Explanation Given: The series ∑ k = 0 ∞ ( x 2 − 1 3 ) k . Result used: (1) Root test: Let ∑ k = 1 ∞ a k be an infinite series with non-negative terms and let ρ = lim k → ∞ a k k . then, (i) If 0 ≤ ρ < 1 , the series converges. (ii) If ρ > 1 , (including ρ = ∞ ) the series diverges. (2) The power series representation of 1 ( 1 − x ) is ∑ k = 0 ∞ x k . (3) The geometric series ∑ k = 0 ∞ r n converges if | r | < 1 and diverges if | r | ≥ 1 . Calculation: Substitute ( x 2 − 1 ) 3 for x in the result 1 ( 1 − x ) = ∑ k = 0 ∞ x k . ∑ k = 0 ∞ ( x 2 − 1 3 ) k = 1 1 − ( x − 1 ) 3 = 1 3 − x 2 + 1 3 = 1 4 − x 2 3 ∑ k = 0 ∞ ( x 2 − 1 3 ) k = 3 4 − x 2 Therefore, the function representation of the given series is 3 4 − x 2 . The interval of convergence of the series ∑ k = 0 ∞ ( x 2 − 1 3 ) k is computed as follows, Consider the Root test ρ = lim k → ∞ | a k | k as follows, ρ = lim k → ∞ | ( x 2 − 1 3 ) k k | = lim k → ∞ | ( x 2 − 1 3 ) | = | x 2 − 1 3 | Since, ρ < 1 , the series ∑ k = 0 ∞<

Calculus, Single Variable: Early Transcendentals (3rd Edition)

3rd Edition

ISBN: 9780134766850

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

Publisher: PEARSON

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Chapter 11.2, Problem 71E

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To find: The function represented by the given series and interval of convergence.

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Chapter 11.2, Problem 70E

Chapter 11.2, Problem 72E

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Chapter 11 Solutions

Calculus, Single Variable: Early Transcendentals (3rd Edition)

Ch. 11.1 - Verify that p3 satisfies p3(k)(a)=f(k)(a), for k =...Ch. 11.1 - Prob. 2QC Ch. 11.1 - Prob. 3QC Ch. 11.1 - Write out the next two Taylor polynomials p4 and...Ch. 11.1 - Prob. 5QC Ch. 11.1 - Prob. 6QC Ch. 11.1 - Suppose you use a second-order Taylor polynomial...Ch. 11.1 - Does the accuracy of an approximation given by a...Ch. 11.1 - The first three Taylor polynomials for f(x)=1+x...Ch. 11.1 - Prob. 4E

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Chapter 11 Solutions