Combining Fourier Series for Step Function: $\phi_{n} = \sin(nx)$

[stunner5000pt]

Find the FOurier Series in terms of \phi_{n} = \sin(nx) of the step function

f(x) = 0 for 0 \leq x \leq \frac{1}{2} \pi [/tex]<br /> f(x) =1 for \frac{1}{2} \pi &amp;lt; x \leq \pi<br /> <br /> now i have no problem finding the series for each branch. But how would i combine them?<br /> <br /> for the 0 to 1/2 pi<br /> \frac{4}{\pi} \sum_{n=1}^{\infty} \sin nx<br /> for the 1/2 pi to pi<br /> \frac{-4}{\pi} \sum_{n=1}^{\infty} \frac{1}{n} \left((-1)^n + \cos(\frac{n \pi}{2}) \right)<br /> <br /> please help me on combining the two! <br /> <br /> Thank you for your help

 

[HallsofIvy]

Strictly speaking, you are not finding the Fourier series for two different functions, you are finding the c_n in \Sum c_n sin(nx) by integrating a single function.
And, I might point out, you have NOT found the Fourier series on x between 0 and \frac{\pi}{2}. f(x)= 0 there so the Fourier series is just 0.
c_n= \frac{2}{\pi}\int_0^\pi f(x)sin(nx)dx= \frac{2}{/pi}\left(\int_0^{\frac{\pi}{2}}0 sin(nx)dx+ \int_{\frac{\pi}{2}}^\pi 1 sin(nx)dx\right)

That is just
c_n= \frac{2}{\pi}\int_{\frac{\pi}{2}}^\pi sin(nx)dx