Newsgroups: sci.math
From: pmontgom@cwi.nl (Peter L. Montgomery)
Subject: Re: Weird Functions
Date: Sat, 2 May 1998 08:03:25 GMT

In article <01bd758c$349fc8c0$1825a3ce@familycomp> 
"Matt Denham" <majora@bendnet.com> writes:
>I assume you know of functions that are of the form x=f(f(f(x))), or:

>The most elementary example of this is:
>
>f(x)=x-1/x
>
>where x is not 0 or 1 (these degenerate to part of their own triad, which
>also includes -1/0).
>
>Can somebody give a function similar to this, but x=f(f(f(f(f(x))))) 
>[pentads of numbers that are cyclic by this function]?
>
	Recall 
                                tan(x) + tan(y)     
                  tan(x + y) =  ----------------
                                1 - tan(x) tan(y)

whenever tan(x), tan(y), tan(x + y) are all defined.
Temporarily leaving t fixed but unspecified, try

                    x + tan(t)
            f(x) = -------------
                    1 - x tan(t)

Then f(tan(x)) = tan(x + t) whenever everything is defined,
This holds for all x, so

         (f^5)(tan(x)) = tan(x + 5*t)

where f^5 denotes the five-fold composition of f.
The tangent function has period pi, so let t be a multiple of pi/5.
That is, tan(t) = +- sqrt(5 +- 2*sqrt(5)).
-- 
        Peter-Lawrence.Montgomery@cwi.nl    San Rafael, California
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From: fredh@ix.netcom.com (Fred W. Helenius)
Newsgroups: sci.math
Subject: Re: Weird Functions
Date: Sat, 02 May 1998 10:08:17 GMT

"Matt Denham" <majora@bendnet.com> wrote:

>I assume you know of functions that are of the form x=f(f(f(x))), <snip>

>The most elementary example of this is:

>f(x)=x-1/x

You mean (x-1)/x, of course.

>where x is not 0 or 1 (these degenerate to part of their own triad, which
>also includes -1/0).
>
>Can somebody give a function similar to this, but x=f(f(f(f(f(x))))) 

Very similar indeed:  let phi = (sqrt(5) + 1)/2, then

f(x) = (phi*x - 1)/x

will do the trick (except for x=0,1,phi,1/phi).

Substituting 1/phi, -phi or -1/phi for phi will also work.

--
Fred W. Helenius	<fredh@ix.netcom.com>

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Von:David G Radcliffe (radcliff@alpha2.csd.uwm.edu)
Betrifft:Re: No f:R -> R with f(f(x)) = g(x) 
Newsgroups:sci.math
Datum:1999/09/08 
 

Philo D. <doozy@earthling.net.nospam> wrote:
: In article <norrisdt.936807309@rintintin.colorado.edu>, Doug Norris
: <norrisdt@rintintin.colorado.edu> wrote: 
:> You're really missing the point of the last person's response.  The very
:> simple function g(x)=-x satisfies your claim. 
: It does not.  There are (discontinuous) functions f : R -> R
: such that f(f(x)) = -x for all real x.

Example:
          /   x+1  if x>0 and ceil(x)  is odd,
          |  -x+1  if x>0 and ceil(x)  is even,
 f(x)  =  |   x-1  if x<0 and floor(x) is odd,
          |  -x-1  if x<0 and floor(x) is even,
          \    0   if x=0. 

-- 
David Radcliffe				radcliff@alpha2.csd.uwm.edu

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Von:Robert Israel (israel@math.ubc.ca)
Betrifft:Re: No f:R -> R with f(f(x)) = g(x) 
Newsgroups:sci.math
Datum:1999/09/08 
 


In article <19990907155921.14021.00003115@ng-cb1.aol.com>,
 seraphsama@aol.com (Seraph-sama) writes:
>  I have found a very simple elementary function g:R -> R and I have proved that
> absolutely NO function f:R -> R, continuous or not, exists such that f(f(x)) =
> g(x). Has anyone ever done this before? Is my result publishable?

E.g. g(x) = x + 2 - 2 exp(x).  It suffices to take any function g such
that 
a) g has exactly one fixed point A (in this case A = 0)
b) there is exactly one point B <> A with g(B) = A 
c) there is some point C with g(C) = B.

No, it isn't publishable.

Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            
Vancouver, BC, Canada V6T 1Z2 


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