# Tetration Forum

Full Version: Universal uniqueness criterion?
You're currently viewing a stripped down version of our content. View the full version with proper formatting.
Pages: 1 2 3 4 5 6
As I now see, even if the uniqueness criterion is valid, the precondition:
$\lim_{n\to\infty} \exp_b^{\circ n}(z)=\infty$ for each $z\in f(S)$
can not be true for any $f$ for $b=e$ (and probably for each $b>e^{1/e}$)!

Michał Misiurewicz [1] showed 1981 that the Julia set of $\exp$ is the whole $\mathbb{C}$ (that is not directly visible from the $e^z$ fractal). The Julia set is the boundary of the set $K$ of all $z$ such that $\lim_{n\to\infty} \exp^{\circ n}(z)=\infty$. That means that the set $K$ and its complement $\mathbb{C}\setminus K$ is dense in $\mathbb{C}$. In every neighborhood of any complex number $z_0$ there is a complex number $z$ such that $\lim_{n\to\infty} \exp^{\circ n}(z)=\infty$ and also a $w$ such that $\lim_{n\to\infty} \exp^{\circ n}(w)\neq\infty$!

And that implies that $f(S)$, which contains an open non-empty set, always contains points $w$ such that $\lim_{n\to\infty} \exp^{\circ n}(w)\neq\infty$.

[1] Michał Misiurewicz (1981). On iterates of e^z. Ergodic Theory and Dynamical Systems, 1 , pp 103-106, doi:10.1017/S014338570000119X
bo198214 Wrote:...
(U2) $\lim_{n\to\infty}\exp_b^{\circ n}(z)= \infty$ for all $z\in f(S)$ ...

$f(\delta(z))=\exp_b^{{\delta_Z(z)}} (f({\delta_S(z)})$
...
$
\infty=\lim_{m\to\infty} \exp_b^{\circ \delta_Z(w_m)}( c)
...$

Do you make any difference between $\exp_b^{\circ p}(q)$ and $\exp_b^{p}(q)$ ?
bo198214 Wrote:...
And that implies that $f(S)$, which contains an open non-empty set, always contains points $w$ such that $\lim_{n\to\infty} \exp^{\circ n}(w)\neq\infty$.
...
Why do you think that this set is open? Wiki says:
In metric topology and related fields of mathematics, a set U is called open if, intuitively speaking, starting from any point x in U one can move by a small amount in any direction and still be in the set U. In other words, the distance between any point x in U and the edge of U is always greater than zero.
http://en.wikipedia.org/wiki/Open_set

The small deviation from a point where the limit is finite allows to get (in the limit) so high walues as you want, because the right hand side of the plot of tetration is pretty scratched. I would replace "open non-empty set" to "non-empty set".
Uniqueness of analytic tetration, scheme of the proof.

Notations.
Let $\mathbb{C}$ be set of complex numbers.
Let $\mathbb{R}$ be set of real numbers.
Let $\mathbb{N}$ be set of integer numbers.
Let $\mathbb{C}^{\prime}=\mathbb{C} \backslash \{ z| z \in \mathbb{R}, z\le 2 \}$
Let $b>1$ be base of tetration.
Assume that there exist analytic tetration $F$ on base $b$, id est,
(0) $F$ is analytic at $\mathbb{C}^{\prime}$
(1) for all $z\in \mathbb{C}^{\prime}$, the relation $F(z+1)=\exp_b(F(z))$ holds
(2) $F(0)=1$
(3) $F$ is real increasing function at $\{z \in \mathbb{R}| z>-2 \}$.

Properties.
From assumption (1) and (2) it follows, that function $F$ has singularity at -2, at -3 and so on.

Consider following Assumption:
There exist entire 1-periodic function $h$ such that
$G(z)=F(z+h(z))$ is also analytic tetration on base $b$.

Then, function $I(z)=z+h(z)$ is not allowed to take values -2, -3, ..
being evaluated at elements of $\mathbb{C}^{\prime}$.

This means that function
$J(z)=I(z) \frac{z+2}{I(z)+2} =(z+h(z)) \frac{z+2}{z+h(z)+2}$
is entire function.

(Weak statement which seems to be true)
Function $J$ cannot grow faster than linear function at infinity in any direction.

Therefore, it is linear function. Therefore, $h=0$

Therefore, there exist only one analytic tetration.
Kouznetsov Wrote:
bo198214 Wrote:...
And that implies that $f(S)$, which contains an open non-empty set, always contains points $w$ such that $\lim_{n\to\infty} \exp^{\circ n}(w)\neq\infty$.
...
Why do you think that this set is open?

Just take a point $w$ of the interior of $S$ such that $f'(w)\neq 0$ (if such a point does not exist then $f$ is constant), then there is a neighborhood $W$ of $w$ such that $f$ is bijective on $W$. That means $f$ and $f^{-1}$ are continuous (because analytic). And the preimage of an open set is open under a continuous map. So the preimage $f(W)$ of $W$ under $f^{-1}$ is open and so we find arbitrary many points $z$ in $f(W)\subseteq f(S)$ such that $\exp^{\circ n}(z)\not\to\infty$.

Kouznetsov Wrote:Do you make any difference between $\exp_b^{\circ p}(q)$ and $\exp_b^{p}(q)$ ?

Oh no its both the same here.
Kouznetsov Wrote:Consider following Assumption:
There exist entire 1-periodic function $h$ such that
$G(z)=F(z+h(z))$ is also analytic tetration on base $b$.

Then, function $I(z)=z+h(z)$ is not allowed to take values -2, -3, ..
being evaluated at elements of $\mathbb{C}^{\prime}$.

But $I(z)$ does not need to be entire, it can also omit the values -2, -3 , ... as arguments.
For an entire function we would know that it can omit at most 1 value (Little Picard).

Edit: Oh I see now $I(z)$ needs not to be entire, but can be continued being entire by $I(z\pm n)=I(z)\pm n$ if $I=f^{-1}\circ g$ can be/is defined on the strip $S$.
Ok, I give it another try:

Proposition.
Let $S=\{z:0<\Re(z)\le 1\}$, $S_\epsilon=\{z:0<\Re(z)<1+\epsilon\}$, $\epsilon>0$ and $D\supseteq S_\epsilon$ being a domain (open and connected) of definition and $H\supseteq S_\epsilon$ being a domain (open and connected) of values for a holomorphic function.
Let $f$ be a holomorphic function on $D$, $G:=f(S_\epsilon)\subseteq H$, such that
(0) $H\subseteq f(D)$.
(1) $f(0)=1$
(2) $f(z+1)=F(f(z))$
(U) $f^{-1}(G)$ has bounded real part.
Then $g=f$ for every other on $D$ holomorphic $g$ satisfying (0), (1), (2), (U).

Proof.
$h(z):=g^{-1}(z)-f^{-1}(z)$ has bounded real part on $G$. We consider $f^{-1}$ and $g^{-1}$ and so $h$ to be holomorphic on the same Riemann surface $G$. $\delta(z):=h(f(z))=g^{-1}(f(z))-z$ is a 1-periodic function, holomorphic on $S_\epsilon$. As $S_\epsilon\supset S$ it can be continued to an entire function, so it has to take on every complex value with at most one exception already on the strip $S$ otherwise it is a constant. Now $\delta(S)=\delta(S_\epsilon)=h(G)$ has bounded real part and hence can not take on every value, so $h(z)=0$ and $g=f$.
bo198214 Wrote:Ok, I give it another try:
..
Bo, Congratulations! You seem to do well. I copypast your theorem into the latex source, but I shall not work on it until I collect all the pieces into the single file.
(I have not yet type the uniqueness of the "Children's factorial" you suggested). While, you generate results faster than I describe them. The draft becomes longer, but does not yet approach to the finish. If you want to work with the text, let me know; I send you the last version. Keep in touch, K.
(10/05/2008, 12:22 AM)Kouznetsov Wrote: [ -> ]Bo, Congratulations! You seem to do well...
The preprint is here: http://www.ils.uec.ac.jp/~dima/PAPERS/2009matan.pdf
Any comments and ctyrics is welcommed.
"We call a region H an initial region of F iff F(z) {is not in} H for all z {in} H..."
does this mean $\exists z : F(z) \notin H$ (there is SOME z such that F(z) is NOT in H) or $\forall z (F(z) \notin H$) (there is NO z such that F(z) IS in H)?
Pages: 1 2 3 4 5 6