THE CAUCHY – SCHWARZ MASTER CLASS - PART 2 pot
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 6 pot
... 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...
THE CAUCHY – SCHWARZ MASTER CLASS - PART 8 pot
... =
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...
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 ... ≥ 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...
THE CAUCHY – SCHWARZ MASTER CLASS - PART 3 pptx
... 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
+...
THE CAUCHY – SCHWARZ MASTER CLASS - PART 4 doc
... 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...
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 ... 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...
THE CAUCHY – SCHWARZ MASTER CLASS - PART 7 pptx
... 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...
THE CAUCHY – SCHWARZ MASTER CLASS - PART 10 pps
... 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...
THE CAUCHY – SCHWARZ MASTER CLASS - PART 11 docx
... 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...