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Topic: fractional iteration of f(x)=1/(1+x)
Replies: 31   Last Post: Feb 7, 2009 11:47 PM

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 matt271829-news@yahoo.co.uk Posts: 2,136 Registered: 1/25/05
Re: fractional iteration of f(x)=1/(1+x)
Posted: Feb 7, 2009 11:47 PM

On Jan 23, 8:09 pm, Matt <matt271829-n...@yahoo.co.uk> wrote:
> On Jan 23, 10:58 am, "alainvergh...@gmail.com"
>
>
>
>
>
> <alainvergh...@gmail.com> wrote:

> > On 23 jan, 11:18, Gottfried Helms <he...@uni-kassel.de> wrote:
>
> > > Am 23.01.2009 08:51 schrieb Gottfried Helms:
>
> > > > Am 22.01.2009 22:19 schrieb Gottfried Helms:
> > > >> Just for exercise I considered the function
>
> > > >>  f(x) = 1/(1+x)
>
> > > >> under iteration and a meaningful description for fractional
> > > >> iteration. I thought, it should be possible to use

>
> > > > For the function f°h(x) with x=1, depending on h I get
>
> > > >   f°h(1) = phi + sqrt(5)*sum(k=1,inf,((2-3*phi)*(phi-1)^h) ^k)
>
> > > > where phi = (sqrt(5)-1)/2 ~ 0.61803...
>
> > > > Here the order of exponents h and k in the term
>
> > > >     ((2-3*phi)*(phi-1)^h) ^k
>
> > > ...
> > > This is shorter expressed by the formula for geometric series:
> > > In
> > >     f°h(1) = phi + sqrt(5)*sum(k=1,inf,((2-3*phi)*(phi-1)^h) ^k)
> > > let
> > >    c = (2-3*phi)*(phi-1)^h

>
> > > then write f1(h) for f°h(1) then with that c
>
> > >    f1(h) = phi + sqrt(5)*c/(1-c)
>
> > > Hmm. Maybe, I'll arrive at Matt's formula this way...
>
> > > --------------------------------------------------------
>
> > > The previous question remains open: is there another
> > > approach to the fractional iterates, which would give
> > > different values? Because this function seems sufficient
> > > simple so that the whereabouts of this "regular iteration
> > > using fixpoint-shift" could be studied more easily than
> > > in the tetration-discussion, where fractional iterates
> > > (computed with this method) usually lead to strongly divergent
> > > series, even with complex coefficients, for which we
> > > then cannot find the values with arbitrary precision.

>
> > > For instance, in the plot in [1] we notice, that the
> > > spiral around the limit case f1(inf) = phi has nearly
> > > equal angles when iterated with a fixed h. But this
> > > is not exact the case (see pic 3).
> > > Could an equi-angular interpolationmethod for fractional
> > > iterates be found?

>
> > > Gottfried Helms
>
> > > [1]http://go.helms-net.de/math/tetdocs/FracIterAltGeom.htm
> > > (updated)

>
> > > Gottfried- Masquer le texte des messages précédents -
>
> > > - Afficher le texte des messages précédents -
>
> > Good morning,
>
> > The problem of continuous iteration (ax +b)/(cx +d)
> > is simply solve:
> > from a direct classification according to the number
> > of fixed points:zero, one, two,infinity and pole or not.

>
> > Here:
> > h(x) =1/(1+x) two fixed points  x1 = (-1+sqrt(5))/2
> >                                 x2 = (-1-sqrt(5))/2
> > and a pole x =-1
> > May be written (h(x)-x1)/(h(x)-x2)=a*(x-x1)/(x-x2) ,
> > for x = 0 ,h(0) = 1   then a = (1+sqrt(5)/(1-sqrt(5))

>
> > And also (h^[n](x)-x1)/(h^[n](x)-x2)=a^n *(x-x1)/(x-x2)
> > just mind 'a'  is real negative!

>
> For f(x) = (a*x + b)/(c*x + d), I get
>
>    f^n(x) = a/c + (b*c - a*d)/c*((c*x - a + alpha)*alpha^(n - 1) -
> (c*x - a + beta)*beta^(n - 1)) / ((c*x - a + alpha)*alpha^n - (c*x - a
> + beta)*beta^n)
>
> with
>
>    alpha = (a + d + sqrt((a - d)^2 + 4*b*c)) / 2
>    beta  = (a + d - sqrt((a - d)^2 + 4*b*c)) / 2

It occurred to me later that this method fails when (a - d)^2 + 4*b*c
= 0, but I only just got round to trying to figure it out. In this
case we have a function of the form

f(x) = (a*x + b^2)/(-c^2*x + a + 2*b*c)

with a solution

f^n(x) = ((a + b*c - b*c*n)*x + b^2*n) / (-c^2*n*x + a + b*c +
b*c*n)