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

[JmsNxn]

Leo, If a function is holomorphic in a neighborhood of \(0\), then it is expandable in a taylor series at \(0\).

If a functions IS NOT holomorphic in a neighborhood of \(0\), then it IS NOT expandable in a taylor series at \(0\).

What Baker is saying, is that there is a divergent series at \(0\)--THEREFORE, it is not holomorphic in a neighborhood of \(0\).

(*) Unless it is an LFT-- by which it wouldn't be a Euclidean mapping

But James, holomorphic refers to differentiable, not the convergent Taylor series, it can have divergent series, as mostly hypergeometric functions do?
Here's another example, consider \(f(z)=\sum_{n\ge0}{_2\mathrm{F}_0(n,a;;c)z^n}\), this function diverges because all its coefficients refers to a divergent series, \(_2\mathrm{F}_0(n,a;;c)=\sum_{k\ge0}{\frac{(a)_k(n)_k}{k!}c^n}\), but still can be continued by \(f(z)=-cze^{\frac{z-1}{c}}\big(\frac{z-1}{c}\big)^{a-1}\Gamma(1-a,\frac{z-1}{c})\), choosing proper a, the function is holomorphic, has a divergent series in some sense but indeed a convergent series(Borel sum).

Ps. what if you don't consider the petals of log⁡(1+z)? because e^z−1 maps Re(z)<−M for very large M into the neighborhood of 0, that is why I defined the branch cuts, so that taking both petals of log⁡(1+z) and a "petal" strip from Re(z)<0∧|Im(z)|<1 produces a holomorphic halfiterate at 0

I wonder if you'd try this?

LEO!

Have you not heard of Cauchy's theorem.

If a function is complex differentiable, it is expandable in a taylor series.

If a taylor series converges, it is complex differentiable!!!!

HOLOMORPHY = TAYLOR SERIES EXPANDABLE