This is a question that has bothered myself and Gottfried Helms a fair amount of late. He has made his case for the following result, but a proof escapes both of us. The question is deceptively simple, but keeps eluding each of my attempts when we get into the finer details.


Let's start by calling:


$$
f(z) = e^{z}-1\\
$$


Then $f$ has a parabolic fixed point at $0$ with multiplier $1$. By which we have:


$$
f(z) = z + O(z^2)\\
$$


There exists an attracting petal, and a repelling petal; and conveniently they are separated by $i\mathbb{R}$. By which, we are only focused on the attracting petal, which is given as $\Re(z) < 0$. Using the standard Ecalle construction of an Abel function, which I'm mostly just following Milnor's *Dynamics in One Complex Variable*. We can construct an Abel function:


$$
\alpha(f(z)) = \alpha(z) + 1\\
$$


This function is $2 \pi i$ periodic, and is holomorphic for $\Re(z) < 0$. We are primarily focused near the boundary point $z = 0$--where the Abel function has a singularity.


Now we can invert this to the left of zero, so that we can construct a super function $\alpha^{-1}(z)$, and by this, we can get to our function in question:


$$
g(z) = \alpha^{-1}(\alpha(z) + 1/2)\\
$$


I will spare the details, but this function is also $2 \pi i$ periodic, and holomorphic for $\Re(z) < 0$. We don't need this though, just that it's holomorphic on a domain to the left of $0$, with $0$ on the boundary. This function satisfies the equation:


$$
g(g(z)) = f(z)\\
$$




---


Gottfried first came to me with this question, and so I am not the most versed in the construction of this series, but it relates to a question which he had investigated on this forum--of the $\sin$ half iterate. But, the essential statement, is that we can construct a formal series:


$$
h(z) = \sum\_{n=1}^\infty d\_n z^n\\
$$


Such that this series satisfies the formal manipulations:


$$
h(h(z)) = \sum\_{n=1}^\infty \frac{z^n}{n!}\\
$$


By which this series has a radius 0 convergence (it's a divergent series).


The question then becomes, is it possible to represent this divergent series, as Euler would represent the divergent series:


$$
\sum\_{n=0}^\infty (-1)^nn!z^n\\
$$


Using the Laplace transform:


$$
\int\_0^\infty \frac{e^{-zx}}{1+x}dx = \sum\_{n=0}^\infty (-1)^nn!z^n\\
$$


Or, rather, could we do something similar, in the same vein that Borel summation is done.




---


This means, we have the function $g(z)$, which is holomorphic, and a boundary value at $0$. We have a divergent/formal series at $z=0$. If we were to take:


$$
\mathcal{B}h(z) = \sum\_{n=0}^\infty d\_n\frac{z^n}{n!}\\
$$


Will this have a non-zero radius of convergence? By which we could Borel sum the divergent series. I know I'm missing something crucial, and every attempt I've made a using traditional arguments fails because $g(z)$ is a rather anomalous function.


Essentially, can we make the same treatment of Euler's series, of this series. By which we'd be able to pull out a bound $d\_n = O(c^n n!)$ for some $c > 0$.



Borel summation and the Abel function of $e^z-1$
So, there's been some developments on the tetration forum, and I came across two papers:


*Introduction to 1-summability and the resurgence theory*, David Sauzin


<https://hal.archives-ouvertes.fr/hal-00860032v1/document#cite.LY>


And:


*The Fixed Point of the Parabolic Renormalization Operator*,
Oscar Lanford III, Michael Yampolsky


<https://arxiv.org/pdf/1108.2801.pdf>


Long story short, both of these papers provide our result. They specifically mention an expansion by Ecalle for the Abel function: $\alpha(f(z)) = \alpha(z) + 1$ that:


$$
\alpha(z) = \frac{1}{z} - A\log(z) + \sum\_{k=0}^\infty b\_k z^k
$$


Where in these papers they prove that $b\_k = O(c^kk!)$. This doesn't directly answer our question, but, through a bit of algebra, you can identify that:


$$
\alpha^{-1}(1/2 + \alpha(z)) = g(z) = \sum\_{k=0}^\infty g\_k z^k
$$


And that the values $g\_k$ have similar bounds to $b\_k$--this is just mule work as far as I'm concerned.


The word they use specifically, is that the coefficients are of Gevrey class 1. This is further identified by a conversation between Will Jagy and Ecalle, on the thread [Does the formal power series solution to $f(f(x))= \sin( x) $ converge?](https://mathoverflow.net/questions/45608/does-the-formal-power-series-solution-to-ffx-sin-x-converge).


This goes to show that we can express the half iterate $g(z)$ in Euler's manner. Numerical evidence has further confirmed this, as much of us at the tetration forum have been evaluating different and more general integrals using this basis.


This answers the question entirely to my satisfaction, and further answers it using work other than my own. I'm just answering it here for the benefit of anyone who comes across this.


