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I am to use Cauchy's Multiplication Theorem and the Binomial Theorem in order to prove

$\exp(x+y)=\exp(x)\exp(y) $

but I have no idea where to begin. All I can think of doing is setting $\exp(x)$ as the sum to infinity of $(x^n)/n!$ and similarly for $\exp(y)$, $(y^n)/n!$

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Well if $x,y\in \mathbb{R}$ then by definition \begin{equation}\exp(x)\exp(y)=(\sum_{k=0}^{\infty}\frac{x^k}{k!})(\sum_{k=0}^{\infty}\frac{y^k}{k!}) \end{equation} The Cauchy's Multiplication Theorem tells as that \begin{equation}\sum_{k=0}^{\infty}\sum_{k=0}^{n}a_kb_{n-k}=\sum_{k=0}^{\infty}a_k\sum_{k=0}^{\infty}b_k\end{equation} when at least one of the two series of the RHS converge absolutely. In our case we have that \begin{equation}\exp(x)\exp(y)=\sum_{n=0}^{\infty}\sum_{k=0}^{n}\frac{x^k}{k!}\frac{y^{n-k}}{(n-k)!}\end{equation} Now because $\binom{n}{k}=\frac{n!}{k!(n-k)!}$, \begin{equation}\exp(x)\exp(y)=\sum_{n=0}^{\infty}\sum_{k=0}^{n}\frac{x^k}{k!}\frac{y^{n-k}}{(n-k)!}=\sum_{n=0}^{\infty}\frac{1}{n!}\sum_{k=0}^{n}\binom{n}{k}x^ky^{n-k}\end{equation} A straightforward application of the binomial theorem yields the resut: \begin{equation}\exp(x)\exp(y)=\sum_{n=0}^{\infty}\frac{1}{n!}(x+y)^n=\exp(x+y)\end{equation}

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    @Mathlete No problem. Have a good day2012-12-01