# What is the shape of parabolic critical orbit?

The parabolic critical orbits of discrete dynamical system form n-th arm stars : which shapes are conjugated with "regular" n-th arm stars

Here are 2 images of parabolic critical orbits for 2 functions from the same family : fm(z)=z^4+mz : : : Using visual analysis and grid ( gray points) one can say that one is "regular" and the second is distorted.

Questions :

1. Is it possible to check the shape ( regular/distorted) of parabolic critical orbit knowing only function ? ( not using visual analysis )
2. How to describe distortion of orbit ?

TIA

Sounds like you need to study the Leau-Fatour flower theorem. Roughly: if $f$ has Taylor expansion $f(z)=z+a z^{n+1}+\cdots$, then the complex number $v$ is a repulsion vector for $f$ at the origin if $n a v^n=1$ and $v$ is an attraction vector if $n a v^n=-1$. More specifically, if an orbit $z_k$ tends to zero, then $z_k\sim v\left/\sqrt[n]{k}\right.$ as $k\to \infty$.

The reason that some orbits tend straight to the origin under the iteration of $f$ and others move along a "distorted" path is simply that some initial values lie right on an attractive vector while most do not.

If, as in your case, $f$ has Taylor expansion $f(z)=\lambda z+a z^{n+1}+\cdots$ where $\lambda$ is a $p^{\text{th}}$ root of unity, then we examine $f^p$.

Let's apply this to your specific examples. In your simple example $f_s(z)=z^4-z$, there are three critical points with arguments $0$, $2\pi/3$, and $-2\pi/3$. Since the coefficient of $z$, namely $-1$, is a 2nd root of unity, we examine $$f_s(f_s(z)) = z-z^4+\left(z^4-z\right)^4$$ to find that the attraction vectors satisfy $-3v^3=-1$. The attraction vectors have the same argments.

In your more complicated example, we have $f_c(z)=z^4-i z$ and there are three critical points with arguments $-\pi/2$, $\pi/6$, and $5\pi/6$. In this case, however, the coefficient of $z$, namely $-i$ is a fourth root of unity, thus we examine $$f_c(f_c(f_c(f_c(z)))) = z-(84+76i)z^{13}+\cdots.$$ The attraction vectors now satisfy $-12(84+76i)v^{12}=-1$ and the resulting arguments are simply different.

Application to images

One other note: a solid understanding of the repulsion vectors allows us to generate very nice images of these types of Julia sets. Here, for example, is the Julia set of $f(z)=z+z^5$: The algorithm is something like so: For each point in a window, iterate $f$ starting from the point say 100 times. Any point that starts inside the bounded portion of the Fatou set will generally be semi-close to the origin. We then classify the point according to it's argument; in this particular case, there are four domains depending on which quadrant the final iterate lies in. The image illustrating just this classification looks like so: The first image is then obtained via a boundary scanning technique applied to the second. More generally, if the iterate $z_n$ of tends to a fixed parabolic point, then the initial seed $z_0$ is classified according to the argument of $z_n-z_0$, the classification being provided by the flower theorem.

As I'm sure you know, since you're asking about parabolic points, an escape time algorithm would take forever to generate that type of image, since the dynamics are so slow there. If you want resolution of $1/100$, it would take roughly $2 \, 10^8$ iterates to move the point $z_0=0.01$ to $z=2$ by iterating $f(z)=z+z^5$.

• Thx for the answer. <br/> "The reason that some orbits tend straight to the origin under the iteration of f and others move along a "distorted" path is simply that some initial values lie right on an attractive vector while most do not."<br/>Initial values are critical points. Do you mean that critical points do not lie on an attractive vectors ? – Adam Apr 14 '13 at 20:38
• Lets take for example function f(z)=z^4-z. It has 6 petals see commons.wikimedia.org/wiki/File:Julia_set_for_f(z)_=_z^4_-z.png. <br/> Taylor expansion of f around z=0 is -z+z^4+8*z^7-44*z^10-56*z^13+1206*z^16-2112*z^19+... <br/>So a=1 and n=3 ? Then attracting vectors v are solutions of v^3=-1. Am I right ? – Adam Apr 14 '13 at 20:59
• @Adam - Yes (to your first question) See the edit. – Mark McClure Apr 14 '13 at 21:02
• And no to your second question, since you're not considering $f(f(z))$. – Mark McClure Apr 14 '13 at 21:03
• Yes. To find p I have to check which root of unity is $\lambda$ the coefficient of z. – Adam Apr 14 '13 at 21:20 critical orbit, attracting and repelling vectors. One can see that cr points do not lie on attracting vectors but tend to them very fast.