Let $P(z)=-z+z^{p+1}$ where $p$ is a positive integer. How many petals does $P$ have at the origin?
From a few examples in my notes, I noticed the following:
- $-z+z^{14}$ has 13 petals.
- $-z+z^3$ has 4 petals.
I'm not able to find a proof of the result above, but I am able to deduce that there are:
- an $2p$ petals if $p+1$ is odd
- if $p+1$ is even then there are $p$ petals.
Could you provide some intuition, or a sketch of the result?
There is a theorem I'm leaning on to get some information:
Let $R$ be a rational map and suppose that $R(z)=z+az^{p+1}+...$ near the origin. Let $\Pi_j$ be the petals of $R$ and for each $j$ let $F_j$ be the component of the Fatou set that contains $\Pi_j$. Then $R^n(z) \to 0$ and $\arg R^n(z) \to 2\pi k/p$ on $F_k$ and $F_0,...,F_{p-1}$ are distinct components of the Fatou set.
But using this theorem, it only shows me that there should always be $p$ petals (ex, there is a petal in between every $\frac{2\pi k_i}{13}, \frac{2\pi k_{i+1}}{13}$).
Besides getting a sketch of the proof, I have one more questions:
- In the even case, there is an overlap between attracting or repelling petals?
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In the parabolic case critical orbit looks like n-th arm star
Arms tend to attracting directions near fixed point. Number of attracting petals is equal to n.
So I have checked critical orbits using this Maxima CAS script. Here are examples for p=2 and p=13
My results are listed below:
so: