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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.

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  • $\begingroup$ Mightn't it make more sense to think about forward differences than about the derivative? $\endgroup$
    – LSpice
    Commented Feb 1, 2019 at 18:45
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    $\begingroup$ 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
    Commented Feb 1, 2019 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
    Commented Feb 2, 2019 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
    Commented Feb 2, 2019 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
    Commented Feb 2, 2019 at 15:10

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