# Exponential Power Series Expansion

1. Sep 16, 2011

### Lucid Dreamer

I want to show that $e^x e^x = e^{2x}$ using a power series expansion. So I start with

$$\sum_{n=0}^\infty \frac{x^n}{n!} \sum_{m=0}^\infty \frac{x^m}{m!}$$
$$\sum_{n=0}^\infty \sum_{m=0}^\infty \frac{x^n}{n!} \frac{x^m}{m!}$$
$$\sum_{n=0}^\infty \sum_{m=0}^\infty \frac{x^{m+n}}{m!n!}$$

I am at a loss of where to go from here. I want to reduce the last expression down to $\sum_{n=0}^\infty \frac{(2x)^n}{n!}$ but I am not sure of how to get rid of one of the summations.

2. Sep 16, 2011

### Lucid Dreamer

I think I got it...in case anyone was interested.

I can use a substitution on the indicies of the double series as follows. Let $m=q$ and $n=p-q$, with the condition that $q \le p$. This gives

$$\sum_{p=0}^\infty \sum_{q=0}^p \frac{x^q}{q!}\frac{x^{p-q}}{(p-q)!}$$
Which one recognizes as a binomial expansion. Where $\frac{(x+x)^n}{n!} = \sum_{k=0}^n \frac{1}{k!(n-k)!} x^k x^{n-k}$. Thus our double series reduces down to

$$\sum_{p=0}^\infty \frac{(x+x)^p}{p!} = e^{2x}$$