{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

This preview shows pages 1–2. Sign up to view the full content.

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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}