Math 128A - HW6 Solutions.pdf

# 25 48 36 16 3 3 10 18 6 1 1 8 8 1 1 6 18 10 3 3 16 36

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- 25 48 - 36 16 - 3 - 3 - 10 18 - 6 1 1 - 8 0 8 - 1 - 1 6 - 18 10 3 3 - 16 36 - 48 25 3 . 10 3 . 12 3 . 14 3 . 18 3 . 24 = 3 . 167 1 . 500 2 . 833 5 . 167 6 . 500 . The first and fifth rows of the matrix are the coefficients of the left and right endpoint formulas. The middle row is the midpoint formula. And the second and fourth rows correspond to the result of 4.1.22(26). Note the reversal of order and change of sign in rows 1,5 and 2,4. Evaluating E ( t ) = L di dt + Ri then gives t 1 . 00 1 . 01 1 . 02 1 . 03 1 . 04 E ( t ) 3 . 54 1 . 91 3 . 22 5 . 51 6 . 83 Exercise 4.1.29. Consider the function e ( h ) = ε h + h 2 6 M, where M is a bound for the third derivative of a function. Show that e ( h ) has a minimum at 3 p 3 ε/M . Solution. Note that e ( h ) is differentiable away from 0 since it is a rational function and is only undefined at the division by 0 caused by h = 0. We assume M > 0. (If f 000 ( x ) 0, the 3-point midpoint scheme is exact, and the motivation for studying the competing effects of truncation error and roundoff error disappears.) To find critical points, note e 0 ( h ) = - ε h 2 + h 3 M. Setting this equal to 0 yields 3 ε = h 3 M h * = 3 p 3 ε/M

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as the only real root. To show it is a minimum we compute the concavity at that point: e 00 ( h ) = 2 ε h 3 + 1 3 M hence e 00 ( h * ) = 2 ε 3 ε/M + 1 3 M = M. Since M is a bound on the third derivative’s magnitude, it must be non-negative, and we assume it is strictly positive. Thus, the function is concave up and h * is a local minimum.
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