algebraic problems on the triangle inequality

... Trang 7Marine pollution The discharge of sewage into the sea off the coast of Bombay has affected the fish population there The deposition of tar-like residues on the beaches on the west and east ... the rivers Luni, Jojari or Bandi Due to the sandy nature of the strata, considerable percolation of the waste water containing these toxic constituents to the saturat- ed zone has occurred The ... released to the atmosphere in varying amounts In 1983, the consumption of coal in India exceeded 100 million tonnes, while crude oil consumption was over 30 million tonnes It is estimated that these

Ngày tải lên: 15/03/2014, 23:20

... two components. In connection with the first assertion of Theorem 3.1, the following open question seems to be interesting Question 2 Is it true that the conclusion of the first part of Theorem ... without the additional assumption on the recession cone 0+D?. In order to derive information about the numbers χ(E w (P)) and χ(E(P)) from the information about the number χ(domF ), one has to ... whether the estimates hold true, or not In our opinion, the parametric affine variational inequality approach can help to study these estimates The following solution existence theorem for monotone

Ngày tải lên: 06/08/2014, 05:20

... organizational pattern is best suited to admin- ister the construction and operation of these works? 3 What is the best means of financing the required organiza- tion, construction, and operation? The ... FREDERICK GuTHEIM, Staff Director Trang 3FOREWORD The Joint Committee on Washington Metropolitan Problems, pur- suant to the terms of House Concurrent Resolution 172, 85th Congress, is conducting ... metropolitan view of the problem Many of the previous studies reflect the political fractionalization of the Washington metropolitan area, and are concerned with only a dictated by the limited interests

Ngày tải lên: 06/03/2014, 15:21

... solution if and only if the following integral condition holds: 0 < 1 0 Theorem 4.2 Suppose that H holds, then problem 1.1 –1.3 has a CΔ ld 0, 1Tpositive solution if and only if the following ... problem1.1 -1.2 , then which implies that u is concave on 0, 1T Combining this with the boundary conditions, we have uΔ0 > 0, uΔ1 < 0 Therefore uΔ0 uΔ1 < 0 So by 10, Theorem 1.115, there ... 0, 1T positive solutions is given by constructing upper and lower solutions and with the maximum principle The nonlinearity f mean that the functions f in1.1 A function xt ∈ C ld 0, 1T∩

Ngày tải lên: 21/06/2014, 20:20

... Chipot We construct the asymptotics of the sharp constant in the Friedrich-type inequality for functions, which vanish on the small part of the boundary Γ ε 1 . It is assumed that Γ ε 1 con- sists ... defined as the completion of the set of functions from the space C ∞ (Ω)bythenorm   Ω (u 2 + |∇u| 2 )dx. The space ◦ H 1 (Ω) is the set of functions from the space H 1 (Ω), with zero trace on ∂Ω. ... that the function u belongs to H 1 (Ω). Consider the following problem: Δu =−λ ε u in Ω, u = 0onΓ ε , ∂u ∂ν = 0on∂Ω \ Γ ε . (3.1) Definit ion 3.1. The function u ∈ H 1 (Ω,Γ ε ) is a solution of

Ngày tải lên: 22/06/2014, 11:20

... ∈ K, then (i)-(ii) ofDefinition 2.1reduce to the definitions of convexity and strong convexity, respectively Remark 2.3 h is strongly monotone with the constantσ > 0 if h is strongly convex ... K, then (i)–(iv) ofDefinition 2.5reduce to the definitions of coerciveness, strong monotonicity, monotonicity, relaxed monotonic-ity, respectively Obviously, theη-coerciveness implies η-monotonicity. ... auxiliary problem for the equilibrium problem (1.1) They then showed that the approximate solutions generated by the auxil-iary problem converge to the exact solution of the equilibrium problem

Ngày tải lên: 22/06/2014, 18:20

... moment, so we postpone this until the next section, and consider the con-ditions onu In addition, it is assumed that u satisfies the boundary conditions in (1.2) These conditions are imposed in, ... impose the condition that As mentioned above, most papers on time scale boundary value problems use the above formulation, but fail to impose any condition on the relationship between the points{ ... just mentioned concern the basic formulation of the linear operator formed from the left-hand side of (1.1), together with the boundary conditions (1.2), so from now on we simply consider the formulation

Ngày tải lên: 22/06/2014, 22:20

... inequalities on f and on the boundary conditions. In Section 3, the a priori bounds from Section 2 are used in conjunction with the non- linear alternative to prove the existence of solutions to the BVPs ... were the works [1, 2, 3, 4, 8, 9, 10]. 2. A priori bounds on solutions In order to apply Theorem 1.8, a priori bounds on solutions to the BVPs are needed. In this section conditions on f and on the ... occurs in (a,σ 2 (b)), then y(t) <R, t ∈ [a,σ 2 (b)] by Lemma 2.2. This concludes the proof.  Johnny Henderson et al. 99 The question now arises on whether the conditions α,β,γ,δ>0canberemovedfrom

Ngày tải lên: 23/06/2014, 00:20

... image inΓ0(K) of the sequence M n =(u n+1,u n ) on Q0(K) by the birational transformation φ K Then the symmetric point of m n+1 with respect to the diagonal x = y lies onΓ0(K) and on the straight ... (1.6)passes throughM0 The quarticsQ(K) are invariant on the action of F, and thus the points M nmove onthe quartic passing throughM0, more precisely on its positive componentQ0(K). The mapF has a geometrical ... interpretation If M ∈ R+ ∗2, letM be the second point ofthe quarticQ(K) which passes through M whose first coordinate is the same as those of M (there is only one such point M because the point at the

Ngày tải lên: 23/06/2014, 00:20

... each other at one end of the route and finish next to each other at the other end, the poset is the Young diagram corresponding to the partition (n, n) of 2n, and a linear extension corresponds ... pawn promotion.) The composer must then ensure that Y is the only such position reachable within the stated number of moves, and the solver must first find the target position Y using the solver’s ... based on a presentation titled “How do I mate thee? Let me count the ways” that I gave the banquet of the conference in honor of Richard Stanley’s 60th birthday; the article is dedicated to him on

Ngày tải lên: 07/08/2014, 08:22

... this further, one may consider imposing additional local constraints on the graph and asking whether the aforementioned bounded decreases Kahn and Kim [6] conjectured that if the graph is triangle-free, ... then the upper bound can be improved to O(∆/ log ∆) Kim [7] proved this with the additional hypothesis that G contains no 4-cycle Soon after, Johansson proved the conjecture Theorem (Johansson ... phrase it only for chromatic number In the next section we present the lower bound in Theorem and the rest of the paper is devoted to the proof of the upper bound The last section describes the minor

Ngày tải lên: 07/08/2014, 21:20

... Maker The free edges of Gk(S) are called available Trang 3The rest of the paper is organized as follows In Section 3 we prove Theorem 1.1 andin Section 4 we prove Theorem 1.2 Finally, in Section ... that the edge that connects v and the other endpoint of that path is still free If he can now close this longer path into a cycle in his next move, then he does so and continues to Phase X Otherwise ... asked there, which of these two values is the correct answer We settle this question as follows: Theorem 1.1 For sufficiently large n we have τ (Hn) = n + 1 Note that, though this only improves the

Ngày tải lên: 07/08/2014, 21:21

... They also conjectured that the constants c0 and C0 in Theorem 2 can be chosen arbitrarily close to each other Conjecture 3 For every graph G and positive constant ε, there exist a positive constant ... q) Trang 3In contrast to the recent results on the connectivity and Hamiltonicity games, so far there has been no progress towards resolving Conjecture 3 In fact, there is no non-trivial graph ... games on graphs that have attracted considerable attention include the diameter game [1], the planarity, colorability, and minor games [6] For more information on positional games, we refer the

Ngày tải lên: 08/08/2014, 14:22

... distance between them. Specifically, the larger the distance, the smaller the collision probability. For one hash table, they first partition the space randomly into high-dimensional cubes. 23 Then, they ... identification We divide the video identification problems into 6 levels based on the noise between the original and the duplicated version video clips. Figure 1.2 illustrates these different level problems ... increasing the search speed, because they only care the performance on precision and recall in current stage. 2.1.2 Video Segmentation and Feature Extraction This module is the main part of the whole

Ngày tải lên: 05/10/2015, 19:02

... depending on the total population of the country and the proportion of the population that is older than 65 The regressions also control for the proportion of the population that ... that long-surviving leaders have longer horizons: the longer the years in office of the leader, the lower is the sharegain of the poor and the greater is the sharegain of the rich Robustness: Other ... long-surviving leaders have longer horizons: the longer the years in office of the leader, the lower is the sharegain of the poor and the greater is the sharegain of the

Ngày tải lên: 29/08/2016, 08:07

... and|.|vdenotes the normalized v-adic norm on the completion k v of k at v Next, since k has characteristic >0, all the valuations v are non-archimedean Then one checks that the norm on A is non-archimedean. ... where the main result isTheorem 3.1 In Section4we consider the same problem, but under the assumption that k is perfect, which gives us finer results, where the main result isTheorem 4.5 Along the ... topology on H1 flat(k,G) Definition The canonical topology on H1(k,G) (resp H1flat(k,G)) is the k s/k- (resp k¯ k-) canonical topology Trang 6It is the same as we define the corresponding topology on H1flat(k,G)

Ngày tải lên: 11/12/2017, 23:22

... According to the Statutes of the IPhOs [1] the theoretical examination consists of three problems of 5 h duration The experimental examination consists of one or two problems of 5 h duration The knowledge ... report on the competition problems given at the 39th International Physics Olympiad held in Hanoi, Vietnam, which consist of three theoretical and one experimental problems The abbreviated solutions ... required to solve the problems is included in the syllabus [2] During the preparation of the problems, the academic committee also paid attention to some other aspects, namely, the relation of physics

Ngày tải lên: 12/12/2017, 06:44

... refer to Section 2 for the precise meaning of the operation Sq? on the domain of the algebraic transfer We next explain the idea of the proof of Theorem 1.1 Let Pk '= H*(BVk) be the polynomial ... The reader who does not wish to follow the invariant theory computation above may be satisfied by the following weaker theorem, and then would not need to read the paper's last 3 sections Theorem ... is not an isomorphism This theorem is proved by observing that, on the one hand, (IF2 P4)2L4 (F2 P4)GL 474 Trang 4THE BEHAVIOR OF THE ALGEBRAIC TRANSFER and on the other hand, Ext4 20 (F F)

Ngày tải lên: 16/12/2017, 01:00

... (CCMP) encryption. By caching the PMK on the station and the AP, both parties can skip 802.1X/EAP authentication (fast) while maintaining a PMK that ensures that the station is allowed on the network ... an 802.1X/EAP authentication by sending frames through its current AP and to the new AP through the wired LAN. Once the handset authenti- cates to the new AP, it then stores the PMK that is created ... is observed, there may be a problem with the network. When such problems do arise, the first place to look is the configu- ration of the APs, controllers, and servers that comprise the WLAN infrastructure. Configuring...

Ngày tải lên: 24/01/2014, 09:20

... Theorem 5. Keep the notation and the assumptions of this theorem. Assume that the parameter w>1 is fixed, as well as the sequences (U n ) n≥1 and (V n ) n≥1 occurring in the definition of Condition ... Σ k is the input alphabet, δ : Q × Σ k → Q is the transition function, q 0 is the initial state, ∆ is the output alphabet and τ : Q → ∆ is the output function. Annals of Mathematics On the ... assumption the sequence a is recurrent there exist at least two oc- currences of the letter a. We then apply the same trick as in the proof of Theorem 3, and we again conclude by applying Theorem...

Ngày tải lên: 16/02/2014, 06:20