Infinite compositions of analytic functions

In mathematics, infinite compositions of analytic functions (ICAF) offer alternative formulations of analytic continued fractions, series, products and other infinite expansions, and the theory evolving from such compositions may shed light on the convergence/divergence of these expansions. Some functions can actually be expanded directly as infinite compositions. In addition, it is possible to use ICAF to evaluate solutions of fixed point equations involving infinite expansions. Complex dynamics offers another venue for iteration of systems of functions rather than a single function. For infinite compositions of a single function see Iterated function. For compositions of a finite number of functions, useful in fractal theory, see Iterated function system.

Although the title of this article specifies analytic functions, there are results for more general functions of a complex variable as well.

Notation[]

There are several notations describing infinite compositions, including the following:

Forward compositions: ${\displaystyle F_{k,n}(z)=f_{k}\circ f_{k+1}\circ \dots \circ f_{n-1}\circ f_{n}(z).}$

Backward compositions: ${\displaystyle G_{k,n}(z)=f_{n}\circ f_{n-1}\circ \dots \circ f_{k+1}\circ f_{k}(z).}$

In each case convergence is interpreted as the existence of the following limits:

${\displaystyle \lim _{n\to \infty }F_{1,n}(z),\qquad \lim _{n\to \infty }G_{1,n}(z).}$

For convenience, set F_n(z) = F_1,n(z) and G_n(z) = G_1,n(z).

One may also write ${\displaystyle F_{n}(z)={\underset {k=1}{\overset {n}{\mathop {R} }}}\,f_{k}(z)=f_{1}\circ f_{2}\circ \cdots \circ f_{n}(z)}$ and ${\displaystyle G_{n}(z)={\underset {k=1}{\overset {n}{\mathop {L} }}}\,g_{k}(z)=g_{n}\circ g_{n-1}\circ \cdots \circ g_{1}(z)}$

Contraction theorem[]

Many results can be considered extensions of the following result:

Contraction Theorem for Analytic Functions^[1] — Let f be analytic in a simply-connected region S and continuous on the closure S of S. Suppose f(S) is a bounded set contained in S. Then for all z in S there exists an attractive fixed point α of f in S such that:

$${\displaystyle F_{n}(z)=(f\circ f\circ \cdots \circ f)(z)\to \alpha .}$$

Infinite compositions of contractive functions[]

Let {f_n} be a sequence of functions analytic on a simply-connected domain S. Suppose there exists a compact set Ω ⊂ S such that for each n, f_n(S) ⊂ Ω.

Additional theory resulting from investigations based on these two theorems, particularly Forward Compositions Theorem, include location analysis for the limits obtained here [1]. For a different approach to Backward Compositions Theorem, see [2].

Regarding Backward Compositions Theorem, the example f_2n(z) = 1/2 and f_2n−1(z) = −1/2 for S = {z : |z| < 1} demonstrates the inadequacy of simply requiring contraction into a compact subset, like Forward Compositions Theorem.

For functions not necessarily analytic the Lipschitz condition suffices:

Theorem^[4] — Suppose ${\displaystyle S}$ is a simply connected compact subset of ${\displaystyle \mathbb {C} }$ and let ${\displaystyle t_{n}:S\to S}$ be a family of functions that satisfies

$${\displaystyle \forall n,\forall z_{1},z_{2}\in S,\exists \rho :\quad \left|t_{n}(z_{1})-t_{n}(z_{2})\right|\leq \rho |z_{1}-z_{2}|,\quad \rho <1.}$$

Define:

$${\displaystyle {\begin{aligned}G_{n}(z)&=\left(t_{n}\circ t_{n-1}\circ \cdots \circ t_{1}\right)(z)\\F_{n}(z)&=\left(t_{1}\circ t_{2}\circ \cdots \circ t_{n}\right)(z)\end{aligned}}}$$

Then ${\displaystyle F_{n}(z)\to \beta \in S}$ uniformly on ${\displaystyle S.}$ If ${\displaystyle \alpha _{n}}$ is the unique fixed point of ${\displaystyle t_{n}}$ then ${\displaystyle G_{n}(z)\to \alpha }$ uniformly on ${\displaystyle S}$ if and only if ${\displaystyle |\alpha _{n}-\alpha |=\varepsilon _{n}\to 0}$.

Infinite compositions of other functions[]

Non-contractive complex functions[]

Results^[5] involving entire functions include the following, as examples. Set

${\displaystyle {\begin{aligned}f_{n}(z)&=a_{n}z+c_{n,2}z^{2}+c_{n,3}z^{3}+\cdots \\\rho _{n}&=\sup _{r}\left\{\left|c_{n,r}\right|^{\frac {1}{r-1}}\right\}\end{aligned}}}$

Then the following results hold:

Theorem E1^[6] — If a_n ≡ 1,

$${\displaystyle \sum _{n=1}^{\infty }\rho _{n}<\infty }$$

then F_n → F is entire.

Theorem E2^[5] — Set ε_n = |a_n−1| suppose there exists non-negative δ_n, M₁, M₂, R such that the following holds:

$${\displaystyle {\begin{aligned}\sum _{n=1}^{\infty }\varepsilon _{n}&<\infty ,\\\sum _{n=1}^{\infty }\delta _{n}&<\infty ,\\\prod _{n=1}^{\infty }(1+\delta _{n})&<M_{1},\\\prod _{n=1}^{\infty }(1+\varepsilon _{n})&<M_{2},\\\rho _{n}&<{\frac {\delta _{n}}{RM_{1}M_{2}}}.\end{aligned}}}$$

Then G_n(z) → G(z) is analytic for |z| < R. Convergence is uniform on compact subsets of {z : |z| < R}.

Additional elementary results include:

Theorem GF3^[4] — Suppose ${\displaystyle f_{k}(z)=z+\rho _{k}\varphi _{k}(z)}$ where there exist ${\displaystyle R,M>0}$ such that ${\displaystyle |z|<R}$ implies ${\displaystyle |\varphi _{k}(z)|<M,\forall k,\ }$ Furthermore, suppose ${\textstyle \rho _{k}\geq 0,\sum _{k=1}^{\infty }\rho _{k}<\infty }$ and ${\textstyle R>M\sum _{k=1}^{\infty }\rho _{k}.}$ Then for ${\textstyle R*<R-M\sum _{k=1}^{\infty }\rho _{k}}$

$${\displaystyle G_{n}(z)\equiv \left(f_{n}\circ f_{n-1}\circ \cdots \circ f_{1}\right)(z)\to G(z)\qquad {\text{ for }}\{z:|z|<R*\}}$$

Theorem GF4^[4] — Suppose ${\displaystyle f_{k}(z)=z+\rho _{k}\varphi _{k}(z)}$ where there exist ${\displaystyle R,M>0}$ such that ${\displaystyle |z|<R}$ and ${\displaystyle |\zeta |<R}$ implies ${\displaystyle |\varphi _{k}(z)|<M}$ and ${\displaystyle |\varphi _{k}(z)-\varphi _{k}(\zeta )|\leq r|z-\zeta |,\forall k.\ }$ Furthermore, suppose ${\textstyle \rho _{k}\geq 0,\sum _{k=1}^{\infty }\rho _{k}<\infty }$ and ${\textstyle R>M\sum _{k=1}^{\infty }\rho _{k}.}$ Then for ${\textstyle R*<R-M\sum _{k=1}^{\infty }\rho _{k}}$

$${\displaystyle F_{n}(z)\equiv \left(f_{1}\circ f_{2}\circ \cdots \circ f_{n}\right)(z)\to F(z)\qquad {\text{ for }}\{z:|z|<R*\}}$$

Theorem GF5^[5] — Let ${\displaystyle f_{n}(z)=z+zg_{n}(z),}$ analytic for |z| < R₀, with |g_n(z)| ≤ Cβ_n,

$${\displaystyle \sum _{n=1}^{\infty }\beta _{n}<\infty .}$$

Choose 0 < r < R₀ and define

$${\displaystyle R=R(r)={\frac {R_{0}-r}{\prod _{n=1}^{\infty }(1+C\beta _{n})}}.}$$

Then F_n → F uniformly for

Example GF1: ${\displaystyle F_{40}(x+iy)={\underset {k=1}{\overset {40}{\mathop {R} }}}\left({\frac {x+iy}{1+{\tfrac {1}{4^{k}}}(x\cos(y)+iy\sin(x))}}\right),\qquad [-20,20]}$[3]

Example GF1:Reproductive universe – A topographical (moduli) image of an infinite composition.

Example GF2: ${\displaystyle G_{40}(x+iy)={\underset {k=1}{\overset {40}{\mathop {L} }}}\,\left({\frac {x+iy}{1+{\tfrac {1}{2^{k}}}(x\cos(y)+iy\sin(x))}}\right),\ [-20,20]}$

Example GF2:Metropolis at 30K – A topographical (moduli) image of an infinite composition.

Linear fractional transformations[]

Results^[5] for compositions of linear fractional (Möbius) transformations include the following, as examples:

Theorem LFT4^[10] — If f_n → f where f is parabolic with fixed point γ. Let the fixed-points of the {f_n} be {γ_n} and {β_n}. If

$${\displaystyle \sum _{n=1}^{\infty }\left|\gamma _{n}-\beta _{n}\right|<\infty \quad {\text{and}}\quad \sum _{n=1}^{\infty }n\left|\beta _{n+1}-\beta _{n}\right|<\infty }$$

then F_n(z) → λ, a constant in the extended complex plane, for all z.

Examples and applications[]

Continued fractions[]

The value of the infinite continued fraction

${\displaystyle {\cfrac {a_{1}}{b_{1}+{\cfrac {a_{2}}{b_{2}+\cdots }}}}}$

may be expressed as the limit of the sequence {F_n(0)} where

${\displaystyle f_{n}(z)={\frac {a_{n}}{b_{n}+z}}.}$

As a simple example, a well-known result (Worpitsky Circle*^[11]) follows from an application of Theorem (A):

Consider the continued fraction

${\displaystyle {\cfrac {a_{1}\zeta }{1+{\cfrac {a_{2}\zeta }{1+\cdots }}}}}$

with

${\displaystyle f_{n}(z)={\frac {a_{n}\zeta }{1+z}}.}$

Stipulate that |ζ| < 1 and |z| < R < 1. Then for 0 < r < 1,

${\displaystyle |a_{n}|<rR(1-R)\Rightarrow \left|f_{n}(z)\right|<rR<R\Rightarrow {\frac {a_{1}\zeta }{1+{\frac {a_{2}\zeta }{1+\cdots }}}}=F(\zeta )}$, analytic for |z| < 1. Set R = 1/2.

Example. ${\displaystyle F(z)={\frac {(i-1)z}{1+i+z{\text{ }}+}}{\text{ }}{\frac {(2-i)z}{1+2i+z{\text{ }}+}}{\text{ }}{\frac {(3-i)z}{1+3i+z{\text{ }}+}}\cdots ,}$ ${\displaystyle [-15,15]}$

Example: Continued fraction1 – Topographical (moduli) image of a continued fraction (one for each point) in the complex plane. [−15,15]

Example.^[5] A fixed-point continued fraction form (a single variable).

${\displaystyle f_{k,n}(z)={\frac {\alpha _{k,n}\beta _{k,n}}{\alpha _{k,n}+\beta _{k,n}-z}},\alpha _{k,n}=\alpha _{k,n}(z),\beta _{k,n}=\beta _{k,n}(z),F_{n}(z)=\left(f_{1,n}\circ \cdots \circ f_{n,n}\right)(z)}$

${\displaystyle \alpha _{k,n}=x\cos(ty)+iy\sin(tx),\beta _{k,n}=\cos(ty)+i\sin(tx),t=k/n}$

Example: Infinite Brooch - Topographical (moduli) image of a continued fraction form in the complex plane. (6<x<9.6),(4.8<y<8)

Direct functional expansion[]

Examples illustrating the conversion of a function directly into a composition follow:

Example 1.^[6][12] Suppose ${\displaystyle \phi }$ is an entire function satisfying the following conditions:

${\displaystyle {\begin{cases}\phi (tz)=t\left(\phi (z)+\phi (z)^{2}\right)&|t|>1\\\phi (0)=0\\\phi '(0)=1\end{cases}}}$

Then

${\displaystyle f_{n}(z)=z+{\frac {z^{2}}{t^{n}}}\Longrightarrow F_{n}(z)\to \phi (z)}$.

Example 2.^[6]

${\displaystyle f_{n}(z)=z+{\frac {z^{2}}{2^{n}}}\Longrightarrow F_{n}(z)\to {\frac {1}{2}}\left(e^{2z}-1\right)}$

Example 3.^[5]

${\displaystyle f_{n}(z)={\frac {z}{1-{\tfrac {z^{2}}{4^{n}}}}}\Longrightarrow F_{n}(z)\to \tan(z)}$

Example 4.^[5]

${\displaystyle g_{n}(z)={\frac {2\cdot 4^{n}}{z}}\left({\sqrt {1+{\frac {z^{2}}{4^{n}}}}}-1\right)\Longrightarrow G_{n}(z)\to \arctan(z)}$

Calculation of fixed-points[]

Theorem (B) can be applied to determine the fixed-points of functions defined by infinite expansions or certain integrals. The following examples illustrate the process:

Example FP1.^[3] For |ζ| ≤ 1 let

${\displaystyle G(\zeta )={\frac {\tfrac {e^{\zeta }}{4}}{3+\zeta +{\cfrac {\tfrac {e^{\zeta }}{8}}{3+\zeta +{\cfrac {\tfrac {e^{\zeta }}{12}}{3+\zeta +\cdots }}}}}}}$

To find α = G(α), first we define:

${\displaystyle {\begin{aligned}t_{n}(z)&={\cfrac {\tfrac {e^{\zeta }}{4n}}{3+\zeta +z}}\\f_{n}(\zeta )&=t_{1}\circ t_{2}\circ \cdots \circ t_{n}(0)\end{aligned}}}$

Then calculate ${\displaystyle G_{n}(\zeta )=f_{n}\circ \cdots \circ f_{1}(\zeta )}$ with ζ = 1, which gives: α = 0.087118118... to ten decimal places after ten iterations.

Theorem FP2^[5] — Let φ(ζ, t) be analytic in S = {z : |z| < R} for all t in [0, 1] and continuous in t. Set

$${\displaystyle f_{n}(\zeta )={\frac {1}{n}}\sum _{k=1}^{n}\varphi \left(\zeta ,{\tfrac {k}{n}}\right).}$$

If |φ(ζ, t)| ≤ r < R for ζ ∈ S and t ∈ [0, 1], then

$${\displaystyle \zeta =\int _{0}^{1}\varphi (\zeta ,t)\,dt}$$

has a unique solution, α in S, with ${\displaystyle \lim _{n\to \infty }G_{n}(\zeta )=\alpha .}$

Evolution functions[]

Consider a time interval, normalized to I = [0, 1]. ICAFs can be constructed to describe continuous motion of a point, z, over the interval, but in such a way that at each "instant" the motion is virtually zero (see Zeno's Arrow): For the interval divided into n equal subintervals, 1 ≤ k ≤ n set ${\displaystyle g_{k,n}(z)=z+\varphi _{k,n}(z)}$ analytic or simply continuous – in a domain S, such that

${\displaystyle \lim _{n\to \infty }\varphi _{k,n}(z)=0}$ for all k and all z in S,

and ${\displaystyle g_{k,n}(z)\in S}$.

Principal example^[5][]

${\displaystyle {\begin{aligned}g_{k,n}(z)&=z+{\frac {1}{n}}\phi \left(z,{\tfrac {k}{n}}\right)\\G_{k,n}(z)&=\left(g_{k,n}\circ g_{k-1,n}\circ \cdots \circ g_{1,n}\right)(z)\\G_{n}(z)&=G_{n,n}(z)\end{aligned}}}$

implies

${\displaystyle \lambda _{n}(z)\doteq G_{n}(z)-z={\frac {1}{n}}\sum _{k=1}^{n}\phi \left(G_{k-1,n}(z){\tfrac {k}{n}}\right)\doteq {\frac {1}{n}}\sum _{k=1}^{n}\psi \left(z,{\tfrac {k}{n}}\right)\sim \int _{0}^{1}\psi (z,t)\,dt,}$

where the integral is well-defined if ${\displaystyle {\tfrac {dz}{dt}}=\phi (z,t)}$ has a closed-form solution z(t). Then

${\displaystyle \lambda _{n}(z_{0})\approx \int _{0}^{1}\phi (z(t),t)\,dt.}$

Otherwise, the integrand is poorly defined although the value of the integral is easily computed. In this case one might call the integral a "virtual" integral.

Example. ${\displaystyle \phi (z,t)={\frac {2t-\cos y}{1-\sin x\cos y}}+i{\frac {1-2t\sin x}{1-\sin x\cos y}},\int _{0}^{1}\psi (z,t)\,dt}$

Example 1: Virtual tunnels – Topographical (moduli) image of virtual integrals (one for each point) in the complex plane. [−10,10]

Two contours flowing towards an attractive fixed point (red on the left). The white contour (c = 2) terminates before reaching the fixed point. The second contour (c(n) = square root of n) terminates at the fixed point. For both contours, n = 10,000

Example.^[13] Let:

${\displaystyle g_{n}(z)=z+{\frac {c_{n}}{n}}\phi (z),\quad {\text{with}}\quad f(z)=z+\phi (z).}$

Next, set ${\displaystyle T_{1,n}(z)=g_{n}(z),T_{k,n}(z)=g_{n}(T_{k-1,n}(z)),}$ and T_n(z) = T_n,n(z). Let

${\displaystyle T(z)=\lim _{n\to \infty }T_{n}(z)}$

when that limit exists. The sequence {T_n(z)} defines contours γ = γ(c_n, z) that follow the flow of the vector field f(z). If there exists an attractive fixed point α, meaning |f(z) − α| ≤ ρ|z − α| for 0 ≤ ρ < 1, then T_n(z) → T(z) ≡ α along γ = γ(c_n, z), provided (for example) ${\displaystyle c_{n}={\sqrt {n}}}$. If c_n ≡ c > 0, then T_n(z) → T(z), a point on the contour γ = γ(c, z). It is easily seen that

${\displaystyle \oint _{\gamma }\phi (\zeta )\,d\zeta =\lim _{n\to \infty }{\frac {c}{n}}\sum _{k=1}^{n}\phi ^{2}\left(T_{k-1,n}(z)\right)}$

and

${\displaystyle L(\gamma (z))=\lim _{n\to \infty }{\frac {c}{n}}\sum _{k=1}^{n}\left|\phi \left(T_{k-1,n}(z)\right)\right|,}$

when these limits exist.

These concepts are marginally related to active contour theory in image processing, and are simple generalizations of the Euler method

Self-replicating expansions[]

Series[]

The series defined recursively by f_n(z) = z + g_n(z) have the property that the nth term is predicated on the sum of the first n − 1 terms. In order to employ theorem (GF3) it is necessary to show boundedness in the following sense: If each f_n is defined for |z| < M then |G_n(z)| < M must follow before |f_n(z) − z| = |g_n(z)| ≤ Cβ_n is defined for iterative purposes. This is because ${\displaystyle g_{n}(G_{n-1}(z))}$ occurs throughout the expansion. The restriction

${\displaystyle |z|<R=M-C\sum _{k=1}^{\infty }\beta _{k}>0}$

serves this purpose. Then G_n(z) → G(z) uniformly on the restricted domain.

Example (S1). Set

${\displaystyle f_{n}(z)=z+{\frac {1}{\rho n^{2}}}{\sqrt {z}},\qquad \rho >{\sqrt {\frac {\pi }{6}}}}$

and M = ρ². Then R = ρ² − (π/6) > 0. Then, if ${\displaystyle S=\left\{z:|z|<R,\operatorname {Re} (z)>0\right\}}$, z in S implies |G_n(z)| < M and theorem (GF3) applies, so that

${\displaystyle {\begin{aligned}G_{n}(z)&=z+g_{1}(z)+g_{2}(G_{1}(z))+g_{3}(G_{2}(z))+\cdots +g_{n}(G_{n-1}(z))\\&=z+{\frac {1}{\rho \cdot 1^{2}}}{\sqrt {z}}+{\frac {1}{\rho \cdot 2^{2}}}{\sqrt {G_{1}(z)}}+{\frac {1}{\rho \cdot 3^{2}}}{\sqrt {G_{2}(z)}}+\cdots +{\frac {1}{\rho \cdot n^{2}}}{\sqrt {G_{n-1}(z)}}\end{aligned}}}$

converges absolutely, hence is convergent.

Example (S2): ${\displaystyle f_{n}(z)=z+{\frac {1}{n^{2}}}\cdot \varphi (z),\varphi (z)=2\cos(x/y)+i2\sin(x/y),>G_{n}(z)=f_{n}\circ f_{n-1}\circ \cdots \circ f_{1}(z),\qquad [-10,10],n=50}$

Example (S2)- A topographical (moduli) image of a self generating series.

Products[]

The product defined recursively by

${\displaystyle f_{n}(z)=z(1+g_{n}(z)),\qquad |z|\leqslant M,}$

has the appearance

${\displaystyle G_{n}(z)=z\prod _{k=1}^{n}\left(1+g_{k}\left(G_{k-1}(z)\right)\right).}$

In order to apply Theorem GF3 it is required that:

${\displaystyle \left|zg_{n}(z)\right|\leq C\beta _{n},\qquad \sum _{k=1}^{\infty }\beta _{k}<\infty .}$

Once again, a boundedness condition must support

${\displaystyle \left|G_{n-1}(z)g_{n}(G_{n-1}(z))\right|\leq C\beta _{n}.}$

If one knows Cβ_n in advance, the following will suffice:

${\displaystyle |z|\leqslant R={\frac {M}{P}}\qquad {\text{where}}\quad P=\prod _{n=1}^{\infty }\left(1+C\beta _{n}\right).}$

Then G_n(z) → G(z) uniformly on the restricted domain.

Example (P1). Suppose ${\displaystyle f_{n}(z)=z(1+g_{n}(z))}$ with ${\displaystyle g_{n}(z)={\tfrac {z^{2}}{n^{3}}},}$ observing after a few preliminary computations, that |z| ≤ 1/4 implies |G_n(z)| < 0.27. Then

${\displaystyle \left|G_{n}(z){\frac {G_{n}(z)^{2}}{n^{3}}}\right|<(0.02){\frac {1}{n^{3}}}=C\beta _{n}}$

and

${\displaystyle G_{n}(z)=z\prod _{k=1}^{n-1}\left(1+{\frac {G_{k}(z)^{2}}{n^{3}}}\right)}$

converges uniformly.

Example (P2).

${\displaystyle g_{k,n}(z)=z\left(1+{\frac {1}{n}}\varphi \left(z,{\tfrac {k}{n}}\right)\right),}$

${\displaystyle G_{n,n}(z)=\left(g_{n,n}\circ g_{n-1,n}\circ \cdots \circ g_{1,n}\right)(z)=z\prod _{k=1}^{n}(1+P_{k,n}(z)),}$

${\displaystyle P_{k,n}(z)={\frac {1}{n}}\varphi \left(G_{k-1,n}(z),{\tfrac {k}{n}}\right),}$

${\displaystyle \prod _{k=1}^{n-1}\left(1+P_{k,n}(z)\right)=1+P_{1,n}(z)+P_{2,n}(z)+\cdots +P_{k-1,n}(z)+R_{n}(z)\sim \int _{0}^{1}\pi (z,t)\,dt+1+R_{n}(z),}$

${\displaystyle \varphi (z)=x\cos(y)+iy\sin(x),\int _{0}^{1}(z\pi (z,t)-1)\,dt,\qquad [-15,15]:}$

Example (P2): Picasso's Universe – a derived virtual integral from a self-generating infinite product. Click on image for higher resolution.

Continued fractions[]

Example (CF1): A self-generating continued fraction.^[5] [4]

${\displaystyle {\begin{aligned}F_{n}(z)&={\frac {\rho (z)}{\delta _{1}+}}{\frac {\rho (F_{1}(z))}{\delta _{2}+}}{\frac {\rho (F_{2}(z))}{\delta _{3}+}}\cdots {\frac {\rho (F_{n-1}(z))}{\delta _{n}}},\\\rho (z)&={\frac {\cos(y)}{\cos(y)+\sin(x)}}+i{\frac {\sin(x)}{\cos(y)+\sin(x)}},\qquad [0<x<20],[0<y<20],\qquad \delta _{k}\equiv 1\end{aligned}}}$

Example CF1: Diminishing returns – a topographical (moduli) image of a self-generating continued fraction.

Example (CF2): Best described as a self-generating reverse Euler continued fraction.^[5]

${\displaystyle G_{n}(z)={\frac {\rho (G_{n-1}(z))}{1+\rho (G_{n-1}(z))-}}\ {\frac {\rho (G_{n-2}(z))}{1+\rho (G_{n-2}(z))-}}\cdots {\frac {\rho (G_{1}(z))}{1+\rho (G_{1}(z))-}}\ {\frac {\rho (z)}{1+\rho (z)-z}},}$

${\displaystyle \rho (z)=\rho (x+iy)=x\cos(y)+iy\sin(x),\qquad [-15,15],n=30}$

Example CF2: Dream of Gold – a topographical (moduli) image of a self-generating reverse Euler continued fraction.

References[]