Question

use the first five terms of the exponential series of e^x and a calculator to approximate e^0.2 to the nearest hundredth

Answer 1

Using the Maclaurin series;

`e^x=1+x+(x^2)/(2!)+(x^3)/(3!)+(x^4)/(4!)+(x^5)/(5!)+...`

Thus;

`\begin{gathered} x=0.2; \\ \\ e^(0.2)=1+(0.2)+((0.2)^2)/(2!)+((0.2)^3)/(3!)+((0.2)^4)/(4!) \\ \\ e^(0.2)\approx1.22 \\ \end{gathered}`

ANSWER: 1.22