Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that R,(x) - 0.]1f(x) =a = 49(2n – 1)!(-1)"-2n.72n +1(х - 49)лn = 12n. 72n + 1. n!(x - 49)"= 0(-1)- 3. 5..... (2n - 1)2". 72n - 1. n!(x - 49)nn = 11(-1)n + 1(x - 49)"2n. 72n. n!n = 1(-1)1:3. 5..... (2n - 1)2n . 72n + 1 . n!(x - 49)"

Answered: Image /qna-images/answer/039b3fb0-3278-4e70-a4da-088f85d8efc9.jpg

Answered: f(x) * a = 49 (2n – 1)! 2". 72n + 1 +…

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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f(x) * a = 49 (2n – 1)! 2". 72n + 1 + (x – 49)" n = 1 1 ㅇ윽+>m.72n+ 1.n! 1 (x - 49)" n = 0 1 1:3.5. .... (2n – 1) -E(-1)^- (x – 49)" 2n. 72n - 1. n! n= 1 1 -E(-1)^ + 1 (x - 49)" + 2". 72n. n! n = 1 S(-1)^3. 5. .... (2n – 1) 2". 72n + 1.n! 1 (x - 49)" n = 1