convergence to matrix exponential I know that
$$\left[1+\frac1n x +O\left(\frac1{n^2}\right)\right]^{tn}\to e^{tx}$$
as $n\to\infty$ for any real numbers $x,t$. And we get other variations that as additional perturbations to the exponent or the inside, etc as long as everything is controlled appropriately.
I want to replace $x$ with a square matrix $A$ though and get
$$\left[I+\frac1n A +O\left(\frac1{n^2}I\right)\right]^{tn}\to e^{tA}$$
as $n\to\infty$.
Am I using the big-O notation acceptably there? It's just some fixed matrix divided by $n^2$ or maybe with some other higher order terms: $O\left(\frac1{n^2}I\right)=A_2/n^2+A_3/n^3+\cdots$ with the $A_j$ being square matrices.
Is this valid? Are there convergence issues if $A$ has a large norm or condition number or something similar?
The only reason I ask is because of large numerical errors I see in Matlab and R, which I imagine is to be expected. R is especially bad at this (at least with the packages I'm using). Matlab does pretty good, but does break down eventually.
Theoretically, it seems fine since the expression on the left will look more and more like the exponential power series (from the binomial expansion and ignoring the higher order terms). The exponential power series converges for any matrix. But maybe there is a potential issue with the higher order terms?
 A: Ok, I found the answer. Also my notation in the question is a bit uncertain. Let $O\left(\frac1{n^2}\right)$ represent an appropriately sized matrix where each component is $O\left(\frac1{n^2}\right)$.
We only need to demonstrate that
$\left[I+\frac1n A +O\left(\frac1{n^2}\right)\right]^{n}\to e^{A}$ as $n\to\infty.$
Performing the expansion gives
$$\left[I+\frac1n A +O\left(\frac1{n^2}\right)\right]^{n}=
\left(I+\frac1n A\right)^n + \left[O\left(\frac1{n^2}\right)\right]^n+\text{other terms}.$$
Clearly $\left[O\left(\frac1{n^2}\right)\right]^n$ is no problem as $n\to\infty$. The other terms left out are not really part of a binomial expansion because we are not guaranteed that the matrices involved commute. That being said, exactly one of the terms bounds the decay as $n\to\infty$. Here are a few of those other terms.
The first term can be written as the summation:
$$\sum_{k=1}^{n-1} \left[O\left(\frac1{n^2}\right)\right]^k\cdot \left(I+\frac1n A\right) \cdot \left[O\left(\frac1{n^2}\right)\right]^{n-1-k}=n\cdot O\left(\frac1{n^2}\right)^k\cdot O(1)\cdot O\left(\frac1{n^2}\right)^{n-1-k}=O\left(\frac1{n^2}\right)^{n-2}.$$
So clearly this term is not a problem.
Each subsequent term can be written similarly as a mix of the factors $I+\frac1n A$ and $O(1/n^2)$. As we increase the number of times $I+\frac1n A$ appears, the decay as $n\to\infty$ becomes slower. The final missing term can be written as:
$$\sum_{k=1}^{n-1} \left(I+\frac1n A\right)^k\cdot O\left(\frac1{n^2}\right) \cdot \left(I+\frac1n A\right)^{n-1-k}=n\cdot O(1)^k\cdot O\left(\frac1{n^2}\right)\cdot O(1)^{n-1-k}=O\left(\frac1{n}\right).$$
Every term in the expansion of $\left[I+\frac1n A +O\left(\frac1{n^2}\right)\right]^{n}$ is $O(1/n)$ except the first, so we conclude that
$$\left[I+\frac1n A +O\left(\frac1{n^2}\right)\right]^{n}=\left(I+\frac1n A\right)^n+O\left(\frac1n\right).$$
The desired conclusion follows immediately.
The power of $t$ isn't a problem either, although to be really careful one would need to reason about continuity or some other careful convergence matters involving matrices, but it is intuitively clear to me that it will all work out now. So this isn't all that rigorous or complete of an answer, but it satisfies me.
