(10/15/2022, 01:17 AM)MphLee Wrote: but I'd like to begin to chew this material... everyone here seems to understand terms and definitions but I'm a total beginner.
What the hell is, formally, a flower?

(10/15/2022, 01:17 AM)MphLee Wrote: Looking at the pictures it seems the image of a path. In particular it seems it has to do with images of the family of paths \[\gamma_n:\mathbb S^1\to \mathbb C\]

(08/28/2022, 11:45 AM)bo198214 Wrote: Just to give some visual impressions I was drawing some Leau-Fatou flowers.
Leau-Fatou flowers occur at parabolic fixed points (i.e. \(f'(z_0)^n=1\) for some integer n>0).
Here we look at the most prominent case \(f'(0)=1\) for the most simple functions, i.e. polynomials.
The Leau-Fatou flower has 2*p petals, where p is the iterative value of f (Ecalle writes it \(p={\rm valit}[f]\)),
which means the index k in the powerseries development of f up to which it is the identity function.
For example \({\rm valit}[z\mapsto z+z^2]=1\), \({\rm valit}[z\mapsto z+z^{10}]=9\).
The petals alternating in attracting/repelling when moving around the fixed point.

So I took a circle around the fixed point and see how it deformed under application of f.
These are the pictures


On each petal there is an Abel function or called Fatou-Coordinate defined, and these functions are typically different on each petal (except the logit is analytic at the fixed point, but in case of polynomials (and entire functions) this is not the case).
The same applies to fractional iterates.
That's why it is interesting to look at the Borel-Summation of the formal powerseries logit, or the formal powerseries of some iterative root.
The Borel-Summation should provide values on a circle around the fixed point, while the circle traverses the different petals.
So if the Borel-summation of a (say) a half iterate returns the a different function on each petal, then there should be breaks visible between the petals.
Maybe the Borel-summation returns only one function and its continuation around the circle - then one would need to find different ways to obtain the other half-iterates.

Just for comparison I put here the Leau-Fatou flower of a function with analytic logit, which has a half-iterate that is analytic at 0.
This function is \(f(x)=-{\rm arccot}(1-{\rm cot}(x))\) it has this power series development at 0:
\[f(x)=x +  x^{2} +  x^{3} + \frac{2}{3} x^{4} - \frac{43}{45} x^{6} - \frac{29}{15} x^{7} - \frac{778}{315} x^{8} - \frac{374}{189} x^{9} + \frac{122}{14175} x^{10} + \frac{782}{225} x^{11} + \frac{3515884}{467775} x^{12} + \frac{45094}{4455} x^{13} + \frac{360268549}{42567525} x^{14} - \frac{1110241}{19348875} x^{15} + O(x^{16}) \]
So it has two petals, but it looks more "round" or convex than the flower of \(z+z^2\):

Also just for comparison the flower of the sine \(x - \frac{1}{6} x^{3} + \frac{1}{120} x^{5} - \frac{1}{5040} x^{7} + \frac{1}{362880} x^{9} + O(x^{10})\) and \({\rm valit}(\sin)=2\) so it has 4 petals:


As final picture "the growth of the flower"   I continuously increase p in \(z+z^p\):

Unfortunately there is a typo in the title, I forgot the r in \(\gamma\) but you know, what I meant!

So hopefully soon, I will continue with the calculation of the Borel-summation.