Question

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

Short answer

The convergence points are:

Step-by-step solution

Given

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

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

Definition of Fourier series

The Fourier series for the function f(x):

$$\begin{gathered} f\left(x\right)=\frac{a_0}{2}+\sum _{n=1}^{\infty }\left(a_n\cos nx+b_n\sin nx\right) \\ {a}_0=\frac{1}{\pi }{\int }_{-\pi }^{\pi }f\left(x\right)dx \\ {a}_n=\frac{1}{\pi }{\int }_{-\pi }^{\pi }f\left(x\right)\cos nxdx \\ {b}_n=\frac{1}{\pi }{\int }_{-\pi }^{\pi }f\left(x\right)\sin nxdx \end{gathered}$$

If f(x)is an even function:

$$\begin{gathered} b_n=0a \\ f\left(x\right)=\frac{a_0}{2}+\sum _{n=1}^{\infty }{a}_n\cos nx \end{gathered}$$

If f(x) is an odd function:

$$\begin{gathered} a_0=a_n \\ =0 \\ f\left(x\right)=\sum _{n=1}^{\infty }{b}_n\sin nx \end{gathered}$$

Sketch the function

The sketch for the given function is shown below.

Use Fourier series 

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

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.

Find the Convergence points

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, $x=-\frac{\pi }{2}$.

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

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, $x=-\pi$.

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

At point,$x=\pi$.

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

At point, $x=-2\pi$.

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

At point, $x=2\pi$.

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

Thus, the convergence points are: