Iterability of exp(x)-1

[bo198214]

exp(x)-1 definately has non-integer iterated; see heirarchies of height 1/2 at http://tetration.org/Combinatorics/Schro...index.html which is listed in the OEIS as A052122. I don't have access to my computer, but it looks like our results for heirarchies of height 1/2 agree. I also have the general solution which checks with OEIS entries for heirarchies of height -2, -1, 1/2, 1, 2, 3 and 4.

The general formula comes from the double binomial expansion,
\( {f^{\circ s}}_n=\sum_{i=0}^{n-1} (-1)^{n-1-i}\left(s\\i\right)\left(s-1-i\\n-1-i\right){f^{\circ i}}_{n} \)
The formula is reliable, I just computed it for the case s=1/2, to exemplify convergence.

However I just looked in Baker's Paper and indeed he states (as a German native I just translate it):

Proposition 17. Let \( F(z)=e^z-1 \); for each real \( \sigma \) let \( F_\sigma(z) \) be the by
\( F(F_\sigma(z))=F_\sigma(F(z)) \)
uniquely determined formal series which has the form
\( F_\sigma(z)=z+\frac{\sigma}{2}z^2+\sum_{m=3}^\infty a_m(\sigma)z^m \).
Then \( F_\sigma(z) \) has a positive radius of convergence if and only if \( \sigma \) is an integer number. \( F_\sigma(z) \) is the \( n \)-th iterate of \( F(z) \) for integer \( \sigma=n>0 \), hence an entire function. \( F_\sigma(z) \) is the inverse series development of \( F_{-n} \) for integer \( \sigma=n<0 \).

So instead just of to numerically verify, can we prove that \( (e^x-1)^{\circ 1/2} \) converges for some \( x>0 \)?