Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) +--- Thread: Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) (/showthread.php?tid=163) Pages: 1 2 3 Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - Ivars - 05/19/2008 I wondered what results can be obtained if we calculate 2^I , 2^(2^I) , etc. $2^I = e^{I*\ln2}$ $2^{(2^I)} = 2^{e^{I*\ln2}}$ $e^{I*\ln2}= cos (I*\ln2)+I*sin(I*\ln2)$ $2^{2^I}= 2^{cos (I*\ln2)+I*sin(I*\ln2)}$ etc up to infinity of power tower. Also, another interesting bases: $\phi^I, \phi^{\phi^I}...$ Golden mean may be interesting since $I/2=sin(I*\ln(\phi))$ $e^I, e^{e^I}, e^{e^{e^I}}, ...$ $\Omega^I,\Omega^{\Omega^I}...$ where $\ln(\Omega)=-\Omega$ And in general a and z as basis. I checked with Pari precision 200: $\Omega^{\Omega^I,...}=0.68000247...$ $\ln2^{(ln2^{(ln2^I)}...} = 0.757558...$ ${1/e}^{1/e^{1/e^{...I}}} = 0.567143...= \Omega$ but $\Omega^{1/\Omega}= 1/e$ so this iteration where it converges leads to numbers whose self roots is base a - for real numbers - as IF there was no I on top.. Like normal tetration Would You know of anyplace where such results would be published, e.g graphically on complex plane? Ivars RE: Calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - bo198214 - 05/20/2008 Ivars Wrote:I wondered what results can be obtained if we calculate 2^I , 2^(2^I) , etc. $2^I = e^{I*\ln2}$ $2^{(2^I)} = 2^{e^{I*\ln2}}$ $e^{I*\ln2}= cos (I*\ln2)+I*sin(I*\ln2)$ There is a little error here: $e^{i\varphi} = \cos(\varphi) + i\sin(\varphi)$ There is no i contained inside the cos and sin. Otherwise I dont know, looks pretty chaotic to me, so why not taking a fractal explorer (for example chaos pro) putting in the formula z_0=I, z=c^z (c is the to be colored point in the plane) and show us the results? So you can see how it behaves at all points in the plane not only your favourite values. RE: Calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - Ivars - 05/20/2008 Oops, thanks I am thinking something else and doing paralel calculations in my head, stupid mistake. Anyway, thanks for the link, I will try the programm You suggested as i of course wanted to know distribution over values of z ( or real a) , but did not know how to get there. Ivars RE: Calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - Ivars - 05/20/2008 I did the fractals for the iteration in this thread, with ChaosPro. As these are the first fractals I make in my life, I hope I have not made some stupid mistake. Anyway, the fractal on complex plane ( unfortunately, I could not map coordinates with the program) exibits very rich behaviour, but as it it module grows very fast, even setting bailout value at 10E200 leaves out many places. I will start with general picture on area 10E12*10E12 and 10E6*10E6, then zoom closer to the center of the picture to Area 5*5, then show very interesting neighboroughoods around $e^{-e}, e^{1/e}, e^{\pi/2}$, (last one in next post). In big scales, interesting to note the three lines making simmetric angles. In smaller, there is a whole lot of creatures. 4*10E12*10E12: [attachment=353] 4*10E6*10E6: [attachment=354] x=[-5;5], y=[-5,5]: [attachment=355] Area around $e^{-e}$ [attachment=356] Area around $e^{1/e}$ [attachment=357] It was worth it. Ivars RE: Calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - Ivars - 05/20/2008 Few more fractal pictures from z^(z^(z^(z^...........(z^I) Very interesing neighboroughood of e^pi/2, with 4 armed spirals. This part seems to be infintely fractal, and quite complex. I wonder of course it what we see is structure of I ( since I^(1/I)= e^(pi/2)) as this map gives numbers whose selfroot is z.At least for real base there is no difference with infinite tetration h??? And if we add few more I on top of infinite power tower, it will not change anything? Fine region around $z=e^{\pi/2}$ [attachment=358] UltraFine region (another 1000 times zoomed) around $z=e^{\pi/2}$ [attachment=359] There also few interesting points where many rays converge to, one of them is x=1.998.. , Y= 1.1972.. Of course, exact values can only be obtained by higher bailout settings. [attachment=360] Also, I add 2 regions near 0 [attachment=361] and -1 , also interesting patterns. [attachment=362] Thanks Henryk for helping me to learn about fractal images.I think this may somehow also help to understand what "imagination" does Ivars RE: Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - Ivars - 05/20/2008 And this little scary thing reminding of embrio in certain stages of development is the large scale repeating structure when program reaches its limits: [attachment=363] Ivars RE: Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - Ivars - 05/20/2008 I added one more I on top, so now iteration becomes: z_0=I^I z=c^z I got a symmetric picture of fish in large scale. [attachment=364] Ivars RE: Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - bo198214 - 05/21/2008 RE: Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - Ivars - 05/21/2008 Hi Henryk But why would a^(a^(a^.......a^I) lead to the same limit as h(a), namely w such that a=w^(1/w) ? The way it goes step by step is totally different because iteration starts as complex number.... So we can map the trajectories, how they converge, whereas in purely real number case we can not. why: $a^{(a^{(a^{...(a^I)}}}= h(a)$??? I do not understand this, and is this even true? On otherhand, it may be a useful tool to see the differences between various real bases a as trajectories will be different. Can You please tell me how can I plot these trajectories for e.g. integer iterations? I am ready to learn another piece of software, perhaps PARI or even SAGE- this map is so simple that programming should not be too difficult. Ivars RE: Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - Ivars - 05/22/2008 Since this is my first fractal map, it is interesting to see another different pattern ending at another important point on real axis: $e^{-\pi/2}$: [attachment=368] Ivars