# Calculating the Buddhabrot

My Question

Why is it that the Buddhabrot visually differs so drastically from the Mandelbrot? Based on articles I've read, such as the one on Wikipedia, it seems that the Buddhabrot should simply be a negative of the Mandelbrot and otherwise visually identical.

My Reasoning
Wikipedia states that in order to render a Buddhabrot, you must iterate (complex) points that aren't in the Mandelbrot set through the Mandelbrot function and trace their paths to escape. The points themselves and their paths supposedly visually compose the Buddhabrot.

Mandelbrot Function: $f(z_{n+1})=z_n^2+c$

This is where my confusion arises; why is it that Buddhabrot renderings show activity in regions that are part of the Mandelbrot Set? By definition, any point who's path/trajectory passes through the Mandelbrot Set is part of the Mandelbrot Set because once the path is in the Set, it cannot escape.

Any clarifications on where my comprehension may be wrong or tips on how to render the Buddhabrot are pleasantly welcomed!

Buddhabrot(tilted) (left), Mandelbrot (right)

That is true for the Julia sets, where all iterations are for the same $c$, but different start points $z_0$, but not for the Mandelbrot set where $c$ varies and the start point is fixed as $z_0=0$.
If some $z_n$ of the sequence $z_{k+1}=z_k^2+c$, $z_0=0$ is inside the bounded component (note that the M~ set itself, the fractal, is only the boundary curve), this does not mean that the sequence of $\tilde c=z_n$, started again at $\tilde z_0=0$, $\tilde z_1=z_n$, is in any way related to the first sequence. Indeed, $z_{n+1}=z_n^2+c$ is different from $\tilde z_2=z_n^2+z_n$.
• I found the flaw in my algorithm, I was plotting the original random $c$ values instead of the more recently calculated $z$ values that describe escape trajectories. Thank you for you great answer though! @LutzL Jul 16, 2017 at 8:09