91Ó°ÊÓ

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

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

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

The Fourier series for the function f(x) :

$$\begin{gathered} \mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\frac{{\mathbf{a}}_{\mathbf{0}}}{\mathbf{2}}\mathbf{+}\underset{\mathbf{n}\mathbf{=}\mathbf{1}}{\overset{\mathbf{∞}}{\mathbf{∑}}}\left(a_{n}cosnx+b_{n}sinnx\right) \\ {\mathit{a}}_{\mathbf{0}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{π}}{\mathbf{∫}}_{\mathbf{-}\mathbf{π}}^{\mathbf{π}}\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathit{d}\mathit{x} \\ {\mathit{a}}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{π}}{\mathbf{∫}}_{\mathbf{-}\mathbf{π}}^{\mathbf{π}}\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathit{c}\mathit{o}\mathit{s}\mathit{n}\mathit{x}\mathit{d}\mathit{x} \\ {\mathit{b}}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{π}}{\mathbf{∫}}_{\mathbf{-}\mathbf{π}}^{\mathbf{π}}\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathit{s}\mathit{i}\mathit{n}\mathit{n}\mathit{x}\mathit{d} \end{gathered}$$

If f(x) is an even function:

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

If f(x) is an odd function:

$$\begin{gathered} {\mathit{a}}_{\mathbf{0}}\mathbf{=}{\mathit{a}}_{\mathbf{n}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0} \\ \mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\underset{\mathbf{n}\mathbf{=}\mathbf{1}}{\overset{\mathbf{∞}}{\mathbf{∑}}}{\mathit{b}}_{\mathbf{n}}\mathit{s}\mathit{i}\mathit{n}\mathit{n}\mathit{x} \end{gathered}$$

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

Use the solution of the example one as follows:

$$\begin{gathered} g\left(x\right)=f(x+\mathrm{π}) \\ g(x-\mathrm{π})=f\left(x\right) \\ g(x-\mathrm{π})=\frac{1}{2}+\frac{2}{\mathrm{π}}\underset{k=2n+1}{\overset{∞}{∑}}\frac{\sin (kx-k\mathrm{π})}{k} \\ =\frac{1}{2}+\frac{2}{\mathrm{π}}\underset{k=2n+1}{\overset{∞}{∑}}\frac{\sin \left(kx\right)\cos \left(k\mathrm{π}\right)-\sin \left(k\mathrm{π}\right)\cos \left(kx\right)}{k} \end{gathered}$$

Calculate further as shown below.

$=\frac{1}{2}+\frac{2}{\mathrm{π}}\underset{k=2n+1}{\overset{∞}{∑}}\frac{\sin \left(kx\right)\cos \left(k\mathrm{π}\right)}{k},\left[\sin \right(k\mathrm{π})=0$ , where $k=2n+1]$

$=\frac{1}{2}-\frac{2}{\mathrm{π}}\underset{k=2n+1}{∑}\frac{\sin \left(kx\right)}{k},\left[\cos \right(k\mathrm{π})=-1$ , where $k=2n+1]$

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