Physics Help Forum Fourier Series of (x)= abs(sin(x))

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 May 2nd 2008, 05:33 PM #1 Junior Member   Join Date: May 2008 Posts: 6 Fourier Series of (x)= abs(sin(x)) ............... Last edited by nairbdm; May 8th 2008 at 02:26 PM.
May 3rd 2008, 02:18 AM   #2
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 Originally Posted by nairbdm I need to find both the complex and the sine/cosine Fourier series for the function f(x)= abs(sin(x)) Any help would be greatly appreciated, Thanks.
Asked and replied to at MHF.

May 3rd 2008, 10:33 PM   #3
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 Originally Posted by nairbdm since the period of abs(sinx) is pi, should the integral for cn be: (1/pi) * integral from -pi/2 to pi/2 of abs(sinx) *e^(-2inx/pi) and if that is correct where should I go from there?
That looks OK. Now you have to integrate. Note |sin x| = -sin x for -pi/2 < x < 0.

May 7th 2008, 06:43 AM   #4
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 Originally Posted by nairbdm okay, now I got: Cn= [(-2in)(e^(-inpi)-e^(inpi))] / ((-2in)^2 + 1) can that simplifty any more?
The exponentials will simplify. Note that $\displaystyle e^{i n \pi} = \left( e^{i \pi} \right)^n$ and $\displaystyle e^{-i n \pi} = \left( e^{- i \pi} \right)^n$ ......

 Tags abssinx, fourier, series

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# serie de fourier abs(sin(x))

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