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Fourier Series Problem

  • Thread starter Ratpigeon
  • Start date
  • #1
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Homework Statement


Use the Fourier series of
f(x) = { 1 |x|<a
{ 0 a<|x|<[itex]\pi[/itex]
for 0<a<[itex]\pi[/itex]
extended as a 2-Pi periodic function for x [itex]\in[/itex]R
to find
[itex]\sum[/itex] Sin2(na)/n2
2. Homework Equations [/b

I got that the fourier series of f(x) was
a/[itex]\pi[/itex]+[itex]\sum[/itex] (2/(m[itex]\pi[/itex]) sin(ma) sin(mx)


The Attempt at a Solution


I'm not sure what to do. I'm guessing that since its meant to use the Fourier series of f; it should be related to f(a), but f(a) is undefined, and the denominator is wrong...
Could anyone please point me in the right direction?
 

Answers and Replies

  • #2
vela
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Your f(x) is an even function, so you should have cosines instead of sines:
$$f(x) = \frac{a}{\pi} + \sum_{m=1}^\infty \frac{2}{m\pi} \sin ma \cos mx.$$ Now if you look at the sum you're trying to evaluate, you need another power of m on the bottom and you want to turn the cosine into a sine. Any idea of how to get that?
 
  • #3
gabbagabbahey
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it should be related to f(a), but f(a) is undefined
The formulas you are likely using for the Fourier Series of a function [itex]f(x)[/itex] require the function to be piecewise smooth and periodic over the Reals, as well as being defined at each discontinuity as equal to the mean value of the one-sided limits, so you require

[tex]f(x=\pm a) = \frac{1}{2} \left( \lim_{x \to \pm a^+}f(x) + \lim_{x \to \pm a^-}f(x) \right) = \frac{1}{2}[/tex]
 
  • #4
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I integrate it from zero to a to get the extra sine term?
 
  • #5
vela
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Sounds good. What do you get?
 
  • #6
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a Pi/2. Thanks for the help :)
 
  • #7
uart
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a Pi/2. Thanks for the help :)
Ok, you're on the right track but not quite there yet. I think there's another term.
 
  • #8
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I don't think there's another term - the 1/2 A0 term goes to zero when you integrate it from
-a to a.
 
  • #9
uart
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I don't think there's another term - the 1/2 A0 term goes to zero when you integrate it from
-a to a.
No actually it doesn't.

BTW. Previously you said you would integrate from zero to pi? Either way will work, but zero to pi is easier.
 
  • #10
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Integrating from zero to a got that it was Pi/2 (a+a^2/Pi); is that right?
Thanks for pointing it out
 
  • #11
uart
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Pi/2 (a+a^2/Pi); is that right?
Almost. Check your "signs".
 
  • #12
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Minus, sorry, thanks - I need sleep...
 
  • #13
uart
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Yep, minus.

And as a nice little "sanity" check, notice that the sum is now zero when [itex]a = \pi[/itex]. :smile:
 
  • #14
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Thank you :)
 

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