For each of the periodic functions in Problems 5.1 to 5.11 , use Dirichlet's theorem to find the value to which the Fourier series converges at$\mathbf{\text{x=0,±π/2,±π,±2π}}$ .

The given function is

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

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

The Fourier series for the function :

$$\begin{gathered} \mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{{\mathbf{a}}_{\mathbf{0}}}{\mathbf{2}}\mathbf{+}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\mathbf{(}{\mathbf{a}}_{\mathbf{n}}\mathbf{}\mathbf{cos}\mathbf{}\mathbf{nx}\mathbf{+}{\mathbf{b}}_{\mathbf{n}}\mathbf{}\mathbf{sin}\mathbf{}\mathbf{nx}\mathbf{)} \\ \mathbf{}\mathbf{}{\mathbf{a}}_{\mathbf{0}}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\pi }}{\mathbf{\int }}_{\mathbf{-}\mathbf{\pi }}^{\mathbf{\pi }}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{dx} \\ \mathbf{}\mathbf{}{\mathbf{a}}_{\mathbf{n}}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\pi }}{\mathbf{\int }}_{\mathbf{-}\mathbf{\pi }}^{\mathbf{\pi }}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{}\mathbf{cos}\mathbf{}\mathbf{nxdx} \\ \mathbf{}\mathbf{}{\mathbf{b}}_{\mathbf{n}}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\pi }}{\mathbf{\int }}_{\mathbf{-}\mathbf{\pi }}^{\mathbf{\pi }}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{}\mathbf{sin}\mathbf{}\mathbf{nxdx} \end{gathered}$$

$$\begin{gathered} \mathbf{If}\mathbf{}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{}\mathbf{is}\mathbf{}\mathbf{an}\mathbf{}\mathbf{even}\mathbf{}\mathbf{function}\mathbf{:}\mathbf{}\mathbf{} \\ {\mathbf{b}}_{\mathbf{n}}\mathbf{=}\mathbf{0} \\ \mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{{\mathbf{a}}_{\mathbf{0}}}{\mathbf{2}}\mathbf{+}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}{\mathbf{a}}_{\mathbf{n}}\mathbf{}\mathbf{cos}\mathbf{}\mathbf{nx} \end{gathered}$$

$$\begin{gathered} \mathbf{If}\mathbf{}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{is}\mathbf{}\mathbf{an}\mathbf{}\mathbf{odd}\mathbf{}\mathbf{function}\mathbf{:}\mathbf{}\mathbf{} \\ {\mathbf{a}}_{\mathbf{0}}\mathbf{=}{\mathbf{a}}_{\mathbf{n}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0} \\ \mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}{\mathbf{b}}_{\mathbf{n}}\mathbf{}\mathbf{sin}\mathbf{}\mathbf{nx} \end{gathered}$$

The sketch for the given function is shown below.

The given function is an odd function (remove the mean value).

When,

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

When, $a_0=1$.

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

Coefficients of $b_n$are $0,\frac{4}{2\tau \tau },0,0,0,\frac{4}{6\tau \tau },0,0,0,.....$where $\left(n>0\right)$.

The expansion is $f\left(x\right)=\frac{1}{2}+\frac{4}{\tau \tau }\sum _{n=2+4m}^{\infty }\frac{\sin nx}{n}$,where $m=0,1,2,......$.

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

And, 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, .

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

At point,localid="1660823675680" $x=-\frac{\mathrm{\pi }}{2}$

localid="1660823682313" $$\begin{gathered} f\left(x\right)=\frac{1}{2}\left[f\left(-\frac{\mathrm{\pi }}{2}+\right)+f\left(-\frac{\mathrm{\pi }}{2}-\right)\right] \\ f\left(x\right)=\frac{1}{2}\left[0+1\right] \\ f\left(x\right)=\frac{1}{2} \end{gathered}$$

At point,localid="1660823690198" $x=\frac{\mathrm{\pi }}{2}$.

localid="1660823696484" $$\begin{gathered} f\left(x\right)=\frac{1}{2}\left[f\left(\frac{\mathrm{\pi }}{2^{+}}\right)+f\left(\frac{\mathrm{\pi }}{2^{-}}\right)\right] \\ f\left(x\right)=\frac{1}{2}\left[0+1\right] \\ f\left(x\right)=\frac{1}{2} \end{gathered}$$

At point,localid="1660823700927" $x=-\mathrm{\pi }$.

localid="1660823707336" $$\begin{gathered} f\left(x\right)=\frac{1}{2}\left[f\left(-{\mathrm{\pi }}^{+}\right)+f\left(-{\mathrm{\pi }}^{-}\right)\right] \\ f\left(x\right)=\frac{1}{2}\left[1+0\right] \\ f\left(x\right)=\frac{1}{2} \end{gathered}$$

At point,localid="1660823714350" $x=-2\mathrm{\pi }$.

localid="1660823722025" $$\begin{gathered} f\left(x\right)=\frac{1}{2}\left[f\left(-2{\mathrm{\pi }}^{+}\right)+f\left(-2{\mathrm{\pi }}^{-}\right)\right] \\ f\left(x\right)=\frac{1}{2}\left[1+0\right] \\ f\left(x\right)=\frac{1}{2} \end{gathered}$$

At point,localid="1660823728112" $x=2\mathrm{\pi }$.

localid="1660823735419" $$\begin{gathered} f\left(x\right)=\frac{1}{2}\left[f\left(2{\mathrm{\pi }}^{+}\right)+f\left(2{\mathrm{\pi }}^{-}\right)\right] \\ f\left(x\right)=\frac{1}{2}\left[1+0\right] \\ f\left(x\right)=\frac{1}{2} \end{gathered}$$

Hence the convergence points are: