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Find a Fourier series to represent xx2 from x=π to x=π. Hence show that, 112122+132142+=π212.
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solution:

Let ,xx2=a02+n=1ancosnx+n=1bnsinnx By Euler's formulae, we have,

a0=1πππ(xx2)dx=1π[x22x33]ππ

=1π[(π22π33)(π22+π33)]=2π23

an=1πππ(xx2)cosnxdx=1π[(xx2)sinnxn(12x)(cosnxn2)+(2)(sinnxn3)]nπ=1π[(12π)cosnπn2(1+2π)cosnπn2]=1π(4πcosnπn2)=4(1)nn2bn=1πππ(xx2)sinnxdx

[cosnπ=(1)n]

=1π[(xx2)(cosnxn)(12x)(sinnxn2)+(2)(cosnxn3)]nπ=1π[(π2π)cosnπn2cosnπn3+(ππ2)cosnπn+2cosnπn3]=1π[2πcosnπn]=2(1)nnxx2=π234n=1(1)nn2cosnx2n=1(1)nnsinnx=π234[cosx12+cos2x22cos3x32+]=π23+4[cosx12cos2x22+cos3x32]+2[sinx1sin2x2+sin3x3]

Putting, x=0,

we get 0=π23+4(112122+132142+)

112122+132142+=π212

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