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Math Help - Fixed point iteration help

  1. #1
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    Fixed point iteration help

    With f defined as f(x)=e^x+6x-5, show the fixed point iteration
    x_{n+1}=\frac{5-e^{x_{n}}}{6}, n=0,1,...
    must converge to x_{*} with x_{0} $\in$ [0,1]

    Had a go at this by using contraction mapping theorem and putting it in the form f(x)=x-g(x), and as f(x)=0 then can work out what g(x) is but I'm not sure it is a contraction so not sure how can know it converges?
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    Re: Fixed point iteration help

    Quote Originally Posted by Bloodzeed View Post
    With f defined as f(x)=e^x+6x-5, show the fixed point iteration
    x_{n+1}=\frac{5-e^{x_{n}}}{6}, n=0,1,...
    must converge to x_{*} with x_{0} $\in$ [0,1]

    Had a go at this by using contraction mapping theorem and putting it in the form f(x)=x-g(x), and as f(x)=0 then can work out what g(x) is but I'm not sure it is a contraction so not sure how can know it converges?
    You must examine the function

    g(x)=\frac{5-e^{x}}{6} on the interval [0,1], and see if it is a contraction or not. What criteria do you have for showing a function is a contraction?
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    Re: Fixed point iteration help

    Quote Originally Posted by Ackbeet View Post
    You must examine the function

    g(x)=\frac{5-e^{x}}{6} on the interval [0,1], and see if it is a contraction or not. What criteria do you have for showing a function is a contraction?
    but here we dont have clue about L, except that it should be between 0 and 1 so how do we prove in general that this function is a contraction?
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    Re: Fixed point iteration help

    Quote Originally Posted by Darkprince View Post
    but here we dont have clue about L, except that it should be between 0 and 1 so how do we prove in general that this function is a contraction?
    Is the function differentiable everywhere? If so, can you put a bound on the derivative in the interval of interest? See your other thread for info about the derivative approach.
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    Re: Fixed point iteration help

    Quote Originally Posted by Ackbeet View Post
    Is the function differentiable everywhere? If so, can you put a bound on the derivative in the interval of interest? See your other thread for info about the derivative approach.
    Hello again at this post Thank you very much, I think I have found another way to solve this exercise if you can review it and comment on how rigorous it is please! (and if it is correct)

    abs(g(x)-g(y))=abs((e^(y)-e^(x)) / 6) = 1/6 * abs ( e^(x) * (e^(y-x)-1))=e^(x)/6 * abs (e^(y-x)-1), since e^(x)=abs(e^(x))

    Then e^(x)/6 * abs (e^(y-x)-1) <= e/6 * abs (e^(y-x)-1) (since x=1 gives e^(x)=1 and x=1 is the maximum of the interval [0,1]

    Now e/6 * abs (e^(y-x)-1) <= e/6 * abs( (e-1) (y-x))= e/6 * (e-1) * abs(x-y) [using that abs(e-1) = e-1 and that abs(y-x)= abs (x-y)

    So finally abs(g(x)-g(y)) <= k abs(x-y) where k= e/6 * (e-1) and 0<k<1


    Hope I found a rigorous, correct proof for contraction of our function, I will really appreciate if you review it!! Thank you very much again!!
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    Re: Fixed point iteration help

    Quote Originally Posted by Darkprince View Post
    Hello again at this post Thank you very much, I think I have found another way to solve this exercise if you can review it and comment on how rigorous it is please! (and if it is correct)

    abs(g(x)-g(y))=abs((e^(y)-e^(x)) / 6) = 1/6 * abs ( e^(x) * (e^(y-x)-1))=e^(x)/6 * abs (e^(y-x)-1), since e^(x)=abs(e^(x))

    Then e^(x)/6 * abs (e^(y-x)-1) <= e/6 * abs (e^(y-x)-1) (since x=1 gives e^(x)=1 and x=1 is the maximum of the interval [0,1]

    Now e/6 * abs (e^(y-x)-1) <= e/6 * abs( (e-1) (y-x))
    Not sure where that's coming from. How do you get that

    \frac{e}{6}\,|e^{y-x}-1|\le\frac{e}{6}\,|(e-1)(y-x)|?

    = e/6 * (e-1) * abs(x-y) [using that abs(e-1) = e-1 and that abs(y-x)= abs (x-y)

    So finally abs(g(x)-g(y)) <= k abs(x-y) where k= e/6 * (e-1) and 0<k<1


    Hope I found a rigorous, correct proof for contraction of our function, I will really appreciate if you review it!! Thank you very much again!!
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    Re: Fixed point iteration help

    Quote Originally Posted by Ackbeet View Post
    Not sure where that's coming from. How do you get that

    \frac{e}{6}\,|e^{y-x}-1|\le\frac{e}{6}\,|(e-1)(y-x)|?

    Consider (e^z)-1 and (e-1)*z. If you sketch both for z in [0,1] you see that (e^z)-1<=(e-1)*z for z in [0,1] So it can be justified by a sketch, don't really know if there is an algebraic manipulation for that. So does the proof considered rigorous?
    Last edited by Darkprince; November 8th 2011 at 11:17 AM.
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    Re: Fixed point iteration help

    To use the derivative approach, all you need do is show that the derivative exists everywhere and is less than 1 in magnitude on the interval. That implies the function is Lipschitz with Lipschitz constant less than one, which implies that the function has a unique fixed point on that interval.
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    Re: Fixed point iteration help

    Quote Originally Posted by Ackbeet View Post
    To use the derivative approach, all you need do is show that the derivative exists everywhere and is less than 1 in magnitude on the interval. That implies the function is Lipschitz with Lipschitz constant less than one, which implies that the function has a unique fixed point on that interval.
    Thank you again, sorry I edited my post, I just found that that inequality can be justified by a sketch as I wrote above. Do you think now is a good proof?

    I think if I can justify the inequality even by a sketch is still justified and I am obtaining an exact L and at the same moment proving directly that it is a contraction!

    But you may have a different opinion (your opinion is obviously "heavier" than mine )
    Last edited by Darkprince; November 8th 2011 at 01:25 PM.
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