<a name="docA001">
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="ch03c001">　Index begin　Index this file
2009-10-24-10-39 start
■　Exercise 3.5 problem statement
　　textbook page 48
（Monotonicity and a Ratio Bound）

Show that if f:[0,1]→(0,∞) is 
non-increasing, then one has

<a name="ch03c003">
2009-10-24-10-49 here
[[
eqn.3.12 is created on 2009-05-22-15-19
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.
Around 2009-05-22, it is practice HTML
math equation period.
]]
<a name="ch03c004">
2009-10-24-10-55 start
As a hint, one might consider the 
possibility of proving a Lagrange type
identity by beginning with a double 
integral on [0,1]*[0,1] whose integrand
is guaranteed to be positive by our 
monotonicity hypothesis.
2009-10-24-10-59 here




<a name="ch03c005">　Index begin　Index this file
2009-10-24-11-07 start
■　Exercise 3.5 hint
　　textbook page 238
Since f is nonnegative and nonINcreasing
one has the integral inequality
  0≦∫[0,1]∫[0,1]{f(x)f(y)(y-x)(f(x)-f(y))}dxdy ---eqn.AJ001

<a name="ch03c006">
since the integrand is nonnegative. One
may now complete the proof by simple 
expansion. Incidentally, this way of
exploiting monotonicity is exceptionally
rich, and several variations on this
theme are explored at length in Chapter 5.
2009-10-24-11-17 stop

<a name="ch03c007">
2009-10-24-11-22 start
　　textbook page 238 line 10 say
[[
Since f is nonnegative and nonDEcreasing
.....
]]
<a name="ch03c008">
LiuHH change to next line
Since f is nonnegative and nonINcreasing
Problem statement use 
[[
f:[0,1]→(0,∞) is non-increasing
]]
and from solving reasoning, it is 
'non-increasing' correct. All three 
CSMC_Errata source not mention this
point. LiuHH change from nonDEcreasing
to nonINcreasing, hope it is right.
2009-10-24-11-29 stop

<a name="CSMC_Errata">
2009-10-28-11-39 start
"All three CSMC_Errata source" are next

2009-02-01-15-22
http://www-stat.wharton.upenn.edu/~steele/Publications/Books/CSMC/CSMC_Errata.pdf
save as CSMC_Errata.pdf

2009-02-01-15-25
http://www-stat.wharton.upenn.edu/~steele/Publications/Books/CSMC/ByronSchmuland.txt
save as CSMC_Errata_ByronSchmuland.txt

2009-02-01-15-27
http://www-stat.wharton.upenn.edu/~steele/Publications/Books/CSMC/LangeListCSMCTypos.pdf
save as CSMC_Errata_LangeListCSMCTypos.pdf
2009-10-28-11-42 stop

<a name="ch03c009">　Index begin　Index this file
2009-10-24-11-33 start
■　Exercise 3.5 solution


Exercise 3.5 is a conditional inequality.
Giving condition is
1. f(x):[0,1]→(0,∞) that is
   f(x)>0 on [0,1]
2. f(x) is non-increasing, that is
   f(x) can be constant or decreasing
   on [0,1]

<a name="ch03c010">
From second point
Let m=0.1, n=0.6, we have n>m
then f(n) ≦ f(m) f(x) decreasing
(n-m)≧0 , f(m)-f(n)≧0, multiply
get (n-m)*(f(m)-f(n))≧0
change n to y, change m to x, get
    (y-x)*(f(x)-f(y))≧0 ---eqn.AJ002

for eqn.AJ002 we assumed the condition
     y>x     ---eqn.AJ003
  f(x)≧f(y) ---eqn.AJ004

<a name="ch03c011">
From first point, we get
    f(x)>0 on x∈[0,1] ---eqn.AJ005
  and
    f(y)>0 on y∈[0,1] ---eqn.AJ006

eqn.AJ002 and eqn.AJ005 and eqn.AJ006
multiply together get
<a name="ch03c012">
 eqn.AJ001 integrand be positive or
 zero, that is
   f(x)f(y)(y-x)(f(x)-f(y))≧0 ---eqn.AJ007

What left is dx and dy.
Both dx and dy are positive, because
integration range
  ∫[0,1]∫[0,1] require dx and dy start
from 0, end at 1, which is positive
increment.

In eqn.AJ001 all terms are positive/zero,
the integration result is positive/zero.

<a name="ch03c013">
expand eqn.AJ007
   f(x)f(y)(y-x)(f(x)-f(y))≧0 ---eqn.AJ007
to
   f(x)f(y)*[y*f(x)-y*f(y)-x*f(x)+x*f(y)]≧0
to
   [f(x)f(y)*y*f(x)-f(x)f(y)*y*f(y)
   -f(x)f(y)*x*f(x)+f(x)f(y)*x*f(y)]≧0
to //move all negative term to other side
    f(x)f(y)*y*f(x)+f(x)f(y)*x*f(y)
   ≧ ---eqn.AJ008
    f(x)f(y)*y*f(y)+f(x)f(y)*x*f(x)
2009-10-24-11-53 here
how to eliminate '+' from eqn.AJ008 ?

<a name="ch03c014">
2009-10-24-15-06 start
    f(x)f(y)*y*f(x)+f(x)f(y)*x*f(y)
   ≧ ---eqn.AJ008
    f(x)f(y)*y*f(y)+f(x)f(y)*x*f(x)

how to eliminate '+' from eqn.AJ008 ?
(eqn.AJ009 and eqn.AJ010 are two
 wondering equation and deleted.)

<a name="ch03c014a">　Index begin　Index this file
2009-10-25-10-30 start
eqn.AJ008 is left two terms ≧ right two terms
need change to left one term ≧ right one term
Because in target eqn.3.12, do not allow
two term addition. Problem is inequality
it is flexible, we can change two terms
change from   different1 + different2
change  to   sameTermLeft+sameTermLeft
which is   2*sameTermLeft
<a name="ch03c014b">
Do same thing for right side
change from   different3 + different4
change  to 2*sameTermRight
then inequality become
   2*sameTermLeft ≧ 2*sameTermRight
cancel '2*', satisfy target eqn.3.12

<a name="ch03c014c">
First step
change from   different1 + different2
change  to   sameTermLeft+sameTermLeft
need alter either one of different1 or
different2, the following discuss how
and caution.
2009-10-25-10-50 stop

<a name="ch03c015">
2009-10-24-15-10 here
We used the following condition
     y>x     ---eqn.AJ003
  f(x)≧f(y) ---eqn.AJ004

if there is no clear border line and
 bring greater than side to smaller value
and/or
 bring   less  than side to bigger value
cause inequality uncertainty.

<a name="ch03c016">
if there is no clear border line and
do the reverse
 bring greater than side to bigger value
and/or
 bring   less  than side to smaller value
enlarge inequality gap, inequality stand
firmly.
<a name="ch03c017">
Next use enlarge inequality gap method
to remove addition in the equation.

A1  f(x)f(y)*y*f(x)+f(x)f(y)*y*f(x)
   ≧ //red term above greater than below. 9810241509
A2  f(x)f(y)*y*f(x)+f(x)f(y)*x*f(y)
   ≧ ---eqn.AJ008
A3  f(x)f(y)*y*f(y)+f(x)f(y)*x*f(x)
   ≧ //blue term above greater than below. 9810241510
A4  f(x)f(y)*x*f(y)+f(x)f(y)*x*f(x)

<a name="ch03c018">　Index begin　Index this file
eqn.AJ008 was A2≧A3, now enlarge gap
push A2 to greater A1
push A3 to smaller A4, result
A1 ≧ A2 ≧ A3 ≧ A4   , then
A1 ≧ A4 is true which is eqn.AJ011 below

A1  f(x)f(y)*y*f(x)+f(x)f(y)*y*f(x)
   ≧ ---eqn.AJ011
A4  f(x)f(y)*x*f(y)+f(x)f(y)*x*f(x)
2009-10-24-16-23 here

<a name="ch03c019">
A1 is 2*f(x)f(y)*y*f(x), removed '+'
A4 can not remove '+' because two terms 
in A4 are different "x*f(y)" and "x*f(x)"
A4 can not write as 2*...
From A3 to A4 reduce value, enlarge 
inequality gap, which is safe. Now we
need to change from A4 to A5
A4  f(x)f(y)*x*f(y)+f(x)f(y)*x*f(x)
A5  f(x)f(y)*x*f(x)+f(x)f(y)*x*f(x)
<a name="ch03c020">
A5 has two identical terms, has factor '2'
BUT
  f(x)≧f(y) ---eqn.AJ004
then A5 ≧ A4, A5 is closer to A2!!
Whether A1≧A5 is true? Now compare A1
with A5

<a name="ch03c021">
A1  f(x)f(y)*y*f(x)+f(x)f(y)*y*f(x)
   ?≧? 
A5  f(x)f(y)*x*f(x)+f(x)f(y)*x*f(x)
We have
     y>x     ---eqn.AJ003
then A1≧A5 is true.
A1 and A5 both remove factor 2, get

    f(x)f(y)*y*f(x)
   ≧ ---eqn.AJ012
    f(x)f(y)*x*f(x)

<a name="ch03c022">
integrate get
  ∫[0,1]{f(x)f(x)}dx * ∫[0,1]{f(y)*y}dy
   ≧ ---eqn.AJ013
  ∫[0,1]{xf(x)f(x)}dx * ∫[0,1]{f(y)}dy

move from numerator to denominator get
  ∫[0,1]{f(x)f(x)}dx / ∫[0,1]{f(y)}dy
   ≧ ---eqn.AJ014
  ∫[0,1]{xf(x)f(x)}dx / ∫[0,1]{f(y)*y}dy

<a name="ch03c023">
both ∫[0,1]{...}dx and ∫[0,1]{...}dy
integrate over whole range [0,1]
both x and y are dummy variable.
Change y to x, get page 48 eqn.3.12.
2009-10-24-16-37 stop



<a name="ch03c024">　Index begin　Index this file
2009-10-24-16-59 start
■　Exercise 3.6 problem statement
　　textbook page 49
（Monotonicity of the Product Defect）

Show that for a pair of monotone sequences
0≦a1≦a2≦...≦an and
0≦b1≦b2≦...≦bn the quantity defined by

<a name="ch03c026">
2009-10-24-17-11 here
are also monotone nondecreasing. 
Specifically, show that for each integer
n=0,1,... one has Dn≦Dn+1.
2009-10-24-17-14 here




<a name="ch03c027">
2009-10-24-17-26 start
■　Exercise 3.6 hint
　　textbook page 238

One expand and factors
  Dn+1-Dn

<a name="ch03c030">
2009-10-24-17-43 here
According to Mitrinovic (1970, p.206) this
elegant observation is due to R.R. Janic.
The interaction between order relation and
quadratic inequality is developed more 
extensively in Chapter 5.
2009-10-24-17-46 stop






<a name="ch03c031">　Index begin　Index this file
2009-10-24-17-48 start
■　Exercise 3.6 solution


Expand Dn+1 as following

<a name="ch03c034">
2009-10-24-18-26 here
where red terms are Dn.
Blue terms cancel "-an+1bn+1"

Black terms and blue terms in eqn.AJ019
and eqn.AJ020 are Dn+1-Dn
which is same as eqn.AJ015.

<a name="ch03c035">
In eqn.AJ015 'n' in "nan+1bn+1" is 
absorbed in eqn.AJ016 right side term.
Summation sum n times, need n "an+1bn+1"

eqn.AJ016 right side summation "an+1bn+1"
is constant, because variable is 'j' and
"an+1bn+1" do not have 'j' factor.

Similarly, eqn.AJ016 left side summation 
"-bn+1" is constant.

<a name="ch03c036">
From eqn.AJ016 to eqn.AJ017 is simple
factor.

Given condition is
 an+1 ≧ aj (j＜n+1)
and
 bn+1 ≧ bj (j＜n+1)
eqn.AJ017 is non negative. Then the start
point 
  Dn+1 - Dn ≧ 0
is true. Problem solved.
2009-10-24-18-53 stop



<a name="ch03c037">　Index begin　Index this file
2009-10-24-18-58 start
■　Exercise 3.7 problem statement
　　textbook page 49
（The Four-Letter Identity via Polarization）

For any real numbers aj, bj, sj, tj,
1≦j≦n, there is an identity due 
independently to Binet and Cauchy 
which states that

<a name="ch03c040">
2009-10-24-19-14 here
This generalizes Lagrange's identity, 
as one can check by setting sj=bj and
tj=aj, but it is much more informative
to know that the Cauchy-Binet identity 
may be obtained as a corollary of the 
much simpler result of Lagrange.

<a name="ch03c041">
In fact, the passage is quite straight-
forward, provided one knows how to 
exploit the polarization transformation
  f(u) → {f(u+v)-f(u-v)}/4 ---eqn.AJ023
This transformation carries the function
u→u2 into the two-variable function
 (u,v)→uv, and it is devilishly effective
at morphing identities with squares into 
new ones where the squares are replaced by
products.

<a name="ch03c042">
To see how this works, check that the
four-variable identity follows from the
two-variable Lagrange identity after two
sequential polarizations. To keep your 
calculation tidy, you may want to use 
the shorthand

 |α β| = αδ-βγ ---eqn.3.14
 |γ δ|

<a name="ch03c043">
and the easy verified identity that follows
from the definition (3.14)

 |α+α' β| =  |α β| + |α' β|  ---eqn.3.15
 |γ+γ' δ|    |γ δ|   |γ' δ|

This shorthand recalls the notation for
the determinant of a two-by-two matrix.
but to solve this problem one does not 
need to know more about determinant than
the two self-evident relations (3.14)
and (3.15)
2009-10-24-19-33 stop





<a name="ch03c044">　Index begin　Index this file
2009-10-24-20-29 start
■　Exercise 3.7 hint
　　textbook page 239
In the suggested shorthand, Lagrange's
identity can be written as

2009-10-24-21-19 here
which is the shorthand version of the
target identity.
2009-10-24-21-20 stop




<a name="ch03c047">
■　Exercise 3.7 solution


Not familiar with polarization 
transformation.
LiuHH skip Exercise 3.7 solution
2009-10-24-21-35 stop



<a name="ch03c048">　Index begin　Index this file
2009-10-25-11-00
■　Product of two sequence-sum

In exercise 3.4, equation 3.11 we see
if in eqn.3.11 set x=1, then eqn.3.11
is product of two sequence-sum
In exercise 3.8, equation 3.17 we use
this product. Here write product of two
sequence-sum in explicit form. Assume
sequence {a} = {a1,a2,a3,a4,a5}
sequence {b} = {b1,b2,b3,b4,b5}
SUM{ai} = a1+a2+a3+a4+a5 ---eqn.AJ027
SUM{bj} = b1+b2+b3+b4+b5 ---eqn.AJ028
<a name="ch03c049">
Multiply out get
SUM{ai}*SUM{bj}= ---eqn.AJ029
 a1*b1+a1*b2+a1*b3+a1*b4+a1*b5
+a2*b1+a2*b2+a2*b3+a2*b4+a2*b5
+a3*b1+a3*b2+a3*b3+a3*b4+a3*b5
+a4*b1+a4*b2+a4*b3+a4*b4+a4*b5
+a5*b1+a5*b2+a5*b3+a5*b4+a5*b5
where red color terms are equal index 
terms. 
<a name="ch03c050">
They can be written as
 a1*b1+a2*b2+a3*b3+a4*b4+a5*b5
or as
 SUM[p=1,5]{apbp} ---eqn.AJ030
Black terms in eqn.AJ029 are cross-index
terms. 
We observe that am*bn = an*bm

<a name="ch03c050a">
2009-10-28-12-15 make up start
if not sequence self product, then 
[[
We observe that am*bn = an*bm
]]
is not true !!

<a name="ch03c050b">
A necessary condition is next
For a-sequence = b-sequence case, that is
for a-sequence * a-sequence case,
(equation 3.11 use a-sequence * a-sequence
 and use b-sequence * b-sequence)
2009-10-28-12-24 make up stop

<a name="ch03c051">
eqn.AJ029 can be written as
//require a-sequence = b-sequence
SUM{ai}*SUM{bj} = eqn.AJ029 = 
  SUM[p=1,5]{apbp}
+2*[a2*b1] //upper triangle merge 
+2*[a3*b1+a3*b2] //to lower triangle
+2*[a4*b1+a4*b2+a4*b3] //generate "2*"
+2*[a5*b1+a5*b2+a5*b3+a5*b4] ---eqn.AJ031
<a name="ch03c052">
Let us see colume one in eqn.AJ031,
//require a-sequence = b-sequence
it is Col1 = b1*SUM[k=2,5]{ak} ---eqn.AJ032
same  Col2 = b2*SUM[k=3,5]{ak} ---eqn.AJ033
same  Col3 = b3*SUM[k=4,5]{ak} ---eqn.AJ034
same  Col4 = b4*SUM[k=5,5]{ak} ---eqn.AJ035

<a name="ch03c053">　Index begin　Index this file
SUM{ai}*SUM{bj} = eqn.AJ029 =
//require a-sequence = b-sequence
  SUM[p=1,5]{apbp}
+2*b1*SUM[k=2,5]{ak}
+2*b2*SUM[k=3,5]{ak}
+2*b3*SUM[k=4,5]{ak}
+2*b4*SUM[k=5,5]{ak} ---eqn.AJ036

<a name="ch03c054">
We can write second summation for bm
as next
//require a-sequence = b-sequence
SUM{ai}*SUM{bj} = eqn.AJ029 = 
  SUM[p=1,5]{apbp} ---eqn.AJ037
+2*SUM[m=1,5-1]{bm*SUM[k=m+1,5](ak)}

<a name="ch03c055">
Final equation is
//require a-sequence = b-sequence
  SUM{ai}*SUM{bj} =  ---eqn.AJ038
  SUM[p=1,5]{apbp} + 2*SUM[1≦m＜k≦5]{bm*ak}
Above is for 5-term sequence, for general
n-term sequence, change 5 to n in eqn.AJ038

<a name="ch03c057">
eqn.AJ039 is the expression used in 
eqn.3.11 (in eqn.3.11 change x to one)
eqn.AJ090 is the expression will be 
used in eqn.3.17
2009-10-25-11-53 stop



<a name="ch03c058">　Index begin　Index this file
2009-10-25-12-19 start
■　Exercise 3.8 problem statement
　　textbook page 50
（Milne and Gauges of Proportionality）

We have seen that the term

<a name="ch03c059">
2009-10-25-12-34 here
provides a natural measure of propor-
tionality for the pair of vectors 
(a1,a2,...,an) and (b1,b2,...,bn). 
But one can think of other measures of 
proportionality that are just as reasonable.
For example, if we restrict our attention
to vectors of positive terms, then one
might equally well use the self-normalized
sum

<a name="ch03c061">
2009-10-25-12-45 here
Develop an identity containing R that will
permit you to prove the inequality of
E.A. Milne

<a name="ch03c063">
2009-10-25-12-58 here
Next, use your identity to show that one
has equality in the bound (3.17) if and 
only if the vectors (a1,a2,...,an) 
and (b1,b2,...,bn) are proportional.
Incidentally, the bound (3.17) was 
introduced by Milne in 1925 to help 
explain the biases inherent in certain 
measurement of stellar radiation.
2009-10-25-13-02 stop






<a name="ch03c064">　Index begin　Index this file
2009-10-25-13-52 start
■　Exercise 3.8 hint
　　textbook page 239

After expanding the two products, one
sees that the difference of the left-
hand side and the right-hand side of
Milne's inequality (3.17) can be
written as a symmetric sum

<a name="ch03c066">
2009-10-25-14-07 here
When each summand is put over the
denominator (ai+bi)(aj+bj), the 
numerator may be simplified, and one
finds that the difference coincides
with the definition (3.16) of R.
2009-10-25-14-10 stop







<a name="ch03c067">　Index begin　Index this file
2009-10-28-12-58 start
When write eqn.AJ039, LiuHH follow eqn.3.11
and ignored the necessary condition that
is eqn.3.11 is for
  sequence_a multiply sequence_a itself
However in eqn.AJ039, LiuHH used two
different named sequence. 
On 2009-10-28-12-13 during proofread,
find that
[[
We observe that am*bn = an*bm
]]
is wrong. Because am*bn = an*bm is not
a given condition, how can I use it?
am*bn = an*bm is true only
if sequence_a = sequence_b !

On 2009-10-28-12-54 add eqn.AJ090
and modify the following 
Exercise 3.8 solution.
2009-10-28-13-10 stop

<a name="ch03c067a">
2009-10-25-14-39 start
■　Exercise 3.8 solution


Our main tool is eqn.AJ039
Our main tool is eqn.AJ090
Product of two sequence-sum
In eqn.3.17, we shall apply
eqn.AJ090 twice.

<a name="ch03c068">
First expand eqn.3.17 right side, 
greater than side G.

<a name="ch03c070">
2009-10-25-15-00 here
When carry out G－L, above red terms
cancel out.
Black terms in eqn.AJ042 and eqn.AJ043
match eqn.AJ041 exactly.


index k and m are dummy variable,
change k to i, change m to j

'2*' in eqn.AJ042 let aibj (was akbm)
expand twice. It become eqn.AJ041
aibj+ajbi

<a name="ch03c071">
'2*' in eqn.AJ043 let
 ajbj(ai+bi)/(aj+bj)
expand twice. It become eqn.AJ041
－ajbj(ai+bi)/(aj+bj)
－aibi(aj+bj)/(ai+bi)


Up to here G－L = eqn.AJ041 is verified.
2009-10-25-15-17 here

<a name="ch03c072">　Index begin　Index this file
Next in eqn.AJ041 do
reduction of fractions to a common 
denominator.

It is easy to see common denominator 
is (ai+bi)*(aj+bj) ---eqn.AJ044

<a name="ch03c073">
Next calculate numerator part

numerator =  // below ---eqn.AJ045
 aibj(ai+bi)*(aj+bj)
+ajbi(ai+bi)*(aj+bj)
-ajbj(ai+bi)*(ai+bi)
-aibi(aj+bj)*(aj+bj)
 =   // below ---eqn.AJ046
 aibj(ai*aj+ai*bj+biaj+bibj)
+ajbi(ai*aj+ai*bj+biaj+bibj)
-ajbj(ai*ai+ai*bi+biai+bibi)
-aibi(aj*aj+aj*bj+bjaj+bjbj)
<a name="ch03c074">
 =  // below ---eqn.AJ047
 aibjaiaj+aibjaibj+aibjbiaj+aibjbibj
+ajbiaiaj+ajbiaibj+ajbibiaj+ajbibibj
-ajbjaiai-ajbjaibi-ajbjbiai-ajbjbibi
-aibiajaj-aibiajbj-aibibjaj-aibibjbj
 =  // below ---eqn.AJ048
 aiaiajbj+aiaibjbj+aiajbibj+aibibjbj
+aiajajbi+aiajbibj+ajajbibi+ajbibibj
-aiaiajbj-aiajbibj-aiajbibj-ajbibibj
-aiajajbi-aiajbibj-aiajbibj-aibibjbj
<a name="ch03c075">
 =  // below ---eqn.AJ049
+aiaibjbj
+ajajbibi
-2aiajbibj

[[
+aiaibjbj (+1)
+ajajbibi (+1)
 aiajbibj (+1+1-1-1-1-1)
 aibibjbj (+1-1)
 aiaiajbj (+1-1)
 aiajajbi (+1-1)
 ajbibibj (+1-1)
]]
2009-10-25-15-54 here

<a name="ch03c076">
numerator =  // below ---eqn.AJ050
  +aiaibjbj+ajajbibi-2aiajbibj
 =(aibj－ajbi)2

Now numerator in eqn.AJ050 is same as
numerator in eqn.3.16

common denominator eqn.AJ044
 (ai+bi)*(aj+bj)
is same as denominator in eqn.3.16 

<a name="ch03c077">　Index begin　Index this file
The only difference is factor 1/2 used
in eqn.3.16 

We can say DEFINE R=(G-L)/2

Exercise 3.8 hint said
[[
the 
numerator may be simplified, and one
finds that the difference coincides
with the definition (3.16) of R.
]]
Problem solved.
2009-10-25-16-21 stop

2009-10-25-18-59 done proofread
2009-10-25-19-11 done spelling check
2009-10-28-13-55 second proofread to here




<a name="ch03c078">
2009-10-26-12-57
■　Why one is not a prime number?

On 2009-09-15, LiuHH uploaded
Prime number and prime decomposition
http://freeman2.com/jsprime2.htm

If uncheck the checkbox at
[[
NOT output to box1, just update internal data. Check [v] here. 
]]
and click [Find Prime] button,
jsprime2.htm build prime number list.
Output to Box 1.

<a name="ch03c079">
At first LiuHH list 1 as first prime 
number, later read online pages and
learned that one is not a prime. 
LiuHH changed code, delete 1 from list.
But, why?

Reading "Yearning for the Impossible"
by John Stillwell, ISBN 1-56881-254-x
page 158 say (meaning, not exact copy)
<a name="ch03c080">
[[
Any integer has prime decomposition,
this prime decomposition is unique.
If we allow one to be a prime number
then 10=2*5 prime decomposition is 
not unique any more, since
  10=2*5=2*5*1=2*5*1*1 ... etc
are all prime decomposition. We want
keep prime decomposition uniqueness,
and demand that one NOT be a prime.
]]
one is NOT a prime and one is NOT a 
non-prime either.
2009-10-26-13-14 stop




<a name="ch04a001">　Index begin　Index this file
2009-10-26-13-30 start
■■Chapter 04: On Geometry and Sums of Squares

■　Distance definition
Distance is an important concept in both
physics and mathematics. Given two points
x=(x1,x2,...,xd) and
y=(y1,y2,...,yd)
<a name="ch04a002">
distance between point x and point y is
ρ(x,y).
Math expression in 2-dimensional case is
  ρ(x,y)=√[(y1-x1)2 + (y2-x2)2] ---eqn.AJ051
Math expression in d-dimensional case is
  ρ(x,y)=√[(y1-x1)2 + (y2-x2)2
          + ... + (yd-xd)2] ---eqn.AJ052

<a name="ch04a003">
If we have no choice, and distance IS 
the expression of eqn.AJ052, we say 
eqn.AJ052 is a distance theorem.

On the other hand,
If we have several choices, and decide
to use eqn.AJ052 as distance formula,
in this case we say 
eqn.AJ052 is a distance definition.

<a name="ch04a004">
Either theorem-view or definition-view
have supporter. If we view distance as
least resistance, then eqn.AJ052 is a 
distance definition.

<a name="ch04a005">
eqn.AJ052 root on Pythagoras theorem.
If we say eqn.AJ052 is defined, that 
mean we have a few distance definition
choices, we choose eqn.AJ052 as our
distance definition. In this case,
Pythagoras theorem become unimportant.
Because We can switch to other distance
formula easily.

<a name="ch04a006">
Either theorem-view or definition-view
we need to make sure that distance in
eqn.AJ052 has shortest value, that is
given any third point z, the distance
ρ(x,z)+ρ(z,y) can not be smaller than
ρ(x,y). In other words we require
  ρ(x,z)+ρ(z,y)≧ρ(x,y) ---eqn.AJ053
stand firmly.
eqn.AJ053 is named triangle inequality.

<a name="ch04a007">
An important thing about eqn.AJ052 is
that how eqn.AJ052 lead us to view
higher dimensional world?
Will eqn.AJ052 give us a higher dimension
picture which comply with our 2D and 3D
experience? or eqn.AJ052 give us a higher
dimension picture contradict with our
2D and 3D experience?
2009-10-26-14-14 here

<a name="ch04a008">
2009-10-26-16-28 start
Please see fig.4.1 and click "Plot5" 
button. There are four unit radius ball
(radius=1) put one in each quadrant and
stay as close to (0,0) as possible. 
Space inside four balls has a smaller
red ball just tangent to all four balls.

<a name="ch04a009">
Each unit ball center at (1,1) or (-1,1)
or (-1,-1) or (1,-1). 
Distance from either center to (0,0) is
 d2=sqrt(1*1+1*1)=sqrt(2) ---eqn.AJ054
Inner tangent ball radius is
 r2= sqrt(2)-1 ---eqn.AJ055
where 2 coincide with two dimension.

<a name="ch04a010">
Now turn to 3-D case, although no drawing,
it is easy to figure out that
Distance from either 3-D unit ball center
to xyz coordinate system center (0,0,0) is
 d3=sqrt(1*1+1*1+1*1)=sqrt(3) ---eqn.AJ056
Inner tangent ball radius is
 r3= sqrt(3)-1 ---eqn.AJ057
where 3 coincide with three dimension.

<a name="ch04a011">
Same reasoning, push to higher dimension
space, the general equation is
Distance from either n-D unit ball center
to n-dimensional coordinate system center
 (0,0,0, ... ,0) is
 dn=sqrt(n) ---eqn.AJ058
Inner tangent ball radius is
 rn= sqrt(n)-1 ---eqn.AJ059

<a name="ch04a012">
Any one object this reasoning? If no one
object above analysis, we will see weird
thing happen.

Just above our 3D space, in 4-D space
 r4= sqrt(4)-1 = 2-1 = 1 ---eqn.AJ060
Inner tangent ball is also unit radius!

<a name="ch04a013">　Index begin　Index this file
■　ride space shuttle goto 9-D space
Let us ride space shuttle goto 9-D space
we find
 r9= sqrt(9)-1 = 3-1 = 2 ---eqn.AJ061
Inner tangent ball radius is two !!

When dimension go higher and higher, 
inner tangent ball radius is bigger 
and bigger.

<a name="ch04a014">
n-D space, inner tangent ball radius is
 rn= sqrt(n)-1
assume n is big enough, we can ignore 1
then for large n
 rn ≈ sqrt(n) ---eqn.AJ062
rn is ball radius. Ball volume is
proportional to radius self multiply
n times,
<a name="ch04a015">
 vn ≈∝ [sqrt(n)]^n = n^(n/2) ---eqn.AJ063
function x^x grow much faster than e^x
function e^x, base number e no change,
function x^x, base number x also grow!
2009-10-26-17-00 stop

<a name="ch04a016">
How to call f(x)=x^x ?
is it pile number?
same thing x pile higher and higher.
2009-10-26-17-09 stop

<a name="ch04a017">　Index begin　Index this file
2009-10-26-20-12 start
■　Distance properties
Distance is an important concept. We
need a mathematics specification for
distance. Let ρ(x,y) be distance 
function. x and y are two points in 
space. Let S be n dimensional space
where points x and y are defined.

<a name="ch04a018">
We require
1. distance be positive or zero. That
   is distance is not a vector.
   distance can not have negative
   value. Math expression is
   ρ(x,y)≧0 for all x, y in S ---eqn.AJ064

<a name="ch04a019">
2. distance be zero, if two points x
   and y coincide. Math expression is
   ρ(x,y)=0 if and only if x=y ---eqn.AJ065

<a name="ch04a020">
3. distance is not a vector, distance
   from x to y and distance from y to
   x are same value. Math expression
   is
   ρ(x,y)=ρ(y,x) for all x,y in S ---eqn.AJ066

<a name="ch04a021">
4. distance between two points x and y
   have shortest length value. any
   third points can not reduce this
   shortest length value.
   Math expression is
   ρ(x,y)≦ρ(x,z)+ρ(z,y) ---eqn.AJ067
     for all x,y,z in S

A pair (S,ρ) with above four properties
is called metric space.

<a name="ch04a022">　Index begin　Index this file
2009-10-26-20-43 here
■　Triangle Inequality for Euclidean
　　Distance
Above four distance properties, first
three properties are easy to understand.
But fourth property is not as evident.
We need to prove eqn.AJ067.
eqn.AJ067 is called Triangle Inequality.

<a name="ch04a023">
Let ρ be a function with two input 
variables, each is defined in
d-dimensional space. ρ output a single
value (real number) We write as
 ρ : Rd*Rd→R ---eqn.AJ068
Function ρ(x,y) is defined by
  ρ(x,y)=√[(y1-x1)2 + (y2-x2)2
  + ... + (yd-xd)2] ---eqn.4.2 (AJ052 too)
<a name="ch04a024">
Above are given condition.
Below ask to prove
   ρ(x,y)≦ρ(x,z)+ρ(z,y) ---eqn.4.3 (AJ067 too)
     for all x,y,z in Rd

<a name="ch04a025">
Key point to prove eqn.4.3 is 
translation property.
Let us goto eqn.4.2, if point x and y
both move same amount w, then x become
x+w, y become y+w. The distance between
x+w and y+w is same as distance between
x and y. Because eqn.4.2 carry out the
difference between x and y, same amount
w cancel in this difference operation.
2009-10-26-21-05 here

<a name="ch04a026">
We want to prove
   ρ(x,y) ?≦? ρ(x,z)+ρ(z,y) ---eqn.4.3
Since distance is relative concept. We
can assign x to be 0 that is (0,0,...,0)
to simplify our work, then eqn.4.3
become
   ρ(0,y) ?≦? ρ(0,z)+ρ(z,y) ---eqn.AJ069
Here, ρ(0,y) and ρ(0,z) are both simpler.
How about ρ(z,y)? can we make it simpler?
Translation property say:
ρ(z,y) and ρ(z-z,y-z) are same thing.
Please see Draw 608. eqn.AJ069 become
   ρ(0,y) ?≦? ρ(0,z)+ρ(0,y-z) ---eqn.AJ070
eqn.AJ070 is not in its simplest form, we 
can still do one more transform, make it
simpler. Assign y to be u+v, assign z=u.
u,v and x,y,z are all vectors in same 
dimensional space.
<a name="ch04a027">
target equation 4.3 become
   ρ(0,u+v) ?≦? ρ(0,u)+ρ(0,v) ---eqn.AJ071
eqn.AJ071 wait for proving. 
Because distance is non-negative, we can
square eqn.AJ071 and use eqn.4.2, we find

<a name="ch04a031">
2009-10-26-21-56 here
Up to here, do we have any question
about eqn.AJ074 ? NO !! No question
at all ! eqn.AJ074 is our old friend
Cauchy's Inequality !

eqn.AJ074 is true, then the starting
eqn.4.3 is true.

This problem link Triangle Inequality
with Cauchy's Inequality. Both are
equivalent.
2009-10-26-22-00 stop

<a name="ch04a032">　Index begin　Index this file
2009-10-27-12-15 start
■　Norm definition

Distance function ρ(x,y) is defined by
  ρ(x,y)=√[(y1-x1)2 + (y2-x2)2
  + ... + (yd-xd)2] ---eqn.4.2 (AJ052 too)
In which, (y1-x1)2 can be generalized to
  (y1-x1)*(n1-m1) or in short as u1*v1
This observation remind us the standard
inner product formula
　　〈u,v〉＝u1*v1＋u2*v2＋...＋ud*vd ---eqn.AJ075
<a name="ch04a033">
Distance function ρ(x,y) can be written
in shorter version as
  ρ(x,y)=〈y-x,y-x〉1/2 ---eqn.AJ076
We understand that both x and y are
vectors in d-dimensional space and
  〈y-x,y-x〉=(y1-x1)2 + (y2-x2)2
  + ... + (yd-xd)2 ---eqn.AJ077

<a name="ch04a034">
eqn.AJ077 can be defined as the length
square of vector y-x. Length of a vector
is called norm of the vector. 
Norm should have the following properties.
1. ∥v∥=0 if and only if v=0 ---eqn.AJ078
   left regular '0' is a pure number.
   right bold 0 is a vector [0,0,...,0]
<a name="ch04a035">
2. ∥αv∥=|α|∥v∥ for all real number α ---eqn.AJ079
   if v=[1,2,3], if α=5, eqn.AJ079 says
   ∥[5*1,5*2,5*3]∥=5*∥[1,2,3]∥
3. ∥u+v∥≦∥u∥+∥v∥ ---eqn.AJ080
   for u and v in same dimensional
   vector space V.

<a name="ch04a036">
If V is a vector space and ∥.∥ is a
norm on V, then (V,∥.∥) is called a
normed linear space.
"Linear" come from  2. ∥αv∥=|α|∥v∥
because α at left side is power one
and α at right side is also power one
2009-10-27-12-57 here

<a name="ch04a037">
v is a vector v=[v1,v2,...,vd]
Inner product
 〈v,v〉＝v1*v1＋v2*v2＋...＋vd*vd ---eqn.AJ081
convert vector [v1,v2,...,vd] to a 
number (length square).
A norm operation ∥v∥ also convert 
vector v to a number (length).

<a name="ch04a038">
Inner product and norm operation are
related. If one is given, the other
is derived.

<a name="ch04a039">
If (V,〈.,.〉) is inner product space 
and is given, then
  ∥v∥=〈v,v〉1/2 ---eqn.AJ082
define a norm on V.
eqn.AJ082 right side is given, left
side is derived. So,
to each given inner product space
(V,〈.,.〉) we can find a related 
normed linear space  (V,∥.∥)

<a name="ch04a040">
On the other hand,
If (V,∥.∥) is a normed linear space
and is given, then 
  ρ(x,y)=∥x-y∥ ---eqn.AJ083
define a metric on V.
Therefore to each given normed linear 
space (V,∥.∥) we can find a related
metric space (V,ρ〈.,.〉)
(metric space = distance measurable
                space)
2009-10-27-13-27 stop

<a name="ch04a041">
2009-10-27-18-18 start
Figure 4.2 is in Textbook page 58.
There are two points X and V. This
is 2-D problem, 
X has coordinate (Xx, Xy)
V has coordinate (Vx, Vy)
X represent a line end, the other end
is O=(0,0).
V represent a point.
P(V) is projection point of V on OX.
Point V is called 'eye'  (by LiuHH)
Point P is called 'foot' (by LiuHH)
<a name="ch04a042">
From 2009-01-28 to 2009-01-31 LiuHH 
wrote a program eyefoot2.htm
http://freeman2.com/eyefoot2.htm
eyefoot2.htm solve foot coordinate
for 2D, 3D up to 12 dimension. The
equation used in eyefoot2.htm is
textbook page 56 line -1 eqn.4.5

<a name="ch04a043">
To solve projection foot point require
straight line go through (0,0) ? If I
have an arbitrary line not go through
(0,0) how do I find projection foot
coordinate?

<a name="ch04a044">
Equation 4.5 based on the assumption
that line go through (0,0), for a line
not go through (0,0) we can parallel
translate line so line do go through
(0,0). This part is easy, we need to
pay attention to two other things for
a correct solution.
<a name="ch04a045">
First, eye point must move same amount
       as line did.
Second, after find projection foot,
      parallel move line, eye point
      and foot point back to what line
      was.
Otherwise, the answer is wrong. Same
reason apply to 3D to higher dimension
space problem.

<a name="ch04a046">　Index begin　Index this file
2009-10-27-18-47 here
■　Projection formula defined
For each v and each x≠[0]=(0,0,...,0)
in d-dimensional space, let P(v) be
the point on the line L={tx:t∈real}
(t is pure number. x, L are vectors)
that is closest to v. Show that one 
has

<a name="ch04a048">
2009-10-27-19-03 here
How to read eqn.4.5 ?
eqn.4.5 is a vector equation,
where x is a vector and x-vector pass
coordinate system original point (0,0)
or (0,0,...,0) for higher dimension 
problem.
〈x,v〉 is a pure number, so does 〈x,x〉
〈x,v〉/〈x,x〉 is a pure number.
〈x,v〉/〈x,x〉 normalize x to be unit 1
〈x,v〉 is cosine between vectors x, v.
<a name="ch04a049">
cosine means PROJECTION of v on x.
This proportional ratio 〈x,v〉/〈x,x〉
multiply to x , get x*{〈x,v〉/〈x,x〉}
Which is the coordinate of projection
point. It is our answer. 
Answer is a vector too.
2009-10-27-19-18 here

<a name="ch04a050">　Index begin　Index this file
2009-10-27-19-24 start
■　Projection formula proved
How to prove eqn.4.5 ?
Please see figure 4.2 again.
point Q is an arbitrary point on line
OX. Distance VQ is greater than VP.
We can set up a function for distance
VQ, and minimize this function to get
VP.
<a name="ch04a051">
Distance VQ squared is
  ρ2(v,tx)=〈v-tx,v-tx〉---eqn.AJ084
in eqn.AJ084,
tx is Q point coordinate,
 v is V point coordinate,
v-tx is distance from Q to V.
both v and x are vectors,
both v and x are given,
t is a pure number, t is a variable.
optimal top is unknown.
v-top*x is shortest distance .
Let us multiply out eqn.AJ084 get

<a name="ch04a052">
  〈v-tx,v-tx〉＝〈v,v〉-2t〈v,x〉+t2〈x,x〉---eqn.AJ085

<a name="ch04a055">
2009-10-27-20-07 here
eqn.4.6 is squared-distance between V, Q.
t varies, then Q moves on line OX. v and
x are both constant. The minimum distance
occur when 
  t=〈v,x〉/〈x,x〉 ---eqn.AJ088
t term in eqn.4.6 vanish. Achieve minimum
value. (if t term not vanish, t term
squared has value greater than zero)

eqn.AJ088 is exactly used in eqn.4.5
Projection formula eqn.4.5 is proved.
2009-10-27-20-18 stop

<a name="ch04a056">
2009-10-27-22-10 start
Projection equation 4.5 is designed 
for line pass through (0,0).
If line NOT pass through (0,0), what
it will be? or what trouble we get?
Please see next equation.

<a name="ch04a058">
2009-10-27-22-17 here
Pass two points define a line. If line OX
not pass (0,0) it must pass another point
let us say the other point is D. Projection
foot coordinate will be eqn.AJ089. If we 
expand eqn.AJ089, result is much complicate
than eqn.4.5.

<a name="ch04a059">
Another application, we start from eqn.4.3
and write it to equivalent form eqn.AJ071
Compare eqn.AJ071 with eqn.4.3, you will 
find how warm a zero is ! Zero let us save
a lot of work !
2009-10-27-22-31 stop

2009-10-28-18-29 done proofread
2009-10-28-19-10 done spelling check

<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