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For an alternating series whose summands are decreasing in magnitude, the true sum S lies between any two successive partial sums: min {SN, SN+1} ≤ S ≤ max {SN, SN+1}. Consider S = Σ n=1 (−1)n+1 n4 Answer: Nmin = 71 Answer: and write S≈ SN = (a) Find the smallest value of N for which the interval bracketing S in line (*) above has length at most 10-6. N SN + SN+1 2 n=1 (b) Using the N found in part (a), approximate S by the midpoint of the interval implicit in line (*). A spreadsheet may be helpful to calculate the sum SN. (−1)n+1 n4

For an alternating series whose summands are decreasing in magnitude, the true sum S lies between any two successive partial sums: min {SN, SN+1} ≤ S ≤ max {SN, SN+1}. Consider S = Σ n=1 (−1)n+1 n4 Answer: Nmin = 71 Answer: and write S≈ SN = (a) Find the smallest value of N for which the interval bracketing S in line (*) above has length at most 10-6. N SN + SN+1 2 n=1 (b) Using the N found in part (a), approximate S by the midpoint of the interval implicit in line (*). A spreadsheet may be helpful to calculate the sum SN. (−1)n+1 n4

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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For an alternating series whose summands are decreasing in magnitude, the true sum S lies between any two successive partial sums: min {SN, SN+1} ≤ S ≤ max {SN, SN+1}. ∞ Consider S = (-1)+1 n4 n=1 Answer: Nmin Answer: " = S≈ (a) Find the smallest value of N for which the interval bracketing S in line (*) above has length at most 10-6. 71 and write * (b) Using the N found in part (a), approximate S by the midpoint of the interval implicit in line (*). A spreadsheet may be helpful to calculate the sum SN. SN + SN+1 2 N SN = Σ n=1 = (−1)n+1 n4
Transcribed Image Text:For an alternating series whose summands are decreasing in magnitude, the true sum S lies between any two successive partial sums: min {SN, SN+1} ≤ S ≤ max {SN, SN+1}. ∞ Consider S = (-1)+1 n4 n=1 Answer: Nmin Answer: " = S≈ (a) Find the smallest value of N for which the interval bracketing S in line (*) above has length at most 10-6. 71 and write * (b) Using the N found in part (a), approximate S by the midpoint of the interval implicit in line (*). A spreadsheet may be helpful to calculate the sum SN. SN + SN+1 2 N SN = Σ n=1 = (−1)n+1 n4
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