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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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    Yeah, I've corrected them. $X$ and $Y$ are matrices instead of real numbers, meaning that binomial theorem doesn't hold in this case.2012-05-10
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    Do you have some reason for thinking that it can be simplified?2012-05-10
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    Just as Marvis did, when the dimensionalites of $X$ and $Y$ are reduced to 1, this expression can be simplified elegantly.2012-05-10
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    Sure, when the variables commute, we know 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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    Thanks, but I forgot an important contraint (NOTE: $x\cdot y \neq y \cdot x$). That means the binomial theorem doesn't hold.2012-05-10
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    @JohnSmith I am confused about the last line of your question. Reals come with multiplications being commutative.2012-05-10
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    Sorry, $x$ and $y$ are not real numbers, you can regard them as matrices, and $xy \neq yx$.2012-05-10
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    Thanks to Marvis's hint, I get it!2012-05-10
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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