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How did we get the power series summation? How did we know to set it up like this.( Circled in pink)

How did we get the power series summation? How did we know to set it up like this.( Circled in pink)

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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How did we get the power series summation? How did we know to set it up like this.( Circled in pink)

Basic Skills » 9-26. Taylor series and interval of convergence a. Use the definition of a Taylor/Maclaurin series to find the first four nonzero tern centered at a. b. Write the power series using summation notation. C. Determine the interval of convergence of the series. 1 9. f(x) = -, a = 1 %3D
Transcribed Image Text:Basic Skills » 9-26. Taylor series and interval of convergence a. Use the definition of a Taylor/Maclaurin series to find the first four nonzero tern centered at a. b. Write the power series using summation notation. C. Determine the interval of convergence of the series. 1 9. f(x) = -, a = 1 %3D
k! k=0 k=0 11.3.9 a. f(x) = 1 so f(1) = 1. f'(x) 3 so f'(1) = -2. f"(x) 24 so f"(1) = 6. f"(x) = – SO %3D %3D %3D f"(1) = -24. So the Taylor series is 15 SO %3D 1- 2(x – 1) + 6 2! (x-1)2 (x- 1)3 24 +.··=1– 2(x – 1) + 3(x – 1)² – 4(x – 1)3 + .. 3! | b. E(-1)*(k+1)(x – 1)*. k=0 (-1)k+1(k + 2)(x – 1)k+1 (-1)k(k + 1)(x – 1)k C. r= lim = |x – 1|. This is less than 1 for -1 <x -1< 1, or 0< I< 2. %3D The series diverges at the endpoints by the Divergence Test, so the interval of convergence is (0, 2). Copyright 2019 Pearson Education, Inc. 2.
Transcribed Image Text:k! k=0 k=0 11.3.9 a. f(x) = 1 so f(1) = 1. f'(x) 3 so f'(1) = -2. f"(x) 24 so f"(1) = 6. f"(x) = – SO %3D %3D %3D f"(1) = -24. So the Taylor series is 15 SO %3D 1- 2(x – 1) + 6 2! (x-1)2 (x- 1)3 24 +.··=1– 2(x – 1) + 3(x – 1)² – 4(x – 1)3 + .. 3! | b. E(-1)*(k+1)(x – 1)*. k=0 (-1)k+1(k + 2)(x – 1)k+1 (-1)k(k + 1)(x – 1)k C. r= lim = |x – 1|. This is less than 1 for -1 <x -1< 1, or 0< I< 2. %3D The series diverges at the endpoints by the Divergence Test, so the interval of convergence is (0, 2). Copyright 2019 Pearson Education, Inc. 2.
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