# 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.

## 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 Suslov (2003).

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

## Relations

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

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

Clearly,

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

## References

• Exton, H. (1983), q-Hypergeometric Functions and Applications, New York: Halstead Press, Chichester: Ellis Horwood, ISBN 0853124914, ISBN 0470274530, ISBN 978-0470274538
• Gasper, G. & Rahman, M. (2004), Basic Hypergeometric Series, Cambridge University Press, ISBN 0521833574
• Ismail, M. E. H. (2005), Classical and Quantum Orthogonal Polynomials in One Variable, Cambridge University Press.
• Jackson, F. H. (1908), "On q-functions and a certain difference operator", Transactions of the Royal Society of Edinburgh, 46, 253-281.