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Compute the Fourier series for $x^3$ and use it to compute the value of $\sum\limits_{n = 1}^{\infty} \frac{1}{n^4}$.

I determined the coefficients of the Fourier series, which are

$$a_0 = \dfrac{\pi^3}{2}; \qquad a_n = \dfrac{6(\pi^2 n^2 - 2)(-1)^n + 12}{\pi n^4}$$

Then, I get

$$x^3 = \dfrac{\pi^3}{4} + \sum\limits_{n = 1}^{\infty} \dfrac{6(\pi^2 n^2 - 2)(-1)^n + 12}{\pi n^4}\cos(nx)$$

If $x = \pi$, then

$$\begin{aligned} \pi^3 &= \dfrac{\pi^3}{4} + \sum\limits_{n = 1}^{\infty} \dfrac{6(\pi^2 n^2 - 2)(-1)^n + 12}{\pi n^4}\cos(n\pi)\\ \dfrac{3\pi^3}{4} &= \sum\limits_{n = 1}^{\infty} \dfrac{6(\pi^2 n^2 - 2)(-1)^n + 12}{\pi n^4}(-1)^n \end{aligned}$$

I'm stuck. It's easy to compute $\sum\limits_{n = 1}^{\infty} \frac{1}{n^2}$, using the Fourier series, but for this type of problem I'm stuck.

Any comments or suggestions? By the way, I know that

$$\sum\limits_{n = 1}^{\infty} \dfrac{1}{n^4} = \dfrac{\pi^4}{90}$$

I need to know how to get there.

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From your identity

\begin{equation*} \frac{3\pi ^{3}}{4}=\sum_{n=1}^{\infty }\frac{6(\pi ^{2}n^{2}-2)(-1)^{n}+12}{ \pi n^{4}}(-1)^{n} \end{equation*}

expanding the right hand side and using the result $\sum_{n=1}^{\infty }\frac{1}{n^{2}}=\frac{\pi^2}{6}$, we get

\begin{eqnarray*} \frac{3\pi ^{4}}{4} &=&\sum_{n=1}^{\infty }\frac{6(\pi ^{2}n^{2}-2)(-1)^{n}+12}{n^{4}}(-1)^{n} \\ &=&6\pi ^{2}\sum_{n=1}^{\infty }\frac{1}{n^{2}}-12\sum_{n=1}^{\infty }\frac{1 }{n^{4}}+12\sum_{n=1}^{\infty }\frac{(-1)^{n}}{n^{4}} \\ &=&\pi ^{4}-12\sum_{n=1}^{\infty }\frac{1}{n^{4}}-12\sum_{n=1}^{\infty } \frac{(-1)^{n-1}}{n^{4}}. \end{eqnarray*}

Now we need to express the alternating series $\sum_{n=1}^{\infty }\frac{ (-1)^{n-1}}{n^{4}}$ in terms of $\sum_{n=1}^{\infty }\frac{1}{n^{4}}$, e.g. as follows

\begin{eqnarray*} \sum_{n=1}^{\infty }\frac{\left( -1\right) ^{n-1}}{n^{4}} &=&\sum_{n=1}^{ \infty }\frac{1}{n^{4}}-2\sum_{n=1}^{\infty }\frac{1}{(2n)^{4}} =\sum_{n=1}^{\infty }\frac{1}{n^{4}}-\frac{1}{2^{3}}\sum_{n=1}^{\infty } \frac{1}{n^{4}}=\frac{7}{8}\sum_{n=1}^{\infty }\frac{1}{n^{4}}. \end{eqnarray*}

Then

\begin{eqnarray*} \frac{3\pi ^{4}}{4} &=&\pi ^{4}-12\sum_{n=1}^{\infty }\frac{1}{n^{4}}-\frac{ 21}{2}\sum_{n=1}^{\infty }\frac{1}{n^{4}} \\ &=&\pi ^{4}-\frac{45}{2}\sum_{n=1}^{\infty }\frac{1}{n^{4}}. \end{eqnarray*}

Solving for $\sum_{n=1}^{\infty }\frac{1}{n^{4}}$ we finally obtain

\begin{equation*} \sum_{n=1}^{\infty }\frac{1}{n^{4}}=\frac{2}{45}\left( \pi ^{4}-\frac{3\pi ^{4}}{4}\right) =\frac{\pi ^{4}}{90}. \end{equation*}

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  • $\begingroup$ What a pure genius. I haven't seen that type of derivation of the alternating series before. Nice one. It will be useful for my computation! :D $\endgroup$ – NasuSama Feb 23 '14 at 23:35
  • $\begingroup$ Many thanks. Glad to help. Google Dirichlet eta function or Alternating zeta function. $\endgroup$ – Américo Tavares Feb 23 '14 at 23:41
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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} \pi x\cot\pars{\pi x}&=1 + \sum_{n = 1}^{\infty}{2x^{2} \over x^{2} - n^{2}} =1 - 2x^{2}\sum_{n = 1}^{\infty}{1 \over n^{2}} - 2x^{4}\sum_{n = 1}^{\infty}{1 \over n^{4}} - 2x^{6}\sum_{n = 1}^{\infty}{1 \over n^{6}} - \cdots \end{align}

$$ \pi x^{1/2}\cot\pars{\pi x^{1/2}}=1 - 2x\sum_{n = 1}^{\infty}{1 \over n^{2}} - 2x^{2}\sum_{n = 1}^{\infty}{1 \over n^{4}} - 2x^{3}\sum_{n = 1}^{\infty}{1 \over n^{6}} - \cdots $$ For $\verts{z} \sim 0$: \begin{align} z\cot\pars{z}&={z \over \tan\pars{z}} \sim {z \over z + z^{3}/3 + 2z^{5}/15} ={1 \over 1 + z^{2}/3 + 2z^{4}/15} \\[3mm]&\sim 1 - \pars{{z^{2} \over 3} + {2z^{4} \over 15}} + \pars{{z^{2} \over 3} + {2z^{4} \over 15}}^{2} \sim 1 - {z^{2} \over 3} - {2z^{4} \over 15} + {z^{4} \over 9} =1 - {z^{2} \over 3} - {z^{4} \over 45} \end{align}

$$ \pi x^{1/2}\cot\pars{\pi x^{1/2}}\sim 1 - {\pi^{2} \over 3}\,x - {\pi^{4} \over 45}\,x^{2} $$

$$ \color{#00f}{\large\sum_{n = 1}^{\infty}{1 \over n^{4}}} = -\,\half\,\pars{-\,{\pi^{4} \over 45}} = \color{#00f}{\large{\pi^{4} \over 90}} $$

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