Suppose $\{p_k\}$ is a sequence of polynomials with $p_k(0)=1$. Let $a_1,a_2,\ldots$ be an enumeration of all of the zeros of the $p_k$. Suppose that $$\prod_{k=1}^\infty p_k(z)$$ converges uniformly on compact subsets of $\mathbb{C}$. Is there some permutation $\sigma:\mathbb{N}\to\mathbb{N}$ such that $$f_\sigma (z)=\prod_{k=1}^\infty \left(1-\frac{z}{a_{\sigma(k)}}\right)$$ converges uniformly on compact subsets of $\mathbb{C}$?

This is a follow-up to Factoring a convergent infinite product of polynomials., in which an example of such $\{p_k\}$ is given along with a permutation $\sigma$ for which the product $f_\sigma (z)$ does $\underline{\text{not converge}}$.

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  • Since you have $a_{\sigma(k)}$ in a denominator, you probably don't want to include $0$ in your enumeration of the zeros, even though it is a zero of all the $p_k$. – Robert Israel Dec 6 at 15:56
  • Thank you. I meant $p_k(0)=1$. – user122916 Dec 6 at 15:57
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    My intuition is telling me that the answer is definitely "no". But I need to think more. – mathworker21 Dec 9 at 3:00
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    I think this is true, but possibly not that easy to prove. It might be provable with a product version of the "polygonal confinement theorem" by Steinitz: sites.math.washington.edu/~morrow/335_17/levy.pdf – Lukas Geyer Dec 11 at 23:45

Let $p$ be any prime number.

Now, $i^{p^s} = 1$ for $s\neq 1$. Now if your statement is correct, we can apply logarithm to both sides and prove that product of prime converges to $0$ which is not true.

Thus there exists no polynomial as such because in that case all the prime factors are nullified.

I have shown you the answer by taking "$s$" dimension polynomial, whose roots are all the prime numbers.

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