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?
It is a non-converging asymptotic series. See Will Jagy's answer on mathoverflow: https://mathoverflow.net/questions/45608/formal-power-series-convergence/46765#46765
There is a singularity at the origin. I would add a little bit to Will Jagy's answer. There are actually four different disconnected half iterate functions, in the four different directions, +real, +imag, -real, -imag, associated with the four Leau flower petals. If we take a point between +real and +imag, we get two different answers. For example, take $x=\exp(\frac{\pi i}{4}) \approx 0.707+0.707i$. Depending on which of the two nearby quadrants we iterate $\sinh$ or $\sinh^{-1}$ towards, we get two slightly different answers for the half iterate, both of which are correct.
$$ \text{half}_1(\exp(\frac{\pi i}{4})) = 0.653259304673551626921 + 0.769356780291275338686i$$ $$ \text{half}_2(\exp(\frac{\pi i}{4})) = 0.653259380995613385458 + 0.769356792332637384377i$$
These two numbers differ by $0.8 \times 10^{-7}$. The first half iterate is the result ofiterating $x \mapsto \sinh(x)$ 40 times $\approx 0.00649983843037054 + 0.268429858003731i$, and then taking the half iterate series, and then iterating $x \mapsto \sinh^{-1}(x)$ 40 times. The second half iterate is the result of first iterating $x \mapsto \sinh^{-1}(x)$ 40 times $\approx0.278230216164981 + 0.0137470522792658i$, and then taking the half iterate and then iterating $x \mapsto \sinh(x)$. Both results are printed accurate to 28 decimal digits, and they both do converge to the printed result, as the iteration count goes to infinity. The half iterate series itself is an asymptotic series with optimal convergence using some number of terms in the series. The result using the half iterate series would match the results of using Abel function solution for the half iterate, if both are used in the same quadrant.
The singularity is at the origin, but the different between the quadrants gets smaller as you get closer to the origin. Besides the singularity at the origin, I suspect there is also a singularity at i, where the $\sinh^{-1}(i)$ has a branch point due to the derivative of $\sinh'(\pi i/2)=0$ where $\sinh(\pi i/2)=i$ .