# 9.10 | Taylor and Maclaurin Series

The Form of a Convergent Power Series
If $$f$$ is represented by a power series $$f(x) = \sum a_n (x-c)^n$$ for all $$x$$ in an open interval $$I$$ containing $$c$$, then
$$a_n = \frac{f^{(n)}(c)}{n!}$$
and
$$f(x) = f(c) + f'(c)(x-c) + \frac{f^{\prime \prime}(c)}{2!} (x-c)^2 + \cdots.$$
Definition of Taylor and Maclaurin Series
If a function $$f$$ has derivatives of all orders at $$x = c$$, then the series
$$\sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!}(x-c)^n = f(c) + f'(c)(x-c) + \frac{f^{\prime \prime} (c)}{2!} (x-c)^2 + \frac{f^{\prime \prime \prime} (c)}{3!} (x-c)^3 + \cdots$$
is called the Taylor series for $$f(x)$$ at $$c$$. Moreover, if $$c = 0$$, then the series is called the Maclaurin series for $$f$$.
Taylor’s Remainder
Let $$P_n(x)$$ be the $$n$$th Taylor Polynomial of $$f$$ centered at $$c$$. The Taylor remainder of $$f$$ is given by
$$R_n(x) = f(x) – P_n(x).$$
Taylor’s Inequality: If $$\left| f^{(n+1)}(x) \right| \leq M$$ on an interval $$I$$ containing $$c$$, then the remainder $$R_n(x)$$ of the Taylor series satisfies the inequality
$$\left|R_n(x) \right| \leq \frac{M}{(n+1)!} \left| x-c \right|^{n+1}$$
on $$I$$.
Convergence of Taylor Series
If $$\lim \limits_{n \rightarrow \infty} R_n = 0$$ for all $$x$$ in the interval $$I$$, then the Taylor series for $$f$$ converges and equal to $$f(x)$$,
$$f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(c)}{n!}(x-c)^n.$$
$$\displaystyle f(x) = \sum_{n=0}^{\infty} c_n x^n$$
Interval of Convergence Radius of Convergence
$$\displaystyle \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n = 1+ x + x^2 + x^3 + \cdots$$ $$(-1,1)$$ 1
$$\displaystyle e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1+ x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$$ $$(-\infty,\infty)$$ $$\infty$$
\displaystyle \begin{aligned} \sin x & = \sum_{n=0}^{\infty} (-1)^n\frac{x^{2n+1}}{(2n+1)!} \\ & = x – \frac{x^3}{3!} + \frac{x^5}{5!} – \frac{x^7}{7!} + \cdots \end{aligned} $$(-\infty,\infty)$$ $$\infty$$
\displaystyle \begin{aligned} \cos x & = \sum_{n=0}^{\infty} (-1)^n\frac{x^{2n}}{(2n)!} \\ & = 1 – \frac{x^2}{2!} + \frac{x^4}{4!} – \frac{x^6}{6!} + \cdots \end{aligned}
$$(-\infty,\infty)$$ $$\infty$$
\displaystyle \begin{aligned} \tan^{-1} x & = \sum_{n=0}^{\infty} (-1)^n\frac{x^{2n+1}}{2n+1} \\ & = x – \frac{x^3}{3} + \frac{x^5}{5} – \frac{x^7}{7} + \cdots \end{aligned}
$$(-1,1]$$ $$1$$
\displaystyle \begin{aligned} \ln(1+ x) & = \sum_{n=1}^{\infty} (-1)^{n-1}\frac{x^n}{n} \\ & = x – \frac{x^2}{2} + \frac{x^3}{3} – \frac{x^4}{4} + \cdots \end{aligned}
$$(-1,1]$$ $$1$$
\displaystyle \begin{aligned} (1+ x)^k &= \sum_{n=0}^{\infty} {k \choose n} x^n \\ &= 1 + k x + \frac{k(k-1)}{2!}x^2 + \cdots \end{aligned}
$$\scriptstyle \begin{cases} [-1,1], & \text{ if } k > -1 \text{ & } k \not \in \mathbb{Z} \\ (-\infty, \infty), & \text{ if } k>-1 \text{ & } k \in \mathbb{Z} \\ (-1,1], & \text{ if } k=-1\\ \end{cases}$$
$$1$$
$$\displaystyle f(x) = \sum_{n=0}^{\infty} c_n x^n$$
Interval of Convergence Radius of Convergence
$$\displaystyle \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$$ $$(-1,1)$$ 1
$$\displaystyle e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$$ $$(-\infty,\infty)$$ $$\infty$$
$$\displaystyle \sin x = \sum_{n=0}^{\infty} (-1)^n\frac{x^{2n+1}}{(2n+1)!}$$ $$(-\infty,\infty)$$ $$\infty$$
$$\displaystyle \cos x = \sum_{n=0}^{\infty} (-1)^n\frac{x^{2n}}{(2n)!}$$
$$(-\infty,\infty)$$ $$\infty$$
$$\displaystyle \tan^{-1} x = \sum_{n=0}^{\infty} (-1)^n\frac{x^{2n+1}}{2n+1}$$
$$(-1,1]$$ $$1$$
$$\displaystyle \ln(1+ x) = \sum_{n=1}^{\infty} (-1)^{n-1}\frac{x^n}{n}$$
$$(-1,1]$$ $$1$$
$$(1+ x)^k = \sum_{n=0}^{\infty} {k \choose n} x^n$$
$$\scriptstyle \begin{cases} [-1,1], \; & \text{ if } k > -1 \text{ & } k \not \in \mathbb{Z} \\ (-\infty, \infty), \; & \text{ if } k>-1 \text{ & } k \in \mathbb{Z} \\ (-1,1], \; & \text{ if } k=-1\\ \end{cases}$$
$$1$$
• $$\displaystyle \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$$

• Interval of Convergence: $$(-1,1)$$
• Radius of Convergence: 1
• $$\displaystyle e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$$

• Interval of Convergence: $$(-\infty,\infty)$$
• Radius of Convergence: $$\infty$$
• $$\displaystyle \sin x = \sum_{n=0}^{\infty} (-1)^n\frac{x^{2n+1}}{(2n+1)!}$$

• Interval of Convergence: $$(-\infty,\infty)$$
• Radius of Convergence: $$\infty$$
• $$\displaystyle \cos x = \sum_{n=0}^{\infty} (-1)^n\frac{x^{2n}}{(2n)!}$$

• Interval of Convergence: $$(-\infty,\infty)$$
• Radius of Convergence: $$\infty$$
• $$\displaystyle \tan^{-1} x = \sum_{n=0}^{\infty} (-1)^n\frac{x^{2n+1}}{2n+1}$$

• Interval of Convergence: $$(-1,1]$$
• Radius of Convergence: $$1$$
• $$\displaystyle \ln(1+ x) = \sum_{n=1}^{\infty} (-1)^{n-1}\frac{x^n}{n}$$

• Interval of Convergence: $$(-1,1]$$
• Radius of Convergence: $$1$$
• $$\displaystyle (1+ x)^k = \sum_{n=0}^{\infty} {k \choose n} x^n$$

• Interval of Convergence:
$$\begin{cases} [-1,1], & k > -1 \text{ & } k \not \in \mathbb{Z} \\ (-\infty, \infty), & k>-1 \text{ & } k \in \mathbb{Z} \\ (-1,1], & k=-1\\ \end{cases}$$
• Radius of Convergence: $$1$$
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 $$-\infty < x < \infty$$.