Padé approximants: Convergence examples

We will first illustrate the Montessus’s theorem with the function:

$\displaystyle f(z) = \frac{z}{4 + z^2} = \frac{z}{(z - 2i)(z + 2i)}$

where $z \in \mathbb{C}$ . This function has two poles at $z = 2i$ and $z = -2i$ .

The corresponding Maclaurin series is:

$\displaystyle \frac{1}{4}z - \frac{1}{16}z^3 + \frac{1}{64}z^5 - \frac{1}{256}z^7 + \frac{1}{1024}z^9 + \mathcal{O}(z^{11})$

The corresponding $P(2,2)$ approximant is (calculations were made according to the linear systems presented in this post):

$\displaystyle P(2,2) = \frac{z}{4 + z^2}$

We see that, in this case, if we set $n = 2$ (the total number of poles) we recover the original function since $P(2,2) = f(z)$ . The graphs of the $f(z)$ , $P(2,2)$ and the corresponding Maclaurin series are presented in the figure below.

Left, graph of

$\displaystyle \frac{z}{4 + z^2}$

and its corresponding $P(2,2)$ . Right, graph of the Maclaurin series

$\displaystyle \frac{1}{4}z - \frac{1}{16}z^3$

in the complex $\mathbb{C}$ -plane. The poles at $-2i$ and $2i$ are clearly visible on the left picture. Hue and brightness are used to display phase and magnitude, respectively.