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 Mizugadro, pentation, Book tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 02/13/2015, 01:21 PM (02/13/2015, 02:25 AM)Kouznetsov Wrote: (02/13/2015, 12:45 AM)tommy1729 Wrote: Im fascinated by the (subset) Taylor(x,ln(x)) (or transseries if you want) used for the parabolic fixpoint. Its ring structure makes solving the zoom equation ( F(f(x)) = f(ax) ) natural. ( polynomials of Taylor(x,ln(x)) are of same type and also Taylor(ax,ln(ax)) is of same type ! ) This needs more attention imho. ..Thank you for your interest, tommy1729. Have you write some similar expansions? Do they work? Quote:Are all parabolic fixpoint expansions (for iterating analytic functions) like this ?? I think so. You may compare it to expansion of superfunction of transferfunction zex(z)=z exp(z) It is described in chapter 12 of book "Суперфункции". The Mathematica code to caluate the expansion is suggested in section 12.2. Quote:.. what about terms ln(x)^7 x^3 then ? You may extract the coefficient from the C++ implementation at http://mizugadro.mydns.jp/t/index.php/E1etf.cin Best regard, Dmitrii The term ln(x)^7 x^3 does not occur in your system because 7>3. You work with P_n(ln(x)) x^n where P_n is a polynomial of degree n. hence my comment about ln(x)^7 x^3. and also the term " subset " of the taylor series (x,ln(x)). This gives me some question mark feelings. --- Could you give a link to the chapter 12 you mentioned ? Maybe its in an earlier post but im unsure. --- I would like to see a proof that all parabolic fixpoint expansions are like this. --- IM not sure if we use / do things the same way. I consider solving the zoom equation F(f(x)) = f(ax) because polynomial/Taylor of P_n(ln(x)) x^n = P*_n(ln(x)) x^n and P_n(ln(ax)) ax^n = P_n(ln(x)+ln(a)) a^n x^n = P°_n(ln(x)) x^n. Therefore we get a ring structure that is completely solvable. Im not sure if its solutions is unique , and think not because the Taylor series is a subset. And even without the Taylor series subset Im not sure. I also notice that an integral or derivative can be of DIFFERENT TYPE. Contrary to Taylor series. But the main issue is convergeance. If im correct the series will only work if x > 1 because of the log parts ? Or is it x > 0 ? If we solve the zoom equation in terms of Taylor we get a formal Taylor with radius zero right ? But does this new series expansion give a nonzero radius then ? ALWAYS ? --- Does this relate to the mittag-leffler expansion ? --- Alot to do. regards tommy1729 Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 02/14/2015, 01:22 AM (02/13/2015, 01:21 PM)tommy1729 Wrote: (02/13/2015, 02:25 AM)Kouznetsov Wrote: Quote:.. what about terms ln(x)^7 x^3 then ? You may extract the coefficient from the C++ implementation at http://mizugadro.mydns.jp/t/index.php/E1etf.cinThe term ln(x)^7 x^3 does not occur in your system because 7>3.Hence, the corresponding coefficient is zero. Please, look: My expansion use ln(x)^m x^n for positive m and negative n. Quote:You work with P_n(ln(x)) x^n where P_n is a polynomial of degree n. hence my comment about ln(x)^7 x^3. and also the term " subset " of the taylor series (x,ln(x)). This gives me some question mark feelings. I doubt, if the expansion with powers of ln(x) and x should be called "taylor series". The Taylor series correspond to expansion of holomorphic function in a point, that is inside the range of holomorphism. Neither logarithm, nor negative powers of the argument are allowed there. Perhaps, this observation resolves your "question mark feelings". Quote: Could you give a link to the chapter 12 you mentioned ? Maybe its in an earlier post but im unsure. Now the book is in "one piece". At the beginning, I had suggested the link http://mizugadro.mydns.jp/t/index.php/Суперфункции does it open from your country? The book can be ordered at https://www.morebooks.de/store/ru/book/С...59-56202-0 and/or loaded at http://www.ils.uec.ac.jp/~dima/BOOK/202.pdf http://mizugadro.mydns.jp/BOOK/202.pdf Chapter 12 begins at page 147 of the printed version. Numeration of pages begins with -3, this is standard for that Editorial. The book is a little bit in Russian, but the formulas are in English, and the codes are in C++ and in Mathematica; hope you understand the last three languages mentioned. Quote:I would like to see a proof that all parabolic fixpoint expansions are like this. Some approaches to this can be found in the book 5+ methods for real analytic tetration Henryk Trappmann, Dmitrii Kouznetsov June 28, 2010. https://bbuseruploads.s3.amazonaws.com/b...2Z1HF1TZ82 Also, some hints are in the article H.Trappmann, D.Kouznetsov. Computation of the Two Regular Super-Exponentials to base exp(1/e). Mathematics of Computation. Math. Comp., v.81 (2012), p. 2207-2227. ISSN 1088-6842(e) ISSN 0025-5718(p) http://www.ams.org/journals/mcom/0000-00...2590-7.pdf http://mizugadro.mydns.jp/PAPERS/2012e1eMcom2590.pdf offprint http://mizugadro.mydns.jp/PAPERS/2011e1e.pdf (preprint) https://bitbucket.org/bo198214/e1e/raw/6...3/main.pdf I do not like that "proof" by Henryk, described there, and I do not include it into the "Суперфункции". I suspect, if you want some accurate "proof that all parabolic fixpoint expansions are like this", then you have to write it by yourself. I think, before to deal with the "general proof", you should consider one or two special cases. Quote:IM not sure if we use / do things the same way. We do not do things in the same way. We do things in different ways. I consider argument z and treat 1/z as small parameter. Using this approach, I suggest the simple algorithms to evaluate superfunctions; I implement them in C++, I plot complex maps of the function and maps of the agreement, I supply them with generators. You consider argument x as small parameter, You suggest the complicated algorithms, You do not implement them in C++ (nor in Fortran), You do not plot the complex maps of superfunctions, You do not plot the maps of the agreement at the substitution of your approximations into the transfer equation. Quote:I consider solving the zoom equation F(f(x)) = f(ax) because polynomial/Taylor of P_n(ln(x)) x^n = P*_n(ln(x)) x^n and P_n(ln(ax)) ax^n = P_n(ln(x)+ln(a)) a^n x^n = P°_n(ln(x)) x^n. Therefore we get a ring structure that is completely solvable. Consider to plot the map of the solution and the map of the agreement. I mean expansion of superfunction at infinity, for large |x|. This expansion provides the efficient algorithm for the evaluation of superfunction. You seem to deal with the Schroeder rquation and the Schroeder functions. In some cases, the superfunction can be expressed through the Schroeder function and vice versa. But it is not a case of tetration to base exp(1/e). Quote:.. But the main issue is convergeance. No convergence. The series are asymptotic. With my expansion, using the transfer equation, for any argument you like, you can get so many decimal digits as you like. In this sense, the solution is exact. We can use it to plot the complex maps. With your expansion, I think, you cannot do the same. Quote:If im correct the series will only work if x > 1 because of the log parts ? Or is it x > 0 ? I do not know about your series. I doubt, if your series "work", in the sense of efficient algorithm to evaluate the superfunction. As for my series, for the evaluation, you should shift the argument to deal with large values of |x|. Note, that in my expansion, the argument appear in NEGATIVE powers. Quote:If we solve the zoom equation in terms of Taylor we get a formal Taylor with radius zero right ? [quote] About radius zero, I think so. About Taylor, I think, "not". [quote] But does this new series expansion give a nonzero radius then ? ALWAYS ? Not. I think, always it does NOT. I think, all the "primary" series in this case have radius of convergence zero. These are typical perturbation series; there is empiric observation that all non–trivial perturbation series diverge. (There is common confusion among physicists: many colleagues believe, that, at small value of the expansion parameter, the perturbation series converge; usually, it costs a lot of efforts in a hard discussion, to convince them, that it is not so.) Quote:Does this relate to the mittag-leffler expansion ? Do you mean http://en.wikipedia.org/wiki/Mittag-Leffler's_theorem ? It may give you some hints, but I do not think you can use it "as is" for tetration, because the tetration is not a meromorphic function. Best regard, Dmitrii. tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 02/15/2015, 05:23 PM Something different. Maybe I should read somewhere again but I wonder how tetration is related to (laser) physics ? My english is not perfect and neither is my russian. Also my knowledge of lasers and electromagnetism is limited. SO ... laser tetration for dummies ? regards tommy1729 Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 02/16/2015, 12:03 AM (02/15/2015, 05:23 PM)tommy1729 Wrote: Maybe I should read somewhere again but I wonder how tetration is related to (laser) physics ?Formalism of superfunctions can be used both, for evaluation of tetration and in (laser) physics. I wanted to use tetration for representation of large numbers; in particular, for the normalisation factor of approximations of the multi-particle wave function; in particular, for the Bose–Einstein condensate. Lasers are used in the preparation. That time, I was surprised, that tetration is not yet developed, so, I upset with mathematicians. I think, they were supposed to do this job in the past century. (Say, in 1950; perhaps, the bolsheviks and nazi just killed the most of Russian and German mathematicians, who could do that). So, I did the computational part by myself for natural tetration. Then Henryk asked me to do the same for various values of base... If you are interested in the history of science, you may trace this process in details by the list of my publications.. Then I had declared, that I can construct a superfunction for any growing holomorphic function. Including the transfer function of a laser amplifier: D.Kouznetsov. Superfunctions for amplifiers. Optical Review, July 2013, Volume 20, Issue 4, pp 321-326. http://link.springer.com/article/10.1007...013-0058-6 Quote:My english is not perfect and neither is my russian. Which language should I convert my Book after to finish the English version? Que idioma habla Usted? あなた　の　くに　は　なに　ですか？ Quote:Also my knowledge of lasers and electromagnetism is limited. The Soviet Veterans often use such arguments in the discussion... Hope, you are not of that kind. Well, indicate, please, the first place in my publications, that you cannot understand, and I shall try to explain it at your level. I mean both, your level in Russian/English, and your level in electromagnetism. And, also, at your level in the general philosophy. While, you may look also at the "TORI axioms". I use them all the time. Let me understand better your level. Have you graduated from some elementary school? Do you know fundamentals of Algebra? How about mathematical analysis? Did you hear about the Cauchi integral? Quote:... laser tetration for dummies ? This is an error. I do not suggest tetration for laser science. I suggest superfunctions for laser science. Perhaps, there is misprint somewhere. Let us correct it. Could you indicate, please: Where did you see "laser tetration"? Could you provide the URL? mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 02/16/2015, 12:16 AM (This post was last modified: 02/16/2015, 12:20 AM by mike3.) (02/16/2015, 12:03 AM)Kouznetsov Wrote: (02/15/2015, 05:23 PM)tommy1729 Wrote: Maybe I should read somewhere again but I wonder how tetration is related to (laser) physics ?Formalism of superfunctions can be used both, for evaluation of tetration and in (laser) physics. I wanted to use tetration for representation of large numbers; in particular, for the normalisation factor of approximations of the multi-particle wave function; in particular, for the Bose–Einstein condensate. Lasers are used in the preparation. That time, I was surprised, that tetration is not yet developed, so, I upset with mathematicians. I think, they were supposed to do this job in the past century. (Say, in 1950; perhaps, the bolsheviks and nazi just killed the most of Russian and German mathematicians, who could do that). So, I did the computational part by myself for natural tetration. Then Henryk asked me to do the same for various values of base... If you are interested in the history of science, you may trace this process in details by the list of my publications.. Then I had declared, that I can construct a superfunction for any growing holomorphic function. Including the transfer function of a laser amplifier: D.Kouznetsov. Superfunctions for amplifiers. Optical Review, July 2013, Volume 20, Issue 4, pp 321-326. http://link.springer.com/article/10.1007...013-0058-6 I am curious: how large are these numbers which occur in this Bose-Einstein condensate physics, which require tetration to write down? And also what the advantage would be of using a smooth and holomorphic tetration function (which is more complicated to evaluate) as opposed to just "iterated-EXP", i.e. notation of the form $\exp_b^n(x)$ for some whole-number $n$, base $b$ and $x$ in the range $[1, b)$, analogous to exponential scientific notation, which does not use the smooth-&-holomorphic extension of the exponential? tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 02/16/2015, 01:05 AM (02/16/2015, 12:03 AM)Kouznetsov Wrote: (02/15/2015, 05:23 PM)tommy1729 Wrote: Maybe I should read somewhere again but I wonder how tetration is related to (laser) physics ?Formalism of superfunctions can be used both, for evaluation of tetration and in (laser) physics. I wanted to use tetration for representation of large numbers; in particular, for the normalisation factor of approximations of the multi-particle wave function; in particular, for the Bose–Einstein condensate. Lasers are used in the preparation. That time, I was surprised, that tetration is not yet developed, so, I upset with mathematicians. I think, they were supposed to do this job in the past century. (Say, in 1950; perhaps, the bolsheviks and nazi just killed the most of Russian and German mathematicians, who could do that). So, I did the computational part by myself for natural tetration. Then Henryk asked me to do the same for various values of base... If you are interested in the history of science, you may trace this process in details by the list of my publications.. Then I had declared, that I can construct a superfunction for any growing holomorphic function. Including the transfer function of a laser amplifier: D.Kouznetsov. Superfunctions for amplifiers. Optical Review, July 2013, Volume 20, Issue 4, pp 321-326. http://link.springer.com/article/10.1007...013-0058-6 Quote:My english is not perfect and neither is my russian. Which language should I convert my Book after to finish the English version? Que idioma habla Usted? あなた　の　くに　は　なに　ですか？ Quote:Also my knowledge of lasers and electromagnetism is limited. The Soviet Veterans often use such arguments in the discussion... Hope, you are not of that kind. Well, indicate, please, the first place in my publications, that you cannot understand, and I shall try to explain it at your level. I mean both, your level in Russian/English, and your level in electromagnetism. And, also, at your level in the general philosophy. While, you may look also at the "TORI axioms". I use them all the time. Let me understand better your level. Have you graduated from some elementary school? Do you know fundamentals of Algebra? How about mathematical analysis? Did you hear about the Cauchi integral? Quote:... laser tetration for dummies ? This is an error. I do not suggest tetration for laser science. I suggest superfunctions for laser science. Perhaps, there is misprint somewhere. Let us correct it. Could you indicate, please: Where did you see "laser tetration"? Could you provide the URL? I want to thank you for your reply. However some comments. I dont know if you feel offended or something but your reply sounds a bit sarcastic and agressive. I hope you are in a good mood. Some quotes : ...so, I upset with mathematicians. I think, they were supposed to do this job in the past century. Which language should I convert my Book after to finish the English version? Que idioma habla Usted? あなた　の　くに　は　なに　ですか？ The Soviet Veterans often use such arguments in the discussion... Hope, you are not of that kind. Have you graduated from some elementary school? Did you hear about the Cauchi integral? (end quotes) Maybe I asked some silly questions , created some misunderstandings etc etc But I hope your not trying to offend me. Maybe you are joking and/or do not know me well. Anyway thanks for the answer. Peace. regards tommy1729 Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 02/16/2015, 01:47 AM (02/16/2015, 01:05 AM)tommy1729 Wrote: .. May be I asked some silly questions ..Do not worry. Ask that you find interesting and related to the topic. It is better to ask (and answer) 10 silly questions, than to miss one deep and important question. I try to answer all your questions. For me, it will be easier, if you ask questions one–by–one: then, the preview of my answer fits the size of the frame. Quote:But I hope your not trying to offend me. I do not try to offend you. I try to answer your questions in the best way I can. Quote:Maybe you are joking and/or do not know me well. .. This is common confusion: Colleagues think, that I am joking, while I am pretty serious. Richard Feynman and Dmitrii Becklemishev had reported similar confusions since past century. http://en.wikipedia.org/wiki/Surely_You'...r._Feynman! http://mizugadro.mydns.jp/t/index.php/Female_logic Yes, I do not know you. But I try to help you in understanding/using of superfunctions. Let me know, if any of my answers is wrong. Consider to answer my questions and formulate your questions. I appreciate any questions on the content of my book: they help me to make it better. Sincerely, Dmitrii. tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 02/16/2015, 01:56 AM thank you No questions anymore for now. regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 02/16/2015, 02:23 AM In your female logic link I detect 3-valued logic. I defend that idea. ( In math mainly ) I had many heated debates about it. regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 02/17/2015, 01:34 PM I would like to point out that Kouznetsov's conjecture that his Taylor(ln(x),x) type functions approximate (asymptotic) the superfunction is trivial. All +real-analytic functions are well approximated by Taylor or laurent series. A stronger version ( under conditions ) was the basic idea for fake function theory. It might however be true that this Taylor(ln(x),x) is a better approximation. regards tommy1729 « Next Oldest | Next Newest »

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