Answer to Question #206423 in Calculus for Oula

Question

Use the exponential series to find an approximation to e-1correct to three significant figures.

 

Answer 1

$e^x=1+\dfrac{x}{1!}+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+...$

$=\displaystyle\sum_{n=0}^{\infin}\dfrac{x^n}{n!} \ \ \text{ for all }x$

$e^{-1}=1+\dfrac{-1}{1!}+\dfrac{(-1)^2}{2!}+\dfrac{(-1)^3}{3!}+\dfrac{(-1)^4}{4!}$

$+\dfrac{(-1)^5}{5!}+\dfrac{(-1)^6}{6!}+\dfrac{(-1)^7}{7!}+\dfrac{(-1)^8}{8!}+...$

$=1-1+\dfrac{1}{2}-\dfrac{1}{6}+\dfrac{1}{24}-\dfrac{1}{120}+\dfrac{1}{720}-\dfrac{1}{5040}$

$+\dfrac{1}{40320}-...$

$\dfrac{1}{40320}\approx0.0000245$

$\approx\dfrac{20160-6720+1680-336+56-1}{40320}\approx\dfrac{14839}{40320}$

$\approx0.368$

$e^{-1}=0.368$