1
$\begingroup$

I'm currently playing arround with my custom fractal renderer and on Math SE in this answer Américo suggested the following function:

$z_{n+2}=z_{n+1}^{3}+c^{z_{n}}$

But to get the first value I'd need $z_{n+1}$, but for this I'd need $z_{n-1}$ and so on. How can I iterate this function?

$\endgroup$
2
  • 1
    $\begingroup$ You can set $z_0$ and $z_1$ to some values and compute the rest of $z_n$ from there. $\endgroup$
    – Regret
    Dec 4, 2014 at 10:38
  • $\begingroup$ Thank you. Add it as an aswer and I'll mark it as correct :) $\endgroup$
    – Sebb
    Dec 4, 2014 at 10:40

1 Answer 1

1
$\begingroup$

$$z_{n+1}=z_n^2+c$$

In Mandelbrot's sequence, $z_0$ is defined to be $0$. The definition of $z_{n+1}$ requires only $z_n$ to be defined, so a single starting value, $z_0$, is enough to define $z_n$ for all $n$.


$$z_{n+2}=z_{n+1}^{3}+c^{z_{n}}$$

In this sequence, to define $z_{n+2}$, you need both $z_{n+1}$ and $z_n$ to be defined. So, to define $z_n$ for all $n$, you need two starting values: $z_0$ and $z_1$. They can be of any value, but changing them will change the rest of the sequence.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .