Understanding Summation with Delta Functions and Exponents in Math

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SUMMARY

The discussion centers on the mathematical expression involving the summation of a delta function, specifically \(\sum_{n=0}^\infty \delta(n - n_0) z^{-n}\), which simplifies to \(z^{-n_0}\). Participants clarify that the delta function \(\delta(n - n_0)\) equals 1 when \(n = n_0\) and 0 otherwise, leading to the conclusion that the summation equals 1 only if \(n_0\) is a positive integer. This understanding resolves the confusion regarding the steps from the summation to the final result.

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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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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?
 
Last edited:
Dick said:
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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