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I have seen this post Fourier series expansion of $\frac{\pi^4}{96}$ and $\frac{\pi^4}{90}$ but it seems to skip some steps that I don't understand. I also looked at some others, but I haven't found one that exactly answers what I need.

Using $f(e^{i\theta}) = |\theta|$. I found the Fourier series to be

$$|\theta| = \frac{\pi}{2} - \frac{2}{\pi} \sum_{k = -\infty}^{\infty} \frac{e^{ik\theta}}{k^2}$$

$c_k = \begin{cases} 0 & \ \mathrm{if} \ k \ \mathrm{is \ even} \\ \frac{-2}{\pi k^2} & \ \mathrm{if} \ k \ \mathrm{ is \ odd} \\ \end{cases}$

So, one could say

$$|\theta| = \frac{\pi}{2} - \frac{2}{\pi} \sum_{k = -\infty}^{\infty} \frac{e^{ik\theta}}{(2k+1)^2}$$

Now, for Parseval's identity, we need $|c_k|^2 = \frac{4}{\pi^2 (2k+1)^4}$ and $\frac{1}{2\pi} \int_{-\pi}^{\pi} |f(e^{i\theta})|^2 d\theta = \frac{\pi^2}{3}$.

So,this means $$\frac{\pi^2}{3} = \sum_{k = -\infty}^{\infty} \frac{4}{\pi^2(2k+1)^4}.$$

So,

$$\frac{\pi^4}{12} = \sum_{k = -\infty}^{\infty} \frac{1}{(2k+1)^4}.$$

From here is where I am having problems. I'm unsure of how to convert this Fourier series to its equivalent starting at $1$. I thought it was just $$\frac{\pi^4}{12} = 2\sum_{k = 1}^{\infty} \frac{1}{(2k+1)^4},$$ but this is of course not true as the result would be $$\frac{\pi^4}{24} = \sum_{k = 1}^{\infty} \frac{1}{(2k+1)^4},$$ which is not the correct answer. (Edit: This part is correct, the answer is just not coming out correctly somehow.) I'm really not sure what the issue is. I did the same process using $f(e^{i\theta}) = \begin{cases} -1 & \ -\pi < \theta < 0 \\ 1 & \ 0 < \theta < \pi \end{cases}$ to prove the series for $\frac{\pi^2}{8}$, but it does not seem to be working here. Does anyone see what's happening?

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  • $\begingroup$ Notice that the terms for $k=0$ and $k=-1$ are the same, the terms for $k=1$ and $k=-2$ are the same, and so on... $\endgroup$
    – Nicolas
    Jun 1, 2021 at 18:16
  • $\begingroup$ First thing I noticed is that there's something odd going on with $k=0$. (Technically, your infiinte sum should be $k=-\infty, -1$ and $k=1,\infty$ as $-2/\pi k^2$ is not defined at $k=0$. ) $\endgroup$
    – daruma
    Jun 1, 2021 at 18:18
  • $\begingroup$ Second thing I noticed is that your Parseval's identity doesn't seem right. You want to evaluate $\frac{1}{2\pi}\int_{-\pi}^{\pi} |f|^2 d\theta$ as this is equal to $\sum c_k^2$. $\endgroup$
    – daruma
    Jun 1, 2021 at 18:20
  • $\begingroup$ @daruma I thought that is why we found the $\frac{\pi}{2}$ term. That is the $c_0$ term found by a separate formula. The second thing you mentioned was a typo. I did indeed do it that way. The result is $\frac{\pi^2}{3}$ $\endgroup$
    – Nolan P
    Jun 1, 2021 at 18:28
  • $\begingroup$ @Nicolas I'm not sure I see the point of that. Can you elaborate? $\endgroup$
    – Nolan P
    Jun 1, 2021 at 18:28

1 Answer 1

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To summarize what happened in the comments, the problem is that you forgot to factor in the $c_0^2$ term when doing your series.

$\pi^2/3=\pi^2/4+2\sum_{k=1}^{\infty}c_k^2$

here we noted that $c_{-k}=c_k$ so we grouped the sum together to get a series from $1$ to $\infty$.

Now, $$\frac{\pi^2}{3}-\frac{\pi^2}{4}=2\sum_{n=1}^{\infty} \frac{4}{\pi^2(2n+1)^4}$$ (here $k=2n+1$ as we omit all the even terms).

So we get $$\frac{\pi^4}{96}=\sum_{n=0}^{\infty}\frac{1}{(2n+1)^4}$$

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  • $\begingroup$ I think you forgot your $\pi^2/4$ which would explain why you were out by a factor of $4$. $\endgroup$
    – daruma
    Jun 1, 2021 at 18:54
  • $\begingroup$ Have you checked the formula $c_k(f)=-\frac{2}{k^2\pi}$? I did not find it but could have flawed my computation. EDIT: It's fine actually, I've missed the factor 2! $\endgroup$
    – Nicolas
    Jun 1, 2021 at 18:57
  • $\begingroup$ Ah I see what you mean now. This totally makes sense. When I did $\frac{\pi^2}{8}$, I had $c_0 = 0$, so I didn't even know I had forgotten it! This definitely cleared it up. Thank you! $\endgroup$
    – Nolan P
    Jun 1, 2021 at 19:05
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    $\begingroup$ Dropped a $\pi^2$? $\endgroup$ Jun 1, 2021 at 19:09

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