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Section A.5 Table of Taylor Expansions
Let \(n\ge \) be an integer. Then if the function \(f\) has \(n+1\) derivatives on an interval that contains both \(x_0\) and \(x\text{,}\) we have the Taylor expansion
\begin{align*}
f(x)&=f(x_0)+f'(x_0)\,(x-x_0)+\dfrac{1}{2!}f''(x_0)\,(x-x_0)^2+\cdots
+\dfrac{1}{n!}f^{(n)}(x_0)\,(x-x_0)^n\\
&\hskip0.5in
+\dfrac{1}{(n+1)!}f^{(n+1)}(c)\,(x-x_0)^{n+1}\qquad\hbox{for some $c$ between
$x_0$ and $x$}
\end{align*}
The limit as \(n\rightarrow\infty\) gives the Taylor series
\begin{align*}
f(x)&=\sum_{n=0}^\infty\dfrac{f^{(n)}(x_0)}{n!}(x-x_0)^n
\end{align*}
for \(f\text{.}\) When \(x_0=0\) this is also called the Maclaurin series for \(f\text{.}\) Here are Taylor series expansions of some important functions.
\begin{align*}
e^x&=\sum_{n=0}^\infty \dfrac{1}{n!}x^n
&&\text{for } -\infty \lt x \lt \infty\\
&=1+x+\dfrac{1}{2}x^2+\dfrac{1}{3!}x^3+\cdots+\dfrac{1}{n!}x^n+\cdots\\
\sin x&=\sum_{n=0}^\infty\dfrac{(-1)^n}{(2n+1)!}x^{2n+1}
&&\text{for } -\infty \lt x \lt \infty\\
&=x-\dfrac{1}{3!}x^3+\dfrac{1}{5!}x^5-\cdots
+\dfrac{(-1)^n}{(2n+1)!}x^{2n+1}+\cdots\\
\cos x&=\sum_{n=0}^\infty\dfrac{(-1)^n}{(2n)!}x^{2n}
&&\text{for } -\infty \lt x \lt \infty\\
&=1-\dfrac{1}{2!}x^2+\dfrac{1}{4!}x^4-\cdots
+\dfrac{(-1)^n}{(2n)!}x^{2n}+\cdots\\
\dfrac{1}{1-x}&=\sum_{n=0}^\infty x^n
&&\text{for } -1\le x \lt 1\\
&=1+x+x^2+x^3+\cdots+x^n+\cdots\\
\dfrac{1}{1+x}&=\sum_{n=0}^\infty(-1)^n x^n
&&\text{for } -1 \lt x\le 1\\
&=1-x+x^2-x^3+\cdots+(-1)^nx^n+\cdots\\
\ln(1-x)&=-\sum_{n=1}^\infty\dfrac{1}{n}x^n
&&\text{for } -1\le x \lt 1\\
&=-x-\half x^2-\dfrac{1}{3}x^3-\cdots-\dfrac{1}{n}x^n-\cdots\\
\ln(1+x)&=-\sum_{n=1}^\infty\dfrac{(-1)^n}{n}x^n
&&\text{for } -1 \lt x\le 1\\
&=x-\half x^2+\dfrac{1}{3}x^3-\cdots-\dfrac{(-1)^n}{n}x^n-\cdots\\
(1+x)^p&=1+px+\dfrac{p(p-1)}{2}x^2+\dfrac{p(p-1)(p-2)}{3!}x^3+\cdots\\
&\hskip0.5in+
\dfrac{p(p-1)(p-2)\cdots(p-n+1)}{n!}x^n+\cdots
\end{align*}
