# Use Cauchy's Multiplication Theorem and the Binomial Theorem to prove $\exp(x+y)=\exp(x)\exp(y)$

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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## 1 Answer

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

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If you could give me some hints and pointers, rather than posting the actual proof, that'd be great. I don't really understand why you're doing what you're doing at each step? – Mathlete Dec 1 '12 at 16:44
Sure I shall add more details if you would like! – Nameless Dec 1 '12 at 16:45
That'd be really useful, thanks! Could you possibly explain where and how you've used Cauchy's multiplication theorem? That's the main part I'm having trouble with. – Mathlete Dec 1 '12 at 16:46
I completely understand that now, thank you very much! – Mathlete Dec 1 '12 at 16:52
@Mathlete No problem. Have a good day – Nameless Dec 1 '12 at 16:53