Explanation Calculation: Consider a function f ( x ) = x 4 − 3 x 2 − 3 and an interval [ 1 , 2 ] . Initial approximation is given as p 0 = 1 . Substitute f ( x ) = 0 and solve for x : f ( x ) = 0 x 4 − 3 x 2 − 3 = 0 x 4 = 3 x 2 + 3 = ( 3 x 2 + 3 ) 1 4 Now, assume that g ( x ) = ( 3 x 2 + 3 ) 1 4 . Recall that p n = g ( p n − 1 ) , by fixed-point iteration method. Now, calculate the value of p 1 by substituting the initial value p 0 = 1 in g ( x ) = ( 3 x 2 + 3 ) 1 4 . p 1 = g ( p 0 ) = ( 3 ( 1 ) 2 + 3 ) 1 4 = ( 3 + 3 ) 1 4 = 1.56508458 Now, calculate absolute difference between the values of p 0 and p 1 . | p 1 − p 0 | = | 1.56508458 − 1 | = | 0.56508458 | > 0.01 The accuracy should be less than 0.01 ( 10 − 2 ) . calculate the value of p 2 by substituting the initial value p 1 = 1.56508458 in g ( x ) = ( 3 x 2 + 3 ) 1 4 . p 2 = g ( p 1 ) = ( 3 ( 1.56508458 ) 2 + 3 ) 1 4 = 1.793572879 Now, calculate absolute difference between the values of p 1 and p 2 . | p 2 − p 1 | = | 1.793572879 − 1.56508458 | = | 0.22849 | > 0.01 The accuracy should be less than 0.01 ( 10 − 2 ) . calculate the value of p 3 by substituting the initial value p 2 = 1.793572879 in g ( x ) = ( 3 x 2 + 3 ) 1 4 . p 3 = g ( p 2 ) = ( 3 ( 1.793572879 ) 2 + 3 ) 1 4 = 1

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ISBN: 9781305253667

Author: Richard L. Burden, J. Douglas Faires, Annette M. Burden

Publisher: Cengage Learning

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Chapter 2.2, Problem 7ES

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The given equation x4−3x2−3=0, p0=1 with fixed point iteration method when the accuracy is less than 10−2.

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Chapter 2.2, Problem 6ES

Chapter 2.2, Problem 8ES

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Chapter 2 Solutions

Numerical Analysis

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The given equation x 4 − 3 x 2 − 3 = 0 , p 0 = 1 with fixed point iteration method when the accuracy is less than 10 − 2 .

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The given equation x4−3x2−3=0, p0=1 with fixed point iteration method when the accuracy is less than 10−2.

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Chapter 2 Solutions