radius of convergence of half iterate of sinh(z)? The half iterate of sinh(z) has a formal power series, centered around z=0.  Does the formal power series for the half iterate converge at the origin? This is equivalent to asking if the half iterate is analytic at the origin.  If it converges, what is the radius of convergence?  Are there any other singularities in the complex plane for the half iterate of the sinh(z)?  The sinh(z) is an entire function, with exponential growth at the real axis, for both positive and negative z.  It has a fixed point of zero, with a fixed point multiplier of 1.
$$\sinh(x)=\frac{\exp(x)-\exp(-x)}{2} = x + \frac{x^3}{6} + \frac{x^5}{120}+ \frac{x^7}{7!} + \frac{x^9}{9!} ....$$
One can generate a formal half iterate of the function, such that half(half(x))=sinh(x).  
$$\text{half}(x) = x + \frac{x^3}{12} + \frac{-x^5}{160}+ \frac{53x^7}{40320} + \frac{-23x^9}{71680} + \frac{92713x^{11}}{1277337600} + ....$$
Such a half iterate would have "half-exponential" growth as abs(real(z)) gets bigger, which is why I was curious if it was entire, since I don't know of any entire "half exponential" functions.  The coefficients of the half iterate of asinh(z) appear to be mostly decreasing, with the 41st coefficient $\approx  0.0000000047072111$.  Does such a half iterate function converge at the fixed point at the origin, or is the convergence only illusory, with the series actually asymptotic, rather than converging?   Another way of generating the half iterate is $\alpha^{-1}(\alpha(z)+\frac{1}{2})$, where $\alpha(z)$ is the abel function.  Presumably, this generate the same half iterate.  What are the half iterate's singularities in the complex plane?
 A: (This is a comment for Sheldon's answer but contains a picture so I made it an answer)      
I'd never expected to see iteration to the same fixpoint using $f(x)$ and $f^{o-1}(x)$ When I learned about repelling and attracting fixpoints that properties were mutually connected to the iteration of the function and the iteration of its inverse.... Strange...
 
Remark 1: the documented markers on the orbits of the simple iterations (the curves) are at iterations $(0),2,6,14,30,62,...,2^k-2,...$ . The interpolation of the orbits is based on linear interpolation (plust slightly enhanced) between $z_{k+200} \ldots z_{k+201}$  (for the sinh-curve) and   $z_{-k-200}  \ldots  z_{-k-201}$ for the asinh-curve. 
Remark 2: the attempt to increase the speed of convergence by using the Newton-iteration was not much successful - we don't get the doubling of correct digits by each iteration ("as usual" with Newton in many other cases). It seems to be a very interesting example how Newton-iteration can get spoiled... (However, it is still a far better method to approach the fixpoint: the markers on the Newton-iteration orbits are only one iteration apart.)
