proof of proposition 1
Đề tài " A stable trace formula III. Proof of the main theorems" pptx
... established a key reduction in the proof of one of the global theorems In this paper, we shall complete the proof of the theorems We shall combine the global reduction of [II] with the expansions that ... With the proof of Local Theorem in hand, we will see that the expansions of Sections 2–5 reduce immediately to two pairs of simple identities The first pair leads directly to a proof of Global ... the proof of the global theorems We have fixed the integers dder and rder in (1. 1) In this section, we shall apply the induction hypotheses (1. 2)– (1. 5) to the terms in the main expansions of [I,...
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Forest-Backed Bonds Proof of Concept Study: FINAL DRAFT - CIRCULATED TO STEERING GROUP ppt
... ($/tonne) Roundwood ($/m3) Sawnwood ($/m3) Wood Pulp ($/tonne) 1, 400 1, 200 1, 000 800 600 400 200 19 61 1966 19 71 1976 19 81 1986 19 91 1996 20 01 source data: FAOSTAT | © FAO Statistics Division 2007 | ... 6% 1% 0% 39,2 71 3% 4% 11 0,557 5% ASIA AND PACIFIC 97,377 57% 5% 15 % 12 % 0% 70,979 12 % 7% 206,705 9% LATAM AND CARRIBEAN 18 4,727 17 % 2% 4% 1% 1% 3 51, 249 2% 1% 5 41, 580 2% Total Protection Area 14 % ... Reserved 10 0 77 93.2 37.8 45 70.9 99.4 76 .1 98 53.2 99 .1 20.5 10 0 Community/Ind 13 0.3 5.9 0 0.6 1. 2 16 .5 31. 3 0.9 0 Individual/Firm 0 0 55 9.3 33 0 80 5.3 46 00 97 10 6.5 56.3 19 .8 7.4 15 10 .2 79.5...
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Đề tài " Proof of the Lov´asz conjecture " docx
... in [15 ] 9 81 ´ PROOF OF THE LOVASZ CONJECTURE H ∗ (Hom (C2r +1 , Kn )) D2 2n − d1 d1 d1 d1 D1 n−2 d1 d1 d1 d1 n−3 D0 d2 d1 q p d1 d1 d1 2r − 2r − 2r ∗,∗ p,q Figure 4 .1: The E1 -tableau, for E1 ⇒ ... of V (C2r +1 ) The sign of this is ( 1) r ; hence γ([σV (C2r +1 ),2r +1 ]) = ( 1) r [σV (C2r +1 ),2r +1 ] 2r 1, n−2 Lemma 4 .13 E1 = rχ0 + r 1 + χn+r +1 , as a Z2 -module Proof 1, 1 , , τ2r +1, 1 can be ... Z2 Z2 Z2 Table 4 .1 2r 1, 2 2r,2 → E1 is Z[d1 ( 1, r )], and 2d1 ( 1, r ) = d1 (2 1, r ) = If 2r + ≥ 9, d : E1 2r,2 2r ,1 we can compute E1 and E1 completely, since H i (Hom+ (C2r +1 , Kn )) vanish...
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Đề tài " A proof of the Kepler conjecture " pdf
... gives a proof of Claim 1. 11 Chapter proves Claim 1. 16 Chapters through give a proof of Claim 1. 13 Chapters through 11 give a proof of Claim 1. 12 Chapters 12 through 14 give a proof of Claim 1. 14 Claim ... consequence of Claims 1. 11 1. 16 Its proof appears in Theorem 5 .11 1. 4 Proofs of the central claims The previous section showed that the main results in the introduction (Theorems 1. 1, 1. 7, and 1. 9) ... hypotheses of Claim 1. 16 are satisfied The conclusion of Claim 1. 16 is the conclusion of Theorem 1. 9 Lemma 1. 20 Assume Claims 1. 11 1. 16 (Theorem 1. 1) holds Then the Kepler conjecture Proof As pointed...
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Đề tài " A proof of Kirillov’s conjecture " potx
... lemmas which are needed for the proof of our “Key Proposition above We also sketch the proof of Theorem 1. 4 In Section we A PROOF OF KIRILLOV’S CONJECTURE 211 recall some facts about distributions ... is G invariant Remark 10 .2 As a corollary to Theorem 10 .1 we obtain another proof of Conjecture 1. 1 The proof follows word for word the proof in [Ber, 5.4], and is omitted 10 .2 Scalar product in ... = Without loss of generality we ˜ can assume that a1 ,1 = Then ˜ λA = (r1 − 1) + (r2 + − 2t2 ) − (r1 + − 2t1 ) − (r2 − 1) = 2(t1 − t2 ) ≤ a ˆ Let A2 ,1 = (ˆi,j ) Then by Lemma 6 .11 we have ai,j...
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báo cáo hóa học: "Design of a complex virtual reality simulation to train finger motion for persons with hemiparesis: a proof of concept study" docx
... Pinky Day 10 Day 60 Angle (deg) Middle 40 20 -20 0.4 0.8 1. 2 1. 5 1. 9 2.3 0.4 0.7 1. 1 1. 4 1. 7 2 .1 Time (sec) Figure Independent Finger Flexion Independent Finger Flexion Left Panel: Depiction of independent ... results from Actual 12 0 Index 60 -60 -12 0 -12 0 12 0 Ring 60 -60 -12 0 -12 0 200 400 Fractionation (deg ) Daily Average Fractionation (deg ) 10 0 http://www.jneuroengrehab.com/content/6 /1/ 28 Target Middle ... # H133E050 011 ) References Authors' contributions SVA participated in the robotic/VR system design, study Target Actual Middle Fractionation (deg) 12 0 Index 60 -60 -12 0 -12 0 12 0 Ring 60 -60 -12 0...
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Báo cáo hóa học: " Pilot proof of concept clinical trials of Stochastic Targeted (STAR) glycemic control" docx
... interests Received: 18 May 2 011 Accepted: 19 September 2 011 Published: 19 September 2 011 Evans et al Annals of Intensive Care 2 011 , 1: 38 http://www.annalsofintensivecare.com/content /1/ 1/38 References ... 35 (10 ) :17 38 -17 48 doi :10 .11 86/ 211 0-5820 -1- 38 Cite this article as: Evans et al.: Pilot proof of concept clinical trials of Stochastic Targeted (STAR) glycemic control Annals of Intensive Care 2 011 ... patients Comput Methods Programs Biomed 2 011 Evans et al Annals of Intensive Care 2 011 , 1: 38 http://www.annalsofintensivecare.com/content /1/ 1/38 Page 12 of 12 44 Hann CE, Chase JG, Lin J, Lotz T,...
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Báo cáo hóa học: " Research Article Proof of One Optimal Inequality for " doc
... for a correction of one part of the proof, and for his improving of the organization of the paper This work was supported by Vega no 1/ 015 7/08 and Kega no 3/7 414 /09 Journal of Inequalities and ... 0g or, in virtue of the definition of p 2p p − 3α α − α 2p 0, b > Let p α be a solution of...
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Proof of heaven, a neurosurgeons journey into the afterlife eben alexander
... The Core 10 What Counts 11 An End to the Downward Spiral 12 The Core 13 Wednesday 14 A Special Kind of NDE 15 The Gift of Forgetting 16 The Well 17 N of 18 To Forget, and to Remember 19 Nowhere ... begun 15 The Gift of Forgetting We must believe in free will We have no choice —ISAAC B SINGER (19 02 19 91) The view of human consciousness held by most scientists today is that it is composed of ... Saturday in 19 75, the rest of the UNC jumpers and I teamed up with some of our friends at a paracenter in eastern North Carolina for some formations On our penultimate jump of the day, out of a D18 Beechcraft...
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Báo cáo toán học: "A Short Proof of the Rook Reciprocity" pps
... purpose of this note is to provide such a proof The knowledgeable reader will recognize that the main idea is borrowed from [4] Proof Observe that d ( 1) R(B; − x − 1) = ( 1) d B rk ( − x − 1) d−k ... the set S of rooks on B is nonempty cancel out of the above sum, because they are counted once for each subset of S, with alternating signs Thus what survives is the set of placements of d non-taking ... − x − 1) d−k d k=0 d B ( 1) k rk (x + d − k)d−k = k=0 B First assume x is a positive integer Add x extra rows to [d] × [d] Then rk (x + d − k)d−k is the number of ways of first placing k rooks...
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Báo cáo toán học: "A short proof of a partition relation for triples" pdf
... Xm 1 ]Y0 , ,Ym 1 = {Z0 ∪ · · · ∪ Zm 1 | Z0 ∈ [X0 ]Y0 ∧ ∧ Zm 1 ∈ [Xm 1 ]Ym 1 }, the most important consequence of which is the fact that [X, Y ]1, 2 is just the set of triples {x0 , y0 , y1 } ... pair of distinct elements z0 and z1 of Z, put 0 if f {z0 , z1 , d0 } = 1, fD {z0 , z1 } = n − if f {z0 , z1 , dn 1} = 1, or n if f {z0 , z1 , d} = for each d ∈ D By Fact ... collection of all n-element chains of X; [X]n = {{x0 , , xn 1 } | x0 , , xn 1 ∈ X ∧ x0 < · · · < xn 1 } And more generally, for two finite sequences of orders X0 , , Xm 1 and Y0 , , Ym 1 ,...
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Báo cáo toán học: " A Purely Combinatorial Proof of the Hadwiger Debrunner (p, q) Conjectur" docx
... factor (1 − h) (1 − 2h) (1 − (p − 1) h) We exaggerate this factor by replacing it by (1 − hp2 /2); if we use Lemma 3, and set h = i/(d + 1) , we obtain i=( q d +1 i= / 1+ ( p d +1 ) (1 − ip2 /(2(d + 1) )) ... such (d + 1) -element set can occur in this way at most N p have at least to N d +1 q d +1 / q d +1 p d +1 / N −d 1 p−d 1 N −d 1 p−d 1 times, which tells us we must different intersecting (d +1) -element ... meet each of the original sets Proof of Lemma If we can find a set of f(d, g) points that meets the convex hull of every subset of g|Q| of the elements of Q, it will meet every one of our polytopes...
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Báo cáo toán học: " A Bijective Proof of Garsia’s q-Lagrange Inversion Theorem" pdf
... Math Soc 305 (19 88), 4 314 65 [16 ] S Ramanujan and L J Rogers, Proof of certain identities in combinatory analysis, Proc London Math Soc 19 (19 19), 211 – 216 [17 ] D Singer, Q-Analogues of Lagrange ... (i 1) |Ti | i =1 Fej q −(j 1) ej j =1 Hence W (T, N )u|N | = (T,N )∈Sk ∞ p k w(Ti )q (i 1) |Ti | p=k (T1 , ,Tk )∈Cp e1 , ,ep 1 i =1 ∞ j =1 p k w(T )q (i 1) |T | u|T | p=k i =1 T ∈C Fej q −(j 1) ej uej = ... journal of combinatorics (19 97), #R26 31 Proof We have |T | q −(i 1) |Ni | Wq (T, N ) = wq (T ) i =1 B(N ) q −(i 1) |Ni | · q = wq (T ) − |T | j=B(N ) +1 (j 1) i =1 B(N ) = wq (T ) |T | B(N ) q −(i 1) |Ni...
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Báo cáo toán học: " Proof of the Alon-Tarsi Conjecture for n = 2r p" docx
... p(p − 1) , (11 ) and combining this with (3), we get p!(p − 1) ! p(p − 1) = (p − 1) !(p − 2)! |NG| = (12 ) Finally, combining (9), (10 ), and (12 ), we have fdels(p) − fdols(p) (p − 1) ! p 1 (p − 1) !(p ... Combinatorica 12 (19 92), 12 5 14 3 [2] A A Drisko, On the number of even and odd latin squares of order p + 1, Adv Math 12 8 (19 97), 20–35 [3] R Huang and G.-C Rota, On the relations of various conjectures ... − 1) !(p − 2)! ≡ ( 1) (p − 1) ! AT(p) = ≡ ( 1) p 1 ≡ ( 1) p 1 (mod p) (13 ) (p − 2)! (mod p) (mod p), since (p − 2)! ≡ (mod p), by Wilson’s theorem Let us record the known cases of the extended Alon-Tarsi...
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Báo cáo toán học: "A Proof of the Two-path Conjecture" pps
... be xi1 , xi2 and yi1 , yi2 with xi1 = yj1 and xi2 = yj2 Using symmetry again, we may assume that i1 < i2 , j1 < j2 , and i1 ≥ j1 With this labeling, again the vertices x0 , , xi1 , yj1 +1 , ... Proof Label the vertices of X as x0 , x1 , , xn , with xi 1 adjacent to xi for ≤ i ≤ n Similarly, label the vertices of Y as y0 , y1, , yn Let s be the number of common vertices; thus ... endpoint of one of the paths We may assume that x0 = yk for some k with k < n Now we consider two cases If yk +1 is not a vertex of X, then we replace the edge xn 1 xn with the edge yk+1x0 to create...
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Báo cáo toán học: "A Concise Proof of the Littlewood-Richardson Rule" pdf
... 37 (19 88), 799–802 [G] V Gasharov, A short proof of the Littlewood-Richardson rule, European J Combin 19 (19 98), 4 51 453 [KN] M Kashiwara and T Nakashima, Crystal graphs for representations of ... Univ Press, Oxford, 19 95 [LR] D E Littlewood and A R Richardson, Group characters and algebra, Phil Trans A 233 (19 34), 99 14 1 [RS] J B Remmel and M Shimozono, A simple proof of the Littlewood-Richardson ... of the qanalogue of classical Lie algebras, J Algebra 16 5 (19 94), 295–345 [L] P Littelmann, A Littlewood-Richardson rule for symmetrizable Kac-Moody algebras, Invent Math 11 6 (19 94), 329–346 [M]...
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Báo cáo toán học: "Matrix-free proof of a regularity characterization" doc
... e tool in the proof of the algorithmic version of Szemer´di’s lemma is Corollary 1. 3 e the electronic journal of combinatorics 10 (2003), #R39 1. 1.2 The matrix proof of Theorem 1. 2 We briefly ... /16 In our proof of Theorem 1. 5, we not appeal to the matrix argument of Section 1. 1.2 We show G2 =⇒ G1 follows directly from an application of the Szemer´di Regularity e Lemma itself Proof of ... dij = d ± 4ε0 The proof of Proposition 2.2 will then be complete upon the proofs of Claims 2.3 and 2.4 the electronic journal of combinatorics 10 (2003), #R39 2.4 .1 Proof of Claim 2.3 Recall...
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Báo cáo toán học: "A Combinatorial Proof of the Sum of q-Cubes" pptx
... bijective proof n q k =1 k 1 − qk 1 q − q k 1 − q k +1 n +1 = + 2 1 q 1 q (2) q The above equation is considered a “q-analogue” because equation (1) can be found by taking limq 1 of equation (2) We ... journal of combinatorics 11 (2004), #R9 (a) If 1 ≥ 1 , there are three subcases: i If 1 is even, then φ(λ, µ) = (λ, 1, µ2 ) ii If 1 is odd and µ2 is even, then φ(λ, µ) = (λ, µ2, 1 ) iii If 1 ... (λ, µ2 − 1, ( 1 + 1) ) (b) If 1 < 1 , there are three subcases: i If 1 is even, then φ(λ, µ) = (µ, 1 , λ2 ) ii If 1 is odd and λ2 is even, then φ(λ, µ) = (µ, λ2 , 1 ) iii If 1 is odd and...
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