poveli1e

2022-01-23

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

Allison Compton

Expert

Let $f\left(x\right)={x}^{2}$ for $x\in \left(-\pi ,\pi \right)$. The results of computing the Fourier coefficients

for , and
Therefore ${|{a}_{n}|}^{2}=\frac{4}{{n}^{4}}$ for $n\in \mathbb{Z},n\ne 0$ and ${|{a}_{0}|}^{2}=\frac{{\pi }^{4}}{9}$
By Plancherel/Parsevals

kumewekwah0

Expert

We have $f\in {L}^{2}\left[-\pi ,\pi \right]$ then
$\frac{1}{2}{A}_{0}^{2}+\sum _{n=1}^{\mathrm{\infty }}\left({A}_{n}^{2}+{B}_{n}^{2}\right)=\frac{1}{\pi }{\int }_{-\pi }^{\pi }{f}^{2}\left(x\right)dx$
Here ${A}_{n}$ and ${B}_{n}$ are Fourier coefficients of f.
Let $f\left(x\right)={x}^{2}$ for $x\in \left(-\pi ,\pi \right)$. Then $f\in {L}^{2}\left[-\pi ,\pi \right]$ and
$f\left(x\right)\sim \frac{{\pi }^{2}}{3}+4\sum _{n=1}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{n}}{{n}^{2}}\mathrm{cos}nx$
Therefore
$\frac{1}{2}{\left(\frac{2{\pi }^{2}}{3}\right)}^{2}+\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{4{\left(-1\right)}^{n}}{{n}^{2}}\right)}^{2}=\frac{1}{\pi }{\int }_{-\pi }^{\pi }{x}^{4}dx$
So
$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{n}^{4}}=\frac{{\pi }^{4}}{90}$

RizerMix

Expert

The series $\sum _{n\ge 1}\frac{1}{{n}^{4}}$ also appears in the following funny way. Consider the operator ${d}^{2}/d{x}^{2}$ on $\left[0,\pi \right]$ s.t. Dirichlet boundary conditions. Find Green's function $G\left(x,s\right)$. Find the normalized in ${L}^{2}\left[0.\pi \right]$ eigen functions ${u}_{n}\left(x\right)$ of the operator; the eigenvalues are ${\lambda }_{n}={n}^{2}$.Evaluate the integral ${\int }_{0}^{\pi }{\int }_{0}^{\pi }{G}^{2}\left(x,s\right)dxds$ in two ways, a) using the explicit formula for $G\left(x,s\right)$ and b) using the representation $G\left(x,s\right)=\sum _{n\ge 1}\frac{{u}_{n}\left(x\right){u}_{n}\left(s\right)}{{\lambda }_{n}}$.After simplifications, you will find $\sum _{n\ge 1}\frac{1}{{n}^{4}}=\frac{{\pi }^{4}}{90}$.

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