Thread Rating:
  • 0 Vote(s) - 0 Average
  • 1
  • 2
  • 3
  • 4
  • 5
Infinite tetration of the imaginary unit
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: 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)


Corrected on 19th-02-08: (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.
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?

OK, Ivars. Thanks ! I corrected.
I hope it's OK now. Please .... check. again

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?
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 !!! Wink


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?
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 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 Sad !

The big problem is to show and correctly demonstrate all this. Unfortunately, at this date, I only have clear, but ... vague, guessings.

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,...,

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.

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<b<e^-e we have this bi-furcation, a periodicity of 2, and expressed in the matrix-formulation 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 tetra-root to be involved, and actually an inverse operation of tetra-root (to be defined and described...).

Hmmm. Maybe it's time to do some more precise descriptions.


*** life is complex: you need to consider its real and its imaginary parts *** Smile
Gottfried Helms, Kassel
To see the trajectories of the partial expressions, if I assume 3-step period even for the convergent case, where f0=f1=f2


convergent - unfurcated

convergent: trifurcated, but in limit f0=f1=f2

tri-furcation , but this converges if infinite height is assumed

tri-furcation without convergence
Gottfried Helms, Kassel
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, top-down, then this gives the coordinates for

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 Helms, Kassel

Possibly Related Threads...
Thread Author Replies Views Last Post
  [MO] Is there a tetration for infinite cardinalities? (Question in MO) Gottfried 10 5,934 12/28/2014, 10:22 PM
Last Post: MphLee
  Remark on Gottfried's "problem with an infinite product" power tower variation tommy1729 4 2,517 05/06/2014, 09:47 PM
Last Post: tommy1729
  Problem with infinite product of a function: exp(x) = x * f(x)*f(f(x))*... Gottfried 5 3,649 07/17/2013, 09:46 AM
Last Post: Gottfried
  Wonderful new form of infinite series; easy solve tetration JmsNxn 1 2,857 09/06/2012, 02:01 AM
Last Post: JmsNxn
  The imaginary tetration unit? ssroot of -1 JmsNxn 2 3,363 07/15/2011, 05:12 PM
Last Post: JmsNxn
  the infinite operator, is there any research into this? JmsNxn 2 3,369 07/15/2011, 02:23 AM
Last Post: JmsNxn
  Tetration and imaginary numbers. robo37 2 3,231 07/13/2011, 03:25 PM
Last Post: robo37
  Infinite Pentation (and x-srt-x) andydude 20 15,314 05/31/2011, 10:29 PM
Last Post: bo198214
  Imaginary zeros of f(z)= z^(1/z) (real valued solutions f(z)>e^(1/e)) Gottfried 91 48,808 03/03/2011, 03:16 PM
Last Post: Gottfried
  Infinite tetration fractal pictures bo198214 15 13,760 07/02/2010, 07:22 AM
Last Post: bo198214

Users browsing this thread: 1 Guest(s)