TITLE: half iterate of $x^2+c$
QUESTION [21 upvotes]: I'm looking for literature on fractional iterates of $x^2+c$, where c>0.  For c=0, generating the half iterate is trivial.
$$h(h(x))=x^2$$
$$h(x)=x^{\sqrt{2}}$$
The question is, for $c>0,$ and $x>1$, when is the half iterate of $x^2+c$ smaller than the half iterate of $x^2$?  We know that the full iterate is always larger, since $x^2+c>x^2$, for $c>0$, and $x>1$. Intuitively, one would think that the half iterate of $x^2+c$ would also always be larger, but I believe I have found some counter examples.
In examining the parabolic case for $c=0.25$, I believe $x=800000000$ is a counter example.  $800000000^{\sqrt{2}} \approx 3898258249628$, but I calculate the half iterate of $f(x)=x^2+0.25$, $h_{x^2+0.25}(800000000) \approx 3898248180100$, which is smaller.  
For $c=0$, this is the equation for the superfunction which can be used to calculate fractional iterations.  $f(x)=x^2$, and $g(x) = f^{o x}$, $g(z) = 2^{2^z}$.  For $c=0.25$, this is the parabolic case, which has been studied a great deal in understanding the mandelbrot set, and the superfunction is entire, and I presume there is a uniqueness criteria.  For $c>0.25$, the problem becomes trickier because $x^2+c$ has complex fixed points, and I am also looking for any literature on unique solutions to calculating real valued fractional iterates for $c>0.25$.   
What I am also interested in is the abel function of $x^2$, which is $\text{abel}(z) = \log_2(\log_2(z))$.   I am interested in the abel function of $x^2$ composed with the superfunction of $x^2+c$.
$$\theta(z)=\text{abel}_{x^2}(\text{superfunction}_{x^2+c}(z))-z$$
As real $z$ increases, if $\theta$ converges to a $1$-cyclic function, as opposed to a constant, then there are counter examples like the one I gave, and sometimes the superfunction is growing slower than $2^{2^z}$, and othertimes it is growing faster, with the two function intersecting each other an infinite number of times.  I'm also wondering if $\theta$ converge to an analytic function?  Any relevant links would be appreciated.
- Sheldon

REPLY [2 votes]: Similar in nature to Cosmas Zachos' answer using Schroeder's method, it is easy to notice that solutions to
$$f^m(x)=(\underbrace{f\circ\dots\circ f}_m)(x)=x^2+c=\psi(x)$$
all satisfy
$$\psi\circ f=f^{m+1}=f\circ\psi$$
and, assuming non-negativity,
$$f=\psi^{-1}\circ f\circ\psi=\psi^{-n}\circ f\circ\psi^n$$
where $\psi^{-1}(x)=\sqrt{x-c}$. For sufficiently large $x$, one can see that
$$|\psi^{-1}(x+\epsilon)-\psi^{-1}(x)|\le\frac12|\epsilon|$$
and hence as long as $f\circ\psi^n$ is known to be large and $g\circ\psi^n$ is within $\epsilon$ of it for some fixed $\epsilon$ and all large $x$, then the error will vanish as $n\to\infty$, so we may define
$$f=\lim_{n\to\infty}\psi^{-n}\circ g\circ\psi^n$$
Notably, defining $g(x)=|x|^{2^{1/m}}$ and considering $x$ so that $|x|<\psi(x)$ suffices.
For $c>0$, numerical computation tends to suggest this is increasing in $n$ i.e. $g$ is an under-approximation of $f$.
Taking $(m,c,x)=(2,0.25,800000000)$, we get
$$f(x)\simeq3898258249628.005$$
which is equivalent to $g(x)$ for the shown places. A table of values:
\begin{array}{c|c|c}x&g(x)&f(x)\\\hline1&1.00000&1.10352\\10&25.95455&25.99583\\100&673.63884&673.65057\\1000&17483.99554&17483.99862\\10000&453789.29806&453789.29886\end{array}
Try it online!