# Tetration Forum

Full Version: Universal uniqueness criterion?
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(06/22/2009, 02:24 PM)bo198214 Wrote: [ -> ].. Well, but $\exp$ *has* more fixed points. In every strip $2\pi i k < \Im(z) < 2\pi i (k+1)$ there is a fixed point of $\exp$.
Henryk, how about to build up "another" holomorphic tetration that goes to other fixed points at $\pm i \infty$?
And what will do such a function at $x+iy$ while $x \rightarrow -\infty$ ?
(06/22/2009, 07:19 PM)Kouznetsov Wrote: [ -> ]
(06/22/2009, 02:24 PM)bo198214 Wrote: [ -> ].. Well, but $\exp$ *has* more fixed points. In every strip $2\pi i k < \Im(z) < 2\pi i (k+1)$ there is a fixed point of $\exp$.
Henryk, how about to build up "another" holomorphic tetration that goes to other fixed points at $\pm i \infty$?
I dont think that there is an initial region connecting other conjugated fixed point pairs than the one closest to the real axis. (Plot the straight line connecting two such fixed points and plot its image under exp. Both lines intersect.)

On the other hand I asked you som time ago to apply your algorithm to other fixed points, but you somehow did not follow that path.
(06/22/2009, 09:46 PM)bo198214 Wrote: [ -> ]I asked you som time ago to apply your algorithm to other fixed points, but you somehow did not follow that path.
Assume that for another tetration $F$, we have some probe function $E(y)=F(iy)$ for real $y$; id est, along the imaginary axis.
Assume, Function $E$ is suposed to go to $L_m$ at infinity and $L_m^*$ at minus infinity; $\log(L_m)=L_m+2\pi i m$, and $m$ is integer constant.
In order to adjust the probe function to some tetration, we need to evaluate the contour integral. If we use the same contour as in the paper http://www.ams.org/mcom/2009-78-267/S002.../home.html
then we need the values at $1+i y$ and $-1+iy$ for real $y$.
It is easy to estimate values at $1+i y$; use estimate $\exp(E(y))$.
As for $-1+i y$; we use estimate $\log(E(y))+2\pi i m$ for positive $y$ and $\log(E(y))-2\pi i m$ for negative $y$. Either we have singularity (jump) at $-1$, or $m=0$.

This explains, why I did not construct such "another tetration", but this is not a proof that this is impossible. Suggest the holomorphic probe function to begin with. Such a function should have some smooth kink of the phase, in order to avoid the jump;
but allow such a jump for the principal branch of its logarithm.
Then we can run the same algorithm, with additional control of the branch of the logarithm. Such a control should recover $F(-1+iy)=\log(F(iy))+i\pi m$,
adding unity or minus unity to $m$ each time when $E(y)$ passes through negative real values. We need the holomorphic kinky probe function, then we can run the algorithm to recover the kinky tetration.
I think about something like $E(y)=
\Re(L_1) +(1-\Re(L_1))/ \cosh(y) + \Im(L_1) \tanh(y)$
bo198214 Wrote:
Tetratophile Wrote:(3) We can repeat (4.2) as many times as needed to get to $D_2$.

No, we can not.
1. The curve may get out of the strip of bijectivity of what indeed happens for . (plot the curves!)
2. may hit , i.e. the extension of overlaps with the original domain. If meets at an angle then meets at an angle . So the image curves rotate around L. This includes also the comment of Ben, that if , i.e. if is real, then then the images do not rotate, and hence one would never reach if it hits in a different angle.
Yeah, I knew it wouldn't work right when I postd it, because of the issues with f's domain of biholomorphy. I edited that many times, but exp still doesn't meet the condition.

Kouznetsov Wrote:Such a function should have some smooth kink of the phase, in order to avoid the jump; but allow such a jump for the principal branch of its logarithm...
I think about something like $E(y)=
\Re(L_1) +(1-\Re(L_1))/ \cosh(y) + \Im(L_1) \tanh(y)$

@Kouznetsov: Can you show me how E(y) would have the behavior as you speculate? Also do the tetrations constructed with other fixed point pairs still satisfy the basic functional equations b[4](z+1) = b^b[4]z and b[4]1=b?
bo198214 Wrote:
Kouznetsov Wrote:Such a function should have some smooth kink of the phase, in order to avoid the jump; but allow such a jump for the principal branch of its logarithm...
I think about something like $E(y)=
\Re(L_1) +(1-\Re(L_1))/ \cosh(y) + \Im(L_1) \tanh(y)$
Can you show me how E(y) would have the behavior as you speculate? Also do the tetrations constructed with other fixed point pairs still satisfy the basic functional equations b[4](z+1) = b^b[4]z and b[4]1=b?
I played it a little bit. We should take an exponential of the expression I wrote and scale the argument; then it reproduces the kink of reasonable size. Of course, such a probe function does not satisfy the tetration equation, but it shows some reasonabe behavior within the strip along the imaginary axis: after the esponentiation, it has just one kink in the upper halfplane and one in the lower halfplane; this is all we need for the function at the first iteration. We may glue it to the asymptotics, and plot the function even before to make the precise evaluation.
@Kouznetsov: how are the other tetrations relevant to our discussion of uniqueness?

@Kouznetsov: Also I just have a quick question to ask about tetration: What is the mathematical reason that the tetrational approaches the fixed points $L$ or $L*$ as you go to $\pm i \infty$? I think instead, that $\lim_{x \rightarrow -\infty} x\pm ci = L\mathrm$ or $L*$, where c is a nonzero real number, should be the limit that corresponds to the fixed point. (when the imaginary part is zero this doesn't work)
Why is this intuitive reasoning incorrect?:
to get from 1 to a non-real number by iteration of exp, you need complex iteration (real iterations always give real numbers)
But then since $L,L*$ are repelling with respect to the exponential, you need infinite negative iterations of exp (positive iterations of log). If you infinitely iterate log on a nonreal number, you get closer to L, why isn't this reflected in the tetrational graph? Bo said something like log is not in the initial region anymore, could you clarify that for me? I have attached a schematic diagram to represent my reasoning:
[attachment=497]
how does the tetrational for base b>exp(1/e) actually behave at large values of real part of non-real arguments?
(06/23/2009, 09:28 PM)Tetratophile Wrote: [ -> ]@Kouznetsov: how are the other tetrations relevant to our discussion of uniqueness?
I expect, all "other" tetrations have additional singularities and cutlines.

Quote:@Kouznetsov: Also I just have a quick question to ask about tetration: What is the mathematical reason that the tetrational approaches the fixed points $L$ or $L*$ as you go to $\pm i \infty$? I think instead, that $\lim_{x \rightarrow -\infty} x\pm ci = L\mathrm$ or $L*$, where c is a nonzero real number, should be the limit that corresponds to the fixed point.
You have no need to type "instead", both should be correct. The reason in not so mathematical, but computational: it is easier to work with a function, holomorphic in wide range, and create all other tetrations, just modifying the argument, if necessary.
In principle, "all animals are equal"; but as soon as you begin to plot graphics, it happens, that "some animals are more equal than others".
For me, "more equal" are animals with wide range of holomorphizm.

Quote:
.. to get from 1 to a non-real number by iteration of exp, you need complex iteration (real iterations always give real numbers)
Yes.

Quote:But then since $L,L*$ are repelling with respect to the exponential, you need infinite negative iterations of exp (positive iterations of log).
Yes.

Quote:If you infinitely iterate log on a nonreal number, you get closer to L, why isn't this reflected in the tetrational graph?
? In the left hand side of the plot of tetration, values approach $L$ in the upper halfplane and $L^*$ in the lower halfplane; this corresponds to the graphic you supply.

If you mean our discussion with Bo about a super-exponential that approaches $L_1$ instead of $L=L_0$, then it is not correct:
In some region, at the iteration of logarithm, we have to add $2 \pi i$ in the upper halfplane and $-2 \pi i$ in the lower halfplane.
The question is, wether we can match them for large enough real part of the argument.
I am not yet successful to construct or plot such a function.
I expect such function to have two additional horizontal cutlines, going to -infinity.

Quote:Bo said something like log is not in the initial region anymore, could you clarify that for me?
As I understand, the initial region is $
C=\{z \in \mathbf{C} : \Re(z)\ge 1 , \|z|\le |L| \}$

if some point $z$ is inside the initial region (not at the margin), then
$\log(z)$ in outside the initial region.
The same about $\log(z)+2 \pi i$ .

Is this that you were asking for?

Quote:I have attached a schematic diagram to represent my reasoning:
how does the tetrational for base b>exp(1/e) actually behave at large values of real part of non-real arguments?
There are pictures at http://www.ils.uec.ac.jp/~dima/PAPERS/2009fractae.pdf
The tetrational shows complicated quasi-periodic behavior.
"bo198214 Wrote:2. may hit , i.e. the extension of overlaps with the original domain. If meets at an angle then meets at an angle . So the image curves rotate around L. This includes also the comment of Ben, that if , i.e. if is real, then then the images do not rotate, and hence one would never reach if it hits in a different angle.

*Fortunately the derivatives at the fixed points L and L* is not a real number, (they are 0.318...+1.337...i) Of course I'm talking about the fixed points of the principal branch of the logarithm.

*What is the problem with the extension overlapping with the original domain?

*What initial regions do all of the known methods of complex extensions of tetrations have? If their initial regions don't deviate too much from each other that their exponentials leave the domain of biholomorphy, we can prove the weaker theorem that all of the known extensions are the same. Can anyone show the (mapping graphs on the complex plane) exponentials and the logarithms of the intiial regions?
(07/04/2009, 11:17 PM)Tetratophile Wrote: [ -> ]*What is the problem with the extension overlapping with the original domain?
*What initial regions do all of the known methods of complex extensions of tetrations have? If their initial regions don't deviate too much from each other that their exponentials leave the domain of biholomorphy, we can prove the weaker theorem that all of the known extensions are the same. Can anyone show the (mapping graphs on the complex plane) exponentials and the logarithms of the intiial regions?
The region G where the tetrtion is biholomorphic is plotted in figure 2 of
http://www.ils.uec.ac.jp/~dima/PAPERS/2009uniabel.pdf
exp(G) is shown at the right (between green and blue curves) and
log(G) is shown at the legt (between the red curve and the black line).
One has no need to evaluate tetration in order to plot this figure.
The region G is filled with lines of constant real part of arctetration and lines of constant imaginary part in figure 1b of
http://www.ils.uec.ac.jp/~dima/PAPERS/2009fractae.pdf
You may extract the figures mentioned and just overlap them.
(07/04/2009, 11:17 PM)Tetratophile Wrote: [ -> ]*Fortunately the derivatives at the fixed points L and L* is not a real number, (they are 0.318...+1.337...i) Of course I'm talking about the fixed points of the principal branch of the logarithm.
Yes, that should be likely for fixed point not on the real axis.

Quote:*What is the problem with the extension overlapping with the original domain?
Thats perhaps not so much the problem, more of a problem is that it is no domain anymore - boundary crosses itself.

Quote:*What initial regions do all of the known methods of complex extensions of tetrations have?

The property of a region to be initial does not depend on the super-function, but just on the function itself. So every initial region of exp can possibly be a domain of definition for tetration. I think/hope every method returns values on the default initial region - the crescent - of all z such that |z|<|L| and such that Re(z)>Re(L). But of course it may still return values on other initial regions.

And still it could be possible (though I dont think so and am close to prove my conjecture) that one tetration is biholomorphic only on one initial region and the other tetration is biholomorphic only on the other initial region.

For a picture of iterated exponentiation have a look at figure 8 in Dmitriis article
http://www.ams.org/mcom/2009-78-267/S002.../home.html
The black lines are the integer iterations of the straight line from L to L*.
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