# Euler function

(Redirected from Euler's function)
Jump to navigation Jump to search

In mathematics, the Euler function is given by

$\phi (q)=\prod _{k=1}^{\infty }(1-q^{k}).$ Named after Leonhard Euler, it is a model 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.

$\phi (e^{-\pi })={\frac {e^{\pi /24}\Gamma \left({\frac {1}{4}}\right)}{2^{7/8}\pi ^{3/4}}}$ $\phi (e^{-2\pi })={\frac {e^{\pi /12}\Gamma \left({\frac {1}{4}}\right)}{2\pi ^{3/4}}}$ $\phi (e^{-4\pi })={\frac {e^{\pi /6}\Gamma \left({\frac {1}{4}}\right)}{2^{{11}/8}\pi ^{3/4}}}$ $\phi (e^{-8\pi })={\frac {e^{\pi /3}\Gamma \left({\frac {1}{4}}\right)}{2^{29/16}\pi ^{3/4}}}({\sqrt {2}}-1)^{1/4}$ Using the Pentagonal number theorem, exchanging sum and integral, and then invoking complex-analytic methods, one derives

$\int _{0}^{1}\phi (q){\text{d}}q={\frac {8{\sqrt {\frac {3}{23}}}\pi \sinh \left({\frac {{\sqrt {23}}\pi }{6}}\right)}{2\cosh \left({\frac {{\sqrt {23}}\pi }{3}}\right)-1}}.$ 