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Math Help - Fixed point method : where is my error?

  1. #1
    MHF Contributor arbolis's Avatar
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    Fixed point method : where is my error?

    I've mostly solved the following problem :
    The equation x^2-x-1=0 has a zero in [-1,0].
    a)Verify that the function g(x)=x^2-1 has 2 fixed points and that the derivative of g in a fixed point has its absolute value greater than 1. Take note that the existence of a fixed point doesn't imply that the derivative evaluated in this point has an absolute value lesser than 1. DONE
    b)Show that
    g(x) satisfies the conditions of existence of a fixed point in [-1,0]. DONE (I showed that g was contractive on this interval)
    c)Chose a
    x_0 closed to the fixed point of g in [-1,0], analyze the convergence of the sequence x_{n+1}=g(x_n). This is where I'm stuck in. My work : I chose x_0=-1. I get x_1=0. Then x_2=-1. It shouldn't be possible! It's not convergent, it will oscillate infinitely! Where is my error? If g is contractive on [-1,0], it must have a fixed point, so what am I doing wrong here?
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi
    Quote Originally Posted by arbolis View Post
    I've mostly solved the following problem :
    The equation x^2-x-1=0 has a zero in [-1,0].
    a)Verify that the function g(x)=x^2-1 has 2 fixed points and that the derivative of g in a fixed point has its absolute value greater than 1. Take note that the existence of a fixed point doesn't imply that the derivative evaluated in this point has an absolute value lesser than 1. DONE
    b)Show that
    g(x) satisfies the conditions of existence of a fixed point in [-1,0]. DONE (I showed that g was contractive on this interval)

    g is not contractive on
    [-1,\,0] : \frac{|g(-0.9)-g(-0.8)|}{|-0.9-(-0.8)|}=\frac{|0.9^2-0.8^2|}{0.9-0.8}=0.9+0.8=1.7>1. To show that g has a fixed point on [-1,\,0], think about the intermediate value theorem.

    c)Choose a x_0 closed to the fixed point of g in [-1,0], analyze the convergence of the sequence x_{n+1}=g(x_n). This is where I'm stuck in. My work : I chose x_0=-1. I get x_1=0. Then x_2=-1. It shouldn't be possible! It's not convergent, it will oscillate infinitely!
    x_0=-1 is not the only possibility, x_0\in [-1,\,0]. You should try to link the fixed point and the limit of the sequence : let's assume that (x_n) has a limit L. What can be said about L ?
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  3. #3
    MHF Contributor arbolis's Avatar
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    g is not contractive on [-1,0]
    Oh... Therefore this website is wrong : Homeomath : fonction contractante, where it states " Si une fonction définie et dérivable sur un intervalle I et telle que pour tout réel x de I on a f ' (x) < 1 alors f est contractante". It should be the absolute value of f'(x). Ok, I will try with the intermediate value theorem.

    What can be said about L ?
    I think L is the negative root of x^2-x-1.
    You should try to link the fixed point and the limit of the sequence
    I still don't know what to chose for x_0. According to my calculator the root of the x^2-x-1 is between -0.62 and -0.61, so it means I should start x_0 close to it? If yes, then how could I have guessed it without a calculator?
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by arbolis View Post
    I think L is the negative root of x^2-x-1.
    OK
    I still don't know what to chose for x_0. According to my calculator the root of the x^2-x-1 is between -0.62 and -0.61, so it means I should start x_0 close to it?
    In fact, whichever x_0 one chooses, the sequence will "oscillate" : x_{2n} will tend to 0 (resp. -1) and x_{2n+1} will tend to -1 (resp. 0). I think that asking you to choose x_0 "closed to the fixed point" is just a means of preventing you from choosing x_0=-1 or x_0=0 which gives a sequence that can be studied very easily.

    Let x_0\in]-1,\,0[ (choose any value but the negative root of x^2-x-1 )
    The sequence (x_n) is neither increasing nor decreasing. (reason : g(x)-x is positive on [-1,\,L[ and negative on ]L,\,0]) In this case, we usually study the two sequences defined by a_n=x_{2n} and b_n=x_{2n+1}. Since (a_n) and (b_n) are monotone*, finding their limits (if they do converge) is not too hard. If these two limits are equal, the sequence (x_n) converges, if they are not equal, (x_n) diverges.

    * a_n=x_{2n}=g(x_{2n-1})=g\circ g(x_{2(n-1)})=g\circ g(a_{n-1}) and, as g is decreasing on [-1,\,0], g\circ g is increasing thus (a_n) is monotone. It'll be increasing if a_0<a_1 and decreasing otherwise. The same goes for (b_n).
    Last edited by flyingsquirrel; May 24th 2008 at 11:51 AM.
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