Evaluate \sum_{n=1}^\infty\frac{1}{n^4} using Parseval's theorem (Fourier series).

poveli1e

poveli1e

Answered question

2022-01-23

Evaluate n=11n4 using Parseval's theorem (Fourier series).

Answer & Explanation

Allison Compton

Allison Compton

Beginner2022-01-24Added 16 answers

Let f(x)=x2 for x(π,π). The results of computing the Fourier coefficients 
an=12πππx2ex dx =2cos(πn)n2=2(1)nn2 
for nZ, n0, and a0=12πππx2ex dx =2cos(πn)n2=2(1)nn2 
Therefore |an|2=4n4 for nZ,n0 and |a0|2=π49 
By Plancherel/Parsevals

kumewekwah0

kumewekwah0

Beginner2022-01-25Added 14 answers

We have fL2[π,π] then
12A02+n=1(An2+Bn2)=1πππf2(x)dx
Here An and Bn are Fourier coefficients of f.
Let f(x)=x2 for x(π,π). Then fL2[π,π] and
f(x)π23+4n=1(1)nn2cosnx
Therefore
12(2π23)2+n=1(4(1)nn2)2=1πππx4dx
So
n=11n4=π490
RizerMix

RizerMix

Expert2022-01-27Added 656 answers

The series n11n4 also appears in the following funny way. Consider the operator d2/dx2 on [0,π] s.t. Dirichlet boundary conditions. Find Green's function G(x,s). Find the normalized in L2[0.π] eigen functions un(x) of the operator; the eigenvalues are λn=n2.Evaluate the integral 0π0πG2(x,s)dxds in two ways, a) using the explicit formula for G(x,s) and b) using the representation G(x,s)=n1un(x)un(s)λn.After simplifications, you will find n11n4=π490.

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