Periodicity

For a periodic composition, we want

\[ p_n(x)=x \]

and thus, keeping our conventions from the general case, we want

\[ \frac{a_nx+bk_n}{k_nx+c_n}=x \]

\[ \left\{ \begin{array}{ll} a_n=c_{n}\\ k_{n}=0 \end{array} \right. \]

One can check that the second condition is sufficient for the first one to be satisfied so we will only considere this one. This second equations is equivalent to

\[ \left(a+c+\sqrt{\rho}\right)^n=\left(a+c-\sqrt{\rho}\right)^n \]

which is equivalent to

\[ \left(1+\frac {\sqrt{\rho}} {a+c}\right)^n=\left(1-\frac {\sqrt{\rho}}{a+c}\right)^n \]

except that it eliminates the solution \(c=-a\), which is a solution of period 2. To continue, we can pose \(z=\frac{\rho}{(a+c)^2}\) and apply the binomial theorem:

\[ \sum_{i=0}^j \begin{pmatrix}j\\i\end{pmatrix} z^{\frac{i-1}{2}}=\sum_{i=0}^j \begin{pmatrix}j\\i\end{pmatrix} (-1)^{i}z^{\frac{i-1}{2}}. \]

Let \(i=2m+1\) and \(k=j-1\), we finally have

\[ \sum_{m=0}^{\lfloor\frac{k}{2}\rfloor}\frac{k!}{(2m+1)!(k-2m)!}z^m=0. \]

Finding solutions in z, we can go back to the rational function via

\[ b=\frac{(z-1)(a^2+c^2)}{4}+\frac{(z+1)ac}{2}. \]

Here is a list of solutions for z:

Period		z
1		1
2	\(a=-c\)	1
3		1	-3
4	\(a=-c\)	1		-1
5		1			\(\pm2\sqrt{5}-5\)
6	\(a=-c\)	1	-3			\(-\frac 13\)

Following solutions can be computed but they become complicated very fast. Also from a certain point, it asks to solve a polynomial of degree of 5 or more, even taking into account solutions that we get from sub-periods.

This is, as far as I know, a complete description of the periodicity of composition of rational linear functions.