# Lagrange interpolation for rational functions

Lagrange interpolation is very useful. I was wondering if there was an equivalent that is not using polynomials but rational functions, one polynomial divided by another. Look at this example:

Say I want a function that passes through the $(x,y)$ points $(3,0),(4,-1),(6,3)$ and $(7,2)$. Using Lagrange interpolation I get $$-\frac{x^3-15x^2+70x-102}{2}$$ but using this interpolation method for rational functions I get the much simpler $(x-3)/(x-5)$. This is a case where the rational function interpolation is a lot simpler than polynomial interpolation.

Is there a method to get $(x-3)/(x-5)$ given the points above, a sort of interpolation method that uses rational functions?

I totally agree that both functions $$f(x)=-\frac{x^3-15x^2+70x-102}{2}$$ $$g(x)=\frac{x-3}{x-5}$$ pass through points $(3,0),(4,-1),(6,3),(7,2)$.
But what does happen if you want to interpolate close to $x=5$ ? $f(5)=1$ but $g(5^{\pm})=\pm \infty$. This makes a small problem, isn(t ?
Quoting Wikipedia page : Lagrange polynomials are used for polynomial interpolation. For a given set of distinct points $x_j$ and numbers $y_j$, the Lagrange polynomial is the polynomial of the least degree that at each point $x_j$ assumes the corresponding value $y_j$.
• I agree with your answer if the goal is to find a function behaving reasonably nicely for the real topology near to the given interpolation points, but that is only one of the many uses of interpolation. it is a very interesting question to ask for the smallest degree rational function taking specified values at specified points, or alternatively asking how many points are needed to determine such a function. As for the example, if I'm working on the projective line $\mathbb P^1$, then the linear fractional transformation $g(x)$ is much easier to analyze than is the cubic polynoimal $f(x)$. – Joe Silverman Dec 31 '16 at 14:40