q-exponential

The term q-exponential occurs in two contexts. The q-exponential distribution, based on the Tsallis q-exponential is discussed in elsewhere.

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, $e_q(z)$ is the q-exponential corresponding to the classical q-derivative while ${{ℰ}}_q(z)$ are eigenfunctions of the Askey–Wilson operators.

The q-exponential is also known as the quantum dilogarithm.

Definition

The q-exponential $e_q(z)$ is defined as

$e_q(z)=\sum _{n=0}^{\mathrm{\infty }}\frac{z^n}{[n]{!}_q}=\sum _{n=0}^{\mathrm{\infty }}\frac{z^n(1-q{)}^n}{(q;q{)}_n}=\sum _{n=0}^{\mathrm{\infty }}{z}^n\frac{(1-q{)}^n}{(1-q^n)(1-q^{n-1})\cdots (1-q)}$

where $[n]{!}_q$ is the q-factorial and

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

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

${\left(\frac{d}{dz}\right)}_q{e}_q(z)=e_q(z)$

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

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

Here, $[n{]}_q$ is the q-bracket. For other definitions of the q-exponential function, see Template:Harvtxt, Template:Harvtxt, and Template:Harvtxt.

Properties

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

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

Addition Formula

The analogue of $\exp (x)\exp (y)=\exp (x+y)$ does not hold for real numbers $x$ and $y$. However, if these are operators satisfying the commutation relation $xy=qyx$, then $e_q(x)e_q(y)=e_q(x+y)$ holds true.

Relations

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

$E_q(z)={}_1{\varphi }_1(\genfrac{}{}{0ex}{}{0}{0};z)={\sum }_{n=0}^{\mathrm{\infty }}\frac{q^{(\genfrac{}{}{0ex}{}{n}{2})}(-z{)}^n}{(q;q{)}_n}={\prod }_{n=0}^{\mathrm{\infty }}(1-q^{n}z)=(z;q{)}_{\mathrm{\infty }}.$

Clearly,

$\underset{q\to 1}{\lim }{E}_q(z(1-q))=\underset{q\to 1}{\lim }\sum _{n=0}^{\mathrm{\infty }}\frac{q^{(\genfrac{}{}{0ex}{}{n}{2})}(1-q{)}^n}{(q;q{)}_n}(-z{)}^n=e^{-z}.$

Relation with Dilogarithm

$e_q(x)$ has the following infinite product representation:

$e_q(x)={(\prod _{k=0}^{\mathrm{\infty }}(1-q^k(1-q)x))}^{-1}.$

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

$\begin{array}{rl}\log e_q(x)& =-\sum _{k=0}^{\mathrm{\infty }}\log (1-q^k(1-q)x)\\ & =\sum _{k=0}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}\frac{(q^k(1-q)x{)}^n}{n}\\ & =\sum _{n=1}^{\mathrm{\infty }}\frac{((1-q)x{)}^n}{(1-q^n)n}\\ & =\frac{1}{1-q}\sum _{n=1}^{\mathrm{\infty }}\frac{((1-q)x{)}^n}{[n{]}_{q}n}\end{array}.$

By taking the limit $q\to 1$,

$\underset{q\to 1}{\lim }(1-q)\log e_q(x/(1-q))={\mathrm{L}\mathrm{i}}_2(x),$

where ${\mathrm{L}\mathrm{i}}_2(x)$ is the dilogarithm.