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.
There are several notations describing infinite compositions, including the following:
Forward compositions:
Fk,n(z)=fk\circfk\circ...\circfn\circfn(z).
Backward compositions:
Gk,n(z)=fn\circfn\circ...\circfk\circfk(z).
In each case convergence is interpreted as the existence of the following limits:
\limn\toF1,n(z), \limn\toinftyG1,n(z).
For convenience, set and .
One may also write
Fn(z)=\underset{k=1}{\overset{n}{R}}fk(z)=f1\circf2\circ … \circfn(z)
Gn(z)=\underset{k=1}{\overset{n}{L}}gk(z)=gn\circgn-1\circ … \circg1(z)
Many results can be considered extensions of the following result:
Let 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, fn(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.[1] For a different approach to Backward Compositions Theorem, see the following reference.[2]
Regarding Backward Compositions Theorem, the example f2n(z) = 1/2 and f2n−1(z) = −1/2 for S = demonstrates the inadequacy of simply requiring contraction into a compact subset, like Forward Compositions Theorem.
For functions not necessarily analytic the Lipschitz condition suffices:
Results involving entire functions include the following, as examples. Set
\begin{align} fn(z)&=anz+cn,2
2+c | |
z | |
n,3 |
z3+ … \\ \rhon&=\supr\left\{\left|cn,r
| ||||
\right| |
\right\} \end{align}
Then the following results hold:
Additional elementary results include:
Example GF1:
F40(x+iy)=\underset{k=1}{\overset{40}{R}}\left(
x+iy | |
1+\tfrac{1 |
{4k}(x\cos(y)+iy\sin(x))}\right), [-20,20]
Example GF2:
G40(x+iy)=\underset{k=1}{\overset{40}{L}}\left(
x+iy | |
1+\tfrac{1 |
{2k}(x\cos(y)+iy\sin(x))}\right), [-20,20]
Results for compositions of linear fractional (Möbius) transformations include the following, as examples:
The value of the infinite continued fraction
\cfrac{a1}{b1+\cfrac{a2}{b2+ … }}
may be expressed as the limit of the sequence where
f | ||||
|
.
As a simple example, a well-known result (Worpitsky's circle theorem[4]) follows from an application of Theorem (A):
Consider the continued fraction
\cfrac{a1\zeta}{1+\cfrac{a2\zeta}{1+ … }}
with
f | ||||
|
.
Stipulate that |ζ| < 1 and |z| < R < 1. Then for 0 < r < 1,
|an|<rR(1-R) ⇒ \left|fn(z)\right|<rR<R ⇒
a1\zeta | |||
|
=F(\zeta)
Example.
F(z)= | (i-1)z | |
1+i+z+ |
(2-i)z | ||
1+2i+z+ |
(3-i)z | |
1+3i+z+ |
… ,
[-15,15]
Example. A fixed-point continued fraction form (a single variable).
fk,n(z)=
\alphak,n\betak,n | |
\alphak,n+\betak,n-z |
,\alphak,n=\alphak,n(z),\betak,n=\betak,n(z),Fn(z)=\left(f1,n\circ … \circfn,n\right)(z)
\alphak,n=x\cos(ty)+iy\sin(tx),\betak,n=\cos(ty)+i\sin(tx),t=k/n
Examples illustrating the conversion of a function directly into a composition follow:
Example 1.[5] Suppose
\phi
\begin{cases}\phi(tz)=t\left(\phi(z)+\phi(z)2\right)&|t|>1\ \phi(0)=0\ \phi'(0)=1\end{cases}
f | ||||
|
\LongrightarrowFn(z)\to\phi(z)
Example 2.
f | ||||
|
\LongrightarrowFn(z)\to
1 | |
2 |
\left(e2z-1\right)
Example 3.
fn(z)=
z | |
1-\tfrac{z2 |
{4n}}\LongrightarrowFn(z)\to\tan(z)
Example 4.
g | ||||
|
\left(\sqrt{1+
z2 | |
4n |
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)=
\tfrac{e\zeta | |
4 |
To find α = G(α), first we define:
\zeta}{4n}}{3+\zeta | |
\begin{align} t | |
n(z)&=\cfrac{\tfrac{e |
+z}\\ fn(\zeta)&=t1\circt2\circ … \circtn(0) \end{align}
Then calculate
Gn(\zeta)=fn\circ … \circf1(\zeta)
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
gk,n(z)=z+\varphik,n(z)
\limn\to\varphik,n(z)=0
gk,n(z)\inS
Source:
\begin{align} gk,n(z)&=z+
1 | |
n |
\phi\left(z,\tfrac{k}{n}\right)\\ Gk,n(z)&=\left(gk,n\circgk-1,n\circ … \circg1,n\right)(z)\ Gn(z)&=Gn,n(z) \end{align}
implies
λ | |||
|
G | ||||
|
n | |
\sum | |
k=1 |
\phi\left(Gk-1,n(z)\tfrackn\right)
eq |
1 | |
n |
n | |
\sum | |
k=1 |
\psi\left(z,\tfrac{k}{n}\right)\sim
1 | |
\int | |
0 |
\psi(z,t)dt,
where the integral is well-defined if
\tfrac{dz}{dt}=\phi(z,t)
λn(z0) ≈
1 | |
\int | |
0 |
\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.
\phi(z,t)=
2t-\cosy | +i | |
1-\sinx\cosy |
1-2t\sinx | |
1-\sinx\cosy |
,
1 | |
\int | |
0 |
\psi(z,t)dt
Example. Let:
g | ||||
|
\phi(z), with f(z)=z+\phi(z).
Next, set
T1,n(z)=gn(z),Tk,n(z)=gn(Tk-1,n(z)),
T(z)=\limn\toTn(z)
when that limit exists. The sequence defines contours γ = γ(cn, 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 Tn(z) → T(z) ≡ α along γ = γ(cn, z), provided (for example)
cn=\sqrt{n}
\oint\gamma\phi(\zeta)d\zeta=\limn\to
c | |
n |
n | |
\sum | |
k=1 |
\phi2\left(Tk-1,n(z)\right)
and
L(\gamma(z))=\limn\to
c | |
n |
n | |
\sum | |
k=1 |
\left|\phi\left(Tk-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
The series defined recursively by fn(z) = z + gn(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 fn is defined for |z| < M then |Gn(z)| < M must follow before |fn(z) − z| = |gn(z)| ≤ Cβn is defined for iterative purposes. This is because
gn(Gn-1(z))
infty | |
|z|<R=M-C\sum | |
k=1 |
\betak>0
serves this purpose. Then Gn(z) → G(z) uniformly on the restricted domain.
Example (S1). Set
f | ||||
|
\sqrt{z}, \rho>\sqrt{
\pi | |
6 |
S=\left\{z:|z|<R,\operatorname{Re}(z)>0\right\}
\begin{align} Gn(z)&=z+g1(z)+g2(G1(z))+g3(G2(z))+ … +gn(Gn-1(z))\ &=z+
1 | \sqrt{z}+ | |
\rho ⋅ 12 |
1 | |
\rho ⋅ 22 |
\sqrt{G | ||||
|
\sqrt{G2(z)}+ … +
1 | |
\rho ⋅ n2 |
\sqrt{Gn-1(z)} \end{align}
converges absolutely, hence is convergent.
Example (S2):
f | ||||
|
⋅ \varphi(z),\varphi(z)=2\cos(x/y)+i2\sin(x/y),>Gn(z)=fn\circfn-1\circ … \circf1(z), [-10,10],n=50
The product defined recursively by
fn(z)=z(1+gn(z)), |z|\leqslantM,
has the appearance
Gn(z)=z\prod
n | |
k=1 |
\left(1+gk\left(Gk-1(z)\right)\right).
In order to apply Theorem GF3 it is required that:
\left|zgn(z)\right|\leC\betan,
infty | |
\sum | |
k=1 |
\betak<infty.
Once again, a boundedness condition must support
\left|Gn-1(z)gn(Gn-1(z))\right|\leC\betan.
If one knows Cβn in advance, the following will suffice:
|z|\leqslantR=
M | |
P |
where P=
infty | |
\prod | |
n=1 |
\left(1+C\betan\right).
Then Gn(z) → G(z) uniformly on the restricted domain.
Example (P1). Suppose
fn(z)=z(1+gn(z))
2}{n | |
g | |
n(z)=\tfrac{z |
3},
\left|Gn(z)
| \right|<(0.02) | |||||||
n3 |
1 | |
n3 |
=C\betan
and
Gn(z)=z
n-1 | |
\prod | |
k=1 |
\left(1+
| |||||||
n3 |
\right)
converges uniformly.
Example (P2).
gk,n(z)=z\left(1+
1 | |
n |
\varphi\left(z,\tfrackn\right)\right),
Gn,n(z)=\left(gn,n\circgn-1,n\circ … \circg1,n\right)(z)=
n | |
z\prod | |
k=1 |
(1+Pk,n(z)),
Pk,n(z)=
1 | |
n |
\varphi\left(Gk-1,n(z),\tfrac{k}{n}\right),
n-1 | |
\prod | |
k=1 |
\left(1+Pk,n(z)\right)=1+P1,n(z)+P2,n(z)+ … +Pk-1,n(z)+Rn(z)\sim
1 | |
\int | |
0 |
\pi(z,t)dt+1+Rn(z),
\varphi(z)=x\cos(y)+iy\sin(x),
1 | |
\int | |
0 |
(z\pi(z,t)-1)dt, [-15,15]:
Example (CF1): A self-generating continued fraction.
\begin{align} Fn(z)&=
\rho(z) | |
\delta1+ |
\rho(F1(z)) | |
\delta2+ |
\rho(F2(z)) | |
\delta3+ |
…
\rho(Fn-1(z)) | |
\deltan |
,\\ \rho(z)&=
\cos(y) | +i | |
\cos(y)+\sin(x) |
\sin(x) | |
\cos(y)+\sin(x) |
, [0<x<20],[0<y<20], \deltak\equiv1 \end{align}
Example (CF2): Best described as a self-generating reverse Euler continued fraction.
G | |||||
|
\rho(Gn-2(z)) | |
1+\rho(Gn-2(z))- |
…
\rho(G1(z)) | ||
1+\rho(G1(z))- |
\rho(z) | |
1+\rho(z)-z |
,
\rho(z)=\rho(x+iy)=x\cos(y)+iy\sin(x), [-15,15],n=30