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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 $$\exp(x)=\sum_{n=0}^{\infty}\frac{x^n}{n!}.$$

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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
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 $\displaystyle\binom nk=\frac{n!}{k!(n-k)!}$.

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