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0 1 2 3 1 1 1 2 0 1 4 9 1 3 4 6 14 6 14 36 14 36 98

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Unformatted text preview: D E F 4 1 = 0 1 2 Using MATLAB, C 4 D = -4 1 E C 6 D = 4 10 E If I fit the best cubic C + Dx + Ex2 + F x3 to those four points, Au = b can be solved directly by u = A-1 b since A is invertible e = b - Au = b - A(A-1 b) = b - b = 0 The error vector e = 0 Proof : By gaussian 1 0 0 0 1 1 0 2 4 0 3 9 1 0 0 0 1 1 0 0 2 0 0 6 1 0 0 0 1 1 0 0 2 0 0 0 F =0 E=1 D = -4 C=4 elimination, 0 4 1 u = -3 8 -4 27 -3 0 4 1 -3 u = 2 6 24 6 C 0 4 D -3 1 = 6 E 2 6 0 F # 4 1 e = 0 1 4 1 = 0 1 0 0 = 0 0 1 1 - 1 1 4 1 - 0 1 # 0 1 2 3 0 0 4 1 1 -4 4 8 1 9 27 0 3 18) u10 u9 = 1 (u1 + u2 + + u9 ) 9 1 = (u1 + u2 + + u10 ) 10 1 1 (u1 + u2 + + u9 ) u10 + 10 10 1 9 u10 + u9 10 10 = = # -1 22) t = 1 2 5 b = 13 17 1 -1 5 C 1 1 = 13 D 1 2 17 Normal Equation AT Au = AT b 1 1 1 1 1 -1 1 2 1 -1 5 1 1 1 13 1 u = -1 1 2 17 2 35 3 2 u= 2 6 42 1 6 -2 35 u= -2 3 42 14 9 = 4 # # The closest line is = 9 - 4x b = A - b u 1 -1 5 9 1 = 1 - 13 4 1 2 17 0 = 0 0 # The error e = 0 because this b is linear combinations of A # 4 Section 2.4 1 1) 1 3 2 2 3 Incidence Matrix Node 1 2 -1 1 Atriangle = 0 -1 -1...
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This note was uploaded on 05/02/2013 for the course MATH 101 taught by Professor Ns during the Fall '12 term at MIT.

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