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How do you calculate an infinite fixed point for $f=x^2+x$, so that $f^{o n}(x)$ never repeats, and doesn't go to infinity and doesn't go to zero? Can such a sequence of points densely cover the entire Cauliflower Julia set boundary? I may have discovered approximations for such points sort of accidentally, but I don't know much about it. Also, any references?

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How is it possible to never repeat , not go to 0 and not go to infinity ? You mean a limit cycle ? – mick Oct 12 '12 at 16:42
@mick, an infinite fixed point cycle.... otherwise it wouldn't densely cover the boundary of the cauliflower. An approximation for such a number: -0.86805350289426250593637022087715 + 1.0327822503323875673886840036798*I. I ran across it trying to solve another post I made on stack exchange, looking for references. I'll look up Milnor's book when I get home. – Sheldon L Oct 12 '12 at 17:11
up vote 2 down vote accepted

(By the way, I prefer $z\mapsto z^2+\frac14$ over $z\mapsto z^2+z$ for aestetical reasons.)

As long as the Julia set $J$ is connected, there is a holomorphic map $\phi$ of the exterior of the Julia set to the exterior of the unit disc and such that $\phi(z^2+z)=\phi(z)^2$. For any irrational angle $e^{i\alpha}$, $\frac\alpha\pi\notin \mathbb Q$, the orbit of $e^{i\alpha}$ under squaring is dense in $S^1$. Provided that $z_0=\lim_{r\to1^+}\phi^{-1}(re^{i\alpha})$ exists, I'd expect that the orbit of $z_0$ under $z\mapsto z^2+z$ is dense in $\partial J$.

Edit: Actually I jumped too short when I first wrote this up - I was thinking of the density of $n\alpha\bmod 2\pi$ instead of $2^n\alpha\bmod 2\pi$. Thus $\frac\alpha{2\pi}$ should not just be irrational; rather it should be a normal number in base 2.

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yeah, I figured it out on the way home. I've also decided that sometimes z^+1/4 is better, and sometimes z^2+z is better, depending on the problem. For this problem, I definitely agree z^2+1/4 is better. – Sheldon L Oct 12 '12 at 22:25

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