THE CAUCHY – SCHWARZ MASTER CLASS - PART 9 potx
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
} +...
THE CAUCHY – SCHWARZ MASTER CLASS - PART 3 pptx
... 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...
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 ... 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...
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 ... 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...
THE CAUCHY – SCHWARZ MASTER CLASS - PART 6 pot
... 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...
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 ... 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...
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 refineme...
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 ... 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...
THE CAUCHY – SCHWARZ MASTER CLASS - PART 10 pps
... 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...
THE CAUCHY – SCHWARZ MASTER CLASS - PART 11 docx
... 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...
THE CAUCHY – SCHWARZ MASTER CLASS - PART 12 ppsx
... 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...
THE CAUCHY – SCHWARZ MASTER CLASS - PART 13 pps
... 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...
THE CAUCHY – SCHWARZ MASTER CLASS - PART 14 pptx
... 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 (...
THE CAUCHY – SCHWARZ MASTER CLASS - PART 15 docx
... 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...
THE CAUCHY – SCHWARZ MASTER CLASS - PART 16 ppt
... 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,...