91Ó°ÊÓ

A Fourier series is defined as an infinite series used in the analysis of periodic functions, in which the terms are constants multiplied by sine or cosine functions of integer multiples of the variable.

Given function in the problem 9

$f\left(x\right)=\left\{\begin{array}{c}x,\\ -2,\end{array}\begin{array}{cc}0& <x<1\\ 1& <x<2\end{array}\right\}$

Here the sine series coefficient

$$\begin{gathered} a_0=0 \\ {a}_n=2\frac{2}{4}\left[{∫}_0^{1}x\sin \frac{n\mathrm{Ï€³\ae }}{2}dx-2{∫}_1^2\sin \frac{n\mathrm{Ï€}x}{2}dx\right] \\ ={\overline{)\left[-\frac{2x}{n\mathrm{Ï€}}\cos \frac{n\mathrm{Ï€}x}{2}+{\left(\frac{2}{n\mathrm{Ï€}}\right)}^2\sin \frac{n\mathrm{Ï€³\ae }}{2}\right]}}_0^1{\overline{)+\frac{4}{n\mathrm{Ï€}}\cos \frac{n\mathrm{Ï€³\ae }}{2}}}_1^2 \\ =-\frac{6}{n\mathrm{Ï€}}\cos \frac{n\mathrm{Ï€}}{2}+{\left(\frac{2}{n\mathrm{Ï€}}\right)}^2\sin \frac{n\mathrm{Ï€}}{2}+\frac{4}{n\mathrm{Ï€}}{\left(-1\right)}^n \end{gathered}$$

Then the series will be

$f_s\left(x\right)=\underset{n=1}{\overset{∞}{∑}}\left[{\left(\frac{2}{n\mathrm{Ï€}}\right)}^2\sin \frac{n\mathrm{Ï€}}{2}-\frac{6}{n\mathrm{Ï€}}\cos \frac{n\mathrm{Ï€}}{2}+\frac{4}{n\mathrm{Ï€}}{\left(-1\right)}^n\right]\sin \frac{n\mathrm{Ï€³\ae }}{2}$

The Cosine series coefficient:

$$\begin{gathered} a_0=2\frac{2}{4}\left[{∫}_0^{1}xdx-2{∫}_1^{2}dx\right]=\frac{1}{2}{x^2}_0^1\overline{)-2x{\overline{)}}_1^2} \\ =-\frac{3}{2} \\ {a}_n=2\frac{2}{4}\left[{∫}_0^{1}x\cos \frac{n\mathrm{Ï€³\ae }}{2}dx-2{∫}_1^2\cos \frac{n\mathrm{Ï€³\ae }}{2}dx\right] \\ =\left[\frac{2x}{n\mathrm{Ï€}}\sin \frac{n\mathrm{Ï€³\ae }}{2}+{\left(\frac{2}{n\mathrm{Ï€}}\right)}^2\cos \frac{n\mathrm{Ï€³\ae }}{2}\right]{\left|{}_0^1-\frac{4}{n\mathrm{Ï€}}\sin \frac{n\mathrm{Ï€³\ae }}{2}\right|}_1^2 \end{gathered}$$

$a_n=\frac{6}{n\mathrm{Ï€}}\sin \frac{n\mathrm{Ï€}}{2}+{\left(\frac{2}{n\mathrm{Ï€}}\right)}^2\left(\cos \frac{n\mathrm{Ï€}}{2}-1\right)a$

Then the Fourier series will be

$f\left(x\right)=\frac{3}{4}+\underset{n-1}{\overset{∞}{∑}}\left[\frac{6}{n\mathrm{Ï€}}\sin \frac{n\mathrm{Ï€}}{2}+{\left(\frac{2}{n\mathrm{Ï€}}\right)}^2\left(\cos \frac{n\mathrm{Ï€}}{2}-1\right)\right]\cos \frac{n\mathrm{Ï€³\ae }}{2}$

The exponential series coefficient

$$\begin{gathered} c_0=\frac{1}{2}\left[{∫}_0^{1}xdx-2{∫}_1^{2}dx\right] \\ =-\frac{3}{4} \\ {c}_n=\frac{1}{2}\left[{∫}_0^{1}x{e}^{-inx\mathrm{Ï€}}dx-2{∫}_1^2{e}^{-inx\mathrm{Ï€}}dx\right] \\ =\frac{1}{2}\left[\left(-\frac{x}{n\mathrm{Ï€}}{e}^{-inx\mathrm{Ï€}}+\frac{1}{n^2{\mathrm{Ï€}}^2}{e}^{-inx\mathrm{Ï€}}\right){\left|{}_0^1+\frac{2}{n\mathrm{Ï€}}{e}^{-in\mathrm{Ï€³\ae }}\right|}_1^2\right] \end{gathered}$$

$$\begin{gathered} c_n=\frac{1}{2}\left[\left(-\frac{{\left(-1\right)}^n}{in\mathrm{Ï€}}+\frac{1}{n^2{\mathrm{Ï€}}^2}\left({\left(-1\right)}^n-1\right)\right)+\frac{2}{n\mathrm{Ï€}}\left(1-{\left(-1\right)}^n\right)\right] \\ =\frac{1}{2}\left[\frac{1}{{\mathrm{Ï€}}^2{\mathrm{n}}^2}\left({\left(-1\right)}^n-1\right)+\frac{i}{n\mathrm{Ï€}}\left(3{\left(-1\right)}^n-2\right)\right] \end{gathered}$$

Then the Fourier series will be

$f_e\left(x\right)=-\frac{3}{4}+\frac{1}{2}\underset{n=-∞}{\overset{∞}{∑}}\left[\frac{1}{{\mathrm{π}}^2{\mathrm{n}}^2}\left({\left(-1\right)}^n-1\right)+\frac{i}{n\mathrm{π}}\left(3{\left(-1\right)}^n-2\right)\right]e^{in\mathrm{π}x}$