# Given $g(h(z))$ is convergent, what can be said about the convergence of $g(z)$ and $h(z)$? [closed]

Consider the iterated function $f^t(z)=f(f(f(...f(z))...))$ where $t \in \mathbb Z$ and $f(z)$ is convergent. Then the iterates of $f(z)$ such as $f^2(z), f^3(z), f^4(z)$ are convergent. Now let $r \in \mathbb Q$ and $s \in \mathbb N$ where $r \times s = t$ and $h(z) = f^r(z)$. Then $h^s(z)=f^t(z)$. Does the convergence of $h^s(z)$ imply the convergence of $h(z)$. I want to prove that the convergence of $f(z)$ implies the convergence of $f^t(z)$.

# Background

Let $f(z)$ and $g(z)$ be holomorphic functions, then the Bell polynomials can be constructed using Faa Di Bruno's formula.

$D^nf(g(z))=$ $\sum_{\pi(n)} \frac{n!}{k_1! \cdots k_n!}$ $(D^kf)(g(z))$ $\left(\frac{Dg(z)}{1!}\right)^{k_1}$ $\cdots \left(\frac{D^ng(z)}{n!}\right)^{k_n}$

A partition of $n$ is $\pi(n)$, usually denoted by $1^{k_1}2^{k_2}\cdots n^{k_n}$ with $k_1+2k_2+ \cdots nk_n=k$; where $k_i$ is the number of parts of size $i$. The partition function $p(n)$ is a decategorized version of $\pi(n)$, the function $\pi(n)$ enumerates the integer partitions of $n$, while $p(n)$ is the cardinality of the enumeration of $\pi(n)$.

Setting $g(z) = f^{t-1}(z)$ results in

$D^n f^t(z) = \sum_{\pi(n)} \frac{n!}{k_1! \cdots k_n!}$ $(D^k f)(f^{t-1}(z))$ $\left(\frac{Df^{t-1}(z)}{1!}\right)^{k_1}$ $\cdots \left(\frac{D^n f^{t-1}(z)}{n!}\right)^{k_n}$

The Taylors series of $f^t(z)$ is derived by evaluating the derivatives of the iterated function at a fixed point $f^t(0)$ by setting $z=0$ and separating out the $k_n$ term of the summation that is dependent on $D^n f^{t-1}(0)$.

$D^n f^t(0) = \sum \frac{n!(D^k f)(0)}{k_1! \cdots k_{n-1}!}$ $\left(\frac{Df^{t-1}(0)}{1!}\right)^{k_1} \cdots$ $\left(\frac{D^n f^{t-1}(0)}{(n-1)!}\right)^{k_{n-1}}$ $+ (D f)(0) D^n f^{t-1}(0)$

The remaining $p(n)-1$ terms of the summation are only dependent on $D^k f^{t-1}(0)$, where $0<k<n$.

Complex Ackermann Function

So for tetration and the Ackermann function in general, the problem of extending them to the complex numbers can be simply reduced to the problem of continuous iteration of functions. The problem of continuous iteration of functions is solved by taking the Taylor series $f^t(z)=\sum_{j=1}^\infty D^j f^t(0) z^j$

Let $f(z) \equiv a \rightarrow z \rightarrow k$.

Theorem. When $f^t(z)$ where $t \in \mathbb{C}$ is defined, then $a \rightarrow b \rightarrow k+1$ where $a,b \in \mathbb{C}$ and $k \in \mathbb{N}$.

\begin{eqnarray} f(1) &=& a \rightarrow 1 \rightarrow k = a\\ f^2(1) &=& f(a) = a \rightarrow a \rightarrow k = a \rightarrow 2 \rightarrow k+1\\ f^3(1) &=& f(a \rightarrow a \rightarrow k) = a \rightarrow (a \rightarrow a \rightarrow k) \rightarrow k = a \rightarrow 3 \rightarrow k+1\\ f^t(1) &=& a \rightarrow t \rightarrow k+1 \\ \end{eqnarray}

Therefore, when $f^t(z)$ where $t \in \mathbb{C}$ is defined, then $a \rightarrow b \rightarrow k+1$ where $a,b \in \mathbb{C}$ is defined. $\bullet$

# Convergence

The fractional iterates of functions are not necessarily analytic, the fractional iterates of $e^z-1$ which has a relatively nice expansion only has a zero radius of convergence. But for a complex Ackermann function all that is needed is for $f(1)$ to have a zero radius of convergence. Let $f(z)=a^z$, then the integerial iterates of $a^z$ are convergent.

Now to tackle the main question, given $g(h(z))$ is convergent, what can be said about the convergence of $g(z)$ and $h(z)$?. Obviously $g(z)=h(z)=\frac{1}{z}$ is a counter-example that if composite of two non-convergent functions are convergent then the individual functions must be convergent. But the given series expansion of $f^t(z)$ has no terms $z^{-k}$ where $k \in \mathbb N$. Consider $r,s$ where $f^r(f^s(z))=f^{r+s}(z)$ and $r+s \in N$, then $f^{r+s}(z)$ is convergent. It would seem odd to me if the convergence of $f^{r+s}(z)$ didn't force the convergence of $f^r(z)$ and $f^s(z)$.

More specifically, does tetration of real numbers $^xa$ where $a,x \in \mathbb R$ exist? Can tetration of real and complex numbers be proven not to exist?

## closed as unclear what you're asking by Mikhail Katz, Alexey Ustinov, Wolfgang, user1688, Stefan KohlJan 19 '16 at 11:25

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• Please clarify your notation: is $f^n(x)=f(f(f(...f(x))...))$ or $f(x)f(x)...f(x)$? – Michael Jan 18 '16 at 19:46
• What type of convergence are you talking about? Pointwise? Uniform? etc. – Arturo Magidin Jan 18 '16 at 20:10
• @ArturoMagidin Uniform convergence would be great, but other modes of convergence are of interest. – Daniel Geisler Jan 18 '16 at 20:31
• My point is: which one are you talking about? You aren't clear. Are you asking about any (fixed) mode of convergence? What? – Arturo Magidin Jan 18 '16 at 22:00
• It still doesn't make sense to speak of a "convergent" function without defining the term. Do you mean analytic (i.e. convergent Taylor series expansion)? In the case of 1/z it has a convergent Laurent series expansion, but it is not analytic at $0$. Is that what you mean? Also, I don't understand the goal of your last part "Convergence", which is most unclear. It is very confusing as $f$ seems to stand for different things. Please take into account my answer below if $f$ is a general parabolic germ, since in that case your last sentence does not hold. [...] – Loïc Teyssier Mar 1 '16 at 20:49

## 1 Answer

If I understand correctly the setting (say $f$ is a germ at $0\in\mathbb C$ of a biholomorphic function), the answer is no in general, but yes generically. The diffeomorphismn $f$ can always be written as the time-$1$ flow $\exp X$ of a formal vector field $X$. The set $$T:=\{t\in\mathbb C : \exp (tX)~ \mathrm{converges}\}$$ is a group containing $\mathbb Z$ with the following properties. Let $f'(0)=\exp (2i\pi\lambda)$ for some $\lambda\in\mathbb C$.

1. If $T$ is not discrete then $X$ is holomorphic and $T=\mathbb C$.
2. If $\lambda$ is not real then $T$ is not discrete (generic case, hyperbolic diffeomorphism).
3. If $\lambda$ is irrational then $T$ is not discrete if and only if $f$ is analytically linearizable (Siegel problem, linearizable on a measure-1 set of irrationals thanks to Brjuno condition).
4. If $\lambda$ is rational then either $f$ is linearizable (i.e. periodic see here, non-generic property) and $T$ is not discrete, or the answer can be read in Écalle-Voronin invariants $\phi$ (a tuple of germs of a biholomorphism) of the parabolic germ $f$ (see also this question and the self-answer of Will Jagy, and most related questions about smooth/analytic fractional iterates). Namely, $t\in T$ if and only if $\phi(\exp(2i\pi t)h)=\exp(2i\pi t)\phi(h)$. The invariant $\phi$ is very hard to compute, though.
• The conditions you list seem to be closely related to the Classification of Fixed Points as explained in Complex Dynamics by Carleson and Gamelin. Super attractor: $f′(0)=0$ Attractor or repellor: $|f′(0)| \ne 1$ uses Schroeder's equation Parabolic rationally neutral: $f′(0)=1$ uses Abel's equation Rationally Neutral: $f′(0) \ne 1$ and $f′(0)$ is a root of unity Irrationally Neutral: $|f′(0)|=1$ and $f′(0)$ is not a root of unity – Daniel Geisler Jan 18 '16 at 21:52
• Yes indeed. Non-discrete $T$ is equivalent to the germ being analytically conjugate to its formal normal form, as defined by the standard methods you're talking about. – Loïc Teyssier Jan 18 '16 at 22:21