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

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SUMMARY

The discussion focuses on deriving the Fourier series for a step function defined as \( f(x) = 0 \) for \( 0 \leq x \leq \frac{1}{2} \pi \) and \( f(x) = 1 \) for \( \frac{1}{2} \pi < x \leq \pi \). The Fourier coefficients are calculated using the formula \( c_n = \frac{2}{\pi} \int_0^\pi f(x) \sin(nx) dx \), leading to \( c_n = \frac{2}{\pi} \int_{\frac{\pi}{2}}^\pi \sin(nx) dx \). The series for each segment is given as \( \frac{4}{\pi} \sum_{n=1}^{\infty} \sin(nx) \) for \( 0 \leq x \leq \frac{1}{2} \pi \) and \( \frac{-4}{\pi} \sum_{n=1}^{\infty} \frac{1}{n} \left((-1)^n + \cos(\frac{n \pi}{2}) \right) \) for \( \frac{1}{2} \pi < x \leq \pi \). The key takeaway is the integration of the single function across the specified intervals to combine the series.

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

f(x) = 0 for [itex]0 \leq x \leq \frac{1}{2} \pi [/tex]<br /> f(x) =1 for [itex]\frac{1}{2} \pi < x \leq \pi[/itex]<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 /> [tex]\frac{4}{\pi} \sum_{n=1}^{\infty} \sin nx[/tex]<br /> for the 1/2 pi to pi<br /> [tex]\frac{-4}{\pi} \sum_{n=1}^{\infty} \frac{1}{n} \left((-1)^n + \cos(\frac{n \pi}{2}) \right)[/tex]<br /> <br /> please help me on combining the two! <br /> <br /> Thank you for your help[/itex]
 
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Strictly speaking, you are not finding the Fourier series for two different functions, you are finding the cn in [itex]\Sum c_n sin(nx)[/itex] by integrating a single function.
And, I might point out, you have NOT found the Fourier series on x between 0 and [itex]\frac{\pi}{2}[/itex]. f(x)= 0 there so the Fourier series is just 0.
[tex]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)[/tex]

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

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