http://www.savagehost.com/consider-the-iteration-show-that-x_n-decreases-monotonically-to-%e2%88%9a2-is-the-rate-linear-superlinear-or-quad.html Just another WordPress weblog Tue, 02 Mar 2010 23:13:58 -0600 http://wordpress.org/?v=2.8 hourly 1 http://www.savagehost.com/consider-the-iteration-show-that-x_n-decreases-monotonically-to-%e2%88%9a2-is-the-rate-linear-superlinear-or-quad.html/comment-page-1#comment-2858 simplici Sat, 04 Jul 2009 15:04:19 +0000 http://www.savagehost.com/consider-the-iteration-show-that-x_n-decreases-monotonically-to-%e2%88%9a2-is-the-rate-linear-superlinear-or-quad.html#comment-2858 1. It is not true that x_n decreases monotonically to √2. It depends on x_0 and x_1. A. if x_1 = x_0 = 0, then x_2 is undefined B. if x_1 = 1, x_0 = 0, then x_2 = 2, which is an increase 2. If both x_n and x_(n-1) are close to but greater than √2 then x_(n+1) = x_n - dx, where dx is positive, so x_(n+1) < x_n Furthermore, if x_n = √2 + e, then dx is approximately: 2e / (2 √2) or e / √2 Because √2 > 1, e / √2 = dx is less than e, so x_(n+1) > √2 As for convergence, if the denominator were 2 x_n, this would just be Newton's method, which converges quadratically, but since the denominator is x_n + x_(n-1) it does not converge quadratically. 1. It is not true that x_n decreases monotonically to √2. It depends on x_0 and x_1.
A. if x_1 = x_0 = 0, then x_2 is undefined
B. if x_1 = 1, x_0 = 0, then x_2 = 2, which is an increase
2. If both x_n and x_(n-1) are close to but greater than √2
then x_(n+1) = x_n – dx, where dx is positive, so x_(n+1) < x_n
Furthermore, if x_n = √2 + e, then dx is approximately:
2e / (2 √2) or e / √2
Because √2 > 1, e / √2 = dx is less than e, so x_(n+1) > √2
As for convergence, if the denominator were 2 x_n, this would just be Newton’s method, which converges quadratically, but since the denominator is x_n + x_(n-1) it does not converge quadratically.

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