Q-exponential

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 other definitions of the q-exponential function, see , , and .

Properties

For real q1, the function e_q(z) is an entire function of z. For q, e_q(z) is regular in the disk |z|.

Note the inverse, ~e_q(z) ~ e_{1/q} (-z) =1.

Addition Formula

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

Relations

For -1, a function that is closely related is E_q(z). It is a special case of the basic hypergeometric series,

: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\to1}E_{q}\left(z(1-q)\right)=\lim_{q\to1}\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) has the following infinite product representation: :e_q(x)=\left(\prod_{k=0}^\infty(1-q^k(1-q)x)\right)^{-1}. On the other hand, \log(1-x)=-\sum_{n=1}^\infty\frac{x^n}{n} holds. When |q|,

:\begin{align} \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]_qn} \end{align}.

By taking the limit q\to 1, :\lim_{q\to 1}(1-q)\log e_q(x/(1-q))=\mathrm{Li}_2(x), where \mathrm{Li}_2(x) is the dilogarithm.

References

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References

1. Zudilin, Wadim. (14 March 2006). "Quantum dilogarithm".
2. (1994-02-20). "Quantum dilogarithm". Modern Physics Letters A.
3. (2011). "Quantum Calculus". Springer.