Suppose that $f(z)=\displaystyle\prod_{k=1}^\infty p_k(z)$ is a convergent product of polynomials $p_k$ such that $p_k(0)=1$. I want to know if I can "factor" $f(z)$ in the following way: if we list the roots of all the $p_k$ as $r_1, r_2, r_3, \ldots$, must the product $\displaystyle\prod_{j=1}^\infty \left( 1-\frac{z}{r_j}\right)$ converge?

I understand that the Weierstrass Factorization Theorem gives a factorization for $f(z)$ that involves exponential terms to ensure convergence, but I am wondering whether knowing only that $\displaystyle\prod_{k=1}^\infty p_k(z)$ converges is enough to conclude that the roots grow fast enough.

  • $\begingroup$ Is the first product assumed to converge pointwise everywhere? And does "convergent" mean convergent to a nonzero value? $\endgroup$ – zhw. Jul 6 '18 at 22:53
  • $\begingroup$ The function $f(z)$ is assumed to be an entire function, so yes, the product converges uniformly on compact subsets of $\mathbb{C}$. $\endgroup$ – user122916 Jul 7 '18 at 4:22
  • $\begingroup$ Just to make sure I understand: if you order the $r_j$'s based on their appearance in the sequence $p_1,p_2,p_3,\dots$, then the fact that $\prod p_k$ converges obviously implies $\prod (1-\frac{z}{r_j})$ converges. But you are asking whether $\prod (1-\frac{z}{r_j})$ converges absolutely, i.e. whether any ordering of the $r_j$'s makes $\prod (1-\frac{z}{r_j})$ converge (conditionally). Is this accurate? $\endgroup$ – mathworker21 Jul 8 '18 at 8:06
  • $\begingroup$ Actually, it is that "obviously implies" part that I am stuck on. Could you explain a bit? $\endgroup$ – user122916 Jul 8 '18 at 14:37

$\prod_{n=1}^\infty (1-\frac{z^2}{n^2})$ converges locally uniformly on $\mathbb{C}$. We can let $p_1 = \prod_{n=1}^{N_1} (1-\frac{z^2}{n^2}), p_2 = \prod_{n=N_1+1}^{N_2} (1-\frac{z^2}{n^2}), p_3 = \prod_{n=N_2+1}^{N_3} (1-\frac{z^2}{n^2})$, etc. for positive integers $N_1 < N_2 < \dots$. We may order the roots of $p_1$ as $-1,-2,...,-N_1,1,2,\dots,N_1$, and order the roots of $p_2$ as $-(N_1+1),\dots,-N_2, (N_1+1),\dots, N_2$, etc. The point is that $\prod_{n=1}^\infty (1+\frac{z}{n})$ goes to infinity if $z \in \mathbb{R}^+$, so if $z \in \mathbb{R}^-$, then $\prod_{n=N_j+1}^{N_{j+1}} (1+\frac{z}{-n})$ will be large for $z \in \mathbb{R}^-$. So we can choose the $N_i$'s to be spaced out enough so that we don't get local uniform convergence. Note exactly what's going on is that at the cutoff of each $p_k$, we are fine, since we multiplied together the $(1+\frac{z}{n})$s with the $(1-\frac{z}{n})$s, but the intermediate product (exactly "half way" between $p_k$ and $p_{k+1}$) is the issue.

  • $\begingroup$ I think the issue is that the degree of the $p_k$'s does not need to be bounded. I think I have a counterexample, and will type it up in an answer below. $\endgroup$ – user122916 Jul 12 '18 at 20:39
  • $\begingroup$ My example doesn't work :). I am trying to turn this argument into an $\epsilon-\delta$ style argument, but I am having issues. In fact, I cannot even convince myself that $\lim_{N\to \infty} \prod_{j=1}^N \left(1-\frac{z}{r_j}\right)$ converges. Any suggestions? I suppose I am looking to more precisely show that the intermediate products "won't really matter." $\endgroup$ – user122916 Jul 12 '18 at 21:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.