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(a) Find the smallest value of N for which the interval bracketing S in line (*) above has length at most 10- Answer: Nmin Answer: (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. = SZ SN + SN+1 2 =

(a) Find the smallest value of N for which the interval bracketing S in line (*) above has length at most 10- Answer: Nmin Answer: (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. = SZ SN + SN+1 2 =

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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2:56 ✓ bartleby.com/questior Q SEARCH a_{N+1} <= 10^-8 1/((N+1) ^ 5) <= 10^-8 (N+1)^5 >= 10^8 ASK Step 2: Find the smallest value We want to find the smallest value of N such that: N+1>= (10^8)^(1/5) N + 1 >= 100 N>= 99 Vo)) LTE+ LTE1 TE2 50% + VX MATH Therefore, the smallest value of N for which the interval bracketing S has length at most 10^-8 is N = 99. 6 Step 3: Find the value of N (b) Using N = 99, we can approximate S by the midpoint of the interval bracketing S: S_X = (S_{99} + S_{100})/2 We can calculate the partial sums S_{99} and S_{100} using a calculator or spreadsheet. Using Excel, we can enter the following formula in cell A1: 000
Transcribed Image Text:2:56 ✓ bartleby.com/questior Q SEARCH a_{N+1} <= 10^-8 1/((N+1) ^ 5) <= 10^-8 (N+1)^5 >= 10^8 ASK Step 2: Find the smallest value We want to find the smallest value of N such that: N+1>= (10^8)^(1/5) N + 1 >= 100 N>= 99 Vo)) LTE+ LTE1 TE2 50% + VX MATH Therefore, the smallest value of N for which the interval bracketing S has length at most 10^-8 is N = 99. 6 Step 3: Find the value of N (b) Using N = 99, we can approximate S by the midpoint of the interval bracketing S: S_X = (S_{99} + S_{100})/2 We can calculate the partial sums S_{99} and S_{100} using a calculator or spreadsheet. Using Excel, we can enter the following formula in cell A1: 000
For an alternating series whose summands are decreasing in magnitude, the true sum Slies between any two successive partial sums: Consider S = ∞ n=1 (*) (−1)²+¹ n.5 Answer: Answer: N min > - S≈ (a) Find the smallest value of N for which the interval bracketing S in line (*) above has length at most 10-⁹. min {SN, SN+1} <S≤ max {SN, SN+1}. (−1)n+1 n5 and write SN N (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 n=1
Transcribed Image Text:For an alternating series whose summands are decreasing in magnitude, the true sum Slies between any two successive partial sums: Consider S = ∞ n=1 (*) (−1)²+¹ n.5 Answer: Answer: N min > - S≈ (a) Find the smallest value of N for which the interval bracketing S in line (*) above has length at most 10-⁹. min {SN, SN+1} <S≤ max {SN, SN+1}. (−1)n+1 n5 and write SN N (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 n=1
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