For which points is there an analytic function through them If $(x_i,y_i)$ is an countable infinite set of points, with $x_i\neq x_j$ when $i \neq j$, does there always exist a sequence of real numbers $a_n$, such that $y_i=\sum_{n>1} a_nx_i^n$ for all $i>0$? Is it unique?
If {${x_i}$}$=\mathbb{N}$, i think its possible, what if {${x_i}$}$=\mathbb{Q}$?
What is required of $(x_i,y_i)$?
And same questions for $a_n\in \mathbb{Q}$?
 A: More generally: for any open subset $\Omega$ of the complex numbers,  any sequence of distinct points $z_n$ in $\Omega$  that has no limit point in $\Omega$,  and any sequence $w_n$ of complex numbers, there is a function $f$ analytic in $\Omega$ with $f(z_n) = w_n$.  This is a standard result of complex analysis that follows from Mittag-Leffler's theorem (see e.g. Rudin, "Real and Complex Analysis", theorem 15.15).   
On the other hand, for a sequence of distinct points $z_n$ that has a subsequence $z_{n_j}$ converging to $p \in \Omega$, the values $f(z_{n})$ must be far from arbitrary: analyticity of $f$ at $p$ requires $f(z_{n_j})$ to converge to $f(p)$, $(f(z_{n_j}) - f(p))/(z_{n_j} - p)$ to converge to $f'(p)$, etc., and the values of $f(z_{n_j})$ determine the function uniquely in the connected component of $\Omega$ containing $p$.
A: One of the ways this could fail for $\{x_i\} = \mathbb{Q}$ is if the power series ended up having a finite radius of convergence.
For example, take $y_i = (1+x_i^2)^{-1}$ for all $x_i$.  To start, if we're searching for a continuous function $f$ so that $f(x_i) = y_i$ then we are forced to take $f(x) = (1+x^2)^{-1}$ since continuous functions on Hausdorff spaces are uniquely determined by their values on dense subsets of their domains.  Now this function happens to be analytic on $|x|<1$, so it has a power series development there,
$$
f(x) = \sum_{k=0}^{\infty} a_k x^k.
$$
However, this power series diverges for $|x| > 1$, so that
$$
y_i \neq \sum_{k=0}^{\infty} a_k x_i^k
$$
for all $x_i$ satisfying $|x_i| > 1$.  Likewise, $f$ has a Laurent series for $|x| > 1$ which diverges for $|x| < 1$.  There is no power series for $f$ which converges on $\mathbb{R} \setminus \{\pm 1\}$.
Basically, given any $\{y_i\}$, we must at least be able to find a continuous function $f$ such that $y_i = f(x_i)$.  Steven's comment gives a clear counterexample.  In addition, we obviously want the function to be analytic if we're going to even think about representing it with a power series.  And, if we do, this power series must converge on all of $\mathbb{R}$, and hence on all of $\mathbb{C}$.  Thus the admissible $\{y_i\}$ you seek are precisely the values on $\mathbb{Q}$ of an entire function $f(x) = \sum_{k=0}^{\infty} a_k x^k$.
