solhw7 - Solutions for Homework 7 Foundations of...

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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 1 ( x ) = 1 3 ( x 3 + 1) Which, if any, of the three roots can you compute with 1 ( x ) and how would you choose x (0) for each computable root? (7.1.b) Consider using the iteration function 2 ( x ) = 3 2 x 1 6 ( x 3 + 1) Which, if any, of the three roots can you compute with 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 1 ( x ) or 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 < < 1 1 < 1 < 1 1 < 2 < 2 1 Consider 1 ( x ) = 1 3 ( x 3 + 1). We have 1 ( x ) = x 2 | x | < 1 | 1 ( x ) | < 1 and so 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 ) 1 ( x ) = x for any root, x of f ( x ). However, since | x | > 1 | 1 ( x ) | > 1 we know that there must exist neighborhoods around and 2 on which 1 ( x ) does not converge to and 2 respectively. We also know this because 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 1 (1) = 2 3 and 1 ( 1) = 0 So after one application of 1 we are in the interval that we know converges to 1 and therefore have If 1 x (0) 1 then x ( k +1) = 1 ( x ( k ) ) 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 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 1 on which 1 ( x ) is a contraction mapping....
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This note was uploaded on 07/25/2011 for the course MAD 5403 taught by Professor Gallivan during the Spring '11 term at University of Florida.

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solhw7 - Solutions for Homework 7 Foundations of...

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