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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?

Mike Battaglia
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    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

2 Answers2

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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.

lhf
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    See also https://math.stackexchange.com/a/1529860/589. – lhf Jul 20 '17 at 17:21
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    So it doesn't seem the interpolating function must be unique. For example, if we sample at the integers, we can add sin(2pix) 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
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    @MikeBattaglia, I don't know. Ask a separate question. – lhf Jul 21 '17 at 17:12
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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?

Rutger Moody
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