06/25/2008, 06:15 AM

Perhaps the following two msgs in sci.math of David Libert are appropriately posted to this thread. I just copy&paste the whole msg to provide the context.

Perhaps we move this post to another place - Henryk?

and a correction, posted later on the day

Perhaps we move this post to another place - Henryk?

Code:

`(David Libert ah170@FreeNet.Carleton.CA )`

Rotwang (sg552@hotmail.co.uk) writes:

> > On 19 Jun, 04:07, Rotwang <sg...@hotmail.co.uk> wrote:

[deletions...]

>> >> Based on the recent discussions it seems that there is no known way to

>> >> define non-integer iteration in general, but I was wondering whether

>> >> anybody here knows whether it can always be defined in the restricted

>> >> setting given here, or alternatively whether there are any no-go

>> >> theorems which show that it can't.

>> >>

>> >> Of particular interest is the case where f (z) = exp (z), for z in {z

>> >> in C | 0 <= im z < 2 pi} since this is an important transformation in

>> >> the book I'm reading. In this case f^t (z) would be a generalisation

>> >> of tetration to non-integer height...

> >

> > Sorry, this is wrong. I mean a generalisation of iterated

> > exponentiation to non-integer height. Does anybody know if such a

> > thing exists?

This question came up here and in other recent threads. I have not read

all of the large volume of writing about this in all threads, so I may have

overlooked something, but from what I have so far seen this has not yet been

answered.

I don't know the answer either, nor have I found one on the Internet.

I will note some things related to the question, some I think I have proven

myself, others found.

First, I Googled on fractional iteration and it does seem to be a big topic.

Tommy in other threads has written about tetration as related to this. I have

previously heard of tetration in making higer iterations of exponentiation possibibly

into the transfinite, along the lines of ordinat notation generalizations of

Ackermann's function. But that Google search did show that this word is also applied

to fractional and continuous interpolations.

I will note some specific interesting looking leads from that search:

a sci.math.research thread from 2005 fractional iteration of functions

http://groups.google.com/group/sci.math.research/browse_thread/thread/b7f7ebcdc67d6bce

the FOM thread Fractional Iteration and Rates of Growth near the top of

http://www.cs.nyu.edu/pipermail/fom/2006-May/thread.html#start

and: http://www.tetration.org/Dynamics/

What I seem to be seeing above and these various threads is it often possible to

define various continuous interpolations but it is hard to say which if any are

canoical. Rotwang above proposed more stringent requirement on a solution, but that

leaves open the question of existence. If that were solved there is also the question

of uniqueness.

A similar point, concerning another case with a more known solution, as I understand

it there are many interpolations of the integer factorial function to the complex

numbers, but if we require log-convexity we have unique existence with the Gamma function.

I will add some related results I may have proven. I don't want to take the time here

to write out proofs. Maybe it would be safer to call these conjectures :-) .

So I think ZF proves that if A is an infinite set which can be well-ordered and

f: A >->> A is an injection, then there are 2^A many g: A >->> A with

f = g o g . (That is function compoistion).

ZF proves for A infinite well-orderable there are 2^A many functions : A -> A, so this

would be as many possible.

Regarding f: R -> R (R the reals), if f is non-decreasing

(this is: x<y -> f(x) <= f(y) ) and continuous

then there are #R many non-decreasing and continous g : R -> R

with f = g o g .

By smooth I mean all finite iterated derivatives exist everywhere and are continous.

For f : R -> R non-decreasing and smooth there are #R many g non-decreasing and smooth

with f = g o g.

On the other hand if f : R -> R is non-increasing, then there are no non-decreasing

g : R -> R with f = g o g, and there are no non-increasing g : R -> R with

f = g o g .

But if f : R -> R is continous and non-increasing, there are #R many g : R -> R

which are piecewise continous on countably many half-open intervals with f = g o g .

It seems a mess to try anything for f non-increasing.

For f non-decreasing, you can keep repeating the .5 construction of new g's above. You

can define dyadic rational exponentiation on these, by using the contruction to take .5 's

and ordinary integer powers to take multiples. The right things commute for this to be

well-defined.

The ordering relations among these are as expected and monotonic, so by taking appropriate

sups or infs of dyadic rational exponents you can extend this to real exponents of iteration.

Unfortunately, if f started as continuous or smooth, you have each of the slices is

respectivelty continous or smooth, but overall operation considered over variable exponent

need not be continuous or smooth in that exponent variable. Because each time you halved

the exponent you made a non-canonical choice of how to do so, and could have taken wildly

different functions.

Indeed none of the functions produced in any of these claims are canonical. The internal

proofs made arbitrary choices along the way, that is why they produced many functions

instead of one.

and a correction, posted later on the day

Code:

`David Libert (ah170@FreeNet.Carleton.CA) writes:`

[deletions...]

> > So I think ZF proves that if A is an infinite set which can be well-ordered and

> > f: A >->> A is an injection, then there are 2^A many g: A >->> A with

> > f = g o g . (That is function compoistion).

I will revise the above, after reconsidering the proof more carefully. For A as above

and f : A >->> A as above, I will define a 2-cycle to be a sibset {x, y} of A

such that x ~= y and f(x) = y and f(y) = x .

The first revision will be if the number of 2-cycles is finite and odd, then

there are no g as above such that f = g o g.

So regarding the remaining cases (finite and even # of 2-cycles or infinitely many 2-cycles) :

Let B = {x in A | f(x) ~= x} .

If B is uncountable then there are 2^#B many g : A >->> A s.t. f = g o g .

If B is finite there are finitely many g : A >->> A s.t. f = g o g .

If B is denumerable (countable and infinite)

then Aleph_0 <= #{ g : A >->> A | f = g o g} <= 2^Aleph_0 .

> > But if f : R -> R is continous and non-increasing, there are #R many g : R -> R

> > which are piecewise continous on countably many half-open intervals with f = g o g .

Also, this statement is claimed for f smooth and non-increasing, then g can be taken

to be piecewise smooth on the same sort of intervals.

Gottfried Helms, Kassel