Timeline for A $q$-analogue of a characterization of polynomials by binomial coefficients
Current License: CC BY-SA 4.0
6 events
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| Jul 16, 2020 at 10:31 | comment | added | Mark Wildon | Anyway, thank you for the very clear answer, which I think is definitive and shows that the generalization (I now realise) I really wanted is not feasible. | |
| Jul 16, 2020 at 10:26 | comment | added | Mark Wildon | Hence if $P(ax+b) = Q(x)$ then $(P([0]_q), \ldots, P([N-1]_q)) \in \langle u_q^{(0)}, \ldots, u_q^{(d)} \rangle$ if and only if $(Q([0]_q), \ldots, Q([N-1]_q)) \in \langle u_q^{(0}), \ldots, u_q^{(d)} \rangle$. For binomial coefficients this is useful, because such affine changes of variable move the evaluation points in a 'natural way'. (E.g. they can be reversed.) But for $q$-binomial coefficients, it's not so useful: e.g, there is no affine transformation sending $[0]_1, [1]_q, [2]_q$ to $[2]_q, [1]_q, [0]_q$ when $q\not= 1$. | |
| Jul 16, 2020 at 10:21 | comment | added | Mark Wildon | Thank you. I think the critical point is that the binomial coefficients $\binom{x}{d}$ are the minimal degree polynomials that interpolate the functions $0 \mapsto 0, 1 \mapsto 0, \ldots, d-1 \mapsto 0, d \mapsto 1$. Thus $(v_0, \ldots, v_{N-1})$ is in the span of $u_1^{(0)}, \ldots, u_1^{(d)}$ if and only if there is a polynomial $P$ of degree $\le d$ such that $P(j) = v_j$ for each $j$. This generalizes to the $q$-binomial coefficients replacing $j \mapsto 0$, $d \mapsto 1$ with $[j] \mapsto 0$ and $[d]_q \mapsto 1$ and $P(j) = v_j$ with $P([j]_q) = v_j$. | |
| Jul 15, 2020 at 16:50 | vote | accept | Mark Wildon | ||
| Jul 14, 2020 at 17:17 | comment | added | Sam Hopkins | See also: mathoverflow.net/questions/218696/q-integer-valued-polynomials | |
| Jul 14, 2020 at 17:14 | history | answered | Fedor Petrov | CC BY-SA 4.0 |