Calculus with Applications (10th Edition)

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

Chapter 12 - Sequences and Series - 12.5 Taylor Series - 12.5 Exercises - Page 647: 13

Answer

$1+4x^2+8x^4+\frac{32}{3}x^6+...+\frac{1}{n!}(4x^2)^n+...$

Work Step by Step

We are given $f(x)=e^{4x^2}$ for $r$ in $(-\infty,\infty)$ The Taylor series for $e^x$ is $1+x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+...+\frac{1}{n!}x^n+...$ The Taylor series for $f(x)=e^{4x^2}$ is $f(x)=1+4x^2+\frac{1}{2!}(4x^2)^2+\frac{1}{3!}(4x^2)^3+...+\frac{1}{n!}(4x^2)^n+...$ $=1+4x^2+8x^4+\frac{32}{3}x^6+...+\frac{1}{n!}(4x^2)^n+...$
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