Timeline for Positivity of a finite sum involving Stirling numbers
Current License: CC BY-SA 3.0
20 events
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| Feb 5, 2018 at 11:23 | history | edited | Wolfgang |
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| Sep 18, 2016 at 10:52 | comment | added | Pietro Majer | Yes, the problem is now reasonably tractable, but my time is over :) | |
| Sep 17, 2016 at 20:19 | comment | added | Tomeu Fiol | @Pietro Majer: Thanks, very nice ! Apologies for not writing explicitly the polynomials $S_n(x)$ - what you call $P_n(x)$ - in my previous reply. It might have saved you some work. Please check the reply to your answer below. | |
| Sep 17, 2016 at 17:23 | answer | added | Pietro Majer | timeline score: 16 | |
| Sep 17, 2016 at 16:54 | comment | added | Pietro Majer | Really, it's the Chebyshev measure of first kind | |
| Sep 17, 2016 at 16:20 | comment | added | Pietro Majer | I think the $a_{n,m}$ are Fourier coefficient of the polynomials $P_n(x)=\big( nx/2 +n/2-1\big)^{\underline n}$ (falling factorial), with respect to a certain symmetric measure on $[-1,1]$ quite concentrated at $\pm 1$. I'll make a computation to reconstruct the measure. Since the $P_n$ are odd/even according to the parity of $n$, and since they are also large at $\pm 1$, this hopefully would explain both positivity and vanishing properties for the $a_{n,m}$. | |
| Sep 17, 2016 at 13:33 | comment | added | Tomeu Fiol | @ T. Amdeberhan: Many thanks for your arguments. I agree with the form you propose for the $b_{n,n-2k-1}$ coefficients (I understand that you meant $n^{n-2k-1}$ instead of $n^{2k+1}$ in your first comment). While having an explicit form for the polynomials $P_k(n)$ that you introduced would be great, it is more than I need. Arguing that all their coefficients must be positive would be enough. | |
| Sep 17, 2016 at 13:23 | comment | added | Tomeu Fiol | @Timothy Chow: Thanks for your ideas. As for knowing that $a_{n,m}=0$ when $n+m$ is even, the reason is the following. I am interested in a family of polynomials $S_n(x)$ over the reals. These polynomials have well-defined parity, $S_{n}(-x)=(-1)^{n+1}S_n(x)$. The $a_{n,m}$ are (up to an overall factor) the coefficients of $S_n$ in terms of Chebyshev polynomials of the first kind $T_m(x)$: $S_n(x) = \sum_m a_{n,m} T_m(x)$. The condition on $a_{n,m}$ then follows from the parity of $S_n, T_m$. | |
| Sep 17, 2016 at 13:08 | comment | added | Tomeu Fiol | @ Steve Huntstman: Thanks for the references. I will look at the techniques in the book you mention, and try to apply them. | |
| Sep 17, 2016 at 3:33 | comment | added | T. Amdeberhan | $b_{n,n-7}=\frac4{45}n^{n-7}\binom{n}4(35n^5+588n^4+4856n^3+20736n^2+30848n+15360)$. As one can see, it gets "uglier". | |
| Sep 17, 2016 at 3:30 | comment | added | T. Amdeberhan | $b_{n,n-5}=\frac1{15}n^{n-5}\binom{n}3(5n^3+48n^2+184n+96)$. | |
| Sep 17, 2016 at 2:28 | comment | added | Timothy Chow | @TomeuFiol : What is your proof that $a_{n,m}=0$ when $n+m$ is even? | |
| Sep 17, 2016 at 1:51 | comment | added | T. Amdeberhan | The fact that $n^m\vert b_{n,m}$ can be proved. In $a_{n,m}$ replace $x=\frac{n}4$. Notice that although $\sum_{j=0}^{n-1}$, in fact $\sum_{j=m}^{n-1}$ since $\binom{2j}{j+m}=0$ when $j<m$. Consequently, $a_{n,m}$ as a polynomial in $x$ has an obvious factor $x^m$. That means $n^m$ factors out up on retreating to $\frac{n}4$ in place of $x$. | |
| Sep 17, 2016 at 1:07 | comment | added | T. Amdeberhan | I am convinced that there is no closed formula in general, unlike some special cases $(n,n−1),(n,n−3)$. Not even a finite recurrence. Perhaps the best we can say is this: $b_{n,n-2k-1}=\alpha_k\times n^{2k+1}\binom{n}{k+1}\times P_k(n)$ for some polynomial $P_k\in\mathbb{Z}[n]$ of degree $2k-1$ and for some positive $\alpha_k\in\mathbb{Q}$. More importantly, each $P_k(y)$ has positive coefficients! | |
| Sep 16, 2016 at 21:05 | comment | added | Timothy Chow | I just noticed that the summand is 0 unless $j\ge m$, so it is trivial that $b_{n,n-1}=n^{n-1}$ and I think it is also easy to prove the formula for $b_{n,n-3}$. But I'm still not sure how to prove that $n^m\mid b_{n,m}$. | |
| Sep 16, 2016 at 20:42 | comment | added | Timothy Chow | Actually, it looks like $b_{n,n-3}=n^{n-2}(n-1)(n+4)/3$ and that $b_{n,m}$ is always divisible by $n^m$. But I don't have a conjectural formula for $b_{n,n-5}$. | |
| Sep 16, 2016 at 20:30 | comment | added | Timothy Chow | If we define $b_{n,m} := 4^{n-1}a_{n,m}$ then $b_{n,m}$ is always an integer, so it might be easier to find a combinatorial interpretation. For example, empirically it seems that $b_{n,n-1} = n^{n-1}$, and $b_{n,n-3}$ appears to factor into small prime factors. | |
| Sep 16, 2016 at 20:11 | comment | added | Steve Huntsman | Have you tried using the representation at mathoverflow.net/questions/34151 for the Stirling numbers and then applying the techniques at math.upenn.edu/~wilf/AeqB.html? | |
| Sep 16, 2016 at 13:48 | review | First posts | |||
| Sep 16, 2016 at 13:52 | |||||
| Sep 16, 2016 at 13:45 | history | asked | Tomeu Fiol | CC BY-SA 3.0 |