Example 6

Find the Fourier series representation of the following periodic function given on its one full period:

$$f(x) = \left \{ \begin{array}{rl} 1, & 0 < x < 2 \\ -1, & ... ...\ 0, & x = -2, 0 {\mbox and} 2.\\ \end{array}\right .$$

Solution:

The function has period 4, and for our calculations we need the half-period $L = 2$.

$a_0 = \displaystyle{\frac{1}{L} \int_{-L}^L f(x) dx = \ \frac{1}{2} \int_{-2}^0 (-1) dx + \frac{1}{2} \int_{0}^2 (1) dx = -1 + 1 = 0 ,}$

$a_n = \displaystyle{\frac{1}{L} \int_{-L}^L f(x) \cos \frac{n\pi x}{L} dx ... ... dx + \ \frac{1}{2} \int_{0}^2 1 \cdot \cos \frac{n\pi x}{2} dx = \ }$

$${ = \biggl. \frac{1}{2} (-\frac{2}{n \pi}) \sin(\frac{n \pi x... ...c{1}{2} \frac{1}{2n \pi} \sin(\frac{n \pi x}{2}) \biggr\vert^{2}_{0} = 0 ,}$$

and

$b_n = \displaystyle{\frac{1}{L} \int_{-L}^L f(x) \sin \frac{n\pi x}{L} dx ... ... dx + \ \frac{1}{2} \int_{0}^2 1 \cdot \sin \frac{n\pi x}{2} dx = \ }$

$${ = \biggl. \frac{1}{2} \frac{2}{n \pi} \cos(\frac{n \pi x}{2... ...2}{n\pi} - \ \frac{2}{n\pi} \cos (n \pi) = \frac{2}{n\pi}(1 - (-1)^n) ,}$$

whence

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

The resulting Fourier series representation for $f(x)$ is the following:

$f(x) = \displaystyle{ \sum_{{\mbox n odd}} \frac{4}{n\pi} \sin (\frac{n \pi ... ...\sin(\frac{\pi x}{2}) + \frac{1}{3} \sin(\frac{3\pi x}{2}) + \ldots) .}$