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
This may be helpful.
Let $$ f(x) = \frac{-1 + \sqrt{1 + 4 x}}{2}, \; \; x > 0 $$ We use a technique of Ecalle to solve for the Fatou coordinate $\alpha$ that solves $$ \alpha(f(x)) = \alpha(x) + 1. $$ For any $x > 0,$ let $x_0 = x, \; x_1 = f(x), \; x_2 = f(f(x)), \; x_{n+1} = f(x_n).$ Then we get the exact $$ \alpha(x) = \lim_{n \rightarrow \infty} \frac{1}{x_n} - \log x_n + \frac{x_n}{2} - \frac{x_n^2}{3} + \frac{13 x_n^3}{36} - \frac{113 x_n^4}{ 240} + \frac{1187 x_n^5}{ 1800} - \frac{877 x_n^6}{ 945} - n. $$ The point is that this expression converges far more rapidly than one would expect, and we may stop at a fairly small $n.$ It is fast enough that we may reasonably expect to solve numerically for $\alpha^{-1}(x).$
We have $$ f^{-1}(x) = x + x^2. $$ Note $$ \alpha(x) = \alpha(f^{-1}(x)) + 1, $$ $$ \alpha(x) - 1 = \alpha(f^{-1}(x)) , $$ $$ \alpha^{-1} \left( \alpha(x) - 1 \right) = f^{-1}(x). $$ It follows that if we define $$ g(x) = \alpha^{-1} \left( \alpha(x) - \frac{1}{2} \right), $$ we get the miraculous $$ g(g(x)) = \alpha^{-1} \left( \alpha(x) - 1 \right) = f^{-1}(x) = x + x^2. $$
I put quite a number of relevant pdfs at BAKER. The host computer for this was down for about a year but has recently been repaired.
EDIT, TUESDAY:
Note that $\alpha$ is actually holomorphic in an open sector that does not include the origin, such as real part positive. That is the punchline here, $\alpha$ cannot be extended around the origin as single-valued holomorphic. So, since we are finding a power series around $0,$ not only are there a $1/z$ term, which would not be so bad, but there is also a $\log z$ term. So the $\ldots -n$ business is crucial.
I give a complete worked example at my question https://mathoverflow.net/questions/45608/formal-power-series-convergence as my answer https://mathoverflow.net/questions/45608/formal-power-series-convergence/46765#46765
The Ecalle technique is described in English in a book, see K_C_G PDF or go to BAKER and click on K_C_G_book_excerpts.pdf The Julia equation is Theorem 8.5.1 on page 346 of KCG. It would be no problem to produce, say, 50 terms of $\alpha(x)$ with some other computer algebra system that allows longer power series and enough programming that the finding of the correct coefficients, which i did one at a time, can be automated. No matter what, you always get the $\alpha = \mbox{stuff} - n$ when $f \leq x.$
As I said in comment, the way to improve this is to take a few dozen terms in the expansion of $\alpha(x)$ so as to get the desired decimal precision with a more reasonable number of evaluations of $f(x).$ So here is a brief version of the GP-PARI session that produced $\alpha(x):$
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