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?

youtalking about? You aren't clear. Are you asking aboutany(fixed) mode of convergence? What? $\endgroup$ – Arturo Magidin Jan 18 '16 at 22:00i.e.convergent Taylor series expansion)? In the case of1/zit 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. [...] $\endgroup$ – Loïc Teyssier Mar 1 '16 at 20:49