From: rusin@washington.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: f(f(f(x))) = x Date: 24 Feb 1995 16:31:38 GMT In article <3iktj5$kad@ixnews3.ix.netcom.com>, david petry wrote: >In >dalarson@students.wisc.edu (david larson) writes: >>... I was wondering if there exist any functions (other than f(x) >>= x) for which f(f(f(x))) = x. [Petry's complex-plane example deleted] >I don't know if any functions that take on only real values for real >arguments satisfy your equation. The original poster did not specify domain or range. The use of composites suggests that domain must equal range, but that's it; then the desired condition on f is that it perform a permutation of order 3 on the domain. In particular, the points in the domain fall into orbits of size 1 or 3; the poster wanted an example in which some 3 points are permuted: x, y, z distinct and f(x)=y, f(y)=z, f(z)=x. Certainly we can do this with real numbers. Take f(x)=x for all x except that f permute {1,2,3} cyclically. The interesting question is whether a _continuous_ such f exists. The answer is no. Assume x, y, and z are real and the domain of f includes an interval with all three points in it. Relabelling suitably, we may assume either xy>z; the two cases are similar so I'll assume the former. Then for any number c in the interval (y,z) we may use the intermediate value theorem twice to find elements w1, w2 in the intervals (x,y) and (y,z) respectively with f(w1)=c, f(w2)=c. Then f is not one-to-one, a contradiction (that is, w1=f(f(f(w1)))=f(f(c))=...=w2.) dave