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I'd like to know what $$\prod_{k=1}^n (1-x^k)$$

evaluates to (assuming there is a simple closed form) and what it "is" in the context of commutative algebra (of which I knew little and recall less).

I'm sure I've seen this in the past but don't know where to place it. LaTeX search doesn't help.

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    $\begingroup$ That's probably the simplest form. It's zero on roots of unity up to degree $n$. $\endgroup$
    – knucklebumpler
    Commented Mar 30, 2011 at 21:39
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    $\begingroup$ What does "what it 'is' in the context of commutative algebra" mean? It is a certain polynomial. I don't know what else there is to say. What do you want to know about it? $\endgroup$
    – Qiaochu Yuan
    Commented Mar 30, 2011 at 21:40
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    $\begingroup$ Looks like a finite mathworld.wolfram.com/q-PochhammerSymbol.html $\endgroup$
    – graveolensa
    Commented Mar 30, 2011 at 21:42
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    $\begingroup$ If it's got a name (e.g., "the Herp-Derp polynomial"), or other stuff that will help me find context for it online. $\endgroup$
    – S Huntsman
    Commented Mar 30, 2011 at 21:43
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    $\begingroup$ @Qiaochu: Assuming it's hard to evaluate in general, are there special values of $n$ for which it's easy to evaluate? $\endgroup$
    – S Huntsman
    Commented Mar 30, 2011 at 21:53

1 Answer 1

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Well, one has $$\prod_{n\geq1}(1-x^k) = \sum_{-\infty\leq n\leq\infty}(-1)^nx^{(3n^2-n)/2}.$$ This is a consequence of Jacobi's triple product identity.

You asked about a finite product, but from this equality you can tell what are the coefficients in the expanded finite product.

The context for this identities is, among others, the theory of partitions. I am sure you will find a proof of this in Andrews' excellent The Theory of Partitions, along with related information.

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  • $\begingroup$ Right, but the OP's question is about the finite product. I guess this tells us what the first $n$ coefficients are. $\endgroup$
    – Qiaochu Yuan
    Commented Mar 30, 2011 at 21:51
  • $\begingroup$ Indeed, I should note that. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Mar 30, 2011 at 21:52
  • $\begingroup$ Great, thanks. I don't have Andrews but I will see if it's in Stanley's books, which I have at home. $\endgroup$
    – S Huntsman
    Commented Mar 30, 2011 at 21:58
  • $\begingroup$ I see that equation 1.30 in Stanley gives the connection with partitions. In fact the reciprocal of the product that is in the title is what actually interested me, so this is quite convenient. Thanks again! $\endgroup$
    – S Huntsman
    Commented Mar 30, 2011 at 22:04
  • $\begingroup$ @SHuntsman I used the mentioned reciprocal to answer one of my questions on "Number of partitions of 2n with no element greater than n" here. Did you have a comparable problem? $\endgroup$
    – draks ...
    Commented Aug 3, 2012 at 20:33

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