128a_su09_hw4_sol

128a_su09_hw4_sol - MATH 128A, SUMMER 2009: HOMEWORK 4...

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MATH 128A, SUMMER 2009: HOMEWORK 4 SOLUTIONS 3.4.11. S ( x ), S 0 ( x ), and S 00 ( x ) must be continuous at x = 1; so 2 = S 1 (1) = S 0 (1) = 2, b = S 0 1 (1) = S 0 0 (1) = - 1, and 2 c = S 00 1 (1) = S 00 0 (1) = - 6. The natural boundary condition requires S 00 ( x ) to be zero at x = 0 and x = 2, so 0 = S 00 0 (0) = 0 and 0 = S 00 1 (2) = 2 c + 6 d . This gives us coefficients ( b,c,d ) = ( - 1 , - 3 , 1). 4.1.6a+8a. At the endpoints, we use the one-sided three-point formulas (equation (4.4) in the section and its mirror-image). At the other points, we use the centered three-point formula (4.5), which is typically more accurate. All error bound formulas involve f (3) ( ξ ) = 8( e 2 ξ - sin(2 ξ )), where ξ is an x -value between the three which are sampled. f (3) is a positive decreasing function on [ - 0 . 3 , 0], so 8( e 2 ξ min - sin(2 ξ min )) is the best possible upper-bound for | f (3) ( ξ ) | . A less careful bound is 8( e 2(0) + 1) = 16. x f ( x ) Approx. f 0 ( x ) Exact f 0 ( x ) Err. Bound (good) Bound (easy) - 0 . 3 - 0 . 27652 - 0 . 06030 - 0 . 03167 0 . 02864 0 . 02969 0 . 05333 - 0 . 2 - 0 . 25074 0 . 57590 0 . 56180 0 . 01410 0 . 01485 0 . 02667 - 0 . 1 - 0 . 16134 1 . 25370 1 . 24012 0 . 01358 0 . 01413 0 . 02667 0 0 1 . 97310 2 . 00000 0 . 02690 0 . 02826 0 . 05333 4.1.28. The fourth-order Taylor polynomials are f ( x 0 - 2 h ) = f ( x 0 ) - 2 hf 0 ( x 0 ) + 4 h 2 2! f 00 ( x 0 ) - 8 h 3 3! f 000 ( x 0 ) + 16 h 4 4! f (4) ( x 0 ) + O ( h 5 ) f ( x 0 - h ) = f ( x 0 ) - hf 0 ( x 0 ) + h 2 2! f 00 ( x 0 ) - h 3 3! f 000 ( x 0 ) + h 4 4! f (4) ( x 0 ) + O ( h 5 ) f ( x 0 + h ) = f ( x 0 ) + hf 0 ( x 0 ) + h 2 2! f 00 ( x 0 ) + h 3 3! f 000 ( x 0 ) + h 4 4! f
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This note was uploaded on 04/06/2010 for the course MATH various taught by Professor Tao/analysis during the Spring '10 term at UCLA.

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128a_su09_hw4_sol - MATH 128A, SUMMER 2009: HOMEWORK 4...

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