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00 For |x| < 1, let f(x) = By the geometric series, we know that E x" for |x|< 1. Notice that dx = - In (1- y) for lyl<1. Use integration, find the Taylor's series for In (1- y) %3D n=0 00 In (1- y) = E (The expression is a function of y and n. Be aware of the sign.) n=0 The radius of the convergence for the Taylor series obtained above is R =

00 For |x| < 1, let f(x) = By the geometric series, we know that E x" for |x|< 1. Notice that dx = - In (1- y) for lyl<1. Use integration, find the Taylor's series for In (1- y) %3D n=0 00 In (1- y) = E (The expression is a function of y and n. Be aware of the sign.) n=0 The radius of the convergence for the Taylor series obtained above is R =

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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y 00 1 dx For |x| < 1, let f(x) 1 By the geometric series, we know that E x" for |x|< 1. Notice that - In (1 - y) for lyl<1. Use integration, find the Taylor's series for In (1 - y) %3D %D 1-X 1-x 1-x n =0 00 In (1- y) = 2 (The expression is a function of y and n. Be aware of the sign.) n= 0 The radius of the convergence for the Taylor series obtained above is R =
Transcribed Image Text:y 00 1 dx For |x| < 1, let f(x) 1 By the geometric series, we know that E x" for |x|< 1. Notice that - In (1 - y) for lyl<1. Use integration, find the Taylor's series for In (1 - y) %3D %D 1-X 1-x 1-x n =0 00 In (1- y) = 2 (The expression is a function of y and n. Be aware of the sign.) n= 0 The radius of the convergence for the Taylor series obtained above is R =
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