Re: How can this tetration be extended?

On Jun 22, 2:24 pm, mike3 <mike4...@xxxxxxxxx> wrote:
I've seen and played with various candidates for an extension of the
tetration (power tower, hyper4, "repeated exponentiation") function to
real towers (the number of exponents.). Most of them focus on real-
number bases, often the ranges 1 < b < e^(1/e), b = e^(1/e), and b > e^
(1/e) (or b > 1). But what about b in the range from 0 to 1? The range
e^-e < b < 1 converges similar to 1 < b < e^(1/e) but oscillates as it
does so. The "regular iteration" and "matrix operator" (Bell/Carleman
matrix power) gives complex values for this tetration at fractional
"tower", and when graphed on a complex plane, it forms a spiral as the
tower is increased continuously from 0, that homes in on the fixed
point. e^-e also appears to converge but at a slower rate.

I haven't posted in the tetration threads in a while. It's
too bad that I missed this thread when mike3 first posted.

Yes, I've thought about tetration in this interval myself,
and the fact that b^x is decreasing on [0,1] results in
the function b^^n for integer n alternately increasing and
decreasing as n-->infinity.

If we were to assume that b^^x for real x is simply an
oscillating real-valued function, we have problems. Such
an oscillating function is not invertible, whereas one of
the desiderata for a tetration function is that it have an
inverse, a superlogarithmic function. Indeed, as mike3
points out, tetration as defined by Robbins is based on
linearizing the _superlog_. Small wonder, then, that when
mike3 tried to use the Robbins method it failed to
converge to a solution.

So I once wondered whether b^^x for b<1, x nonintegral
real ought to be imaginary, just as mike3 has done. If
b^^x were a complex spiral, then the function could be 1-1
and still have a superlog inverse. And sure enough, we see
that the Gottfried Helms matrix method gives complex
values for b^^x.

The way I sometimes think about it is as follows:

-- The base 0, for tetration, acts a little like the
number -1 for exponentiation. We already know that as far
as tetration is concerned, 0^0=1 (since lim x->0 x^x=1),
and so we see that 0^^n=0 for n odd and 1 for n even, just
as (-1)^n=(-1) for n odd and 1 for n even.

-- The interval (0,1) for tetration acts a little like the
interval (-1,0) for exponentiation. Exponentation in the
interval (-1,0) spirals around the complex axis before it
converges to 0 while tetration in (0,1) spirals around the
complex axis, and it usually converges as well (though
mike3 has already mentioned the exceptions).

-- The base 1 for tetration acts a little like the number
0 for exponentiation. We have 1^^n=1 for positive n, just
as 0^n=0 for positive n.

-- The interval (0,1] for exponentiation can be said to
correspond to the interval (1,eta] for tetration. For
exponentiation in (0,1] converges to either 0 or 1, while
tetration in (1,eta] always converges as well.

-- Finally, the interval (eta,infinity) for tetration
corresponds to the (1,infinity) for exponentiation since
in either case, we have increasing without bound.

Of course, these correspondences aren't exact. Still, the
Tetration Forum thread to which mike3 links below also
compares the tetration of numbers less than 1 to the
exponentiation of negative numbers.

And so it does appear that defining b^^n for b<1 and
n nonintegral real to be nonreal complex appears to
satisfy the desiderata of most tetration experts.
.

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