# two fixed points, same fractional iteration

Suppose a function f(z) has two fixed points, one repelling, and the other attracting. Call the repelling fixed point f(-1)=-1, and the attracting fixed point f(+1)=+1. I'm interested in functions where the fractional iterates are the same, developed from either fixed point.

We can generate fractional iterates, $g_{-1}(z)=f^{oz}$ from the Schroeder function of f(z) developed around the fixed point of -1, and also from the fixed point of +1, $g_{+1}(z)=f^{oz}$. For what functions "f" will the two fixed points agree on their fractional iterates, such that $g_{-1}(z)=g_{+1}(z+k)$, where "k" is a constant?

The only case I can find that works is $f(z)=\frac{z+c}{1+cz}$, where $0<|c|<1$, and the inverse function is $f^{-1}(z)=\frac{z-c}{1-cz}$. Then $g(z)=\tanh(z\tanh^{-1}(c))$, which is derived using the tangent angle sum equation. Are there any other functions f with symmetrical fractional iterates from both fixed points, or is this function family of functions the only functions with symmetrical fractional iterates from both fixed points?

I know of one other case, iterating z^2, involving a super-attracting fixed point of zero, and a repelling fixed point of 1.

• It appears that this formula works for other values of c, with $|c|>1$. For example, $c=\sqrt 5$, leads to the Fibonacci ratio, see math.stackexchange.com/questions/589841/… So perhaps the formula works of $c \ne 1$ and $c \ne -1$. Dec 7, 2013 at 21:03
• In favorites. Tommy and myself have discussed a generalized question like this. And very recent I wondered about your 2 solutions of half-iterates for x - 1/x. I think they are related similarly ( by a moebius transform ). But that is based on Visuals and gut feeling , may be wrong.
– mick
Dec 22, 2014 at 22:26
• Yeah, I remember at the time when I posted this, that it was a Moebius transformation. There is one other "sort of" solution, the parabolic case with $f(x)=\frac{x}{1-x}$, which has only one fixed point, but both the repelling and attracting Leau-Fatou leaves have the same Abel function. Dec 23, 2014 at 8:50

## 1 Answer

A complete answer to this question can be found in the paper

MR0729400 S. Dubuc, Étude théorique et numérique de la fonction de Karlin-McGregor, J. Analyse Math. 42 (1982/83), 15–37.

and references in it. Shortly speaking there are no other rational functions with this property. But of course there are functions on $[-1,1]$, and they are completely described in the references to this paper.

• Thanks! I need to lookup that paper. Mar 31, 2015 at 0:37