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I was reading about the Baker-Hausdorff formula. And there was a proof of it. While I understood (in general) how it was proven ,there was an instance that got me asking how does this identity (the one I wrote below) is true:

$$\Sigma_{n=0}^\infty\Sigma_{m=0}^\infty \frac{(-1)^m(aA)^{n+m}}{n!m!}B$$

This is equal to:

$$\Sigma_{n=0}^\infty\Sigma_{d=0}^n\frac{(-1)^da^nA^{n-d}}{d!(n-d)!}BA^d$$

where A,B are matricies

imbAF
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  • If you use $k$ instead of $n$ in the second expression, it follows from the substitution $k=n+m$ and $d=n$. The point is to make the value of $n+m$ your main index and then parametrize the pairs of numbers which add to it using $d$. – anon Apr 07 '22 at 19:08

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