1
$\begingroup$

So I'm trying to find the number of superattracting period-$n$ orbits in the family $z\rightarrow z^2+c$ for $n = 1,2,3,4,5,6$.

I think I found an algorithm to compute this.

$0 \rightarrow c \rightarrow c^2 + c \rightarrow (c^2+c)^2+c \rightarrow ((c^2+c)^2+c)^2+c\rightarrow (((c^2+c)^2+c)^2+c)^2 + c \rightarrow ((((c^2+c)^2+c)^2+c)^2 + c)^2 + c$

So I know that Period 1 has a superattracting point at $c = 0$. To find the superattracting orbit for the Period-2 bulb, you solve for $c^2 + c = 0$, which yields $c = 0,-1$. How would one solve for $c$ for the higher periods? Would you have to use Mathematica?

$\endgroup$
1
$\begingroup$

Methods :

  • all centers for given period $p$
  • one center $c_p$ for given period p near given point c

For all centers

...

$\endgroup$
0
$\begingroup$

Mu-ency's enumeration of features page details combinatorial methods for finding the number of superattracting period-$n$ orbits:

We start by considering the lemniscates as polynomials set equal to zero, and solving them, i.e. "finding the roots". Each root gives the nucleus of some mu-atom. Since the $n$th lemniscate is of order $2^{n-1}$, it follows that for each period $n$ there are $2^{n-1}$ roots.

However, a root of period $3$ (for example) also shows up as a root of period $6$, $9$, etc. but we don't count it that way. It only counts as a root of period $3$. So, from $2^{n-1}$ we must subtract the number of roots of all lower periods that are divisors of $n$. This gives the following formula for the number of Mu-Atoms of period $n$: $$N(n) = 2^{n-1} - \sum_{0 < f < n \\n \equiv 0 \mod f} N(f)$$

... This is Sloane's sequence A000740.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.