Roots of z^z^z+1 (pictures in MSE) - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Computation (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=8) +--- Thread: Roots of z^z^z+1 (pictures in MSE) (/showthread.php?tid=1254) Roots of z^z^z+1 (pictures in MSE) - Gottfried - 02/07/2020 Dear friends -  I've contributed in a thread in MSE https://math.stackexchange.com/questions/1415029/solve-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.    RE: Roots of z^z^z+1 (pictures in MSE) - Daniel - 02/08/2020 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 Shell Thron boundary. So  $^nz$  would be related to an iterated Shell Thron boundary. RE: Roots of z^z^z+1 (pictures in MSE) - Gottfried - 02/08/2020 (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" ) RE: Roots of z^z^z+1 (pictures in MSE) - Daniel - 02/09/2020 Check out my page on the fractal of the exponential map at -1. The map of $^z(-1)$ is self similar as you zoom in.