# Golden spirals in the Mandelbrot set?

The Mandelbrot set is defined by iterations of $f_c(z) = z^2 + c$. When plotted in the parameter plane, images coloured by various methods are full of logarithmic spirals, which occur due to the asymptotic self-similarity of the Mandelbrot set around Misiurewicz points.

The Golden spiral is a particular logarithmic spiral, which expands by a factor of $\phi^4$ each time it turns through $2 \pi$ (where $\phi = \frac{1}{2}\left(\sqrt{5} + 1\right)$ is the Golden ratio).

The spirals in the Mandelbrot set are associated with the multiplier of the Misiurewicz point at the center. The Misiurewicz point is strictly pre-periodic: $f_c^{q + p}(0) = f_c^{q}(0) ; 0 < p \in \mathbb{N}, 0 < q \in \mathbb{N}$ and its multiplier is the derivative of its periodic cycle: $$m = \prod_{k=q}^{q+p-1} 2 f_c^k(0)$$ By construction, all Misiurewicz points (and thus their multipliers) are algebraic numbers, which may or may not be relevant (I don't know enough to be sure).

The question is: is there a Golden spiral in the Mandelbrot set?

Equivalently: is there a Misiurewicz point whose multiplier $m$ satisfies $\arg m \equiv \frac{\pi \log |m|}{2 \log \phi} \mod 2\pi$?

Some animated examples of non-Golden logarithmic spirals in the Mandelbrot set, coloured using exterior distance estimation, can be found in my blog post.

This question was originally posed by Fraktalist on fractalforums.org.

• @Adam afaict the (English version of the) book just states 2 facts that are already known: 1. the golden spiral is a particular logarithmic spiral; and 2. some logarithmic spirals are in the Mandelbrot set. This however does not answer the question of whether there is a golden spiral in the Mandelbrot set. – Claude May 3 '18 at 21:26

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.