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Fixed Point Iteration Convergence

  1. Oct 17, 2011 #1
    1. The problem statement, all variables and given/known data
    Consider the system
    x = [itex]\frac{1}{\sqrt{2}}[/itex] * [itex]\sqrt{1+(x+y)^2}[/itex] - 2/3
    y = x = [itex]\frac{1}{\sqrt{2}}[/itex] * [itex]\sqrt{1+(x-y)^2}[/itex] - 2/3

    Find a region D in the x,y-plane for which a fixed point iteration

    xn+1 = [itex]\frac{1}{\sqrt{2}}[/itex] * [itex]\sqrt{1+(x_n + y_n)^2}[/itex] - 2/3

    yn+1 = [itex]\frac{1}{\sqrt{2}}[/itex] * [itex]\sqrt{1+(x_n - y_n)^2}[/itex] - 2/3

    is guaranteed to converge to a unique solution for any (x0,y0)[itex]\in[/itex]D

    a) State clearly what properties this region must have
    b) find a region with these properties and show it has these properties

    2. Relevant equations

    Seen above



    3. The attempt at a solution
    Not really sure where to start.
    I don't know, in general, what properties are required.
     
  2. jcsd
  3. Oct 18, 2011 #2
    Try to find bounds on x_(n+1) in terms of x_n ,y_n ( for instance, when it's less than x_n).
     
  4. Oct 18, 2011 #3
    I'm confused as to where that leads :(.

    Also, I realized there is a typo. There shouldn't be an "x =" in the second line.
     
  5. Oct 18, 2011 #4
    Here is what I tried:

    Assume xn<yn.

    Then,

    yn+1 = [itex]\frac{1}{\sqrt{2}}[/itex]*[itex]\sqrt{1+(x_n-y_n)^2}[/itex]
    = [itex]\frac{1}{\sqrt{2}}[/itex] * 1 - 2/3
    = [itex]\frac{3-2\sqrt{2}}{3\sqrt{2}}[/itex]

    So, y_n+1 bounded above by [itex]\frac{3-2\sqrt{2}}{3\sqrt{2}}[/itex] ?
     
    Last edited: Oct 18, 2011
  6. Oct 18, 2011 #5
    amath.colorado.edu/courses/5600/2005fall/Tests/.../final-96ans.pdf

    This problem seems similar (#7) but I don't quite understand it.
     
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