Euler function

${\displaystyle \varphi (q)=\prod _{k=1}^{\mathrm{\infty }}(1-q^k),|q|<1.}$

Named after Leonhard Euler, it is a model example of a q-series and provides the prototypical example of a relation between combinatorics and complex analysis.

Properties

The coefficient ${\displaystyle p(k)}$ in the formal power series expansion for ${\displaystyle 1/\varphi (q)}$ gives the number of partitions of k. That is,

${\displaystyle \frac{1}{\varphi (q)}=\sum _{k=0}^{\mathrm{\infty }}p(k)q^k}$

where ${\displaystyle p}$ is the partition function.

The Euler identity, also known as the Pentagonal number theorem, is

${\displaystyle \varphi (q)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}(-1{)}^n{q}^{(3n^2-n)/2}.}$

${\displaystyle (3n^2-n)/2}$ is a pentagonal number.

The Euler function is related to the Dedekind eta function as

${\displaystyle \varphi (e^{2\pi i\tau })=e^{-\pi i\tau /12}\eta (\tau ).}$

The Euler function may be expressed as a q-Pochhammer symbol:

${\displaystyle \varphi (q)=(q;q{)}_{\mathrm{\infty }}.}$

The logarithm of the Euler function is the sum of the logarithms in the product expression, each of which may be expanded about q = 0, yielding

${\displaystyle \ln (\varphi (q))=-\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}\frac{q^n}{1-q^n},}$

which is a Lambert series with coefficients -1/n. The logarithm of the Euler function may therefore be expressed as

${\displaystyle \ln (\varphi (q))=\sum _{n=1}^{\mathrm{\infty }}{b}_n{q}^n}$

where ${\displaystyle b_n=-\sum _{d|n}\frac{1}{d}=}$ -[1/1, 3/2, 4/3, 7/4, 6/5, 12/6, 8/7, 15/8, 13/9, 18/10, ...] (see OEIS A000203)

On account of the identity ${\displaystyle \sigma (n)=\sum _{d|n}d=\sum _{d|n}\frac{n}{d}}$ , where ${\displaystyle \sigma (n)}$ is the sum-of-divisors function, this may also be written as

${\displaystyle \ln (\varphi (q))=-\sum _{n=1}^{\mathrm{\infty }}\frac{\sigma (n)}{n}{q}^n}$.

Also if ${\displaystyle a,b\in {\mathbb{R}}^{+}}$ and ${\displaystyle ab={\pi }^2}$, then

${\displaystyle a^{1/4}{e}^{-a/12}\varphi (e^{-2a})=b^{1/4}{e}^{-b/12}\varphi (e^{-2b}).}$