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Math Help - Convergence of fixed point iteration question

  1. #1
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    Convergence of fixed point iteration question

    Let p>0 and  x = \sqrt{p+\sqrt{p+\sqrt{p+ \cdots }}} , where all the square roots are positive. Design a fixed point iteration x_{n+1} = F (x_{n}) with some F which has x as a fixed point. We prove that the fixed point iteration converges for all choices of initial guesses greater than -p+1/4.



    Let x_{n+1}=F(x_{n})= \sqrt{p+x_{n}} so x is a fixed point for F since F(x)=x.
    Now let g(x)=\sqrt{p+x}. We have g'(x)=\frac{1}{2 \sqrt{p+x}} .

    I can see that for   x > -p + 1/4 , we have that g'(x) <1.

    From there I am not sure how to proceed to obtain convergence for  x_{0} >  -p +\frac{1}{4} .
    Last edited by math8; April 19th 2010 at 01:55 PM.
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  2. #2
    MHF Contributor chisigma's Avatar
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    Such a problem recently has been 'attacked' in...

    http://www.mathhelpforum.com/math-he...nvergence.html

    Here the sequence is defined as...

    x_{n+1} = \sqrt {p+x_{n}} \rightarrow \Delta_{n}= x_{n+1}-x_{n} = \sqrt{p + x_{n}} - x_{n} = f(x_{n}) (1)

    The fixed point is the x_{0} for which is f(x_{0})=0 so that is...

    x_{0} = \frac{1 + \sqrt{1 + 4 p}}{2} (2)

    ... and that means that it must be p > - \frac{1}{4}...

    Kind regards

    \chi \sigma
    Last edited by chisigma; April 20th 2010 at 04:58 AM.
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