hw8 - Solutions of Selected Problems in HW8 3.5.7 In order...

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Solutions of Selected Problems in HW8 3.5.7 In order to solve this problem, let us recall the following facts; if { u 1 , ··· , u n } is an or- thonormal basis for V and x = c 1 u 1 + ··· + c n u n , then c i = < x , u i > . Moreover, by the Parseval’s equality, the norm of x can be computed as || x || 2 = c 2 1 + ··· + c 2 n . In this problem, we have 4 = < x , u 1 > = c 1 and 0 = < x , u 2 > = c 2 . Since || x || = 5, we get c 2 1 + c 2 2 + c 2 3 = 25 Therefore c 1 = 4 ,c 2 = 0 and c 3 = ± 3. 3.5.27 (a) < 1 ,x > = R 1 - 1 1 · xdx = 1 2 x 2 | 1 - 1 = 0 . (b) || 1 || = ( R 1 - 1 1 · 1 dx ) 1 2 = 2 . || x || = ( R 1 - 1 x · xdx ) 1 2 = ( 1 3 x 3 | 1 - 1 ) 1 2 = 6 3 . (c) Normalize two vectors 1 and x as u 1 = 1 / || 1 || = 1 2 = 2 2 and u 2 = x/ || x || = 3 6 x = 6 2 x . Then { u 1 , u 2 } is an orthonormal set. The least squares approximation for x 1 3 with respect to { u 1 , u 2 } is p = < x 1 3 , u 1 > u 1 + < x 1 3 , u 2 > u 2 . Now,
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This note was uploaded on 07/17/2008 for the course MATH 571 taught by Professor Kim during the Spring '08 term at Ohio State.

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hw8 - Solutions of Selected Problems in HW8 3.5.7 In order...

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