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Math Help - Poisson summation formula

  1. #1
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    Poisson summation formula

    I just have no idea how to apply Poisson summation formula. Can someone give some help?
    a) Let \tau be fixed with Im(\tau)>0. Apply the Poisson summation formula to f(z)=(\tau +z)^{-k} where k \geq 2 to obtain
    \sum_{n=-\infty}^{\infty} \frac {1}{(\tau+n)^k}= \frac{(-2\pi i)^k}{(k-1)!} \sum_{m=1}^{\infty} m^{k-1} e^{2 \pi im \tau}
    b) Does this formula still hold whenever \tau is any complex number that is not an integer?
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  2. #2
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    Quote Originally Posted by namelessguy View Post
    I just have no idea how to apply Poisson summation formula. Can someone give some help?
    a) Let \tau be fixed with Im(\tau)>0. Apply the Poisson summation formula to f(z)=(\tau +z)^{-k} where k \geq 2 to obtain
    \sum_{n=-\infty}^{\infty} \frac {1}{(\tau+n)^k}= \frac{(-2\pi i)^k}{(k-1)!} \sum_{m=1}^{\infty} m^{k-1} e^{2 \pi im \tau}
    b) Does this formula still hold whenever \tau is any complex number that is not an integer?
    As for how to apply Poisson sommation formula, everything is given in the text: they even give the function to use. The formula says:
    \sum_{n\in\mathbb{Z}} f(n)=\sum_{n\in\mathbb{Z}} \widehat{f}(n),
    provided the second sum is absolutely convergent. And if the Fourier transform is defined as \widehat{f}(\xi)=\int e^{-2i\pi x\xi}f(x)dx.

    The left-hand side is exactly what you want, so you have to check that the right-hand side matches the formula you're given. In other words, you have to compute the Fourier transform of f.

    This can be tedious, but there are tricks:
    first, notice that f(x)=\frac{(-1)^{k-1}}{(k-1)!}\frac{d^{k-1}}{dx^{k-1}}\left(\frac{1}{\tau+x}\right), hence \widehat{f}(\xi)=\frac{(-1)^{k-1}}{(k-1)!}(2i\pi\xi)^{k-1}\mathcal{F}\left(\frac{1}{\tau+x}\right)(\xi). You have to be careful here, since x\mapsto \frac{1}{\tau+x} is not integrable, but only square-integrable.
    Then you can notice that (by easy integration) \mathcal{F}^{-1}\left(H(\xi)e^{2i\pi \xi\tau}\right)(x)=\frac{1}{-2i\pi(\tau+x)} where H is the Heavyside function ( H(x)=1 if x\geq 0 and H(x)= 0 else). As a consequence, using Fourier inversion formula (in L^2), \mathcal{F}\left(\frac{1}{\tau+x}\right)(\xi)=(-2i\pi)H(\xi)e^{2i\pi \xi\tau}.
    We deduce \widehat{f}(\xi)=\frac{(-1)^{k-1}}{(k-1)!}(2i\pi\xi)^{k-1}(-2i\pi)e^{2i\pi\xi\tau}H(\xi) in L^2 (hence for almost every \xi), and hence for every \xi because both functions are continuous (even for \xi=0 because k\geq 2).

    As a consequence, \widehat{f}(n)=\frac{(-2i\pi)^k}{(k-1)!}n^{k-1}e^{2i\pi n\tau} if n> 0 and 0 otherwise.

    The series \sum_n \widehat{f}(n) is absolutely convergent because |e^{2i\pi n\tau}|=e^{{\rm Re}(2i\pi n\tau)}=e^{-2\pi n {\rm Im}(\tau)} and {\rm Im}(\tau)>0. That's why we can apply the Poisson sommation formula. And it gives exactly what is expected.

    By the way, if {\rm Im}(\tau)\leq 0, the series would diverge.
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  3. #3
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    Thanks a lot for your help Laurent. I really need to review this Fourier Transform materials. I wasn't able to apply this concept to show that \sum_{n=-\infty}^{\infty} \frac {1}{(\tau+n)^2}=\frac{\pi^2}{sin^2(\pi \tau)}. I plug in k=2 on both sides of the formula, but got nowhere.
    I then tried an alternative way suggested by my friend. I integrate f(z)=\frac{\pi cot \pi z}{(\tau + n)^2} over the circle \mid z \mid=R_N. I almost got the result (still working on showing cot \pi z is bounded in that circle).
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  4. #4
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    Quote Originally Posted by namelessguy View Post
    Thanks a lot for your help Laurent. I really need to review this Fourier Transform materials. I wasn't able to apply this concept to show that \sum_{n=-\infty}^{\infty} \frac {1}{(\tau+n)^2}=\frac{\pi^2}{sin^2(\pi \tau)}. I plug in k=2 on both sides of the formula, but got nowhere.
    For this, you should compute \sum_{m=1}^\infty m x^m where x=e^{2i\pi\tau}. This sum is obtained by differentiating \sum_{m=0}^\infty x^m=\frac{1}{1-x}.
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