<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">,<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="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="docA06">
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="docA07">
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 here

<a name="ch01d001">　Index begin　Index this file
2009-08-23-12-09 start

■　Exercise 1.14 problem statement
　　textbook page 16
（A Cauchy Triple and Loomis-Whitney）
Here is a generalization of Cauchy's Inequality
that has as a corollary a discrete version of 
the Loomis-Whitney inequality. a result which
in the continuous case provides a bound on the
volume of a set in terms of the volumes of the 
projections of that set onto lower dimensional
subspaces. The discrete Loomis-Whitney inequality
(1.26) was only recently developed, and it has
applications in information theory and the theory
of algorithms.

<a name="ch01d002">
(a) Show that for any nonnegative aij,bjk,cki
    with 1≦i,j,k≦n one has the triple product
    inequality

<a name="ch01d003">
2009-08-23-12-36 here
(b) Let A denote a finite set of points in Z3
    and let Ax, Ay, Az denote the projections
    of A onto the corresponding coordinate
    planes that are orthogonal to the x, y,
    or z-axes. Let |B| denote the cardinality 
    of a set B⊂Z3 and show that the
    projections provide an upper bound on the
    cardinality of A:
      |A|≦|Ax|½ |Ay|½ |Az|½  -----eqn.1.26
2009-08-23-12-46 here

<a name="ch01d004">　Index begin　Index this file
2009-08-23-13-05 start
■　Exercise 1.14 hint
　　textbook page 230 and 231
More often than one might like to admit,
tidiness is important in problem solving, 
and here the hygienic use of parentheses 
can make the difference between success
and failure. One just carefully computes.

<a name="ch01d011">
2009-08-23-14-37 here
This proof of the triple product bound (1.25)
follows Tiskin (2002). Incidentally, the 
corollary (1.26) was posed as a problem
on the 33rd International Mathematical
Olympiad (Moscow, 1992). More recently,
Hammer and Shen (2002) note that the
corollary may be obtained as an application
of Kolmogorov complexity. George (1984, p.243)
outline a proof of continuous Loomis-Whitney
inequality, a result which can be used to 
give a third proof of the discrete bound (1.26)
2009-08-23-14-44 here





<a name="ch01d012">　Index begin　Index this file
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="ch01d013">
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 here

2009-08-23-16-07 start
The following repeat hint work and fill in
detail steps.

<a name="ch01d015">
2009-08-23-16-14 here
Above equation left hand side is start point.
∑[i,j,k=1 to i,j,k=n] is in fact short write
for three sums
∑[i=1 to i=n] ∑[j=1 to j=n] ∑[k=1 to k=n] 
page 231 line 3 equation re-group and let
j stand out, j is different from i,k.
Left hand side is symmetry to a,b,c and i,j,k.
It is possible to re-group and let i or k
stand out. Textbook hint choose j.

page 231 line 3A equation is hint equation
page 231 line 3B equation is same as 3A 
But 3B use both [...] and {...} to indicate
∑[i,k=1 to i,k=n] {...} cover range clearly.

<a name="ch01d016">
[...] in page 231 line 3B equation is a dot 
product of two sequences
seq.1 √(ai1),√(ai2),√(ai3) ... √(ain)
seq.2 √(b1k),√(b2k),√(b3k) ... √(bnk)
Next apply Cauchy's Inequality to seq.1 and
seq.2 get the next equation.

<a name="ch01d018">
2009-08-23-16-47 here
Change from line 3B to line 4B is a result
of applying Cauchy's Inequality. We do not
just apply Cauchy Inequality once. If n=5
we apply Cauchy Inequality 5*5=25 times !
Because application take place inside of
∑[i,k=1 to i,k=n], i sum 1 to 5 and
 k sum 1 to 5. Total sum 25 times.
line 3B equation aij½ and bjk½ both have
square root. square root disappeared in 
line 4B.  aij and bjk are free from '½'
<a name="CauchySquareRoot">
Cauchy put '½' out side of ∑[j=1 to j=n]
Please see Cauchy Inequality below.

Right hand side ak squared, summed
then whole sum take square root. That is
"put '½' out side of ∑[j=1 to j=n]".

<a name="ch01d019">
Next step, re-group again.
Our final goal is eqn.1.25
In line 4B, both {∑[j=1 to j=n]aij}½
and {∑[j=1 to j=n]bjk}½ are close to
final goal. Hint choose {∑[j=1 to j=n]bjk}½
to stand out and we get next re-group 
line 5B equation.

<a name="ch01d021">
2009-08-23-17-17 here
In line 5B equation, we will apply Cauchy
Inequality again. Before that, identify two
sequences as following.
seq.3 √(ck1),√(ck2),√(ck3) ..... √(ckn)
seq.4 √(∑[j]a1j),√(∑[j]a2j) ..... √(∑[j]anj)
line 5B ∑[i=1,n][.....] is simply seq.3 dot
product with seq.4.
Now seq.3 and seq.4 are in hand, how can we
not invite Cauchy again?

<a name="ch01d023">
2009-08-23-17-51 here
Here explain 
seq.4 √(∑[j]a1j),√(∑[j]a2j) ..... √(∑[j]anj)
operation.

Page 231 line 5B, right end sum over j only
(∑[j=1,j=n]aij)^½

Page 231 line 7B, right end summation sum
over i and j.
(∑[i,j=1;i,j=n]aij)^½

<a name="ch01d024">
Both take square root after summation
page 231 line 7B is Cauchy Inequality greater
than side. It is seq.4 dot product with seq.4
This square operation removed square root from
line 5B right end summation.
seq.4 dot product with seq.4 sum over i
therefore page 231 line 7B right end 
summation sum over i and j.
Cauchy Inequality take square root after
sum over i. This action get page 231 
line 7B right end
(∑[i,j=1;i,j=n]aij)^½
<a name="ch01d025">
page 231 line 7B has a happy news.
Right end summation
(∑[i,j=1;i,j=n]aij)^½
not mix with k term any more. Right end 
summation can be moved out of ∑[k] which
is next re-qroup step.

<a name="ch01d027">
2009-08-23-18-20 here
Now identify two sequences as following
seq.5 √(∑[j]bj1),√(∑[j]bj2) ..... √(∑[j]bjn)
seq.6 √(∑[i]c1i),√(∑[i]c2i) ..... √(∑[i]cni)
seq.5 dot product with seq.6 is exact
 ∑[k]{...} in page 231 line 8B equation
Apply Cauchy Inequality to seq.5 and seq.6
get next inequality greater than side.

<a name="ch01d029">
2009-08-23-18-32 here
It is magic ! 
Up to here aij, bjk, cki are separated!!
2009-08-23-18-36 what is going on ??!!
accident created curve text !! 
2009-08-23-18-38 stop

<a name="ch01d030">
2009-08-23-21-16 start
Extra work.
Exercise 1.14 did not ask 
What condition the equality in eqn.1.25
come true?
Equality condition is that 
Two sequences in Cauchy's Inequality are
proportional.
When solve exercise 1.14 we applied Cauchy 
Inequality three times, then it must be
each time, two sequences are proportional.

<a name="ch01d031">
Let us see first pair sequences
seq.1 √(ai1),√(ai2),√(ai3) ... √(ain)
seq.2 √(b1k),√(b2k),√(b3k) ... √(bnk)
If require seq.1 and seq.2 are proportional,
Matrix aij and matrix bjk should be
relate to each other in some way.

Second pair sequences
seq.3 √(ck1),√(ck2),√(ck3) ..... √(ckn)
seq.4 √(∑[j]a1j),√(∑[j]a2j) ..... √(∑[j]anj)
If require seq.3 and seq.4 are proportional,
Elements of matrix cki should be 
Matrix aij row sum.

<a name="ch01d032">
Third pair sequences
seq.5 √(∑[j]bj1),√(∑[j]bj2) ..... √(∑[j]bjn)
seq.6 √(∑[i]c1i),√(∑[i]c2i) ..... √(∑[i]cni)
If require seq.5 and seq.6 are proportional,
Matrix bjk column sum and matrix cki row sum 
should have a kind of transpose relation..

But the process is not unique.
page 231 line 3B left hand side is symmetric to
a,b,c. We can re-group let either one stand out.
From line 4B to line 5B, we can let aij 
meet Cauchy, we can also let bjk to meet
Cauchy.

<a name="ch01d033">
The equality condition in eqn.1.25 is 
complicate. eqn.1.25 use '≦' not use '＜'

LiuHH guess that 
The equality condition in eqn.1.25 is that
Matrix a,b,c are all rank n=1 (just a number)
 OR if rank n>1
Either one of a,b,c matrix is all zero
matrix.
2009-08-23-21-55 stop



<a name="ch01d034">　Index begin　Index this file
2009-08-24-11-31 start

■　Exercise 1.15 problem statement
　　textbook page 17
（An Application to Statistical Theory）
<a name="ch01d035">
If p(k;θ)≧0 for all k∈D and θ∈Θ and if
  ∑[k∈D]{p(k;θ)}=1 for all θ∈Θ  -----eqn.1.27
then for each θ∈Θ one can think of 
  Mθ={p(k;θ);k∈D}  ---eqn.AE01
as specifying a probability model where
p(k;θ) represents the probability that we
"observe k" when the parameter θ is the
true "state of nature". If the function
g:D→R satisfies
  ∑[k∈D]{g(k)p(k;θ)}=θ for all θ∈Θ ---eqn.1.28
then g is called an unbiased estimator 
of the parameter θ. Assuming that D is 
finite and p(k;θ) is a differentiable 
function of θ.
<a name="ch01d036">
Show that one has the lower bound
  ∑[k∈D]{[g(k)-θ]^2*p(k;θ)}≧1/I(θ) ---eqn.1.29

where I:Θ→R is defined by the sum
  I(θ)=∑[k∈D]{[pθ(k;θ)/p(k;θ)]^2*p(k;θ)} ---eqn.1.30

where pθ(k;θ)=∂p(k;θ)/∂θ. The quantity
defined by the left side of the bound
(1.29) is called the variance of the 
unbiased estimator g, and the quantity
I(θ) is known as the Fisher information
at θ of the model Mθ
The inequality (1.29) is known as the 
Cramer-Rao lower bound. and it has 
extensive application in mathematical
statistics.
2009-08-24-12-01 stop



<a name="ch01d038">　Index begin　Index this file
2009-08-24-12-20 start
■　Exercise 1.15  hint
　　textbook page 231
If we differentiate the identity (1.27)
and (1.28) we find for all θ∈Θ that
  ∑[k∈D]{pθ(k;θ)}=0  -----eqn.AE02
and
  ∑[k∈D]{g(k)*pθ(k;θ)}=1  -----eqn.AE03
<a name="ch01d039">
Consequently, we have the identity
 1=1-θ*0
  =∑[k∈D]{[g(k)-θ]*pθ(k;θ)}

<a name="ch01d041">
2009-08-24-12-54 here
which yields the Cramer-Rao inequality (1.29)
when we apply Cauchy's inequality to this
sum of brackted terms.

  The derivation of the Cramer-Rao inequality
may be the most significant application of
the 1-trick in all of applied mathematics.
It has been repeated in hundreds of papers
and books.
2009-08-24-12-57 stop







<a name="ch01d042">　Index begin　Index this file
2009-08-24-14-41 start
■　Exercise 1.15 solution

Exercise 1.15 hint already solved this 
problem. LiuHH fill detail steps as 
following.
Start point is eqn.1.27 and eqn.1.28
eqn.1.27 says the total sum of probability
is one.
eqn.1.28 says unbiased estimator g(k)
multiply by probability at k and sum over
all possible k, get the true state of 
nature θ.
<a name="ch01d043">
differentiate the identity (1.27) and 
(1.28) with respect to θ, we find for
all θ∈Θ that
  ∑[k∈D]{pθ(k;θ)}=∂1/∂θ=0  -----eqn.AE02
and
  ∑[k∈D]{g(k)*pθ(k;θ)}=∂θ/∂θ=1  -----eqn.AE03

<a name="ch01d044">
eqn.AE03 - θ*eqn.AE02
 =1-θ*0　　　//where this ZEROed θ go? 
 = 1
 =∑[k∈D]{[g(k)-θ]*pθ(k;θ)}
next, equal to eqn.AE04
Here we insert a multiply-by-ONE. This 
ONE is red color terms. Re-group and get
eqn.AE05

<a name="ch01d045">
Cauchy's inequality need two sequences.
Refer to eqn.AE05, find two sequences as
below.
seq.1
　　[g(1)-θ]√[p(1;θ)],
　　[g(2)-θ]√[p(2;θ)],
　　 ... 
　　[g(n)-θ]√[p(n;θ)]
<a name="ch01d046">
seq.2
　　pθ(1;θ)/√[p(1;θ)],
　　pθ(2;θ)/√[p(2;θ)],
　　 ... 
　　pθ(n;θ)/√[p(n;θ)]

seq.2 actually used
　　√[p(1;θ)]*[pθ(1;θ)/p(1;θ)],
　　√[p(2;θ)]*[pθ(2;θ)/p(2;θ)],
　　 ... 
　　√[p(n;θ)]*[pθ(n;θ)/p(n;θ)]
eqn.1.30 preserve the ratio pθ(k;θ)/p(k;θ).

Now we apply Cauchy's inequality to above
seq.1 and seq.2
seq.1 dot product with seq.2 get eqn.AE05

<a name="ch01d047">
Cauchy's inequality less than side do not
equal to one directly. Here eqn.AE05 equal
to one, this equality has nothing to do 
with Cauchy.

Cauchy's inequality greater than side is
seq.1 dot product with seq.1 take square root
seq.2 dot product with seq.2 take square root
then multiply two square root. (see Cauchy)

<a name="ch01d048">
seq.1 dot seq.1 is next

<a name="ch01d050">
2009-08-24-15-57 here
Cauchy's inequality is
 seq.1 dot seq.2 ≦
 √{ [seq.1 dot seq.1]*[seq.2 dot seq.2] }

replace with equation number, get
 1 = eqn.AE05 ≦ √{ eqn.AE06 * eqn.AE07 }

eqn.AE05 start from 1, eqn.AE05 equal to 1.

<a name="ch01d051">
Cauchy's inequality become
 √{ eqn.AE06 * eqn.AE07 } ≧ 1
Square this equation, get
    eqn.AE06 * eqn.AE07   ≧ 1
rewrite as next line
    eqn.AE06  ≧ 1/eqn.AE07 ---eqn.AE08

eqn.AE08 is same as eqn.1.29
Problem solved.
2009-08-24-16-12 stop

Chapter one is done.




<a name="ch02a001">　Index begin　Index this file

2009-08-25-09-32 start
■■Chapter 02: The AM-GM Inequality

■　Arithmetic mean and geometric mean

'mean' procedure is to take average of two
or multiple numbers. Now take the simplest
example. Let a1 and a2 be two numbers in
consideration.

<a name="ch02a002">
Arithmetic mean = AM = (a1+a2)/2 ---eqn.AE09
Geometric mean  = GM = √(a1*a2) ---eqn.AE10

'Arithmetic' in 'Arithmetic mean' come from
the fact that +,-,*,/ four operations are
basic arithmetic operations.

<a name="ch02a003">
'Geometric' in 'Geometric mean' come from
the fact that in the right triangle, draw
a perpendicular line from right angle to 
hypotenuse. This line divide right triangle
to two similar and smaller right triangles.
Shorter section of hypotenuse multiply with
longer section of hypotenuse equal to square
of perpendicular line.
2009-08-25-10-15 here.
So perpendicular line is equal to square
root of shorter section multiply with 
longer section. This is an average by 
square root and appears in geometry.

<a name="ch02a004">
eqn.AE09 and eqn.AE10 are for two elements
in a sequence. What to do if a sequence 
contain three or more elements?
For three elements sequence do we get
AM = (a1+a2+a3)/3 ---eqn.AE11
GM = √(a1*a2*a3)  ---eqn.AE12NO
?
<a name="ch02a005">
Although square root is a frequently 
used function. But eqn.AE12NO has 
trouble. If our sequence has all length
elements. Then a1 and a2 and a3 are all
length. a1*a2*a3 rise to length third 
power--volume. √(a1*a2*a3) change
dimention to length power 1.5. But 
<a name="ch02a006">
addition in eqn.AE11 let whole equation
keep length unit (power 1.0)
length power 1.5 and length power 1.0 
can not be compared.
"Two acre is greater than one mile" is
a meaningless statement.
In three variable case, Geometric mean
take cubic root for three elements product.
<a name="ch02a007">
In n variable case, 
Arithmetic mean let n-sum divide by n
AM = (a1+a2+ ... +an)/n ---eqn.AE13
Geometric mean take power of 1/n for n 
elements product.
GM = (a1*a2* ... *an)1/n  ---eqn.AE14
Each operation, return elements product to
its original physics unit.

<a name="ch02a008">
Change eqn.AE12NO to next correct equation
GM = (a1*a2*a3)1/3  ---eqn.AE12
eqn.AE12 (length) can be compared with 
eqn.AE11 (length)
AM = (a1+a2+a3)/3 ---eqn.AE11

2009-08-25-10-42 stop

<a name="ch02a009">　Index begin　Index this file
2009-08-25-13-01 start
■　AM-GM Inequality -- start point.
Given one sequence a1, a2, a3 ... an 
we have two average method.
AM = (a1+a2+ ... +an)/n ---eqn.AE13
GM = (a1*a2* ... *an)1/n  ---eqn.AE14
The result should be different. 
Which one get greater value? 
What condition give us AM=GM ?
<a name="ch02a010">
Start from easiest two elements sequence 
a1, a2. It is easy to use numerical
value to find out the answer. First 
assume all elements are non-negative. 
Avoid imaginary result.
<a name="ch02a011">
Let a1=3 and a2=12
AM = (a1+a2)/2 = 7.5
GM = (a1*a2)1/2 = 6
Numerical result suggest AM > GM.
Is it true in general? When AM=GM ?

<a name="ch02a012">
Textbook use a systematic approach.
Given sequence a1, a2, a3,  ... an

First show that AM≧GM is valid for 
n be 2k where k is a positive integer.
for example, n=2, 4, 8, 16, 32 etc 
are true. (n=3,5 etc are not covered)

<a name="ch02a013">
Second show that AM≧GM is valid for 
n be any positive integer.
Second step is called "Cauchy leap 
forward and fall back method"

Third show that AM≧GM is valid for 
1/n be any rational number that is 
smaller pos. integer/greater pos. integer
(n=π,√5 etc are not covered)

<a name="ch02a014">
Fourth show that AM≧GM is valid for 
n be any real number that is whole real
axis include n=π and n=e=2.71828182845...
2009-08-25-13-31 here

<a name="ch02a015">　Index begin　Index this file

■　GM≦AM two dimensional case
Let x and y be any real number, then
 (x-y)*(x-y)≧0  ---eqn.AE15 
(See figure, please click [Draw 603])

is true. Expand eqn.AE15, get
 x*x + y*y ≧ 2*x*y
or
 xy ≦ x2/2 + y2/2 ---eqn.AE16
This is textbook page 19 line 3 eqn.2.1
eqn.AE16 is called 'humble bound'.
Cross term xy replaced with equal term x*x, y*y

<a name="ch02a016">
Now let x and y be any non-negative number,
and replace x with √x, replace y with √y
Put √x and √y to eqn.AE16, we get
 √x√y ≦ (√x)2/2 + (√y)2/2
re-arrange, get
two dimensional AM-GM inequality
     √(xy)     ≦ (x + y)/2 ---eqn.AE17
Geometric mean ≦ Arithmetic mean
or
 4√(xy) ＜ 2*x + 2*y ---eqn.AE18 for x≠y
This is textbook page 19 eqn.2.2
Sequence elements weight, coefficient 1/2 or
power 1/2, is same as sequence dimension 2.

<a name="ch02a017">
eqn.AE18 has a redundant factor 2.
It help us explain eqn.AE18 better.
Let x and y be the length of a rectangle 
two sides. Rectangle has four sides, x,
x and y, y. Rectangle perimeter length
is 2*x+2*y which is right hand side of
eqn.AE18
<a name="ch02a018">
Rectangle area is xy. √(xy) has dimension
of length. √(xy) is the one side average
length of rectangle. It has four sides.
Rectangle averaged total perimeter length
is 4√(xy). eqn.AE18 tell us this 
averaged total perimeter length is 
  less than or equal to the 
  actual total perimeter length.

<a name="ch02a019">
From
 √(xy) ≦ (x + y)/2 ---eqn.AE17
if we require rectangle total perimeter 
length be constant, that is x+y=C, then
variable area √(xy) ≦ (C)/2 constant
square x=y=C/2 has largest area
 √(xy) =C/2 ≦ (x + y)/2 = (C/2+C/2)/2
'largest' is maximum, when reach maximum,
inequality '≦' become equality '=' C/2=C/2

<a name="ch02a020">
From
 √(xy) ≦ (x + y)/2 ---eqn.AE17
4√(xy) ≦ (2x + 2y) ---eqn.AE18
if we require rectangle total area xy
be constant, that is xy=A, then
constant √(A)≦(x+y)/2 variable perimeter
square x=y=√A has smallest perimeter 
4√(xy)=4√A ≦ (2√A + 2√A) = 4√A
'smallest' is minimum, when reach minimum,
inequality '≦' become equality '=' 4√A=4√A

2009-08-25-14-11 stop

<a name="ch02a021">　Index begin　Index this file
2009-08-25-17-23 start
■　GM≦AM is valid for n=2k k=any positive integer
Cauchy leap forward and fall back method
proved AM-GM inequality has next form

<a name="ch02a022">
2009-08-25-17-59 here
Before meet Cauchy for n=any positive integer,
First show GM≦AM is valid for n=2k
where k=any positive integer.

From eqn.AE15 to eqn.AE17 it showed 
for n=2k and k=1 (that is for n=2)
GM≦AM is true. Next case,
<a name="ch02a023">
for n=2k and k=2 (that is for n=4)
it is easy to extend the proof.
n=4 use four element sequence 
[a1,a2,a3,a4]
Apply two dimensional AM-GM inequality
three times to four element sequence

<a name="ch02a025">
2009-08-25-18-17 here
First time apply two dimensional AM-GM
inequality. Eqn.2.4 left term (a1a2a3a4)¼
is less than eqn.2.4 middle term
[(a1a2)½+(a3a4)½]/2

Second time apply two dimensional AM-GM
inequality. Eqn.2.4 middle red term and
eqn.2.4 right red term participate.

<a name="ch02a026">
Third time apply two dimensional AM-GM
inequality. Eqn.2.4 middle blue term and
eqn.2.4 right blue term participate.

Up to here, eqn.2.4
four dimensional AM-GM inequality is done.

Repeat the same process, it is easy to
conclude that 
AM-GM inequality is true for n=2,4,8,16,32 ...
that is true for n=2k and k is any positive
integer. 

When n=5 or 6 or 7 above procedure do not
support these non-2k n's.

Sequence elements weight, coefficient 1/n or
power 1/n, is same as sequence dimension n.

2009-08-25-18-28 stop

<a name="ch02a027">　Index begin　Index this file

2009-08-25-19-07 start
■　GM≦AM is valid for n=any positive integer
Cauchy leap forward and fall back method

Textbook page 20 introduce 
GM≦AM is valid for n=2k k=any positive integer
Textbook page 21 introduce 
GM≦AM is valid for n=any positive integer.
Textbook page 101 line 11 said
"Cauchy's leap-forward fall-back induction"
Textbook page 249 line 5 said
"Cauchy's argument (page 20) ....."
"fall-back step, one choose k ....."

Based on above text sections, LiuHH guessed
page 21 method is called
"Cauchy leap forward and fall back method".
2009-08-25-19-24 here.

<a name="ch02a028">
Assume we have a sequence with n elements
and n is NOT equal to 2k k=any positive integer
For numerical example, assume n=5.
Assume [a1,a2,a3,a4,a5]=[2,5,7,8,9]
n=5>4=2^2 , and n=5<8=2^3, n=5 is not 
covered by previous section.

<a name="ch02a029">
Cauchy leap forward method is to pad dummy
sequence elements from a6, a7 to a8

Since n=5<8=2^3, 8 is next higher integer
which IS 2k k=any positive integer=3. 
Dummy element a8 reached 2k=23 requirement.
Assign what value to a6, a7, a8 ?
We have sequence [2,5,7,8,9] in hand. 
<a name="ch02a030">
One trial is to assign
a6=a7=a8=[2+5+7+8+9]/5=6.2=A ---eqn.AE19
That is to assign arithmetic mean to dummy
elements a6,a7,a8.
Numerical value eqn.AE19 is special case.
Next use symbol equation for general case

<a name="ch02a032">
2009-08-25-19-59 here
Original numerical sequence is
[2,5,7,8,9]
New numerical sequence is
[2,5,7,8,9 , 6.2,6.2,6.2]

Original symbol sequence is
[a1,a2,a3,a4,a5]
New symbol sequence is
[a1,a2,a3,a4,a5 , α6,α7,α8]
(note: dummy α6,α7,α8 IS a6,a7,a8)

<a name="ch02a033">
Apply eight dimensional AM-GM inequality
(not in this page, figure out in your brain.)
to above new created symbol sequence.
Above pad dummy, Cauchy leap forward part.
Below cut dummy, Cauchy fall back section.
remember A = [a1+a2+a3+a4+a5]/5 ---eqn.AE19

<a name="ch02a035">
2009-08-25-20-35 here
re-write above equation
[(a1a2a3a4a5)1/8]*A3/8 ≦ A
or (fall back, cut dummy right here.)
(a1a2a3a4a5)1/8 ≦ A5/8
Whole equation take 8/5 power get
(a1a2a3a4a5)1/5 ≦ A
<a name="ch02a036">
Put A back, get
(a1a2a3a4a5)1/5 ≦ [a1+a2+a3+a4+a5]/5
This is a proof for GM≦AM is valid for n=5.
In above analysis, if change '8' to 2k
and change '5' to 'n'. Then it is same as
textbook derivation.  2k and 'n' notation
is general for any positive integer n.

Up to here, still
sequence elements weight, coefficient 1/n or
power 1/n, is same as sequence dimension n.

The following dimension is NOT weight
2009-08-25-20-50 stop

<a name="ch02a038">　Index begin　Index this file
2009-08-26-12-00 start
■　GM≦AM is valid for n=any rational number
rational number means a number which is a
result of integer_A / integer_B
require integer_B not equal to zero.

eqn.2.6 is valid for n=any positive integer.
eqn.2.6 do not cover n=7/3 or n=12/5 etc.
7/3 and 12/5 are rational numbers.
Is it possible to extend AM-GM Inequality
to n=any rational number? 
This is done in textbook page 22.
<a name="ch02a039">
Assume that p1, p2 ... pn are nonnegative
rational numbers that sum to one. 
p1+p2+...+pn=1 ---eqn.AE20 
Numerical example is 1/7, 1/7, 2/7, 3/7.
For a non-negative real sequence a1, a2,
 ... an {'non-negative' avoid complex
number jump out of magician's hat.)
<a name="ch02a040">
show that next equation is true.
AM-GM Inequality is general for any
 rational weights
a1p1a2p2...anpn ≦ p1a1+p2a2+...+pnan ---eqn.2.7
ALERT: p1+p2+...+pn=1 ---eqn.AE20

Now use above numerical example, we have
p1=1/7, p2=1/7, p3=2/7, p4=3/7
They sum to one
p1+p2+p3+p4=1/7+1/7+2/7+3/7=1
<a name="ch02a041">
Re-write eqn.2.7 for this numerical example
a1p1a2p2a3p3a4p4 ≦ p1a1+p2a2+p3a3+p4a4
become
a11/7a21/7a32/7a43/7 ≦  ---eqn.AE21
    (1/7)*a1+(1/7)*a2+(2/7)*a3+(3/7)*a4
eqn.AE21 is different from eqn.2.6, because
eqn.2.6 has common power 1/n at less than side
eqn.2.6 has common multiplier 1/n at greater than side
Actually eqn.AE21 present no problem at all,
<a name="ch02a042">
we can re-write eqn.AE21 as next
("lots of repetition" textbook page 22, line 11)
a11/7a21/7a31/7a31/7a41/7a41/7a41/7
 ≦ (1/7)*a1+(1/7)*a2   ---eqn.AE22
   +(1/7)*a3+(1/7)*a3+(1/7)*a4+(1/7)*a4+(1/7)*a4
all a3 repeat terms arrange in red color,
all a4 repeat terms arrange in blue color,
Sequence elements weight, coefficient
or power 1/7,1/7,2/7,3/7, is NOT same
as sequence dimension n=4. compare

<a name="ch02a043">
Now eqn.AE22 satisfy eqn.2.6 and
GM≦AM is valid for n=any positive integer
apply to pj=rational number.
In above numerical example, 
change '4' to 'n'
change '7' to 'M'
change '2/7' to 'ki/M'
change '3/7' to 'kj/M'
<a name="ch02a044">
then it is same as textbook derivation.
and
AM-GM Inequality is general for any
 rational weights (page 22, eqn.2.7)
a1p1a2p2...anpn ≦ p1a1+p2a2+...+pnan ---eqn.2.7
ALERT: p1+p2+...+pn=1 ---eqn.AE20

2009-08-26-13-07 stop

<a name="ch02a045">　Index begin　Index this file

2009-08-26-15-42 start
■　GM≦AM is valid for n=any real number
Previous section, proved that 
GM≦AM is valid for n=any rational number
rational numbers are isolated points on 
real axis.
On real axis, irrational numbers dominate.

<a name="ch02a046">
irrational numbers are those can not be 
expressed with rational numbers. 
Example: 
√2=1.4142135623730951...
√3=1.7320508075688772...
PI=3.141592653589793...
E=2.718281828459045... etc.

<a name="ch02a047">
irrational numbers must be infinite long,
because whenever they stopped, then that
finite length number is a rational number!
For example, 
PI=3.141592653589793... is infinite long
But
P0=3.141592653589793    is finite long
we can write P0 as
P0=3141592653589793/1000000000000000
  =integer_A/integer_B=rational number

<a name="ch02a048">
Irrational numbers are infinite long,
then, how can we extend from
GM≦AM is valid for n=any rational number
to
GM≦AM is valid for n=any real number
?
This is discussed in textbook page 22.
It is also simple, but exhaust!
<a name="ch02a049">
Brief speak, we chase to infinite!
Do as following. 
Let sequence elements weight be
pj(t), j=1,2,3 ... and t=1,2,3 ... 
j is finite, t goto infinite.
j is sequence element index, finite number.
t is number of digits used for a special
  "finite-length irrational number".
Numerical example,
assume three elements sequence (j=1,2,3)
[3, 7, 11]
assume t=10
p1(t)=p1(10)=0.5773502691 (1/√3 10 digit)
p2(t)=p2(10)=0.3183098861 (1/PI 10 digit)
p3(t)=p3(10)=1-p1(10)-p2(10)
            =0.1043398448 (10 digit)
<a name="ch02a050">
then problem is rational for
3^(p1)*7^(p2)*11^(p3)
≦
3*(p1)+7*(p2)+11*(p3)
numerical result is
4.499123315487912
≦
5.1079583028

<a name="ch02a051">
Paste
[[
p1=0.5773502691
p2=0.3183098861
p3=0.1043398448
pow(3,p1)*pow(7,p2)*pow(11,p3)
3*p1+7*p2+11*p3
]]
to box3 in complex2.htm#calculator
http://freeman2.com/complex2.htm#calculator
click "test box3 command, output to box4"
get above numerical result.

<a name="ch02a052">
Up to here, it is t=10 digits solution.
Repeat same process, let t goto 100, 1000
and approach to infinity. This procedure
let us approach to irrational number solution.
BUT, this procedure is not complete, because
we can not argue the equality condition for
a chase to infinity method.

<a name="ch02a053">
AM-GM Inequality is general for any
 irrational weights (page 23, eqn.2.9)
a1p1a2p2...anpn ≦ p1a1+p2a2+...+pnan ---eqn.2.9
ALERT: p1+p2+...+pn=1 ---eqn.AE20

Prof. J. Michael Steele still reach out 
his helping hand for completeness.
Please see next section.
2009-08-26-16-50 stop

<a name="ch02a054">　Index begin　Index this file
2009-08-27-11-31 start
■　GM≦AM perfect proof by Prof. George Polya
Above sections prove AM-GM inequality start 
from two dimensional case eqn.AE17 or eqn.2.2
then generalize to
AM-GM Inequality is general for any
irrational weights (page 23, eqn.2.9)
Above proof is building brick style. 
Second proof generalize first  proof,
third  proof generalize second proof ... etc.
Eventually, we have one hand hold irrational
weights, but the other hand is empty for not
knowing equality condition.

<a name="ch02a055">
The next proof has nothing to do with above
building bricks. 
Next proof is stand alone.
Next proof is complete.
Next proof is done by Prof. George Polya.
Next proof is covered in textbook page 23
to page 25.

Please see next two figures first
2009-08-27-11-54 here

<a name="ch02a058">
2009-08-27-11-59 here
Fig 603 red lines are the same as textbook
page 23 fig.2.1

Fig 603 red lines equation is next
  1+x ≦ exp(x) for all real x ---eqn.2.8 page 23

Fig 603 blue line equation is next
  y(x)=x*x ---eqn.AE23

<a name="ch02a059">
eqn.AE23 is memtioned in textbook page 23 
line 7 . Textbook stated that
Cauchy used  y(x)=x*x≧0
Polya  used  1+x ≦ exp(x) ---eqn.2.8
Both masters get great result from simple
equations.

Fig 603 red lines show up on textbook 
page 23 fig.2.1
Fig 604 is actually used for Polya's proof.

<a name="ch02a060">
Fig 603 use 1+x≦exp(x) ---eqn.2.8  page 23
Fig 604 use x≦exp(x-1) ---eqn.2.10 page 24

In eqn.2.8, change of variable
let x=y-1  ---eqn.AE24
then 1+x≦exp(x) become 1+y-1≦exp(y-1)
become y≦exp(y-1) 
change dummy variable y to x get eqn.2.10

<a name="ch02a061">
Our target is to prove
AM-GM Inequality is general for any
 REAL weights (page 23, eqn.2.9)
a1p1a2p2...anpn ≦ p1a1+p2a2+...+pnan ---eqn.2.9
where
a1,a2,...,an are non-negative real numbers
and they are elements of a sequence.
p1,p2,...,pn are positive real numbers
with constraint p1+p2+...+pn=1 ---eqn.AE20
<a name="ch02a062">
Please pay special attention here,
At starting point, we 
do not require 1/pj's to be 2^k, 
do not require 1/pj's to be positive integer, 
do not require 1/pj's to be rational/irrational.
Just require p1,p2,...,pn are positive real
numbers. This is fundamental difference between
Polya's proof and previous many generalization.

<a name="ch02a063">　Index begin　Index this file

■　Why require p1+p2+...+pn=1 ? ---eqn.AE20
It is easy to understand from dimensional
analysis.
Assume sequence in discussion is an array
of length data. That is a1,a2,...,an are 
all length.
p1,p2,...,pn MUST BE pure numbers.
because in a1p1, p1 is a power of length a1
A power MUST BE pure number, or else we
can not explain what is the meaning of
length rise to mass power.
<a name="ch02a064">
Right hand side of eqn.2.9 is 
p1a1+p2a2+...+pnan
pure number p1 * length a1 plus same terms
result is still length.
Left hand side of eqn.2.9 is 
a1p1a2p2...anpn
Base a1,a2,...,an are all length,
ONLY IF p1+p2+...+pn=1
the left hand side is length rise to power
one which is still length.
<a name="ch02a065">
left hand side length ≦ right hand side length
only length compare with length.
length can not compare with area, volume etc.

For physics consistant reason, 
WE MUST INSIST  p1+p2+...+pn=1
2009-08-27-12-57 stop

<a name="200908301802">
2009-08-30-18-02 start
On the other hand, if you see the following
problem
[[
Please prove: if x, y, z are real numbers, 
with x, y, z > 0 and xyz = 1 then
 x + y + z ≦ x2 + y2 + z2
]]
In this case, x and y and z can not be any
physics quantity, because 
 length can not be ≦ length_squared
Here, x, y, z must be pure number.
2009-08-30-18-08 stop

<a name="ch02a066">　Index begin　Index this file
2009-08-27-14-03 start
■　Polya's proof, change of variable x→x-1
Our target equation is eqn.2.9
Polya's proof start at eqn.2.8
 1+x≦exp(x) ---eqn.2.8  page 23
then change of variable to get
 x≦exp(x-1) ---eqn.2.10 page 24
Both valid for all real x.
See x≦exp(x-1) click, click [Draw 604]

<a name="ch02a067">
We can start from eqn.2.8 and modify ak
to the form 1+xk to fit eqn.2.8.
We can also start from eqn.2.10 and not 
modify ak. Textbook suggest reader explore
both ways (page 24 line 7).
Textbook give us guide line for eqn.2.10
method.

<a name="ch02a068">
eqn.2.10 is valid for all real x. Now
sequence element ak is real, we have
 x=ak and
 ak≦exp(ak-1) ---eqn.AE25 page 24 line 13
See ak≦exp(ak-1) click, click [Draw 604]
eqn.AE25 rise to pk power
 akpk≦[exp(ak-1)]pk
 akpk≦exp(akpk-pk) ---eqn.AE26 page 24 line 13
<a name="ch02a069">
We multiply eqn.AE26 from k=1 to k=n get
 ∏[k=1,n]{akpk}≦∏[k=1,n]{exp(akpk-pk)} ---eqn.AE27
(eqn.AE27 right hand side has one further 
 step at one pace below see eqn.2.11)

Define G=eqn.AE27 left hand side 
 G=a1p1a2p2...anpn ---eqn.AE28
G is geometric mean in target inequality.

<a name="ch02a070">
Define R(a1,a2,...,an)=eqn.AE27 right hand side 
 R=∏[k=1,n]{exp(akpk-pk)}
 R=exp(∑[k=1,n]{akpk-pk})
 R=exp(∑[k=1,n]{akpk}-∑[k=1,n]{pk})
Here ∑[k=1,n]{pk}=1, R reduce to
 R=exp(∑[k=1,n]{akpk}-1) ---eqn.2.11 page 24

<a name="ch02a071">
eqn.AE27 tell us that 
geometric mean G is bounded by R
 G ≦ R ---eqn.AE27 simplified
R is exp(arithmetic mean - 1) so
R is NOT arithmetic mean.

geometric mean ≦ arithmetic mean
is not established yet.
2009-08-27-14-59 here

<a name="ch02a072">　Index begin　Index this file

■　How R relate to arithmetic mean A?
Compare eqn.2.11 with eqn.2.10
 R=exp(∑[k=1,n]{akpk}-1) ---eqn.2.11
 x≦exp(x-1) ---eqn.2.10
exp() match exp(), '-1' match '-1', 
See x≦exp(x-1) click, click [Draw 604]
<a name="ch02a073">
then !! LET
 x=AM=∑[k=1,n]{akpk} ---eqn.AE29
jump out of one's mind!
eqn.2.10 change to next 
 AM≦exp(AM-1)  //AM=arithmetic mean
that is
 ∑[k=1,n]{akpk}≦exp(∑[k=1,n]{akpk}-1) ---eqn.AE30
that is
 AM≦R
<a name="ch02a074">
eqn.AE27 tell us GM≦R  (simplified)
eqn.AE30 tell us AM≦R
Our goal is to establish relation between
AM and GM. Go a big loop, we get both
AM and GM are not greater than R !!
What can we conclude !?
2009-08-27-15-28 here

<a name="ch02a075">　Index begin　Index this file
2009-08-27-16-40 start
■　Professor Steele played magic
Although both GM≦R and AM≦R, we can write
max{GM, AM}≦R ---eqn2.12 simplified, page 24

normalization played magic before. change
from adding to multiplying. Can we get
help here? smaller compare with smaller?
Find out the "bigger smaller"? Just try.
<a name="ch02a076">
Define 
 A = p1a1+p2a2+...+pnan ---eqn.AE31
 A = Arithmetic mean
Define
 αk = ak/A ---eqn.AE32
That is take Arithmetic mean as unit length.

<a name="ch02a077">
Re-visit eqn.AE27
replace ak with αk get the following

<a name="ch02a080">
2009-08-27-17-35 here
eqn.AE33 is textbook page 25 line 12 equation.
eqn.AE33 exp() function has sum ∑ over k, but
denominator A (Arithmetic mean) is not a
function of k, move A out of sum ∑, get 
eqn.AE34. Refer to definition of eqn.AE31
eqn.AE34 simplify to 1.
<a name="ch02a081">
Re-write eqn.AE33 less than side and greater
than side (now value 1) get eqn.AE35.
eqn.AE35 numerator is geometric mean G
eqn.AE35 denominator is Arithmetic mean A
eqn.AE35 says GM ≦ AM
MAGIC !!
<a name="ch02a082">
remember p1+p2+...+pn=1
eqn.AE35 denominator A use it.
remember weights pk can be ANY positive
real number. no integer, rational limit.
2009-08-27-17-53 stop

<a name="ch02a083">　Index begin　Index this file
2009-08-27-19-10 start
■　Equality GM=AM is easy.
eqn.AE35 give us GM≦AM.
When the equality exist?
Our starting point is
 x≦exp(x-1) ---eqn.2.10 for all real x
We say ak is real, so get
 ak≦exp(ak-1) ---eqn.AE25 page 24 line 13
<a name="ch02a084">
We turn to normalization, define
 αk = ak/A ---eqn.AE32
Then key starting point is
 (ak/A)≦exp((ak/A)-1) ---eqn.AE36
From eqn.AE36 and from Figure 604
(goto Figure 604, click [Draw 604] button)
<a name="ch02a085">
we know equality condition occur at point
(x,y)=(1,1)
In eqn.AE36, 'x' is 'ak/A'
In eqn.AE36, 'y' is 'exp((ak/A)-1)'
If the point (1,1) is on the curve? try
let  x=ak/A=1
get  y=exp((ak/A)-1)=exp(1-1)=1
True !!
<a name="ch02a086">
So equality condition is
 x=ak/A=1 ---eqn.AE37
eqn.AE37 says ak=A for any k.
Then from k=1 to k=n, we have
 a1=a2=...=an=A=Arithmetic mean ---eqn.AE38
eqn.AE38 is equality condition !!
<a name="ch02a087">
We chase to infinity and can not draw
any conclusion for equality condition.
Now Professor George Polya's perfect proof
and Professor Michael Steele's magic hand
bring an easy solution to us.

Thank you! Professor George Polya
Thank you! Professor Michael Steele
2009-08-27-19-39 stop

<a name="ch02a088">
2009-08-27-21-16 start
■　A tiny problem for reader
■　A generalization for dimension consistency
In previous section
Why require p1+p2+...+pn=1 ? ---eqn.AE20
explained the reason why.

Suppose in another application, we have 
a1p1a2p2...anpn ≦ p1a12+p2a22+...+pnan2 ---eqn.AE39
Can you explain that we must insist
 p1+p2+...+pn=2  ---eqn.AE40
?
2009-08-27-21-24 stop

<a name="ch02a089">
2009-08-28-11-46 done proofread.
2009-08-28-11-46 start
If pj in eqn.AE39 is probability,
then we need to work at a condition
 100% probability = 2
Is this right? 
LiuHH put a question mark here.
2009-08-28-11-49 stop



<a name="ch02a090">
2010-04-08-17-23 start
Above ask
[[
 100% probability = 2
Is this right? 
]]
Answer is WRONG !!

After read 
Chapter 08: The Ladder of Power Means
know the answer.
<a name="ch02a091">
 100% probability = 1 not 2
eqn.AE39 less than side is still
regular GM. But the GM value 
take square like greater than
side each sequence element 
did -- take square.
2010-04-08-17-28 stop



<a name="20090831">
2009-08-30-19-40 start
'Update 2009-08-31' add
200908301802
and 'docB??' below
and program sum i from 1 to n
for ∑[i=1,n]{i},   ∑[i=1,n]{i*i}
and ∑[i=1,n]{i^3}, ∑[i=1,n]{i^4}
2009-08-30-19-45 stop

from 2009-08-31-08-30 to 2009-08-31-09-18
added "∑ i^"[box] allow arbitrary power.
([box] value)

<a name="docB01">
2009-08-30-09-18 start
LiuHH is mathematics admirer and outsider.
LiuHH is programming admirer and outsider.
So both math work and programming work have
inaccurate statement or low efficient code.
Next explain 1+2+3+4+5+...+n calculation.
For symmetry matrix, store lower triangle
elements is enough for calculation. To get
data from lower triangle matrix need target
row number diagonal element location.
For rank 1 sub-matrix, location=1
For rank 2 sub-matrix, location=1+2
For rank 3 sub-matrix, location=1+2+3
.....
Above start form first element location=1
<a name="docB02">
Below start form first element location=0
For rank 1 sub-matrix, location=1    -1=0
For rank 2 sub-matrix, location=1+2  -1=2
For rank 3 sub-matrix, location=1+2+3-1=5
.....
Please see example
In tute0009.htm need to create a sequence
for diagonal elements location. 
<a name="docB03">
For rank-10 matrix, the sequence is
0,2,5,9,14,20,27,35,44,54 (start from 0)
or
1,3,6,10,15,21,28,36,45,55 (start from 1)
In the calculation loop, use previous data
add current rank n. One addition operation
finish one rank. 
(Better than three operations for one rank)

<a name="docB04">
Next goto tute0010.htm
Here find a button 'summation factorial'
(wrong math term, but LiuHH do not know the
 right math term)
This button calculate one rank value directly.
No previous value as base for new value. In 
this case it is better to use formula to find
the value of 'summation factorial'. 
<a name="docB05">
For example rank-10
1+2+3+4+5+6+7+8+9+10 = 10*11/2 = 55
When write tute0010.htm, LiuHH  copy code from
tute0009.htm, still do nine addition for 55.
This is low efficient code. Because equation
n*(n+1)/2 that is 10*(10+1)/2
one addition (10+1)
one multiplication 10*(10+1)
one division 10*(10+1)/2
three operations get answer for one rank.
<a name="docB06">
For rank 1000, tute0010.htm add 1000 times
but 1000*(1000+1)/2 still three operations.
Now (2009-08-14) LiuHH's computer can not
input Chinese. Both tutc0010.htm (Chinese)
and tute0010.htm (English) not update.
(I can not create Chinese document)
Just make a note here in tute0011.htm
<a name="docB07">
If you want duplicate LiuHH's code
be alert !! 
Code may not be efficient!

If you want use LiuHH's math reasoning,
be alert !! 
Reasoning may not be correct!
2009-08-30-09-52 stop


<a name="docB08">
2009-08-30-12-28 start
Now explain the equation 
  n*(n+1)/2 ---eqn.AE41
eqn.AE41 is 'summation factorial' formula.
For rank n=4, we have the following matrix

[ 11 　1 　2 　3] 
[ 1 　12 　4 　5]  ---eqn.AE42
[ 2 　4 　13 　6] 
[ 3 　5 　6 　14] 

<a name="docB09">
put n=4 into eqn.AE41, get
Lower triangle element numbers
 = 4*(4+1)/2 = 10 (red plus blue)
It can be explained as following.
 n*(n+1)/2 has two part
 n*n/2  and  n/2
Rank n matrix has n*n elements.
Half of it is n*n/2 
n*n/2 is the red color elements in eqn.AE42
<a name="docB10">
But we want whole diagonal elements!
n*n/2 do not cover the blue elements in
eqn.AE42
Rank n matrix has n diagonal elements.
Half of it (take care the blue elements)
is n/2
Lower triangle element numbers is 
 n*n/2 + n/2 = (n*n+n)/2 = n*(n+1)/2 
for complete diagonal.
eqn.AE41 is good for any rank number n.
2009-08-30-12-45 stop

<a name="20090901">
2009-09-01-08-03 start
"Update 2009-09-01" add equation for
∑ i^(5) from i=1 to i=n which is
n*n*n*n*n*n/6+n*n*n*n*n/2+5*n*n*n*n/12-n*n/12
2009-09-01-08-05 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