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Summation help

by FrogPad
Tags: summation
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FrogPad
#1
Apr22-07, 07:02 PM
P: 837
I don't see how the following works:

[tex] \sum_{n=0}^\infty \delta ( n - n_0 ) z^{-n} = z^{-n_0} [/tex]

I am missing the steps from [itex] \sum_{n=0}^\infty \delta ( n - n_0 ) z^{-n} [/itex] to [itex] z^{-n_0} [/itex].

If I try this step by step:
[tex] \sum_{n=0}^\infty \delta ( n - n_0 ) z^{-n} = \sum_{n=0}^\infty \delta ( n - n_0 ) z^{-n_0} = z^{-n_0} \sum_{n=0}^\infty \delta ( n - n_0 ) [/tex]

Now, how is [itex] \sum_{n=0}^\infty \delta ( n - n_0 ) [/itex] equal to 1. I don't get that.

Thanks
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Dick
#2
Apr22-07, 07:30 PM
Sci Advisor
HW Helper
Thanks
P: 25,228
delta(n-n0) is equal to 1 if n=n0 and zero otherwise. So the only way the sum could be nonzero is if n0 is a positive integer. Is n0 a positive integer?
FrogPad
#3
Apr22-07, 07:45 PM
P: 837
Quote Quote by Dick View Post
delta(n-n0) is equal to 1 if n=n0 and zero otherwise. So the only way the sum could be nonzero is if n0 is a positive integer. Is n0 a positive integer?
:) - wow, i've been looking at this crap for too long. I can't believe I missed that.

Thanks man :) Yeah, n0 is a positive integer.

time for a break...


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