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Question first:

Show that if $s_1 < s_2 < \dots$ is an increasing sequence of positive integers and $P(x)$ is a nonzero polynomial then we cannot have $$ P(x) \equiv \prod_{j=1}^\infty (1 - x^{s_j}) $$ as formal series.

The right-hand side really means $\lim_{N \to \infty} \prod_{j=1}^N (1 - x^{s_j})$ where the notion of convergence is as described in this math.SE answer by Bill.

The intuition is that if we take $x \to 1^-$ then the left-hand side tends to zero at a slower rate than the right-hand side, since $P$ has the root $1$ with only finite multiplicity, while every factor on the RHS tends to zero. However, since formal and functional power series aren't actually the same thing (see link above), I'm not sure how to make this precise, even after formally inverting the $1-x$ factors on the left-hand side (by multiplying both sides by $(1+x+x^2+\dots+)^k$).

Does anyone know the correct incantations?

(For context, this came up in a solution to a recent IMO proposal. Since it was for high school students I swept the convergence issues under the rug, but myself I'd like to know exactly what the right thing to do is.)

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    $\begingroup$ If they are equal then the constant term of $P(x)$ is $1$, so it can be factored as $\prod_{i=1}^m (1 - \alpha_iX)$ over its reciprocal roots in $\mathbf C$. Every formal power series in $1 + X{\mathbf C}[[X]]$ has a unique decomposition as a product $\prod_{k\geq 1} (1 + a_kX^k)$ with $a_k \in \mathbf C$ (you can solve for the $a_k$'s recursively). In particular, $P(X)$ arises like this with a product having finitely many terms, so it is not also such a product with infinitely many terms. $\endgroup$
    – KConrad
    Commented Jul 7, 2016 at 20:33
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    $\begingroup$ Why must $P(X)$ be a finite product? For example, $P(X) := 1+X+X^2$ doesn't appear to be such a finite product, unless I'm missing something. (It appears $P(X) = (1+X)(1+X^2)(1-X^3)\dots$.) $\endgroup$
    – Evan Chen
    Commented Jul 7, 2016 at 20:48
  • $\begingroup$ Ah, I see my error: the unique decomposition I described has terms $1 + a_kX^k$ with distinct $k$, but when you factor a polynomial over its reciprocal roots you will have parts all of degree 1, e.g., $1-X^2 = (1-X)(1+X)$ is not a decomposition into the form $\prod_{k \geq 1} (1 + a_kX^k)$. $\endgroup$
    – KConrad
    Commented Jul 7, 2016 at 23:48

2 Answers 2

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Equivalently, we'll show that we cannot have

$$\frac{1}{P(x)} = \frac{1}{\prod_{j=1}^{\infty} (1 - x^{s_j})}$$

as formal power series. The idea is that the LHS has a pole of finite order at $x = 1$ while the RHS has an essential singularity at $x = 1$. Precisely, the coefficients on the LHS have asymptotic growth a polynomial times an exponential. On the other hand, the coefficients of the RHS can be shown to have growth both strictly larger than any polynomial (by truncating the product) and strictly smaller than any exponential (by comparing to the growth rate in the case where $s_j = j$, which is known). So the two rates of growth can't match.

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Just define the RHS as $\sum c_ix^i$, where $$ c_i = \sum_{\substack{i_1 < \dots < i_k \\ s_{i_1} + \dots s_{i_k} = i}} (-1)^k. $$ Then $c_i$ is a finite sum. It's clear that $|c_i| \le p(i) \ll (1+\epsilon)^i$ for all $\epsilon > 0$, where $p(i)$ is the partition function. Therefore, the RHS as defined converges absolutely for all $|x| < 1.$ This allows us to substitute values $x = 1 - \delta_i$ for $\delta_i \rightarrow 0$, and compute the RHS as written. As $x \rightarrow 1$, the RHS $\rightarrow 0$, so $P(x) \rightarrow 0$, so $P(1) = 0.$ Let $P(x) = (1-x)Q(x).$ Considering only values $|x| < 1$ still, we can multiply both sides by $(1+x+\dots)$ formally, and continue taking limits as $x \rightarrow 1.$ Repeating this argument, we can finish.

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