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:

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 in the following reference.^[4] For a different approach to Backward Compositions Theorem, see the following reference.^[5]

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:

Infinite compositions of other functions

Non-contractive complex functions

Results 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:

Additional elementary results include:

Example GF1: ${\displaystyle F_{40}(x+iy)={\underset {k=1}{\overset {40}{\mathop {R} }}}\left({\frac {x+iy}{1+{\tfrac {1}{4^{k}}}(x\cos(y)+iy\sin(x))}}\right),\qquad [-20,20]}$^[9]

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

Linear fractional transformations

Results^[8] for compositions of linear fractional (Möbius) transformations include the following, as examples:

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*^[14]) 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.^[8] A fixed-point continued fraction form (a single variable).

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

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

Direct functional expansion

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

Example 1.^[7][15] 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.^[7]

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

Example 3.^[6]

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

Example 4.^[6]

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

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.

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

Source:^[8]

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

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

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).

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

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

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

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

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

Continued fractions

Example (CF1): A self-generating continued fraction.^[8]

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

Example (CF2): Best described as a self-generating reverse Euler continued fraction.^[8]

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

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

See also

• Generalized continued fraction

References