9.7 | Taylor Polynomials and Approximations


Definitions of \(n\)th Taylor Polynomial and \(n\)th Maclaurin Polynomial
If \(f\) has \(n\) derivatives at \(c\), then the polynomial
$$
P_n(x) = \sum_{i=0}^n \frac{f^{(i)}(c)}{i!}(x-c)^i
$$
is called the \(n\)th Taylor polynomial for \(f\) at \(c\).
If \(c=0\), then
$$
P_n(x) = \sum_{i=0}^n \frac{f^{(i)}(0)}{i!}x^i
$$
is also called the \(n\)th Maclaurin polynomial for \(f\)
Taylor’s Theorem If \(f\) has derivatives up to order \(n+1\) in an interval \(I\) containing \(c\), then for each \(x\) in \(I\), there exists a number \(z\) between \(x\) and \(c\) such that
$$
\begin{aligned}
f(x) & = f(c) + f'(c)(x-c) + \frac{f^{\prime \prime}(c)}{2}(x-c)^2 + \cdots + \frac{f^{(n)}(c)}{n!}(x-c)^n + R_n(x) \\
& = P_n(x) + R_n(x)
\end{aligned}
$$
where
$$
R_n(x) = \frac{f^{(n+1)}(z)}{(n+1)!}(x-c)^{n+1}
$$

Find the first, second, third, fourth, fifth, and sixth degree Maclaurin polynomial for
$$
f(x) = e^x.
$$


The \(n^{th}\)-degree Maclaurin polynomial is given by:
$$
M_n(x) = \sum_{k = 0}^n \frac{f^{(k)}(0)}{k!}x^k
$$

\(\displaystyle k\) \(\displaystyle f^{(k)}(x)\) \(\displaystyle \frac{f^{(k)}(0)}{k!}x^k\)
0 \(e^x\) \(1\)
1 \(e^x\) \(x\)
2 \(e^x\) \(\frac{1}{2}x^2\)
3 \(e^x\) \(\frac{1}{3!}x^3\)
4 \(e^x\) \(\frac{1}{4!}x^4\)
5 \(e^x\) \(\frac{1}{5!}x^5\)
6 \(e^x\) \(\frac{1}{6!}x^6\)

Using the table on the left, we obtain:
$$
\begin{aligned}
M_1(x) & = 1 + x \\
M_2(x) & = 1 + x + \frac{x^2}{2} \\
M_3(x) & = 1 + x + \frac{x^2}{2} + \frac{x^3}{3!} \\
M_4(x) & = 1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + \frac{x^4}{4!}\\
M_5(x) & = 1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!}\\
M_6(x) & = 1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \frac{x^6}{6!}\\
\end{aligned}
$$

E 9.7 Exercises