# Tetration Forum

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Inspired by Kouznetsov's consideration I investigated a bit more in this direction and found interesting results:

1.
Let F be holomorphic on the right half plane, and let F(z+1)=zF(z),
F(1)=1, further let F be bounded on the strip 1<=Re(z)<2, then F is
the gamma function.

2.
$\exp_b$ is also determined as the only function that satisfies $b^{x+1}=bb^x$, $b^1=b$ and is bounded on the strip 1<=R(z)<2 (or 0<=R(z)<1).

3. I started a thread in sci.math.research and it turned out that the same criterion also makes the Fibonacci function
$F(z)=\frac{\phi^z - (1-\phi)^z}{\sqrt{5}}$, $\phi=\frac{1+sqrt(5)}{2}$ the unique extension of $F(n+1)=F(n)+F(n-1)$, $F(0)=0$, $F(1)=1$.

So I really would guess that this criterion (which is a slightly weaker demand than Kouznetsov's criterion) also implies uniqueness for tetration, at least for base $b>e^{1/e}$.
Maybe its not even difficult to prove.
Where are these proven? How are these proven? By whom?

So by "bounded" are you referring to the fact that Kouznetsov's extension requires that the limit towards $i\infty$ is finite? Is this the same as saying that this limit exists?

Andrew Robbins
Also, I admire the criticism of the people in that thread, but given the right adjectives, there is really no problem with what was said. I think that if you replace "entire" with "holomorphic over domain D" (where D is the complex plane minus any singularities or branch cuts) and suffix "unique" with "unique up to branch choice" (or state some property that only the principal branch has) then we might get better responses, and certainly more clarity.

Andrew Robbins
andydude Wrote:So by "bounded" are you referring to the fact that Kouznetsov's extension requires that the limit towards $i\infty$ is finite? Is this the same as saying that this limit exists?

His assumption is that the limit is the fixed point in the upper complex halfplane, not only finite. I mentioned somewhere already that for $1 with regular tetration this is no more true (i.e. has no limit), however it is still bounded. So perhaps this is a good generalization.

Quote:Where are these proven? How are these proven? By whom?

Ok, lets start with the uniqueness of $b^x$ (I dont know whether is somewhere written):

Proposition.Let $b>1$ then $f(z)=b^z=e^{\ln(b)z}$ is the only holomorphic solution, defined on the right halfplane $\Re(z)\ge 0$ [this condition is not necessary, I just include it to emphasize on non-entire functions], of the equations $f(z+1)=bf(z)$, $f(0)=1$ which is bounded on the strip $S$ given by $0\le \Re(z)<1$.

Proof
We know that every other solution must be of the form $g(z)=f(z+p(z))$ where $p$ is a 1-periodic holomorphic function (this can roughly be seen by showing periodicity of $h(z)=f^{-1}(g(z))-z$).

In this case this means:
$f(z+p(z))=b^{z+p(z)}=b^{p(z)}b^z=b^{p(z)}f(z)=:q(z)f(z)$
where $q$ is also a 1-periodic function. As $g$ (and $f$) is bounded on $S$, $q$ must be bounded too.
As $q$ is periodic it can be continued from $S$ to the whole plane $\mathbb{C}$ and is hence an entire holomorphic function, which is still bounded. By Liouville $q$ must be constant:
$g(z)=qf(z)$. And now we apply $g(0)=1$ and see that $q=1$.

I will describe the proof for the Gamma function in another post, I found it in a German complex analysis book: Reinhold Remmert, "Funktionnentheorie", Springer, 1995. As reference is given: H. Wielandt 1939.

And the proof for the Fibonacci function is given in the sci.math.research thread by Waldek Hebisch (key points), and a bit fleshed out by me.
Very nice. I followed your proof up until "By Liouville", what does this mean? Is this the name of a theorem?
andydude Wrote:Very nice. I followed your proof up until "By Liouville", what does this mean? Is this the name of a theorem?

Theorem of Liouville
Ok, now the promised proof for the uniqueness of the gamma function, which was point 1. here. I nearly only translate it from the mentioned book:

We consider the function $v=F-\Gamm$ on the right half plane, it also satisfies $v(z+1)=zv(z)$. Hence $v$ has a meromorphic continuation to $\mathbb{C}$. Poles can be at most at 0, -1, -2 , ....
As $v(1)=0$ we have $\lim_{z\to 0} zv(z)=0$, hence $v$ has a holomorphic continuation to 0 and also to to each $-n$, $n\in\mathbb{N}$ by $v(z+1)=zv(z)$.

$v(z)$ restricted to $1\le \Re(z)<2$ is bounded because $\Gamma$ is bounded there. Then $v(z)$ is also restricted on $S$ given by $0\le \Re(z)\le 1$. $q(z)$ defined by $v(z)v(1-z)$ is bounded on $S$ because $v(z)$ and $v(1-z)$ have the same values on $S$. Now $q(z+1)=-q(z)$, hence $q$ is bounded on whole $\mathbb{C}$ and by Liouville $q(z)=q(1)=0$. Hence $v\equiv 0$ and $F\equiv\Gamma$.
Perhaps the following two msgs in sci.math of David Libert are appropriately posted to this thread. I just copy&paste the whole msg to provide the context.
Perhaps we move this post to another place - Henryk?

Code:
(David Libert ah170@FreeNet.Carleton.CA ) Rotwang (sg552@hotmail.co.uk) writes: > > On 19 Jun, 04:07, Rotwang <sg...@hotmail.co.uk> wrote:   [deletions...] >> >> Based on the recent discussions it seems that there is no known way to >> >> define non-integer iteration in general, but I was wondering whether >> >> anybody here knows whether it can always be defined in the restricted >> >> setting given here, or alternatively whether there are any no-go >> >> theorems which show that it can't. >> >> >> >> Of particular interest is the case where f (z) = exp (z), for z in {z >> >> in C | 0 <= im z < 2 pi} since this is an important transformation in >> >> the book I'm reading. In this case f^t (z) would be a generalisation >> >> of tetration to non-integer height... > > > > Sorry, this is wrong. I mean a generalisation of iterated > > exponentiation to non-integer height. Does anybody know if such a > > thing exists?   This question came up here and in other recent threads.  I have not read all of the large volume of writing about this in all threads, so I may have overlooked something, but from what I have so far seen this has not yet been answered.   I don't know the answer either, nor have I found one on the Internet. I will note some things related to the question, some I think I have proven myself, others found.   First, I Googled on  fractional iteration  and it does seem to be a big topic. Tommy in other threads has written about tetration as related to this.  I have previously heard of tetration in making higer iterations of exponentiation possibibly into the transfinite, along the lines of ordinat notation generalizations of Ackermann's function.  But that Google search did show that this word is also applied to fractional and continuous interpolations.   I will note some specific interesting looking leads from that search: a sci.math.research thread from 2005   fractional iteration of functions http://groups.google.com/group/sci.math.research/browse_thread/thread/b7f7ebcdc67d6bce the FOM thread  Fractional Iteration and Rates of Growth     near the top of http://www.cs.nyu.edu/pipermail/fom/2006-May/thread.html#start and:    http://www.tetration.org/Dynamics/   What I seem to be seeing above and these various threads is it often possible to define various continuous interpolations but it is hard to say which if any are canoical. Rotwang above proposed more stringent requirement on a solution, but that leaves open the question of existence.  If that were solved there is also the question of uniqueness.   A similar point, concerning another case with a more known solution, as I understand it there are many interpolations of the integer factorial function to the complex numbers, but if we require log-convexity we have unique existence with the Gamma function.   I will add some related results I may have proven.  I don't want to take the time here to write out proofs.  Maybe it would be safer to call these conjectures  :-) .   So I think ZF proves that if A is an infinite set which can be well-ordered and f: A >->> A   is an injection,   then there are  2^A  many   g: A >->> A   with f = g o g  .   (That is function compoistion).   ZF proves for A infinite well-orderable there are  2^A many functions : A -> A, so this would be as many possible.   Regarding   f: R -> R   (R the reals),  if  f is  non-decreasing (this is:   x<y ->  f(x) <= f(y) )   and continuous then there are #R many  non-decreasing and continous  g : R -> R with   f = g o g .   By smooth I mean all finite iterated derivatives exist everywhere and are continous.   For  f : R -> R   non-decreasing and smooth  there are  #R many g  non-decreasing and smooth with  f = g o g.   On the other hand  if  f : R -> R  is non-increasing,  then there are no non-decreasing g : R -> R  with  f = g o g,   and there are no non-increasing g : R -> R  with f = g o g .   But if  f : R -> R  is continous and non-increasing,  there are #R many  g : R -> R which are piecewise continous on countably many half-open intervals with  f = g o g .   It seems a mess to try anything for f non-increasing.   For f non-decreasing, you can keep repeating the .5 construction of new g's above.  You can define dyadic rational exponentiation on these, by using the contruction to take .5 's and ordinary integer powers to take multiples.  The right things commute for this to be well-defined.   The ordering relations among these are as expected and monotonic, so by taking appropriate sups or infs of dyadic rational exponents you can extend this to real exponents of iteration.   Unfortunately, if f started as continuous or smooth, you have each of the slices is respectivelty continous or smooth, but overall operation considered over variable exponent need not be continuous or smooth in that exponent variable.  Because each time you halved the exponent you made a non-canonical choice of how to do so, and could have taken wildly different functions.   Indeed none of the functions produced in any of these claims are canonical.  The internal proofs made arbitrary choices along the way, that is why they produced many functions instead of one.

and a correction, posted later on the day
Code:
David Libert (ah170@FreeNet.Carleton.CA) writes:   [deletions...] > >   So I think ZF proves that if A is an infinite set which can be well-ordered and > > f: A >->> A   is an injection,   then there are  2^A  many   g: A >->> A   with > > f = g o g  .   (That is function compoistion).   I will revise the above, after reconsidering the proof more carefully.  For A as above and    f : A >->> A   as above,  I will define a 2-cycle to be a sibset  {x, y} of A such that    x ~= y   and    f(x) = y     and     f(y) = x .   The first revision will be if the number of 2-cycles is finite and odd,  then there are no   g   as above such that    f =  g o g.   So regarding the remaining cases  (finite and even # of 2-cycles  or  infinitely many 2-cycles) :   Let  B  =  {x in A |   f(x)  ~= x} .   If B is uncountable then there  are  2^#B  many  g : A >->> A   s.t.    f = g o g .   If B is finite  there are finitely many  g : A >->> A   s.t.   f = g o g .   If B is denumerable  (countable and infinite)   then     Aleph_0 <=  #{ g : A >->> A |  f = g o g}    <=   2^Aleph_0 . > >   But if  f : R -> R  is continous and non-increasing,  there are #R many  g : R -> R > > which are piecewise continous on countably many half-open intervals with  f = g o g .   Also, this statement is claimed for f smooth and non-increasing, then g can be taken to be piecewise smooth on the same sort of intervals.
We have the usual conditions
(1) $f$ is a holomorphic function defined on $C'=\mathbb{C}\setminus (-\infty,-2]$
(2) $f(0)=1$
(3) $f(z+1)=b^{f(z)}$, $b>1$.
(4) $f$ is real on $(-2,\infty)$ and strictly increasing.

The following criterion should pick a unique function out of the possible solutions for the above conditions:
(U1) $f$ is bounded on the strip $S:=\{ x+iy| 0< x\le 1, y\in\mathbb{R}\}$.
(U2) The values of $f$ fill the complex plane: $f(D)=\mathbb{C}$, $D:=\{x+iy| -2.

Proof attempt:

Let $g$ be some function that also satisfies (1), (2), (3),(4) and (U1), (U2). We put appropriate cuts on the domain of $f$ such that we have an inverse $f^{-1}$ which always maps reals to reals.

$f^{-1}(g(z))-z$ is a holomorphic and 1-periodic function. We know that if a 1-periodic holomorphic function is bounded on $S$ then it is already a constant.
So if we could show that $\delta(z):=f^{-1}(g(z))$ has a bounded real part on $S$ then would
$e^{\delta(z)-z}=e^{\Re(\delta(z))+i\Im(\delta(z))+\Re(z)+i\Im(z)}=e^{\Re(\delta(z))+\Re(z)}e^{i\left(...\right)}$
be bounded and 1-periodic on $S$, hence $e^{\delta(z)-z}=c$ hence $\delta(z)-z=d$ hence $\delta(z)=z+d$ for some $d\in\mathbb{C}$. But of course $d=0$ because $0=f^{-1}(1)=f^{-1}(g(0))=\delta(0)=d$.

So we still need show that $f^{-1}(g(z))$ has bounded real part on $S$. We know already that $g(S)$ is bounded.

By condition (U2) we divide the area $g(S)$ into images of $f$ of $S_n = \{ x+iy| n, i.e.

$g(S)\subseteq \mathbb{C}= f(D) =\bigcup_{n=-2}^{\infty} f(S_n)$.

Because $g(S)$ is bounded, it is a subset of a compact set, which can be covered by merely finitely many $f(S_n)$:

$g(S)\subseteq f(S_{n_1})\cup f(S_{n_2})\cup \dots \cup f(S_{n_k})$

so $f^{-1}(g(S))\subseteq S_{n_1}\cup \dots\cup S_{n_k}$ which has bounded real part.
I am not so convinced whether $f$ is invertible (it is usually not injective on the complex plane) without getting trouble. So here is the next revision/attempt:

bo198214 Wrote:We have the usual conditions
(1) $f$ is a holomorphic function defined on $C'=\mathbb{C}\setminus (-\infty,-2]$
(2) $f(0)=1$
(3) $f(z+1)=b^{f(z)}$, $b>1$.
(4) $f$ is real on $(-2,\infty)$ and strictly increasing.
The following criterion should pick a unique function out of the possible solutions for the above conditions:
(U1) $f$ is bounded on the strip $S:=\{ x+iy| 0< x\le 1, y\in\mathbb{R}\}$.
(U2) $\lim_{n\to\infty}\exp_b^{\circ n}(z)= \infty$ for all $z\in f(S)$. ($\lim_{n\to\infty} z_n=\infty$ is just an abbreviation for the more save $\lim_{n\to\infty}1/z_n=0$.)

Proof attempt:

Let $g$ be some function that also satisfies (1), (2), (3),(4) and (U1), (U2).
Then $g(z)=f(\delta(z))$, where $\delta(z)-z$ is a 1-periodic function (This is easily provable for bijective $f$ and $g$ but I am not 100% sure yet for arbitrary $f$ and $g$, but it should go through.)

As in the previous proof we just need to show that $\delta$ has bounded real part on $S$.

We know already that $g(S)=f(\delta(S))$ is bounded.
We can write
$f(\delta(z))=\exp_b^{{\delta_Z(z)}} (f({\delta_S(z)})$
where $\delta_Z(z)\in\mathbb{Z}$ is the next lower integer (floor) of the real part of $\delta(z)$ and $\delta_S(z):=\delta(z)-\delta_Z(z)\in S$.

$\delta_S(S)\subseteq S$ hence $f(\delta_S(S))\subseteq f(S)$ is bounded by (U1) and $\lim_{n\to\infty}\exp_b^{\circ n}(z)=\infty$ for every $z\in f(\delta_S(S))$ by (U2).

If the real part of $\delta$ would be unbounded then also $\delta_Z$ would be unbounded on $S$. We choose a sequence $z_n$ such that $\delta_Z(z_n)$ is strictly increasing (and hence unbounded).

The sequence $f(z_n)$ is bounded and hence has a converging subsequence. Let $v_n$ be a subsequence of $z_n$ such that $\lim_{n\to\infty} f(v_n)=d$.

The sequence $f(\delta_S(v_n))$ is bounded and hence has a converging subsequence. Let $w_n$ be a subsequence of $v_n$ such that $c_n:= f(\delta_S(w_n))\in f(S)$ converges, let $c:=\lim_{n\to\infty}c_n$.
We still have $w_n\in S$, $\lim_{n\to\infty} f(w_n)=d$ and $\delta_Z(w_n)$ is strictly increasing.

$c$ is element of the closure $\overline{f(S)}$. (U2) holds also on $\overline{f(S)}=f(\overline{S})$ by (3).
Thatswhy
\begin{align*}
\infty=\lim_{m\to\infty} \exp_b^{\circ \delta_Z(w_m)}( c)=\lim_{m\to\infty} \lim_{n\to\infty} \exp_b^{\circ \delta_Z(w_m)}(c_n)=\lim_{n\to\infty} \exp_b^{\circ \delta_Z(w_n)}(c_n)= \lim_{n\to\infty} f(w_n)=d
\end{align*}

in contradiction to $d$ being finite.
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