# Why the cardioid shows up in the Mandelbrot set?

How does the main cardioid appear in the Mandelbrot set? I also wonder why something "weird" happens at a point with coordinate 2 on the actual coordinate line, I mean why is the point a sort of bounding point? (this question has already been raised, but I have not found a clear answer)

• For your second question, do you understand that iterating a point of radius $>2$ only increases its radius, so the sequence of radii of the iterates of any such point diverges to infinity? Feb 6, 2020 at 7:44
• So why doesn't this happen, for example, with numbers > 1, but < 2? Feb 6, 2020 at 8:43
• It does, for most such numbers. Notice that except for a small region to the left (in the negative real direction) and some tiny filaments near $\pm \mathrm{i}$, most points in the annulus you describe are not points of the Mandelbrot set. Feb 6, 2020 at 8:51
• Well, how to prove that as soon as Z(n) becomes more than 2, the sequence will start tending to infinity? Feb 6, 2020 at 9:26

The main cardioid is the set of $$c$$ such that $$f_c(z)=z^2+c$$ has a fixed point $$f_c(z)=z$$ which is attracting, meaning $$|f_c'(z)|<1$$. You can probably redo that computation yourself.

A non obvious theorem imply that if $$f_c$$ has such an attracting fixed point, then $$c$$ is in the Mandelbrot set. More precisely, that theorem says that if $$f_c$$ has an attracting fixed point, then the sequence $$(z_n)$$ defined by $$z_{n+1}=z_n^2+c$$ and $$z_0=0$$ must converge to that fixed point, so in particular is bounded.

Not responding to the cardioid question. Responding to the $$|z| > 2$$ implies divergence question.

Let $$c = z_0 = x+\mathrm{i} y$$ be the starting point for an iteration with the distance between $$z_0$$ and the origin greater than $$2$$. That is, $$|z_0| = \sqrt{x^2 + y^2} > 2$$. It is convenient to also represent this point in polar coordinates, $$z_0 = |z_0| \mathrm{e}^{\mathrm{i} \theta_0}$$, where $$\theta$$ is the angle on the plane to the point, measured anticlockwise from the positive real axis.

Applying the iteration for the Mandelbrot set, we get the point $$z_1 = z_0^2 + c = |z_0|^2 \mathrm{e}^{\mathrm{i} 2\theta_0} + |z_0| \mathrm{e}^{\mathrm{i} \theta_0}$$ This is the sum of two complex numbers. The first has polar angle $$2\theta_0$$ and the second has polar angle $$\theta_0$$. Let's be a little lazy and not try to figure out how those two angles are related, but just try to find out the range of things that could happen using the triangle inequality.

If $$2\theta_0$$ and $$\theta_0$$ point in opposite directions (are antiparallel), the sum gives the shortest length it can: $$\text{shortest case}: |z_1| \geq |z_0|^2 - |z_0| = (|z_0| - 1) |z_0| \text{.}$$ Since $$|z_0| > 2$$, this is the product of a number bigger than $$1$$ with $$|z_0|$$, so is a number bigger than $$|z_0|$$.

If $$2\theta_0$$ and $$\theta_0$$ point in the same direction (are parallel), the sum gives the longest length it can: $$\text{longest case}: |z_1| \leq |z_0|^2 + |z_0| = (|z_0| + 1) |z_0| \text{.}$$ Since $$|z_0| > 2$$, this is the product of a number bigger than $$3$$ with $$|z_0|$$, so is a number bigger than $$|z_0|$$.

If $$2\theta$$ and $$\theta$$ point in less special directions, the length is between the two above, so is still a number bigger than $$|z_0|$$.

What we have shown is that if your starting point has $$|z_0| > 2$$, then $$|z_1| > |z_0| > 2$$. We can continue the analysis and discover $$|z_2| > |z_1|$$, then $$|z_3| > |z_2|$$. This tells us we have an increasing sequence of radii. It is not quite enough to show that the sequence of radii diverge. For that we have to show that the amounts of increase don't eventually decrease to zero. (If the radii have to increase by increasing amounts, the radii will ultimately diverge to infinity.) So let's look at that iteration more carefully.

To remind, $$|z_{i+1}| > |z_i| > 2$$ for $$0 \leq i \leq n$$. Suppose $$|z_n| = 2+\varepsilon$$, where $$\varepsilon > 0$$, so we have a name for how far above $$2$$ we are at this step. As above, we have $$(|z_n| - 1)|z_n| < |z_{n+1}| \text{.}$$ We examine \begin{align*} (|z_n| - 1)|z_n| &= (1+\varepsilon)(2+\varepsilon) \\ &= 2 + 3\varepsilon +\varepsilon^2 \text{.} \end{align*} When $$\varepsilon$$ is small, its square is very small, so when the radius is just above $$2$$, we triple the distance above $$2$$ on every iteration. When $$\varepsilon$$ is large, its square is larger than its triple, so when the radius is far above $$2$$, we square the distance above $$2$$ on every iteration. In any case, for a starting point with radius greater than $$2$$, the sequence of radii of its iterates increases, more and more rapidly, diverging to infinity.

• I didn't quite understand how you got "shortest case" and "longest case". What did these inequalities result from? Feb 7, 2020 at 8:16
• @Artyom : "If $2\theta_0$ and $\theta_0$ point in opposite directions (are antiparallel), the sum gives the shortest length it can" and "If $2\theta_0$ and $\theta_0$ point in the same direction (are parallel), the sum gives the longest length". All three inequalities are the triangle inequality. Feb 7, 2020 at 14:41
• Excuse me for my lack of judgment, but I still can't figure out how you went from the first iterative formula to the triangle inequalities. Please explain. Feb 7, 2020 at 15:06
• Are you reading what's written? "This is the sum of two complex numbers. The first has polar angle $2\theta_0$ and the second has polar angle $\theta_0$." I've added a diagram, but I can't fathom why it is necessary. Feb 7, 2020 at 15:35
• Thank you very much. And yes, I read what you wrote, but my problem was not in the angels, but in a banal misunderstanding of how you operated with vectors (modules of complex numbers), and it turned out to be simple. Thank you again, all the best. Feb 7, 2020 at 15:49

Watch this video regarding the carioid question: