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Finding the Fourier Sine Series

  1. Nov 3, 2013 #1
    Find the Fourier SIne Series for f(x) = x on -L < x < L (Full Fourier)

    Ok, so my issue is in calculating the coefficients for the sine and cosine parts, more so an interpretation. So I have calulated the sine and cosine series to this point:

    let An: Cosine series Bn: sine series:

    An = l/(n[itex]\pi[/itex])) [ sin(n[itex]\pi[/itex] + sin(-n[itex]\pi[/itex])] - l/(n2[itex]\pi[/itex]2) [sin(n[itex]\pi[/itex]) - sin(-n[itex]\pi[/itex])]

    Bn = l/(n[itex]\pi[/itex])) [ cos(n[itex]\pi[/itex] + cos(-n[itex]\pi[/itex])] - l/(n2[itex]\pi[/itex]2) [cos(n[itex]\pi[/itex]) - cos(-n[itex]\pi[/itex])]

    My questions are these: How do I interpret the relationship between cos(n[itex]\pi[/itex]) and cos(-n[itex]\pi[/itex])? and the same with the sine values?

    My take is this: for the sine function, since all of the sine coeffcients have a [itex]\pi[/itex] in them then it doesn't matter what value I select for "n", they will all go to 0.

    As for the cosine: Since cos(n[itex]\pi[/itex]) and cos(-n[itex]\pi[/itex]) end up going to the same value, just in opposite directions, I will have 2cos(ncos(n[itex]\pi[/itex]) since cos(n[itex]\pi[/itex]) = cos(-n[itex]\pi[/itex]) in values.
  2. jcsd
  3. Nov 3, 2013 #2


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    Yes. ##\sin n\pi = 0## for any integer ##n##, so just leave those terms out of your series.

    I think you mean ##\cos(n\pi) + \cos(-n\pi) = 2\cos(n\pi)## and ##\cos(n\pi) - \cos(-n\pi) = 0##. Also sometimes you will see ##\cos(n\pi)=(-1)^n##.
  4. Nov 3, 2013 #3

    Yes that actually brings up anouther question I had with regards to ##\cos(n\pi)=(-1)^n##

    I noticed that they expressed a couple of final results using that expression of (-1)n. Why is that?
  5. Nov 3, 2013 #4


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    I don't understand your question. They are equal.
  6. Nov 3, 2013 #5
    It's ok I figured it out......there was another example with and extra (-1) that they didn't include initially so: (-1)(-1)n = (-1)n+1
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