Combinatorial Identity

**BrownianMan** · November 2nd 2012, 06:38 PM

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?

**girdav** · November 3rd 2012, 07:18 AM

Use Cauchy product and Chu-Vandermonde equality.

**BrownianMan** · November 3rd 2012, 12:34 PM

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.

**BrownianMan** · November 4th 2012, 09:59 AM

Anyone?

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