<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="ch04b001">　Index begin　Index this file
2009-10-29-13-10 start
■　Cauchy's inequality as by-product
Let us re-visit eqn.4.6 .
eqn.4.6 has variable t to second power,
eqn.4.6 is a quadratic curve, represent
point V (fixed) to point Q (moving on OX)
dustance_squared. Please see figure 4.2.

<a name="ch04b002">
The minimum distance occur when 
  t=〈v,x〉/〈x,x〉 ---eqn.AJ088
t term in eqn.4.6 vanish. In this case,
eqn.4.6 simplify to

<a name="ch04b003">
2009-10-29-13-34 here
eqn.4.7 right side no variable t.
eqn.4.7 left side has variable t, but
left side also has "min", "min" lock
t to a specific value eqn.AJ088.
Over all, eqn.4.7 is a constant, minimum
distance_squared between point V and Q.

<a name="ch04b004">
What can we get from eqn.4.7?
eqn.4.7 left side is minimum distance-
squared, which is a non-negative value.
(not allow negative distance). Then
eqn.4.7 right side is also non-negative.
eqn.4.7 right side denominator is positive.
eqn.4.7 right side numerator must be non-
negative. 
<a name="ch04b005">
That is
  〈v,v〉〈x,x〉－〈v,x〉2≧0 ---eqn.AK001
Re-write eqn.AK001 as
  〈v,x〉2≦〈v,v〉〈x,x〉
Re-write eqn.AK001 one step more
  〈v,x〉≦〈v,v〉1/2*〈x,x〉1/2 ---eqn.AK002
eqn.AK002 is Cauchy's inequality !!

<a name="ch04b006">
We start from proving projection formula
eqn.4.5, end up with prove Cauchy's
inequality as by-product ! Surprise !

Cauchy's inequality is basically a re-
statement of -1 ≦ cosθ ≦ +1 which is 
universally true, so we see Cauchy every
where, that is what it should be.

<a name="ch04b007">
eqn.4.7 let us know why the defect 
〈v,v〉〈x,x〉－〈v,x〉2 is not zero?
From figure 4.2 we get the answer,
defect rise because point V do not 
wait in line OX (for next service)

<a name="ch04b008">
The flip side story is that if we want 
〈v,v〉〈x,x〉－〈v,x〉2 = 0 ---eqn.AK003
if and only if point V step into the
waiting line OX. In this case line OV
have no component perpendicular to OX.
In other words, line OV and line OX are
same line, their difference is just a 
length ratio constant.
2009-10-29-14-12 here

<a name="ch04b009">　Index begin　Index this file
2009-10-29-14-15 start
■　Projection formula proved -- second method
We proved the minimum distance occur when 
  t=〈v,x〉/〈x,x〉 ---eqn.AJ088
We started from minimize the distance
equation eqn.AJ084
We can use different reason to prove the
same equation eqn.AJ088.
<a name="ch04b010">
Please see figure 4.2 again.
When minimum distance occurred, 
vector VP is perpendicular to vector OX
Not use minimization concept, 
use perpendicular concept, can we get 
same answer? Try.

<a name="ch04b011">
Point P (Q same) move on line OX.
Point P coordinate is t*x
t is a pure number, x is vector.

vector VP is from point P to point V
vector VP is vector v minus vector p
vector VP is  v - tx
<a name="ch04b012">
(remember as
  vector difference = end point minus start point
 when start point is [0]=(0,0, ... 0)
 then
  vector difference = end point minus [0]
  vector difference = end point
)

<a name="ch04b013">
vector OX is x - [0] = x
where [0] is coordinate (0,0, ... 0)
We deliberately choose line OX pass [0]
jus attempt to get this simplification
from x - d to x - [0] = x . Now 
vector x perpendicular to vector v - tx
<a name="ch04b014">
we write the mathematical expression
  〈x,v - tx〉=0  ---eqn.AK004
In eqn.AK004, '=0' mean two vectors are
perpendicular to each other. That is
 cos(90 degree)=0  ---eqn.AK005
Alert! eqn.AK004 use perpendicular concept
       Not use minimization concept,
       eqn.AK004 do not have t square, 
       do not have minimum value as goal.

<a name="ch04b015">
In eqn.AK004, t is only unknown, 
expand eqn.AK004 to
  〈x,v〉 - 〈x,tx〉=0 
(t is pure number, can be moved out
 of 〈.,.〉) or
  〈x,v〉 - t〈x,x〉=0 
and solve for t get
  t=〈v,x〉/〈x,x〉 ---eqn.AJ088

Projection formula proved -- second method
is done.
2009-10-29-14-40 stop

<a name="ch04b016">　Index begin　Index this file
2009-10-29-20-43 start
■　A bound for the product of two linear
　　forms. (textbook page 59, problem 4.4)

Show that for all real uj, vj and xj
1≦j≦n, one has the following upper bound
for a product of two linear forms:

<a name="ch04b018">
2009-10-29-21-01 here
Problem say "product of two linear forms"
eqn.4.8 left side is product of two sum.
Both should be linear form. We should
conclude immediately that
[uj]=[u1,u2,...,un]
and
[vj]=[v1,v2,...,vn]
are two constant sequences. and
[xj]=[x1,x2,...,xn]
is one variable sequence.

<a name="ch04b019">
What special point eqn.4.8 has? Left
side two summations each contain one
constant coefficient and one variable.
Right side of eqn.4.8 
constant [uj] and [vj] are
separated from variable [xj]

<a name="ch04b020">
First step, let us apply Cauchy Inequality
to eqn.4.8 left side two terms. We get

<a name="ch04b021">　Index begin　Index this file
2009-10-29-21-31 here
In eqn.AK006, red terms is one application
of Cauchy's inequality. Black terms is
second application. 

eqn.AK006 has √(∑x*x) twice. 
Re-write eqn.AK006 as next

<a name="ch04b023">
2009-10-29-21-41 here
Now compare eqn.4.8 with eqn.AK007
Both equation left side are the same.
Both equation right side right end 
term are the same.
Difference is { ... } term. Now write
only term in { ... } as below

<a name="ch04b026">
2009-10-29-22-10 here
eqn.4.8 and eqn.AK007 difference part
is eqn.AK008 and eqn.AK009 red text part.
The difference is a Cauchy inequality.
eqn.AK009 is smaller than eqn.AK008,
so eqn.4.8 is sharper than eqn.AK007.
2009-10-29-22-14 stop

<a name="ch04b026a">
2009-10-31-08-33 start
"sharper" mean inequality equation 
greater than side is closer to less
than side. Smaller gap, smaller error
and still keep correct inequality sign.
2009-10-31-08-35 stop

<a name="ch04b027">　Index begin　Index this file
2009-10-30-10-57 start
■　Reflection analysis
Before prove eqn.4.8, we need to know
more about projection and reflection.
Please see figure 4.3
Please click ==> Plot7 button.
Given line L=OX pass through coordinate
original point (0,0).
Given point V not on line OX.
<a name="ch04b028">
We can find V projection on OX, which
is point P on line OX. Point P must 
satisfy the following inequality
  ∥OP∥≦∥OV∥ ---eqn.4.9
eqn.4.9 is simply |cos α|≦1 .
cosine function is projection function!

<a name="ch04b029">
We already defined projection P(V) at
eqn.4.5 and proved projection formula
eqn.4.5 at eqn.4.6 by minimization
concept. Second proof use perpendicular
concept. 
Above proved projection formula eqn.4.5
but not prove ∥OP∥≦∥OV∥ . 
<a name="ch04b030">
Below use eqn.4.5 as our starting point
and show ∥OP∥≦∥OV∥.

<a name="ch04b031">
2009-10-30-11-27 here
In eqn.Ak010, left equal sign is 
application of eqn.4.5
Middle equal sign is expansion of 
middle normed term and cancel one 
∥x∥.
Right "less than equal" sign is 
application of Cauchy's inequality.
<a name="ch04b032">
Cauchy's inequality say
  |〈x,v〉| ≦ ∥x∥*∥v∥ ---eqn.Ak011
re-write eqn.Ak011 as 
  |〈x,v〉|/∥x∥ ≦ ∥v∥ ---eqn.Ak012
it is still Cauchy's inequality. We can
not just know Cauchy's front view, we
must also know Cauchy's side view.

<a name="ch04b033">
eqn.Ak010 achieved our goal,
  ∥OP∥≦∥OV∥ ---eqn.4.9
[ O is (0,0) and omitted in eqn.Ak010 ]
2009-10-30-11-37 stop

<a name="ch04b034">　Index begin　Index this file
2009-10-30-13-32 start
■　Equation for reflection R(V) of point V
In figure 4.3, we need find equation for
vector OR, which is reflection of point V
in the line OX.
point V coordinate is given,
point X coordinate is given,
Projection point P is given by eqn.4.5
<a name="ch04b035">
How to find reflection R(V) coordinate
from above given information?
Go to figure 4.3
Vector OR is combined by vector OP plus
vector PR
vector OP is eqn.4.5
vector PR is negative of vector PV
vector PV is vector OV minus vector OP,
<a name="ch04b036">
then vector PR is 
negative of vector OV minus vector OP
which is
vector OP minus vector OV = vector PR
Now
Vector OR = vector OP + vector PR
Vector OR = vector OP + vector OP - vector OV
Vector OR = 2*vector OP - vector OV
<a name="ch04b037">
Point O is (0,0), we can drop O and
write mathematics equation for
reflection R(V) of point V as
  R(V) = 2*P(V) - V ---eqn.Ak013
This is textbook page 60 line 12 equation.
In eqn.Ak013, independent variable is
point V.
Both projection P(V) and reflection R(V)
are dependent variables.

<a name="ch04b038">　Index begin　Index this file
■　Reflection has length preserve property
In figure 4.3 we can argue that
right triangle OPR and right triangle OPV
have common side OP, same right angle and
PV=PR, both right triangle are identical, 
so both hypotenuse are same length. Which 
prove reflection has length preserve property.

<a name="ch04b039">
Textbook use algebraic reasoning, instead 
of geometry reasoning. We follow textbook 
and do as following.
First show that OP dot product OP is same
as OP dot product OV.

<a name="ch04b042">
2009-10-30-14-18 here
Both eqn.AK014 and eqn.AK015 right side
are the same, so we conclude that
  〈P(v),P(v)〉=〈P(v),v〉 ---eqn.AK016

Our goal is
Reflection has length preserve property
we need eqn.AK016 to achieve our goal.

<a name="ch04b043">　Index begin　Index this file
Reflection length square is
  ∥R(V)∥2 = ∥R(V)∥∥R(V)∥
  = // apply eqn.Ak013
  〈2P(v)-v,2P(v)-v〉
  = // expand inner product
  4〈P(v),P(v)〉-4〈P(v),v〉+〈v,v〉
  = // apply eqn.AK016, cancel left two terms
  0+〈v,v〉
  = // same term inner product is norm square
  ∥v∥2

<a name="ch04b044">
Up to here, we proved
  ∥R(V)∥2 = ∥v∥2 ---eqn.AK017
Because length can not be a negative
number, eqn.AK017 is same as
  ∥R(V)∥ = ∥v∥ ---eqn.AK018
eqn.AK018 is an important result, it
state that 
Reflection has length preserve property.
2009-10-30-14-35 here

<a name="ch04b045">　Index begin　Index this file
■　Prove eqn.4.8
Now let us back to textbook page 59, 
problem 4.4
To prove eqn.4.8 do as following.

First of all, above analysis work
with one coefficient sequence V and
one variable sequence X.

<a name="ch04b046">
However, problem 4.4 use two coefficient 
sequence U, V and one variable sequence X.
The following, U and V are two different
points.

<a name="ch04b047">
We use Cauchy's inequality and use 
reflection length preserve property.
Start with reflection of U, R(U), 
inner product with V,
 Cauchy's inequality tell us
  〈R(u),v〉≦∥R(u)∥∥v∥
  =  // reflection R(u) length preserved
  ∥u∥∥v∥
The net result is textbook page 61
  〈R(u),v〉≦∥u∥∥v∥ ---eqn.4.11

<a name="ch04b048">
Next start from 〈R(u),v〉 again, but
use different equation,
  〈R(u),v〉
  = // eqn.Ak013
  〈2P(u)-u,v〉
  = // expansion
  2〈P(u),v〉-〈u,v〉
  = // eqn.4.5

<a name="ch04b050">
2009-10-30-15-19 here
In eqn.AK019 , left term is vector x
inner product (dot product) with
vector v, other coef. like 〈x,u〉
and ∥x∥2 are just pure number.
Finally get next equation.
  〈R(u),v〉= // ---eqn.AK020
  [2〈x,u〉〈x,v〉/∥x∥2]-〈u,v〉
eqn.AK020 is a pure number, 
inner product convert two vectors
to a pure number.

<a name="ch04b051">
Now combine eqn.4.11 with eqn.AK020
we find
  [2〈x,u〉〈x,v〉/∥x∥2]-〈u,v〉
   ≦∥u∥∥v∥ ---eqn.AK021
Re-arrange eqn.AK021 get
  〈x,u〉〈x,v〉  // ---eqn.AK022
  ≦[(〈u,v〉+∥u∥∥v∥)*∥x∥2]/2
<a name="ch04b052">
Please open second browser window and
goto eqn.4.8
Compare eqn.4.8 with eqn.AK022
they are SAME equation !!
textbook page 59, problem 4.4 solved.
2009-10-30-15-45 stop

<a name="ch04b053">　Index begin　Index this file
2009-10-31-16-56 start
■　The Light Cone Inequality
Textbook page 62 to 65 introduce
light cone inequality.

Light cone is a physics term used in
relativity theory. One important
theory is
In the universe, no particle can
gain speed faster than light speed.
<a name="ch04b054">
This theory tell us that not only we
can not travel faster than light speed,
we can not send signal faster than 
light speed either. When time and
space link together, we must make sure
that distance measurement not exceed
the distance traveled by light in
related time interval. 

<a name="ch04b055">
Let us put distance traveled by light 
and distance measurement together into
one sequence as following
define x_a to be
x_a=[t*c,x1,x2,...,xd] ---eqn.AK023A
define y_a to be
x_a=[u*c,y1,y2,...,yd] ---eqn.AK023B
where c=light speed≈3*10^8 meter/second
      t=u=time in seconds
      xi=yi=distance component in meter
<a name="ch04b056">
and define time-space array product
to be // alert ! not a common usage
  [x_a,y_a]=  ---eqn.AK024
  tc*uc+x1y1+x2y2+...,+xdyd
Whether eqn.AK024 is a convenient choice?
If measured distance =
   distance traveled by light
how can we tell from eqn.AK024 expression?
eqn.AK024 definition can not answer this 
question!

<a name="ch04b057">　Index begin　Index this file
■　Define Lorentz product
Dutch physicist Hendrick Antoon Lorentz
(1853-1928) (textbook page 62 line -7)
introduced one expression and answered
above question.
For time-space array
 x=(tc; x1,x2,...,xd)
 y=(uc; y1,y2,...,yd)
<a name="ch04b058">
define Lorentz product as next
 [x,y]=tc*uc－x1*y1－...－xd*yd ---eqn.4.14
compare with vector inner product
 〈x,y〉=＋x1*y1＋...＋xd*yd ---eqn.AK025
where c is light speed,
      t and u are both time.
<a name="ch04b059">
Please be alert about the difference
between eqn.4.14 and eqn.AK025.

eqn.4.14   has time factor t, u
eqn.AK025  no time factor
eqn.4.14   has length limit tc, uc
eqn.AK025  no length limit
eqn.4.14   assign length measurement
          be negative, tc be positive.
eqn.AK025 length measurement positive.
<a name="ch04b060">
Back to the question
If measured distance =
   distance traveled by light
how can we tell from Lorentz product?
Now set [x,y] to [x,x]
 [x,x]=tc*tc－x1*x1－...－xd*xd ---eqn.AK026
Answer:
If Lorentz product＞0 length measurement＜tc
If Lorentz product＝0 length measurement＝tc
If Lorentz product＜0 length measurement＞tc

<a name="ch04b061">　Index begin　Index this file
■　Light Cone Inequality formula
Based on Lorentz product definition
eqn.4.14, we find a
Generalized Cauchy Inequality which
use the following transformation
matrix (for 3-D case)
[tc,x1,x2,x3]*  ---eqn.AK027
[1  0  0  0][uc]
[0 -1  0  0][y1]
[0  0 -1  0][y2]
[0  0  0 -1][y3]
Compare!!
Regular Cauchy Inequality use matrix
expression which has all eigenvalues
to be one (no -1)

<a name="ch04b062">
eqn.AK027 and eqn.4.14 are same thing.
Because Lorentz product use a 
transformation matrix with negative
eigenvalue -1, -1, -1, then Cauchy's
inequality reverse greater then/less
 than direction.
that is 
Light Cone Inequality:
  √([x,x][y,y])≦[x,y] ---eqn.4.15 page 63
  √([a,a][b,b])≧[a,b] ---Cauchy's inequality

<a name="ch04b063">
Above two equations reverse inequality
direction and yet both correct? WHY !?

Lorentz product is assigned by human
   being.
Cauchy's inequality use natural
   arithmetic rule.

Above light cone inequality reasoning
is different from textbook.
LiuHH's reasoning could be WRONG !!

Please read textbook for better
explanation.
2009-10-31-18-05 stop



<a name="ch04b064">　Index begin　Index this file
2009-11-03-10-29 start
■　Sequence products
Assume we have 3-dimensional problem
sequence A = [a1,a2,a3]
sequence B = [b1,b2,b3]
Product of sequence A with sequence B
is defined to be
(a1+a2+a3)*(b1+b2+b3)
 =a1*(b1+b2+b3)
 +a2*(b1+b2+b3)
 +a3*(b1+b2+b3)
<a name="ch04b065">
That is
  SeqProd =  ---eqn.AK028
  a1*b1+a1*b2+a1*b3
 +a2*b1+a2*b2+a2*b3
 +a3*b1+a3*b2+a3*b3
We can write eqn.AK028 in matrix form
  SeqProd =  ---eqn.AK029
 [a1,a2,a3]*
 [1  1  1][b1]
 [1  1  1][b2]
 [1  1  1][b3]
<a name="ch04b066">
If we multiply out eqn.AK029, we get 
eqn.AK028. First step multiply is
  SeqProd =  ---eqn.AK030
 [a1,a2,a3]*
 [b1+b2+b3]
 [b1+b2+b3]
 [b1+b2+b3]
in eqn.AK030, [a1,a2,a3] is row vector
and
 [b1+b2+b3]
 [b1+b2+b3]
 [b1+b2+b3]
is a column vector. Second step multiply
get eqn.AK028.

<a name="ch04b067">　Index begin　Index this file

■　Sequence is vector components
eqn.AK028 is one method of two sequence
multiplication. 
If the sequence is vector components in 
Cartesian coordinates, there is cosine
function involved in multiplication. 

Two vector components in same coordinate
have cos(0) coefficient in the product.
For example x axis component
 ax*bx*cos(0)
which is
 ax*bx*1

<a name="ch04b068">
Two vector components in different coordinate
have cos(90) coefficient in the product.
For example x,y axis component
 ax*by*cos(90)
which is
 ax*bx*0
In the case
 sequence elements = vector components
we need different product rule.

<a name="ch04b069">　Index begin　Index this file
■　Vector inner product
In vector form it is
  V1DotV2 =  ---eqn.AK031
 [a1,a2,a3]*
 [cos(0)  cos(90) cos(90)][b1]
 [cos(90) cos(0)  cos(90)][b2]
 [cos(90) cos(90) cos(0) ][b3]
Simplify to
  V1DotV2 =  ---eqn.AK032
 [a1,a2,a3]*
 [1  0  0][b1]
 [0  1  0][b2]
 [0  0  1][b3]
<a name="ch04b070">
When multiply out, we get
  V1DotV2 =  ---eqn.AK033
 a1b1+a2b2+a3b3
eqn.AK033 is valid only if 
sequence elements = vector components

eqn.AK033 is called vector inner product,
        also called vector  dot  product.
(there is vector cross product rule)

vector inner product answer is a pure 
number, not a vector any more.
(vector cross product result is another
 vector)

<a name="ch04b071">
■　Vector length
If vector A dot product with itself,
the answer is
  V1DotV1 =  ---eqn.AK034
 a1a1+a2a2+a3a3
eqn.AK034 is a wonderful result !
Because in 3-dimensional geometry, 
point A distance_square from original
point (0,0,0) is exact a1a1+a2a2+a3a3.
(please review Pythagoras theorem)
which represent vector length_squared.
Since vector start at (0,0,0), and
vector end at (a1,a2,a3)

<a name="ch04b072">
We can define distance of point A from
(0,0,0) as
 RNorm = sqrt(V1DotV1) that is
 RNorm = √[a1a1+a2a2+a3a3] ---eqn.AK035
Here "Norm" represent vector length.

<a name="ch04b073">
We can define distance of point A from
point B as DistBA
 = sqrt((V1-V2)Dot(V1-V2)) ---eqn.AK036
Explicit form of eqn.AK036 is
 DistBA = ---eqn.AK037
 √[(a1-b1)2+(a2-b2)2+(a3-b3)2]
If in eqn.AK037 we set b1=b2=b3=0 for
 (0,0,0) coordinate original point,
then eqn.AK037 reduce to eqn.AK035
2009-11-03-11-28 stop

<a name="ch04b074">　Index begin　Index this file

2009-11-03-13-06 start
■　Move to complex number
Above we defined vector length
 RNorm = √[a1a1+a2a2+a3a3] ---eqn.AK035
eqn.AK035 is valid for real number 
sequence. Assume real sequence R1
 R1=[r1,r2,r3]
 R1=[1, 3, 5] ---eqn.AK038
Substitute eqn.AK038 to eqn.AK035 for
a numerical result, we find
 RNorm = √[1*1+3*3+5*5] = √(35) = 5.916079783099616
Vector [1, 3, 5] length is 5.916079783099616

<a name="ch04b075">
We always want to extend mathematical
calculation to more general case. Let
us try complex number.
Assume complex sequence C1
 C1=[c1,c2,c3]
 C1=[1+2i, 3+4i, 5+6i] ---eqn.AK039
Substitute eqn.AK039 to eqn.AK035 for
a complex test run, we find
 CNorm = √[(1+2i)*(1+2i)+(3+4i)*(3+4i)+(5+6i)*(5+6i)]
 CNorm = √[(-3+4i)+(-7+24i)+(-11+60i)]
 CNorm = √[-21+88i]
<a name="ch04b076">
 CNorm = 5.893682627418461+i*7.465620865857989
We get sequence C1 norm is a complex
number, but, a norm is length, which 
should be a real non-negative number!

What can we do?
Do we accept complex norm is complex?
Do we insist complex norm is real?
Mathematician choose complex norm is real.
Then how can we get real out of complex?

<a name="ch04b077">　Index begin　Index this file
■　Complex inner product
Real number inner product has real-world
application. That is in a given coordinate
system, vector components are all real
number.

How about complex inner product? Do we
see any coordinate system use complex 
number as component value? No! It can
not be so!, because one complex number
is actually combined by one real and
one imaginary.

<a name="ch04b078">
We define complex inner product rule 
just for academic reason.

How can we get real out of complex?
The answer is complex conjugate. For
example 1+2i multiply by its conjugate
1-2i get (1+2i)*(1-2i)=1*1-(2i)*(2i)
=1-(-4)=5
[ recall (a+b)*(a-b)=a*a-b*b ; i*i=-1]
<a name="ch04b079">
Complex number a+bi, its conjugate is
a-bi, NOT the other way, NOT -a+bi
To find norm for a complex sequence,
we need to modify the real equation
 RNorm = √[a1a1+a2a2+a3a3] ---eqn.AK035
to
 CNorm = √[c1c1j+c2c2j+c3c3j] ---eqn.AK040
Here use super_j for conJugate.

<a name="ch04b080">
Complex norm come from complex inner
product. 
For complex sequence C=[c1,c2,c3]
and complex sequence D=[d1,d2,d3]
Define Complex inner product as 
following
 〈c,d〉 = c1d1j+c2d2j+c3d3j ---eqn.AK041
When we talk about norm, set D sequence
to C sequence, then √〈c,c〉 eqn.AK041 
become CNorm eqn.AK040.
Alert,  if c≠d, then 〈c,d〉 is complex,
however if c＝d, then 〈c,d〉 is real.

<a name="ch04b081">
Now let us test
C1=[1+2i, 3+4i, 5+6i] ---eqn.AK039
with CNorm eqn.AK040, we get
 CNorm eqn.AK040 =
 CNorm = √[(1+2i)*(1-2i)+(3+4i)*(3-4i)+(5+6i)*(5-6i)]
 CNorm = √[(5+0i)+(25+0i)+(61+0i)]
 CNorm = √[(91+0i)]
 CNorm = 9.539392014169456+0i
Use complex conjugate for complex 
inner product definition eqn.AK041
we get real number norm 9.539 for 
a complex sequence!

<a name="ch04b082">
If complex norm were complex, 
we can NOT introduce inequality to
complex number. 
Because 1*1=1>0, i*i=-1<0 ! square
 is not positive any more, we can
 not compare complex number.

Now complex norm is real number, we
CAN introduce inequality to complex
number. We get more job to do.
Shouldn’t we inequalitist be happy?
2009-11-03-14-10 stop

<a name="ch04b083">　Index begin　Index this file

2009-11-03-15-50 start
■　Complex conjugate? which one?
We said define Complex inner product
as following
 CDot1 = √[c1d1j+c2d2j+c3d3j] ---eqn.AK041
how about switch conjugate to following
 CDot2 = √[c1jd1+c2jd2+c3jd3] ---eqn.AK042
AK042 take conjugate for 1st sequence C
AK041 take conjugate for 2nd sequence D
<a name="ch04b084">
When we find norm, we set first sequence
equal to second sequence. Then eqn.AK042
and eqn.AK041 make no difference. But
how about non-norm application? what is
the difference? 
If 1st sequence C is different from 2nd
sequence D, two answers are conjugate to
each other.

<a name="ch04b085">
OK, then 
who take conjugate for 1st sequence? (ans. physicists)
who take conjugate for 2nd sequence? (ans. mathematician)

2009-11-01-15-08 LiuHH accessed
http://mathstat.carleton.ca/~ckfong/C5.pdf

C5.pdf page 1 of 7 has
[[
<a name="ch04b086">
Remark: In (5.1),
 〈x,y〉 = √[x1y1j+x2y2j+x3y3j] ---eqn.(5.1)
 it is not clear why we prefer to take
 complex conjugates of components of y
 instead of components of x.
 Actually this is more or less due to
 the tradition of mathematics, rather
 than our preference.
 (Physicists have a different tradition!)
]]

<a name="ch04b087">
On 2009-11-02 LiuHH write code to
http://freeman2.com/complex2.htm
add complex array dot product function
cdotf(array1,array2)
In this function, LiuHH set complex 
conjugate to array2. Because when take
norm, array1=array2, let array1 be 
original unchanged array, and alter
array2 to complex conjugate. This
arrangement is more natural.
2009-11-03-16-20 here

<a name="ch04b088">　Index begin　Index this file
2009-11-03-16-40 start
■　Compare results
on
2009-05-30-16-26 LiuHH accessed
http://www.math.umd.edu/~hck/Hermitian.pdf
save as complex_inner_product_math.umd.edu_~hck_Hermitian.pdf
page 1 of 3 has 
[[
<a name="ch04b089">
The vector space Cn has a standard
 inner product, 〈u, v〉 = u*v. Recall
 u* = uT,j so another formula is
〈u, v〉 = uT,j v. So for example
〈[1 + i, 2 − i], [3 − 2i, 1 + i]〉
 = (1 − i)(3 − 2i) + (2 + i)(1 + i)
 = 1 − 5i + 1 + 3i = 2 − 2i.
You can compute this in Matlab as
 dot([1+i, 2-i], [3-2i, 1+i]).
]]
<a name="ch04b090">
LiuHH note:
uT=u transposed
uj=u complex conjugate
uT,j has both.

<a name="ch04b091">
in 
http://freeman2.com/complex2.htm#box03
In Box3 paste next
[[
chkbxA01.checked=chkbxA03.checked=0;
chkbxA02.checked=1 //use ';' separate numbers
array1='1+i;2-i'
array2='3-2i;1+i'
cdotf(array1,array2) 
]]
<a name="ch04b092">
then click
"test box3 command, output to box4"
button, output is
[[
cdotf(array1,array2) 
2,2
]]
"2,2" is "2+2i"

<a name="ch04b093">
http://www.math.umd.edu/~hck/Hermitian.pdf
answer is "2-2i"

complex2.htm  use a_sequence * b_seq_conjugate
Hermitian.pdf use a_seq_conjugate * b_sequence
two answer are conjugate to each other.

2009-11-03-16-40 start
<a name="ch04b094">
2009-11-03-20-05 stop !!
During this four hours, spend about 
three hour debug why complex2.htm 
get wrong answer?
Finally found that input formula
contains unicode minus sign
"2-i" is actually "2&#8722;i"
ANSI '-' and unicode '−' are different
complex2.htm can not process "2&#8722;i"
2009-11-03-20-10 stop 

<a name="ch04b095">　Index begin　Index this file
2009-11-04-14-22 start
■　Complex inner product example
Let V be a complex vector space, for
example Cd for dimension=d.
A three-dimension complex vector space
C3 has element like
  V1=[1+2i, 3+4i, 5-6i]
The set of complex valued continuous 
function on [0,1] also form a complex
vector space.
2009-11-04-14-29 here (why require [0,1]?)

<a name="ch04b096">
We require function domain to be [0,1]
that is a normalized domain. Any domain
[a,b] can be transformed to [0,1] 
A numerical example of complex valued
continuous function on [0,1] is next

<a name="ch04b097">
2009-11-01-16-50 LiuHH accessed
http://www.fen.bilkent.edu.tr/~franz/la05/compl.pdf
save as complex_inner_product_fen.bilkent.edu.tr_compl.pdf
page 2/2 example
Following is a copy from compl.pdf
[[
We can also define a dot product on function
spaces such as the vector space V of all 
polynomials with complex coefficients by
putting

2009-11-04-15-15 here
 ＝ 1 - (1+3i)/2 +i = (1-i)/2 ---eqn.AK046

<a name="ch04b099">
and

]]
Above is a copy from compl.pdf
2009-11-04-15-26 here
Above is numerical example of complex 
valued continuous function on [0,1]

<a name="ch04b100">　Index begin　Index this file
■　Complex inner product five rules
A function on VxV defined by the mapping
 (a,b) → 〈a,b〉 ∈ C
is a complex inner product.
We say that 
 (V, 〈.,.〉) is a complex inner product space
provided that the pair (V, 〈.,.〉) has five
basic properties. The first four properties
parallel to those required in real inner 
product space.

<a name="ch04b101">
p.1, 〈v,v〉≧0 for all v∈V
p.2, 〈v,v〉＝0 if and only if v＝0
p.3, 〈αv,w〉＝α〈v,w〉 for all α∈C and v,w∈V
p.4, 〈u,v+w〉＝〈u,v〉+〈u,w〉 for all u,v,w∈V
We assume next p.5 property
p.5, 〈v,w〉＝〈w,v〉j for all v,w∈V
<a name="ch04b102">
textbook use over-bar for complex conjugate.
But text editor can not create over-bar.
In this file, use superj as complex conjugate
of super.
2009-11-04-15-53 here

<a name="ch04b103">
V is a complex vector space.
 v∈V  means  vector v (lowercase) belong to
complex vector space V (uppercase)
Elements of vector v are all complex numbers.

p.3, 〈αv,w〉＝α〈v,w〉 for all α∈C and v,w∈V
say α is a complex number (α∈C)
 v,w are two complex vectors (v,w∈V)
Please pay attention to p.3
p.3 require 〈αv,w〉＝α〈v,w〉
not require 〈αv,w〉＝αj〈v,w〉

<a name="ch04b104">　Index begin　Index this file

■　Sesquelinear map
http://mathstat.carleton.ca/~ckfong/C5.pdf
page 2 of 7 introduce "sesquelinear map"
[[
Definition. By an inner product on a
complex vector space we mean a device of
assigning to each pair of vectors x and
 y a complex number denoted by 〈x, y〉,
 such that
<a name="ch04b105">
the following conditions are satisfied:
(C1) 〈x, y〉 ≥ 0, and //LiuHH: 〈x, x〉≥0
     〈x, x〉 = 0 if and only if x = 0.
(C2) 〈y, x〉 = 〈x, y〉j.

<a name="ch04b106">
(C3) The inner product is a
     "sesquelinear map", i.e.
〈a1x1 + a2x2, y〉 = a1〈x1, y〉 + a2〈x2, y〉 ---eqn.AK048
〈x, b1y1 + b2y2〉 = b1j〈x, y1〉 + b2j〈x, y2〉. ---eqn.AK049
(Actually the second identity of (C3)
 above is the consequence of the first,
 together with (C2).
]]
2009-11-04-16-11 stop

<a name="ch04b107">　Index begin　Index this file
Attention: Graph Cauchy inequality,
complex number, drawing program
2009-11-05-10-03 start
■　Cauchy-Schwarz for a complex inner
　　product
(textbook page 65, problem 4.6)
Let v,w be complex numbers in complex 
inner product space (V,〈.,.〉). 
Show that
  |〈v,w〉|≦〈v,v〉1/2〈w,w〉1/2 ---eqn.4.18
Furthermore, show that v≠0 then one has
equality in the bound (4.18) if and only
if w=λv for some complex number λ.

<a name="ch04b108">
We can not compare complex number 
directly. However for complex number 
real properties, such as norm or dot
product, we can do comparison. We have
two complex numbers v and w. Let us 
create a new complex number v-w and 
find its dot product.
  0≦〈v-w,v-w〉 //cdotf is real number≧0
   = //expand 〈v-w,v-w〉 get next line
   〈v,v〉+〈w,w〉-〈v,w〉-〈w,v〉
<a name="ch04b109">
   　 //rule 5 applied, special to complex
   = //superj is complex conjugate
   〈v,v〉+〈w,w〉-{〈v,w〉+〈v,w〉j}
   = //〈v,w〉+〈v,w〉j = 2Re〈v,w〉
   〈v,v〉+〈w,w〉-2Re〈v,w〉 ---eqn.AK050
we get
  0≦〈v,v〉+〈w,w〉-2Re〈v,w〉 ---eqn.AK051
re-write as
  Re〈v,w〉≦[〈v,v〉+〈w,w〉]/2 ---eqn.4.19
We use Magic adding to multiplying,
normalization. Define
   vn=v/〈v,v〉1/2 ---eqn.AK052
   wn=w/〈w,w〉1/2 ---eqn.AK053
Where supern indicate normalization.
<a name="ch04b110">
In eqn.4.19 use eqn.AK052 and eqn.AK053
Magic change eqn.4.19 to
  Re〈v,w〉≦〈v,v〉1/2〈w,w〉1/2 ---eqn.4.20
Our goal is eqn.4.18 (=eqn.AK054)
  |〈v,w〉|≦〈v,v〉1/2〈w,w〉1/2 ---eqn.AK054
eqn.4.20 and eqn.AK054 are very different.
Because Re〈v,w〉≦|〈v,w〉| and |〈v,w〉|≧0
but Re〈v,w〉 ＜＝＞ 0 all possibility exist.
eqn.4.20 can not imply eqn.AK054 !!
2009-11-05-10-50 here

<a name="ch04b111">　Index begin　Index this file
■　Saved by a self improvement
The following is a copy of textbook 
page 66.
The saving grace of inequality (4.20) is
that it is of the self-improving kind.
If we exploit the generality appropriately,
we can derive an apparently stronger
inequality.

<a name="ch04b112">
If we write 〈v,w〉=ρeiθ ---eqn.AK055
with ρ＞0 and if we set u=e-iθv ---eqn.AK056
then the properties of the complex inner 
product give us the identity
  〈u,u〉=〈v,v〉 ---eqn.AK057
and
  〈u,w〉=Re〈u,w〉=|〈v,w〉| ---eqn.AK058
<a name="ch04b113">
so the real part bound (4.20) for the pair
u and w gives us
  |〈v,w〉|=Re〈u,w〉≦〈u,u〉1/2〈w,w〉1/2
          =〈v,v〉1/2〈w,w〉1/2 ---eqn.AK059
The out side terms yield the complex 
Cauchy-Schwarz inequality in the precisely
form we expected. So the bound (4.20) was
strong enough after all.

Above is a copy of textbook page 66. 
All v_over_tilde change to 'u'.
2009-11-05-11-07 stop

<a name="ch04b114">　Index begin　Index this file
2009-11-05-13-44 start
■　Numerical value help understand
LiuHH read above 
"Saved by a self improvement" section
several times, but not really understand.
2009-11-04 LiuHH wrote code lines to
calculate test points, draw on fig2.5
drawing board, read figure, this method
really help LiuHH understand. Do you
want to do same thing?. Code is next
<a name="ch04b115">
[[
v='1+2i';//define v with a real complex number
w='3-4i';//define w
vw=cdotf(v,w);//vw has length rho, angle theta
pvw=cpolr(vw);//show up rho and theta
pu=cpolr(v);  //change v to polar
pu[1]=pu[1]-pvw[1]; //rotate reverse theta degree
u=cxryi(pu); //create new v => u, |u|=|v|
cdotf(u,u)   //page 66, line -10
cdotf(v,v)   //expect cdotf(v,v)=cdotf(u,u)
uw=cdotf(u,w)//u dot w get real only
uw //expect uw=cgetr(uw)=cabsf(vw)
cgetr(uw)//uw is real only, cgetr get real
cabsf(vw) //9811041957
v+';'+w+';'+vw+';'+u+';'+uw //five points for plotting
]]
<a name="ch04b116">
Paste above 14 lines to
http://freeman2.com/complex2.htm#box03
then click
"test box3 command, output to box4"
button. Goto box 4 copy last line
1+2i;3-4i;-5,10;1.3416407864998738,-1.788854381999832;11.180339887498949,-8.881784197001252e-16
it is five points use ';' as separator. 
<a name="ch04b117">
Paste five complex numbers to
http://freeman2.com/tute0014.htm#fig2.5
box 1, In
"Box 1 complex separator: [ ]newline [ ]';' [ ]',' [ ]tab"
click ';' checkbox. then 
click "plot3" button.
Figure for five points will show up
at one pace below.
Point A is point v
Point B is point w, pivot point, not move.
Point C is point vw = cdotf(v,w)
Point D is point u=e-iθv ---eqn.AK056
Point E is point uw=〈u,w〉=Re〈u,w〉=|〈v,w〉| ---eqn.AK058

<a name="ch04b118">
Key point in this figure is that
We steer point A to D (rotate v to u)
then point C follow rotate to point E.
point C to point E eliminate imaginary
part, whole complex number C=〈v,w〉
become real E=〈u,w〉. 
"make 〈v,w〉 real" is our main goal.

<a name="ch04b119">　Index begin　Index this file

■　Explain the whole story
　　as following.
Original problem is
Let v,w be complex numbers, find 
  |〈v,w〉|≦〈v,v〉1/2〈w,w〉1/2 ---eqn.4.18
but we find
  Re〈v,w〉≦〈v,v〉1/2〈w,w〉1/2 ---eqn.4.20
where Re〈v,w〉 is real part of
  vw = cdotf(v,w) = -5+10i
<a name="ch04b120">
that is
  Re〈v,w〉=-5
We want |-5+10i| = 11.180
we do not want Re〈v,w〉=-5
We want the whole vector (-5+10i) length
we do not want vector projection on x axis.

<a name="ch04b121">
〈v,w〉=-5+10i, which has phase angle
2.0344439357957027 rad.
If we reverse rotate 〈v,w〉=-5+10i an angle
of negative of 2.0344439357957027 rad
then vector 〈v,w〉=-5+10i rotate to
vector 11.180+0i , new vector's projection
on x axis is its whole length 11.180 !!
"make 〈v,w〉 real" is our main goal.

<a name="ch04b122">
Next, how can we rotate 〈v,w〉?
let s=〈v,w〉, write complex numbers v,w,s 
in polar form (p_something = polar form)
 v='1+2i' ==> pv=2.236*exp(1.107)
 w='3-4i' ==> pw=5*exp(-0.9273)
 s=-5+10i ==> ps=11.180*exp(2.034)
complex conjugate of w is
 wj='3+4i' ==> pwj=5*exp(+0.9273)
<a name="ch04b123">
then
 s=〈v,w〉
 ps_length = pv_length*pwj_length=2.236*5=11.180
 ps_phase  = pv_phase +pwj_phase =1.107+0.9273=2.034
We want to rotate s, both pv and pwj
contribute s rotation. Now we hold 
w unchange, pwj_phase keep constant, 
<a name="ch04b124">
rotate v (pv) only, then 
 ps_phase=pv_phase+CONSTANT_pwj_phase 
formula tell us that 
 ps_phase_change  = pv_phase_change
We want ps_phase change to zero rad.
so that vector s has no imaginary
component. 
<a name="ch04b125">
We choose phase_change to be -ps_phase,
 then
 ps_phase_change  = pv_phase_change
become
 ps_phase - ps_phase = pv_phase - ps_phase
get
 ps_zero_rad_phase = pv_phase - ps_phase

<a name="ch04b126">　Index begin　Index this file
■　Create second problem
With this understanding, next we create
second problem.
Second problem's answer is different
from first problem's answer, with
better and complete result.
Let w='3-4i'
define u replace v, such that
polar form of u = polar form of v - ps_phase
(v is given for first problem
 u is fabricated for second problem)

<a name="ch04b127">
Next three lines achieve this goal.
pu=cpolr(v);  //u start = v (polar form)
pu[1]=pu[1]-pvw[1]; //rotate reverse theta degree
u=cxryi(pu); //create new v => u, |u|=|v|

Explain above three lines as below.
pu=cpolr(v);  
is u start = v and it is in polar form.

<a name="ch04b128">
Next line is rotation - ps_phase rad.
pu[1]=pu[1]-pvw[1]; //'s' is 'vw', s=〈v,w〉
  //in polar expression pu[0]=vector length
  //and pu[1]=vector phase angle

after rotation, recover real+imag*i form
u=cxryi(pu); 
Now u value is 1.3416-1.7888i
which is point D in tute0014.htm#fig2.5

<a name="ch04b129">
We just found u and w is given, now 
we create 〈u,w〉, code line is
uw=cdotf(u,w)//u dot w get real only
uw //uw value is 11.180 + zero (e-16)
11.180339887498949,-8.881784197001252e-16

we know 〈u,w〉 has only real component.
  〈u,w〉=Re〈u,w〉=|〈v,w〉| ---eqn.AK058
is true,
left equality come from 〈u,w〉 has only 
real component
right equality come from vector 〈v,w〉
and vector 〈u,w〉 just rotate, their
vector length are the same value.

<a name="ch04b130">
Next see
  |〈v,w〉|=Re〈u,w〉≦〈u,u〉1/2〈w,w〉1/2
          =〈v,v〉1/2〈w,w〉1/2 ---eqn.AK059
eqn.AK059 first line equality come 
from above mentioned eqn.AK058.
eqn.AK059 first line inequality come
from Cauchy-Schwarz inequality. In 
this part Re〈u,w〉 is actually |〈u,w〉|
because 〈u,w〉 has no imaginary part.
"make 〈v,w〉 real 〈u,w〉" is our main
goal.

<a name="ch04b131">
eqn.AK059 first line to second line come
from the fact that u vector and v vector
are same length, they just rotate apart.
When take norm, both u and v have same
length.
eqn.AK059 beginning and ending form
  |〈v,w〉|≦〈v,v〉1/2〈w,w〉1/2 ---eqn.AK060
<a name="ch04b132">
This is our answer.
Both v and w are complex number. We
can not compare v and w for inequality.
Only their real property (norm, length)
can be compared.

Hope above analysis will help you 
understand.
2009-11-05-15-06 stop



<a name="ch04b133">
2009-11-06-14-39 start
Real number Cauchy inequality is
   〈a,b〉 ≦〈a,a〉1/2〈b,b〉1/2 ---eqn.AK061
Complex number Cauchy inequality is
  |〈v,w〉|≦〈v,v〉1/2〈w,w〉1/2 ---eqn.AK060
Can we write complex number Cauchy
inequality as
   〈v,w〉 ≦〈v,v〉1/2〈w,w〉1/2 ---eqn.AK062
?
eqn.AK061 not use absolute value,
why eqn.AK060 add absolute value?
Please think. 
Answer is in this page.
You need to dig it out ! 
2009-11-06-14-46 stop

<a name="ch04b134">
■　Cauchy inequality in one dimension
2009-11-07-10-31 start
On 2009-11-06-20-17 LiuHH review 
numerical example found 
Cauchy inequality is actually an
equality !! Why?
v='1+2i';
w='3-4i';
vw=cdotf(v,w)='-5+10i'
are not colinear,
why |-5+10i|=√(|1+2i|*|3-4i|)=11.1803
Above v='1+2i'; and w='3-4i'; represent
sequence V has just one element '1+2i'
sequence W has just one element '3-4i'

<a name="ch04b135">
This is complex Cauchy inequality.
Go to real　Cauchy inequality to find
hint.
If real sequence A=[2] has one element.
If real sequence B=[5] has one element.
This is one dimensional problem. In two
and multiple dimensional problem, we 
need pay attention to whether sequence
A and B are colinear? If A and B are
colinear, then Cauchy inequality become
equality.
<a name="ch04b136">
In one dimensional problem, how can A 
and B are not colinear? They have no
second dimension to become non-colinear.
If A=[2] and B=[5] , their Cauchy 
inequality is
 2*5 ≦ √(2*2) * √(5*5)
it is simply 10=10 !!
<a name="ch04b137">
For one dimensional problem, Cauchy's
inequality become equality (identity).

The following is a code for numerical
example of two-elements array complex
Cauchy inequality, at the end, output
shows inequality.
<a name="ch04b138">
Please go to
http://freeman2.com/complex2.htm#box03
In Box3 paste next
[[
//2009-11-07-10-26
v1='1+2i';//V-seq. 1st element
v2='2-3i';//V-seq. 2nd element
w1='3-4i';//W-seq. 1st element
w2='5-i'; //W-seq. 2nd element
v12=v1+';'+v2;//build V seq.
w12=w1+';'+w2;//build W seq.
vw=cdotf(v12,w12);//V seq. dot W seq.
pvw=cpolr(vw);//change vw to polar
pv1=cpolr(v1);//change v1 to polar
pv2=cpolr(v2);//change v2 to polar
pu1=pv1;//create U seq. 1st element
pu2=pv2;//create U seq. 2nd element
pu1[1]=pu1[1]-pvw[1];//rotate U element
pu2[1]=pu2[1]-pvw[1];//this is key point
u1=cxryi(pu1);//change polar U to x+iy
u2=cxryi(pu2);//
u12=u1+';'+u2;//build U seq. for cdotf()
uw=cdotf(u12,w12);//real due to rotation
vv=cdotf(v12,v12);//real due to self prod.
ww=cdotf(w12,w12);//real due to self prod.
vn=csqrt(vv);//sqrt of self dot is length
wn=csqrt(ww);//length of W sequence
vnwn=cmulf(vn,wn)//greater than side vn*wn
vnwn//print greater than side value
uw  //print less than side value
//this time no surprise, they are inequal
//2009-11-07-10-29
]]
<a name="ch04b139">
then click
"test box3 command, output to box4"
button, output is
[[
vnwn//print greater than side value
30.298514815086232,0
uw  //print less than side value
8.544003745317534,1.7763568394002505e-15
]]
<a name="ch04b140">
You can change
v1='1+2i';
v2='2-3i';
w1='3-4i';
w2='5-i'; 
value and check the result.

<a name="ch04b141">
Next two lines
pu1[1]=pu1[1]-pvw[1];//rotate U element
pu2[1]=pu2[1]-pvw[1];//this is key point
indicate, Explain the whole story
is true for two dimensional case and
multiple dimensional case.
2009-11-07-11-21 stop

　Index begin　Index this file



<a name="docB001">
2009-10-31-09-10 start
Program "A bound for the product of
 two linear forms" box 4 output has
the following meaning
Top four lines are four numbers.
Line #1 is eqn.4.8 less than side
 ∑ujxj∑vjxj
Line #2 is eqn.4.8 great than side
 {[∑ujvj+√(∑uj2∑vj2)]*∑xj2}/2
<a name="docB002">
Line #3 is eqn.AK007 great than side
 [√(∑uj2∑vj2)]*∑xj2
Line #4 is arbitrary/curious test value
 [∑ujvj]*∑xj2
<a name="docB003">
eqn.4.8 say #1≦#2 (also #2≦#3)
eqn.AK007 say #1≦#3
but, #1 ≦＝≧ #4 
no equation relate #1 and #4
2009-10-31-09-26 stop

2009-10-31-09-35 done proofread
2009-10-31-09-46 done spelling check

2009-11-06-15-15 done proofread
2009-11-06-15-42 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