Is there a simple way to interpolate smoothly between levels of a complex-valued continued fraction?

Question

I have two complex numbers, $a = x_1 + y_1 i$ and $b = x_2 + y_2 i$. These serve as inputs to an infinite continued fraction of the form $f_n = a + \frac{b}{f_{n - 1}}$, with $f_1 = a$. Thus the first few levels of the continued fraction are:

• $f_1 = a$
• $f_2 = a + \frac{b}{a} = \frac{a^2 + b}{a}$
• $f_3 = a + \frac{b}{a + \frac{b}{a}} = \frac{a^3 + 2 a b}{a^2 + b}$
• $f_4 = a + \frac{b}{a + \frac{b}{a + \frac{b}{a}}} = \frac{a^4 + 3 a^2 b + b^2}{a^3 + 2 a b}$
• $f_5 = a + \frac{b}{a + \frac{b}{a + \frac{b}{a + \frac{b}{a}}}} = \frac{a^5 + 4 a^3 b + 3 a b^2}{a^4 + 3 a^2 b + b^2}$

For most values of $a$ and $b$, the values of $f_n$ spiral around a fixed point $M = \frac{a + \sqrt{a^2 + 4 b}}{2}$, getting ever closer as $n$ increases.

With only the points or the points-and-vectors visible, the spiral isn't always obvious, but connecting the points with circular arcs makes it more apparent:

However, while it isn't difficult to construct circular arcs with positional continuity ($C^0$ & $G^0$, so there are no gaps between points) and tangent continuity ($G^1$, so the tangents of the circular arcs which meet at each point are the same), that is as far as the continuity goes. The piecewise arcs do not make up a smooth function which interpolates meaningfully between the points.

I want to find a function $f(x) = z$ with a real-valued input $x$ and complex-valued output $z$ which passes through $a$ at $f(1)$ and all subsequent points at their respective index values, so $f_n = f(n)$ and which ideally is infinitely-differentiable (smooth) everywhere, except at $f(0)$ which is undefined due to division by zero.

Is that even possible? If so, how do I find that function? If not, are there alternatives?

Update: While struggling to understand MathWonk's answer, I realized that I don't need to understand all the underlying mechanics if WolframAlpha can handle it for me. The key insight that MathWonk provided is that $z ↦ a + \frac{b}{z}$ can be rewritten as $z ↦ \frac{a z + b}{1 z + 0}$, which is the standard form of a linear fractional transformation $z ↦ \frac{a z + b}{c z + d}$ with $c = 1$ and $d = 0$. This corresponds to the $2 × 2$ matrix $\left(\begin{matrix}a & b \\ 1 & 0\end{matrix}\right)$. The left column of the $n$th integer power of that matrix corresponds to the numerator and denominator of the $n$th level of my continued fraction. Asking WolframAlpha to evaluate the expression $((a, b), (1, 0))^x$ produces a complicated output matrix which, using the knowledge that the left two entries in the output matrix are the numerator and denominator of the continued fraction at input-value $x$, simplifies to $\frac{(λ_1 - λ_2) λ_2^t}{λ_1^t - λ_2^t} + λ_1$, where $λ_1 = \frac{a + \sqrt{a^2 + 4 b}}{2}$ and $λ_2 = \frac{a - \sqrt{a^2 + 4 b}}{2}$. Plotting that expression in the complex plane gives the following graph, which is exactly what I wanted:

I'll probably keep trying to figure out how the linear-algebra magic works under the hood to produce that answer, since I like to know what's actually going on to produce useful solutions, but it's nice to have a concrete result!

Answer 1

Here is a broad outline of a solution. The linear-fractional transformation $z\to z'=T(z)=a+ \frac{b}{z}$ can also be expressed in the format of a $2\times 2$ matrix equation, and the iterates of this transformation then become expressed as consecutive integer powers of a matrix $M$. linear-fractional maps as matrices

In more detail, you will find that the numerator and denominator of your iterated function are the two components of the quantity $M^n V$ where $V=(1,0)$ is a $2\times 1$ column vector. Sample shown below.

The next step is to compute those matrix powers. This can be done by consulting powers of a general 2x2 matrix

In particular, use this formula, contained therein:

Finally,the interpolation you seek boils down to interpolation of integer powers of the eigenvalues $\alpha, \beta$. For that final step you can simply use the identity $ \alpha^n= e^{n \ln (\alpha)}$ in which $n$ is now free to be non-integer. Likewise for $\beta$.