Examples

Taylor polynomial approximations

Let $f(x)=\sin x$ and let $T_n(x)$ be the Taylor polynomials expanded around $c=0$.

By considering the alternating series error bound, find the first $n$ for which $T_n(0.02)$ must have error less than ${10}^{-6}$.

Solution

Write the Maclaurin series of $\sin x$ because we are expanding around $c=0$:

$$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots =\sum _{n=0}^{\infty }{(-1)}^n\frac{x^{2n+1}}{(2n+1)!}$$

This series is alternating, so the AST error bound formula applies (“Next Term Bound”):

$$|E_n|\le a_{n+1}$$

Find smallest $n$ such that $a_{n+1}\le {10}^{-6}$, and then we know:

$$|E_n|\le a_{n+1}\le {10}^{-6}\gg \gg |E_n|\le {10}^{-6}$$

Plug $x=0.02$ in the series for $\sin x$:

$$a_{2n+1}=\frac{{(0.02)}^{2n+1}}{(2n+1)!}$$

Solve for the first time $a_{2n+1}\le {10}^{-6}$ by listing the values:

$$\begin{array}{c}\frac{{0.02}^1}{1!}=0.02,\frac{{0.02}^3}{3!}\approx 1.33\times {10}^{-6},\\ \\ \frac{{0.02}^5}{5!}\approx 2.67\times {10}^{-11},\dots \end{array}$$

The first time $a_{2n+1}$ is below ${10}^{-6}$ happens when $2n+1=5$.

This is NOT the same $n$ as in $T_n$. That $n$ is the highest power of $x$ allowed.

The sum of prior terms is $T_4(0.02)$.

Since $T_4(x)=T_3(x)$ because there is no $x^4$ term, the final answer is $n=3$.

Taylor polynomials to approximate a definite integral

Approximate ${\displaystyle {\int }_0^{0.3}{e}^{-x^2}dx}$ using a Taylor polynomial with an error no greater than ${10}^{-5}$.

Solution

Plug $u=-x^2$ into the series of $e^u$:

$$\begin{array}{c}{e}^u=\sum _{n=0}^{\infty }\frac{u^n}{n!}\\ \\ \gg \gg e^{-x^2}=\sum _{n=0}^{\infty }\frac{{(-x^2)}^n}{n!}\gg \gg \sum _{n=0}^{\infty }{(-1)}^n\frac{1}{n!}{x}^{2n}\end{array}$$

Find an antiderivative by terms:

$$\begin{array}{c}\int \sum _{n=0}^{\infty }{(-1)}^n\frac{1}{n!}{x}^{2n}dx\\ \\ \sum _{n=0}^{\infty }{(-1)}^n\frac{1}{n!}\frac{x^{2n+1}}{2n+1}\\ \\ \gg \gg x-\frac{1}{1!(3)}{x}^3+\frac{1}{2!(5)}{x}^5-\frac{1}{3!(7)}{x}^7+\cdots \end{array}$$

Plug in bounds for definite integral:

$$\begin{array}{c}{\int }_0^{0.3}\sum _{n=0}^{\infty }{(-1)}^n\frac{1}{n!}{x}^{2n}={\sum _{n=0}^{\infty }{(-1)}^n\frac{1}{n!}\frac{x^{2n+1}}{2n+1}|}_0^{0.3}\\ \\ \gg \gg \sum _{n=0}^{\infty }{(-1)}^n\frac{1}{n!}\frac{{(0.3)}^{2n+1}}{2n+1}\\ \\ \gg \gg 0.3-\frac{{(0.3)}^3}{3}+\frac{{(0.3)}^5}{10}-\frac{{(0.3)}^7}{42}+\cdots \end{array}$$

Notice alternating series pattern. Apply error bound formula, “Next Term Bound”:

$$\frac{{(0.3)}^5}{10}\approx 2.43\times {10}^{-4},\frac{{(0.3)}^7}{42}\approx 5.21\times {10}^{-6}$$

So we can guarantee an error less than $5.21\times {10}^{-6}$ by summing the first terms through ${\displaystyle \frac{{(0.3)}^5}{10}}$:

$$0.3-\frac{{(0.3)}^3}{3}+\frac{{(0.3)}^5}{10}=0.291243$$