128ahw3sum10 - MATH 128A, SUMMER 2010, HOMEWORK 3 SOLUTION...

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MATH 128A, SUMMER 2010, HOMEWORK 3 SOLUTION BENJAMIN JOHNSON Homework 3: Due Tuesday, July 6 2.3; 2, 12a, 19 2.4; 1b, 6b, 10 2.5; 1d, 6, 9 section 2.3 2. Let f ( x ) = - x 3 - cos x and p 0 = - 1. Use Newton’s method to find p 2 . Could p 0 = 0 be used? solution: We have f 0 ( x ) = - 3 x 2 + sin x . So p 1 = - 1 - f ( - 1) f 0 ( - 1) = - 1 - 1 - cos( - 1) - 3+sin( - 1) - 0 . 880333. So p 2 = - 0 . 880333 - f ( - 0 . 880333) f 0 ( - 0 . 880333) ≈ - 0 . 865684. p 0 = 0 could not be used for Newton’s method because f 0 (0) = 0. 12. Use all three methods in this section to find solutions to within 10 - 7 for the following problems. a. x 2 - 4 x + 4 - ln x = 0 for 1 x 2 and for 2 x 4 solution: To do this exercise, I used the “Java Programs” online interface at http://www.as.ysu.edu/~faires/Numerical-Analysis/DiskMaterial/ programs/Java/JavaPrograms.htm For Newton’s Method, I used initial points 1.5 and 3: I P0 F0 ==================================================== 1 1.406720935 0.010718631 2 1.412369957 0.000039953 3 1.412391172 0.000000001 4 1.412391172 0.000000000 No of Iterations 4 Approximate solution 1.412391172 with F(p) = 0.000000000 Tolerance 1.0E-7 I P0 F0 ==================================================== 1 3.059167373 0.003692746 2 3.057106055 0.000004476 3 3.057103550 0.000000000 4 3.057103550 -0.000000000 No of Iterations 4 Approximate solution 3.057103550 Date : July 6, 2010. 1
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2 BENJAMIN JOHNSON with F(p) = -0.000000000 Tolerance 1.0E-7 For the Secant Method I used initial points 1.5 and 1.6, and 3.0 and 3.1: I Pn F(Pn) ========================================== 2 1.399400417 0.024675988 3 1.414190660 -0.003384813 4 1.412406598 -0.000029051 5 1.412391154 0.000000035 6 1.412391172 -0.000000000 No of Iterations 6 Approximation solution P = 1.412391172 with F(P) = -0.000000000 Tolerance 1.0E-7 I Pn F(Pn) ========================================== 2 3.055647080 -0.002600623 3 3.057067614 -0.000064221 4 3.057103581 0.000000055 5 3.057103550 -0.000000000 No of Iterations 5 Approximation solution P = 3.057103550 with F(P) = -0.000000000 Tolerance 1.0E-7 For the Method of False Position, I used starting points 1 and 2, and 2 and 4;
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128ahw3sum10 - MATH 128A, SUMMER 2010, HOMEWORK 3 SOLUTION...

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