Applications of Series — Question 5

Question 5 – Solution

Step 1: Substitute into the exponential series.

$$e^{-x^2}=1-x^2+\frac{x^4}{2!}-\frac{x^6}{3!}+\frac{x^8}{4!}-\cdots.$$ For $0\le x\le0.5$, the alternating term magnitudes decrease.

Step 2: Integrate through the $x^6$ term.

$$\begin{align*} I&\approx\left[x-\frac{x^3}{3}+\frac{x^5}{10}-\frac{x^7}{42}\right]_0^{0.5}\\&=0.5-\frac{0.5^3}{3}+\frac{0.5^5}{10}-\frac{0.5^7}{42}\\&=\boxed{0.4612723214}. \end{align*}$$

Step 3: Bound the integrated remainder.

The omitted integrand has magnitude at most $x^8/4!$, and the integrated alternating remainder satisfies $$|R|\le\int_0^{0.5}\frac{x^8}{4!}\,dx=\frac{0.5^9}{9\cdot4!}<9.05\times10^{-6}.$$ Thus the estimate is accurate to within $0.00000905$.