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Fourier Series and the first term 
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#1
Mar2112, 09:15 AM

P: 29

I wasn't really sure where to post this because I am covering this in 2 classes (Math and Physics). Figured this would be my best bet.
The fourier series of some Function is [itex]a_{0}/2+etc...[/itex]. I've looked in several textbooks but none explain why the [itex]1/2[/itex] is there, and not in any of the other terms of the summation. I do have a homework problem concerning this, but my professor said it's ok to not explain this part of the Fourier series. I'm intrigued by this now, so I'd like to know. 


#2
Mar2112, 09:40 AM

Mentor
P: 18,036

Remember that the general term of a Fourier series is
[tex]<f,e_i>[/tex] where the [itex]e_i[/itex] is normalized (has norm 1) We want to define [tex]a_n=\frac{1}{\pi}\int_{\pi}^\pi f(x)\cos(nx)dx[/tex] If n>0, then this holds. The [itex]1/\pi[/itex] comes from normalizing the cosine. That is, the above is actually equal to [tex]<f,\cos nx>[/tex] but cos(nx) does not have norm 1, but rather pi. So we must divide by pi to normalize. We want the formula for a_{n} to hold for n=0 as well. But in this case, we have [tex]<f,1>[/tex] and the 1 is not normalized and has norm 2pi. So in order to normalize the thing, we must divide by 2pi. Division by pi is already taken care of in the definition of a_{n}, so we must also divide by 2. 


#3
Mar2112, 09:51 AM

P: 29

So, why doesn't the 2 tag along with the rest of the terms? In this case, it seems like there is a piecewise function under conditions n=0 and n>0.
Sorry if I'm asking a dumb question. 


#4
Mar2112, 10:10 AM

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PF Gold
P: 4,500

Fourier Series and the first term
Integrating from pi to pi:
The function f=1 yields 2pi The function f=cos^{2}(nx) yields pi. 


#5
Mar2112, 03:38 PM

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#6
Mar2112, 06:07 PM

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PF Gold
P: 4,500




#7
Mar2512, 10:59 PM

P: 27

I watched a video by MIT's OCW for 18.03 when I was doing this a while back, and the prof derived the Fourier series and explained the 1/2 quite well



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