Combinatorial Identity

Oct 2009
114
2
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?
 
Jul 2009
678
241
Rouen, France
Use Cauchy product and Chu-Vandermonde equality.
 
Oct 2009
114
2
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.
 
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