Let $a\in \mathbb R$ and $f(x)$ be and infinitely differentiable function on an interval $I$ containing $a.$ Then the one-dimensional Taylor series of $f$ around $a$ is given by \[ f(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+\cdots \]

which can be written in the most compact form: \[ f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n. \]

Taylor's theorem gives an approximation of a $k$-times differentiable function around a given point by a $k$-th order Taylor polynomial.

For example, the best linear approximation for $f(x)$ is $$f(x)\approx f(a)+f'(a)(x-a).$$ This linear approximation fits $f(x)$ with a line through $x=a$ that matches the slope of $f$ at $a.$

For a better approximation we can add other terms in the expansion. For instance, the best quadratic approximation is $$f(x)\approx f(a)+f'(a)(x-a)+\frac12 f''(a)(x-a)^2.$$

Suppose that a function $f$ is analytic throughout a disk $|z -z_0|< R,$ centered at $z_0$ and with radius $R.$ Then $f(z)$ has the power series representation \begin{eqnarray}\label{seriefunction} f(z)=\sum_{n=0}^{\infty} a_n(z-z_0)^n,\quad |z-z_0|<R, \qquad \qquad (1) \end{eqnarray} where $ \, a_n=\dfrac{f^{(n)}(z_0)}{n!},\quad n=0,1,2,\ldots $

That is, series (1) converges to $f(z)$ when $z$ lies in the stated open disk.