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?
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.