Infinite tetration of the imaginary unit
#8
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<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.

Gottfried

*** life is complex: you need to consider its real and its imaginary parts *** Smile
Gottfried Helms, Kassel
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Messages In This Thread
Infinite tetration of the imaginary unit - by GFR - 02/10/2008, 12:09 AM

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