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Fourier Series Troubles

  1. Apr 6, 2012 #1
    1. The problem statement, all variables and given/known data
    Let [itex]f(x) = x^2 on [-\pi,\pi][/itex]. Computer the Fourier Coefficients of the 2π-periodic extension of f. Use Dirichlet's Theorem to determine where the Fourier Series of f converges. Use the previous two conclusions to show that [itex]\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}[/itex]


    2. Relevant equations
    My book (baby Rudin) uses
    [itex]c_n = \frac{1}{2\pi} \int_{-\pi}^\pi f(x) e^{-inx} dx[/itex]
    as the equation for the coefficients of the series.


    3. The attempt at a solution
    My main problem is the last part, but I'm sure it's rooted in another piece of the puzzle. I've found that
    [itex]c_n = \frac{2(-1)^n}{n^2}[/itex]
    for n ≠ 0 via Maple and
    [itex]c_0 = \frac{\pi^2}{3}[/itex]
    by computation.

    My book doesn't actually have anything I can see called "Dirichlet's Theorem", but I found a paper that said if f(x) was continuous then the series would converge to that point, otherwise it would converge to the midpoint of the jump discontinuity. Is this right?
    If so, then the series ought to converge at every point, even the endpoints, because x^2 is continuous and even. So then we'd have, because [itex]\exp(i n \pi) = (-1)^n[/itex], that'd cancel the (-1)^n in the series and thus given the equation
    [itex]\frac{\pi^2}{3} + 2\sum_{n=1}^\infty \frac{1}{n^2} = \pi^2 = f(\pi)[/itex]
    but that doesn't work out too well.


    Where'd I go wrong?

    This is the first of four almost identical problems, but if I can do one I can do them all. Thanks for any input.
     
  2. jcsd
  3. Apr 6, 2012 #2

    Dick

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    Check your value for c0. I think it's wrong.
     
  4. Apr 6, 2012 #3
    I'm positive it's correct.
    [itex]\frac{1}{2\pi} \left(\frac{\pi^3}{3} - \frac{(-\pi)^3}{3}\right) = \frac{1}{2\pi} * \frac{2\pi^3}{3} = \frac{\pi^2}{3}[/itex]
     
  5. Apr 6, 2012 #4

    Dick

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    Oh yeah, you are right. Now I'm thinking you forgot to sum over the negative values of n as well. How about that?
     
  6. Apr 6, 2012 #5
    Yea, I think that's totally the problem. In the book they kind of glossed over that the negatives disappeared at some point and a coefficient of 2 popped out front. But the negative sums would be the same as the positive sums here because c_n is even.

    Thanks for guiding me to my error.
     
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