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Attempt: I computed the complex Fourier Series of $f(x) = e^{x + 2πxi}$

$f(x) = ∑_n(e^{2in}−e^2)e^{inx−in−1}/(2π)(in−2iπ−1)$ where $n$ is any integer.

How do I use this to calculate $∑_{n>0}1/(1+π^2n^2)$? I tried equating f(1), f(-1), f(0) but they didn't work.

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    $\begingroup$ Is it $\sum_{k=1}^{\color{red}\infty}\frac{(-1)^\color{red}{n}}{1+k^2 \pi^2}$ ? $\endgroup$ – Claude Leibovici Jul 23 '17 at 4:12
  • $\begingroup$ A bounty for a question already solved multiple times on the site (and in every textbook on the subject)? Hmmm... $\endgroup$ – Did Jul 25 '17 at 16:53
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    $\begingroup$ @Did Don't worry, he accepted his own answer. $\endgroup$ – adfriedman Jul 25 '17 at 19:38
  • $\begingroup$ @adfriedman ...Which does not even solve the question (as you noted). Strange character, I must admit. $\endgroup$ – Did Jul 25 '17 at 19:44
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Using $f(x)=\cosh(x/\pi)$, the Fourier coefficients are given by

\begin{align} a_n &= \frac{1}{\pi}\int_{-\pi}^{\pi} \cosh(x/\pi) \cos(n x)\,dx = 2 \sinh(1) \frac{(-1)^n}{1+\pi^2n^2}\\ b_n &= \frac{1}{\pi}\int_{-\pi}^{\pi} \cosh(x/\pi) \sin(n x)\,dx = 0 \end{align}

So on $[-\pi,\pi]$, \begin{align} f(x) = \cosh(x/\pi)&= \frac{1}{2}a_0 + \sum_{n=1}^{\infty} a_n \cos(n x) + \sum_{n=1}^{\infty} b_n \sin(n x) \\ &= \sinh(1) + 2\sinh(1)\sum_{n=1}^{\infty} \frac{(-1)^n}{1+\pi^2n^2}\cos(nx) \end{align} at $x=0$ and with some rearranging we have

\begin{align} \sum_{n=1}^{\infty} \frac{(-1)^n}{1+\pi^2n^2} &= \frac{\cosh(0) - \sinh(1)}{2\sinh(1)} = \frac{1 - \sinh(1)}{2\sinh(1)}\\ &= \frac{1}{2}\big( \mathrm{csch}(1) - 1\big) \end{align} giving $$\boxed{\sum_{n=1}^{\infty} \frac{(-1)^n}{1+\pi^2n^2} = \frac{1}{2}\left( \frac{1+2e-e^2}{e^2-1}\right)}$$

Motivation As to why I chose $\cosh(x/\pi)$, I recalled that the Fourier coefficients for $e^x$ were something like ''$\frac{\cos(n\pi)}{1+n^2}$'', so it made sense to try and compute coefficients after some kind of scaling. $e^{x/\pi}$ turned out to work quite well for the cosine terms, but not for the sine terms. The same turned out to be true of $e^{-x/\pi}$, so the hope was that some linear combination of the two would cancel out all of the sine terms. I took the two most obvious linear combinations, $\sinh(x/\pi)$ and $\cosh(x/\pi)$, where the latter turned out to work.

Update To address the change of question description, we will solve the more general form first. We just established that for $x\in [-\pi,\pi]$,

$$\cosh(x/\pi) = \sinh(1) + 2\sinh(1) \sum_{n=1}^{\infty} \frac{(-1)^n}{1+\pi^2n^2}\cos(nx)$$ which may be arranged to \begin{align} \sum_{n=1}^{\infty} \frac{(-1)^n}{1+\pi^2n^2}\cos(nx) &= \frac{1}{2}\left(\frac{\cosh(x/\pi)}{\sinh(1)} - 1\right) \end{align}

Of course $\cos(n\pi)=(-1)^n$, so evaluating at $x=\pi$ gives $$\sum_{n=1}^{\infty} \frac{1}{1+\pi^2n^2} = \sum_{n=1}^{\infty} \frac{(-1)^n\cos(n\pi)}{1+\pi^2n^2} = \frac{1}{2}\left(\frac{\cosh(\pi/\pi)}{\sinh(1)} - 1\right) = \frac{1}{2}\big(\coth(1) - 1\big)$$ after expanding the definition of $\coth(1)$ and a bit of algebra you get $$\boxed{\sum_{n=1}^{\infty} \frac{1}{1+\pi^2n^2} = \frac{1}{e^2-1}}$$

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  • $\begingroup$ The OP has changed the question a bit: the $(-1)^n$ has been replaced by $1$. Why not modify your answer accordingly and get the bounty? $\endgroup$ – Alex M. Jul 29 '17 at 9:46
  • $\begingroup$ @AlexM. I guess he did. I have updated accordingly $\endgroup$ – adfriedman Jul 30 '17 at 10:28
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    $\begingroup$ @Jeffrey, I strongly suggest that you unaccept your own answer (and delete it, to recover some reputation points), and accept this one instead. $\endgroup$ – Alex M. Jul 30 '17 at 11:41
  • $\begingroup$ Good answer although it was chaotic question. (+1) $\endgroup$ – user90369 Aug 1 '17 at 9:39

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