09/11/2022, 11:36 AM
(09/10/2022, 12:08 PM)tommy1729 Wrote: Just like exp(x) - 1 has 2 solutions that fundamentally disagree ( no 1 periodic function unites them )
Two things:
- There are two types of multiplicities here: One comes from the petals - for \(f(x)=x+a_{n+1}x^{n+1} + O(n+2)\) you always have 2n petals with (typically different) Abel functions, which then provide 2n solutions for all fractional/real/complex iterates. But all these solutions have multiplier 1, i.e. are of the form \(f^{\circ t}(x)=x+ta_{n+1}x^{n+1} + O(n+2)\) asymptotically at 0.
However in this thread we are looking at fractional iterates with multiplier being an n-th root of unity! In the case \(f(x)=e^x-1\) is n=1, which means we can not take an iterational root with multiplier say -1, or i. We can do this for example for \(f(x)=x+x^3\), which has n=2, i.e. taking a half iterate with multiplier -1 is possible here:
\[ f^{\circ \frac{1}{2}|2} = - x - \frac{1}{2} x^{3} + \frac{3}{8} x^{5} - \frac{9}{16} x^{7} + \frac{133}{128} x^{9} - \frac{507}{256} x^{11} + \frac{3399}{1024} x^{13} - \frac{6261}{2048} x^{15} - \frac{283347}{32768} x^{17} + \frac{4149645}{65536} x^{19} + O(x^{21}) \]
- Of course they are united by a 1-periodic function - every two Abel functions of a function f that are suitably overlapping are "united" by a 1-periodic function.
But I made some extra effort to even show that numerically, please have a look here.
PS: Calling something "almost trivial" with only a vague understanding of the problem bears a bad taste ...