a simple idea ... maybe considered before ?
let us approximate exp(x) with polynomials of degree n.
let us call that exp_n(x).
now we can solve for f_n(f_n(x)) = exp_n(x)
if n is large enough and then f_n(x) being a polynomial of degree < n.
( we ignore the terms of degree > n in f_n(f_n(x)) )
now if lim f_n(x) exists we got f(x) with f(f(x) = exp(x).
....
headscratch ...
regards
tommy1729
tommy1729 Wrote:headscratch ...
That doesn't work. Suppose
 = x^2)
then
) = x^4)
. Chopping off the second series gives
 = 0)
which is false.
andydude Wrote:tommy1729 Wrote:headscratch ...
That doesn't work. Suppose then . Chopping off the second series gives which is false.
well , thanks for your reply andy.
but i dont agree , and i assume you misunderstood what i meant.
in your example there is nothing to chop off ?
the degree of f(f(x)) = 4 just as desired.
what is ignored is more something like this
( with some imagination , example not so good )
f(x) = a + b x^4 + (x^17) / 17!
f(f(x)) = polynomial of degree 16 + ' some terms of degree > 16 '
' those terms of degree > 16 ' are caused by the x^17 part so they are dropped.
they are pretty small for some x afterall since i took x ^17 / 17!
that is important , what we ignore must be relatively small.
but exp(x) has a fast converging taylor series so *that* should probably not be a problem.
however in the OP i assumed a limit , that might be trickier.
but i believe that when taking the limit in a good way , we have chance at succeeding ...
i hope this clarifies a bit ...
regards
tommy1729
tommy1729 Wrote:i hope this clarifies a bit ...
Nope. Still head scratching.
andydude Wrote:Nope. Still head scratching.
Me too.
Most polynomials have no polynomial half iterate, so how can your suggestion work?