Question

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$\mathbf{x}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{\pm }\mathbf{\pi }\mathbf{/}\mathbf{2}\mathbf{,}\mathbf{\pm }\mathbf{\pi }\mathbf{,}\mathbf{\pm }\mathbf{2}\mathbf{\pi }$ .

Short answer

x	$f\left(x\right)$
0	$0$
$-\frac{\tau \tau }{2}$	$0$
$\frac{\tau \tau }{2}$	$\frac{\tau \tau }{2}$
$-\tau \tau$	$\frac{\tau \tau }{2}$
$\tau \tau$	$\frac{\tau \tau }{2}$
$-2\tau \tau$	$0$
$2\tau \tau$	$0$

The convergence points are:

Step-by-step solution

Given

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

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

Definition of Fourier series

The Fourier series for the function$\mathbf{f}\left(x\right)$ :

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

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

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

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

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

Sketch the function

The sketch for the given function is shown below.

 Step 4: Find the Coefficients

Coefficients are given below.

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

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

Calculate more as shown below.

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

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

The expansion is

$f\left(x\right)=\frac{\mathrm{\pi }}{4}-\frac{1}{\mathrm{\pi }}\sum _{n=1+2m}^{\infty }\frac{1}{n^2}\left[{\left(-1\right)}^n-1\right]\cos nx+\sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n+1}}{n}\sin nx,wherem=0,1,2,.....$

Use Fourier series and find the convergence points

The Fourier series converges to $f\left(x\right)\to$ 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, $x=-\frac{\mathrm{\pi }}{2}$.

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

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

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

At point, .

role="math" localid="1659358955699" $$\begin{gathered} f\left(x\right)=\frac{1}{2}\left[f\left(-{\mathrm{\pi }}^{+}\right)+f\left(-{\mathrm{\pi }}^{-}\right)\right] \\ f\left(x\right)=\frac{1}{2}\left[0+\mathrm{\pi }\right] \\ f\left(x\right)=\frac{\mathrm{\pi }}{2} \end{gathered}$$

At point, .

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

Therefore, the convergence points are:

x	$f\left(x\right)$
0	0
$-\frac{\tau \tau }{2}$	0
$\frac{\tau \tau }{2}$	$\frac{\tau \tau }{2}$
$-\tau \tau$	$\frac{\tau \tau }{2}$
$\tau \tau$	$\frac{\tau \tau }{2}$
$-2\tau \tau$	0
$2\tau \tau$	0