FS9

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Find the Fourier series for f(t) = \begin{cases}\frac{4}{\pi}t & 0 \le t < \frac{\pi}{2},\\ \frac{-4}{\pi}t & \frac{-\pi}{2} \le t \le 0\end{cases}

This is the general Fourier Series:

f(x) = {a_0\over 2} + \sum_{n=1}^\infty \left[a_n\cos({2n\pi x \over T}) + b_n\sin({2n\pi x\over T})\right]\,

a_n = {2\over T}\int_c^{c+T} f(x) \cos({2n\pi x\over T})\,dx\,

b_n = {2\over T}\int_c^{c+T} f(x) \sin({2n\pi x\over T})\,dx\,

So the given function can be replaced by its Fourier expansion:

f(t) = {a_0\over 2} + \sum_{n=1}^\infty\left[a_n \cos\left(\frac{2n\pi x}{\pi}\right) + b_n \sin\left(\frac{2n\pi x}{\pi}\right)\right]\,

a_0 = {2\over\pi} \int_{\frac{-\pi}{2}}^0 \frac{-4}{\pi}t\,dt + {2\over\pi} \int_0^\frac{\pi}{2} \frac{4}{\pi} t\,dt=2\,

a_n = \frac{2}{\pi} \int_{-\pi\over 2}^0 \frac{-4}{\pi}t \cos(2n t)\,dt + \frac{2}{\pi}\int_0^\frac{\pi}{2} \frac{4}{\pi}t\cos(2n t)\,dt = \frac{4((-1)^n-1)}{\pi^2 n^2}\,

b_n = \frac{2}{\pi}\int_{\frac{-\pi}{2}}^0 \frac{-4}{\pi}t \sin(2n t)\,dt + \frac{2}{\pi} \int_0^{\frac{\pi}{2}} \frac{4}{\pi}t \sin(2n t)\,dt=0\,

So the solution is

f(t) = 1+\sum_{n=1}^\infty \left[ \frac{4}{n^2\pi^2} ((-1)^n-1) \cos(2n t)\right]\,

= 1+\sum_{n=1}^\infty \left[ \frac{-4\cdot 2}{(2n-1)^2\pi^2} \cos(2(2n-1) t)\right]\,

= 1-\frac{8}{\pi^2}\sum_{n=1}^\infty \frac{\cos\left[2(2n-1)t\right]}{(2n-1)^2}\,

Fourier Series

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