Chapter 11 - Section 11.10 - Taylor and Maclaurin Series - 11.10 Exercises - Page 772: 50

$\frac{1}{\sqrt[10] e}=e^{-0.1}\approx 0.90483$

Given:$\frac{1}{\sqrt[10] e}=e^{-0.1}$ The series for $e^{x}=\Sigma_{n=0}^{\infty}\frac{x^{n}}{n!}$ Put $x=-0.1$ $e^{x}=\Sigma_{n=0}^{\infty}(-1)^{n}\frac{(0.1)^{n}}{n!}$ Here, $a_{n}=(-1)^{n}\frac{(0.1)^{n}}{n!}$ So, $s_{2}=a_{0}+a_{1}+a_{2}+a_{3} \approx 0.90483$ Thus, $\frac{1}{\sqrt[10] e}=e^{-0.1}\approx 0.90483$