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[edit]

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), Suslov (2003) and Cieslinski (2011).

Properties[edit]

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}.

Addition Formula[edit]

If {\displaystyle xy=qyx}, {\displaystyle e_{q}(x)e_{q}(y)=e_{q}(x+y)} holds.

Relations[edit]

For {\displaystyle -1<q<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)=\;_{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,

{\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}.~}

Relation with Dilogarithm[edit]

{\displaystyle e_{q}(x)} has the following infinite product representation:

{\displaystyle e_{q}(x)=\left(\prod _{k=0}^{\infty }(1-q^{k}(1-q)x)\right)^{-1}.}

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

{\displaystyle \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}}.}

By taking the limit {\displaystyle q\to 1},

{\displaystyle \lim _{q\to 1}(1-q)\log e_{q}(x/(1-q))=\mathrm {Li} _{2}(x),}

where {\displaystyle \mathrm {Li} _{2}(x)} is the dilogarithm.

References[edit]

• 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.
• Ismail, M. E. H. & Zhang, R. (1994), “Diagonalization of certain integral operators,” Adv. Math. 108, 1–33.
• Ismail, M.E.H. Rahman, M. & Zhang, R. (1996), Diagonalization of certain integral operators II, J. Comp. Appl. Math. 68, 163-196.
• Jackson, F. H. (1908), “On q-functions and a certain difference operator”, Transactions of the Royal Society of Edinburgh, 46, 253-281.