Understanding Summation with Delta Functions and Exponents in Math

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The discussion centers on the mathematical expression involving the summation of a delta function and its relationship to exponents. The key point is that the summation \sum_{n=0}^\infty \delta(n - n_0) z^{-n} simplifies to z^{-n_0} when n_0 is a positive integer. Participants clarify that the delta function equals 1 only when n equals n_0, making the sum non-zero solely under this condition. The confusion arises from understanding why \sum_{n=0}^\infty \delta(n - n_0) equals 1, which is confirmed when n_0 is indeed a positive integer. The conversation concludes with a realization of the oversight and a light-hearted acknowledgment of the complexity involved.
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I don't see how the following works:

\sum_{n=0}^\infty \delta ( n - n_0 ) z^{-n} = z^{-n_0}

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

If I try this step by step:
\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 )

Now, how is \sum_{n=0}^\infty \delta ( n - n_0 ) 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...
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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