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 Dmitrii Kouznetsov's Tetration Extension bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 05/23/2008, 10:09 AM Kouznetsov Wrote:Quote:Quote:I have plotted the only one, $F$ such that $F(0)=1$. At $b=\sqrt{2}$, for example, $\lim_{x \rightarrow \infty} F_{\sqrt{2}}(x+{\rm i}y)=2$; $\lim_{x \rightarrow -\infty} F_{\sqrt{2}}(x+{\rm i}y)=4$. This contradicts $F(z^\ast)=F(z)^\ast$?No. $F_{\sqrt{2}}(x+{\rm i}y)$ approaches the limiting values at any real $y$. (in spite the cutlines!) Oh that was my mistake, I thought you are speaking about the limits for imaginary infinity. Quote:I tried to post the code as "source" together with the picture, but it was not accepted... Ya, give me the extension of the file so I can add it to the accepted file types. Otherwise you just can append a .txt to the end of the filename, then its accepted if not too big. Suffix .gz also always works up to 1MB. Quote:I know that usually the "self-made" codes dislike to run at another computer, so, let us do it with few steps, reproducing the figures one by one. Let us begin with very simple code. Please, reproduce first Figure 1 from the source posted at http://en.citizendium.org/wiki/Image:Exa...nLog01.png please, tell me how does it run at your computer and send me (or post) the resulting picture. It is supposed to be identical with the eps file I got at my computer. Gosh, you are really doing that in C! Creating eps in C, its unbelievable. But yes the code compiles with -lm and produce the same (at least visually) image. Quote:Yes, but they seem to be singular.. What example would you suggest to begin with? I would begin with what I am best with: tetration by regular iteration. Quote:"Equal to mine" is tetration that has no singularities at the right hand side of the complex halfplane. But only if your uniqueness conjecture is correct. Ok this demand can reduced to having no singularities in the strip $S$ of $0<\Re(z)\le 1$. Quote: I did not see any map of real and imaginary parts, nor those of modulus and phase. Tell me if I am wrong. Hm, indeed I think we have only real plots available, as I am not familiar with complex function visualization. However most tetrations considered here are given as a powerseries developed at 0. So its not so easy to continue them to the whole strip $S$. However for the regular tetration I have an iterative formula so this should be possible to make a plot similar to yours. And as I already explained, it is periodic along the imaginary axis, hence has no limit at $i\infty$. Does that mean that it is different to your's? Which value does your $F_{\sqrt{2}}$ assume at $i\infty$? Is it the nearest complex fixed point of $\sqrt{2}^z$? Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 05/23/2008, 02:15 PM bo198214 Wrote:... I would begin with what I am best with: tetration by regular iteration.It does not click.. May be I find it later.. Quote:Quote:"Equal to mine" is tetration that has no singularities at the right hand side of the complex halfplane. But only if your uniqueness conjecture is correct. Ok this demand can reduced to having no singularities in the strip $S$ of $0<\Re(z)\le 1$. You are right, the strip is enough. Quote:Hm, indeed I think we have only real plots available, as I am not familiar with complex function visualization. If we work together, soon you will be. Quote:However most tetrations considered here are given as a powerseries developed at 0. So its not so easy to continue them to the whole strip $S$. However for the regular tetration I have an iterative formula so this should be possible to make a plot similar to yours. Go ahead! I hope you can reproduce the period. Quote:And as I already explained, it is periodic along the imaginary axis, hence has no limit at $i\infty$. Does that mean that it is different to your's? No, that does not. The analytic tetrations should coincide. Does your soluiton show period $\frac{2\pi \rm i}{ \ln\big(\ln(\sqrt{2})\Big)+\ln(2)}\approx 17.1431 \!~\rm i$? Quote:Which value does your $F_{\sqrt{2}}$ assume at $i\infty$? Is it the nearest complex fixed point of $\sqrt{2}^z$? The previous code fails at $b \le exp(1/\rm e)$, because it needs such a value. I made another code for $b=\sqrt{2}$; it does not assume the limiting value. But it knows the asymptotic. Well, I see, you are ready for figure 2. I am cleaning up the code. While, you can read the description I have posted. Do you already have writing access at citizendium? For other users who watch only the last post, I repeat the url: en.citizendium.org/wiki/Usermitrii_Kouznetsov/Analytic_Tetration P.S. At the preview, part of url appears as "smile"; should be U s e r : D m i t r i i _ K o u z n e t s o v without spaces bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 05/23/2008, 03:47 PM (This post was last modified: 11/20/2008, 09:28 AM by bo198214.) Kouznetsov Wrote:bo198214 Wrote:... I would begin with what I am best with: tetration by regular iteration.It does not click.. May be I find it later.. Regular iteration is definitely something you have to learn about. The usual starting article for doing so is G. Szekeres, Regular iterations of real and complex functions. Acta Math. 100 (195, 203-258 However I think it is written somewhat cumbersome. If you speak French I advice you to read Jean Écalle, Théorie des invariants holomorphes, Publ. Math. Orsay No. 67–7409 Lars Kinderman has a very good collection of literature for the topic. Quote:Quote:And as I already explained, it is periodic along the imaginary axis, hence has no limit at $i\infty$. Does that mean that it is different to your's? No, that does not. The analytic tetrations should coincide. Does your soluiton show period $\frac{2\pi \rm i}{ \ln\big(\ln(\sqrt{2})\Big)+\ln(2)}\approx 17.1431 \!~\rm i$? Hm, not exactly, what I derived some posts earlier would give merely $\frac{2\pi i}{\ln(\ln(\sqrt{2}))}$, yours is $\frac{2\pi i}{\ln(\ln(2))}$. Btw. there are infinitely many analytic functions that have no singularity on right halfplane: If $F$ has no singularity there, also $F(x+c\sin(2\pi x))$, $0, is a superexponential with no singularities on the right halfplane and any 1-periodic function $p$ with $p(x)>-x$ in $0 would do. More interesting seems here the boundedness on the strip $0<\Re(z)\le 1$, which seems to be an universal uniqueness criterion. Quote:I made another code for $b=\sqrt{2}$; it does not assume the limiting value. But it knows the asymptotic. ah, ok. Quote:Do you already have writing access at citizendium? Hm, I signed up but did not get any response yet ... Quote:For other users who watch only the last post, I repeat the url: en.citizendium.org/wiki/Usermitrii_Kouznetsov/Analytic_Tetration P.S. At the preview, part of url appears as "smile"; should be U s e r : D m i t r i i _ K o u z n e t s o v without spaces You can include clickable urls into your post with the construction: Code:[url=http://....]how the link appears[/url] Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 05/23/2008, 04:35 PM bo198214 Wrote:Quote:No, that does not. The analytic tetrations should coincide. Does your soluiton show period $\frac{2\pi \rm i}{ \ln\big(\ln(\sqrt{2})\Big)+\ln(2)}\approx 17.1431 \!~\rm i$? Hm, not exactly, what I derived some posts earlier would give merely $\frac{2\pi i}{\ln(\ln(\sqrt{2}))}$, yours is $\frac{2\pi i}{\ln(\ln(2))}$.Bo, namely for $b=\sqrt{2}$ my expression can be simplified to yours. Quote:Btw. there are infinitely many analytic functions that have no singularity on right halfplane: If $F$ has no singularity there, also $F(x+c\sin(2\pi x))$, $0, is a superexponential with no singularities on the right halfplane and any 1-periodic function $p$ with $p(x)>-x$ in $0 would do. I disagree. At some smooth contour $u$ in your strip, the contour $v=u+c \sin(2\pi u)$ enters the left halfplane and crosses the cut $v<-2$. You have no need to evaluate any tetration in order to see it. Corresponding contour $F(v)$ is not continuous. Fig.2 is ready at http://en.citizendium.org/wiki/Tetration...rameters00. Try to run it. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 05/23/2008, 05:52 PM Kouznetsov Wrote:Bo, namely for $b=\sqrt{2}$ my expression can be simplified to yours. Yes you are right, again stupid mistake of mine. For $b=\sqrt{2}$ the fixed point is $a=2$ and the period is $\frac{2\pi i}{\ln(\ln(a))}=\frac{2\pi i}{\ln(\ln(2))}\approx -17.1431*I$. Good, so seems quite as if tetration by regular iteration gives the same as tetration with your approach. As it is too expensive to compute the contours, I give here pictures of the regular tetration as 3d plot, absolute value over the complex plane, here it is: over -1-I*10 ... +1+I*10     over 0 ... 1+I*20     One can clearly see the period of -17.14 along the imaginary axis. Quote:Quote:Btw. there are infinitely many analytic functions that have no singularity on right halfplane: If $F$ has no singularity there, also $F(x+c\sin(2\pi x))$, $0, is a superexponential with no singularities on the right halfplane and any 1-periodic function $p$ with $p(x)>-x$ in $0 would do. I disagree. At some smooth contour $u$ in your strip, the contour $v=u+c \sin(2\pi u)$ enters the left halfplane and crosses the cut $v<-2$. You have no need to evaluate any tetration in order to see it. Corresponding contour $F(v)$ is not continuous. You are right. As sin is not bounded in the real part on the strip it can go to arbitrary negative values. Quote:Fig.2 is ready at http://en.citizendium.org/wiki/Tetration...rameters00. Try to run it. ok, that works too (but as a c++ program, which I wasnt aware of). Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 05/23/2008, 11:03 PM Quote:As it is too expensive to compute the contours, ... One can clearly see the period of -17.14 along the imaginary axis. How long does it take, to plot $F(z)-F(z+T)$ ? Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 05/24/2008, 05:36 AM (This post was last modified: 05/24/2008, 06:29 AM by Kouznetsov.) Henryk, I hope you agree with Figure 2. Let us move to Figure 3. Please, run testfile of the fast plotter; tell me if the code for figure http://en.citizendium.org/wiki/Image:Con...xample.jpg runs well at your computer. You will need functions ado and conto for Figure 3. If it works, try fig.3 http://en.citizendium.org/wiki/AnalyticT...seSqrt2v00. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 05/24/2008, 09:43 AM (This post was last modified: 05/24/2008, 09:44 AM by bo198214.) Kouznetsov Wrote:Quote:As it is too expensive to compute the contours, ... One can clearly see the period of -17.14 along the imaginary axis. How long does it take, to plot $F(z)-F(z+T)$ ?     This is not in scale and took 1562 seconds. You see the difference is in the range of error. But I mean it was already the theoretic result, that it is periodic with $2\pi i/\ln(\ln(2))$ so no need to wonder. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 05/24/2008, 09:53 AM Kouznetsov Wrote:Henryk, I hope you agree with Figure 2. As long as you didnt have explained it I can not agree Quote:Let us move to Figure 3. Please, run testfile of the fast plotter; tell me if the code for figure http://en.citizendium.org/wiki/Image:Con...xample.jpg runs well at your computer. You will need functions ado and conto for Figure 3. If it works, try fig.3 http://en.citizendium.org/wiki/AnalyticT...seSqrt2v00. Yes works. If I understand that right, you have a certain procedure that creates the contour plot from a mesh of function values (I guess by somehow interpolating the gaps), at least it takes not additional computation to approach the function argument $z$ while approach a certain value $|f(z)|$. So then give me the next step of your Dmitrii-makes-contourplot-of-tetration-tutorial. Btw. if you allow me a very little criticism, your code is in the complete spirit of assembler, as if the world hadnt changed since that time Kouznetsov Fellow Posts: 151 Threads: 9 Joined: Apr 2008 05/24/2008, 11:24 AM (This post was last modified: 05/24/2008, 11:26 AM by Kouznetsov.) bo198214 Wrote:Kouznetsov Wrote:How long does it take, to plot $F(z)-F(z+T)$ ?This is not in scale and took 1562 seconds. You see the difference is in the range of error. But I mean it was already the theoretic result, that it is periodic with $2\pi i/\ln(\ln(2))$ so no need to wonder.Thanks, Henryk. I think, at least for $\ln(b)<1/\rm e$, our tetrations coincide. I still wonder why do you direct the axis down (at the top it is $\sim 5\times 10^{-10}$ and at the bottom $\sim 160\times 10^{-10}$), but anyway, I believe, you have many correct decimal digits. « Next Oldest | Next Newest »

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