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 Mizugadro, pentation, Book tommy1729 Ultimate Fellow Posts: 1,354 Threads: 328 Joined: Feb 2009 01/14/2015, 01:21 AM In your paper on pentation you define a2,a3. But what is a1 ? regards tommy1729 Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 01/14/2015, 04:08 AM (01/14/2015, 01:21 AM)tommy1729 Wrote: In your paper on pentation you define a2,a3. But what is a1 ?Perhaps, a1 is just unity. Sorry if I forgot to specify it. tommy1729 Ultimate Fellow Posts: 1,354 Threads: 328 Joined: Feb 2009 01/15/2015, 01:12 PM (This post was last modified: 01/15/2015, 01:20 PM by tommy1729.) Why isnt pentation defined as $pent(z) = sexp^{[z]}(x_0) =$ lim $sexp^{[n]}( L' ^z slog^{[n]}(x_0))$ Do you really believe lim $sexp^{[n]}(f(L,L',z)) =$ lim $sexp^{[n]}( L' ^z slog^{[n]}(x_0))$ where $f$ is a simple elementary function ? Afterall approximating slog^[n] with an elementary function seems wrong/divergent ? Lim $sexp^{[n]}(x+y)$ is usually very different from lim $sexp^{[n]}(x)$ even if $y$ is small or getting smaller with growing $n$. Also its not defined as lim $sexp^{[n]}(L' ^{z-n})$. I assume its (your def of pentation in the paper) meant as an acceleration of lim $sexp^{[n]}(L' ^{z-n})$. If not that would appeal weird and dubious to me. Is that acceleration really a big improvement ? Still reading and thinking , I dont have much time ... regards tommy1729 Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 01/15/2015, 03:09 PM First, thank, you, Tommy, for your interest in that article. I try to answer your questions, although I do not understand some notations. (01/15/2015, 01:12 PM)tommy1729 Wrote: Why isnt pentation defined as $pent(z) = sexp^{[z]}(x_0) =$ lim $sexp^{[n]}( L' ^z slog^{[n]}(x_0))$I think, your sexp is equivalent of my tet, and your slog is equivalent of my ate. The natural pentation pen in my book and in my article is defined through the following way: Some integer $M>1$ is chosen. $\varepsilon=\exp(kz)$ $f(z)=L_{e,4,0} + \sum_{m=0}^{M-1} a_m \varepsilon^m$ Where $L_{e,4,0}\approx -1.850354529$ is zeroth fixed point of the 4th Ackermann to base e, id est, real solution $x$ of equation $\mathrm{tet}(x)=x$. Increment $k$ and coefficients $a$ are determined by substitution of the asymptotic supertetration $F(z)=f(z)+O(\varepsilon^M)$ into the transfer equation $F(z+1)=\mathrm{tet}(F(z))$ I define $F_n(z)=\mathrm{tet}^n(f(z-n))$ $F(z)=\lim_{n\rightarrow \infty} F_n$ $\mathrm{pen}(z)=F(x_1+z)$ where $x_1$ is real solution of equation $F(x_1)=1$ My claim is, that resulting pen does not depend on the $M$ chosen. I define it in my way, because it seems to work. Quote:Do you really believe lim $sexp^{[n]}(f(L,L',z)) =$ lim $sexp^{[n]}( L' ^z slog^{[n]}(x_0))$ where $f$ is a simple elementary function ? No. I do not understand the notation. I would say, $\mathrm{tet}(z)=\mathrm{pen}(1+\mathrm{ape}(z))$ where $\mathrm{ape}=\mathrm{pen}^{-1}$ Is it that you mean? Quote:Afterall approximating slog^[n] with an elementary function seems wrong/divergent ? It depends on the skills of the colleague who makes the approximation. After to construct the approximation, the researcher should substitute it into the equation, calculate the residual and plot the map of the agreement. Then we'll see, how many decimal digits may it keep for any given argument. If the numerical test passes, we may ask Henryk to write the long and complicated proof for some "Aequationes Mathematicae". Quote:Lim $sexp^{[n]}(x+y)$ is usually very different from lim $sexp^{[n]}(x)$ even if $y$ is small or getting smaller with growing $n$. Also its not defined as lim $sexp^{[n]}(L' ^{z-n})$. What is L' ? I doubt if these equations are useful for the evaluation of tetration. Usually, such a limits are very slow to converge; I have not enough patience "to press a key, to have a tea". Quote:I assume its (your def of pentation in the paper) meant as an acceleration of lim $sexp^{[n]}(L' ^{z-n})$. If not that would appeal weird and dubious to me. I do not see any equivalent of this formula in my article. Nor in my book.. Quote: Is that acceleration really a big improvement ? Hmm... As soon, as anybody reproduces the complex maps from my Book or from my articles with any alternative algorithms, it will be possible to compare the efficiency. There is no reason to compare acceleration of a car to the acceleration of the unmovable rock. The only, we may say, that the car moves, while the rock does not. However, one can refer to the theory of relativity, quantum mechanics, etc., but this does not look serious. I repeat my old statement: Until now, nobody could calculate and plot complex maps of tetration (nor pentation) with any algorithm, faster or simpler than those I had presented. As soon, as you or anybody else present any algorithm, that does the same, we can plot the maps of the agreement and compare the algorithms. Tommy, if you do not have much time, then, do not collect many questions in a bunch. Ask as soon as you have formulated one first question. For me, it will be also easier to answer. Now, the preview does not fit one screen.. Best regard, Dima. tommy1729 Ultimate Fellow Posts: 1,354 Threads: 328 Joined: Feb 2009 01/16/2015, 12:31 AM Ok I understand now. But I fear I have to tell you your construction is not new. It's just a classical fixpoint method used on an nonstandard function. Your claim that it does not depend on M is thereby true and easy to prove. To give a big hint : suppose an analytic function f has a fixpoint at 0 with f ' (0) = Y > 1. THEN for any real k , as n goes to oo : pent(z) = sexp^[z](x_0) = f^[n] ( Y^(z-k) slog^[n+k](x_0) ). Or said differently for any distinct pair reals k_1,k_2 : f^[n] ( Y^(z-k_1) slog^[n+k_1](x_0) ) = f^[n] ( Y^(z-k_2) slog^[n+k_2](x_0) ). At least if both sides converge. Once you can see that , you will understand. SECOND HINT : plug in koenigs function. *** For those who can still follow , the real question is what if sexp(x-t) = x is a parabolic fixpoint ? Then how do we get GOOD convergeance for sexp(x-t)^[z]. Maybe thats not so hard either , but it seems a logical followup question. *** --- Again for those who still follow , My answer is illuminating but the convergeance speedup is not yet understood. --- I wonder about Eremenko's Conjecture regarding pentation. I rediscovered Eremenko's Conjecture as a kid , guess that explains it. regards tommy1729 MphLee Fellow Posts: 95 Threads: 7 Joined: May 2013 02/09/2015, 08:19 PM (01/11/2015, 05:01 PM)Kouznetsov Wrote: Quote:..I can't understand half of the contents.. Ask questions. I made also account at http://dmitriikouznets.livejournal.com for questions. But if with formulas, here seems to be easier. You may use also http://math.eretrandre.org/hyperops_wiki...Kouznetsov Thank you very much for the kindness, so I'm happy to accept your invite and ask you something! I wonder if you were able to explain to me some of your last results and achievements about tetration and the state of the research in laymen's terms... if is possible. Given my poor knowledge of dynamics and basic analysis is pretty painful for me to go through your papers without "losing my mind" : (1) for example how do you achieve uniqueness in easy words? I mean, what are the properties that the solution must satisfy and that allow us to derive/(imply) the uniqueness? (2) what is/are your algorithm/s for tetration? For the second question I did my best but I was never able to go through the overwhelming amount of formulas and find something.. Looking at Mizugadro's page of Tetration it says that three different algorithms are used for bases belonging to three different domains [A]$b\in ]1;e^{1/e}[$, [B]$b=e^{1/e}$ and [C]$b\in ]e^{1/e},\infty[$ (let's skip the complex cases) [A] For this case Mizugadro says that the related tetration function ${\rm tet}_b$( aka $1$-based supefunction) is constructed with regular iteration at smallest of the real fixed points $L$ of the function $\log_b$, even if I don't know much about how the regular iteration method works, it says that the function should be of the form $\displaystyle F(z) = L+\sum_{n=1}^{N} a_n {e^{kzn}} + o({e^{kxN}})$ ...but I'm lost before finding the values for $k$ and for the coefficients $a_n$ in the case of tetration... references of Mizugadro sends me to "D.Kouznetsov, H.Trappmann. Portrait of the four regular super-exponentials to base sqrt(2). Mathematics of Computation, 2010, v.79, p.1727-1756. " but the link is broken and  D.Kouznetsov. (2009). Solutions of $F(z+1)=\exp(F(z))$ in the complex $z$-plane. is really too complex for me... [B] For the case $b=e^{1/e}$ is says that a modified formula is needed $\displaystyle F(z)=\mathrm e\cdot\left(1-\frac{2}{z}\left( 1+\sum_{m=1}^{M} \frac{P_{m}\big(-\ln(\pm z) \big)}{(3z)^m} +\mathcal{O}\!\left(\frac{|\ln(z)|^{m+1}}{z^{m+1}}\right) \right) \right)$ But since I can't understand how to evaluate the coefficents $c_{m,n}$ of the polynomials $P_m$... I searched it in the reference [2] Trappmann-Kouznetsov-Computation of the Two Regular Super-Exponentials to base exp(1/e). Mathematics of computation, 2012 February 8. In this paper you (and Trappmann) say that we have to apply the regular iteration to the function $h(x)=e^x -1$ and since it is conjugated to $f(x)=(e^{1/e})^x$ by the function $\tau(x)=e(x+1)$ iterating $h$ gives us the iterates of $f$(hence tetration too) because $(\tau h \tau^{-1})^n =\tau h^n \tau^{-1}$. Later the text shows some known old methods (Lévy's, Newton limit formula, Fatou/Walker's ) and then (pag 9) the new expansion $\displaystyle F(z)=\mathrm e\cdot\left(1-\frac{2}{z}\left( 1+\sum_{m=1}^{M} \frac{P_{m}\big(-\ln(\pm z) \big)}{(3z)^m} +\mathcal{O}\!\left(\frac{|\ln(z)|^{m+1}}{z^{m+1}}\right) \right) \right)$ So if I'm not lost we have that for $\eta=e^{1/e}$ and $F(X_1)=1$ the following is true ${\rm tet}_{\eta}(z)=F(z+X_1)$ But again, what are the coefficents $c_{m,n}$? [C] Ok, here with Cauchi integral, "iterated Cauchi algorithm" I'm totally lost. What I've got is (reading Natural tet at Mizugadro) that superfunctions are identified up to a $1$-periodic function $\theta(x)$ but inside the equivalence class of supefunctions (up to the equivalence relation given by $f(x)=g(x+\theta(x))$ we can achieve uniqueness looking for the asymptotic approach to the fixed point (of the transfer function?) so, for example, also ${\rm tet}(z+\theta(x))$ is a superfunction of exp but it is not... the right one? If you could explain to me some of these points I would be infinitely grateful to you. Quote: Quote:About Pentation: l have to notice that long ago, in 2006, Rubtsov and Romerio [1] were able to compute a first approximation of the fixed point $L_{e,4,0}=-1.850354529...$. Their first approximation where denoted by $\sigma=-1.84140566...$ and $pent_e (-\infty)=sln(\sigma)={{}^\sigma}e=\sigma...$ (where pent is computed using their approximation). .. [1] Rubtsov, Romerio - Notes on Hyper-Operations, Progress Report -NKS forum III, Final review 3, 2006. http://math.eretrandre.org/tetrationforu...hp?aid=222 Thank you for the link, I add it to the article http://mizugadro.mydns.jp/t/index.php/Superfunction Do you understand, how do they calculate the tetration? I think, my algorithms are more efficient, because Rubtsov and Romerio do not present any complex map of tetration, nor pentation. Well, actually they didn't. Those values for sigma (the fixedpoint of natural tetration) are obtained with the fixed-point iteration method using approximations of natural superlogarithm (inverse of natural tetration) (linear and cubic). The linear approximation uses $x+1$ ($-1) while the Cubic approximation is defined as follow (See [i]"Andrew Robbins - Solving for the Analytic Piecewise Extension of Tetration and the Super-logarithm[i]") $SE_3(x)=-{2\over 13}x^3+{3\over 13}x^2+{12\over 13}x-1$ if $-1 $SE_3(x)=\exp(SE_3(x-1))$ if $0 So using the the linear approximation we get $\sigma_1\simeq -1.84140566043696...$ using the cubic one $\sigma_3\simeq -1.8497049770580847...$ so if we denote your approximation with $\sigma_K$ we have that $\sigma_3-\sigma_K \simeq -0.00064931.. ..$ close enough! MathStackExchange account:MphLee tommy1729 Ultimate Fellow Posts: 1,354 Threads: 328 Joined: Feb 2009 02/09/2015, 10:35 PM mainly @ mphlee I think Kouznetsov's results are mainly alternative computational methods , rather then new solutions or theoretical ideas. 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. I realise that does not completely answer your questions , so apologies for that. Im also sorry to lack large amounts of enthousiasm at this point. As for the case base > eta , I expressed doubt and skepticism relating the Cauchy integral method ( or whatever its called ). Im also somewhat annoyed by pseudocode algorithms rather than math notation , what makes analysis harder imho. The Cauchy integral method seems TO ME like an over or under determined set of equations. regards tommy1729 Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 02/10/2015, 12:07 AM Thank you, MphLee, for your interest. (02/09/2015, 08:19 PM)MphLee Wrote: .. I wonder if you were able to explain to me some of your last results..Yes, I shall try. Quote:(1) for example how do you achieve uniqueness in easy words? I mean, what are the properties that the solution must satisfy and that allow us to derive/(imply) the uniqueness? I postulate the holomorphism in so wide range as I can. I postulate the specific behaviour at infinity. These postulates allow the efficient evaluation. Quote:(2) what is/are your algorithm/s for tetration? For the second question I did my best but I was never able to go through the overwhelming amount of formulas and find something.. Looking at Mizugadro's page of Tetration it says that three different algorithms are used for bases belonging to three different domains [A]$b\in ]1;e^{1/e}[$, [B]$b=e^{1/e}$ and [C]$b\in ]e^{1/e},\infty[$ (let's skip the complex cases) Yes. Let us consider these three cases one by one. Quote:[A] For this case Mizugadro says that the related tetration function ${\rm tet}_b$( aka $1$-based supefunction) is constructed with regular iteration at smallest of the real fixed points $L$ of the function $\log_b$, even if I don't know much about how the regular iteration method works, it says that the function should be of the form $\displaystyle F(z) = L+\sum_{n=1}^{N} a_n {e^{kzn}} + o({e^{kxN}})$ ...but I'm lost before finding the values for $k$.. Let us find you. What happens, it you substitute the representation above into the transfer equation $F(z+1)=\exp_b(F(z))$ ? Please, express the left hand side and the right hand side as series with respect to the small parameter $\varepsilon=\exp(kz)$ and collect terms with the same power of $\varepsilon$. You may compare the result with general formulas (6.3) and (6.4) of my book "Суперфункции" for the specific transfer function $T=\exp_b$ Quote:and for the coefficients $a_n$ in the case of tetration... references of Mizugadro sends me to "D.Kouznetsov, H.Trappmann. Portrait of the four regular super-exponentials to base sqrt(2). Mathematics of Computation, 2010, v.79, p.1727-1756. " but the link is broken Please, type the source that indicates the wrong link and indicate, which ULR does not work. I shall try to fix it. Then we'll consider the other your questions. Sincerely. Dmitrii. Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 02/10/2015, 12:18 AM P.S., addition to my previous message: Quote:Quote:references of Mizugadro sends me to "D.Kouznetsov, H.Trappmann. Portrait of the four regular super-exponentials to base sqrt(2). Mathematics of Computation, 2010, v.79, p.1727-1756. " but the link is broken Please, type the source that indicates the wrong link and indicate, which ULR does not work. I shall try to fix it. .. I just found and corrected one misprint in one reference [1] in the article http://mizugadro.mydns.jp/t/index.php/Tetration May be, you mean that. Now it seems to work. If any other link does not work, let me know as soon as you find it. Sincerely, Dmitrii. MphLee Fellow Posts: 95 Threads: 7 Joined: May 2013 02/10/2015, 12:12 PM Ok, let's try it. LEFT SIDE Substituting it in $F(z+1)$ it gives $\displaystyle F(z) = L+\sum_{n=1}^{N} a_n {\varepsilon^n} + o(\varepsilon^N)$ $\displaystyle F(z+1) = L+\sum_{n=1}^{N} a_n {e^{k(z+1)n}} + o({e^{k(z+1)N}})$ $\displaystyle F(z+1)=L+\sum_{n=1}^{N}(a_n{e^{kn}})\varepsilon^n+ o(\varepsilon^N{e^{kN}})$ Let's pretend for a second that I know how the little-o notation works (even if I don't even know what it is), now we have a new series that instead of the coefficents $a_n$ has the coefficents $b_n=a_n e^{kn}$ that looks really similar to the one we started with $\displaystyle F(z+1)=L+\sum_{n=1}^{N}b_n\varepsilon^n+ o(\varepsilon^N{e^{kN}})$ LEFT --- RIGHT SIDE Now I substitute it in $T(F(z))$ $\displaystyle \exp_b(F(z)) = \exp_b(L+\sum_{n=1}^{N} a_n {\varepsilon^n} + o(\varepsilon^N))$ Im not really good with series but maybe I could use the definition of exponentiation (for T=exp_b) and obtain this $\displaystyle b^{ F(z)} =e^{\ln(b)\cdot F(z)}= \exp_b(L+\sum_{n=1}^{N} a_n {\varepsilon^n} + o(\varepsilon^N))$ $\displaystyle \sum_{n=0}^{\infty}{\ln(b)^i\over i!}F(z)^i= \sum_{n=0}^{\infty}{\ln(b)^i\over i!}\left (L+\sum_{n=1}^{N} a_n {\varepsilon^n} + o(\varepsilon^N) \right)^i$ At this point with my poor knowledge of the subject I can't continue, so looking at your book it says that series has the property that (6.4) $\displaystyle T(F(z)) = L+T^{[1]}(L)\sum_{n=1}^{\infty}a_n {\varepsilon^n} +{T^{[2]}(L)\over 2}\left(\sum_{n=1}^{\infty}a_n {\varepsilon^n}\right )^2+...$ The summation in (6.4) seems an infinite series because I saw the dots “...”, so if I understand well the formulas, assuming that the sequence 1,2,6... is the factorial and with transfer function $T=\exp_b$ (with $T^{[n]}=T^{[n]}(L)$) the final form should be something like that $\displaystyle T(F(z)) =L+\sum_{i=1}^{\infty}{T^{[i]}(L)\over i!}\left(\sum_{n=1}^{\infty}a_n {\varepsilon^n}\right )^i$ RIGHT --- So the question is how can be the left side the same as the right side? $\displaystyle F(z+1)=L+\sum_{n=1}^{N}b_n\varepsilon^n+ o(\varepsilon^N{e^{kN}})=L+\sum_{i=1}^{\infty}{T^{[i]}(L) \over i!}\left(\sum_{n=1}^{\infty}a_n {\varepsilon^n}\right )^i=T(F(z))$ So what the book is probably trying to say is that is possible to manipulate algebraically the RIGHT side in order to obtain a series of the form $\displaystyle T(F(x))=L+\sum_{n=1}^{\infty}\tau_n\varepsilon^n$ with the coefficients $\tau_n$ defined in the formulas (6.5), (6.6) and (6.7) of your book $\tau_1=T'(L)$ $\tau_2=T'(L)a_2+T^{[2]}(L)/2$ $\tau_3=T'(L)a_3+T^{[2]}(L)a_2+T^{[3]}(L)/6$ and probably $\tau_n=T'(L)a_n+T^{[2]}(L)a_{n-1}+...+T^{[n-1]}(L)a_2+T^{[n]}(L)/n!$ I don't know the rules that makes you able to turn the series on the RIGHT in a series with coefficients $\tau_n$ but if is possible and we assume that $T(F(z))=F(z+1)$ then we should have that $F(z+1)=L+b_1\varepsilon+b_2\varepsilon^2+b_3\varepsilon^3+...=L+ {{\tau}}_1 \varepsilon + {\tau}_2 \varepsilon^2 + {\tau}_3 \varepsilon^3+...=T(F(z))$ and thus $b_n=a_n e^{kn}={\tau}_n$. This makes us able to find $k$ $b_1=a_1 e^{k}=T'(L)=\exp'_b(L)=\ln(b)b^L$ so $k=\ln({\ln(b)b^L\over a_1})$ Is this correct? And since we have $k$ and $a_1$ the other $a_n$ are given by your formulas (6.8 ), (6.9) and (6.10)... Thank you alot for the effort and your help. I'm really curious about this but I'm also sorry if is so hard, I'm a donkey in analysis and im not good with powerseries as well. MathStackExchange account:MphLee « Next Oldest | Next Newest »

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