Series Table of Contents Complex Analysis Laurent Series


Taylor series

For real functions

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:
\[ f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n. \]

Recall that, in real analysis, 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 $$f(x)\approx f(a)+f'(a)(x-a)+\frac12 f''(a)(x-a)^2.$$

The following applet shows the partial sums of the Taylor series for a given function. Drag the slider to show more terms of the series. Drag the point a or change the function.

For complex functions

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, \end{eqnarray} where \begin{eqnarray} a_n=\frac{f^{(n)}(z_0)}{n!},\quad n=0,1,2,\ldots \end{eqnarray} That is, series (\ref{seriefunction}) converges to $f(z)$ when $z$ lies in the stated open disk.

Every complex power series (\ref{seriefunction}) has a radius of convergence. Analogous to the concept of an interval of convergence for real power series, a complex power series (\ref{seriefunction}) has a circle of convergence, which is the circle centered at $z_0$ of largest radius $R \gt 0$ for which (\ref{seriefunction}) converges at every point within the circle $|z-z_0|=R.$ A power series converges absolutely at all points $z$ within its circle of convergence, that is, for all $z$ satisfying $|z - z_0| \lt R,$ and diverges at all points $z$ exterior to the circle, that is, for all $z$ satisfying $|z-z_0| \gt R.$ The radius of convergence can be:

  1. $R = 0$ (in which case (\ref{seriefunction}) converges only at its center $z = z_0$),
  2. $R$ a finite positive number (in which case (\ref{seriefunction}) converges at all interior points of the circle $|z - z_0| = R),$ or
  3. $R = \infty$ (in which case (\ref{seriefunction}) converges for all $z$).

The radius of convergence can be calculated using the ratio test of convergence. For example, if:

  1. $\displaystyle \lim_{n\rightarrow \infty} \left| \frac{a_{n+1}}{a_n}\right| = L\neq 0,$ the radius of convergence is $R=\dfrac{1}{L}$;
  2. $\displaystyle \lim_{n\rightarrow \infty} \left| \frac{a_{n+1}}{a_n}\right|= 0,$ the radius of convergence is $R=\infty$;
  3. $\displaystyle \lim_{n\rightarrow \infty} \left| \frac{a_{n+1}}{a_n}\right|= \infty,$ the radius of convergence is $R=0.$

Dynamic exploration

Use the following applet to explore Taylor series representations and its radius of convergence which depends on the value of $z_0.$

On the left side of the applet below, a phase portrait of a complex function is displayed. On the right side, you can see the approximation of the function through it's Taylor polynomials at the blue base point $z_0.$ The complex function, the base point $z_0,$ the order of the polynomial (vertical slider) and the zoom (horizontal slider) can be modified.

f(z) =

You can also select one function of the following list:

Maclaurin series

A Taylor series with centre $z_0=0$ \[ f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}z^n \] is referred to as Maclaurin series.

Some important Maclaurin series are:

\begin{eqnarray*} \displaystyle \frac{1}{1-z}&=& \sum_{n=0}^{\infty} z^n, \quad |z|\lt 1; \\ \displaystyle e^z &=& \sum_{n=0}^{\infty} \frac{z^n}{n!}, \quad |z|\lt \infty;\\ \displaystyle \sin z &=& \sum_{n=0}^{\infty} (-1)^n\frac{z^{2n+1}}{(2n+1)!}, \quad |z|\lt \infty;\\ \displaystyle \cos z &=& \sum_{n=0}^{\infty} (-1)^n \frac{z^{2n}}{(2n)!}, \quad |z|\lt \infty;\\ \displaystyle \sinh z &=& \sum_{n=0}^{\infty} \frac{z^{2n+1}}{(2n+1)!}, \quad |z|\lt \infty;\\ \displaystyle \cosh z &=& \sum_{n=0}^{\infty} \frac{z^{2n}}{(2n)!}, \quad |z|\lt \infty; \end{eqnarray*}

Exercise: Find the Maclaurin series expansion of the function \[ f(z)=\frac{z}{z^4+9} \] and calculate the radius of convergence.

Note: The applet was originally written by Aaron Montag using CindyJS. The source can be found at GitHub.

Laurent Series