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) If ∑ a k converges, then lim k → ∞ a k = 0 and if lim k → ∞ a k ≠ 0 , then the series diverges. Given: The function, 1 ( 1 − 10 x ) 2 . Calculation: Obtain Maclaurin series of 1 ( 1 − 10 x ) 2 . Integrate the function 1 ( 1 − 10 x ) 2 . ∫ f ( x ) d x = ∫ 1 ( 1 − 10 x ) 2 d x = ∫ ( 1 − 10 x ) − 2 d x = ( 1 − 10 x ) − 2 + 1 − 2 + 1 ⋅ 1 − 10 Simplify further, ∫ f ( x ) d x = ( 1 − 10 x ) − 1 1 ⋅ 1 10 = 1 10 ⋅ 1 ( 1 − 10 x ) Substitute 10 x for x in 1 1 − x = 1 + x + x 2 + x 3 + ⋯ , 1 1 − ( 10 x ) = 1 + ( 10 x ) + ( 10 x ) 2 + ( 10 x ) 3 + ⋯ = 1 + 10 x + ( 10 ) 2 x 2 + ( 10 ) 3 x 3 + ⋯ ( 10 x ) k + ⋯ = ∑ k = 0 ∞ ( 10 x ) k Multiply with 1 10 on both sides, 1 10 ⋅ 1 1 − ( 10 x ) = 1 10 ⋅ ∑ k = 0 ∞ ( 10 x ) k Differentiate the series 1 10 ⋅ ∑ k = 0 ∞ ( 10 x ) k to get back the original series. d d x ( 1 10 ⋅ ∑ k = 0 ∞ ( 10 x ) k ) = 1 10 ⋅ ∑ k = 1 ∞ k ( 10 x ) k − 1 ⋅ 10 = 1 10 ⋅ ∑ k = 1 ∞ 10 k ⋅ ( 10 x ) k − 1 = ∑ k = 1 ∞ k ( 10 x ) k − 1 Therefore, the required Maclaurin series is ∑ k = 1 ∞ k ( 10 x ) k − 1 . Obtain the interval of convergence of the series. Let a k = k ( 10 x ) k − 1 . Then, a k + 1 = ( k + 1 ) ( 10 x ) ( k + 1 ) − 1 . Use the Ratio test to compute the value | a k + 1 a k |

Calculus: Early Transcendentals (2nd Edition)

2nd Edition

ISBN: 9780321947345

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

Publisher: PEARSON

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Chapter 9, 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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Calculus: Early Transcendentals (2nd Edition)

Ch. 9.1 - Suppose you use a second-order Taylor polynomial...Ch. 9.1 - Does the accuracy of an approximation given by a...Ch. 9.1 - The first three Taylor polynomials for f(x)=1+x...Ch. 9.1 - Prob. 4E Ch. 9.1 - How is the remainder Rn(x) in a Taylor polynomial...Ch. 9.1 - Explain how to estimate the remainder in an...Ch. 9.1 - Linear and quadratic approximation a. Find the...Ch. 9.1 - Linear and quadratic approximation a. Find the...Ch. 9.1 - Linear and quadratic approximation a. Find the...Ch. 9.1 - Linear and quadratic approximation a. Find the...

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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 determines 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.

Chapter 9 Solutions

Additional Math Textbook 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 determines 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.

Chapter 9 Solutions

Additional Math Textbook 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 .