= Solution
Let $\beta=\mu[A]+\mu[B]$ and $\gamma=\mu[A+B]$, where all three <matrices> are symmetric. Submultiplicativity and the triangle inequality give
$$
\|[e^{xB},A]e^{xA}\|_2
\leq2\|A\|_2e^{x(\mu[A]+\mu[B])},
$$
and similarly the other <commutator> term is bounded by $2\|B\|_2e^{x\beta}$. Apply part (a) also to $A+B$ in the integral from part (b). The outer factor one-half cancels these twos, leaving
$$
\|F(t)-e^{t(A+B)}\|_2
\leq(\|A\|_2+\|B\|_2)\int_0^t e^{(t-x)\gamma+x\beta}\,dx.
$$
The <exponential divided difference> evaluates this integral. Thus
$$
\boxed{\|F(t)-e^{t(A+B)}\|_2\leq
(\|A\|_2+\|B\|_2)
\frac{e^{t\beta}-e^{t\gamma}}{\beta-\gamma}}
$$
when $\beta\ne\gamma$, and
$$
\boxed{\|F(t)-e^{t(A+B)}\|_2\leq
(\|A\|_2+\|B\|_2)\,te^{t\gamma}}
$$
when they coincide. The second expression is both the direct equal-exponent integral and the continuous limit of the first. The <Rayleigh-Ritz variational principle> also gives $\gamma\leq\beta$, although the integral computation does not require a strict inequality. These are valid coarse norm bounds; the cancellation between the two products can make the actual small-step error substantially smaller.
Back to article page
