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Complex Fourier Series Problem

  1. Dec 13, 2016 #1
    Hi, I'm starting to studying fourier series and I have troubles with one exercises of complex fourier series with
    f(t) = t:
    $$t=\sum_{n=-\infty }^{\infty } \frac{e^{itn}}{2\pi }\int_{-\pi}^{\pi}t\: e^{-itn} dt$$
    $$t=\sum_{n=-\infty }^{\infty } \frac{cos(tn)+i\, sin(tn)}{2\pi }\int_{-\pi}^{\pi}t\: e^{-itn} dt$$
    $$t=\sum_{n=-\infty }^{\infty } \frac{cos(tn)+i\, sin(tn)}{2\pi }\: (2i)(\frac{\pi cos(\pi n)}{n}-\frac{sin(\pi n)}{n^{2}})$$
    $$t=\sum_{n=-\infty }^{\infty } \left ( \frac{sin(tn)sin(n\pi )}{n^{2}\pi }-\frac{sin(nt)cos(n\pi )}{n} \right )+i\left (\frac{ cos(tn)cos(\pi n)}{n}-\frac{cos(nt)sin(n\pi )}{n^{2}} \right )$$
    Because the imaginary part is a odd function only remains the term with n=0
    so:
    $$t=\sum_{n=-\infty }^{\infty } \left ( \frac{sin(tn)sin(n\pi )}{n^{2}\pi }-\frac{sin(nt)cos(n\pi )}{n} \right )+\lim_{n\rightarrow 0}\, \, i\left (\frac{ cos(tn)cos(\pi n)}{n}-\frac{cos(nt)sin(n\pi )}{n^{2}} \right )$$
    Because the real part is a even function we can transform it into this:
    $$t=2\sum_{n=1 }^{\infty } \left ( \frac{sin(tn)sin(\pi n )}{n^{2}\pi }-\frac{sin(nt)cos(n\pi )}{n} \right )+\lim_{n\rightarrow 0}\, \, \left ( \frac{sin(tn)sin(n\pi )}{n^{2}\pi }-\frac{sin(nt)cos(n\pi )}{n} \right)+$$
    $$+\lim_{n\rightarrow 0}\, \, i\left (\frac{ cos(tn)cos(\pi n)}{n}-\frac{cos(nt)sin(n\pi )}{n^{2}} \right )$$
    the first limit is 0 and in the sum we can delete the term with contains ##sin(\pi n )## and get:
    $$t=-2\sum_{n=1 }^{\infty }\frac{sin(nt)cos(n\pi )}{n}+\lim_{n\rightarrow 0}\, \, i\left (\frac{ cos(tn)cos(\pi n)}{n}-\frac{cos(nt)sin(n\pi )}{n^{2}} \right )$$
    $$t=-2\sum_{n=1 }^{\infty }(-1)^{n}\frac{sin(nt)}{n}+\lim_{n\rightarrow 0}\, \, i\left (\frac{ cos(tn)cos(\pi n)}{n}-\frac{cos(nt)sin(n\pi )}{n^{2}} \right )$$

    this is right if the limit is equal to 0 but is undefined so where is the error?
     
  2. jcsd
  3. Dec 13, 2016 #2
    Ok well, sorry just ignore what I said I thought the limit is with n to infinity.
     
  4. Dec 13, 2016 #3

    Charles Link

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    Homework Helper

    Suggestion is to treat the n=0 case separately at the beginning, so that you don't have these false n's (with n=0) showing up in the denominator. The n=0 term integrates to zero on the right integral at the very top.
     
  5. Dec 13, 2016 #4
    .-. You are right, That's the way to banish the term n=0.
    until the 3 step the n=0 case works fine but after expanding all,the 4 step ruined everything.
    other question:
    why expanding things make mistakes? , I mean is algebra, you have some example where expanding make mistakes?
    thanks =D
     
  6. Dec 14, 2016 #5

    nrqed

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    Gold Member

    Well, the limit is not undefined. In the limit n goes to zero, ##\cos n \pi/n ## goes to 1/n plus a correction of order n and higher which go to zero, and ##sin(n\pi)/n^2 ## goes to ##\pi/n## plus corrections of order n and higher. So the two terms you had cancel out.

    if you don't like this limiting procedure, then the answer is that the integration and the limit ##n ## goes to zero do not commute. Strictly speaking, your integral of ##t e^{-itn}## is valid only for ##n \neq 0##. For n=0 you have to set n=0 before doing the integration. But I prefer to think of the integral as being correct for any n and that the limit n->0 is perfectly ok.
     
  7. Dec 14, 2016 #6
    Thanks , I guess I have to keep solving exercises
     
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