a. Explanation Given: The geometric series S = ∑ k = 0 ∞ r k , where | r | < 1 and its sequence of partial sums S n = ∑ k = 0 n r k . Result used: The geometric series ∑ n = 1 ∞ a r n − 1 = a + a r + a r 2 + ⋯ is convergent if | r | < 1 and its sum is a 1 − r , where a ≠ 0 is the first term and r is the common ratio of the series. The sum of the finite sum of the geometric series is ∑ k = 0 n − 1 a r k = a 1 − r n 1 − r . Calculation: Compute the value N ( r ) , such that | S − S N ( r ) | < 10 − 4 , at r = − 0.9 . Here, the geometric series ∑ k = 0 ∞ ( − 0.9 ) k with first term of the series is a = 1 and common ratio is | r | = 0.9 < 1 . Compute the sum of the geometric series by the Result stated above as follows, S = 1 1 + 0.9 = 1 1.9 That is, S = 1 1.9 . Obtain the value of S n = ∑ k = 0 n r k . S n = 1 − ( − 0.9 ) n 1 + 0.9 = 1 − ( − 0.9 ) n 1.9 Suppose n = 80 , S 80 = 1 − ( − 0.9 ) 80 1.9 . | S − S 80 | = 1 × 10 − 4 . Suppose n = 81 , S 80 = 1 − ( − 0.9 ) 81 1.9 . | S − S 81 | = 9.3 × 10 − 5 . Therefore, n = N ( r ) = 81 . Compute the value N ( r ) , such that | S − S N ( r ) | < 10 − 4 , at r = − 0.7 . Here, the geometric series ∑ k = 0 ∞ ( − 0.7 ) k with first term of the series is a = 1 and common ratio is | r | = 0.7 < 1 . Compute the sum of the geometric series by the Result stated above as follows, ∑ k = 0 ∞ ( − 0.7 ) k = 1 1 + 0.7 = 1 1.7 That is, ∑ k = 0 ∞ ( − 0.9 ) k = 1 1.7 . Obtain the value of S n = ∑ k = 0 n r k . S n = 1 − ( − 0.7 ) n 1 + 0.9 7 = 1 − ( − 0.7 ) n 1.7 Suppose n = 23 , S 23 = 1 − ( − 0.7 ) 23 1.7 . | S − S 23 | = 1.1 × 10 − 4 Suppose n = 24 , S 24 = 1 − ( − 0.7 ) 24 1.7 . | S − S 24 | = 7.9 × 10 − 5 Therefore, n = N ( r ) = 24 . Compute the value N ( r ) , such that | S − S N ( r ) | < 10 − 4 , at r = − 0.5 . Here, the geometric series ∑ k = 0 ∞ ( − 0.5 ) k with first term of the series is a = 1 and common ratio is | r | = 0.5 < 1 . Compute the sum of the geometric series by the Result stated above as follows, ∑ k = 0 ∞ ( − 0.5 ) k = 1 1 + 0.5 = 1 1.5 That is, ∑ k = 0 ∞ ( − 0.5 ) k = 1 1.5 . Obtain the value of S n = ∑ k = 0 n r k . S n = 1 − ( − 0.5 ) n 1 + 0.5 = 1 − ( − 0.5 ) n 1.5 Suppose n = 11 , S 11 = 1 − ( − 0.5 ) 11 1.5 . | S − S 11 | = 1.6 × 10 − 4 Suppose n = 12 , S 12 = 1 − ( − 0.5 ) 12 1.5 . | S − S 12 | = 8.1 × 10 − 5 Therefore, n = N ( r ) = 12 . Compute the value N ( r ) , such that | S − S N ( r ) | < 10 − 4 , at r = − 0.2 . Here, the geometric series ∑ k = 0 ∞ ( − 0.2 ) k with first term of the series is a = 1 and common ratio is | r | = 0.2 < 1 . Compute the sum of the geometric series by the Result stated above as follows, ∑ k = 0 ∞ ( − 0.2 ) k = 1 1 + 0.2 = 1 1.2 That is, ∑ k = 0 ∞ ( − 0.2 ) k = 1 1.2 . Obtain the value of S n = ∑ k = 0 n r k . S n = 1 − ( − 0.2 ) n 1 + 0.2 = 1 − ( − 0.2 ) n 1.2 Suppose n = 4 , S 4 = 1 − ( − 0.2 ) 4 1.2 . | S − S 4 | = 2.7 × 10 − 4 Suppose n = 5 , S 5 = 1 − ( − 0.2 ) 5 1.2 . | S − S 5 | = 5.3 × 10 − 5 Therefore, n = N ( r ) = 5 . Compute the value N ( r ) , such that | S − S N ( r ) | < 10 − 4 , at r = − 0.2 . The geometric series f ( r ) = ∑ k = 0 ∞ r k at r = 0 is 1. Obtain the value of S n = ∑ k = 0 n r k . S n = 1 − ( 0 ) n 1 + 0 = 1 1 Suppose n = 0 , S 0 = 1 b. To determine To sketch: The graph of N ( r ) for various values of r . c. To determine To describe: The rate of convergence of the geometric series depend on r.

Calculus: Early Transcendentals (2nd Edition)

2nd Edition

ISBN: 9780321947345

Author: William L. Briggs, Lyle Cochran, Bernard Gillett

Publisher: PEARSON

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Chapter 8.3, Problem 100E

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To complete: The given table showing the smallest value of n, calling it N(r), such that |S−Sn|<10−4, for various value of r.

To determine

To sketch: The graph of N(r) for various values of r.

To determine

To describe: The rate of convergence of the geometric series depend on r.

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Chapter 8.3, Problem 99E

Chapter 8.4, Problem 1E

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Explain the conditions under which the Radius of Convergence of the Power Series is a "finite positive real number" r>0

Chapter 8 Solutions

Calculus: Early Transcendentals (2nd Edition)

Ch. 8.1 - Define sequence and give an example.Ch. 8.1 - Suppose the sequence {an} is defined by the...Ch. 8.1 - Suppose the sequence {an} is defined by the...Ch. 8.1 - Prob. 4E Ch. 8.1 - Prob. 5E Ch. 8.1 - Given the series k=1k, evaluate the first four...Ch. 8.1 - The terms of a sequence of partial sums are...Ch. 8.1 - Consider the infinite series k=11k. Evaluate the...Ch. 8.1 - Explicit formulas Write the first four terms of...Ch. 8.1 - Explicit formulas Write the first four terms of...

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The given table showing the smallest value of n , calling it N ( r ) , such that | S − S n | < 10 − 4 , for various value of r .

Solution Summary: The author explains the value of the geometric series N(r) for various values of r.

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To complete: The given table showing the smallest value of n, calling it N(r), such that |S−Sn|<10−4, for various value of r.

To sketch: The graph of N(r) for various values of r.

To describe: The rate of convergence of the geometric series depend on r.

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Chapter 8 Solutions

The given table showing the smallest value of n , calling it N ( r ) , such that | S − S n | &lt; 10 − 4 , for various value of r .

Solution Summary: The author explains the value of the geometric series N(r) for various values of r.

Videos

To complete: The given table showing the smallest value of n, calling it N(r), such that |S−Sn|<10−4, for various value of r.

To sketch: The graph of N(r) for various values of r.

To describe: The rate of convergence of the geometric series depend on r.

Want to see the full answer?

Chapter 8 Solutions

The given table showing the smallest value of n , calling it N ( r ) , such that | S − S n | < 10 − 4 , for various value of r .