Function to calculate the period of a Mandelbrot Set point

Question

I'm trying to implement a program to calculate the period of a Mandelbrot Set point.

I'm using the multipliers of periodic points as explained here:

https://rlbenedetto.people.amherst.edu/talks/mhc_ug14.pdf

I've already implemented a first version but it's not detecting well all periods inside the bulbs.

I'm using this formula:

where "phi prime" is 2*z and "phi N" the Nth iteration of the Mandelbrot formula z²+c.

UPDATE:

I want to calculate the cycle of a point. Basically I want to build this image.

I'm not a mathematician but what I understood from the λ function, "n" is the minimum period/cycle of the orbit for the complex point "a".

The way to get the minimum "n" is by using brute force iterating "n" from 1 to MAX_PERIOD you want to check.

There is another explanation here for this method to get the period.

LINKS:

• The program I'm implementing.

Answer 1

As the comments already have stated, the value of $\lambda$ is not the period of the point $a$, but rather, it refers to a property of the dynamics of $a$; specifically, whether the $n$-cycle that $a$ belongs to is an attractive, repelling, or indifferent one. What we mean by these terms is this:

• Attractive: points in the attraction basin--loosely speaking, some neighborhood of the cycle of periodic points--will have a forward orbit whose limit will tend toward the cycle. This is sort of analogous to the concept of a stable equilibrium in physics, where small perturbations away from the equilibrium will still return to that equilibrium.
• Repelling: points in a neighborhood of the cycle are repelled from the cycle, even though the cycle itself is periodic. This is like an unstable equilibrium (e.g., balancing a pencil on its tip is impossible but there is a theoretical point where the pencil's center of gravity is directly above the balancing point).
• Indifferent: the dynamics of this case are complex--it's neither attracting nor repelling. The ways in which orbits in a neighborhood of a cycle behave are not all the same. See how many known classifications and types of indifferent fixed-points are there? for more information.

That said, all of this has nothing to do with the period of the cycle itself. A naive calculation of the period would be done by calculating a sufficiently large number of iterations $\{z_0, z_1, \ldots \}$ and then comparing $z_n$ against $z_{n-k}$ for $k \in \{1, 2, \ldots, m\}$ where $m$ is the largest period you want to detect. If you can find that $|z_n - z_{n-k}| < \epsilon$ for sufficiently large $n$, this suggests a cycle of length $k$ and you would confirm by looking at $$\{|z_n - z_{n-k}|\}_{n=n'}^{n'+k}$$ and seeing that these are all sufficiently small. There are more sophisticated approaches, but this is the elementary way.

Answer 2

What I do to create an image like the one you link, for $f_c(z) = z^2 + c$:

• start iteration from $z_0 := 0$, with $m := \infty$
• for each $n = 1, 2, 3, ...$ in order
  • calculate $z_n := f_c(z_{n-1})$
  • if $|z_n| < m$
    • set $m := |z_n|$
    • use Newton's method to solve $w = f_c^{\circ n}(w)$ with initial guess $w^{(0)} := z_n$ (this may fail to converge, in which case continue with the next $n$), the steps are $$w^{(i+1)} := w^{(i)} - \frac{f_c^{\circ n}(w^{(i)}) - w^{(i)}}{{f_c^{\circ n}}'(w^{(i)}) - 1}$$
    • calculate the derivative of the cycle $\lambda := {f_c^{\circ n}}'(w)$
    • if $|\lambda| < 1$, then the cycle is attractive and $c$ is within a hyperbolic component of period $n$, stop (success). $\lambda$ may used as "interior coordinates" within the hyperbolic component. $w$ and $n$ can be used for interior distance estimation.

The point of using Newton's method is to accelerate the computation of $w$, a point in the limit cycle attractor. Computing $w$ just by iterating $f_c$ could take many 1000s of iterations, especially when $\lambda$ is close to $1$.

I have no complete proof of correctness (but this doesn't mean I think it is incorrect; the images seem plausible). It relies on the "atom domains" surrounding each hyperbolic component of a given period.

It also relies on the cycle reached by Newton's method being the same cycle as the limit cycle approached by iteration: this is true for the quadratic Mandelbrot set because there is only one finite critical point, $0$ ($\infty$ is a fixed point) and each attracting or parabolic cycle has a critical point in its immediate basin (see https://math.stackexchange.com/a/3952801), which means there can be at most one attracting or parabolic cycle.

For an implementation in C99 you can see my blog post at https://mathr.co.uk/blog/2014-11-02_practical_interior_distance_rendering.html

Answer 3

Here is "my" algorithm:

• choose parameter c
• compute critical orbit ( forward orbit of critical point ) and find period of it's limit cycle:

Steps:

• start with critical point: z = z0 = 0.0
• make n forward iterations of z0 to (possibly) fall into attracting cycle now z = zn
• make n forward iterations of zn to compute attracting cycle
• check backwards whether the last iterate z has already been visited before

Max iteration and precision ( epsilon ) might need to be adjusted Here is the code for complex quadratic polynomialand for the other polynomials

Comparison of "my" and Claude's methods:

"my"	Claude's
compute z,n	compute z,n,lambda
gives only attracting z	gives all z, also repelling
checks dynamics by using forward iteration	uses numerical method for finding roots
need very small epsilon and very large n for points near boundary	can use bigger epsilon and smaller n