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Math Help - Chaos Theory: Linear Mappings

  1. #1
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    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?

    Please help,
    Katy
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  2. #2
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    Quote Originally Posted by harkapobi View Post
    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?

    Please help,
    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.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    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.

    Please help,
    Katy
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  4. #4
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    I've come up with this answer, could someone please let me know if its right or wrong please?

    If a=1, f(x)=x, then all points are prime period 1 points.

    If a=-1, f(x)=-x then x=0 is a prime period 1 point, and all other points are prime period 2 because f(f(x))=x.

    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.

    And there are no other possiblilities are there? Is this enough to show that there are only prime period 1 and 2 points for a linear mapping?
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