TITLE: Series related to the Mandelbrot set
QUESTION [5 upvotes]: 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: 

For what $c$ are $\color{blue}{the\ blue\ terms}$ small enough that $z(c)=z_\infty(c)$, 
How if at all is the convergence of $(2)$ connected with the convergence of the iteration $z_{n+1}=z_n^2+c$? 
I would also to like to know whether there is a closed form for $z(c)$ [see comment]  
What is the radius of convergence of $(2)$?

REPLY [6 votes]: 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*}

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 $|c| \le \dfrac{1}{4}$.

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



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


Animation of successive iterations of the Mandelbrot Set: