Q-exponential

Not to be confused with the Tsallis q-exponential.

In combinatorial mathematics, a q-exponential is a q-analog of the exponential function, namely the eigenfunction of a q-derivative. There are many q-derivatives, for example, the classical q-derivative, the Askey–Wilson operator, etc. Therefore, unlike the classical exponentials, q-exponentials are not unique. For example, ${\displaystyle e_{q}(z)}$ is the q-exponential corresponding to the classical q-derivative while ${\displaystyle {\mathcal {E}}_{q}(z)}$ are eigenfunctions of the Askey–Wilson operators.

The q-exponential is also known as the quantum dilogarithm.^[1][2]

Definition

The q-exponential ${\displaystyle e_{q}(z)}$ is defined as

${\displaystyle e_{q}(z)=\sum _{n=0}^{\infty }{\frac {z^{n}}{[n]!_{q}}}=\sum _{n=0}^{\infty }{\frac {z^{n}(1-q)^{n}}{(q;q)_{n}}}=\sum _{n=0}^{\infty }z^{n}{\frac {(1-q)^{n}}{(1-q^{n})(1-q^{n-1})\cdots (1-q)}}}$

where ${\displaystyle [n]!_{q}}$ is the q-factorial and

${\displaystyle (q;q)_{n}=(1-q^{n})(1-q^{n-1})\cdots (1-q)}$

is the q-Pochhammer symbol. That this is the q-analog of the exponential follows from the property

${\displaystyle \left({\frac {d}{dz}}\right)_{q}e_{q}(z)=e_{q}(z)}$

where the derivative on the left is the q-derivative. The above is easily verified by considering the q-derivative of the monomial

${\displaystyle \left({\frac {d}{dz}}\right)_{q}z^{n}=z^{n-1}{\frac {1-q^{n}}{1-q}}=[n]_{q}z^{n-1}.}$

Here, ${\displaystyle [n]_{q}}$ is the q-bracket. For other definitions of the q-exponential function, see Exton (1983), Ismail & Zhang (1994), and Cieśliński (2011).

Properties

For real ${\displaystyle q>1}$, the function ${\displaystyle e_{q}(z)}$ is an entire function of ${\displaystyle z}$. For ${\displaystyle q<1}$, ${\displaystyle e_{q}(z)}$ is regular in the disk ${\displaystyle |z|<1/(1-q)}$.

Note the inverse, ${\displaystyle ~e_{q}(z)~e_{1/q}(-z)=1}$.

Addition Formula

The analogue of ${\displaystyle \exp(x)\exp(y)=\exp(x+y)}$ does not hold for real numbers ${\displaystyle x}$ and ${\displaystyle y}$. However, if these are operators satisfying the commutation relation ${\displaystyle xy=qyx}$, then ${\displaystyle e_{q}(x)e_{q}(y)=e_{q}(x+y)}$ holds true.^[3]

Relations

For ${\displaystyle -1<q<1}$, a function that is closely related is ${\displaystyle E_{q}(z).}$ It is a special case of the basic hypergeometric series,

${\displaystyle E_{q}(z)=\;_{1}\phi _{1}\left({\scriptstyle {0 \atop 0}}\,;\,z\right)=\sum _{n=0}^{\infty }{\frac {q^{\binom {n}{2}}(-z)^{n}}{(q;q)_{n}}}=\prod _{n=0}^{\infty }(1-q^{n}z)=(z;q)_{\infty }.}$

Clearly,

${\displaystyle \lim _{q\to 1}E_{q}\left(z(1-q)\right)=\lim _{q\to 1}\sum _{n=0}^{\infty }{\frac {q^{\binom {n}{2}}(1-q)^{n}}{(q;q)_{n}}}(-z)^{n}=e^{-z}.~}$

Relation with Dilogarithm

${\displaystyle e_{q}(x)}$ has the following infinite product representation:

${\displaystyle e_{q}(x)=\left(\prod _{k=0}^{\infty }(1-q^{k}(1-q)x)\right)^{-1}.}$

On the other hand, ${\displaystyle \log(1-x)=-\sum _{n=1}^{\infty }{\frac {x^{n}}{n}}}$ holds. When ${\displaystyle |q|<1}$,

${\displaystyle {\begin{aligned}\log e_{q}(x)&=-\sum _{k=0}^{\infty }\log(1-q^{k}(1-q)x)\\&=\sum _{k=0}^{\infty }\sum _{n=1}^{\infty }{\frac {(q^{k}(1-q)x)^{n}}{n}}\\&=\sum _{n=1}^{\infty }{\frac {((1-q)x)^{n}}{(1-q^{n})n}}\\&={\frac {1}{1-q}}\sum _{n=1}^{\infty }{\frac {((1-q)x)^{n}}{[n]_{q}n}}\end{aligned}}.}$

By taking the limit ${\displaystyle q\to 1}$,

${\displaystyle \lim _{q\to 1}(1-q)\log e_{q}(x/(1-q))=\mathrm {Li} _{2}(x),}$

where ${\displaystyle \mathrm {Li} _{2}(x)}$ is the dilogarithm.