B BrownianMan Oct 2009 114 2 Nov 2, 2012 #1 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?

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?

girdav Jul 2009 678 241 Rouen, France Nov 3, 2012 #2 Use Cauchy product and Chu-Vandermonde equality.

B BrownianMan Oct 2009 114 2 Nov 3, 2012 #3 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. Last edited: Nov 3, 2012

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.