THE CAUCHY – SCHWARZ MASTER CLASS - PART 2 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 } +...

Ngày tải lên: 14/08/2014, 05:20

... triangle with the traditional labelling of Figure 6 .2, the law of cosines tells us that a 2 = b 2 + c 2 −2bc cos α. Show that this law implies the area formula a 2 =(b − c) 2 +4A tan(α /2) , then show ... 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...

Ngày tải lên: 14/08/2014, 05:20

... = n  k=1  k/n (k−1)/n ya k dy = 1 2n 2 n  k=1 (2k − 1)a k , (8 .26 ) 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 ... 1)a k ∞  n=k 1 n 2 ≤ ∞  k=1 (2k − 1)a k ∞  n=k  1 n − 1 2 − 1 n + 1 2  = ∞  k=1 2k − 1 k − 1 2 a k =2 ∞  k=1 a k and, when we insert this bound in the...

Ngày tải lên: 14/08/2014, 05:20

... 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 ... ≥ 2 and if a j ≥ 0 for all 1 ≤ j ≤ n, then there exists a constant λ = λ(a,p) such that a j ∈ [(λ − δ 1 2 ) 2/ p , (λ + δ 1 2 ) 2/ p ] for all j =1, 2, ,n. (9 .20 ) In othe...

Ngày tải lên: 14/08/2014, 05:20

... B.C.). The classic identity 1=cos 2 (α + β)+sin 2 (α + β) permits one to deduce that (a 2 1 + a 2 2 )(b 2 1 + b 2 2 ) equals (a 1 b 1 + a 2 b 2 ) 2 +(a 1 b 2 − a 2 b 1 ) 2 . Fig. 3.1. In the right ... Q n (a 1 ,a 2 , ,a n ; b 1 ,b 2 , ,b n ) that is given by the difference of the squares of the two sides of Cauchy s inequality, then Q n equals (a 2 1 +...

Ngày tải lên: 14/08/2014, 05:20

... w≤ ˜ v, ˜ v 1 2 w, w 1 2 = v, v 1 2 w, w 1 2 . The outside terms yield the complex Cauchy Schwarz inequality in the precisely the form we expected, so the bound (4 .20 ) was strong enough after all. The ... e 1 e 1 + x 2 , e 2 e 2 ···= ··· x n = x n , e 1 e 1 + x n , e 2 e 2 + ···+ x n , e n e n . Exercise 4.9 (Gram–Schmidt Implies Cauchy Sch...

Ngày tải lên: 14/08/2014, 05:20

... 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 ... using the AM- GM inequality, and when it is applied to the additive bound (5.5), one finds  n  k=1 a 2 k  1 2  mM n  k=1 b 2 k  1 2 ≤ 1 2  n  k=1 a 2 k...

Ngày tải lên: 14/08/2014, 05:20

... have  t 0 x 2 f(x) dx = f(t)t 2 +1 2 +1 − 1 2 +1  t 0 x 2 +1 f  (x) dx = f(t)t 2 +1 2 +1 + 1 2 +1  t 0 x 2 +1 |f  (x)|dx (7.8) ≥ 1 2 +1  t 0 x 2 +1 |f  (x)|dx. By the hypothesis (7.7) the ... it is often applied in proba- bility theory. For example, if we take f(u)=e −u 2 /2  √ 2 then it gives the lower bound e −t 2 /2 2t √ 2 ≤ 1 √ 2  ∞ t e −u 2 /2 du...

Ngày tải lên: 14/08/2014, 05:20

... y) 1 y 2 dy =Γ (2 )Γ(1 − 2 )for0<λ<1 /2. (10 .25 ) As a consequence, one finds that the evaluation of the integral (10.8) yields the famous functional equation for the Gamma function, Γ (2 )Γ(1 2 )= π sin ... n  m n  2 ∞  n=1 ∞  m=1 b 2 n m + n  n m  2 , so, when we consider the first factor on the right-hand side we see ∞  m=1 ∞  n=1 a 2 m m + n  m n  2...

Ngày tải lên: 14/08/2014, 05:20

... 1 72 Hardy’s Inequality and the Flop terms of A n and A n−1 , then we have ∆ n = A 2 n − 2A n a n = A 2 n − 2A n  nA n − (n −1)A n−1  =(1− 2n)A 2 n +2( n −1)A n A n−1 , but unfortunately the ... T N equals s 1 a 2 1 −s N+1 (a 1 +a 2 +···+ a n ) 2 + N  n =2 s n  2( a 1 +a 2 +···+ a n−1 )a n +a 2 n  and, since s N+1 (a 1 + a 2 + ···+ a n ) 2 ≥ 0, we at last find...

Ngày tải lên: 14/08/2014, 05:20