<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="ch10b001">　Index begin　Index this file
2010-04-10-11-48 start
■　Problem 10.2
Suppose that the constant C 
satisfies

<a name="ch10b003">
for all pairs of sequences
of real numbers {am} and {bn}.
Show that C≧π.
2010-04-10-12-01 stop

<a name="ch10b004">
2010-04-10-12-12 start
The summation in eqn.10.10 is
infinity summation. If we use
diverge sequence
  lim[n→∞]bn＞0 ---eqn.BF001
then eqn.10.10 two side are
  infinity ＜ C*infinity 
There is no point to study this
infinity inequality.

<a name="ch10b005">
We require both {am} and {bn}
converge at infinity limit.
It is necessary that
  lim[m→∞]am=0 ---eqn.BF002A
  lim[n→∞]bn=0 ---eqn.BF002B

<a name="ch10b006">　Index begin　Index this file
We also need a parameter ε
build in both {am} and {bn}
so that we have a handle to
work with. Let us define
  an=bn=n-ε-1/2 ---eqn.BF003
{am} and {bn} are identical
sequences.
<a name="ch10b007">
Put eqn.BF003 to eqn.10.10 
greater than side, we find

<a name="ch10b008">
eqn.BF004 right side has
m summation and n summation,
m and n are dummy variable.
We can change m to n.
Both summation are identical.
Product of square root of two 
identical term is this term 
to power one.
eqn.BF004 become next

<a name="ch10b010">
textbook Page 157 line 26
say For any nonnegative decreasing
function f:[0,∞]→Real, we have the 
integral bound //simplified equation.
  ∑[n=1,∞]f(n)≦∫[x=0,∞]f(x)dx ---eqn.BE123
We will do order of magnitude 
analysis below. We ask only
approximately equal to, so we say

<a name="ch10b012">
2010-04-10-13-01 here
Next, verify integration in 
eqn.10.11B
  ∫[x=1,∞]{x(-1-2*ε)}dx
 =∫[x=1,∞]d{x(-1-2*ε+1)}/(-1-2*ε+1)
 =[∞(-1-2*ε+1) - 1(-1-2*ε+1)]/(-1-2*ε+1)
 =[0-1]/(-2*ε)
 =1/(2*ε) ---eqn.BF005
<a name="ch10b013">
Above calculation indicate that
Hilbert's inequality greater
than side is in the order of
1/(2*ε).

Next see Hilbert's inequality 
less than side.
2010-04-10-13-10 stop

<a name="ch10b014">
2010-04-10-14-52 start
eqn.10.10 can be written
symbolically as
  less_than_side＜  ---eqn.BF006
  C*greater_than_side
Greater than side is in the order 
of 1/(2*ε). eqn.BF006 can be
written as
  less_than_side＜C/(2*ε) ---eqn.BF007
<a name="ch10b015">
We target at C≧π, eqn.BF007
become (expected)
  less_than_side＜π/(2*ε) ---eqn.BF008
Let us see eqn.10.10 less than 
side

<a name="ch10b017">
We want to prove that
when ε approach to zero,
eqn.BF009 approach to π/(2*ε).
We lay out this result as a
lemma.

<a name="ch10b018">　Index begin　Index this file
2010-04-10-15-15 here
■　Double Sum Lemma
　　Textbook page 160, line 15

<a name="ch10b019">
We start from integral version 
of eqn.BF010. Target equation
change to next.

<a name="ch10b020">
The term (x+y) tie x,y two
variables together. Hard to do 
integral separation. We can 
change variable as following. 
Let
  u=y/x ---eqn.BF012
to eliminate y, use
  y=u*x ---eqn.BF013
<a name="ch10b021">
eqn.BF011 denominator 
  x(ε+1/2)*y(ε+1/2)*(x+y)
 =x(ε+1/2)*(u*x)(ε+1/2)*(x+u*x)
 =x(2ε+1)*u(ε+1/2)*x*(1+u) ---eqn.BF014
Alert, x(2ε+1) come from x(ε+1/2)
and (u*x)(ε+1/2) two sources.

<a name="ch10b022">
First key point is that addition
of two variables (x+y) change to
multiplication of two variables
x*(1+u), integral separation is
possible now.

<a name="ch10b023">　Index begin　Index this file
Second key point is that y=u*x
give us
  dy=d(u*x) ---eqn.BF015
then
  dy=u*dx+x*du ---eqn.BF016
eqn.BF011 use dx*dy, we have
  dx*dy=dx*u*dx+dx*x*du ---eqn.BF017
dx*x*du enter the following
equation, but dx*u*dx drop out.
Because dx*dx is higher order
smaller than dx. (if dx=1.e-4
then dx*dx=1.e-8＜＜1.e-4)
<a name="ch10b024">
From above analysis rewrite 
eqn.BF011 as next

<a name="ch10b027">
But, not so good news about 
eqn.BF018 is that variable u 
lower bound is u=1/x ! See
  u=y/x ---eqn.BF012
y lower bound is 1, y replaced
by u with the rule u=y/x , and
y=1 change to u=1/x. Then, the
integration of u is influenced
by x ! Complicated calculation.
<a name="ch10b028">　Index begin　Index this file
This Double Sum Lemma analysis
is approximation every where. 
It is OK to add one more. Now
change u integral
 from ∫[u=1/x,u=∞] //hard calculate
   to ∫[u=0,  u=∞] //easy calculate
added ∫[u=0, u=1/x] !! //trouble
<a name="ch10b029">
Now we are working at eqn.10.10
less than side. 
Less than side add few positive
value ?! Inequality could be
reversed !! That is a damage to 
our inequality !! ('damage' used
in textbook page 160, line-4)
U integration ∫[u=0, u=1/x] add
how much to less than side?
Let's take a look.

<a name="ch10b032">
2010-04-10-17-00 here
think why eqn.BF020 has errata 
change?

<a name="ch10b033">　Index begin　Index this file
2010-04-10-17-53
textbook page 160 line-2 equation
right end use the term x(+ε-1/2) 
errata page change to  x(-ε-1/2) 
eqn.BF021 is to verify which one
is correct. 
x-(-ε+1/2) left '-' from u=1/x
x-(-ε+1/2) right '-' from u(+1-ε-1/2)
LiuHH found textbook is correct.

<a name="ch10b034">
2010-04-10-18-05 here
eqn.BF020 left '0＜' is because
the integrand and du are all
positive. The integration result
is positive.
eqn.BF020 right '＜' is because
integrand change from (1+u)
to 1, dropped positive u from
denominator.

<a name="ch10b035">
eqn.BF020 right end come from
eqn.BF019 square bracket term.
Let us go back to eqn.BF019.

<a name="ch10b036">
eqn.BF019 use lower bound u=1/x
eqn.BF022 use lower bound u=0
Since x≧1, eqn.BF022 has greater
integral value than eqn.BF019.
eqn.BF019 is equality for I(ε) 
eqn.BF022 is equality for I(ε) 
too ! Because eqn.BF022 used
big O() to absorb the approximate
values. big O() even absorb a
negative sign ! (is this right?)
eqn.BF022 u-integration lower
bound reduce to u=0 added extra
positive value. For equality to
be true. big O must be negative
value. big O() is blue integral
in eqn.BF022. 



<a name="ch10b037">
Why eqn.BF022 blue integral
integrand is x-ε-3/2 ? 
It is because
  x-ε-3/2=x+ε-1/2*x-1-2*ε ---eqn.BF023
where x+ε-1/2 come from eqn.BF020
 and  x-1-2*ε come from eqn.BF019

<a name="ch10b038">　Index begin　Index this file
eqn.BF022 blue integral result
is (this is big O value)

<a name="ch10b039">
2010-04-10-19-00 here
2010-04-11-11-49 insert start
eqn.BF024 is big O value. We
still have one more calculation.
eqn.BF022 left end x integration
separated from u integration.
Next evaluate left end x 
integration.

<a name="ch10b041">
We require ε approach to zero 
but not equal to zero.
For non-zero ε, 1/(∞2ε) is 1/∞
which evaluate to zero.
The result of eqn.BF025 is
1/(2ε) which contribute to next
eqn.BF026 first factor right to
equality sign.
2010-04-11-12-08 insert stop

<a name="ch10b042">
2010-04-10-19-00 here
Textbook say above result is

2010-04-10-19-17 here
<a name="ch10b043">　Index begin　Index this file
eqn.BF026 is target equation
eqn.10.10 less than side
estimation.
target equation greater than 
side is in the order of 1/(2*ε).
Now put less than and greater
than into one equation.

2010-04-10-19-50 stop
not see numerator ε in eqn.BF024
tripped at eqn.BF027 Red was O(1)
eight hour delayed eqn.BF028

<a name="ch10b045">
2010-04-11-04-40 start
In eqn.BF027, ε approach to zero
but ε is not zero. Whole equation
multiply by ε, we get eqn.BF028.
Left term and right term cancel ε.
Red term get a ε. When ε goto zero, 
red term (was big O) become zero.
Also damage (big O) go to zero. 
eqn.BF028 left term factor u-ε-1/2 
become u-1/2. eqn.BF028 reduce to

<a name="ch10b047">
In eqn.BF029 cancel factor 1/2.
Left side has value π, please see 
eqn.10.9. 

<a name="ch10b048">
At this point (after refer to
eqn.10.9 get value π)
If we consider eqn.BF029 less
than side only, we conclude 
Double Sum Lemma
If we consider eqn.BF029 both
side, we solved Problem 10.2
2010-04-11-04-52 stop



<a name="ch10b049">　Index begin　Index this file
2010-04-11-13-50 start
■　Exercise 10.1 problem statement
　　textbook page 162
（Guaranteed Positivity）

Show that for any real numbers
a1,a2,...,an one has

<a name="ch10b052">
Obviously the second inequality
implies the first, so the bound
(10.16) is mainly a hint which
makes the link to Hilbert's
inequality. As a better hint,
one might consider the possibility
of representing 1/λj as an
integral.
2010-04-11-14-04 stop




<a name="ch10b053">　Index begin　Index this file
2010-04-11-14-06 start
■　Exercise 10.1 hint
　　textbook page 267

The proof fits in a single line:
just note

<a name="ch10b055">
2010-04-11-14-18 from
http://www-stat.wharton.upenn.edu/~steele/Publications/Books/CSMC/ByronSchmuland.txt
include
[[
267     5    There are two typos in the displayed equation.
             (comma after = under the sum, x_j should be x^j)
             but the whole solution has to be reworked since
             \int_0^1 x^{\lambda_j}\,dx=1/(1+\lambda_j), not 1/\lambda_j.
             Use 1/\lambda_j=\int_0^\infty exp(-\lambda_jx)\,dx instead.
]]

<a name="ch10b056">
2010-04-11-14-23 continue
and integrate over [0,1]. For the
general case, one naturally uses
the representation 
  1/λj≠∫[x=0,1]xλjdx ---eqn.BF031

<a name="ch10b057">
This problem is a reminder that
there are many circumstance
where dramatic progress is made
possible by replacing a number
(or a function) with an appropriate
integral. Although this example 
and the one given by Exercise 7.5
page 116, are simple, the basic
theme has countless variations.
Some of these variation are quite
deep.
2010-04-11-14-31 stop

<a name="ch10b059">
2010-04-11-17-46 start
■　Exercise 10.1 discussion

Compare eqn.BF030 with eqn.10.16 
need to relate

2010-04-11-18-06 here
<a name="ch10b061">
Verify eqn.BF032 and eqn.BF033.

<a name="ch10b062">
Both eqn.BF032 and eqn.BF033
right side equality are correct.
(LiuHH change eqn.BF032 left
 equality to red inequality)
To solve Exercise 10.1, use
eqn.BF033.



<a name="ch10b063">　Index begin　Index this file
2010-04-11-18-29 here
■　Exercise 10.1 solution


The following solve for general
case eqn.10.17 which cover the
special case eqn.10.16

<a name="ch10b064">
eqn.BF033 move λj from denominator
to power of 2.718281828459045...
Change λj to λj+λk get

<a name="ch10b066">
In eqn.BF037 sum of j and sum 
of k are separated. Also j and
k are dummy variable. eqn.BF037
right side is the square of one
summation. Square of real number
is non-negative.

eqn.10.17 is proved. eqn.10.16 
is special case of eqn.10.17.
Exercise 10.1 is done.
2010-04-11-19-02 stop



<a name="ch10b067">　Index begin　Index this file
2010-04-12-10-51 start
■　Exercise 10.2 problem statement
　　textbook page 163
（Insertion of a Fudge Factor）

There are many ways to continue
the theme of Exercise 10.1, and
this exercise is one of the most
useful. It provides a generic
way to leverage an inequality
such as Hilbert's.

<a name="ch10b068">
Show that if the complex array
{ajk:1≦j≦m,1≦k≦n} satisfies
the bound
  |∑[j,k]ajkxjyk|
 ≦M∥x∥2∥y∥2 ---eqn.10.18
then one also has the bound
  |∑[j,k]ajkhjkxjyk|
 ≦αβM∥x∥2∥y∥2 ---eqn.10.19
provided that the factors hjk have
<a name="ch10b069">
an integral representation of
the form
  hjk=∫[D]fj(x)gk(x)dx ---eqn.10.20
for which for all j and k one
has the bound
  ∫[D]|fj(x)|2dx≦α2 ---eqn.10.21A
and
  ∫[D]|gk(x)|2dx≦β2 ---eqn.10.21B
2010-04-12-11-05 stop





<a name="ch10b070">
2010-04-12-11-09 start
■　Exercise 10.2 hint
　　textbook page 267

We substitute, switch order, 
apply the bound (10.18), switch 
again, and finish with Cauchy's 
inequality to find

<a name="ch10b074">
2010-04-12-12-50 here
The bound αβM∥x∥2∥y∥2 now 
follows from the assumption 
(10.21)
2010-04-12-12-52 stop



<a name="ch10b075">
2010-04-12-14-12 start
■　Exercise 10.2 discussion
eqn.10.20 has hjk definition.
Substitute eqn.10.20 to 
eqn.BF038 left side, result
is eqn.BF038 right side. hjk 
definition carries integration
which show up at eqn.BF038 
right side. 

<a name="ch10b076">
Most discussion write in between
equations. From eqn.BF038 to 
eqn.BF043. Still think how to 
switch ∑ and ∫.
2010-04-12-15-24 stop



<a name="ch10b077">
■　Exercise 10.2 solution


Why eqn.BF041 has 
reverse inequality?
How to switch ∑ and ∫ ?

Exercise 10.2 is NOT DONE
2010-04-12-15-25 stop
2010-04-12-16-39 delete eqn.BF040 and stop



<a name="ch10b078">　Index begin　Index this file
2010-04-12-17-12 start
■　Exercise 10.3 problem statement
　　textbook page 163
（Max. Version of Hilbert's Inequality）

Show that for every pair of 
sequences of real numbers
{an} and {bn} one has

<a name="ch10b080">
and show that 4 may not be
replaced by a smaller constant.
2010-04-12-17-27 stop





<a name="ch10b081">
2010-04-12-17-29 start
■　Exercise 10.3 hint
　　textbook page 267

To mimic our proof of Hilbert's
inequality we take λ＞0 and use
the analogous splitting to find

<a name="ch10b084">
By Cauchy's inequality the square
of the double sum is bounded by 
the product of the sum given by

<a name="ch10b085">
and the corresponding sum
containing {bn2}

<a name="ch10b086">
If we take λ=1/4, we have

<a name="ch10b089">　Index begin　Index this file
so, to complete the proof we
only need to note that the {bn2}
sum satisfies an exact analogous
bound.

<a name="ch10b090">
Finally, the usual "stress testing"
method shows that 4 cannot be
replaced with a smaller value.
After setting 
  an=bn=n-ε-1/2 ---eqn.BF051
one checks that

<a name="ch10b093">　Index begin　Index this file
Peeking ahead and take
  K(x,y)=1/max(x,y) ---eqn.BF054
in Exercise 10.5, one finds
that the constant 4 in the
bound (10.3) is perhaps best
understood when interpreted as
the integral

2010-04-12-19-04 stop




<a name="ch10b095">
2010-04-13-11-21 start
■　Exercise 10.3 solution


In Problem 10.1, textbook use 
Cauchy's Inequality for countable 
set to prove Hilbert's Inequality.
<a name="ch10b096">
Here, Max. Version of Hilbert's 
Inequality, still use same method.
If we use
  αs=am/√(m+n) ---eqn.BE115
  βs=bn/√(m+n) ---eqn.BE116
we can not reach our goal.
Textbook use eqn.10.4A and 
eqn.10.4B as starting point. 
That is how we get eqn.BF044. 
<a name="ch10b097">
Red term in eqn.BF044 is one 
and λ is under our control. 
From eqn.BF044 to eqn.BF045
it is simple splitting. Please
pay attention to eqn.BF045 
denominator, it is √ * √

<a name="ch10b098">　Index begin　Index this file
Two summations in eqn.BF045 are
similar and separated. am series 
and bn series are formulated 
in eqn.BF046 and eqn.BF047.
The summation which contain
1/max(m,n) is single out as
eqn.BF048
<a name="ch10b099">
In web page tute0037.htm, start 
from ch10a048 to ch10a068
textbook tell us how to find λ 
value. The best value is 
  λ=1/4  ---eqn.BE126
eqn.BF048 applied λ=1/4.

<a name="ch10b100">
To solve max(m,n), eqn.BF048
split sum 
from ∑[n=1,∞] 
 to  ∑[n=1,m] + ∑[n=m+1,∞] 
∑[n=1,m] has max(m,n)=m and
∑[n=m+1,∞] has max(m,n)=n.
We have simpler eqn.BF048 right
side equation.
<a name="ch10b101">
In ∑[n] summation operation,
m is constant, m can be moved 
out of n summation. 
2010-04-13-12-05 here

<a name="ch10b102">
2010-04-13-12-18 made eqn.BF049
●●change from 'no ≦' to 'ok ＝'

2010-04-13-12-20 here, 
why eqn.BF050 is true?

<a name="ch10b103">　Index begin　Index this file
2010-04-13-12-38 refer to
eqn.BE123
  ∑[n=1,∞]{1/√n}≦∫[x=0,∞]dx/√x ---eqn.BF056
find proof !

<a name="ch10b104">
Write eqn.BF049 left summation
  ∑[n=1,m]{1/√n}
 ≦∫[x=0,m]dx/√x
 =∫[x=0,m]d(√x√x)/√x
 =∫[x=0,m]2*√x*d(√x)/√x
 =∫[x=0,m]2*d(√x)
 =2*(√m-√0)
 =2*√m
then we have
  ∑[n=1,m]{1/√n}≦2*√m ---eqn.BF057

<a name="ch10b105">
Write eqn.BF049 right summation
  ∑[n=m+1,∞]{n(-3/2)}
 ≦∫[x=m,∞]x(-3/2)dx
 =∫[x=m,∞]d[x(+1-3/2)]/(+1-3/2)
 =∫[x=m,∞]d[x(-1/2)]/(-1/2)
 =[∞(-1/2)-m(-1/2)]/(-1/2)
 =[0-1/√m]/(-1/2)
 =2/(√m)
then we have
  ∑[n=m+1,∞]{n(-3/2)}≦2/(√m) ---eqn.BF058

<a name="ch10b106">
eqn.BF057 and eqn.BF058 support
eqn.BF050 left inequality. But 
look like strict inequality '＜'
instead of '≦'. If m=1, 
eqn.BF049 ∑[n=1,1]{1/√n} = 1
eqn.BF050 2√m = 2*1 = 2 get
eqn.BF049 ＜ eqn.BF050
2010-04-13-13-05 stop
2010-04-13-14-00 start

<a name="ch10b107">　Index begin　Index this file
Above analysis give us

<a name="ch10b108">
Same story for b sum eqn.BF047.
Put both result back to
eqn.BF045 get greater than 
side coefficient 4*4. After 
take square root in eqn.10.22
we get coefficient 4 as 
required.
2010-04-13-14-11 here

<a name="ch10b109">
To prove that coefficient 4
is the best value. textbook
use "stress testing", 
Exercise 10.3 specify sequences
{an} and {bn} as 
"for every pair of sequences 
 of real numbers", then set
  an=bn=n-ε-1/2 ---eqn.BF051
still satisfy requirement.
<a name="ch10b110">
When ε approach zero and n≧1
1/√n is a positive monotone 
decrease function. Satisfy
eqn.BE123 requirement.
Again refer to eqn.BE123
  ∑[n=1,∞]{n-ε-1/2}
 ≦∫[x=0,∞]{x-ε-1/2dx} ---eqn.BF059
Evaluate eqn.BF059 integration
as following

<a name="ch10b112">
2010-04-13-14-51 wonder, why infinity ?
in ∫[x=1,∞]xpdx, p=-1 is border.
p＜-1 get finite, p≧-1 get infinity.
for nearly zero ε, -ε-1/2＞-1.
2010-04-13-14-57 LiuHH ●●change
from set
  an=bn=n-ε-1/2 ---eqn.BF051
to set
  an=bn=n-1-2ε ---eqn.BF061
then the result of eqn.BF025 is
valid. eqn.BF052 would be written
as

How to evaluate O(1) ?
2010-04-13-15-07 stop

<a name="ch10b114">
2010-04-13-17-22 start
From eqn.BF052 to eqn.BF053 look
like it is a simple multiplication
by 4. This 4 possibly come from
eqn.BF050. Liu,Hsinhan is not
sure how to get eqn.BF053. We have
one summation ∑[n=1,∞]an=1/(2ε)
Why two summation
∑[m=1,∞]∑[n=1,∞]aman/max(m,n)
is simply multiply by 4?

<a name="ch10b115">　Index begin　Index this file
Hilbert's Inequality never see 
aman, only see ambn and amam
For example eqn.10.1, eqn.10.10.
Why aman show up at eqn.BF053
LiuHH thought aman should show 
up at eqn.10.1 and eqn.10.10
proving process.
<a name="ch10b116">
How Cauchy's Inequality for 
countable set convert aman?
Please see eqn.10.2
Cauchy Inequality should contain 
aman, please see textbook page 14 
eqn.1.22, 2αβ is 2aman and 2xy is 
2bmbn.

<a name="ch10b117">
LiuHH wish junior, senior level
reader also write study notes.
So that LiuHH can learn more.
LiuHH is just sophomore level
and target at freshman level.

Exercise 10.3 is NOT DONE
2010-04-13-17-46 stop



<a name="ch10b118">　Index begin　Index this file
2010-04-14-11-42 start
■　Power integration is simple
Liu,Hsinhan target at freshman 
level reader. Advanced reader
can take a break.

We see few power integrations.
List them as following

<a name="ch10b125">　Index begin　Index this file
2010-04-14-12-11 done collect 6 eqn.
Now we pay attention how to 
change from left integral to
right integral. Above six
equations has two groups.

eqn.BF035 integral variable
locate at power. 1-eqn-group.

Other five integral have their
variable locate at base. They
are 5-eqn-group.

<a name="ch10b126">
Now observe 5-eqn-group 
(exclude eqn.BF035)
All left integrand has power p
and right integrand has power p+1.
For example, eqn.BF060 
 left integrand has power   -ε-1/2
right integrand has power +1-ε-1/2
left side has 'dx'
right side has 'd(x1-ε-1/2)'

<a name="ch10b127">
Why move 'd' from d single_x to
d all_x then all_x get power + 1?
The reason is simple, very simple.

If x is length, left integration
x-ε-1/2dx has physics dimension of
length to power '-ε-1/2' plus one.
This power one come from length x 
in 'dx'
<a name="ch10b128">
Right integration has d(x1-ε-1/2)
It is length to power '1-ε-1/2'.
Physics dimension is balanced.
Differentiation symbol 'd' do not
carry Physics dimension. 'd' just
say take very very tiny amount.

<a name="ch10b129">
Why right integration has division
of '1-ε-1/2' ?
This is elementary calculus rule
  d[x^n]=n*d[x^(n-1)] ---eqn.BF063
In order to be able go backward
from right integration to left 
integration with same value.
The forward calculation divide
a '1-ε-1/2', let backward factor
(1-ε-1/2) cancel forward 1/(1-ε-1/2)
to one.
2010-04-14-12-35 here.

<a name="ch10b130">　Index begin　Index this file
If problem is ∫x-1dx
we get
  ∫x-1dx=∫d[x-1+1]/(-1+1) ---eqn.BF064
divide by zero !!
∫x-1dx is special integration
its answer is
  ∫x-1dx=log(x) ---eqn.BF065

Above is 5-eqn-group 
      variable is at base.
<a name="ch10b131">
Next  1-eqn-group
      variable is at power.
Any power must be pure number.
If not, we have trouble.
What is the meaning of two
meters (length) to power of
three seconds (time)? This is
non-sense statement.

<a name="ch10b132">
Power must be pure number.
That is eqn.BF035 left side
e-x*λj has pure number '-x*λj'
If x is length, x*λ is pure number
then λ is 1/length !
1/λ is 1/(1/length)=length 

<a name="ch10b133">
eqn.BF035 left side is 
pure number e-x*λj multiply
by length dx
eqn.BF035 right side is 
1/λ=1/(1/length)=length 
eqn.BF035 has physics dimension
consistency !!

<a name="ch10b134">
In eqn.BF035, how denominator λ
is born ?
We know the calculus rule for
exponential function is that
  d[e^y]/dy = e^y ---eqn.BF066
or
  d[e^y] = e^y * dy ---eqn.BF067
<a name="ch10b135">　Index begin　Index this file
now eqn.BF035 left side is
e-x*λjdx
Simple !! multiply by ONE
this ONE is next
  1=-λj/(-λj) ---eqn.BF068
<a name="ch10b136">
For constant λ,
  λ*dx=d(λx) ---eqn.BF069
then we have
  e-x*λjd(-λjx)/(-λj) ---eqn.BF070
the term e-x*λjd(-λjx) match
d[e^y] = e^y * dy rule and
denominator λ is born.
2010-04-14-12-56 stop

2010-04-14-17-52 done proofread
2010-04-14-18-18 done spelling check



<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