# Euler function

(Redirected from Euler's function)
Modulus of phi on the complex plane, colored so that black=0, red=4
For other meanings, see List of topics named after Leonhard Euler.

In mathematics, the Euler function is given by

$\phi(q)=\prod_{k=1}^\infty (1-q^k).$

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

## Properties

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

$\frac{1}{\phi(q)}=\sum_{k=0}^\infty p(k) q^k$

where $p(k)$ is the partition function of k.

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

$\phi(q)=\sum_{n=-\infty}^\infty (-1)^n q^{(3n^2-n)/2}.$

Note that $(3n^2-n)/2$ is a pentagonal number.

The Euler function is related to the Dedekind eta function through a Ramanujan identity as

$\phi(q)= q^{-\frac{1}{24}} \eta(\tau)$

where $q=e^{2\pi i\tau}$ is the square of the nome.

Note that both functions have the symmetry of the modular group.

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

$\phi(q)=(q;q)_\infty$

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:

$\ln(\phi(q))=-\sum_{n=1}^\infty\frac{1}{n}\,\frac{q^n}{1-q^n}$

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

$\ln(\phi(q))=\sum_{n=1}^\infty b_n q^n$

where

$b_n=-\sum_{d|n}\frac{1}{d}=$ -[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 following identity,

$\sum_{d|n} d = \sum_{d|n} \frac n d$

this may also be written as

$\ln(\phi(q))=-\sum_{n=1}^\infty \frac{q^n}{n} \sum_{d|n} d$

## Special values

The next identities come from Ramanujan's lost notebook, Part V, p. 326.

$\phi(e^{-\pi})=\frac{e^{\pi/24}\Gamma\left(\frac14\right)}{2^{7/8}\pi^{3/4}}$
$\phi(e^{-2\pi})=\frac{e^{\pi/12}\Gamma\left(\frac14\right)}{2\pi^{3/4}}$
$\phi(e^{-4\pi})=\frac{e^{\pi/6}\Gamma\left(\frac14\right)}{2^{{11}/8}\pi^{3/4}}$
$\phi(e^{-8\pi})=\frac{e^{\pi/3}\Gamma\left(\frac14\right)}{2^{29/16}\pi^{3/4}}(\sqrt{2}-1)^{1/4}$