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? 



Properties of rational functions 
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 $d-e$.


