# LUCAS WILLEMS

A 26 year-old student passionate about maths and programming

# Mandelbrot set symmetry

## Article

Mandelbrot set is the set of all sequences defined as follow :

$$c \in \mathbb{C} \qquad u_c(n+1) = u_c(n)^2 + c \qquad u_c(0) = 0$$

with a convergent module. This set can be represented as follow with, in black, values of $$c$$ for which module converge, and in black, those for which module diverge :

This representation allows us to see that Mandelbrot set seems to be symmetric about the real axis. But how to prove it ? It is what I'm going to show you.

Summary

## Proof

First, we have to prove that $$u_{\overline c}(n) = \overline{u_c(n)}$$. To do it, we are going to use induction with the following statement :

$$\forall n \in \mathbb{N} \quad H_n : u_{\overline c}(n) = \overline{u_c(n)}$$

that leads us to the following reasoning :

Bases :
For $$n = 0$$, $$u_{\overline c}(0) = 0 = \overline{u_c(0)}$$. So, $$H_0$$ holds.

Induction steps :
For $$n+1$$ :

$$u_{\overline c}(n+1) = u_{\overline c}(n)^2 + \overline c$$And, as we assume $$H_m$$ holds :

$$u_{\overline c}(n+1) = \overline{u_c(n)}^2 + \overline c = \overline{u_c(n)^2} + \overline c = \overline{u_c(n)^2 + c} = \overline{u_c(n+1)}$$

Consequently, $$H_n \Rightarrow H_{n+1}$$.

Conclusion :
$$H_n$$ holds $$\forall n \in \mathbb{N}$$.

Now, we can easily prove what is interesting us. We just have to notice that $$|a| = |\overline a|$$ and so $$|\overline{u_c(n)}| = |u_c(n)| = |u_{\overline c}(n)|$$, that allows us to write :

$$\lim_{n \to +\infty} |u_c(n)| = \lim_{n \to +\infty} |u_{\overline c}(n)|$$

In other words, we have just proved that the convergence of the module of the sequence with $$c$$ is the same than with $$\overline c$$, where $$c$$ and $$\overline c$$ are 2 complex numbers symmetric about the real axis, what proves the symmetry.