What is q-calculus ?



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<q<1 :

\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)


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.


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<r<1

\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}


This is one of the most famous books for undergraduates :

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


3 thoughts on “What is q-calculus ?

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s