a. Explanation Given: The function defined by the geometric series f ( x ) = ∑ k = 0 ∞ x 2 k . Result used: The sum of the geometric series is ∑ k = 0 ∞ a r k = a 1 − r if | r | < 1 and does not exists if | r | ≥ 1 . Calculation: From the Result, the geometric series defined only by f ( x ) = ∑ k = 0 ∞ x 2 k when the ratio x 2 is less than 1, which means | x | < 1 . Compute the function f ( x ) = ∑ k = 0 ∞ x 2 k at x = 0 . Here, | x | < 1 and the first term a = 1 . Thus, by Result stated above f ( 0 ) = 1 1 − 0 2 . f ( 0 ) = 1 Compute the function f ( x ) = ∑ k = 0 ∞ x 2 k at x = 0.2 . Here, | x | < 1 and the first term a = 1 . Thus, by Result stated above f ( 0.2 ) = 1 1 − 0.2 2 . f ( 0.2 ) = 1 1 − 0.4 = 1 0.6 = 10 6 That is, f ( 0 b. To determine To find: The domain of f .

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 96E

To determine

To evaluate: The values of f(0),f(0.2),f(0.5),f(1) and f(1.5) if possible.

To determine

To find: The domain of f.

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Chapter 8.3, Problem 95E

Chapter 8.3, Problem 97E

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Calculus: Early Transcendentals (2nd Edition)

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The values of f ( 0 ) , f ( 0.2 ) , f ( 0.5 ) , f ( 1 ) and f ( 1.5 ) if possible.

Solution Summary: The author evaluates the geometric series f(x)=displaystyle undersetk=0overset