- #1

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

- Thread starter mhill
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- #1

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[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

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).

- #3

matt grime

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Perhaps you meant to integrate the RHS over the real line?

- #4

Defennder

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

http://img136.imageshack.us/img136/3605/impulsetrainqr8.jpg [Broken]

Here FT refers to fourier transform and FS fourier series. As for how "it can be shown", I have no idea.

http://img136.imageshack.us/img136/3605/impulsetrainqr8.jpg [Broken]

Here FT refers to fourier transform and FS fourier series. As for how "it can be shown", I have no idea.

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- #5

matt grime

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- #6

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[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.

- #7

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[tex]

\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

[tex]

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

[/tex]

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'

- #8

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[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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