# Properties of rational fractions

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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Asked and answered simultaneously on MO mathoverflow.net/questions/86723/… –  David Speyer Jan 26 '12 at 23:19
@DavidSpeyer: Could your answer on MO be translated into a precalculus-level answer here? –  Isaac May 2 '12 at 7:20
Certainly the statement could. Given $2d+1$ pairs $(x_i, y_i)$, there will usually be is a unique degree $d$ rational function passing through them. More precisely, such a function will exist UNLESS there is some integer $e>0$ and some subset of $2d+1-e$ of the points which lie on a rational function of degree $d-e$. Whether the argument can be simplified to precalc level is less clear. –  David Speyer May 2 '12 at 13:17

## 2 Answers

I don't have a "pre-calculus" simplification either, but I see some things not mentioned in here or on the MO answers, so I suppose I can throw my stuff in...

Consider the $(m,n)$ rational function

$$R(x)=\frac{P(x)}{Q(x)}=\frac{p_0+p_1 x+\cdots+p_m x^m}{q_0+q_1 x+\cdots+q_n x^n}$$

Although there are $m+n+2$ coefficients $p_k,q_k$ indicated in $R(x)$, in fact only $m+n+1$ conditions are needed to uniquely determine it, since one can easily cancel out a common factor from the numerator and the denominator.

Ostensibly, having $m+n+1$ points that your $R(x)$ should pass through ($R(x_j)=y_j, j=1,\dots,m+n+1$), should be enough, and that one only needs to solve the homogeneous system of linear equations $P(x_j)=y_j Q(x_j)$. Consider the example of fitting a $(1,1)$ rational function $\dfrac{a+bx}{c+dx}$ to the following points:

$$\begin{array}{c|ccc}x_j&0&1&2 \\\hline\\y_j&1&2&2\end{array}$$

which yields the set of equations

\begin{align*} a+0\cdot b&=1\cdot(c+0\cdot d)\\ a+1\cdot b&=2\cdot(c+1\cdot d)\\ a+2\cdot b&=2\cdot(c+2\cdot d) \end{align*}

Solving this set of equations and substituting into the expression for the $(1,1)$ rational function yields the function $R(x)=\dfrac{2x}{x}$, or $R(x)=2$ after canceling out common factors. It can now be seen that this function passes through the second and third points, but certainly not the first. (In the parlance of rational interpolation, the point $(0,1)$ is termed an inaccessible point, and the two other points are said to be in special position.)

As it turns out, things simplify if we only consider rational functions with relatively prime numerators and denominators (i.e. $P(x)$ and $Q(x)$ have no common polynomial factors of positive degree). The applicable theorem, then, is

If the coefficients of a rational function $R(x)$ satisfy the $m+n+1$ linear equations determined by the given points, then the interpolation problem has no inaccessible points, if and only if the equivalent rational function $R^\ast(x)$, with relatively prime numerator and denominator also satisfy those $m+n+1$ equations.

See Bulirsch and Stoer for a more detailed discussion, and these two articles for reliable algorithms for constructing rational interpolants and/or detecting which subset of a given set of points is in special position.

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Given $2d+1$ pairs $(x_i,y_i)$, there will usually be a unique degree $d$ rational function passing through them. More precisely, such a function will exist UNLESS there is some integer $e>0$ and some subset of $2d+1−e$ of the points which lie on a rational function of degree $d−e$. Whether the argument can be simplified to precalc level is less clear. – David Speyer May 2 at 13:17

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