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cot4501fa07_midterm2_solutions

# cot4501fa07_midterm2_solutions - Numerical Analysis for...

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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 ( α ) = 0 for some α I . For a given x n I , define x n +1 = x n f ( x n ) f ( x n ) . Then there exists a point ξ n between α and x n such that ( α x n +1 ) = 1 2 ( α x n ) 2 f ′′ ( ξ n ) f ( x n ) . (a) [5 points] If ξ n α and x n α as n → ∞ , what does the ratio R n = α x n +1 ( α x n ) 2 converge to? The ratio R n = α x n +1 ( α x n ) 2 converges to lim n →∞ R n = 1 2 lim n →∞ f ′′ ( ξ n ) f ( x n ) = 1 2 f ′′ ( α ) f ( α ) . (b) [5 points] Evaluate R n (in the limiting case as n → ∞ ) for f ( x ) = 2 e βx 1. For f ( x ) = 2 e βx 1, f ( x ) = 2 βe βx and f ′′ ( x ) = 2 β 2 e βx . Therefore, lim n →∞ R n = 1 2 f ′′ ( α ) f ( α ) = β 2 . (c) [10 points] The Newton Error Formula can be used to approximately “peek ahead” at the solution f ( α ) = 0 when we are getting close to convergence. Imagine that you are executing Newton’s method and are at time step n . You think you are somewhat close to the true answer. In order to get a better estimate ˆ α ( n ) of the true answer α, you replace ξ n by x n in the Newton Error Formula to get an estimate ˆ α ( n ). Using this approximation in the above Newton’s Error Formula, find a solution for ˆ α ( 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 [( α x n +1 ) = 1 2 ( α x n ) 2 f ′′ ( ξ n ) f ( x n ) ] to find an approximate value ˆ α ( n ) for the true root α by substituting x n for the unknown ξ n . Therefore α ( n ) x n +1 ) = 1 2 α ( n ) x n ) 2 f ′′ ( x n ) f ( x n ) . 1

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Define S n = 1 2 f ′′ ( x n ) f ( x n ) . Now α ( n ) x n +1 ) = S n α ( n ) x n ) 2 = S n α ( n ) 2 α ( n ) x n + x 2 n ) . Grouping terms, we get S n ˆ α ( n ) 2 (2 S n x n + 1)ˆ α ( n ) + ( S n x 2 n + x n +1 ) = 0 . This is a quadratic equation in ˆ α ( n ). Solving for ˆ α ( n ), we get ˆ α ( 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 ˆ α ( n ) = (2 S n x n + 1) ± 4 S n T n + 1 2 S n . It is not clear at this point whether we should take the positive or negative root. It is also not clear whether 4 S n T n + 1 0. If the latter is not true, then the estimate ˆ α ( n ) becomes complex which is unacceptable.
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cot4501fa07_midterm2_solutions - Numerical Analysis for...

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