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

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 amy666 Posts: 1,859 Registered: 6/27/07
Re: Iteration formula transformation
Posted: Feb 12, 2009 7:22 AM

> On Feb 11, 8:05 pm, amy666 <tommy1...@hotmail.com>
> wrote:

> > > It's well known and straightforward that if f(x)
> =
> > > phi^-1(1 + phi(x))
> > > for some function phi, then f^n(x) = phi^-1(n +
> > > phi(x)), where "f^n"
> > > denotes iteration of f. But say we have an

> expression
> > > for f^n that we
> > > found some other way and we know is a valid
> > > continuous function
> > > iteration, then how to recover phi? As an

> alternative
> > > to guessing the
> > > correct form followed by some trial-and-error, I
> > > found

> >
> > >     phi(x) = Integral dx/g(0,x)
> >
> > > where g(n,x) = d/dn f^n(x).
> >
> > > Probably nothing new, but kind of cute I thought.
> >
> > ??
> >
> > 2*x is the nth iterate of x+2.
> >
> > phi(x) = x + 2 = integral dx / d/d0 f^0(x) ???
> >
> > i dont get it.
> >
> > some examples plz

>
> Very simple example:
>
> f(x) = x + 2
> f^n(x) = x + 2*n (= nth iterate of f(x))
> g(n,x) = d/dn f^n(x) = 2
> phi(x) = Integral dx/g(0,x) = Integral dx/2 = x/2 + C
> (C is an
> arbitrary constant)
> phi^-1(x) = 2*(x - C)
> phi^-1(n + phi(x)) = x + 2*n = f^n(x)
>
> Another example:
>
> f(x) = a*x/(x + a) (a is any constant)
> f^n(x) = a*x/(n*x + a)
> g(n,x) = d/dn f^n(x) = -a*x^2/(n*x + a)^2
> phi(x) = Integral dx/g(0,x) = Integral -a*dx/x^2 =
> a/x + C
> phi^-1(x) = a/(x - C)
> phi^-1(n + phi(x)) = a*x/(n*x + a) = f^n(x)

hmm

how do you work out f(x) = e*x ?