# 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 $a_n\in \mathbb{Q}$?

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you need $x_i \neq x_j$ for $i \neq j$. –  user29743 Apr 25 '12 at 0:37
For $\{x_i\}=\mathbb{Q}$, consider using $y_i=0$ for all negative $x_i$ and $y_i=1$ for all positive $x_i$. More generally, if the set $x_i$ has any limit points then $y_i$ is constrained to be continuous at those limit points. –  Steven Stadnicki Apr 25 '12 at 0:39
I retracted my wrong answer from before. A friend explained why the proof is correct for the case where the domain is the natural numbers. –  user29743 Apr 25 '12 at 1:29

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 $p$.

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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 $f(x) = \sum_{k=0}^{\infty} a_k x^k$.

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