Question

In each of the following problems you are given a function on the interval $\mathbf{-}\mathbf{\pi }\mathbf{<}\mathbf{x}\mathbf{<}\mathbf{\pi }$. Sketch several periods of the corresponding periodic function of period$\mathbf{2}\mathit{\tau }\mathit{\tau }$ . Expand the periodic function in a sine-cosine Fourier series,

role="math" localid="1659239194875" $\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{\{}\begin{array}{l}\mathbf{1}\mathbf{,}\mathbf{}\mathbf{-}\mathbf{\pi }\mathbf{<}\mathbf{x}\mathbf{<}\frac{\mathbf{\pi }}{\mathbf{2}}\mathbf{and}\mathbf{}\mathbf{0}\mathbf{<}\mathbf{x}\mathbf{<}\frac{\mathbf{\pi }}{\mathbf{2}}\\ \mathbf{0}\mathbf{,}\mathbf{}\mathbf{-}\frac{\mathbf{\pi }}{\mathbf{2}}\mathbf{<}\mathbf{x}\mathbf{<}\mathbf{0}\mathbf{}\mathbf{and}\mathbf{}\frac{\mathbf{\pi }}{\mathbf{2}}\mathbf{<}\mathbf{x}\mathbf{<}\mathbf{\pi }\end{array}$

Short answer

The resultant expansion is $f\left(x\right)=\frac{1}{2}+\frac{4}{\mathrm{\pi }}\sum _{n=2+4m}^{\infty }\frac{\sin nx}{n}wherem=0,1,2-....$

Step-by-step solution

Given data

The given function is $\mathrm{f}\left(\mathrm{x}\right)=\{\begin{array}{l}1,-\mathrm{\pi }<\mathrm{x}<\frac{\mathrm{\pi }}{2}\mathrm{and}0<\mathrm{x}<\frac{\mathrm{\pi }}{2}\\ 0,-\frac{\mathrm{\pi }}{2}<\mathrm{x}<0\mathrm{and}\frac{\mathrm{\pi }}{2}<\mathrm{x}<\mathrm{\pi }\end{array}$.

Concept of Fourier series

An infinite sum of sines and cosines is used to represent the expansion of a periodic function f(x) into a Fourier series.

The orthogonality relationships of the sine and cosine functions are used in Fourier series.

Sketch the points

The sketch for the function is shown below.

Find the coefficients of the series

The fundamental period is $\mathit{\tau }\mathit{\tau }$.

Remove the mean value to make the function odd, so ${\mathit{A}}_{\mathbf{n}}\mathbf{=}\mathbf{0}\mathbf{.}$.

The rest of the coefficients are given below.

$$\begin{gathered} {\mathrm{A}}_0=\frac{1}{\mathrm{\pi }}\left[\int -{\mathrm{\pi }}^{-\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\mathrm{dx}+{\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\mathrm{dx}\right]=1 \\ {\mathrm{B}}_{\mathrm{n}}=\frac{1}{\mathrm{\pi }}\left[{\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\sin \left(\mathrm{nx}\right)\mathrm{dx}+{\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\sin \left(\mathrm{nx}\right)\mathrm{dx}\right] \\ {\mathrm{B}}_{\mathrm{n}}=-\frac{1}{\mathrm{n\pi }}\left[\cos \left(\mathrm{nx}\right)\overline{){}_{-\mathrm{\pi }}^{\raisebox{1ex}{$-\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}+\cos \left(\mathrm{nx}\right)\overline{){}_0^{\raisebox{1ex}{$-\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right] \\ {\mathrm{B}}_{\mathrm{n}}=-\frac{1}{\mathrm{n\pi }}\left[2\cos \frac{\mathrm{n\pi }}{2}-\left(1+{\left(-1\right)}^{\mathrm{n}}\right)\right] \\ \mathrm{The}\mathrm{coefficient}\mathrm{has}\mathrm{the}\mathrm{series}(\mathrm{starting}\mathrm{with}\mathrm{n}=1). \\ {\mathrm{B}}_{\mathrm{n}}:0,\frac{4}{2\mathrm{\pi }},0,0,0,\frac{4}{6\mathrm{\pi }},0,0,0,\frac{4}{10\mathrm{\pi }},.... \\ \mathrm{Thus},\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{2}+\frac{4}{\mathrm{\pi }}\sum _{\mathrm{k}=2+4\mathrm{n}}^{\infty }\frac{\sin \left(\mathrm{kx}\right)}{\mathrm{k}}\mathrm{m}=0,1,2,3,.... \end{gathered}$$