Calculus MacLaurin series

Let $\displaystyle f(x)=\frac{e^{-x^2}-1}{x}$ . We have $\displaystyle e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots$ , so $\displaystyle e^{-x^2}=1-x^2+\frac{x^4}{2!}-\frac{x^6}{3!}+\dots$ and $\displaystyle f(x)=-x+\frac{x^3}{2!}-\frac{x^5}{3!}+\frac{x^7}{4!}-\dots$ . The terms of this series have alternating signs and decreasing absolute values (when $|x|\le1$ ); therefore, the sum from the $n$ th term to infinity is bounded by the $n$ th term. For example, let $\displaystyle g(x)=-x+\frac{x^3}{2!}-\frac{x^5}{3!}$ . Then $\displaystyle |f(x)-g(x)|=\left|\frac{x^7}{4!}+\dots\right|\le\frac{x^ 7}{4!}$ .

Now, suppose $\displaystyle\varepsilon=\sup_{x\in[0,1/5]}|f(x)-g(x)|$ . Then $|\int_0^{1/5} f(x)\,dx-\int_0^{1/5} g(x)\,dx|\le\int_0^{1/5}|f(x)-g(x)|\,dx\le\varepsilon/5$ . Our goal is to make $\varepsilon/5<10^{-6}$ . One can check that $\displaystyle\frac{(1/5)^7}{5\cdot 4!}<10^{-6}$ , but $\displaystyle\frac{(1/5)^5}{5\cdot 3!}>10^{-6}$ , so we have to approximate f(x) with a partial sum of three terms, i.e., with g(x).

Calculating $\int_0^{1/5}g(x)\,dx$ gives −0.019801778, which is −0.019802 when rounded to the 6th digit.