## Calculus 8th Edition

${f^{2n}(0)}=\frac{(2n)!}{n!}$
${e^{x}}=\Sigma_{n=0}^\infty\frac{x^{n}}{n!}$ $f(x)={e^{x^{2}}}=\Sigma_{n=0}^\infty\frac{(x^{2})^{n}}{n!}=\Sigma_{n=0}^\infty\frac{x^{2n}}{n!}$ 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, ${f^{2n}(0)}=\frac{(2n)!}{n!}$