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, $$e_q(z)$$ is the q-exponential corresponding to the classical q-derivative while  $$\mathcal{E}_q(z)$$ 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}^\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 $$[n]!_q$$ 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, , and.

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,


 * $$E_{q}(z)=\;_{1}\phi_{1}\left({\scriptstyle{0\atop 0}}\, ;\,z\right)=\sum_{n=0}^{\infty}\frac{q^{\binom{n}{2}}(-z)^{n}}{(q;q)_{n}}=\prod_{n=0}^{\infty}(1-q^{n}z)=(z;q)_\infty. $$

Clearly,
 * $$\lim_{q\to1}E_{q}\left(z(1-q)\right)=\lim_{q\to1}\sum_{n=0}^{\infty}\frac{q^{\binom{n}{2}}(1-q)^{n}}{(q;q)_{n}}

(-z)^{n}=e^{-z} .~ $$

Relation with Dilogarithm
$$e_q(x)$$ has the following infinite product representation:
 * $$e_q(x)=\left(\prod_{k=0}^\infty(1-q^k(1-q)x)\right)^{-1}. $$

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


 * $$\begin{align}

\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]_qn} \end{align}.$$

By taking the limit $$q\to 1$$,
 * $$\lim_{q\to 1}(1-q)\log e_q(x/(1-q))=\mathrm{Li}_2(x), $$

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