# simplify this expression

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. – John Smith May 10 '12 at 2:41
Do you have some reason for thinking that it can be simplified? – Gerry Myerson May 10 '12 at 2:45
Just as Marvis did, when the dimensionalites of $X$ and $Y$ are reduced to 1, this expression can be simplified elegantly. – John Smith May 10 '12 at 2:48
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? – Gerry Myerson May 10 '12 at 4:06

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. – John Smith May 10 '12 at 2:31
@JohnSmith I am confused about the last line of your question. Reals come with multiplications being commutative. – user17762 May 10 '12 at 2:33
Sorry, $x$ and $y$ are not real numbers, you can regard them as matrices, and $xy \neq yx$. – John Smith May 10 '12 at 2:35
Thanks to Marvis's hint, I get it! – John Smith May 10 '12 at 7:02
+1 for this answer. You might consider up-voting the question. (So far I'm the only one who's done so.) – Michael Hardy May 10 '12 at 12:51

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! – John Smith May 10 '12 at 5:42