hmwk_4_2010_solutions_rev

# 1 24 f 1 h 57 f 1 1 33 f 2 h 10 f 1 2 a 2 63 5 f h 2

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(1) 0 + - 24 f 1 h + 57 f (1) 1 + - 33 f 2 h + 10 f (1) 2 a 2 = - 63 . 5 f 0 h 2 + - 24 . 25 f (1) 0 h + 22 f 1 h 2 + - 68 f (1) 1 h + 41 . 5 f 2 h 2 + - 12 . 75 f (1) 2 h a 3 = 32 . 75 f 0 h 3 + 11 . 75 f (1) 0 h 2 + - 8 f 1 h 3 + 38 f (1) 1 h 2 + - 24 . 75 f 2 h 3 + 7 . 75 f (1) 2 h 2 4

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a 4 = - 8 f 0 h 4 + - 2 . 75 f (1) 0 h 3 + 1 f 1 h 4 + - 10 f (1) 1 h 3 + 7 f 2 h 4 + - 2 . 25 f (1) 2 h 3 a 5 = 0 . 75 f 0 h 5 + 0 . 25 f (1) 0 h 4 + 0 f 1 h 5 + 1 f (1) 1 h 4 + - 0 . 75 f 2 h 5 + 0 . 25 f (1) 2 h 4 We can then group f 0 , f (1) 0 , f 1 , f (1) 1 , f 2 , and f (1) 2 terms together. However, we must also account for the fact that the interpolation function is g(x) = a 0 + a 1 x + a 2 x 2 + a 3 x 3 + a 4 x 4 + a 5 x 5 , so we must multiply each equation a i by x i for i = 0, 1, 2, 3, 4, 5. This grouping procedure results in the following equation: f ( x ) = - 18 + 57 h x + - 63 . 5 h 2 x 2 + 32 . 75 h 3 x 3 + - 8 h 4 x 4 + 0 . 75 h 5 x 5 · f 0 + - 9 h + 24 x + - 24 . 25 h x 2 + 11 . 75 h 2 x 3 + - 2 . 75 h 3 x 4 + 0 . 25 h 4 x 5 · f (1) 0 + 9 + - 24 h x + 22 h 2 x 2 + - 8 h 3 x 3 + 1 h 4 x 4 · f 1 + - 18 h + 57 x + - 68 h x 2 + 38 h 2 x 3 + - 10 h 3 x 4 + 1 h 4 x 5 · f (1) 1 + 10 + - 33 h x + 41 . 5 h 2 x 2 + - 24 . 75 h 3 x 3 + 7 h 4 x 4 + - 0 . 75 h 5 x 5 · f 2 + - 3 h + 10 x + - 12 . 75 h x 2 + 7 . 75 h 2 x 3 + - 2 . 25 h 3 x 4 + 0 . 25 h 4 x 5 · f (1) 2 If we let the bracketed terms above be g 1 , g 2 , g 3 , g 4 , g 5 , and g 6 , we note that this equation is of the form: f ( x ) = g 1 f 0 + g 2 f (1) 0 + g 3 f 1 + g 4 f (1) 1 + g 5 f 2 + g 6 f (1) 2 Therefore, these six g functions are our basis functions. Let h = 1, plot each of these functions, and display all six plots in Figure 1. Since we let h = 1, our three interpolation points (h, 2h, 3h) equal 1, 2, and 3 respectively. We can therefore see that these functions meet all the necessary requirements. g 1 = 1 at x=1, equals zero at x=2 and x=3, and has a slope of zero at all three points. g 2 has a slope of 1 at x=1, has a slope of zero at x=2 and x=3, and equals zero at all three interpolation points. g 3 = 1 at x=2, equals zero at x=1 and x=3, and has a slope of zero at all three points. g 4 has a slope of 1 at x=2, has a slope of zero at x=1 and x=3, and equals zero at all three interpolation points. g 5 = 1 at x=3, equals zero at x=1 and x=2, and has a slope of zero at all three points. Finally, g 6 has a slope of 1 at x=3, has a slope of zero at x=1 and x=2, and equals zero at all three interpolation points. In summary, the interpolation functions have appropriate functional values and derivatives and thus match the necessary requirements. 5
(a) Basis g1 (b) Basis g2 (c) Basis g3 (d) Basis g4 (e) Basis g5 (f) Basis g6 Figure 1: Plots for the six basis functions.

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• Fall '08
• Westerink,J
• Derivative, Zagreb, Highways in Croatia, Histone H3, xvec, EAvec

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