Question

Find the three Fourier series in problem9 and 10.

Short answer

$$\begin{gathered} f_{s}x=\sum _{n=1}^{\infty }\left[{\left(\frac{2}{n\mathrm{\pi }}\right)}^2\sin \frac{n\mathrm{\pi }}{2}-\frac{6}{n\mathrm{\pi }}\cos \frac{n\mathrm{\pi }}{2}+\frac{4}{n\mathrm{\pi }}{\left(-1\right)}^n\right]\sin \frac{n\mathrm{\pi x}}{2} \\ {f}_{c}x=\frac{3}{4}+\sum _{n=1}^{\infty }\left[\frac{6}{n\mathrm{\pi }}\sin \frac{n\mathrm{\pi }}{2}+{\left(\frac{2}{n\mathrm{\pi }}\right)}^2\left(\cos \frac{n\mathrm{\pi }}{2}-1\right)\right]\cos \frac{n\mathrm{\pi }}{2} \\ {f}_{e}x=-\frac{3}{4}+\frac{1}{2}\sum _{-\infty }^{\infty }\left[\frac{1}{{\mathrm{\pi }}^2{\mathrm{n}}^2}\left({\left(-1\right)}^n-1\right)+\frac{i}{n\mathrm{\pi }}\left(3{\left(-1\right)}^n-2\right)\right]e^{in\mathrm{\pi }x} \end{gathered}$$

Step-by-step solution

Meaning of the Fourier series

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 parameter

Given function in the problem 9

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

Find the Fourier series for the sine series in Problem 9

Here the sine series coefficient

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

Then the series will be

$f_s\left(x\right)=\sum _{n=1}^{\infty }\left[{\left(\frac{2}{n\mathrm{\pi }}\right)}^2\sin \frac{n\mathrm{\pi }}{2}-\frac{6}{n\mathrm{\pi }}\cos \frac{n\mathrm{\pi }}{2}+\frac{4}{n\mathrm{\pi }}{\left(-1\right)}^n\right]\sin \frac{n\mathrm{\pi x}}{2}$

Find the Fourier series for the cosine series in Problem 10

The Cosine series coefficient:

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

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

Then the Fourier series will be

$f\left(x\right)=\frac{3}{4}+\sum _{n-1}^{\infty }\left[\frac{6}{n\mathrm{\pi }}\sin \frac{n\mathrm{\pi }}{2}+{\left(\frac{2}{n\mathrm{\pi }}\right)}^2\left(\cos \frac{n\mathrm{\pi }}{2}-1\right)\right]\cos \frac{n\mathrm{\pi x}}{2}$

Find the Fourier series for the exponential series in Problem 10

The exponential series coefficient

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

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

Then the Fourier series will be

$f_e\left(x\right)=-\frac{3}{4}+\frac{1}{2}\sum _{n=-\infty }^{\infty }\left[\frac{1}{{\mathrm{\pi }}^2{\mathrm{n}}^2}\left({\left(-1\right)}^n-1\right)+\frac{i}{n\mathrm{\pi }}\left(3{\left(-1\right)}^n-2\right)\right]e^{in\mathrm{\pi }x}$