Illustrating the Leau-Fatou flowers

[bo198214]

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.

[JmsNxn]

HOLY FUCK, BO!

You have a paper and a half here!

HOLY FUCK! 

Just remember to thank me when I explain how Borel sums work!

[Gottfried]

Wonderful! What a pleasure to see this! 

We seem to have a much inspired time here currently!  

Regards -

Gottfried

[JmsNxn]

Hey, Bo. I thought I'd add an equivalent expression to the Leau Flower theorem, which is just my way of phrasing The result through Milnor's lens.

If \(\mathcal{N}\) is a neighborhood of \(0\); then:

\[
\bigcap_{n=0}^\infty f^{\circ n}(\mathcal{N}) \cup f^{\circ -n}(\mathcal{N}) =\mathcal{U}\\
\]

And the set \(\mathcal{U}\) is a neighborhood of \(0\). Which is just a fancy way of doing the petal theorem.

[bo198214]

We seem to have a much inspired time here currently!  

Absolutely!

\[
\bigcap_{n=0}^\infty f^{\circ n}(\mathcal{N}) \cup f^{\circ -n}(\mathcal{N}) =\mathcal{U}\\
\]

In this post I just want to (before we start with the Borel summation) showing the different half iterates of \(f(x)=e^x-1\) - the classical way!
Two neighbouring petals are overlapping and one can compare the values in the overlapping area.
There one can get half-iterate on the attracting petal by following the iterates of \(f\) and the half-iterate on the repelling petal by following the iterates of \(f^{-1}\). In our case \(f(x)=e^x-1\) the attractive petal is some neighbourhood of 0 without  \((0,\infty)\) and the repelling petal is some neighbourhood of 0 without \( (-\infty,0)\).
These are the trajectories from 0+0.5i:
   
The regular half iterates on the attractive and repelling petals can be defined as
\begin{align}
f^{\circ \frac{1}{2}}_-(z)  &= \lim_{n\to\infty} f^{\circ - n}(h_{20}(f^{\circ n}(z)))\\
f^{\circ \frac{1}{2}}_+(z) &= \lim_{n\to\infty} f^{\circ n}(h_{20}(f^{\circ -n }(z)))
\end{align}
where \(h_{20}\) the formal (non-converging) powerseries of the half iterate truncated to 20 (though 2, i.e. \(z+\frac{1}{4}z^2\) would be sufficient, imho).

So from a circle of radius 3 I took an arc not containing the positive real axis for \(f^{\circ \frac{1}{2}}_-\) and an arc not containing the negative real axis for \(^{\circ \frac{1}{2}}_+\) and these are the images of the arcs under the half iterates.
   
While I could take close to the full arc for the blue curve (\(-0.9\pi..0.9\pi\)) I only could use a smaller part of the arc (\(0.4\pi..1.6\pi\)) for the red curve because the exponentials just increase too much numerically to the right.
But one can really see that both differ a lot! It is not as tiny a difference as with the half iterates of \(\sqrt{2}^x\) (on the other hand we didn't look at the complex plane there).

[JmsNxn]

We seem to have a much inspired time here currently!  

Absolutely!

\[
\bigcap_{n=0}^\infty f^{\circ n}(\mathcal{N}) \cup f^{\circ -n}(\mathcal{N}) =\mathcal{U}\\
\]

In this post I just want to (before we start with the Borel summation) showing the different half iterates of \(f(x)=e^x-1\) - the classical way!
Two neighbouring petals are overlapping and one can compare the values in the overlapping area.
There one can get half-iterate on the attracting petal by following the iterates of \(f\) and the half-iterate on the repelling petal by following the iterates of \(f^{-1}\). In our case \(f(x)=e^x-1\) the attractive petal is some neighbourhood of 0 without  \((0,\infty)\) and the repelling petal is some neighbourhood of 0 without \( (-\infty,0)\).
These are the trajectories from 0+0.5i:

The regular half iterates on the attractive and repelling petals can be defined as
\begin{align}
f^{\circ \frac{1}{2}}_-(z)  &= \lim_{n\to\infty} f^{\circ - n}(h_{20}(f^{\circ n}(z)))\\
f^{\circ \frac{1}{2}}_+(z) &= \lim_{n\to\infty} f^{\circ n}(h_{20}(f^{\circ -n }(z)))
\end{align}
where \(h_{20}\) the formal (non-converging) powerseries of the half iterate truncated to 20 (though 2, i.e. \(z+\frac{1}{4}z^2\) would be sufficient, imho).

So from a circle of radius 3 I took an arc not containing the positive real axis for \(f^{\circ \frac{1}{2}}_-\) and an arc not containing the negative real axis for \(^{\circ \frac{1}{2}}_+\) and these are the images of the arcs under the half iterates.

While I could take close to the full arc for the blue curve (\(-0.9\pi..0.9\pi\)) I only could use a smaller part of the arc (\(0.4\pi..1.6\pi\)) for the red curve because the exponentials just increase too much numerically to the right.
But one can really see that both differ a lot! It is not as tiny a difference as with the half iterates of \(\sqrt{2}^x\) (on the other hand we didn't look at the complex plane there).

Super fascinating! I really love these explanations by you, because they are reinforcing many things I've read, but never quite "clicked with". I've always focused too much on geometric cases; and I always thought I got the parabolic case, but I wasn't confident enough to speak on it. These last few weeks of your comments and visual descriptions have really made me see it clearer.

Thanks a lot for the graphs too, I really fucking love these Leau Flower Petal graphs. Milnor is so stingy with the visual arguments, lol.

Great having you active again, bo!

Regards, James

[bo198214]

Super fascinating! I really love these explanations by you, because they are reinforcing many things I've read, but never quite "clicked with". I've always focused too much on geometric cases; and I always thought I got the parabolic case, but I wasn't confident enough to speak on it. These last few weeks of your comments and visual descriptions have really made me see it clearer.

I am glad and that's super motivating too.

Also it cleared up things for me. For example I always thought that the parabolic case is (in a quite direct manner) the merge of hyperbolic fixed points. But this is surely not true when considering the number of petals which is 2*(n-1) while the number of perturbed fixed points is n. (Only for the case n=2 the numbers would equal.) Also depending on the perturbation the fixed points would come from different directions and have different multipliers. I made a picture of the perturbed Leau-Fatou flowers for the case n=5, \(f(x)=x+x^5 - c^5\) where I moved c around 0, \(|c|=0.4\) or \(|c|=0.2\):
       
The 5 fixed points are equally spaced on the circle with radius 0.4 or 0.2.
Red fixed point means repelling, green means attracting.

PS: the left picture is like a symbol of the outer chaotic world, while the right picture looks like a state of meditation ... perhaps we just have to find the parabolic fixed point inside ourselves!

[Gottfried]

PS: the left picture is like a symbol of the outer chaotic world, while the right picture looks like a state of meditation ... perhaps we just have to find the parabolic fixed point inside ourselves!

  At first view I thought the right one would give a nice toy for childs, a Babuschka in calm, while the left the Babuschka is running a 100 m hurdle line... Now - you've seen a state of meditation... so, why not that latter one

Amazing -

Gottfried

[bo198214]

  At first view I thought the right one would give a nice toy for childs, a Babuschka in calm, while the left the Babuschka is running a 100 m hurdle line... Now - you've seen a state of meditation... so, why not that latter one

Haha, I can see your appeal to the matryoshka doll!

[bo198214]

So just for Tommy

Just like exp(x) - 1 has 2 solutions that fundamentally disagree ( no 1 periodic function unites them ) 

I show how the two Abel functions of \(e^x-1\) are connected by a 1-periodic function.
Normally it is clear that if you have two Abel functions \(\alpha_+\) and \(\alpha_-\) of \(f\) that are defined on a suitable region where you can take the inverse and which includes \(z\) and \(z+1\) that for \(\tau=\alpha_-\circ \alpha^{\circ -1}_+\):
\[ \tau(z+1)=\alpha_-(\alpha^{\circ -1}_+(z+1))=\alpha_-(f(\alpha^{\circ -1}_+(z)))=\alpha_-(\alpha^{\circ -1}_+(z))+1=\tau(z)+1 \]
Hence \(\theta(z)=\tau(z)-z\) is a 1-periodic function, and you can express \(\alpha_-(z)=\alpha_+(z+\theta(z))\)

In case of two adjacent petals the above conditions are satisfied and we can compute \(\tau\).
Lets do that for \(f(x)=e^x-1\):
We calculate the inverse Abel function as before:

\begin{align}
f^{\circ \frac{1}{2}}_+(z) &= \lim_{n\to\infty} f^{\circ n}(h_{20}(f^{\circ -n }(z)))
\end{align}
where \(h_{20}\) the formal (non-converging) powerseries of the half iterate truncated to 20 (though 2, i.e. \(z+\frac{1}{4}z^2\) would be sufficient, imho).

only that we change the \(\frac{1}{2}\) to \(t\). 
\[ \alpha_+^{\circ -1}(t) = f^{\circ t}_+(z_0) \]


The Abel function can be defined similarly:
\begin{align}
\alpha_-(z)=\lim_{n\to\infty} \alpha_{20}(f^{\circ n}(z)) - n
\end{align}
where
\begin{align}
\alpha_{20}(z)=\frac{1}{3}\log(-z)-\frac{2}{x} - \frac{1}{36}x + \frac{1}{540}x^{2} + \frac{1}{7776}x^{3} - \frac{71}{435456}x^{4} + \frac{8759}{163296000}x^{5} + \frac{31}{20995200}x^{6} - \frac{183311}{16460236800}x^{7} + \frac{23721961}{6207860736000}x^{8} + \dots
\end{align}

So with these ingredients we can calculate \(\tau(t)=\alpha_-(\alpha_+^{\circ -1}(t))=\alpha_-( f^{\circ t}_+(z_0))\) where I choose \(z_0=3i\) and let t run from 0 to 2 and show the curve \(\tau(t)\) in the complex plane: