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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
=a)
and
=b)
with equal real part and with
-\Im(a)>2\pi )
. One needs to show that then there is always a pair of points
)
and
)
with equal real part and with
-\Im(c_1)=2\pi)
.
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.
)>\Re(a)\forall t\in (0,1))
, then this is a consequence of the Jordan curve theorem. We have
 < \Im(a+2\pi i) < \Im(b) < \Im(b+2\pi i) )
. The closed Jordan curve
has the point
+2\pi i\approx a+2\pi i)
in its interior and the point
+2\pi i\approx b+2\pi i)
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?
Please remember to stay hydrated.
ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\
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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.
*has* more fixed points. In every strip
there is a fixed point of
?
