Math 128A - HW7 Solutions.pdf

# Solution replacing h by h 3 in 1 and multiplying by 3

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Solution. Replacing h by h 3 in (1) and multiplying by 3 yields 3 M = 3 N h 3 + K 1 3 h 3 + K 2 3 h 2 9 + · · · . (2) Subtracting (1) from (2) and dividing by 2 gives a new approximation of M in which the linear term K 1 h has been eliminated: M = N 2 ( h ) + e K 2 h 2 + e K 3 h 3 + · · · , N 2 ( h ) = 3 N ( h 3 ) - N ( h ) 3 - 1 . Repeating, we can eliminate the quadratic term: 9 M = 9 N 2 h 3 + e K 2 9 h 2 9 + e K 3 9 h 3 27 + · · · , M = N 3 ( h ) + b K 3 h 3 + b K 4 h 4 + · · · , N 3 ( h ) = 9 N 2 ( h 3 ) - N 2 ( h ) 9 - 1 . Expressing N 3 ( h ) in terms of N ( h ) gives the desired result: M = 27 N ( h 9 ) - 12 N ( h 3 ) + N ( h ) 16 + O ( h 3 ) . Exercise 4.2.10. Suppose that N ( h ) is an approximation to M for every h > 0 and that M = N 2 ( h ) + K 2 h 2 + K 4 h 4 + K 6 h 6 + · · · , for some constants K 2 , K 4 , K 6 , . . . . Use the values N ( h ), N ( h 3 ) , and N ( h 9 ) to produce an O ( h 6 ) approximation to M . Solution. Let N 2 ( h ) = N ( h ) and relabel the K j ’s so that M = N 2 ( h ) + K 2 h 2 + K 4 h 4 + K 6 h 6 + · · · . Replacing h by h 3 and multiplying by 3 2 = 9 yields 9 M = 9 N 2 h 3 + 9 K 2 h 3 2 + 9 K 4 h 3 4 + · · · . Subtracting the first equation from the substituted one (as we did in 9. ) and dividing by 8 gives a new approximation of M in which the quadratic term K 2 h 2 has been eliminated: M = N 4 ( h ) + e K 4 h 4 + e K 6 h 6 + · · · , N 4 ( h ) = 9 N 2 ( h 3 ) - N 2 ( h ) 9 - 1 . Repeating, we can eliminate the quartic term by multiplying by 3 4 = 81:

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81 M = 81 N 4 h 3 + e K 4 81 h 4 81 + e K 6 81 h 6 3 6 + · · · , M = N 6 ( h ) + O ( h 6 ) , N 6 ( h ) = 81 N 4 ( h 3 ) - N 4 ( h ) 81 - 1 . Expressing N 6 ( h ) in terms of N 2 ( h ) = N ( h ) gives the desired result: 80 N 6 ( h ) = 81 N 4 h 3 - N 4 ( h ) 8 · 80 N 6 ( h ) = 81 9 N 2 h 9 - N 2 h 3 - 9 N 2 h 3 - N 2 ( h ) M = 729 N ( h 9 ) - 90 N ( h 3 ) + N ( h ) 640 + O ( h 6 ) . Exercise 4.3.2. Approximate the following integrals using the Trapezoidal rule. (a) Z 0 . 25 - 0 . 25 (cos x ) 2 dx Solution. Use b - a 2 ( f ( a ) + f ( b )). We get 0 . 25 - ( - 0 . 25) 2 (cos 2 ( - 0 . 25) + cos 2 (0 . 25)) 0 . 25[0 . 938791 + 0 . 938791] 0 . 469396 Exercise 4.3.4. Find a bound for the error in Exercise 2 using the error formula, and compare this to the actual error. (a) Z 0 . 25 - 0 . 25 (cos x ) 2 dx Solution. The error bound for the Trapezoidal rule is - h 3 12 f 00 ( ξ ) for some ξ ( a, b ). In our case, we have h = 0 . 5 and f 0 ( x ) = - 2 cos x sin x = - sin(2 x ), and f 00 ( x ) = - 2 cos 2 x . Since | f 00 ( x ) | ≤ 2, we obtain the error bound 0 . 5 3 12 · 2 0 . 0208333.
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• Spring '08
• Rieffel
• Trapezoid, Romberg's method

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