The rationally indifferent case sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 03/18/2015, 08:47 AM (This post was last modified: 03/18/2015, 09:07 AM by sheldonison.) This thread is a summary of the rationally indifferent case, where $z \mapsto \lambda z + kz^2 + ...$, where $|\lambda|$=1. The method of Jean Ecalle works for the Abel function for $\lambda=1$, for the parabolic case; see for example: http://mathoverflow.net/questions/179736...l-iterates For the indifferent case, with an irrational multiplier, there is the well known Schroeder equation, but for the indifferent case with a rational multiplier, Ecalle's method for $f^{[\circ n]}$ seems to be all that I have been able to find information for; but I'm still searching for references. For example, see my three year old question on overflow for the 5-periodic case for $z \mapsto \exp(\frac{2\pi i}{5})z + z^2\;\;\;$ http://mathoverflow.net/questions/93411/...-and-repel I have posted an equation whose convergence is unproven: math eretrandre exp(-e) here, for the 2-periodic case, for exp(-e)~=0.0660. I also used it for the simpler 2-periodic solution for $z \mapsto -z + z^2$. This would also work for the 5-periodic case above. Kouznetsov's method also works for the bipolar tetration solution for the 5-periodic case, b~=1.96513877 + 0.441243326i. My own bipolar complex base Kneser solution requires two Schroeder solutions and two theta mappings, so it cannot work for rationally indifferent periodic cases since there is no Schroeder function solution. One could also consider an indifferent case where $\lambda=\exp(2\pi i \alpha)$ multiplier, with $\alpha$ well approximated by a rational number for example very close to 1/5, and $\alpha$ is not a Brjuno_number. Surely such a problem has been considered .... My understanding of Jean Yoccoz's Field's medal is that for such a $\lambda$ multiplier, the Schroeder function does not converge. So then for such a $\lambda$ multiplier, then it would seem that all that is possible is the merged Fatou solution. - Sheldon « Next Oldest | Next Newest »