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Sequence Convergence & Fixed Point Theorem

  1. May 8, 2013 #1
    1. The problem statement, all variables and given/known data

    Let g(x)= (2/3)*(x+1/(x^2)) and consider the sequence defined by pn= g(pn-1) where n≥1

    a) Determine the values of p0 [itex]\in[/itex] [1,2] for which the sequence {pn} from 0 to infinity converges.

    b) For the cases where {pn} converges (if any), what is the rate of convergence?


    2. Relevant equations

    http://en.wikipedia.org/wiki/Fixed-point_theorem

    Fixed Point Theorem

    attachment.php?attachmentid=58589&stc=1&d=1368037754.png


    3. The attempt at a solution

    For part a, my answer is that ANY point p0 between 1 and 2 will converge, because the sequence satisfies the fixed point theorem.


    g(x) exists on [1,2] and is continuous
    g'(x) = (2/3)(1-2/(x^3)) exists and is continuous on [1,2]

    There is a positive constant k<1 for which |g'(x)|≤k

    By plotting g'(x), i found k = 2/3

    Therefore, the fixed point theorem is satisfied, and so should the answer be "any value of p0 from 1 to 2 will cause the sequence to converge"?

    For part b, I am not sure how to find the rate of convergence. I heard that you have to take the taylor series expansion, but I am not sure. Any help? Thanks
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. May 8, 2013 #2

    haruspex

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    Yes, that shows it converges throughout the interval.
    For the rate, let p be the fixed point and x be small. If the rate of convergence is a then g(p+x) is approximately p + ax, i.e. the error gets multiplied by factor a at each step. Take the Taylor expansion of g at p, just the first two terms, and see what that simplifies to.
     
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