review3s - Numerical Analysis/Fall...

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Unformatted text preview: Numerical Analysis/Fall 2009/Review #3/Solutions (1) (a) f (1) =- 5 < 0, (3) = 5 > 0, so by the Intermediate Value Theorem, f has a root in [1 , 3]. Since f ( x ) = 2 x + 1 > 0 on [1 , 3], f is increasing, and the root is unique. (b) 1.00000 3.00000 2.00000 2.00000 3.00000 2.50000 2.00000 2.50000 2.25000 2.00000 2.25000 2.12500 2.12500 2.25000 2.18750 2.18750 2.25000 2.21875 2.18750 2.21875 2.20313 Final estimate is 2 . 203125. (c) ln 2 × 10 6 ln 2 ’ 20 . 9, so at least 21 steps would be required. (2) (a) 2., 2.200000000, 2.192592593, 2.192582404 f ( x ) = 2 x + 1, so f ( p ) 6 = 0, p is a simple root, and convergence is quadratic. (b) f ( x ) =- e- x , so the iterating formula is g ( x ) = x- f ( x ) f ( x ) = x + 1 . Thus p i = i + 1. The problem is that f has no roots, and Newton’s method does not converge. (3) L ( x ) = ( x- 1)( x- 2)( x- 3)- 6 L 1 ( x ) = x ( x- 2)( x- 3) 2 L 2 ( x ) = x ( x- 1)( x- 3)- 2 L 3 ( x ) = x ( x- 1)( x- 2) 6 L ( x ) = 1 2 x 3- 1 2 x 2- x + 2 (4) x y...
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This note was uploaded on 10/10/2010 for the course MATH 01:640:111 taught by Professor Carey during the Spring '10 term at Saint Louis.

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review3s - Numerical Analysis/Fall...

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