8
$\begingroup$

It is not hard to find easy formulas for the derivative of the function $\binom{x}{k}$, for instance it is not too hard to see (for $k$ an integer) that

$\frac{d}{dx} \binom{x}{k} = \sum_{i=1}^k \frac{1}{i} \binom{x-i}{k-i}.$

Now I was wondering if there is a nice q-analog of this formula. So if I assume that $\frac{d}{dx}$ is still the right thing to so and we write $\frac{d}{dx} \binom{x}{k}_q$ as a linear combination of $\binom{x-i}{k-i}_q$'s, then we have

$\frac{d}{dx} \binom{x}{1}_q = \frac{q^x \log(q)}{q-1} \binom{x-1}{0}_q.$

$\frac{d}{dx} \binom{x}{2}_q = \frac{q^x \log(q)}{q(q^2-1)} \left[ 2q \binom{x-1}{1}_q + \binom{x-2}{0}_q \right].$

$\frac{d}{dx} \binom{x}{3}_q = \frac{q^x \log(q)}{(q^2-1)(q^3-1)} \cdot \left[ 3(q^2-1) \binom{x-1}{2}_q + (q-1) \binom{x-2}{1}_q+ q^{-3} (2q^x+q^3-2q^2-q) \binom{x-3}{0}_q \right].$

This seems to turn out ugly. For $q \rightarrow 1$ it does not even seem to have the right limit for $k=3$ which is a property that a "nice q-analog" should have. So at this point I concluded that this is not the right generalization (if there is any). The derivative of $\binom{x}{k}$ is used very often, so I feel like that someone should have thought about this before me.

$\endgroup$
  • $\begingroup$ Mightn't it make more sense to think about forward differences than about the derivative? $\endgroup$ – LSpice Feb 1 at 18:45
  • 5
    $\begingroup$ There is such a thing as a $q$-derivative; have you tried that? That said, I'm not very keen on the meaningfulness of Gaussian binomial coefficients with non-integers on the top -- those lead to non-integer exponents, and at that point it's not clear what ring we are working in anymore. Puiseaux series? $\endgroup$ – darij grinberg Feb 1 at 20:18
  • $\begingroup$ Why forward difference? Originally, I was wondering about $\left( \binom{x}{k} - \binom{y}{k} \right) / (x-y)$ for the difference of $x$ and $y$ being very small. That is the derivative. For forward differences it is much easier. Googling gave me q-derivative, but that does not seem to give something nice. All of that said, I am very open to not looking at the derivative. My original question is more accurately described by looking at $\left( \binom{x}{k}_q - \binom{y}{k}_q \right)/(\binom{x}{1}_q - \binom{y}{1}_q)$ for $|x-y|$ small. $\endgroup$ – Ratio Bound Feb 2 at 14:03
  • $\begingroup$ Not sure if this is meaningful, but for a family $\mathcal{F}$ of $k$-sets in $\{ 1, \ldots, n \}$, it is sometimes convenient to write $|\mathcal{F}| = \binom{x}{k}$ for comparing different families. If we look at $k$-spaces of $\mathbb{F}_q^n$ instead, then writing $|\mathcal{F}| = \binom{x}{k}_q$ seems to be a convenient thing to do as well. $\endgroup$ – Ratio Bound Feb 2 at 14:21
  • $\begingroup$ Maybe $\left( \binom{x}{k}_q - \binom{y}{k}_q \right)/\left( \binom{x}{1}_q - \binom{y}{1}_q \right)$ is indeed the right thing instead of what I suggested first. For $k=2$ we obtain $\frac{2}{q+1} \binom{x-1}{1}_q + \frac{1}{q^2+q} \binom{x-2}{0}_q$ as limit, which looks good. For $k=3$, we obtain $\frac{3}{q^2+q+1} \binom{x-1}{2}_q + \frac{q^2+q-2}{q(q^2-1)(q^2+q+1)} \binom{x-2}{1}_q + \frac{1}{q^2 (q^2+q+1)} \binom{x-3}{0}_q$. And here the limit for $q \rightarrow 1$ does the right thing. So I suppose that I have nearly answered my own question (a bit too fast, should not have asked). $\endgroup$ – Ratio Bound Feb 2 at 15:10

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.