Math128_hw1

Math128_hw1 - Math 128A, Spring 2007 Homework 1 Solution...

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Unformatted text preview: Math 128A, Spring 2007 Homework 1 Solution handout Evaluating the first 11 terms of the sequence given by a = 100 ln(101 / 100) and a n = 100(1 /n- a n- 1 ) yields ˆ a = 0 . 995033 ˆ a 1 = 0 . 496691 ˆ a 2 = 0 . 330853 ˆ a 3 = 0 . 248017 ˆ a 4 = 0 . 198348 ˆ a 5 = 0 . 165241 ˆ a 6 = 0 . 142563 ˆ a 7 = 0 . 029448 ˆ a 8 = 9 . 555183 ˆ a 9 =- 944 . 407156 ˆ a 10 = 94450 . 715592 These diverge from a n = R 1 100 x n / (100 + x ) dx because the computer is working with approximations of real numbers. When the computer approximates a , it produces a value ˆ a = a + , where is some small error. If we feed this through the recurrence relation, we get ˆ a n = a n + (- 100) n so this small initial error is amplified by the recurrence relation. 3.1.10 Graphing e- x- sin( x ), it’s easy to see that [ a, b ] = [0 , π/ 2] contains the smallest positive root. After n iterations, the error is bounded by b- a 2 n +1 = π 2 n +2 so it’s sufficient to find n satisfying π 2 n +2 < 10- 10...
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This note was uploaded on 05/07/2008 for the course MATH 128A taught by Professor Rieffel during the Spring '08 term at Berkeley.

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Math128_hw1 - Math 128A, Spring 2007 Homework 1 Solution...

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