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I'm working on a problem (specifically, I'm using an exam paper without course notes to prepare for a course starting in September),

Define the partition function $P(q)$ and give its infinite product expression.

Wikipedia has

The generating function for $p(n)$ is given by $$\sum_{n=0}^\infty p(n)x^n=\prod_{k=1}^\infty\left(\frac{1}{1-x^k}\right)$$

But how does this allow us to evalute $p(n)$?

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    $\begingroup$ $\frac{1}{1-x^k} = 1+x^k+x^{2k}+x^{3k}+\cdots$. As an example, to find the number of partitions of $0,1,2,3,4$, expand $(1+x+x^2+x^3+x^4+\cdots)(1+x^2+x^4+\cdots)(1+x^3+\cdots)(1+x^4+\cdots)(1+\cdots)$ to get $1+x+2x^2+3x^3+5x^4+\cdots$ so the answers are $1,1,2,3,5$ respectively $\endgroup$
    – Henry
    Commented Jun 30, 2022 at 22:17
  • $\begingroup$ @Henry But we want to find a function of $n$ and now we're working with only $x$? And when I put $x=0$ into $1+x+2x^2+\dots$ I get $1$ (as expected), but when I put $x=1$ in it diverges. What am I missing? $\endgroup$
    – mjc
    Commented Jun 30, 2022 at 22:26
  • $\begingroup$ When you put $x=1$ into $\sum_{n=0}^\infty p(n)x^n$ you get $\sum_{n=0}^\infty p(n)$ which counts every partition of every natural number... it is $\infty$, as it ought to be. $\endgroup$
    – Brian Moehring
    Commented Jun 30, 2022 at 22:30
  • $\begingroup$ @BrianMoehring OK; so, back to my original question: how does this allow us to evaluate $p(n)$ for a particular $n$? $\endgroup$
    – mjc
    Commented Jun 30, 2022 at 22:32
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    $\begingroup$ Formal way: take the $n$th derivative of the sum or product at $x=0$ and divide that by $n!$; clearly with the sum that gives $p(n)$, and doing it to the product gives the numerical value. Easy method: look at the coefficient of $x^n$ in the expansion. $\endgroup$
    – Henry
    Commented Jun 30, 2022 at 22:37

1 Answer 1

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From comments as requested:

$\frac{1}{1-x^k} = 1+x^k+x^{2k}+x^{3k}+\cdots$.

As an example, to find the number of partitions of $0$, $1$, $2$, $3$, and $4$, expand $(1+x+x^2+x^3+x^4+\cdots)(1+x^2+x^4+\cdots)(1+x^3+\cdots)(1+x^4+\cdots)(1+\cdots)$ to get $1+x+2x^2+3x^3+5x^4+\cdots$ so the answers are $1$, $1$, $2$, $3$, and $5$ respectively.

  • The formal way to use generating functions: take the $n$th derivative of the sum or product at $x=0$ and divide that by $n!$; clearly with the sum that gives $p(n)$, and doing it to the product gives the numerical value.

  • Easy method: look at the coefficient of $x^n$ in the expansion.

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