JmsNxn's new workout of the growthrate of coeffs in halfiterate of exp(z)-1f°0.5(z)

[Gottfried]

It seems very likely that James has now hit the nail for the proof of summability of the fractional iterates of \( \exp(z)-1 \).   

Going with this now many days recently, it came to me that I have this nice wedding-picture at the wall, where my older brother and my sister-in-law had taken a photo and had let it convert to an acryl-painting on linnen having a wooden frame. 
See this:
     

They gave me the honor of citing a topic of my webpage - namely the halfiteration-thingy. At one edge of this wooden block-frame my then hobby of the challenge of the Erdoes-statement is taken:

   

Seems to be solved now...  

Thanks James for your work!

Community -greetings:

Gottfried

[bo198214]

Seems to be solved now...  

Thanks James for your work!

Hip, Hip, Hurray!

[JmsNxn]

I'm very flattered; but this is definitely an Ecalle/Baker level result that you just have to dig for Alexandre Eremenko used to always tell me to read more Baker--as he answered most of my questions with odd results from Baker, lol.

Thank you for putting me on this problem though. I've always wanted to use Mellin transform/Borel/Euler/Fourier stuff on tetration, and this problem has definitely helped approach this problem for the parabolic case--which I always thought would be untenable. And since it's so easy in the geometric case, I always knew there was a way to do it for the parabolic case--Especially because \(\eta \uparrow \uparrow z\) has a beautiful representation with Mellin transforms:

\[
\begin{align}
\vartheta(x) &= \sum_{k=0}^\infty \left(\eta^{\eta^{\cdots k+1\,\text{times}\,^\eta}}\right) \frac{(-x)^k}{k!} = \eta - \eta^\eta x + \eta^{\eta^\eta}x^2/2! -\eta^{\eta^{\eta^\eta}}x^3/3!\,+\,...\\
\Gamma(1-z) \cdot \eta \uparrow \uparrow z &= \sum_{k=0}^\infty \left(\eta^{\eta^{\cdots k+1\,\text{times}\,^\eta}}\right)\frac{(-1)^k}{k!(k+1-z)} + \int_1^\infty \vartheta(-x)x^{-z}\,dx\\
\end{align}
\]

Which is valid, at least, for \(\Re(z) > 0\).

I think we've just opened up a bunch of new techniques that I'm familiar with; but now we talk about the Abel function, rather than the inverse Abel function--and I think most of the heavy lifting's already done!