Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - Review - Exercises - Page 826: 62



Work Step by Step

${e^{x}}=\Sigma_{n=0}^\infty\frac{x^{n}}{n!}$ $f(x)={e^{x^{2}}}=\Sigma_{n=0}^\infty\frac{(x^{2})^{n}}{n!}=\Sigma_{n=0}^\infty\frac{x^{2n}}{n!}$ We can sat that the coefficient of $x^{2n}$ is in power series of $f(x)$ is $\frac{1}{n!}$. Thus, $\frac{f^{2n}(0)}{2n!}=\frac{1}{n!}$ Hence, ${f^{2n}(0)}=\frac{(2n)!}{n!}$
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