Brieskorn manifold

In mathematics, infinite compositions of analytic functions (ICAF) offer alternative formulations of 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.

Notation

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

Forward compositions:

$F_{k,n}(z)=f_k\circ f_{k+1}\circ \cdots \circ f_{n-1}\circ f_n(z)$

Backward compositions:

$G_{k,n}(z)=f_n\circ f_{n-1}\circ \cdots \circ f_{k+1}\circ f_k(z).$

Convergence is interpreted as the existence of $\underset{n\to \mathrm{\infty }}{lim}{F}_{1,n}(z)$ and $\underset{n\to \mathrm{\infty }}{lim}{G}_{1,n}(z).$

For convenience, set F_n(z) = F_1,n(z) and G_n(z) = G_1,n(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 Template:Overline of S. Suppose f(Template:Overline) is a bounded set contained in S. Then

$F_n(z)=f\circ f\circ \cdots \circ f(z)\to \alpha ,$

the attractive fixed point of f in S, for all z in Template:Overline.

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(z)} converges uniformly on compact subsets of S to a constant function F(z) = λ.

Backward (outer or left) Compositions Theorem. {G_n(z)} 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} converge to γ.

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.

Infinite compositions of other functions

General analytic functions

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

$\begin{array}{cc}{f}_n(z)& =a_{n}z+c_{n,2}{z}^2+c_{n,3}{z}^3+\cdots \\ {\rho }_n& =\underset{r}{sup}\left\{{\left|c_{n,r}\right|}^{\frac{1}{r-1}}\right\}\end{array}$

Then the following results hold:

Theorem E1. If a_n ≡ 1,

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

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:

$\begin{array}{cc}\sum _{n=1}^{\mathrm{\infty }}{\epsilon }_n& <\mathrm{\infty },\\ \sum _{n=1}^{\mathrm{\infty }}{\delta }_n& <\mathrm{\infty },\\ \prod _{n=1}^{\mathrm{\infty }}(1+{\delta }_n)& <M_1,\\ \prod _{n=1}^{\mathrm{\infty }}(1+{\epsilon }_n)& <M_2,\\ {\rho }_n& <\frac{{\delta }_n}{R{M}_1{M}_2}.\end{array}$

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

Theorem GF3. Let {f_n} be a sequence of complex functions defined on S = {z : |z| < M}. Suppose there exists a non-negative sequence {β_n} such that

$C\sum _{n=1}^{\mathrm{\infty }}{\beta }_n<M,$

$|f_n(z)-z|<C{\beta }_n,z\in S.$

Set $R=M-C\sum _{n=1}^{\mathrm{\infty }}{\beta }_n>0$. Then G_n(z) → G(z) for |z| < R, uniformly on compact subsets.

Theorem GF4. Let f_n(z) = z(1+g_n(z)), analytic for |z| < R₀, with |g_n(z)| ≤ Cβ_n,

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

Choose 0 < r < R₀ and define

$R=R(r)=\frac{R_0-r}{\prod _{n=1}^{\mathrm{\infty }}(1+C{\beta }_n)}.$

Then F_n → F uniformly for |z| ≤ R. Furthermore,

$|F^{\prime }(z)|\le \prod _{n=1}^{\mathrm{\infty }}(1+\frac{R_0}{r}C{\beta }_n)$.

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 $z\ne \beta =\underset{n\to \mathrm{\infty }}{lim}{\beta }_n$, 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

$\begin{array}{cc}\sum _{n=1}^{\mathrm{\infty }}|{\gamma }_n-{\beta }_n|& <\mathrm{\infty }\\ \sum _{n=1}^{\mathrm{\infty }}n|{\beta }_{n+1}-{\beta }_n|& <\mathrm{\infty }\end{array}$

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

Examples & applications

Continued fractions

The value of the infinite continued fraction

$\frac{a_1}{b_1+\frac{a_2}{b_2+\dots }}$

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

$f_n(z)=\frac{a_n}{b_n+z}.$

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

Consider the continued fraction

$\frac{a_1\zeta }{1+\frac{a_2\zeta }{1+\dots }}$

with

$f_n(z)=\frac{a_n\zeta }{1+z}.$

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

$|a_n|<rR(1-R)\Rightarrow |f_n(z)|<rR<R\Rightarrow \frac{a_1\zeta }{1+\frac{a_2\zeta }{1+\dots }}=F(\zeta )$, analytic for |z| < 1.

Set R = 1/2.

Direct functional expansion

An example illustrating the conversion of a function directly into a composition follows:

Suppose that for |t| > 1, $\phi (tz)=t(\phi (z)+\phi (z{)}^2)$, an entire function with φ(0) = 0, φ′(0) = 1. Then $f_n(z)=z+\frac{z^2}{t^n}\Rightarrow F_n(z)\to \phi (z)$.

Example. $f_n(z)=z+\frac{z^2}{2^n}\Rightarrow F_n(z)\to \frac{1}{2}(e^{2z}-1)$

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

$G(\zeta )=\frac{\frac{e^{\zeta }}{4}}{3+\zeta +\frac{\frac{e^{\zeta }}{8}}{3+\zeta +\frac{\frac{e^{\zeta }}{12}}{3+\zeta +\dots }}}$

To find α = G(α), first we define:

$\begin{array}{cc}{t}_n(z)& =\frac{\frac{e^{\zeta }}{4n}}{3+\zeta +z}\\ f_n(\zeta )& =t_1\circ t_2\circ \cdots \circ t_n(0)\end{array}$

Then calculate $G_n(\zeta )=f_n\circ \cdots \circ f_1(\zeta )$ with ζ = 1, which gives: α = 0.087118118... to ten decimal places after ten iterations.

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

$f_n(\zeta )=\frac{1}{n}\sum _{k=1}^n\phi (\zeta ,\frac{k}{n}).$

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

$\zeta ={\int }_0^1\phi (\zeta ,t)dt$

has a unique solution, α in S, with $\underset{n\to \mathrm{\infty }}{lim}{G}_n(\zeta )=\alpha$.

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 $g_{k,n}(z)=z+{\phi }_{k,n}(z)$ analytic - or simply continuous - in a domain S, such that

$\underset{n\to \mathrm{\infty }}{lim}{\phi }_{k,n}(z)=0$

for all k and z in S imply $g_{k,n}(z)\in S$.

Example 1

$g_{k,n}(z)=z+\frac{k}{n^2}f(z).$

Now, set $T_{1,n}(z)=g_{1,n}(z)$ and $T_{k,n}(z)=g_{k,n}(T_{k-1,n}(z))$. If $\underset{n\to \mathrm{\infty }}{lim}{T}_{n,n}(z)=T(z)$ exists, the initial point z has moved to a new position, T(z), in a fashion described above (for large values of n, $g_{k,n}(z)\approx z$). It is not difficult to show that f(z) = αz + β, α ≥ 0 implies $T_{n,n}(z)\to e^{\frac{\alpha }{2}}z+b\beta$. A byproduct of this derivation is the following representation:

$\underset{n\to \mathrm{\infty }}{lim}\prod _{k=1}^n(1+\frac{2k}{n^2}x)=e^x,x\in 𝐑.$

And of course, if f(z) ≡ c, then

$T(z)=z+c{\int }_0^{1}tdt.$

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 2

$g_n(z)=z+\frac{c_n}{n}\phi (z),$

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

$T(z)=\underset{n\to \mathrm{\infty }}{lim}{T}_n(z)$

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) $c_n=\sqrt{n}$. If c_n ≡ c > 0, then it seems apparent - though not rigorously proven for many cases - that T_n(z) → T(z), a point on the contour γ = γ(c, z). It is easily seen that

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

and

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

when these limits exist.

These concepts are marginally related to active contour theory in image processing.

Self-replicating series & products

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

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

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

Example (S1): Set

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

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

$\begin{array}{cc}{G}_n(z)& =z+g_1(z)+g_2(G_1(z))+g_3(G_2(z))+\cdots +g_n(G_{n-1}(z))\\ & =z+\frac{1}{\rho \cdot 1^2}\sqrt{z}+\frac{1}{\rho \cdot 2^2}\sqrt{G_1(z)}+\frac{1}{\rho \cdot 3^2}\sqrt{G_2(z)}+\cdots +\frac{1}{\rho \cdot n^2}\sqrt{G_{n-1}(z)}\end{array}$

converges absolutely, hence is convergent.

Products

The product defined recursively by $f_n(z)=z(1+g_n(z))$, |z| ≤ M, have the appearance

$G_n(z)=z\prod _{k=1}^n(1+g_k(G_{k-1}(z))).$

In order to apply theorem (GF3) it is required that $|z\cdot g_n(z)|\le C{\beta }_n$ where

$\sum _{k=1}^{\mathrm{\infty }}{\beta }_k<\mathrm{\infty }.$

Once again, a boundedness condition must support

$|G_{n-1}(z)\cdot g_n(G_{n-1}(z))|\le C{\beta }_n.$

If one knows Cβ_n in advance, setting |z| ≤ R = M/P where

$\prod _{n=1}^{\mathrm{\infty }}(1+C{\beta }_n)=P$

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

Example (P1): Suppose that $f_n(z)=z(1+g_n(z))$ where $g_n(z)=\frac{z^2}{n^3}$, observing after a few preliminary computations, that |z| ≤ 1/4 implies |G_n(z)| < 0.27. Then

$|G_n(z)\cdot \frac{G_n(z{)}^2}{n^3}|<(0.02)\frac{1}{n^3}=C{\beta }_n$

and

$G_n(z)=z\cdot \prod _{k=1}^{n-1}(1+\frac{G_k(z{)}^2}{n^3})$

converges uniformly.

References

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