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Please show that for all $x,y\in\mathbb{R}$,

$e^{x+y} - e^xe^y = \lim_{k\to\infty} \sum_{n=1}^k \sum_{j = 0}^n\left(\frac{x^{k+j}}{(k+j)!}\frac{y^{n-j}}{(n-j)!} + \frac{x^j}{j!}\frac{y^{k+n-j}}{(k+n-j)!}\right)\;.$

(This is from a homework assignment)

I'm thinking of a way to proceed by induction, but it is not clear to me how to utilize it for this problem. Can someone work out this problem so I can ask questions from it?

2 Answers 2

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Hint: The Taylor expansion for $e^x$ is

$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!} =\lim_{M\to \infty}\sum_{n=0}^{M}\frac{x^n}{n!} \; .$

So you can find the Taylor expansions for $e^{x+y}$ and $e^y$ as well. The rest is a matter of combining the series together to obtain $e^{x+y}-e^xe^y$.

See Martin's worked out solution.

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    @user89$1$7 Perhaps if you showed us your work, someone could point out what to do next.2011-04-01
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I've tried to do some computations. Although I did not obtain the required formula, I am posting it here - hopefully someone can either spot a mistake in what I did or modify the method to obtain the above result.

$\sum\limits_{i=0}^k \sum\limits_{i=0}^k \frac{x^i}{i!} \frac{y^j}{j!}= \sum\limits_{m=0}^{2k} \sum\limits_{\substack{i+j=m\\0\le i,j \le k}} \frac{x^i}{i!} \frac{y^j}{j!}= \sum\limits_{m=0}^k \sum\limits_{\substack{i+j=m\\0\le i,j \le k}} \frac{x^i}{i!} \frac{y^j}{j!}+ \sum\limits_{m=k+1}^{2k} \sum\limits_{\substack{i+j=m\\0\le i,j \le k}} \frac{x^i}{i!} \frac{y^j}{j!} $

The first sum:

$\sum\limits_{m=0}^k \sum\limits_{\substack{i+j=m\\0\le i,j \le k}} \frac{x^i}{i!} \frac{y^j}{j!}= \sum\limits_{m=0}^k \sum\limits_{i=0}^m \frac{x^iy^{m-i}}{i!(m-i)!}= \sum\limits_{m=0}^k \frac1{m!} \sum\limits_{i=0}^m \binom mi x^iy^{m-i}= \sum\limits_{m=0}^k \frac{(x+y)^m}{m!} $

The second sum:

$\sum\limits_{m=k+1}^{2k} \sum\limits_{\substack{i+j=m\\0\le i,j \le k}} \frac{x^i}{i!} \frac{y^j}{j!}= \sum\limits_{n=1}^k \sum\limits_{\substack{i+j=k+n\\0\le i,j \le k}} \frac{x^i}{i!} \frac{y^j}{j!}= \sum\limits_{n=1}^k \sum\limits_{i=n}^k \frac{x^i}{i!} \frac{y^{n+k-i}}{(n+k-i)!} $

The difference $e^xe^y-e^{x+y}$ is the limit of the last sum.

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    Yeah, I suppose something like that is going on. Since in the end, the goal is to show that the limit vanishes for $k \to \infty$, I don't think it really matters a lot what you compare it to. It's like choosing between $1-1=0$ and $3-3=0$.2011-04-01