Fixed Point Iteration Convergence

  • Thread starter Scootertaj
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  • #1
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Homework Statement


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

Homework Equations



Seen above



The Attempt at a Solution


Not really sure where to start.
I don't know, in general, what properties are required.
 

Answers and Replies

  • #2
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Try to find bounds on x_(n+1) in terms of x_n ,y_n ( for instance, when it's less than x_n).
 
  • #3
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Try to find bounds on x_(n+1) in terms of x_n ,y_n ( for instance, when it's less than x_n).
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.
 
  • #4
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Try to find bounds on x_(n+1) in terms of x_n ,y_n ( for instance, when it's less than x_n).
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:
  • #5
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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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