Question

(a) Given $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathbf{}\left(\pi -x\right)\mathbf{/}\mathbf{2}$on$\left(0,\pi \right)$ , find the sine seriesof period$\mathbf{2}\mathbf{\pi }$ for $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}$.

(b) Use your result in (a) to evaluate$\underset{}{\overset{}{\mathbf{\sum }\mathbf{1}\mathbf{/}{\mathbf{n}}^{\mathbf{2}}}}$ .

Short answer

Part a) the sine series is $f\left(x\right)=\underset{}{\overset{}{\sum \frac{\sin \left(nx\right)}{n}}}$

Part (b) the sum is $\sum _{n=1}^{\infty }=\frac{{\mathrm{\pi }}^2}{6}$

Step-by-step solution

Given information

The given function is $f\left(x\right)=\left(\mathrm{\pi }-\mathrm{x}\right)/2$ and the sum is $\underset{}{\overset{}{\sum 1/n^2}}$

Meaning of the Fourier Series and Definition of Parseval theorem.

A Fourier series is an infinite sum of sines and cosines expansion of a periodic function. The orthogonality relationships of the sine and cosine functions are used in the Fourier Series.

Parseval's theoremstates thata signal's energy can be represented as the average energy of its frequency components.

Part (a) Find the sine series

The sine series coefficients are

$$\begin{gathered} b_n=\frac{2}{\mathrm{\pi }}{\int }_0^{\mathrm{\pi }}\frac{1}{2}\left(\mathrm{\pi }-\mathrm{x}\right)\sin \left(nx\right)dx \\ ={\int }_0^{\mathrm{\pi }}\sin \left(nx\right)dx-\frac{1}{\mathrm{\pi }}{\int }_0^{\mathrm{\pi }}x\sin \left(nx\right)dx \\ =-{\frac{\cos \left(nx\right)}{n}}_0^{\mathrm{\pi }}{\left|-\frac{1}{\mathrm{\pi }}\left[-\frac{x}{n}\cos \left(nx\right)+\frac{1}{n^2}\sin \left(nx\right)\right]\right|}_0^{\mathrm{\pi }} \\ =-\frac{{\left(-1\right)}^n-1}{n}-\frac{1}{\mathrm{\pi }}\frac{-\mathrm{\pi }{\left(-1\right)}^{\mathrm{n}}}{n} \end{gathered}$$

Further solving

$b_n=\frac{1}{n}$

The function is then:

$f\left(x\right)=\sum _{n=1}^{\infty }\frac{\sin \left(nx\right)}{n}$

Part (b) Evaluate the sum∑1/n2

First find the average of the square of the function over the interval:

$$\begin{gathered} {\left|f\left(x\right)\right|}^2=\frac{1}{\mathrm{\pi }}{\int }_0^{\mathrm{\pi }}{\left(\frac{\mathrm{\pi }-\mathrm{x}}{2}\right)}^{2}dx \\ =\frac{1}{4\mathrm{\pi }}{\int }_0^{\mathrm{\pi }}\left({\mathrm{\pi }}^2-2{\mathrm{\pi x\_x}}^2\right)dx \\ =\frac{1}{4\mathrm{\pi }}{\overline{)\left({\mathrm{\pi }}^2\mathrm{x}-{\mathrm{\pi x}}^2+\frac{1}{3}{\mathrm{x}}^3\right)}}_0^{\mathrm{\pi }} \\ =\frac{{\mathrm{\pi }}^2}{12} \end{gathered}$$

By Parseval theorem:

$$\begin{gathered} \frac{{\mathrm{\pi }}^2}{12}=\frac{1}{2}\sum _{n=1}^{\infty }\frac{1}{n^2} \\ \sum _{n=1}^{\infty }\frac{1}{n^2}=\frac{{\mathrm{\pi }}^2}{6} \end{gathered}$$

Part (a) the sine series is $f\left(x\right)=\sum _{n=1}^{\infty }\frac{\sin \left(nx\right)}{n}$

Part (b) the sum is $\sum _{n=1}^{\infty }\frac{1}{n^2}=\frac{{\mathrm{\pi }}^2}{6}$