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GregA
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[SOLVED] Proof of convergence
Let \alpha[/tex] be a fixed point of x = g(x) and let (x_n) be the sequence generated by the fixed point iteration scheme. Using the first two terms of the Taylor series for<br /> g(x) about \alpha[/tex] we can get an approximation for g(x_n):<br /> g(x_n) = g(\alpha) + (x_n - \alpha)g&amp;#039;(\alpha)[/tex]&lt;br /&gt; (Assuming the terms in the sequence are close to α we have neglected non-&lt;br /&gt; linear terms.)&lt;br /&gt; &lt;br /&gt; Show that |x_n - \alpha| = |x_0 - \alpha||g&amp;amp;#039;(\alpha)|^{n}[/tex] for all n \geq 1 hence show that the sequence (x_n) converges to \alpha[/tex] for g&amp;amp;amp;amp;#039;(x)&amp;amp;amp;amp;lt; 1&amp;amp;lt;br /&amp;amp;gt; &amp;amp;lt;h2&amp;amp;gt;Homework Equations&amp;amp;lt;/h2&amp;amp;gt;&amp;amp;lt;br /&amp;amp;gt; &amp;amp;lt;h2&amp;amp;gt;The Attempt at a Solution&amp;amp;lt;/h2&amp;amp;gt;&amp;amp;lt;br /&amp;amp;gt; I&amp;amp;amp;#039;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&amp;amp;amp;#039;s happening if I can.&amp;amp;lt;br /&amp;amp;gt; If I suppose that my function g(x) = \sqrt{x+1} then the value of a fixed point \alpha = \frac{1+\sqrt{5}}{2}[/tex]&amp;amp;amp;lt;br /&amp;amp;amp;gt; Now if I let x_0 = 2 and set n = 1 then x_1 = \sqrt{2+1} and I am under the impression that I can now show:&amp;amp;amp;lt;br /&amp;amp;amp;gt; |\sqrt{3} - \alpha| = |2 -\alpha||\frac{1}{2}(\frac{1}{\sqrt{\alpha+1}})|[/tex]&amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;gt; \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}})|&amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;gt; But this is false!...How am I misinterpreting the given statement or what have I done wrong?&amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;gt; Please don&amp;amp;amp;amp;amp;#039;t prove the problem for me.
Homework Statement
Let \alpha[/tex] be a fixed point of x = g(x) and let (x_n) be the sequence generated by the fixed point iteration scheme. Using the first two terms of the Taylor series for<br /> g(x) about \alpha[/tex] we can get an approximation for g(x_n):<br /> g(x_n) = g(\alpha) + (x_n - \alpha)g&amp;#039;(\alpha)[/tex]&lt;br /&gt; (Assuming the terms in the sequence are close to α we have neglected non-&lt;br /&gt; linear terms.)&lt;br /&gt; &lt;br /&gt; Show that |x_n - \alpha| = |x_0 - \alpha||g&amp;amp;#039;(\alpha)|^{n}[/tex] for all n \geq 1 hence show that the sequence (x_n) converges to \alpha[/tex] for g&amp;amp;amp;amp;#039;(x)&amp;amp;amp;amp;lt; 1&amp;amp;lt;br /&amp;amp;gt; &amp;amp;lt;h2&amp;amp;gt;Homework Equations&amp;amp;lt;/h2&amp;amp;gt;&amp;amp;lt;br /&amp;amp;gt; &amp;amp;lt;h2&amp;amp;gt;The Attempt at a Solution&amp;amp;lt;/h2&amp;amp;gt;&amp;amp;lt;br /&amp;amp;gt; I&amp;amp;amp;#039;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&amp;amp;amp;#039;s happening if I can.&amp;amp;lt;br /&amp;amp;gt; If I suppose that my function g(x) = \sqrt{x+1} then the value of a fixed point \alpha = \frac{1+\sqrt{5}}{2}[/tex]&amp;amp;amp;lt;br /&amp;amp;amp;gt; Now if I let x_0 = 2 and set n = 1 then x_1 = \sqrt{2+1} and I am under the impression that I can now show:&amp;amp;amp;lt;br /&amp;amp;amp;gt; |\sqrt{3} - \alpha| = |2 -\alpha||\frac{1}{2}(\frac{1}{\sqrt{\alpha+1}})|[/tex]&amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;gt; \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}})|&amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;gt; But this is false!...How am I misinterpreting the given statement or what have I done wrong?&amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;gt; Please don&amp;amp;amp;amp;amp;#039;t prove the problem for me.
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