# Show that for $0<x<\pi$ ; $x(\pi-x)=\frac{\pi^2}{6}-\big(\frac{\cos2x}{1^2}+\frac{\cos4x}{2^2}+\frac{\cos6x}{3^2}+…\big)$

This a question from Fourier Series:

Show that for $$0
$$x(\pi-x)=\frac{\pi^2}{6}-\big(\frac{\cos2x}{1^2}+\frac{\cos4x}{2^2}+\frac{\cos6x}{3^2}+.....\big)$$

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:
$$a_{0}=-\frac{\pi^2}{3}$$
$$a_{k}=\frac{4(-1)^{k+1}}{k^2},\forall k=1,2,3....$$
$$b_k=\frac{2(-1)^{k+1}}{k},\forall k=1,2,3,....$$

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.,

• Substituting $x=0,\pi$, the given expression is true. So you may change the interval into [$0,\pi$]. – 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 $$x=\pi$$

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: 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:

$$\sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}6$$

• And what about the interval being open interval? – Esha Sep 20 at 16:28
• How is the function $x(\pi-x)$ even ? $f(-x)=-x(\pi+x)$ which is not equal to $x(\pi-x)$ . – Esha Sep 20 at 16:33
• @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. – Jean-Claude Arbaut Sep 20 at 16:35
• 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 ? – Esha Sep 20 at 16:43
• Also, $f$ is not even . What about that ? – Esha Sep 20 at 16:44