1 pgs ts đỗ văn lưu pgs ts phan huy khải 2000 giải tích lồi nhà xuất bản khoa học và kỹ thuật hà nội

Device independent playground investigating and opening up a quantum black box

Ngày tải lên : 09/09/2015, 11:23

... =1+ x a+b c 24 x =1 a+b c 25 z =1 a b+c 10 /3 52/9 26 z=0 a b+c - 40/9 27 z =1 a b+c 28 x =1 a+b c 29 z =1 a b+c 10 /3 52/9 30 z=0 a b+c 18 /5 11 4/25 31 y =1 a+b c 14 /5 32 z=0 a b+c 18 /5 11 4/25 33 y =1 ... only 16 possibilities: A0 = 1, A1 = 1, B0 = 1 and B1 = 1 Consider the CHSH expression, CHSH = A0 B0 + A0 B1 + A1 B0 − A1 B1 , (2 .15 ) = A0 (B0 + B1 ) + A1 (B0 − B1 ) (2 .16 ) Note that (B0 + B1 ... = ax a+c b 12 z = ax a+c b 13 y =1+ z a b+c - 40/9 14 y =1+ x a+b c 10 /3 52/9 15 x=0 a+b c 16 y =1 a+b c - 40/9 17 x=0 a+b c 18 z=0 a b+c 19 z =1 a b+c 9/2 20 z=0 a b+c 16 /5 26/5 21 z =1 a b+c 9/2...

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Jensen’s Inequality

Ngày tải lên : 18/10/2013, 03:20

... r g Sk +1 h i f f IE 1 + r,k +1 gSk +1 jF k = 1 + r,k IE + r g Sk +1 jF k f 1 + r,k IE g + r Sk +1 jF k f 1 + r,k g IE + r Sk+1jF k = 1 + r,k g Sk ; So f 1 + r,k ... 0; 1; : : : ; n: Example 7 .1 (Stopped Process) Figure 7.3 shows our familiar 3-period binomial example Deﬁne ! = Then !1 = T; !1 = H: S HH = 16 S2^ ! ! = if if : S2 HT = S1 T = S1 ... f0; 1; : : : ; ng k The set f kg is in F k , so the set f k + 1g = f kgc is also in F k We compute h i h IE Yk +1 ^ jF k = IE If kgY + If k+1gYk +1 jF k = If kgY + If k+1gIE Yk +1 jF...

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Báo cáo hóa học: " On Minkowski’s inequality and its application" doc

Ngày tải lên : 20/06/2014, 22:20

... 2 011 , 2 011 : 71 http://www.journalofinequalitiesandapplications.com/content/2 011 /1/ 71 Page of 1/ s s(p−t)/(s−t) 1/ s (f (x)+g(x)) dx ≥ p s f (x)dx s + g (x)dx × f t (x)dx (1: 4) 1/ t t(s−p)/(s−t) 1/ t ... Inequalities and Applications 2 011 , 2 011 : 71 http://www.journalofinequalitiesandapplications.com/content/2 011 /1/ 71 ˜ ˜ ˜ Wi (K +L )1/ (n−i) ≤ Wi (K )1/ (n−i) + Wi (L )1/ (n−i) , ˜ with equality if and ... doi :10 .10 16/00 01- 8708(88)90077 -1 Gardner, RJ: Geometric Tomography Cambridge University Press, New York (19 96) Lutwak, E: Dual mixed volumes Pacific J Math 58, 5 31 538 (19 75) doi :10 .11 86 /10 29-242X-2 011 -71...

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Báo cáo hóa học: " Extension of Hu Ke’s inequality and its applications Jing-Feng Tian" pptx

Ngày tải lên : 20/06/2014, 22:20

... m n m r =1 j =1 Arj (1 − er + es ) Asi s =1 i =1 m s =1 i =1 r =1 j =1 n m m n s =1 i =1 j =1 r =1 = n λj Arj (1 − er + es ) n m Asi ≥ λ Arjj (1 − er Asi n = A 1 s1 s =1 n λj + es ) − 1 1 Ar1 (1 − er + ... n m n m r =1 j =1 Arj (1 − er + es ) Asi s =1 i =1 n n − 1 − er + e s ) A 1 A 1 (1 s1 r1 ≥ λj m j=2 s =1 r =1 m n λ A 1 Arjj (1 s1 × m n n × s =1 r =1 j=2 − 1 Aλ r1 = λj − er + e s ) s =1 r =1 j=2 n λj ... Asjj A 1 (1 r1 n m j=2 λj × m r =1 n n × λ Asjj A 1 (1 r1 λ A 1 Arjj (1 − er + es ) s1 s =1 r =1 j=2 n n λj − er + e s ) ( 21) s =1 r =1 n − 1 Aλ r1 = m j=2 λj × m r =1 Aλ s1 − s =1 n λ Arjj er + r =1 n...

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Báo cáo hóa học: " A new interpretation of Jensen’s inequality and geometric properties of -means" pptx

Ngày tải lên : 21/06/2014, 00:20

... http://www.journalofinequalitiesandapplications.com/content/2 011 /1/ 48 Page of 15 and hence ψ(ϕ 1 ( (1 − t)u + tv)) ≤ (1 − t)ψ(ϕ 1 (u)) + tψ(ϕ 1 (v)) for all t Î (0, 1) This means that ψ ∘ -1 is convex Conversely, if ψ ∘ -1 is convex, we ... λc,ξs1 ,ξs2 (x) = s2 λc,ϕ,ψ (x) + − s2 (s1 = 0) (3) and λc,ξs1 ,ξs2 (x) = s2 s − s1 − s1 s1 λc,ϕ,ψ (x) − s1 1 s1 (s1 = 0) (4) for each x Î I\{c} If s1 = 0, then it is trivial by (3) that λc,ξs1 ... Applications 2 011 , 2 011 :48 http://www.journalofinequalitiesandapplications.com/content/2 011 /1/ 48 Page of 15 Then u Î (I)\{d} and d Î ((I))∘, hence (λc,ϕ,ψ ◦ ϕ 1 )(u) = (ψ ◦ ϕ 1 )(u) − (ψ ◦ ϕ 1 )(d)...

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Báo cáo hóa học: " On improvements of the Rozanova’s inequality" potx

Ngày tải lên : 21/06/2014, 01:20

... Appl 16 2, 317 –3 21 (19 91) doi :10 .10 16/0022-247X ( 91) 9 015 2-P 14 Zhao, CJ, Cheung, WS: Sharp integral inequalities involving high-order partial derivatives J Inequal Appl (2008) Article ID 5 714 17 15 ... Zaved Mat 12 5, 75–80 (19 75) doi :10 .11 86 /10 29-242X-2 011 -33 Cite this article as: Zhao and Cheung: On improvements of the Rozanova’s inequality Journal of Inequalities and Applications 2 011 2 011 :33 ... 2 011 , 2 011 :33 http://www.journalofinequalitiesandapplications.com/content/2 011 /1/ 33 Page of Then a b 0 D1 D2 f x(s, t) p(s, t) p(s, t)g D1 D2 x(s, t) D1 D2 p(s, t) a b 0 D1 D2 p(s, t)g ≤w D1...

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Báo cáo hóa học: " Research Article The Best Lower Bound Depended on Two Fixed Variables for Jensen’s Inequality with Ordered Variables" pot

Ngày tải lên : 21/06/2014, 07:20

... let p1 , p2 , , pn be positive real numbers such that p1 p1 a1 p2 a2 p ··· p pn an p a 11 a22 · · · ann ≥ Qi ai , ak ··· 2. 21 ak 1 p2 , Q i aRk k 2.22 with equality for a1 Remark 2 .11 For p1 ··· ... 3 .14 where ··· pn xn pk 1 xk 1 Rk xk , pk xk Yk pk xk Rk p1 x1 Xk ··· X p1 x1 p2 x2 ··· , 3 .15 pn xn The inequality 3 .14 follows from Lemma 2.2, since xk , X ∈ Yk , Xk and Rk Xk Yk Rk xk X 3 .16 ... p1 p1 a1 p2 a2 ≥ Qi ··· p p p2 ··· 2 .10 pn Then, p pn an − a 11 a22 · · · ann Rk ak − Qi Qi / Qi Rk Rk R / Qi Rk ak k 2 .11 , Journal of Inequalities and Applications with equality for ··· a2 a1...

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Báo cáo sinh học: " Research Article Improvement and Reversion of Slater’s Inequality and Related Results" pdf

Ngày tải lên : 21/06/2014, 17:20

... page φ y2 − φ y1 φ x2 − φ x1 ≤ x2 − x1 y2 − y1 3. 21 with x1 ≤ y1 , x2 ≤ y2 , x1 / x2 , y1 / y2 Since by Theorem 3.6, Γt is log-convex, we can set in 3. 21 : φ x log Γx , x1 t, x2 s, y1 u, and y2 ... exp M1 ,1 exp 2u − u u 1 , log xi u 1 Pn Mu 1 x u / 0, 1, Pn log2 d − Pn M2 log x 2Pn log M0 x − 2d 2Pn log d − log M0 x Pn d log2 d n u 1 i pi xi u 1 duMu 1 x u P n Mu x − d u n i n i 1 − dM 1 ... results can also be given for 2 .10 and 2 .15 Namely, suppose that φ /ψ has inverse function, then from 2 .10 and 2 .15 we have ξ φ ψ ξ φ ψ 1 1 φ d ψ d 1/ Pn 1/ Pn n i pi n i pi xi − d φ xi − 1/ Pn...

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Báo cáo hóa học: " Research Article On the Strengthened Jordan’s Inequality" pot

Ngày tải lên : 21/06/2014, 22:20

... no 4, pp 422–423, 19 69 [12 ] J P Williams, “Problems and solutions: solutions of advanced problems: 5642,” The American Mathematical Monthly, vol 76, no 10 , pp 11 53 11 54, 19 69 [13 ] J.-L Li, “On ... n +1 π − (2x) (n ≥ 2) x π π (3 .13 ) (3 .14 ) hold with equality if and only if x = π/2 Furthermore, the constants (π − 2)/π and 2/π in (3 .13 ), as well as the constants 2/(nπ n +1 ) and (π − 2)/π n +1 ... (2 .13 ) is slightly stronger than inequality (2 .15 ) In the case when x ∈ (0,x0 ], inequality (2 .13 ) is slightly stronger than inequality (1. 2), where 7π + 240π − 2880 x0 = 12 π 1/ 2 ≈ 1. 2 (2 .16 )...

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Báo cáo hóa học: " Research Article A Generalized Wirtinger’s Inequality with Applications to a Class of Ordinary Differential Equations" docx

Ngày tải lên : 22/06/2014, 02:20

... k 1/ 2 T n dt 1/ 2 x t − nr ˙ dt T n 1 x t ˙ dt n2 κ2 T x t ˙ dt, 2 .10 that is, ˙ ¨ ⇒ T x ≤ nκT x ˙ x ≤ nκ x ¨ From 2 .1 and T |x ˙ t |2 dt 2 .11 0, we have ¨ ˙ 2π x ≤ T x 2 .12 Combining 2 .11 and ... inequality, one may see 15 and the references therein Now, we are ready to prove our main results We first give the proof of Theorem 1. 3 Proof of Theorem 1. 3 From 1. 1 and Definition 1. 1, for all t, u ∈ ... Applications, vol 14 , no 8, pp 7 01 715 , 19 90 T Furumochi, “Existence of periodic solutions of one-dimensional differential-delay equations,” Tohoku Mathematical Journal, vol 30, no 1, pp 13 –35, 19 78 S Chapin,...

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Báo cáo hóa học: " Research Article A Hilbert’s Inequality with a Best Constant Factor" doc

Ngày tải lên : 22/06/2014, 02:20

... > 1, 1/ p 1/ q 1, r > 1, 1/ r 1/ s 1, t ∈ 0, , − min{r, s} t min{r, s} ≥ λ > p q − min{r, s} t, such that ∞ > ∞ np 1 t 2t−λ /r 1 an > 0, ∞ > ∞ nq 1 t 2t−λ /s 1 bn > 0, n n then ∞ ∞ am bn m n 1m ... we have K o1 K o1 O

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Báo cáo hóa học: " Research Article Jensen’s Inequality for Convex-Concave Antisymmetric Functions and Applications" pptx

Ngày tải lên : 22/06/2014, 02:20

... assume that x1 ∈ 1, and x2 ∈ 0, The equation of the straight line through points x1 , f x1 , 0, is y Since f is convex on 1, and x1 < p1 x1 f p1 x1 p1 p2 x2 p2 f x1 x x1 p2 x2 / p1 ≤ 2 .1 p2 ≤ 0, ... f x1 p1 x1 x1 p1 p2 x2 p2 2.2 S Hussain et al It is enough to prove that f x1 p1 x1 x1 p1 p2 x2 p1 f x1 ≤ p2 p1 p2 f x2 p2 2.3 which is obviously equivalent to the inequality f x1 f x2 ≤ x1 x2 ... , xn 1 i xi − xi , i 1, , n If xi xn 1 i ≤ and A2 xi , xn Gn Gn ≥ n i A2 xi , xn A2 xi , xn 1 i pi pn 1 i 1 i pi xi pn 1 i xn 1 i / pi pn ≤ 1/ 2, i 1, , n, then 1 i 1/ 2Pn ≥ An An 1 i ,...

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Báo cáo hóa học: " Research Article Diamond-α Jensen’s Inequality on Time Scales" pdf

Ngày tải lên : 22/06/2014, 03:20

... k 1g 2N 2k 1 1/ 2N 1 α2k g 2k ln N k α2k 2k ln g t ∇t 1 α 2k ln g 2k 2N − N 1 k g 2N − ln 1 α 2N ln 1 α 2k 1 α 2k k g 1/ 2N 1 k N 1 ln ⎝ 3.20 1 α 2k N k ⎛ g 2k 1/ 2N 1 k N α2 g 2k k g 2k 1 α ... 1 α 2k 1/ 2N 1 ⎞ ⎠ k We conclude that ln α N 1 k k g ⎛ ≥ ln ⎝ 2k N 1 1−α 2N − k N k k 12 g 2k 1/ 2N 1 N α2 g 2k k g 2k k 1/ 2N 1 1−α ⎞ 3. 21 ⎠ k On the other hand, N 1 α 2k g 2k k 1 α N N 1 2k ... 2N ln α 1 ≥ 2N 2N g 2N 1 α 1 N 1 k k g 2N − α ln t Δt k t ∇t 1 N k k 12 g 2N − 1 α 1, b 2N , and g : {2k : k ln g t ♦α t 2N − α α 2N ln g t Δt 2N 1 N 1 k k ln g 2N − N 1 k α2k N 1 ln 2N −...

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Báo cáo hóa học: " Research Article A Reﬁnement of Jensen’s Inequality for a Class of Increasing and Concave Functions" pptx

Ngày tải lên : 22/06/2014, 03:20

... tx1 f q,t,x1 qt x1 t,x1 θ∈ 0 ,1/ 1 q tx1 x− σ2 − q tx1 x1 qtx1 − t x1 , 2b 1 2.8 1 − qtx1 q − q t2 x1 − q tx1 21 1 − qtx1 x − θσ − qtx1 Maximizing over x1 , we get 1 t x1 −f x − θσ θ∈ 0 ,1/ 1 q ... 1 α 2x1 qtx1 − qtx1 1 α 2x1 1 α 1 α 2x α qtx1 1 α 1 1 1 1 α α− Ye Xia 13 Maximizing the last expression over q, we get − α / 2x1 Maximizing the result over x1 , we get 1 α 2a 1 2 .10 Also, ... x − θσ θ∈ 0 ,1/ 1 q tx1 θ∈ 0 ,1/ 1 q tx1 −f x − θσ − qtx1 x − θσ − qtx1 k 2x1 qtx1 − qtx1 k 2x1 k − qtx1 k k 1 1 1 1 Maximizing the last expression over q and t, we get 1/ k x1 / k −bk/ k...

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Báo cáo hóa học: " Research Article Some New Results Related to Favard’s Inequality" pot

Ngày tải lên : 22/06/2014, 03:20

... Δu f 1/ v−u 2 .12 Journal of Inequalities and Applications Proof An equivalent form of 1. 12 is ϕ y2 − ϕ y1 ϕ x2 − ϕ x1 ≤ , x2 − x1 y2 − y1 2 .13 where x1 ≤ y1 , x2 ≤ y2 , x1 / x2 , and y1 / y2 ... Theorem 1. 1 Let f and p be nonnegative and integrable functions on a, b , with then for < r < s < t, r, s, t / 1, one has t−r Ds s s 1 ≤ t−s Dr r r 1 Dt t t 1 b p a x dx 1, s−r 1. 3 Remark 1. 2 For ... s is log-convex, and Γα s s s 1 Γα α s α sB s 1, α s − f t tα dt log f t tα dt Γα 1 α 1 H α 1 − H α log α α log α 2 f s t tα dt , s / 0, 1, 1 α 1 α f t tα dt , 3 .16 f t tα dt f t tα dt log f t...

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Báo cáo hóa học: " Research Article Hilbert’s Type Linear Operator and Some Extensions of Hilbert’s Inequality" docx

Ngày tải lên : 22/06/2014, 06:20

... x 1 ε x 1 ε ∞ ∞ t ( 1 ε)/2 dt dx 1/ x Amin {1, t } + B max {1, t } ∞ x 1 ε (2 .13 ) t ( 1 ε)/2 dt dx Amin {1, t } + B max {1, t } 1/ x t ( 1 ε)/2 dt dx Amin {1, t } + B max {1, t } For x ≥ 1, we get 1/ x ... ( 1 ε)/2 dt Amin {1, t } + B max {1, t } 1/ x ( 1 ε)/2 t At + B 1/ x dt ≤ B = 1 B − (1 + ε)/2 x ≤ 1/ 4 x B t ( 1 ε)/2 dt (2 .14 ) 1 (1+ ε)/2 (setting < ε < 1/ 2) Thus ∞ 0< ≤ B ≤ B x 1 ε ∞ ∞ 1/ x t ( 1 ε)/2 ... y } x ∞ 1/ 2 dt + t At + B A 1+ (1/ 2) AB B 1/ 2 √ AB =√ (setting t = 1/ u) dx t 1/ 2 dt Amin {1, t } + B max {1, t } 1/ 2 A/B ∞ A/B 1/ 2 dt t A + Bt 1 1/ 2 dt + √ t 1+ t AB 1 1/ 2 u du + √ 1+ u AB ∞...

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Báo cáo hóa học: " Research Article Hilbert’s Type Linear Operator and Some Extensions of Hilbert’s Inequality" doc

Ngày tải lên : 22/06/2014, 06:20

... x 1 ε x 1 ε ∞ ∞ t ( 1 ε)/2 dt dx 1/ x Amin {1, t } + B max {1, t } ∞ x 1 ε (2 .13 ) t ( 1 ε)/2 dt dx Amin {1, t } + B max {1, t } 1/ x t ( 1 ε)/2 dt dx Amin {1, t } + B max {1, t } For x ≥ 1, we get 1/ x ... ( 1 ε)/2 dt Amin {1, t } + B max {1, t } 1/ x ( 1 ε)/2 t At + B 1/ x dt ≤ B = 1 B − (1 + ε)/2 x ≤ 1/ 4 x B t ( 1 ε)/2 dt (2 .14 ) 1 (1+ ε)/2 (setting < ε < 1/ 2) Thus ∞ 0< ≤ B ≤ B x 1 ε ∞ ∞ 1/ x t ( 1 ε)/2 ... y } x ∞ 1/ 2 dt + t At + B A 1+ (1/ 2) AB B 1/ 2 √ AB =√ (setting t = 1/ u) dx t 1/ 2 dt Amin {1, t } + B max {1, t } 1/ 2 A/B ∞ A/B 1/ 2 dt t A + Bt 1 1/ 2 dt + √ t 1+ t AB 1 1/ 2 u du + √ 1+ u AB ∞...

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Báo cáo hóa học: " Research Article New Strengthened Carleman’s Inequality and Hardy’s Inequality" ppt

Ngày tải lên : 22/06/2014, 06:20

... obtain ∞ λn +1 a 1 aλ2 · · · aλn n 1/ Λn ∞ = n =1 λn +1 λ 1/ Λn λ λ c 11 c22 · · · cnn n =1 ∞ λn +1 λ λ λ c 11 c22 · · · cnn ≤ n =1 ∞ = λm cm am m =1 ∞ c1 a1 1 c2 a2 λ2 · · · cn an λn 1/ Λn n 1/ Λn λm cm ... results ∞ n =1 an Theorem 1. 3 (see [3, Theorem 1] ) Let an ≥ 0(n ∈ N) and < ∞ 1/ n a1 a2 · · · a n

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Báo cáo hóa học: " Research Article New Strengthened Carleman’s Inequality and Hardy’s Inequality" docx

Ngày tải lên : 22/06/2014, 06:20

... obtain ∞ λn +1 a 1 aλ2 · · · aλn n 1/ Λn ∞ = n =1 λn +1 λ 1/ Λn λ λ c 11 c22 · · · cnn n =1 ∞ λn +1 λ λ λ c 11 c22 · · · cnn ≤ n =1 ∞ = λm cm am m =1 ∞ c1 a1 1 c2 a2 λ2 · · · cn an λn 1/ Λn n 1/ Λn λm cm ... results ∞ n =1 an Theorem 1. 3 (see [3, Theorem 1] ) Let an ≥ 0(n ∈ N) and < ∞ 1/ n a1 a2 · · · a n

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Báo cáo hóa học: "Research Article On the Precise Asymptotics of the Constant in Friedrich’s Inequality for Functions " potx

Ngày tải lên : 22/06/2014, 11:20

... ⎩εn/2 1 + δ(ε)εn−2 1/ 2 1/ 2 + o (lnε) 1/ 2 + δ(ε)| lnε| + o εn/2 1 + δ(ε)εn−2 1/ 2 1/ 2 , , if n = 2, if n > 2, (3.42) as ε→0 Finally, due to (3.36), (3.37), and (3.38) we get that 1 ≤ ϕ(ε), − 1 1 ... Russian Mathematical Surveys, vol 44, no 3, pp 19 5 19 6, 19 89, translation in Uspekhi Matematicheskikh Nauk, vol 44, no 3(267), pp 15 7 15 8, 19 89 [11 ] R R Gadyl’shin and G A Chechkin, “A boundary ... Princeton, NJ, USA, 19 51 [15 ] R R Gadyl’shin, “On the asymptotics of eigenvalues for a periodically ﬁxed membrane,” St Petersburg Mathematical Journal, vol 10 , no 1, pp 1 14 , 19 99, translation...

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