Can every rational function be represented in barycentric form?

Question

This article about polynomial interpolation claims that (it is known that) every rational function may be represented in barycentric form: $$r(x)=\frac{\sum_{j=0}^N\frac{w_j}{x-x_j}y_j}{\sum_{j=0}^N\frac{w_j}{x-x_j}}$$ What about the rational function $r(x,y)=\frac{xy}{x^2+y^2}$? But even if I accept that we are talking only about univariate functions, I would still like to know a bit more about the cited statement:

1. Is this statement strictly true for $\mathbb R$, i.e. for any rational function $r(x)=\frac{p(x)}{q(x)}$ with $p,q\in\mathbb R[x]$ (polynomials with real coefficients) there are $w_j, x_j, y_j \in\mathbb R$ representing $r(x)$?
2. Is this statement strictly true for $\mathbb C$, i.e. for any rational function $r(x)=\frac{p(x)}{q(x)}$ with $p,q\in\mathbb C[x]$ (polynomials with complex coefficients) there are $w_j, x_j, y_j \in\mathbb C$ representing $r(x)$?
3. For $r(x)=\frac{p(x)}{q(x)}$ and $N=\operatorname{deg}p+\operatorname{deg}q$, is the set of $x_j\in\mathbb C$ for which appropriate $w_j,y_j\in\mathbb C$ exist open and dense in $\mathbb C^{N+1}$?
4. Where can I learn more about this topic, since when is this known, ... ?

Answer 1

I found a derivation in one of the papers from 1997 of Jean-Paul Berrut cited in Barycentric Lagrange Interpolation. Jean-Paul Berrut, Lloyd N. Trefethen. The claim is that any rational function $r(x)=\frac{p(x)}{q(x)}$ with $\max(\operatorname{deg}p,\operatorname{deg}q)\leq n$ for $n+1$ distinct interpolation points $x_j$ with $q(x_j)\neq 0$ can be written in barycentric form $$r(x)=\frac{\sum_{j=0}^n\frac{u_j}{x-x_j}r_j}{\sum_{j=0}^n\frac{u_j}{x-x_j}}$$ for $r_j:=r(x_j)$ and suitable weights $u_j$. The derivation is actually quite simple:

Let $l(x)=(x-x_0)(x-x_1)\dots(x-x_n)$ and $w_j=\left.\frac{x-x_j}{l(x)}\right|_{x=x_j}$. The barycentric Lagrange formula represents the denominator as $q(x)=l(x)\sum_{j=0}^n\frac{w_j}{x-x_j}q_j$. For $\tilde{p}(x):=l(x)\sum_{j=0}^n\frac{w_j}{x-x_j}q_jr_j$ we have $$\frac{\tilde{p}(x)}{q(x)}=\frac{\sum_{j=0}^n\frac{w_j}{x-x_j}q_jr_j}{\sum_{j=0}^n\frac{w_j}{x-x_j}q_j}=\frac{\sum_{j=0}^n\frac{u_j}{x-x_j}r_j}{\sum_{j=0}^n\frac{u_j}{x-x_j}}$$ with $u_j:=w_jq_j$, and $\tilde{p}(x)=p(x)$ follows from $\operatorname{deg}p\leq n$ and $q_j=q(x_j)\neq 0$. We have $p(x_j)=q_jr_j$, because $q_j=q(x_j)\neq 0$. So $\tilde{p}(x)$ coincides with $p(x)$ at $n+1$ points, and both are polynomials of degree less than $n+1$, so they are identical.

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The article Some New Aspects of Rational Interpolation. Claus Schneider, Wilhelm Werner from 1986 uses the barycentric form for rational interpolation and highlights its advantages, but doesn't seem to waste time or space to state or prove such a basic fact about the barycentric form. It might indeed have been stated and proved for the first time by Jean-Paul Berrut, maybe because people at that time grew tired of basic folklore results which practitioners knew but didn't publish, because the results were not new or non-trivial enough.

A good place to find more information about rational interpolation might be the reference section of this article about rational interpolation, i.e. the articles

1. Recent developments in barycentric rational interpolation. Jean–Paul Berrut, Richard Baltensperger and Hans D. Mittelmann.
2. Barycentric rational interpolation with no poles and high rates of approximation. Michael S. Floater, Kai Hormann.