Applications of Series — Question 1

Question 1 – Solution

Step 1: Write the cubic polynomial.

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

Step 2: Evaluate at $x=0.1$.

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

Step 3: Bound the error.

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.