Euler function

Domain coloring plot of ϕ on the complex plane

Euler function $\varphi (x)$.

In mathematics, the Euler function is given by

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

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

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

where $p$ is the partition function.

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

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

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

The Euler function is related to the Dedekind eta function as

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

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

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

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

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

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

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

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

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

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

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

Special values

The next identities come from Ramanujan's Notebooks:

$\varphi (e^{-\pi })=\frac{e^{\pi /24}\mathrm{\Gamma }\left(\frac{1}{4}\right)}{2^{7/8}{\pi }^{3/4}}$

$\varphi (e^{-2\pi })=\frac{e^{\pi /12}\mathrm{\Gamma }\left(\frac{1}{4}\right)}{2{\pi }^{3/4}}$

$\varphi (e^{-4\pi })=\frac{e^{\pi /6}\mathrm{\Gamma }\left(\frac{1}{4}\right)}{2^{11/8}{\pi }^{3/4}}$

$\varphi (e^{-8\pi })=\frac{e^{\pi /3}\mathrm{\Gamma }\left(\frac{1}{4}\right)}{2^{29/16}{\pi }^{3/4}}(\sqrt{2}-1{)}^{1/4}$

Using the Pentagonal number theorem, exchanging sum and integral, and then invoking complex-analytic methods, one derives

${\int }_0^1\varphi (q)\mathrm{d}q=\frac{8\sqrt{\frac{3}{23}}\pi \sinh \left(\frac{\sqrt{23}\pi }{6}\right)}{2\cosh \left(\frac{\sqrt{23}\pi }{3}\right)-1}.$

References

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