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Assignment 12 Solutions

Assignment 12 Solutions - MATH 138 Calculus 2 Solutions to...

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Unformatted text preview: MATH 138 Calculus 2, Solutions to Assignment 12 1: Estimate the following values so the error is ≤ 1 1000 . (a) ln(5 / 4) Solution: ln(1 + x ) = x- 1 2 x 2 + 1 3 x 3- 1 4 x 4 + ··· , so we have ln ( 5 4 ) = 1 4- 1 2 · 4 2 + 1 3 · 4 3- 1 4 · 4 4 + ··· ∼ = 1 4- 1 2 · 4 2 + 1 3 · 4 3 = 1 4- 1 32 + 1 192 = 43 192 with error E ≤ 1 4 · 4 4 = 1 1024 < 1 1000 by the A.S.T. (b) 5 √ e Solution: e x = 1 + x + 1 2! x 2 + 1 3! x 3 + 1 4! x 4 + ··· , so we have 5 √ e = e 1 5 = 1 + 1 5 + 1 5 2 2! + 1 5 3 3! + 1 5 4 4! + ··· ∼ = 1 + 1 5 + 1 5 2 2! + 1 5 3 3! = 1 + 1 5 + 1 50 + 1 750 = 916 750 with error E = 1 5 4 4! + 1 5 5 5! + 1 5 6 6! + ··· = 1 5 4 4! 1 + 1 5 · 5 + 1 5 2 · 5 · 6 + 1 5 3 · 5 · 6 · 7 + ··· ≤ 1 5 4 4! 1 + 1 5 2 + 1 5 4 + 1 5 6 + ··· = 1 5 4 4! 1 1- 1 25 = 1 5 4 4! 25 24 = 1 13200 where we used the C.T. and the formula for the sum of a geometric series. 2: Approximate the value of the definite integral Z 1 p 4 + x 3 dx so the error is at most 1 1000 ....
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Assignment 12 Solutions - MATH 138 Calculus 2 Solutions to...

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