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Series identity

by mhill
Tags: identity, series
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mhill
#1
Jun11-08, 08:01 AM
P: 193
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]
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matt grime
#2
Jun11-08, 08:05 AM
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Clearly it isn't (and I assume that you mean to have an x in the exponent on the LHS).
matt grime
#3
Jun11-08, 09:15 AM
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Perhaps you meant to integrate the RHS over the real line?

Defennder
#4
Jun11-08, 10:37 AM
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P: 2,616
Series identity

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.
matt grime
#5
Jun11-08, 01:10 PM
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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)...
rbj
#6
Jun11-08, 03:11 PM
P: 2,251
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.
mhill
#7
Jun12-08, 02:49 AM
P: 193
rbj and matt were right only this

[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'
mhill
#8
Jun12-08, 05:13 AM
P: 193
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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