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We have found the following Maclaurin series. x4n 142n (2n)! Now we can use this to find the Maclaurin series of the given function, treating the term 13x as a constant and using the rule can = cªn. f(x) = 13x cos 05 (1141+²2) cos(4x²) = (-1)^- n = 0 = 13x ‹ Ë (−1)”. n = 0 = - Σ n = 0 (-1)". x4n 142n (2n)! 4n+1 142 (2n)! X

We have found the following Maclaurin series. x4n 142n (2n)! Now we can use this to find the Maclaurin series of the given function, treating the term 13x as a constant and using the rule can = cªn. f(x) = 13x cos 05 (1141+²2) cos(4x²) = (-1)^- n = 0 = 13x ‹ Ë (−1)”. n = 0 = - Σ n = 0 (-1)". x4n 142n (2n)! 4n+1 142 (2n)! X

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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Step 2 We have found the following Maclaurin series. Now we can use this to find the Maclaurin series of the given function, treating the term 13x as a constant and using the rule can 1 f(x) = 13x cos (x²) 14 00 = 13× Σ (1)". 142n (2n)! tan n = 0 = cos(-1⁄4ײ) = [ (-1)^ COS n = 0 ∞ A n = 0 (-1)" An+1 +An 142n (2n)! X 14²n (2n)! X =cΣªn. C an
Transcribed Image Text:Step 2 We have found the following Maclaurin series. Now we can use this to find the Maclaurin series of the given function, treating the term 13x as a constant and using the rule can 1 f(x) = 13x cos (x²) 14 00 = 13× Σ (1)". 142n (2n)! tan n = 0 = cos(-1⁄4ײ) = [ (-1)^ COS n = 0 ∞ A n = 0 (-1)" An+1 +An 142n (2n)! X 14²n (2n)! X =cΣªn. C an
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