Question

Find the value of $e^{-1/5}$ correct to four places of decimal.

Answer 1

$e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+...$ to $\infty$ (i)
putting $x=(-\frac{1}{5})$ in (i), we get $e^{-1/5}=1-\frac{1}{5}+\frac{1}{2!}{(-\frac{1}{5})}^2+\frac{1}{3!}{(-\frac{1}{5})}^3+\frac{1}{4!}{(-\frac{1}{5})}^4+...$ to $\infty$
$\Rightarrow$ $e^{-1/5}=1-\frac{2}{10}+\frac{2^2}{2!}.\frac{1}{{10}^2}-\frac{2^3}{2!}.\frac{1}{{10}^3}+\frac{2^4}{2!}.\frac{1}{{10}^4}+...$
$\Rightarrow$ $e^{-1/5}=1-0.200000+0.200000-0.001333+0.0000066$
$\Rightarrow$ $e^{-1/5}=0.8187$ (correct to 4 decimal places)