I have two complex numbers, $a = x_1 + y_1 i$ and $c = x_2 + y_2 i$. These serve as inputs to a quadratic map $f_n = f_{n - 1}^2 + c$, with $f_0 = a$. Thus the first few iterations of the map are:
- $f_0 = a$
- $f_1 = a^2 + c$
- $f_2 = (a^2 + c)^2 + c$
- $f_3 = ((a^2 + c)^2 + c)^2 + c$
- $f_4 = (((a^2 + c)^2 + c)^2 + c)^2 + c$
Which can be expanded to:
- $f_0 = a$
- $f_1 = a^2 + c$
- $f_2 = a^4 + 2 a^2 c + c^2 + c$
- $f_3 = a^8 + 4 a^6 c + 6 a^4 c^2 + 2 a^4 c + 4 a^2 c^3 + 4 a^2 c^2 + c^4 + 2 c^3 + c^2 + c$
- $f_4 = a^{16} + 8 a^{14} c + 28 a^{12} c^2 + 4 a^{12} c + 56 a^{10} c^3 + 24 a^{10} c^2 + 70 a^8 c^4 + 60 a^8 c^3 + 6 a^8 c^2 + 2 a^8 c + 56 a^6 c^5 + 80 a^6 c^4 + 24 a^6 c^3 + 8 a^6 c^2 + 28 a^4 c^6 + 60 a^4 c^5 + 36 a^4 c^4 + 16 a^4 c^3 + 4 a^4 c^2 + 8 a^2 c^7 + 24 a^2 c^6 + 24 a^2 c^5 + 16 a^2 c^4 + 8 a^2 c^3 + c^8 + 4 c^7 + 6 c^6 + 6 c^5 + 5 c^4 + 2 c^3 + c^2 + c$
For a subset of values of $a$ and $c$ where $\left|1 - \sqrt{1 - 4 c}\right| ≤ 1$, the values of $f_n$ spiral around a fixed point $M = \frac{1 - \sqrt{1 - 4 c}}{2}$, getting ever closer as $n$ increases. Extending the definition backwards to negative values of $n$ with the relation $f_n = \sqrt{f_{n + 1} - c}$, the negative-$n$ points spiral towards the fixed point $m = \frac{1 + \sqrt{1 - 4 c}}{2}$.
I want to find a closed-form function $F(t) = z$ with a real-valued input $t$ and complex-valued output $z$ which passes through $a$ at $F(0)$ and all subsequent points at their respective index values, so $f_n = F(n)$, and which ideally is infinitely-differentiable (smooth) everywhere (or almost everywhere).
Is that even possible? If so, how do I find that function? If not, are there alternatives?
Define the mapping $\,\phi(z) := z^2+c\,$ with fixed points $\,m\,$ and $\,M\,$ which famously occurs in the definition of the Mandelbrot set. The problem of how to numerically interpolate the iterated function (also referred to as iterated composition) of $\,\phi(z)\,$ is just one special case in the area of complex dynamics.
To begin, define a recursive sequence with
$$ a_0 := a,\qquad a_n = \phi^{\circ n}(a) := \phi(a_{n-1}). \tag1 $$
Define notation $\, q := 2m\,$ and verify that $\, c = m\, M, \; 1 = m+M .\,$ Let
$$ T(x) := q\,x(1-x),\;\; K(x) := m - q\,x,\;\; K^{\circ -1}(x) = \frac{m-x}q. \tag2 $$
Verify that
$$ x = K^{\circ -1}\circ K(x) := K^{\circ -1}(K(x)) $$
and
$$ T(x) = K^{\circ -1}\circ\phi\circ K(x) := K^{\circ -1}(\phi(K(x))). \tag3 $$
In other words, the quadratic mapping $\,\phi\,$ is conjugate via the linear map $\,K\,$ and its inverse function $\,K^{\circ -1}\,$ to a logistic map $\,T\,$ with fixed points at $\,0\,$ and $\,\infty.\,$
From Schröder's equation there is a formal power series expansion
$$ g(z) := z + c_2\,z^2 + c_3\,z^3 + c_4\,z^4 + \cdots\tag4 $$
such that the functional equation
$$ q\,g(z) = g\circ T(z) := g(T(z)) \tag5 $$
is satisfied where the coefficients $\,c_2,\,c_3,\,c_4,\,\dots\,$ are determined by the functional equation. Substitute definition $(4)$ into equation $(5)$ and equate coefficients to solve for $\,c_2,\,c_3,\,c_4.\,$
The result is $$ c_2 = \frac{1}{q\!-\!1}, c_3 = \frac{2q}{(q\!-\!1)(q^2\!-\!1)}, c_4 = \frac{q+5q^3}{(q\!-\!1)(q^2\!-\!1)(q^3\!-\!1)}. \tag6 $$
Note that for the exceptional case of $\,q=1,\,m=M=1/2\,$ these coefficients are all $\,\infty,\,$ $T(z) = z(1-z)\,$ and $\,\phi(z) = z^2+1/4.\,$ This corresponds to the map $\,x\to x-x^2\,$ which has a different convergence behavior as in several MSE questions such as MSE 2861768.
Apply the functional equation $(5)$ iteratively to get
$$ q^n\,g(z) = g(T^{\circ n}(z)),\;\; T^{\circ n}(z) = g^{\circ -1}(q^n g(z)). \tag7 $$
Use equation $(3)$ to get
$$ \phi^{\circ n}(x) = K\circ T^{\circ n}\circ K^{\circ-1}(x) := K(T^{\circ n}(K^{\circ-1}(x))). \tag8 $$
Now the $\,n\,$ can be replaced by any real number. For example, with $\,n = 1/2,\,$ define
$$ \theta(m \!+\!z) := m \!+\!u\,z \!+\!\frac{z^2}{u+u^2} \!-\!\frac{2z^3}{u^2(1+u)^2(1+u^2)} + \cdots \tag9 $$
where $\, u := \sqrt{q}.\,$ Then $\,\phi(z) = \theta^{\circ2}(z) := \theta\circ\theta(z) := \theta(\theta(z)).\,$ Verify that
$$ \theta^{\circ2}(m \!+\!z) = m + 2mz + z^2 = (m \!+\!z)^2 + c. \tag{10} $$
However, a more direct approach is to assume an Ansatz of
$$ \phi^{\circ n}(m\!+\!z) = m + c_1\,x + c_2\,x^2 + c_3\,x^3 + \dots \tag{11}$$
where $\,c_1,\,c_2,\,c_3,\,\dots\,$ depend on $\,m\,$ and $\,n\,$ only. This formal power series must satisfy the equation
$$ \phi\circ\phi^{\circ n}(m\!+\!z) := \phi(\phi^{\circ n}(m\!+\!z)) = \phi^{\circ n+1}(m\!+\!z). \tag{12}$$
Substitute definition $(11)$ into equation $(12)$ and equate coefficients to solve for $\,c_1,\,c_2,\,c_3.\,$ The result is that
$$ c_1 = q^n, \; c_2 = \frac{q^n(q^n-1)} {q(q-1)}, \; c_3 = \frac{2q^n(q^n-1)(q^n-q)} {q^2(q-1)(q^2-1)}. \tag{13} $$
As expected, when $\,n=1\,$ equation $(13)$ reduces to
$$ c_1 = 2m,\quad c_2 = 1,\quad c_3 = 0,\quad\dots\, \tag{14} $$
agreeing with equation $(10)$. Also, with $\,n=1/2\,$ equation $(13)$ agrees with the half iterate equation $(9)$.
The question asks for a closed-form function for the $\,\phi^{\circ n}(x)$ of equation $(8)$
The answer to this is no, except in rare cases. An alternative is given by using the power series of $\,g(z)\,$ as given in equation $(4)$ or directly the one in equation $(11)$.