(08/05/2021, 04:36 PM)Leo.W Wrote: And do you already solved the noncontructable cases? I'm like, stuck in the very initial case \( f(z)=-z(1-z) \) at \( f(0)=0,f'(0)=-1 \), I wonder if anyone had already successfully construct a second iterative root?
Though if you assume the second iterative root has a Taylor series at z=0, quickly sooner you'll notice it's not helpful, contradicting the fundamental laws of iterated functions.
Regards
Leo
This is certainly computable. Start by taking \( f^{\circ 2}(z) \) which has a fixed point \( f(0)=0,\,f'(0) = 1 \). From here we would use Ecalle's construction of the abel function. It will not be holomorphic in a neighborhood of zero--you are correct; zero will be on the boundary. The function \( f^{\circ 2}(z) \) has an "attracting petal" and a "repelling petal". Where a Petal will be a domain in \( \mathbb{C} \) in which;
\(
\lim_{n\to\infty} f^{\circ 2n}(z) = 0\,\,\text{for the attracting petal}\\
\lim_{n\to\infty} f^{\circ 2n}(z) \,\text{diverges for the repelling petal}\\
\)
The value \( 0 \in \overline{\mathcal{P}_{\pm}} \) where \( \mathcal{P}_{\pm} \) is the attracting/repelling petal respectively.
I'm going to follow Milnor here and give you a quick sketch on how to construct the Abel function. I highly suggest reading John Milnor's Dynamics in one complex variable. This is found at about page 104; and continues to page 120 or so. I'm only giving a rough rough sketch here; as the arguments are very intricate.
To begin; our function:
\(
f^{\circ 2}(z) = g(z) = z + a z^2+...\\
\)
The \( 2 \) means we only have 2 petals. One repelling; and one attracting.
We first make the substitution \( w = \frac{-1}{2az^2} = \frac{c}{z^2} \); and concern ourselves with the attracting petal; where we now have a fixed point at \( w = \infty \). I won't walk through the whole proof; just the exact construction method. We're going to find a mapping \( \psi(w) \) such that,
\(
g\circ \psi(w) = \sqrt{\frac{c}{w}}(1+a\frac{c}{w} + o(1/w))\\
\)
Then taking,
\(
F(w) = \phi \circ g \circ \psi = w + 1 + o(1)\\
F(w) = w + 1 + O(\frac{1}{sqrt{|w|}})\\
\)
By composing with the map \( \phi(z) = c/z^2 \). We've succesfully "changed our variables" to the best possible situation. A large section of the petal will now be mapped to \( \Re(w) > R \) for large enough \( R > 0 \). Choose a point \( \hat{w} \) in this half plane, and,
\(
\beta(w) = \lim_{k\to\infty} F^{\circ k}(w) - F^{\circ k}(\hat{w})\\
\)
satisfies the Abel equation,
\(
\beta(F(w)) = \beta(w) + 1\\
\lim_{w \to \infty} \frac{\beta(w)}{w} = 1\\
\)
And now; we simply change our variables back to \( z \) and we've successfully constructed an abel function \( \alpha \) on the attractive petal. A similar procedure can be done with the repelling petal by taking the inverse \( g^{-1} \).
I suggest reading up on Ecalle's method; it's really quite a beautiful construction. There are many fascinating questions. Luckily; the logistic function is rather simple to handle; more complex functions produce much more difficult questions. Specifically; classifying petals; how many there are; how many are repelling/attracting etc...
Anyway, for our case here; we can construct a unique Abel function on \( \mathcal{P}_{\pm} \) calling \( \alpha_\pm \)--which is two abel functions. Then, locally we can define,
\(
h_{\pm}(z) = \alpha^{-1}_{\pm}(\alpha_{\pm}(z) + 1/4)\\
\)
Where,
\(
h_+(h_+(z)) = f(z) \,\,\text{for}\,\,z \in \mathcal{P}_+\\
h_-(h_-(z)) = f(z) \,\,\text{for}\,\,z \in \mathcal{P}_-\\
\)
I hope this helps. You should note that \( \mathcal{P}_+ \cap \mathcal{P}_- = \emptyset \). And that these two sets are separated by the julia set.
Regards, James
I made an error with how we find the square root of \( f \). It's too late to fix now; but \( f \) may have two repelling petals ; which would mean we'd have to take another change of variable. The same result pretty much stands; there just might be a change of variable at the end. It'll just amount to a bunch of minus signs is all really.