f_2(x) = f(f(x))
f_3(x) = f(f(f(x)))
etc.

The question is how to define f_n(x) when n is not integer. I have seen
the fractional fourier transform, and it gives a good idea of what I am
talking about. Has there been any work on this allready? I am
considering writing a paper but need refferences.

It seems there are some very serious open questions relating to this
subject. One question I find personally very interesting, relates to
how many iterative function solutions a particular function should
have. For instance take f(f(x)) = e^x. The function e^x has no fixed
point on the real line but an infinity of them in C. Does each new
fixed point create its own solution for an iteration function? Are
there solutions of the iteration function outside of the fixed points?

For instance f(x) = 6 + 2x - x^2, has two fixed points x = 3 and x =
-2. Using the method I alluded to in the first post, you can generate
two series solutions to f(x) from the fixed points. So does the
function above have more then one 2 solutions for the iteration
function, or does it have exactly 2?
Of course this could have been answered years ago in other papers for
all I know...

Consider the iterates of f(z); typically there will be not only fixed
points but period k fixed points. Given that f(z) has a period two fixed
point, now consider the iterates of g(z) such that g(z) = f(f(z)). Both
of the period two fixed points for f(z) serve as period one fixed points
for g(z).

Cris Moore suggested an interesting test for validating a continuously
iterated function. It’s easy to kludge together a function that is
faithful to the Lyapunov exponent of a fixed point in the fixed point’s
neighborhood; the trick is for it to also be faithful to the Lyapunov
exponent of other fixed points in their neighborhoods. I had some
positive but not conclusive results with my own work when I modeled the
Taylor series from a fixed point of the continuous iteration of a very
low entropy exponential function c^z with the value of c near to 1. This
takes an incredible amount of computational power to do right since as z
moves away from the fixed point the Taylor series must model the fixed
point’s Lyapunov exponent, further out model chaotic behavior, and then
as z takes even larger values the Taylor series must settle down and
model a second fixed point and its Lyapunov exponent.

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.

I think it may be appropriate to extend your question about whether a
solution is associated with each fixed point to ask if a solution is
associated with every set of period k fixed points. What about solutions
for when k goes to infinity? What about solutions for strange attractors?

Good luck,
Daniel

[1] R. Aldrovandi and L. P. Freitas,
     Continuous iteration of dynamical maps,
     J. Math. Phys. 39, 5324 (1998)

You've picked up some 'stuff' for your study about fractional
iteration fonctions.
You've written "one question I find personaly interesting relates to
how many iterative functions solutions a particular function should
have " .I am mainly concerned with real ,continuous functions.
I want to show two examples.
    1°)f^[2](x)= x  ,just have to write a symmetrical {f,x} expression
, like x*f+f+x =3 ...Is there a generalization to [n] ?

    2°) f(x) = 2x +1 , f^[1/2](x)= ? the iterated line (ax+b)[r]  
gives two roots f^[1/2](x)=-xsqrt(2)-1-sqrt(2);
          f^[1/2](x)= xsqrt(2)-1+sqrt(2) ,
          compute f^[1/3](x)!!!
Please do tell me about the significant results you arrived at.
Honestly I think we still have to work in order to clarify and to make
continuous iterated fractions more 'catchable',

1. Yes, each fixed point creates its own solution for an iteration
function.

2. Yes, there are solutions of the iteration function outside of the
fixed points; in particular there are infinitely many real analytic
flows into which e^x embeds on the reals.

By "solution" you seem to mean a smooth function, domain unspecified.
If you compute this solution by computing power series at a fixed point
(so that what you're computing is actually the germ of a solution, but
never mind) then I think you need to know whether the power series
converges before you can assert that there is a solution (defined somewhere
besides the fixed point of course).

For example I looked for a solution to  g o g = x - x^2  of the form
g(x) = x + a2 x^2 + a3 x^3 + ... ; the rational coefficients seem to
grow too rapidly to permit convergence (I went as far as the coefficient of
x^100 and it appeared  log(|a_n|)  grew faster than linearly). So if
this series has a zero radius of convergence, in what sense is it a solution?

Your specific quadratic may be different. I looked instead at the function
f(x) =  4x - 3x^2  so I could avoid surds, looking for a power-series
solution starting  g(x) = 2x + a2 x^2 + ... . I "solved" this in a very
round-about way. First I found the coefficients a3, a4, ..., a_75 in turn.
Factoring numerators and denominators and looking for a pattern, I found
that   a_{n+1} / a_n = (3 n^2 - 6)/(4n^2 + 6n + 2)  so the series appears to
converge for |z| < 4/3. In particular the value  g(x)  would be the limit of
the partial sums  S_n  which satisfy a recurrence
  (4n^2 + 6n + 2) S_{n+1} = ((4+3x)n^2 + 6n+(2-6x)) S_n - 3x(n^2 - 2) S_n .
Maple solves this recurrence relation and I can massage the answer to be  S_n =
   (2*x)*(   hypergeom([1, 1-2^(1/2), 1+2^(1/2)],[3/2, 2],3/4*x) +
 (2/Pi)^(1/2)*(sin(Pi*(-1+2^(1/2)))/4)*n^(-3/2)*((3/4)*x)^n
 *hypergeom([1, n+1-2^(1/2), n+1+2^(1/2)],[n+2, n+3/2],3/4*x)
 *(GAMMA(n+1-2^(1/2))*GAMMA(n+1+2^(1/2))/GAMMA(2*n+3)*4^n*n^(3/2)*4/Pi^(1/2) ) )
The first line is free of  n's  and the last line tends to  1  as  n->oo;
Maple did not succeed in evaluating the limit of the rest as  n --> oo, except
numerically, and for the few  x  I tried it seems this limit is zero, leaving
  g(x) = 2*x*hypergeom([1, 1-2^(1/2), 1+2^(1/2)],[3/2, 2],3/4*x) .
Indeed, this  g  has the correct Taylor-series coefficients (through x^75).
I have to say I didn't expect a "closed form" solution. Computing numerical
values of  g(g(x)) we check we do indeed get  4x-3x^2  but only for  
x < 1.07  or so, after which the function  g  starts to decrease.

Well, this is a lot of guesswork but it seems to nominate a specific
function to be a solution to the functional equation; I suppose someone who
uses the hypergeometric functions regularly could prove  g o g = 4x - 3x^2.
I want only to observe that the function  g  numerically takes the value
g(1) = 1.3226017298723... rather than  g(1) = 1,  so this is not the
same as the solution you would construct from the fixed point of f at 1.

Since you asked about solving  g o g = exp  and seem to be interested in
complex-analytic solutions, I will just point out that there is no
such solution defined everywhere in the complex plane, as can be deduced
from thinking about the order of entire functions; see Polya,
"On an integral function of an integral function",  
Journal London Math. Soc. 1 (1926), p. 12-15.

f(x) = c, F(n,x) = c, c is a constant belonging to C. No other
conditions on the iterate.

I was thinking also about piecewise continuous functions p(x) of the
various trivial solutions to see if they are amenable to solution. I
was using Mathematica to define them and play around. For instance:

p(x) = x, if x > 1
2x + 1, if 0 < x <= 1
x^3, if x <= 0

This function has overlap in the values that the function produces
[p(3) = p(1)]. If I could find a general solution or general method the
results might be helpfull in approximating other more general functions
also. This is just an idea currently... It looks like that if the
piecewise functions do not have overlap the solution for the total
iterate will be the set of piecewise iterate solutions, otherwise the
iterations of p(x) become much harder. So for now this method would
only work for functions that are allways increasing or allways
decreasing and can be modeled as piecewise of the trivial solutions. Of
course don't take any of this to be definite except the solution for
f(x) = c for the iterate (sorry).

The classification of fixed points states that complex dynamics (the
dynamics of the complex plane) can result in hyperbolic, irrationally
neutral, rationally neutral, parabolic rationally neutral, and
superattracting fixed points. They all have distinct dynamical
pro