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Does Fourier series of x^2 converge?

  1. May 4, 2012 #1
    I'm trying to show that the Fourier series of [itex] f(x)=x^2[/itex] converges and I can't. Does anybody know if it actually does converge? (I'm assuming that [itex] f(x)=x^2[/itex] for [itex] x\in [-\pi,\pi][/itex]).
    The Fourier Series itself is [itex] \displaystyle\frac{\pi^2}{3}+4\sum_{n=1}^\infty \frac{(-1)^n}{n^2}\cos nx [/itex]
    I tried using Dirichlet's test but it wasn't working for me, though that may be because I'm doing something wrong.
  2. jcsd
  3. May 4, 2012 #2
    Isn't it easy to show absolute convergence?? Since [itex]\sum \frac{1}{n^2}[/itex] converges

    Do you also want to show that it converges to [itex]x^2[/itex]?? There are many theorems out there that give you that, so it depends on what you have seen.
  4. May 4, 2012 #3
    I was just looking to see if it converged. So yes, you're absolutely right. I was going well off-beam with my attempt.

    And about converging to [itex]x^2[/itex]; I guess it does so uniformly since the function is continuous on the circle and the Fourier series converges absolutely.

    Thank you!
  5. May 4, 2012 #4
    It's not because a function is continuous and because the Fourier series converges absolutely, that you can have uniform convergence (I think).
    Here, I think you can infer uniform convergence from the Weierstrass M-test.
  6. May 4, 2012 #5
    No I think it is. I quote Corollary 2.3 from "Fourier Analysis" by Stein and Shakarchi -

    "Suppose that [itex]f [/itex] is a continuous function on the circle and that the Fourier series of [itex]f [/itex] is absolutely convergent, [itex]\sum_{n=-\infty}^\infty |\hat{f}(n)|<\infty. [/itex] Then, the Fourier series converges uniformly to [itex]f[/itex], that is
    [itex]\displaystyle \lim_{N\to\infty}S_N(f)(\theta)=f(\theta) [/itex] uniformly in [itex]\theta. [/itex]
  7. May 4, 2012 #6
    Oh ok, I did not know that result. Nice!!
  8. May 4, 2012 #7
    Yeah it's pretty sweet. It's not too restrictive.
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