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I want to know how to simplify the following expression by using the fact that $\sum_{i=0}^\infty \frac{X^i}{i!}=e^X$. The expression to be simplified is as follows:

$\sum_{i=0}^{\infty} \sum_{j=0}^i \frac{X^{i-j}}{(i-j)!} \cdot \frac{Y^j}{j!}\;,$ where $X$ and $Y$ are square matrices (not commutative). (That is, $X\cdot Y \neq Y \cdot X$).

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    Sure, when the variables commute, we $k$now it can be simplified. But do you have some reason for thinking it can be simplified in the case that actually interests you, when the variables don't commute?2012-05-10

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Multiply and divide the innermost term by $\displaystyle i!$ and use binomial theorem. Move the mouse over the gray area to get the answer.

This gives us $\displaystyle \sum_{i=0}^{\infty} \frac1{i!} \sum_{j=0}^{i} \frac{x^{i-j} y^j i!}{(i-j)!j!} = \sum_{i=0}^{\infty} \frac1{i!} \sum_{j=0}^{i} \binom{i}{j} x^{i-j} y^j = \sum_{i=0}^{\infty} \frac{(x+y)^i}{i!} = e^{x+y}$ where $\displaystyle \sum_{j=0}^{i} \binom{i}{j} x^{i-j} y^j = (x+y)^i$ from binomial theorem.

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    +1 for this answer. You might consider up-voting the question. (So far I'm the only one who's done so.)2012-05-10
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Even if $X$ and $Y$ don't commute, it's still true that this expression is equal to $e^X e^Y$; it's just not true that this is equal to $e^{X+Y}$.

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    Yeah, I think so. Thanks!2012-05-10