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Unformatted text preview: r k 6 = 0, do (a) k =r T k p k p T k Ap k , (b) x k +1 = x k + k p k , (c) r k +1 = r k + k Ap k , (d) k = r T k +1 Ap k p T k Ap k , (e) p k +1 =r k +1 + k +1 p k . (f) k k + 1. 1 2 Suppose that the kth iterate generated by the algorithm is not the solution point x * . Prove that r T k r i = 0 , i = 0 , 1 , . . . , k1 span { r , r 1 , . . . , r k } = span r , Ar , . . . , A k r , span { p , p 1 , . . . , p k } = span r , Ar , . . . , A k r , p T k Ap i = 0 , i = 0 , 1 , . . . , k1 . Therefore the sequence { x k } converges to x * in at most n steps....
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This note was uploaded on 12/27/2011 for the course MATH 104b taught by Professor Ceniceros,h during the Fall '08 term at UCSB.
 Fall '08
 Ceniceros,H
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