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I know this question has been widely answered here, but without using Fourier analysis. Also there is a video referring to this trick but I want to use a different Fourier series.

First of Parseval's Theorem states: $\displaystyle{\dfrac{1}{\pi}\int_{-\pi}^{\pi}[f(x)]^2 = [a_0]^2+\sum_{n=1}^{\infty}[a_n]^2+[b_n]^2}$.

I calculate $a_0$ like $\frac{1}{2\,\pi}\,\int_{-\pi}^{\pi}f(x)\,\mathrm{dx}$ instead of $\frac{1}{\pi}\,\int_{-\pi}^{\pi}f(x)\,\mathrm{dx}$, so no need for $\frac{1}{2}\,a_0$

Here I'd like to involve the series of $\cosh(x)$ whose partition I have been asked for in a prior question.

It should be: $\displaystyle{\cosh(x) = \underbrace{\dfrac{\sinh(\pi)}{\pi}}_{a_0}+\sum_{n = 1}^{\infty}\underbrace{\color{red}{2}\,\dfrac{1}{\pi}\,\dfrac{\sinh(\pi)}{1+n^2}\,(-1)^n\,\cos(n\,x)}_{a_n}}$

Plugging those terms into the original theorem:

$$\begin{align} &\dfrac{1}{\pi}\int_{-\pi}^{\pi}\cosh^2(\pi) = \left(\dfrac{\sinh(\pi)}{\pi}\right)^2+\sum_{n=1}^{\infty}\left(\color{red}{2}\,\dfrac{1}{\pi}\,\dfrac{\sinh(\pi)}{1+n^2}\,(-1)^n\right)^2 \\\\ & 1+\dfrac{\sinh(2\,\pi)}{2\,\pi}= \dfrac{\sinh^2(\pi)}{\pi^2}\,\left[1+\sum_{n=1}^{\infty}\color{red}{4}\,\left(\dfrac{1}{1+n^2}\right)^2\right]\\\\ &\sqrt{\left(\dfrac{\pi^2}{\sinh^2(\pi)}+\dfrac{\sinh(2\,\pi)\,\pi}{2\,\sinh^2(\pi)}-1\right)\,\color{red}{\dfrac{1}{4}}} = \sum_{n=1}^{\infty}\dfrac{1}{1+n^2} \quad ? \end{align}$$

Actually the value is coming close to the approximated sum, but it's not exactly the same result...

Edit

I fixed the Fourier Series highlighting the missing term in red. On the other hand I eradicated some factors, not sure about making it worse. It's still differing from the approximation.

approximation $\approx 1,0767$


Therefore it works exactly like suggested by Stefan Lafon in the remarks: setting $x=\pi$

$\begin{align} &\cosh(\pi) = \dfrac{\sinh(\pi)}{\pi}+\displaystyle{\sum_{n=1}^{\infty}2\,\dfrac{1}{\pi}\,\dfrac{\sinh(\pi)}{1+n^2}}\\\\ &\Rightarrow \quad \dfrac{\cosh(\pi)\,\pi}{2\,\sinh(\pi)}-\dfrac{1}{2} =\displaystyle{\sum_{n=1}^{\infty}\dfrac{1}{1+n^2}} \end{align}$

It just remains a mystery why the method on top fails

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    $\begingroup$ I don't think you should be using Parseval. Rather, use the Fourier series expansion that you have for $\cosh(x)$, and evaluate at $x=\pi$. $\endgroup$ Jul 25, 2021 at 21:32

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Note that when you did Parseval, you had (ignoring other pieces in the expression)

$$\sum_{n=1}^{\infty} \frac{1}{(1+n^2)^{\color{red}2}}.$$

Note the exponent in red. This is not $\displaystyle\sum_{n=1}^{\infty} \frac{1}{1+n^2}$ and is the source of the issues you are having.

Per Stefan Lafon (for posterity and the purposes of having an answer to this question): simply evaluate your Fourier series at $x=\pi$ to get the correct expression which is

$$ \frac{\pi}{2} \operatorname{coth}(\pi) - \frac{1}{2}.$$

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  • $\begingroup$ I thought taking the square root would have fixed the issue but apparently it's changing the sum. Anyway the other method has worked. $\endgroup$
    – Leon
    Jul 27, 2021 at 19:36
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    $\begingroup$ Recall that $(a^2 + b^2)^{1/2} \neq a + b$ in general. This is just an extension of that fact. $\endgroup$ Jul 27, 2021 at 19:43

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