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

log1mzovSome 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…

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About goobypl5

pizza baker, autodidact, particle physicist
This entry was posted in Math, Mathematica and tagged , , , , . Bookmark the permalink.

2 Responses to Pade approximants: a case study (i)

  1. Pingback: Pade approximants: a case study (ii) | randomgooby

  2. Pingback: Pade approximants: a case study (iii) | randomgooby

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