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Topic: Iteration formula transformation
Replies: 14   Last Post: Feb 12, 2009 12:45 PM

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 alainverghote@gmail.com Posts: 812 Registered: 5/31/08
Re: Iteration formula transformation
Posted: Feb 11, 2009 1:15 PM

On 11 fév, 17:03, Gottfried Helms <he...@uni-kassel.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,

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