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Series identity

  1. Jun 11, 2008 #1
    for every sequence of numbers a_n E_n is this identity correct ?

    [tex] \sum_{n= -\infty}^{\infty}a_n e^{2\pi i E_{n}}= \sum_{n= -\infty}^{\infty}a_n \delta (x-E_{n}) [/tex]
  2. jcsd
  3. Jun 11, 2008 #2

    matt grime

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    Clearly it isn't (and I assume that you mean to have an x in the exponent on the LHS).
  4. Jun 11, 2008 #3

    matt grime

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    Perhaps you meant to integrate the RHS over the real line?
  5. Jun 11, 2008 #4


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    That looked a little familiar to me so I dug out my signals notes and found the following:


    Here FT refers to fourier transform and FS fourier series. As for how "it can be shown", I have no idea.
  6. Jun 11, 2008 #5

    matt grime

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    I suspect I know who this poster is, and the same advice I've given repeatedly still applies: have you tried it for any examples? Eg a_i=0 for i=/=0 and a_0=1, E_0=0, then the LHS is 1 and the RHS is d(x) (d for delta)...
  7. Jun 11, 2008 #6


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    This is true

    [tex] \sum_{n=-\infty}^{+\infty} e^{i 2 \pi n x} = \sum_{k=-\infty}^{+\infty} \delta(x-k) [/tex]

    but to generalize it with arbitrary coefficients (that are placed on both sides) is not a true equality.
  8. Jun 12, 2008 #7
    rbj and matt were right only this

    \sum_{n=-\infty}^{+\infty} e^{i 2 \pi n x} = \sum_{k=-\infty}^{+\infty} \delta(x-k)
    [/tex] (1)

    is correct , however my question is if using Fourier analysis we could generalized to an identity

    \sum_{n=-\infty}^{+\infty}a_{n} e^{i 2 \pi n x} = \sum_{k=-\infty}^{+\infty} b_{n}\delta(x-k)

    where the a_n and b_n are related by some way , this is interesting regarding an article of Functional equation for Dirichlet series, using (1) the author was able to proof the functional equation for Riemann Zeta, my idea was to develop a functional equation for almost every dirichlet series to see where they have the 'poles'
  9. Jun 12, 2008 #8
    If we have in the general case

    [tex] \sum_{n=-\infty}^{+\infty}b_{n} e^{i 2 \pi n x} = D A(x) [/tex]

    Where A(x) is the partial sum of a_n and D is the derivative operator , in case A(x)=[x] we recover usual delta identity , then i believe we can calculate b_n by the Fourier integral

    [tex] b_n = \int_{0}^{1} dx DA(x) e^{-2i\pi x} [/tex]
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