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periods of tetration around real axis

Overview

Tetration by Escape Tetration by Period
Tetration by Escape Tetration by Period

My name is Daniel Geisler , welcome to my web site dedicated to promoting research into the question of what lies beyond exponentiation. Tetration is defined as iterated exponentiation but while exponentiation is essential to a large body of mathematics, little is known about tetration due to its chaotic properties. The standard notation for tetration is \(^{1}a=a,  ^{2}a=a^a,  ^{3}a=a^{a^a},\) and so on. Mathematicians have been researching tetration since at least the time of Euler but it is only at the end of the twentieth century that the combination of advances in dynamical systems and access to powerful computers is making real progress possible.

The big question in tetration research is how can tetration be extended to complex numbers. How do you compute numbers like \(^{.5}2\), and \(^{\pi i}e\) ? This web site will show how to compute these and other problems. See the Tetration page for a one page overview of extending tetration to the complex numbers. A one page overview of my research can be found at Math Equation.

Paper in peer review

A New Kind of Science: Open Problems & Projects - Page 33
How can one extend recursive function definitions to continuous numbers? What is the continuous analog of the Ackermann function? The symbolic forms of the Ackermann function with a fixed first argument seem to have obvious interpretations for arbitrary real or complex values of the second argument. But is there a general way to extend these kinds of recursive definitions to continuous cases? Given a way to do this, how does it apply to recursive definitions like those on page 130? What happens to all the irregularities when one is between integer values? Or is it only possible to find simple continuous generalizations to functions that show fundamentally simple behavior? Can this be used as a characterization of when the behavior is simple? | | NKS page: 906 | Fields: functional analysis; recursive function theory

My paper The Existence and Uniqueness of the Taylor Series of Iterated Functions has been rejected by the Annals of Mathematics. While I disagree with a number of issues in the referee's report, I think it may also contain important perspectives on iterated functions that folks in the tetration research community have been unaware of. Referee’s report on The Existence and Uniqueness of the Taylor Series of Iterated Functions.

From Complex to Banach space

The following is a new result based on an expansion of a theorem from my recent paper that is an algebraic expansion of $f^t(z)$ with $t \in \mathbb{N}$. It is my attempt to extend $f(z)$ and thus $f^t(z)$ from the complex plane to Banach space. The expression $D^k f^t(f_0) = \sum_{j=0}^{k-1} \sigma(k) \lambda^j$ is due to the repeated application of geometrical progressions. Of course the standard summation for geometrical progressions breaks down when considering the roots of unity. This simple fact is reflected in the Classification of Fixed Points, in fact much of the Classification of Fixed Points can be worked out from the preceding equation including topological conjugacy and the functional equations. This is the source of both Schroeder's and Abel's functional equations. The only difference between the proof in the complex plane to Banach space is to realize that $D^k f^t(f_0)$ can possess a much wider variety of symmetries. All linearizations are the result of different symmetries simplifying $D^k f^t(f_0)$.

The $n^{th}$ Derivative

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} \label{eq:FaaDiBruno} $ 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 \begin{eqnarray} 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} \end{eqnarray} The Taylor series of $f^t(z)$ is derived by evaluating the derivatives of the iterated function at a fixed point $f^t(f_0)$ by setting $z=0$ and separating out the $k_n$ term of the summation that is dependent on $D^n f^{t-1}(f_0)$. \begin{eqnarray} D^n f^t(f_0) = & \\ & \sum \frac{n!(D^k f)(f_0)}{k_1! \cdots k_{n-1}!} \left(\frac{Df^{t-1}(f_0)}{1!}\right)^{k_1} \cdots \left(\frac{D^n f^{t-1}(f_0)}{(n-1)!}\right)^{k_{n-1}}\\ & +(D f)(f_0) D^n f^{t-1}(f_0) \end{eqnarray} and rewriting $(D^k f)(f_0)$ as $f_k$. \begin{eqnarray} D^n f^t(f_0) = & \\ & \sum \frac{n! f_k}{k_1! \cdots k_{n-1}!} \left(\frac{Df^{t-1}(f_0)}{1!}\right)^{k_1} \cdots \left(\frac{D^{n-1} f^{t-1}(f_0)}{(n-1)!}\right)^{k_{n-1}} \\ & + \lambda D^n f^{t-1}(f_0) \end{eqnarray} The remaining $p(n)-1$ terms of the summation are only dependent on $D^k f^{t-1}(f_0)$, where $0\gt k \gt n $. Let this partial summation be written as $\sigma(n)$ with $\sigma(0)=0$ and $\sigma(1) = 1$. $ \sigma(n)=\sum \frac{n! f_k}{k_1! \cdots k_{n-1}!} \left(\frac{Df^{t-1}(f_0)}{1!}\right)^{k_1} \cdots \left(\frac{D^{n-1} f^{t-1}(f_0)}{(n-1)!}\right)^{k_{n-1}} $ Rewriting the $p(n)-1$ terms of the summation as $\sigma(n)$ will help in writing a proof by general induction. For $n \gt 1$ $ D^n f^t(f_0)=\sigma(n) + \lambda D^n f^{t-1}(f_0) $
The Taylor series of an iterated function $f^t(z)$ can be constructed given a fixed point where $t \in \mathbb{N}$.
Proof. Assume the given fixed point is at zero. The Taylor series of $f^t(z)$ can be constructed for some positive value of $R$ where $0 \lt |z| \lt R$ if and only if $D^n f^t(f_0)$ can be constructed for every $n \geq 0$. Prove by strong induction.
Basis Steps:
Case $n=0$. By definition $D^0 f^t(f_0) = f_0$, so $D^0 f^t(f_0)$ can be constructed.
Case $n=1$. From the first derivative, $D^1 f^t(f_0) = \lambda^t$, so $D^1 f^t(f_0)$ can be constructed.
Induction Step:
Case. Assume that $D^k f^t(f_0)$ can be constructed for all $k$, where $0 \leq k < n$. Using the Dynamical Recurrance equation, $D^k f^t(f_0)=\sigma(k) + \lambda D^k f^{t-1}(f_0)$. The function $\sigma(k)$ in only dependent on $D^0 f(f_0), \ldots, D^k f(f_0)$ and $D^k f^t(f_0), \ldots, D^{(k-1)} f^t(f_0)$. By the strong induction hypothesis, $\sigma(k)$ can be constructed. Therefore the Dynamical Recurrance equation can be reduced to a geometrical progression based on $\lambda$ that can be represented by a summation. \begin{eqnarray} D^k f^t(f_0) = \sum_{j=0}^{k-1} \sigma(k) \lambda^j \end{eqnarray} This completes the induction step that $D^n f^t(f_0)$ can be constructed for all whole numbers $n$.
The Taylor series for $f^t(z)$ is
$ f^t(z) = \sum_{n=0}^\infty \sum_{j=0}^{n-1} \frac{\sigma(n)}{n!} \lambda^j (z-f_0)^n $
$\blacksquare$

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