Clarification on the Weierstrass factorization theorem Given any entire function $f(z)$, with zeros on $\{a_n\}$, it must be of the form
$$f(z) = z^m e^{g(z)} \prod_1^\infty E_{p_n}(z/a_n)$$
where the $E_{p_n}$ denote elementary factors, and $g$ is entire.
Is this factorization unique or is there a counterexample? Can we always choose the $p_n$ so that the same zeros $a_n$ give the same values of $p_n$?
 A: It is unique, since $m$, and each elementary factor are completely determined by (the growth rate of) $f$. Uniqueness of $g$ is then a consequence.
If by $p_n$ you mean the polynomial in the exponential factor of elementary factor, and your question is whether you can take the same $p_n$ as you change $f$, then no, you cannot, since the exponential factor depends on the growth rate of $f$.
A: It is not unique. Recall that
$$E_p(z) = (1-z)\exp\left( z + \frac{z^2}2 + \cdots + \frac{z^p}p \right).$$
For a trivial example, let $f(z) = 1-z$, with $a_1 = 1$. Then
$$
f(z) = z^0 e^{-z} \underbrace{(1-z)\exp(z)}_{=E_1(z)}
= z^0 e^{-z-z^2/2} \underbrace{(1-z)\exp(z+\frac{z^2}2)}_{=E_2(z)}.
$$
In general you have some wiggle room in how you choose $p_n$.
(If the zeros satisfy some additional properties, you can get uniqueness by choosing ''$p$ as small as possible''. Rudin calls this canonical products).

More details added The $E_p$:s are really just fudge factors to make the infinite product converge, and I'm not sure what you mean by choosing them as small as possible. Sometimes (depending on the zero set of $f$) it's possible to choose $p_n$ indepently of $n$, and in this case it makes sense to take $p_n$ as small as possible. Doing this will result in what many authors call canonical products and these are unique.
In general, however, $p_n$ will have to depend on $n$ and increasing or decreasing a few $p_n$:s here and there won't affect the convergence of the infinite product.
