I noticed that for the (real) function $f(x)=\frac{1}{1-x}$, $f(f(f(x)=f^3(x)=x$ for all real $x$.
This surprised me, and I was naturally curious about rational functions that elicit the identity after 4 or 5 or, in general, $n$ iterations.
I looked at rational functions of the form $f(x)=\frac{1}{z-x}$ for $z\in\mathbb{R}$ and wondered for which $z$ did $f^n(x):=f(f(...f(f(x))...))=x$.
After playing around for a bit on Wolfram and Desmos, I came up with the following conjecture: If $n\in\mathbb{N}$ and $n\ge2$ and$f(x)=\frac{1}{2\cos(\frac{\pi}{n})-x}$, then $f^n(x)=x$.
Note that I do not claim that $f$ is the only rational function of the form $f(x)=\frac{1}{z-x}$ where $z\in\mathbb{R}$.
So I have a few questions:
1) Is this conjecture true?
2) How would someone go about proving it, if it is true?
3) Is this a well-known theorem or a somewhat-immediate corollary or special case of a well-known theorem? If so, what is that theorem?
4) How would I find all $z\in\mathbb{R}$ such that if $f(x)=\frac{1}{z-x}$, then $f^n(x)=x$? Is such a thing easy to do?
5) What field of math would ask questions like this?
The conjecture seems to be true and if it is, it could be proved by mathematical induction.I checked it for up to $n=4$ and it worked. There are other values for $z$ for example for $n=3$, we also have $z=-1$ and for $n=4$ we have $z=\pm \sqrt 2$ and $z=0$
Note that with $f^3 =f$ we have also $f^5=f$ and $f^7=f$,....
You find these type of questions in discrete dynamical systems.