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Unformatted text preview: Solutions for Homework 7 Foundations of Computational Math 1 Fall 2010 Problem 7.1 Let f ( x ) = x 3 − 3 x + 1. This polynomial has three distinct roots. (7.1.a) Consider using the iteration function φ 1 ( x ) = 1 3 ( x 3 + 1) Which, if any, of the three roots can you compute with φ 1 ( x ) and how would you choose x (0) for each computable root? (7.1.b) Consider using the iteration function φ 2 ( x ) = 3 2 x − 1 6 ( x 3 + 1) Which, if any, of the three roots can you compute with φ 2 ( x ) and how would you choose x (0) for each computable root? (7.1.c) For each of the roots you identified as computable using either φ 1 ( x ) or φ 2 ( x ), apply the iteration to find the values of the roots. (You need not turn in any code, but using a simple program to do this is recommended.) Solution: We can evaluate f ( x ) = x 3 − 3 x + 1 to get a broad idea of where the roots are located if we find an appropriate sign pattern to the values of f . We have f ( − 2) = − 8 + 6 + 1 = − 1 f ( − 1) = − 1 + 3 + 1 = 3 f (0) = 1 f (1) = 1 − 3 + 1 = − 1 f (2) = 8 − 6 + 1 = 3 So we have 3 distinct roots and the open intervals that contain them are − 2 < α < − 1 − 1 < α 1 < 1 1 < α 2 < 2 1 Consider φ 1 ( x ) = 1 3 ( x 3 + 1). We have φ ′ 1 ( x ) = x 2  x  < 1 →  φ ′ 1 ( x )  < 1 and so φ 1 ( x ) is a contraction mapping for  x  < 1 and converges to a unique fixed point. We must verify that this fixed point is a root of f ( x ). We have f ( x ) = x 3 − 3 x + 1 f ( x ∗ ) = 0 → 0 = ( x ∗ ) 3 − 3 x ∗ + 1 → 3 x ∗ = ( x ∗ ) 3 + 1 → x ∗ = 1 3 ( ( x ∗ ) 3 + 1 ) ∴ φ 1 ( x ∗ ) = x ∗ for any root, x ∗ of f ( x ). However, since  x  > 1 →  φ ′ 1 ( x )  > 1 we know that there must exist neighborhoods around α and α 2 on which φ 1 ( x ) does not converge to α and α 2 respectively. We also know this because φ 1 ( x ) cannot be a contraction mapping on an interval containing more than one fixed point. To see that we can extend the interval of convergence beyond − 1 < x (0) < 1 note that φ 1 (1) = 2 3 and φ 1 ( − 1) = 0 So after one application of φ 1 we are in the interval that we know converges to α 1 and therefore have If − 1 ≤ x (0) ≤ 1 then x ( k +1) = φ 1 ( x ( k ) ) → α 1 This interval can be extended on the rightside by noting that if 1 < x (0) < 3 √ 2 then x (1) < 1 and therefore convergence to α 1 would follow. Similar extension could be considered on the left. The reasoning can be extended further by finding values of x (0) where − 1 ≤ x (2) ≤ 1 etc. These extensions on left and right are essentially trying to iterate to get the bound on the interval around α 1 on which φ 1 ( x ) is a contraction mapping....
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 Spring '11
 Gallivan
 Calculus, Numerical Analysis, Convergence, Mathematical analysis, Metric space

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