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Find the Maclaurin series of the function. f(x) = ln (1 – 5x) Choose the Maclaurin series. In (1 – 5x) In (1 – 5x) = In (1 – 5x) == -Σ In (1 – 5x) = Σ n=l == h=1 Σ Σ n=1 n=1 5"x" η (-1)n-15¹ xn n 5nxh 5n (−1)n-1 x5n 5n Identify the interval on which the series is valid.

Find the Maclaurin series of the function. f(x) = ln (1 – 5x) Choose the Maclaurin series. In (1 – 5x) In (1 – 5x) = In (1 – 5x) == -Σ In (1 – 5x) = Σ n=l == h=1 Σ Σ n=1 n=1 5"x" η (-1)n-15¹ xn n 5nxh 5n (−1)n-1 x5n 5n Identify the interval on which the series is valid.

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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I need to know what the expansion is valid for, thanks

 

 

Find the Maclaurin series of the function. f(x) = ln (1 - 5x) Choose the Maclaurin series. In (1 - 5x) = In (1 - 5x) = In (1 – 5x) Σ == n=1 ∞ Σ n=1 n=1 ∞ 5" xn n (−1)n−15″ xn n 5n xn 5n (−1)n-1x5n In (1 - 5x) = - 5n n=1 Identify the interval on which the series is valid.
Transcribed Image Text:Find the Maclaurin series of the function. f(x) = ln (1 - 5x) Choose the Maclaurin series. In (1 - 5x) = In (1 - 5x) = In (1 – 5x) Σ == n=1 ∞ Σ n=1 n=1 ∞ 5" xn n (−1)n−15″ xn n 5n xn 5n (−1)n-1x5n In (1 - 5x) = - 5n n=1 Identify the interval on which the series is valid.
ln (1 − 5x) = Σ n=1 In (1 - 5x) = ln (1 − 5x) = -Σ n=1 Incorrect n 5n xn 5n ∞ 】 (−1)n-¹x³n 5n n=1 Identify the interval on which the series is valid. (Give your answer as an interval in the form (*, *). Use the symbol ∞ for infinity, U for combining intervals, and an appropriate type of parenthesis "(",")", "[","]" depending on whether the interval is open or closed. Enter Ø if the interval is empty. Express numbers in exact form. Use symbolic notation and fractions where needed.) The expansion valid for:
Transcribed Image Text:ln (1 − 5x) = Σ n=1 In (1 - 5x) = ln (1 − 5x) = -Σ n=1 Incorrect n 5n xn 5n ∞ 】 (−1)n-¹x³n 5n n=1 Identify the interval on which the series is valid. (Give your answer as an interval in the form (*, *). Use the symbol ∞ for infinity, U for combining intervals, and an appropriate type of parenthesis "(",")", "[","]" depending on whether the interval is open or closed. Enter Ø if the interval is empty. Express numbers in exact form. Use symbolic notation and fractions where needed.) The expansion valid for:
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