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Use substitution to find the Taylor series at x 0 of the function 4 sin (-x). Choose the correct Taylor series. F (-1)"(4x)?n + 1 (2n + 1)! 43x3 45,5 = 4x - 3! O A. E %3D 5! n=0 - 4x2n + 1 4x 4x° 4x- 3! О в. %3D (2n + 1)! 5! n=0 4(-1)"(x)²n + 1 4x 4x = 4x - 3! 4x5 Oc. Ос. (2n + 1)! 5! n=0 5 4(-1)"(-x)²n*1 4x3 4x5 O D. E - 4x + 3! (2n + 1)! 5! n=0 +

Use substitution to find the Taylor series at x 0 of the function 4 sin (-x). Choose the correct Taylor series. F (-1)"(4x)?n + 1 (2n + 1)! 43x3 45,5 = 4x - 3! O A. E %3D 5! n=0 - 4x2n + 1 4x 4x° 4x- 3! О в. %3D (2n + 1)! 5! n=0 4(-1)"(x)²n + 1 4x 4x = 4x - 3! 4x5 Oc. Ос. (2n + 1)! 5! n=0 5 4(-1)"(-x)²n*1 4x3 4x5 O D. E - 4x + 3! (2n + 1)! 5! n=0 +

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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Question 6
Use substitution to find the Taylor series at x = 0 of the function 4 sin (-x). Choose the correct Taylor series. 00 (- 1)"(4x)²n - O A. E + 1 43x3 45x5 = 4x - 3! (2n + 1)! n=0 5! 00 - 4x2n +1 4x 4x° О в. 2 = - 4x - 3! (2n + 1)! 5! n= 0 4(-1)"(x)²n- (2n + 1)! 00 4x Ос. = 4x - 3! 5! n= 0 00 4(-1)"(-x)2n- 4x° 4x5 O D. E = - 4x + (2n + 1)! 3! 5! n=0
Transcribed Image Text:Use substitution to find the Taylor series at x = 0 of the function 4 sin (-x). Choose the correct Taylor series. 00 (- 1)"(4x)²n - O A. E + 1 43x3 45x5 = 4x - 3! (2n + 1)! n=0 5! 00 - 4x2n +1 4x 4x° О в. 2 = - 4x - 3! (2n + 1)! 5! n= 0 4(-1)"(x)²n- (2n + 1)! 00 4x Ос. = 4x - 3! 5! n= 0 00 4(-1)"(-x)2n- 4x° 4x5 O D. E = - 4x + (2n + 1)! 3! 5! n=0
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