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.

\[\begin{align*} e^x&=\sum_{n=0}^\infty \dfrac{1}{n!}x^n &&\text{for } -\infty \lt x \lt \infty\\ &=1+x+\dfrac{1}{2}x^2+\dfrac{1}{3!}x^3+\cdots+\dfrac{1}{n!}x^n+\cdots\\ \sin x&=\sum_{n=0}^\infty\dfrac{(-1)^n}{(2n+1)!}x^{2n+1} &&\text{for } -\infty \lt x \lt \infty\\ &=x-\dfrac{1}{3!}x^3+\dfrac{1}{5!}x^5-\cdots +\dfrac{(-1)^n}{(2n+1)!}x^{2n+1}+\cdots\\ \cos x&=\sum_{n=0}^\infty\dfrac{(-1)^n}{(2n)!}x^{2n} &&\text{for } -\infty \lt x \lt \infty\\ &=1-\dfrac{1}{2!}x^2+\dfrac{1}{4!}x^4-\cdots +\dfrac{(-1)^n}{(2n)!}x^{2n}+\cdots\\ \dfrac{1}{1-x}&=\sum_{n=0}^\infty x^n &&\text{for } -1\le x \lt 1\\ &=1+x+x^2+x^3+\cdots+x^n+\cdots\\ \dfrac{1}{1+x}&=\sum_{n=0}^\infty(-1)^n x^n &&\text{for } -1 \lt x\le 1\\ &=1-x+x^2-x^3+\cdots+(-1)^nx^n+\cdots\\ \ln(1-x)&=-\sum_{n=1}^\infty\dfrac{1}{n}x^n &&\text{for } -1\le x \lt 1\\ &=-x-\frac{1}{2} x^2-\dfrac{1}{3}x^3-\cdots-\dfrac{1}{n}x^n-\cdots\\ \ln(1+x)&=-\sum_{n=1}^\infty\dfrac{(-1)^n}{n}x^n &&\text{for } -1 \lt x\le 1\\ &=x-\frac{1}{2} x^2+\dfrac{1}{3}x^3-\cdots-\dfrac{(-1)^n}{n}x^n-\cdots\\ (1+x)^p&=1+px+\dfrac{p(p-1)}{2}x^2+\dfrac{p(p-1)(p-2)}{3!}x^3+\cdots\\ &\hskip0.5in+ \dfrac{p(p-1)(p-2)\cdots(p-n+1)}{n!}x^n+\cdots \end{align*}\]

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