
Re: Iteration formula transformation
Posted:
Feb 11, 2009 1:15 PM


On 11 fév, 17:03, Gottfried Helms <he...@unikassel.de> wrote: > Am 11.02.2009 15:50 schrieb Matt: > > > > >> but how do you find the derivative of f^n(x) wrt to n > >> without having the fractional iterate before? > > > You don't. I don't claim that this formula helps to find the > > fractional iterate in the first place. > > I see. > > > iterates. All the formula does is provide a mechanical way of > > transforming a known f^n(x) into the form f^n(x) = phi^1(n + phi(x)). > > Yes, got it. Thanks! > > Gottfried
Bonsoir Matt and Gottfried,
I am sorry I mistook about 'your g(r,x)' function. The given formula is right and very interesting: Let us write g(r,x) = f[r](x) ,r real phi(g(r,x)) = phi(x) + r derivation with r , g'r(r,x)*phi'(g(r,x)) = 1 for r =0 ,g(0,x)=x , so g'r(0,x)*phi'(x) = 1 And your given formula phi(x) = integral(dx/g'r(0,x)) is simpler than mine: phi(x)=integral(g'r(r,x)dx/g'x(r,x))
Amicalement, Alain

