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Poisson's summation formula

  1. Jun 13, 2009 #1
    I am trying to understand the derivation of the Poisson's sum formula. Wikipedia's article is like crosswords to me. I checked mathworld's take on it. It looked simple, but it stated that the equation is derived from a more general result without stating or proving that general result. Here's the link. Can you please tell me how that generalisation is derived? thank u.
    link:

    http://mathworld.wolfram.com/PoissonSumFormula.html
     
  2. jcsd
  3. Jun 13, 2009 #2
    can you at least tell me under which category in mathematics does this topic lie so that I know what kind of book I'm searching for.?
     
  4. Jun 13, 2009 #3
    It is very easy to derive the formula using Fourier series.
     
  5. Jun 13, 2009 #4
    Consider the functions:

    [tex]e_{k}(x) = \exp(2 \pi i k x)[/tex]

    For functions defined on the interval from n-1/2 to n + 1/2 with n some arbitrary integer, we define the inner product:

    [tex]\langle f,g\rangle = \int_{n - 1/2}^{n +1/2} f(x)g^{*}(x)dx[/tex]

    Then we see that:

    [tex]\langle e_k, e_r\rangle = \delta_{k,r}[/tex]

    where [itex]\delta_{k,r}[/itex] is the Kronecker delta, which is zero if the indices are differet and 1 if the indices are equal. This means that the [itex]e_k[/itex] form an orthonormal basis of functions. It can be shown that these functions also form a complete set which then implies that we can expand any function f as:


    [tex]f(x) = \sum_{k=-\infty}^{\infty} \langle f, e_{k}\rangle e_{k}(x) [/tex]

    [Note that this is completly analogous to how you can expand any vector in terms of basis vectors. If V is a 3d vector and ex, ey and ez are the unit vectors in the x, y, and z, directions, then V = <V,ex>ex +<V,ey> + <V,ez>ez. The inner products <V,ei> are, of course, the components of V in the ith direction.]

    At x = n, we have

    [tex]e_{k}(n)= \exp(2 \pi i k n) = 1[/tex],

    so we have:


    [tex]f(n) = \sum_{k=-\infty}^{\infty} \langle f, e_k\rangle =
    \sum_{k=-\infty}^{\infty}\int_{n-1/2}^{n+1/2} f(x) \exp(-2 \pi i k x) dx[/tex]


    Now, sum both sides over n from minus infinity to infinity:

    [tex]\sum_{n=-\infty}^{\infty}f(n) = \sum_{n=-\infty}^{\infty}\sum_{k=-\infty}^{\infty}\int_{n-1/2}^{n+1/2} f(x) \exp(-2 \pi i k x) dx[/tex]


    Interchange the summations over n and k:

    [tex]\sum_{n=-\infty}^{\infty}f(n) = \sum_{k=-\infty}^{\infty}\sum_{n=-\infty}^{\infty}\int_{n-1/2}^{n+1/2} f(x) \exp(-2 \pi i k x) dx[/tex]


    Summing over n yields

    [tex]\sum_{n=-\infty}^{\infty}\int_{n-1/2}^{n+1/2}f(x)\exp(-2\pi i k x) dx=\int_{-\infty}^{\infty}f(x)\exp(-2\pi i k x)dx=\hat{f}(k)[/tex]

    So, we have:

    [tex]\sum_{n=-\infty}^{\infty}f(n)=\sum_{k=-\infty}^{\infty}\hat{f}(k)[/tex]
     
    Last edited: Jun 14, 2009
  6. Jun 14, 2009 #5
    Count thank you greatly but I cannot understand through the language of the proof can you use latex? please? :)
     
  7. Jun 14, 2009 #6

    I've edited the posting.
     
  8. Jun 14, 2009 #7
    "For functions defined on the interval from n-1/2 to n + 1/2 with n some arbitrary integer, we define the inner product:

    LaTeX Code: \\langle f,g\\rangle = \\int_{n - 1/2}^{n +1/2} f(x)g^{*}(x)dx

    Then we see that:

    LaTeX Code: \\langle e_k, e_r\\rangle = \\delta_{k,r}

    where LaTeX Code: \\delta_{k,r} is the Kronecker delta, which is zero if the indices are differet and 1 if the indices are equal. This means that the LaTeX Code: e_k form an orthonormal basis of functions. It can be shown that these functions also form a complete set which then implies that we can expand any function f as:"


    I did not understand something here, what is the nature of g(x)? is it an arbitrary function or not? And how exactly did you deduce tht it is a kronecker's delta? the product of the two exponentials indeed will be 1 if the indices are the same but how will it always be zero if they arent? it could be exp (2pix (k-n)) for example..
     
  9. Jun 16, 2009 #8
    I do not understand how <ek , er> results in a delta function. Isnt the integrand going to be e^(2pi i k) * e^(-2pi i r). While this is equal to 1 if k = r, it isnt necessarily 0 if r isnt equal to k. How did u interpret it as delta? please clarify.
     
  10. Jun 16, 2009 #9

    Redbelly98

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    Staff Emeritus
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    Equivalently, the integrand is e^[2 pi i (k-r)].

    (k-r) is an integer, and the integral is done over a range of 1 (from n-½ to n+½). So the integral is done over an integer number of periods of the function e^[2 pi i (k-r)]. Therefore, the integral is zero.
     
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