I am trying to understand the classification of points in the Mandelbrot set. There are an infinite number of baby Mandelbrots, each associated with a defined set of landing rays.
There are the pre periodic Misiurewicz branching points, which also have landing rays. What are the rest of the points in the Mandelbrot set called? Are they called chaotic? Also, is this set of remaining uncountably infinite? What is known about landing rays for these remaining points?
Dynamical Systems – Classification of Mandelbrot Set Points
complex-dynamicsdynamical systemsfractals
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As noted by Sheldon, the good starting point must be a critical point. There is indeed a theorem that says that if you have an attracting cycle, then at least one critical point must belong to its attraction basin.
Roughly, the idea of the proof is as follows : around an attracting fixed point, there is a linearizing coordinate $\phi$ such that $f(\phi(z))=\lambda \phi(z)$. That coordinate is defined only in a neighborhood of the attracting fixed point, however using the functional equation it satisfies it is possible to prolonge it until you meet a critical point. Now if you have a cycle instead of a fixed point, just replace $f$ by $f^p$ to get the same result.
The Mandelbrot set is of mathematical interest because in complex dynamics, the global behaviour of the dynamics is generally ruled by the dynamics of the critical points. Thus knowing the dynamics of the critical points give you information on all of the dynamics. In the simplest family $z^2+c$, there is only 1 critical point (0) and so it is natural to look at what happens to its dynamics depending on $c$.
If you want to generalize the notion of Mandelbrot set for, say, cubic polynomials $z^3 + az+b$, you would have to look at the behaviour of two critical points, and so not only would you get a set in $\mathbb{C}^2$, you would also need to make a choice in your definition : are you looking at parameters where both critical points are attracted to a cycle, or one of them, or none ?
In your case, there is only one critical point, so your set is a reasonable analogue of the Mandelbrot set.
EDIT : note that the definition of the Mandelbrot set does not use attracting cycles, but depends on whether or not the critical point goes to infinity. It is conjectured (it's one of the most important conjecture in the field) that the interior of the Mandelbrot set is exactly composed of parameters for which the critical point is attracted to a cycle. However it is well known that in the boundary of the Mandelbrot set, you have no attracting cycles.
EDIT 2 : One of the most interesting features of the Mandelbrot set is that its boundary is exactly the locus of bifurcation, i.e. the set of parameters for which the behaviour of the dynamics changes drastically. If you choose any holomorphic family of holomorphic maps $f_\lambda$, you can also define the locus of bifurcation for this family. It has been proved that this set is either empty or contains copies of the Mandelbrot set.
I am very new to algebraic number theory, so this may be complete garbage...
The first few iterations of $f_c^n(0)$, with their degree, are:
$$\begin{aligned} 0 && -\infty\\ c && 1 \\ c^2 + c &&2 \\ c^4 + 2c^3 + c^2 + c &&4 \\ c^8+4c^7+6c^6+6c^5+5c^4+2c^3+c^2+c&&8\\\vdots&&2^{n-1}\end{aligned}$$
Thus, Misiurewicz points $M_{q,p}$ of preperiod $q > 0$ and period $p > 0$ are roots of the monic integer-coefficient polynomial $$f_c^{p+q}(0) - f_c^q(0)$$ and therefore each $c = M_{q,p}$ is an algebraic integer, with $\deg M_{q,p} \le 2^{p + q - 1}$. The algebraic integers are closed under addition and multiplication, therefore the multiplier $$m =\prod_{n=q}^{p+q-1} 2 f_c^n(0)$$ is also an algebraic integer.
The desired multiplier is $(i \phi)^t, t\in \mathbb{R}$. $0 < t$ because Misiurewicz points are repelling, $|m|>1$. $i\phi$ is an algebraic integer of degree $4$, with minimal polynomial $x^4+3x^2+1$. Define $w(\cdot)$ to be the number of non-zero coefficients of the minimal polynomial of the multiplier, then $w(i\phi) = 3$.
Here is a table of some $w(M_{q,p})$:
q \ p 1 2 3 4 5 6
2 2 3 7 13 31 55
3 3 4 13 25 61
4 6 9 22 49
5 13 15 49
6 28 33
7 59
Conjecture 1: $\deg M_{q,1} = 2^{q-1}-1$
Conjecture 2: $w(M_{q,1}) = 2^{q-1}-(q-2)$
Conjecture 3: $w(M_{q,p}) = O(\deg M_{q,p})$
Claim: $w(M_{q,p}) = 3$ occurs only for $M_{3,1}$ and $M_{2,2}$, and their multipliers have degrees $3$ and $2$ respectively.
Now it is enough (is it?) to consider lowest-terms $\frac{a}{b} = t \in \mathbb{Q}$. Now suppose there is an $m$ such that $$m^{\frac{1}{a}} = (i \phi)^{\frac{1}{b}}$$ The right hand side has minimal polynomial $x^{4b}+3x^{2b}+1$ which also has $3$ non-zero coefficients, and similarly the left hand side has the same number of non-zero coeffients as $m$. Both are monic.
Therefore, the only possible choices for $m$ are the $m$ corresponding to each $M_{3,1}$ or $M_{2,2}$. The first implies:
$$x^{3a} - 4x^{2a} + 16 = 0 = x^{4b} + 3x^{2b} + 1 $$
So $3a=4b$, ie $a = 4n, b = 3n, n > 0$ but because $a$ and $b$ are coprime that means $n = 1$ and the equations boil down to:
$$x^{12} - 4x^{8} + 16 = 0 = x^{12} + 3x^{6} + 1$$
Rearranging gives:
$$4x^{8} + 3x^{6} - 15 = 0$$
But this has no solutions in the algebraic integers due to the leading $4$.
Similarly, for $M_{2,2}$ we get
$$x^{2a} - 8x^{1a} + 32 = 0 = x^{4b} + 3x^{2b} + 1$$
where $2a = 4b$ implies $a = 2$ and $b = 1$ so
$$x^4 - 8x^2 + 32 = 0 = x^4 + 3x^2 + 1$$
Rearranging gives $11 x^2 - 31 = 0$ but this has no solution in the algebraic integers due to the leading $11$.
Therefore there is no such $m$, and thus there are no Golden spirals in the Mandelbrot set.
Best Answer
boundary points of Mandelbrot set ( c-points ) :
Siegel points
parabolic points = root points of hyperbolic components = biaccesible points ( landing points of 2 rays )
Misiurewicz points
Chaotic points
Cremer points