THE CAUCHY – SCHWARZ MASTER CLASS - PART 8 pot
THE CAUCHY – SCHWARZ MASTER CLASS - PART 8 pot
... 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 refinemen...
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THE CAUCHY – SCHWARZ MASTER CLASS - PART 2 pot
... focus the appli- cation of an inequality on the point (or the region) where the inequality is most effective. For example, in the derivation of the AM-GM inequal- ity from the bound 1 + x ≤ e x , the ... least the following: • “Can you solve your problem in a special case?” 28 The AM-GM Inequality Pursuit of a Principle By the hypothesis on the left-hand side of the...
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THE CAUCHY – SCHWARZ MASTER CLASS - PART 6 pot
... 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 representation (6.7), the ... the Gauss–Lucas Theorem. Exercise 6.12 (The Gauss–Lucas Theorem) Show that for any complex polynomial P (z)=a 0 + a 1 z + ···+ a n z n , the roots of the derivative P (z) are...
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THE CAUCHY – SCHWARZ MASTER CLASS - PART 9 potx
... 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 ... proofs of his bound (9.37). In the first of these he called on the Cauchy Binet formula [see (3.7), page 49], and the second he used the AM-GM inequality which he wrote in...
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THE CAUCHY – SCHWARZ MASTER CLASS - PART 3 pptx
... worth the attention of the world’s mathematicians at the dawn of the 20th century. The prob- lems were wisely chosen, and they have had a profound influence on the development of mathematics over the ... characterization of the case of equality for the n-dimensional Cauchy inequality. Problem 3.1 (On Equality in Cauchy s Bound) Show that if (b 1 ,b 2 , ,b n ) =0then equal...
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THE CAUCHY – SCHWARZ MASTER CLASS - PART 4 doc
... 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 ... a Plan If the Cauchy Schwarz Master Class were to have a final exam, then the light cone inequality would provide fertile ground for the develop- ment of good problems. O...
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THE CAUCHY – SCHWARZ MASTER CLASS - PART 5 doc
... 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 ... connec- tion to probability theory, and it has many applications in other areas of mathematics. Nevertheless, the probabilistic interpretation of the bound 5 Consequences of...
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THE CAUCHY – SCHWARZ MASTER CLASS - PART 7 pptx
... 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 ... replaced by 1. The essence of the challenge is therefore to beat the naive immediate application of Schwarz s inequality. Taking the Hint If we want to apply the pat...
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THE CAUCHY – SCHWARZ MASTER CLASS - PART 10 pps
... and some of these are quite brief. Nevertheless, for the connoisseur of tech- niques for exploiting Cauchy s inequality, this proof of Hilbert’s inequal- ity is a sweet victory. Finally, there is ... Compensating Difficulties The two sums on the right-hand side of the naive bound (10.3) diverge, but the good news is that they diverge for different reasons. In a sense, the first fac...
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THE CAUCHY – SCHWARZ MASTER CLASS - PART 11 docx
... y 2 )/2. In the first two examples the flop was achieved with help from Cauchy s inequality or Schwarz inequality, but the basic idea is obviously quite general. In the next problem (and in several of the ... 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...
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