Thread Rating:
  • 0 Vote(s) - 0 Average
  • 1
  • 2
  • 3
  • 4
  • 5
Roots of z^z^z+1 (pictures in MSE) [update 8'2022]
#1
Dear friends - 

I've contributed in a thread in MSE https://math.stackexchange.com/questions...olve-xxx-1   on the question for finding of solutions for z^z^z = -1 . The question is of 2015 and because of lots of chaotic numerical difficulties I left the question with finding three roots using the log(log(z^z^z))=log(log(-1)), exposed in a Newton-fractal.                    
This year I came back to the question, partially overcame the numerical chaos in Pari/GP and have now a discussion worth to mention it also here.    

A finding is, that the separation of the complex values z^z^z+1 into real and imaginary components gives families of continuous curves of zero values, and the lines occur with some periodicity.     
Because roots of the full complex values of the function are only there where both components are zero, we find the roots on a discrete lattice-style set of surely infinite points.        

See the overlay of the contour-plots of real and imaginary components in the neighbourhood of a known root at about 5.277+11.641i . The roots are on the intersections of the white curves.   

[Image: LBsQ0.png]
Gottfried Helms, Kassel
Reply
#2
Note since , that 
 

 
where  is related to the Shell Thron boundary. So 
 
would be related to an iterated Shell Thron boundary.
Daniel
Reply
#3
(02/08/2020, 01:45 AM)Daniel Wrote: Note since , that 
 

 
where  is related to the Thom Schell boundary. So 
 
would be related to an iterated Thom Schell boundary.

Yes, that nice has come to me tonight too :-) Things like this might become "my darling" one day ...                            

Anyway, there is one more observation which made me think again.                

I've verbally written about a "lattice" when I saw the curves of the zero-values intersecting. While the painted intersection interestingly seem to show nearly orthogonal angle at the intersections (is it so?) it moreover seems  that each pair of curves can only have *one* intersection - but which is then no more a "lattice".               
Is this so? What consequences does it have - for instance for a better root-finding procedure than the Newton-algorithm which has this nasty numerical chaos at many coordinates?         


(Btw. the two authors's names are (D.) "Shell" and (?.) "Thron" )
Gottfried Helms, Kassel
Reply
#4
Check out my page on the fractal of the exponential map at -1. The map of is self similar as you zoom in.
Daniel
Reply
#5
Update for the initial posting.   



Just played around with this problem (main question in MSE: https://math.stackexchange.com/questions...olve-xxx-1 ) again, and by a sudden idea - reflecting my "invention" of the iterated-branched-logarithm in the "periodic-points" thread - I got now apparently the complete solution for the finding of roots in \( \,^3z +1 \), and the generalizations for the finding of roots in \( \,^3z \pm 1 \) and \( \,^4z \pm 1 \); the scheme is so simple that its further generalization to \( \,^mz \pm 1 \) is simply an extension of a couple of lines of code.



I've not yet time to set this up in mathjaxform, to make this visible directly here in the forum-box, but I upload   the complete file where


  • I've put together my long answer to the initial problem in MSE (2015, update 2020) pg 1-8     
  • with the new appendix pg 9-13 containing my solution which I found yesterday & today. This is heuristic, and perhaps spurious cases missing - don't know how to handle the problem to prove to really have the exhaustion of all the possible roots.                


Here is the link (to possibly later updated versions) https://go.helms-net.de/math/tetdocs/_ot...ration.pdf , but also uploaded it here.
.pdf   roots_of_xxx+1_V4_fullexploration.pdf (Size: 1.34 MB / Downloads: 25)
Criticism and constructive ideas are much welcome.

Update 2.9.22 - the method is not yet strong enough to generalize easily to \( \,^4z \pm 1\) and higher; after I found missings in the \( \,^5z + 1 \) case, a sharper look at the assumption of attractivity of the formula 4.1b (in the above linked-to edition) shows an over-generalization of the findings with the \( \,^3z \pm 1 \) case. I'll have to consider now whether -and then how- this problem with attractiveness/contractiveness of the iteration can be fixed at all. So - hold on, you all nice readers...  hope this need not be dismissed in whole.

Gottfried
Gottfried Helms, Kassel
Reply
#6
(08/28/2022, 12:00 PM)Gottfried Wrote: Update for the initial posting.   

Just played around with this problem (main question in MSE: https://math.stackexchange.com/questions...olve-xxx-1 ) again, and by a sudden idea - reflecting my "invention" of the iterated-branched-logarithm in the "periodic-points" thread - I got now apparently the complete solution for the finding of roots in \( \,^3z +1 \), and the generalizations for the finding of roots in \( \,^3z \pm 1 \) and \( \,^4z \pm 1 \); the scheme is so simple that its further generalization to \( \,^mz \pm 1 \) is simply an extension of a couple of lines of code.

I've not yet time to set this up in mathjaxform, to make this visible directly here in the forum-box, but I upload   
  • the basic file where I've put together my long answer to the initial problem in MSE (2015, update 2020) pg 1-8                                
  • the appendix pg 9-12 only containing my solution which I found yesterday & today. This is heuristic, and perhaps spurious cases missing - don't know how to handle the problem to prove to really have the exhaustion of all the possible roots.                  

Criticism and constructive ideas are much welcome.

Gottfried
 

Just flipping through this initially. This is super cool. The day we start relating "roots of tetration" and "Number theory" we can finally get some traction. The way we are just flippantly talking about \(^m z \pm 1\)--it's going to be so valuable. Not much else to add, but dope as hell, Gottfried Big Grin

Imagine we get an Euler's theorem--no one ever (way back in the day) guessed that \(z^m \pm 1=0\) would ever just become \(z^m + 1=0\) -- thru a "change of variables" Wink
Reply


Possibly Related Threads…
Thread Author Replies Views Last Post
  On n-periodic points of the exp() - A discussion with pictures and methods Gottfried 1 3,762 06/10/2020, 09:34 AM
Last Post: Gottfried
  Funny pictures of sexps bo198214 45 111,156 08/27/2009, 06:52 PM
Last Post: bo198214
  Pari-TTY - update Gottfried 0 4,399 10/13/2007, 04:02 PM
Last Post: Gottfried



Users browsing this thread: 1 Guest(s)