If We Use The Notation M N And T N For The Composite Midpoint And Trapezoidal Rules In Class We Saw That B 5 Mm 1 (312.25 KiB) Viewed 37 times
If we use the notation M(n) and T(n) for the composite midpoint and trapezoidal rules, in class we saw that b 5* (= Mm) +(mexe ) h2 f(x) dx = M(n) + (b − a). f"(e) and 24 = ° s(e) de =T(n) – (6-0) "C). h2 f"(e 12 (a) Take a combination of these two equations to eliminate the error term, and write the integral as a combination of M(n) and T(n). Then write a function called betterint that takes as input the function, bounds and n, that calls mid and trap and outputs the combination that should give a better answer. I don't need to see the code, just the formula. 2 (b) Given em dx e2 – 1, evaluate with n= 100, 200, 400 using the midpoint and trape- zoidal rules, and find the errors. By taking the ratio of the errors, verify that the errors are behaving as expected, going down by a factor of four. S = >
2 ſe (c) Given ex dx = e2 – 1, evaluate using betterint with n= 100, 200, 400 and find the errors. What is the ratio of errors? Given that if the error is proportional to hk, halving the step size should reduce the error by a factor of 2k, can you guess the order of our new method?
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