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The problem states:

Expand the function $f(x)=\sin{\frac{5x}{6}}$ into a Fourier series on $(-\pi, \pi)$, then find the sum $$S=\sum_{n=1}^{+\infty} \frac{(-1)^n}{36n^2-25}.$$

I was able to find the series fairly easily, and it's $$S(x)=\frac{36}{\pi}\sum_{n=1}^{+\infty} \frac{(-1)^{n+1}n}{36n^2-25}\sin{nx}.$$

It's apparent that the sum I'm supposed to find is connected to the series in some way. More precisely, the derivative of $$G(x)=\frac{36}{\pi}\sum_{n=1}^{+\infty}\frac{(-1)^n}{36n^2-25}\cos{nx}$$ is $S(x)$ (because $\frac{(-1)^{n+1}n}{36n^2-25}\sin{nx}$ uniformly converges by Dirichlet's test). And $G(0) $ is the sum that I need.

However, I'm unable to put the finishing touches on the problem, because if I try integrating $S(t)$ from $0$ to $x$, I get $$T(x)=\frac{36}{\pi}\sum_{n=1}^{+\infty}\frac{(-1)^n}{36n^2-25}(1-\cos{nx})=\frac{36}{\pi} (S-G(x)).$$

If I were to pick a different constant than $0$, I'd have a similar problem, so I can't derive the result directly.

Is there a way to bypass the issue of there not being a constant $c$ for which $\cos{nc}$ for all $n \in \mathbb{N}$?

P.S. Full disclosure: this question is from an exam, and it's already been asked (Finding the sum of Fourier series of a function $f(x)=\sin(\frac{5x}{6})$) and even though it was unanswered, I'm confident mine warrants an answer because a lot more effort was put into the formulation of the issue than the last one (no offense to the last asker), and my question is fairly abstract, as the answer to this question would give me a good idea on the methodology behind finding sums based on Fourier series of functions.

EDIT: I've just noticed that I can't deduce uniform convergence on $(-\pi, \pi)$ from Dirichlet's test the way I wanted to, as the sums $|\sum_{k=1}^{n} \sin{kx}|$ are not bounded because zero is in the interval, so by putting $x_{n}=\frac{1}{n}$, I get $|\sum_{k=1}^{n} \frac{k}{n}|$, which diverges as $n \to \infty$. So I'm unsure about whether the $S_{n}(x)$ (the sequence of partial sums) uniformly converges as well.

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    $\begingroup$ $G(x)=\frac{36}{\pi}\sum_{n=1}^{+\infty}\frac{(-1)^n}{36n^2-25}\cos{nx},x \in [0,2\pi]$ is the only one function such that $G' = f$ and $\int_0^{2\pi} G(x)dx= 0$. Let $G_0(x) = \int_0^x f(t)dt$ then $$G(x) = G_0(x) - \frac{1}{2\pi}\int_0^{2\pi} G_0(x)dx$$ $\endgroup$ – reuns May 27 '17 at 22:13
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    $\begingroup$ And $\sum_{n=1}^{+\infty}\frac{(-1)^n}{36n^2-25}\cos{nx}$ converges absolutely and hence uniformly to $G$ $\endgroup$ – reuns May 27 '17 at 22:17
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As an alternative approach, you may notice that $$\begin{eqnarray*} S = \frac{1}{10}\sum_{n\geq 1}(-1)^n\left(\frac{1}{6n-5}-\frac{1}{6n+5}\right)&=&\frac{1}{10}\sum_{n\geq 1}(-1)^n\int_{0}^{1}\left(x^{6n-6}-x^{6n+4}\right)\,dx\\&=&\frac{1}{10}\int_{0}^{1}\frac{x^{10}-1}{1+x^6}\,dx\\&=&\frac{1}{10}\left(\frac{1}{5}-\int_{0}^{1}\frac{1+x^4}{1+x^6}\,dx\right)\\&=&\frac{1}{10}\left(\frac{1}{5}-\int_{0}^{1}\frac{1+x^4}{(1+x^2)(1-x^2+x^4)}\,dx\right)\\ &=&\frac{1}{10}\left(\frac{1}{5}-\left[\frac{2}{3}\arctan(x)+\frac{1}{3}\arctan\frac{x}{x^2-1}\right]_{0}^{1}\right)\\&=&\frac{1}{10}\left(\frac{1}{5}-\frac{\pi}{3}\right)=\color{red}{\frac{1}{50}-\frac{\pi}{30}}.\end{eqnarray*}$$

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