anonymous
  • anonymous
Fourier Series Help!
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[f(x)= 0 -3\le x<-1, 3 -1\le x < 1, 0 1 \le x < 3\]
anonymous
  • anonymous
what's the question?
anonymous
  • anonymous
Solve for 3 non zero terms

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anonymous
  • anonymous
remind me, we have to integrate between these points, right?
anonymous
  • anonymous
Yes with formulas for ao, an, bn
anonymous
  • anonymous
\[ f(x) = a0 + \sum_{n=1}^{\infty}(an \cos nx +bn \sin nx)\]
anonymous
  • anonymous
Not familiar with it in that form but it looks similar yeah
anonymous
  • anonymous
so the question is : 0 + \[\int\limits_{-3}^{-1} (an \cos n x + bn \sin nx) dx \] for the first term right?
anonymous
  • anonymous
let me look
anonymous
  • anonymous
then 3 - (the integral between -1 and 1) , right?
anonymous
  • anonymous
for the second term.
anonymous
  • anonymous
\[ao= 1/2L \int\limits_{-L}^{L} f(x) dx\]
anonymous
  • anonymous
\[an= 1/L \int\limits_{-L}^{L} f(x)\cos* n \pi x/L dx\]
anonymous
  • anonymous
lol, yes! For the first one the interval is between [-3,-1) so you'll integrate and get the following form:\[= \int\limits_{-3}^{-1}a0 dx + \int\limits_{-3}^{-1}\sum_{n=1}^{\infty}(an \cos nx + bn \sin nx)dx\]
anonymous
  • anonymous
That is for solving a0 right? I'm not used to seeing sigma or it done with all the integrals combined.
anonymous
  • anonymous
\[=2 a0 +\sum_{n=1}^ {\infty} an \int\limits_{-3}^{-1} \cos nx dx + \sum_{n=1}^{\infty}bn \int\limits_{-3}^{-1}\sin nx dx\]
anonymous
  • anonymous
lol, that's the general form of fourier's series ^_^
anonymous
  • anonymous
then you can integrate cos nx and sin nx and treat n as csts :), then you'll get the answer
anonymous
  • anonymous
Hmm alright. Thanks
anonymous
  • anonymous
separate them and solve each one alone then combine them and you'll get: \[\int\limits_{-3}^{-1}\cos nx dx\] and then integrate : \[\int\limits_{-3}^{-1}\sin nx dx\] find the answer and combine it with : = 2ao + _____ + _____ ( the answers you'll get after integrating)
anonymous
  • anonymous
same thing goes with the other zeros ^_^ good luck! hope I made it easier for you :)
anonymous
  • anonymous
A little :) Thanks
anonymous
  • anonymous
lol, np ^_^ good luck.

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