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Fixed point iteration, locally convergent

  1. Feb 14, 2015 #1
    1. The problem statement, all variables and given/known data
    For which of them will the corresponding fixed point iteration xk+1 = g(xk) be locally convergent to the solution xbar in [0, 1]? (The condition to check is whether |g'(xbar)| < 1.)
    A) 1/x2 -1
    B)...
    C)...

    compute xbar to within absolute error 10-4.
    2. Relevant equations
    3. The attempt at a solution
    I
    don't think I understand the question in general terms, but this is what I think I have to do.
    The solution xbar is 1 because g(1)=0. It's not locally convergent because |g'(1)|=-2

    Since xbar isn't locally convergent I can't computer xbar. But calculating it would look like:
    k | xk
    1 1
    2 -2
    3 .25
    4 128

    Can you please help me understand what this problem is asking for, or similar examples of this problem, or what I'm doing wrong. Thank you
     
  2. jcsd
  3. Feb 14, 2015 #2

    mfb

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    That is not a fixed point as your point changes (from 1 to 0) and the next iteration becomes impossible because you would have to divide by zero.
    What does "fixed point" mean? What is fixed?
    You'll have to check that again once you find the right xbar.
    Oh sure you can, just not directly with the given iteration.
     
  4. Feb 14, 2015 #3
    I'm guessing a fixed point is a point that can't change. My teacher said xbar is the solution, but I'm not sure of what and how to get it.
     
  5. Feb 14, 2015 #4

    mfb

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    It does not change if you apply g to it, so the next iteration starts with the same value again. What does that mean in terms of equations?
     
  6. Feb 14, 2015 #5
    Is it 0 because g(0)=0
     
  7. Feb 14, 2015 #6

    mfb

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    ##g(0)=\frac{1}{0^2}-1## is undefined.
     
  8. Feb 14, 2015 #7
    Sorry I was thinking 2x for no reason.

    I'm looking at the graph and all the values of x seem to have different y's. I'm not sure what I could put into g(x) that wouldn't change.

    Where does the 1+51/2/2 come from in the attached picture.
     

    Attached Files:

    Last edited: Feb 14, 2015
  9. Feb 14, 2015 #8
    I did fixed point iteration, but it's definitely not converging. I wish I knew what xbar is and how to get it.
     
  10. Feb 14, 2015 #9

    SammyS

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    Here's that image:


    capture-png.79090.png
    I would have never guessed this from the OP.

    Apparently, ##\displaystyle\ \frac{1+\sqrt{5}}{2} \ ## is xbar and is a solution to the equation, g(x) = x .
     
  11. Feb 14, 2015 #10
    That's the root of g'(x), but my problem doesn't have any roots, and my problem has an interval. So I have no idea.
     
  12. Feb 15, 2015 #11

    mfb

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    Did you draw a sketch? You can get a rough idea where the fixed point is, and solve the equation numerically.
     
  13. Feb 15, 2015 #12
    Yes I graphed it. I don't know what the I 'm looking for though. I tried the roots of the of f(x) and g'(x). And points close to the intersection of the graphs, but nothing is converging when I do it numerically.

    edit:
    I just tried it with another g(x) and it's converging to the root of f(x). So I'm pretty sure the root of f(x) is what I'm looking for and the original g(x) is not convergent.
     
    Last edited: Feb 15, 2015
  14. Feb 15, 2015 #13

    mfb

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    What is f(x)?

    It is not convergent, right, but it still has a fixed point. Can you write down the equation for the fixed point? What has to be true at this point? We said it in words, but you'll need the equation.
     
  15. Feb 15, 2015 #14

    SammyS

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    I take it that the image you posted, (and what I responded to), was for an example with the function ## \displaystyle \ g(x)=1+\frac{1}{x} \ ##. For this example problem, ## \displaystyle \ x_\text{bar}=\frac{1+\sqrt{5}}{2} \ ## because for that value of xbar, g(xbar) = xbar . In other words, xbar is a root of the equation, g(x) = x. (It's one of two roots.)

    xbar is not a root (zero) of g'(x). Note that ## \displaystyle \ g'(x)=-\frac{1}{x^2} \ ## which has no root. However, as stated in the example, ## \displaystyle \ \left|g'(x_\text{bar})\right|<1 \, ,\ ## so the fixed point iteration gives a sequence converging to xbar for the example problem.

    It's not entirely clear to me what the problem is that you're asked to solve. I guess that you're to do similarly for ## \displaystyle \ \frac{1}{x^2}-1 \ ## (or is it ## \displaystyle \ \frac{1}{x^2-1}\, ? \ ##) . It's likely the former since that gives xbar in [0, 1] .

    Are you calling the function f(x) or calling it g(x) for your problem?
     
  16. Feb 15, 2015 #15
    I figured it out. Thank for the help
     
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