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Euler's Partition Product Is a q-Series Identity by Vicfred

tags: mathematics, modular forms, q-series, partitions

The reciprocal of the Euler product generates the partition numbers. The formulas are more useful when each symbol has a job rather than merely decorating the theorem.

Notation

A modular form \(f\) on the upper half-plane transforms predictably under fractional linear maps and has a Fourier expansion in \(q=e^{2\pi i\tau}\). Its coefficients often encode arithmetic.

$$ \sum_{n\ge0}p(n)q^n=\prod_{m\ge1}(1-q^m)^{-1} $$

The definition determines which expressions are legal; only then does the identity become meaningful. An equality in \(\mathcal A\) may change ambient meaning, so I keep \(\mathsf D\) separate from \(\mathsf C\).

$$ (q;q)_\infty=\prod_{m\ge1}(1-q^m) $$

Stress the formula

The computation below is not a second theorem. It is a checksum for the definitions and a place to inspect the difficult LaTeX at full size.

$$ \frac1{(q;q)_\infty}=1+q+2q^2+3q^3+5q^4+7q^5+\cdots $$

Interpretation

The formula is reusable precisely because it says which pieces are structural and which belong only to the worked example.

$$ \begin{aligned} \mathsf{D}\;&:\quad \sum_{n\ge0}p(n)q^n=\prod_{m\ge1}(1-q^m)^{-1},\\[5pt] \mathsf{C}\;&:\quad (q;q)_\infty=\prod_{m\ge1}(1-q^m). \end{aligned} $$

Limit of the argument

Weight, level, character, and cusp conditions are part of the definition. A formal \(q\)-series is not automatically a modular form.

$$ \boxed{\begin{gathered} \text{compact conclusion}\\[-2pt] (q;q)_\infty=\prod_{m\ge1}(1-q^m) \end{gathered}} $$

With the dependency made explicit, the same pattern can be recognised safely in nearby problems. A changed hypothesis should now be easy to spot.

This article was posted on Tue 02 July 2013. Facts and circumstances may have changed since publication.
Please contact me before jumping to conclusions if something seems wrong or unclear.