Iterations of $f(x)=\dfrac{ax+b}{cx+d}$ Consider $f(x)=\dfrac{ax+b}{cx+d}$, where $c\neq0$ and $f(x)$ is not equal to a constant. Is it necessarily true that $f^{[n]}(x)=f(x)$ for some natural number $n > 1$?
 A: Consider $f(x)=\frac{x}{x+1}$. Then $f^{[n]}(x)=\frac{x}{nx+1}$.
If this is true for some $n$ (and it is for $n=1$) then 
$$\begin{align}f^{[n+1]}(x)&=\frac{\frac{x}{x+1}}{n\frac{x}{x+1}+1}\\
&=\frac{x}{nx+(x+1)}\\
&=\frac{x}{(n+1)x+1}\end{align}$$ And so by induction $f^{[n]}(x)=\frac{x}{nx+1}$ for all $n\in\mathbb{N}$. This shows $f^{[n]}$ will never equal $f$ for $n\in\mathbb{N}$.
A: Hint: for each function $f$ consider the corresponding matrix $$\begin{pmatrix} a & b\\ c & d\end{pmatrix}.$$
(Assume that matrices normalized so that $\det \begin{pmatrix} a & b\\ c & d\end{pmatrix} = 1$. The group of such matrices is denoted by $SL(2, \mathbb{R})$; the set of such maps $f$ is isomorphic to the group $SL(2, \mathbb{R})$. See http://en.wikipedia.org/wiki/SL2%28R%29 .)
Note that the matrix corresponding to the composition of functions $f$ and $g$ equals the product of matrices corresponding to functions $f$ and $g$. Thus the problem reduces to finding a matrix $A$ such that $A^n \neq A$ for every $n > 1$. We can choose 
$$A = \begin{pmatrix} \cos \psi & \sin \psi\\ -\sin \psi & \cos \psi\end{pmatrix},$$
where, $\psi$ is such that $\psi/\pi \notin {\mathbb Q}$; e.g. $\psi = 1$. Then 
$$A = \begin{pmatrix} \cos (\psi n) & \sin (\psi n)\\ -\sin (\psi n)& \cos (\psi n)\end{pmatrix} \neq A.$$
A: Let $f(x)=1/(x+1)$. Then $f(f(x))=(x+1)/(x+2)$, and $f(f(f(x)))=(2x+3)/(3x+5)$, the next few being
$$\frac{3x+5}{5x+8},\frac{5x+8}{8x+13},\frac{8x+13}{13x+21},\frac{13x+21}{21x+34}.$$
As this series is continued, one gets coefficients which are adjacent Fibonacci numbers. This can be shown via induction. So no higher iteration of this $f(x)$ is the starting function $f(x)$, and in fact no terms in the series of iterations are equal as functions.
