For each of the periodic functions in Problems5.1to 5.11, use Dirichlet's theorem to find the value to which the Fourier series converges at $\mathit{x}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{\pm }\mathit{\pi }\mathbf{/}\mathbf{2}\mathbf{,}\mathbf{}\mathbf{\pm }\mathit{\pi }\mathbf{,}\mathbf{}\mathbf{\pm }\mathbf{2}\mathit{\pi }$.

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

The given points are $x=0,\pm \frac{\pi }{2},\pm \pi ,\pm 2\pi$.

The Fourier series for the function f(x):

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

If f(x)is an even function:

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

If f(x)is an odd function:

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

The sketch for the given function is shown below.

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

Coefficients of $a_n$are given below.

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

Coefficients of$a_n$are stated below.

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

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

Coefficients of $b_n$are shown below

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

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

The expansion is $f\left(x\right)=\frac{1}{4}+\frac{1}{\pi }\left[\sum _{n=1+4m}^{\infty }\frac{\cos nx}{n}-\sum _{n=3+4m}^{\infty }\frac{\cos nx}{n}+\sum _{n=1+2m}^{\infty }\frac{\sin nx}{n}+2\sum _{n=2+4m}^{\infty }\frac{\sin nx}{n}\right]$ where

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

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

The Fourier series converges to $\frac{1}{2}\left[f\left(x^{+}\right)+f\left(x^{-}\right)\right]\to$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, $x=-\frac{\pi }{2}$

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

At point, $x=\frac{\pi }{2}$

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

At point, $x=-\pi$

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

At point, $x=\pi$

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

At point, $x=-2\pi$

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

At point, $x=2\pi$

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

Hence, the convergence points are: