05/23/2011, 09:01 PM (This post was last modified: 05/23/2011, 10:37 PM by sheldonison.)
The recent posts on the Taylor series for the superfunctions of \( \eta=exp(1/e) \) reminded me that I want to post my theory, that as imag(z) increases, the lower and upper superfunctions at eta converge towards each other, plus a constant. Here, sexp(z) is the lower superfunction, with sexp(0)=1, and cheta(z) is the upper superfunction, normalized so that cheta(0)=2e.
\( \text{sexp}_{\eta}(z)=\text{cheta}(z+\theta(z)+ k )
\). Where \( \theta(z) \) is a 1-cyclic function, which quickly decays to zero as imag(z) increases. Then, the constant "k" is 5.0552093131039000 + 1.0471975511965977*I. And then we have for any real number x,
\( \lim_{z \to i\infty}\text{sexp}_{\eta}(x+z)=\text{cheta}(x+z+k) \)
- Sheldon
05/23/2011, 09:49 PM (This post was last modified: 05/23/2011, 09:51 PM by bo198214.)
(05/23/2011, 09:01 PM)sheldonison Wrote: Where \( \theta(z) \) is a 1-cyclic function, which quickly decays to zero as imag(z) increases. Then, the constant "k" is 5.0552093131039000 + 1.0471975511965977*I.
Without looking at numerical values I at least want to mention, if theta is 1-cyclic and bounded (which would follow from decay towards ioo) then theta would be constant, because every bounded entire function must be constant.
05/23/2011, 11:26 PM (This post was last modified: 05/24/2011, 08:28 AM by sheldonison.)
(05/23/2011, 09:49 PM)bo198214 Wrote:
(05/23/2011, 09:01 PM)sheldonison Wrote: Where \( \theta(z) \) is a 1-cyclic function, which quickly decays to zero as imag(z) increases. Then, the constant "k" is 5.0552093131039000 + 1.0471975511965977*I.
Without looking at numerical values I at least want to mention, if theta is 1-cyclic and bounded (which would follow from decay towards ioo) then theta would be constant, because every bounded entire function must be constant.
PS: you have some typos in the thread title.
Yeah, if the 1-cyclic function is a sum, \( \theta(z)=\sum_{n=1}^{\infty}a_n \exp(2n\pi zi) + b_n \exp(-2n\pi zi) \), then all of the b_n terms must be zero, so it decays to zero as imag(z) increases. The theta(z) function is a Kneser mapping function, which is the sum of the a_n terms, with singularities for integer values of z. It is defined if imag(z)>0, or (if imag(z)=0 and z is not an integer). Then the sexp(z) function for imag(z)<0 is defined by a Schwarz reflection. And, just like sexp base e, the sexp_eta(z) winds up with singularities for integers <= -2, and the singularity cancels for integers>-2.
I was curious to see if anyone else had noticed this. Unfortunately, I don't know how to prove it. edit: In a prevoius post Henryk pointed out that if you have a sickle between the two conjugated fixed points, then that is a required condition. I'll have to try to understand that better ....
I have also numerically calculated the terms for theta(z), and verified that \( \text{sexp}_{\eta}(z)=\text{cheta}(z+\theta(z)+ k) \). At imag(z)=I, the upper harmonics have already decayed enough that the graph looks very sinusoidal. Here is the graph of theta(z) at imag(z)=1, from -1+i to 1+i, where the amplitude of the main harmonic has decayed to 0.00017. Closer to the real axis, as I have I previously posted, the graph has has progressively more and more high frequency terms, due to the singularities at integer values of z.
- Sheldon
05/24/2011, 01:06 PM (This post was last modified: 05/24/2011, 02:25 PM by bo198214.)
Ah, ok, I see \( \theta \) is not entire.
Interesting conjecture. It seems also confirmed by the pictures (made by Dmitrii) on page 8 in the attached "Computation of the Two Regular Super-Exponentials to base exp(1/e)".
I look forward to reading Dimitrii's paper. Apparently, it must still be pre-publication, as all I can see is, "You are not the submitter of that submission".
I originally thought I would calculate theta(z) by generating a Kneser Riemann mapping, which would require some modifications to my Kneser.gp code, that I haven't had time to try yet. So yesterday I calculated \( \theta(z) \), the easier way, using the equation.
\( \theta(z)=\text{cheta}^{[-1]}(\text{sexp}_\eta(z))-z \). With 100 terms, I get double precision accurate results for imag(z)>=0.04, for the equation generating
\( \text{sexp}_\eta(z) = \text{cheta(z+\theta(z)) \)
(05/24/2011, 02:18 PM)sheldonison Wrote: Apparently, it must still be pre-publication, as all I can see is, "You are not the submitter of that submission".
Oh, dear, I uploaded it to the arxiv. But it takes some time there until it is public.
So, I attached the file in the original post. It is currently under review in "Mathematics of Computation".
05/25/2011, 04:03 AM (This post was last modified: 05/26/2011, 01:02 PM by sheldonison.)
(05/24/2011, 01:06 PM)bo198214 Wrote: Ah, ok, I see \( \theta \) is not entire.
Interesting conjecture. It seems also confirmed by the pictures (made by Dmitrii) on page 8 in the attached "Computation of the Two Regular Super-Exponentials to base exp(1/e)".
Dimitrii noticed the similarity too, page 20. "Outside the positive part of the real axis, F3(z) approaches e at |z|->infinity in a similar way as F1 does."
- Sheldon