Timeline for Sum of $q$-binomial coefficients
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 26, 2022 at 14:21 | comment | added | aleph | @PeterTaylor If I calculated correctly, Nate's answer implies that $ \binom{2n}{n}_q > \binom{2n}{n-a}_q \cdot q^{a^2} $, which means that the sum of the other coefficients is $ 2 \sum_{a=1}^n \binom{2n}{n-a}_q < 2 \binom{2n}{n}_q \sum_{a=1}^n q^{-a^2} < 2 \binom{2n}{n}_q \sum_{a=1}^\infty q^{-a^2} $. Since $ 2 \sum_{a=1}^\infty q^{-a^2} < 1 $ for any $ q \geqslant 3 $, the middle coefficient is larger than the sum of all the others when $ q \geqslant 3 $. For $ q = 2 $ this is perhaps not true, but taking into account the residues I am interested in, this case also holds. | |
| Feb 26, 2022 at 10:51 | comment | added | Mark Wildon | The identity $\sum_{i=0}^n \binom{2m}{i}_q (-1)^i = (1-q)(1-q^3) \ldots (1-q^{2m-1})$ is correct: see for instance (3.3.8) in Andrews' textbook, The theory of partitions. Of course if $n$ is odd then the left-hand side is zero because $\binom{n}{i}_q = \binom{n}{n-i}_q$. | |
| Feb 25, 2022 at 17:41 | comment | added | Peter Taylor | I don't think Nate's answer quite gets you that. $[n]_q!$ is a polynomial of degree $\frac{n(n-1)}{2}$, so $\binom{n}{k}_q$ has degree $(n-k)k$. If you fix $n$ then simply because the polynomials are of higher degree there will be some $q_0$ such that for $q > q_0$ the middle coefficient(s) exceed the sum of the others, but with the exceptions of $n \in \{ 0, 1, 2, 3, 4, 5, 7 \}$ that $q_0$ seems to be more than 1 (although taking the sums over residues of the question instead, the cutoff appears empirically to approach 1 from below without reaching it). | |
| Feb 25, 2022 at 9:34 | vote | accept | aleph | ||
| Feb 25, 2022 at 9:33 | vote | accept | aleph | ||
| Feb 25, 2022 at 9:34 | |||||
| Feb 25, 2022 at 9:32 | comment | added | aleph | Yes, I had in mind prime powers, although it should be true for all reals $ \geqslant 1 $ (by Nate's answer below). | |
| Feb 25, 2022 at 0:10 | comment | added | Peter Taylor | @LeechLattice, I think it would be more accurate to say "in some contexts in which $q$-analogues are used..." There are other contexts where $q$ is complex; as I understand the history, in the first applications of $q$-analogues it was likely to be a root of unity. | |
| Feb 24, 2022 at 19:10 | history | edited | YCor |
edited tags
|
|
| Feb 24, 2022 at 16:23 | answer | added | Nate | timeline score: 4 | |
| Feb 24, 2022 at 16:18 | comment | added | LeechLattice | @PeterTaylor In the context of $q$-analogues, the number $q$ means the size of a finite field, so it's a prime power. | |
| Feb 24, 2022 at 16:12 | comment | added | Peter Taylor | Quick experimentation turns up the counterexample $n=11$, $\ell=3$, $q=-1$ where the $h$ which maximises is $h = 1 \not\equiv 5 \pmod 3$. However, given the background I suspect that you want to add an assumption that $q$ is an integer $> 0$. | |
| Feb 24, 2022 at 15:56 | history | asked | aleph | CC BY-SA 4.0 |