06/08/2011, 01:18 PM

(06/08/2011, 09:55 AM)bo198214 Wrote: It is the iteration at infinity.

To see whats happening there one moves the fixpoint at infinity to 0, by conjugating with 1/x:

in thread tid 403 I , Bo , Mike and Ben already discussed moving fixpoints

in particular - as the title says - f(f(x)) = exp(x) + x.

this thread seems similar.

however , i consider this slightly different.

exp(x) + x has a " true " fixpoint at oo.

with " true " i mean that it touches the id(x) line.

in this case of x^2 + 1 , i would not call oo a fixpoint , but rather say that Bo uses so-called " linearization " :

g^[-1](f^[n](g(x))) = [g^[-1](f(g(x))]^[n]

Quote:.

you take lim n -> oo ... and then i see no n.

so this needs a correction.

Quote:Again similar to the Schröder case, we have an alternative expression:

If we even roll back our conjugation with we get:

.

i dont get how you arrive at this ... g and f on both sides ?

Quote:This may lead into a new way of computing fractional iterates of exp, because we just approximate exp(x) with polynomials and approximate the half-iterate of exp with the half-iterate of these polynomials.

hmm ... i wonder if (x^2 + 1)^[h] is analytic at x = 0 for small h.

the reason is that lim h-> 0 (x^2 + 1)^[h] = abs(x) and abs(x) is not analytic at 0.

im not sure about being analytic elsewhere either , though maybe levy böttcher schröder imply so ( i still do not know enough about them ).

also approximating exp with polynomials might be troublesome ; does the n'th approximation of the half-iterate converge when n -> oo ? do we really get an analytic function at n = oo ?

as a bad example that does not make sense in the case of exp(x) but gives an idea of what i mean ( i dont have a good one for the moment ) for instance if an n'th polynomial satisfies f(x) = f(-x) we are in trouble.

polynomials also have zero's and those zero's will need to drift towards oo fast if the sequence of polynomials wants to approximate exp(x) well.

so i suggest working with the (n^2)'th polynomial approximations , rather then compare the n'th with the (n+1)'th.

despite much comment , dont get me wrong , i like this idea.

regards

tommy1729