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 \( \frac{\2pi i}{L}\approx 4.447+1.058i \). I you go to \( \Im(\infty) \) 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 \( \text{sexp}_b(z) \) 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 bases\( \\[15pt]

{1<a<e^{e^{-1}} } \)really 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 \( \\[20pt]

{^{\frac{1}{x}}a} \)...

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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