q-exponential

Not to be confused with the Tsallis q-exponential.

In combinatorial mathematics, the q-exponential is a q-analog of the exponential function, namely the eigenfunction of the q-derivative

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 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, $~e_q(z) ~ e_{1/q} (-z) =1$ .

For $q<1$ , a function that is closely related is

e_q(z) = E_q(z(1-q)).

Here, $E_q(t)$ is a special case of the basic hypergeometric series:

E_q(z) = \;_{1}\phi_0 (0;q,z) = \prod_{n=0}^\infty \frac {1}{1-q^n z} ~.

F. H. Jackson (1908), On q-functions and a certain difference operator, Trans. Roy. Soc. Edin., 46 253-281.

Exton, H. (1983), q-Hypergeometric Functions and Applications, New York: Halstead Press, Chichester: Ellis Horwood, 1983, ISBN 0853124914, ISBN 0470274530, ISBN 978-0470274538

Gasper G., and Rahman, M. (2004), Basic Hypergeometric Series, Cambridge University Press, 2004, ISBN 0521833574