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
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...