Properties of rational functions

Question

Hi everyone. It is well known that a polynomial of degree $n$ is completely determined by $n+1$ points. Now, is there any similar result for rational functions?

Answer 1

This answer builds on Joe Silverman's, and uses the same notation. He writes "as your conditions $F(x_i)=c_i$ ... are independent".

Suppose, for $1 \leq e \leq d$, that there do not exist any $2d+1-e$ of the points which can by interpolated by a rational function of degree $d-e$. Than I claim the conditions are independent.

Proof Suppose that $F(x_i) = c_i$ and $F'(x_i) = c_i$. Let $F(x) = \sum a_i x^i / \sum b_i x^i$ as in the previous answer. Let $C$ be the curve in $\mathbb{P}^1 \times \mathbb{P}^1$ cut out by $$y_1 \left( \sum a_i x_1^i x_2^{d-1} \right) = y_2 \left(\sum b_i x_1^i x_2^{d-1} \right)$$ where $(x_1:x_2)$ are homogenous coordinates on the first $\mathbb{P}^1$ and $(y_1:y_2)$ are homogenous coordinates on the second. Define $C'$ similarly.

Then the curves $C$ and $C'$ meet at the $2d+1$ points $(x_i:1) \times (c_i : 1)$. However, a computation in $H^{\bullet}(\mathbb{P}^1 \times \mathbb{P}^1)$ shows that the intersection product $C \cdot C' = 2d$. The only way that this can happen is that $C$ and $C'$ have a common component. In particular, $C$ has more than one component.

Now, $C$ has degree $(1,d)$ in $\mathbb{P}^1 \times \mathbb{P}^1$. So, if it has more than one component, then one of them is of degree $(1,d-e)$ and the others are $e$ lines of the form $x=\mathrm{constant}$. There can be $e$ points on the vertical lines (since the $x_i$ are distinct); that leaves $2d+1-e$ points on the component of degree $(1,e)$. In other words, $2d+1-e$ points which are interpolated by a rational function of degree $d-e$.

Answer 2

A rational function of degree $d$ (of 1 variable) has the form $$F(x) = \frac{a_dx^d+a_{d-1}x^{d-1}+\dots+a_0}{b_dx^{d}+b_{d-1}x^{d-1}+\dots+b_0},$$ where one should view the coefficients $[a_d,\ldots,a_0,b_d,\ldots,b_0]$ as homogeneous coordinates in the projective space $\mathbb{P}^{2d+1}$. Fixing a value $F(x_i)=c_i$ gives a linear equation in the coefficients that determines a hyperplane in $\mathbb{P}^{2d+1}$. Since $2d+1$ independent hyperplanes intersect in exactly one point, it follows that as long as your conditions $F(x_i)=c_i$ with $1\le i\le 2d+1$ are independent, then they determine a unique rational function of degree less than or equal to $d$. (The degree can be smaller than $d$ if it turns out that $a_d=b_d=0$.)

Answer 3

I recently got an e-mail about this question, so I'm going to take the opportunity to write up the answer in an elementary way.

By a "rational function of degree $d$", I mean a function $f(x)/g(x)$ where $f$ and $g$ are relatively prime and $\max(\deg f, \deg g) = d$. By a rational function of degree $\leq d$, I mean that $\deg f$, $\deg g \leq d$. Here are the main claims:

Uniqueness: Given $2d+1$ pairs $(x_1, y_1)$, $(x_2, y_2)$, .., $(x_{2d+1}, y_{2d+1})$ with distinct $x$ values, they can be interpolated by at most one rational function of degree $\leq d$.

Existence: Suppose that there does not exist any $2d+1-e$ element subset of our pairs which can be interpolated by a rational function of degree $\leq d-e$, for $1 \leq e \leq d$. Then there is a unique rational function of degree $\leq d$ interpolating them.

Proof of claim about uniqueness: Suppose to the contrary that we had two rational functions $f_1(x)/g_1(x)$ and $f_2(x)/g_2(x)$ with $f_1(x_i)/g_1(x_i) = f_2(x_i)/g_2(x_i)$ for all $i$, and $\deg f_1$, $\deg f_2$, $\deg g_1$, $\deg g_2 \leq d$.

Then the polynomial $f_1(x) g_2(x) - f_2(x) g_1(x)$ has degree $\leq 2d$ and vanishes at the $2d+1$ points $x_i$, so it is identically $0$ and we conclude that $f_1(x) g_2(x) = f_2(x) g_1(x)$. So $f_1(x)/g_1(x) = f_2(x)/g_2(x)$. $\square$

Proof of claim about existence Write $f(x) = f_d x^d + \cdots + f_1 x + f_0$ and $g(x) = g_d x^d + \cdots + g_1 x + g_0$. Consider the linear equations for the $f_j$ and $g_j$ given by $$f(x_j) = g(x_j) y_j \qquad (\ast).$$ This is $2d+1$ linear equations in $2d+2$ variables, so there is a nonzero solution; let $f(x)$ and $g(x)$ be the polynomials that solve the equations $(\ast)$.

We would like to claim that the rational function $f(x)/g(x)$ takes the value $y_j$ at $x_j$, but there is a problem. If $f(x_j) = g(x_j) = 0$, then $(\ast)$ is satisfied, but $\lim_{x \to x_j} \tfrac{f(x)}{g(x)}$ need not equal $y_j$.

Let $e$ be the number of $x_j$ for which $f(x_j) = g(x_j) = 0$. Then $f(x)/g(x)$ is actually a rational function of degree $\leq d-e$ which interpolates the $2d+1-e$ other pairs. This contradicts our assumption, so we don't have $f(x_j) = g(x_j) = 0$ for any $j$ and we are done. $\square$

As an example to show that we need the extra condition in the existence case, take $d=1$ with $y_1 = y_2 \neq y_3$. A rational function of degree $\leq 1$ is always either injective or constant, so the condition that $f(x_1)/g(x_1) = f(x_2)/g(x_2)$ forces $f(x)/g(x)$ to be a constant function, and then it can't interpolate $(x_3, y_3)$.