Let's call a real-valued continuous function an n-th identity root when it generates the identity function after exactly n compositions with itself. For example f(x)=-x is a 2nd identity root because f(f(x))=x.

1. The function f(x)=1-1/x is a 3rd identity root. Unfortunately it is undefined at x=0. Are there identity roots for n>2 which are defined for all real numbers?

2. For a given real number c and n>1, give an example of an n-th identity root which is defined for all real numbers except c. How many such roots exist?

1) no

given n>2
suppose it is possible and there exists an  f that is an an n-th identity root defined for all real numbers
 f must be a bijection
 f must be strictly increasing or strictly decreasing
 let f'(x)=f(f(x)
 f' is strictly increasing and is an n'-th identity root for som n'>1
 f' must not be the identity, so for some x f'(x) > x or f'(x) < x.
 wlog f'(x)>x  (if < replace < for > in remaining)
 f'(f'(x)>f'(x)>x (strictly increasing, apply f to both sides)
 f'(f'(f'(x)))>x
 and so on. in particular, let f" = f' composed with itself n' times
 f"(x)>x thus f' is not an n'-th identity root which is a contradition

2) I'll just look at n>2
  Here is an extension of 1-1/x to simply add some linear pieces into the "circle"
if x<c : f(x) = (c+n-2)-1/(x-c)
if c<x<c+1 : f(x) = (c+1)-1/(x-c)
if c+1<=x<=c+n-2 : f(x) = x-1
if c+n-2 < x : f(x) = (c+n-2)-1/(x-(c+n-3))

There are infinite such roots, in fact an uncountable number.  In fact, An even larger number than the number of reals.  oodles of infinities basically.

Posted by Joel on 2006-11-07 20:33:02