Expand $\cos(x)$ in sinus-series in the interval $(0,\pi/2)$ and use the result to calculate $$\sum_{n=1}^{\infty}\frac{n^2}{(4n^2-1)^2}.$$

Since we only need the odd terms in the Fourier expansion, we know that $a_n=0$, so we only need $b_n$ in order to find the sine expansion.

In my book, there is a chapter on "Fourier series on intervals", and there they state that:

If $f$ is a piecewise smooth function on $[0,l]$, then the Fourier sine expansion is given by

$$f(x)=\sum_{n=1}^{\infty}b_n\sin\left(\frac{\pi n x}{l}\right),\tag{1}$$


$$b_n=\frac{2}{l}\int\limits_{0}^{l}f(x)\sin\left(\frac{\pi n x}{l}\right).\tag{2}$$

So we should be able to use only $(1)$ and $(2)$ with $l=\pi/2$ to solve this problem. So

\begin{align} b_n=\frac{4}{\pi}\int\limits_{0}^{\pi/2}\cos(x)\sin(2nx)dx = \frac{4(\sin(\pi n)-2n)}{\pi(1-4n^2)}\tag{3}. \end{align}

This gives

$$\cos(x)=\frac{8}{\pi}\sum_{n=1}^{\infty}\frac{n}{4n^2-1}\sin(2nx), \quad \forall \ n\in\mathbb{N}\tag{4}$$

2 questions remain:

  1. I used wolfram alpha to calculate the integral in $(3)$. What methods do I use to do this? Repeated integration by parts and solve for the integral?
  2. How do I calculate the desired sum?
  • $\begingroup$ The function to expand is probably not the regular $\cos$. Instead, consider the function that coincides with $\cos$ on $(0, \frac \pi 2)$, that is odd and is $\pi$ periodic. $\endgroup$ – Stefan Lafon Jan 25 at 16:30
  • $\begingroup$ But they explicitly state that I'm to expand $\cos(x)$. So I am pretty convinced that $\cos(x)$ is the regular cos that is to be expanded. I might have misunderstood your comment. $\endgroup$ – Parseval Jan 25 at 17:30

As posted by the OP, two steps remain:

  1. Perform the integral:

$$ b_n=\frac{4}{\pi}\int\limits_{0}^{\pi/2}\cos(x)\sin(2nx)dx \\ = \frac{1}{i \pi}\int\limits_{0}^{\pi/2}e^{ix(1+2n)}- e^{ix(1-2n)}+e^{-ix(1-2n)}-e^{-ix(1+2n)}dx \\ = \frac{1}{i \pi} \left(\frac{e^{ix(1+2n)}}{i(1+2n)}|_{0}^{\pi/2} - \frac{e^{ix(1-2n)}}{i(1-2n)}|_{0}^{\pi/2} + \frac{e^{-ix(1-2n)}}{-i(1-2n)}|_{0}^{\pi/2} -\frac{e^{-ix(1+2n)}}{-i(1+2n)}|_{0}^{\pi/2} \right)\\ = \frac{1}{i \pi} \left(\frac{i(-1)^n - 1}{i(1+2n)} - \frac{i(-1)^n - 1}{i(1-2n)} + \frac{-i(-1)^n - 1}{-i(1-2n)} - \frac{-i(-1)^n - 1}{-i(1+2n)} \right)\\ = \frac{1}{- \pi (1 - 4 n^2)} \left( -4n(i(-1)^n - 1) - 4n (-i(-1)^n - 1) \right)\\ = \frac{-8n}{\pi(1-4n^2)} \\ = \frac{4(\sin(\pi n)-2n)}{\pi(1-4n^2)} $$ where the last step only indicates the "desired" solution by WolframAlpha, where anyway $\sin(\pi n) = 0$ .

2 Calculating the desired sum.

From OP's last result, $$ \cos(x)=\frac{8}{\pi}\sum_{n=1}^{\infty}\frac{n}{4n^2-1}\sin(2nx), $$ multiply this formula with $\cos(x) $ and integrate, using the integral which was just derived:

$$ \frac{\pi}{4} = \int\limits_{0}^{\pi/2}\cos^2(x) dx = \frac{8}{\pi} \sum_{n=1}^\infty \frac{n}{4n^2 -1} \int\limits_{0}^{\pi/2}\cos(x)\cdot\sin(2nx) dx = \\ \frac{8}{\pi} \sum_{n=1}^\infty \frac{n}{4n^2 -1} \cdot \frac{2 n}{4n^2 -1}\\ = \frac{16}{\pi} \sum_{n=1}^\infty \frac{n^2}{(4n^2 -1)^2} $$ so finally

$$ \sum_{n=1}^\infty \frac{n^2}{(4n^2 -1)^2} = \frac{\pi^2}{64} $$

  • $\begingroup$ Wonderful, thanks a lot! $\endgroup$ – Parseval Jan 28 at 12:46

Consider the function $f(x) = \cos(x) \cdot {\rm{sign}}(x)$ in the interval $(-\pi, \pi)$. This function is identical to $\cos(x)$ in the interval $(0, \pi/2)$ so the expansion will be correct in that interval.

You want $f(x) = \sum_{m=1}^\infty b_m \sin(mx)$.

Integrating with $\sin(nx)$ gives

$$ \int\limits_{-\pi}^{\pi}\sin( n x)\sin(nx)dx = \pi $$ $$ \int\limits_{-\pi}^{\pi}\sin( m x)\sin(nx)dx = 0 \quad (m\ne n) $$ and

$$ \int\limits_{-\pi}^{\pi}\cos(x)\cdot {\rm{sign}}(x) \cdot \sin(nx)dx = 2 \int\limits_{0}^{\pi}\cos(x)\cdot \sin(nx)dx = 2 \frac{n (1+\cos(\pi n))}{n^2 - 1} \tag{*} $$

So you have $$ \cos(x) = \frac{2}{\pi} \sum_{n=1}^\infty \frac{n (1+\cos(\pi n))}{n^2 - 1} \sin(nx) $$ where the coefficients are nonzero for even n.

Substituting $n$ with $2 m$ gives $$ \cos(x) = \frac{8}{\pi} \sum_{m=1}^\infty \frac{m}{4m^2 -1} \sin(2 m x) $$

No use the integral in (*) again: multiply the last formula with $\cos(x) $ and integrate:

$$ \frac{\pi}{2} = \int\limits_{0}^{\pi}\cos^2(x) dx = \frac{8}{\pi} \sum_{m=1}^\infty \frac{m}{4m^2 -1} \int\limits_{0}^{\pi}\cos(x)\cdot\sin(2mx) dx = \frac{32}{\pi} \sum_{m=1}^\infty \frac{m^2}{(4m^2 -1)^2} $$ so finally

$$ \sum_{m=1}^\infty \frac{m^2}{(4m^2 -1)^2} = \frac{\pi^2}{64} $$

  • $\begingroup$ Thanks for a nice solution and I'll give you an upvote. However this answer fails to provide an answer that aids in helping me to understand my mistake. I'd rather learn where I went wrong in my own attempt then see a customized soluton that would only work for this particular problem. For example, I've never had to deal with the $sgn(x)$ function and there is not a single example in the book where they use a method like this to solve a similar problem. This problem is solveable using standard given formulae and I want to know why it did not work this time. $\endgroup$ – Parseval Jan 25 at 19:29
  • $\begingroup$ If you want it without that symmetry condition, you would have to set $\cos(x) = \sum_{m=1}^\infty b_m \sin(2mx)$ which we know (in hindsight) from the solution above. It is not clear to me a priori that the odd terms $\sin((2m+1)x)$ should not be present in the series at all - do you have an argument? Then, you can integrate $\int_0^{\pi/2} \sin(2mx) \sin(2nx) dx = 0$ for $m \ne n$ and everything works as above. (continued) $\endgroup$ – Andreas Jan 25 at 19:59
  • $\begingroup$ (continued) The point is that in general, $\int_0^{\pi/2} \sin(mx) \sin(nx) dx $ is not zero and therefore the calculation of the coefficients does not work straightforward if you cannot exclude the odd values of m right from the start. $\endgroup$ – Andreas Jan 25 at 19:59
  • $\begingroup$ I've been working on this problem a bit further now, I'll edit the answer and show you what I mean by applying the formulas straight on. Stand by. $\endgroup$ – Parseval Jan 25 at 20:52
  • $\begingroup$ Please see my edit now Andreas. Thanks for the time and effort you put in, I apreciate it. Oh and the answer $\pi^2/64$ is indeed correct. $\endgroup$ – Parseval Jan 25 at 21:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.