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?
 A: 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.
A: 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?
