# Combinatorial identity

1. Nov 4, 2012

### 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.

2. Nov 4, 2012

### haruspex

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:
$B_{k}(x)=\sum_{n\geq 0}\binom{n}{k}\frac{x^{n}}{n!} = \sum_{n\geq k}\binom{n}{k}\frac{x^{n}}{n!} = \sum_{n\geq 0}\binom{n+k}{k}\frac{x^{n+k}}{(n+k)!} = x^k\sum_{n\geq 0}\frac{x^{n}}{n!k!} = \frac{x^k}{k!}\sum_{n\geq 0}\frac{x^{n}}{n!} = \frac{x^k}{k!}e^x$

3. Nov 4, 2012

### BrownianMan

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

4. Nov 4, 2012

### haruspex

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

5. Nov 5, 2012

### BrownianMan

You're right. My mistake.

Thanks.

6. Nov 5, 2012