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

E 9.10 Exercises