# 10.3 | Finding and Using Taylor Series

Find the Taylor series representation for $$\displaystyle f(x) = \frac{1}{1+x}$$ at $$x=2$$.

\begin{aligned} f(x) & = \frac{1}{1+x} = \frac{1}{3+(x-2)} \\ & = \frac{1}{3 \left[ 1 + \left( \frac{x-2}{3} \right) \right]} = \frac{1}{3} \frac{1}{1-\left(-\frac{x-2}{3} \right)} \\ & = \sum_{n=0}^{\infty} (-1)^n \frac{(x-2)^n}{3^{n+1}}, \; \; \left|-\frac{x-2}{3} \right|<1 \end{aligned} where

• we change the form of $$f(x)$$ so that it includes the expression $$(x-2)$$,
• and then apply the formula
$$\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n, |x| <1$$ replacing $$x$$ with $$-\frac{x-2}{3}$$.

So the Taylor series representation for $$\displaystyle f(x) = \frac{1}{1+x}$$ at $$x=2$$ is
$$\sum_{n=0}^{\infty} (-1)^n \frac{(x-2)^n}{3^{n+1}},$$
and the interval of convergence is $$-1 < x < 5$$.

Find the Maclaurin series for $$\displaystyle f(x) = x^2 \sin 2x$$.

\begin{aligned} f(x) & = x^2 \sin 2x = x^2 \sum_{n=0}^{\infty} (-1)^n \frac{(2x)^{2n+1}}{(2n+1)!} \\ & = \sum_{n=0}^{\infty} (-1)^n \frac{2^{2n+1} x^{2n+3}}{(2n+1)!} \end{aligned}
where

• we apply the formula
$$\sin x = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}$$
replacing $$x$$ with $$2x$$,
• distribute the exponent $$2n+1$$ over $$2x$$,
• and multiply in $$x^2$$

So the Maclaurin series for $$\displaystyle f(x) = x^2 \sin 2x$$ is
$$\sum_{n=0}^{\infty} (-1)^n \frac{2^{2n+1} x^{2n+3}}{(2n+1)!},$$
and the interval of convergence is $$-\infty < x < \infty$$.

Find the Taylor series for $$\displaystyle f(x) = e^{-2x}$$ centered at $$3$$.

Observe that,
\begin{aligned} f(x) & = e^{-2x} = e^{-2(x-3) -6} = e^{-6}e^{-2(x-3)} \\ & = e^{-6}\sum_{n=0}^{\infty} \frac{(-2(x-3))^n}{n!} \\ & = \sum_{n=0}^{\infty} \frac{(-1)^n2^n(x-3)^n}{e^6 n!} \end{aligned}
where

• rewrite $$f(x)$$ so that it includes the expression $$(x-3)$$
• apply the formula
$$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$$
replacing $$x$$ with $$2(x-3)$$,
• distribute the exponent $$n$$ over $$2(x-3)$$,
• and multiply in $$e^{-6}$$

So the Taylor series for $$\displaystyle f(x) = e^{-2x}$$ centered at $$3$$ is
$$\sum_{n=0}^{\infty} \frac{(-1)^n2^n(x-3)^n}{e^6 n!},$$
and the interval of convergence is $$-\infty < x < \infty$$.