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07/05/2009, 09:13 PM
(This post was last modified: 07/05/2009, 09:15 PM by bo198214.)
(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.

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(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.

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07/06/2009, 08:56 AM
(This post was last modified: 07/06/2009, 09:01 AM by bo198214.)
(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

is injective and connects two points

and

with equal real part and with

. One needs to show that then there is always a pair of points

and

with equal real part and with

.

This sounds very plausible but I couldnt prove it except for certain simple shapes of

, e.g. convex.

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07/06/2009, 09:20 AM
(This post was last modified: 07/06/2009, 09:30 AM by bo198214.)
(07/06/2009, 08:56 AM)bo198214 Wrote: One needs to show that then there is always a pair of points and with equal real part and with .

This is equivalent to that

and

intersect.

If

only extend to the right, i.e.

, then this is a consequence of the Jordan curve theorem. We have

. The closed Jordan curve

has the point

in its interior and the point

in its exterior. Hence there must be an intersection of

and

(as

does not pass [a,b] for

.)

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06/26/2022, 08:49 AM
(This post was last modified: 06/26/2022, 08:49 AM by Catullus.)
I do not understand why there must be that kind of boundedness with the universal uniqueness criterion. Why?

ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ

Please remember to stay hydrated.

Sincerely: Catullus

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(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 ...

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(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.

Catullus, it's okay to accept when something is outside of your purview, and beyond your scope. But it's also a tad rude to bring up old threads and ask a question like this. This thread more than already answers the question you are asking.