Re: fractional iteration of functions

qmagick_at_yahoo.com
Date: 01/30/05 

Date: Sun, 30 Jan 2005 15:30:04 +0000 (UTC)

Alain et al.

I am sorry that I cannot at this point release what results I have
arrived at in this public forum. First, I need to do alot of work and
read what everyone else has done to make sure I am not stealing anyone
elses ideas or saying anything blatantly stupid. Second, in the event
that I have come up with something original, I do want to get credit of
course. If I have not come up with anything, I will be the first to
tell you.

To give a hint at what I have found, check out this paper:
L. Smolarek, The formula of fractional iteration of function
differentiable at its fixed point. Zeszyty Nauk. Mat. Uniw. Gdans. 7
(1987), 99-107

Just looking at the paper's title I get the feeling that Mr. Smolarek
might have discovered something similar to my own results. If anyone
here has read this paper please let me know his methodology/results so
I can be for sure... I have allready emailed Mr. Smolarek, but he has
not given back any response as of yet. His paper is in Polish so that
might also be a problem (I don't speak Polish!). Maybe we can all
pressure him to tell us.

Alain. I would like to answer your two questions with some definiteness
so here goes. f(f(x)) = x, does not have just one solution for the
iterate of the function f. In any event, it is not the iterate we are
looking for but the set of functions with periods of two intersecting y
= x. I am using the term iterate to mean a continuous iteration
solution of a function f(x) with continuous variables n,x over some
range for the function F(n,x). F is the "iterate" of f(x). F has to
have the following properties:

1. F(0,x) = x
2. F(1,x) = f(x)
3. F(m, F(n, x)) = F(n, F(m, x)) = F(n+m, x)
4. n is over at least one continuous range including n = 0, and n = 1.
x must also be continuous over a given range for each n so defined.

So what I am saying is that the functional equation f(f(x)) = x has an
infinity of solutions. Any iterate with a period 2 solution along y = x
will suffice, and there are an infinity of them that I do not yet know
how to classify. This is a guesse though. Sorry. I could definitely be
wrong on this one. Maybe I will work on it. I would just be happy with
a general solution for iterates of functions though...

In answer to question 2*) above. The following functions have iterate
solutions for the whole of the complex plane over suitably defined
regions that are easily seen (x,n belong to C):

1. f(x) = x, F(n,x) = x, no need for explaining I hope.
2. f(x) = x + a, F(n,x) = x + na, dito of above. Above is special case
of current where a = 0.
3. f(x) = ax + b F(n,x) = a^n x + (a^n-1)b/(a-1), a != 1. If a = 1 use
above formula.
4. f(x) = (ax+b)/(cx+d), F(n,x) = (M_00 x + M_01)/(M10 x + M_11)

M_00 is the 00 component of the Matrix

| a b | n
| c d |

or the matrix formed from the coefficients raised to the continuous
power n and evaluated for 00. I think this is just the Moebius group. I
imagine you could generalize this result...

5. f(x) = ax^p, F(n,x) = a^((p^n-1)/(p-1)) x^(p^n), a != 0, p != 1.
This function is rather boring if you ask me. It just grows insanely
fast or decays insanely slowly.

These are what I am calling the trivial solutions. Whether they are
trivial or whether they are true, or allready found out yet I do not
know and do not too much care. Try them out yourself to see if they
meet the conditions for iterate solutions I have presented as
conditions 1-4. They should, but if they do not let me know! I will
then have to recant my inexcusable error.

So for the example Alain has given,
f(x) = 2x + 1 and find f^[1/3](x),
a = 2, b = 1, n = 1/3. This gives:

F(1/3, x) = 2^(1/3) x + (2^(1/3) - 1) 1 / (2 - 1)
= 2^(1/3) x + (2^(1/3) - 1)

Now for a test. F(1/3, F(1/3, F(1/3, x))) should equal 2x + 1. Lets
break it up into two parts. First, F(1/3, F(1/3,x)) should equal
F(2/3,x) right?

F(2/3, x) = 2^(2/3) x + (2^(2/3) - 1) 1 / (2 - 1)
= 2^(2/3) x + (2^(2/3) - 1)

F(1/3, F(1/3, x)) = 2^(1/3) [(2^(1/3) x + (2^(1/3) - 1)] + (2^(1/3) -
1)
= 2^(2/3) x + (2^(2/3) - 1)
= F(2/3, x)

So far so good. Now to finish.

F(1/3, F(2/3, x)) = 2^(1/3)[2^(2/3) x + (2^(2/3) - 1)] + (2^(1/3) - 1)
= 2x + 1 = F(1,x)

by the way, F(1/2,x) gives the same formula Alain gave, which is a good
thing.

There you go! As I said, this is a "trivial" solution but you should do
it on your own to make sure the solution meets the conditions 1-4. If
you do the formulas above enough you will see the trick in each one
that makes them work. Let me know if these are original or not, but I
doubt it very seriously.

Well, now for the bad news, or just a recap of what everyone has said
and what I have observed to date. Mr. Geisler has said there are a
number of "methods" for finding the iterates of various functions. From
what little I have read, this seems definitely to be true. I believe
they can roughly be catagorized in the following list:

I. Interpolation.
Using a computer create the functions f(x), f(f(x)), etc. and
interpolate the results from F(n,x) to F(n+1,x) for what the function
is for F(m,x) where n < m < n + 1. The down side of this is that it is
not analytic and merely a very serious "fudge".

II. Pattern recognition.
Take a function f(x) and iterate it over and over again. If you find a
pattern in F(n,x) elevate n to continuous and check whether it meets
conditions 1-4. This is how the trivial solutions above were generated
but it might work for other functions.Can not know until you try.

III. Coefficient Solutions.
Use various series expansions of f(x) and iterate them. Create another
expansion for F(n,x). Equate coefficients This technique is interesting
but has the drawback that it can not be shown to be rigorously the
actuall iterate solution, in fact, it probably is not. One other
approach in this method is making the cooeficients in F(n,x) functions
of n. Then one can produce what is called a "flow" if I remember
correctly. This method also has the drawback that it it is
computationally insane and not rigorously analytic.

IV. Taylor or Laurent Series expansions.
Finding the Taylor series or Laurent series expanded around a fixed
point or otherwise. This is what I am also doing (as well as Geisler).
What are the problems? Most functions can not be expressed as a series,
or that we know that a iterate of one can! Aside from that, the method
seems a very good one. Yet again I can't say exactly what I am doing
as of yet. What is true is that if f(x) is significantly well behaved
there is a chance this method could come out correctly for a large
class of functions, like f(z) = z^2 - 1 and so on.

Finally I want to respond to Mr. Geisler:

Daniel Geisler Jan 29, 2:00 pm
>Since my own work is based on finding the Taylor series of iteration
>functions from a fixed point, assuming the existence of a fixed point
>and that the function is smooth, the very existence of a solution
>indicates that a solution is associated with every fixed point. If
there
>is a solution for each fixed point it may be that we are only talking
>about one solution appearing as many under translation from one >fixed
point to another.

This is an intrigueing idea. Let me summarise it and tell me if I am
somewhat correct. For a function f(x) there exist a set of iterate
solutions generated via the fixed points in a one-to-one and onto
correspondence.

f(p0) = p0
...
f(p_n) = p_n

Make a set p = { p0, p1, ... , p_n } where the set can be infinite or
not. Solve the iterates of f(x) for each p_m and label them F_m (n, x).
I allready see the difference. I am very sorry. What you were talking
about were points x0, x1, ... x_k such that
f(x0) = x1
f(x1) = x2
etc.
f(x_k) = x0

Or a k period. It seems then that if F(n,x) is the iterate of f(x) then
the function F would have to traverse all the points from 0 to k
continuously The fixed points would just be cycles with period 0 or 1
depending how you look at or define it. I do not see also why you could
not have cycles of non-integral period. Of course other points would
not belong in a period at all. This brings to mind all sorts of
Topological questions that I did not even begin to think of. Perhaps
chaos is just a non-periodic set of points with boundaries? I really
have no idea. Requires alot more thought. What about those non-integral
periods though huh? I think I still did not get the whole import of
your idea though...

I would very much like to hear about how Mr. Geisler's method works if
it is allready published, just to make sure it is not the same as my
own. I also want to note that solving the iterates of functions is not
necessarily related to more general functional equations. For instace,
if we knew what the iterate is of e^x, we would not necessarily know
how to solve f(f(x)) + f(x) = e^x. This is just one example but shows
that "functional equations" and "function iterates" should be treated
as two seperate, albeit related topics.

Plus, I hope Mr. Geisler if you don't mind that I use the idea of Mr.
More to test the functions I generate. Is his idea allready published
or should I only use it only for my own confidence tests?

Sincerely to everyone,
-- NPC