General strategy of error bound of matrix exponential I want to ask General strategy of the error bound of the matrix exponential.
For example, suppose, $A, B$ are finite dimension $n \times n$ matrices with complex coefficients. Using Baker–Campbell–Hausdorff formula or simple calculation. One can show that ($x \geq 0$)
$e^{Ax}e^{Bx}-e^{Bx}e^{Ax}=[A,B]x^2+O(|x^3|).$
But the thing I want is that the clear bounds. I want to find the $c$ such that
$||e^{Ax}e^{Bx}-e^{Bx}e^{Ax}-[A,B]x^2|| \leq c|x^3|.$ Here $c$ should be function of $A,B.$
The first method I want to try is to use Taylor's theorem. But the computation is so tedious. Is there any way that I can find the clear expression for $c$?
More generally, given any matrix expression in terms of parameters $x$, one can easily see the first few terms,  but is there a more elegant way to control the error bounds?
Or in the simplest case, what is the error bounds for Baker–Campbell–Hausdorff formula $e^{Ax}e^{Bx}$ to $n$ order. I mean the clear expression of the constant before $|x|^n$ not just the big O notation.
 A: Here is a method that gives a simple explicit formula in this case and should give similar ones in more general cases.
Fix a matrix norm that is submultiplicative, such as the largest singular value. Assume $x \leq 1$.
We have
$$ e^{Ax} e^{Bx}  - e^{Bx} e^{Ax} - [A,B]x^2 = \sum_{n,m=0}^{\infty}  \frac{A^n B^m }{ n!m!} x^{n+m}  -  \sum_{n,m=0}^{\infty}  \frac{B^m. A^n  }{ n!m!} x^{n+m} - (AB-BA) x^2  = \sum_{0 \leq n, m \leq \infty, n+m\geq 3}\frac{A^n B^m }{ n!m!} x^{n+m} -\sum_{0 \leq n, m \leq \infty, n+m\geq 3}\frac{ B^mA^n }{ n!m!} x^{n+m}  $$
so $$ \left|  e^{Ax} e^{Bx}  - e^{Bx} e^{Ax} - [A,B]x^2\right| = \left| \sum_{0 \leq n, m \leq \infty, n+m\geq 3}\frac{A^n B^m }{ n!m!} x^{n+m} -\sum_{0 \leq n, m \leq \infty, n+m\geq 3}\frac{ B^mA^n }{ n!m!} x^{n+m}\right| \leq \sum_{0 \leq n, m \leq \infty, n+m\geq 3}\frac{|A|^n |B|^m }{ n!m!} x^{n+m} +\sum_{0 \leq n, m \leq \infty, n+m\geq 3}\frac{ |B|^m |A|^n }{ n!m!} x^{n+m} \leq 2\sum_{0 \leq n, m \leq \infty}\frac{ |B|^m|A|^n }{ n!m!}x^3 = 2e^{ |A| + |B|} x^3. $$
