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Unformatted text preview: Numerical Analysis for Computing COT 4501 Spring 2005 Midterm II Solutions April 21, 2009 1. [35 points] Newton Error Formula: Let f C 2 ( I ) be given, for some interval I R , with f ( ) = 0 for some I . For a given x n I , define x n +1 = x n f ( x n ) f ( x n ) . Then there exists a point n between and x n such that ( x n +1 ) = 1 2 ( x n ) 2 f ( n ) f ( x n ) . (a) [5 points] If n and x n as n , what does the ratio R n = x n +1 ( x n ) 2 converge to? The ratio R n = x n +1 ( x n ) 2 converges to lim n R n = 1 2 lim n f ( n ) f ( x n ) = 1 2 f ( ) f ( ) . (b) [5 points] Evaluate R n (in the limiting case as n ) for f ( x ) = 2 e x 1. For f ( x ) = 2 e x 1, f ( x ) = 2 e x and f ( x ) = 2 2 e x . Therefore, lim n R n = 1 2 f ( ) f ( ) = 2 . (c) [10 points] The Newton Error Formula can be used to approximately peek ahead at the solution f ( ) = 0 when we are getting close to convergence. Imagine that you are executing Newtons method and are at time step n . You think you are somewhat close to the true answer. In order to get a better estimate ( n ) of the true answer , you replace n by x n in the Newton Error Formula to get an estimate ( n ). Using this approximation in the above Newtons Error Formula, find a solution for ( n ) in terms of x n +1 ,x n ,f ( x n ) and f ( x n ). You are asked to use the Newton Error Formula [( x n +1 ) = 1 2 ( x n ) 2 f ( n ) f ( x n ) ] to find an approximate value ( n ) for the true root by substituting x n for the unknown n . Therefore ( ( n ) x n +1 ) = 1 2 ( ( n ) x n ) 2 f ( x n ) f ( x n ) . 1 Define S n = 1 2 f ( x n ) f ( x n ) . Now ( ( n ) x n +1 ) = S n ( ( n ) x n ) 2 = S n ( ( n ) 2 2 ( n ) x n + x 2 n ) . Grouping terms, we get S n ( n ) 2 (2 S n x n + 1) ( n ) + ( S n x 2 n + x n +1 ) = 0 . This is a quadratic equation in ( n ). Solving for ( n ), we get ( n ) = (2 S n x n + 1) radicalbig (2 S n x n + 1) 2 4 S n ( S n x 2 n + x n +1 ) 2 S n = (2 S n x n + 1) radicalbig 4 S n ( x n x n +1 ) + 1 2 S n . Defining T n def = f ( x n ) f ( x n ) , we can rewrite the Newton update equation as x n x n +1 = T n from which we get that ( n ) = (2 S n x n + 1) 4 S n T n + 1 2 S n ....
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This note was uploaded on 01/22/2012 for the course COT 4501 taught by Professor Davis during the Spring '08 term at University of Florida.
 Spring '08
 Davis

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