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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.] Find the associated radius of convergence R. F(x) = 5 cos(x), a- 137 Step 1 The Taylor series formula is F(a) + F'(a)(x - a) + ex- a)2 + f"(a), 2)3 + 4a) x - a)* + .. The function, f(x) = 5 cos(x), has derivatives F'(x) = -5 sin(x) -5 sin(z) F"(x) =-5 cos(x) |-5 cos(z) f"(x) = 5 sin(x) 5 sin(x)|- and F(4)(x) = 5 cos(x) 5 cos (r) Step 2 With a = 137, f(13#) = -5v -5 F'(13m) = 의 F"(137) = 5 V f"(137) = 0 , and f(4)(13m) = 5- Step 3 Therefore, the Taylor Series begins -5 -s + 0v 의 이 (x-13m) + (x- 13m)2 + (x - 13n)3 + (x - 137) +.... 3! 41 Step 4 Thus, the general formula is Submit Skip (you cannot come back)

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.] Find the associated radius of convergence R. F(x) = 5 cos(x), a- 137 Step 1 The Taylor series formula is F(a) + F'(a)(x - a) + ex- a)2 + f"(a), 2)3 + 4a) x - a)* + .. The function, f(x) = 5 cos(x), has derivatives F'(x) = -5 sin(x) -5 sin(z) F"(x) =-5 cos(x) |-5 cos(z) f"(x) = 5 sin(x) 5 sin(x)|- and F(4)(x) = 5 cos(x) 5 cos (r) Step 2 With a = 137, f(13#) = -5v -5 F'(13m) = 의 F"(137) = 5 V f"(137) = 0 , and f(4)(13m) = 5- Step 3 Therefore, the Taylor Series begins -5 -s + 0v 의 이 (x-13m) + (x- 13m)2 + (x - 13n)3 + (x - 137) +.... 3! 41 Step 4 Thus, the general formula is Submit Skip (you cannot come back)

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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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.] Find the associated radius of convergence R. f(x) = 5 cos(x), a = 137 Step 1 The Taylor series formula is f(a) + f'(a)(x - a) + (x - a)2 + ax - a)3 + 4(a)x- a)4 + ... The function, f(x) - 5 cos(x), has derivatives F'(x) = -5 sin(x) -5 sin(r) F"(x) = -5 cos(x) -5 cos (1) f"(x) = 5 sin(x) 5 sin(x)|, and F(4) (x) = 5 cos(x) 5 cos (r) Step 2 With a = 13I, f(137) = -5 F'(137) = 0 F"(137) = 5 v f"(137) = 0 v and f(4)(137) = 5 -5 Step 3 Therefore, the Taylor Series begins -5V 의 이 (x (x - 137)2 + 13п)3 + -5 V (x- 137)4 + ... 137) 2! 4! Step 4 Thus, the general formula is Skip (you cannot come back)
Transcribed Image Text: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.] Find the associated radius of convergence R. f(x) = 5 cos(x), a = 137 Step 1 The Taylor series formula is f(a) + f'(a)(x - a) + (x - a)2 + ax - a)3 + 4(a)x- a)4 + ... The function, f(x) - 5 cos(x), has derivatives F'(x) = -5 sin(x) -5 sin(r) F"(x) = -5 cos(x) -5 cos (1) f"(x) = 5 sin(x) 5 sin(x)|, and F(4) (x) = 5 cos(x) 5 cos (r) Step 2 With a = 13I, f(137) = -5 F'(137) = 0 F"(137) = 5 v f"(137) = 0 v and f(4)(137) = 5 -5 Step 3 Therefore, the Taylor Series begins -5V 의 이 (x (x - 137)2 + 13п)3 + -5 V (x- 137)4 + ... 137) 2! 4! Step 4 Thus, the general formula is Skip (you cannot come back)
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