# Math Help - Chaos Theory: Linear Mappings

1. ## Chaos Theory: Linear Mappings

Hi, I have the following question

Prove that each linear mapping $f: R \rightarrow R$ (R is the set of real numbers) has only periodic points with prime period 1 or 2.

How do I go about this? I know a linear mapping has the form f(x)=ax for some a. Do I need to split it into cases for different values of a?

Katy

2. Originally Posted by harkapobi
Hi, I have the following question

Prove that each linear mapping $f: R \rightarrow R$ (R is the set of real numbers) has only periodic points with prime period 1 or 2.

How do I go about this? I know a linear mapping has the form f(x)=ax for some a. Do I need to split it into cases for different values of a?

Katy
We have to be a little bit careful here. Because you are talking about linear mappings in the sense of "chaos theory", "periodic" refers to composition, not f(x+ p)= f(x) as in "periodic" trig functions.

Saying that a function, f, has "period 1" means that f(x)= x for all x.
Saying that it has "period 2" means that f(f(x))= x for all x.

For what a is f(x)= ax= x? If a is not that value, is there a value that makes [tex]f(f(x))= a(ax)= a^2x= x[tex]?

Suppose a not either of those values. Is it possible that can "periodic with period n"- that is that $f^n(x)= x$? Here, " $f^n(x)$ means the composition: f(f(f(...f(x)...))). And, for f(x)= ax, that is $a^nx$.

3. Originally Posted by HallsofIvy
We have to be a little bit careful here. Because you are talking about linear mappings in the sense of "chaos theory", "periodic" refers to composition, not f(x+ p)= f(x) as in "periodic" trig functions.

Saying that a function, f, has "period 1" means that f(x)= x for all x.
Saying that it has "period 2" means that f(f(x))= x for all x.

For what a is f(x)= ax= x? If a is not that value, is there a value that makes [tex]f(f(x))= a(ax)= a^2x= x[tex]?

Suppose a not either of those values. Is it possible that can "periodic with period n"- that is that $f^n(x)= x$? Here, " $f^n(x)$ means the composition: f(f(f(...f(x)...))). And, for f(x)= ax, that is $a^nx$.
I'm a little confused. I thought that the question was saying that the periodic points of a linear mapping only have prime period 1 or 2. So f(x)=x for some point x, not all x? Isn't a period 1 point the point were x intersects the identity mapping?

For example if a =-1, then f(x)= -x which has one prime period 1 point at 0 and infinitely many prime period 2 points because f(f(x))= x. Is this right?

I know how to prove that there are prime period 1 and 2 points but I dont see how to prove that there arent any higher prime period points.

For any other value of a, f(x)= ax has the fixed point x=0 which has prime period 1. $f^n(x)=a^nx$ (for n=2,3,4...) has only one fixed point at x=0, which is a prime period 1 point and therefore not prime period n.