Show, by integrating the Maclaurin series for f(x) = that for |x| < 1, 1.3. 5... (2n – 1) x2n+1 2.4 · 6 .… · (2n) 2n + 1 sinx = x+ > n=1

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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Show, by integrating the Maclaurin series for f(x) =
that for |x| < 1,
1.3. 5... (2n – 1) x2n+1
2.4 · 6 .… · (2n) 2n + 1
sinx = x+ >
n=1
Transcribed Image Text:Show, by integrating the Maclaurin series for f(x) = that for |x| < 1, 1.3. 5... (2n – 1) x2n+1 2.4 · 6 .… · (2n) 2n + 1 sinx = x+ > n=1
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