pset4 - MIT OpenCourseWare http/ocw.mit.edu 18.085...

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MIT OpenCourseWare http://ocw.mit.edu 18.085 Computational Science and Engineering I Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .
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18.085 - Mathematical Methods for Engineers I Prof. Gilbert Strang Solutions - Problem Set 4 Section 2.3 4 0 1 0 7) b = 1 x = 1 A = 1 1 0 2 1 2 1 3 1 3 Au = b Normal Equation A T Au = A T b 1 0 4 1 1 1 1 1 1 1 1 1 1 1 0 1 2 3 1 2 0 1 2 3 0 u = 1 3 1 4 6 6 6 14 u = 4 1 14 6 � � 6 u = 20 6 4 4 3 = 1 Nearest Line, C + Dx = 3 x # 8) p = Au 1 0 = 1 1 3 1 2 1 1 3 3 = 2 1 0 For y = 3 x at x = 0 , y = 3 0 = 3 at x = 1 , y = 2 at x = 2 , y = 1 at x = 3 , y = 0 1
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Those four values do lie on the line C + Dx # l = b p 4 1 3 2 = 0 1 1 0 1 1 1 1 = A T e = 0 1 1 1 0 1 1 1 = 0 1 2 3 1 0 1 Verified that A T e = 0 # 12) Parabola C + Dx + Ex 2 A b 1 ⎛⎬ 0 0 ⎛⎬ 4 = C D 1 1 1 1 1 2 4 0 E 1 3 9 1 Normal Equation A T Au = A T b C D = E 1 0 0 4 1 1 1 1 1 1 1 1 1 1 1 1 C D = E 0 1 2 3 0 1 2 3 1 2 4 0 0 1 4 9 0 1 4 9 1 3 9 1 4 6 14 6 14 36 6 4 14 36 98 10 Using MATLAB, C D = 4 4 1 E Cubic C + Dx + Ex 2 + Fx 3 C D E F 1 0 0 0 4 1 1 1 1 1 = 1 2 4 8 0 1 3 9 27 1 2
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If I fit the best cubic C + Dx + Ex 2 + Fx 3 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 elimination, 1 0 0 0 4 0 1 1 1 u = 3
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