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Find a Fourier series to represent x−x2 from x=−π to x=π. Hence show that, 112−122+132−142+……=π212.
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written 3.0 years ago by |
solution:
Let ,x−x2=a02+∑∞n=1ancosnx+∑∞n=1bnsinnx By Euler's formulae, we have,
a0=1π∫π−π(x−x2)dx=1π[x22−x33]π−π
=1π[(π22−π33)−(π22+π33)]=−2π23
an=1π∫π−π(x−x2)cosnxdx=1π[(x−x2)sinnxn−(1−2x)(−cosnxn2)+(−2)(−sinnxn3)]n−π=1π[(1−2π)cosnπn2−(1+2π)cosnπn2]=1π(−4π⋅cosnπn2)=−4(−1)nn2bn=1π∫π−π(x−x2)sinnxdx
[∵cosnπ=(−1)n]
=1π[(x−x2)(−cosnxn)−(1−2x)(−sinnxn2)+(−2)(cosnxn3)]n−π=1π[(π2−π)cosnπn−2cosnπn3+(−π−π2)cosnπn+2cosnπn3]=1π[−2π⋅cosnπn]=−2(−1)nn∴x−x2=−π23−4−∑n=1(−1)nn2cosnx−2−∑n=1(−1)nnsinnx=−π23−4[−cosx12+cos2x22−cos3x32+……]=−π23+4[cosx12−cos2x22+cos3x32−……]+2[sinx1−sin2x2+sin3x3−……]
Putting, x=0,
we get 0=−π23+4(112−122+132−142+……)
⇒112−122+132−142+……=π212.
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