<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="ch09e001">　Index begin　Index this file
2010-04-02-11-08 start
■　Exercise 9.15 problem statement
　　textbook page 154
（An l2 Interpolation Bound）

Let cjk 1≦j≦m, 1≦k≦n be an
array of nonnegative real
numbers for which one has the
implication

<a name="ch09e003">
2010-04-02-11-33 here
//  Why  require  1≦p≦2 ?
Show that for all 1≦p≦2 one
then has the bound

<a name="ch09e004">
where
  q=p/(p-1) ---eqn.BE003
and
  M=max|cjk| ---eqn.BE004
2010-04-02-11-45 stop




<a name="ch09e005">　Index begin　Index this file
2010-04-02-11-49 start
■　Exercise 9.15 hint
　　textbook page 266

One can proceed barehanded, but 
it is also instructive to apply
the result of the preceding 
exercise. 
<a name="ch09e006">
From the hypothesis we have 
  ∥Tx∥2≦∥x∥2 ---eqn.BE005
and from the definition of M
one finds
  ∥Tx∥∞≦M∥x∥1 ---eqn.BE006
Since the linear system

<a name="ch09e008">
has 
  θ＝(2-p)/p ∈ [0,1] ---eqn.BE008
as its unique solution, the
bound (9.43) is indeed a
corollary of Exercise 9.14.
2010-04-02-12-05 stop


<a name="ch09e009">　Index begin　Index this file
2010-04-02-15-55 start
■　Exercise 9.15 discussion

Here present a numerical example
for eqn.BE001,BE002 and eqn.9.43
cjk is a m×n matrix
xk is a n×1 column vector
We use all positive elements
and drop absolute sign.

<a name="ch09e010">
Let [C] be
      [.1  .2]
  [C]=[.3  .4] ---eqn.BE009
      [.5  .6]
we have m=3, n=2 ---eqn.BE010
(m=3 rows, n=2 columns)

then c11=.1, c12=.2, ---eqn.BE011
 and c21=.3, c22=.4, ---eqn.BE012
 and c31=.5, c32=.6. ---eqn.BE013

<a name="ch09e011">
 Set [x]=[7] ---eqn.BE014
         [8] 
that is x1= 7 ---eqn.BE015
    and x2= 8 ---eqn.BE016
//this is lowercase x 2*1 vector
//here is uppercase X 3*1 vector

<a name="ch09e012">
eqn.BE001 [X]=[C][x] is a
 (3*2)(2*1)=(3*1) column vector

 [C][x] become
  [.1  .2] [7] [ .1*7+.2*8 ] [2.3]
  [.3  .4]*   =[ .3*7+.4*8 ]=[5.3] ---eqn.BE017
  [.5  .6] [8] [ .5*7+.6*8 ] [8.3]

<a name="ch09e013">
Since [X]=[C][x]  ---eqn.BE018
//eqn.BE018 = eqn.BE001
we have
      [2.3]
  [X]=[5.3] ---eqn.BE019
      [8.3]
//this is uppercase X 3*1 vector
//here is lowercase x 2*1 vector

<a name="ch09e014">　Index begin　Index this file
Next see eqn.BE002
Less than side has the following
value //∑[j=1,m] and m=3
  ∑[j=1,3]Xj2
   =2.3*2.3+5.3*5.3+8.3*8.3
   =102.27 ---eqn.BE020
<a name="ch09e015">
eqn.BE002 greater than side has 
the following value //∑[k=1,n] n=2
  ∑[k=1,2]xk2=7*7+8*8=113 ---eqn.BE021
then
  102.27 ≦ 113  ---eqn.BE022
satisfy assumption eqn.BE002.
2010-04-02-16-50 here

Above are assumptions
<a name="ch09e016">
Below is conclusion.
We need calculate eqn.9.43
p has restriction 1≦p≦2 
//  Why  require  1≦p≦2 ?
We assume
  p=1.8 ---eqn.BE023
<a name="ch09e017">
Holder's inequality require
  1/p + 1/q = 1 ---eqn.BA037
with p=1.8, find q
  1/1.8 + 1/q = 1
  1/q = 1 - 1/1.8
  q=2.25 ---eqn.BE024

<a name="ch09e018">
Refer to Exercise 9.14
Interpolation two end given
conditions are
  ∥Tx∥t0≦M0∥x∥s0 ---eqn.9.40A
  ∥Tx∥t1≦M1∥x∥s1 ---eqn.9.40B
//T is m×n matrix, elements are cjk
We need to identify these two
conditions first.

<a name="ch09e019">　Index begin　Index this file
Exercise 9.15 hint say
[[
From the hypothesis we have 
  ∥Tx∥2≦∥x∥2 ---eqn.BE005
]]
<a name="ch09e020">
Compare eqn.9.40A with eqn.BE002
eqn.BE002 whole equation take
square root, then eqn.BE002 
is 2-norm equation (for p-norm
definition see eqn.9.11, set 
p=2 get 2-norm) we find
eqn.BE002 less than side 
<a name="ch09e021">
p-norm p number t0=2 ---eqn.BE025
eqn.BE002 greater than side 
p-norm p number s0=2 ---eqn.BE026
eqn.9.40A equivalent M0 in 
eqn.BE002 is M0=1 ---eqn.BE027
(t0, s0, M0 are from eqn.9.40A)

<a name="ch09e022">
That is
  ∥Tx∥t0≦M0∥x∥s0 ---eqn.9.40A
reduce to
  ∥Tx∥2 ≦1*∥x∥2 ---eqn.BE028
eqn.BE028 = sqrt(eqn.BE002)

Interpolation work has two
ends, above is one end.
<a name="ch09e023">
Next is second end.

Exercise 9.15 hint say
[[
from the definition of M
one finds
  ∥Tx∥∞≦M∥x∥1 ---eqn.BE006
]]

<a name="ch09e024">　Index begin　Index this file
Definition of M is next
  M=max|cjk| ---eqn.BE004
Our cjk in hand is
      [.1  .2]
  [C]=[.3  .4] ---eqn.BE009
      [.5  .6]
  M=max{.1, .2, .3,
        .4, .5, .6}=0.6 ---eqn.BE029
(Before verify eqn.9.43, we need
get the value of M.)

<a name="ch09e025">
  ∥Tx∥∞ is //∥Tx∥ is eqn.BE017
  [2.3]
  [5.3]
  [8.3]∞

<a name="ch09e026">
Infinite norm is the maximum
absolute value element, in
above numerical example
  ∥Tx∥∞=8.3 ---eqn.BE030

<a name="ch09e027">
  ∥x∥1 is //[x] is eqn.BE014
  [7]
  [8]1
One-norm is the sum of all
elements absolute values, in 
above numerical example
  ∥x∥1= |7| + |8| =15 ---eqn.BE031

<a name="ch09e028">
Above numerical analysis 
change
  ∥Tx∥∞≦M∥x∥1 ---eqn.BE006
to
  8.3≦0.6*15=9

<a name="ch09e029">　Index begin　Index this file
LiuHH's puzzle begin here
Exercise 9.15 hint say
[[
from the definition of M
one finds
  ∥Tx∥∞≦M∥x∥1 ---eqn.BE006
]]
<a name="ch09e030"> //puzzle solved here
One-norm and infinity-norm are
very different! Why they stay
in same equation?
Why 
from the definition of M
  M=max|cjk| ---eqn.BE004
one finds
  ∥Tx∥∞≦M∥x∥1 ---eqn.BE006
?  //puzzle solved here
2010-04-02-17-44 here

<a name="ch09e031">
M in eqn.BE004 is pick out the 
maximum value one element from 
all m*n=6 elements.
M in eqn.BE006 is pick out the 
maximum value one element from 
(3*1) vector ∥Tx∥ then 
divide by the sum of two
elements in (2*1) x vector.
Two M selection are different.

<a name="ch09e032">
Why 
from the definition of M
  M=max|cjk| ---eqn.BE004
one finds
  ∥Tx∥∞≦M∥x∥1 ---eqn.BE006
? //puzzle solved here
2010-04-02-17-53 stop and think

<a name="ch09e033">　Index begin　Index this file
2010-04-02-18-58 write code
//[[ code start
var c1=[[.1, .2],[.3, .4],[.5, .6]];
var x1=[7,8]
//you can change above number.
//if you do not know programming
//do not change below.
var m=c1.length;
var n=c1[0].length;
var o=x1.length;
if(n!=o){alert('n!=o')}
var i0,i1,i2;
var mmax=0; //eqn.BE004 M value
for(i0=0;i0<m;i0++)
for(i1=0;i1<n;i1++)
if(c1[i0][i1]>mmax)mmax=c1[i0][i1];
var y1=Array(m);
for(i0=0;i0<m;i0++)
{y1[i0]=0;
for(i1=0;i1<n;i1++)
y1[i0]+=c1[i0][i1]*x1[i1];}
var normInf=0; //∥Tx∥_∞ 
for(i0=0;i0<m;i0++)
if(abs(y1[i0])>normInf)normInf=abs(y1[i0]);
var normOne=0; //∥x∥_1
for(i1=0;i1<n;i1++)
normOne+=abs(x1[i1]);
normInf //echo print variable value
mmax*normOne //echo print value
//eqn.BE006 is normInf≦mmax*normOne
//
//complex2.htm#calculator require
//each line present a '='. If one
//line no '=', complex2.htm will
//echo print variable value.
//code 'else' can not stay alone
//in a line. Must write as
// else { dummy=0; 
//if you copy from source file
//(not copy from screen) please
//change "&lt;" to "<" and
//change "&gt;" to ">" 
//2010-04-02-23-01
//]] code stop

2010-04-02-19-17 get correct output

<a name="ch09e034">
change x1=[7,8] to [2,1]
to [-.1, .2]
always get 
normInf ＜ mmax*normOne

<a name="ch09e035">
Calculator page is next (local)
http://freeman2.com/complex2.htm#calculator
Box3, input JS command
Box4, output for Box3
click button is
"test box3 command, output to box4"
2010-04-02-19-26 stop

<a name="ch09e036">
2010-04-02-21-32 start
Exercise 9.15 hint say
[[
from the definition of M
one finds
  ∥Tx∥∞≦M∥x∥1 ---eqn.BE006
]]
<a name="ch09e037"> puzzle begin here
eqn.BE006 is correct. It is not
easy to understand from one-norm
and infinity-norm view point.
It is much easier to digest if
we expand eqn.BE006.

<a name="ch09e038">　Index begin　Index this file
Go back to the numerical example
      [.1  .2]
  [C]=[.3  .4] ---eqn.BE009
      [.5  .6]

Maximum element in [C] is 0.6
In eqn.BE006,
  M=0.6 ---eqn.BE032

<a name="ch09e039">puzzle begin here
 Set [x]=[7] ---eqn.BE014
         [8] 

 [C][x] become ([C][x] is Tx)
  [.1  .2] [7] [ .1*7+.2*8 ] [2.3]
  [.3  .4]*   =[ .3*7+.4*8 ]=[5.3] ---eqn.BE017
  [.5  .6] [8] [ .5*7+.6*8 ] [8.3]

 M∥x∥1 become [ .6*7+.6*8 ]
[ blue line is  0.6*(7+8),
  7+8 is ∥x∥1 one-norm.]

<a name="ch09e040">
Please compare red three lines
and blue one line.
Red is three elements of Tx
Blue is one element of M∥x∥1
7 and 8 are from x vector which
is arbitrary, also 7,8 are common 
to all four lines (red 3, blue 1)
7,8 do not determine inequality
greatness/smallness.

<a name="ch09e041">puzzle begin here
To see picture clearly, set 
x=[1, 1], then red and blue 
become the following
  [ .1+.2 ]
  [ .3+.4 ]
  [ .5+.6 ]

  [ .6+.6 ]
<a name="ch09e042">
Blue .6 come from the maximum one
in red. Blue line has two max.
Red lines only one line has one
maximum 0.6. The compete rule is
select a maximum value from red
and compare with blue. Red can
be at most equal blue, if red 
has one line both are maximum 0.6.
Otherwise blue is always greater 
than red.

<a name="ch09e043">　Index begin　Index this file
"Select a maximum value from red
three lines" this action is 
∥Tx∥∞ infinite-norm.

(7+8) in
 M∥x∥1=[.6*7+.6*8]=0.6*(7+8)
is ∥x∥1 one-norm

<a name="ch09e044"> puzzle begin here
Up to here explained
[[
from the definition of M
one finds
  ∥Tx∥∞≦M∥x∥1 ---eqn.BE006
]]
2010-04-02-21-59 stop

<a name="ch09e045">
2010-04-03-10-04 start
Now back to
[[
Interpolation two end given
conditions are
  ∥Tx∥t0≦M0∥x∥s0 ---eqn.9.40A
  ∥Tx∥t1≦M1∥x∥s1 ---eqn.9.40B
]]

<a name="ch09e046">
First end condition is
  ∥Tx∥2≦∥x∥2 ---eqn.BE005
  t0=2 ---eqn.BE025
  s0=2 ---eqn.BE026
  M0=1 ---eqn.BE027

<a name="ch09e047">
Second end condition is
  ∥Tx∥∞≦M∥x∥1 ---eqn.BE006
  t1=∞ ---eqn.BE033
  s1=1 ---eqn.BE034
  M1=M=0.6 ---eqn.BE032

<a name="ch09e048">　Index begin　Index this file
Exercise 9.14 conclusion is
[[
<a name="ch09d154">
for each 0≦θ≦1, one has the
bound
  ∥Tx∥t≦Mθ∥x∥s ---eqn.9.41
for all x in Realn. where
  Mθ=M1θM01-θ ---eqn.BD051
and where s and t are given by
eqn.9.42A and eqn.9.42B.
]]
Exercise 9.14 eqn.9.41 match 
Exercise 9.15 eqn.9.43 

<a name="ch09e049">
We arbitrary assigned
  p=1.8 ---eqn.BE023
  q=2.25 ---eqn.BE024

s-norm in eqn.9.41 is  //s≡p
p-norm in eqn.9.43 p=1.8
t-norm in eqn.9.41 is  //t≡q
q-norm in eqn.9.43 q=2.25

<a name="ch09e050">
Find θ from eqn.9.42A (s=p=1.8)
  1/s = θ/s1 + (1-θ)/s0 ---eqn.9.42A
  1/1.8 = θ/1 + (1-θ)/2
  1/1.8 = θ/2 + 1/2
  θ = 2*(-1/2+1/1.8)
  θ = 0.1111111=1/9 ---eqn.BE035
Exercise 9.14 require 0≦θ≦1
Now 0≦0.1111111≦1, OK
2010-04-03-10-48 here

<a name="ch09e051">
Find θ from eqn.9.42B (t=q=2.25)
  1/t = θ/t1 + (1-θ)/t0 ---eqn.9.42B
  1/2.25 = θ/∞ + (1-θ)/2
  1/2.25 =  0  + 1/2-θ/2
  θ = 2*(1/2-1/2.25) =0.1111111=1/9

Both calculation get same θ=1/9
related t=q=2.25
  and   s=p=1.8
2010-04-03-10-56 here

<a name="ch09e052">
2010-04-03-11-08 start
eqn.9.43 M power is (2-p)/p
We have p=1.8
  (2-p)/p=0.1111111=1/9 ---eqn.BE036
With
  M=0.6 ---eqn.BE032
<a name="ch09e053">　Index begin　Index this file
From eqn.9.43
  Mθ=M(2-p)/p=M1/9
  Mθ=0.61/9
  Mθ=pow(0.6,1/9)
  Mθ=0.9448223083691852 ---eqn.BE037

<a name="ch09e054">
Above find Mθ from M(2-p)/p
Below find Mθ from M1θM01-θ
We have
  M0=1 ---eqn.BE027
  M1=0.6 ---eqn.BE032
From eqn.BD051
  Mθ=M1θM01-θ ---eqn.BD051
  Mθ=0.61/9*11-1/9
  Mθ=0.61/9
  Mθ=0.9448223083691852 ---eqn.BE038
eqn.BE037 and eqn.BE038 give same Mθ

<a name="ch09e055">
All data are in hand, finally 
verify eqn.9.43 numerically

eqn.9.43 less than side is
  {∑[j=1,m](|Xj|q)}1/q
and
      [2.3]
  [X]=[5.3] ---eqn.BE019
      [8.3]
<a name="ch09e056">
Calculation for [X] q-norm is
next (t=q=2.25)
  [X]_q-norm=(2.3^2.25 ---eqn.BE039
    +5.3^2.25+8.3^2.25)^(1/2.25)
Javascript code is
  [X]_q-norm=pow(pow(2.3,2.25)+pow(5.3,2.25)+pow(8.3,2.25),(1/2.25));
Numerical answer is
  [X]_q-norm=9.700420405668534 ---eqn.BE040

<a name="ch09e057">
Next, eqn.9.43 greater than 
side is
  M(2-p)/p*{∑[k=1,n](|xk|p)}1/p
and
  [x]=[7] ---eqn.BE014
      [8] 
<a name="ch09e058">　Index begin　Index this file
Calculation for [x] p-norm is
next (s=p=1.8)
  [x]_p-norm  ---eqn.BE041
    =(7^1.8+8^1.8)^(1/1.8)
Javascript code is
  [x]_p-norm=pow(pow(7,1.8)+pow(8,1.8),(1/1.8));
Numerical answer is
  [x]_p-norm=11.042592979385444 ---eqn.BE042

<a name="ch09e059">
eqn.9.43 whole equation is
  [X]_q-norm≦Mθ*[x]_p-norm
and 
  Mθ=0.94482 ---eqn.BE037/eqn.BE038
<a name="ch09e060">
Final over all check :
  [X]_q-norm≦Mθ*[x]_p-norm
//Uppercase [X] 3*1 vector
//lowercase [x] 2*1 vector
  9.70042 ≦ 0.94482*11.04259
or
  9.70042 ≦ 10.433259 ---eqn.BE043
is this right?
Look OK !
A long numerical check !!
2010-04-03-11-46 stop


<a name="ch09e061">　Index begin　Index this file
2010-04-03-13-07 start
■　Exercise 9.15 solution


Exercise 9.15 is an application
of Exercise 9.14. and Exercise 
9.14 is Interpolation for 
Positive Matrices. Interpolation
need two end conditions.

<a name="ch09e062">
Exercise 9.15 assume the truth
of eqn.BE002 which represent
first end condition, we get
eqn.BE005

Problem defined M value, from
which we get the second end 
condition eqn.BE006

<a name="ch09e063">
Exercise 9.14 say if given two 
end condition eqn.9.40A & B
then the interpolation result
is eqn.9.41

<a name="ch09e064">
Exercise 9.15 target equation
eqn.9.43 should be same as
eqn.9.41.

Identify first end condition 
is easy.
Identify second end condition 
is not straight forward.
However, it is till manageable.
2010-04-03-13-36 here

<a name="ch09e065">
The heart calculation of 
Exercise 9.15 is next few
lines. Start from
  1/s = θ/s1 + (1-θ)/s0 ---eqn.9.42A
where
  s0=2 ---eqn.BE026
  s1=1 ---eqn.BE034
Interpolation two end condition
s0-norm s0 number and
s1-norm s1 number are 
determined by problem statement.

<a name="ch09e066">　Index begin　Index this file
Exercise 9.14 use s same as
Exercise 9.15 use p, so s=p
In eqn.9.42A, 1/s become 1/p
  1/p = θ/1 + (1-θ)/2
  1/p = θ/2 + 1/2
  2/p = θ   + 1
  θ = 2/p - 1
  θ = 2/p - p/p
<a name="ch09e067">
  θ = (2- p)/p ---eqn.BE044
Result of heart calculation 
of Exercise 9.15 is eqn.BE044

<a name="ch09e068">
eqn.BE044 tell us that
Exercise 9.14 eqn.9.41 Mθ and
Exercise 9.15 eqn.9.43 M(2-p)/p
are one thing.

Exercise 9.15 solved.
2010-04-03-13-53 stop



2010-04-03-18-53 done proofread

<a name="ch09e069">
2010-04-03-19-48 start
What is the meaning of eqn.BE007?
  (1/p, 1/q)=θ(1/1, 1/∞)
      + (1－θ)(1/2, 1/2) ---eqn.BE045
eqn.BE045 is simplified eqn.BE007

<a name="ch09e070">
(1/2, 1/2) is hint indicated
first end equation eqn.BE005
greater than side, less than
side two p-norm 1/p numbers.

<a name="ch09e071">　Index begin　Index this file
(1/1, 1/∞) is hint indicated
second end equation eqn.BE006
greater than side, less than
side two p-norm 1/p numbers.

<a name="ch09e072">
(1/p, 1/q) is interpolated
equation eqn.9.41 two p-norm
 1/p numbers.

eqn.BE007 is actually two 
equations.

<a name="ch09e073">
First equation is
  1/p=θ(1/1) + (1－θ)(1/2) ---eqn.BE046
  1/p=θ + 1/2 －θ/2
  1/p=θ/2 + 1/2
  2/p=θ + 1
  θ = -1+2/p
  θ = (-p+2)/p ---eqn.BE047

<a name="ch09e074">
Second equation is
  1/q=θ(1/∞) + (1－θ)(1/2) ---eqn.BE048
but 
  1/p + 1/q = 1 ---eqn.BA037
so
  1/q=1-1/p
<a name="ch09e075">
then
  1-1/p = θ(1/∞) + (1－θ)/2
  1-1/p = 0 + (1－θ)/2
  2-2/p = (1－θ)
  θ = 1-2+2/p
  θ = -1+2/p
  θ = (-p+2)/p ---eqn.BE049

<a name="ch09e076">
eqn.BE044, eqn.BE047, eqn.BE049
are all the same. They are
result of heart calculation.
2010-04-03-20-11 stop

2010-04-03-20-24 done spelling check
2010-04-03-20-58 upload above notes

<a name="ch09e077">　Index begin　Index this file
2010-04-04-11-01 start
Why require 1≦p≦2 ? //answer
For interpolation, We have
two end conditions
  ∥Tx∥t0≦M0∥x∥s0 ---eqn.9.40A
  ∥Tx∥t1≦M1∥x∥s1 ---eqn.9.40B
<a name="ch09e078">
First one
  ∥Tx∥t0≦M0∥x∥s0 ---eqn.9.40A
reduce to
  ∥Tx∥2 ≦1*∥x∥2 ---eqn.BE028
<a name="ch09e079">
Second one 
  ∥Tx∥t1≦M1∥x∥s1 ---eqn.9.40B
change to //why
  ∥Tx∥∞≦M*∥x∥1 ---eqn.BE006

<a name="ch09e080">
Our target equation is
  ∥Tx∥t≦Mθ∥x∥s ---eqn.9.41
Above is Exercise 9.14 style
If use   Exercise 9.15 style
t⇒q and s⇒p, we find target
<a name="ch09e081">
  ∥Tx∥q≦Mθ∥x∥p ---eqn.BE050
Two end conditions are
  ∥Tx∥2 ≦1*∥x∥2 ---eqn.BE028
  ∥Tx∥∞≦M*∥x∥1 ---eqn.BE006
eqn.BE050 is interpolation of
eqn.BE028 and eqn.BE006, so
  1≦p≦2 ---eqn.BE051 //answer why
  2≦q≦∞ ---eqn.BE052
  M≦Mθ≦1 ---eqn.BE053

<a name="ch09e082">　Index begin　Index this file
Our numerical example has
  1≦1.8≦2  //p= 1.8 see eqn.BE023
  2≦2.25≦∞ //q=2.25 see eqn.BE024
  θ = 0.1111111=1/9 ---eqn.BE035

<a name="ch09e083">
For  M≦Mθ≦1 ---eqn.BE053
get  0.6≦0.9448223≦1 
//0.6 see eqn.BE029
//0.9448223 see eqn.BE037
//1 see eqn.BE027
2010-04-04-11-35 stop

========= Chapter nine end here =========




<a name="ch10a001">　Index begin　Index this file
2010-04-05-12-25 start
■■Chapter 10: Hilbert's Inequality 
　　and Compensating Difficulties

Web page (local)
http://freeman2.com/tute0009.htm
has topic "Hilbert's Inequality
 and Schur Constant" which is
related to Chapter ten.
<a name="ch10a002">
tute0009.htm only demonstrate 
that Schur Constant is maximum 
eigenvalue of Hilbert matrix.
tute0009.htm did not prove that
Schur Constant is PI.

<a name="ch10a003">　Index begin　Index this file
■　Compare Hilbert with Cauchy

Hilbert's Inequality and
Cauchy's Inequality are similar.
List them below for comparison.

<a name="ch10a004">
Both Cauchy and Hilbert equation
right side use a-seq. and b-seq.
2-norms. Hilbert add constant C.
Cauchy and Hilbert equation left
side are different.
Cauchy  use one summation but
Hilbert use two summations.
Whether it is possible that
Cauchy  use two summations?

<a name="ch10a005">
Below is generalized Cauchy's 
Inequality. Key point is it has
cross product, that is it has 
ai bj with i!=j 
[M]=symmetric matrix

<a name="ch10a006">　Index begin　Index this file
The concept of "Generalized 
Cauchy's Inequality" come from
Exercise 1.7 problem statement
　　textbook page 14
（Flexibility of Form）
[[
<a name="ch10a007">
Assume x, y, α, β are all real 
numbers, please prove
  (5αx+αy+βx+3βy)2  -----page 14 eqn.1.22
 ≦(5α2+2αβ+3β2)(5x2+2xy+3y2)
Please explain eqn.1.22 is 
Cauchy~Schwarz Inequality eqn.1.16. 
For this problem, you need design 
a special dot product 〈.,.〉.
]]
<a name="ch10a008">
Liu,Hsinhan do not know if some
author use the name "Generalized 
Cauchy's Inequality", but LiuHH
use this term here in tute*.htm
sreies.
Google search for "Generalized 
Cauchy's Inequality" 14 results.
No one use it same as LiuHH did.
2010-04-07-20-01

<a name="ch10a009">
(5αx+αy+βx+3βy) in eqn.1.22
can be written as

<a name="ch10a011">
Second step multiplication get
  x*(5*α+1*β)+y*(1*α+3*β)
  = 5*α*x+β*x+α*y+3*β*y ---eqn.BE102
eqn.BE102 is eqn.1.22 less than
side term.
2010-04-05-13-14 stop

<a name="ch10a012">　Index begin　Index this file
2010-04-05-14-41 start
Please observe eqn.1.22
eqn.1.22 less than side use
square matrix in eqn.BE101
  [x, y] [ 5 　 1 ] [α] ---eqn.BE103
         [ 1 　 3 ] [β]

<a name="ch10a013">
eqn.1.22 greater than side use
same square matrix in eqn.BE101 
twice, once for [α,β],
  [α, β] [ 5 　 1 ] [α] ---eqn.BE104
         [ 1 　 3 ] [β]

<a name="ch10a014">
and once for [x,y]
  [x, y] [ 5 　 1 ] [x] ---eqn.BE105
         [ 1 　 3 ] [y]

<a name="ch10a015">
The key point to observe is that
eqn.1.22 three matrix multiplication
use SAME matrix
         [ 5 　 1 ]  ---eqn.BE106
         [ 1 　 3 ] 
or write as [ 5  1 ; 1  3 ]
';' indicate change line.

<a name="ch10a016">
Matrix [ 5  1 ; 1  3 ] keep same
inequality direction as regular
Cauchy's Inequality, because
Matrix [ 5  1 ; 1  3 ] has all
positive eigenvalues
5.414213562373096 
2.5857864376269046 

<a name="ch10a017">　Index begin　Index this file
Since eqn.AC01 is Generalized 
Cauchy's Inequality, how to 
change 
from Generalized Cauchy's Ineq.
 to  regular Cauchy's Inequality?
The answer is the square matrix.
<a name="ch10a018">
If we change from
matrix [ 5  1 ; 1  3 ] 
to
matrix [ 1  0 ; 0  1 ] 
or better view as
         [ 1 　 0 ]  ---eqn.BE107
         [ 0 　 1 ] 
for all three application of
matrix multiplication in eqn.AC01
the result is regular Cauchy's 
Inequality eqn.1.7

<a name="ch10a019">
Matrix [ 1  0 ; 0  1 ] has two
eigenvalues 1 and 1, both are
positive.
Matrix [ 1  0 ; 0  1 ] has all
zero off-diagonal elements.

<a name="ch10a020">
What is the difference between
Generalized Cauchy's Inequality
and Hilbert's Inequality

<a name="ch10a021">
Generalized Cauchy's Inequality
use same matrix for all three 
applications.
Hilbert use [1/(m+n)] as less 
than side matrix and use
identity matrix twice at 
greater than side.
<a name="ch10a022">
Cauchy use three identical
       square matrices.
Hilbert use two different
       matrices at three
       applications.

<a name="ch10a023">　Index begin　Index this file
Use simplest 3×3 Hilbert matrix
to explain (instead of ∞ size)
Avoid use footnote am and bn
change to [α,β,γ] and [x,y,z].
(next is 3×3 Hilbert matrix)

<a name="ch10a025">
In eqn.BE108 
"1/2" come from 1/(m+n)=1/(1+1)
Since  "1/2" sit at (m,n)=(1,1)
"1/3" come from 1/(m+n)=1/(1+2)
Since  "1/3" sit at (m,n)=(1,2)
"1/3" come from 1/(m+n)=1/(2+1)
Since  "1/3" sit at (m,n)=(2,1)
etc. Hilbert matrix is symmetry.
<a name="ch10a026">
Matrix multiplication first step
is next //see begin

<a name="ch10a027">
Carry out vector dot product 
between red vector and blue 
vector. Before dot product,
two vectors has six elements.
After dot product, all six
merge to one number. Let red 
vector do the same for all
three rows, get

<a name="ch10a029">
eqn.BE110 has no square matrix,
eqn.BE110 has one row vector
[ α   β   γ ] and the other one 
is column vector. Carry vector 
dot product again get a single
number
<a name="ch10a030">
  α*(x/2 + y/3 + z/4)
 +β*(x/3 + y/4 + z/5)
 +γ*(x/4 + y/5 + z/6)
 =
  α*x/2 + α*y/3 + α*z/4
 +β*x/3 + β*y/4 + β*z/5
 +γ*x/4 + γ*y/5 + γ*z/6 ---eqn.BE111

<a name="ch10a031">
eqn.BE111 is a double sum
eqn.BE111 is Hilbert's Inequality
less than side ∑[m=1,3]∑[n=1,3]
If let upper bound 3 goto
infinity, and let α,β,γ array
size goto infinity, let x,y,z 
array size goto infinity we get
eqn.10.1 less than side exactly.
Here α,β,γ array is a-sequence
 and x,y,z array is b-sequence
am and bn are used in eqn.10.1

Above is Hilbert's less than side
<a name="ch10a032">
Below is Hilbert's greater than 
side (next is identity matrix)

<a name="ch10a033">　Index begin　Index this file
After expansion, get
  α*α+β*β+γ*γ ---eqn.BE113
Same story for 
  x*x+y*y+z*z ---eqn.BE114

Change α, β, γ to am
Change x, y, z to bn to get
eqn.10.1 expression.
2010-04-05-16-53 stop

<a name="ch10a034">
2010-04-05-18-43 start
Why Hilbert's Inequality eqn.10.1
has a constant C＝π and Cauchy's 
Inequality eqn.1.7 do not have a
constant C? (why Cauchy use C=1)
(Liu,Hsinhan self study raised many
 questions. Self ask / self answer
 may not be correct ! ALERT !!)



<a name="ch10a035">
If generalized Cauchy's Inequality 
use Hilbert's matrix at all three
summations, since Hilbert's matrix
has all positive eigenvalues (but
ill conditioned), then generalized 
Cauchy inequality is true without
use C＞1.

<a name="ch10a036">
Now, it is Hilbert (not Cauchy),
Hilbert change Cauchy's greater
than side two matrices to identity
matrices. Off diagonal elements
are all zero. See eqn.BE112.
Dropped many positive terms, it
cause Cauchy's greater than side
reduce to smaller than less than
<a name="ch10a037">
side. In order to keep its greater
position, Hilbert added a constant
C＝π＞1 to lift the should-be-great
but falling term to stand great.
Is this right story?
2010-04-05-19-05 here

<a name="ch10a038">　Index begin　Index this file
■　Problem 10.1 
(Hilbert's Inequality)

Show that there is a constant C
such that for every pair of
sequences of real numbers {an} 
and {bn} one has

2010-04-05-19-12 stop

<a name="ch10a040">
2010-04-05-19-42 start
There are many method to prove
Hilbert's Inequality. Textbook
start from Cauchy's inequality
to prove Hilbert's Inequality.

<a name="ch10a041">
Hilbert's Inequality less than
side use double summation
∑[m=1,∞]∑[n=1,∞]
Hilbert's Inequality greater
than side use single summation
∑[m=1,∞] and ∑[n=1,∞]

<a name="ch10a042">
On the other hand, Generalized 
Cauchy's Inequality use double
sum at both side, see eqn.AC01.
Regular Cauchy Inequality use
single sum at both side.

<a name="ch10a043">　Index begin　Index this file
Hilbert and Cauchy are not 
compatible. Textbook use set
concept to solve this problem.

<a name="ch10a044">
If S is any countable set and
{αs} and {βs} are collections
of real numbers indexed by S,
then Cauchy's Inequality can
be written as

<a name="ch10a045">
Hilbert use matrix element 
index m,n as matrix element.
At location (m,n), matrix 
element has value 1/(m+n).
Because Cauchy's inequality
work with two sequences and
1/(m+n) show up at both seq.
We need build sequence for
Cauchy.

<a name="ch10a046">
Textbook say if naively use 
the next two sequences for 
Cauchy
  αs=am/√(m+n) ---eqn.BE115
  βs=bn/√(m+n) ---eqn.BE116
They will fit Cauchy's less
than side, but greater than
side has value infinity. 
<a name="ch10a047">
Who worry less than side is 
smaller than infinity? The
reason we get infinity at
greater than side is that
greater than side two terms
one has   ∑[m=1,∞]{1/m} the
other has ∑[n=1,∞]{1/n}
both have infinity value.

<a name="ch10a048">　Index begin　Index this file
Textbook suggest us to use 
the next two sequences for 
Cauchy inequality.

<a name="ch10a050">
2010-04-05-20-39 here
where 
  s=(m,n) ---eqn.BE117
  λ＞0    ---eqn.BE118
Here λ is a parameter we can
control. //see value compare
Now substitute eqn.10.4A and
eqn.10.4B into eqn.10.2 we 
find

2010-04-05-21-00 stop

<a name="ch10a053">　Index begin　Index this file
2010-04-06-09-45 start
The first factor in eqn.BE120 
contains am2 which is not a
function of n, am2 can be moved
out of ∑[n=1,∞] summation.

<a name="ch10a055">
Generalized Cauchy use double
summation, regular Cauchy use 
single summation. eqn.BE121
change from generalized Cauchy
to regular Cauchy. If we can
adjust λ value such that

<a name="ch10a057">
Hilbert's Inequality greater 
than side is bounded by finite
number, then the proof of 
Hilbert's Inequality will be 
done.

For any nonnegative decreasing
function f:[0,∞]→Real, we have
the integral bound

<a name="ch10a059">
2010-04-06-10-15 here
We need put eqn.BE121 red term
into eqn.BE123. Red term sum
over n, so n is variable.
eqn.BE123 right side use x as
variable. In red term we change
n to x, get the following

<a name="ch10a061">
Above x integration equation,
we treat m as constant. Change
variable, let
  x=m*y ---eqn.BE124
absorbed m, x change to y.
Integration upper/lower bound
not change, because for x=m*y 
x=0 ⇒ y=0 and x=∞ ⇒ y=∞.

<a name="ch10a063">　Index begin　Index this file
After change variable x=my, we
change from x related constant
Bλ to y related constant Cλ.

The integration in eqn.10.6B
is convergent if
  0＜λ＜1/2 ---eqn.BE125
We define Cλ as following

2010-04-06-10-59 stop

<a name="ch10a065">
2010-04-06-13-40 start
We want find
smallest value of greater side
(largest value of smaller side)
The integration of eqn.10.7 is
the greater than side factor of
eqn.BE120. We try to minimize
Cλ where λ is free variable.

<a name="ch10a066">
Mathematica and Maple programs
give us a quick answer. The
integration of eqn.10.7 has a
closed form solution, it is

<a name="ch10a068">　Index begin　Index this file
In sin(2πλ), λ is a variable.
Also 0＜λ＜1/2 is required. Then
  λ=1/4  ---eqn.BE126
give us
  sin(2πλ)=sin(2π/4)
          =sin(π/2)=1 ---eqn.BE127
Sin function maximum value is 1.
Denominator sin has maximum,
we find eqn.10.8 has minimum
at λ=1/4. 
<a name="ch10a069"> verify C＝π
Then what is Cλ at λ=1/4?

<a name="ch10a070">
The proof of eqn.10.8 can
be found in textbook written
by Bak and Newman (1997) and
Cartan (1995).

<a name="ch10a071">
Replace the red term in 
eqn.BE121 with minimum value π
Do similar replacement for 
second term in eqn.BE120. Then 
eqn.BE119≦new_eqn.BE120 is the 
square of Hilbert's Inequality.

Problem 10.1 solved.
2010-04-06-14-24 stop



<a name="ch10a072">　Index begin　Index this file
2010-04-06-18-44 start
■　Quarter Circle Lemma 
　(textbook page 161)

For all m≧1 we have the bound

<a name="ch10a074">
2010-04-06-18-59 here
Please click "Draw fig1001" 
button below.

<a name="ch10a075">
2010-04-06-19-01 start
Equation 10.13 has two parameters
m and n. It is summation for n
from n=1 to n=infinity. m is 
constant.
In fig.10.1, horizontal x axis
is measured by √m (not by m !). 
Vertical y axis is measured by 
√n (not by n !). 
<a name="ch10a076">
Red quarter circle has radius √m 
which is a constant. Moving point
p move on vertical dash line. 
p is not shown in fig.10.1 When
p has a specific n value, p is 
marked with C and D.
<a name="ch10a077">　Index begin　Index this file
In Figure 10.1, points A,B,C,D
has the coordinates
  A=(0,0)    ---eqn.BE128
  B=(√m, 0)  ---eqn.BE129
  C=(√m, √[n-1]) ---eqn.BE130
  D=(√m, √n) ---eqn.BE131
<a name="ch10a078">
When p move on vertical dash line
each time increase one step,
represent n value increase by one.
Remember √n is y axis scale.
Triangle ∆ACD area and shaded
triangle ∆AEF area can be 
calculated. We are interested
at the sum of shaded triangle 
areas which is bounded by
quarter circle √m*√m*PI/4.

<a name="ch10a079">
Figure 10.1 show one typical
∆ACD and ∆AEF. 
First, let us calculate area 
of ∆ACD 
Base of ∆ACD is CD, we know
  C=(√m, √[n-1]) ---eqn.BE130
  D=(√m, √n) ---eqn.BE131
then
  CD=√n - √[n-1] ---eqn.BE132
<a name="ch10a080">
Height of ∆ACD is AB. We know
  A=(0,0)    ---eqn.BE128
  B=(√m, 0)  ---eqn.BE129
then
  AB=√m-0=√m ---eqn.BE133
Area of ∆ACD is
  Area(∆ACD)=AB*CD/2
  Area(∆ACD)=√m*(√n-√[n-1])/2 ---eqn.BE134

<a name="ch10a081">
Second, let us calculate area 
of shaded ∆AEF.
Because ∆AEF is similar to ∆ACD
Find two triangle's area ratio.
shaded ∆AEF has one side AF on
red circle. Circle radius is
constant. So choose side AF 
and side AD. To find ratio of 
AD to AF, find AD and AF first.
<a name="ch10a082">　Index begin　Index this file
  AF=radius=AB=√m ---eqn.BE135
Next see AD. 
From Pythagoras theorem,
  AD^2 = AB^2 + BD^2 ---eqn.BE136
  AD^2 = (√m)^2 + (√n)^2
  AD^2 =    m   +   n
  AD   = √(m+n) ---eqn.BE137
<a name="ch10a083">
The ratio AD/AF is
  AD/AF = √(m+n)/√m ---eqn.BE138
eqn.BE138 is line ratio, for
area ratio, we square eqn.BE138
get
  ∆ACD/∆AEF=(AD/AF)^2
  ∆ACD/∆AEF=(m+n)/m ---eqn.BE139
or
  ∆AEF/∆ACD=m/(m+n) ---eqn.BE140
<a name="ch10a084">
We already found
  Area(∆ACD)=√m*(√n-√[n-1])/2 ---eqn.BE134
then eqn.BE134 and eqn.BE140
give us
  Area(∆AEF)=[m/(m+n)]
       *√m*(√n-√[n-1])/2 ---eqn.BE141
Textbook use An for Area(∆AEF)
Write in better math equation
we have

<a name="ch10a086">
2010-04-06-19-51 here
We need simplify the term
[√n－√(n－1)]
If f(x)=√x, then
  d[f(x)]/dx=d[√x]/dx
    =1/[2√x] ---eqn.BE142
or
  d[√x]=dx/[2√x] ---eqn.BE143
Integrate eqn.BE143 from x=n-1
to x=n get

<a name="ch10a088">
Up to here eqn.BE144 is an
equality equation. Right side
is same as left side, not 
simplified. But right side
integration let us to observe
that n and n-1 are constants
(n are integration limits)
and 1/√x is monotone decrease.
<a name="ch10a089">
We can create an inequality.
Change the integrand from 1/√x
to 1/√n, then integration
become ∫[x=n-1,n]{dx/√n}. To
compare, see next two lines
 ∫[x=n-1,n]{dx/√x} is greater
 ∫[x=n-1,n]{dx/√n} is smaller
dx in "dx/√n" say x is variable
√n in "dx/√n" say n not vary
<a name="ch10a090">
 ∫[x=n-1,n]{dx/√n} simplify to
  {∫[x=n-1,n]dx}/√n ---eqn.BE145
  = [n-(n-1)]/√n = 1/√n
then
  √n－√(n－1)＞(1/2)/√n ---eqn.BE146
(1/2) come from eqn.BE144
If n=5, we get √5－√(5－1)＞(1/2)/√5 
0.2360679 ＞ 0.2236067 (0.23 ＞ 0.22)

<a name="ch10a091">
eqn.10.14 change to next

<a name="ch10a092">　Index begin　Index this file
Here An is just one
shaded triangle. When point P 
move along dash vertical line
push n to infinity, the total
shaded area is the summation
of An.
<a name="ch10a093">
Red circle radius is √m
Red circle area is PI*√m*√m
Quarter area is PI*m/4
The factor "m/4" cancel from
quarter area and eqn.10.15 
the result is eqn.10.13.

Cancel red, black is eqn.10.13.
Quarter Circle Lemma is done.
2010-04-06-20-37 stop

<a name="ch10a095">　Index begin　Index this file
2010-04-07-12-00 start
■　Hilbert's Inequality second proof

There are many method to prove
Hilbert's Inequality. Textbook
provided Quarter Circle Lemma
pave the way for second proof.
<a name="ch10a096">
Second proof reference is
(a) The Cauchy-Schwarz Master Class
(b) An Elementary Proof of Hilbert's
    Inequality 
    by Krzysztof Oleszkiewicz
    The American Mathematical Monthly,
    Vol.100 No.3 (Mar.1993) pp.276-280

<a name="ch10a097">
2010-03-06-18-22 Liu,Hsinhan accessed
http://www.myoops.org/twocw/nctu/upload/fourier/supplement/hilbert%20inequality.pdf
hilbert_elementary_proof_myoops.org.pdf

Second proof still start from
Cauchy end at Hilbert. Please
see Hilbert's Inequality
<a name="ch10a098">
We start from eqn.10.1 less than 
side. Key point is how to split 
less than side. We write

<a name="ch10a102">
2010-04-07-12-50 here
Quarter Circle Lemma eqn.10.13
tell us that the red term and
the blue term in eqn.BE151 
are both less that PI. Therefore
eqn.BE151 is less than the next

<a name="ch10a104">
We find
  eqn.BE149 ＜ eqn.BE152
take square root of it, the
result is Hilbert's Inequality
eqn.10.1
Second proof is done.

<a name="ch10a105">　Index begin　Index this file
If you compare first proof and
second proof. Main difference
is that Prof. J. Michael Steele
tell us why we take power_1/4.
The second proof use power_1/4
directly and no mention why.
2010-04-07-13-00 stop

2010-04-08-11-06 done proofread



<a name="ch10a106">
2010-04-08-14-37 start
Verify eqn.10.9 numerically.
Please goto
http://freeman2.com/integral.htm#NumIntegral
in g(x) box fill in
 1/(1+x)/sqrt(x) ---eqn.BE153
this is Javascript code for the
integrand of eqn.10.9
<a name="ch10a107">
In steps box fill in 200
x-bgn and x-end use 0 and 1000
In "Newton-Cotes formula n= "
click choose 6
Then click "integral g(x) A"
answer is
[[
From x=0 to x=1000
integration value is Infinity
]]

<a name="ch10a108">
Because x-bgn=0, the function
 1/(1+x)/sqrt(x) ---eqn.BE153
has divide by zero trouble.

<a name="ch10a109">
To solve this problem, change 
variable as following.
Let
  x=√x*√x ---eqn.BE154
rewrite eqn.BE153 as
  1/(1+√x*√x)/√x ---eqn.BE155
<a name="ch10a110">
Integration has a 'dx', all
together become
  d(√x*√x)/(1+√x*√x)/√x ---eqn.BE156
Now,
  d(√x*√x)=2*√x*d(√x) ---eqn.BE157
eqn.BE156 become
  [2*√x*d(√x)]/(1+√x*√x)/√x ---eqn.BE158
<a name="ch10a111">
In eqn.BE158, we cancel √x
from numerator and denominator
the result is
  [2*d(√x)]/(1+√x*√x) ---eqn.BE159
<a name="ch10a112">
We are afraid of 1/√x in 
eqn.BE156, but we are not
afraid of 1/(1+√x*√x) in 
eqn.BE159
When x approach zero,
 1/√x shoot to infinity 
    (then answer infinity)
 1/(1+√x*√x) still finite.
<a name="ch10a113">
eqn.BE159 and eqn.10.9 are
identical, eqn.BE159 is
well behaved. 
Write √x as y, eqn.BE159
become
  [2*d(y)]/(1+y*y) ---eqn.BE160
y is dummy variable, change
y to x, get
  [2*dx]/(1+x*x) ---eqn.BE161

<a name="ch10a114">
Now fill in the following 
data 
g(x) box : 2/(1+x*x)
(program ask NOT fill in dx)

steps box : 800
x-bgn : 0
x-end : 1000
<a name="ch10a115">
answer is
[[
From x=0 to x=1000
integration value is 3.139578121643638
]]
3.13957 is near PI=3.141592653589793
answer 3.13957 is acceptable.
eqn.10.9 is reasonable.
2010-04-08-15-02 stop

2010-04-08-17-10 done spelling check

<a name="ch10a116">
2010-04-10-22-04 start
Start from eqn.BE144 to eqn.BE146
prove an inequality with help
from integration. The next is
to prove same inequality using
simple algebra without using
integration. Start from 
conclusion go backward, see if
reach a truth.
<a name="ch10a117">
Start from CSMC page 161, 
line -4, here n≧1.
  √n-√(n-1) ＞ 1/(2√n) ---eqn.BE162
Next change '＞' to '?＞?'
indicate uncertain.
Move √n from right side 
denominator to left side 
numerator.
  √n√n-√n√(n-1) ?＞? 1/2
  n-√n√(n-1) ?＞? 1/2
<a name="ch10a118">
Move square root to right
side.
  n-1/2 ?＞? √n√(n-1)
n≧1, so n-1/2＞0.
Square whole equation
  (n-1/2)*(n-1/2) ?＞? n*(n-1)
  n*n-n+1/4 ?＞? n*n-n
Cancel n*n-n what left is
  1/4 ?＞? 0
<a name="ch10a119">
0.25 is greater than zero, 
that is certainly true. Then
the beginning eqn.BE162 is
true.
This proof do not use 
integration.
2010-04-10-22-14 stop



<a name="docB001">　Index begin　Index this file
2010-04-09-21-32 start
"Update 2010-04-14" change [Draw fig1001]
from unit circle to radius-ten circle
Put C on y-axis=8 and D on y-axis=9
Still not to scale √m and √n, but 
better than unit circle.
<a name="docB002"> //Draw program
"Update 2010-04-14" add [COLOR Oles.pie]
button. Let user see to-scale picture.
"Update 2010-04-14" add alert "Program 
environment is MSIE 6.0, please use MSIE"
2010-04-09-21-38 stop

<a name="docB003">
"Update 2010-04-14" add 
simple algebra proof without 
using integration.
2010-04-10-22-21

<a name="docB004">
2010-04-14-18-22 start
"Update 2010-04-14" improved
Oleszkiewicz's pie 
report detail and accept color
code '#6dfde9' or 'fabca6' at
R:b and e two boxes.
2010-04-14-18-25 stop

<a name="docB005">
2010-04-19-12-30 start
"Update 2010-04-19" made two 
changes.
First, in Oleszkiewicz's pie 
add m=n alert. If set m=n,
Data output add two lines 
[[
Ole.pie/m=0.38684046862414894,meet PI/8 at m=n=infinity.
m=n,PI/8=0.39269908169872414,Ole.pie/m:PI/8=ratio below
]]
<a name="docB006">
If m is finite, all triangles
always have a gap from circle.
Ole.pie always ＜ Sector area
If m is finite, n can be ＞ m.
m → infinity, Ole.pie → Sector
m → infinity, n≦m. limit n=m

<a name="docB007">
"Update 2010-04-19" 2nd change
added
Square [pie1] [pie2] ; circle [pie3] [pie4]
First version of Oleszkiewicz's 
pie use black line only. Can not
see the gradient at high n end.
Second version added color, let
color change show gradient.
<a name="docB008">
Oleszkiewicz's pie do not have
uniform color gradient. Now add
[pie1] to [pie4] buttons for
uniform color gradient.
[pie1] [pie2] draw square 
[pie3] [pie4] draw circle 
Color go different direction.
2010-04-19-12-51 stop

<a name="docB009">
2010-04-20-10-08 start
"Update 2010-04-20" change
"Update 2010-04-19" 'pie4' function.
"Update 2010-04-19" 'pie3' draw color
start from circle center to outer.
'pie4' draw color start from circle
outer to center. But user can achieve
this function by switch begin and end
color code.

"Update 2010-04-20" change 'pie4' 
function. Color start from angle=0
to angle=180 degree. //Draw program
2010-04-20-10-18 stop

<a name="docB010">
2010-04-25-14-18 start
"Update 2010-04-25" add code for
[pie4] button, allow pole or singular
selection. Earlier code has only
singular, new code add pole code.
2010-04-25-14-20 stop


<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