Rational difference equation

One approach[7] to developing the transformed variable ${\displaystyle x_{t}}$, when ${\displaystyle ad-bc\neq 0}$, is to write

${\displaystyle y_{t+1}=\alpha -{\frac {\beta }{y_{t}}}}$

where ${\displaystyle \alpha =(a+d)/c}$ and ${\displaystyle \beta =(ad-bc)/c^{2}}$ and where ${\displaystyle w_{t}=y_{t}-d/c}$.

Further writing ${\displaystyle y_{t}=x_{t+1}/x_{t}}$ can be shown to yield

${\displaystyle x_{t+2}-\alpha x_{t+1}+\beta x_{t}=0.}$

This approach[8] gives a first-order difference equation for ${\displaystyle x_{t}}$ instead of a second-order one, for the case in which ${\displaystyle (d-a)^{2}+4bc}$ is non-negative. Write ${\displaystyle x_{t}=1/(\eta +w_{t})}$ implying ${\displaystyle w_{t}=(1-\eta x_{t})/x_{t}}$, where ${\displaystyle \eta }$ is given by ${\displaystyle \eta =(d-a+r)/2c}$ and where ${\displaystyle r={\sqrt {(d-a)^{2}+4bc}}}$. Then it can be shown that ${\displaystyle x_{t}}$ evolves according to

${\displaystyle x_{t+1}=\left({\frac {d-\eta c}{\eta c+a}}\right)\!x_{t}+{\frac {c}{\eta c+a}}.}$

The equation

${\displaystyle w_{t+1}={\frac {aw_{t}+b}{cw_{t}+d}}}$

can also be solved by treating it as a special case of the more general matrix equation

${\displaystyle X_{t+1}=-(E+BX_{t})(C+AX_{t})^{-1},}$

where all of A, B, C, E, and X are n × n matrices (in this case n = 1); the solution of this is[9]

${\displaystyle X_{t}=N_{t}D_{t}^{-1}}$

where

${\displaystyle {\begin{pmatrix}N_{t}\\D_{t}\end{pmatrix}}={\begin{pmatrix}-B&-E\\A&C\end{pmatrix}}^{t}{\begin{pmatrix}X_{0}\\I\end{pmatrix}}.}$