THE CAUCHY – SCHWARZ MASTER CLASS - PART 6 pot

... b 2 2 ) 1 2 +(a 2 3 + a 2 4 ) 1 2 (b 2 3 + b 2 4 ) 1 2 ≤ (a 2 1 + a 2 2 + a 2 3 + a 2 4 ) 1 2 (b 2 1 + b 2 2 + b 2 3 + b 2 4 ) 1 2 , which is Cauchy s inequality for n = 4. Extend this argument to obtain Cauchy s ... exam- ple, by Cauchy s inequality for n = 2 applied twice, one has a 1 b 1 + a 2 b 2 + a 3 b 3 + a 4 b 4 = {a 1 b 1 + a 2 b 2 } +...

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... Exercise 3. 8. Roots and Branches of Lagrange’s Identity Joseph Louis de Lagrange (1 73 6–1 8 13) developed the case n =3of the identity (3. 4) in 17 73 in the midst of an investigation of the geom- etry ... mathematical address of all time. In his lecture, Hilbert de- scribed 23 problems which he believed to be worth the attention of the world’s mathematicians at the dawn...

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... bonus. The numer- ator on the right-hand side of the identity (4. 7) must also be positive, and this observation gives us yet another proof of the Cauchy Schwarz inequality. There are even two further ... that the case n =1 of Bessel’s inequality is equivalent to the Cauchy Schwarz inequality. Exercise 4. 11 (Gram–Schmidt and Products of Linear Forms) Use the Gram–Schmidt p...

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... A}. Exercise 5. 3 (Cauchy Schwarz and the Cross-Term Defect) If u and v are elements of the real inner product space V for which on has the upper bounds u, u≤A 2 and v, v≤B 2 , then Cauchy s inequality ... inequality may be of help. Exercise 5. 5 (The P´olya–Szeg˝o Converse Restructured) The converse Cauchy inequality (5. 7) is expressed with the aid of bounds on the ra...

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... function, the left-hand side of the bound (6. 16) is identically equal to zero, but in general the bound (6. 16) asserts a more subtle relation. More precisely, it tells us that the left-hand side is ... exploiting the hypothesis f  (·) ≥ 0 by noting that it implies that the integrand f  (·) is nondecreasing. In fact, our hypothesis contains no further information, so the re...

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... may suspect that the fundamental theorem of calculus will somehow help. This is The Cauchy- Schwarz Master Class, so here one may not need long to think of applying the 1-trick and Schwarz s inequality ... factor on the right- hand side of the bound (7. 4) would be replaced by 1. The essence of the challenge is therefore to beat the naive immediate application of Schwa...

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... one has the bound max 1<k≤n k−1  j=1 1 |x σ(k) − x σ(j) | ≥ 1 8 n log n. 1 28 The Ladder of Power Means Despite the differences in the two forms (8. 18) and (8. 19), the defini- tion (8. 19) should ... ( 189 6–1 981 ) observed in his lectures on the geome- try of numbers that the limit representation (8. 2) for the geometric mean can be used to prove an elegant refineme...

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... central core of the classical theory of inequal- ities, and we have already seen three of these: the Cauchy Schwarz inequality, the AM-GM inequality, and Jensen’s inequality. The quartet is completed ... S p = n  k=1 a k b p k for p>0. (9. 37) Rogers gave two proofs of his bound (9. 37). In the first of these he called on the Cauchy Binet formula [see (3.7), page 49] , and...

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... 0wehave  ∞  m=1 a 2 m ()  1 2  ∞  n=1 b 2 n ()  1 2 = ∞  n=1 1 n 1+2 ∼  ∞ 1 dx x 1+2 = 1 2 . (10. 11) Closing the Loop To complete the solution of Problem 10. 2, we only need to show that the corresponding sum for the left-hand side of Hilbert’s inequality (10. 10) is asymptotic ... that all the quantities in the inequality (10. 10) tend to infinity as  → 0. This pa...

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... negativity of the last term, the proof of the preflop inequality (11. 14) is complete. Finally, we know already that the flop will take us from the inequality (11. 14) to the inequality (11. 9) of our ... some of the clutter may be removed by setting A n =(a 1 + a 2 + ···+ a n )/n. Also, if we consider the term-by- term differences ∆ n between the summands in the preflop ineq...

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... E 2 2 (x 1 ,x 2 ,x 3 ). (12. 10) Now the “remarkable identity” (12. 6) springs into action. The assertion (12. 9) says for three variables what the inequality (12. 8) says for two, therefore (12. 6) tells us ... first group (12. 12) follow from the induction hypothesis H n and the identity (12. 6). All of the inequalities of H n have come to us for free, except for one. If we writ...

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... vectors, then in matrix notation the relation (13. 7) says that α ∈ H(β)=⇒ α = Dβ (13. 10) where D is the doubly stochastic matrix defined by the sums (13. 8). Now, to complete the solution of the first ... Day-to-Day Example The final challenge addresses a typical example of the flood of prob- lems that one can solve — or invent — with help from the tools devel- oped in this chap...

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... provide the single most common source. Such bounds lie behind the two introductory examples (14. 2) and (14. 3), and, although these are particularly easy, they still point toward an important theme. Linear ... challenge problem. In fact, these may be the best known of van der Corput many inequalities, even though they are notably less subtle than the bound (14. 17). Exercise 14. 5 (...

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... Exercise 4.2. The derivative on the left is equal to ∇f(x), u which is bounded by ∇f(x)u = ∇f(x) by the Cauchy Schwarz inequality. On the other hand, the derivative on the right is equal ... + n  k=1 p 2 k + n  k=1 1/p 2 k , (14.44) and then we estimate the last two terms separately. By the 1-trick and the hypothesis p 1 + p 2 + ···+ p n = 1, the first of these...

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... of the few articles to advocate the inductive approach to Cauchy s inequality that is favored in this chapter. The Cram´er–Rao inequality of Exercise 1.15 illustrates one way that the Cauchy Schwarz ... 101 Cauchy, Augustin-Louis, 10 Cauchy Binet identity, 49 Cauchy Schwarz inequality, 8 as accidental corollary, 57 cross term defects, 83 geometric proof, 58 self-generalization,...

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