Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 12 - Sequences and Series - Chapter Review - Review Exercises - Page 662: 49


For $r$ in $(-\infty,\infty)$, the Taylor series for $f(x)=2x^3e^{-3x}$ is $=2x^3-6x^4+9x^5-9x^6+...+\frac{2x^3(-3x)^n}{n!}+...$

Work Step by Step

We are given $f(x)=2x^3e^{-3x}$ for $r$ in $(-\infty,\infty)$ The Taylor series for $e^x$ is $1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...+\frac{1}{n!}x^n+...$ With $c=2x^3$, the Taylor series for $f(x)=2x^3e^{-3x}$ is $f(x)=2x^3+2x^3.(-3x)+2x^3\frac{(-3x)^2}{2!}+2x^3\frac{(-3x)^3}{3!}+...+2x^3\frac{(-3x)^n}{n!}+...$ $=2x^3-6x^4+9x^5-9x^6+...+\frac{2x^3(-3x)^n}{n!}+...$
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