Illustrating the Leau-Fatou flowers

[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).