Error Bounds on Remainder Term of Multiplication of Two Matrix Exponentials Let $A, B \in \mathbb{C}^{n \times n}$
$$\begin{align*}
e^A e^B &= \left(I + A + \frac{A^2}{2!} + \dots \right)\left(I + B + \frac{B^2}{2!} + \dots \right) \\
&= I + A + B + \frac{A^2 + 2AB + B^2}{2!} + R
\end{align*}$$
How can I calculate the bound on $\lVert R \rVert$, where $\lVert \cdot \rVert$ may be any induced (subordinate) matrix norm?
I've searched for Taylor's Theorem on matrix functions and end up with the Fréchet derivative. But I'm not sure how to apply it for this problem.
 A: If $A,B$ are not comparable, then $A^2,B^2,AB$ are also not comparable. Thus there is no reason to group them.
You must consider  $e^{tA}e^{tB}$ where $t$ is a small real number. Then $||R(t)- (A^3+3A^2B+3AB^2+B^3)t^3/6||=o(t^3)$ when $t$ tends to $0$.
In other words, for $t$ small enough, $||R(t)||\leq ||A^3+3A^2B+3AB^2+B^3|||t|^3/3\leq (||A||+||B||)^3|t|^3/3$.
EDIT. Answer to @ obareey . The previous result uses standard result about the Taylor formula; unfortunately, it is very difficult to find an explicit $t_0$ s.t. $|t|<t_0$ implies the above inequality concerning $R(t)$. If you want an explicit bound, then we must consider the whole series that defines the exp; we can proceed as follows: since the exp function is normally-convergent over $M_n(\mathbb{C})$, one has
$||R||\leq \sum_{i\geq 3}(||A||+||B||)^i/i!=e^{||A||+||B||}-1-(||A||+||B||)-(||A||+||B||)^2/2=e^{\theta(||A||+||B||)}(||A||+||B||)^3/6$ where $\theta\in(0,1)$.
Conclusion: $||R||\leq e^{||A||+||B||}(||A||+||B||)^3/6$. The trouble lies in the fact that $e^{||A||+||B||}$ may be a great number.
