Maybe the solution at z=0 to f(f(z))=-z+z^2

[Leo.W]

maybe a silly comment but

why not consider h(z) = - z + z^2 , g(z) = h(h(z))

and then consider the super and abel of g(z) ?

regards

tommy1729

Hey tommy, the reason was discussed way earlier at https://math.eretrandre.org/tetrationfor...318&page=4
I can repeat lol

Abel part:
It'll be infelicitous for us to compute the Abel function for a nonconstructible type, we call a such function \(f\). Let's say \(q\) is the smallest positive integer making \(f^q(z)\) is constructible, or having \(1\) as its multiplier.
The Abel function \(\alpha_f(z)\) we iterate to converge is actually a scaled version of different branches of the same Abel function of \(f^q\): \(\alpha_{f^q}(z)\), it has the property \(\alpha_{f^q}(f^q(z))=\alpha_{f^q}(z) + 1\), but never \(\alpha_{f^q}(f(z))=\alpha_{f^q}(z) + \tfrac{1}{q}\) since this Abel function works officially for the regularized root of \(f\).

* We call \(g\) as a regularized root of \(f\) (a function having multiplier \(1\)) if \(g\) also has \(1\) as its multiplier.
And every such function \(f\) has at least 1 regularized root. We can write \(g =(f^q)^{\tfrac{1}{q}}\)


Let's look at this example \(f(z)=-z+z^2\).
We first calculate \(f^2(z)=z-2z^3+z^4\), that's good, we have \(q=2\) and a totally closed form of the inverse function, making the computation easier as fuck. Then we calculate the Abel function \(\alpha_{f^2}(z)=\frac{1}{4z^2}+\frac{1}{4z}+\frac{11}{8}\log(z)+O(z)\) around \(z=0\), and we use iteration to make it converge by \(\alpha_{f^2}(z)=\lim_{k\to\infty}{\alpha_{f^2}(f^{2k}(z))-k}\). Whoa- it just converges as expected!
But when we do computation and find the inverse abel, the results just got us cold as it
a) it's multivalued as fuck
b) it'll oscillate among branches, for example we won't have \(\alpha_{f^2}(f(z))=\alpha_{f^2}(z)+\tfrac{1}{2}\) but totally 2 different branches.
c) its result is exactly the fucking scaled abel function of \((f^2)^{\tfrac{1}{2}}(z)=z-z^3+\tfrac{1}{2}z^4-\frac{3}{2}z^5+O(z^6)\), this is what we call the regularized root of \(f^2\).

* so for b), we have \(\alpha_{f^2}((f^2)^{\tfrac{1}{2}}(z))=\alpha_{f^2}(z)+\tfrac{1}{2}\) exactly
Any Abel function generalized in this way, terminating in the regularized root case, that is so different from our initial \(f\).

Superfunction part:
It can be calculated in the same procedure in Abel part, and we still gets into the blackhole of regularized root.
My "P Method" was to combine in the simplistic way the 2(or N) different branch of a superfunction, and it's successful at many cases, but not this nonconstructible case. In this case the merged superfunction either diverge to total discontinuities or converge to 0 for all noninteger \(z\).
Besides, if you really constructed a superfunction, it must be oscillative since the multiplier is -1, or generally nonconstructible, this superfunction would contradicts with the Abel function you construct, then you can't combine the 2 functions to create iterations. If you really wanna do so, computing a nice enough Abel function is the eternal nightmare.

And the point is their combination gives only polynomial, or Taylor series for iterations instead, like I computed, that \(f^{1/2}(z)\) just has no such series, but an multivalued asymptotic expansion. You may notice about the wierd \(\pi\) appearing in the formula of f, in fact it's descended from \(\log(i)\), which is multivalued.