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

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

I'm having some trouble with this one. Does anyone have any hints?

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- Nov 2nd 2012, 06:38 PMBrownianManCombinatorial Identity
Let $\displaystyle B_{k}(x)=\sum_{n\geq 0}\binom{n}{k}\frac{x^{n}}{n!}$. Show that

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

I'm having some trouble with this one. Does anyone have any hints? - Nov 3rd 2012, 07:18 AMgirdavRe: Combinatorial Identity
Use Cauchy product and Chu-Vandermonde equality.

- Nov 3rd 2012, 12:34 PMBrownianManRe: Combinatorial Identity
I did that and I got $\displaystyle B_{k}(x)B_l(x)=\binom{n}{k+l}\frac{(2x)^{n}}{n!}=B _{k+l}(2x)$.

I'm not sure where the $\displaystyle \frac{1}{2^{k+l}}\binom{k+l}{l}$ comes from. - Nov 4th 2012, 09:59 AMBrownianManRe: Combinatorial Identity
Anyone?