A long time ago I asked a series of questions about finding a matrix as function of a function that by matrix multiplication will "move forward" in the variable and calculate future function values for us automatically in the following sense:
$${\bf v_k = M}^k{\bf v_0}$$
So that $$f(k\cdot\Delta) = {\bf v_k}_1 = {\bf M}^k{\bf v_0}_1$$
For some function $t\to f(t)$.
I managed to find several ways to do this for a couple of families, rotation matrices maybe being the simplest one for $f(t) = \cos(t)$ and $${\bf M} = \begin{bmatrix}\cos(\Delta)&\sin(\Delta)\\-\sin(\Delta)&\cos(\Delta)\end{bmatrix} \Rightarrow {\bf M}^k = \begin{bmatrix}\cos(k\Delta)&\sin(k\Delta)\\-\sin(k\Delta)&\cos(k\Delta)\end{bmatrix}$$
Is there some way we can derive differentiation or integration in a similar fashion?
My own work is limited to concluding that a naive integration approximator will be $$\frac{1}{\Delta}\left(\sum_{i=0}^k {\bf M}^i\right){\bf v_0}$$
And in the same fashion we can approximate derivative:
$$\frac{1}{\Delta}\left( {\bf M}^{k+1} - {\bf M}^{k}\right){\bf v_0}$$
Or maybe a midpoint-derivative, as it is in numerical analysis often preferred as it gives better results:
$$\frac{1}{2\Delta}\left( {\bf M}^{k+1} - {\bf M}^{k-1}\right){\bf v_0}$$
But can we derive it as one matrix somehow? It feels like this $\Delta$ will be intimately coupled with how to do this somehow...