Summation help

  1. 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.

  2. jcsd
  3. Dick

    Dick 25,913
    Science Advisor
    Homework Helper

    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: Apr 22, 2007
  4. :) - 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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