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 (3rd Edition)

3rd Edition

ISBN: 9780134763644

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

Publisher: PEARSON

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Chapter 11, Problem 33RE

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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 (3rd Edition)

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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 11 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

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.

Chapter 11 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 .