# How can the lyapunov exponents for the Mandelbrot Set be computed?

I am trying to find a way to calculate the Lyapunov exponents of the Mandelbrot set. There are some very nice diagrams that you can find on Flickr of a plot of the Lyapunov exponents of the Mandelbrot set, leading to a much higher detail in the structure of the set.

I would like to do this as well and have found the definition of the Lyapunov Exponent but I cannot find a good paper or documentation on how to calculate the Lyapunov Exponent for the M-Set. Especially the derivative of the function $z[n+1] = z^n + c$ with regards to the starting position $x_0$ is what I cannot figure out.

Do I have to use an approximation for this derivative? If yes, which one? If no, how can these exponents be calculated for the Mandelbrot set.

Many thanks!

Lukas M.

PS: I am a hobby mathematician, but mainly an engineer. A description in terms of an algorithm would greatly help. Thank you for your consideration.

• You can't define the derivative for a discrete process. That does not mean there is no equivalent to the Lyapunov exponent, but it certainly means that you can't use the definition you have at hand. – Raskolnikov Jul 31 '13 at 12:15
• Other point, the formula to generate the Mandelbrot set is $z_{n+1}=z_n^2+c$. – Raskolnikov Jul 31 '13 at 12:31
• @Raskolnikov you said you cannot define the derivative for a discrete process. then again the definition you provide for the maximal lyapunov exponent requires a derivative. – Lukas Mosser Jul 31 '13 at 13:37
• Yes, but that is not the derivative of the iterated process, but the derivative of the map $f$ which is being iterated. And that is something different altogether. But I maybe misunderstood what you meant initially. – Raskolnikov Jul 31 '13 at 13:39

$$\lambda_f(z_0) = \lim_{n\rightarrow\infty} \frac{1}{n} \sum_{i=0}^{n-1} \left ( \ln\left|f'(z_i)\right | \right )$$
$$f'(x) = \frac{d}{dz}f_c(z) = 2z$$ means first derivative of f with respect to z