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

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} ~.$$

References

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

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