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Math Help - Fourier series

  1. #1
    MHF Contributor arbolis's Avatar
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    Fourier series

    I posted a very similar exercise a few days ago. I think this one is a bit easier.
    Let f(t)=0 if - \pi < t \leq 0, 1 if 0<t \leq \pi.
    1)Calculate \hat f (n) for all n \in \mathbb{Z}.
    2)Calculate \sum _{n=-\infty}^{\infty} |\hat f (n)|^2.
    3)Study the pointwise convergence of the Fourier series of f.

    Attempt: 1)For all n \neq 0, I get that \hat f(n)=\frac{i(-1)^n}{2 \pi n}. For n=0, \hat f(0)=\frac{1}{2} which is consistent with the average of the right and left limit of the function at t=0.
    2) \sum _{n=-\infty}^{\infty} |\hat f (n)|^2 =||f||^2 = \lim _{N \to \infty} ||S_N f||^2+ ||f- S_N f||^2=\frac{1}{2 \pi} \int _{-\pi}^{\pi} |f(t)|^2 dt. So I think I should calculate the Fourier series of f.
    It gave me f(t)=\frac{1}{2}+ \sum _{n=-\infty}^{-1} \frac{i(-1)^n}{2 \pi n} \left [  \cos (tn) - i \sin (tn)  \right ] + \sum _{n=1}^{+\infty} \frac{i(-1)^n}{2\pi n} \left [  \cos (tn) - i \sin (tn)  \right ].
    So that ||S_N f||^2 =\frac{1}{2\pi} \int _{- \pi}^{\pi} \left | \sum _{n=-N} ^{n=N} _{n \neq 0} \frac{i(-1)^n}{2 \pi n} + \frac{1}{2} \right | ^2.
    The algebra really get messy... Too messy. I suppose there is an easier way to solve the question.
    3)Maybe I should use the Dirichlet–Dini Criterion (that I found in Convergence of Fourier series - Wikipedia, the free encyclopedia). But we never studied it, so I'm guessing there's another way.


    Edit: For 2), if I plug directly the expression of \hat f (n) into the infinite series, it diverges since I fall over 2 harmonic series.
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  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    For 2) use Parseval's Identity

    \displaystyle \sum_{n=-\infty}^{\infty}|\hat{f}(n)|^2=\frac{1}{2\pi}\int_  {-\pi}^{\pi}|f(x)|^2dx=\int_{-\pi}^{0}|0|^2dx+\int_{0}^{\pi}|1|^2dx=...

    Anywhere the f(x) is differentiable the series will converge point wise. So for this series everywhere but 0.
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  3. #3
    MHF Contributor arbolis's Avatar
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    Oh... now it seems so simple!
    Did you forget a 1/(2 pi ) factor in the right side?
    So that \sum_{n=-\infty}^{\infty}|\hat{f}(n)|^2=\frac{1}{2\pi} \int _{0}^{\pi} dx=\frac{1}{2}, or I'm missing something and \sum_{n=-\infty}^{\infty}|\hat{f}(n)|^2=\pi?
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  4. #4
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    You are correct it got lost in translation.
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  5. #5
    MHF Contributor arbolis's Avatar
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    Thanks for all. My exam (which is worth 100% of the grade for this course) starts in about 11 hours now. I'm very stressed and will reread my whole course and stop to do exercises now.
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