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Universal uniqueness criterion?
#11
As I now see, even if the uniqueness criterion is valid, the precondition:
for each
can not be true for any for (and probably for each )!

Michał Misiurewicz [1] showed 1981 that the Julia set of is the whole (that is not directly visible from the fractal). The Julia set is the boundary of the set of all such that . That means that the set and its complement is dense in . In every neighborhood of any complex number there is a complex number such that and also a such that !

And that implies that , which contains an open non-empty set, always contains points such that .

[1] Michał Misiurewicz (1981). On iterates of e^z. Ergodic Theory and Dynamical Systems, 1 , pp 103-106, doi:10.1017/S014338570000119X
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#12
bo198214 Wrote:...
(U2) for all ...


...

Do you make any difference between and ?
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#13
bo198214 Wrote:...
And that implies that , which contains an open non-empty set, always contains points such that .
...
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".
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#14
Question 
Uniqueness of analytic tetration, scheme of the proof.

Notations.
Let be set of complex numbers.
Let be set of real numbers.
Let be set of integer numbers.
Let
Let be base of tetration.
Assume that there exist analytic tetration on base , id est,
(0) is analytic at
(1) for all , the relation holds
(2)
(3) is real increasing function at .

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

Consider following Assumption:
There exist entire 1-periodic function such that
is also analytic tetration on base .

Then, function is not allowed to take values -2, -3, ..
being evaluated at elements of .

This means that function

is entire function.

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

Therefore, it is linear function. Therefore,

Therefore, there exist only one analytic tetration.
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#15
Kouznetsov Wrote:
bo198214 Wrote:...
And that implies that , which contains an open non-empty set, always contains points such that .
...
Why do you think that this set is open?

Just take a point of the interior of such that (if such a point does not exist then is constant), then there is a neighborhood of such that is bijective on . That means and are continuous (because analytic). And the preimage of an open set is open under a continuous map. So the preimage of under is open and so we find arbitrary many points in such that .

Kouznetsov Wrote:Do you make any difference between and ?

Oh no its both the same here.
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#16
Kouznetsov Wrote:Consider following Assumption:
There exist entire 1-periodic function such that
is also analytic tetration on base .

Then, function is not allowed to take values -2, -3, ..
being evaluated at elements of .

But 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 needs not to be entire, but can be continued being entire by if can be/is defined on the strip .
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#17
Ok, I give it another try:

Proposition.
Let , , and being a domain (open and connected) of definition and being a domain (open and connected) of values for a holomorphic function.
Let be a holomorphic function on , , such that
(0) .
(1)
(2)
(U) has bounded real part.
Then for every other on holomorphic satisfying (0), (1), (2), (U).

Proof.
has bounded real part on . We consider and and so to be holomorphic on the same Riemann surface . is a 1-periodic function, holomorphic on . As 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 otherwise it is a constant. Now has bounded real part and hence can not take on every value, so and .
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#18
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.
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#19
(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.
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#20
"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 (there is SOME z such that F(z) is NOT in H) or ) (there is NO z such that F(z) IS in H)?
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