Suppose I have a system of recursive functions of discrete time
$$\mathbf{x}(t+1) = \mathbf{A}\mathbf{x}(t)$$
where $\mathbf{A}$ is a square matrix. How can I find a system of first order ordinary differential equations
$$\mathbf{\dot{x}}(t) = \mathbf{B}\mathbf{x}(t)$$
so that, sharing the initial state $\mathbf{x}(0)$, every solution $(t, \textbf{x})$ to the former is also a solution to the latter?
Letting $\mathbf{B} = \mathbf{A}$ doesn’t work.
If $D$ is differential operator, then
$$ D{\bf x}(t)= \dot{\bf x}(t) \tag{1} $$
Moreover,
$$ e^D {\bf x}(t) = {\bf x}(t + 1) \tag{2} $$
In you case the operator $D$ is just a matrix ${\bf B}$, so that
$$ e^{\bf B} = {\bf A} $$