# Chaos Theory: period points

• Feb 20th 2010, 06:25 AM
harkapobi
Chaos Theory: period points
Hi, I have this question

Let f: $R \rightarrow R$ (R is the set of real numbers) and let $f^2 = f(f(x))$ intersect the graph of the identity mapping id in just one point (call it p). How many points of period-1 does f have? In how many points does the graph of f intersect that of id?

I know that the point p is a fixed point of $f^2$ and I know that if f has a fixed point then it is also a fixed point of $f^2$, so f can have at most one fixed point. But I just can't seem to get my head around this question.

Yes, if p is a point of period 1 for f, then it is a fixed point for f and so for $f^2$. Thus, if $f^2$ has only one fixed point, then so does f. The only thing you seem to have left undone is the number of points in which the graph of f intersects the graph of the identity function. But if p lies on both y= f(x) and y= x, then y= f(p)= p and p is a fixed point. Since f has only one fixed point, in how many points can y= f(x) intersect y= x?
So is there no way that there can be no fixed points for f and a fixed point for $f^2$?