Infinite compositions of analytic functions

Notation

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

Forward compositions: F_k,n(z) = f_k ∘ f_k+1 ∘ ... ∘ f_n−1 ∘ f_n.

Backward compositions: G_k,n(z) = f_n ∘ f_n−1 ∘ ... ∘ f_k+1 ∘ f_k

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

lim n → ∞ F 1 , n ( z ) , lim n → ∞ 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 F n ( z ) = R n k = 1 f k ( z ) = f 1 ∘ f 2 ∘ ⋯ ∘ f n ( z ) and G n ( z ) = L n k = 1 g k ( z ) = g n ∘ g n − 1 ∘ ⋯ ∘ g 1 ( z )

Contraction theorem

Many results can be considered extensions of the following result:

Contraction Theorem for Analytic Functions. 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 F n ( z ) = ( f ∘ f ∘ ⋯ ∘ f ) ( z ) → α , where α is the attractive fixed point of f in S.

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.

Forward (inner or right) Compositions Theorem. {F_n} converges uniformly on compact subsets of S to a constant function F(z) = λ.Backward (outer or left) Compositions Theorem. {G_n} converges uniformly on compact subsets of S to γ ∈ Ω if and only if the sequence of fixed points {γ_n} of the {f_n} converges to γ.

Additional theory resulting from investigations based on these two theorems, particularly Forward Compositions Theorem, include location analysis for the limits obtained here [2]. For a different approach to Backward Compositions Theorem, see [3].

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. Suppose S is a simply connected compact subset of C and let t n : S → S be a family of functions that satisfies ∀ n , ∀ z 1 , z 2 ∈ S , ∃ ρ : | t n ( z 1 ) − t n ( z 2 ) | < ρ | z 1 − z 2 | , ρ < 1. Define:Then F n ( z ) → β ∈ S uniformly on S . If α n is the unique fixed point of t n then G n ( z ) → α uniformly on S if and only if | α n − α | = ε n → 0 .

Non-contractive complex functions

Results involving entire functions include the following, as examples. Set

f n ( z ) = a n z + c n , 2 z 2 + c n , 3 z 3 + ⋯ ρ n = sup r { | c n , r | 1 r − 1 }

Then the following results hold:

Theorem E1. If a_n ≡ 1, ∑ n = 1 ∞ ρ n < ∞ then F_n → F, entire.Theorem E2. Set ε_n = |a_n−1| suppose there exists non-negative δ_n, M₁, M₂, R such that the following holds: ∑ n = 1 ∞ ε n < ∞ , ∑ n = 1 ∞ δ n < ∞ , ∏ n = 1 ∞ ( 1 + δ n ) < M 1 , ∏ n = 1 ∞ ( 1 + ε n ) < M 2 , ρ n < δ n R M 1 M 2 . Then G_n(z) → G(z), analytic for |z| < R. Convergence is uniform on compact subsets of {z : |z| < R}.

Additional elementary results include:

Theorem GF3. Suppose f n ( z ) = z + ρ n φ ( z ) where there exist R , M > 0 such that | z | < R implies | φ ( z ) | < M . Furthermore, suppose ρ k ≥ 0 , ∑ k = 1 ∞ ρ k < ∞ and R > M ∑ k = 1 ∞ ρ k . Then for R ∗ < R − M ∑ k = 1 n ρ k G n ( z ) ≡ ( f n ∘ f n − 1 ∘ ⋯ ∘ f 1 ) ( z ) → G ( z )  for  { z : | z | < R ∗ } Theorem GF4. Suppose f n ( z ) = z + ρ n φ ( z ) where there exist R , M > 0 such that | z | < R and | ζ | < R implie | φ ( z ) | < M and | φ ( z ) − φ ( ζ ) | < r | z − ζ | . Furthermore, suppose ρ k ≥ 0 , ∑ k = 1 ∞ ρ k < ∞ and R > M ∑ k = 1 ∞ ρ k . Then for R ∗ < R − M ∑ k = 1 n ρ k F n ( z ) ≡ ( f 1 ∘ f 2 ∘ ⋯ ∘ f n ) ( z ) → F ( z )  for  { z : | z | < R ∗ } Theorem GF5. Let f n ( z ) = z + z g n ( z ) , analytic for |z| < R₀, with |g_n(z)| ≤ Cβ_n, ∑ n = 1 ∞ β n < ∞ . Choose 0 < r < R₀ and define R = R ( r ) = R 0 − r ∏ n = 1 ∞ ( 1 + C β n ) . Then F_n → F uniformly for |z| ≤ R. Furthermore, | F ′ ( z ) | ≤ ∏ n = 1 ∞ ( 1 + R 0 r C β n ) .

Example GF1: F 40 ( x + i y ) = R 40 k = 1 ( x + i y 1 + 1 4 k ( x cos ⁡ ( y ) + i y sin ⁡ ( x ) ) ) , [ − 20 , 20 ]

Example GF2: G 40 ( x + i y ) = L 40 k = 1 ( x + i y 1 + 1 2 k ( x cos ⁡ ( y ) + i y sin ⁡ ( x ) ) ) , [ − 20 , 20 ]

Linear fractional transformations

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

Theorem LFT1. On the set of convergence of a sequence {F_n} of non-singular LFTs, the limit function is either:

(a) a non-singular LFT,

(b) a function taking on two distinct values, or

(c) a constant.

In (a), the sequence converges everywhere in the extended plane. In (b), the sequence converges either everywhere, and to the same value everywhere except at one point, or it converges at only two points. Case (c) can occur with every possible set of convergence.

Theorem LFT2. If {F_n} converges to an LFT, then f_n converge to the identity function f(z) = z.Theorem LFT3. If f_n → f and all functions are hyperbolic or loxodromic Möbius transformations, then F_n(z) → λ, a constant, for all z ≠ β = lim n → ∞ β n , where {β_n} are the repulsive fixed points of the {f_n}.Theorem LFT4. 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 ∑ n = 1 ∞ | γ n − β n | < ∞ and ∑ n = 1 ∞ n | β n + 1 − β n | < ∞ then F_n(z) → λ, a constant in the extended complex plane, for all z.

Continued fractions

The value of the infinite continued fraction

a 1 b 1 + a 2 b 2 + ⋯

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

f n ( z ) = a n b n + z .

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

Consider the continued fraction

a 1 ζ 1 + a 2 ζ 1 + ⋯

with

f n ( z ) = a n ζ 1 + z .

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

| a n | < r R ( 1 − R ) ⇒ | f n ( z ) | < r R < R ⇒ a 1 ζ 1 + a 2 ζ 1 + ⋯ = F ( ζ ) , analytic for |z| < 1. Set R = 1/2.

Example. F ( z ) = ( i − 1 ) z 1 + i + z   +   ( 2 − i ) z 1 + 2 i + z   +   ( 3 − i ) z 1 + 3 i + z   + ⋯ , [ − 15 , 15 ]

Direct functional expansion

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

Example 1. Suppose ϕ is an entire function satisfying the following conditions:

{ ϕ ( t z ) = t ( ϕ ( z ) + ϕ ( z ) 2 ) | t | > 1 ϕ ( 0 ) = 0 ϕ ′ ( 0 ) = 1

Then

f n ( z ) = z + z 2 t n ⟹ F n ( z ) → ϕ ( z ) .

Example 2.

f n ( z ) = z + z 2 2 n ⟹ F n ( z ) → 1 2 ( e 2 z − 1 )

Example 3.

f n ( z ) = z 1 − z 2 4 n ⟹ F n ( z ) → tan ⁡ ( z )

Example 4.

g n ( z ) = 2 ⋅ 4 n z ( 1 + z 2 4 n − 1 ) ⟹ G n ( z ) → 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. For |ζ| ≤ 1 let

G ( ζ ) = e ζ 4 3 + ζ + e ζ 8 3 + ζ + e ζ 12 3 + ζ + ⋯

To find α = G(α), first we define:

t n ( z ) = e ζ 4 n 3 + ζ + z f n ( ζ ) = t 1 ∘ t 2 ∘ ⋯ ∘ t n ( 0 )

Then calculate G n ( ζ ) = f n ∘ ⋯ ∘ f 1 ( ζ ) with ζ = 1, which gives: α = 0.087118118... to ten decimal places after ten iterations.

Theorem FP2. Let φ(ζ, t) be analytic in S = {z : |z| < R} for all t in [0, 1] and continuous in t. Set f n ( ζ ) = 1 n ∑ k = 1 n φ ( ζ , k n ) . If |φ(ζ, t)| ≤ r < R for ζ ∈ S and t ∈ [0, 1], then ζ = ∫ 0 1 φ ( ζ , t ) d t has a unique solution, α in S, with lim n → ∞ G n ( ζ ) = α .

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 g k , n ( z ) = z + φ k , n ( z ) analytic or simply continuous – in a domain S, such that

lim n → ∞ φ k , n ( z ) = 0 for all k and all z in S,

and g k , n ( z ) ∈ S .

Principal example

g k , n ( z ) = z + 1 n ϕ ( z , k n ) G k , n ( z ) = ( g k , n ∘ g k − 1 , n ∘ ⋯ ∘ g 1 , n ) ( z ) G n ( z ) = G n , n ( z )

implies

λ n ( z ) ≐ G n ( z ) − z = 1 n ∑ k = 1 n ϕ ( G k − 1 , n ( z ) k n ) ≐ 1 n ∑ k = 1 n ψ ( z , k n ) ∼ ∫ 0 1 ψ ( z , t ) d t ,

where the integral is well-defined if d z d t = ϕ ( z , t ) has a closed-form solution z(t). Then

λ n ( z 0 ) ≈ ∫ 0 1 ϕ ( z ( t ) , t ) d t .

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. ϕ ( z , t ) = 2 t − cos ⁡ y 1 − sin ⁡ x cos ⁡ y + i 1 − 2 t sin ⁡ x 1 − sin ⁡ x cos ⁡ y , ∫ 0 1 ψ ( z , t ) d t

Example. Let:

g n ( z ) = z + c n n ϕ ( z ) , with f ( z ) = z + ϕ ( z ) .

Next, set 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

T ( z ) = lim n → ∞ 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) c n = 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

∮ γ ϕ ( ζ ) d ζ = lim n → ∞ c n ∑ k = 1 n ϕ 2 ( T k − 1 , n ( z ) )

and

L ( γ ( z ) ) = lim n → ∞ c n ∑ k = 1 n | ϕ ( T k − 1 , n ( z ) ) | ,

when these limits exist.

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

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 g n ( G n − 1 ( z ) ) occurs throughout the expansion. The restriction

| z | < R = M − C ∑ k = 1 ∞ β k > 0

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

Example (S1). Set

f n ( z ) = z + 1 ρ n 2 z , ρ > π 6

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

G n ( z ) = z + g 1 ( z ) + g 2 ( G 1 ( z ) ) + g 3 ( G 2 ( z ) ) + ⋯ + g n ( G n − 1 ( z ) ) = z + 1 ρ ⋅ 1 2 z + 1 ρ ⋅ 2 2 G 1 ( z ) + 1 ρ ⋅ 3 2 G 2 ( z ) + ⋯ + 1 ρ ⋅ n 2 G n − 1 ( z )

converges absolutely, hence is convergent.

Example (S2): f n ( z ) = z + 1 n 2 ⋅ φ ( z ) , φ ( z ) = 2 cos ⁡ ( x / y ) + i 2 sin ⁡ ( x / y ) , > G n ( z ) = f n ∘ f n − 1 ∘ ⋯ ∘ f 1 ( z ) , [ − 10 , 10 ] , n = 50

Products

The product defined recursively by

f n ( z ) = z ( 1 + g n ( z ) ) , | z | ⩽ M ,

has the appearance

G n ( z ) = z ∏ k = 1 n ( 1 + g k ( G k − 1 ( z ) ) ) .

In order to apply Theorem GF3 it is required that:

| z g n ( z ) | ≤ C β n , ∑ k = 1 ∞ β k < ∞ .

Once again, a boundedness condition must support

| G n − 1 ( z ) g n ( G n − 1 ( z ) ) | ≤ C β n .

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

| z | ⩽ R = M P where P = ∏ n = 1 ∞ ( 1 + C β n ) .

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

Example (P1). Suppose f n ( z ) = z ( 1 + g n ( z ) ) with g n ( z ) = z 2 n 3 , observing after a few preliminary computations, that |z| ≤ 1/4 implies |G_n(z)| < 0.27. Then

| G n ( z ) G n ( z ) 2 n 3 | < ( 0.02 ) 1 n 3 = C β n

and

G n ( z ) = z ∏ k = 1 n − 1 ( 1 + G k ( z ) 2 n 3 )

converges uniformly.

Example (P2).

g k , n ( z ) = z ( 1 + 1 n φ ( z , k n ) ) , G n , n ( z ) = ( g n , n ∘ g n − 1 , n ∘ ⋯ ∘ g 1 , n ) ( z ) = z ∏ k = 1 n ( 1 + P k , n ( z ) ) , P k , n ( z ) = 1 n φ ( G k − 1 , n ( z ) , k n ) , ∏ k = 1 n − 1 ( 1 + P k , n ( z ) ) = 1 + P 1 , n ( z ) + P 2 , n ( z ) + ⋯ + P k − 1 , n ( z ) + R n ( z ) ∼ ∫ 0 1 π ( z , t ) d t + 1 + R n ( z ) , φ ( z ) = x cos ⁡ ( y ) + i ⋅ y sin ⁡ ( x ) , ∫ 0 1 ( z π ( z , t ) − 1 ) d t , [ − 15 , 15 ] :

Continued fractions

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

F n ( z ) = ρ ( z ) δ 1 + ρ ( F 1 ( z ) ) δ 2 +   ρ ( F 2 ( z ) ) δ 3 +   ⋯   ρ ( F n − 1 ( z ) ) δ n , ρ ( z ) = cos ⁡ ( y ) cos ⁡ ( y ) + sin ⁡ ( x ) + i sin ⁡ ( x ) cos ⁡ ( y ) + sin ⁡ ( x ) , [ 0 < x < 20 ] , [ 0 < y < 20 ] , δ k ≡ 1

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

G n ( z ) = ρ ( G n − 1 ( z ) ) 1 + ρ ( G n − 1 ( z ) ) −   ρ ( G n − 2 ( z ) ) 1 + ρ ( G n − 2 ( z ) ) − ⋯ ρ ( G 1 ( z ) ) 1 + ρ ( G 1 ( z ) ) −   ρ ( z ) 1 + ρ ( z ) − z , ρ ( z ) = ρ ( x + i y ) = x cos ⁡ ( y ) + i y sin ⁡ ( x ) , [ − 15 , 15 ] , n = 30