Báo cáo toán học: "Product and puzzle formulae for GLn Belkale-Kumar coefficients" potx

... we don’t.) For t = 1 this is the original BK -puzzle P, and for all t the resulting total shape is a triangle. Consider now the BK -puzzle at t = 0: all the s-edges have collapsed, and each (i, ... G = GL n case. the electronic journal of combinatorics 18 (2011), #P76 4 Product and puzzle formulae for GL n Belkale-Kumar coefficients Allen Knutson ∗ Department of Mathematics...

Ngày tải lên: 08/08/2014, 14:23

... possibl e to perform all relevant control and configuration com- mands for every station. The baseband board was used for the real-time tests of the MIMAX baseband processor in several setups. First, ... perform the combining and processing of the complex antenna signal in the digital baseband. Since complete transmitter and receiver are required for each path, the resulting power...

Ngày tải lên: 20/06/2014, 21:20

Ngày tải lên: 05/08/2014, 15:20

... 5)  23. 2 The lower bound For a graph G, a complete graph K and vertex set S ⊆ V (G), we say that S is (G, +v, K) maximal if and only if K  G[S] and K ⊆ G[S ∪ {v}], for every vertex v ∈ V (G) ... , 9}] ∼ = C 9 . Graphs F 1 and F 2 are shown in Figures 1 and 2 . Our computational approach is based on the following lemmas and observations. Lemma 1. For r  2 and positive i...

Ngày tải lên: 08/08/2014, 11:20

... w 1 = 1 and w i ≤ 1 + max{w 1 , w 2 , . . . , w i−1 } for all 2 ≤ i ≤ d. We call them restricted growth words here. See [16, 17 , 19] and [6, 8] for some of their combinatorial properties and applications. ... structural features of S and T . Section 4 describes the place-action structure of T and the original motivation for our work. Our main results are proven in Sections 5...

Ngày tải lên: 08/08/2014, 12:23

... is not true for Lorentzen and Nilsen’s formulation. We have solved the LP with inequalities (2) and (4) using the above described iterative procedure for 1 ≤ I ≤ 15, 5 ≤ J ≤ 20 (for J ≤ 4 the ... between these bounds and the best upper bounds known. For example the lower bound for M(15, 10) in [1] is 958, the lower bound using Lorentzen and Nilsen’s formulation is 962 and t...

Ngày tải lên: 07/08/2014, 06:20

Ngày tải lên: 05/08/2014, 15:20

... na h ´ Eireann in Cork (Ireland). Further, I would like to thank Stephanie Halbeisen for many hints and fruitful discussions, and the referee for numerous extremely helpful suggestions and remarks on an ... integers N,with(n, m) ∈ E if n = m and for some x, y ∈ N we have n = x + y and m = x · y . Notice that for all n ∈ N,(n, n +1)∈ E (this is just because n =1· n and n +...

Ngày tải lên: 07/08/2014, 08:20

... right-continuous, and thus, is continuous. By (11) and (14), we have ηk ≤ f u (0, x) ≤ k, and so −(1 −η) k ≤ f u (0, x) −k ≤ 0 for every 10 Figure 1: Infinite beam on nonlinear and non-uniform elastic ... C q (R) if u ∈ C q (R) for q = 0, 1, 2, . . (F2) f(u, x) ·u ≥ 0, and f u (u, x) ≥ 0 for every u, x ∈ R. (F3) For every υ ≥ 0, sup x∈R, |u|≤υ   ∂ q f ∂u q (u, x)   <...

Ngày tải lên: 20/06/2014, 20:20

... exists m ≥ 0 such that β(t) ≥ ϕ ∗ for all t ∈ I − and β  (t) = m ≥ max y ≤max I β f(t, β(t), y) for a.a. t ∈ I 0 . Therefore, α and β are lower and upper solutions for problem (1).  Example 5 ... that β (t) ≥ k(t) for all t ∈ I − and β  (t) = m ≥ max y ≤max I β(t) f(y) for all t ∈ I 0 . (25) So we deduce from (24) and (25) that α and β are lower and upper solutions...

Ngày tải lên: 20/06/2014, 20:20