# 218 hw1 - 34375 0 375 2 = 0 359375 f(0 359375 = 00262 Root...

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ME 218 Homework 1 Solution Problem # 1 f (0 . 6) = 0 . 36 f (0 . 75) = - 0 . 375 A root is bracketed by the interval. Iteration 1 midpoint = 0 . 6+0 . 75 2 = 0 . 675 f (0 . 675) = - 0 . 0412 Root is between 0.6 and 0.675 Iteration 2 midpoint = 0 . 6+0 . 675 2 = 0 . 6375 f (0 . 6375) = . 1509 Root is between 0.6375 and 0.675 Iteration 3 midpoint = 0 . 6375+0 . 675 2 = 0 . 65625 f (0 . 65625) = 0 . 05273 Root is between 0.65625 and 0.675 Problem # 2 f (0) = - 1 f (1) = 1 . 123 A root is bracketed by the interval. midpoint = 0+1 2 f (0 . 5) = 0 . 3307 Root is between 0 and 0.5 midpoint = 0+0 . 5 2 = 0 . 25 f (0 . 25) = - 0 . 2866 Root is between 0.25 and 0.5 1

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midpoint = 0 . 25+0 . 5 2 = 0 . 375 f (0 . 375) = 0 . 03628 Root is between 0.25 and 0.375 midpoint = 0 . 25+0 . 375 2 = 0 . 3125 f (0 . 3125) = - 0 . 121899 Root is between 0.3125 and 0.375 midpoint = 0 . 3125+0 . 375 2 = 0 . 34375 f (0 . 34375) = - 0 . 04195 Root is between 0.34375 and 0.375 midpoint =
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Unformatted text preview: . 34375+0 . 375 2 = 0 . 359375 f (0 . 359375) =-. 00262 Root is between 0.359375 and 0.375 midpoint = . 359375+0 . 375 2 = 0 . 3671 f (0 . 3671) = 0 . 01688 Root is between 0.359375 and 0.3671875 midpoint = . 359375+0 . 3671875 2 = 0 . 36328 The maximum error of the bisection method is | b-a | 2 n where b and a are the initial boundaries of the bisection interval, and n is the step number. Here, the error is:-1 128 = . 0078125. This can also be computed using the current bisection interval: . 3671875-. 359375 = 0 . 0078125 After 7 steps, this meets the exit criteria of x error < . 01. 2...
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218 hw1 - 34375 0 375 2 = 0 359375 f(0 359375 = 00262 Root...

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