# Analytic “Lagrange” interpolation for a countably infinite set of points?

Suppose I have a finite set of points on the real plane, and I want to find the univariate polynomial interpolating all of them. Lagrange interpolation gives me the least-degree polynomial going through all of those.

Is there an analogous construct for a countably infinite, sparse set of points on the real plane, instead using analytic functions and power series?

There is obviously some difficulty in forming a perfect analogy, as Lagrange interpolation yields the "lowest degree" polynomial interpolating the points, whereas there is no such thing as a "lowest degree" power series. However, perhaps there is some generalized measure of the complexity of a power series that is decently workable, and which restricts to the lowest-degree polynomial in the finite case.

If so, how does this work? Is there an easy way to obtain the nth coefficient of the power series from the points?

• Your countably infinite sparse set $A$ of points needs to be so sparse that no sequence of distinct points from $A$ has a finite limit. In that case, there is an entire (i.e., analytic in the whole plane) interpolant. That follows from theorems of Weierstrass and Mittag-Leffler (the former gives an entire function that vanishes at exactly the points in $A$, and the latter gives a function with poles at exactly the points of $A$ and with prescribed principal parts at the poles). – Andreas Blass Jul 20 '17 at 17:17
• do you know about Runge's phenomenon ? – G Cab Jul 20 '17 at 17:24
• So suppose you have a function which is 0 at every nonzero integer, and 1 at x=0. Is there no way to interpolate this and get something like the sinc function? – Mike Battaglia Jul 20 '17 at 23:00
• Never mind, misinterpreted - thanks – Mike Battaglia Jul 21 '17 at 16:26

There is this theorem:

Given two sequences $z_n$ and $w_n$ of complex numbers such that $|z_n| \to \infty$, there exists a holomorphic function $f$ such that $f(z_n) = w_n$ for all $n$.

It is a consequence of the Weierstrass factorization theorem and the Mittag-Leffler theorem.

See this question.

• So it doesn't seem the interpolating function must be unique. For example, if we sample at the integers, we can add sin(2*pi*x) to the result to obtain another valid interpolant. Is there some condition we can place on the interpolant to obtain a simplest one, in some sense? – Mike Battaglia Jul 21 '17 at 17:11
• @MikeBattaglia, I don't know. Ask a separate question. – lhf Jul 21 '17 at 17:12

I was trying something like this myself. Not 100% sure if this is what you mean.

Let $$\{ a_i \}$$ be the infinite sequence you want to interpolate by polynomials in $$t$$. We construct series of $$n$$-th degree polynomials $$X^n(t)$$ such that : $$\forall i \le n : X^n(i) = a_i$$ like this:

$$X^0(t)= \frac{a_0}{0!0!}$$

$$X^1(t)= \frac{a_0}{0!0!} - (t-0)\{\frac{a_0}{0!1!} -\frac{a_1}{1!0!}\}$$

$$X^2(t) = \frac{a_0}{0!0!} - (t-0)\{\frac{a_0}{0!1!} -\frac{a_1}{1!0!} - (t-1)\{ \frac{a_0}{0!2!} - \frac{a_1}{1!1!} + \frac{a_2}{2!0!} \} \}$$

$$X^3(t) = \frac{a_0}{0!0!} - (t-0)\{\frac{a_0}{0!1!} -\frac{a_1}{1!0!} - (t-1)\{ \frac{a_0}{0!2!} - \frac{a_1}{1!1!} + \frac{a_2}{2!0!} - (t-2)\{ \frac{a_0}{0!3!} -\frac{a_1}{1!2!} +\frac{a_2}{2!1!} - \frac{a_3}{3!0!} \} \} \}$$

The idea of course is that every $$X^n(t)$$ is 'cut off' at some point when we fill in an integer $$p < n$$, resulting in the polynomial $$X^p(p)$$ for which we know the relation holds.

This would lead to the general formula :

$$\begin{array}{l} X^n(t) = \\ \frac{a_0}{0!0!} \\- (t-0)\{\frac{a_0}{0!1!} -\frac{a_1}{1!0!} \\- (t-1)\{ \frac{a_0}{0!2!} - \frac{a_1}{1!1!} + \frac{a_2}{2!0!} \\- (t-2)\{ \frac{a_0}{0!3!} -\frac{a_1}{1!2!} +\frac{a_2}{2!1!} - \frac{a_3}{3!0!}\\ - (t-3)\{ \frac{a_0}{0!4!} -\frac{a_1}{1!3!} +\frac{a_2}{2!2!} - \frac{a_3}{3!1!} +\frac{a_4}{4!0!} \\ \vdots\\ -(t-n+1)\{ \frac{a_0}{0!n!} -\frac{a_1}{1!(n-1)!} + \cdots \cdots \cdots \cdots\\ +(-1)^{n-2} \frac{a_{n-2}}{(n-2)!2!} +(-1)^{n-1} \frac{a_{n-1}}{(n-1)!1!} +(-1)^n \frac{a_n}{n!0!} \}\\ \cdots \} \} \} \} \\ \end{array}$$

Which can be verified with the help of the formula $$\sum_{k=0}^n (-1)^k\binom{n}{k}=0$$

Now I assumed that the data points were equally spaced (say every $$y=a_i$$ can be found at $$x=i$$). If this is not the case we could regard the parameter $$t$$ as a parameter on a 2-dimensional curve $$\left( X^n(t), Y^n(t)\right)$$.

I hope if someone reads this they can verify the above result. Does this procedure have a name?