Solving $A_n A_{n+1}=A_{n}+2 A_{n+1}$ to disagree with a question

Question

Solving $$A_n A_{n+1}=A_{n}+2 A_{n+1} \tag1$$ to disagree with a question attached below:

Let us re-write *1) as $$A_{n+1}(A_n-2)=A_n \tag 2$$ Let $A_n-2=B_n/B_{n-1}$ in (2) to simplify and to get $$B_{n+1}-2B_{n-1}+B_n=0 \tag3$$ Let $B_n=x^n$, we get $x^2+x-2=0 \implies x=1,-2.$ We get $$B_n=P +Q (-2)^n \tag4$$. Finally we get $$A_n=\frac{P+Q(-2)^n}{P+Q(-2)^{n-1}}+2=\frac{3P}{P+Q(-2)^{n-1}}\tag5$$ The solution of (1) given by (5) is algebraic of exponential but not not periodic.

Can one justify the attached question containing the recurrence relation (1)?

Edit: Let $Q/P=R$, to write $$A_n=\frac{3}{1+R(-2)^{n-1}}.$$

Answer 1

This is a Ricatti recurrence:

$\begin{align*} w_{n + 1} = \frac{a w_n + b}{c w_n + d} \end{align*}$

with $c \ne 0$ and $a d - b c \ne 0$. There are several ways to solve them. Brand "A Sequence Defined by a Recurrence Relation", AMM 62:7 (1955), pp 489-492 goes as follows.

Define:

$\begin{align*} y_{n + 1} &= \alpha - \frac{\beta}{y_n} \\ \alpha &= a + d \\ \beta &= a d - bc \end{align*}$

Replacing $y_n = x_{n + 1} / x_n$ gives now:

$\begin{align*} x_{n + 2} - \alpha x_{n + 1} + \beta x_n &= 0 \end{align*}$

We need two starting values, pick $x_0 = 1$ for convenience giving $x_1 = y_0$, and you are set.

Another road is to recognize the recurrence as a Möbius tranformation:

$\begin{align*} w_{n + 1} &= \frac{a w_n + b}{c w_n + d} \end{align*}$

It turns out those compose just like $2 \times 2$ matrices multiply, so if you define:

$\begin{align*} M &= \pmatrix{a & b \\ c & d} \\ M^n &= \pmatrix{a^{(n)} & b^{(n]} \\ c^{(n)} & d^{(n)}} \end{align*}$

then:

$\begin{align*} w_n &= \frac{a^{(n)} w_0 + b^{(n)}}{d^{(n)} w_0 + d^{(n)}} \end{align*}$

Yet another way is given by Mitchell "An Analytic Ricatti Solution for Two-Target Discrete-Time Control", Journal of Economic Dynamics and Control 24:4 (2000), pp 615-622.

Define the auxilliary sequence:

$\begin{align*} x_n &= \frac{1}{1 + \eta w_n} \end{align*}$

to get:

$\begin{align*} x_{n + 1} &= \frac{(d \eta - c) x_n + c} {b \eta^2 - (a - d) \eta - c) x_n + a \eta + c} \end{align*}$

Picking $\eta$ so that $b \eta^2 - (a - d) \eta - c = 0$, this is a linear recurrence. Bonus is that it is first order, so it can be solved even if the coefficients aren't constant.