<a name="docA001">　Index begin　Index this file
2009-06-08-19-10 start
This file http://freeman2.com/tute0008.htm
is Freeman's reading notes. Although Freeman
always keep correct view point. But Freeman's
capability is limited, plus no one proofread
this file. Then you can still find wrong view
point. When read, please put question mark as
often as possible. If you suspect any view 
point wrong, please ask a math expert near by.

<a name="docA002">,<a name="textbook">
This file is a note for reading inequality
book written by Professor J. Michael Steele
The Cauchy-Schwarz Master Class  　★★★★★
Below use 'textbook' as abbreviation.
Freeman also read web pages online, and will 
indicate the source URL at discussion point.

<a name="docA003">
Freeman study mechanical engineering. 
Engineering mathematics do not teach 
inequality. Above book is first inequality
book. First time read, it was very hard.
Although high school time learned
  a*a + b*b >= 2*a*b
But this little knowledge do not help.

<a name="docA004">
This file follow textbook chapter section
order, but not continuous, Freeman skip 
those uncertain sections/problems.

This file first function is to learn 
inequality. Second function is to learn
how to use html code to write math 
equations.

<a name="docA005">　Index begin　Index this file
On 2009-01-27-10-08 Freeman accessed
the next page
http://www.sftw.umac.mo/~fstitl/10mmo/inequality.html
save as sftw.umac.mo_text_math_eqn_good.htm
Above page is the main reference for html
math equation.

In order to let reader to build  html math
equation, previous file tute0007.htm page 
end has math symbol and internal code.

This file third function is to display how
to draw curves in web page. The main engine
is XYGraph v2.3 - Technical Figures

Thank you for read Freeman's inequality page.
2009-06-08-19-47 stop

<a name="docA006">
2009-08-23-15-00 start
□　Exercise 1.14 solution

Exercise 1.14 problem (a) is solved by hint
Exercise 1.14 problem (b) is out of LiuHH's
reach. What is "cardinality of a set B⊂Z3" ?
Without clear understand the definition, 
LiuHH is unable to solve problem (b)
LiuHH major in mechanical engineering.
LiuHH is mathematics admirer and outsider.

<a name="docA007">
To solve problems in
The Cauchy-Schwarz Master Class
LiuHH's goal is to solve 50% of the exercise
problems. In reality, solve 40% is good enough.
Please do not be surprise that LiuHH skip some
text section, skip some problems.
2009-08-23-15-12 stop

<a name="ch14d001">　Index begin　Index this file
2010-08-12-12-17 start
■　Exercise 14.7 problem statement
　　textbook page 223
（Rademacher-Menchoff with Weight）

<a name="ch14d002">
Let  ψ1,ψ2,...,ψn be real-valued
functions for which

2010-08-12-12-53 stop





<a name="ch14d006">　Index begin　Index this file
2010-08-12-15-32 start
■　Exercise 14.7 hint
　　textbook page 282

For any A⊂{1,2,...,n} we have

<a name="ch14d008">
2010-08-12-15-51 here
where the last inequality comes 
from applying the hypothesis
(14.39) where
  yj=cj if j∈A, and  ---eqn.BT001
  yj=0 if j NOT∈ A ---eqn.BT002
//Seq. yj and seq. cj relation.
//seq. cj in A, cj size |A|≦n
//Seq. yj in {1,2,...,n}, yj size n

<a name="ch14d009">
Next, if we replace ai by ciψi(x) 
in the real-variable inequality 
(14.26) and integrate, we find

<a name="ch14d013">
which is slightly stronger
than the target inequality 
(14.40)
2010-08-12-16-26 stop

<a name="ch14d015">　Index begin　Index this file
2010-08-12-17-31 start
■　Exercise 14.7 solution


Exercise 14.7 hint start from
eqn.14.63 It is integral of
square of sum of a sequence.
<a name="ch14d016">
This is a partial sequence.
Because "A⊂{1,2,...,n}" mean
partial. Equality in eqn.14.63
come from square operation and
eqn.14.38. If j=k, then ajk=1
thst is demand by eqn.14.38A.
<a name="ch14d017">
Square of sum of a sequence 
must change from (∑[j∈A])2 
to (∑[j∈A])(∑[k∈A]) so that
it cover the general case,
allow different index j≠k be
multiplied.

<a name="ch14d018">
eqn.14.63 less than side has
different index multiplication
and same index multiplication
sum together.
eqn.14.63 greater than side NO
different index multiplication
greater than side has only sum
of same index multiplication
<a name="ch14d019">
Hypothesis (14.39) promise there
is a constant C such that fewer
term sum of same index multipli-
cation multiply C has greater
value than more terms product
sum.

<a name="ch14d020">　Index begin　Index this file
Inequality in eqn.14.63 come
from applying eqn.14.39
Up to here we talk about partial
sum in subset A⊂{1,2,...,n}
eqn.14.26 allow us link
max[1≦k≦n]{partial sum}_square
≦ceiling(log2(n))*{total_sum}_square

<a name="ch14d021">
Why "AA≦" line ci missing?
Why "BB≦" line ci reappear?
2010-08-12-18-23

<a name="ch14d022">
eqn.14.26 both side have ai.
Since we replace ai by ciψi(x) 
then ciψi(x) should present at 
"AA≦" line

<a name="ch14d023">
Exercise 14.7 continue from 
Problem 14.4 which LiuHH was 
unable to understand. More
question is in "B∈B", what is 
the difference between two B's?
B has document at textbook
page 218 upper half page.

<a name="ch14d024">
Liu,Hsinhan need more time to 
think.
Exercise 14.7 is NOT DONE
2010-08-12-18-31 stop



<a name="ch14d025">　Index begin　Index this file
2010-08-12-19-41 start
■　Exercise 14.8 problem statement
　　textbook page 223
（Functions with Geometric Dependence）

<a name="ch14d026">
If the constant ρ satisfies
  0＜ρ＜1 ---eqn.BT004
and the sequence of functions {ψj}
satisfies

<a name="ch14d028">
2010-08-12-20-00 here
for all 1≦j,k≦n, then there is 
a constant M depending only on ρ
such that partial sums
  Sk(x)= ψ1+ψ2+...+ψk ---eqn.BT006
1≦k≦n 
satisfy the maximal inequality

2010-08-12-20-11 stop




<a name="ch14d030">　Index begin　Index this file
2010-08-12-21-27 start
■　Exercise 14.8 hint
　　textbook page 282

//One ●●change, many occurrence.
//suggest change from ρ-|j-k| to ρ+|j-k|
//key point is that 0＜ρ＜1 ---eqn.BT004
// 1/(1-x)=1+x+x^2+x^3+... for 0＜x＜1
// |j-k| are integers.
//2010-08-16-13-55

<a name="ch14d031">
From the splitting

<a name="ch14d037">
Given the inequality (14.64), 
the conclusion of Exercise 14.8 
with the value
  M=(1+ρ)/(1-ρ) ---eqn.BT011
follows from Exercise 14.7.
2010-08-12-22-20 stop



<a name="ch14d038">　Index begin　Index this file
2010-08-12-23-33 start
■　Exercise 14.8 solution


Exercise 14.8 is special.
Exercise 14.8 problem statement
say sequence of functions {ψj(x)}
no one word about number sequence
yj (not a function).

<a name="ch14d039">
Exercise 14.8 hint do just the 
reverse.
Exercise 14.8 hint say number 
sequence yj and no one word 
about seq. of functions {ψj(x)}

<a name="ch14d040">
At the end, 14.8 hint says 
"follows from Exercise 14.7"
Go to Exercise 14.7, it says
Let ψ1,ψ2,...,ψn be real-valued
functions for which eqn.14.38 
indicate that function ψi(x) is 
not orthonormal, since no word 
say ajk=0, eqn.14.38B is not 
required to be zero.

<a name="ch14d041">
Go back to Problem 14.4
It require function φi(x) be
orthonormal.

<a name="ch14d042">
Problem not require orthonormal is 
not compatible with orthonormal-
required problem. Exercise 14.7
still titled "Rademacher-Menchoff 
with Weight"
The word "with Weight", does it
mean to drop orthonormal condition?

<a name="ch14d043">
Rademacher-Menchoff related problems
and exercise need more time to think.
Exercise 14.8 is NOT DONE
2010-08-12-23-56 stop

<a name="ch14d044">　Index begin　Index this file
2010-08-13-09-28 start
Exercise 14.7 use two sequences
{yi} and {cj}. If two are
identical, no need use two. One 
is enough. Difference between
{yi} and {cj} is eqn.BT001 and 
eqn.BT002
<a name="ch14d045">
{yi} is in complete set
{cj} is in subset A
subset=A⊂{1,2,...,n}=complete set
{yi} has n elements
{cj} has |A| elements, |A|≦n
<a name="ch14d046">
Sum {yi} n elements is OK
Sum {cj} |A| elements is OK
Sum {cj} n elements is wonder !
eqn.14.63 sum {cj} |A| elements, 
but, 
eqn.14.40 sum {cj} n elements
How to explain ?

<a name="ch14d047">
Exercise 14.7 problem statement
say: for all real c1,c2,...,cn.
Exercise 14.7 hint say
For any A⊂{1,2,...,n} we have
eqn.14.63 where cj is summed 
within subset A, |A|≦n
<a name="ch14d048">
Problem statement say size cn
Hint say size c|A|. How to 
explain the difference?
2010-08-13-10-00 here

<a name="ch14d049">　Index begin　Index this file
Problem 14.4 (Rademacher-Menchoff
 Inequality) say
[[
Given that the functions
  φk:[0,1] to complex 1≦k≦n ---eqn.BR032
are orthonormal.
]]
<a name="ch14d050">
Orthonormal is combination of 
orthogonal and normalized.

orthonormal definition is 
j=k: ∫[x=0,1]φjφkdx=1 //normalized
j≠k: ∫[x=0,1]φjφkdx=0 //orthogonal
//above ---eqn.BT012
<a name="ch14d051">
Exercise 14.7 link to
Rademacher-Menchoff with Weight
define
j=j: ∫[x=0,1]ψjψjdx=1 //normalized
j≠k: ∫[x=0,1]ψjψkdx=ajk 
<a name="ch14d052">
Exercise 14.7 problem statement
did not say ajk=0 for j≠k
How can we conclude orthogonal
condition?
2010-08-13-10-16 stop



<a name="ch14d053">　Index begin　Index this file
2010-08-13-10-23 start
■　Exercise 14.9 problem statement
　　textbook page 224
（The Subset Lower Bound）

<a name="ch14d054">
Show that for complex numbers
z1,z2,...,zn one has

<a name="ch14d055">
2010-08-13-10-37 here
and show that the constant factor
1/π cannot be replaced by a larger
one. The qualitative message of 
this cancellation story is that
<a name="ch14d056">
there is always some subset with
a sum whose modulus is a large
fraction of the sum of all the
moduli. For a hint one might
consider the special subset Sθ
defined in Figure 14.1.
2010-08-13-10-42 stop

<a name="ch14d057">　Index begin　Index this file
2010-08-13-16-56 start
■　Exercise 14.9 hint
　　textbook page 283

From the definition of Sθ 
one finds

<a name="ch14d061">
2010-08-13-17-38 here
So indeed, there must exist 
some value θ* for which f(θ*)
is at least as large as the 
last sum. By taking 
{zk=exp(ik2π/N):0≦k≦N}
<a name="ch14d062">　Index begin　Index this file
for large N one can show that 
the constant 1/π cannot be
improved. This argument follows
W.W. Bledsoe(1970), Mitrinovic 
(1970, p.331) notes that similar 
results were obtained earlier by
D.Z. Djokovic.
2010-08-13-17-44 stop


<a name="ch14d063">
2010-08-13-19-25 start
■　Exercise 14.9 solution


<a name="ch14d064">
eqn.BT013 "def=" is similar to 
eqn.14.41 greater than side.
Difference is that 
f(θ) may be max[I⊂{1,2,...,n}]
f(θ) may not be max[I⊂{1,2,...,n}]
while eqn.14.41 greater than side
pin point to max[I⊂{1,2,...,n}]

<a name="ch14d065">
"DD=" is whole set Sθ rigid rotate
-θ angle. Sθ arrow is +θ angle
from real axis. Rotate -θ angle 
move Sθ arrow point to positive
real axis direction. 
Why rotate? LiuHH is still 
thinking. why1 why2 why3
<a name="ch14d066">
Although Sθ arrow point to positive
real axis, but the net sum left
to "EE≧" may not be real only.
A complex number with non-zero 
imaginary, we have
abs(complex)≧Real(complex)
For example 
abs(3+4i)=5＞3=real(3+4i)
Above explain "EE≧" 

<a name="ch14d067">　Index begin　Index this file
"FF=" rewrite complex number 
real part as 
complex modulus*cos(phase angle)
Phase angle has zk contribution
arg(zk) and rotation exp(-θ) 
contribution. Net result is
θ-arg(zk) or -θ+arg(zk) it does
not matter, since cos(+θ)=cos(-θ)

<a name="ch14d068">
"GG=" remove outer absolute sign.
Because complex modulus |zk| is
nonnegative and angle θ-arg(zk) 
is within +90 to -90 degree. 
Cosine value is nonnegative for
sure. No need outer absolute 
sign. Please see figure 14.1, 
click [Draw fig.14.1] button

<a name="ch14d069">
Next integrate eqn.BT013 two ends
from θ=0 to θ=2PI. Divide integral
value by angle 2PI, result is 
average of f(θ) over 2PI period.

<a name="ch14d070">
"II≧" is a copy of "EE≧", it come
with eqn.BT013.

<a name="ch14d071">
"JJ=" separate θ dependent variable
and non-θ dependent constants. Do
calculation separately.
Why "JJ=" change from ∑[zk∈Sθ]
to ∑[k=1,n] why from partial 
sum to complete sum?
Why "JJ=" change from ∫[θ=0,2PI]
range 2PI to range 1*PI?
[arg(zk)+π/2]-[arg(zk)-π/2]=π
why still be equality? why1 why2 why3
<a name="ch14d072">　Index begin　Index this file
"KK=" is integration of
cos(θ-arg(zk))*dθ ---eqn.BT301
here θ is variable, arg(zk) is
constant (no θ).
  d[constant]=0 ---eqn.BT302
we add a zero=d[-arg(zk)] to dθ
result is // ---eqn.BT303
cos(θ-arg(zk))*d(θ-arg(zk))
that is
  ∫d[sin(θ-arg(zk))] ---eqn.BT304

<a name="ch14d073">
Upper bound give us 
sin[ arg(zk)+π/2 -arg(zk) ]
=sin[+π/2]=1 ---eqn.BT305
Lower bound give us
sin[ arg(zk)-π/2 -arg(zk) ]
=sin[-π/2]=-1 ---eqn.BT306
<a name="ch14d074">
"KK=" step integral result is
+1-(-1)=2. This numerator 2 
cancel "II≧" side denominator 2
End up with 1/π and sum of |zk|
See "KK=" right side.

<a name="ch14d075">
Why 
"for large N one can show that 
the constant 1/π cannot be
improved."
LiuHH still thinking.
Puzzles are here. why1 why2 why3
Exercise 14.9 is NOT DONE
2010-08-13-20-17 stop

<a name="ch14d076">
2010-08-13-23-25 include
LangeListCSMCTypos.pdf
17. Page 283, line 7. 
Why is it possible to include 
all points rather than just 
those in the half plane Sθ? 
The limits of integration on 
the integral also do not make 
sense.



2010-08-13-22-14 start
<a name="ch14d077">　Index begin　Index this file
■　Exercise 14.10 problem statement
　　textbook page 224
（A Domination Principle）

<a name="ch14d078">
If the complex numbers an satisfies
the bound 
  |an|≦An ---eqn.BT015
1≦n≦N, then the complex array
  {ynr: 1≦n≦N, 1≦r≦R} ---eqn.BT016
satisfies the bounds

2010-08-13-22-32 stop




<a name="ch14d080">
2010-08-13-22-35 start
■　Exercise 14.10 hint
　　textbook page 283

If L and R denotes the left and
right sides of the target bound
(14.42), then by squaring and
change order, one finds the
representation

<a name="ch14d084">
so our hypothesis gives us
L≦R. The bound (14.42) provide
a generic example of a class of
inequalities called majorant 
principles, and the treatment 
given here follows Theorem 4
of Montgomery (1994, p. 132)
2010-08-13-23-01 stop


<a name="ch14d085">
2010-08-14-09-51 start
■　Exercise 14.10 solution


L is defined to be less than 
side of eqn.14.42, that is
  L=∑[r=1,R]∑[s=1,R] ---eqn.BT019
   |∑[n=1,N]anynrynsj|2
//complex conjugate of c is cj

<a name="ch14d086">
Complex square rule say
complex number c1 square is
 c1*c1_conjugate correct
 c1*c1 wrong
Now |∑[n=1,N]anynrynsj|2 ---eqn.BT020
is a square term, equivalent 
c1 is ∑[n=1,N]anynrynsj ---eqn.BT021
equivalent c1_conjugate is
   ∑[n=1,N]anjynrjyns ---eqn.BT022
<a name="ch14d087">　Index begin　Index this file
eqn.BT021 and eqn.BT022 use same
index variable n which cause
confusion, we can not see unlike
index term. Since variable n is 
dummy, change eqn.BT021 index
from n to m as next
   ∑[m=1,N]amymrymsj ---eqn.BT023
So that cross product is visible.
<a name="ch14d088">
L is eqn.BT022 * eqn.BT023 It 
has four summation sign, four
variables are r, s, n, m. This 
step explain "MM=" in eqn.BT017 

<a name="ch14d089">
"MM=" line has four summation
sign. y's participate all four
summations. But two a's has 
only n and m index. 
"NN=" step let ∑n and ∑m stand
in front, they influence every
terms in L formula. Put amanj 
after ∑n, ∑m and before ∑r ∑s
<a name="ch14d090">
Look like
∑n ∑m amanj ∑r ∑s 4_y's here
When we do 4_y's calculation,
2_a's stay out side. Let picture
be more clear.
<a name="ch14d091">
"PP=" step regroup 
from ∑r ∑s 4_y's 
to   ∑r 2_yr ∑s 2_ys
Result is ∑n ∑m amanj multiply
∑r ymrynrj and multiply ∑s ymsjyns
<a name="ch14d092">　Index begin　Index this file
"QQ=" step is to combine two
like terms to square of one.
∑r ymrynrj and ∑s ymsjyns are 
like terms, variable r and s
are dummy variable. If change
s to r, then two multiplied
terms are complex conjugate 
to each other, see next
∑r ymrynrj multiply ∑r ymrjynr 
<a name="ch14d093">
We get square term |∑r ymrynrj|2
in "QQ=" line.
//complex conjugate of c is cj
//Review complex square rule

<a name="ch14d094">
Above is calculation for eqn.14.42
left side L. We do same thing for
right side R get eqn.BT018 

<a name="ch14d095">
eqn.BT018 and "QQ=" line both let
amanj and AmAn stand out clearly.
Compare eqn.BT018 and "QQ=" line 
refer to given condition
  |an|≦An ---eqn.BT015
<a name="ch14d096">
We conclude that right side R
is greater than or equal to
left side L of target eqn.14.42
Exercise 14.10 is done
2010-08-14-10-50 stop



<a name="ch14d097">　Index begin　Index this file
2010-08-14-13-29 start
■　Exercise 14.11 problem statement
　　textbook page 225
（An Inequality of P. Enflo）

<a name="ch14d098">
Show that for vectors 
um 1≦m≦M and vn 1≦n≦N, 
in the inner product space Cd 
one has the bound

2010-08-14-13-48 stop





<a name="ch14d100">
2010-08-14-14-00 start
■　Exercise 14.11 hint
　　textbook page 284

The most direct proof just require
a big piece of paper and a timely
application of Cauchy's inequality.
<a name="ch14d101">
First expand the squares |〈um,vn〉|2 
in terms of the vector components
umj 1≦j≦d and vnk 1≦k≦d. Next,
change the order of summation so 
that the double sum over j and k
is outermost, and only now apply
Cauchy's inequality. Finally, 
<a name="ch14d102">　Index begin　Index this file
within each of the two resulting
rooted expressions, you change 
the order of summation within
each of the braces and reinterpret
the sums innermost sums as inner
products.
<a name="ch14d103">
This solution amplifies the remark
of Montgomery (1994, p. 144) that
manipulations like those used in
the solution of Exercise 14.10 can
be used to prove Enflo's inequality.
<a name="ch14d104">
An alternative solution may be based
on the observation that the functions
  φn,m(x,y)=e(mx)e(ny) ---eqn.BT025
//define e(mx)=exp(2πimx); i=√(-1)
are orthonormal on the space [0,1]2
One then introduce the function

<a name="ch14d106">
and exploits the fact that the 
integral of |f(x,y)|2 over [0,1]2
gives the left side of Enflo's 
inequality.
2010-08-14-14-25 stop


<a name="ch14d107">　Index begin　Index this file
2010-08-14-18-13 start
■　Exercise 14.11 solution


eqn.BT024 say that there are two
groups vectors, 
One is u group, it has M elements.
Two is v group, it has N elements.
M+N vectors are in complex_d_dim
space. Create a numerical example.
<a name="ch14d108">
Real number belong to complex. For 
simplicity assume complex_d_dim is
real_2_dimension.
Assume u group has M=3 elements
{u}={u1, u2, u3}  ---eqn.BT027
as next //real as special complex
{u}={[3.1,3.2], [3.3,3.4],
     [3.5,3.6]} ---eqn.BT028
Vector u1=[3.1,3.2] ---eqn.BT029
number u11=3.1 ---eqn.BT030
number u12=3.2 ---eqn.BT031

<a name="ch14d109">
Assume v group has N=4 elements
{v}={v1, v2, v3, v4} ---eqn.BT032
 as next
{v}={[4.1,4.2], [4.3,4.4],
     [4.5,4.6], [4.7,4.8]} ---eqn.BT033
Vector v3=[4.5,4.6] ---eqn.BT034
number v31=4.5 ---eqn.BT035
number v32=4.6 ---eqn.BT036
<a name="ch14d110">
Double foot note 31, 
3 mean vector count as third.
1 mean first number element.

<a name="ch14d111">
eqn.BT024 less than side is dot
product of one u element and
one v element, sum all possible
combination.
<a name="ch14d112">　Index begin　Index this file
eqn.BT024 greater than side is 
square root of
[[
dot product of one u element and
still one u element, sum all 
possible combination multiply
with similar case for v group.
]]
<a name="ch14d113">
This is different from Cauchy's
inequality.
eqn.BT024 greater than side is
dot product of two from same 
group but different vector.
Cauchy's inequality greater than 
side is dot product of two from 
same vector.

<a name="ch14d114">
Expand the squares |〈um,vn〉|2 
m=1,n=1: |〈u1,v1〉|2 
  =[3.1,3.2]dot[4.1,4.2]
  = 3.1*4.1+3.2*4.2 = 26.15 ---eqn.BT037
square real get 683.8225
If square complex, here need
complex*complex_conjugate

<a name="ch14d115">
In symbolic equation, we have
∑m∑n[um1vn1+um2vn2]2 ---eqn.BT038
For this simple example, each
vector has only two elements.
Write [um1vn1+um2vn2] in third
summation as ∑m∑n∑j[umjvnj]2 ---eqn.BT039

<a name="ch14d116">
How to expand the square?
2010-08-14-19-03 stop

<a name="ch14d117">　Index begin　Index this file
2010-08-15-09-53 start
[[
First expand the squares |〈um,vn〉|2 
in terms of the vector components
umj 1≦j≦d and vnk 1≦k≦d.
]]
<a name="ch14d118">
Expand the square as next line
∑m∑n∑j[umjvnj]*∑μ∑ν∑k[uμkvνk] ---eqn.BT040

<a name="ch14d119">
//drop ∑m∑n∑j∑k[umjvnjumkvnk] ---eqn.BT340
//drop ∑m∑n∑j∑k[umjumjvnkvnk] ---eqn.BT341
//drop ∑m∑n∑j[umjvnj]*∑s∑t∑k[uskvtk]
//drop ∑m∑n∑j[umjvnj*umjvnj]
<a name="ch14d120">
//drop four sum ∑m∑n |∑j∑k[umjvnk]|2
//drop four sum ∑m∑n |∑j∑k[umjvnk]
//drop                *∑s∑t[umsvnt]|
//drop new 4 sum ∑j∑k∑s∑t{∑m∑n[umjvnkumsvnt]}

<a name="ch14d121">
Next,
change the order of summation so 
that the double sum over j and k
is outermost, 
∑j∑k {∑m∑n∑μ∑ν[umjvnjuμkvνk]} ---eqn.BT041

<a name="ch14d122">　Index begin　Index this file
only now apply Cauchy's inequality.
Cauchy's first sequence is
  ∑m∑μ[umjuμk] ---eqn.BT042
Cauchy's second sequence is
  ∑n∑ν[vnjvνk] ---eqn.BT043
<a name="ch14d123">
Cauchy's inequality give us
  ∑m∑n∑μ∑ν[umjvnjuμkvνk]
 ≦{∑m∑μ[umjuμk]2}½
 *{∑n∑ν[vnjvνk]2}½ ---eqn.BT044

<a name="ch14d124">
This is equation for one j and
one k. For total j,k we have
∑j∑k ( ∑m∑n∑μ∑ν[umjvnjuμkvνk])
 ≦   // ---eqn.BT045
∑j∑k ( {∑m∑μ[umjuμk]2}½
      *{∑n∑ν[vnjvνk]2}½ )

<a name="ch14d125">　Index begin　Index this file
Finally, 
within each of the two resulting
rooted expressions, you change 
the order of summation within
each of the braces and reinterpret
the sums innermost sums as inner
products.
<a name="ch14d126">
 ∑m∑n∑μ∑ν (∑j∑k[umjvnjuμkvνk])
 ≦ //  ---eqn.BT046
 {∑m∑μ (∑j∑k [umjuμk])2}½
*{∑n∑ν (∑j∑k [vnjvνk])2}½ 

<a name="ch14d127">
reinterpret as
 ∑m∑n |〈um,vn〉|2
 ≦ // ---eqn.BT047
 {∑m∑μ |〈um,uμ〉|2 }½
*{∑n∑ν |〈vn,vν〉|2 }½ 

<a name="ch14d128">
LiuHH do not have confidence
for above calculation. Worry
could have error, prefer to 
say Exercise 14.11 is NOT DONE
2010-08-15-11-18



<a name="ch14d129">　Index begin　Index this file
2010-08-15-13-32 start
■　Exercise 14.12 problem statement
　　textbook page 225
（Selberg's Inequality）

<a name="ch14d130">
Prove that if x and y1,y2,...,yn 
are elements of a real or complex
product space, then we have

<a name="ch14d132">
Selberg's inequality can sometimes
be used as a replacement for
the orthonormality identity 
(14.22) or Bessel's inequality
(4.29) when the elements 
y1,y2,...,yn are only 
approximately orthogonal.
<a name="ch14d133">
Techniques for relaxing the 
requirement of orthonormality 
have important consequences 
throughout probability, number
theory, and combinatorics.
2010-08-15-14-08 stop





<a name="ch14d134">　Index begin　Index this file
2010-08-15-14-10 start
■　Exercise 14.12 hint
　　textbook page 284

<a name="ch14d135">
One always has
  〈z,z〉≧0 ---eqn.BT048
so if we set
  z=x-(c1y1+c2y2+...+cnyn) ---eqn.BT049
We find for all cj, 1≦j≦n that

<a name="ch14d140">
simple algebra bring us to
the inequality (14.43). This 
argument is based on the classic
exposition of E. Bombieri (1974).
2010-08-15-15-00 stop



<a name="ch14d141">
2010-08-15-16-20 start
■　Exercise 14.12 solution


<a name="ch14d142">
Problem given that x and y1,y2,
...,yn are complex numbers
(complex cover real, real do 
not cover complex) eqn.BT050 
suggest that cj j=1 to n are 
complex.

<a name="ch14d143">
Start from
  〈z,z〉≧0 ---eqn.BT048
Apply eqn.BT049 to get
  [x-(c1y1+c2y2+...+cnyn)] ---eqn.BT054
 *[xj-(c1jy1j+c2jy2j+...+cnjynj)]≧0

<a name="ch14d144">　Index begin　Index this file
Define Complex inner product as 
following
 〈c,d〉 = c1d1j+c2d2j+c3d3j ---eqn.AK041
Left side 'd' do not use super j.
right side 'd' must use super j.
complex conjugate of c is cj
eqn.AK041 explains that eqn.BT048
use 〈z,z〉≧0
not 〈z,zj〉≧0
eqn.BT054 is expanded 〈z,z〉
eqn.BT054 need use super j clearly.
<a name="ch14d145">
Multiply out eqn.BT054, result is
eqn.BT050
In eqn.BT050, can not use 〈x,xj〉
can not use cjckj〈yj,ykj〉
Blue super j is complex conjugate,
complex inner product take complex 
conjugate one more time cause error.

<a name="ch14d146">
LiuHH wonder for a while, why in
cjckj〈yj,yk〉, yk not take complex
conjugate? Above is study notes
and thinking notes.
2010-08-15-16-55 here

<a name="ch14d147">
Humble bound has link at
tute0010.htm#9808131533
and tute0033.htm#ch09a005

<a name="ch14d148">
If a,b are both real number, expand
 (a-b)*(a-b)≧0 get
 a*a+b*b ≧ 2*a*b  ---eqn.AD18
This is called "humble bound"
Textbook page 19, line 6.

<a name="ch14d149">　Index begin　Index this file
In eqn.AD18, 
change 'a' to cj
change 'b' to ck
Divide eqn.AD18 by two get 
eqn.BT051
Use eqn.BT051 replace cjckj
in last term of eqn.BT050
the result is eqn.BT052 

<a name="ch14d150">
Up to here, coefficient cj is
unspecified. We have freedom
to choose cj value.
The choice of cj , and the 
definition of eqn.BT053 
is to use cj denominator to
cancel same factor in "SS+"
and "TT+" two terms.
<a name="ch14d151">
After cancellation of 
∑[k=1,n] |〈yj,yk〉|
Simple algebra after eqn.BT052 
is next //2010-08-15-17-26 here

<a name="ch14d153">
2010-08-15-17-54 here
Define 'sumAB' as eqn.14.43 
less than side expression. 
eqn.BT055 is in the following
simplified form
  0 ≦〈x,x〉-sumAB-sumAB
    +sumAB/2+sumAB/2 ---eqn.BT056
<a name="ch14d154">
where -sumAB+sumAB/2+sumAB/2 add
to zero, what left is
  0 ≦〈x,x〉-sumAB ---eqn.BT057
Move sumAB to less than side, 
the result is target equation 
eqn.14.43 exactly.
2010-08-15-18-01 stop



<a name="ch14d155">　Index begin　Index this file
2010-08-16-10-21 start
Numerical verify eqn.BT010

The following code //2010-08-16-15-10
get incorrect output. 
Typo point is that pow(ρ,-abs(k-j))
Corrected equation: pow(ρ,+abs(k-j))
Corrected code is here

//<a name="ch14d156">
//2010-08-16-11-00 code
var j,k,n,sumN,sumMax,ρ;
n=3;
ρ=.25; //0＜ρ＜1 ---eqn.BT004
sumN=[];
sumN[0]=0;
sumMax=[0,0];
var ans0=ans1=ans2='';
//<a name="ch14d157">
//9908161120 switch j,k
for(k=1;k<=n;k++)
  {
sumN[k]=0;
for(j=1;j<=n;j++) 
sumN[k]+=pow(ρ,-abs(k-j)) //typo line
//'+abs()' OK, but '-abs()' typo
if(sumMax[1]<sumN[k]) {dummy=0;
sumMax[0]=k; //[else] button
sumMax[1]=sumN[k];    }
ans1+='k='+k+', sumN[k]='+sumN[k]+'\n'
  }
//<a name="ch14d158">
ans0=''
+'Given n='+n+', ρ='+ρ+'\n'
+'At k='+sumMax[0]+' maxSum='+sumMax[1]+'\n'
ans0 //eqn.BT010 less than side sum
//2010-08-16-11-14 code
//Above is eqn.BT010 less than summation
//<a name="ch14d159">
//below is eqn.BT010 greater than sum
//require j in Z (+/- integer) next code
//sum from j=1 to j=n100 for +integer
//then double (2*sum100) add negative
//integer. Then add pow(ρ,0) for j=0 case
//sum100 has j=-n100 to j=0 to j=+n100
//2010-08-16-11-35 notes
//2010-08-16-11-24 code
//<a name="ch14d160">　Index begin　Index this file
var n100,sum100;
n100=100
sum100=0
for(j=1;j<=n100;j++)
sum100+=pow(ρ,-abs(j)); //typo line
sum100=2*sum100+pow(ρ,0)
sum100 //eqn.BT010 greater than side sum
//
var sum101; //9908161329 pow(ρ,+abs(j))
n100=100
sum101=0
for(j=1;j<=n100;j++)
sum101+=pow(ρ,+abs(j)); //correct
sum101=2*sum101+pow(ρ,0)
sum101 //eqn.BT010 greater than side sum
//
//<a name="ch14d161">
var onePρoneMρ=(1+ρ)/(1-ρ);
onePρoneMρ  //eqn.BT010 (1+ρ)/(1-ρ)
//2010-08-16-11-28
ans1=(ans1+'').replace(/,/g,'\n'); //9908161114
ans1 //eqn.BT010 less than side detail
//copy/paste the above code to
//http://freeman2.com/complex2.htm#calculator
//box 3, click [test box3 command, output to box4]


<a name="ch14d162">
2010-08-16-09-31 start proofread
tute0052.htm reached eqn.BT010
Do not know how to prove eqn.BT010
write program code to verify this
equation. But code can not reach
the equality shown in eqn.BT010.
//  0＜ρ＜1 ---eqn.BT004
2010-08-16-12-12 stop


//<a name="ch14d163">
//Numerical verify eqn.BT010. Next 
//is corrected code. Key change is 
//from typo  pow(ρ,-abs(k-j))
//to correct pow(ρ,+abs(k-j))
//typo show up at textbook 
//page 282 bottom half page.
//2010-08-16-11-00 code
//2010-08-16-14-59 correct code
var j,k,n,sumN,sumMax,ρ;
n=3; ρ=.25; //0＜ρ＜1 ---eqn.BT004
//You can change above two numbers
//n be integer and 0＜ρ＜1. Output
// maxSum ≦ sum101 = onePρoneMρ
sumN=[];
sumN[0]=0;
sumMax=[0,0];
var ans0=ans1=ans2='';
//<a name="ch14d164">
//9908161120 switch j,k
for(k=1;k<=n;k++)
  {  //pow(ρ,-abs(k-j)) is typo
sumN[k]=0; //+abs(k-j)  is correct
for(j=1;j<=n;j++) //2010-08-16-15-00
sumN[k]+=pow(ρ,+abs(k-j))
if(sumMax[1]<sumN[k]) {dummy=0;
sumMax[0]=k;
sumMax[1]=sumN[k];    }
ans1+='k='+k+', sumN[k]='+sumN[k]+'\n'
  }
//<a name="ch14d165">　Index begin　Index this file
ans0=''
+'Given n='+n+', ρ='+ρ+'\n'
+'At k='+sumMax[0]+' maxSum='+sumMax[1] //+'\n'
ans0 //eqn.BT010 less than side sum
//2010-08-16-11-14 code
//Above is eqn.BT010 less than summation
//<a name="ch14d166">
//below is eqn.BT010 greater than sum
//require j in Z (+/- integer) next code
//sum from j=1 to j=n100 for +integer
//then double (2*sum100) add negative
//integer. Then add pow(ρ,0) for j=0 case
//sum101 has j=-n100 to j=0 to j=+n100
//2010-08-16-11-35 notes
//2010-08-16-11-24 code
//<a name="ch14d167">
var sum101; //9908161329 pow(ρ,+abs(j))
n100=100
sum101=0
for(j=1;j<=n100;j++)
sum101+=pow(ρ,+abs(j))
sum101=2*sum101+pow(ρ,0)
sum101 //eqn.BT010 greater than side sum
//<a name="ch14d168">
var onePρoneMρ=(1+ρ)/(1-ρ);
onePρoneMρ  //eqn.BT010 (1+ρ)/(1-ρ)
//2010-08-16-11-28
ans1=(ans1+'').replace(/,/g,'\n'); //9908161114
ans1 //eqn.BT010 less than side detail
//copy/paste the above code to
//http://freeman2.com/complex2.htm#calculator
//box 3, click [test box3 command, output to box4]
//2010-08-16-15-05 done correct code


<a name="ch14d169">　Index begin　Index this file
2010-08-16-15-33 start
■　∑[j=-inf,+inf]x|j|=(1+x)/(1-x)
　　for 0＜x＜1

After more time think, solved one 
puzzle.
Explain eqn.BT010 as next.
Less than side
  ∑[j=1,n]ρ|j-k| ---eqn.BT058
is summation j over n terms.
<a name="ch14d170">
There are k=n different flavors.
Output "ans1" list n different 
flavors. eqn.BT010 less than side 
choose maximum sum of n flavors.
eqn.BT010 greater than side sum
j from -infinity to +infinity.
<a name="ch14d171">
Inequality in eqn.BT010 has very
simple reason. When all terms
are positive, how can less than 
side n sum beat the infinite sum? 

<a name="ch14d172">
Equality in eqn.BT010 is also
simple. But need a little work.
The basic observation is next

<a name="ch14d173">
Given 0<x<1 //9908161312
(1+x)/(1-x) = what

First simple relation
1/(1-x) = // ---eqn.BT059
  1 +x +x^2 +x^3 +x^4 +x^5+.....

<a name="ch14d174">　Index begin　Index this file
x/(1-x) = what?
Because
1/(1-x) - x/(1-x) = 1
    = (1-x)/(1-x) ---eqn.BT060
then
x/(1-x) = 1/(1-x) - 1 ---eqn.BT061

<a name="ch14d175">
Now come to
(1+x)/(1-x) = 1/(1-x) + x/(1-x)
            = 1/(1-x) + [1/(1-x) - 1]
(1+x)/(1-x) = 1/(1-x)
            + 1/(1-x)
            - 1 ---eqn.BT062
(1+x)/(1-x) = 1 +x +x^2 +x^3 +x^4 +x^5 +.....
            + 1 +x +x^2 +x^3 +x^4 +x^5 +.....
            - 1 ---eqn.BT063
(1+x)/(1-x) = 1/x +x^2 +x^3 +x^4 +x^5 +.....
            + 1/x +x^2 +x^3 +x^4 +x^5 +.....
            + 1 ---eqn.BT064

<a name="ch14d176">
Red line contribute to
 j=|+1|, |+2|, ... |+infinity|
Blue line contribute to
 j=|-1|, |-2|, ... |-infinity|
Black +1 contribute to j=0
x^0=1. j is power of x (x=ρ)

<a name="ch14d177">
Above explained for j∈Z 
from j=-infinity to j=+infinity
eqn.BT010 equality.
2010-08-16-16-13 stop

<a name="ch14d178">　Index begin　Index this file
2010-08-16-19-31 start
Above used simple relation
1/(1-x) = // ---eqn.BT059
  1 +x +x^2 +x^3 +x^4 +x^5+.....
Is it possible to show eqn.BT059
is true?
Yes, it is easy. 
<a name="ch14d179">
Define
  p = 1 +x +x^2 +x^3 +x^4 +x^5
      +..... to infinity ---eqn.BT065
Multiply x to both side.
 xp = x +x^2 +x^3 +x^4 +x^5 +x^6
      +..... to infinity ---eqn.BT066

<a name="ch14d180">
Above two equation red part 
are identical. Then we write
  p = 1 + xp ---eqn.BT067
Solve for p, get
  p = 1/(1-x) ---eqn.BT068

Use eqn.BT065 to replace p in
eqn.BT068, we get eqn.BT059.
2010-08-16-19-39 stop

2010-08-16-20-12 done first proofread
2010-08-17-10-48 done second proofread
2010-08-17-11-03 done spelling check

========= Chapter fourteen end here =========

========= Whole book CSMC  end here =========

<a name="ch14d201">
2010-08-17-12-55 start
■■Study notes conclusion
This study notes is reading of
The Cauchy-Schwarz Master Class
J. Michael Steele ★★★★★
ISBN 978-0-521-54677-5

<a name="ch14d202">
Study notes start from
http://freeman2.com/tute0007.htm
to
http://freeman2.com/tute0052.htm
Notes written by Liu,Hsinhan.
LiuHH back ground is

<a name="ch14d203">
1970-July graduate from Mechanical
Engineering Department, National
Cheng Kung University, Tainan, 
Taiwan, Republic of China. Receive 
bachelor degree.
http://www.me.ncku.edu.tw/
(Cheng Kung = Success in Chinese)

<a name="ch14d204">
1983-July graduate from Mechanical
Engineering Department, University
of Iowa, Iowa City, Iowa, U.S.A.
Receive ms degree.
http://www.engineering.uiowa.edu/

<a name="ch14d205">
LiuHH math background is average 
among engineers. That is weak from
math major view point.

<a name="ch14d206">
2008-11-03-20-15 click purchase
2008-11-06-13-11 receive CSMC.
Around 2008 to 2009 spring, LiuHH 
hope to study calculus and find 
how to prove several theorems.
2009 April or May decide to study
inequality. It is a risky decision
<a name="ch14d207">
because I know nothing about math
inequality. My study and learn
process is not very easy, however
I do learn many new material.
Each chapter has a new topic.

<a name="ch14d208">
Just read CSMC I feel not enough,
always hope to get more reference.
Goto online find related paper.
Those pages from journal or math
society target at advanced reader
they do not explain basic material
it is hard for me to read.
<a name="ch14d209">
On the other hand, those pages for
high school student are elementary
I do not learn new material.
Few pages are in the middle level
but the explanation is minimum.
<a name="ch14d210">
Very often, I go around a big loop
on Internet, finally come back to
CSMC, this book give me the best
help. For example, 
Newton's inequality tute0043.htm
Doubly Stochastic Matrix tute0047
http://freeman2.com/jsmajor2.htm
The program jsmajor2.htm totally
rely upon CSMC instruction.
2010-08-17-15-00 here

<a name="ch14d211">
2010-08-17-19-00 start
CSMC has many exercises and each
exercise has a solution. LiuHH
call it as exercise hint. LiuHH
never learn inequality and weak
in mathematics. Frankly speak I
read hint as lecture material.
<a name="ch14d212">
At beginning of study, I do not
even know that Cauchy inequality
treat two data sequences, AM-GM 
inequality treat one sequence, 
power mean inequality treat one 
data seq. and one pure number
coefficient sequence. All of 
<a name="ch14d213">
these understanding gradually
clear after read, think and do
exercise. Most other mathematics
textbook give numerical answer
of odd number problems. After
comparison, CSMC exercise hint
is really a great help to self 
study reader.

<a name="ch14d214">
Another special point about 
The Cauchy-Schwarz Master Class
is that whole book do not have
numerical problem. From begin
to end all challenge problems
and exercises present symbolic
calculation. Numerical problem
is special case, not general
case and not qualified as proof.
<a name="ch14d215">
Symbolic calculation is general
cover all possible numerical
values. Symbolic calculation is
proof work. Therefore CSMC is a
proof book. Great! CSMC teach
me how to prove, I write code
to do numerical calculation.

<a name="ch14d216">
I receive book on 2008-11-06
and start write reading notes 
on 2009-06-09. In seven month
period, I wrote simpler study 
notes not intended upload to 
Internet. Compare those simple
notes with uploaded tute*.htm
the quality/completeness ratio
is about 1:10. In other words,
<a name="ch14d217">
write study notes for public
reading, drive me write better
and work hard to understand
more. If there are too many
NOT DONE, it does not look 
nice. Now review the overall
<a name="ch14d218">
done and NOT DONE it is about
70% done and 30% NOT DONE. For
me it is possibly the limit of
my capability. I hope continue
study mathematics. In the future
NOT DONE percentage could be
reduced. 

<a name="ch14d219">
If I found CSMC typos I mark
●●change, suggest Professor
J. Michael Steele to review.
But how about typos I created
in my page? No one proofread 
for me, there are certainly
error some where. For example
<a name="ch14d220">
in tute0049.htm#ch14a262 I said
we can not compare complex 
number, but I made exactly this
error at eqn.BQ026, eqn.BQ027
in tute0049.htm#ch14a033
This is my typing error.

<a name="ch14d221">
There is concept error too.
tute0044.htm#ch12c106 has wrong
argument, correction is made at
tute0044.htm#ch12c113c
I can discuss those errors I
found, I can not tell you those
error I still do not know. When
you read, remember: 
<a name="ch14d222">
no body proofread tute*.htm, 
you should be highly suspect 
any thing you do not feel 
comfortable. Ask a mathematics 
expert near by.

<a name="ch14d223">
In twenty two months, CSMC is 
first book I read. I learned 
many new material. 
Liu,Hsinhan sincerely thank 
Professor J. Michael Steele 
for his wonderful work.
<a name="ch14d224">
Liu,Hsinhan also thank my
dear reader. Hope you like
my study notes and hope you
like my programs. I hope to
write more, if possible, in
the future. Goodby.
Liu,Hsinhan 2010-08-17-20-04




<a name="Copyright">　Index begin　Index this file
2009-06-19-10-48
If you are interested in inequality,
suggest you buy the book
The Cauchy-Schwarz Master Class
written by Prof. J. Michael Steele
The Cauchy-Schwarz Master Class is
this web page's textbook.

To buy textbook, that is to show thanks
to Prof. J.M. Steele's great work.
and it is also respect copyright law.
Thank you. Freeman 2009-06-19

The Cauchy-Schwarz Master Class
J. Michael Steele ★★★★★
ISBN 978-0-521-54677-5
2009-06-19-10-56