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# 4.3 - = 2 C-2 D-4-2 2 C-D-2-1 0 2 C D(1 2 C 2 D(2 = 20 D 20...

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Math 2940 Solutions, Fall 2011 Section 4 . 3 4 ) The expression we have to minimize is E ( C, D ) = ( C + 0 D - 0) 2 + ( C + D - 8) 2 + ( C + 3 D - 8) 2 + ( C + 4 D - 20) 2 . Taking partial derivatives, ∂E/∂C = 2 C + 2( C + D - 8) + 2( C + 3 D - 8) + 2( C + 4 D - 20) = 8 C + 16 D - 72 and ∂E/∂D = 0 + 2( C + D - 8) + 2( C + 3 D - 8)(3) + 2( C + 4 D - 20)(4) = 16 C + 52 D - 224 . The normal equations are 4 C + 8 D = 36 and 8 C + 26 D = 112. 6 ) b · a a · a a = (36 / 4) a = (9 , 9 , 9 , 9) = p . Note b - p = ( - 9 , - 1 , - 1 , 11) and this dots to 0 with a . The shortest distance is ( - 9) 2 + ( - 1) 2 + ( - 1) 2 + 11 2 = 204 . 12 ) (a) Here, as in problem 6, a T a is just the dot product of a with itself. This is m . Similarly a T b = m i =1 b i so ˆ x = 1 m m i =1 b i . (b) e = b - ˆ xb and || e || 2 = m i =1 ( b i - ˆ x ) 2 and the standard deviation is the square root of this last quantity. (c) e = b - ˆ xa = (1 , 2 , 6) - 3(1 , 1 , 1) = ( - 2 , - 1 , 3). This dots to 0 with (3 , 3 , 3). The projection matrix P = aa T a T a = 1 3 1 1 1 1 1 1 1 1 1 . 22 ) We need to consider E ( C, D ) = ( C - 2 D - 4) 2 + ( C - D - 2) 2 + ( C - - 1) 2 + ( C + D - 0) 2 + ( C + 2 D ) 2 so ∂E/∂C = 2( C - 2 D - 4) + 2( C - D - 2) + 2( C + 1) + 2( C + D ) + 2( C + 2 D ) = 10

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Unformatted text preview: = 2( C-2 D-4)(-2)+2( C-D-2)(-1)+0+2( C + D )(1)+2( C +2 D )(2) = 20 D +20 so we need to solve 5 C = 5, 10 D =-10 so C = 1 and D =-1. The line of best ﬁt is 1-t . 26 ) We are trying to solve C + D = 0, C + E = 1, C-D = 3 and C-E = 4, that is 1 1 1 1 1-1 1-1 = 1 3 4 . So A T A = 1 1 1 1 1 0-1 0 1-1 1 1 1 1 1-1 1-1 = 4 0 0 0 2 0 0 0 2 . Now A T ~ b = 8-3-3 So solving A T A~x = 8-3-3 gives ~x = 2-3 / 2-3 / 2 . The best plane is 2-(3 / 2) x-(3 / 2) y which at (0 , 0) gives 2, the average of 0, 1, 3 and 4....
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