# What is q-calculus ?

q-DERIVATIVE

Here is a nice diversion for anyone who knows what is the derivative of a simple function $f(x)$. The modern theory of differential and integral calculus began in the XVIIth century with the works of Newton and Leibniz. As it is well known, the derivative of a function $f(x)$ with respect to the variable x is by definition :

$\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$

Now, let us consider the following expression :

$\displaystyle f'(x) = \lim_{q \to 1} \frac{f(qx)-f(x)}{qx-x}$

Of course, this is not valid when $q=1$ or $x=0$ but otherwise this alternative formula is equivalent to the usual derivative. You can convince yourself by writing $\frac{f(x+(q-1)x)-f(x)}{(q-1)x}$, the term $(q-1)x$ playing the role of $h$.

At the beginning of the XXth century, F.H. Jackson studied this modified derivative and many of its consequences. The key concept is the q-derivative operator defined as follows when $0 :

$\displaystyle (D_q f)(x) = \frac{f(qx)-f(x)}{qx-x}$

This q-derivative can be applied to functions not containing $0$ in their domain of definition. Then it reduces to the ordinary derivative when $q$ goes to $1$ :

$\displaystyle \lim_{q \to 1} (D_q f)(x) = f'(x)$

Example : Compute the q-derivative of $x^2+2x+1$

$D_q(x^2+2x+1) = \frac{[(qx)^2+2(qx)+1]-[x^2+2x+1]}{qx-x} = (1+q)x+2$

One can easily check that the q-derivative operator is linear :

$\displaystyle D_q (f+g) = D_q f + D_q g$

$\displaystyle D_q (\lambda f) = \lambda D_q f$

The product rule is slighlty modified but it approaches the usual product rule when q goes to 1 :

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

q-INTEGERS

There is an intriguing relation between arithmetics and the q-derivative. Indeed, if we compute $D_q x^n = \frac{(qx)^n-x^n}{qx-x} = \frac{q^n-1}{q-1} x^{n-1}$ then we notice an analogy with the usual derivation formula $\frac{d(x^n)}{dx} = n x^{n-1}$. It can be made explicit by defining q-integers :

$\displaystyle \{n\}_q = 1 + q + ... + q^{n-1}$

$\displaystyle \{0\}_q = 0$

As you may remember from the courses on geometric series, we have $1 + q + ... + q^{n-1} = \frac{q^n-1}{q-1}$. Our previous computation thus becomes

$\displaystyle D_q x^n = \{n\}_q x^{n-1}$

We notice that q-integers coincide with usual integers when $q=1$, that is to say :

$\displaystyle \{n\}_1 = n$

Example : Recompute the q-derivative of $x^2+2x+1$ by using q-integers

$D_q(x^2+2x+1) = D_q(x^2)+2D_q(x)+D_q(1)$

$D_q(x^2+2x+1) = \{2\}_q x+2\{1\}_q+0 = \{2\}_q x+2$

Here is an application of q-integers to the general Leibniz rule :

$\displaystyle (D_q^n (f g))(x) = \sum_{k=0}^{n} {\binom{n}{k}}_q (D_q^k f)(q^{n-k} x) (D_q^{n-k} g)(x)$

where we have defined the q-binomial coefficient :

$\displaystyle {\binom{n}{k}}_q = \frac{\{n\}_q!}{\{k\}_q! \{n-k\}_q!}$

and the q-factorial :

$\displaystyle \{n\}_q! = \prod_{k=1}^{n} \{k\}_q$, $\{0\}_q! = 1$.

q-INTEGRAL

As we can expect, the q-derivative has a reciprocal operation which is the indefinite q-integral. It is given by the Jackson integral :

$\displaystyle \int f(x) d_q x = (1-q) x \sum_{k=0}^{\infty} q^k f(q^k x)$

If $F_q(x)$ denotes the q-integral of $f(x)$ then we can check that :

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

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

$\displaystyle (D_q F_q)(x) = - \left( \sum_{1}^{\infty} q^k f(q^k x) - q^0 f(q^0 x) - \sum_{1}^{\infty} q^k f(q^k x) \right)$

$\displaystyle (D_q F_q)(x) = f(x)$

Example : Compute $\int_0^1 x^2 d_q x$

$\int_0^1 x^2 d_q x = \left[ (1-q) x \sum_{0}^{\infty} q^k f(q^k x) \right]_0^1$

$\int_0^1 x^2 d_q x = (1-q) 1 \sum_0^\infty q^k (q^k 1)^2 - 0$

$\int_0^1 x^2 d_q x = (1-q) \sum_0^\infty (q^3)^k$

$\int_0^1 x^2 d_q x = \frac{1-q}{1-q^3}$ since $\sum_0^\infty r^k = \frac{1}{1-r}$ when $0

$\int_0^1 x^2 d_q x = \frac{1}{1+q+q^2}$

$\int_0^1 x^2 d_q x = \frac{1}{\{3\}_q}$

which is the q-analog of the usual integral $\int_0^1 x^2 d x = \frac{1}{3}$

COMPLEMENTS

This is one of the most famous books for undergraduates :

Victor Kac and Pokman Cheung, Quantum Calculus, Springer, 2001