If $f(x) = 0$ has infinitely many solutions for Taylor series of $f$, does this uniquely determine $f$? Euler's original proof that $\sum_n 1/n^2 = \pi^2/6$ seemed to implicitly rely on an assumption that the series $1 - x/3! + x^2/5! + \ldots$ was the only series that had $\{(n \pi)^2\}_{n=1}^{\infty}$ as the roots.  Is this true in general? I.e. if I have countable infinite many non-zero roots $x_n$ of $f$ such that $\sum_n 1/x_n$ converges absolutely, and $0$ is not a root of $f$, then if $f$ has a Taylor series at $0$ that converges everywhere, is $f$ (i.e. the Taylor series of $f$ at $0$) uniquely determined up to a scalar multiple, just by knowing the roots?  If $f(0) = 1$, is the Taylor series of $f$ at $0$ always given by the usual formulas for polynomial coefficients in terms of reciprocals of the roots, assuming that the constant term of the polynomial is $1$?  What can be said about a possible Taylor series for $f$ at $0$, if $f(0) = 1$ and $\sum_n 1/x_n$ diverges and/or does not converge absolutely (or is there no possible Taylor series with non-zero radius of convergence)?
 A: Maybe Weierstraß-Factorization-Theorem is something for you. 
It says that for every sequence $a_n$ with $|a_n|\to \infty$ you find an entire function which do have a zero at every $a_n$ and are zero only there.
So in fact after knowing every zero (with multiplicity) two entire functions only differ by a $\exp(g(z))$ term, where $g$ is entire.
It says: (taken from Wikipedia)

Let $f$ be an entire function, and let $\{a_n\}$ be the non-zero zeros of $f$ repeated according to multiplicity; suppose also that $f$ has a zero at $z = 0$ of order $m \geq 0$ (a zero of order $m = 0$ at $z = 0$ means $f(0) \neq 0$). Then there exists an entire function $g$ and a sequence of integers $\{p_n\}$ such that
  $$f(z)=z^m e^{g(z)} \prod_{n=1}^\infty E_{p_n}\!\!\left(\frac{z}{a_n}\right)$$
  where 
  $$E_n(z) = \begin{cases} (1 -z) & \text{if }n=0, \\ (1-z)\exp \left( \frac{z^1}{1}+\frac{z^2}{2}+\cdots+\frac{z^n}{n} \right) & \text{otherwise}. \end{cases}$$

A: It's not true that you can uniquely extract a function from its zeroes when it is not a polynomial. E.g. $\sin x$ and $e^x \sin x$ have identical zeroes. 
