Applications of Series — Question 1

*Question 1 Approximate $e^{0.1}$ using the cubic Maclaurin polynomial for $e^x$. Then use the Lagrange remainder to give a rigorous upper bound for the absolute error.

*Question 1 – Solution

$$P_3(x)=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}.$$

$$\begin{align*} P_3(0.1)&=1+0.1+\frac{0.1^2}{2}+\frac{0.1^3}{6}\\&=1+0.1+0.005+0.0001666667\\&=\boxed{1.1051666667}. \end{align*}$$

Taylor’s theorem gives $$R_3(0.1)=\frac{e^c(0.1)^4}{4!}\quad\text{for some }0<c<0.1.$$ Because $e^c\le e^{0.1}$, $$|R_3(0.1)|\le\frac{e^{0.1}(0.1)^4}{24}<4.61\times10^{-6}.$$ Thus $e^{0.1}$ lies within $0.00000461$ of the stated approximation.