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,
is the q-exponential corresponding to the classical q-derivative while
are eigenfunctions of the Askey–Wilson operators.
The q-exponential is also known as the quantum dilogarithm.[1][2]
Definition
The q-exponential
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)))}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f224e5c169bd7d785bc27d1cb5c515c5416330d1)
where
is the q-factorial and
![{\displaystyle (q;q)_{n}=(1-q^{n})(1-q^{n-1})\cdots (1-q)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/352c9ce4fb89dad19bf549784ea3e9e2c0a6158f)
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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/11dad1ef314348eda0c93effeaf6a67d529b0303)
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}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c5454c7904c36ab46a23fd93c086675aa4f7f618)
Here,
is the q-bracket.
For other definitions of the q-exponential function, see Exton (1983), Ismail & Zhang (1994), Suslov (2003) harvtxt error: no target: CITEREFSuslov2003 (help) and Cieśliński (2011).
Relations
For
, a function that is closely related is
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 }.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/56a6aea7a04fa7e3772ec2f34973157cfe569785)
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}.~}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0a795619eef7485331dc44748925636b4adb7538)
Relation with Dilogarithm
has the following infinite product representation:
![{\displaystyle e_{q}(x)=\left(\prod _{k=0}^{\infty }(1-q^{k}(1-q)x)\right)^{-1}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/248cf42ca82ec385730f281d1d25aef37e5803a2)
On the other hand,
holds.
When
,
![{\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)).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/adeebd93d418e5c68eada4e8607dcf515f21a37f)
By taking the limit
,
![{\displaystyle \lim _{q\to 1}(1-q)\log e_{q}(x/(1-q))=\mathrm {Li} _{2}(x),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/91e1d34e69d61f3d23283a471906beade1915e8d)
where
is the dilogarithm.