FS5

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Find the Fourier series for f(x) = \begin{cases}-1 & -3 \le x < 0\\ 1&0<x\le 3\end{cases} on [-3,3]\,

a_k = \frac{1}{3} \int_{-3}^3 f(x) \cos(\frac{k\pi x}{3})\,dx = \frac{1}{3} \int_{-3}^0 -\cos(\frac{k\pi x}{3})\,dx + \frac{1}{3}\int_0^3 \cos(\frac{k \pi x}{3})\,dx\,

= \frac{1}{k\pi} \left[\sin(\frac{k\pi x}{3})\right]_{-3}^0 - \frac{1}{k\pi}\left[\sin(\frac{k\pi x}{3})\right]_0^3=0\,

a_0 = \frac{1}{3}\int_{-3}^0 -\,dx + \frac{1}{3}\int_0^3 \,dx = -1 + 1 = 0\,

b_k = \frac{1}{3}\int_{-3}^3 f(x) \sin(\frac{k\pi x}{3})\,dx = \frac{1}{3}\int_{-3}^0 -\sin(\frac{k\pi x}{3})\,dx + \frac{1}{3}\int_0^3 \sin(\frac{k\pi x}{3})\,dx \,

= \frac{1}{k\pi} \left[\cos(\frac{k\pi x}{3})\right]_{-3}^0 - \frac{1}{k\pi}\left[\cos(\frac{k\pi x}{3})\right]_0^3\,

= \frac{1}{k\pi} - \frac{(-1)^k}{k\pi} - \frac{(-1)^k}{k\pi} + \frac{1}{k\pi} = \frac{2-2(-1)^k}{k\pi}\,\,\,k=1,2,3,...\,

= \frac{4}{(2k+1)\pi},\,\,\,k=0,1,2,...\,

So the Fourier series is:

f(x) = \sum_{k=0}^\infty \frac{4}{(2k+1)\pi}\sin\left(\frac{(2k+1)\pi x}{3}\right)\,


Fourier Series

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