Explanation Result used: ( 1) Ratio test: Let ∑ k = 1 ∞ a k be an infinite series and let r = lim k → ∞ | a k + 1 a k | . If r < 1 , then the series ∑ k = 1 ∞ a k is converges. (2) Every infinite sum of nonzero constant diverges. Given: The function, 1 1 − x 2 . Calculation: Obtain Maclaurin series of 1 1 − x 2 . Substitute x 2 for x in 1 1 − x = 1 + x + x 2 + x 3 + ⋯ , 1 1 − ( x 2 ) = 1 + ( x 2 ) + ( x 2 ) 2 + ( x 2 ) 3 + ⋯ = 1 + x 2 + x 4 + x 6 + ⋯ + x 2 k + ⋯ = ∑ k = 0 ∞ x 2 k Therefore, the required Maclaurin series is ∑ k = 0 ∞ x 2 k . Obtain the interval of convergence of the series. Let a k = x 2 k . Then, a k + 1 = x 2 k + 2 . Use the Ratio test to compute the value | a k + 1 a k | . | a k + 1 a k | = | x 2 k + 2 x 2 k | Take lim k → ∞ on both sides, lim k → ∞ | a k + 1 a k | = lim k → ∞ | x 2 k + 2 x 2 k | = lim k → ∞ | x 2 | = | x 2 | Since | r | < 1 , the series ∑ k = 0 ∞ x 2 k converges as | x 2 | < 1

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, Problem 29RE

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The Maclaurin series and interval of convergence for the given function by using the geometric series ∑k=0∞xk=11−x, for |x|<1.

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Chapter 11, Problem 28RE

Chapter 11, Problem 30RE

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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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The Maclaurin series and interval of convergence for the given function by using the geometric series ∑ k = 0 ∞ x k = 1 1 − x , for | x | < 1 .

Solution Summary: The author explains the Maclaurin series and interval of convergence for the given function by using the geometric series displaystyle

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The Maclaurin series and interval of convergence for the given function by using the geometric series ∑k=0∞xk=11−x, for |x|<1.

Want to see the full answer?

Chapter 11 Solutions

The Maclaurin series and interval of convergence for the given function by using the geometric series ∑ k = 0 ∞ x k = 1 1 − x , for | x | &lt; 1 .

Solution Summary: The author explains the Maclaurin series and interval of convergence for the given function by using the geometric series displaystyle

Videos

The Maclaurin series and interval of convergence for the given function by using the geometric series ∑k=0∞xk=11−x, for |x|<1.

Want to see the full answer?

Chapter 11 Solutions

The Maclaurin series and interval of convergence for the given function by using the geometric series ∑ k = 0 ∞ x k = 1 1 − x , for | x | < 1 .