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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 11, 2009 3:06 PM

> > 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) = 2*x = integral dx / d/d0 f^0(x) ???
>
> i dont get it.
>
> some examples plz
>
> regards
>
> tommy1729

( corrected typo )