Consider the function \(f(t):=t^2\ \ (-\pi\leq t\leq \pi)\), extended to all of \({\mathbb R}\) periodically with period \(2\pi\). Developping \(f\) into a Fourier series we get \[t^2 ={\pi^2\over3}+\sum_{k=1}^\infty {4(-1)^k\over k^2}\cos(kt)\qquad(-\pi\leq t\leq \pi).\] If we put \(t:=\pi\) here we easily find \(\zeta(2)={\pi^2\over6}\). For \(\zeta(4)\) we use Parseval's formula \[\|f\|^2=\sum_{k=-\infty}^\infty |c_k|^2\] . Here \[\|f\|^2={1\over2\pi}\int_{-\pi}^\pi t^4\>dt={\pi^4\over5}\] and the \(c_k\) are the complex Fourier coefficients of \(f\). Therefore \(c_0={\pi^2\over3}\) and \(|c_{\pm k}|^2={1\over4}a_k^2={4\over k^4}\) \(\ (k\geq1)\). Putting it all together gives \(\zeta(4)={\pi^4\over 90}\).

Source - http://math.stackexchange.com/questions/28329/nice-proofs-of-zeta4-pi4-90

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestThis was an easy proof , but nicely done :)

Log in to reply

I think he forgot to mention the source . :D

http://math.stackexchange.com/questions/28329/nice-proofs-of-zeta4-pi4-90

Log in to reply

Remember You just recently posted a solution to my question. I don't know whether you realize or not that it contained a proof that \(\zeta{(4)} = \dfrac{{\pi}^{4}}{90} \).

Log in to reply

Yes , You are right . But without showing the result of that integral using a different way we can't prove it. Can we ?

Log in to reply

I can't understand what you are trying to say here kindly clarify your statement.

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

\( cos(x) = \dfrac{e^{ix}+e^{-ix}}{2} \)

Log in to reply

Log in to reply

Log in to reply

Log in to reply