9.10 | Taylor and Maclaurin Series | Aori Nevo's Course Webpage

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 $.

\(\displaystyle f(x) = \sum_{n=0}^{\infty} c_n x^n \)	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
\(\displaystyle e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}\)	\((-\infty,\infty)\)	\( \infty \)
\(\displaystyle \sin x = \sum_{n=0}^{\infty} (-1)^n\frac{x^{2n+1}}{(2n+1)!} \)	\((-\infty,\infty)\)	\(\infty\)
\(\displaystyle \cos x = \sum_{n=0}^{\infty} (-1)^n\frac{x^{2n}}{(2n)!} \)	\((-\infty,\infty)\)	\(\infty\)
\(\displaystyle \tan^{-1} x = \sum_{n=0}^{\infty} (-1)^n\frac{x^{2n+1}}{2n+1} \)	\((-1,1]\)	\(1\)
\(\displaystyle \ln(1+ x) = \sum_{n=1}^{\infty} (-1)^{n-1}\frac{x^n}{n} \)	\((-1,1]\)	\(1\)
\( (1+ x)^k = \sum_{n=0}^{\infty} {k \choose n} x^n \)	\(\scriptstyle \begin{cases} [-1,1], \; & \text{ if } k > -1 \text{ & } k \not \in \mathbb{Z} \\ (-\infty, \infty), \; & \text{ if } k>-1 \text{ & } k \in \mathbb{Z} \\ (-1,1], \; & \text{ if } k=-1\\ \end{cases}\)	\(1\)