Trying a bit of combinatorics this winter break, and I don't understand a certain claim.
The claim is that for each $k$ there is a unique polynomial $P_k(x)$ of degree $k$ whose coefficients are in $\mathbb{Q}(q)$, the field of rational functions, such that $P_k(q^n)=\binom{n}{k}_q$ for all $n$.
Here $\binom{n}{k}_q$ is the $q$-binomial coefficient. I guess what is mostly troubling me is that $P_k(q^n)$ is a polynomial in $q^n$. I'm sure it's obvious, but why is the claim true? Thanks.
Best Answer
I think it could go something like this (but I'll bet there's also an inductive proof or even a more direct and beautiful formula): $$ [n]_q = \sum_{i=0}^{n-1}\;q^i = \frac{1-q^n}{1-q} = f(q^n) \qquad\text{for}\qquad 0 < n \in \mathbb{N} \qquad\text{and}\qquad f(x) = \frac{1-x}{1-q} \in \mathbb{Q}(q)[x] $$ Now if we note that $[0]_q=0$ and that $[a+b]_q=[a]_q+q^a[b]_q$ for $a,b\in\mathbb{N}$, then for $a=n-i$ and $b=1+i$, it follows that $[n-i]_q=[n]_q-q^{n-i}[1+i]_q$ for $0 \leq i \leq k < n$, so that $$ \binom{n}{k}_q = \prod_{i=0}^{k-1}\frac{[n-i]_q}{[1+i]_q} = \prod_{i=0}^{k-1}\frac{[n]_q-q^{n-i}[1+i]_q}{[1+i]_q} = P_k(q^n) %\qquad\text{for}\qquad %[m]_q! = \prod_{h=1}^{m} [h]_q $$ for $$ P_k(x) =\prod_{i=0}^{k-1}\frac{f(x)-x\cdot c_i q^{-i}}{c_i} =\prod_{i=0}^{k-1}\left(\frac{f(x)}{c_i}-\frac{x}{q^i}\right) \in \mathbb{Q}(q)[x] $$ since each $c_i=[1+i]_q\in\mathbb{Q}(q)$, and that, furthermore, $deg(P_k)=k$ since each term in the above product is linear. This simplifies to $$ P_k(x) = \prod_{i=1}^{k} \frac{1 - x (1+q-q^{1-i})}{1-q^i}. $$