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Math Help - Combinatorial Identity

  1. #1
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    Combinatorial Identity

    Let B_{k}(x)=\sum_{n\geq 0}\binom{n}{k}\frac{x^{n}}{n!}. Show that

    B_{k}(x)B_{l}(x)=\frac{1}{2^{k+l}}\binom{k+l}{l}B_  {k+l}(2x).

    I'm having some trouble with this one. Does anyone have any hints?
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  2. #2
    Super Member girdav's Avatar
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    Re: Combinatorial Identity

    Use Cauchy product and Chu-Vandermonde equality.
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    Re: Combinatorial Identity

    I did that and I got B_{k}(x)B_l(x)=\binom{n}{k+l}\frac{(2x)^{n}}{n!}=B  _{k+l}(2x).

    I'm not sure where the \frac{1}{2^{k+l}}\binom{k+l}{l} comes from.
    Last edited by BrownianMan; November 3rd 2012 at 01:51 PM.
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    Re: Combinatorial Identity

    Anyone?
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