Question

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="1659236419546" $\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{\{}\begin{array}{cc}\mathbf{0}\mathbf{,}& \mathbf{-}\mathbf{\pi }\mathbf{<}\mathbf{x}\mathbf{<}\mathbf{0}\\ \mathbf{-}\mathbf{1}\mathbf{,}& \mathbf{0}\mathbf{<}\mathbf{x}\mathbf{<}\frac{\mathbf{\pi }}{\mathbf{2}}\\ \mathbf{1}\mathbf{,}& \frac{\mathbf{\pi }}{\mathbf{2}}\mathbf{<}\mathbf{x}\mathbf{<}\mathbf{\pi }\end{array}$

Short answer

The answer of the given function is

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

Step-by-step solution

Given

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

The concept of the Fourier series for the function f(x)

The Fourier series for the function $\mathit{f}\left(x\right)$:

$$\begin{gathered} \mathbf{f}\left(x\right)\mathbf{=}\frac{{\mathbf{a}}_{\mathbf{0}}}{\mathbf{2}}\mathbf{+}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\left(a_n\cos \mathrm{nx}+b_n\sin \mathrm{nx}\right) \\ \mathbf{}\mathbf{}\mathbf{}{\mathbf{a}}_{\mathbf{0}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\pi }}{\mathbf{\int }}_{\mathbf{-}\mathbf{\pi }}^{\mathbf{\pi }}\mathbf{f}\left(x\right)\mathbf{dx} \\ \mathbf{}\mathbf{}\mathbf{}{\mathbf{a}}_{\mathbf{0}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\pi }}{\mathbf{\int }}_{\mathbf{-}\mathbf{\pi }}^{\mathbf{\pi }}\mathbf{f}\left(x\right)\mathbf{cos}\mathbf{}\mathbf{nxdx} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{b}}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\pi }}{\mathbf{\int }}_{\mathbf{-}\mathbf{\pi }}^{\mathbf{\pi }}\mathbf{f}\left(x\right)\mathbf{sin}\mathbf{}\mathbf{nxdx} \\ \mathbf{If}\mathbf{}\mathbf{f}\left(x\right)\mathbf{}\mathbf{is}\mathbf{}\mathbf{an}\mathbf{}\mathbf{}\mathbf{even}\mathbf{}\mathbf{}\mathbf{function}\mathbf{:} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{b}}_{\mathbf{n}}\mathbf{=}\mathbf{0} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{f}\left(x\right)\mathbf{=}\frac{{\mathbf{a}}_{\mathbf{0}}}{\mathbf{2}}\mathbf{+}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}{\mathbf{a}}_{\mathbf{n}}\mathbf{cos}\mathbf{}\mathbf{nx} \\ \mathbf{If}\mathbf{}\mathbf{f}\left(x\right)\mathbf{}\mathbf{is}\mathbf{}\mathbf{an}\mathbf{}\mathbf{odd}\mathbf{}\mathbf{function}\mathbf{:} \\ \mathbf{}\mathbf{}\mathbf{}{\mathbf{a}}_{\mathbf{0}}\mathbf{=}{\mathbf{a}}_{\mathbf{n}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0} \\ \mathbf{f}\left(x\right)\mathbf{=}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}{\mathbf{b}}_{\mathbf{n}}\mathbf{}\mathbf{sin}\mathbf{}\mathbf{nx} \end{gathered}$$

From the given information

The sketch of the given function shown below.

Coefficients of $a_n$:

$$\begin{gathered} {\mathrm{a}}_{\mathrm{n}}=\frac{1}{\mathrm{\pi }}{\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}\mathrm{f}\left(\mathrm{x}\right)\cos \mathrm{nxdx} \\ =\frac{1}{\mathrm{\pi }}\left[-{\int }_0^{\frac{\mathrm{\pi }}{2}}\mathrm{cosnxdx}+{\int }_{\frac{\mathrm{\pi }}{2}}^{\mathrm{\pi }}\mathrm{cosnxdx}\right] \\ =\frac{1}{\mathrm{n\pi }}\left[-{\left[\sin \mathrm{nx}\right]}^{\frac{\mathrm{\pi }}{2}}0+{\left[\sin \mathrm{nx}\right]}_{\frac{\mathrm{\pi }}{2}}^{\mathrm{\pi }}\right] \\ =-\frac{2}{\mathrm{n\pi }}\sin \left(\frac{\mathrm{n\pi }}{2}\right) \end{gathered}$$

Calculate Coefficients of  bn

Coefficients of $b_n$ are given below.

$$\begin{gathered} b_n=\frac{1}{\mathrm{\pi }}{\int }_{-\mathrm{\pi }}^{\mathrm{\pi }}f\left(x\right)\sin nxdx \\ =\frac{1}{\mathrm{\pi }}\left[-{\int }_0^{\frac{\mathrm{\pi }}{2}}s\mathrm{in}nxdx+{\int }_{\frac{\mathrm{\pi }}{2}}^{\mathrm{\pi }}\sin nxdx\right] \\ =\frac{1}{\mathrm{n\pi }}\left[{\left[\cos nx\right]}_0^{\frac{\mathrm{\pi }}{2}}-{\left[\cos nx\right]}_{\frac{\mathrm{\pi }}{2}}^{\mathrm{\pi }}\right] \\ =\frac{1}{\mathrm{n\pi }}\left[2\cos \left(\frac{n\mathrm{\pi }}{2}\right)-\left\{1+{\left(-1\right)}^n\right\}\right] \\ Hencetheexpansionisf\left(x\right)=-\frac{2}{\mathrm{\pi }}\left[\sum _{n=1+4m}^{\infty }\frac{\cos nx}{n}-\sum _{n=3+4m}^{\infty }\frac{\cos nx}{n}\right]-\frac{4}{\mathrm{\pi }}\left[\sum _{n=2+4m}^{\infty }\frac{\sin nx}{n}\right]wherem=0,1,2-- \end{gathered}$$