Posts: 291
Threads: 67
Joined: Dec 2010
(12/30/2010, 12:50 PM)Gottfried Wrote: (12/29/2010, 10:15 PM)JmsNxn Wrote: Thank you. Just one final question, how did you generate this Taylor series? Is there a closed form expression?
James, as far as I see this is a graph which was produced by Dimitri Kousnetzov, who also posted here in the forum (you may use this link to find all all posts of him ) and to a certain extend explained his method here. But there is also a published paper of him where he describes this in detail (I've never understood it, btw, because I seem to lack some basic knowledge about cauchyintegrals and riemannmappings, but for a student of mathematics this may be completely familiar). I think his article is also in our (the tetrationforum's) database of literature (see the related message litrefdb in the forum)
And have a happy new year 
Gottfried
I doubt I'll understand it too, I'm only in highschool.
(12/30/2010, 02:34 PM)sheldonison Wrote: (12/30/2010, 12:50 PM)Gottfried Wrote: (12/29/2010, 10:15 PM)JmsNxn Wrote: Thank you. Just one final question, how did you generate this Taylor series? Is there a closed form expression? James, as far as I see this is a graph which was produced by Dimitri Kousnetzov There is no known closed form for the Taylor series for tetration. I generated the Taylor series with the kneser.gp program I wrote. I also posted the mathematical equations behind the algorithm here, http://math.eretrandre.org/tetrationforu...hp?tid=487
The basic idea using base e here, where L is the fixed point such that , , is that if
and and etc.
This can be used to develop a complex valued entire superfunction such that for all values of z.
The problem is that the superf is complex valued, not real valued. A 1cyclic mapping is used to convert this function to an analytic real valued tetration. The 1cyclic theta mapping is equivalent to the Riemann mapping in Kneser's algorithm, although convergence is not proven.
http://math.eretrandre.org/tetrationforu...hp?tid=487
The Taylor series is generated via a unit circle Cauchy integral.
 Sheldon
That's a really clever way of extending it. I wish I thought of it
Posts: 761
Threads: 118
Joined: Aug 2007
12/31/2010, 10:42 AM
(This post was last modified: 12/31/2010, 10:42 AM by Gottfried.)
(12/30/2010, 02:34 PM)sheldonison Wrote: http://math.eretrandre.org/tetrationforu...hp?tid=487
The Taylor series is generated via a unit circle Cauchy integral.
 Sheldon Hi Shel 
it's near new year and ususally I've much time to rethink some more complicated matter in that days. So I thought I'd try again to understand the procedere of the/of your Knesersolution. I've already downloaded your Kneser.gp and have run it two or three times, but what I am really lacking is the entrypoint, the key idea, the door, the basic rationale in which way it differs from the simple regular powerseries approach. (I've also read the paper of Kneser and first time got stuck when he introduced that regions and his handling with that...)
I'm just posting this msg although today I'm just peeking into the forum. I'll have the time to go into the matters possibly tomorrow in the evening or the day after tomorrow.
A "wish you happy new year" to everyone 
Gottfried
Gottfried Helms, Kassel
Posts: 761
Threads: 118
Joined: Aug 2007
(12/25/2010, 06:07 PM)JmsNxn Wrote: I was wondering if anyone had the means by which to reproduce this graph? And also, why this is not the accepted extension for tetration of rational values? A short step back to your opening msg here.
 ) did you think of contacting the author (Dimitri Kousnetzov) ?
 ) on the " And also, why (...)" . I cannot really give an answer; one of the last remarks about it I remember was, that Henryk and Dimitri worked together for a new article and Henryk mentioned something about a problem which came up but could not yet be solved/lacked a proof. I expected, they would post some proceedings here in the forum.
Another "Happy new year" 
Gottfried
Gottfried Helms, Kassel
Posts: 1,389
Threads: 90
Joined: Aug 2007
(12/25/2010, 06:07 PM)JmsNxn Wrote: And also, why this is not the accepted extension for tetration of rational values?
Because it is not theoretically safe.
There is no proof that the Dmitrii's (and also Sheldon's) procedure of computing the sexp even converges.
There is however a theoretic uniqueness criterion and a proven procedure that exactly produces the corresponding sexp/slog. I found that out recently and give links/references perhaps later. However Dmitrii's and also Sheldon's way is simpler to calculate and seem to compute exactly this unique sexp/slog.
Posts: 368
Threads: 44
Joined: Sep 2009
01/06/2011, 05:08 AM
(This post was last modified: 01/06/2011, 05:13 AM by mike3.)
(01/06/2011, 03:02 AM)bo198214 Wrote: (12/25/2010, 06:07 PM)JmsNxn Wrote: And also, why this is not the accepted extension for tetration of rational values?
Because it is not theoretically safe.
There is no proof that the Dmitrii's (and also Sheldon's) procedure of computing the sexp even converges.
There is however a theoretic uniqueness criterion and a proven procedure that exactly produces the corresponding sexp/slog. I found that out recently and give links/references perhaps later. However Dmitrii's and also Sheldon's way is simpler to calculate and seem to compute exactly this unique sexp/slog.
What is this proven algorithm, and does it provide any clues to finding the explicit (or close enough, e.g. if in terms of, say, infinite sums) coefficients of the Taylor series for the tetrational function, i.e. what is in
?
Posts: 1,389
Threads: 90
Joined: Aug 2007
(01/06/2011, 05:08 AM)mike3 Wrote: What is this proven algorithm
Michael Yampolski referred me to one of his articles [1].
I think you need a really good background in holomorphic dynamics to understand the article. Before that you should understand the construction of Shishikura [2] of the Fatou coordinates/Abel function where he uses the measurable Riemann theorem. Because Gaidashev gives a *constructive* measurable Riemann theorem, i.e. one where there is an algorithm to calculate it, and uses it to calculate the Fatou coordinates, if I understood that properly.
The Kneser construction is very specific, it constructs the (i.e. the unique) Fatou coordinates for exp (which I showed in [3]) with the (plain) Riemann theorem. Then Jaydfox used an algorithm for the Riemann mapping theorem to compare Kneser's with Andrew's solution.
Just wanted to show the similarity and that it seems we can not circumvent the (measurable) Riemann mapping theorem, which gives no closed form coefficients; which was regarding your question:
Quote:, and does it provide any clues to finding the explicit (or close enough, e.g. if in terms of, say, infinite sums) coefficients of the Taylor series
[1] Gaidashev, D., & Yampolsky, M. (2007). Cylinder renormalization of Siegel disks. Exp. Math., 16(2), 215–226.
[2] Shishikura, M. (2000). Bifurcation of parabolic fixed points. In Lei, Tan, The Mandelbrot set, theme and variations. Cambridge: Cambridge University Press. Lond. Math. Soc. Lect. Note Ser. 274, 325363 (2000).
[3] Trappmann, H., & Kouznetsov, D. (2010). Uniqueness of holomorphic Abel functions at a complex fixed point pair. Aequationes Math.
Posts: 368
Threads: 44
Joined: Sep 2009
(01/15/2011, 07:06 AM)bo198214 Wrote: (01/06/2011, 05:08 AM)mike3 Wrote: What is this proven algorithm
Michael Yampolski referred me to one of his articles [1].
I think you need a really good background in holomorphic dynamics to understand the article. Before that you should understand the construction of Shishikura [2] of the Fatou coordinates/Abel function where he uses the measurable Riemann theorem. Because Gaidashev gives a *constructive* measurable Riemann theorem, i.e. one where there is an algorithm to calculate it, and uses it to calculate the Fatou coordinates, if I understood that properly.
Ah. But I guess my thoughts that I've had that the Kneser, etc. tetrational function is the "best" tetrational function are likely correct:
(01/15/2011, 07:06 AM)bo198214 Wrote: The Kneser construction is very specific, it constructs the (i.e. the unique) Fatou coordinates for exp (which I showed in [3]) with the (plain) Riemann theorem. Then Jaydfox used an algorithm for the Riemann mapping theorem to compare Kneser's with Andrew's solution.
So would this mean that Andrew's algorithm and Kneser's construction yield the same result?
(01/15/2011, 07:06 AM)bo198214 Wrote: Just wanted to show the similarity and that it seems we can not circumvent the (measurable) Riemann mapping theorem, which gives no closed form coefficients; which was regarding your question:
Quote:, and does it provide any clues to finding the explicit (or close enough, e.g. if in terms of, say, infinite sums) coefficients of the Taylor series
However, I didn't think Riemann's theorem had something to do with whether or not one could give a formula of some kind for the mapping or a function related to the mapping, like . Just that a mapping exists. It is true that pretty much all Riemann mappings cannot be expressed with "elementary" functions, however. Also, I thought that a "closed form" always meant a finitary expression in terms of some "standard" set of operators, e.g. the elementary functions (i.e. functions formed by finitary composition on the first three Ackermann operators and their inverses), and nonelementary standard special functions (hypergeometric, gamma, zeta, etc.). So an infinite sum, product, etc. expansion of the Taylor coefficients would not be "closed form".
But if such an expansion truly does not exist, this would seem to seriously hamper the analysis and even the usefulness of the tetrational function (note how, e.g. complex factorial is very "useful" and has "deep" connections to other areas of mathematics  but the complex tetrational function may simply be too bizarre to be either of those). Or perhaps the formula exists, but it requires new kinds of primitive functions.
At the very least, it seems to be pointing at the idea that the tetration is a kind of function that really has no precedent, an exotic, "alien" kind of function that is really a whole lot different from the conventionally accepted set of special functions. Is this a fair assessment?
Posts: 1,365
Threads: 333
Joined: Feb 2009
(01/15/2011, 07:06 AM)bo198214 Wrote: Just wanted to show the similarity and that it seems we can not circumvent the (measurable) Riemann mapping theorem, which gives no closed form coefficients;
??
no closed form coefficients ? Riemann mapping theorem states there is a holomorphic function !?
so we have a taylor series with its coefficients.
and although not trivial , those are computable not ?
i am either confused or disagreeing.
regards
tommy1729
Posts: 1,389
Threads: 90
Joined: Aug 2007
(01/15/2011, 12:26 PM)mike3 Wrote: So would this mean that Andrew's algorithm and Kneser's construction yield the same result?
Check out Jayd's findings yourself here, it seems numerically giving the same results.
Quote:At the very least, it seems to be pointing at the idea that the tetration is a kind of function that really has no precedent, an exotic, "alien" kind of function that is really a whole lot different from the conventionally accepted set of special functions. Is this a fair assessment?
Well the discussing of the (existence and uniqueness) of perturbed Fatou coordinates (i.e. Abel function at a fixed point pair) is relatively recent in holomorphic dynamics. I think it was not discovered before *because* we have no nice coefficient description. Its a different/nonclassic way which enables to tackle the problem.
(01/15/2011, 09:29 PM)tommy1729 Wrote: no closed form coefficients ? Riemann mapping theorem states there is a holomorphic function !?
so we have a taylor series with its coefficients.
and although not trivial , those are computable not ?
maybe but thats not called a closed form if you can somehow compute it.
Posts: 1,365
Threads: 333
Joined: Feb 2009
(12/25/2010, 06:07 PM)JmsNxn Wrote:
I was wondering if anyone had the means by which to reproduce this graph? And also, why this is not the accepted extension for tetration of rational values?
Sorry to be a newb and ask so many questions
At the moment the image is absent or damaged. Well at least at my pc here.
