Explanation Replace the value of N by 2 N in the error bound to check the correctness of the given statement. Then determine the actual value of the integral and the actual errors and observe if the actual error seems to decrease by 1 16 if N is doubled Given: ∫ 0 π sin x d x Calculation: If we replace N by 2 N the error bound becomes K 4 ( b − a ) 5 180 ( 2 N ) 4 = 1 16 ( K 4 ( b − a ) 5 180 N 4 ) Hence we see that since N is raised to the fourth power, the error bound decreases by 1 16 if N is increased to 2 N . The exact value of the integral is ∫ 0 π sin x d x = [ − cos x ] 0 π = 2 Δ x = π 4 and the endpoints are 0 , π 4 , 2 π 4 , 3 π 4 , π Hence S 4 = 1 3 π 4 [ sin 0 + 4 sin ( π 4 ) + 2 sin ( 2 π 4 ) + 4 sin ( 3 π 4 ) + sin π ] ≈ 2.004560 Similarly, S 8 = 1 3 π 8 [ sin 0 + 4 sin ( π 8 ) + 2 sin ( 2 π 8 ) + ... + 4 sin ( 7 π 8 ) + sin π ] ≈ 2.0002692 S 16 = 1 3 π 16 [ sin 0 + 4 sin ( π 16 ) + 2 sin ( 2 π 16 ) + ...

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ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 8.9, Problem 56E

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To explain why the error bound for $S_{N}$ decreases by $\frac{1}{16}$ if $N$ is increased to $2 N$ and compute the actual error in $S_{N}$ for $\int_{0}^{π} \sin x d x$ for $N = 4, 8, 16, 32, 64$ . And to determine if the actual error seem to decrease by $\frac{1}{16}$ if $N$ is doubled

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Chapter 8.9, Problem 55E

Chapter 8.9, Problem 57E

Chapter 8 Solutions

Calculus - Standalone book

Ch. 8.1 - Prob. 1PQ Ch. 8.1 - Prob. 2PQ Ch. 8.1 - Prob. 3PQ Ch. 8.1 - Prob. 1E Ch. 8.1 - Prob. 2E Ch. 8.1 - Prob. 3E Ch. 8.1 - Prob. 4E Ch. 8.1 - Prob. 5E Ch. 8.1 - Prob. 6E Ch. 8.1 - Prob. 7E

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To explain why the error bound for S N decreases by 1 16 if N is increased to 2 N and compute the actual error in S N for ∫ 0 π sin x d x for N = 4 , 8 , 16 , 32 , 64 . And to determine if the actual error seem to decrease by 1 16 if N is doubled

Solution Summary: The author explains why the error bound for S N decreases by 1 16 if N is increased to 2 N, and determines the actual value of the integral and actual errors.

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To explain why the error bound for $S_{N}$ decreases by $\frac{1}{16}$ if $N$ is increased to $2 N$ and compute the actual error in $S_{N}$ for $\int_{0}^{π} \sin x d x$ for $N = 4, 8, 16, 32, 64$ . And to determine if the actual error seem to decrease by $\frac{1}{16}$ if $N$ is doubled

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Chapter 8 Solutions

To explain why the error bound for S N decreases by 1 16 if N is increased to 2 N and compute the actual error in S N for ∫ 0 π sin x d x for N = 4 , 8 , 16 , 32 , 64 . And to determine if the actual error seem to decrease by 1 16 if N is doubled

Solution Summary: The author explains why the error bound for S N decreases by 1 16 if N is increased to 2 N, and determines the actual value of the integral and actual errors.

Concept explainers

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To explain why the error bound for SN decreases by 116 if N is increased to 2N and compute the actual error in SN for ∫0πsinxdx for N=4,8,16,32,64. And to determine if the actual error seem to decrease by 116 if N is doubled

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Chapter 8 Solutions

To explain why the error bound for $S_{N}$ decreases by $\frac{1}{16}$ if $N$ is increased to $2 N$ and compute the actual error in $S_{N}$ for $\int_{0}^{π} \sin x d x$ for $N = 4, 8, 16, 32, 64$ . And to determine if the actual error seem to decrease by $\frac{1}{16}$ if $N$ is doubled