Example 5

Sketch several periods of the given function and find its Fourier series:

$$f(x) = \left \{ \begin{array}{rl} -1, & -\pi < x < 0 \\ x,... ..., & x = -\pi, 0 {\mbox and} \pi.\\ \end{array}\right .$$

Solution:

The graph of $f$ is sketched in Figure 4.

The values of $f$ at the endpoints of the intervals do not affect the value of the integrals. Thus,

$a_0 = \displaystyle{\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) dx = \ \frac{1}{\pi} \int_{-\pi}^{0} (-1) dx + \ \frac{1}{\pi} \int_{0}^{\pi} x dx =}$

$= \displaystyle{ \biggl. \frac{1}{\pi} (-\pi) + \frac{1}{\pi} \frac{x^2}{2} \biggr\vert^{\pi}_0 = -1 + \frac{\pi}{2}}.$

Calculation of $a_n$ involves integration by parts:

$a_n = \displaystyle{ \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos nx dx = \... ... (-1)\cos nx dx + \ \frac{1}{\pi} \int_{0}^{\pi} x \cos nx dx = \ }$

$= \displaystyle{\biggl. -\frac{1}{\pi} \frac{1}{n} \sin nx \biggr\vert^0_{-\... ...\biggl[\cos nx \biggr]_0^{\pi} = \frac{1}{\pi n^2}\biggl[(-1)^n - 1 \biggr]. }$

Taking into account that $\cos 0 = 1$, $\sin n\pi = \sin (-n\pi) = 0$ and $\cos n \pi = (-1)^n$, we finally obtain

$$a_n = \displaystyle{ \left \{ \begin{array}{cl} \displaystyle... ... 0 & {\mbox for} n {\mbox even.} \\ \end{array}\right . }$$

Similarly, we calculate the coefficients $b_n$:

$b_n = \displaystyle{ \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin nx dx = \... ... (-1)\sin nx dx + \ \frac{1}{\pi} \int_{0}^{\pi} x \sin nx dx = \ }$

$= \displaystyle{\biggl. -\frac{1}{\pi}\biggl(- \frac{1}{n} \cos nx \biggr) \... ...\vert^{\pi}_{0} = \ \frac{1}{n\pi} \biggl(\cos 0 - \cos(-n\pi)\biggr) + }$

$+ \displaystyle{ \frac{1}{\pi}\biggl[\frac{1}{n^2} \sin n\pi - \frac{\pi}{n} \cos n\pi - 0 + 0 \biggr] = \ }$

$= \displaystyle{ \frac{1}{n\pi}\biggl( 1 - (-1)^n\biggr) + \frac{1}{\pi}\b... ...os n\pi \biggr) = \ \frac{1}{n\pi}\biggl( 1 - (-1)^n - \pi (-1)^n \biggr). }$

Thus,

$$b_n = \displaystyle{ \left \{ \begin{array}{cl} \displaystyl... ...}} & {\mbox for} n {\mbox even.} \\ \end{array}\right . }$$

To ensure that the Fourier series converges to $f(x)$ at every point $x$, we need to redefine $f$ at the points of discontinuity $\pm 2n\pi, n = 0,1,2, \ldots$ to the average value $f_{av} = -0.5$. Note that at the points $\pm (2n + 1)\pi, n = 0,1,2, \ldots$, $f(x) = f_{av} = 0$, and no change is required.

With $f(x)$ redefined, we have

$f(x) = \displaystyle{ - \frac{1}{2} + \frac{\pi}{4} + \sum_{{\mbox n \... ...ac{2}{\pi n^2} \cos nx + (\frac{2}{n\pi} + \frac{1}{n}) \sin nx \biggr) + }$

$+ \displaystyle{ \sum_{{\mbox n even}} (- \frac{1}{n}) \sin nx = - \frac... ...\frac{\pi}{4} - \frac{2}{\pi} \cos x \ + (\frac{2}{\pi} + 1) \sin x - }$

$${ - \frac{1}{2} \sin 2x - \frac{2}{9\pi} \cos 3x \ + (\frac{2}{3\pi} + \frac{1}{3}) \sin 3x - \frac{1}{4} \sin 4x - \ldots }$$