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Suppose $X$ and $Y$ are $n\times n$ matrices (assume they are symmetric if you want), is there a reasonable formula for

$ Z=\frac{d}{dt}|_{t=0} e^{X+tY}? $

Obviously if we assume that $[X,Y]=0$ then there is a simple answer.
I assume if there is an reasonable closed form then it should follow from the Baker-Campbell-Hausdorff identity but I am having a bit of a hard time at it.

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For each $k\in\mathbb{N}$, let $X_k$ be the set of all non-commutative monomials of degree $k$ on two variables. Write $p_{0,k}$ for the monomial $p_{0,k}(x,y)=x^k$.

$ \frac{e^{X+tY}-e^X}t=\sum_{k=0}^\infty\frac{(X+tY)^k-X^k}{tk!}=\sum_{k=1}^\infty\frac{\sum_{p\in X_k}p(X,tY)-X^k}{tk!} =\sum_{k=1}^\infty\sum_{p\in X_k\setminus\{p_{0,k}\}}\frac{p(X,tY)}{tk!} $ The only monomials in $X_k$ that will survive after $t\to0$ are those where there is only one $tY$, i.e. $p_{1,k}(x,y)=yx^{k-1}$, $p_{2,k}(x,y)=xyx^{k-2}$, etc.

So $ \lim_{t\to0}\frac{e^{X+tY}-e^X}t=\lim_{t\to0}\sum_{k=1}^\infty\sum_{p\in X_k\setminus\{p_{0,k}\}}\frac{p(X,tY)}{tk!}=\lim_{t\to0}\sum_{k=1}^\infty\sum_{p\in \{p_{1,k},\ldots,p_{k,k}\}}\frac{p(X,tY)}{tk!}\\ =\sum_{k=1}^\infty\frac{YX^{k-1}+XYX^{k-2}+X^2YX^{k-3}+\cdots+X^{k-1}Y}{k!} =\sum_{m=0}^\infty\sum_{n=0}^\infty\frac{X^mYX^n}{(m+n+1)!} $

(Edit: thanks to Robert for the suggestion for a nicer expression)

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    That's definitely nicer. Thanks, Robert!2012-06-25