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Proof of convergence

  1. Mar 8, 2008 #1
    [SOLVED] Proof of convergence

    1. The problem statement, all variables and given/known data
    Let [itex] \alpha[/tex] be a fixed point of [tex]x = g(x)[/tex] and let [tex](x_n)[/tex] be the sequence generated by the fixed point iteration scheme. Using the first two terms of the Taylor series for
    [tex]g(x)[/tex] about [itex]\alpha[/tex] we can get an approximation for [tex]g(x_n)[/tex]:
    [itex]g(x_n) = g(\alpha) + (x_n - \alpha)g'(\alpha)[/tex]
    (Assuming the terms in the sequence are close to α we have neglected non-
    linear terms.)

    Show that [itex]|x_n - \alpha| = |x_0 - \alpha||g'(\alpha)|^{n}[/tex] for all n [tex] \geq 1[/tex] hence show that the sequence [tex](x_n)[/tex] converges to [itex]\alpha[/tex] for [tex]g'(x)< 1[/tex]

    2. Relevant equations

    3. The attempt at a solution
    I'm not so worried about the proving second part (looks obvious if the first part is true) but before trying to prove the first I want to try a couple of examples and see whats happening if I can.
    If I suppose that my function [tex] g(x) = \sqrt{x+1}[/tex] then the value of a fixed point [itex] \alpha = \frac{1+\sqrt{5}}{2}[/tex]
    Now if I let [tex]x_0 = 2[/tex] and set n = 1 then [tex]x_1 = \sqrt{2+1}[/tex] and I am under the impression that I can now show:
    [itex]|\sqrt{3} - \alpha| = |2 -\alpha||\frac{1}{2}(\frac{1}{\sqrt{\alpha+1}})|[/tex]
    [tex]\Rightarrow |\sqrt{3} - \frac{1+\sqrt{5}}{2}| = |2 - \frac{1+\sqrt{5}}{2}||\frac{1}{2}(\frac{1}{\sqrt{\frac{1+\sqrt{5}}{2}+1}})|[/tex]
    But this is false!...How am I misinterpreting the given statement or what have I done wrong?
    Please don't prove the problem for me.
    Last edited: Mar 8, 2008
  2. jcsd
  3. Mar 8, 2008 #2


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    You are right, it is not true- it is only approximately true. After all, you neglected powers higher than the first in the Taylor expansion.

    Proving [itex]|x_n- \alpha|=|x_0- \alpha||g'(\alpha)^n[/itex], to first order, by induction, is pretty easy if you replace [itex]g(x_n)[/itex] and [itex]g(\alpha)[/itex] in [itex]g(x_n) = g(\alpha) + (x_n - \alpha)g'(\alpha)[/itex] by [itex]x_{n+1}[/itex] and [itex]\alpha[/itex].
    Last edited: Mar 8, 2008
  4. Mar 9, 2008 #3
    Cheers for that HallsofIvy, it was the presence of an equals sign, not approximately equals that was throwing me; I couldn't see why it should have been true :redface:
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