hmwk_4_2010_solutions_rev

# A 2 h 3 a 3 h 2 4 a 4 h 3 5 a 5 h 4 g 2 h f 1 f 1 a 2

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a 2 h + 3 · a 3 h 2 + 4 · a 4 h 3 + 5 · a 5 h 4 g (2 h ) = f 1 = f 1 = a 0 + 2 · a 1 h + 4 · a 2 h 2 + 8 · a 3 h 3 + 16 · a 4 h 4 + 32 · a 5 h 5 g (1) (2 h ) = f (1) 1 = f (1) 1 = a 1 + 4 · a 2 h + 12 · a 3 h 2 + 32 · a 4 h 3 + 80 · a 5 h 4 g (3 h ) = f 2 = f 2 = a 0 + 3 · a 1 h + 9 · a 2 h 2 + 27 · a 3 h 3 + 81 · a 4 h 4 + 243 · a 5 h 5 g (1) (3 h ) = f (1) 2 = f (1) 2 = a 1 + 6 · a 2 h + 27 · a 3 h 2 + 108 · a 4 h 3 + 405 · a 5 h 4 3

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The constraint equations may be written in matrix form AX = B. 1 h h 2 h 3 h 4 h 5 0 1 2 h 3 h 2 4 h 3 5 h 4 1 2 h 4 h 2 8 h 3 16 h 4 32 h 5 0 1 4 h 12 h 2 32 h 3 80 h 4 1 3 h 9 h 2 27 h 3 81 h 4 243 h 5 0 1 6 h 27 h 2 108 h 3 405 h 4 a 0 a 1 a 2 a 3 a 4 a 5 = f 0 f (1) 0 f 1 f (1) 1 f 2 f (1) 2 We can then multiply each derivative equation by h so that each column in A is the same order with regards to h: 1 h h 2 h 3 h 4 h 5 0 h 2 h 2 3 h 3 4 h 4 5 h 5 1 2 h 4 h 2 8 h 3 16 h 4 32 h 5 0 h 4 h 2 12 h 3 32 h 4 80 h 5 1 3 h 9 h 2 27 h 3 81 h 4 243 h 5 0 h 6 h 2 27 h 3 108 h 4 405 h 5 a 0 a 1 a 2 a 3 a 4 a 5 = f 0 h · f (1) 0 f 1 h · f (1) 1 f 2 h · f (1) 2 Since all columns of A now have common elements (h raised to some power), we can then group them with our unknowns by factoring them out of the A matrix and adding them to the X vector as follows: 1 1 1 1 1 1 0 1 2 3 4 5 1 2 4 8 16 32 0 1 4 12 32 80 1 3 9 27 81 243 0 1 6 27 108 405 a 0 a 1 h a 2 h 2 a 3 h 3 a 4 h 4 a 5 h 5 = f 0 h · f (1) 0 f 1 h · f (1) 1 f 2 h · f (1) 2 Calculating the inverse of matrix A in Matlab yields the following: A - 1 = - 18 - 9 9 - 18 10 - 3 57 24 - 24 57 - 33 10 - 63 . 5 - 24 . 25 22 - 68 41 . 5 - 12 . 75 32 . 75 11 . 75 - 8 38 - 24 . 75 7 . 75 - 8 - 2 . 75 1 - 10 7 - 2 . 25 0 . 75 0 . 25 0 1 - 0 . 75 0 . 25 Now solving for the unknowns by multiplying A - 1 by the modified B vector and dividing by the h terms in the modified X vector yields the following six equations: a 0 = - 18 f 0 - 9 hf (1) 0 + 9 f 1 - 18 hf (1) 1 + 10 f 2 - 3 hf (1) 2 a 1 = 57 f 0 h + 24 f
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• Fall '08
• Westerink,J
• Derivative, Zagreb, Highways in Croatia, Histone H3, xvec, EAvec

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