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 Dmitrii Kouznetsov's Tetration Extension bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 05/24/2008, 11:39 AM Kouznetsov Wrote:Thanks, Henryk. I think, at least for $\ln(b)<1/\rm e$, our tetrations coincide. I too think so, and above they can not coincide because regular iteration is not defined without fixed point, and development at non-real fixed points does not give real values on the real axis. Though I have to mention that sometimes differences occur at the level of $10^{-15}$ or even $10^{-24}$ especially if you compare regular iterations at different fixed points. Quote:I still wonder why do you direct the axis down (at the top it is $\sim 5\times 10^{-10}$ and at the bottom $\sim 160\times 10^{-10}$), Because I am just too lazy to properly orientate the drawing . Instead was more focussed to let it look nice, haha. Quote: but anyway, I believe, you have many correct decimal digits. I can arbitrarily increase the precision, it just takes longer then. Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 05/24/2008, 12:08 PM (This post was last modified: 05/24/2008, 12:09 PM by Kouznetsov.) bo198214 Wrote:Kouznetsov Wrote:Henryk, I hope you agree with Figure 2.As long as you didnt have explained it I can not agree I have updated the tertation for small b . The minimal of quasi-periods determines the periodicity; the side of longest quasiperiod (negarive real part) has cuts of analyticity. Please, indicate the first statement you consider as "not explained". Quote:So then give me the next step of your Dmitrii-makes-contourplot-of-tetration-tutorial. Please, enter into my code, extract few values and compare with yours. I expect you will see disagreement or order of one percent, even at point $z=0$. This is because I did not adjust well the misplacement along the real axis. Assume, my code evaluates some function $f(z)$. Solve numerically equation $f(x_0)=1$ and compare function $F(z)=f(x_0+z)$ to your tetration. Quote:your code is in the complete spirit of assembler, as if the world hadnt changed since that time You have no need to work much inside the "conto" function; the only, may be, suppress the message about "Copyleft"; it will run even faster. (just type // at the beginning of line). Map your tetration with my conto plotter. Ask questions if any doubt. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 05/26/2008, 07:01 AM (This post was last modified: 05/26/2008, 07:02 AM by bo198214.) Kouznetsov Wrote:bo198214 Wrote:Kouznetsov Wrote:Henryk, I hope you agree with Figure 2.As long as you didnt have explained it I can not agree I have updated the tertation for small b . The minimal of quasi-periods determines the periodicity; the side of longest quasiperiod (negarive real part) has cuts of analyticity. Please, indicate the first statement you consider as "not explained". Good with the accompanying text it becomes clear. One minor correction: In the left part of fig. 2 its not $T$ but $\Im(T)$ as the period is purely imaginary. Did you btw play around with other complex fixed points, differently from the one nearest the real axis? For the distribution of fixed points I even created a small "movie" here. Quote:Please, enter into my code, extract few values and compare with yours. But you didnt give me your code yet! I have the code for fig. 2 and the example of the contourplot. Quote:You have no need to work much inside the "conto" function; the only, may be, suppress the message about "Copyleft"; it will run even faster. (just type // at the beginning of line). Map your tetration with my conto plotter. Attached is the somewhat ameteurish result for base $sqrt{2}$. The red lines are $|f|=1,2,4$. Attached Files   expbs2_20x20_2.eps (Size: 24.44 KB / Downloads: 343) Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 05/26/2008, 09:03 AM (This post was last modified: 05/26/2008, 11:22 AM by Kouznetsov.) bo198214 Wrote:... In the left part of fig. 2 its not $T$ but $\Im(T)$ as the period is purely imaginary.Yes. Thanks! Quote:Did you btw play around with other complex fixed points, differently from the one nearest the real axis? [quote] Not yet. We shold calculate tetrations with various asymptotic behavior. [quote] For the distribution of fixed points I even created a small "movie" here. Good! Could you trace them? A movie would be good for evolution of the map of tetration as ln(b) increases. (Yet, I have only 3 stapshots: b=sqrt(2),b=2,b=e). Quote:But you didnt give me your code yet! Oops... I was sure I did. Now I see, I forgot. Sorry. Check now, codes for fig.3a and fig.3b should be there. Quote:Attached is the somewhat ameteurish result for base $sqrt{2}$. The red lines are $|f|=1,2,4$. Why do blue lines pass through integer values at the real axis? (You may reduce a little bit up-to-last argument in the conto function, and increase a little bit the last argument; contour lines will come closer to singulatity). Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 05/27/2008, 03:58 PM Henryk, I have posted the Gauss-legendre quadrature formulas at http://en.citizendium.org/wiki/GauLegExample/code. Could you check it? I hope, it is the last brick you need for evaluation of the contour integral. By the way, what happened with your application to the http://en.citizendium.org? I wanted you to post there maps of tetration for different fixed points... Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 05/28/2008, 08:58 AM bo198214 Wrote:.. If you speak French I advice you to read Jean Écalle, Théorie des invariants holomorphes, Publ. Math. Orsay No. 67–7409I agree to read French. Could you send please the pdf? bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 10/09/2008, 10:21 PM (This post was last modified: 10/09/2008, 10:25 PM by bo198214.) Just to throw something in, I computed the image of the Cartesian grid for the matrix power super exponential. This can directly be compared with Figure 7 in D.Kouznetsov. Analytic solution of F(z+1)=exp(F(z)) in complex z-plane. Mathermatics of Computation, in press (200, and it looks quite similar:     The picture shows the image of lines with constant real part. The thick red lines are -1,0,1,2. Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 10/10/2008, 01:17 AM (This post was last modified: 10/10/2008, 07:03 AM by bo198214.) bo198214 Wrote:Just to throw something in, I computed the image of the Cartesian grid for the matrix power super exponential. This can directly be compared with Figure 7 in D.Kouznetsov. Analytic solution of F(z+1)=exp(F(z)) in complex z-plane. Mathermatics of Computation, in press (200...     1. I attach the overlap of the figures mentioned. The deviation (if any) is smaller than width of lines; the agreement at least with 2 significant decimal digits. 2. Is the superexponential implemented in the Matrix Power software? Can we cite it? (for some reason, the image does not load by itself, I have to clisk the url...) Moderator's Update: Image now visible Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 11/20/2008, 01:31 AM (This post was last modified: 11/20/2008, 02:29 PM by Kouznetsov.) bo198214 Wrote:Thats always the difficulty with those numerical comparison. If the difference is small does that mean that they are equal?Henryk asked me to calcilate my tetration with displaced basic interval. In the paper http://www.ils.uec.ac.jp/~dima/PAPERS/2008analuxp99.pdf , the basic algorithm evaluates tetration($z$) in the range $|\Re(z)|<1$; precisely (14 digits) in the range $|\Re(z)|<0.8$; I use it for $|\Re(z)| \le 0.5$, and bring to this interval other values using the recurrent relation. Then, I shifted the interval; the new implementation evaluates the tetration for $-1.5\le \Re(z) \le 0.5$. I run the test; it is precise for $-1.25\le \Re(z) \le 0.25$. I use it for $-1\le \Re(z) \le 0$, bringing other cases to this intergal. The plots for old algorithm and for the new one look identical. Then I plot the difference $f=f(z)=\mathrm{(old-new)}*10^{14}$.     The grid covers the range $-10\le\Re(z)\le 10$ $-4\le\Im(z)\le 4$ with unity step. Levels $\Re(f)=-2$ are shown with thick pink lines Levels $\Re(f)=-1$ are shown with thin pink lines Levels $\Re(f)= 1$ are shown with thin green lines Levels $\Re(f)= 2$ are shown with thck green lines Levels $\Im(f)=-2$ are shown with thick red lines Levels $\Im(f)=-1$ are shown with thin red lines Levels $\Im(f)= 1$ are shown with thin blue lines Levels $\Im(f)= 2$ are shown with thck blue lines If variation of function esceed 10 per step of the mesh, then the line is not drawn; so, the right hand side of the figure in vicinity of the real axix left blank; but there, the deviation between these two functions is larger than 1.e-14 and, perhaps, even larger than unity; the only the plotter cannot identify the position of lines, tey are too dence. I see the jumps of the difference at the integer and at half-integer values of the real part; the jumps are at the level of $10^{-14}$. Each of the implementations has its own errors, and they can be revealed comparing two finctions. I attribute the humps at the half-integer values to the old algorithm, and jumps at the integer values to the new algorithm. Comparing these two algorithms, I cannot say that one of them is somehow better than another one. Conclusion: In order to see from numerical evaluation, that two functions are not the same, you should plot them in the complex plane. It is better than to compare the Tailor series. « Next Oldest | Next Newest »

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