91Ó°ÊÓ

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

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$\mathbf{f}\left(x\right)$ :

$$\begin{gathered} \mathbf{f}\left(x\right)\mathbf{=}\frac{{\mathbf{a}}_{\mathbf{0}}}{\mathbf{2}}\mathbf{+}\underset{\mathbf{n}\mathbf{=}\mathbf{1}}{\overset{\mathbf{∞}}{\mathbf{∑}}}\left(a_n\cos \mathrm{nx}+b_n\sin \mathrm{nx}\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{a}}_{\mathbf{0}}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{\mathbf{π}}{\mathbf{∫}}_{\mathbf{-}\mathbf{π}}^{\mathbf{π}}\mathbf{f}\left(x\right)\mathbf{dx} \end{gathered}$$

$$\begin{gathered} {\mathbf{a}}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{π}}{\mathbf{∫}}_{\mathbf{-}\mathbf{π}}^{\mathbf{π}}\mathbf{f}\left(x\right)\mathbf{cosnxdx} \\ {\mathbf{b}}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{π}}{\mathbf{∫}}_{\mathbf{-}\mathbf{π}}^{\mathbf{π}}\mathbf{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{sinnxdx} \end{gathered}$$

If f$\left(x\right)$is an even function:

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

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

The sketch for the given function is shown below.

Coefficients are given below.

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

$$\begin{gathered} Calculatefurtherasfollows: \\ {a}_n=\frac{1}{\mathrm{π}}{∫}_{-\mathrm{π}}^{\mathrm{π}}f\left(x\right)\cos nxdx \\ =\frac{1}{\mathrm{π}}{∫}_0^{\mathrm{π}}f\left(x\right)\cos nxdx \\ =\frac{1}{\mathrm{π}}{\left[\frac{x}{n}\sin nx+\frac{1}{n_2}\cos nx\right]}_0^{\mathrm{π}} \\ =\frac{1}{n^2\mathrm{π}}\left[{\left(-1\right)}^n-1\right] \end{gathered}$$

Calculate more as shown below.

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

$$\begin{gathered} =-\frac{\mathrm{Ï€}}{n\mathrm{Ï€}}{\left(-1\right)}^n \\ So,=\frac{{\left(-1\right)}^{n+1}}{n} \end{gathered}$$

The expansion is

$f\left(x\right)=\frac{\mathrm{π}}{4}-\frac{1}{\mathrm{π}}\underset{n=1+2m}{\overset{∞}{∑}}\frac{1}{n^2}\left[{\left(-1\right)}^n-1\right]\cos nx+\underset{n=1}{\overset{∞}{∑}}\frac{{\left(-1\right)}^{n+1}}{n}\sin nx,wherem=0,1,2,.....$

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

And, the Fourier series converges to $\frac{1}{2}\left[f\left(x^{+}\right)+f\left(x^{-}\right)\right]$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, $x=0$.

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

At point, $x=-\frac{\mathrm{Ï€}}{2}$.

role="math" localid="1659358793562" $$\begin{gathered} f\left(x\right)=\frac{1}{2}\left[f\left(-\frac{\mathrm{Ï€}}{2^{+}}\right)+f\left(-\frac{\mathrm{Ï€}}{2^{-}}\right)\right] \\ f\left(x\right)=\frac{1}{2}\left[0+0\right] \\ f\left(x\right)=0 \end{gathered}$$

At point, $x=\frac{\mathrm{Ï€}}{2}$.

role="math" localid="1659358858343" $$\begin{gathered} f\left(x\right)=\frac{1}{2}\left[f\left(-\frac{\mathrm{Ï€}}{2^{+}}\right)+f\left(-\frac{\mathrm{Ï€}}{2^{-}}\right)\right] \\ f\left(x\right)=\frac{1}{2}\left[\mathrm{Ï€}+0\right] \\ f\left(x\right)=\frac{\mathrm{Ï€}}{2} \end{gathered}$$

At point, .

At point, .

Therefore, the convergence points are: