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527solns9 - 642:527 SOLUTIONS ASSIGNMENT 9 FALL 2007...

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642:527 SOLUTIONS: ASSIGNMENT 9 FALL 2007 Section 9.9: 2 (a) See text solution. (d) The formula is (23) of the text. We have ( e 1 , e 1 ) (= e 1 · e 1 ) = 14, ( e 2 , e 2 ) = 5, ( e 3 , e 3 ) = 70, ( e 1 , u ) = 8, ( e 2 , u ) = 5, and ( e 3 , u ) = 10, so u = (4 / 7) e 1 + e 2 + (1 / 7) e 3 . 12 (d) See text solution. (f) Summary of the Gram-Schmidt process: Given v 1 , . . ., v n we define ˆ e 1 = v 1 / bardbl v 1 bardbl . Now, once we have defined ˆ e 1 , . . . ˆ e k we define ˆ e k +1 by a two-step process: e k +1 = v k +1 k summationdisplay i =1 ( ˆ e i , v k +1 ) ˆ e i , ˆ e k +1 = e k +1 bardbl e k +1 bardbl . So for the given vectors v 1 = (1 , 1 , 1), v 2 = (1 , 0 , 1), v 3 = (1 , 1 , 0) we have ˆ e 1 = v 1 bardbl v 1 bardbl = 1 radicalbig ( v 1 , v 1 ) v 1 = 1 3 (1 , 1 , 1); e 2 = v 2 − ( ˆ e 1 , v 2 ) ˆ e 1 = (1 , 0 , 1) 2 3 (1 , 1 , 1) = 1 3 (1 , 2 , 1) , ˆ e 2 = e 2 bardbl e 2 bardbl = 1 6 (1 , 2 , 1); e 3 = v 3 − ( ˆ e 1 , v 3 ) ˆ e 1 − ( ˆ e 2 , v 3 ) ˆ e 2 = (1 , 1 , 0) 2 3 (1 , 1 , 1) + 1 6 (1 , 2 , 1) = 1 2 (1 , 0 , 1) , ˆ e 3 = e 3 bardbl e 3 bardbl = 1 2 (1 , 0 , 1); Section 9.10: 2 (a) See text solution. (c) With u = (3 , 0 , 1 , 4 , 1) we have ( ˆ e 1 , u ) = 5 / 5, ( ˆ e 2 , u ) = 6 / 6, ( ˆ e 3 , u ) = 4, so the best approximation is 3 summationdisplay i =1 ( ˆ e i , u ) ˆ e i = (1 , 0 , 2 , 0 , 0) + (2 , 0 , 1
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