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 \(-\infty < x < \infty \).

E 10.3 Exercises