# Properties of rational functions

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?

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For any fixed nonnegative integers $a$ and $b$, a rational function with numerator of degree $\leq a$ and denominator of degree $\leq b$ is uniquely determined by $a+b+1$ values. But I'm not sure whether we can fit such a function for any $a+b+1$ given value; it is purely a uniqueness-if-exists assertion. – darij grinberg Jan 26 '12 at 14:38
Thanks Darij. If you find or remember anything else, please let me know. – Hebert Jan 26 '12 at 15:44

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 $d-e$.

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Interesting. Thank you David. – Hebert Jan 27 '12 at 8:36

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 $a_d=b_d=0$.)

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Thanks. I see that it can be generalized to the case when the numerator and the denominator have different degrees, right? – Hebert Jan 26 '12 at 15:46
That should be "independent hyperplanes", not "distinct hyperplanes". Also note that independence of the conditions is not as simple as it is in the polynomial case. For polynomials of degree $\le d$, any $d$ distinct $x_i$ will do. For rational functions with numerator and denominator of degrees $\le d_1$ and $d_2$ respectively, independence depends not only on the $x_i$ but also on the $c_i$, e.g. if more than $d_1$ of the $c_i$ are $0$ they will be dependent. – Robert Israel Jan 26 '12 at 16:55
@Robert Isreal: Thanks, I made the correction. And as you say, the independence is not automatic. – Joe Silverman Jan 26 '12 at 18:22
@Herbert: The situation that I've described handles that case, too, since there's no requirement that $a_d$ or $b_d$ be nonzero, nor a requirement that the numerator and denominator have no common factors. So the family of rational functions that I wrote down includes all rational functions of degree at most $d$. – Joe Silverman Jan 26 '12 at 18:25