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# review2s - (4 T = ±-1 3-1 ² c = ± 4 3 1 ² estimates are...

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Numerical Analysis/Fall 2009/Review #2/Solutions (1) x y xy x 2 0 2 0 0 1 1 1 1 2 2 4 4 3 8 24 9 6 13 29 14 4 6 13 6 14 29 y = 0 . 4 + 0 . 9 x. (2) (a) Z π - π sin mx cos nx dx = 1 2 Z π - π sin( m + n ) x + sin( m - n )) dx = - 1 2 cos( m + n ) x m + n + cos( m - n ) x m - n π - π = 0 since cos is an even function. (b) Since f is an even function, all the b k are zero. πa 0 = π (look at R π - π f ( x ) dx ). For k > 0 πa k = Z π - π f ( x ) cos kx dx = 2 Z π 0 x cos kx dx = 2 x sin kx k + cos kx k 2 π 0 = 2 k 2 (cos - 1) = 2 k 2 (( - 1) k - 1) . (3) Note that a b = 1 3 3 2 . F 4 ( a ) = 3 i 1 - i F 4 ( b ) = 3 1 + 2 i - 1 1 - 2 i F 4 ( a b ) = 9 - 2 + i - 1 - 2 - i F 4 ( a ) × F 4 ( b ) =

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Unformatted text preview: (4) T = ±-1 3-1 ² , c = ± 4 3 1 ² estimates are ± ² , ± 4 3 1 ² , ± 1-1 3 ² , ± 13 9 ² . (5) (a) 7 (b) 5 (6) The condition number of A is 4 . 0001 × 60001 ’ 240000, which is high. In this case, the residual being close to zero does not mean that the estimate is close to a solution....
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review2s - (4 T = ±-1 3-1 ² c = ± 4 3 1 ² estimates are...

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