91Ó°ÊÓ

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

The given points are $x=0,Â\pm \frac{\mathrm{Ï€}}{2},Â\pm \mathrm{Ï€},Â\pm 2\mathrm{Ï€}$.

The Fourier series for the function f(x):

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

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:

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

The sketch for the given function is shown below.

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

Coefficients of $a_n$are given below.

$$\begin{gathered} a_0=\frac{1}{\mathrm{π}}{∫}_{-\mathrm{π}}^{\mathrm{π}}f\left(x\right)dx \\ =\frac{1}{\mathrm{π}}{∫}_0^{\frac{\mathrm{π}}{2}}dx \\ =\frac{1}{\mathrm{π}}[x{]}_0^{\frac{\mathrm{π}}{2}} \\ =\frac{1}{2} \end{gathered}$$

Coefficients of$a_n$are stated below.

$$\begin{gathered} a_n=\frac{1}{\mathrm{π}}{∫}_{-\mathrm{π}}^{\mathrm{π}}f\left(x\right)\cos nxdx \\ =\frac{1}{\mathrm{π}}{∫}_0^{\frac{\mathrm{π}}{2}}\cos nxdx \\ =\frac{1}{n\mathrm{π}}[\sin nx{]}_0^{\frac{\mathrm{π}}{2}} \\ =\frac{1}{n\mathrm{π}}\sin \left(\frac{n\mathrm{π}}{2}\right) \end{gathered}$$

Coefficients of role="math" localid="1664290158734" $a_n$are $\frac{1}{\mathrm{Ï€}},0,-\frac{1}{3\mathrm{Ï€}},0,\frac{1}{5\mathrm{Ï€}},0,-\frac{1}{7\mathrm{Ï€}}$.

Coefficients of $b_n$are shown below

$$\begin{gathered} b_n=\frac{1}{\mathrm{π}}{∫}_{-\mathrm{π}}^{\mathrm{π}}f\left(x\right)\sin nxdx \\ =\frac{1}{\mathrm{π}}{∫}_0^{\frac{\mathrm{π}}{2}}\sin nxdx \\ =-\frac{1}{n\mathrm{π}}[\cos nx{]}_0^{\frac{\mathrm{π}}{2}} \\ =\frac{1}{n\mathrm{π}}\left[1-\cos \left(\frac{n\mathrm{π}}{2}\right)\right] \end{gathered}$$

Coefficients of $b_n$are $\frac{1}{\mathrm{Ï€}},\frac{2}{2\mathrm{Ï€}},\frac{1}{3\mathrm{Ï€}},0,\frac{1}{5\mathrm{Ï€}},\frac{2}{6\mathrm{Ï€}},\frac{1}{7\mathrm{Ï€}}$.

The expansion is $f\left(x\right)=\frac{1}{4}+\frac{1}{\mathrm{π}}\left[\underset{n=1+4m}{\overset{∞}{∑}}\frac{\cos nx}{n}-\underset{n=3+4m}{\overset{∞}{∑}}\frac{\cos nx}{n}+\underset{n=1+2m}{\overset{∞}{∑}}\frac{\sin nx}{n}+2\underset{n=2+4m}{\overset{∞}{∑}}\frac{\sin nx}{n}\right]$ where

$m=0,1,2,.....$

The Fourier series converges to $f\left(x\right)→$At all points where f is continuous.

The Fourier series converges to $\frac{1}{2}\left[f\left(x^{+}\right)+f\left(x^{-}\right)\right]→$at all points where f is discontinuous.

Therefore the series converges to the average value of the right and left limits at a point of discontinuity.

At point, x = 0

$$\begin{gathered} f\left(x\right)=\frac{1}{2}\left[f\left(0^{+}\right)+f\left(0^{-}\right)\right] \\ f\left(x\right)=\frac{1}{2}[0+1] \\ f\left(x\right)=\frac{1}{2} \end{gathered}$$

At Point, $x=-\frac{\mathrm{Ï€}}{2}$

$$\begin{gathered} f\left(x\right)=\frac{1}{2}\left[f\left(-\frac{\mathrm{Ï€}}{2^{+}}\right)+f\left(-\frac{\mathrm{Ï€}}{2^{-}}\right)\right] \\ f\left(x\right)=\frac{1}{2}[0+0] \\ f\left(x\right)=0 \end{gathered}$$

At point, $x=\frac{\mathrm{Ï€}}{2}$

$$\begin{gathered} f\left(x\right)=\frac{1}{2}\left[f\left({\frac{\mathrm{Ï€}}{2}}^{+}\right)+f\left(\frac{\mathrm{Ï€}}{2}-\right)\right] \\ f\left(x\right)=\frac{1}{2}[0+1] \\ f\left(x\right)=\frac{1}{2} \end{gathered}$$

At point, $x=-\mathrm{Ï€}$

$$\begin{gathered} f\left(x\right)=\frac{1}{2}\left[f\left(-{\mathrm{Ï€}}^{+}\right)+f\left(-{\mathrm{Ï€}}^{-}\right)\right] \\ f\left(x\right)=\frac{1}{2}[0+0] \\ f\left(x\right)=0 \end{gathered}$$

At point, $x=\mathrm{Ï€}$

$$\begin{gathered} f\left(x\right)=\frac{1}{2}\left[f\left({\mathrm{Ï€}}^{+}\right)+f\left({\mathrm{Ï€}}^{-}\right)\right] \\ f\left(x\right)=\frac{1}{2}[0+0] \\ f\left(x\right)=0 \end{gathered}$$

At point, $x=-2\mathrm{Ï€}$

$$\begin{gathered} f\left(x\right)=\frac{1}{2}\left[f\left(-2{\mathrm{Ï€}}^{+}\right)+f\left(-2{\mathrm{Ï€}}^{-}\right)\right] \\ f\left(x\right)=\frac{1}{2}[0+1] \\ =\frac{1}{2} \end{gathered}$$

At point, $x=2\mathrm{Ï€}$

$$\begin{gathered} f\left(x\right)=\frac{1}{2}\left[f\left(2{\mathrm{Ï€}}^{+}\right)+f\left(2{\mathrm{Ï€}}^{-}\right)\right] \\ f\left(x\right)=\frac{1}{2}[1+0] \\ f\left(x\right)=\frac{1}{2} \end{gathered}$$

Hence, the convergence points are: