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Math Help - Proving limit goes to neg. infinity with fixed points.

  1. #1
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    Proving limit goes to neg. infinity with fixed points.

    The function f is defined by

    f(x)=((x^3)+1)/3

    has three fixed points, say alpha, beta and gamma where
    -2<alpha<-1, 0<Beta<1, 1<gamma<2.

    For arbitrarily chosen x_1, define {x_n} by setting x_n+1=f(x_n).

    If x_1 < alpha, prove that x_n --> neg infinity as n ---> infinity.


    Right now my thoughts is to first show that this is a mono decreasing sequence. Then I want to show that every |x_n - x_n+1 | >= d for some d. Then it should be easy to show that the sequence goes to negative infinity. My problem is showing that |x_n - x_n+1| >= d. I can't seem to figure out the correct way to say this mathematically. It is possible that this current route wont work at all and that's why I can't show it mathematically. Thanks for any assistance.


    P.S How do you guys put the mathematical symbols into the forum message?

    Thanks!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Dagger2006 View Post
    The function f is defined by

    f(x)=((x^3)+1)/3

    has three fixed points, say alpha, beta and gamma where
    -2<alpha<-1, 0<Beta<1, 1<gamma<2.

    For arbitrarily chosen x_1, define {x_n} by setting x_n+1=f(x_n).

    If x_1 < alpha, prove that x_n --> neg infinity as n ---> infinity.


    Right now my thoughts is to first show that this is a mono decreasing sequence. Then I want to show that every |x_n - x_n+1 | >= d for some d. Then it should be easy to show that the sequence goes to negative infinity. My problem is showing that |x_n - x_n+1| >= d. I can't seem to figure out the correct way to say this mathematically. It is possible that this current route wont work at all and that's why I can't show it mathematically. Thanks for any assistance.


    P.S How do you guys put the mathematical symbols into the forum message?

    Thanks!
    Let $$x_0 be the negative fixed point, and put x_1=x_0-\varepsilon_1 with \varepsilon_1>0.

    Now show that:

    x_2=x_0-\varepsilon_2

    where \varepsilon_2>\varepsilon_1>0

    (mathematical typesetting on MHF uses a LaTeX a add-in to the bulletin board software, you can find further information on how to use it >>here<< )

    CB
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