<a name="docA01">
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="docA02">
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="docA03">
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="docA04">
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="docA05">
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="ch01c001">　Index begin　Index this file
2009-08-04-15-25 start

■　Exercise 1.8 problem statement
　　textbook page 14
（Doing the Sums）
To effectively use Cauchy Inequality, one needs
to estimate upper bounds. Please verify the 
following four real number upper bound equations.

2009-08-04-16-10 stop


<a name="ch01c006">　Index begin　Index this file
2009-08-04-16-48 start
■　Exercise 1.8  hint
　　textbook page 228
For above four equations, apply Cauchy Inequality
to each of them and estimate the summed upper
bounds.

Exercise 1.8 problem A, use geometric progression.
（pay attention to the domain 0≦x＜1）
1 + x^2 + x^4 + x^6 + ... = 1/(1-x*x) -----eqn.AD01

<a name="ch01c007">
Exercise 1.8 problem B, use Euler's famous formula

<a name="ch01c008">
2009-08-04-17-13 here
Exercise 1.8 problem B, you can also use 
telescoping identity
（textbook use "≦", but 1/k/k < 1/k/(k-1)　where
　k≧2，so page 228 line 29 should be strictly
 not equal. LiuHH 2009-08-04-17-16 ）

<a name="ch01c009">
2009-08-04-17-34 here
Exercise 1.8 problem C use next integral
equation.

<a name="ch01c011">
2009-08-04-17-45 here
Exercise 1.8 problem D use the explicit sum
for the squares of the binomial formula.

<a name="ch01c012">
2009-08-04-17-54 here
page 229 line 5 equation can be proved by 
classic counting method. For example, from
n male and n female select n person to form
a committee. Page 229 line 5 equation middle
term (between two '=') count select k member
from n male (then select n-k member from
female) then sum from k=0 to k=n. Page 229 
line 5 equation right side term, direct count
select n member from 2n candidate. Two methods
are the same.
2009-08-04-18-01 stop

<a name="ch01c013">　Index begin　Index this file
2009-08-04-19-48 start
■　Exercise 1.8 solution


Exercise 1.8 equation A, B, C, D 
four  left hand side, all have ak
four right hand side, all have √〔∑ak2〕
This is common term for all four problems.
We need pay attention to its companion term.

Cauchy Inequality handle two number sequences.
We need to identify these two sequences first.
▣　Exercise 1.8 equation A use next two 
sequences.
Seq.1 is common term ak ，k=0 to ∞，k is index
Seq.2 is unique seq. xk ，k=0 to ∞，k is power
where 0≦x＜1 is a rule.
<a name="ch01c014">
Exercise 1.8 equation A left hand side is dot
product of two sequences.
Cauchy Inequality original form is next.

<a name="ch01c015">
2009-08-04-20-18 here
Exercise 1.8 equation A right hand side is 
two sequences length product.
Seq.1 ak become right hand side √〔∑ak2〕 -----eqn.AD01
Seq.2 xk become right hand side √〔∑x2k〕； 0≦x＜1
where ∑x2k from k=0 to k=∞ get the following eqn.
1 + x^2 + x^4 + x^6 + ... = 1/(1-x*x) -----eqn.AD02
Substitute eqn.AD02 right hand side to Cauchy
Inequality and take square root get the answer
page 14 line 25 eqn.A
2009-08-04-20-39 stop

<a name="ch01c016">
2009-08-05-09-18 start
▣　Exercise 1.8 equation B has two sequences
Seq.1 is common term ak ，k=1 to n，k is index
Seq.2 is unique seq. 1/k，k=1 to n，k is integer.
equation A use k=0 to k=∞
equation B use k=1 to k=n
equation B cannot use k=0 to k=∞，
because equation B has 1/k
when k=0, 1/k become infinite, can not handle.
If equation B use from k=1 to k=∞， also get
trouble, because ∑1/k from k=1 to k=∞ has
infinite value.
equation B start from k=1 to finite k=n this
is a reasonable problem.

<a name="ch01c017">
Exercise 1.8 equation B left hand side is dot
product of two sequences.
Exercise 1.8 equation B right hand side is 
two sequences length product.
Sequence 1 ∑ak length is
 √〔a12+a22+...an2〕 -----eqn.AD01
This length apply to Cauchy Inequality directly.
Sequence 2 ∑1/k length is
 √〔(1/1)^2 + (1/2)^2 + (1/3)^2 +...+ (1/n)^2〕 -----eqn.AD03
Problem right hand side is a simple √2 ，
therefore would above equation
(1/1)^2 + (1/2)^2 + (1/3)^2 +...+ (1/n)^2
be less than 2 ？ that is whether
 1 + 1/4 + 1/9 + 1/16 + ... + 1/(n*n) ?＜? 2
Subtract one from both side, get
 1/(2*2)+1/(3*3)+1/(4*4)+...+1/(n*n) ?＜? 1 -----eqn.AD04
<a name="ch01c018">
The following, do not use Euler's formula
The following, use telescoping identity
We can observe one fact.
（next five lines cause absolute inequality）
 1/(2*2)=0.25 < 0.5=1/(2*1) = 1/[2*(2-1)]
 1/(3*3)=1/9  < 1/6=1/(3*2) = 1/[3*(3-1)]
 1/(4*4)=1/16 <1/12=1/(4*3) = 1/[4*(4-1)]
...
 1/(n*n) <           1/[n*(n-1)]
We can write as following
 ∑{1/[k*k]} < ∑{1/[k*(k-1)]}
（ k=2 to k=n ， k is variable, n is constant）
Greater than side can be re-written as following.

<a name="ch01c020">
2009-08-05-10-16 here
page 228 line 29 aux.1
 -1/2 and +1/(3-1) cancel
 -1/3 and +1/(4-1) cancel
 -1/4 and +1/(5-1) cancel
.....
finally, left only first term +1/(2-1) and
last term -1/n
so, equation
 ∑{1/[k*k]} < ∑{1/[k*(k-1)]}
become
∑{1/[k*k]} < +1/(2-1) - 1/n = 1 - 1/n

After expand equation, keep first term, keep
last term, drop all middle terms. It is called 
telescoping identity.

<a name="ch01c021">
Summation symbol ∑ from k=2 sum to k=n 
 ∑{1/[k*k]} value is less than 1 - 1/n
 ∑{1/[k*k]} sum to infinity, its value approach 1 

Add one to both side of inequality equation, get
Summation symbol ∑ from k=1 sum to k=n 
 ∑{1/[k*k]} sum to infinity, its value approach 2 
eqn.AD04 is true.
eqn.AD03 summation value is less than √2

Exercise 1.8 equation B proved to be true.
Absolute inequality come from telescoping
 identity.
2009-08-05-10-33 stop

<a name="ch01c022">
2009-08-07-11-57 start
▣　Exercise 1.8 equation C has two sequences
Seq.1 is common term ak ，k=1 to n，
　　　k is index
Seq.2 is unique seq. 1/√(n+k)，k=1 to n，
　　　k is integer.
n bound k from above.

Sequence 1 length is eqn.AD01
Sequence 2 length square is the sum of all
elements squared.
∑[1/√(n+k)]*[1/√(n+k)]  （k=1 to n）
Sequence 2 length square is 
∑[1/(n+k)]  （k=1 to n）
<a name="ch01c023">
Hint suggest consider integral equation.

<a name="ch01c024">
2009-08-07-12-11 here
How to prove page 229 line 2 left equation ?
If carry out integration, get new problem.
How to prove
1/(n+k) ?＜? log[(n+k)/(n+k-1)]
This thinking is a little bit complicate.
Back to starting point page 229 line 2 left
equation.
Do not change right side (do not integrate)
Re-write left side as the following.

<a name="ch01c025">
2009-08-07-12-25 here
Above equation left integral value is one.
 1/(n+k) multiply by one, not change its
value. Integration variable is x. (n+k) 
is constant. (n+k) is upper limit of 
integration. Move (n+k) to inside of 
integral sign, get the following

<a name="ch01c026">
2009-08-07-12-29 here
Page 229 line 2 left equation not easy to
compare greater than/less than relation. 
But page 229 line 2 aux.2 equation is 
easy to compare. Because most are the 
same, the only difference is two 
denominators. Both n and k are integers
greater than or equal to one.
Integration dx start from x=n+k-1 to x=n+k ,
dx is positive. To compare greater than
or less than reduce to next
1/(n+k) ?＜? 1/x
Since x and n+k are both positive, re-write
as
x ?＜? (n+k) -----eqn.AD05
（here get absolute not equal）
(n+k) is integral upper limit, so eqn.AD05
is true for sure. Go back to original eqn.
must be true.

<a name="ch01c027">
page 229 line 2 left eqn. is from x=n+k-1 
to x=n+k Within this section both n and k 
are constant. x is integral variable.
When consider whole domain, k is variable
k start from k=1 to k=n 
page 229 line 2 left eqn. repeat n times.
Sum all n equations get next

<a name="ch01c028">
2009-08-07-12-50 here
Equality for page 229 line 2 right side 
equation is simple. Right side integration
result
log(2n) - log(n) = log(2n/n) = log(2)
Exercise 1.8 equation C sequence 2 length
square is ∑[1/(n+k)]
From page 229 line 2 both right and left 
side. We can conclude that 
seq.2 length square is less than  log(2)
seq.2 length        is less than √log(2)

<a name="ch01c029">
To solve Exercise 1.8 equation C apply 
Cauchy Inequality to two sequences 

Seq.1 dot seq.2 get page 14 line 27 
equation C left hand directly.
Seq.1 length multiply seq.2 length get 
page 14 line 27 equation C right hand 
side. Done this proof.

Please pay attention to that equation C
is absolute inequality. It come from
eqn.AD05.
2009-08-07-13-11 stop

<a name="ch01c030">
2009-08-08-13-34 start
▣　Exercise 1.8 equation D has two sequences.
Seq.1 is common term ak ，k=0 to n，
　　　k is index
Seq.2 is unique seq. It is sum of binomial
of n and k, k=0 to n

<a name="ch01c031">
k and n are integers，n is upper bound of k 
The following use "bino(n,k)" represent
binomial coefficients of eqn.AD06

Seq.1 length is same as before eqn.AD01.
Seq.2 length square is the sum of elements
squares.
bino(n,0)*bino(n,0) + bino(n,1)*bino(n,1)
+ ..... + bino(n,n)*bino(n,n)
that is
∑[bino(n,k)*bino(n,k)]  （k=0 to n）
Above line can be written as page 229 
line 5 left side expression.
page 229 line 5 equation is textbook hint.
2009-08-08-14-10 here

<a name="ch01c032">
It is not very easy to prove page 229 
line 5 equation. Textbook use classic 
counting method. LiuHH attempted to
expand page 229 line 5 equation without
success.

2009-08-07-22-59 found page online
http://citeseerx.ist.psu.edu/viewdoc/download;jsessionid=0A9ADE926DC711D48836D9C5E5B9950D?doi=10.1.1.38.6614&rep=rep1&type=pdf
Binomial Sums and Identities, Volker Strehl
save as binoSum1.pdf This file, page 5 
to 6 (total 13 pages) prove page 229 
line 5 equation. He did not expand
directly. He use indirect method. Say 
equation left side and right side both
have same property to draw equality
conclusion. Reader can access his page.

<a name="ch01c033">
To solve Exercise 1.8 equation D apply
Cauchy Inequality to two sequences.
Seq.1 dot seq.2 get page 14 line 27 eq.D
left hand side.
Seq.1 length multiply seq.2 length get 
right hand side.
Seq.1 length is the same as above three
problems.
Seq.2 length is unique to the current
problem. From hint done Exercise 1.8
equation D
2009-08-08-14-37 stop

<a name="ch01c034">
2009-08-08-16-34 start
2009-08-07-22-46 access
http://matwbn.icm.edu.pl/ksiazki/aa/aa86/aa8612.pdf
Factors of sums of powers of binomial coefficients
by Neil J. Calkin (Clemson, S.C.)
save as binoSum0.pdf This file page 1 
(total 10 pages) say
[[
It is possible to show (Wilf, personal 
communication, using techniques in [8])
that for 3≦ power ≦9 there is no closed
form for fn,b as a sum of a fixed number
of hypergeometric terms.
]]
2009-08-19-21-01 LiuHH note:
[[
In Neil J. Calkin paper, fn,b was written
as fn,a . LiuHH change 'a' to 'b' avoid
confuse with two following 'a'.
Please see page 229 line 5 left side 
expression. Change power '2' to 'a'
then it is same as fn,a 2009-08-19-21-11
]]

Please see page 229 line 5 eqn. Power 2
change to 3 then no closed form solution, 
only numerical solution available.
2009-08-08-16-40 stop



<a name="ch01c035">　Index begin　Index this file
2009-08-08-19-33 start

■　Exercise 1.9 problem statement
　　textbook page 15
（Beating the Obvious Bounds）
Many math problems need bounds more accurate
than Cauchy Inequality bound. Following is 
an example. Given real sequence aj, j=1,2,...,n 
Please prove the following inequality is true.

2009-08-08-19-54 here
If apply Cauchy Inequality directly, above
equation greater than side coefficient is
2n, not (n+2). When n increase, page 15 
line 8 bound is nearly half that of Cauchy
Inequality.


<a name="ch01c037">　Index begin　Index this file
■　Exercise 1.9  hint
　　textbook page 229
Use T to represent page 15 line 8 left 
hand side. Expand T get

<a name="ch01c038">
2009-08-08-20-08 here
Where set S has the following property
for all (j,k) 1≦j＜k≦n and j+k is even.

From basic inequality 2ajak≦aj2+ak2， we get

<a name="ch01c039">
2009-08-08-20-31 here
where ns is the number of (j,k) pair in S
with j=s or k=s . We have ns ≦ └(n-1)/2┘
therefore

2009-08-08-20-52 stop





<a name="ch01c040">　Index begin　Index this file
2009-08-09-18-58 start
■　Exercise 1.9 solution

Let's see Cauchy Inequality first

<a name="ch01c041">
2009-08-09-19-05 here
Above is general equation. Problem given
equation can be written as sum of two 
Cauchy Inequalities. First Cauchy change
all 'b' to '1' as next

<a name="ch01c042">
2009-08-09-19-15 here
eqn.AD07 use two sequences
Seq.1 is old friend ak ，k=1 to n，
　　　k is index
Seq.2 is unique seq. It is 1, 1, 1, ... 1
　　　repeat n times.
This is one trick.
When elements of ak are all equal, then
seq.1 and seq.2 are in proportion and
first Cauchy Inequality eqn.AD07 become
equality.

<a name="ch01c043">
Next see second Cauchy Inequality, change
all 'b' to (-1)j as the following

<a name="ch01c044">
2009-08-09-19-45 here
eqn.AD08 use two sequences
Seq.1 is old friend ak ，k=1 to n，
　　　k is index
Seq.2 is unique seq. -1, 1, -1 ... +/-1
　　　repeat n times.
This is also one trick.
When sequence ak elements absolute value
are equal and reverse sign from neighbor
to neighbor. Seq.1 and seq.2 are in 
proportion and second Cauchy Inequality 
eqn.AD08 become equality.

<a name="ch01c045">
If above two Cauchy Inequalities are two
separate problems, then each has chance to
become equality. Now, merge two problems
with addition operation to form one problem
Clearly sequence ak can not be proportion
to 1, 1, 1 ... 1 and to -1, 1, -1 ... +/-1
at same time.
Exercise 1.9 is absolute inequality by nature.
2009-08-09-19-58 here
hint said﹕
"Use T to represent page 15 line 8 left 
 hand side."
<a name="ch01c046">
The following use matrix method to separate
common factor a1, a2 ... an
and unique coefficients.

<a name="ch01c050">
2009-08-09-20-37 here
Let's see eqn.AD09 matrix off-diagonal
elements (j not= k)
Element at row 1 column 2 is
(j,k) = (1,2) = +1-1 = 0
Element at row 1 column 3 is (red '2')
(j,k) = (1,3) = +1+1 = 2
All elements with index j+k = even get value 2
All elements with index j+k =  odd get value 0

<a name="ch01c051">
All elements with index j+k = even 
(j,k) bring a value of 2
(k,j) bring a value of 2 again
therefore
all j+k = even get value 4  (j＜k)
all j+k =  odd get value 0  (j＜k)

hint require (j,k)∈S that is
 S is in the set of "j+k = even"

<a name="ch01c052">
Above is eqn.AD09 matrix off-diagonal
   elements (j not= k)
Below is eqn.AD09 matrix  on-diagonal
   elements (j = k)

page 15 line 8 left absolute square expand
get a1*a1 and right absolute square expand
get a1*a1 add together result coefficient 2.
Other on-diagonal terms have same reasoning.
Up to here, explained page 229 line 13
equation.
2009-08-09-20-53 stop

<a name="980809a">
2009-08-09-22-38 create matrix for testing
2
0 2
2 0 2
0 2 0 2
2 0 2 0 2
0 2 0 2 0 2
2 0 2 0 2 0 2
0 2 0 2 0 2 0 2
Copy paste above lower triangle matrix to
http://freeman2.com/tute0009.htm#ch01tb03
box 9 and calculate eigenvalue.
Any rank n should get maximum eigenvalue
be less than n+2
2009-08-09-22-42

<a name="ch01c053">
2009-08-10-04-24 start
eqn.AD09 has two no-zero matrix. Explain
as th following.
Seq.1 a1, a2, a3, ... an,
Seq.2 a1, a2, a3, ... an,
are two identical sequences. Multiply get
 a1 of seq.2 multiply seq.1 from a1 to an
 a2 of seq.2 multiply seq.1 from a1 to an
.....
 an of seq.2 multiply seq.1 from a1 to an
This procedure create eqn.AD09 all one matrix.

<a name="ch01c054">
Now see
Seq.3 -a1, a2,-a3, ... +/-an,
Seq.4 -a1, a2,-a3, ... +/-an,
are two identical sequences. Multiply get
-a1 of seq.4 multiply seq.3 from -a1 to +/-an
-a2 of seq.4 multiply seq.3 from -a1 to +/-an
.....
-an of seq.4 multiply seq.3 from -a1 to +/-an
<a name="ch01c055">
Now explain from index j,k
Seq.3 use index j, j=1,2,3 ... n for -a1 ... +/-an
Seq.4 use index k, k=1,2,3 ... n for -a1 ... +/-an
If j=k (+)*(+)=(+) and (-)*(-)=(+) always (+)
This is diagonal elements of +/- matrix.

If j NOT= k 
if both j and k are  odd,
  for example (-a1)*(-a3) = +a1a3
if both j and k are even,
  for example (+a2)*(+a4) = +a2a4
here odd*odd=(+), even*even=(+)
<a name="ch01c056">
if j and k one is odd, the other is even
for example (-a1)*(+a4) = -a1a4
or (+a2)*(-a7) = -a2a7
here odd*even=(-)
when j NOT= k odd*odd=(+), even*even=(+),
              odd*even=(-)
This condition create eqn.AD09 +/- matrix.

eqn.AD09 all one matrix add +/- matrix get
2/0 matrix. Eqn.AD09 and hint at page 229 
line 13 equation are identical.
2009-08-10-05-03 stop

<a name="ch01c057">
2009-08-10-10-17 start
How to explain
〔〔
where ns is the number of (j,k) pair in S
with j=s or k=s . We have ns ≦ └(n-1)/2┘
〕〕
？
<a name="ch01c058">
Please see color matrix below.

<a name="ch01c059">
2009-08-10-10-20 here
Page 229 line 16 equation left '≦' 
represent take Cauchy Inequality for 
aj and ak two sequences.

Page 229 line 16 equation right '≦' 
explain as following.

eqn.AD10 matrix upper left 3*3 sub-matrix
rank=3, s=3 red 2 is one (j,k) pair.
Red 2 is (1,3) or (3,1)， One of j or k 
should be 3.
（comply with hint said "j=s or k=s",
 s=n=rank 3 ）
one pair ＝ 1 ＝ ns ≦ └(n-1)/2┘ ＝ └(3-1)/2┘
　　　　　　　＝ └1┘ ＝ 1
<a name="ch01c060">
Similarly
rank=4, s=4 blue 2 is one (j,k) pair.
Blue 2 is (2,4) or (4,2)， One of j or k 
should be 4
（comply with hint said "j=s or k=s",
 s=n=rank 4 ）
one pair ＝ 1 ＝ ns ≦ └(n-1)/2┘ ＝ └(4-1)/2┘
　　　　　　　＝ └1.5┘ ＝ 1
<a name="ch01c061">
Similarly

rank=5, s=5 gold 2 is two (j,k) pair.
Gold 2 is pair one (1,5) or (5,1) or
Gold 2 is pair two (3,5) or (5,3)，
One of j or k should be 5
（comply with hint said "j=s or k=s",
 s=n=rank 5 ）
Two pairs ＝ 2 ＝ ns ≦ └(n-1)/2┘
　　　　　　 ＝ └(5-1)/2┘　＝ └2┘ ＝ 2
Keep same reasoning, support page 229
line 16 equation right side '≦'.

<a name="ch01c062">
Key point for page 229 line 19 equation 
is that (n-1)/2 drop all fraction part.
（compare with [four drop, five enter]
 "count 5 and higher fractions as one,
  count 4 and lower  fractions as zero"）.

If (n-1)/2=3.0 get 3
if (n-1)/2=3.5 get 3， drop 0.5
In the expression 2*└(n-1)/2┘ 
if n=odd, for example n=7
　　　　　2*└(7-1)/2┘ = 2*3 = 6 < 7=n
if n=even, for example n=6
　　　　　2*└(6-1)/2┘ = 2*2 = 4 < 6=n
<a name="ch01c063">
This procedure make sure　2*└(n-1)/2┘ ＜ n
Add 2 at both side,
make sure　2+2*└(n-1)/2┘ ＜ n+2
Multiply sequence length to both side
make sure　page 229 line 19 equation
to be true.

Up to here done exercise 1.9 .
2009-08-10-11-05 stop

Exercise 1.9 is not hard, but not easy
to understand. Matrix decomposition and
drop all fraction part, if not find out
with my hand, I may put a question mark
for long time.
2009-08-10-11-12 stop

<a name="9808131533">
2009-08-13-15-33 start
During proofread, get a question. Problem 
start point is page 15 line 8 equation.
Expand left hand side, get page 229 line 13
equation. Here we find cross terms ajak 
and j NOT= k. Finally we reached page 15
line 8 equation. Here no cross terms. 
During deriving process, where change
cross terms ajak to non cross term ajaj ?
Answer is at 
"from basic inequality 2ajak≦aj2+ak2"
This step change all cross terms to 
non cross terms.

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.
2009-08-13-16-04 stop



<a name="ch01c064">　Index begin　Index this file
2009-08-10-11-55 start

■　Exercise 1.10 problem statement
　　textbook page 15
（Schur's Lemma -- The R and C Bound）
For each two dimension array 
 (cjk: 1≦j≦m, 1≦k≦n) and
for a pair of one dimension array
 (xj: 1≦j≦m) and (yk: 1≦k≦n) 
ALERT: sequence xj and yk
 size different ! m≠n
We have the following inequality.

<a name="ch01c066">
2009-08-10-12-12 here
Where R is maximum row sum and C is
maximum column sum. Define as below

<a name="ch01c067">
2009-08-10-12-22 here
Textbook equation 1.23 is called
Schur's Lemma. This equation is possibly
Schur's second most famous equation.
If row number equal to column number
n=m and if all diagonal elements are
one and all off-diagonal elements are
zero. Then Schur's Lemma become Cauchy
Inequality.

（LiuHH note: consider rectangular matrix
and square matrix, Schur's Lemma is 
general and Cauchy Inequality is special
case.）
2009-08-10-12-32 stop


<a name="ch01c068">　Index begin　Index this file
2009-08-10-14-26 start
■　Exercise 1.10 hint
　　textbook page 229
Re-write sequence cjkxjyk as
 [|cjk1/2|*|xj|]*[|cjk1/2|*|yk|]
then apply Cauchy Inequality twice, get

2009-08-10-14-31 here
Two terms inside of '{' and '}' replace
with R and C respectively.








<a name="ch01c070">　Index begin　Index this file
2009-08-10-14-34 here
■　Exercise 1.10 discussion

Whenever see cjkxjyk it is better to use
matrix view point to solve problem. 
Exercise 1.10 state clearly this is 
rectangle matrix, not square matrix.
Following use 3x5 matrix as an example
and explain page 15 line 18 equation.

<a name="ch01c072">
2009-08-10-15-11 here
Matrix elements have two index, for
example cjk 
j is for cjk's row location, above 1≦j≦3
k is for cjk's col location, above 1≦k≦5
Now use eqn.AD11 to explain 
"page 15 line 18 R equation"

<a name="ch01c073">
2009-08-10-15-22 here
Page 15 line 18 R equation find the
maximum value for j=1, j=2, j=3 
When j=1 ，k sum from k=1 to k=5， that is
when j=1 sum c11+c12+c13+c14+c15
　　　　　= 6.6+1.1+1.2+0.5+1.6 = 11
when j=2 sum c21+c22+c23+c24+c25
　　　　　= 1.1+7.8+0.4+2.1+1.9 = 13.3
when j=3 sum c31+c32+c33+c34+c35
　　　　　= 1.2+0.4+8.5+2.3+1.7 = 14.1
Now three summation values are in hand
maximum row sum R is maximum in the set
 [11, 13.3, 14.1] we get R=14.1

<a name="ch01c074">
Above find maximum row sum R 
Below find maximum col sum C 

Following use eqn.AD11 to explain 
"page 15 line 18 C equation"

<a name="ch01c075">
2009-08-10-15-37 here
Please click review example eqn.AD11.
page 15 line 18 C equation require
find maximum column sum for k=1, k=2,
 k=3, k=4, k=5 cases.
When k=1 , j sum from j=1 to j=3, that is
When k=1 sum c11+c21+c31
　　　　　= 6.6+1.1+1.2 = 8.9
When k=2 sum c12+c22+c32
　　　　　= 1.1+7.8+0.4 = 9.3
When k=3 sum c13+c23+c33
　　　　　= 1.2+0.4+8.5 = 10.1
When k=4 sum c14+c24+c34
　　　　　= 0.5+2.1+2.3 = 4.9
When k=5 sum c15+c25+c35
　　　　　= 1.6+1.9+1.7 = 5.2

<a name="ch01c076">
Now five column sums are in hand
max. column sum C is the maximum value
in the set [8.9,9.3,10.1,4.9,5.2] 
we get C=10.1

For this example matrix eqn.AD11 
max.   row  sum  R=14.1
max. column sum  C=10.1
2009-08-10-15-50 here

<a name="ch01c077">　Index begin　Index this file
■　Exercise 1.10 solution

Exercise 1.10 hint already solved the problem.
Please refer to discussion for help.
2009-08-10-15-55 stop



<a name="ch01c078">　Index begin　Index this file
2009-08-10-17-50 start

■　Exercise 1.11 problem statement
　　textbook page 15
（Schwarz's Argument in an Inner Product Space）
Let （V，〈.,.〉） represent inner product space.
Let v, w are elements of （V，〈.,.〉）
Refer to next quadratic equation in t∈R
 p(t) = 〈v+tw, v+tw〉 -----eqn.AD12
Pay attention to above equation, it has non
negative value. Use your knowledge about
quadratic equation and prove page 8 eqn.1.16
＜v,w＞ ≦ ＜v,v＞0.5＜w,w＞0.5 ---page 8 eqn.1.16
which is Cauchy Inequality. In your process, 
determine under what condition inequality
become equality. Then Schwarz's argument 
（page 11） can be applied to prove Cauchy
Inequality.
2009-08-10-18-10 here


<a name="ch01c079">　Index begin　Index this file
■　Exercise 1.11 hint
　　textbook page 230
Schwarz's argument (page 11) just do a 
little modification, then it apply to
generalized Cauchy Inequality. But there
is visual difference. First, we use problem
given condition and quadratic equation p(t)
to get
 0 ≦ p(t) = 〈v,v〉 + 2t〈v,w〉 + t*t〈w,w〉
 p(t) discriminant is D=B*B-AC
 = 〈v,w〉2-〈v,v〉*〈w,w〉
we require D ≦ 0， otherwise p(t) would 
get two real roots, then some section in 
the domain may cause p(t) become negative
which is not allowed.
2009-08-10-18-26 here

<a name="ch01c080">　Index begin　Index this file
■　Exercise 1.11 solution

Please refer to
quadratic discriminant>0 what trouble?
Theorem: Every inner product satisfies the
 Cauchy-Schwarz inequality.
Exercise 1.7 generalized Cauchy Inequality
Similar problem repeat many times. 
Freeman skip this Exercise 1.11
2009-08-10-18-34 stop



<a name="ch01c081">　Index begin　Index this file
2009-08-11-18-50 start

■　Exercise 1.12 problem statement
　　textbook page 16
（Example of a Self-generalization）
Let V be vector space. Let 〈.,.〉 be inner
product for V .
Assume x1, x2, ..... xn and y1, y2, ..... yn
are elements of V. Please prove that we can
find vector version of Cauchy Inequality
as below.

<a name="ch01c084">
2009-08-11-19-18 here
Please pay attention to that if we let
n=1 then equation 1.24 become inner space
Cauchy-Schwarz Inequality. On the other 
hand, if we let n be any positive integer,
but require that vector space V be real 
number space, and require inner product 
to be 〈x, y〉 = xy 
then equation 1.24 become regular Cauchy
Inequality.
2009-08-11-19-25 here


<a name="ch01c085">　Index begin　Index this file
2009-08-11-19-50 start
■　Exercise 1.12 hint
　　textbook page 230
Define a new inner product space (V[n],[.,.])
and require
V[n]={v1, v2, ..... vn : vj∈V, 1≦j≦n }
define [v,w]=∑[j=1,n]〈vj, wj〉， where
v = (v1, v2, ..... vn) and
w = (w1, w2, ..... wn) 
after confirm [v,w] is an effective 
inner product. We can see equation 1.24 
is Cauchy-Schwarz Inequality in inner 
product space [.,.] 
2009-08-11-20-01 stop

2009-08-11-20-28 start
<a name="ch01c086">　Index begin　Index this file
■　Exercise 1.12 discussion

Please compare Cauchy Inequality vector
version and non vector version.

Sequence element of non vector version is
a number,
Sequence element of     vector version is
a vector.
Sequence of a vector version is in fact
a matrix. 
A 5x5 matrix is a 'sequence'.
Cauchy Inequality always ask two sequences.
Following use two 5x5 matrix to explain.

<a name="ch01c089">
2009-08-11-20-38 here
Vector version of Cauchy Inequality eqn.1.24
left hand side is ∑[j=1,n]〈xj, yj〉
dot product of two different vector xj and yj
When j=1 above seq.A and seq.B calculation
is next
seq.A [1.1, 2.1, 3.1, 4.1, 5.1]T and
seq.B [1.6, 2.6, 3.6, 4.6, 5.6]T dot product
Here use row vector and transpose sign 'T'
for a column vector.
When j=1, two vector dot result is a number
1.1*1.6+2.1*2.6+3.1*3.6+4.1*4.6+5.1*5.6=65.8

<a name="ch01c090">
Vector version of Cauchy Inequality eqn.1.24
right hand side is 
√{∑[j=1,n]〈xj, xj〉}*√{∑[j=1,n]〈yj, yj〉}
dot product of two same vector xj and xj
yj is the same as xj.

When j=1 above seq.A calculation is next
seq.A [1.1, 2.1, 3.1, 4.1, 5.1]T and
seq.A [1.1, 2.1, 3.1, 4.1, 5.1]T dot product
dot result is a number
seq.B [1.6, 2.6, 3.6, 4.6, 5.6]T and
seq.B [1.6, 2.6, 3.6, 4.6, 5.6]T dot product
dot result is a number
Multiply two numbers and take square root
get √(58.05*74.8) = 65.895
This is the value of equation 1.24 right
hand side when j=1.
Work for j=1 to j=n finish Vector version
of Cauchy Inequality.

<a name="ch01c091">

Please pay attention to the following
1. this file use 'dot product' and 
   'inner product' equivalently.
2. problem indicate that matrix can be
   square, rectangular or one line matrix.
   A vector version 'sequence' has m column
   vectors. 
   when define [v,w]=∑[j=1,n]〈vj, wj〉 
   Each column vector can have n elements
   (or other number of elements)
　　when n=1 （m＞1） equation 1.24 become
   inner product Cauchy-Schwarz Inequality.

<a name="ch01c092"> (9808221410)
2009-08-22-14-10 deleted
2009-08-11-21-17 here


<a name="ch01c093">　Index begin　Index this file
■　Exercise 1.12 solution

Please refer to hint and discussion.
2009-08-11-21-26 stop

2009-08-11-23-11 start
Exercise 1.12 has a little bit confusion.
LiuHH distinguish m and n.

 m is number of column vectors in a matrix.
 n is number of elements in a vector.
 m not need be equal to n 
Problem statement and hint both not 
distinguish m and n
Problem say "let n=1 and then ....."
is it m and n become one at same time?
A little bit confuse.
2009-08-11-23-18 stop



<a name="ch01c094">　Index begin　Index this file

2009-08-12-08-23 start
■　Exercise 1.13 problem statement
　　textbook page 16
（Application of Cauchy's Inequality to an Array）
If {ajk: 1≦j≦m, 1≦k≦n} is real number
array, please prove

<a name="ch01c096">
2009-08-12-08-46 here
Please explain that equality condition
is that if and only if there exist a
 αj and a βk， with 1≦j≦m, 1≦k≦n 
we can always find ajk = αj + βk
2009-08-12-08-52 here


<a name="ch01c097">　Index begin　Index this file
2009-08-12-08-56 here
■　Exercise 1.13  hint
　　textbook page 230

◎　2009-08-12-12-28 start
◎　Problem give ajk ，hint start from xjk？
◎　What is the relation between them？
◎　Please see page 230 line 18 equation 
◎　and exercise 1.13 discussion, constant
◎　result. (it is not a constant matrix)
◎　2009-08-12-12-35 stop

<a name="ch01c098">
Although xjk is two index matrix, 
if we treat { xjk : 1≦j≦m, 1≦k≦n}
as vector with length m*n , then we can
apply Cauchy Inequality to xjk and use
 one trick. xjk = xjk*1. We get the
following inequality.

<a name="ch01c099">
2009-08-12-09-15 here
Consider the following relation.

<a name="ch01c101">
2009-08-12-09-36 here
then eqn.14.46 left hand side upper
bound is T2， Right hand side can be
written as

<a name="ch01c102">
2009-08-12-09-59 here
Cauchy Inequality upper bound eqn.14.46 
change to problem asked inequality.

Under what condition inequality become
equality? eqn.14.46 applied Cauchy's
Inequality. Equality condition for
Cauchy Inequality is that two array 
are in proportion. Exercise 1.13 use
one trick to solve problem. One array
used is constant [1, 1, 1, ..... 1]
then the other array xjk must be
constant array if equality is required.
<a name="ch01c103">
Let xjk ＝constant＝c where 1≦j≦m, 1≦k≦n
We can use αj = c+rj and βk=ck for ajk .
Above procedure is Theorem one in van Dam
(1998) his method use matrix theory and
instructive corollaries.
2009-08-12-10-18 stop

<a name="ch01c104">　Index begin　Index this file
2009-08-12-11-33 start
■　Exercise 1.13 discussion

First find out the meaning of page 230
line 18 equation.
 ajk is problem original matrix
 xjk is page 230 line 18 transformed matrix
Following use 3x5 numerical matrix to 
observe detail.

<a name="ch01c105">
row sum rj=∑[k=1,n]ajk, column sum ck=∑[j=1,m]ajk
[ 　1.1 　1.2 　1.3 　1.4 　1.5 　] r1= 6.5 ---matrix C
[ 　2.1 　2.2 　2.3 　2.4 　2.5 　] r2=11.5 ---eqn.AD15
[ 　3.1 　3.2 　3.3 　3.4 　3.5 　] r3=16.5
ck＝6.3 　6.6 　6.9 　7.2 　7.5　k=1 to 5, red text not matrix
◎　T=∑rj=6.5+11.5+16.5=34.5
◎　T=∑ck=6.3+6.6+6.9+7.2+7.5=34.5
◎　This equal step length matrix get
◎　constant result, special case.
◎　Non-equal step length matrix not get
◎　constant result. 9808121928

Matrix elements are {ajk : 1≦j≦m, 1≦k≦n}
here m=3, n=5 three row five column matrix.
Based on page 230 line 19 left equation
calculate as following.

<a name="ch01c106"> (9808130950)
◎　2009-08-13-09-50 Deleted this calculation,
◎　Because eqn.AD15 is equal step length matrix
◎　output are all the same -2.3 ，
◎　Suspect and write computer program
◎　If change
  aa=[[1.1,2.2,-4.3,-2.4,1.8],[-2.1,6.2,-0.3,5.4,-2.9],[-3.6,2.4,5.3,4.4,-1.5]];
◎　to
//aa=[[1.1,2.2,-4.3,-2.4,1.8],[-2.1,6.2,-0.3,5.4,-2.9],[-3.6,2.4,5.3,4.4,-1.5]];
◎　run program, result is the same as deleted
◎　result. 2009-08-13-09-56 stop
<a name="ch01c107"> del
<a name="ch01c108"> del
<a name="ch01c109"> del
2009-08-12-12-17 here
Here use numerical matrix find that 
page 230 line 18 equation
get constant matrix.

<a name="ch01c110">
2009-08-12-12-43
ajk is original arbitrary matrix
xjk is constant matrix
Why apply Cauchy Inequality to constant matrix?
Why not apply Cauchy Inequality to original
matrix? Still in thinking.
2009-08-12-12-46

[[
[ 　1.1 　1.2 　1.3 　1.4 　1.5 　]  ---matrix C
[ 　2.1 　2.2 　2.3 　2.4 　2.5 　]  ---eqn.AD15
[ 　3.1 　3.2 　3.3 　3.4 　3.5 　]
]]
<a name="ch01c111">
//begin, begin, begin, 2009-08-12-14-59
//Next line start calculation code.
var mm=3,nn=5;   //matrix C dimension is 3x5
var rr=[0,0,0];  //matrix C has three rows
var cc=[0,0,0,0,0];  //matrix C has five columns
var aa=[[1.1,1.2,1.3,1.4,1.5],[2.1,2.2,2.3,2.4,2.5],[3.1,3.2,3.3,3.4,3.5]];
var xx=[[1.1,1.2,1.3,1.4,1.5],[2.1,2.2,2.3,2.4,2.5],[3.1,3.2,3.3,3.4,3.5]];
//above aa is special case,
//next  aa is arbitrary matrix. 9808121532
  aa=[[1.1,2.2,-4.3,-2.4,1.8],[-2.1,6.2,-0.3,5.4,-2.9],[-3.6,2.4,5.3,4.4,-1.5]];
var i,j,k;
for(j=0;j < mm;j++){
rr[j]=0;  //find row sum
for(k=0;k < nn;k++)rr[j]+=aa[j][k];}
for(k=0;k < nn;k++){
cc[k]=0;  //find column sum
for(j=0;j < mm;j++)cc[k]+=aa[j][k];}
for(j=0;j < mm;j++){ //page 230 line 18 equation
for(k=0;k < nn;k++)xx[j][k]=aa[j][k]-rr[j]/nn-cc[k]/mm;}
for(j=0;j < mm;j++){ //page 230 line 18 equation x_jk value
xx[j]; // this line no '=', print variable value
i=0;}  // complex2.htm special rule, '}' need a dummy '='
rr // this line no '=', print variable value
cc // print this way is complex2.htm added function
//done,done,done 2009-08-12-15-13 above line done.
<a name="ch01c112">
Paste above command to complex2.htm#box03
http://freeman2.com/complex2.htm#box03
box 3, click "test box3 command, output to box4"
output to box4.

<a name="ch01c113">
2009-08-12-15-22 run program, learned that the 
following "constant matrix" is special case.
[[
2009-08-12-12-17 here
Here use numerical matrix find that 
page 230 line 18 equation
get constant matrix.
]]
change aa value, result is not constant matrix.
2009-08-12-15-23  （9808121523）

<a name="ch01c114">　Index begin　Index this file
2009-08-12-16-21 start
■　Exercise 1.13 solution

Hope to solve several problems
Why use page 230 line 18 equation?
Substitute page 230 line 18 equation into
eqn.14.46 right hand, then how to get
page 230 line 22 equation?
Exercise 1.13 is still thinking. 
No solution at this moment.
2009-08-12-16-26 stop

<a name="ch01c115">
2009-08-12-17-20 start

<a name="ch01c116">
2009-08-12-17-35 here
expand eqn.AD16 square term

2009-08-12-17-50 here
How to convert eqn.AD17 to page 230 line 22
equation?
2009-08-12-17-58 thinking

<a name="ch01c117">
2009-08-13-16-54
Following is numerical check eqn.14.46 right
hand side and page 230 line 22. Get equal
value. Confirmed equality. 9808131741
//begin, begin, begin, 9808131655
//Next line start calculation code.
var mm=3,nn=5;   //matrix C dimension is 3x5
var rr=[0,0,0];  //matrix C has three rows
var cc=[0,0,0,0,0];  //matrix C has five columns
var aa=[[1.1,1.2,1.3,1.4,1.5],[2.1,2.2,2.3,2.4,2.5],[3.1,3.2,3.3,3.4,3.5]];
var xx=[[1.1,1.2,1.3,1.4,1.5],[2.1,2.2,2.3,2.4,2.5],[3.1,3.2,3.3,3.4,3.5]];
//above aa is special case,
//next  aa is arbitrary matrix. 9808121532
  aa=[[1.1,2.2,-4.3,-2.4,1.8],[-2.1,6.2,-0.3,5.4,-2.9],[-3.6,2.4,5.3,4.4,-1.5]];
var i,j,k;
for(j=0;j < mm;j++){
rr[j]=0;  //find row sum
for(k=0;k < nn;k++)rr[j]+=aa[j][k];}
for(k=0;k < nn;k++){
cc[k]=0;  //find column sum
for(j=0;j < mm;j++)cc[k]+=aa[j][k];}
for(j=0;j < mm;j++){ //page 230 line 18 equation
for(k=0;k < nn;k++)xx[j][k]=aa[j][k]-rr[j]/nn-cc[k]/mm;}
//
//9808131655 evaluate eqn.14.46 right hand side
//if same value as page 230 line 22 equation
var mnx=0; //eqn.14.46 that is page 230
for(j=0;j < mm;j++){ // line 17 right hand side
for(k=0;k < nn;k++)mnx+=xx[j][k]*xx[j][k];}
mnx=mnx*mm*nn; //done eqn.14.46 right hand side
//that is page 230 line 17 right hand side
//
var tt=0; //page 230 line 19 right T=∑∑a_jk
for(j=0;j < mm;j++){ //evaluate T=∑∑a_jk 
for(k=0;k < nn;k++)tt+=aa[j][k];}
var mnt=0; //page 230 line 22 value
for(j=0;j < mm;j++){
for(k=0;k < nn;k++)mnt+=aa[j][k]*aa[j][k];}
mnt=mnt*mm*nn; //page 230 line 22 left end
var mr=0; //page 230 line 22 -m*∑r*r
for(j=0;j < mm;j++){mr+=rr[j]*rr[j];}
var nc=0; //page 230 line 22 -n*∑c*c
for(k=0;k < nn;k++){nc+=cc[k]*cc[k];}
//mnt=mnt-mr-nc+2*tt*tt; //done page 230 line 22
mnt=mnt-mm*mr-nn*nc+2*tt*tt; //9808131735 correction
mnx //print page 230 line 17 right side value
mnt //print page 230 line 22 value
mnx-mnt //difference should be zero 9808131717stop
//done,done,done 9808131717 above line done.
//check, check, check ∑∑xx=-∑∑aa .9808131955
var uu=0; //uu=∑∑xx_jk
for(j=0;j < mm;j++){ //9808131958
for(k=0;k < nn;k++)uu+=xx[j][k];}
uu+tt  //should be zero 9808131959

<a name="ch01c118">
2009-08-13-17-20 answer
mnx //print page 230 line 17 right side value
1738.0300000000002
mnt //print page 230 line 22 value
1738.0300000000002
mnx-mnt //difference should be zero 9808131717stop
0
2009-08-13-17-37

2009-08-13-20-00 get answer
uu+tt  //should be zero 9808131959
1.7763568394002505e-15

<a name="ch01c119">　Index begin　Index this file
2009-08-13-17-52
page 230 line 17 right hand side value
 = 
page 230 line 22 value
Math expression is complicate, after numerical
check, make sure equality exist. Then think
carefully. Conclusion: EQUAL.

Put eqn.AD17 back to the envelope of
 eqn.AD16, that is mn∑∑[...]，
mn∑∑[ ajk2 ] is common term to both
page 230 line 17 right hand side and 
page 230 line 22 . This is very clear.

<a name="ch01c120">
Put two rj terms in eqn.AD17 to mn∑∑[...]
result is

<a name="ch01c125">
2009-08-13-18-49 here
Above put two rj terms in eqn.AD17
back to mn∑∑[...] 
put two ck terms in eqn.AD17 back to 
mn∑∑[...] use same reason. Omit.

Last work is to put right most term in
eqn.AD17 back to mn∑∑[...] get the
following.

<a name="ch01c128">
2009-08-13-19-06 here
eqn.AD17 aux.7 ＝ 2 * T * T

After above analysis, get conclusion
eqn.14.46 right hand side and page 230
line 22 equation are indeed equal.
2009-08-13-19-10

<a name="ch01c129">　Index begin　Index this file
Two ∑ in eqn.AD17 aux.7 are both T , please
see definition of T
This definition is to add all elements in a
matrix. For numerical example, add matrix C
all elements together, this value is T.

 ∑[j=1,m]rj in eqn.AD17 aux.7
r1 is row one sum value
r2 is row two sum value
r3 is row three sum value
∑[j=1,m]rj add above three values,
that is total sum T !
Reason for ∑[k=1,n]ck is the same. 
No need to repeat.
2009-08-13-19-35

<a name="ch01c130">
2009-08-13-20-02 start
Hint said
[[
then eqn.14.46 left hand side upper
bound is T2， 
]]
Below explain that eqn.14.46 left hand
side value is equal to T2 . If not take 
square then ∑∑xjk = -∑∑ajk = -T
page 230 line 18 link ∑∑xjk and ∑∑ajk .
row sum rj=∑[k=1,n]ajk -----eqn.ad19
col sum ck=∑[j=1,m]ajk -----eqn.ad20
<a name="ch01c131">
page 230 line 18 equation is
xjk = ajk-rj/n-ck/m -----eqn.ad21
where m is total row number， 1≦j≦m,
   n is total column number， 1≦k≦n.
Sum in both row and column for eqn.ad21
get
∑∑xjk = ∑∑ajk-∑∑rj/n-∑∑ck/m -----eqn.ad22

<a name="ch01c132">
∑∑ajk is page 230 line 19 right side
defined T.　What is T？
-∑∑rj/n sum for j and sum for k get
-∑∑rj/n = -∑[j=1,m]rj∑[k=1,n]/n 
-∑∑rj/n = -T*(n/n) = -T -----eqn.ad23
About ∑[j=1,m]rj=T please see﹕what is T？
∑[k=1,n]/n = n/n = 1 this is easiest part.
add five  times for 1/5 get one;
add eight times for 1/8 get one.

Similarly -∑∑ck/m = -T -----eqn.ad24
Put eqn.ad23 and eqn.ad24 to eqn.ad22 get
∑∑xjk = T-T-T = -T -----eqn.ad25
therefore [∑∑xjk]^2 = T*T -----eqn.ad26

<a name="ch01c133">
Substitute eqn.ad26 into eqn.14.46 left
hand side.
Substitute page 230 line 22 equation into
eqn.14.46 right hand side.
T2 in eqn.ad26 and 2*T2 in eqn.AD17 aux.7 
cancel one T2. When finish this procedure,
we get problem required equation， It is 
the same as the following brief version.

<a name="ch01c134">
2009-08-13-20-52 here
Almost give up. Now finally done.
2009-08-13-20-53 stop

<a name="ch01c135">
2009-08-13-22-25 start
Problem ask to explain the equality 
condition is if and only if there exist 
a αj and a βk， with 1≦j≦m, 1≦k≦n 
we can always find ajk = αj + βk
In the analysis, only one place use Cauchy
Inequality. Condition for Cauchy Inequality
to become equality is that two sequences
are in proportion. Exercise 1.13 used one
trick. One sequence is [1, 1, 1, ..... 1] 
the other sequence must be all elements
equal to a constant （L in below）.

<a name="ch01c136">
 xjk entered Cauchy's Inequality, ALERT !!
 ajk asked for equality analysis. ALERT !!
 xjk and ajk are related at page 230 line 18
It is the same as next brief equation
xjk = ajk-rj/n-ck/m -----eqn.ad21
Let all xjk equal to constant L （one trick）
ajk-rj/n-ck/m = L
ajk = L +rj/n+ck/m -----eqn.ad27
<a name="ch01c137">
Isn't problem ask for αj and βk ？
require αj = L +rj/n -----eqn.ad28
require βk = ck/m  -----eqn.ad29
Solve problem !
2009-08-13-22-51 stop

<a name=20091217>
2009-12-17-11-09 start
Update 2009-12-17 change all tute*.htm
(from tute0007.htm to tute0023.htm)
first:
Correct 'Limit' link from '#docA06'
to '#docA006'
second:
Change Javascript index to read from
jslist1e.js so that update jslist1e.js
then update ALL tute*.htm.
2009-12-17-11-23 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