Given a recursive homographic sequence : $$ a_{n+1} = f(a_n) $$

where $f$ is an homographic function : $$x \longmapsto \dfrac{ax+b}{cx+d}$$

Where $a,b,c,d \in \mathbb{C}$ such that : $ad-bc \neq 0$

The usual way to solve this type of recurrence relation is to define a sequence $(V_n)_{n \in \mathbb{N} }$ , such that :

Case 1 : $(U_n)$ has two fixed points ($\alpha$,$\beta$)

$$\forall n \in \mathbb{N}: V_n=\frac{U_n- \alpha}{U_n - \beta}$$

Then $(V_n)$ is a geometric sequence with common ration $r=\frac{c\beta +d}{c\alpha +d}$

Note $U_0$ has to be different than $\alpha$ or $\beta$ otherwise $(U_n)$ is stationary.

Case 2 : $(U_n)$ only admits one fixed point ($\alpha$)

$$\forall n \in \mathbb{N} : V_n=\frac{1}{U_n-\alpha}$$

Then $(V_n)$ is an arithmetic sequence with common difference $d=\frac{2c}{a+d}$

Note $U_0$ has to be different than $\alpha$ otherwise $(U_n)$ is stationary.

My questions :

How would one come up with such a sequence $(V_n)$ ?

What are the fixed points role in deriving the sequence $(V_n)$ ?

What is the intuition behind the sequence ($V_n$) and its intepretation ?

Also , In one of my attemps to solve a specific homographic recurrence relation , I've found that in the case where $f$ admits two fixed points , the common ratio $r$ equals the ratio of the eigen values of the matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix} $ Why ?

I'd like to note that i did my research on the subject but i could not find any satisfactory answer , it appears that all the documents i've consulted did not motivate the solution and rather stated it as is .

Any piece of documentation would be greatly appreciated !


1 Answer 1


As you noticed, given a homographic function (Moebius transformaion) $$f: x \longmapsto \dfrac{ax+b}{cx+d}$$ where $a,b,c,d \in \mathbb{C}$ such that : $ad-bc \neq 0,$ There are only two cases.

Case 1. The function $f$ has two fixed points. This case is equivalent under a suitable homography to fixed points $0$ and $\infty$ which is $ x \mapsto rx $ for some $r$. In more detail, Suppose that $f_1, f_2$ are the two fixed points and $s$ is any other number. By the property of homographies, there is a unique one $g$ which maps $(0, \infty, 1)$ (in that order) to $(f_1, f_2, s)$. Conjugate $f$ by $g$ to get the map $h: x \mapsto g^{-1}(f(g(x))).$ Now notice that $$h(0) = g^{-1}(f(g(0))) = g^{-1}(f(f_1)) = g^{-1}(f_1) = 0$$ which implies that $0$ is a fixed point of $h$. Similarly for $\infty$.

Case 2. The function $f$ has one fixed point. This case is equivalent under a suitable homography to fixed point $\infty$ which is $ x \mapsto \delta+x $ for some $\delta$.

  • $\begingroup$ Where can i learn more about this , I apologize for not being able to follow the answer you provided. Per say why do fixed points imply such homographies ? $\endgroup$ Dec 17, 2022 at 18:06

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