Can every rational function be represented in barycentric form?

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 $\mathbb C^{N+1}$?
• In Barycentric Lagrange Interpolation. Jean-Paul Berrut, Lloyd N. Trefethen., the earliest references seem to be from 1997 of Jean-Paul Berrut. This is a case of an author citing another paper of himself as reference, so "it is known" might only refer to a very small set of people... – Thomas Klimpel Feb 21 '15 at 13:59
• It's not clear, in your displayed equation, what $y_j$ denotes. – John Hughes Feb 21 '15 at 14:10
• @JohnHughes You mean because $y_j$ is not just a coefficient, but also identical to $r(x_j)$? Maybe you have a good point, and it could be related to my troubles understanding the cited statement. – Thomas Klimpel Feb 21 '15 at 14:21
• Actually, I assumed that $y_i$ denoted a variable. I think Trefethen and Berrut use $f_i$ for this, indicating that it's the value that you're trying to interpolate with some function $f$. Now that I think about it, $y_i$ is a perfectly good name too. :) I think if you read the first few pages of the paper cited by Thomas Klimpel, you'll find that the answer to item 1 is "yes". I suspect that "2" is also "yes", and that the proof's identical. Finally, the last page of the cited paper contains a bunch of historical remarks that should address item 4. – John Hughes Feb 21 '15 at 17:36

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.