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 $\mathcal_q(z)$ are eigenfunctions of the Askey-Wilson operators.

Definition

The ''q''-exponential $e_q(z)$ is defined as
:$e_q(z)= \sum_^\infty \frac = \sum_^\infty \frac = \sum_^\infty z^n\frac$

where _q is the ''q''-factorial and
:$(q;q)_n=(1-q^n)(1-q^)\cdots (1-q)$

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

:$\left(\frac\right)_q e_q(z) = e_q(z)$

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\right)_q z^n = z^ \frac
= q z^.

Here, 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, $~e_q(z) ~ e_ (-z) =1$.

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

:$E_(z)=\;_\phi_\left(\, ;\,z\right)=\sum_^\frac=\prod_^(1-q^z)=(z;q)_\infty.$

Clearly,
:$\lim_E_\left(z(1-q)\right)=\lim_\sum_^\frac (-z)^=e^ .~$

Relation with Dilogarithm

$e_q(x)$ has the following infinite product representation:
:$e_q(x)=\left(\prod_^\infty(1-q^k(1-q)x)\right)^.$
On the other hand, $\log(1-x)=-\sum_^\infty\frac$ holds.
When $, q, <1$,
:$\log e_q(x)=-\sum_^\infty\log(1-q^k(1-q)x)=\sum_^\infty\sum_^\infty\frac=\sum_^\infty\frac=\frac\sum_^\infty\frac.$
By taking the limit $q\to 1$,
:$\lim_(1-q)\log e_q(x/(1-q))=\mathrm_2(x),$
where $\mathrm_2(x)$ is the dilogarithm.

In physics

The Q-exponential function is also known as the quantum dilogarithm.

References

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{{DEFAULTSORT:Q-Exponential
Q-analogs
Exponentials