Roots of z^z^z+1 (pictures in MSE) [update 8'2022]
#3
(02/08/2020, 01:45 AM)Daniel Wrote: Note since \( ^3 z+1=0 \), that 
\( ^3 z=-1 \) 
\( ^4 z=1/z \)
\( ^5z=z^{1/z} \) 
where \( z^{1/z} \) is related to the Thom Schell boundary. So 
\( ^nz \) 
would be related to an iterated Thom Schell boundary.

Yes, that nice \( \;^5 z=z^{1/z} \) 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
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RE: Roots of z^z^z+1 (pictures in MSE) - by Gottfried - 02/08/2020, 09:01 AM

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