First Mandelbrot steps

Define the function $f_c(z)=z^2+c$ where $z$ and $c$ are complex numbers and $c$ is a constant. For fixed $c$ , we consider the orbit of $z=0$ under iteration of $f_c$ . That is, we consider the sequence of iterates $0, f_c(0), f_c(f_c(0)), f_c(f_c(f_c(0))), \dots$ .

Expressed recursively: Where $z_0=0$ , and $z_{n+1}=z_n^2+c$ , we consider the sequence of iterates $z_0, z_1, z_2, z_3, \dots$ .

We say that a sequence is bounded if there exists a real number $M>0$ such that $|z_n|\leq M$ for all integers $n$ . And a sequence is unbounded if it is not bounded, that is, if its values go to infinity. Here are some examples of $z_{n+1}=z_n^2+c$ for different values of $c$ .

For $c=0$ , the orbit is fixed: $0, 0^2+0=0, 0^2+0=0, \dots$
For $c=1$ , the orbit grows unbounded in the real numbers: $0, 0^2+1=1, 1^2+1=2, 2^2+1=5, 5^2+1=26, \dots$
For $c=i$ , the orbit is eventually (quickly) periodic, oscillating between $-1+i$ and $-i$ : $0, 0^2+i=i, i^2+i=-1+i, (-1+i)^2+i = -i, (-i)^2+i=-1+i, \dots$
For $c=-1$ , the orbit oscillates between -1 and 0: $0, 0^2-1=-1, (-1)^2-1=0, \dots$
For $c=-i$ , the orbit oscillates between $-1-i$ and $i$ : $0, 0^2-i=-i, (-i)^2-i=-1-i, (-1-i)^2-i=i, i^2-i=-1-i, \dots$
For $c=1+i$ , the orbit is unbounded: $0, 0^2+1+i=1+i, (1+i)^2+1+i = 1+3i, (1+3i)^2+1+i = -7+7i, \dots$

It can be shown that the orbit of $z_0=0$ under $z_{n+1}=z_n^2+c$ is bounded if and only if $|z_n|\leq2$ for all $n$ . (Used in this way, the value 2 is called the orbit’s escape radius.) This justifies the claim above that the orbits for $c=1$ and $c=1+i$ are unbounded.

The following program lets us view the orbits of 0 under $f_c(z)=z^2+c$ . Use the sliders to fix $c$ by setting its real part cr and its imaginary part ci. Note that an orbit’s first two values are 0 and $c$ . Use iterations to set the number of iterates. Whenever an orbit escapes the gray disk $z\leq 2$ , we stop iterating since the orbit proves unbounded.

Iterates of 0 under $f_c(z)=z^2+c$

The Mandelbrot set $M$ comprises the set of complex numbers $c$ such that the orbit of 0 under $f_c(z)=z^2+c$ is bounded. We’ve seen that $M$ is not empty since it contains $0$ , $i$ , $-1$ , and $-i$ , and that its complement $\overline M$ is not empty since $1$ and $1+i$ do not belong to $M$ . You can use the program to visualize the examples given above. By using the sliders, it’s not hard to find values of $c$ for which the orbit of 0 appears bounded at first, only to escape as the number of iterations is increased. Such values of $c$ belong to $\overline M$ , the Mandelbrot set’s complement.

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Michael Laszlo

It is not a something, but not a nothing either

First Mandelbrot steps