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A set of positive integers $d_1, \dots, d_n$ describe a n-dimensional tetrahedron $T$ with the vertices $$ (0,\dots,0), (1/d_1,0,\dots,0), (0, 1/d_2,\dots,0), \dots, (0,\dots,1/d_n).$$ Let $L_T(t)$ be the Ehrhart (quasi-)polynomial of $T$, i.e. the number of integer lattice points in $tT$.

I understand that typically everything beyond the first, second and last coefficients of $L_T(t)$ are rather difficult to determine in general. Is there an approximation possible for the other coefficients? As this for a practical problem, I'm not so much interested in an upper and lower bound, but rather a good "guess".

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Actually, $L_T(t)$ is not a polynomial (it's an honest quasipolynomial except for very special cases of $d_1, \dots, d_n$) and already the second "coefficient" is nontrivial, as is the last "coefficient". (The leading coefficient is a constant--the volume of T.) I'm not sure what exactly you're after in terms of approximating this, but here is one fact I'm aware of: let's write the quasipolynomial as $L_T(t) = c_{n-1}(t) \, t^{n-1} + \dots + c_0(t)$; then each $c_j(t)$ is periodic and so we can ask for its average value; let's call that $a_j$. Thus $a_{n-1} \, t^{n-1} + \dots + a_0$ is a polynomial that somehow encodes how $L_T(t)$ behaves "on average." It turns out this polynomial is a close relative of the Bernoulli polynomials (it's essentially what's known as a Bernoulli-Barnes polynomial); see, e.g., http://math.sfsu.edu/beck/ccd.html, p. 151.

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  • $\begingroup$ I've just found in an earlier paper that for pairwise co-prime $d_1, \dots, d_n$ all the coefficients (except $c_0$) are constant. This would be a possible restriction for me. Would the Bernoulli polynomials still helpful in this case? $\endgroup$
    – Jiro
    Feb 25, 2016 at 22:30
  • $\begingroup$ In terms of approximation, what I'm trying right now, is to replace the $T$ with an equilateral tetrahedron $T_e$ such that $d_e = d'_1 = \dots = d'_n$, where $d_e = \sqrt[n]{\prod_{i=1}^n d_i}$. The equilateral tetrahedron is very simple to scale and the leading coefficient is equal as the volumes of $T$ and $T_e$ are equal. I'm hoping that if the $d_i$'s are not too far away from $d_e$, the error won't be too big. Anyway to provide evidence that this doesn't go wrong horribly? $\endgroup$
    – Jiro
    Feb 25, 2016 at 22:30
  • $\begingroup$ Yes, the pairwise co-prime case will still involve Bernoulli-Barnes polynomials--in fact, they give all coefficients in this case except the last one. $\endgroup$
    – matthias beck
    Feb 26, 2016 at 23:15
  • $\begingroup$ I see how this could work. Just a quick question, in your book p.151, (8.4), is this really $B_d^A(-t)$ or is it $B_d^A(-n)$? $\endgroup$
    – Jiro
    Feb 27, 2016 at 16:20
  • $\begingroup$ Yes, that's a typo: it should be $-n$, as you said. (Thanks for catching it!) $\endgroup$
    – matthias beck
    Feb 28, 2016 at 19:08

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