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Math Help - question on indices of a generating function

  1. #1
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    question on indices of a generating function

    I want to find the generating function for an=n(n-1).

    \sum_{n=0}^{\infty }n(n-1)\frac{x^n}{n!}
    =\sum_{n=0}^{\infty }\frac{x^n}{(n-2)!}
    =x^2\sum_{n=0}^{\infty }\frac{x^{n-2}}{(n-2)!}
    =x^2\sum_{k=-2}^{\infty }\frac{x^{k}}{k!}
    =x^2\sum_{k=0}^{\infty }\frac{x^{k}}{k!}
    =x^2e^x

    Am I justified in the second to last step, ignoring the summands for k=-1 and k=-2?
    If so, is there a reason why I can ignore the first two terms?
    Last edited by Yuuki; May 29th 2013 at 06:09 PM.
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  2. #2
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    Re: question on indices of a generating function

    Is there such a thing as (-2)! or (-1)!?
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  3. #3
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    Re: question on indices of a generating function

    No, but I wasn't sure if (-2)! and (-1)! not existing allowed me to ignore them.
    So because (-2)! and (-1)! don't exist, we can't evaluate them, hence it's okay to ignore them?
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  4. #4
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    Re: question on indices of a generating function

    Only in this case, because you can see if you plug n=0 and n=1 into your original sum that those terms are 0 anyway.

    Really what you should have written is

    \displaystyle \begin{align*} \sum_{n = 0}^{\infty} n(n-1)\frac{x^n}{n!} &= 0 + 0 + \sum_{n = 2}^{\infty} n(n-1)\frac{x^n}{n!} \\ &= \sum_{n = 2}^{\infty} n(n-1)\frac{x^n}{n!} \\ &= \sum_{n = 2}^{\infty} \frac{x^n}{(n-2)!} \\ &= x^2 \sum_{n=2}^{\infty} \frac{x^{n-2}}{(n-2)!} \\ &= x^2\sum_{k=0}^{\infty}\frac{x^k}{k!} \\ &= x^2\,e^x  \end{align*}

    It doesn't make sense to put into your counter values which will give undefined terms.
    Last edited by Prove It; May 29th 2013 at 06:13 PM.
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  5. #5
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    Re: question on indices of a generating function

    Thank you very much, it now makes a lot of sense.
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