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Fourier series for (sin(x))^2 
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#1
Apr2705, 10:12 PM

P: 525

Hey guys i was working on an algorithm for one of my CS classes that included working out the fourier series for the function f(x) = (sin(x))^2. it's been a few years since i've done anything like this, so I did some googling to refresh my memory of how to determine the fourier coefficients, and after some integrals and calculations, came up with:
f(x) = (1/2) + (1/2)cos(2x) + (1/2)sin(2x) I graphed this along with f(x) = (sin(x))^2 and it looked different from it. I was wondering if someone can give me a correct answer to compare with what i came up with, or help with giving me a rough walkthru of the process. Thanks again. 


#2
Apr2705, 10:38 PM

HW Helper
P: 4,124

The answer you got is incorrect.
The two identities you need are: (sin(x))^2 + (cos(x))^2 =1 and cos(2x) = 2(cos(x))^2  1 Use both of these to solve for (sin(x))^2 in terms of cos(2x). 


#3
Apr2805, 09:41 AM

Sci Advisor
HW Helper
P: 3,031

f(x) = f(x) All the component functions must also be even functions. One of yours is not. You can often take advantage of symmetry to eliminate performing many of the integrals that have to be calculated to decompose a function. 


#4
Apr2805, 02:45 PM

P: 998

Fourier series for (sin(x))^2
No integrals at all required here. You can either use the identities learningphysics gave, or use the exponential form of sine:
[tex]\sin{x} = \frac{e^{ix}e^{ix}}{2i}[/tex] and just square it (and then convert back to sines and cosines). 


#5
Apr2805, 08:19 PM

Sci Advisor
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P: 3,031




#6
Apr2805, 09:41 PM

P: 998

Indeed, I wasn't criticizing your reply at all
If you want to find the fourier series for x^n, for example, that symmetry can help a lot! 


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