Báo cáo toán học: "An Inequality Related to Vizing’s Conjecture" docx
... ≤|D|, and the desired inequality (2) follows. References [1] Bert Hartnell and Douglas F. Rall, Domination in Cartesian Products: Vizing’s Conjecture, in Domination in Graphs—Advanced Topics edited by ... electronic journal of combinatorics 7 (2000), #N4 2 Theorem 1 For any graphs G and H, γ(G)γ(H) ≤ 2γ(G H). Proof. Let D be a dominating set of G H. It is sufficient to show that γ(G)γ...
Ngày tải lên: 07/08/2014, 06:20
... q-Identities related to overpartitions and divisor functions Amy M. Fu Center for Combinatorics, LPMC Nankai University, Tianjin 300071, P.R. China Email: ... K. Dilcher, Some q-series identities related to divisor functions, Discrete Math., 145(1995) 83–93. [5] A.M.FuandA.Lascoux,q-identities from Lagrange and Newton interpolation, Adv. Appl. Math., ... 25(1984) 377–379. [10] J. Zeng...
Ngày tải lên: 07/08/2014, 13:21
... Theor. Comput. Sci., 12(2):47–64, 2010. the electronic journal of combinatorics 18(2) (2011), #P18 7 Continued fractions related to (t, q)-tangents and variants Helmut Prodinger Department of Mathematics 7602 ... this continued fra ction expansions stops, since s r (z) = 0. Replacing z by −z we get zs 0 (−z) s −1 (−z) = z a 0 − z a 1 − z a 2 − z . . . . This translates into a continued fr...
Ngày tải lên: 08/08/2014, 14:23
Báo cáo toán học: "An extension of Scott Brown''''s invariant subspace theorem: K-spectral sets " ppsx
Ngày tải lên: 05/08/2014, 09:46
Báo cáo toán học: "Graph Color Extensions: When Hadwiger’s Conjecture and Embeddings Help" doc
... contract to K 5 ,ands +3 colors are both necessary and sufficient to extend the coloring of G[P ]toallofG.The same construction, beginning with a copy of K 4 , creates a graph that does not contract to ... G. A. Dirac, Map colour theorems related to the Heawood color formula, J. London Math. Soc. 31 (1956), 460-471. [12] G. A. Dirac, Map colour theorems related to the Heawood colo...
Ngày tải lên: 07/08/2014, 07:21
Báo cáo toán học: "A Bijective Proof of Borchardt’s Identity" docx
... C B (p, q). Hence case (3.32) with respect to z is equivalent to case (3.31) with respect to z, and we have just shown this case to be impossible. Pictorially, z is obtained from z by swapping ... τ(i)isb i .Thei th row sum is equal to p i and the i th column sum is equal to q i . The weight of the board is equal to (σ), the sign of σ. An illustration of the board B 1 corresp...
Ngày tải lên: 07/08/2014, 08:20
Báo cáo toán học: "On a Generalization of Meyniel’s Conjecture on the Cops and Robbers Game" potx
... Now, the cops move to new positions. At this moment there is no cop in A, so the robber is able to move to any vertex of X in her turn; thus to complete the proof, we need to show that there ... larger than s. Proof. (i) To show connectivity one has to prove that every element of Z 1+k(s+1) 2 can be written as a linear combination of members of S, which is equivalent to the ma...
Ngày tải lên: 08/08/2014, 12:23
Báo cáo toán học: " An introduction to 2-fuzzy n-normed linear spaces and a new perspective to the Mazur-Ulam problem" docx
... vec- tors. Then, for any f 3 , . . . , f n−1 , Ψ( f 1 ) − Ψ(f 0 ), . . . , Ψ(f n−1 ) − Ψ(f 0 ) are linearly dependent. and λf = {(λx, µ) : (x, µ) ∈ f} where λ ∈ S. The linear space (X) is said to ... Provisional PDF corresponds to the article as it appeared upon acceptance. Fully formatted PDF and full text (HTML) versions will be made available soon. An introduction to 2-fuzzy n-norme...
Ngày tải lên: 20/06/2014, 21:20
Báo cáo toán học: "An abstract Kato inequality for generators of positive operators semigroups on Banach lattices " doc
Ngày tải lên: 05/08/2014, 10:20
Báo cáo toán học: "An Area-to-Inv Bijection Between Dyck Paths and 312-avoiding" ppsx
... recurrence, it is necessary to understand why A i (q)=q i−1 C i−1 (q)C n−i (q). Since a path in A i first touches the diagonal at (i, i), it must go from (0, 1) to (i − 1,i) without touching the diagonal ... been shown to be C i−1 and thus have weight C i−1 (q). To these paths, we must add the i − 1 squares just above the diagonal from (0, 0) to (i, i). Thus the part of the paths fro...
Ngày tải lên: 07/08/2014, 06:22