I am learning Fourier Series and I cannot figure this question out,

Find the value of the infinite sum,

$1 + \frac{1}{2} - \frac{1}{4} - \frac{1}{5} + \frac{1}{7} + \frac{1}{8} - \frac{1}{10} - \frac{1}{11} + ... $

I am guessing that you have to say that the sum of the series is $f(x)$ and that $f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n cos(\frac{n\pi x}{l}) b_n sin(\frac{n\pi x}{l})$

and that $a_n = \frac{1}{l} \int_{-l}^{l} f(x) cos(\frac{n\pi x}{l})dx$

$b_n = \frac{1}{l} \int_{-l}^{l} f(x) sin(\frac{n\pi x}{l})dx$

So then I tried to say that

$\frac{1}{2} = a_1 = \frac{1}{l} \int_{-l}^{l} f(x) cos(\frac{\pi x}{l})dx * cos(\frac{\pi x}{l})$

But I am not too sure here or even sure that I am going in the right direction. Does anyone know how to proceed with this problem?

  • $\begingroup$ As the terms are designed, I would take the sum of some Fourier series in $\frac{\pi]{3}$. Let's say $\sum_{n=1}^\infty \frac{\sin nx}{n}$. Now you have to find the exact value of the sum of this series :-) $\endgroup$ – Nicolas FRANCOIS Nov 14 '16 at 2:08
  • 2
    $\begingroup$ Just as a quick note so you can check your answer, we can show the following: First note that we have the partial sums of the series as $$\lim_{m\to\infty}\sum_{k=0}^m \left(\frac{1}{1+6k}+\frac{1}{2+6k}-\frac{1}{4+6k}-\frac{1}{5+6k}\right)$$ We apply the Dominated Convergence Theorem now $$=\int_0^1\left\{ \lim_{m\to\infty}\sum_{k=0}^m\left(x^{6k}+x^{6k+1}-x^{6k+3}-x^{6k+4}\right)\right\}$$ The integral can be expanded in terms of the arctangent function $$=\int_0^1\left(\frac{1}{x^2-x+1}\right) = \frac{2\pi}{3\sqrt{3}}$$ $\endgroup$ – Brevan Ellefsen Nov 14 '16 at 2:30
  • $\begingroup$ Shouldn't the integral be $\int_0^1 \frac{1+x}{1+x^2+x^4}{\rm d}x$ ? $\endgroup$ – Nicolas FRANCOIS Nov 14 '16 at 2:39
  • $\begingroup$ @NicolasFRANCOIS were you referring to my post? If so, note that$$\sum_{k=0}^m\left(x^{6k}+x^{6k+1}-x^{6k+3}-x^{6k+4}\right) = \frac{1-x^{6 m+6}}{x^2-x+1}$$ At which point we take limits by the Dominated Convergence Theorem $\endgroup$ – Brevan Ellefsen Nov 14 '16 at 2:53
  • $\begingroup$ @BrevanEllefsen Brevan, I suggest that you post your comment as a solution. $\endgroup$ – Mark Viola Nov 14 '16 at 3:31

This is not exact science, there is a bit of experimental mathematics here.

As I pointed out in my comment, the shape of the terms suggests working with the sum $$S(x)=\sum_{n=1}^\infty \frac{\sin nx}{x}$$ and take the value at $\frac{\pi}{3}$.

Problem is : what is the function $S$ ? Doing a little research with Wolfram Alpha and Geogebra, you can infer that the function $$f:x\mapsto \begin{cases}\frac{\pi-x}{2}&\text{ if }x\in]0,\pi] \\ \frac{x+\pi}{2}&\text{ if }x\in[-\pi,0[\end{cases}$$ prolonged by $2\pi$-periodicity could do the trick.

$f$ is continuously differentiable by parts, so its Fourier series converges at every continuity point towards $f$. The coefficients are : $$b_n=\frac{2}{\pi}\int_0^\pi \frac{x+\pi}{2}\sin nx{\rm d}x = \frac{1}{n}$$ so $f(x)=S(x)$ except at points of the form $2k\pi$.

Taking the value at $\frac{\pi}{3}$, you find : $S(\pi/3)=\frac{\pi}{3}$, but also $$S(3)=\sum_{n=1}^\infty \frac{\sin n\frac\pi3}{n} = \frac{\sqrt3}{2}(1+\frac12-\frac14-\frac15+\frac17+\frac18-\dots)$$ so your sum evaluates to $\frac2{\sqrt3}\frac\pi3=\frac{2\pi}{3\sqrt3}$, as mentionned by Brevan.

  • $\begingroup$ Oh, yes, you're correct, I forgot to multiply by $\sin\frac{\pi}{3}$ !!! $\endgroup$ – Nicolas FRANCOIS Nov 14 '16 at 3:21
  • $\begingroup$ ah yes, that would do it! Minor mistake. Otherwise, good answer. Nothing beats a heuristic approach in my book :) $\endgroup$ – Brevan Ellefsen Nov 14 '16 at 3:23

Now that Nicolas has posted, and on the request of Dr. MV, here is a solution using more elementary techniques.

We start by expressing the series in any way we can with summation notation so we can start manipulating the series. One trivial way is the following $$\lim_{m\to\infty}\sum_{k=0}^m \left(\frac{1}{1+6k}+\frac{1}{2+6k}-\frac{1}{4+6k}-\frac{1}{5+6k}\right)$$ We now want to simplify the sum. One way we can do this is by expressing the inside as an integral on the domain $[0,1]$, getting the following $$= \lim_{m\to\infty}\sum_{k=0}^m\left\{\int_0^1\left(x^{6k}+x^{6k+1}-x^{6k+3}-x^{6k+4}\right)dx\right\}$$ We now interchange the summation and integral, and then interchange the limit and integral. This is justified by the Dominated Convergence Theorem. We now have the following
$$= \int_0^1\left\{\lim_{m\to\infty}\sum_{k=0}^m\left(x^{6k}+x^{6k+1}-x^{6k+3}-x^{6k+4}\right)\right\}dx$$ $$= \int_0^1\left\{\lim_{m\to\infty}\frac{1-x^{6 m+6}}{x^2-x+1}\right\}dx$$ $$= \int_0^1\frac{dx}{x^2-x+1}$$ We now evaluate this integral to get
$$\left.\frac{2}{\sqrt{3}}\arctan\left(\frac{2 x-1}{\sqrt{3}}\right)\right|_0^1 = \color{red}{\frac{2\pi}{3\sqrt{3}}}$$

  • $\begingroup$ @Dr.MV Here you go :) $\endgroup$ – Brevan Ellefsen Nov 14 '16 at 3:47

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