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7 Polygons with Rational Vertices In this section we will establish formulas for the number of integer points in any rational convex polygon and its integral dilates. A natural first step is to fix a triangulation of the polygon P, which reduces our problem to that of counting integer points in rational triangles. However, this procedure merits some remarks. After counting lattice points in the triangles, we need to put those back together to form the polygon. But then we need to take care of the overcounting on line segments (where the triangles meet).

Note that P itself is a face of P, corresponding to the degenerate hyperplane Rd ,1 and the empty set ∅ is a face of P, corresponding to a hyperplane that does not meet P. The (d − 1)-dimensional faces are called facets, the 1dimensional faces edges, and the 0-dimensional faces vertices of P. Vertices are the “extreme points” of a polytope. A convex d-polytope has at least d + 1 vertices. A convex d-polytope with exactly d + 1 vertices is called a d-simplex. Every 1-dimensional convex polytope is a 1-simplex, namely, a line segment.

7. ♣ Prove that a rational convex polytope can be described by a system of linear inequalities and equations with integral coefficients. 8. 3) of the Eulerian numbers for all integers 1 ≤ k ≤ d, namely: (a) A (d, k) = A (d, d + 1 − k) ; (b) A (d, k) = (d − k + 1)A (d − 1, k − 1) + kA (d − 1, k) ; d A (d, k) = d! ; (c) k=0 k (−1)j (d) A (d, k) = j=0 d+1 (k − j)d . 9. 6); namely, for d ≥ 0, 1 (1−z)d+1 = k≥0 d+k d zk . 10. 7): For t, k ∈ Z and d ∈ Z>0 , (−1)d −t + k d = t+d−1−k . 11. The Stirling numbers of the first kind, stirl(n, m), are defined through the finite generating function n x(x − 1) · · · (x − n + 1) = stirl(n, m) xm .

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