91Ó°ÊÓ

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

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):

$$\begin{gathered} f\left(x\right)=\frac{a_0}{2}+\underset{n=1}{\overset{∞}{∑}}\left(a_n\cos nx+b_n\sin nx\right) \\ {a}_0=\frac{1}{\mathrm{π}}{∫}_{-\mathrm{π}}^{\mathrm{π}}f\left(x\right)dx \\ {a}_n=\frac{1}{\mathrm{π}}{∫}_{-\mathrm{π}}^{\mathrm{π}}f\left(x\right)\cos nxdx \\ {b}_n=\frac{1}{\mathrm{π}}{∫}_{-\mathrm{π}}^{\mathrm{π}}f\left(x\right)\sin nxdx \end{gathered}$$

If f(x)is an even function:

$$\begin{gathered} b_n=0a \\ f\left(x\right)=\frac{a_0}{2}+\underset{n=1}{\overset{∞}{∑}}{a}_n\cos nx \end{gathered}$$

If f(x) is an odd function:

$$\begin{gathered} a_0=a_n \\ =0 \\ f\left(x\right)=\underset{n=1}{\overset{∞}{∑}}{b}_n\sin nx \end{gathered}$$

The sketch for the given function is shown below.

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

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, $x=-\frac{\mathrm{Ï€}}{2}$.

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

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, $x=-\mathrm{Ï€}$.

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

At point,$x=\mathrm{Ï€}$.

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

At point, $x=-2\mathrm{Ï€}$.

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

At point, $x=2\mathrm{Ï€}$.

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

Thus, the convergence points are: