## Pade approximants: a case study (i)

This post was inspired by a lecture series of Carl Bender [1].

## The case

I will consider the function $f$ on the complex plane, defined by

$f(z) = \frac{\log(1-z)}{z}$

Have a look at a complex plot of it

Some properties:

• $f$ has a a logarithmic singularity at $z = 1$.
• $f$ is analytic in $\mathbb{C}\setminus[1,\infty)$
• $f$ has a logarithmic cut on the line $(1,\infty)$

For $|z|<1$ the following series representation holds

$f(z) = -\sum\limits_{k=0}^\infty \frac{z^k}{k+1}$

Claim: The series sucks in practice!

## The study (i)

Once upon a time some guy called Padé came along, being unnerved by the wackness of Taylor-like series, and proposed what is now known as Padé approximants. His idea was to find polynomials $P_n(z)$ and $P_m(z)$ of degree $n$ and $m$, such that

$P^n_m(z )\equiv \frac{P_n(z)}{P_m(z)} = f(z) + \text{corrections}$.

In a similar way of Taylor polynomials approximating $f$, he simply tried rational expressions.

Q: How does one find these these polynomials? A: By comparing to the Taylor series.

Example 1: Find $P^1_2(z )$. Start with an ansatz:

$P^1_2(z) = \frac{a_0+a_1z}{1+b_1z+b_2z^2}$.

where I already put the lowest order coefficient in the denominator to $1$, which is a convention that removes the ambiguousness of the polynomial parameters when considering ratios. So we have four parameters, which we can match to a Taylor polynomial of order three. To make progress, we expand our ansatz in powers of $z$

$P^1_2(z)=(a_0+a_1z)(1-(b_1z+b_2z^2)+(b_1 zb_2 z^2)^2-b_1^3 z^3) +\dots$

where I already dropped terms that will be of fourth power or higher in $z$. This expression should be matched to $-1-\frac{z}{2}-\frac{z^2}{3}-\frac{z^3}{4}$ order by order in $z$. A little algebra yields

$P^1_2(z)=\underbrace{a_0}_{=-1}+z\underbrace{(a_1+b_1)}_{=-\frac{1}{2}} + z^2 \underbrace{\biggl(\frac{b_1}{2}+b_2\biggr)}_{=-\frac{1}{3}}+z^3\underbrace{\biggl(\frac{b_1}{3}+\frac{b_2}{2}\biggr)}_{=-\frac{1}{4}} + \dots$.

This set of equations has a simple solution, resulting in

$P^1_2(z)=\frac{-1+\frac{z}{2}}{1-z+\frac{z^2}{6}}$

A quick fireprobe: this approximant suggests the approximate value $\frac{9}{13}$ for $\log 2$. This isn’t too bad, in fact you can add it to your stash of quick and dirty calculation library. Students are always impressed by this sort of witchcraft. I dare you to find an easier-to-remember value from the series representation…

Example 2: The Taylor polynomials are contained within the Padé approximants as

$P^n_0(z) = -\sum\limits_{k=0}^n \frac{z^k}{k+1}$.

To be continued… I’m lacking blogging time atm…