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 Infinite tetration of the imaginary unit Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 02/14/2008, 07:54 PM [quote=Gottfried] [update] The nonconvergent trifurcation of partial evaluation with b=1.8 *i. The actual fixpoint is repelling [attachment=250] Excuse me for a question without going deep into those beautiful explorations- have You switched the axis (real-> imaginary) as point seems to have moved exactly if You had. Ivars Gottfried Ultimate Fellow Posts: 755 Threads: 115 Joined: Aug 2007 02/14/2008, 08:13 PM Ivars Wrote:[quote=Gottfried] Excuse me for a question without going deep into those beautiful explorations- have You switched the axis (real-> imaginary) as point seems to have moved exactly if You had. Ivars Hmm, funny, the point t = 0.385429071712 + 0.465126311482*I works well with b=1.8*I b^t = 0.385429071712 + 0.465126311482*I Did you get another one, using lambert-w? Gottfried Gottfried Helms, Kassel Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 02/14/2008, 08:25 PM No, no, just graphically that fixpoint looked to be symmetric to usual value if infinite tetration of i =infinite[4]i = 0.438283+i*0.3605924, but it is of course not once You gave a numerical value. Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 02/14/2008, 09:12 PM (This post was last modified: 02/14/2008, 09:14 PM by Ivars.) Quote:Your question concerning the selfroot, if intended as the b-solution of y = b^y, is crucial. Either this solution is exclusively given by b = selfrt(y) = y^(1/y), and in this case we have to explain why the "yellow zone" appears, or the b-solution of y = b^y is not the selfroot alone, but it is accompanied by other "functions" or branches. In the second case the explicitation (extraction of b) in y = b^y would have (at least) two branches, the selfroot and the perimeter of the yellow zone. Its inverse, as I see it, must have (... at least) four branches, as it is shown by a "wild" graphical inversion. GFR I just thought that given the fact that each nth odd root of a real y has 1 real and n-1 conjugate complex values, we can also study substitution: y= nth odd root of X; Then selfroot becomes b= (nth odd root of X ) of (nth odd root of X). If n =3 we automatically get at least complex 4 values of b in addition to 1 real: 1) b = (a+id ) 3rdroot X) of (( a+id) 3rdroot X) 2)b = (a-id ) 3rdroot X) of (( a+id) 3rdroot X) 3) b = (a+id ) 3rdroot X) of (( a-id) 3rdroot X) 4) b = (a-id ) 3rdroot X) of (( a-id) 3rdroot X) 5) b= real 3rd root(X) of real 3rd root of X. for n=5, we will have more additonal complex selfroots for b if n=7, etc. Since any real y is nth odd root of something, those n-1 complex values of y for EACH n are always there, ready for usage, if needed. Odd numbers are good as these roots does not involve negative numbers, but even in case of a negative root, it can always be expressed as multiplication of 2 imaginary subroots, so in the end there is also 1 real root and n imaginary roots = odd number totally: e.g square root of X is: + sgrt x - sgrt x = -i*4th real root(X)* -i* 4th real root(X) = +i*4th real root(X)*i* 4th real root(X) so again, we have 3 roots, whose combinations create different selfroots of Y=sqrt(X) (somehow). sgrtx -I*4th real root(X) +i* 4th real root(X) Of course, this can be extended infinitely, but may that is the whole point? That every finite real number is a finite root of some other number, meaning it has infinite number of invisible complex "companions". Ivars Gottfried Ultimate Fellow Posts: 755 Threads: 115 Joined: Aug 2007 02/14/2008, 11:51 PM (This post was last modified: 02/15/2008, 12:40 AM by Gottfried.) Ioannis kindly gave an approximation for the first of the two questions, Quote:First: what is the value b = x*i, 1.7 < x < 1.75, where convergence suddenly disappears? which seemed to be improvable, so I use that here. At b~ 1.71290*I there seem to be a limit-case; from the plot it seems, the trajectory connects then the three tri-furcation-branches, which would then give an "aequator"-trajectory (but the current value may not be a good enough approximation to the true limit-case)     The points of the three branches do not match, so the question is then, whether in the limit the equator is a smooth or even continuous line... New question to investigate. Btw b/2 is 0.85645*I , a number which looks somehow familiar to me, but don't have a clue yet. Gottfried Gottfried Helms, Kassel Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 02/15/2008, 05:35 PM I think this belongs here as well , as a result involving only I: One more interesting value is h(i*(e^(pi/2))) = h(I*(I^(1/I)). And it is: h(i*i^(1/i)) = h(i^((1/i)+1)) = h( i^(1+i)/i)) = 0,213934198848366+I*0,213934198848366 So that form is a+i*a, and Arg is pi/4, so in exponential form: h( i^(1+i)/i))= +-0,30254864546678200*e^(I*pi/4). May be I have mixed some sign for imaginary part. Ivars Gottfried Ultimate Fellow Posts: 755 Threads: 115 Joined: Aug 2007 06/19/2011, 09:22 PM (This post was last modified: 06/20/2011, 12:11 AM by Gottfried.) Today I reviewed this thread and thought I could add some more info here. First, I'd say, that the base b0~1.71290*I as suggested by Ioannis (Galidakis) in the thread sci.math (200 indeed provides an orbit which is an "aequator". I call it aequator since it seems, that it doesn't diverge to infinity nor to a triple of separated fixpoints nor to one fixpoint.(Compare the first pictures in this thread which show the behave of bases greater and smaller than b0) This base b0 is thus one other interesting complex constant in the study of iteration of exponentials; perhaps we may call it "aequator-constant". The computation due to Ioannis (translated to Pari/GP): Code:b=I*solve(x=1,2,abs(LW(-log(b*I)))-1)  \\LW is the lambert-W-function (implemented using wikipedia-code)    \\   b=1.7129360403744179818*I t = h(b) \\ 0.3920635 + 0.4571543*I    by h()-function x=0 data = vectorv(3*210,r, x=b^x); That this base provides an aequator is also backed by another new observation: if the initial point x0, from where the iteration starts, is nearer to the fixpoint, say x0=0.5 then we still get an aequator-line, and even more smooth - the orbit approximates a circle around the fixpoint: without approximating or diverging. Here are pictures for some other starting points of iteration. (The base is always the same). The complex point x0=1.0+1.0*I was taken: The complex point x0=0.9+0.9*I was taken: The complex point x0=0.7+0.7*I was taken: The complex point x0=0.5+0.5*I was taken: Now it was interesting, whether the original line, x0=0 is also an equator. I computed a lot of iterates, but the empirical result is inconclusive. There are three critical places, where (for higher bases) the entries to the divergence to the triple of oscillating fixpoints reside. I even guess, that we get something like a snowflake-curve, not smooth like with the other startingpoints x0. The complex point x0=0.0+0.0*I (the "original"/reference) was taken: Here I took two computations (and two colors) for the plot because of need of lots of tons of iterates to get a clue of the shape in the sparse regions. The blue points are the first about two-thousand iterates, with an implicte float-accuracy of 400 decimal digits. We see, that there are very sparse regions. After that I switched to 800 and 1200 digits precision and went up to 80000 iterates - no clue, whether the errors accumulate to something horrible. From that 80000 iterates I deleted all in the inner area so I took only the points in the sparse areas into the plot (red dots). It seems, the values are not completely messed, since in principle the blue and red points join reasonably to one curve. Hmm... what does this tell me? For instance: is that aequator through x0=0 asymptotically dense? Is it indeed an aequator at all? Is it enclosed in some disk? For another instance: what does this mean in regard to fractional iteration? Assume an orbit with a starting value x0 inside the limiting aequator (through x0=0), say through x0=0.5+0.5*I which seems to fill a smooth, closed line densely. Are also all fractional iterates on that curve? Gottfried Gottfried Helms, Kassel sheldonison Long Time Fellow Posts: 623 Threads: 22 Joined: Oct 2008 06/20/2011, 05:27 AM (This post was last modified: 06/20/2011, 05:50 AM by sheldonison.) (06/19/2011, 09:22 PM)Gottfried Wrote: The complex point x0=0.0+0.0*I (the "original"/reference) was taken .... Here I took two computations (and two colors) for the plot because of need of lots of tons of iterates to get a clue of the shape in the sparse regions. The blue points are the first about two-thousand iterates, with an implicte float-accuracy of 400 decimal digits. We see, that there are very sparse regions. After that I switched to 800 and 1200 digits precision and went up to 80000 iterates - no clue, whether the errors accumulate to something horrible. From that 80000 iterates I deleted all in the inner area so I took only the points in the sparse areas into the plot (red dots). It seems, the values are not completely messed, since in principle the blue and red points join reasonably to one curve. Hmm... what does this tell me? For instance: is that aequator through x0=0 asymptotically dense? Is it indeed an aequator at all? Is it enclosed in some disk? For another instance: what does this mean in regard to fractional iteration? Assume an orbit with a starting value x0 inside the limiting aequator (through x0=0), say through x0=0.5+0.5*I which seems to fill a smooth, closed line densely. Are also all fractional iterates on that curve? GottfriedHey Gottfried, Coincidently, I've been thinking about the same problem -- the behavior of bases on the Shell-Thron boundary. The base was chosen to be on the Shell-Thron Boundary, with a real period~=2.9883. Starting with x0=0 probably generates a fractal. My conjecture is for bases on the Shell Thron boundary, there is an analytic superfunction with a real period, whose structure depends on what the continued fraction representation of the real period is. As long as the period is a real number (with an infinite continued fraction representation), then I suspect the superfunction is analytic. If the period is a rational number, then I don't think there is an analytic superfunction. For example, this base, with a real period=3, probably doesn't have an analytic superfunction, developed from the neutral fixed point, because starting with a point near L, and iterating the function x=B^x three times, doesn't get you back to the initial starting point. Base= 0.030953557167612060 + 1.7392241043091316i L= 0.39294655583435517 + 0.46203078407110528i Back to your plot, starting with x0=0 generates a fractal, which represents the lower boundary of the real periodic superfunction. The structure of the fractal appears to depends on the continued fraction of the real period, where $\text{period} = 2\pi i /\log(\log(L))$. Anyway, most of this is pure conjecture, and perhaps others have done more work on these complex bases on the Shell-Thron boundary. - Sheldon Gottfried Ultimate Fellow Posts: 755 Threads: 115 Joined: Aug 2007 06/20/2011, 06:11 AM Here is an overlayplot nearer to the x0=0 , x1=1, x2= b0,... orbit. I took x0=0.98 and x0=0.99 as initial points (so defined h=0 there). The plot shows approximation to the fractal(?) equator of the x0=0-orbit. Gottfried Helms, Kassel Gottfried Ultimate Fellow Posts: 755 Threads: 115 Joined: Aug 2007 06/20/2011, 06:20 AM Hi Sheldon - good morning! Ahhh... "Shell-Thron-boundary": nice to see a relation to something known. I'll look at it again in the D. Shell-article. I'll take a look at your base and fixpoint later after the coffee ... Gottfried Gottfried Helms, Kassel « Next Oldest | Next Newest »

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