Proving the Combinatorial Identity for B_k(x)B_l(x)

[BrownianMan]

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? I've tried using Cauchy product and Chu-Vandermonde equality but I get $B_{k}(x)B_l(x)=\sum_{n\geq 0}\binom{n}{k+l}\frac{(2x)^{n}}{n!}=B_{k+l}(2x)$.

I'm not sure where the $\frac{1}{2^{k+l}}\binom{k+l}{l}$ comes from.

I'm also suppose to take the x^n/n! coefficient of both sides and show which binomial coefficient identity this gives.

 

[haruspex]

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? I've tried using Cauchy product and Chu-Vandermonde equality but I get $B_{k}(x)B_l(x)=\sum_{n\geq 0}\binom{n}{k+l}\frac{(2x)^{n}}{n!}=B_{k+l}(2x)$.

I'm not sure where the $\frac{1}{2^{k+l}}\binom{k+l}{l}$ comes from.

I'm also suppose to take the x^n/n! coefficient of both sides and show which binomial coefficient identity this gives.

Probably most useful to you is to find the error in what you have done, but for that you'll need to post your working. (And a statement or URL for any existing result your rely on.)
But fwiw, I get:
$B_{k}(x)=\sum_{n\geq 0}\binom{n}{k}\frac{x^{n}}{n!} = \sum_{n\geq k}\binom{n}{k}\frac{x^{n}}{n!}<br /> = \sum_{n\geq 0}\binom{n+k}{k}\frac{x^{n+k}}{(n+k)!}<br /> = x^k\sum_{n\geq 0}\frac{x^{n}}{n!k!}<br /> = \frac{x^k}{k!}\sum_{n\geq 0}\frac{x^{n}}{n!} = \frac{x^k}{k!}e^x$

 

[BrownianMan]

I don't see how Bk(x) = x^k/k! * e^x. I checked with specific values and it doesn't hold.

 

[haruspex]

I don't see how Bk(x) = x^k/k! * e^x. I checked with specific values and it doesn't hold.

What values? Seems ok for k = 0, 1, 2.

 

[BrownianMan]

What values? Seems ok for k = 0, 1, 2.

You're right. My mistake.

Thanks.

 

[haruspex]

OK, good. Did that help you solve the OP?

 

[BrownianMan]

Yes, I solved it. Thanks again!