Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Section 11.10 - Taylor and Maclaurin Series - 11.10 Exercises - Page 772: 70

Answer

$x+\frac{x^{2}}{2}+\frac{x^{3}}{3}+...$

Work Step by Step

$y=e^{x}ln(1+x)$ $ln(1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}+....+\frac{x^{n}}{n}$ and $e^{x}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+...$ $e^{x}ln(1+x)=(1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+..)(x-\frac{x^{2}}{2}+\frac{x^{3}}{3}+....+\frac{x^{n}}{n})$ $\approx x+\frac{x^{2}}{2}+\frac{x^{3}}{3}+...$
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