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I try to simplify the proof of uniqueness of holomorphic tetration by bo198214.
Let and and
Let
Let be entire 1periodic function.
Let .
According to the Picard’s Little Theorem [1],
either is trivial function (constant),
or or such that .
Let .
Let
Assume, is not constant.
How to prove that at least one of these two statements is true: ,
?
[1] Weisstein, Eric W. ”Picard’s Little Theorem.” From MathWorld–A Wolfram Web Resource. http://mathworld.wolfram.com/PicardsLittleTheorem.html
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11/04/2008, 12:00 PM
(This post was last modified: 11/05/2008, 11:35 AM by bo198214.)
But if you had this lemma, how would you continue with the uniqueness of tetration?
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bo198214 Wrote:... if you had this lemma, how would you continue with the uniqueness of tetration? Did not you found the tail01.tex, or do you just want to have it posted on the forum?
I go to copypast it here; I'll convert it from latex and copyedit a bit.
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11/05/2008, 09:07 AM
(This post was last modified: 11/05/2008, 11:35 AM by bo198214.)
Kouznetsov Wrote:bo198214 Wrote:... if you had this lemma, how would you continue with the uniqueness of tetration? Did not you found the tail01.tex, or do you just want to have it posted on the forum?
I go to copypast it here; I'll convert it from latex and copyedit a bit.
Dmitrii, a forum is not for private communaction, either post it that everybody has a chance to understand or dont post at all.
PS: I just saw that my previous post was nonsense, I changed it accordingly.
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11/06/2008, 03:19 PM
(This post was last modified: 11/10/2008, 09:47 AM by Kouznetsov.)
Lemma is not yet ready, but Bo asked me about pics of slightly modified tetration.
I consider the simplest possible modification of tetration
where
Such modified tetration satisfies the tetration equaitons,
,
For , the function is plotted in the complex plane
It is almost identical function, but the defiation is seen, if the imaginary part becomes of larget than 3.
Now, the plot of modified tetration:
The grid shown occupies the range [10,10], [4,4] with step unity.
Levels of integer values of and
Levels of integer values of are plotted.
Below I show the zoomin of the part of the previoous figure:
There are cutlines there. The structure in the upper part folloes the topology of tetration, but it is reduced in size and strongly deformed.
For highest harmonics like or , added to the argument of tetration in definition of mofified tetratio "tem", the
structures of cuts is smaller, denser and approach closer to the real axis.
If some function satisfies equations
,
and for some , the function differ from tetration for at least , id est,
.
then function is not holomorphic in
In such a way, any deformation of tetration (even small) breaks its continuity.
See also
http://en.citizendium.org/wiki/Tetration
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11/11/2008, 08:34 AM
(This post was last modified: 11/11/2008, 08:36 AM by Kouznetsov.)
bo198214 Wrote:But if you had this lemma, how would you continue with the uniqueness of tetration?
First, I repeat my Lemma: Let
is entire 1periodic funcrion
for some
Then
(end of Lemma 1).
This lemma does not requre any knowledge about properties of tetration.
Now, how do I apply it:
Assume some fixed base and let within the range of holomorphism of tetration, id est, in the complex plane except some set of measure zero.
This set includes one line , and, at ,
additional horizontal cut lines that correspond to the periodicity of tetration.
Assume there exist some function which is also holomorphic within some region ,
which includes at least some vicinity of the segment
.
Assume also, that function is holomorphic at least in and
and is not tetration . (Tetration satisfies the equations above)
Then, there exist function such that in vicinity of the segment
function can be expressed as follows:
and, in some vicinity of the same segment,
We need this expression only in vicinity of the segment ; therefore, we have no need to specify, which of we mean.
There exist only one holomorphic extension of a function in a domain of trivial topology.
Therefore, we can extend the function outside of the range of definition.
Consider function
Assuming that and are not identical in the segment , function should not be identically zero. This function also allows the holomorphic extension. Consider behavior of function in the complex plane. This extension should be periodic, satisfying conditions of my Lemma.
Therefore, function should take values from set inside domain . Function has singularities at these points.
Therefore, function has singularities in the domain . With these singulatiries,
function does not satisfy the criterion in the definition of tetration.
In such a way, there exist only one tetration, that is strictly increasing function in the segment .
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04/12/2009, 09:58 PM
(This post was last modified: 04/12/2009, 11:25 PM by sheldonison.)
Now that I can understand it, this is an amazing graph, showing the fractal copies of the main tetration curve distorted and duplicated with singularities into each vertical strip of the function.
In the tetration curve, the imaginary value for f(z) at the real axis is equal to zero for all z>=2. For the solution based on the fixed point, is the imaginary value of f(z) nonzero everywhere outside of the real axis? Otherwise it would seem likely that there would be other singularities, outside of the real axis.
 Sheldon
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sheldonison Wrote:..
In the tetration curve, the imaginary value for f(z) at the real axis is equal to zero for all z>=2. For the solution based on the fixed point, is the imaginary value of f(z) nonzero everywhere outside of the real axis?
.. What does it mean, "based on the fixed point"? Do you mean the case of other base?
For base , there are two real fixed points.
We may use any of them to buildup the "real" superexponential.
Both resulting superexponentials have imaginary periods.
If such a superexponential has zero somewhere (for example, at minus unity),
then it has countable set of zeros.
sheldonison Wrote:..
Otherwise it would seem likely that there would be other singularities, outside of the real axis. Yes, at , the superexponential F such that F(0)=1 is periodic; hence, just translate the singularity at 2 for the period, and you get the singularity outside the real axis. See the picture for b=sqrt(2) at
http://en.citizendium.org/wiki/Image:Hol...rt2v01.jpg
It shows one additional singularity and the corresponding cut line.
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04/13/2009, 01:03 PM
(This post was last modified: 04/13/2009, 07:19 PM by sheldonison.)
Kouznetsov Wrote:sheldonison Wrote:..
In the tetration curve, the imaginary value for f(z) at the real axis is equal to zero for all z>=2. For the solution based on the fixed point, is the imaginary value of f(z) nonzero everywhere outside of the real axis?
...
Otherwise it would seem likely that there would be other singularities, outside of the real axis. What does it mean, "based on the fixed point"? Do you mean the case of other base? I was refering to the base e solution.
Khoustenov earlier wrote:
Kouznetsov Wrote:
where In your equation, you have the original tet(z) and you also have tem(z)=tet(J(z)). When I said "solution based on a fixed point", I was referring to tet(z), the "correct" solution, rather than the modified tem(z) equation, with the extra singularities.
For tet(z), is the only contour line where at the real axis, for z>2?
Just realized this cannot be true, since . So when the imaginary part >=pi, contours where the imaginary component is zero arise.
For the modified equation tem(z)=tet(J(z)), the fractal copies of the real axis (including the singularities) occur where , and these occur where there is a contour of . Wherever there is a contour line for outside of the real axis, a distorted copy of the original tet(z) real axis gets generated.
All of this seems to imply that the contours of (outside the real axis) have much more to do with the particular onecyclic function, where , then with tet(z) itself.
 Sheldon
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sheldonison Wrote:In the tetration curve, the imaginary value for f(z) at the real axis is equal to zero for all z>=2. For the solution based on the fixed point, is the imaginary value of f(z) nonzero everywhere outside of the real axis?
...
I was refering to the base e solution.
..
For tet(z), is the only contour line where at the real axis, for z>2? No.
The part of the real axis is not the only contour where .
There are many of them in the right hand side of the complex plane.
Both real and imaginary parts of have huge values in vicinity of the real axis. Many times the imaginary part passes through the values that are integer factors of
Each line , at the unity translation , produces the line ,
These lines form the complicated structure.
The fractal of lines is shown in fig.2 at
http://www.ils.uec.ac.jp/~dima/PAPERS/2009fractal.pdf
sheldonison Wrote:Just realized this cannot be true, since . So when the imaginary part >=pi, contours where the imaginary component is zero arise.
Yes.
sheldonison Wrote:For the modified equation tem(z)=tet(J(z)), the fractal copies of the real axis (including the singularities) occur where , and these occur where there is a contour of . Wherever there is a contour line for outside of the real axis, a distorted copy of the original tet(z) real axis gets generated. Yes. In the paper mentioned, the example with modified tetration is also considered.
However, we have no need to modify the tetration in order to have lines
outside the real axis, but these lines are at .
sheldonison Wrote:... the contours of (outside the real axis) have much more to do with the particular onecyclic function, where , then with tet(z) itself. Yes.
The multiple reproduction of reduced and deformed patterns is not a specific property of a tetration.
Take any other function with beautiful pattern in the complex plane, modify its argument with function J , and you get the fractal of the modified patterns at the plot.
