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I'm doing a homework for my analysis class, and a problem says the following:

Let $f : \mathbb{T}^1 \to \mathbb{C}$ be the function $$f(\theta) = \begin{cases} 1 & 0 < \theta < \pi \\ 0 & - \pi < \theta < 0 \end{cases}$$ Compute the Fourier transform of $f$. Finally, it says to use the fact that the Fourier transform is an isometry from $L^2 \left( \mathbb{T}^1 \right) \to \ell^2$ to show that $\sum_{k = 1}^{\infty} k^{-2} = \pi^2 / 6$.

I've computed the Fourier transforms of $f$ as $$\hat{f}(k) = \begin{cases} \frac{1}{2} & k = 0 \\ 0 & \textrm{$k \neq 0$ even} \\ - \frac{i}{\pi k} & \textrm{$k$ odd} \end{cases}$$ If I understand correctly, what it wants me to do next is observe that since $f \mapsto \hat{f}$ is a unitary operator from $L^2 \left( \mathbb{T}^1 \right)$ to $\ell^2$, we have $$\pi = \| f \|_{L^2}^2 = \sum_{k \in \mathbb{Z}} \left| \hat{f}(k) \right|^2 = \frac{1}{4} + \frac{1}{\pi^2} \sum_{\textrm{$k$ odd}} \frac{1}{k^2}$$

Now there's an example in the text that does something with a different function to get that $\sum_{\textrm{$k \geq 1$} odd} k^{-2} = \frac{\pi^2}{8}$, then does some arithmetic trickery to get that $\sum_{k = 1}^{\infty} k^{-2} = \frac{\pi^2}{6}$. But whenever I do this calculation on the $f$ shown above, I get different numbers. I've already sunk a lot of time into this. Do I have an arithmetic error somewhere that I need to iron out, or am I on the wrong track?

Thanks.

EDIT: In this text, the Fourier transform is defined as $$\hat{f}(k) = (2 \pi)^{-1} \int_{- \pi}^{\pi} f(\theta) e^{-k i \theta} \mathrm{d} \theta$$

AJY
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  • I get $\hat f(0)=\sqrt{\pi/2}$. – Angina Seng Oct 28 '19 at 06:56
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    Well, $\frac{1}{4}\cdot \frac{\pi^2}{6}=\sum_{k\geq1}\frac{1}{4}\cdot \frac{1}{k^2}=\sum_{k\geq2\text{ even}}\frac{1}{k^2}$. Therefore, $\sum_{k\geq1\text{ odd}}\frac{1}{k^2}=\frac{\pi^2}{6}-\frac{\pi^2}{24}=\frac{\pi^2}{8}$. – conditionalMethod Oct 28 '19 at 06:57
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    There are different ways to define the Fourier Transform, one of them is unitary and Parseval's identity holds in the form you used. The other one is not unitary and there is a correction coefficient $=1/2\pi$ on the right hand side. I think this might be the source of the confusion. – GReyes Oct 28 '19 at 07:18
  • @conditionalMethod I understand that, but getting to $\pi^2/8$ or $\pi^2/6$ is the problem – AJY Oct 28 '19 at 07:24
  • @GReyes As defined in the text, we have $\hat{f} (k) = (2 \pi)^{-1} \int_{-\pi}^{\pi} f(\theta) e^{-k i \theta} \mathrm{d} \theta$ – AJY Oct 28 '19 at 07:26
  • I guess you forgot the exponential.. – GReyes Oct 28 '19 at 07:28
  • @GReyes fixed it – AJY Oct 28 '19 at 07:29
  • Look at this https://www.johndcook.com/blog/fourier-theorems/ – GReyes Oct 28 '19 at 07:40
  • @GReyes I understand there are different normalizations of the Fourier transform, but I don’t feel like I’m any closer to an answer to my question – AJY Oct 28 '19 at 07:42
  • Refer also to the frequent OP – user Oct 28 '19 at 11:26
  • @AJY Not only the transform itself, but the Parseval's identity changes its form, as indicated in that link. – GReyes Oct 28 '19 at 14:28
  • @user I found that, but it didn’t have the solution the problem asks for – AJY Oct 28 '19 at 14:32
  • @GReyes Are you telling me that the stated equality is false? Because I just pulled out a second book on Fourier analysis off of my desk and it had the same formula in it, i.e. $f(\theta) = \sum_{k \in \mathbb{Z}} \left( \frac{1}{2 \pi} \int_{- \pi}^{\pi} f(\phi) e^{- i k \phi} \mathrm{d} \phi \right) e^{i k \theta}$. Given that $\left{ e^{i k \theta} \right}_{k \in \mathbb{Z}}$ is an orthonormal basis for $L^2 \left( \mathbb{T}^1 \right)$, the isometry equation follows. Now can you please address my actual question? – AJY Oct 28 '19 at 16:29
  • Ok, here is the thing: actually you are using a non-unitary version, the one with coefficient $(2\pi)^{-1}$ in front of the integral. In this version, Parseval's identity is: $|f|^2/(2\pi)=\sum_k|c_k|^2$. Then your left hand side should be $1/2$ instead of $\pi$ and your right hand side is $1/4+\frac{1}{\pi^2}\sum(1/k^2)$ which is equal to $1/2$ because $\sum(1/k^2)=\pi^2/4$ (twice as much as $\pi^2/8$) – GReyes Oct 28 '19 at 17:51
  • @GReyes Can you tell me what part of the comment I just posted was wrong? That’s the Fourier series formula I’ve found in multiple texts. – AJY Oct 28 '19 at 17:59
  • Sure. The only problem with this definition is that Parseval’s identity has an extra $1/2\pi$ in front of the square of the norm of the original. – GReyes Oct 28 '19 at 19:17
  • @GReyes Find my error in the comment I posted two hours ago, 4 back from this one – AJY Oct 28 '19 at 19:19
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    Here it is: $e^{ik\theta}$ is an orthogonal basis, but is is orthonormal with respect to the dot product $\langle v,w\rangle=\frac{1}{2\pi}\int_{-\pi}^{\pi}v(s)\bar w(s),ds$ and then the corresponding $L^2$-norm of $f$ contains that factor as well. The sum of the squares of the projections=Fourier coefficients (with respect to the above dot product) equals the norm of the function, that is what Parseval's identity is about. I honestly do not know how to better explain this. – GReyes Oct 28 '19 at 23:39
  • @GReyes I understand your point now. Thank you for your patience. – AJY Oct 29 '19 at 18:31

1 Answers1

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Thank you to @GReyes for his help. He pointed out that the correct formula for $\| \cdot \|_{L^2}$ should've been $$\| f \|_{L^2} = (2 \pi)^{-1} \int_{- \pi}^{\pi} |f(\theta)|^2 \mathrm{d} \theta .$$

With this fix, the solution to the problem is as follows:

\begin{align*} \hat{f}(k) & = (2 \pi)^{-1} \int_{- \pi}^{ \pi} f(\theta) e^{- i k \theta} \mathrm{d} \theta \\ & = (2 \pi)^{-1} \int_{0}^{\pi} e^{- i k \theta} \mathrm{d} \theta . \end{align*} If $k = 0$, then $e^{- i k \theta} = 1$ for all $\theta$, so we know that $\hat{f}(0) = \frac{1}{2}$. But if $k \neq 0$, then \begin{align*} \hat{f}(k) & = (2 \pi)^{-1} \int_{0}^{\pi} e^{- i k \theta} \mathrm{d} \theta \\ & = (2 \pi)^{-1} \left[ \frac{e^{- i k \theta}}{- i k} \right]_{\theta = 0}^{\pi} \\ & = (2 \pi)^{-1} \frac{1}{- i k} \left( e^{- i k \pi} - 1 \right) \\ & = \frac{i}{2 \pi k} \left( (-1)^{k} - 1 \right) \\ & = \begin{cases} 0 & \textrm{$k \neq 0$ even} \\ - \frac{i}{\pi k} & \textrm{$k$ odd} \end{cases} \end{align*} In summary, $$\hat{f}(k) = \begin{cases} \frac{1}{2} & k = 0 \\ 0 & \textrm{$k \neq 0$ even} \\ - \frac{i}{\pi k} & \textrm{$k$ odd} \end{cases}$$

Now we use Parseval's theorem. We can see that $\| f \|_{L^2}^2 = (2 \pi)^{-1} \int_{- \pi}^{\pi} |f(\theta)|^2 \mathrm{d} \theta = \frac{1}{2}$. Now we're going to use the fact that $\| f \|_{L^2}^2 = \sum_{k \in \mathbb{Z}} \left| \hat{f}(k) \right|^2$ to find $\sum_{k = 1}^{\infty} k^{-2}$. Note that if we can show that $\sum_{\textrm{$k \geq 1$ odd}} k^{-2} = \pi^2 / 8$, the remainder of the proof will follow from the text's argument (see (5.4.21), page 194).

By Parseval's theorem, we have that \begin{align*} \| f \|_{L^2}^2 & = \sum_{k \in \mathbb{Z}} \left| \hat{f}(k) \right|^2 \\ = \frac{1}{2} & = \frac{1}{4} + \sum_{\textrm{$k \in \mathbb{Z}$ odd}} \frac{1}{\pi^2 k^2} \\ & = \frac{1}{4} + \frac{2}{\pi^2} \sum_{\textrm{$k \geq 1$ odd}} k^{-2} \\ \Rightarrow \frac{1}{4} & = \frac{2}{\pi^2} \sum_{\textrm{$k \geq 1$ odd}} k^{-2} \\ \Rightarrow \frac{\pi^2}{8} & = \sum_{\textrm{$k \geq 1$ odd}} k^{-2} . \end{align*} Finally, following the text's lead, we observe that \begin{align*} \sum_{k = 1}^{\infty} k^{-2} & = \sum_{\textrm{$k \geq 1$ odd}} k^{-2} + \sum_{\textrm{$k \geq 1$ even}} k^{-2} \\ & = \sum_{\textrm{$k \geq 1$ odd}} k^{-2} + \sum_{\ell = 1}^{\infty} (2 \ell)^{-2} \\ & = \sum_{\textrm{$k \geq 1$ odd}} k^{-2} + \frac{1}{4} \sum_{\ell = 1}^{\infty} \ell^{-2} \\ \Rightarrow \frac{3}{4} \sum_{k = 1}^{\infty} k^{-2} & = \sum_{\textrm{$k \geq 1$ odd}} k^{-2} \\ & = \frac{\pi^2}{8} \\ \Rightarrow \sum_{k = 1}^{\infty} k^{-2} & = \frac{4}{3} \frac{\pi^2}{8} \\ & = \frac{\pi^2}{6} . \end{align*}

AJY
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