Evaluate ∑n=1∞1n4 using Parseval's theorem (Fourier series).

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Evaluate ∑n=1∞1n4 using Parseval&#039;s theorem (Fourier series).

Allison Compton · Accepted Answer

Let&amp;nbsp;f(x)=x2&amp;nbsp;for&amp;nbsp;x∈(−π,π). The results of computing the Fourier coefficients&amp;nbsp;an=12π∫−ππx2e∈x&amp;nbsp;dx&amp;nbsp;=2cos⁡(πn)n2=2(−1)nn2&amp;nbsp;for&amp;nbsp;n∈Z,&amp;nbsp;n≠0, and&amp;nbsp;a0=12π∫−ππx2e∈x&amp;nbsp;dx&amp;nbsp;=2cos⁡(πn)n2=2(−1)nn2&amp;nbsp;Therefore&amp;nbsp;|an|2=4n4&amp;nbsp;for&amp;nbsp;n∈Z,n≠0&amp;nbsp;and&amp;nbsp;|a0|2=π49&amp;nbsp;By Plancherel/Parsevals

kumewekwah0 · Accepted Answer

We have f∈L2[−π,π] then  12A02+∑n=1∞(An2+Bn2)=1π∫−ππf2(x)dx  Here An and Bn are Fourier coefficients of f.  Let f(x)=x2 for x∈(−π,π). Then f∈L2[−π,π] and  f(x)∼π23+4∑n=1∞(−1)nn2cos⁡nx  Therefore  12(2π23)2+∑n=1∞(4(−1)nn2)2=1π∫−ππx4dx  So  ∑n=1∞1n4=π490

RizerMix · Accepted Answer

The series ∑n≥11n4 also appears in the following funny way. Consider the operator d2/dx2 on [0,π] s.t. Dirichlet boundary conditions. Find Green&#039;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 ∫0π∫0πG2(x,s)dxds in two ways, a) using the explicit formula for G(x,s) and b) using the representation G(x,s)=∑n≥1un(x)un(s)λn.After simplifications, you will find ∑n≥11n4=π490.

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

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