# q-derivative

In mathematics, in the area of combinatorics, the q-derivative, or Jackson derivative, is a q-analog of the ordinary derivative, introduced by Frank Hilton Jackson. It is the inverse of Jackson's q-integration

## Definition

The q-derivative of a function f(x) is defined as

$\left(\frac{d}{dx}\right)_q f(x)=\frac{f(qx)-f(x)}{qx-x}.$

It is also often written as $D_qf(x)$. The q-derivative is also known as the Jackson derivative.

Formally, in terms of Lagrange's shift operator in logarithmic variables, it amounts to the operator

$D_q= \frac{1}{x} ~ \frac{q^{d~~~ \over d (\ln x)} -1}{q-1} ~,$

which goes to the plain derivative, →ddx, as q→1.

It is manifestly linear,

$\displaystyle D_q (f(x)+g(x)) = D_q f(x) + D_q g(x)~.$

It has product rule analogous to the ordinary derivative product rule, with two equivalent forms

$\displaystyle D_q (f(x)g(x)) = g(x)D_q f(x) + f(qx)D_q g(x) = g(qx)D_q f(x) + f(x)D_q g(x).$

Similarly, it satisfies a quotient rule,

$\displaystyle D_q (f(x)/g(x)) = \frac{g(x)D_q f(x) - f(x)D_q g(x)}{g(qx)g(x)},\quad g(x)g(qx)\neq 0.$

There is also a rule similar to the chain rule for ordinary derivatives. Let $g(x) = c x^k$. Then

$\displaystyle D_q f(g(x)) = D_{q^k}(f)(g(x))D_q(g)(x).$

The eigenfunction of the q-derivative is the q-exponential eq(x).

## Relationship to ordinary derivatives

Q-differentiation resembles ordinary differentiation, with curious differences. For example, the q-derivative of the monomial is:

$\left(\frac{d}{dz}\right)_q z^n = \frac{1-q^n}{1-q} z^{n-1} = [n]_q z^{n-1}$

where $[n]_q$ is the q-bracket of n. Note that $\lim_{q\to 1}[n]_q = n$ so the ordinary derivative is regained in this limit.

The n-th q-derivative of a function may be given as:

$(D^n_q f)(0)= \frac{f^{(n)}(0)}{n!} \frac{(q;q)_n}{(1-q)^n}= \frac{f^{(n)}(0)}{n!} [n]_q!$

provided that the ordinary n-th derivative of f exists at x=0. Here, $(q;q)_n$ is the q-Pochhammer symbol, and $[n]_q!$ is the q-factorial. If $f(x)$ is analytic we can apply the Taylor formula to the definition of $D_q(f(x))$ to get

$\displaystyle D_q(f(x)) = \sum_{k=0}^{\infty}\frac{(q-1)^k}{(k+1)!} x^k f^{(k+1)}(x).$

A q-analog of the Taylor expansion of a function about zero follows:

$f(z)=\sum_{n=0}^\infty f^{(n)}(0)\,\frac{z^n}{n!} = \sum_{n=0}^\infty (D^n_q f)(0)\,\frac{z^n}{[n]_q!}$