# Combinatorial Identity

• November 2nd 2012, 06:38 PM
BrownianMan
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?
• November 3rd 2012, 07:18 AM
girdav
Re: Combinatorial Identity
Use Cauchy product and Chu-Vandermonde equality.
• November 3rd 2012, 12:34 PM
BrownianMan
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.
• November 4th 2012, 09:59 AM
BrownianMan
Re: Combinatorial Identity
Anyone?