Half-iterate exp(z)-1: hypothese on growth of coefficients

[Leo.W]

Hey, Leo.

I'm disagreeing with you--and until I see concrete evidence I will continue to do so. This is not the function in question at all. You are talking about a different function entirely.

You have developed a Taylor series about \(0\). This is a question about the divergent series that Gottfried has developed. You have made a function \(g\) that is holomorphic near zero (which I'm not sure about, as I'm pretty sure this contradicts Baker--Which Gottfried mentions). There exists no half iterate that is holomorphic near zero. So you've made a mistake, or Baker has made a mistake. We cannot expand an iteration in a neighborhood of a parabolic fixed point, unless that function is an LFT. Bo and myself even saw this recently in the Karlin Mcgregor paper, where it is written in the beginning.

By your logic, we've now solved \(f(z) = e^{z}-1\) and found a function \(f^{\circ t}(z)\) which is holomorphic about \(z = 0\). That just doesn't happen. Because you are brushing the paths of the Julia set and the attracting/repelling petal. The abel function cannot be expanded in a neighborhood of zero. And through common transformations, we can turn your iteration into an abel function, and now we have an abel function which is holomorphic for \(z \neq 0\) and \(|z| < \delta\). That doesn't happen. Abel functions are not holomorphic like this at parabolic points.

I can't speak to the accuracy of this graph, because I do not understand how you are constructing it. Additionally, a graph doesn't mean holomorphy. Just because it looks holomorphic, doesn't make it holomorphic. What is far more likely, is that you have expanded an asymptotic series, that converges fairly well, and iterated functional relationships to express how close of an asymptotic it is.

This is equivalent to saying that you've found a holomorphic iteration \(\exp^{\circ t}_{\eta}(z)\) such that this expression is holomorphic in \(z\) about \(e\). It's well known this isn't possible.

I apologize, but until I see some form of hard evidence I'll continue to disagree with you. Baker himself has a paper about no Taylor expansion for a half iterate of \(e^z -1\) about \(z=0\). Straight from the horses mouth. See the top answer here https://mathoverflow.net/questions/4347/...ar-and-exp

So unless you want to contradict baker, I'm sorry.

No need to feel sorry, James bro. It's nice to exchange ideas with brilliant guys~

I'd continue to contradict baker. So please show me evidences that showing there ain't a single function that's holomorphic about \(z=0\).
It's insufficient if you say that the Taylor series diverges or there can't be one single Taylor expansion at z=0. It exists and always exists, as we can compute it. A function has a divergent series don't mean the function doesn't exist, for example the series \(f(z)=\sum_{n=0}^{\infty}{n!z^n}\), it is simply the expansion of the function \(\frac{1}{z}e^{-1/z}\mathrm{Ei}(\frac{1}{z})\) at z=0, choosing the right branch cut and asymptotic expansion at z=0 will also give you \(1+z+2!z^2+3!z^3+\cdots\). And beyond that we have ways to analytic continue the series, at least you can't say it's not holomorphic because its series diverge, since there are tons of analytic continuations, especially the hypergeometric functions \(_p\mathrm{F}_q\).
But okay, maybe the definition of the term "holomorphic" diverges?

Yes, you're right, the photo doesnt speak a single truth, I only see a photo as a hint or something that showing there's probabilities, about the existance. From my main branch cut of half iterate \(g\) of \(e^z-1\), p1 shows \(g(g(z))-e^z+1\), because it's multivalued only one branch cut won't cover the whole plane, but still covers around z=0. p2 shows its derivative, and thus hinting it's analytic, real-to-real. 
and *if, just if, you brush off the multivaluedness, you're also banishing the existence of the Julia solution, abel solution and so on, because they're all multivalued, that would be nonsense for us to do researches into flows and iterations.

Why can't we expand parabolic iterations?
I think I have also a similar question to this, for example, it's known that \(f(z)=z+\frac{1}{z}\) has parabolic fixed point at infinity, then why \(f^2(z)=z+\frac{2}{z}+O(z^{-3})\) is parabolic and works and still holomorphic at infinity(what I mean is by conjugation it's still holomorphic at parabolic points)?
And why this 2nd iteration's asymptotic expansion, can't satisfy that \(f^2(z)=f^2(\frac{1}{z})\)?
It's just because the information is localized, and yet the series can't show that \(f^2(z)=f^2(\frac{1}{z})\)
I think there's something wrong, The paper will claim in a singlevalued style, and also it claims

However, there are two half-iterates (or associated Fatou coordinates \(\alpha(e^z - 1) = \alpha(z) + 1\)) that are holomorphic with very large domains. One is holomorphic on the complex numbers without the ray \(\left[ 0,\infty \right)\) along the positive real axis, the other is holomorphic on the complex numbers without the ray \(\left(- \infty,0\right]\) along the negative real axis. And both have the formal power series of the half-iterate \(f(z)\) as asymptotic series at 0.

Only taking singlevaluedness will cause so many limitations when you want a single iteration. Expanding the asymptotic series and Taylor series is equivalent if they're differentiable and has no poles or essential singularities and not at infinity, just because Taylor series is only a special type of wider series. I don't see any reason why it can't be holomorphic around z=0.

I'll just say, there's no holomorphic iterates, if singlevalued, and there won't even be any holomorphic inverse functions, because all inverse functions are multivalued or not-holomorphic except for linear functions, how will you invert z^-1 with holomorphic functions? The singlevaluedness is a limitation, so is the holomorphity. You can't get an iteration always holomorphic, but at least at some points, or in some domains they do, that's enough.
I can show you how to get a main branch cut of it if you're interested. And please show me some evidence that there's not a single half iterate(multivalued or partially-holomorphic). I'll be very appreciative.

@bo: Thank you bo, seems helpful. I think the asymp be proved very soon, looking forward~