
#1
Apr2207, 07:02 PM

P: 838

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 



#2
Apr2207, 07:30 PM

Sci Advisor
HW Helper
Thanks
P: 25,175

delta(nn0) 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?




#3
Apr2207, 07:45 PM

P: 838

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


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