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

[JmsNxn]

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. Not to be harsh leo, but your solution is an asymptotic expansion. Unless you want to rewrite a hundred years of complex dynamics, you can't be right.