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

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Math 128A, Spring 2007 Homework 1 Solution handout Evaluating the first 11 terms of the sequence given by a 0 = 100 ln(101 / 100) and a n = 100(1 /n - a n - 1 ) yields ˆ a 0 = 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 = 1 0 100 x n / (100 + x ) dx because the computer is working with approximations of real numbers. When the computer approximates a 0 , it produces a value ˆ a 0 = a 0 + 0 , where 0 is some small error. If we feed this through the recurrence relation, we get ˆ a n = a n + ( - 100) n 0 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 I.e. n > log 2 π + 10 log 2 10 - 2 32 . 8708. 3.2.2 Running the MATLAB code 1
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x0 = 0; % initial guess x1 = x0 - (x0^3-x0^2-x0-1) / (3*x0^2-2*x0-1); while abs( x1 - x0 ) >= 10^-6 x0 = x1; x1 = x0 - (x0^3-x0^2-x0-1) / (3*x0^2-2*x0-1) end x1 produces the output x1 = 1.8393 3.2.3 (a) With f ( x
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