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 wondering how to prove the following question:

In any unital Banach algebra, we have $\exp(x+y)=\exp(x)\exp(y)$, if $xy=yx$, where $$\exp(x)=\sum_{n=0}^{\infty}\frac{x^n}{n!}.$$

share|improve this question
3  
Do you know how to prove it when $x$ and $y$ are real numbers? Then try to imitate the proof. –  Davide Giraudo Feb 9 '13 at 16:20

1 Answer 1

up vote 1 down vote accepted

$$\exp(x+y)= \sum_{n=0}^\infty \frac{(x+y)^n}{n!}=\\ = \sum_{n\ge 0} \frac{x^n+\binom n1 x^{n-1}y+\dots+\binom n{n-1} xy^{n-1}+y^n}{n!}$$ Now commutativity of $x$ and $y$ is used in there (counting e.g. $xxxyxx$ and $xyxxxx$ together).

Then reorder the sums, and use $\displaystyle\binom nk=\frac{n!}{k!(n-k)!}$.

share|improve this answer

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.