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GregA
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[SOLVED] Proof of convergence
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]
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 what's 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.
Homework Statement
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]
Homework Equations
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 what's 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.
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