elementary superfunctions

[bo198214]

What does mathematica say about
\( f(x)=\frac{x-1}{x+1} \)?
This is strictly increasing and has no real fixed point.

It gives

\(
F(x)=\frac{\left(-\frac{1}{2}-\frac{i}{2}\right)^x(i C-1) + \left(-\frac{1}{2}+\frac{i}{2}\right)^x(i C+1)}{\left(-\frac{1}{2}-\frac{i}{2}\right)^x(i+ C)+\left(-\frac{1}{2}+\frac{i}{2}\right)^x(i-C) }
\)

Well we can do better:
\( f^{\circ u}(z)=\frac{\cos(\frac{\pi}{4}u)z-\sin(\frac{\pi}{4}u) }{\sin(\frac{\pi}{4}u)z+\cos(\frac{\pi}{4}u)} \)
more detailed information in this article

To be complete I will also give Gottfried's solution for \( f(z)=\frac{1}{z+1} \) which has two real fixed points at \( \pm\frac{1}{2}\sqrt{5}-\frac{1}{2} \), and singularity at -1:
\( f^{\circ u}(z)=\frac{\operatorname{fib}_u-\operatorname{fib}_{u-1}z}{\operatorname{fib}_{u+1}-\operatorname{fib}_{u}z} \)
\( \operatorname{fib}_u=\frac{\phi^u-(1-\phi)^u}{\sqrt{5}} \)
\( \phi=\frac{1+\sqrt{5}}{2} \)