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

[Leo.W]

I'd continue to contradict baker. So please show me evidences that showing there ain't a single function that's holomorphic about \(z=0\).

You can have a look into the paper of Baker.
It has something to do with that the parabolic fixed point is a point of non-normality (Julia set) and that the Julia set is a perfect set - means it has no isolated points.
I attach the file here for archiving purpose.

The general theory there is anyways that there are these Leau-Fatou-flower petals around a fixed point of multiplier 1, where you can define Abel-Function (or also called Fatou-Coordinates), hence you have fractional iterates there. And these are *asymptotically* analytic in the fixed point, i.e. the powerseries coefficients converge to those of the formal solution at the fixed point. But the normal thing is that these Abel-Functions/Fatou-Coordinates are different from each other (and hence the iterates). Only in exceptional cases (e.g. LFTs) they agree.

Thank you, bo. I read the paper.

In fact, I don't think my idea contradicts Baker's.

First, when you take singlevalued functions into consideration, like I said, is a limitation for iterations, thus there aint any iterations for any functions except for linear functions and their conjugacies, and because only singlevalued holomorphic invertible functions are linear, only linear functions have iterations. If you banish holomorphity, then you can include rational linear functions.
I also knew some other theorems, one claims: if a singlevalued function f has a n-periodic fixed point a, alternatively \(f(a)\ne a,f^2(a)\ne a,\cdots,f^{n-1}(a)\ne a, f^n(a)=a\), then there lies no n-th iteration root of f. This theorem is enough to prove that \(e^z-1\) has no singlevalued half-iterate.
Just don't ingore the multivaluedness. Because these statements are not puissant when it comes to the real world where almost all functions are multivalued.

Second, as in the paper he claims the statement with the assumption \(f(z),f^2(z),\cdots,f^n(z)\) are all singlevalued, and then applied the theorem to \(f(z)=e^z-1\). But what if applied to \(f^{-1}(z)=\log(z+1)\)? In the inverting case, it doesnt satisfy the assumption anymore, and we only ask for \((f^{-1})^{-\frac{1}{2}}\). This is the way to generate a half-iterate of f.

I understand there's fatal contradictions about parabolic iterations especially involves \(\forall z\in\mathbb{U}_0, f^n(z)\to0\rightarrow (f^n)'(0)\to0\) but \((f^n)'(0)=1\)
This is a classic contradict, I think there're plenties of ways to explain this.
1) This is because the iterations converge so slowly. In other cases when multiplier isn't 1, all points in a neighborhood of a fixed point converges to the fixed point after t iterations, they converge at a exponential speed \(O(s^t)\). But multiplier=1 implies they converge at a very slow speed about \(O(t^{1-\delta})\) for some real \(delta>0\).
2) \((f^n)'(0)=1\) doesnt imply \(\lim_{n\to\infty}{(f^n)'(0)}=1\), all points in that neighborhood, by our converging speed would still culminate at the fixed point, this is pretty like another contradict:
All functions \(f_n(x)=\left\{\begin{matrix}0& x<0\\nx& x\in[0,\frac{1}{n}]\\1& x>\frac{1}{n}\end{matrix}\right.\) are continuous, but as n goes to infinity, it isnt continuous anymore.

I'd follow on. btw I wanna mention up about the iterative logarithm or Julia function, about its multivaluedness.
Now consider an iteration \(f^t\) generated by the schroder function, with multiplier s that \(|s|\ne1,0\), ignore the multivaluedness of schroder function or its inverse, the Julia function must be multivalued since \(f^t(z)=\sigma^{-1}(s^t\sigma(z))\) implies \(f^t(z)=f^{t+2k\pi i\log(s)^{-1}}(z)\), k any integer.
by petals, there should lies halfiterate of a function whose multiplier is -1, me and James discussed earlier. Then in this case, the iteration logarithm gets at a fixed point \(J(f(L))=J(L)=-J(L)=J(L)f'(L)\), showing \(J(L)=0\), then the logarithm fails in a neighborhood of L because \(f^t(z)\sim f^{t+2k}(z)\), the Julia function then should have 2 branch cuts at L.
Also, you can't define julia's function only by \(j(f)=jf'\), to distinct a julia function, you'd need to define an initial value, and claim it's not multiplied by any of the form \(\theta(\alpha(z))\) where theta is 1-periodic and alpha is the abel funct.

But the normal thing is that these Abel-Functions/Fatou-Coordinates are different from each other (and hence the iterates). Only in exceptional cases (e.g. LFTs) they agree.

Please lemme know what is this and why~