carmywarserver class part 2
THE CAUCHY – SCHWARZ MASTER CLASS - PART 2 pot
... one has a1 b1 + a2 b2 + a3 b3 + a4 b4 = {a1 b1 + a2 b2 } + {a3 b3 + a4 b4 } 1 1 ≤ (a2 + a2 ) (b2 + b2 ) + (a2 + a2 ) (b2 + b2 ) 2 4 1 ≤ (a2 + a2 + a2 + a2 ) (b2 + b2 + b2 + b2 ) , 4 which is ... α+β α+β (2. 26) and, for a typical corollary, show that one also has the more timely bound x2004 y + xy 20 04 ≤ x2005 + y 20 05 Exercise 2. 4 (A Canadian Challenge) Participants in the 20 02 Canadian ... to 2k The average A is listed 2k − n times in the padded sequence {αi }, so, when we apply inequality (2. 5) to {αi }, we find 2k −n a1 a2 · · · an ·A 1/2k ≤ a1 + a2 + · · · + an + (2k − n)A 2k...
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THE CAUCHY – SCHWARZ MASTER CLASS - PART 3 pptx
... Pythagoras (circa 497 B.C.) The classic identity = cos2 (α + β) + sin2 (α + β) permits one to deduce that (a2 + a2 )(b2 + b2 ) equals 2 (a1 b1 + a2 b2 )2 + (a1 b2 − a2 b1 )2 Fig 3.1 In the right light, ... needs multiplication to verify the identity of Diophantus, (a1 b1 + a2 b2 )2 = (a2 + a2 )(b2 + b2 ) − (a1 b2 − a2 b1 )2 , 2 (3.10) yet multiplication does not suggest how such an identity might ... We Know The simplest nontrivial case of Lagrange’s identity is (a2 + a2 )(b2 + b2 ) = (a1 b1 + a2 b2 )2 + (a1 b2 − a2 b1 )2 , 2 and, since polynomials may be substituted for the reals in this...
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THE CAUCHY – SCHWARZ MASTER CLASS - PART 4 doc
... (x + y + z) ≤ x2 + y + (b) Show for < x ≤ y ≤ z that √ y2 + z2 ≤ x + y2 + z2 + x2 + z (y − x )2 + (z − x )2 (c) Show for positive x, y, z that √ ≤ x2 + y + z + x 2 + y 2 + z 2 This list can ... centered at the origin Fig 4.1 This arrangement of = 22 + circles in [ 2, 2] 2 has a natural generalization to an arrangement of 2d + spheres in [ 2, 2] d This general arrangement then provokes a question ... inequality Problem 4 .2 (Triangle Inequality for Euclidean Distance) Show that the function ρ : Rd × Rd → R defined by ρ(x, y) = (y1 − x1 )2 + (y2 − x2 )2 + · · · + (yd − xd )2 (4 .2) satisfies the triangle...
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THE CAUCHY – SCHWARZ MASTER CLASS - PART 5 doc
... 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 this ... a2 and b1 ≤ b2 imply ≤ (a2 − a1 )(b2 − b1 ), and when this is unwrapped, we find a1 b2 + a2 b1 ≤ a1 b1 + a2 b2 , Consequences of Order 79 which is precisely the rearrangement inequality (5. 12) ... Show that if < m = x1 ≤ x2 ≤ · · · ≤ xn = M < ∞ then for nonnegative weights with p1 + p2 + · · · + pn = one has n n pj xj j=1 pj j=1 xj ≤ 2 2 (5 .21 ) √ where µ = (m + M ) /2 and γ = mM This bound...
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THE CAUCHY – SCHWARZ MASTER CLASS - PART 6 pot
... implies the area formula a2 = (b − c )2 + 4A tan(α /2) , then show how Jensen’s inequality implies that in any triangle one has √ a2 + b2 + c2 ≥ (a − b )2 + (b − c )2 + (c − a )2 + 3A This bound is known ... bound M x2 − f (¯) ≤ ¯ x n pk k=1 M x2 − f (xk ) k where we have set x = p1 x1 + p2 x2 + · · · + pn xn , and this bound is easily ¯ rearranged to yield n pk f (xk ) −f (¯) ≤ x k=1 M n pk x2 − 2 x ... log 2) visible, the graph is not drawn to scale.) convexity properties of f , so we just differentiate twice to find f (x) = − ex 2( 1 + ex )3 /2 and f (x) = − (1 + ex )−3 /2 ex + (1 + ex )−5 /2 e2x...
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THE CAUCHY – SCHWARZ MASTER CLASS - PART 7 pptx
... note that for any ≤ t < ∞ we have t t f (t)t2α+1 − x2α+1 f (x) dx 2 + 2 + t f (t)t2α+1 + x2α+1 |f (x)| dx = 2 + 2 + t x2α+1 |f (x)| dx ≥ 2 + x2α f (x) dx = (7.8) By the hypothesis (7.7) ... to the splitting xα+β+1 |f (x)| = {x (2 +1) /2 |f (x)|1 /2 } {x (2 +1) /2 |f (x)|1 /2 } we find the nice intermediate bound ∞ (1 + α + β )2 I ≤ ∞ x2α+1 |f (x)| dx x2β+1 |f (x)| dx Now we see how we can ... h such that h ≤ f (x)/D(x) one has h F (x) ≥ h u2 |f (u) |2 du = (x + t )2 |f (x + t) |2 dt h (x + t )2 |f (x) − t D(x) |2 dt ≥ ≥ hx2 {f (x) − h D(x) }2 , or, a bit more simply, we have 1 F (x) ≥ h...
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THE CAUCHY – SCHWARZ MASTER CLASS - PART 8 pot
... Means 129 Problem 8.4 (Termwise Bounds for Carleman’s Summands) Show that for positive real numbers ak , k = 1, 2, , one has (a1 a2 · · · an )1/n ≤ e 2n2 n (2k − 1)ak for n = 1, 2, , (8 .22 ) ... (ny) dy = yak dy = k=1 (k−1)/n 2n2 n (2k − 1)ak , (8 .26 ) k=1 so, in view of the general bound (8 .25 ) and the identity (8 .23 ), the proof of the first inequality (8 .22 ) of the challenge problem is ... Exercise 8 .2 (Harmonic Means and Recognizable Sums) Suppose x1 , x2 , , xn are positive and let S denote their sum Show that we have the bound S n2 S S ≤ + + ··· + (2n − 1) 2S − x1 2S − x2 2S −...
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THE CAUCHY – SCHWARZ MASTER CLASS - PART 9 potx
... convex φ and positive weights w1 , w2 , , wn one has φ w1 x1 + w2 x2 + · · · + wn xn w1 + w2 + · · · + wn w1 φ(x1 ) + w2 φ(x2 ) + · · · + wn φ(xn ) ≤ w1 + w2 + · · · + wn (9.31) Consider the ... real root of the 146 H¨lder’s Inequality o equation n n p /2 aj nλ − 2 p 1−p +n aj j=1 = (9 .22 ) j=1 Since a quadratic equation A 2 + 2Bλ + C = has a real root if and only if AC ≤ B , we see that ... inequality which he wrote in the form w1 +w2 +···+wn w1 x1 + w2 x2 + · · · + wn xn w1 + w2 + · · · + wn xw1 xw2 · · · xwn ≤ n where the values w1 , w2 , ,wn are assumed to be positive but which...
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THE CAUCHY – SCHWARZ MASTER CLASS - PART 10 pps
... (t) k k=1 a2 k =2 k=1 n k a2 k k=1 Hilbert’s Inequality and Compensating Difficulties 165 (c) Combine the preceding observations to conclude that n n ≤ 2 ak n a2 k k=1 k=1 k a2 k (10 .24 ) k=1 This ... Compensating Difficulties In the specific case of f (x) = m2λ x2λ (m + x)−1 , we therefore find ∞ m m+n n n=1 ∞ 2 ≤ m2λ dx = m + x x2λ ∞ 1 dy, (10.6) (1 + y) y 2 where the last equality comes from the change ... (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 2 ∞ ∞ b2 n n m+n m n=1 m=1 2 , so, when we consider the first factor on the right-hand side we see ∞ ∞ m a2 m...
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THE CAUCHY – SCHWARZ MASTER CLASS - PART 11 docx
... a2 − sN +1 (a1 + a2 + · · · + an )2 + sn 2( a1 + a2 + · · · + an−1 )an + a2 n n =2 and, since sN +1 (a1 + a2 + · · · + an )2 ≥ 0, we at last find N n=1 (a1 + a2 + · · · + an ) n N 2 sn (a1 + a2 ... “humble bound” and note that if we replace An An−1 by (A2 + A2 ) /2 we have n n−1 ∆n ≤ (1 − 2n)A2 + (n − 1) A2 + A2 n n n−1 = (n − 1)A2 − nA2 n−1 n After a few dark moments, we now find that we ... now try to write ∆n just in 1 72 Hardy’s Inequality and the Flop terms of An and An−1 , then we have ∆n = A2 − 2An an n = A2 − 2An nAn − (n − 1)An−1 n = (1 − 2n)A2 + 2( n − 1)An An−1 , n but unfortunately...
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THE CAUCHY – SCHWARZ MASTER CLASS - PART 12 ppsx
... 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 action ... 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 x3 ... second bound ( 12. 10) To make the task clear we first rewrite the bound ( 12. 10) in longhand as x1 + x2 + x3 {x1 x2 x3 } ≤ x1 x2 + x1 x3 + x2 x3 ( 12. 11) This bound is trivial if x1 x2 x3 = 0, so...
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THE CAUCHY – SCHWARZ MASTER CLASS - PART 13 pps
... we just take α = (α1 , 2 ) = (ρ + σ, ρ − σ) and take β = (β1 , 2 ) = (ρ + τ, ρ − τ ) There is no loss of generality in assuming α1 ≥ 2 , β1 ≥ 2 , and α1 + 2 = β1 + 2 ; moreover, no loss in ... such that max(x1 , x2 , , xn ) ≤ (x1 + x2 + · · · + xn )/k, show that one has n j=1 k2 ≤ (n − k) + + xj k + x1 + x2 + · · · + xn (13 .21 ) Majorization and Schur Convexity 20 5 Exercise 13.3 (A ... numbers x1 , x2 , , xn such that m xk = k=1 m n n xk + δ (13 .22 ) k=1 where δ ≥ 0, show that the sum of squares has the lower bound n x2 ≥ k k=1 n n xk + k=1 δ2n m(n − m) (13 .23 ) This refinement...
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THE CAUCHY – SCHWARZ MASTER CLASS - PART 14 pptx
... or its square |SM (P ) |2 If we try brute force, we will need an n-term analog of the ¯ familiar formula |c1 + c2 |2 = |c1 |2 + |c2 |2 + 2Re {c1 c2 }, and this calls for 21 2 Cancellation and Aggregation ... have |C(k)| ≤ log2 (k) ≤ log2 (n) and C(k) ⊂ B, 22 0 Cancellation and Aggregation so the double sum bound (14 .25 ) gives us max |a1 + a2 + · · · + ak |2 ≤ log2 (n) 1≤k≤n aj , (14 .26 ) B∈B j∈B and ... Cancellation and Aggregation 20 9 we first note that a1 z1 + a2 z2 + · · · + an zn = a1 S1 + a2 (S2 − S1 ) + · · · + an (Sn − Sn−1 ) = S1 (a1 − a2 ) + S2 (a2 − a3 ) + · · · + Sn−1 (an−1 − an...
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THE CAUCHY – SCHWARZ MASTER CLASS - PART 15 docx
... b2 − xb2 y b1 = xa2 y a2 (xa1 −a2 + y a1 −a2 − xb1 −a2 y b2 −a2 − xb2 −a2 y b1 −a2 ) = xa2 y a2 (xb1 −a2 − y b1 −a2 )(xb2 −a2 − y b2 −a2 ) ≥ 0, since b1 − a2 ≥ b2 − a2 = a1 − b1 ≥ Lee (20 02) notes ... |2 ≤ + (¯1 ν1 + 2 2 )µ1 ν1 + (µ1 ν1 + 2 2 )¯1 ν1 µ ¯ ¯ ¯ ¯ µ and, by expansion, this is the same as ¯ ¯ |µ1 |2 + |ν1 |2 + |µ1 ν1 |2 + | 2 2 |2 + 2Re {µ1 ν1 2 2 } ¯ ¯ ≤ + 2| µ1 ν1 |2 + 2Re ... α sin β = b2 + b2 a2 + a2 , , we find the Pythagorean path to the identity of Diophantus: = cos2 (α + β) + sin2 (α + β) = (a1 b1 + a2 b2 )2 + (a1 b2 − a2 b1 )2 (b2 + b2 )(a2 + a2 ) 2 Solution...
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THE CAUCHY – SCHWARZ MASTER CLASS - PART 16 ppt
... 144, 146, 21 5, 21 9, 22 7, 23 1 defined, 22 6 first used, 12 refinement, 20 5, 28 8 Abel inequality, 20 8, 22 1 Abel, Niels Henrik, 20 8 Acz´l, J., 28 7 e additive bound, 4, 9, 66, 106 and H¨lder, 137, 26 4 o ... 154, 22 7, 26 3, 26 7 defined, 22 6 first used, 12 grand champion, 26 6 stability in H¨lder inequality, 145 o of AM-GM inequality, 35 steepest ascent, 67 Stef˘nescu, D., 26 2 ¸ a stress testing, 26 8 Stromberg, ... inequality, 28 5 Magiropoulos, M., 28 6 majorant principles, 28 4 majorization, 191 Maligranda, L., vi, 26 4, 28 8, 28 9 marriage lemma, 20 6 Marshall, A., 27 7 Matouˇek, J., vi, 28 5 s McConnell, T.R., 27 7 Meng,...
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MySql High Availability Class Part 1
... 1 -22 1. 12 Chapter Summary 1 -23 INTRODUCTION TO MYSQL HIGH AVAILABILITY 2- 1 2. 1 Overview 2- 1 2. 2 What is High Availability? 2- 2 2. 2.1 ... Availability defined 2- 2 2. 2 .2 Downtime 2- 4 2. 2.3 Scheduled vs Unplanned Downtime 2- 5 2. 2.4 Business Impact 2- 6 2. 3 Overview of Availability ... 2- 9 2. 4.1 Linux-HA 2- 9 2. 4 .2 Windows Clustering 2- 9 2. 4.3 Veritas Cluster Agent 2- 9 2. 4.4 Red Hat Cluster Suite 2- 9 2. 5 Heartbeat...
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MySql High Availability Class Part 2
... distribution prohibited Copyright© 20 10, Oracle and/or its affiliates [NDBD] hostname=1 92. 168 .2. 1 [NDBD] hostname=1 92. 168 .2. 1 [MYSQLD] hostname=1 92. 168 .2. 1 04 20 Original Lab 4-G, Step Find a second ... [NDB_MGMD] hostname=1 92. 168 .2. 1 (Server “A” Virtual IP Address) datadir=/var/lib/mysql-cluster [NDBD DEFAULT] NoOfReplicas =2 datadir=/var/lib/mysql-cluster [NDBD] hostname=1 92. 168 .2. 1 (Server “A” Virtual ... hostname=1 92. 168 .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=1 92. 168 .2. 1pp...
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Tổng hợp part 1 question for beginners class
... Hà Nội: Số 89 /27 Đại Cồ Việt SDT: 04 665788 92; Số 2/ 850 Láng SDT: 04 325 95447 Hồ Chí Minh: Số 42 Nguyễn Phi Khanh, Quận SDT: 08 36031143_ Số 1/100 hẻm 56 Lữ Gia, Quận 11 SDT: 08 36 020 443 Working ... Hà Nội: Số 89 /27 Đại Cồ Việt SDT: 04 665788 92; Số 2/ 850 Láng SDT: 04 325 95447 Hồ Chí Minh: Số 42 Nguyễn Phi Khanh, Quận SDT: 08 36031143_ Số 1/100 hẻm 56 Lữ Gia, Quận 11 SDT: 08 36 020 443 • Possibly ... Hà Nội: Số 89 /27 Đại Cồ Việt SDT: 04 665788 92; Số 2/ 850 Láng SDT: 04 325 95447 Hồ Chí Minh: Số 42 Nguyễn Phi Khanh, Quận SDT: 08 36031143_ Số 1/100 hẻm 56 Lữ Gia, Quận 11 SDT: 08 36 020 443 • (For...
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Class Notes in Statistics and Econometrics Part 1 pdf
... Random Variables 199 20 3 21 0 21 2 21 6 21 8 22 1 22 6 23 5 Chapter Random Matrices 9.1 Linearity of Expected Values 9 .2 Means and Variances of Quadratic Forms in Random Matrices 24 5 24 5 24 9 Chapter 10 The ... Chapter 22 Specific Datasets 563 x CONTENTS 22 .1 22 .2 22. 3 22 .4 22 .5 Cobb Douglas Aggregate Production Function Houthakker’s Data Long Term Data about US Economy Dougherty Data Wage Data 563 580 5 92 ... element only, and ∅ for the empty set ∞ (2. 2.9) n=1 ∞ (2. 2.10) n=1 ∞ ,2 = n ,2 = n 0, = n 0, + = n n=1 ∞ n=1 Answer ∞ ,2 n (2. 2.11) , = (0, 2] n (2. 2. 12) n=1 ∞ 0, n =∅ 0, + n = [0, 1] n=1 n=1...
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Class Notes in Statistics and Econometrics Part 2 pptx
... E[(x − E[x] )2 ] var[x] = E[(x − µ )2 ] = E[x2 − 2x(E[x]) + (E[x] )2 ] (3.10.17) 2 = E[x ] − 2( E[x]) + (E[x]) = E[x2 ] − (E[x] )2 = E[x2 − 2xµ + 2 ] = E[x2 ] − 2 2 + 2 = E[x2 ] − 2 • b point ... t2 x2 t3 x3 + + ··· 2! 3! t2 t3 (3.10 .21 ) mx (t) = E[etx ] = + t E[x] + E[x2 ] + E[x3 ] + · · · 2! 3! d t2 (3.10 .22 ) mx (t) = E[x] + t E[x2 ] + E[x3 ] + · · · dt 2! d2 (3.10 .23 ) mx (t) = E[x2 ... Answer Pr[t = 0] = (1 − θ )2 ; Pr[t = 1] = 2 (1 − θ); Pr[t = 2] = 2 Therefore an application of (3.10.1) gives E[t3 ] = 03 · (1 − θ )2 + 13 · 2 (1 − θ) + 23 · 2 = 2 + 6 2 Theorem 3.10.1 Jensen’s...
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