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 Infinite tetration of the imaginary unit GFR Member   Posts: 174 Threads: 4 Joined: Aug 2007 02/10/2008, 12:09 AM (This post was last modified: 02/10/2008, 11:44 AM by GFR.) We know that the infinite tetration of (applied to) a base b can be found by the pricipal branch of the Lambert Function, by: y = W(-ln(b))/(-ln b) = h(b). If we "force" this formula, applying it to base i (the imaginary unit), we get: h(i) = W(-ln(i))/(-ln(i)) = W(-i * Pi/2) / (-i * Pi/2) , i. e. : h(i) = ioo = 0.438283.. + i * 0.360592.. See also: http://mathworld.wolfram.com/PowerTower.html formula 18 It is interesting to remember that: h(e^(Pi/2)) = (e^(Pi/2))oo = {-i,+i} and also that: i2 = e^(-Pi/2) GFR Corrected on 19th-02-08: (e^(Pi/2))oo = -i (Thanks, Ivars !) We get +i with another branch of the W formula (W(-1), instead of W(0)). I check again. Ivars Long Time Fellow    Posts: 366 Threads: 26 Joined: Oct 2007 02/10/2008, 09:52 AM (This post was last modified: 02/10/2008, 09:53 AM by Ivars.) Was not the proper value for h(e^pi/2)=h(i^(1/i))=h(-i^(-1/i)=-i, or, at least, 2 values-+i? Ivars GFR Member   Posts: 174 Threads: 4 Joined: Aug 2007 02/10/2008, 11:35 AM OK, Ivars. Thanks ! I corrected. I hope it's OK now. Please .... check. again GFR Ivars Long Time Fellow    Posts: 366 Threads: 26 Joined: Oct 2007 02/10/2008, 01:28 PM I am more wondering about inverse operation- how do we get e^pi/2 out of -i - that is self root of -i - but self root of - i - what does it does? If tetration of e^(pi/2) creates hypervolumes from base e^(pi/2), than self root starts from that hypervolume -i and ? When we do infinite tetration of e^(pi/2), we can do it step by step, like a limit when number of steps-> infinity- which is of course a speed -up operation ( limit taking) but than we should be able to come back also somehow step by step- but there is none- just selfroot of - i =-i^(1/-i) =e^(pi/2) in one jump. That was the reason I suggested a step dI, hypersurface dI=lnI/I=-pi/2 etc- because differentiation is also a limit operation , so that by infinite differentation of -I we would reach e^(pi/2). But that seems a little out of usual approach. Are there any other? The reason I thought pentation must lead to e^-pi was, on one hand, the idea that MAYBE pentation of e^(pi/2) would lead to further updimensioning of hypervolume - I, which might be somewhat special case ( getting -i from e^(pi/2) seems special to me.) So I thought, Ok, pentation of (e^pi/2) would lead to hypervolumer -i^2 =I^2 = -1= e^-I*pi=e^I*pi. Then sextation would be -I^3= I=e^(-3pi/2) and heptation -I^4 = 1 = e^(-2pi*I). But may be it is not so simple. I think tetration may be special in this sequence as it kind of connects the "normal" operations with hyperoperations. It works with both, in the middle. If so , there has to be inverse superslow operations of similarly important place. e.g 1/2 -ation? The first slower operation between addition and zeration? GFR Member   Posts: 174 Threads: 4 Joined: Aug 2007 02/10/2008, 08:53 PM Halfation? We thought about that . I am preparing, in collaboration with KAR, a thread concerning zeration. It needs a good introduction, because I know that the subject had ... some problems in the framework of Wikipedia. To-night, I shall work on that. I have always problems in following your hyperdimensional mappings. I shall try harder !!! GFR Ivars Long Time Fellow    Posts: 366 Threads: 26 Joined: Oct 2007 02/10/2008, 09:10 PM (This post was last modified: 02/11/2008, 08:04 PM by Ivars.) GFR, I was thinking about Your idea that infinite tetration might be complex in bigger area and You must be perfectly right. It must be complex everywhere where branches of W and ln does not cancel out perfectly. So, in principle, it could be complex even over all real arguments, converging , oscillating, or diverging. I am not sure about negative arguments, though. Nor about complex arguments. Nor about tripleton (I forgot the name of those unalgebraic extensions of complex numbers) , quaternion or octonion or sedenion arguments. Also, what type of operation is self root if it gives inverse of tetration? It is kind the only exactly defined unique root, meaning one x can have only one selfroot? GFR Member   Posts: 174 Threads: 4 Joined: Aug 2007 02/11/2008, 10:15 PM Infinite tetrates may be real or, indeed ...., complex. My guess is that tetraton (y = bx) is one multivalued real and/or complex "function" of x, depending on the value of base b > 0. Oscillations of y = bx, for constant b, should be produced, at b < 1, by a complex y, multivalued "function" of a real variable x, which appear as such in a "real" projection on the yx plane. But, we must find such a "function", the smoothness of which would appear in this real projection (continuous and infinite-time derivable, continuity class Coo). Outside the yellow zone, the oscillations vanish for x -> oo. Inside the yellow zone, they remain persistent at x = +oo (sorry for my ... non standarization!). 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. The beginning of this analysis is the study of y = b^(b^y)), equivalent to y = b^y. But, I don't see how to proceed, in an exhaustive mathematical way. I found a candidate for the additional branches (yellow zone), but I cannot justify why it may be so. Nevertheless, it seems to work ! The big problem is to show and correctly demonstrate all this. Unfortunately, at this date, I only have clear, but ... vague, guessings. GFR Gottfried Ultimate Fellow     Posts: 758 Threads: 117 Joined: Aug 2007 02/12/2008, 09:23 AM (This post was last modified: 02/12/2008, 01:12 PM by Gottfried.) Hmm. Holidays, time to speculate a bit about other aspects of tetration. I played around a bit with an extension of base i to multiples of i. What I got is, that we get bi-or multifurcations, if we observe the intermediate steps of b, b^b, b^b^b, b^b^b^b^b,..., [update] for some b = a*i (b purely imaginary) I got always tri-furcation, and for b = i it makes also sense to separate the partial expressions into three groups. [/update] So for some b the partial evaluation of y= b^...^b^b^b^b gives periodically "partial fixpoints", say f0 = ...(b^b^b^(b^b^b)) f1 = ...(b^b^b^(b^b^b^(b))) f2 = ...(b^b^b^(b^b^b^(b^b))) For some b the three points happen to converge to the same value, where each one follows a certain trajectory, for other b they stabilize very fast to their distinct individual values. Also for not purely imaginary b one gets different multi-furcation with various lengthes of periods. What does this mean? First, in the inverse view, (in that of mapping u->b , where b = exp(u/exp(u))) does it mean, that the map u->b produces a non-continuous complex plane for b? (other wording: there are infinitely many values b in the complex plane, which cannot be expressed by u when b=exp(u/exp(u))) But is this true? Another approach: if we have periodicity then we might try to compute values for the whole period, say b^b^b^(b^b^b) = a^a - but this obviously wrong in iteration. The matrix-approach may help here. A height of tetration is represented by a power of the Bb-matrix. Tb°h=V(1)~* Bb^h [,1] .......// [,1] means: second column of Bb^h Then, periodicity occurs in collected powers of B (=Bb here): Tb°h=V(1)~* (B*B*B)*(B*B*B)*....(B*B*B) =V(1)~ * (B^3)^(h/3) But since B^3 has not the form of a B-matrix (especially the important second column does *no more* provide the coefficients of an exponential series), we cannot readily reformulate this into a^a^a ....; B^3 is another *type* of operator. For bases 0

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