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 eta as branchpoint of tetrational sheldonison Long Time Fellow Posts: 630 Threads: 22 Joined: Oct 2008 06/07/2011, 05:19 PM (This post was last modified: 06/07/2011, 07:24 PM by sheldonison.) (06/02/2011, 02:04 PM)bo198214 Wrote: .... Ok, going with the base on the upper halfplane from e to sqrt(2). As long as we are not landed back on the real axis, the fixpoints are not conjugate, so the tetration value will not be real. But this is nothing new. Now when on the real axis: the Kneser method is not applicable to two real fixpoints. So we must take the value there as limit from above. The kneser tetration $b\mapsto b[4]p$ is real on the real axis $b>\eta$, which implies that $\overline{b} [4] p = \overline{b[4]p}$ (conjugation). So approaching from above or below is just conjugate to each other. So what one need to show imho is that the imaginary part will not tend to zero when approaching the real axis at $b<\eta$. I wonder whether Sheldon could supply some pictures of that, with his fourier algorithm. ....I made a picture of four different superfunction for B=sqrt(2). The first two are the two in your paper, with Dimitrii, generated from the lower fixed point L=2, and the upper fixed point, L=4. The third is the "new function", I posted about here. The fourth is the hypothetical Perturbed Fatout mapping, which is a merged upper/lower fixed point function. As imag(z) grows, this merged upper/lower Perturbed Fatou mapping gets arbitrarily close to the new function I generated, which is approaching the upper fixed point L=4 function, minus 1.05i, which is half the difference between the two periods. But, as imag(z) gets smaller and smaller, it approaches the lower fixed point function L=2 function. The way I drew it, the merger is occurring at imag(z)~=8.6i, where we have one of the gentle transitions from -4 to -2. Here, the magnitude of the two theta(z) functions would be comparable, but both are very small, with an first harmonic amplitude of ~1E-25. I would like to calculate the two theta(z) functions for Fatou(z), for a couple of harmonics, at least. Both theta(z) would have singularities, but there is a large overlap where both theta(z) functions are analytic. As you pointed out, this merged upper/lower Perturbed Fatou function wouldn't have Im(F(z))=0, at the real axis or any other horizontal line. I thought this might help the conversation. It would take me a day or two to calculate the upper and lower theta(z) values to generated this new merged upper/lower fixed point Fatou function. The plots have imag(z)=negative on the left, and imag(z)=positive on the right. $\text{Fatou_{\sqrt{2}}(z)=\text{Usexp_{\sqrt{2}}(z+\theta_U(z)+k)$ $\text{Fatou_{\sqrt{2}}(z)=\text{Lsexp_{\sqrt{2}}(z+\theta_L(z))$ $\Im(k)=(\text{Period}_U -\text{Period}_L)/2$ $\theta_U(z)$ decays to zero as imag(z) increases, with singularities at the real axis at integer values of z. $\theta_L(z)$ decays to zero as imag(z) decreases. It must also have singularities, as imag(z) increases, at approximately imag(z)~=18.1, but I don't understand that as well. - Sheldon     bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 06/07/2011, 05:47 PM (This post was last modified: 06/07/2011, 05:50 PM by bo198214.) Wow, Sheldon, thats such a great overview! Only one clarification: What means ~=? I guess it means not equal. but then your "new function" Im(z)=8.6 must read "Im(f(z))=0" instead of ~=0? It must be symmetric about the real axis. But I am really amazed how quickly you develop your code for new functions! Say, is your "new function" analytic continuable through base eta? sheldonison Long Time Fellow Posts: 630 Threads: 22 Joined: Oct 2008 06/07/2011, 05:54 PM (This post was last modified: 06/07/2011, 08:59 PM by sheldonison.) (06/07/2011, 05:47 PM)bo198214 Wrote: Wow, Sheldon, thats such a great overview! Only one clarification: What means ~=? I guess it means not equal. but then your "new function" Im(z)=8.6 must read "Im(f(z))=0" instead of ~=0? It must be symmetric about the real axis.I have to update that typo in my picture. Im(f(z))~=0 except at the real axis. For the NewFunc (Kneser mapping), Im(f(z))~=0 means that Im(f(z)) is wobbling very close to zero, but not equal to zero. Quote:But I am really amazed how quickly you develop your code for new functions! Say, is your "new function" analytic continuable through base eta? I haven't calculated the Fatou function yet, but now that I have drawn the pictures, it should be easy to calculate (easy compared to a Kneser Riemann mapping), and convergence precision should improve much much quicker than the Kneser Riemann mapping. It will still be an iterative mapping, iteratively generating each of the two theta(z) functions from the other superfunction, and the other theta(z) function. Because there is so much overlap where both theta(z) functions are analytic, no intermediate "sexp" function will be required to generate the mapping. Yes, the Kneser mapping "newfunc", for B-2, which will have singularities at z=-2,-3,-4 ..... Thanks for the compliments. - Shel sheldonison Long Time Fellow Posts: 630 Threads: 22 Joined: Oct 2008 06/08/2011, 02:14 PM (This post was last modified: 06/08/2011, 04:04 PM by sheldonison.) (06/07/2011, 05:47 PM)bo198214 Wrote: But I am really amazed how quickly you develop your code for new functions!Ok, here are the theta(z) values, for the Fatou(z) function,     I used these two equations, which are equal. $\text{Fatou_{\sqrt{2}}(z)=\text{Usexp_{\sqrt{2}}(z+\theta_U(z)+k)$ $\text{Fatou_{\sqrt{2}}(z)=\text{Lsexp_{\sqrt{2}}(z+\theta_L(z))$ Next, calculate thetau via this equation, near imag(z)=2i, by doing a Fourier series over a unit length, starting with thetaL(z)=0. $\theta_U(z)+k=\text{Usexp}^{-1}(\text{Lsexp(z+\theta_L(z))-z$ Now, using the theta_u from above, near imag(z)=15i, by doing a Fourier series over a unit length. $\theta_L(z)=\text{Lsexp}^{-1}(\text{Usexp(z+k+\theta_U(z))-z$ Repeat one more time, and the two series are consistent, and accurate to more than 67 decimal digits accuracy. Be careful to remember that thetaU decays to zero as imag(z) increases, and thetaL decays to zero as imag(z) decreases. $\theta_U(z) = \sum_{n=1}^{\infty} a_n\exp(2n\pi i z)$ and $\theta_L(z) = \sum_{n=1}^{\infty} a_n\exp(-2n\pi i z)$. I verified that there is a very large overlap where $\text{Usexp(z+\theta_U(z)+k)=\text{Lsexp}(z+\theta_L(z))$, accurate to 67 decimal digits, which is the accuracy of the superfunctions that I used. Theta(z) results are posted at 8.5715741i, half the average of the two Periods, where the two theta(z) values have equal magnitudes, and both are equally small. Note the a1 terms have nearly identical magnitudes, with a1_U ~= -conj(a1_L), to 50 digits accuracy. This is where imag(f(z)) goes from 4- to 2+ as z goes from -inf to +inf. At the real axis, thetaU has a singularity, at integer values of z. At the real axis, thetaL has very small terms, with a1~=-2E-50. ThetaL has a singularity, that I don't understand as well, around imag(z) = average of the two Periods. This method would also probably work for complex bases, where one fixed point is repelling, and the other fixed point is attracting. This might provide a mechanism to verify Mike's results for complex bases, posted here.. Also, this post is closely related to my attracting fixed point thread, where I also posted theta(z) mappings for B=sqrt(2). Code:ThetaU(z), starting with the constant "k"; the superfunctions used in these calculations were 67 digits accurate k=5.284046911275929509562319765 + 1.046500431344003802826235228*I a1= 5.132787355776188711993056404 E-27 - 3.625724477536479451525596757 E-25*I a2= 1.511792461497588143794162538 E-50 - 6.197254624020296328476658294 E-49*I a3= 4.741105856335653880260054426 E-74 - 1.519169304975871323751503497 E-72*I a4= 1.570350059077682233462666141 E-97 - 4.321977227564835332037557804 E-96*I a5= 5.396718213010231485424958665 E-121 - 1.334210621916187903869149050 E-119*I For ThetaL, I drove the "k" constant to zero. KL=~0=-1.3080789967888603754 E-68 + 2.880161851 E-67*I a1= -5.132787355776188711993056404 E-27 - 3.625724477536479451525596757 E-25*I a2=  8.268182523502499158985049574 E-51 + 2.060888780809025676827314260 E-49*I a3= -3.994533474813316741919755528 E-75 - 1.060362818161862613179451797 E-73*I a4= -2.431964709922672038589803344 E-99 + 4.894555977076092256343445907 E-98*I a5=  5.436928486837832888855701860 E-123 - 2.185866273307809932369827910 E-122*I mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 06/09/2011, 04:14 AM @sheldonison: Thanks for the coeffs. I'd like to see if I could make a color graph. What is the starting point you use for the upper regular superfunction? (i.e. the value at 0) sheldonison Long Time Fellow Posts: 630 Threads: 22 Joined: Oct 2008 06/09/2011, 02:30 PM (This post was last modified: 06/09/2011, 02:48 PM by sheldonison.) (06/09/2011, 04:14 AM)mike3 Wrote: @sheldonison: Thanks for the coeffs. I'd like to see if I could make a color graph. What is the starting point you use for the upper regular superfunction? (i.e. the value at 0)Hey Mike, $\text{Usexp}_{\sqrt{2}}(z)$ is developed from the L=4 fixed repelling point, with $\text{Usexp}_{\sqrt{2}}(0)=5.767053253764297762019157833944$. $\text{Lsexp}_{\sqrt{2}}(z)$ is developed from the L=2 fixed attracting point, with f(-1)=0, f(0)=1, f(1)=sqrt(2), and the "merger" of the two functions takes place at half the period of the LSexp=8.57i. In practice, one could literally say that for imag(z)<=8.57i Fatou(z)=Lsexp(z), and for imag(z)>=8.57i Fatou(z)=Usexp(z+k), ignoring any of the theta(z) terms, because the two functions are so very nearly identical where they merge. - Sheldon mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 06/09/2011, 10:11 PM (This post was last modified: 06/09/2011, 10:49 PM by mike3.) (06/09/2011, 02:30 PM)sheldonison Wrote: (06/09/2011, 04:14 AM)mike3 Wrote: @sheldonison: Thanks for the coeffs. I'd like to see if I could make a color graph. What is the starting point you use for the upper regular superfunction? (i.e. the value at 0)Hey Mike, $\text{Usexp}_{\sqrt{2}}(z)$ is developed from the L=4 fixed repelling point, with $\text{Usexp}_{\sqrt{2}}(0)=5.767053253764297762019157833944$. $\text{Lsexp}_{\sqrt{2}}(z)$ is developed from the L=2 fixed attracting point, with f(-1)=0, f(0)=1, f(1)=sqrt(2), and the "merger" of the two functions takes place at half the period of the LSexp=8.57i. In practice, one could literally say that for imag(z)<=8.57i Fatou(z)=Lsexp(z), and for imag(z)>=8.57i Fatou(z)=Usexp(z+k), ignoring any of the theta(z) terms, because the two functions are so very nearly identical where they merge. - Sheldon That is odd, since in the complex-base case you dug up that thread on, the "merger" looks to occur right along the real axis, just as it does for, say, base $e$. This makes me suspicious if this function is really the limit of the complex-base function for $b < \eta$. Unless, of course, the graphs are deceptive. Which is possible, given how "subtle" the curve differences are. And half the period, or half the average of the two periods? sheldonison Long Time Fellow Posts: 630 Threads: 22 Joined: Oct 2008 06/09/2011, 11:34 PM (This post was last modified: 06/09/2011, 11:58 PM by sheldonison.) (06/09/2011, 10:11 PM)mike3 Wrote: (06/09/2011, 02:30 PM)sheldonison Wrote: (06/09/2011, 04:14 AM)mike3 Wrote: @sheldonison: Thanks for the coeffs. I'd like to see if I could make a color graph. What is the starting point you use for the upper regular superfunction? (i.e. the value at 0)Hey Mike, $\text{Usexp}_{\sqrt{2}}(z)$ is developed from the L=4 fixed repelling point, with $\text{Usexp}_{\sqrt{2}}(0)=5.767053253764297762019157833944$. $\text{Lsexp}_{\sqrt{2}}(z)$ is developed from the L=2 fixed attracting point, with f(-1)=0, f(0)=1, f(1)=sqrt(2), and the "merger" of the two functions takes place at half the period of the LSexp=8.57i. In practice, one could literally say that for imag(z)<=8.57i Fatou(z)=Lsexp(z), and for imag(z)>=8.57i Fatou(z)=Usexp(z+k), ignoring any of the theta(z) terms, because the two functions are so very nearly identical where they merge. - Sheldon That is odd, since in the complex-base case you dug up that thread on, the "merger" looks to occur right along the real axis, just as it does for, say, base $e$. This makes me suspicious if this function is really the limit of the complex-base function for $b < \eta$. Unless, of course, the graphs are deceptive. Which is possible, given how "subtle" the curve differences are. And half the period, or half the average of the two periods?At half the period of the Lower Superfunction, all three functions, LSexp(z), Usexp(z+k), and Fatou(z) are nearly equal, differing by approximately 1E-25. At the real axis, the Fatou(z) differs from LSexp(z) by 1E-50. At imag(z) =(Uperiod+Lperiod)/2, the average of the two periods, we have Fatou(z) differing from Usexp(z+k) by about 1E-50. I actually have calculated all three of these functions to >100 decimal digits accuracy, to verify my results! If we had developed the two superfunctions of sqrt(2) on their other real valued section, at half their respective imaginary periods, than the merger between the two functions would have occurred at the real axis. But that's not what comes out of the limit equations, where we get the primary real axis of the two superfunctions. Base e is entirely different, since it involves a Kneser Riemann mapping, as opposed to the much simpler merger for Fatou(z). If you have a Kneser Riemann mapping, then the merger occurs at the real axis, which will have singularities at z=-2,-3,-4..... - Sheldon mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 06/10/2011, 12:20 AM (This post was last modified: 06/10/2011, 12:29 AM by mike3.) Now I was able to generate the graph! Yessss!!!! I give two graphs: the first one is the function as given, the second is its conjugate applied to the conjugate of z, i.e. $\bar{F(\bar{z})}$, which would be equivalent to "continuity from above" at the branch cut from $b = \eta$ to $b = -\infty$ in the extended Kneser/disturbed-Fatou tetrational. The second graph is there to better illustrate the continuation of the progression seen in that post about the complex-base attracting-repelling solution that I posted. main:     conj-conj:     The scale is -40 to +40 (WOW!) on both axes. Zero is in the center. Wow. It is soooo close to the attracting regular iteration on the real line... I could see why the regular seems so good now as an extension of tetration, and why it seems like it would create an "analytic" tetrational in the base when concatenated with the Kneser iteration at $b > \eta$. The singularity/branch point of tetration at base $\eta$ must be incredibly mild. Weeeeeeeeeeerd -- tetration seems once again to be the weirdest complex function I've ever seen. I notice something interesting here: the regions of "fractal" structure (where it gets really huge) cross across the imaginary axis. I noticed I was having a heck of a time trying to generate something like this with my continuum sum method. Perhaps that is why. There would have been huge values along the imaginary axis and so the "periodic approximation" I was using didn't work due to the ill behavior. sheldonison Long Time Fellow Posts: 630 Threads: 22 Joined: Oct 2008 06/10/2011, 01:54 PM (This post was last modified: 06/10/2011, 01:55 PM by sheldonison.) (06/10/2011, 12:20 AM)mike3 Wrote: Now I was able to generate the graph! Yessss!!!! .... The scale is -40 to +40 (WOW!) on both axes. Zero is in the center. Wow. It is soooo close to the attracting regular iteration on the real line... I could see why the regular seems so good now as an extension of tetration, and why it seems like it would create an "analytic" tetrational in the base when concatenated with the Kneser iteration at $b > \eta$. The singularity/branch point of tetration at base $\eta$ must be incredibly mild. Weeeeeeeeeeerd -- tetration seems once again to be the weirdest complex function I've ever seen.Looks great! Thanks for making the graph. I'm not sure I understand all of your comments though... regular tetration at a base>eta would be a complex superfunction, with no real valued real axis, right? - Sheldon « Next Oldest | Next Newest »

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