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# solexam2f10 - Foundations of Computational Math I Exam 2...

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Unformatted text preview: Foundations of Computational Math I Exam 2 Take-home Exam Open Notes, Textbook, Homework Solutions Only Due beginning of Class Wednesday, December 1, 2010 Question Points Points Possible Awarded 1. Iterative Methods 25 for Ax = b 2. Iterative Methods 30 for Ax = b 3. Nonlinear Equations 30 4. Nonlinear Equations 25 Total 110 Points Name: Alias: to be used when posting anonymous grade list. 1 Problem 1 (25 points) Suppose B ∈ R n × n is a symmetric positive definite tridiagonal matrix of the form B = parenleftbigg D r L L T D b parenrightbigg where n = 2 k , D r and D b are diagonal matrices of order n/ 2 and L is a lower triangular matrix with nonzeros restricted to its main diagonal and its first subdiagonal. Assume that Bx = b can be solved using Jacobi’s method, i.e., the iteration converges acceptably fast. Partition each iterate x i into the top half and bottom half, i.e., x i = parenleftBigg x ( top ) i x ( bot ) i parenrightBigg 1.a . Assume an initial guess x is given and identify what information, i.e., the pieces of x i for 0 ≤ i ≤ j , determines the values found in the vectors x ( top ) j and x ( bot ) j for any j > 0. 1.b . Can the relationships from 1.a be used to design an iteration that approximates the solution essentially as well but only requires half of the work of Jacobi’s method? 1.c . Relate your new method from 1.b to applying Gauss-Seidel to solve Bx = b starting from the same initial guess x . Solution: The system is of the form Bx = b parenleftbigg D r L L T D b parenrightbiggparenleftbigg x ( top ) x ( bot ) parenrightbigg = parenleftbigg b ( top ) b ( bot ) parenrightbigg Given an initial guess x = parenleftBigg x ( top ) x ( bot ) parenrightBigg the key fact about Jacobi is that it splits into two independent sequences . The Jacobi iteration exploiting the partitioning given is parenleftBigg x ( top ) k +1 x ( bot ) k +1 parenrightBigg = parenleftbigg D − 1 r L D − 1 b L T parenrightbigg parenleftBigg x ( top ) k x ( bot ) k parenrightBigg + parenleftbigg ˜ b ( top ) ˜ b ( bot ) parenrightbigg = parenleftBigg D − 1 r Lx ( bot ) k D − 1 b L T x ( top ) k parenrightBigg + parenleftbigg ˜ b ( top ) ˜ b ( bot ) parenrightbigg 2 It is easily seen from writing out the partition matrix vector operations defining each step that the information from earlier steps that defines x ( top ) j and x ( bot ) j flows as follows: Sequence 1: x ( top ) → x ( bot ) 1 → x ( top ) 2 → x ( bot ) 3 ··· Sequence 2: x ( bot ) → x ( top ) 1 → x ( bot ) 2 → x ( top ) 3 ··· So if we only follow one of them we halve the number of operations and since we know by assumption that Jacobi is converging we would construct an estimate x i = parenleftBigg x ( top ) i x ( bot ) i − 1 parenrightBigg x i = parenleftBigg x ( top ) i − 1 x ( bot ) i parenrightBigg depending on the choice of sequence....
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solexam2f10 - Foundations of Computational Math I Exam 2...

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