Newton Method - Contents 9 Newtons Method 9.1 146 Solving...

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Contents 9 Newton’s Method 146 9.1 Solving Equations with Error Analysis Using Newton’s Method . . . . . . . . . . . . . . . . . 146
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CONTENTS 2
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Chapter 9 Newton’s Method 9.1 Solving Equations with Error Analysis Using Newton’s Method Read the Gold Notes, V.A Newton’s Method in the Special Case. Definition of accuracy to K decimal places: If E is the error in an approximation then the approxi- mation will be considered accurate to K decimal places if | E | < 0 . 5 × 10 - K . Example : Solve sin x - x 4 = 0 with an error less than ± 1 × 10 - 3 . Step 0 : Find the number of roots and suitable intervals each containing exactly one root. One method is to write f ( x ) = sin x - x 4 and use graph sketching techniques to sketch f ( x ). (You are not allowed to plot points.) Since f ( x ) is a di±erential function on R , it is continuous on R . Hence the intermediate value theorem applies. Use this theorem to test and reduce the size of each interval that your graph shows containing a root. Also make sure f 0 ( x ) 6 = 0 in each of the above intervals. For this problem another method works. Write sin x = x 4 and graph sin x and x 4 on the same coordinate system. The roots of sin x - x 4 = 0 occur where the two graphs intersect. If the graphs do not intersect the equation has no roots (an example is ln x - x = 0). It is important to know that sin x < x if x > 0. This result implies sin x - x = 0 has only one root. ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± ± - π r 1 r 2 π 2 r 3 π y = x 4 x y = sin x y - π 2
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147 9.1 Solving Equations with Error Analysis Using Newton’s Method The sketch shows there are exactly three roots with the following properties (a) r 2 = 0 (by inspection) (b) r 3 ( π 2 ) (c) r 1 ( - π, - π 2 ) (d) r 1 = - r 3 (by symmetry; i.e., both functions are odd). We need only obtain r 3 to determine all three roots. To find a suitable interval compute f (2) 0 . 409 and f (3) ≈ - 0 . 608 f (2) and f (3) have opposite signs implies r 3 [2 , 3] by the intermediate value theorem. Next we show f 0 ( x ) 6 = 0 in [2 , 3]. f 00 ( x ) = - sin x f 0 ( x ) is decreasing in [2 , 3]. Also f 0 (2) ≈ - 0 . 666. f 0 (2) < 0 and f 0 ( x ) decreasing in [2 , 3] implies f 0 ( x ) < 0 in [2 , 3]; i.e., f 0 ( x ) 6 = 0 in [2 , 3]. Step 1 : Determine constants M 1 and M 2 so that | f 0 ( x ) | = ± ± ± cos x - 1 4 ± ± ± > M 1 > 0 x [2 , 3] and | f 00 ( x ) | = | - sin x | < M 2 x [2 , 3] . In Step 0 we saw that
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This note was uploaded on 09/22/2011 for the course ECON 101 taught by Professor Mr.tull during the Spring '11 term at De La Salle University.

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Newton Method - Contents 9 Newtons Method 9.1 146 Solving...

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