Q-exponential

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

The analogue of ${\displaystyle \exp(x)\exp(y)=\exp(x+y)}$ does not hold for real numbers ${\displaystyle x}$ and ${\displaystyle y}$. However, if these are operators satisfying the commutation relation ${\displaystyle xy=qyx}$, then ${\displaystyle e_{q}(x)e_{q}(y)=e_{q}(x+y)}$ holds true.^[3]

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,

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

${\displaystyle e_{q}(x)}$ 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 ${\displaystyle |q|<1}$,

${\displaystyle {\begin{aligned}\log e_{q}(x)&=-\sum _{k=0}^{\infty }\log(1-q^{k}(1-q)x)\\&=\sum _{k=0}^{\infty }\sum _{n=1}^{\infty }{\frac {(q^{k}(1-q)x)^{n}}{n}}\\&=\sum _{n=1}^{\infty }{\frac {((1-q)x)^{n}}{(1-q^{n})n}}\\&={\frac {1}{1-q}}\sum _{n=1}^{\infty }{\frac {((1-q)x)^{n}}{[n]_{q}n}}\end{aligned}}.}$

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.

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