# Continuation of a system of recursive finite differences

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.

• The existing answers are technically correct; you do in fact want to find a square matrix $\mathbf{B}$ where $e^\mathbf{B} = \mathbf{A}$. However, this is not necessarily trivial. Sometimes there will be no solution; other times there will be infinitely many. Did you have any particular $\mathbf{A}$ in mind? – Chad Groft Mar 9 '18 at 20:16
• @ChadGroft I supposed finding the matrix logarithm of $\mathbf{A}$ would find me $\mathbf{B}$. The system that inspired the question was $\mathbf{y}(t + 1) = \begin{bmatrix} -0.7 & 0.05 \\ 0.7 & -0.05 \end{bmatrix} \mathbf{y}(t)$. – holomenicus Mar 11 '18 at 3:04
• Yeah, that matrix is singular, so it can't possibly have a logarithm. – Chad Groft Mar 11 '18 at 7:24

If $D$ is differential operator, then
$$D{\bf x}(t)= \dot{\bf x}(t) \tag{1}$$
$$e^D {\bf x}(t) = {\bf x}(t + 1) \tag{2}$$
In you case the operator $D$ is just a matrix ${\bf B}$, so that
$$e^{\bf B} = {\bf A}$$