Re: Old chestnut: f(f(x))=exp(x)

david petry wrote:
The problem of finding a solution to f(f(x)) = exp(x) has come up in
this newsgroup before. I've suggested a possible solution which I
believe will be the "best and most natural" solution, if we can prove
that the Taylor series for the solution of f(f(x)) = exp(x)-1 has a
non-zero radius of convergence.

http://groups.google.com/group/sci.math/msg/3c85c5acfbd2ee49?hl=en&;

Partly inspired by your post I noticed that, defining h(x) = exp(x) -
1, one can solve

f(h(x)) = h(f(x)) (1)

for f as a power series expansion - i.e. f(x) = c1*x + c2*x^2 + c3*x^3
+ ... - simply by equating powers of x and solving for c1, c2, c3, ...
in turn. Since, with h^n(x) defined sensibly, f(x) = h^n(x) satisfies
(1) for all real n, we should get a family of solutions for f, each
corresponding to h^n(x) for a different real n (*subject to
convergence*, and possibly some other things not rigorously proved!).

Anyway, it seems to work. As far as I can see, c1 = 1 is forced, and
then there is a free choice for c2 which determines all the remaining
coefficients. (There is possibly a singular solution for c1 = -1 too,
which I have not investigated.) Rather remarkably, setting c2 = n/2
seems to yield f(x) = h^n(x) for "any" given n (same provisos as
before).

Some number-crunching experiments I have done show that convergence is
very convincing for small |x| and |n|, and that all the functions seem
very nicely behaved in those regions - making an nice smooth
transition, for example, from n = -1 (when f(x) = h^(-1)(x) = log(x +
1)), through n = 0 (when f(x) = h^(0)(x) = x) to n = 1 (when f(x) =
h^(1)(x) = exp(x) - 1).

So, we have a way of expressing the general iteration of exp(x) - 1 as
a power series, subject to the convergence provisos. If there was some
way to represent the iteration of exp(x) in terms of the iteration of
exp(x) - 1 then we could use this method to get a nice power series for
exp^n(x) for "any" real n. Anyone got any ideas whether this can be
done? Maybe it's a forlorn hope... I guess someone has been down this
road before...

.

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