Unformatted text preview:  pp n  ≤ b 1a 1 2 n a condition which is suﬃcient to ensure that the error is smaller than 5 × 1011 is b 1a 1 2 n ≤ 5 × 1011 or n ≥ 35 . • If we wish to compute the cube root of 5, we could proceed by ﬁnding the zeros of f ( x ) = x 35. Let the initial interval be deﬁned by a = 0 and b = 3 and calculate an approximation for 5 1 / 3 using 4 steps of the bisection. n 1 2 3 4 5 a n 1 . 5 1 . 5 1 . 5 1 . 6875 b n 3 3 2 . 25 1 . 875 1 . 875 p n 1 . 5 2 . 25 1 . 875 1 . 6875 1 . 78125 f ( a n )f ( b n ) + + + + + f ( p n )+ ++ The ﬁnal interval is [1 . 6875 , 1 . 78125] and p 6 = 1 . 734375 ≈ p with guaranteed error smaller than (1 . 781251 . 6875) / 2 = 0 . 046875. MATH2051 Practical 1 20101010...
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This note was uploaded on 02/09/2011 for the course MATH 2051 taught by Professor Dr.a.wacher during the Fall '11 term at Durham.
 Fall '11
 Dr.A.Wacher
 Numerical Analysis

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