a generalization of aromaticity hückel apos s 4n 2 rule

... Al-Salam and A Verma, On quadratic transformations of basic series, SIAM J Math Anal., 15 (1984), 414– 421 [3] G Gasper, Summation, transformation, and expansion formulas for bibasic series, Trans ... and quartic summation and transformation formulas, Canad J Math., 42 (1990), 1 27 [5] G Gasper and M Rahman, Basic Hypergeometric Series, Encyclopedia of Mathematics and its Applications, 35 ... have [P ]0 = In general, polynomials p,q arising in bibasic hypergeometric summation have several different representations in terms of falling bibasic factorials From all possibilities, we shall...

Ngày tải lên: 07/08/2014, 06:20

... of diagram squares of rank at least m − In particular, π avoids all patterns of Am if and only if every diagram square is of rank at most m − the electronic journal of combinatorics 9 (2) (20 03), ... component consisting of all diagram squares of rank zero.) It was shown in [10, Lemma 2. 2] that the rank of a square (i, j) which belongs to the diagram of a permutation in Sn is at least i + j − ... from B the diagram squares having rank at most m − and the pairs (i, j) ∈ Om (π) Proof For any m ≥ 2, let D be the set of all diagram squares of rank at most m − The proof is based on the following...

Ngày tải lên: 07/08/2014, 07:21

... head of at least qds /2 non-free escaping paths As by definition of safeness, X has size qds /2, this would imply that the number of non-free escaping pairs is at least (qds /2) 2 , which is larger ... choices for each of u2 , u3 , , us+1 , thus each v ∈ X is the first vertex of at most ds escaping paths This shows that v is on at most ds + msds−1 escaping paths Recall that since the robber was ... elements, and property (2) ensures that |Ns (x)| = |Ns (x)| ≥ qds That is, every x ∈ X is the head of at least qds distinct escaping pairs We say that an escaping pair (x, y) is free if all shortest...

Ngày tải lên: 08/08/2014, 12:23

... the parsing problem is solvable, and this class includes x-bar grammars Let us review the offline parsable grammars Let G be a unification grammar with a set of rules R, a set of terminals T, and ... Context-freeness of Natural Language Linguistics and Philosophy 8(3), 333-343 Such a grammar is desirable because it allows us to assert once that an English VP consists of a verb followed by a suitable list ... (Boisen et al., 1988) This grammar now has about 700 rules and considerable syntactic coverage, so it represents a serious test of our sort system We have found that the sort system is a natural...

Ngày tải lên: 21/02/2014, 20:20

... Oregon and Oregon State, I was watching a show called Dangerous Pursuits on TLC A deranged man had commandeered a bus, and was smashing squad cars and turfing lawns all over Beverly Hills It was damned ... library basement to watch “the Classics”—Johnny-speak for highlights of last season s games Before the lights are dimmed, the head coach says a few (hundred) words As he makes his slow, steady septuagenarian s ... games than any five NFL coaches, who still insisted that his quarterbacks call their own plays “Why not?” says Gags “These guys are a hell of a lot smarter than I am.” He is dumb as a fox Gagliardi...

Ngày tải lên: 24/03/2014, 02:21

... that quaternion addition is associative and commutative 48 The sum of a scalar and a vector is a quaternion with that scalar and that vector, and conversely, a quaternion is the sum of its scalar ... the level of the north road; and similarly in Ex 22 24 A massless ring P is attached to several elastic strings that pass respectively through smooth rings at A, B, C, · · · and are attached to ... plane triangle ABC, whose sides opposite A, B, C, are a, b, c, show by squaring BC = AC − AB, that as a2 = b2 + c2 − 2bc cos A; also from V BCCA = V CAAB = V ABBC show that a : b : c = sin A...

Ngày tải lên: 28/06/2014, 19:20

Ngày tải lên: 05/08/2014, 15:20

Ngày tải lên: 05/08/2014, 15:20

... of the arrangement H A A The combinatorial object associated to an arrangement A is the intersection (semi)lattice LA It is the set of subspaces V of Cn such that ∅ = V = H for H∈B some subset ... can be easily computed using a result about characteristic n polynomials of hyperplane arrangements (see Orlik & Terao [11, Theorem 2. 69]) or more generally subspace arrangements (Athanasiadis ... Proposition 2. 2 [21 ] Let A be an arrangement of a ne hyperplanes in Cn Assume G ≤ Gln (C) is a finite subgroup that leaves A invariant Let ˆ be an additional element that is larger than any element...

Ngày tải lên: 07/08/2014, 06:20

... 1 )s l (i − s) (s − l)! (x + 2s + 2) 2i− 2s (2x + m + s) i s (2x + 2v + 1)j +s 2v+m−1 (−x − 2i − l + s − 1)i s (x + 2s − j + 2) j (x + m + 2j − s + 2 )s (m + 3j − 3s) (4 .22 ) We have to show that (4 .22 ) ... identities of the previous sections (see the paragraph after the proof of Theorem 10 for an account of this relationship) Special cases of this identity appeared previously in the paper [2] of Andrews ... it is seen that this means to check (i+l) /2 s= l (2i − 3s + l) (i − 2s + l + 1 )s l (x + 2s + 1)2i− 2s (−2x−2i−l+2n s 1 )s i (i − s) (s − l)! (−x − 2i − l + s) i s × (−x − 2i − l + s) j (x + 2s − j...

Ngày tải lên: 07/08/2014, 06:20

... observation that S is a Sidon set in G: an equality s1 − s2 = s1 − s2 with s1 = s2 , s1 = s1 creates a rectangle s1 , 0, s2 , s2 + s2 = s1 + s1 It is easy to verify, however, that the cardinality ... p − 2, provided |A| + |B| ≥ p + 2? Do similar estimates hold when A and B are subsets of an arbitrary finite group? If some of the answers are negative, what are the best possible estimates of ... to any s ∈ S there correspond at least m + n − q pairs (a, b) (with a ∈ A, b ∈ B) such that a + b = s, and for any such pair we have (a, b) ∈ R Totally, we have at least |S| (m + n − q) pairs (a, ...

Ngày tải lên: 07/08/2014, 06:20

... Chen, J Pach, M Szegedy, G Tardos: Delaunay graphs of point sets in the plane with respect to axis-parallel rectangles, in: Proc 19th annual ACM-SIAM symposium on Discrete algorithms (SODA) 20 08, ... G d has chromatic number at most k d , since it is the union of d subgraphs, each of which is k-colorable In particular, all graphs in Bd are 2d -colorable We show that this bound is sharp, which ... which is somewhat surprising since Cartesian products of bipartite graphs are bipartite More generally, let f (G; d) = maxG∈G d χ(G) We show that if G is the class of all k-colorable graphs, then...

Ngày tải lên: 07/08/2014, 21:21

... → kx is called an automorphism transform of Gn (Ai ), and Gn (Ai ) is called an automorphism cyclic graph The set Si is called an automorphism parameter set The number k is called a special element ... in terms of Definition and the set of representatives of equivalence classes is M = {2, 14, 4, 5, 8} Compute all A2 -colored chains starting at a ∈ M and we obtain an A2 -colored chain ≺ 2 ≺ ≺ ... partition of Ai In fact, an equivalence class is an orbit of Ai under the action of H Lemma Let k be a special element of order s of the automorphism cyclic graph Gn (Ai ) For an arbitrary a ∈ Ai ,...

Ngày tải lên: 08/08/2014, 11:20

... there exists a2 ∈ A2 , , an ∈ An so that h (a2 , , an ) = 0, then we get f (a, a2 , , an ) = h (a2 , , an ) = 0, which proves the assertion Otherwise h |A2 × ×An ≡ By the division algorithm ... inductive assumption we know that there exist a1 ∈ A1 \ {a} , a2 ∈ A2 , , an ∈ An so that g (a1 , , an ) = Hence f (a1 , a2 , , an ) = (a1 − a) · g (a1 , , an ) = 0, which proves the assertion ... degree is at most |A| − 1, and for all a ∈ A it takes value of Hence f ≡ and the coefficient of x |A| −1 equals But it is also the same as the the left hand side of the asserted equality Theorem (Coefficient...

Ngày tải lên: 08/08/2014, 11:20

... Using similar arguments, it is possible to prove that (16) is satisfied also with the choices of IV ) This assures that in all cases I) − −IV ), S is a semifield spread set Hence, S defines a semifield ... a presemifield is invariant up to isotopy and up to the transpose operation Proof Let S1 and S2 be two presemifields with associated spread sets of linear maps S1 and S2 , respectively If S1 is ... non–singular matrices over Fqn and σ ∈ Aut(Fqn )) φ such that S2 = S1 , where S1 and S2 are the semifield spread sets of matrices associated with S1 and S2 , respectively Starting from an Fq –linear set...

Ngày tải lên: 08/08/2014, 12:23

... Plasma, Saliva and Urine Int J Clin Pharmacol Ther 20 02, 40(6) :24 9 -25 5 Zahran FM, Ardawi MSM, Al-Fayez SF: Carboxyhaemoglobin concentrations in smokers of sheesha and cigarettes in Saudia Arabia BMJ ... conclusion) that "sharing a waterpipe mouthpiece poses a serious risk of transmission of communicable diseases, including tuberculosis and hepatitis" This statement is wrongly attributed to researchers ... tobacco manufacturers to produce various aromatic compounds [3,14] Moreover, as far as tar is concerned, all specialists know that what makes the latter hazardous and carcinogenous is not its quantities...

Ngày tải lên: 11/08/2014, 08:20

... (1980) Omran Kouba Department of Mathematics Higher Institute for Applied Sciences and Technology P.O Box 31983, Damascus, Syria omran kouba@hiast.edu.sy Mathematical Reflections (20 10) ... n σ(k) = 1+ax2 2 xf (x) dx we conclude that σ(k) 2 = log(1 + a) n2 + ak 1 2a • Starting from lim n→∞ n n k=1 ϕ(k) = 2, k π which can be proved in the same way as (3), we conclude that, for every ... )| < This ends the proof of Theorem Applications • It is known that Euler s totient function ϕ has very erratic behaviour, but on the mean we have the following beautiful result, see [2, 18.5],...

Ngày tải lên: 15/08/2014, 16:55

... Karatama s sense Summability allows for a huge variety of processes For instance, an I(0) process is always S( 0) And, the usual I(1) process, a random walk, is S( 1) These are common processes ... series can be seen as particular cases of summable processes But, maybe more important, the concept of summability controls the balancedness of non-linear regressions and generalizes the standard ... almost all the cases Also, the mean and the standard deviation lengths of the intervals are quite reasonable and improves as the sample size grows Overall, we can say that the methodology that...

Ngày tải lên: 14/11/2015, 07:33

... in the most approved fashion Mr Bullsom shook hands as though it were a sacrament; Mary, who was annoyed, did not smile at all "This is all quite unnecessary, Lord Arranmore," she said, while ... uncle was signalling for a cab I shall not change my mind, and I am sorry that you spoke to uncle about it at all." "It is a serious matter to me, Miss Scott," Lord Arranmore said, gravely "And ... you are often in London, Lord Arranmore?" she asked shyly "A great deal too often," he answered "We read about your beautiful parties at Enton," she said, with a sigh "It is such a lovely place."...

Ngày tải lên: 06/07/2014, 05:20

... possessed of such a thing No; I have a sense of justice, that is all a sort of weakness for seeing the scales held fairly Now, don't you think it is reasonable that she should accept this money ... pledges Speak in the House with the same passion and the same eloquence as when you sowed hope in the heart of those suffering thousands Some one must break away from this musty routine of Party ... a glass of wine A small boy shouting his number attracted his attention He called him, and was handed a card "Lord Arranmore!" "You can show the gentleman here," Brooks directed Arranmore came...

Ngày tải lên: 06/07/2014, 05:20