How to solve a recurrence equation like this:
$(a_{n+1},b_{n+1})=(3a_n+5b_n,a_n+3b_n)$
with the initial condition:
$(a_1,b_1)=(3,1)$
The only way I can think of to solve an equation like this is to use generating functions. However, since there are actually two variables ($a_n$ and $b_n$), I don't know if generating functions applies on the question.
Any help or hint would be appreciated. Thanks in advance!
Best Answer
We have a system of homogeneous first order linear difference equations. Define $A \in \mathbb{R}^{2 \times 2}$ by $$ A = \begin{pmatrix} 3 & 5 \\ 1 & 3 \end{pmatrix}$$ Then we have the following difference equation: $$\begin{equation} \begin{pmatrix} a_{n+1} \\ b_{n+1} \end{pmatrix} = \begin{pmatrix} 3 & 5 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} a_n \\ b_n \end{pmatrix}, \quad n \in \mathbb{N} \end{equation}$$ A solution is given by $$\begin{pmatrix} a_n \\ b_n \end{pmatrix} = A^{n-1} \begin{pmatrix} a_1 \\ b_1 \end{pmatrix}$$ Can you find a nice expression for $A^{n-1}$?