Calculus: Early Transcendentals 8th Edition

$e^{-x^{4}}$
Given: $\Sigma_{n=0}^{\infty}{(-1)^{n}}\dfrac{x^{4n}}{n!}$ As we know $e^{x}=\Sigma_{n=0}^{\infty}\dfrac{x^{n}}{n!}$ and $e^{-x}=\Sigma_{n=0}^{\infty}(-1)^{n}\dfrac{x^{n}}{n!}$ Thus, $e^{-x^{4}}=\Sigma_{n=0}^{\infty}{(-1)^{n}}\dfrac{x^{4n}}{n!}$