Question

Sketch several periods of the corresponding periodic function of period$\mathbf{2}\mathit{\pi }$. Expand the periodic function in a sine-cosine Fourier series.

$\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{\{}\begin{array}{l}\mathbf{1}\mathbf{,}\mathbf{}\mathbf{-}\mathbf{\pi }\mathbf{<}\mathbf{x}\mathbf{<}\mathbf{0}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\\ \mathbf{0}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathbf{0}\mathbf{<}\mathbf{x}\mathbf{<}\mathbf{\pi }\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\end{array}$

Short answer

The answer of the given function is $f\left(x\right)=\frac{1}{2}-\frac{2}{\pi }\left(\frac{\sin x}{1}+\frac{\sin 3x}{3}+\frac{\sin 5x}{5}---\right)$.

Step-by-step solution

Given

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

The sketch and the result of the given function are shown below.

$f\left(x\right)=\frac{1}{2}-\frac{2}{\pi }\left(\frac{\sin x}{1}+\frac{\sin 3x}{3}+\frac{\sin 5x}{5}---\right)$

Concept of Fourier series

The Fourier series for the function f(x) :

$$\begin{gathered} \mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\frac{{\mathbf{a}}_{\mathbf{0}}}{\mathbf{2}}\mathbf{+}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\left(a_{n}cosnx+b_{n}sinnx\right) \\ {\mathit{a}}_{\mathbf{0}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\pi }}{\mathbf{\int }}_{\mathbf{-}\mathbf{\pi }}^{\mathbf{\pi }}\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathit{d}\mathit{x} \\ {\mathit{a}}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\pi }}{\mathbf{\int }}_{\mathbf{-}\mathbf{\pi }}^{\mathbf{\pi }}\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathit{c}\mathit{o}\mathit{s}\mathit{n}\mathit{x}\mathit{d}\mathit{x} \\ {\mathit{b}}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\pi }}{\mathbf{\int }}_{\mathbf{-}\mathbf{\pi }}^{\mathbf{\pi }}\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathit{s}\mathit{i}\mathit{n}\mathit{n}\mathit{x}\mathit{d} \end{gathered}$$

If f(x) is an even function:

$$\begin{gathered} {\mathit{b}}_{\mathbf{n}}\mathbf{=}\mathbf{0}\mathit{a} \\ \mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\frac{{\mathbf{a}}_{\mathbf{0}}}{\mathbf{2}}\mathbf{+}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}{\mathit{a}}_{\mathbf{n}}\mathit{c}\mathit{o}\mathit{s}\mathit{n}\mathit{x} \end{gathered}$$

If f(x) is an odd function:

$$\begin{gathered} {\mathit{a}}_{\mathbf{0}}\mathbf{=}{\mathit{a}}_{\mathbf{n}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0} \\ \mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}{\mathit{b}}_{\mathbf{n}}\mathit{s}\mathit{i}\mathit{n}\mathit{n}\mathit{x} \end{gathered}$$

Calculate with the help of Fourier series

Given function is $f\left(x\right)=\frac{1}{2}-\frac{2}{\pi }\left(\frac{\sin x}{1}+\frac{\sin 3x}{3}+\frac{\sin 5x}{5}---\right)$.

Use the solution of the example one as follows:

$$\begin{gathered} g\left(x\right)=f(x+\pi ) \\ g(x-\pi )=f\left(x\right) \\ g(x-\pi )=\frac{1}{2}+\frac{2}{\pi }\sum _{k=2n+1}^{\infty }\frac{\sin (kx-k\pi )}{k} \\ =\frac{1}{2}+\frac{2}{\pi }\sum _{k=2n+1}^{\infty }\frac{\sin \left(kx\right)\cos \left(k\pi \right)-\sin \left(k\pi \right)\cos \left(kx\right)}{k} \end{gathered}$$

Calculate further as shown below.

$=\frac{1}{2}+\frac{2}{\pi }\sum _{k=2n+1}^{\infty }\frac{\sin \left(kx\right)\cos \left(k\pi \right)}{k},\left[\sin \right(k\pi )=0$ , where $k=2n+1]$

$=\frac{1}{2}-\frac{2}{\pi }\sum _{k=2n+1}\frac{\sin \left(kx\right)}{k},\left[\cos \right(k\pi )=-1$ , where $k=2n+1]$

$$\begin{gathered} =\frac{1}{2}-\frac{2}{\pi }\left(\frac{\sin x}{1}+\frac{\sin 3x}{3}+\frac{\sin 5x}{5}---\right) \\ =f\left(x\right) \end{gathered}$$