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I've been working on a program that draws fractal images, and I was struck by a question that came to mind.

It is clear that the Mandelbrot fractal contains infinitely many copies of itself, but I've been wondering, is it a countable or uncountable infinity?

Vivian River
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3 Answers3

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This is similar to the mathstack question, mandelbrot-bulbs-countable. It turns out that every bulb/Cardioid has a hyperbolic center that is the solution of an algebraic equation; and algebraic numbers are countable. The hyperbolic centers are all zeros of the following sequence of equations: $f_1=x$, $f_2=x^2+x$, $$f_{n}=(f_{n-1})^2+x$$ $$f_{n}=0$$

The roots of each of the $f_n$ equations are the hyperbolic centers. The main Cardioid of every mini-mandelbrot also has a main hyperbolic center, which represent a subset of these countable algebraic numbers.

For example, consider the roots of $f_4=0$. Two of these zeros are $x=-0.156520166833755\pm1.03224710892283i$, which are the hyperbolic center of the largest period 4 mini-mandelbrot, and its conjugate $f_4 = x^8 + 4x^7 + 6x^6 + 6x^5 + 5x^4 + 2x^3 + x^2 + x$. Another zero is -1.94079980652948 which is a smaller period 4 mini-mandelbrot. The other zeros are period=4 and period=2 bulbs, as well as the period=1 main Cardioid at x=0.

Another example is $f_3=x^4 + 2x^3 + x^2 + x$. One of the zeros of $f_3=0$ is $-1.75487766624669$, which is the hyperbolic center of the main Cardioid of the largest period 3 mini-mandelbrot.

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Sheldon L
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  • Can you point me to where I can learn about why the Mandelbrot fractal has this property that the hyperbolic centers are zeroes to a series of equations? Is the converse true, that every solution to every equation in the series is the center of a bulb or cardioid? – Vivian River Apr 13 '14 at 02:54
  • Converse is true. see: Complex Dynamics by Lennart Carleson, Chapter 8.1, Quadratic polynomials, the Mandelbrot set. Super attracting points have derivative of zero. These are centers of repeating components, $f^n=f^0$, where the derivative is zero. Turns out all bulbs/cardioids have such a center point where the derivative is zero, which means that center is asolution of f^n(0)=0. – Sheldon L Apr 13 '14 at 12:40
  • Also, see this mathstack question; http://math.stackexchange.com/questions/491279/mandelbrot-boundary/499238#499238 Each Period n component has a period "n" attracting fixed point. If the fixed point is zero (at the center), it is super-attracting. The boundary of the component is where the fixed point changes from attracting to neutral, or parabolic. – Sheldon L Apr 13 '14 at 14:33
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Every copy has a central cardioid with finite, positive area, disjoint from all the other copies' central cardioids. The areas must add up to a finite number; thus, there can't be uncountably many of them, as a sum of uncountably many positive numbers can't be finite.

user2357112
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As @Sheldon suggested, I looked at the "Mandelbulb" countable question, and the answer given there is awesome, some I'm adding it here as "Community WIKI".

I like the way the answers above provide so much new information, but I love the simplicity of this explanation.

I don't know if you have a precise definition of "bulb", but it's reasonable to expect that any bulb ought to contain a sufficiently small ball. Any ball contains a point with rational coordinates, because the latter are dense. Assuming bulbs are disjoint, this lets us define an injection from bulbs to rational points, which are countable, therefore there cannot be uncountably many bulbs.

Vivian River
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