourier series for f (x) = sin x in −p < x < p.
[JNTU 2002]
Solution The function is defined in S= (−p, p) by
f (x) = sin x (1)
Here
(i) x∈ S ⇒ −x ∈ S and (ii) f ( −x) =
sin(−x) =−sin x=−f ( x) ∀ x ∈ S.
∴ f is an odd function in S and hence a
0
=a
n
= 0
for all n.
The Fourier series of the function is given by
1
sinsin
n
n
xbnx
∞
=
=
∑
(2)
where
[]
=
=−−+
∫
∫
0
0
2
sinsin
1
cos(1)cos(1)
n
bxnxdx
nxnxdx
p
p
p
p
(3)
0
1sin(1)sin(1)
0,1
11
sin0 for(1),(1)or0
nxnx
n
nn
ppnn
p
p
pp
−+
⎡⎤
=−=≠
⎢⎥
−+
⎣⎦
==−−∵
(4)
n = 1
==−
⎛⎞⎡⎤
=−=−−−0=
⎜⎟
⎢⎥
⎝⎠⎣⎦
∫∫
1
00
0
21
sinsin(1cos2)
1sin211
(0)(sin2)1
22
bxxdxxdx
x
x
pp
p
pp
pp
pp
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