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Apologies if this question has been asked here before. The Mandelbrot set is the set of complex numbers $c$ for which the iteration $z_{n+1}(c)=z_n (c)^2+c$ with $z_0(c)=0$ does not diverge (we write $z_\infty(c)=\lim\limits_{n\rightarrow\infty}z_n(c)$ where it converges). Thus every $z_n(c)$ can be written as a polynomial in $c$ of order $2^{n-1}$, so we define:

$$z_n(c)=\sum_{j=0}^\infty \beta_{n,j}c^j$$

where $\beta_{n.j}$ is only nonzero for $j\le2^{n-1}$. Using the Mandelbrot iteration formula and the Cauchy product, we get:

$$\beta_{n+1,j}=\begin{cases}\sum\limits_{k=0}^j\beta_{n,k}\beta_{n,j-k}&j\ne1\\2\beta_{n,0}\beta_{n,1}+1&j=1\end{cases}\tag{1}$$

Now $\beta_{n,0}=0\;\;\forall n\ge0$ and $\beta_{n,1}=1\;\;\forall n\ge1$ and using $(1)$ we can show that $\beta_{n+1,2}=\beta_{n,1}^2=1\;\;\forall n\ge1$, that $\beta_{n+1,3}=2\;\;\forall n\ge2$ and in general that $\beta_{n,j}=\beta_j\;\;\forall n\ge j$ where:

$$\beta_j=\begin{cases}0&j=0\\1&j=1\\\sum\limits_{k=0}^j \beta_k\beta_{j-k}&j\ge2\end{cases}$$

(I calculated the first few terms $0,1,1,2,5,14,42,132,429,...$ by hand, but the OEIS would not give me a general formula). Thus the first $n+1$ terms of $z_n(c)$ are $\beta_i c^i$ as seen here in red:

$$z_0(c)=\color{#ff0000}{0}$$ $$z_1(c)=\color{#ff0000}{0+c}$$ $$z_2(c)=\color{#ff0000}{0+c+c^2}$$ $$z_3(c)=\color{#ff0000}{0+c+c^2+2c^3}+\color{#0000ff}{c^4}$$ $$z_4(c)=\color{#ff0000}{0+c+c^2+2c^3+5c^4}+\color{#0000ff}{6c^5+6c^6+4c^7+c^8}$$

and so on. Thus as $n\rightarrow\infty$ the first $n+1$ terms of $z_n(c)$ will tend to the following series:

$$z(c)=\sum_{j=0}^\infty\beta_j c^j\tag{2}$$

However, because of the extra $2^{n-1}-n$ terms in $z_n(c)$ (shown in blue), I assume that $\{z_n(c)\}$ will not always converge to $z(c)$, especially since the Mandelbrot set is not a disc (and power series converge within discs), but I do not know of any way of describing the extra terms which will enable a connection to be made between the convergence of the Mandelbrot iteration and the convergence of $(2)$.

My questions are the following:

  1. For what $c$ are $\color{blue}{the\ blue\ terms}$ small enough that $z(c)=z_\infty(c)$,
  2. How if at all is the convergence of $(2)$ connected with the convergence of the iteration $z_{n+1}=z_n^2+c$?
  3. I would also to like to know whether there is a closed form for $z(c)$ [see comment]
  4. What is the radius of convergence of $(2)$?
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    $\begingroup$ I thought I'd put this as a comment rather than an edit, just in case anyone else arrived at the same series as me and didn't know a general formula for $z(c)$. In my question I asked for a closed form for $z(c)$, but I have since found that the numbers $\beta_j$ are Catalan numbers and that a closed form for $z(c)$ is $\frac{1-\sqrt{1-4c}}{2c}$ as can be seen here. $\endgroup$ – Anon Jan 6 '17 at 1:25
  • $\begingroup$ Can you send me computed terms? TIA $\endgroup$ – Adam Feb 12 '18 at 18:02
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  • Despite expanding finite iterations, there's a series maps the exterior of a unit disk to the exterior of the $M$ set.

    \begin{align*} \psi(z) &= z+\sum_{m=0}^{\infty} \frac{\beta_{0,m+1}}{z^m} \\ z_{n} &= \sum_{m=0}^{\infty} \beta_{n,m} z^{2^n-m} \\ &= z^{2^{n}}+o(1) \end{align*}

    enter image description here

    See John H. Ewing, Glenn Schober, The area of the Mandelbrot Set

  • Strictly speaking, we won't say the iteration converges when the final states are oscillating. Hence, not all the points on the Mandelbrot set converge to a limit. All points in $n$-periodic cycles $(n>1)$ or chaotic bands are bounded, so they belong to the Mandelbrot set but do not converge.

  • If you insist for how interior of $M$ is mapped to its final states. Please see derivation about period $1$ cycle and its final state below:

    Snapshot I, snapshot II and snapshot III

    In conclusion, $$c=\frac{re^{i\phi}}{2}-\frac{r^2e^{2i\phi}}{4} \mapsto z=\frac{re^{i\phi}}{2}$$ where $(r,\phi) \in [0,1] \times [0,2\pi)$

  • For period one cycle, $z$ and $c$ can be expressed in quadratic: $$z^2-z+c=0$$

    Taking the branch enclosing the super-attractive point $c=0$, \begin{align} z &= \frac{1-\sqrt{1-4c}}{2} \\ &= \sum_{k=1}^{\infty} \frac{(-1)^{n-1}c^{n}}{2^{2n-1}} \binom{\frac{1}{2}}{n} \\ &= c+c^2+2c^4+5c^6+14c^8+42c^{10}+\ldots \end{align}

    which has the radius convergence of $\dfrac{1}{4}$.

    The boundary of the mapping:

    $$c=\frac{e^{i\theta}}{4} \mapsto z=\frac{1-\sqrt{|\sin \frac{\theta}{2}|+|\sin \frac{\theta}{2}|^2}}{2}- \frac{i\operatorname{sgn} (\sin \theta) \sqrt{|\sin \frac{\theta}{2}|-|\sin \frac{\theta}{2}|^2}}{2}$$

  • The iteration and series agree only when $|c|=\dfrac{1}{4}$.

    Below is the comparison between the exact iterations (blue) and the series summation (red):

    enter image description here

    The green boundaries refer pre-image and image enclosed for period two cycle.

    enter image description here

  • Animation of successive iterations of the Mandelbrot Set:

    enter image description here

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  • $\begingroup$ I'm a little confused about what you're driving at here. $\endgroup$ – Anon Dec 18 '16 at 7:43
  • $\begingroup$ If you starting with $c\in M$, the final state will be periodic when $c$ falls on the periodic disk, i.e. $z_{n}=z_{n+p}$ for some $p\in \mathbb{N}$. The worst case is $c$ is a Misiurewicz point, it'll be chaotic. Usually, we maps exterior of $M$ set and observe the asymptotic behaviour of the "final state". Your notation seem to be come from the relevant papers. $\endgroup$ – Ng Chung Tak Dec 18 '16 at 7:51
  • $\begingroup$ Apologies for my obtuseness, but since the powers of $z$ in your series are negative (i.e. a Laurent series), I'm not sure exactly how it connects to my problem. $\endgroup$ – Anon Dec 18 '16 at 8:33
  • $\begingroup$ Your approach is only do-able for periodic disks but not chaotic band. By the way, your series (interior) may even has no radius of convergence at which $c$ is a Misiurewicz point (and it "walks" randomly/chaotically). By the way, all the baby Mandelbrot sets "distribute" like Cantor set. The mapping according to the paper maps onto the exterior boundary of $M$ set (which is connected) when $|z|=1$ though it converges very very slowly. $\endgroup$ – Ng Chung Tak Dec 18 '16 at 8:47
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    $\begingroup$ Thank you for your answer, especially the edits. They helped me to understand part of my question. However my main question is about the series in my question and about how its convergence relates to $z\rightarrow z^2+c$, so my main question still remains unanswered. $\endgroup$ – Anon Jan 6 '17 at 1:33

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