Quiz 24B Solution Guyang Cao December 2 2024 1 Question 1 Solution Let f t log sin t we have f t 1 sin2 1 which means the error bound is part of 100000 times this bound is 47 48 also allowed sin2 1 1 1 1 sin t cos t and then f t 1 sin2 t Therefore f 1 1 2 24 1 2 1 3 0 0004708 which means the integer 2 Question 2 Solution In this problem we evaluate the precision of Composite Simpson s rule and Joe s rule 1 By the code above we have Simpson s result be 3 1415926530 which has 10 correct digits and Joe s result be 3 1435688003 which only has 3 correct digits Therefore x 10 9 also allowed y 3 2
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