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Unformatted text preview: Comment on Program 3 Foundations of Computational Math 1 Fall 2011 1 The Spectral Radii 1.1 Jacobi and GaussSeidel Methods From homework we have that A = T = 1 . . . . . . . . . 1 1 . . . . . . 1 1 . . . . . . . . . . . . . . . . . . . . . 1 1 . . . . . . 1 1 . . . . . . . . . 1 j = 2 cos j, = n + 1 q j = ( sin( j ) , sin(2 j ) , . . ., sin( nj ) ) T and that G J =  1 T . We therefore have ( G J ) =  2 cos n  = 2 cos 1 where k = k . Since the matrix is tridiagonal ( G gs ) = 2 ( G J ) The number of expected steps for Jacobi given an initial error vector e (0) with norm = bardbl e (0) bardbl and a desired error of d = bardbl e (0) bardbl follows d d ( G J ) d (log 10 e d log 10 e ) log 10 ( G J ) and half of that for GaussSeidel. We have the following spectral radii for the values of and n of interest. 1 n cos 1 ( G J ) ( G gs ) 2 100 0.99951628 0.9995163 0.9990328 2 1000 0.99999508 0.9999951 0.9999902 2 2000 0.99999877 0.9999988 0.9999975 2 10000 0.99999995 1.0000000 0.9999999 3 100 0.99951628 0.6663442 0.4440146 3 1000 0.99999508 0.6666634 0.4444401 3 2000 0.99999877 0.6666658 0.4444433 3 10000 0.99999995 0.6666666 0.4444444 4 100 0.99951628 0.4997581 0.2497582 4 1000 0.99999508 0.4999975 0.2499975 4 2000 0.99999877 0.4999994 0.2499994 4 10000 0.99999995 0.5000000 0.2500000 Note that for = 2 the methods are expected to converge very slowly, if at all, in practice. Also note that is more important than n in determining the radii. So we would not expect significant variation in the convergence as a function of n for a given . 1.2 Symmetric GaussSeidel Method In this section we prove the the basic statements in the notes about Symmetric GaussSeidel applied to a symmetric positive definite matrix. For Symmetric GaussSeidel, if A = A T and D negationslash = 0 we have A = D L L T M = ( D L ) D 1 ( D L T ) = A + LD 1 L T G = I M 1 A = ( D L T ) 1 L ( D L ) 1 L T Note that Le 1 = 0 and Me 1 = Ae 1 . If we assume further that A is symmetric positive definite, i.e., v negationslash = 0 v T Av > 0 then it is easy to see that the diagonal elements of A must be positive, i.e., D is symmetric positive definite and diagonal. It also follows that D 1 exists, is symmetric positive definite, and diagonal. M is also symmetric positive definite and has a Cholesky factorization M = CC T , C = ( D L ) D 1 / 2 . The form of C follows simply from M = ( D L ) D 1 ( D L T ). It is easy to see that C is lower triangular with positive diagonal elements. Therefore, M = CC T is the Cholesky factorization of M which is therefore symmetric positive definite....
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This note was uploaded on 11/27/2011 for the course MAD 5403 taught by Professor Gallivan during the Spring '09 term at FSU.
 Spring '09
 gallivan

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