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Combinatorial identity

  1. Nov 4, 2012 #1
    Let [itex]B_{k}(x)=\sum_{n\geq 0}\binom{n}{k}\frac{x^{n}}{n!}[/itex]. Show that

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

    I'm having some trouble with this one. Does anyone have any hints? I've tried using Cauchy product and Chu-Vandermonde equality but I get [itex]B_{k}(x)B_l(x)=\sum_{n\geq 0}\binom{n}{k+l}\frac{(2x)^{n}}{n!}=B_{k+l}(2x)[/itex].

    I'm not sure where the [itex]\frac{1}{2^{k+l}}\binom{k+l}{l}[/itex] comes from.

    I'm also suppose to take the x^n/n! coefficient of both sides and show which binomial coefficient identity this gives.
     
  2. jcsd
  3. Nov 4, 2012 #2

    haruspex

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    Probably most useful to you is to find the error in what you have done, but for that you'll need to post your working. (And a statement or URL for any existing result your rely on.)
    But fwiw, I get:
    [itex]B_{k}(x)=\sum_{n\geq 0}\binom{n}{k}\frac{x^{n}}{n!} = \sum_{n\geq k}\binom{n}{k}\frac{x^{n}}{n!}
    = \sum_{n\geq 0}\binom{n+k}{k}\frac{x^{n+k}}{(n+k)!}
    = x^k\sum_{n\geq 0}\frac{x^{n}}{n!k!}
    = \frac{x^k}{k!}\sum_{n\geq 0}\frac{x^{n}}{n!} = \frac{x^k}{k!}e^x[/itex]
     
  4. Nov 4, 2012 #3
    I don't see how Bk(x) = x^k/k! * e^x. I checked with specific values and it doesn't hold.
     
  5. Nov 4, 2012 #4

    haruspex

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    What values? Seems ok for k = 0, 1, 2.
     
  6. Nov 5, 2012 #5
    You're right. My mistake.

    Thanks.
     
  7. Nov 5, 2012 #6

    haruspex

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    OK, good. Did that help you solve the OP?
     
  8. Nov 5, 2012 #7
    Yes, I solved it. Thanks again!
     
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