# Tetration Forum

Full Version: Universal uniqueness criterion?
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(06/22/2009, 09:46 PM)bo198214 Wrote: [ -> ]
(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.

What about using Kneser's approach to produce the alternate tetration functions?
(07/05/2009, 08:57 PM)BenStandeven Wrote: [ -> ]What about using Kneser's approach to produce the alternate tetration functions?

There is no initial region connecting an alternative fixed point pair.
The image of the straight line connecting a secondary fixed point pair overlaps with itself.
Secondary fixed points lie in a range with imaginary part greater or less than pi.
The vertical line connecting a pair is longer than 2*pi.
This means the image revolves more than once around 0 with constant radius, hence overlapping itself.

I tried to construct different connecting lines of a secondary fixed point pair and failed. I believe there is no initial region connecting two secondary fixed points.
(07/05/2009, 09:13 PM)bo198214 Wrote: [ -> ]
(07/05/2009, 08:57 PM)BenStandeven Wrote: [ -> ]What about using Kneser's approach to produce the alternate tetration functions?

There is no initial region connecting an alternative fixed point pair.
The image of the straight line connecting a secondary fixed point pair overlaps with itself.
Secondary fixed points lie in a range with imaginary part greater or less than pi.
The vertical line connecting a pair is longer than 2*pi.
This means the image revolves more than once around 0 with constant radius, hence overlapping itself.

I tried to construct different connecting lines of a secondary fixed point pair and failed. I believe there is no initial region connecting two secondary fixed points.

Yeah, that's right; the path would have to pass through a point with imaginary value 2 pi, and also through its conjugate. Then the other side of the region would intersect itself at the exponential of that point.
(07/06/2009, 12:53 AM)BenStandeven Wrote: [ -> ]the path would have to pass through a point with imaginary value 2 pi, and also through its conjugate. Then the other side of the region would intersect itself at the exponential of that point.

Say the curve $\gamma: [0,1]\to \mathbb{C}$ is injective and connects two points $\gamma(0)=a$ and $\gamma(1)=b$ with equal real part and with $\Im(b)-\Im(a)>2\pi$. One needs to show that then there is always a pair of points $c_1=\gamma(t_1)$ and $c_2=\gamma(t_2)$ with equal real part and with $\Im(c_2)-\Im(c_1)=2\pi$.

This sounds very plausible but I couldnt prove it except for certain simple shapes of $\gamma$, e.g. convex.
(07/06/2009, 08:56 AM)bo198214 Wrote: [ -> ]One needs to show that then there is always a pair of points $c_1=\gamma(t_1)$ and $c_2=\gamma(t_2)$ with equal real part and with $\Im(c_2)-\Im(c_1)=2\pi$.

This is equivalent to that $\gamma$ and $\gamma+2\pi i$ intersect.
If $\gamma$ only extend to the right, i.e. $\Re(\gamma(t))>\Re(a)\forall t\in (0,1)$, then this is a consequence of the Jordan curve theorem. We have $\Im(a) < \Im(a+2\pi i) < \Im(b) < \Im(b+2\pi i)$. The closed Jordan curve $[a,b]\cup \gamma$ has the point $\gamma(0+\epsilon)+2\pi i\approx a+2\pi i$ in its interior and the point $\gamma(1-\epsilon)+2\pi i\approx b+2\pi i$ in its exterior. Hence there must be an intersection of $\gamma+2\pi i$ and $\gamma$ (as $\gamma+2\pi i$ does not pass [a,b] for $t\in (0,1)$.)
I do not understand why there must be that kind of boundedness with the universal uniqueness criterion. Why?
(06/26/2022, 08:49 AM)Catullus Wrote: [ -> ]I do not understand why there must be that kind of boundedness with the universal uniqueness criterion. Why?

I don't understand the question ... why? Because this is how this proof works ...
(06/27/2022, 05:15 PM)bo198214 Wrote: [ -> ]
(06/26/2022, 08:49 AM)Catullus Wrote: [ -> ]I do not understand why there must be that kind of boundedness with the universal uniqueness criterion. Why?

I don't understand the question ... why? Because this is how this proof works ...

Lmao. Bo is out here with the patience of a saint.