Re: fractional iteration of functions

From: Dave Rusin (rusin_at_math.niu.edu)
Date: 01/30/05 

Date: Sun, 30 Jan 2005 15:30:06 +0000 (UTC)

In article <ctdgh5$s1d$1@news.ks.uiuc.edu>, <qmagick@yahoo.com> wrote:

>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?

Or does it have any at all?

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.

dave