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How it looks (i.θ)ₐ
#14
(04/23/2015, 04:52 PM)marraco Wrote: That was my dumb mistake. I didn't needed sexp(10), but sexp(I*10).
Maracco,
Beautiful graphs! Also, I*10 should work just fine! sexp(10*I)=0.318128684402243 + 1.33723930629532*I As I mentioned earlier, Kneser.gp doesn't implement tetration for bases<eta, and I strongly discourage using sexp(z) for bases<eta, since it does do something that I thought was interesting at the time, but I forget what; so you probably got that error using a base<eta.

Quote:¿Do you think that the tiny spirals at the end of the lines ͥ·ˣa (as x→∞) are accurate, or just a numerical artifact?

The spirals are not artifacts. As imag(z) increases, the function goes toward's Koenig's solution, which has the spirals. For base(e), Koenig's solution is periodic with period . I you go to with slope orthogonal to the Period, than the spirals will go away.

Quote:¿Did you attempted to find the base for which the lines turn into an "ellipse" (or something close)?. I guess a=e^(e^-1); that would "explain" the change of state.

Yes, base eta=e^(1/e) is the transition. In fact, the family of functions for tetration of various bases is analytic to the right of eta, but there is a mild singularity at eta itself. I generated a very accurate Taylor series for the first derivative of centered around b=2; see complex base tetration thread#15 The series begins with a0=0.889364954621 which is sexp'(z) for b=2.

Kneser.gp includes an implementation of sexpeta(z), which is for b=exp(1/e), which approaches e as imag(z) goes to infinity. I now know that the analytic function, extended from bases>eta, to bases<eta is not real valued at the real axis for bases<eta; this function is not included in Kneser.gp

Quote:Now, if the basesreally turn into ellipses, then it should be easy to find an algebraic expression for tetration to real exponents, or at least an important insight for ...

It sounds as though you would be interested in Koenig's solution for bases<eta (or equivalently; JmsNxn's solution) I would recommend you write your own Koenig's solution Smile It is much much much easier than generating a full blown Kneser Riemann mapping solution, but it is a required step in the process. Traditionally, these Tetration functions for bases<eta are imaginary periodic, developed from the attracting fixed point, using Koenig's solution. I just checked; in kneser.gp the imaginary periodic superfunction for bases<eta is included as the function "superf2(z)", and the variable with the associated period is "Period2". superf2(z) is the function you are probably interested in for real bases<eta; it is Koenig's solution.
- Sheldon
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Messages In This Thread
How it looks (i.θ)ₐ - by marraco - 04/18/2015, 11:20 PM
RE: How it looks (i.θ)ₐ - by sheldonison - 04/19/2015, 02:40 PM
RE: How it looks (i.θ)ₐ - by marraco - 04/19/2015, 08:40 PM
RE: How it looks (i.θ)ₐ - by sheldonison - 04/19/2015, 11:19 PM
RE: How it looks (i.θ)ₐ - by marraco - 04/20/2015, 02:35 AM
RE: How it looks (i.θ)ₐ - by Gottfried - 04/20/2015, 07:53 AM
RE: How it looks (i.θ)ₐ - by marraco - 04/21/2015, 02:34 AM
RE: How it looks (i.θ)ₐ - by marraco - 04/21/2015, 03:20 AM
RE: How it looks (i.θ)ₐ - by sheldonison - 04/21/2015, 08:23 AM
RE: How it looks (i.θ)ₐ - by marraco - 04/23/2015, 04:52 PM
RE: How it looks (i.θ)ₐ - by JmsNxn - 04/23/2015, 11:15 PM
RE: How it looks (i.θ)ₐ - by sheldonison - 04/23/2015, 11:20 PM
RE: How it looks (i.θ)ₐ - by marraco - 04/26/2015, 12:50 AM
RE: How it looks (i.θ)ₐ - by sheldonison - 04/26/2015, 05:08 AM
RE: How it looks (i.θ)ₐ - by marraco - 04/20/2015, 03:46 AM
RE: How it looks (i.θ)ₐ - by marraco - 04/23/2015, 02:21 AM
RE: How it looks (i.θ)ₐ - by marraco - 04/25/2015, 07:52 PM



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