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Compute f(x) f'(x) f"(x) f" (1) = = = = f(iv)(x) = 1 1+x² 2x (1+x²)² f(x) = Σ k=0 (6x²-2) (x²+1) ³ 24x(x²-1) (+1)* 4 1. 24 (5x4-10x²+1) (x²+1)³ - ƒ(v)(z) = − 240x(x² − 3)(3x²-1) (x²+1) 6 f(0) 1²k f'(0) f"(0) = 1 We see that for the odd terms f(2+1)(0) = and we also see that for the even derivatives f(2k) (0) = Hence the Taylor series for f centered at 0 is given by = 0 = -2 f" (0) = 0 f(iv) (0) = 24 f(") (0) = 0

Compute f(x) f'(x) f"(x) f" (1) = = = = f(iv)(x) = 1 1+x² 2x (1+x²)² f(x) = Σ k=0 (6x²-2) (x²+1) ³ 24x(x²-1) (+1)* 4 1. 24 (5x4-10x²+1) (x²+1)³ - ƒ(v)(z) = − 240x(x² − 3)(3x²-1) (x²+1) 6 f(0) 1²k f'(0) f"(0) = 1 We see that for the odd terms f(2+1)(0) = and we also see that for the even derivatives f(2k) (0) = Hence the Taylor series for f centered at 0 is given by = 0 = -2 f" (0) = 0 f(iv) (0) = 24 f(") (0) = 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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f(2)= Compute f(x) f'(x) f"(x) f"(z) 1 1+z² f(0) (1) = f(x) = = || f(iv) (r) = || 1 1+x² 2x (1+x²)² k=0 (6x²-2) (x²+1) ³ |_24x(x²-1) (x²+1)* 4 24(5x¹ − 10x²+1) (x²+1) ³ 240x(x²-3)(3x²-1) (x²+1) 6 f(0) 2² f'(0) f"(0) f™ (0) We see that for the odd terms f(2+1) (0) = and we also see that for the even derivatives f(2k) (0) = Hence the Taylor series for f centered at 0 is given by f(0) (0) = 1 = 0 = -2 = f(iv) (0) = 24 0 = 0 4.
Transcribed Image Text:f(2)= Compute f(x) f'(x) f"(x) f"(z) 1 1+z² f(0) (1) = f(x) = = || f(iv) (r) = || 1 1+x² 2x (1+x²)² k=0 (6x²-2) (x²+1) ³ |_24x(x²-1) (x²+1)* 4 24(5x¹ − 10x²+1) (x²+1) ³ 240x(x²-3)(3x²-1) (x²+1) 6 f(0) 2² f'(0) f"(0) f™ (0) We see that for the odd terms f(2+1) (0) = and we also see that for the even derivatives f(2k) (0) = Hence the Taylor series for f centered at 0 is given by f(0) (0) = 1 = 0 = -2 = f(iv) (0) = 24 0 = 0 4.
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