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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?

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It is not considered polite to simply post imperative sentences as your question - it isn't conducive to people wanting to help you. You should explain where this question is coming from, what you have tried, etc. Also, the question sounds like homework - if it is, you should add the "homework" tag to your question. –  Zev Chonoles Mar 31 '11 at 7:42
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The OP has improved the wording of the question, and I've added the homework tag and formatted the question -- I think that removes the reasons for the close votes. @user8917: If you intend to ask questions here in the future, please look at how I did the formatting (by clicking on "edit") so you can try to do it yourself next time. –  joriki Mar 31 '11 at 10:17

2 Answers 2

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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Thank you, I tried it out but am lost in this. can you please show how to do the problem? –  mary Apr 1 '11 at 3:53
    
I runned through the algebra, but am not getting the answer above. Can someone please show how this? –  mary Apr 1 '11 at 4:47
    
@user8917 Perhaps if you showed us your work, someone could point out what to do next. –  Martin Sleziak Apr 1 '11 at 9:56

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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you made a small typo at the end, the sum over $i$ should be running from $n$ to $k$ since $n$ is smaller than $k$. –  Raskolnikov Apr 1 '11 at 12:32
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@Raskolnikov Thanks a lot. Now, after your correction, I guess that if I compared the sum with $\sum\limits_{m=0}^{2k} \frac{(x+y)^m}{m!}$ instead of $\sum\limits_{m=0}^k \frac{(x+y)^m}{m!}$, I would get something which would be pretty close to the formula given by OP. (I am too lazy to do all the compatations again.) –  Martin Sleziak Apr 1 '11 at 12:52
    
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 $3-3=0$. –  Raskolnikov Apr 1 '11 at 13:19

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