Fourier series

[Papapetros Vaggelis]

Let \(\displaystyle{\delta\in\left(0,\pi\right)}\) and \(\displaystyle{f:\mathbb{R}\longrightarrow \mathbb{R}}\) be a \(\displaystyle{2\,\pi}\) -



periodical function given by

\(\displaystyle{f(x)=\begin{cases}1\,\,\,\,\,,-\delta\leq x\leq \delta\\
0\,\,\,\,\,,\,\,\delta<\left|x\right|\leq \pi \end{cases}}\)





Calculate the \(\displaystyle{\rm{Fourier}}\) series of \(\displaystyle{f}\) and prove that

1. \(\displaystyle{\sum_{n=1}^{\infty}\dfrac{\sin\,(n\,\delta)}{n}=\dfrac{\pi-\delta}{2}}\)



2. \(\displaystyle{\sum_{n=1}^{\infty}\dfrac{\sin^2\,(n\,\delta)}{n^2\,\delta}=\dfrac{\pi-\delta}{2}}\)



3. \(\displaystyle{\sum_{n=1}^{\infty}\dfrac{1}{\left(2\,n-1\right)^2}=\dfrac{\pi^2}{8}}\)

[Tolaso J Kos]

Hello Vaggelis. Here is a solution to this nice exercise.

First of all we expand \( f \) in Fourier series. The function is a piecewise discontinuous function yet it can be expanded into Fourier Series. Let us compute the coefficients:

\( \require{cancel} {\color{gray} \blacksquare} \;\; \displaystyle a_0= \frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\, {\rm d}t= \frac{1}{\pi}\left [ \cancelto{0}{\int_{-\pi}^{-\delta}0 \, {\rm d}t}+ \int_{-\delta}^{\delta}\, {\rm d}t+ \cancelto{0}{\int_{\delta}^{\pi}0 \, {\rm d}t} \right ]=\frac{2\delta}{\pi} \)

\( \require{cancel} {\color{gray} \blacksquare} \;\; \displaystyle a_n= \frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\cos nt \, {\rm dt}= \frac{1}{\pi} \left [ \cancelto{0}{\int_{-\pi}^{-\delta}0 \, {\rm d}t} + \int_{-\delta}^{\delta}\cos nt \, {\rm d}t + \cancelto{0}{\int_{\delta}^{\pi} 0 \, {\rm d}t} \right ]=\frac{2\sin \delta n}{\pi n} \)

\( {\color{gray} \blacksquare} \;\; \displaystyle b_n = \frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\sin nt \, {\rm d}t = 0 \)

Hence the Fourier series of \( f \) is :

$$f(x)= \frac{\delta}{\pi}+ \sum_{n=1}^{\infty}\frac{2\sin \delta n}{\pi n}\; \cos nx$$

a) For \( x =0 \) the above formula gives us:

$$1= \frac{\delta}{\pi}+ \sum_{n=1}^{\infty}\frac{2\sin n\delta}{\pi n}\Rightarrow \sum_{n=1}^{\infty}\frac{\sin n \delta}{n} = \frac{\pi-\delta}{2}$$

b) By making use of Parseval's identity we have that:

$$4\sum_{n=1}^{\infty}\frac{\sin^2 n \delta}{\pi^2 n^2}+ \frac{2\delta^2}{\pi^2}= \frac{2\delta}{\pi}\Rightarrow \sum_{n=1}^{\infty}\frac{\sin^2 n \delta}{n^2 \delta}= \frac{\pi-\delta}{2}$$

c) Taking advantage of question b) and subbing \( \delta=\frac{\pi}{2} \) we get the desired result since \( \sin^2 \frac{\pi n}{2} \) is \( 1 \) at odd and \( 0 \) at even.

Some comments:
The series \( \displaystyle \sum_{n=1}^{\infty}\frac{1}{\left ( 2n-1 \right )^2}= \sum_{n=0}^{\infty}\frac{1}{\left ( 2n+1 \right )^2} \) is a special case of the \( \lambda \) function, that is \( \lambda(2) \). The \( \lambda \) function is defined as:

$$\lambda(s)= \sum_{n=0}^{\infty}\frac{1}{\left ( 2n+1 \right )^s}, \; \mathfrak{Re}(s)>1$$

and it has a closed form given by

$$\zeta(s)= \sum_{n=1}^{\infty}\frac{1}{(2n)^s}+ \sum_{n=0}^{\infty}\frac{1}{(2n+1)^s} \Leftrightarrow \zeta(s)= \frac{1}{2^s}\zeta(s) + \sum_{n=0}^{\infty}\frac{1}{(2n+1)^s} \Leftrightarrow$$
$$\Leftrightarrow \sum_{n=0}^{\infty}\frac{1}{(2n+1)^s} =(1-2^{-s})\zeta(s) \tag{1}$$

And one more thing (may be off topic but I think it's worth it). Using the closed formula we derived for the \( \lambda \) function one can find the functional equation of \( \eta \) Dirichlet, that is of the function \( \displaystyle \eta(s)= \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^s} , \; \mathfrak{Re}(s)>1 \). Indeed, since the sum converges absolutely that means we can re-arrange its terms by splitting it up into even and odd terms.

$$\begin{aligned}
\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^s} &=\left ( 1+\frac{1}{3^s}+ \frac{1}{5^s}+\cdots \right )- \left ( \frac{1}{2^s}+ \frac{1}{4^s}+ \frac{1}{6^s}+\cdots \right ) \\
&= \sum_{n=0}^{\infty}\frac{1}{(2n+1)^s}- \sum_{n=1}^{\infty}\frac{1}{(2n)^s}\\
&=\sum_{n=0}^{\infty} \frac{1}{(2n+1)^s}- \frac{1}{2^s} \sum_{n=1}^{\infty}\frac{1}{n^s}\\
&\overset{(1)}{=}\left ( 1-2^{-s} \right )\zeta(s)- \frac{1}{2^s}\zeta(s)\\
&=\left ( 1-2^{1-s} \right )\zeta(s)
\end{aligned}$$