Mizugadro, pentation, Book MphLee Long Time Fellow Posts: 373 Threads: 29 Joined: May 2013 02/10/2015, 12:48 PM (This post was last modified: 02/10/2015, 12:49 PM by MphLee.) (02/09/2015, 10:35 PM)tommy1729 Wrote: For instance the solutions for bases between 1 and eta are unique. So basicly these are traditional fixpoint methods. I recently wrote about parabolic fixpoints and I assume your aware of the koenigs function. Truncating Taylor series by polynomials or other functions can lead to alternative ways of computation. mmhh... to be honest I don't know what is the Koenigs function. My knowledge is almost negative in this field (or immaginary xd). BTW I'm reading about it right now.. It actually says that if $f$ is a function with some properties (holomorphic mapping of the unit disck to itself... and so on) then there exist an unique holomorphic solution to the schroeder's equation $S(f(x))=f'(0)\cdot S(x)$ with $S(0)=0$ and $S'(0)=1$. Then this unique solution is called Koenigs function. So if the conditions of the Koenings theorem are satisfied by some function we are interested in it should give us an unique holomorphic solution to a schroeder equation, and since its conjugated with the abel equation we will get an unique Abel function as well... is this the point? When I have time I'll check the post you are talking about (parabolic fixedpoints) MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 02/10/2015, 02:40 PM (This post was last modified: 02/10/2015, 02:42 PM by Kouznetsov.) (02/10/2015, 12:12 PM)MphLee Wrote: Ok, let's try it. .. .. $k=\ln({\ln(b)b^L\over a_1})$ Is this correct?I think, you are doing in the correct way. I hope, you made no errors. Keep in mind, that $L$ is fixed point, id est, $b^L=L$ Do you have any software at your computer to check your deduction? I used Maple; then I found that it is not so good, http://en.wikisource.org/wiki/Maple_and_Tea and Mathematica does this better. In addition, I use C++ to plot graphics and the complex maps, and you can do the same. You may use my algorithms and codes, and you may write your own codes, it is better for your education. Keep in mind, that all codes have bugs. If you get some tens coefficients with Mathematica and use them in the C++ code, the resulting implementation may return of order of 14 significant figures, and allow to plot the complex maps in the real time. You may load the examples from http://mizugadro.mydns.jp/t/index.php/Category:Book http://mizugadro.mydns.jp/t/index.php/Category:C%2B%2B Try to write your own algorithms. Run the tests. Substitute the functions you get into the equations they are supposed to satisfy. Plot the residuals. Compare your calculus to my calculus. Try to make your algorithms more precise, than my ones. Try to make your algorithms shorter and simpler, than my ones. Try to make your algorithms faster, than my ones. Try to make your algorithms more general, than my ones. Citius, Altius, Fortius! Quote: .. I'm really curious about this but I'm also sorry if is so hard.. Do not worry. Keep doing. If you see any error in my book, let me know immediately. Ask a question as soon as you can formulate it. Your questions help me to organise better the English version of the Book. Sincerely, Dmitrii. Gottfried Ultimate Fellow Posts: 898 Threads: 130 Joined: Aug 2007 02/11/2015, 09:04 PM (This post was last modified: 02/11/2015, 09:25 PM by Gottfried.) Hello Dmitrii - I still wish to learn/understand your specific method. This time I followed the link to your wiki http://mizugadro.mydns.jp/t/index.php/Tetration and looked at the entry for the tetration to base b=exp(1/e). I do not understand how you arrive at the polynomials, and especially how at the set of power-series in the table with the increasing m. However, I could find the coefficients of the first row (m=0) by my standard procedure, and my method allows 12,16 or 20 correct digits for them. So if I knew, how our methods are related, and if they come out to be identical (at least for that exact base), then you could take something useful from my method, at least for the more accurate computation of the m=0 coefficients. The way how I arrive at that many digits is some well experienced method of Noerlund-summation of alternating divergent series. Unfortunately not many people are used to such methods and thus are unable to apply them if their projected coefficients stem from formal expressions which represent divergent series. Or from functions, whose formal expressions as powerseries have zero-range of convergence, but have taylor-coefficients with alternating signs - in the latter case they can in many nontrivial cases be summed for an actual given argument x or z outside the range of convergence. What I'm doing, in short, is using the concept of Carlemanmatrices. Carlemanmatrices are invented for the formal manipulation of composition (and thus also iteration) of functions which have a power series representation (=are analytic). For the given function f(x), the exponential with base exp(1/e) and its conjugation to some g(x) which is defined to satisfy f(x)=g(x/L-1)+1)*L (with L its fixpoint) the Carleman-matrix G is of special simple form (triangular with units in the diagonal) which allows easily computation of its (Matrix-)logarithms and thus arbitrary fractional powers by G^h = Exp(h * Log(G)) and where G^h contains then the coefficients of the power-series for the h'th iterate (g°h(x/L-1)+1)*L even for fractional or complex iteration-height h. Tetration in your terminology is then the evaluation at x=1 where the iterationheight h is given as the p+qi - argument in your description. If I leave -in my procedures - the iteration-height parameter h indetermined in the resulting two-variable power series, but evaluate for the case x=1 I arrive at the coefficients c very similar to that in your article, but can supply many more digits precision - however only because of the technique of Noerlund-summation for the series where the x=1 - argument is involved/evaluated. If this is interesting for you then ask for more information (I can also supply the Pari/GP-toolbox for that computations). See here for instance the approximations of the first few coefficients c (after my carlemanmatrix/Noerlundsummation - concept). With powerseries of 64 terms I get in the rows 59 to 64 the partial sums of the summations for the c-coefficients, when that summations are understood as summing columns of data. Here are the values: Code:...:  ... 59:  1.000000000000000  0.6110954537716500   -0.2317026144766614   0.09178128765991140    -0.03756492168250494    0.01577372205052486 60:  1.000000000000000  0.6110954537716508   -0.2317026144767069   0.09178128766078397    -0.03756492169126225    0.01577372210673970 61:  1.000000000000000  0.6110954537716513   -0.2317026144767332   0.09178128766130164    -0.03756492169661943    0.01577372214214125 62:  1.000000000000000  0.6110954537716514   -0.2317026144767481   0.09178128766160907    -0.03756492169989380    0.01577372216440198 63:  1.000000000000000  0.6110954537716515   -0.2317026144767563   0.09178128766179094    -0.03756492170189427    0.01577372217838127 64:  1.000000000000000  0.6110954537716517   -0.2317026144767610   0.09178128766189732    -0.03756492170311400    0.01577372218715045 ...: ... Where your coefficients were given as Code:m    c_{m,0}    c_{m,1}    c_{m,2}       c_{m,3}    c_{m,4} 0       1       0.61061   -0.23171       0.09225    -0.03757 1       0       0.69521    0.41315      -0.16027     0.07007 2       0      -0.57851    0.18323       0.49162    -0.15216 3       0       0.64730   -0.62933      -0.51128     0.51372 4       0      -0.84098    1.23261       0.42470    -0.97551 5       0       1.19090   -2.12653      -0.06895     1.57684 The data at m=0 match suspiciously well, however I've no further idea yet how to reproduce the rows at m=1,m=2 and so forth. Also, to exclude the possibility of a pure incidence/of a "false positive", it would perhaps be good if I could see more of the coefficients in that first row of your table... Gottfried Helms, Kassel Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 02/12/2015, 12:27 AM (02/11/2015, 09:04 PM)Gottfried Wrote: http://mizugadro.mydns.jp/t/index.php/Tetration and looked at the entry for the tetration to base b=exp(1/e). I do not understand how you arrive at the polynomials, and especially how at the set of power-series in the table with the increasing m.Thank you Gottfried for your interest. The coefficients in the table are calculate using the primitive fit, quick and dirty. Quote: However, I could find the coefficients of the first row (m=0) by my standard procedure, and my method allows 12,16 or 20 correct digits for them. Can you evaluate tetration to the complex base? Quote:So if I knew, how our methods are related, and if they come out to be identical (at least for that exact base), then you could take something useful from my method, at least for the more accurate computation of the m=0 coefficients. I think, first we should check, that your tetration has the same behaviour as my one. Can you reproduce any figures from my book "Суперфункции" or from the article about "holomorphic ackermanns" with your algorithm? I mean for some definite, fixed value of base. Quote: .. Code:...:  ... 59:  1.000000000000000  0.6110954537716500   -0.2317026144766614   0.09178128765991140    -0.03756492168250494    0.01577372205052486 ...Where your coefficients were given as Code:m    c_{m,0}    c_{m,1}    c_{m,2}       c_{m,3}    c_{m,4} 0       1       0.61061   -0.23171       0.09225    -0.03757 1       0       0.69521    0.41315      -0.16027     0.07007 2       0      -0.57851    0.18323       0.49162    -0.15216 3       0       0.64730   -0.62933      -0.51128     0.51372 4       0      -0.84098    1.23261       0.42470    -0.97551 5       0       1.19090   -2.12653      -0.06895     1.57684 The data at m=0 match suspiciously well, however I've no further idea yet how to reproduce the rows at m=1,m=2 and so forth. Also, to exclude the possibility of a pure incidence/of a "false positive", it would perhaps be good if I could see more of the coefficients in that first row of your table... Why do you think that it is suspiciously well? Only 3 decimal digits. ".. precision of such evaluation is not so high (perhaps, only two or three decimal digits are significant in the estimates above).." These coefficients are estimated from the "general" fit, formula (17.1) in book "Суперфунцкии". You may extract the implementation from the generator of figure 17.1 in the same book. Then you may differentiate it so many times as you like. I doubt, if more coefficients have any sense: the precision of the fit is poor. It is barely sufficient to plot the camera-ready version of figure 17.1, but if you zoom-in the figure with scaling factor of order of 1000, you'll see the defects. For the precise evaluation of coefficients you ask, the tetration should be already implemented for the complex base, as complex tet(complex b, complex z); Do you have the efficient algorithm for the evaluation? Could you implement it in C++? Then I think, we can evaluate the coefficients of the expansion with some 14 digits. The idea is simple: You evaluate the derivatives of tetration for base $b$ at some loop around the point of interest, $t_n(b,z) = \mathrm{d}^n \mathrm{tet}_b(z) / \mathrm {d} z^n$ and $t_n(b) =t_n(b,0)$ For example, $b_k=\exp(1/\mathrm{e}) + r~ \exp(2 \pi \mathrm{i}k/N)$ for some $r\approx 0.2$ and $N=1024$ Then, you may calculate the coefficients through derivatives of these $t_n$ at zero as the Cauchi integral, approximating it as a discrete sum with respect to $k$ Gottfried Ultimate Fellow Posts: 898 Threads: 130 Joined: Aug 2007 02/12/2015, 04:10 PM (This post was last modified: 02/12/2015, 04:17 PM by Gottfried.) Hi Dmitrii again - thanks for your answer! (02/12/2015, 12:27 AM)Kouznetsov Wrote: (02/11/2015, 09:04 PM)Gottfried Wrote: http://mizugadro.mydns.jp/t/index.php/Tetration and looked at the entry for the tetration to base b=exp(1/e). I do not understand how you arrive at the polynomials, and especially how at the set of power-series in the table with the increasing m.Thank you Gottfried for your interest. The coefficients in the table are calculate using the primitive fit, quick and dirty.Well, here I not even know what you might mean with "primitive fit". Can I find this in your Tori-wiki? If I find it I'd try this using Pari/GP to see myself, trying to use greater precision, whether my approach can be finetuned to arrive at your values at all. Quote:Quote: However, I could find the coefficients of the first row (m=0) by my standard procedure, and my method allows 12,16 or 20 correct digits for them. Can you evaluate tetration to the complex base? Well, this would only be the next step, I'm afraid. If I cannot reproduce your values in the simple (=real) case, why invest more work to implement the even more difficult complex bases... My problem is, just to find some firm way whether at all, and then how, I could reproduce your easiest coefficients with the procedures of my own language. Only if this is successful I'd go further and try complex bases too. (There will still be enough work for me to understand the method for the remaining bases, where you use Cauchy-integral and Riemann-mapping: still I've not yet the slightest idea what this really and practically is and how it is implemented - I'll still have a large corpus to study ...) (Snipping the rest of your answer here, just focusing on the first two aspects) Kind regards - Gottfried Ah, p.s.: instead of the word "preliminary" in the exp(1/e)-section - did you possibly mean "provisorical"? Gottfried Helms, Kassel tommy1729 Ultimate Fellow Posts: 1,859 Threads: 402 Joined: Feb 2009 02/13/2015, 12:45 AM 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. Are all parabolic fixpoint expansions (for iterating analytic functions) like this ?? what about terms ln(x)^7 x^3 then ? regards tommy1729 Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 02/13/2015, 02:04 AM (02/12/2015, 04:10 PM)Gottfried Wrote: .. (02/11/2015, 09:04 PM)Gottfried Wrote: http://mizugadro.mydns.jp/t/index.php/Tetration and looked at the entry for the tetration to base b=exp(1/e). I do not understand how you arrive at the polynomials, and especially how at the set of power-series in the table with the increasing m... Well, here I not even know what you might mean with "primitive fit". Can I find this in your Tori-wiki?Yes. http://mizugadro.mydns.jp/t/index.php/Fit1.cin This file is cited at http://mizugadro.mydns.jp/t/index.php/Fi...0bx10d.png which is cited at http://mizugadro.mydns.jp/t/index.php/Tetration Today, I had some crash at Mizugadro. I have recovered it, but I did not find its reason. For the case if any similar happen again, there is copy at http://en.citizendium.org/wiki/File:Tetreal10bx10d.png Quote:If I find it I'd try this using Pari/GP to see myself, trying to use greater precision, whether my approach can be finetuned to arrive at your values at all. I think, first we should compare our results for some fixed value of base, for example, for natural tetration , or for b=sqrt(2). Then we may consider the derivatives with respect to base. Quote:.. Can you evaluate tetration to the complex base? .. Well, this would only be the next step, I'm afraid. If I cannot reproduce your values in the simple (=real) case, why invest more work to implement the even more difficult complex bases... My approach implies complex base and complex argument. You deal with the primitive fit, instead of to compare with advanced approximations, that are declared to provide 14 digits. If you want to deal with the Taylor expansion, you may begin with the Taylor expansion of natural tetration at zero. It is shown in figure 14.6 of my book and also loaded to TORI at http://mizugadro.mydns.jp/t/index.php/File:Vladi04.jpg http://mizugadro.mydns.jp/t/index.php/Vladinaiv49.cin Quote:Ah, p.s.: instead of the word "preliminary" in the exp(1/e)-section - did you possibly mean "provisorical"? Many times I saw word "preliminary" in the scientific texts, and I think, I know its meaning. But I do not know the meaning of word "provisorical". So, type "preliminary". I think, in the way I had mentioned, we may get of order of 14 digits for some tens by tens coefficients. However, first, tetration of the complex argument and complex base should be evaluated with similar precision. Fist, efficient algorithms of the evaluation and the tests, verifications. Then, any expansion you like. Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 02/13/2015, 02:25 AM (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 Gottfried Ultimate Fellow Posts: 898 Threads: 130 Joined: Aug 2007 02/13/2015, 10:09 AM (This post was last modified: 02/13/2015, 12:27 PM by Gottfried.) Hello Dmitrii - (02/13/2015, 02:04 AM)Kouznetsov Wrote: Quote:If I find it I'd try this using Pari/GP to see myself, trying to use greater precision, whether my approach can be finetuned to arrive at your values at all. I think, first we should compare our results for some fixed value of base, for example, for natural tetration , or for b=sqrt(2). Then we may consider the derivatives with respect to base.Well, here I see no problem; as you write that you use for that the regular tetration (which employs power series centered around the lower fixpoint (which is real)) - and that is nicely expressible by the Carlemanmatrix-algebra. So here we have automatically identity because the methods match even formally. That simple-to-formulate Carlemanmatrix-approach ( for the base 1

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