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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) = i[4]oo = 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))[4]oo = {i,+i}
and also that:
i[4]2 = e^(Pi/2)
GFR
Corrected on 19th0208: (e^(Pi/2))[4]oo = i (Thanks, Ivars !)
We get +i with another branch of the W formula (W(1), instead of W(0)). I check again.
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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
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OK, Ivars. Thanks ! I corrected.
I hope it's OK now. Please .... check. again
GFR
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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?
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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. Tonight, I shall work on that.
I have always problems in following your hyperdimensional mappings. I shall try harder !!!
GFR
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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?
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Infinite tetrates may be real or, indeed ...., complex.
My guess is that tetraton (y = b[4]x) is one multivalued real and/or complex "function" of x, depending on the value of base b > 0. Oscillations of y = b[4]x, 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 infinitetime 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 bsolution 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 bsolution 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
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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 bior 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 trifurcation, 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 multifurcation 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 noncontinuous 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 matrixapproach may help here.
A height of tetration is represented by a power of the Bbmatrix.
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 Bmatrix (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<b<e^e we have this bifurcation, a periodicity of 2, and expressed in the matrixformulation by the operator B^2
f0 = V(1)~ (B^2)^h // h integer
f1 = V(1)~ B*(B^2)^h // h integer
So here seems the tetraroot to be involved, and actually an inverse operation of tetraroot (to be defined and described...).
Hmmm. Maybe it's time to do some more precise descriptions.
Gottfried
*** life is complex: you need to consider its real and its imaginary parts ***
Gottfried Helms, Kassel
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02/12/2008, 01:37 PM
(This post was last modified: 02/14/2008, 05:16 PM by Gottfried.)
To see the trajectories of the partial expressions, if I assume 3step period even for the convergent case, where f0=f1=f2
Gottfried
convergent  unfurcated
convergent: trifurcated, but in limit f0=f1=f2
trifurcation , but this converges if infinite height is assumed
trifurcation without convergence
Gottfried Helms, Kassel
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02/13/2008, 03:39 PM
(This post was last modified: 02/15/2008, 06:31 PM by Gottfried.)
[update]
The nonconvergent trifurcation of partial evaluation with b=1.8 *i. The actual fixpoint is repelling
Data: for the partial evaluation read from left to right, topdown, then this gives the coordinates for
\(
\begin{matrix} {rrr}
1 & b & b^b \\
 & & \\
b^{b^b} & b^{b^{b^b}} & b^{b^{b^{b^b}}} \\
 & & \\
\ldots & \ldots & \ldots \\
\text { } & & \\
& \text{the data are:} & \\
& & \\
1.0000 & 1.8000*I & 0.029026+0.051555*I \\
0.93538+0.071130*I & 0.092461+1.5470*I & 0.045888+0.080842*I \\
0.89836+0.10796*I & 0.13746+1.4245*I & 0.057241+0.10054*I \\
0.87334+0.13111*I & 0.16534+1.3498*I & 0.065447+0.11492*I \\
0.85502+0.14708*I & 0.18474+1.2989*I & 0.071635+0.12595*I \\
0.84094+0.15873*I & 0.19920+1.2619*I & 0.076432+0.13469*I \\
0.82975+0.16753*I & 0.21049+1.2339*I & 0.080221+0.14179*I \\
0.82066+0.17436*I & 0.21959+1.2121*I & 0.083256+0.14765*I \\
0.81314+0.17976*I & 0.22710+1.1946*I & 0.085712+0.15256*I \\
 & & \\
\ldots & \ldots & \ldots \\
 & & \\
trajectory0 & trajectory1 & trajectory2 \\
\end{matrix}
\)
Another better view, comparing 4 different bases:
It would be nice to locate two spcial coordinates.
First: what is the value b = x*i, 1.7 < x < 1.75, where convergence suddenly disappears?
Second: what is the value b=x*i, 1.75<x<1.8, where the trajectories give a traight line  if this occurs at all. It seems so, since at x=1.8 the trajectory is continuously left bound and x=1.75 the curve is continuously right bound.
Gottfried
Gottfried Helms, Kassel
