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q-exponential

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Not to be confused with the Tsallis 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.

The q-exponential is also known as the quantum dilogarithm.

Definition

The q-exponential e q ( z ) {\displaystyle e_{q}(z)} is defined as

e q ( z ) = ∑ n = 0 ∞ z n [ n ] q ! = ∑ n = 0 ∞ z n ( 1 − q ) n ( q ; q ) n = ∑ n = 0 ∞ z n ( 1 − q ) n ( 1 − q n ) ( 1 − q n − 1 ) ⋯ ( 1 − q ) {\displaystyle e_{q}(z)=\sum _{n=0}^{\infty }{\frac {z^{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 {\displaystyle !_{q}} is the q-factorial and

( q ; q ) n = ( 1 − q n ) ( 1 − q n − 1 ) ⋯ ( 1 − q ) {\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

( d d z ) q e q ( z ) = e q ( z ) {\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

( d d z ) q z n = z n − 1 1 − q n 1 − q = [ n ] q z n − 1 . {\displaystyle \left({\frac {d}{dz}}\right)_{q}z^{n}=z^{n-1}{\frac {1-q^{n}}{1-q}}=_{q}z^{n-1}.}

Here, [ n ] q {\displaystyle _{q}} is the q-bracket. For other definitions of the q-exponential function, see Exton (1983), Ismail & Zhang (1994), and Cieśliński (2011).

Properties

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} .

Addition Formula

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

Relations

For − 1 < q < 1 {\displaystyle -1<q<1} , a function that is closely related is E q ( z ) . {\displaystyle E_{q}(z).} It is a special case of the basic hypergeometric series,

E q ( z ) = 1 ϕ 1 ( 0 0 ; z ) = ∑ n = 0 ∞ q ( n 2 ) ( − z ) n ( q ; q ) n = ∏ n = 0 ∞ ( 1 − q n z ) = ( z ; q ) ∞ . {\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,

lim q → 1 E q ( z ( 1 − q ) ) = lim q → 1 ∑ n = 0 ∞ q ( n 2 ) ( 1 − q ) n ( q ; q ) n ( − z ) n = e − z .   {\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

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

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

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

log ⁡ e q ( x ) = − ∑ k = 0 ∞ log ⁡ ( 1 − q k ( 1 − q ) x ) = ∑ k = 0 ∞ ∑ n = 1 ∞ ( q k ( 1 − q ) x ) n n = ∑ n = 1 ∞ ( ( 1 − q ) x ) n ( 1 − q n ) n = 1 1 − q ∑ n = 1 ∞ ( ( 1 − q ) x ) n [ n ] q n . {\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}}{_{q}n}}\end{aligned}}.}

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

lim q → 1 ( 1 − q ) log ⁡ e q ( x / ( 1 − q ) ) = L i 2 ( x ) , {\displaystyle \lim _{q\to 1}(1-q)\log e_{q}(x/(1-q))=\mathrm {Li} _{2}(x),}

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

References

  1. Zudilin, Wadim (14 March 2006). "Quantum dilogarithm" (PDF). wain.mi.ras.ru. Retrieved 16 July 2021.
  2. Faddeev, L.d.; Kashaev, R.m. (1994-02-20). "Quantum dilogarithm". Modern Physics Letters A. 09 (5): 427–434. arXiv:hep-th/9310070. Bibcode:1994MPLA....9..427F. doi:10.1142/S0217732394000447. ISSN 0217-7323. S2CID 119124642.
  3. Kac, V.; Cheung, P. (2011). Quantum Calculus. Springer. p. 31. ISBN 978-1461300724.
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