solhw7 - Solutions for Homework 7 Foundations of...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

{[ snackBarMessage ]}

Page1 / 16

solhw7 - Solutions for Homework 7 Foundations of...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online