Question

. Determine the Fourier series for the periodic

function:

f (x) =  −

2

,

when

−π <

x

< 0

+

2

,

when 0

< x

< π

which is periodic outside this range of

period 2π.

Answer 1

$\text{Given the function: } f(x) = \begin{cases} -2, \text{ when } & - \pi < x < 0 \\ 2, \text{ when } & 0 < x < \pi\end{cases} \\ \text{The fourier series will be of the form: } \\ \displaystyle \frac12a_0 + \sum_{n = 1}^{\infty} [a_n \cos nx + b_n \sin nx] \\ \text{Where: } \\ \displaystyle a_0 = \frac1\pi \int_{-\pi}^{\pi} f(x) dx = \frac1\pi \int_{-\pi}^{0} f(x) dx +\frac1\pi \int_{0}^{\pi} f(x) dx = \frac1\pi \int_{-\pi}^{0} -2 \,dx + \frac1\pi \int_{0}^{\pi} 2\, dx \\ \quad = \left[ \frac{-2x}{\pi} \right]_{-\pi}^{0} + \left[ \frac{2x}{\pi} \right]_{0}^{\pi} = 0 \\ \displaystyle a_n = \frac1\pi \int_{-\pi}^{\pi} f(x)\cos nx \, dx = \frac1\pi \int_{-\pi}^{0} f(x)\cos nx \,dx +\frac1\pi \int_{0}^{\pi} f(x)\cos nx \, dx \\ \quad = \frac1\pi \int_{-\pi}^{0} -2\cos nx \,dx + \frac1\pi \int_{0}^{\pi} 2\cos nx\, dx \\ \quad = \left[ \frac{-2\sin nx}{n\pi} \right]_{-\pi}^{0} + \left[ \frac{2\sin nx}{n\pi} \right]_{0}^{\pi} = \frac{4 \sin n \pi}{n \pi} = 0 \,\, \forall \, n\in \N \\ \displaystyle b_n = \frac1\pi \int_{-\pi}^{\pi} f(x)\sin nx \, dx = \frac1\pi \int_{-\pi}^{0} f(x)\sin nx \,dx +\frac1\pi \int_{0}^{\pi} f(x)\sin nx \, dx \\ \quad = \frac1\pi \int_{-\pi}^{0} -2\sin nx \,dx + \frac1\pi \int_{0}^{\pi} 2\sin nx\, dx \\ \quad = \left[ \frac{2\cos nx}{n\pi} \right]_{-\pi}^{0} + \left[ \frac{-2\cos nx}{n\pi} \right]_{0}^{\pi} = \frac{2}{n\pi} - \frac{2\cos n\pi}{n\pi} - \frac{2\cos n\pi}{n\pi} + \frac{2}{n\pi} \\ \quad = \frac{4 - 4\cos n\pi}{n\pi} \\ \displaystyle b_n = \begin{cases} 0 & \text{ when n is even} \\ \frac{8}{n\pi} & \text{ when n is odd} \end{cases} \\ \text{we can just write $n$ as $n = 2k - 1$ where $k \in \N$} \\ \text{So thefore the fourier series for $f(x)$ is given as: } \\ \displaystyle f(x) = \frac8\pi \sum_{k = 1}^{\infty} \frac{\sin (2k-1)x}{2k-1}$