This a question from Fourier Series:

Show that for $0<x<\pi$

First of all, the interval given is an open interval i.e. $(0,\pi)$, but I have read that Fourier Series are only applicable for closed intervals. Then how can I solve this question using Fourier Series ?
Finding out the Fourier coefficients considering the interval $[-\pi,\pi]$, I have calculated the following:
$a_{k}=\frac{4(-1)^{k+1}}{k^2},\forall k=1,2,3....$
$b_k=\frac{2(-1)^{k+1}}{k},\forall k=1,2,3,....$

But I think all this not useful in this question as the interval given is different. Also, even if I consider the Fourier series in the interval $[0,\pi]$, then also I will not be able to write the equality ($=$) sign in the equality to be shown in the question because equality means the series converges to the function $x(\pi-x)$ and for convergence of the Fourier series, the initial assumption is that the given function is a periodic function of periodicity $2\pi$. But here the function $x(\pi-x)$ is defined over a period of $\pi$.

Can anyone help me out here ? I will be highly grateful.,

  • $\begingroup$ Substituting $x=0,\pi$, the given expression is true. So you may change the interval into [$0,\pi$]. $\endgroup$ – cosmo5 Sep 20 at 15:44

Hint: You can use :Fourier expansion of $f(x)=|x|$ which is:

$$|x|=\frac{\pi}2-\frac{4}{\pi}\big(\frac{ cos x}{1^2}+\frac{ cos 3x}{3^2}+ . . . \frac{ cos (2n-1)x}{(2n+1)^2}+ . . .\big)$$; $(-\pi≤x≤\pi)$

$f(x)=|x|=\frac{\pi^2}8$ for $x=±\pi$ or $x=0$

And that of function $f(x)=x^2$ which is:

$$x^2=\frac{\pi^2}{3}-4\big[\frac{cos x}{1^2}-\frac{cos 2x}{2^2}+\frac{cos 3x}{3^2}-\frac{cos 4x}{4^2} . . .\big]$$; $(-\pi, \pi)$

$f(x^2)=\frac{\pi^2}6$ for $x=\pi$

| cite | improve this answer | |

This is a standard application of Fourier series. See https://en.wikipedia.org/wiki/Fourier_series for the basics.

To compute the Fourier series of a given function, you first need a periodic function. For the Fourier series to involve only cosine terms, you need the function to also be even. Note that the period need not be $2\pi$.

Let $f$ be a $\pi$-periodic function, defined on $[0,\pi]$ by $f(x)=x(\pi-x)$. $f$ is even, because the function $x\to x(\pi-x)$ is symmetric with respect to $x=\pi/2$.

Here is a plot of $f$, showing $6$ periods:

enter image description here

Then the cosine Fourier coefficients are, for $n>0$:

$$a_n=\frac{2}{\pi}\int_0^{\pi} f(x)\cos(2nx)\,\mathrm dx=\frac{2}{\pi}\int_0^{\pi}x(\pi-x)\cos(2nx)\,\mathrm dx$$

Now, two integrations by parts:

$$a_n=\frac{2}{\pi}\left[x(\pi-x)\frac{\sin (2nx)}{2n}\right]_0^\pi-\frac{2}{\pi}\int_0^{\pi}(\pi-2x)\frac{\sin(2nx)}{2n}\,\mathrm dx\\=-\frac{2}{\pi}\int_0^{\pi}(\pi-2x)\frac{\sin(2nx)}{2n}\,\mathrm dx\\=\frac{2}{\pi}\left[(\pi-2x)\frac{\cos (2nx)}{4n^2}\right]_0^\pi+\frac{2}{\pi}\int_0^{\pi}2\frac{\cos(2nx)}{3n^2}\,\mathrm dx\\=\frac{2}{\pi}\left[(\pi-2x)\frac{\cos (2nx)}{4n^2}\right]_0^\pi\\=-\frac{1}{n^2}$$

The sine coefficients are $b_n=0$ since the function $f$ is even.

Last, the constant coefficient:

$$a_0=\frac{2}{\pi}\int_0^\pi f(x)\,\mathrm dx=\frac{2}{\pi}\int_0^\pi x(\pi-x)\,\mathrm dx\\=\frac{2}{\pi}\left[\pi\frac{x^2}2-\frac{x^3}{3}\right]_0^\pi=\frac{2}{\pi}\cdot\frac{\pi^3}{6}=\frac{\pi^2}{3}$$

Now, the function $f$ is continuous and piecewise $C^1$, hence the series converges everywhere to the function, hence, for all $x\in[0,\pi]$,

$$x(\pi-x)=\frac{a_0}{2}+\sum_{n=1}^\infty a_n\cos(2n x)=\frac{\pi^2}{6}-\sum_{n=1}^\infty \frac{\cos(2n x)}{n^2}$$

Note that for $x=0$, you get the classic series:


| cite | improve this answer | |
  • $\begingroup$ And what about the interval being open interval? $\endgroup$ – Esha Sep 20 at 16:28
  • $\begingroup$ How is the function $x(\pi-x)$ even ? $f(-x)=-x(\pi+x)$ which is not equal to $x(\pi-x)$ . $\endgroup$ – Esha Sep 20 at 16:33
  • $\begingroup$ @Esha $f$ is a $\pi$-periodic function. It's only equal to $x(\pi-x)$ on $[0,\pi]$ (you take this part as the period, and you complete the function by "copying" on each interval $[k\pi,(k+1)\pi]$). I proved a bit more than what was asked, namely the equality holds on the closed interval: it's not a problem. $\endgroup$ – Jean-Claude Arbaut Sep 20 at 16:35
  • $\begingroup$ So the open interval $(0,\pi))$ is not a problem ? Fourier series for open and closed interval is the same ? And Fourier series can be found for open intervals too ? $\endgroup$ – Esha Sep 20 at 16:43
  • $\begingroup$ Also, $f$ is not even . What about that ? $\endgroup$ – Esha Sep 20 at 16:44

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.