carmywarserver class part 1
THE CAUCHY – SCHWARZ MASTER CLASS - PART 2 pot
... (a1 a2 · · · ak ) 1/ k = k =1 (a1 c1 a2 c2 · · · ak ck )1/ k (c1 c2 · · · ck )1/ k k =1 ∞ ≤ a1 c1 + a2 c2 + · · · + ak ck k(c1 c2 · · · ck )1/ k k =1 ∞ = ∞ ak ck k =1 j=k , j(c1 c2 · · · cj )1/ j (2 .17 ) ... e(ak /A) 1 = 1, (2 .13 ) unless A A and we always have ak ≤ e(ak /A) 1 , A so we see that one also has a1 A p1 a2 A p2 ··· an A n pn < exp pk k =1 ak A 1 = 1, (2 .14 ) unless ak = A for all k = 1, 2, ... c1 c2 · · · cj 1 = j j 1 and c1 c2 · · · cj = (j + 1) j , so division gives us the explicit formula cj = (j + 1) j =j 1+ j j 1 j j From this formula and our original bound (2 .17 ) we find ∞ ∞ (a1...
- 18
- 220
- 0
THE CAUCHY – SCHWARZ MASTER CLASS - PART 3 pptx
... (a1 , a2 , , an ) and (b1 , b2 , , bn ) that the quantity n n aj bj + 2x j =1 aj bk 1 j
- 14
- 336
- 0
THE CAUCHY – SCHWARZ MASTER CLASS - PART 4 doc
... to d d (uj + vj )2 ≤ j =1 d u2 + j d 1/ 2 u2 j j =1 j =1 d 1/ 2 vj j =1 j =1 and this in turn may be simplified to the equivalent bound d d u j vj ≤ j =1 1/2 d u2 j j =1 1/2 vj j =1 vj , + On Geometry and ... satisfy the On Geometry and Sums of Squares 71 triangular system of linear relations x1 = x1 , e1 e1 x2 = x2 , e1 e1 + x2 , e2 e2 ··· = ··· xn = xn , e1 e1 + xn , e2 e2 + · · · + xn , en en Exercise ... upper bound for a product of two linear forms: n n vj xj ≤ uj xj j =1 j =1 n n j =1 n u2 j u j vj + j =1 1/2 n vj j =1 1/2 x2 (4.8) j j =1 The charm of this inequality is that it leverages the presence...
- 22
- 271
- 0
THE CAUCHY – SCHWARZ MASTER CLASS - PART 5 doc
... nonnegative real numbers a1 , a2 , b1 , and b2 satisfy max{a1 , a2 } ≥ max{b1 , b2 } and a1 + a2 = b1 + b2 , then for nonnegative x and y, one has xb1 y b2 + xb2 y b1 ≤ xa1 y a2 + xa2 y a1 (5.22) Prove ... inequality (5 .12 ) 80 Consequences of Order aj an 1 an ··· bσ(n 1) bσ(n) ··· bτ (n 1) bτ (n) ak a1 a2 bσ (1) bσ(2) ··· bσ(j) ··· bσ(k) bτ (1) bτ (2) ··· bτ (j) ··· bτ (k) Fig 5 .1 An interchange ... function n n aθ+x bθ−x j j fθ (x) = j =1 aθ−x bθ+x , j j x ∈ R j =1 If we set θ = 1, we see that f1 (0 )1/ 2 gives us the left side of Cauchy’s inequality while f1 (1) 1/2 gives us the right side Show...
- 14
- 240
- 0
THE CAUCHY – SCHWARZ MASTER CLASS - PART 6 pot
... value V = (1+ r1 ) (1+ r2 ) · · · (1+ rn ) of our investments after n years must satisfy the bounds n (1 + rG )n ≤ (1 + rk ) ≤ (1 + rA )n , (6.23) k =1 where rG = (r1 r2 · · · rn )1/ n and rA = (r1 + r2 ... before in Exercise 2 .11 that for nonnegative aj and bj , j = 1, 2, , n one has superadditivity of the geometric mean: (a1 a2 · · · an )1/ n +(b1 b2 · · · bn )1/ n ≤ {(a1 + b1 )(a2 + b2 ) · · · ... n n pk f (xk ) − f k =1 n pk xk k =1 = µ j =1 n pj pk (xj − xk )2 (6 .16 ) k =1 Context and a Plan This result is from the same famous 18 85 paper of Otto Ludwig H¨lder o (18 59 -19 37) in which one finds...
- 18
- 243
- 0
THE CAUCHY – SCHWARZ MASTER CLASS - PART 7 pptx
... + 1) I = xα+β +1 |f (x)| dx Thus, if we then apply Schwarz’s inequality to the splitting xα+β +1 |f (x)| = {x(2α +1) /2 |f (x) |1/ 2 } {x(2β +1) /2 |f (x) |1/ 2 } we find the nice intermediate bound ∞ (1 ... 1) 2 = dx t t ≤ f (x) dx 1 1 dx, f (x) (7 .14 ) so, by our hypothesis we find t c 1 t−2 (t − 1) 2 ≤ 1 dx, f (x) and when we let t → ∞ we find the bound ∞ dx c 1 ≤ f (x) (7 .15 ) Since we were challenged ... for p1 + p2 + · · · + pn = the bound (7. 21) reduces to exactly to the Integral Intermezzo 11 5 classic AM-GM bound, n n apk ≤ k k =1 pk ak (7.22) k =1 Incidentally, the integral analog (7. 21) of...
- 15
- 212
- 0
THE CAUCHY – SCHWARZ MASTER CLASS - PART 8 pot
... has n k =1 n 1/ s pk xs k ≤ 1/ t pk xt k (8 .10 ) k =1 Finally, show that then one has equality in the bound (8 .10 ) if and only if x1 = x2 = · · · = xn 12 4 The Ladder of Power Means Fig 8 .1 If x > ... the bound of Case I gives us n pk x−t k 1/ t n pk x−s k ≤ k =1 1/ s k =1 Now, when we take reciprocals we find n pk x−s k n 1/ s pk x−t k ≤ k =1 1/t , k =1 1 so when we substitute xk = yk , we get ... weighted sums of reciprocals: · · · xpn n ≤ p1 x1 + p2 x2 + · · · + pn xn ≤ xp1 xp2 p1 p2 + + ··· + x1 x2 p1 p2 + + ··· + x1 x2 pn , xn pn xn (8 .15 ) (8 .16 ) Going to Extremes The last of the power...
- 15
- 183
- 0
THE CAUCHY – SCHWARZ MASTER CLASS - PART 9 potx
... formula (1/ s, 1/ t) = θ (1/ s1 , 1/ t1 ) + (1 − θ) (1/ s0 , 1/ t0 ) the bound (9. 41) automatically recaptures the inequality (9.24) from Challenge Problem 9.6; one only needs to set t1 = 1, s1 = 1, M1 = ... relation 1/ p + 1/ q = By splitting and by H¨lder’s inequality we find o m m n n q (cjk xp )1/ p (cjk yj )1/ q k cjk xk yj = j =1 k =1 j =1 k =1 m n 1/ p m n cjk xp k ≤ 1/ q q cjk yj j =1 k =1 , (9.26) j =1 k =1 and ... n 1 + = p q r n 1/ r ⇒ j =1 n 1/ p ap j ≤ (aj bj )r 1/ q bq j j =1 j =1 (b) Given p, q, and r are bigger than 1, show that if 1 + + = 1, p q r then one has the triple produce inequality n n 1/ p j=1...
- 20
- 249
- 0
THE CAUCHY – SCHWARZ MASTER CLASS - PART 10 pps
... we see that as → we have ∞ a2 ( ) m ∞ b2 ( ) n m =1 ∞ = n =1 n =1 ∞ ∼ n1+2 dx = (10 .11 ) x1+2 Closing the Loop To complete the solution of Problem 10 .2, we only need to show that the corresponding ... the Plan When we apply Cauchy’s inequality (10 .2) to the pair (10 .4), we find ∞ ∞ am bn m+n m =1 n =1 ∞ ∞ ≤ a2 m m m+n n m =1 n =1 2λ ∞ ∞ b2 n n m+n m n =1 m =1 2λ , so, when we consider the first factor ... explicit: ∞ 1 π dy = 2λ (1 + y) y sin 2πλ for < λ < 1/ 2 (10 .8) Now, since sin 2πλ is maximized when λ = 1/ 4, we see that the smallest value attained by Cλ with < λ < 1/ 2 is equal to ∞ C = C1/4 = 1 √...
- 11
- 166
- 0
THE CAUCHY – SCHWARZ MASTER CLASS - PART 11 docx
... (x)) dx = ak (11 .17 ) n 1 and, since ϕ(x)/x is nondecreasing, we also have the bound n 1/ n ak = exp k =1 −ϕ(n) n n ≤ exp n 1 −ϕ(x) x dx (11 .18 ) When we sum the relations (11 .17 ) and (11 .18 ), we then ... bound in our summation by parts inequality (11 .11 ), we find N n =1 (a1 +a2 +· · ·+an ) n N ≤4 n =1 (a1 +a2 +· · ·+an ) an , (11 .13 ) n and this is almost the inequality (11 .10 ) that we wanted to prove ... k=n ≤ k2 ∞ k=n ∞ k(k − 1) = k=n 1 − k 1 k = ≤ , n 1 n and, since s1 = + s2 ≤ + 1/ (2 − 1) = 2, we see that ∞ k=n ≤ k2 n for all n ≥ (11 .12 ) Hardy’s Inequality and the Flop 17 1 Now, when we use this...
- 12
- 245
- 0
THE CAUCHY – SCHWARZ MASTER CLASS - PART 12 ppsx
... assume x1 x2 x3 = We can then divide our bound by (x1 x2 x3 )2 to get 1 1 + + x1 x2 x1 x3 x2 x3 ≤ 1 1 + + x1 x2 x3 which may be expanded and simplified to 1 1 1 + + ≤ + + x1 x2 x1 x3 x2 x3 x1 x2 ... symmetric functions, the target bound (12 .15 ) may be written more systematically as E0 (1/ x1 , 1/ x2 , , 1/ xn )E2 (1/ x1 , 1/ x2 , , 1/ xn ) ≤ E1 (1/ x1 , 1/ x2 , , 1/ xn ), and one now sees that this ... assertions: E0 (x1 , x2 , x3 )E2 (x1 , x2 , x3 ) ≤ E1 (x1 , x2 , x3 ), E1 (x1 , x2 , x3 )E3 (x1 , x2 , x3 ) ≤ (12 .9) E2 (x1 , x2 , x3 ) (12 .10 ) Now the “remarkable identity” (12 .6) springs into...
- 13
- 180
- 0
THE CAUCHY – SCHWARZ MASTER CLASS - PART 13 pps
... chapter Problem 13 .5 Given x, y, z ∈ (0, 1) such that max(x, y, z) ≤ (x + y + z)/2 < 1, (13 .19 ) show that one has the bound 1+ x 1 x 1+ y 1 y 1+ z 1 z ≤ + (x + y + z) − (x + y + z) 2 (13 .20) If this ... sum n f (x1 , x2 , , xn ) = φ(xk ) (13 .17 ) k =1 is Schur convex Thus, for α, β ∈ (a, b)n with α ≺ β one has n n φ(αk ) ≤ k =1 φ(βk ) (13 .18 ) k =1 Orientation To familiarize the bound (13 .18 ), one ... 19 2 Majorization and Schur Convexity corresponding ordered values, {α[j] } and {β[j] }, we could just as well write the chain (13 .1) as (1, 1, 1, 1) ≺ (0, 1, 1, 2) ≺ (1, 3, 0, 0)...
- 17
- 158
- 0
THE CAUCHY – SCHWARZ MASTER CLASS - PART 14 pptx
... cn−h n =1 h=0 N +H H H = cn−j n =1 j=0 N +H cn−k ¯ H k=0 H 1 H |cn−s | + Re = n =1 s=0 cn−s cn−t ¯ s=0 t=s +1 N H 1 H N +H |cn | + Re = (H + 1) cn−s cn−t ¯ n =1 s=0 t=s +1 n =1 N H 1 H N +H ≤ (H + 1) |cn ... j =1 k =1 yj , (14 .39) j =1 for any n real numbers y1 , y2 , , yn , then we also have k max 1 k≤n n cj ψj (x) dx ≤ C log2 (4n) j =1 c2 k (14 .40) k =1 for all real c1 , c2 , , cn Exercise 14 .8 ... one has the bound M N | um , | m =1 n =1 M M ≤ 1/ 2 | um , uµ | m =1 µ =1 N N 1/ 2 | v n , vν | n =1 ν =1 Exercise 14 .12 (Selberg’s Inequality) Prove that if x and y1 , y2 , , yn are elements of...
- 18
- 164
- 0
THE CAUCHY – SCHWARZ MASTER CLASS - PART 15 docx
... factors n Dn +1 − Dn = n aj bj + nan +1 bn +1 − bn +1 j =1 n j =1 n aj − an +1 j =1 bj j =1 n aj (bj − bn +1 ) + an +1 = n (bn +1 − bj ) j =1 (an +1 − aj )(bn +1 − bj ) ≥ = j =1 According to Mitrinovi´ (19 70, p 206) ... µ2 ν2 ) 1 1 + ( 1 1 + µ2 ν2 ) 1 1 µ ¯ ¯ ¯ ¯ µ and, by expansion, this is the same as ¯ ¯ | 1 |2 + | 1 |2 + | 1 1 |2 + |µ2 ν2 |2 + 2Re { 1 1 µ2 ν2 } ¯ ¯ ≤ + 2| 1 1 |2 + 2Re { 1 1 µ2 ν2 ... L ≡ | 1 |2 + | 1 |2 + |µ2 ν2 |2 ≤ + | 1 1 |2 , but the substitution |µ2 ν2 |2 = (1 − | 1 |2 ) (1 − | 1 |2 − |ν3 |2 ) gives us L = + | 1 1 |2 + |ν3 |2 (| 1 |2 − 1) ≤ + | 1 1 |2 since | 1 |2 ≤...
- 59
- 235
- 0
THE CAUCHY – SCHWARZ MASTER CLASS - PART 16 ppt
... Schwarz inequality, 10 , 11 centered, 11 5 pointwise proof, 11 5 Schwarz’s argument, 11 , 63 failure, 13 6 in inner product space, 15 in light cone, 63 Schwarz, Hermann Amandus, 10 Selberg inequality, ... Russian edition, 19 49), Dover Publications, New York Index 1- trick, 11 0, 14 4, 14 6, 215 , 219 , 227, 2 31 defined, 226 first used, 12 refinement, 205, 288 Abel inequality, 208, 2 21 Abel, Niels Henrik, ... Yacovlevich, 10 , 19 0, 285 and AM-GM inequality, 11 5 Chebyshev contact, 76 vis-a-vis Schwarz, 11 Buzano, M.L., 286 Cai, T., vi, 246 cancellation, origins of, 210 Carleman inequality, 11 8 Carleson proof, 17 3...
- 22
- 336
- 0
MySql High Availability Class Part 1
... 1- 15 1. 9 .1 MySQL Community Web Page 1- 16 1. 10 Installing MySQL 1- 18 1. 11 Installing the "world" database 1- 22 1. 12 Chapter Summary 1- 23 ... 1- 1 1. 1 Learning Objectives 1- 1 1. 2 The MySQL Overview 1- 2 1. 2 .1 MySQL Partners 1- 3 1. 3 MySQL Products 1- 4 1. 3 .1 MySQL ... Enterprise 1- 9 1. 6 Supported Operating Systems 1- 11 1.7 MySQL Certification Program 1- 12 1. 8 Training Curriculum Paths 1- 13 1. 9 MySQL Website...
- 214
- 359
- 0
MySql High Availability Class Part 2
... Availability 11 Unauthorized reproduction or distribution prohibited Copyright© 2 010 , Oracle and/or its affiliates [NDBD] hostname =19 0 .14 5.45 .15 [MYSQLD] hostname =19 2 .16 8.2 .1 [MYSQLD] hostname =19 0 .14 5.45 .15 ... distribution prohibited Copyright© 2 010 , Oracle and/or its affiliates [NDBD] hostname =19 2 .16 8.2 .1 [NDBD] hostname =19 2 .16 8.2 .1 [MYSQLD] hostname =19 2 .16 8.2 .1 04 20 Original Lab 4-G, Step Find a ... hostname =19 2 .16 8.2 .1 (Replace IP with inet address obtained in step 1. 4) ns a r t n o an Revision s Add the hostname to the [NDBD] section also ) de i m u o [NDBD] g c ent G n i hostname =19 2 .16 8.2.1pp...
- 14
- 276
- 0
Tổng hợp part 1 question for beginners class
... SDT: 08 360 311 43_ Số 1/ 100 hẻm 56 Lữ Gia, Quận 11 SDT: 08 36020443 • Do you like (to go to/going to) parties? (Why?/Why not?) • Do you often go to parties? • Do you usually have these parties at ... Minh: Số 42 Nguyễn Phi Khanh, Quận SDT: 08 360 311 43_ Số 1/ 100 hẻm 56 Lữ Gia, Quận 11 SDT: 08 36020443 • (For university students) What are your favourite classes/courses/subjects at university? • ... 360 311 43_ Số 1/ 100 hẻm 56 Lữ Gia, Quận 11 SDT: 08 36020443 Sport/Exercise In the questions below, the word "sport" might be replaced by the word, "exercise" for some questions • Do you like any particular...
- 102
- 906
- 3
Class Notes in Statistics and Econometrics Part 1 pdf
... Known 10 99 10 99 11 08 11 11 111 6 11 26 11 32 11 46 11 52 11 65 Chapter 54 .1 54.2 54.3 54 Dynamic Linear Models Specification and Recursive Solution Locally Constant Model The Reference Model 11 69 11 69 11 75 ... Measures 10 36 10 37 10 38 10 38 10 42 10 43 10 43 10 44 10 45 10 50 Chapter 49 Distributed Lags 49 .1 Geometric lag 49.2 Autoregressive Distributed Lag Models 10 51 1062 10 63 Chapter 50 .1 50.2 50.3 10 73 10 73 10 76 ... Distribution 5 .10 The Lognormal Distribution 5 .11 The Cauchy Distribution 13 9 13 9 14 6 14 8 15 4 15 8 16 4 16 5 16 6 17 2 17 4 17 4 Chapter Sufficient Statistics and their Distributions 6 .1 Factorization...
- 88
- 363
- 0
Class Notes in Statistics and Econometrics Part 2 pptx
... instance, we could use the following set of bit strings for the eight horses: 0, 10 , 11 0, 11 10, 11 110 0, 11 110 1, 11 111 0, 11 111 1 Show that the the expected length of the message you send to your friend ... · · · + pmkm , then 3 .11 ENTROPY 10 3 H(p1 , , pn ) = H(w1 , , wn ) + + w1 H(p 11 /w1 + · · · + p1k1 /w1 ) + + w2 H(p 21 /w2 + · · · + p2k2 /w2 ) + · · · + + wm H(pm1 /wm + · · · + pmkm /wm ... (0, 1/ 36], in (1/ 36,3/36], in (3/36,6/36], in (6/36 ,10 /36], in (10 /36 ,15 /36], in (15 /36, 21/ 36], in ( 21/ 36,26/36], in (26/36,30/36], 10 in (30/36,33/36], 11 in (33/36,35/36], and 12 in (35/36 ,1] ...
- 75
- 303
- 0
Tìm thêm: xác định các nguyên tắc biên soạn khảo sát các chuẩn giảng dạy tiếng nhật từ góc độ lí thuyết và thực tiễn khảo sát chương trình đào tạo của các đơn vị đào tạo tại nhật bản khảo sát chương trình đào tạo gắn với các giáo trình cụ thể xác định thời lượng học về mặt lí thuyết và thực tế điều tra đối với đối tượng giảng viên và đối tượng quản lí điều tra với đối tượng sinh viên học tiếng nhật không chuyên ngữ1 khảo sát thực tế giảng dạy tiếng nhật không chuyên ngữ tại việt nam nội dung cụ thể cho từng kĩ năng ở từng cấp độ xác định mức độ đáp ứng về văn hoá và chuyên môn trong ct mở máy động cơ lồng sóc hệ số công suất cosp fi p2 đặc tuyến hiệu suất h fi p2 đặc tuyến mômen quay m fi p2 đặc tuyến dòng điện stato i1 fi p2 động cơ điện không đồng bộ một pha sự cần thiết phải đầu tư xây dựng nhà máy thông tin liên lạc và các dịch vụ phần 3 giới thiệu nguyên liệu chỉ tiêu chất lượng theo chất lượng phẩm chất sản phẩm khô từ gạo của bộ y tế năm 2008