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 ) {displaystyle e_{q}(z)} is the q-exponential corresponding to the classical q-derivative while E q ( z ) {displaystyle {mathcal {E}}_{q}(z)} are eigenfunctions of the Askey-Wilson operators. 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 ) {displaystyle e_{q}(z)} is the q-exponential corresponding to the classical q-derivative while E q ( z ) {displaystyle {mathcal {E}}_{q}(z)} are eigenfunctions of the Askey-Wilson operators. The q-exponential e q ( z ) {displaystyle e_{q}(z)} is defined as where [ n ] q ! {displaystyle _{q}!} is the q-factorial and is the q-Pochhammer symbol. That this is the q-analog of the exponential follows from the property where the derivative on the left is the q-derivative. The above is easily verified by considering the q-derivative of the monomial Here, [ n ] q {displaystyle _{q}} is the q-bracket.For other definitions of the q-exponential function, see Exton (1983), Ismail & Zhang (1994), Suslov (2003) and Cieslinski (2011). For real q > 1 {displaystyle q>1} , the function e q ( z ) {displaystyle e_{q}(z)} is an entire function of z {displaystyle z} . For q < 1 {displaystyle q<1} , e q ( z ) {displaystyle e_{q}(z)} is regular in the disk | z | < 1 / ( 1 − q ) {displaystyle |z|<1/(1-q)} . Note the inverse,   e q ( z )   e 1 / q ( − z ) = 1 {displaystyle ~e_{q}(z)~e_{1/q}(-z)=1} . If x y = q y x {displaystyle xy=qyx} , e q ( x ) e q ( y ) = e q ( x + y ) {displaystyle e_{q}(x)e_{q}(y)=e_{q}(x+y)} holds.