Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose I pick a collection $A \subset \mathbb{C}$ of points in the complex plane and attempt to construct a "polynomial" with those roots via, $$f(z):=\Pi_{\alpha \in A} (z-\alpha).$$

If $A$ is finite, we get a polynomial.

If $A=\{n\pi:n \in \mathbb{Z}\}$, according to Euler we get $f(x)=\sin(x)$. Edit: this example is not right as Qiaochu has pointed out; see his answer for more details.

What about other subsets of the complex plane? Other countable subsets without accumulation points? Countable sub with accumulation points like $A=\{1/n:n \in \mathbb{Z}\}$? Uncountable subsets?? When does the product converge, and if it does how does the spatial distribution of $A$ effect the properties of $f$?

This question was motivated by the question here: Determining the density of roots to an infinite polynomial

share|cite|improve this question
up vote 8 down vote accepted

That is not what Euler's product expansion of the sine looks like. It is very much supposed to be in the form $$\frac{\sin z}{z} = \prod_{n \ge 1} \left( 1 - \frac{z^2}{\pi^2 n^2} \right).$$

The product you've written down does not converge for $A = \{ n \pi \}$ unless $z \in A$. Indeed, its factors don't go to $1$, which is a necessary condition exactly analogous to the condition for infinite series that the terms need to go to $0$. In fact one can switch between infinite sums and infinite products using the logarithm, which can be used to prove the following.

(First I need to mention that the theorem below requires the convention wherein a product which tends to $0$ is said to diverge. This is because the logarithm of such a product diverges to $-\infty$.)

Theorem: Let $a_n \in \mathbb{C}$ be a sequence such that $\sum |a_n|^2$ converges. Then $\prod (1 + a_n)$ converges if and only if $\sum a_n$ converges.

Sketch. Use the fact that $\log (1 + a_n) = a_n + O(|a_n|^2)$.

So we can make sense of the "infinite polynomial"

$$\prod_{\alpha \in A} \left( 1 - \frac{z}{\alpha} \right)$$

for countable $A$ such that $\sum_{\alpha \in A} \frac{1}{|\alpha|^2}$ and $\sum_{\alpha \in A} \frac{1}{\alpha}$ both converge. See also the Weierstrass factorization theorem.

Note that by the identity theorem, the zeroes of a holomorphic function are isolated, so if you want your product to be holomorphic with $A$ as its zero set, $A$ needs to be discrete.

Infinite sums and products do not behave well for uncountably many terms, the basic reason being the following.

Theorem: Let $S$ be an uncountable set of positive real numbers. Then for any positive real $r$, there is a finite subset of $S$ whose sum is greater than $r$.

(In other words, no sum with uncountably many terms can converge absolutely.)

Proof. The sets $S_{\epsilon} = \{ s : s \in S, s > \epsilon \}$ for $\epsilon$ a positive rational are a countable collection of sets whose union is $S$. Since a countable union of countable sets is countable, it follows that there exists $\epsilon$ such that $S_{\epsilon}$ is uncountable. Then the result is clear.

share|cite|improve this answer
Thanks for the correction, Qiaochu. Is that first theorem intended just to show that the euler example is wrong, or are you also saying it has stronger implications to the broader question? – Nick Alger Jun 21 '12 at 5:34
@Nick: well, it furnishes a pretty large class of examples. – Qiaochu Yuan Jun 21 '12 at 5:35
Ok, I see you're still editing. Interesting that you cant have an "uncountable polynomial", whereas you can have a linear operator with uncountable spectrum. Non-point spectrum, but still strange. – Nick Alger Jun 21 '12 at 5:38
So, thinking about the first theorem you wrote and considering the Riemann series theorem, I think unconditionally convergent "infinite polynomials" can be identified as the spectra of trace-class operators. – Nick Alger Jun 21 '12 at 5:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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