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

[Leo.W]

Interestingly the iterative logarithm of \(e^x-1\) follows a similar pattern.
The iterative logarithm is an indicator whether all iterates at the fixed point are analytic, the defining equation is \(j(f(x))=f'(x)j(x)\).
It also satisfies \(j(x)=\frac{\partial f^{\circ t}(x)}{\partial t}\big|_{t=0}\).
It is called iterative logarithm because if you consider it a functional mapping f to j, i.e. \(j=\text{logit}[f]\) then you have \(\text{logit}[f^{\circ t}] = t\;\text{logit}[f]\).

Anyways this is the pattern:


with the advantage that it doesn't depend on t, which we took as t=1/2.

The half iterate of \(\log(z+1)\) has a similar asymptotic expansion at zero.

There are two half iterates at \(0\), they are holomorphic for \(g^-(z)\) for \(\Re(z) < 0\) and \(g^+(z)\) is holomorphic for \(\Re(z) > 0\) (which is developed using the attracting petal of \(\log(z+1)\)). Both functions produce the same asymptotic series at \(0\). And we are given \(g^+(g^+(z)) = g^{-}(g^{-}(z))= e^z-1\) while being two different solutions. But if you expand the asymptotic series about \(z \approx 0\) they produce the same asymptotic series. Per, again, Baker.

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@Leo

Also, I have no idea what you are talking about Leo. If a function isn't holomorphic at \(0\), there is no taylor series. If a function is holomorphic at \(0\), there is a taylor series. You are trying to say that it's holomorphic at 0 is in direct contradiction to baker. I literally just showed you the paper.

So the halfiterate of e^z-1 is holomorphic at 0? Maybe my sentences are a lil bit tangled up, I apologize.
I'm not saying holomorphic at 0 is in direct contradiction to baker. I read the paper, it just says about holomorphity in the whole plane, not at a single point. Also it says the series diverge, but diverge is not the same as not holomorphic. Also the halfiterate can be analytic continued through 0, so to the whole plane.
Does "holomorphic" mean the same to both of us? I meant the analytic continued version has the same series(asymp or even Taylor) at 0.
I think the 2 different solutions should be different branch cuts of the same halfiterate. Plus there're more halfiterates.
I'm very curious now about how 

There are two half iterates at \(0\), they are holomorphic for \(g^-(z)\) for \(\Re(z) < 0\) and \(g^+(z)\) is holomorphic for \(\Re(z) > 0\) (which is developed using the attracting petal of \(\log(z+1)\)).

is derived? maybe you could give me a hint

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 \(Re(z)<0\wedge |Im(z)|<1\) produces a holomorphic halfiterate at 0