Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\exp(y)$, $(y^n)/n!$

share|improve this question

1 Answer 1

up vote 3 down vote accepted

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}

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.