Find the fourier series of a piecewise continuous function

Question

Consider the function $f(x) = \begin{cases} 0 & & x \in [-\pi, 0) \\ 1 & & x \in [0, \frac{\pi}{2})\\ 0 & & x \in [\frac{\pi}{2}, \pi)\\ \end{cases}$ Calculate its fourier series

I am unsure if this is correct but i have:

$a_0=\frac{1}{2\pi} \int_{-\pi}^{\pi}f(x)dx=\frac{1}{2\pi} \int_{0}^{\frac{\pi}{2}}1dx=\frac{1}{2\pi}[x]_{0}^{\frac{\pi}{2}}=\frac{1}{4}$

$a_n=\frac{1}{\pi} \int_{-\pi}^{\pi}f(x)\cos(nx)dx=\frac{1}{\pi} \int_{0}^{\frac{\pi}{2}}\cos(nx)dx=\frac{1}{\pi}[\frac{1}{n}\sin(nx)]_{0}^{\frac{\pi}{2}}=\frac{1}{n\pi}$

$b_n=\frac{1}{\pi} \int_{-\pi}^{\pi}f(x)\sin(nx)dx=\frac{1}{\pi} \int_{0}^{\frac{\pi}{2}}\sin(nx)dx=\frac{1}{\pi}[\frac{1}{n}(-\cos(nx))]_{0}^{\frac{\pi}{2}}=\frac{-1}{n\pi}$

So we have $f(x)=\frac{1}{4}+\sum_{n=0}^{\infty}[\frac{1}{n\pi} \cos(nx) - \frac{1}{\pi} \sin(nx)]$

Is this correct?

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Answer 1

Looks OK to me, although you lost the 1/n term in the sine series.

As long as you can do the integration, you can find the Fourier series of any piecewise continuous function exactly as you have done here.

Answer 2

The setup is alright but the evaluations are not correct. For example $$ \dfrac{1}{\pi} \int_0^{\pi/2} \sin(nx) \; dx = \dfrac{1 - \cos(n \pi/2)}{n \pi} \neq - \dfrac{1}{n \pi} $$

Answer 3

It should be $$\\ { a }_{ 0 }=\frac { 1 }{ 2 } \\ { a }_{ n }=\frac { \sin { \left( \frac { \pi n }{ 2 } \right) } }{ n\pi } \\ { b }_{ n }=\frac { 1-\cos { \left( \frac { \pi n }{ 2 } \right) } }{ n\pi } \\ \\ \\ $$