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Differential Equations Solutions 174

Differential Equations Solutions 174 - 1 = α j h 2 µ...

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184 Chapter 30. Solutions: Case Study: Fast Solvers and Sylvester Equations (b) The algorithm fails if L ( i, i ) + R ( j, j ) = 0 for some value of i and j . The main diagonal elements of triangular matrices are the eigenvalues of the matrix, so it is necessary and sufficient that L and R have no common eigenvalues. (c) If AU + UB = C , then WLW U + UYRY = C . Multiplying on the left by W and on the right by Y , we obtain LU + UR = C . CHALLENGE 30.3. The algorithm of Challenge 2(a) reduces to U ( i, j ) = F ( i, j ) / ( L ( i, i ) + R ( j, j )) for i, j = 1 , . . . , n , which requires n 2 additions and divi- sions. CHALLENGE 30.4. (a) Recall the identities sin( a ± b ) = sin a cos b ± cos a sin b. If we form B x times the j th column of V , then the k th element is v k 1 ,j + 2 v k,j v k +1 ,j h 2 = α j h 2 sin ( k 1) n + 1 + 2 sin kjπ n + 1 sin
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Unformatted text preview: + 1 ¶ = α j h 2 µ − sin kjπ n + 1 cos jπ n + 1 + cos kjπ n + 1 sin jπ n + 1 + 2 sin kjπ n + 1 − sin kjπ n + 1 cos jπ n + 1 − cos kjπ n + 1 sin jπ n + 1 ¶ = α j h 2 µ 2 − 2 cos jπ n + 1 ¶ sin kjπ n + 1 = 1 h 2 µ 2 − 2 cos jπ n + 1 ¶ v k,j = λ j v k,j . Stacking these elements we obtain B x v j = λ j v j . (b) This follows by writing the k th component of Vy . CHALLENGE 30.5. See the website for the programs. The results are shown in Figures 30.1 and 30.2. All of the algorithms give accurate results, but as n gets large, the efficiency of the fast algorithm of Challenge 4 becomes more apparent....
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