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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, < x 1 3 , u 1 > = R 1 - 1 2 2 x 1 3 dx = 0 and < x 1 3 , u 2 > = R 1 - 1 6 2 x · x 1 3 dx = 3 6 7 Therefore the approximation is p = 0 · 2 2 + 3 6 7 · 6 2 x = 9 7 x.
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