I want to find a closed form of $a_i$ sequence below:

$$a_i = \alpha a_{i-1} + \beta b_{i-1}$$ $$b_i = \gamma a_{i-1} + \delta b_{i-1}$$

I did some expansions and arrived to (assuming I have no errors) something that looks simpler:

$$a_i = f(i) + \sum_{j=1}^i j\delta^ja_j$$

But solving the second term is something I can not figure out as well.

  • 2
    $\begingroup$ Do you have specific values for $\alpha, \beta, \gamma, \delta$? There are general methods to solve such recurrences. Your recurrence is of the form $\mathbf{x}_{i} = A\mathbf{x}_{i-1}$, where $A = \begin{bmatrix} \alpha & \beta \\ \gamma & \delta \end{bmatrix}$ and $\mathbf{x}_{n} = \begin{bmatrix} a_{n} \\ b_{n}\end{bmatrix}$. You can show that this implies that $\mathbf{x}_{i} = A^{i}\mathbf{x}_{0}$ for all $i\in \mathbb{N}$. So if you have learnt about how to find powers of a matrix, you could go from here. $\endgroup$ – Minus One-Twelfth Mar 2 '19 at 1:20
  • 1
    $\begingroup$ You can also try to eliminate one of the variables and end up with a second-order linear recurrence for the other, which you can then solve if you have learnt about how to solve these (for some examples and information about this, see here for instance). For example, see the example from slide 43 here. $\endgroup$ – Minus One-Twelfth Mar 2 '19 at 1:22

Your problem can be written like this:

$$ \mathbf{x}_i = \begin{bmatrix} a_i \\ b_i \end{bmatrix} = \mathbf{A} \mathbf{x}_{i-1} \hspace{20mm} \mathbf{A} = \begin{bmatrix} \alpha & \beta \\ \gamma & \delta \end{bmatrix} $$

Assume you know $\mathbf{x}_0$. Then $\mathbf{x}_i = \mathbf{A}^i \mathbf{x}_0$.

Here are the first few values of $\mathbf{A}^n$. $$ \mathbf{A}^0 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\ \mathbf{A}^1 = \begin{bmatrix} \alpha & \beta \\ \gamma & \delta \end{bmatrix} \\ \mathbf{A}^2 = \begin{bmatrix} \alpha^2 + \beta\gamma & \alpha\beta + \beta\delta \\ \alpha\gamma + \gamma\delta & \beta\gamma + \delta^2 \end{bmatrix} \\ $$

I believe that for arbitrary $\alpha, \beta, \gamma, \delta$, you will need to compute $\mathbf{A}^n$ explicitly. See this post on how to calculate square matrix powers. If your matrix $\mathbf{A}$ satisfies a recurrence relation, then you can obtain a closed-form solution.


The defining matrix $A$ has trace $T = \alpha + \delta$ and determinant $D = \alpha \delta - \beta \gamma .$ The Cayley-Hamilton Theorem says $$ A^2 - TA + DI = 0, $$ $$ A^2 = TA - DI. $$ In particular, $$ a_{i+2} = T a_{i+1} - D a_i $$


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