<a name="docA01">
2009-06-08-19-10 start
This file http://freeman2.com/tute0007.htm
is Freeman's reading notes. Although Freeman
always keep correct view point. But Freeman's
capability is limited, plus no one proofread
this file. Then you can still find wrong view
point. When read, please put question mark as
often as possible. If you suspect any view
point wrong, please ask a math expert near by.
<a name="docA02">
This file is a note for reading inequality
book written by ProfessorJ. MichaelSteeleThe 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, this 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="ch00a001">Index
2009-06-09-10-09 start
■■Chapter zero: basic material
2009-06-11-16-24 start
Provide a chapter zero for basic material.
chapter zero theorem one
Square of any real number
greater than equal to zero.
This is a definition of product operation,
no need to proof.
<a name="ch00a002">
■ chapter zero theorem two
Two real number sequences
a1, a2, a3, ... an
b1, b2, b3, ... bn
absolute value of sum of element product
less than or equal to
sum of absolute value of element product
for example
<a name="ch00a003">
sequence 1: -0.65, 0.83, 0.97
sequence 2: 0.99, 0.07,-0.92
first step is sum of element product
-0.65*0.99 + 0.83*0.07 + 0.97*(-0.92)
= -1.4778
last step is to take absolute value 1.4778
compare with
sum of absolute value of element product
|-0.65*0.99|+|0.83*0.07|+|0.97*(-0.92)|
=1.594
here last step is sum.
<a name="ch00a004">
Above result indicate
last step take absolute value get smaller value,
first step take absolute value get greater value.
1.4778 ≦ 1.594
because
last step take absolute value allow positive
and negative elements cancel each other. Get
smaller sum. On the other hand
first step take absolute value, when come to
sum step, no cancellation. Get greater value.
This is easy to understand. If sequence has two
elements, its equation is
|a1*b1+a2*b2| ≦ |a1*b1|+|a2*b2|
Equation for arbitrary number of elements is
（2009-06-11-16-45 here）
<a name="ch00a006">
In the future, store all basic material in
chapter zero.
2009-06-11-17-00 stop
<a name="ch00a007">
2009-06-13-16-18 start
This file use the following notation:
Use I for integer, negative, zero, positive.
Use I0+ for non-negative integer
Use I+ for positive integer
<a name="ch00a008">
Use R for real number
Use Rn for n th demension real space
Use R0+ for non negative real
Use R+ for positive real
<a name="ch00a009">
Use Ra for rational number
Use Ri for irrational number
2009-06-13,16,27 here (future add more)
<a name="ch00a010">
Use C[a,b] for continuous function (C)
Domain for real function is
variable x : start from a<=x to x<=b
include end points.
x∈[a,b] indicate variable x stay within
[x=a, x=b] , "∈" is "inside of"
[a,b] means x can equal to left end point a
and x can equal to right end point b
(a,b] means x can not equal to left end point a
and x can equal to right end point b
[a,b) means x can equal to left end point a
and x can not equal to right end point b
(a,b) means x can not equal to left end point a
and x can not equal to right end point b
function f(x)= 1/x not allow x=0
function f(x)= 1/(x-1) not allow x=1
domain for variable x is (-∞,1)∪(1,+∞)
"∪" indicate union of two sections
(-∞,1)∪(1,+∞) means x can never reach -∞
x not allow to be 1, x can never reach +∞
2009-06-13-17-52 stop
<a name="ch01a001">Index
2009-06-09-10-40 start
■■Chapter 01: Starting with Cauchy
■ Cauchy inequality ~ n=1 and n=2
(equation for infinite sequence eq.1.7)
assume two real number sequences
a1, a2, a3, ... an
b1, b2, b3, ... bn
each elements can be positive/negative
the following inequality exist.
a1b1+a2b2+a3b3+ ... +anbn ≦
√（a12+a22+a32+ ... +an2）＊
√（b12+b22+b32+ ... +bn2） ---eq.AA01
<a name="ch01a002">
right side of inequality are sum of squares
then take square root, which is positive number.
If sequence has positive and negative numbers,
left side of inequality has cancellation, then
left side is smaller than right side. But if
sequence has all positive number, whether above
inequality eq.AA01 still hold? Because equation
is true for general case, we can start from
simple case, if it is not true for simple case,
then inequality eq.AA01 is not true in general.
(best case is conditional true)
Simplest case is one sequence one element
sequence a has one element a1
sequence b has one element b1<a name="ch01a003">
eq.AA01 become
a1b1 ？≦？ √（a12）＊√（b12） ---eq.AA02
"？≦？" indicate uncertain inequal relation
square and square root cancel. eq.AA02
left side and right side are the same.
If a1 and b1 have same sign, the result is
equality. If they have opposite sign, result
is inequality. Negative number less than
positive number, it has no value to discuss.
If both side are positive, then it has value
to study.
Below is second simple case
sequence a has two elements a1 and a2
sequence b has two elements b1 and b2<a name="ch01a004">
eq.AA01 become
a1b1+a2b2 ？≦？ √（a12+a22）＊√（b12+b22） ---eq.AA03
Square whole equation to eliminate the
square root, but if it is safe to square
whole equation?, for example -3＜+2 after
square, get 9 NOT＜ 4. This case, negative
less than positive is already inequal. No
need to square to find the inequality.
When both side are positive, square will
not get surprise. Square eq.AA03 get
<a name="ch01a005">
[a1b1+a2b2]*[a1b1+a2b2]
？≦？
（a12+a22）＊（b12+b22）
expand
a12b12+a22b22+2*a1b1a2b2
？≦？
a12b12+a22b22+a22b12+a12b22<a name="ch01a006">
cancel same term get
2*a1b1a2b2 ？≦？ a22b12+a12b22
move to same side
０ ？≦？ a22b12+a12b22 - 2*a1b1a2b2
right hand side is a perfect square
０ ？≦？ (a2b1 - a1b2)2 ---eq.AA04
<a name="ch01a007">
Real sequence operation get real, real square
is non-negative.
eq.AA04 is true, "？≦？" change to "≦"
０ ≦ (a2b1 - a1b2)2
Back to starting eq.AA03 is true.
(Next equation not use "？≦？")
a1b1+a2b2 ≦ √（a12+a22）＊√（b12+b22） ---eq.AA03
Up to here eq.AA01 is true for n=1 and n=2
Whether eq.AA01 is true for n>2?
2009-06-09-11-44 here
<a name="ch01a008">Index
■ Cauchy's Inequality～proof by induction
Below proof by induction
First assume eq.AA01 is true for n, next
find out if it is still true for n+1
a1b1+a2b2+a3b3+ ... +anbn +an+1bn+1
＝ //sum already true n term plus n+1 term
[a1b1+a2b2+a3b3+ ... +anbn] +an+1bn+1
//although it is n+1 term sum, view it as
//sum of two terms. assume n term is
//already true, apply it, get next "≦"
<a name="ch01a009">
≦ √（a12+a22+a32+ ... +an2）
＊√（b12+b22+b32+ ... +bn2） + an+1bn+1
＜＝√（a12+a22+a32+ ... +an2+an+12）
＊√（b12+b22+b32+ ... +bn2+bn+12）
above "＜＝" use two term sum (see next line)
c1d1+c2d2 ≦ √（c12+c22）＊√（d12+d22） ---eq.AA03
Up to here, proof by induction is done.
<a name="ch01a010">
The difficulty of above proof is to treat
√（a12+...+an2） as a single term c1
√（b12+...+bn2） as a single term d1
an+1 as c2
bn+1 as d2
and
<a name="ch01a011">
use two term sum formula for n+1 term sum.
although √（a12+...+an2） locate inside of √ .
But this term will get squared c12，remove √ ，
then sum with c22 achieve perfect result.
（a12+a22+a32+ ... +an2+an+12）
2009-06-09-12-28 stop
<a name="ch01a012">Index
■ Cauchy's Inequality～proof by
quadratic discriminant
2009-06-09-14-10 start
Above is textbook page 2 to 3, use
induction method.
Below is teacher Lai's inequality lecture
notes, proof by quadratic discriminant.
Next is access record
2009-05-20-21-37
http://web.tcfsh.tc.edu.tw/jflai/math4/%E4%B8%8D%E7%AD%89%E5%BC%8F.pdf
web.tcfsh.tc.edu.tw_980520a.pdf
<a name="ch01a013">
Assume two real number sequences
a1, a2, a3, ... an
b1, b2, b3, ... bn
every element can be positive or negative,
two sequences have the following relation
（a12+a22+a32+ ... +an2）＊
（b12+b22+b32+ ... +bn2）
≧
（a1b1+a2b2+a3b3+ ... +anbn）2 ---eq.AA05
<a name="ch01a014">
eq.AA05 and eq.AA01 are same thing
square eq.AA01 exchange left/right and reverse
inequal sign, get eq.AA05.
Define a quadratic equation f(x) as following
f(x)=(a1x-b1)2+(a2x-b2)2+.....+(anx-bn)2 ---eq.AA06A
expand squares
f(x)=
(a12*x*x-2*a1b1*x+b12)+(a22*x*x-2*a2b2*x+b22)
+.....+(an2*x*x-2*anbn*x+bn2) ---eq.AA06B
collect same power of x, get
f(x)=（a12+a22+a32+ ... +an2）*x*x
-2*(a1b1+a2b2+a3b3+ ... +anbn)*x
+（b12+b22+b32+ ... +bn2） ---eq.AA06C
<a name="ch01a015">
eq.AA06 is sum of real numbers, therefore f(x)≧0
and f(x) curve not below x axis. discriminant of
eq.AA07 ≦0 If quadratic equation is A*x*x+B*x+C=0
discriminant is B*B-4*A*C, next is discriminant
[-2*(a1b1+a2b2+a3b3+ ... +anbn)]2
-4*（a12+a22+a32+ ... +an2）
*（b12+b22+b32+ ... +bn2） ≦0 ---eq.AA07
eq.AA07 delete '4', move to same side, get
eq.AA05 Which is to be proved.
2009-06-09-14-39 here
<a name="ch01a016">Index
■ Cauchy's Inequality～proof by geometry
Above is teacher Lai's proof by quadratic
discriminant
Below is proof by geometry （Draw 602 ==602）
Here treat one number sequence as one vector.
Draw a 2D vector need two components. This
discussion is limited to two elements sequence.
Blue line: vector one coordinates (a1,a2)
to simplify analysis, let Blue line coincide
with x axis, then a2=0
Vector one length = √(a1*a1+a2*a2)= a1 ---eq.AA08
Red solid line: vector two coordinates (b1,b2)
Vector two length = √(b1*b1+b2*b2) ---eq.AA09
<a name="ch01a017">
Red dash line: Vector two component along
Vector one.
Red dash line length
= √(b1*b1+b2*b2)*cos(angle) ---eq.AA10
("angle" indicate angle from blue to red)
red solid length multiply by blue length =
√(b1*b1+b2*b2) * √(a1*a1+a2*a2)
red dash multiply by blue (dot product) =
a1*b1+a2*b2
Cauchy inequality say
Blue vector length multiply by red solid
vector length
≧
Blue vector length multiply by red dash
vector length
<a name="ch01a018">
that is
√(a1*a1+a2*a2) * √(b1*b1+b2*b2)
≧ a1*b1+a2*b2 ---eq.AA11
Because vector component is always shorter
than or equal to vector length itself.
Cauchy's inequality (2D problem) is true.
From geometry view point, Cauchy's inequality
is in fact a restatement of
-1 ≦ cos(angle) ≦ 1。
3D problem is the same. Two vectors in 3D
space is in fact a 2D problem. Because 2D
is a sub-space of 3D. Same reason apply to
higher dimension space.
Two vectors can form only 2D problem.
Please click the following "Draw 602" button.
2009-06-09-15-24 stop
<a name="ch01a020">
■ Graph 601
2009-06-09-15-29 start
Below is "Draw 601", this is text book
page 117 exercise 7.7
All graphic file "Draw 601" has detail
document to explain the meaning of wx??
//define curve 1 x=f1(t) and y=g1(t)
wx01="t";
wx02="exp(-t*t/2)";
Start from "Draw 602" document is omitted.
During draft work exercise 7.7 is only one
graph which use all five curves. Therefore
keep it as "Draw 601" of this file.
Graph parameter table allow at most five
curve definitions, please see "Modify 601"
After done graph, then Freeman understand
the meaning of exercise 7.7 Complicated
integration value can be estimated its
lower bound at domain left end function
value and first derivative value. Discuss
at future due time.
2009-06-09-15-40 stop
<a name="ch01a022">Index
2009-06-09-16-55 start
■ Addition change to multiplication
Many learned simple inequality in high
school such as
0≦(a-b)*(a-b) ---eq.AA12
after expansion, it becomes
0 ≦ a*a - 2*a*b + b*b
then
2*a*b ≦ a*a + b*b
If replace a with √a and replace b with
√b get the following
2*√a*√b ≦ √a*√a + √b*√b
that is
√(a*b) ≦ (a + b)/2 ---eq.AA13
Left hand side is geometric mean (GM)
Right hand side is arithmetic mean (AM)
<a name="ch01a023">
From right triangle ⊿ABC right angle
point B take height line to the opposite
side. Mark the foot as D,
then similar triangle side ratio give us
AD : BD = BD : DC
which is BD*BD = AD*DC
or BD = √(AD*DC)
BD is called geometric mean (GM) of AD
and DC.
Above sequence a has single element a
and sequence b has single element b
<a name="ch01a024">
Below sequence a has two element a1,a2
and sequence b has two element b1,b2
0 ≦ (a1-b1)2 + (a2-b2)2
after expansion, it becomes
0 ≦ (a12-2*a1b1+b12)+(a22-2*a2b2+b22)
then
2*(a1b1+a2b2) ≦ (a12+a22)+(b12+b22)
that is
a1b1+a2b2 ≦ 0.5*(a12+a22)+0.5*(b12+b22)
---eq.AA14
This is a result for two element sequence.
Multiple element sequence and infinite many
element sequence follow same pattern. Textbook
has infinite many element sequence equation.
Outer frame
0
1
;
Inner frame
0
1
If change text size
need change width.
Width of above equation
<a name="ch01a026">
Compare with Cauchy Inequality
（see eq.AA01）
k=∞
∑
k=1
a k
b k
≦
[
k=∞
∑
k=1
a k2
]
1/2
[
k=∞
∑
k=1
b k2
]
1/2
page 5, eq.1.7
width of above equation
<a name="ch01a027">
2009-06-09-18-00 here
Eq.1.6 and eq.1.7 are similar.
Left hand side are the same
(for positive sequences)
eq.1.6 right hand side are two terms addition.
eq.1.7 right hand side are two terms multiplication.
eq.1.6 right hand side term has coefficient 0.5
eq.1.7 right hand side term has power 0.5
Is it possible change one to another?
<a name="ch01a028">
■ Magic adding to multiplying, normalization
There is a magic process, change addition to
multiplication. That is textbook page 5 main
topic.
The method is normalization.
Use coefficient 0.5 plus coefficient 0.5 = 1
to eliminate addition process.
To simplify discussion, use two elements
sequence as an example.
elements of a sequence should be treated
as components of a vector
sequence a : a1, a2 length is √（a12+a22） ---eq.AA15
sequence b : b1, b2 length is √（b12+b22） ---eq.AA16
<a name="ch01a029">
require their length all be one,
adjust as below
let c1 = a1/√（a12+a22） ---eq.AA17
let c2 = a2/√（a12+a22） ---eq.AA18
let d1 = b1/√（b12+b22） ---eq.AA19
let d2 = b2/√（b12+b22） ---eq.AA20
Can not divide by zero? If sequence a length
is zero, it must have all zero elements,
page 4, eq.1.6 become
0≦ 0 + b sequence square sum
page 5, eq.1.7 become
0≦ 0 * b sequence square sum
Both are correct, no need to prove.
Only when both a sequence and b sequence
are non-zero length, then it has value to
discuss them. We consider only non-zero
sequence. divide by length is safe.
<a name="ch01a030">
After normalization, c sequence and
d sequence have following relation.
（Alert: sequence c and sequence d
both are length one.）
k=∞
∑
k=1
|c k
* d k|
≦
0.5
k=∞
∑
k=1
c k2
+
0.5
k=∞
∑
k=1
d k2
page 4, eq.1.6b
width of above equation
<a name="ch01a031">
2009-06-09-18-40 here
Inequality smaller end remove absolute sign,
that is OK, because this removal let its value
become even smaller.
Inequality greater end remove absolute sign,
that is TROUBLE, because this removal let
greater end value become smaller, then whole
equation's inequality is uncertain.
Textbook remove absolute sign, here follow
text step.
Refer to eq.AA17 to eq.AA20 page 4, eq.1.6b
become
k=∞
∑
k=1
ak
√[∑(aj2）]
*
bk
√[∑(bj2）]
≦
0.5 * 1
+
0.5 * 1
page 5 line 13 eq.
width of above equation
<a name="ch01a032">
2009-06-09-18-54 here
Above page 5 line 13 eq. is a key point.
Equation right side two '1' are results of
normalization.
Equation right side two 0.5 sum to one,
eliminate addition process.
Equation left side denominator is length
of original vectors. It has no relation
with sum index k. Vector length can be
moved out side of sum symbol. Move Vector
length to right hand side. Its result is
page 5 eq.1.7 Cauchy's inequality.
Up to here
Magic adding to multiplying is done.
2009-06-09-19-00 stop
<a name="ch01a033">Index
2009-06-09-19-31 start
■ When Cauchy inequality become equal? I
Go back to see teacher Lai's quadratic equation
f(x)=（a12+a22+a32+ ... +an2）*x*x
-2*(a1b1+a2b2+a3b3+ ... +anbn)*x
+（b12+b22+b32+ ... +bn2） ---eq.AA07
<a name="ch01a034">
"discriminant B*B-4*A*C≦0 ", in which
"<0" cause inequality
"<0" cause f(x) no real root
"<0" cause f(x) curve not intersect with
x axis.
If discriminant B*B-4*A*C=0, cause equality
equation. Under what condition B*B-4*A*C=0
come true? If there is repeated roots, then
B*B-4*A*C=0. "has (repeated) root" allow us
to assign f(x)=0
(the opposite case is B*B-4*A*C<0 , f(x) not
intersect with x axis, we can not ask f(x)=0)
<a name="ch01a035">
Step back to eq.AA06
f(x)=(a1x-b1)2+(a2x-b2)2+.....+(anx-bn)2 ---eq.AA06
We require f(x)=0 (for B*B-4*A*C=0)
eq.AA06 is sum of real number square. equal
to zero?! then must require each square term
equal to zero, then get
a1x-b1=a2x-b2=.....=anx-bn=0 ---eq.AA21
solve for x, get
x=b1/a1=b2/a2=.....=bn/an ---eq.AA22
this x is ratio of two vector length.
<a name="ch01a036">
state eq.AA22 mathematically, it is
two sequences each corresponding element pair
ratio is constant.
Elements of a sequence should be treated
as components of a vector， State from vector
view point:
If two vectors (number sequences) point to
same direction, then Cauchy's inequality
become equality. Two vectors must be on
same line. They can point to same direction,
or point to opposite direction. They can
have different length.
(If point to opposite direction, absolute
value of two side of inequality are the
same. One positive, one negative)
2009-06-09-20-04 stop
<a name="ch01a037">
2009-06-09-21-47 start
From geometry illustration, it is clear
that when two vectors (sequences) component
ratio are constant, Cauchy's inequality
become equality.
Please click Draw 602 button
Cauchy inequality greater value side is red
solid line multiply by blue solid line.
Cauchy inequality smaller value side is red
dash line multiply by blue solid line.
If red solid line coincide with blue solid
line, projection of red on blue is same as
red solid line itself. In this case, Cauchy
inequality become equality.
2009-06-09-21-56 stop
<a name="ch01a038">
2009-06-10-09-56 start
Below is a simple program calculate two
side values of Cauchy inequality. For test
reason, you can fill in random numbers by
click "random" button. Default generate
five numbers for each sequence. You can
change it.
2009-06-10-09-59 stop
Sequence has
numbers
Cauchy
When sequence 1 proportional to sequence 2, Cauchy inequality
become equality. Please give
=
Box 2 change
2009-06-12-14-50 start
■ Sequence Quadratic Equation
Below is a simple test program. Output
a quadratic curve (red color)
Input first number sequence to Box 5
Input second number sequence to Box 6
then click "Modify 603 Too" button.
Do not click "Draw 603" button
Do not click "Modify 603" button
Because no data in data bank. (If use
data bank, you can not change its value)
Red curve is quadratic curve calculated
from Box 5 and Box 6 see eq.AA06C
Blue curve is quadratic curve calculated
from Box 5 and Box 5 (identical seq.)
Two side of an inequality have different
values, if move one side to the other side.
New equation is written as ≧0 or =0 or
≦0. Red curve is Cauchy inequality, red
curve not tangent not cross x axis.
Blue curve is Cauchy inequality, blue
curve tangent x axis. If function curve
meet (tangent or cross) x axis, the cross
point is where f(x)=0 (User can not control
blue curve)
If two sequence are proportional, curve
tangent x axis at x=proportional ratio
If two sequence are identical, curve
tangent x axis at x=1 (blue curve)
If two sequence are proportional, but
opposite sign, curve tangent x axis at
x<0 and -x=proportional ratio.
2009-06-12-15-04 here
<a name="ch01a043">Index
■ Graph 603, Drawing Board
Box 5
Box 6
Do not click "Draw 603" and "Modify 603" button
<a name="ch01a044">
2009-06-12-15-06 start
The main purpose to write "Modify 603 Too"
is for its three dash curves. Dash curves
do not read data from box 5,6.
Green dash line is conditional inequality.
When variable t is in [-1.4, +1.4} curve has
negative value, below x axis. Otherwise, curve
has positive value and above x axis.
Other four curves do not cross x axis.
Silver (gray) curve not tangent to x axis.
It is strict inequality. Any point on silver
curve can never equal to zero. One side is
always greater than the other side.
Both black curve and blue curve tangent x
axis. Black curve is relative flat. Blue
curve is relative narrow. Although both
tangent to x axis, but black curve every
where close to x axis, blue curve move
away from x axis quickly.
black curve is easier to become equality.
blue curve is harder to become equality.
2009-06-12-15-22 stop
<a name="ch01a045">
■ Goodbye Zero Nine
2009-06-12-15-31 start
Javascript use float number for calculation.
Answer may contain long string of '000000'
or '999999', for example
2.00000000000001
-25.999999999999996
On 2009-06-12-11-58 write a function
function bye09(in09) //9806121158
To use this function, call as following
coef2=bye09(coef2);
where coef2 is a number. After call
2.00000000000001 change to 2
-25.999999999999996 change to -26
It is handy, you can use it too.
If number is 1.23111111111 there is no
change.
function bye09(in09) handle output number
for better looking. Do not call bye09()
during calculation, that is just slow down
process. If expect answer to be irrational
Do not call bye09().
If expect answer to be integer or short
decimal number like 1.2. In these case
you can call bye09().
2009-06-12-15-40 stop
2009-06-17-19-05
bye09() process only one number at a time.
If you have a string of several numbers.
Do not put number string as input argument
bye09() will send string back immediately.
This file has sample code at time stamp
'9806171858' it is a right point to call.
'9806171830' is a wrong point to call.
2009-06-17-19-09 stop
<a name="ch01a047">Index
2009-06-13-10-00 start
■ Short symbol, vector dot product
Cauchy inequality equation use three times
vector dot product. If write complete equation,
it is a complicate work. Above equation used
sum ∑, already simplified. Whether it is
possible to simplify further? If we make a new
symbol and a set of rules for it, everybody
follow this symbol and rule, then further
simplification is possible. Do the following.
Sequence 1: a1, a2, a3, ... an ---eq.AA23
Sequence 2: b1, b2, b3, ... bn ---eq.AA24
the sum of their elements product is
a1*b1+a2*b2+a3*b3+.....+an*bn ---eq.AA25
First simplification is to use ∑(ak*bk)
for above equation. Understand that index k
start from k=1 to k=n
The next simplification, drop index k .
Require sequence 1 and sequence 2 product
sum be written as ＜a,b＞ .
<a name="ch01a048">
When reader see ＜a,b＞ then write a equation
in his/her brain as following
a1*b1+a2*b2+a3*b3+.....+an*bn
No need to print long equation. Allow edit
work to be maximum simplified.
<a name="ch01a049">
Use sequence product sum symbol ＜a,b＞
rewrite Cauchy inequality equation as below
Use a for sequence 1 （eq.AA23）
Use b for sequence 2 （eq.AA24），
Cauchy inequality equation is
＜a,b＞≦＜a,a＞0.5 *＜b,b＞0.5 text page 7 eq.1.13
in which someTerm0.5 means square root for
someTerm.
2009-06-13-10-50 stop
<a name="ch01a050">
2009-06-13-11-50 start
■ Rule for vector dot product
Assume vector v and vector w are both real
elements vectors. Use R for real number.
(relatively, it is not complex 1+2i ，not
quarternion [12;3,4,5]）
Use d for space demension, an integer. We
say
method 1: vector v=[1.2, 3.4, 5.6] is a real
number vector in 3D space.
method 2: symbol v∈Rd （d=3）
Define Ｖ≡R3≡real number 3D vector space
method 3: symbol v∈Ｖ
Above three methods are same, but one after
one more simplified.
<a name="ch01a051">
Vector v∈Ｖ and vector w∈Ｖ dot product is
＜v,w＞=v1*w1+v2*w2+v3*w3+.....+vn*wn ---eq.AA26
The following is dot product rule.
(Text page 7)
⑴ ＜v,v＞≧0 ---eq.AA27
for all v∈Ｖ
⑵ ＜v,v＞=0 ---eq.AA28
if and only if v=0=[0,0,0,....,0]
⑶ ＜αv,w＞=α＜v,w＞ ---eq.AA29
α∈R, for all v,w∈Ｖ
⑷ ＜u,v+w＞=＜u,v＞+＜u,w＞ ---eq.AA30
for all u,v,w∈Ｖ
⑸ ＜v,w＞=＜w,v＞ ---eq.AA31
for all v,w∈Ｖ
<a name="ch01a052">
Rule ⑴ ask: vector dot itself is a non-negative
number. vector dot itself is defined to be
vector length square. A length is non-
negative. Rule ⑴ follows.
Rule ⑵ ask: non-zero length vector can not say
itself is zero length. Only vector with
self dot equal to zero can be zero length.
Rule ⑶ say: two vectors dot product, if any one
change length, the dot result change value
with same ratio.
<a name="ch01a053">
Rule ⑷ say: two vectors dot product, if any one
is a composite of 3rd and 4th vectors.
Then the other one dot with 3rd and 4th
then sum to same result. In other words
sum first then dot or dot first then sum
get same result. Operation order do not
change answer.
Rule ⑸ say: two vectors dot product, there is
no master/follower difference. They are
on equal foot.
Above rules reflect the fact of our living
environment. Write down as mathematic rules
to serve our needs.
2009-06-13-12-40 stop
2009-06-13-16-28 start
<a name="ch01a054">Index
■ Define weighted vector dot product
Define Ｖ≡Rn≡ n dimensional real space,
n is a positive integer.
（Ｖ definition changed. Previous define V
to be 3D real vector space.）
If n=3, then the equation below apply to
3D space.
Require weighted factor wj to be positive
real， Mathematical expression is next
{wj∈R+; j=1,2, ..., n}
require a∈Ｖ ; b∈Ｖ
define ＜a,b＞ with weighted factor wj
The equation is next
a1*b1*w1+a2*b2*w2+.....+an*bn*wn ---eq.AA32
<a name="ch01a055">
that is ∑(aj*bj*wj) ---eq.AA33
eq.AA33 is called inner product of vector
a and vector b with weighted sum. Start
from j=1 to j=n .
Under some situation, weight function wj
represent probability pj. In this case
∑{pj}=1 ---eq.AA34
2009-06-13-17-10 here
<a name="ch01a056">
■ Define function product integral
Above eq.AA32 is two number sequence elements
product first, then sum . On the other hand,
integration is also a sum process. Below change
number sequence to function f(x) and g(x) and
define integral of dot product of two functions.
Require f(x), g(x) be define in V=C[a,b]
In which, C means continuous function. [a,b]
indicate independent variable x can vary from
a<=x to x<=b .
Define function product integral as below.
<a name="ch01a057">
<f,g> =
x=b
∫
x=a
f(x) g(x) dx
---text page 8, line 3
width of above equation
2009-06-13-17-29 here
<a name="ch01a058">
Text page 8, line 3 equation do not have
weight function. We can add weight function.
Require w(x) : [a,b] --> R be a continuous
function. The meaning of "[a,b] --> R" is
to require function w(x) input variable x
vary in the section [x=a, x=b]. a <= b
Function w(x) output value in all real axis.
[a,b] is part of real axis,
R is whole real axis.
2D curve input (independent variable) is
x axis.
2D curve output (dependent variable) is
y axis.
"part of real axis" and "whole real axis"
are two different thing. Do not mix up.
For example, if function is for distance
traveled in one minute.
input is time t∈[0,1]（"part of real axis"）
Output is distance L∈[0, one light minute]
（To me, "one light minute" is already
"whole real axis", it is 18000000 km）
Main point is
"part of real axis" and "whole real axis"
are two different things, even physics meaning
are not the same. (time and distance are
different)
<a name="ch01a059">
Define two functions with weight function
product integral as following.
<f,g> =
x=b
∫
x=a
f(x) g(x) w(x) dx
---text page 8, line 7
width of above equation
Text page 8, line 7 equation will be used
in the future.
2009-06-13-18-15 stop
<a name="ch01a060">
■ Sequence sum change to integral
2009-06-14-11-33 start
Above talk about integration ∫ ，
Further above talk about summation ∑ ，
and many text book say
number sequence sum can be calculated by
integration method. For example
<a name="ch01a062">
2009-06-14-11-56 here
Number sequence summation and integration
are very different, how can we use integration
to solve sequence summation problem?
Integration has function [example f(x)=1/x]
Integration has differential symbol dx
Integration has integral symbol ∫
Integration has lower/upper limit.
All of these are not used in sequence summation.
Plus
<a name="ch01a063">
Sequence summation is addition one by one.
Sequence summation is a one dimensional problem.
Integration is function value in y axis multiply
by step length dx in x axis.
Integration is a two dimensional problem.
How can we replace 1D problem with a 2D problem?
Those who think more, will find out above
questions.
We can explain as following （2009-06-14-12-16 here）
2009-06-14-12-23 here
In ∫dx/x we can slice integration area to
stripes. Require
Between x=1 to x=2 function value be 1/1
Between x=2 to x=3 function value be 1/2
.....
Between x=n-1 to x=n function value be 1/(n-1)
Original integration ∫dx/x become
∫dx/1 + ∫dx/2 + ∫dx/3 +.....+ ∫dx/(n-1)
Function 1/x change to constants
1/1, 1/2, 1/3, .....
Constant can be moved out of integration
symbol. Then integration become ∫dx
Between x=1 to x=2 value is (∫dx=1) * 1/1
Between x=2 to x=3 value is (∫dx=1) * 1/2
.....
Up to here, integration become sequence
summation.
Consider the opposite process. Sequence
summation can change to integration too.
Summation (1/1) + (1/2) + (1/3) +.....
no dx ?! Yes ! there is, above equation
is the same as below
(1/1)*1 + (1/2)*1 + (1/3)*1 + .....
Here "*1" IS "*dx" !
From x=2 to x=3 ， ∫dx value is 1，
so sequence one term (1/2)*1
is (1/2)*∫dx
that is ∫(1/2)*dx
because (1/2) is a constant.
Now sequence summation can change to
integration. Integration step length
dx=1 allow us explain smoothly.
This dx=1 can be defined as one meter,
or can be defined as one millimeter.
If data is dense, sequence summation
gradually close to integration value.
Sequence summation is step shape curve
integration function is a smooth curve.
Above numerical approach may have slight
error (computer integration use step/
stripe summation) but, sequence summation
and integration have same characteristic
behavior.
Sum from x=1 to x=∞ for 1/x answer is ∞
Integral x=1 to x=∞ for ∫dx/x also get ∞
If integral to infinity, step length is
one centimeter? one meter? one kilo-meter?
no difference, relative to infinity large,
these step size are all infinity small.
Use integration to solve sequence summation
is a common technique.
2009-06-14-12-53 stop
<a name="ch01a064">Index
■ Dot product proof Cauchy Ineq. I
2009-06-14-16-57 start
Cauchy inequality equation discussed
without using vector dot rule.
After that, discussed vector dot rule.
The following use vector dot rule to
prove Cauchy inequality equation. This
is textbook page 8 and 9 main topic.
Define Ｖ≡Rn≡ n dimensional
real vector space. n is a positive integer.
Define ＜.,.＞ be dot product of two vectors
both defined in space Ｖ. See rules
Vector v∈Ｖ and vector w∈Ｖ has the following
relation
<a name="ch01a065">
＜v,w＞ ≦ ＜v,v＞0.5＜w,w＞0.5 ---page 8, eq.1.16
Vector equation eq.1.16 and sequence equation
eq.1.13 are the same.
If both v and w are not zero length, eq.1.16
equal relation happen
＜v,w＞ ＝ ＜v,v＞0.5＜w,w＞0.5 ---eq.AA37
if and only if
v = λw ---eq.AA38
require that λ is not zero.
<a name="ch01a066">
■ Alert v = λw
2009-06-14-17-55 here
Math new-hand may not pay attention to
v = λw In fact we should be alert and
can not ignore the meaning of v = λw
Given v∈Ｖ and w∈Ｖ , so v and w are both
n dimensional real vectors. But λ is not
（not see λ∈Ｖ） λ is a real number, not a
vector. Assume discuss 3D problem.
Assume λ=0.5， assume
vector v1=[1,2,3]
vector w1=[4,5,6]
<a name="ch01a067">
v1 and w1 NO v1 = λ*w1 relation, because
two vectors component ratio
1/4 NOT= 2/5 NOT= 3/6
then v1 = λ*w1 is not true.
Only if all component ratio have same value
then v1 = λ*w1 become true.
If use random number generator to build v∈Ｖ
and build w∈Ｖ , probability of the existence
of relation v = λw is nearly zero. This kind
hard-to-see equation come alive in front of
our eye, we should pay special attention to
it.
<a name="ch01a068">
Equation v = λw require that vector v and
vector w have same component ratio. View
from geometry, it is a condition for
vector v and vector w be collinear and
point to same direction. If ratio is a
negative number. It is collinear and
point to opposite direction.
Two vector collinear and same direction
Cauchy inequality become equality, for
example 5=5.
Two vector collinear and opposite direction
Cauchy inequality stay inequality. But two
side have same absolute value. For example
-5 ＜ +5.
<a name="ch01a069">
Please goto Modify 603 Too test the
following sample sequence
In Box 5 input 1 2 3
In Box 6 input 3 6 9
Because 1/3 = 2/6 = 3/9 two sequences are
collinear and same direction.
Both red curve and blue curve tangent to x
axis. (Blue ratio = 1:1, red ratio = 1:3)
Next try
In Box 5 input 1 2 3
In Box 6 input -3 -6 -9
Because -1/3 = -2/6 = -3/9 two sequences are
collinear and in opposite direction. Red
curve tangent at x=-3
"Modify 603 Too" control only red curve.
Other four curves can be changed at
function 2 x(t)
function 2 y(t)
.....
function 5 x(t)
function 5 y(t)
boxes.
2009-06-14-18-47 stop
<a name="ch01a070">Index
■ Dot product proof Cauchy Ineq. II
2009-06-15-08-53 start
Now back to
＜v,w＞ ≦ ＜v,v＞0.5＜w,w＞0.5 ---page 8 eq.1.16
Use vector dot product rule to prove
Cauchy Inequality.
Vector dot product rule 1 (textbook page 7)
⑴ ＜v,v＞≧0 ---eq.AA27
eq.1.16 has two vectors v, w ，two vectors
can combine to one vector v-w ，substitute
this one vector to rule 1 get
＜v-w,v-w＞≧0 ---eq.AA39
<a name="ch01a071">
Treat left side "v-w" as one vector, treat right
side "v-w" as two vectors addition.
Use rule 4
⑷ ＜u,v+w＞=＜u,v＞+＜u,w＞ ---eq.AA30
re-write eq.AA39 as the following
（change part is red）
＜v-w,v-w＞=＜v-w,v＞+＜v-w,-w＞
Use rule 5
⑸ ＜v,w＞=＜w,v＞ ---eq.AA31
re-write again
＜v-w,v-w＞=＜v,v-w＞+＜-w,v-w＞
Use rule 4
＜v-w,v-w＞=＜v,v＞+＜v,-w＞+＜-w,v＞+＜-w,-w＞
<a name="ch01a072">
Use rule 3
⑶ ＜αv,w＞=α＜v,w＞ ---eq.AA29
change to
＜v-w,v-w＞=＜v,v＞-＜v,w＞-＜w,v＞+(-1)*(-1)＜w,w＞
=＜v,v＞-＜v,w＞-＜w,v＞+＜w,w＞
=＜v,v＞+＜w,w＞-＜v,w＞-＜w,v＞
Use rule 5
＜v-w,v-w＞=＜v,v＞+＜w,w＞-＜v,w＞-＜v,w＞
=＜v,v＞+＜w,w＞-2*＜v,w＞
<a name="ch01a073">
Based on eq.AA39 ＜v-w,v-w＞≧0 get
＜v,v＞+＜w,w＞-2*＜v,w＞≧0
that is
＜v,v＞+＜w,w＞≧2*＜v,w＞
that is
2*＜v,w＞≦＜v,v＞+＜w,w＞
that is
＜v,w＞≦0.5*＜v,v＞+0.5*＜w,w＞
---page 9 eq.1.17
2009-06-15-09-59 here
<a name="ch01a074">
Page 9 eq.1.17 and page 4 eq.1.6 are the same.
Both are half way Cauchy inequality, greater
value side is sum of two vector length.
Not whole way Cauchy inequality, greater value
side is product of two vector length.
If v or w is zero vector, for example [0,0,0]
Page 8 eq.1.16 ＜v,w＞ ≦ ＜v,v＞0.5＜w,w＞0.5
become 0 ≦ 0 ，it is correct for sure. We
need to consider both v and w are non-zero
vector. Under this condition, vector length
can be used to divide other number (1/0 not
occur) Then we can normalize vector v and
w . Please see Magic adding to multiplying.
Change page 8 eq.1.16 to page 9 line -10
＜v,w＞ ≦ √(＜v,v＞)*√(＜w,w＞) ---eq.AA40
Above is to use dot product to prove Cauchy
Inequality.
2009-06-15-10-33 stop
<a name="ch01a075">
■ When Cauchy inequality become equal? II
2009-06-15-10-58 start
If v or w is zero vector, Cauchy Inequality
become equality.
0 = 0*√(＜w,w＞)
0 = √(＜v,v＞)*0
Below, rule out zero vector condition. Then
both vector v and w can be normalized get
v1 and w1.
We require Cauchy Inequality
＜v,w＞ ≦ √(＜v,v＞)*√(＜w,w＞) -----eq.AA40
become equality
＜v,w＞ = √(＜v,v＞)*√(＜w,w＞)
<a name="ch01a076">
This requirement apply to normalized v1 and w1
＜v1,w1＞ = √(＜v1,v1＞)*√(＜w1,w1＞)
＜v1,w1＞ = 1*1 = 1
Vector v1 and w1 dot value cos(angle) = 1 ?
Must be angle=zero, so vector v1 and w1
are identical. Both length are one. Their
angle is zero.
Vector v1 and w1 are in same direction, then
before normalization,
Vector v and w are in same direction too.
(dot product operation do not rotate either
vector)
This conclusion is answer.
<a name="ch01a077">
Vector v and w in same
direction get equality.
Two vector are in same direction, it means
that two vector component ratio are the same.
Assume v =[1, 2, 3]
Assume w =[3.3, 6.6, 9.9]
component ratio 1/3.3 = 2/6.6 = 3/9.9
Then vector v and vector w are in same
direction.
Cauchy Inequality become equality.
Above is textbook page 9 material.
2009-06-15-11-23 here
<a name="ch01a078">
Please go to
■ Cauchy inequality calculation
and random number
In box 1 and 2 input
1, 2, 3
3.3, 6.6, 9.9
then click "Cauchy inequality" button
or click "Cauchy" button. Output to
box 3.
<a name="ch01a079">
Please go to
■ Graph 603, Drawing Board
In box 5 and 6 input
1, 2, 3
3.3, 6.6, 9.9
then click "Modify 603 Too" button. If
you use MSIE browser, you will see five
curves. Only red curve reflect the data
in box 5 and 6. Red curve tangent to x
axis, indicate there is repeated root.
discriminant is zero, two sequences are
proportional. The proportional ratio is
red point tangent point value. For example
above data get tangent at x=3.3 .
2009-06-15-11-46 stop
<a name="ch01a080">
2009-06-15-16-23 start
■ Can not compare complex number
Above is real number Cauchy inequality.
Below is complex number Cauchy inequality.
We know 1>0 is a matter of course, no need
to prove. Then, how about
imaginary i=√(-1) > 0
If we make it a rule i=√(-1) > 0
when take square operation, strange result
show up.
Real 1*1 = 1 > 0 OK, but
Imaginary i*i =-1 > 0 ？NO!
We can not say greater than or less than
relation for imaginary/complex number. But
complex number has properties expressed in
real numbers. These real number properties
can be used to compare greater than or less
than relation.
<a name="ch01a081">
Freeman's another web page complex2.htm has
function definition cabsf('m+ni')=√(m*m+n*n)>=0
It calculate the norm (or the length) of a
complex number 'm+ni' in which both m, n are
real, only i is imaginary. For example, define
a = 1+2i
b = 3-4i
|a| = √(1*1+2*2) = √(5)
|b| = √(3*3+(-4)*(-4)) = √(25)
We can not say: b > a ， but
we can say: |b| > |a|
<a name="ch01a082">
On the other hand, if two complex number
have same norm, then their phase angle
can be compared. For example
a = 1+2i
b = 2-1i
∠(a) = cargf('1+2i') = 1.1071487177940904
∠(b) = cargf('2-1i') =-0.4636476090008061
We can say﹕∠(a) > ∠(b)
Phase angle has principle value range,
cosine function principle value range is
from -180 degree (-PI/2 rad.)
to +180 degree (+PI/2 rad.)
Alert: angle +190 become -170
Compare phase angle may cause puzzle.
Above two examples, complex number output
real number, we can compare only real number.
<a name="ch01a083">Index
■ Complex number Cauchy Inequality
2009-06-15-16-54 here
Below is complex number Cauchy Inequality
We can compare only complex norm.
2009-06-06-21-41 access
http://www.emis.de/journals/JIPAM/images/010_03_JIPAM/010_03.pdf
save as emis.de-etc-010_03.pdf
Below is reading record.
Theorem 2.3
Define As = (a1, . . . , an)
define Bs = (b1, . . . , bn)
to be two complex number sequences.
Cauchy-Bunyakovsky-Schwarz’s inequality
short as (CBS)−inequality, is next
（Alert: ak, bk are all complex）
2009-06-15-17-30 here
Define A* be complex conjugate for A
(Complex multiplication use 'Ｘ' )
If A = 1+2i then A* = 1-2i ，
complex number reverse its imaginary
sign to form its conjugate.
<a name="ch01a085">
Prove (CBS)−inequality as following.
Let λ be a complex number, create next equation.
<a name="ch01a087">
2009-06-15-18-48 here
Substitute eq.AA43 λ0 to λ in eq.AA42
eq.AA42 first term ∑|ak|2 not change
eq.AA42 second term change as following
|
∑
a k
b k
∑
|
b k
|
2
|
2
k=n
∑
k=1
|
b k
|
2
-----eq.AA44
width of above equation
<a name="ch01a088">
2009-06-15-19-03
Numerator and denominator have a non-zero
common term ∑|bk|2 ，which is cancelled
Then eq.AA42 second term become
<a name="ch01a090">Index
2009-06-15-19-22 here
eq.AA42 third term use λ conjugate λ*，this
conjugate become eq.AA46 left half numerator
conjugate. λ denominator ∑|bk|2 is square of
absolute of a complex number, a real number.
But λ numerator ∑ a k b k is complex, when λ
change to conjugate， numerator also change
to conjugate. eq.AA46 left side numerator
and right side ∑ a k b k are conjugate each
other. Product [ ∑ak bk ] [ ∑ak bk ]* is a
real number. Its value is same as eq.AA45
numerator |∑ a k b k|2<a name="ch01a091">
Re-write eq.AA42 as following
＝
k=n
∑
k=1
|
a k
|
2
+
|∑
a k
b k
|
2
∑
|
b k
|
2
－2*Re
|∑
a k
b k
|
2
∑
|
b k
|
2
---eq.AA47
width of above equation
<a name="ch01a092">
2009-06-15-20-06 here
'Re' in eq.AA47 is redundant, because both
numerator and denominator take absolute value,
which can be real only. (if λ not take λ0
value then 'Re()' is necessary)
eq.AA47 is right hand side of eq.AA41
Left hand side of eq.AA41 is a perfect
square, then eq.AA47 ≧ 0, get next
k=n
∑
k=1
|
a k
|
2
－
|∑
a k
b k
|
2
∑
|
b k
|
2
≧０
---eq.AA48
width of above equation
<a name="ch01a093">
2009-06-15-20-20 here
Take common-denominator for eq.AA48 get
∑ | a k |2 ∑ | b k |2 - |∑ a k b k|2≧０
This is final complex number (CBS)−inequality.
emis.de theorem 2.3
In which a k and b k are all complex.
2009-06-15-20-26 stop
Change direction, same thing
|∑akbk|2≦∑|ak|2 ∑|bk|2
This form is closer to emis.de theorem 2.3
2009-06-15-20-34 stop
<a name="ch01a094">
■ "c+di" Ｘ "c-di" vs. |"c+di" Ｘ "c+di"|
2009-06-15-21-42 start
Above has one section
[[
eq.AA46 left side numerator
and right side ∑ a k b k are conjugate each
other. Product [ ∑ak bk ] [ ∑ak bk ]* is a
real number. Its value is same as eq.AA45
numerator |∑ a k b k|2
]]
Below explain why.
<a name="ch01a095">
Assume complex ak = c+di
its conjugate ak* = c-di
in which both c, d are real, i=√(-1)
Compare the value of [c+di] Ｘ [c-di]
and absolute value of [c+di] Ｘ [c+di]
whether they are the same?
The following use '*' as multiplication
not as complex conjugate.
[c+di] * [c-di] = c*c - (di)*(di)
= c*c - d*d*i*i
= c*c - d*d*(-1)
= c*c + d*d ---eq.AA49
<a name="ch01a096">
See the other term
absolute value of [c+di] * [c+di]
Carry out multiplication first.
[c+di] * [c+di] = c*c + 2*c*d*i + di*di
= c*c - d*d + 2*c*d*i
Its absolute value is
real (c*c - d*d) square plus
imaginary (2*c*d) square, then
take the square root of the sum.
Second term
= √[(c*c - d*d)*(c*c - d*d) + (2*c*d)*(2*c*d)]
= √[c*c*c*c - 2*c*c*d*d + d*d*d*d + 4*c*d*c*d]
= √[c*c*c*c + 2*c*c*d*d + d*d*d*d]
= √[(c*c+d*d)*(c*c+d*d)]
= c*c+d*d ---eq.AA50
<a name="ch01a097">
From eq.AA49 and eq.AA50 it is clear that
value of [c+di] Ｘ [c-di] and absolute
value of [c+di] Ｘ [c+di] are the same
Therefore in eq.AA42 second term and
third term cancel each other is a correct
operation. （see eq.AA47）
2009-06-15-22-02 stop
<a name="ch01a098">Index
■ Why define λ this way?
2009-06-16-10-05 start
λ in eq.AA42 is an arbitrary complex number
We can assign a value to λ . This is a key
step, it is also an experience.
Based on eq.AA42，how can one find out the
λ0 in eq.AA43？
eq.AA42 first term and second term are
both absolute value, real number. But
third term is a complex number. Real
number can be compared for inequality.
Complex number can not.
（In order to find out this magic λ0 ,
ignore Re() in eq.AA42 ）
eq.AA42 term two has λ , term three has
λ*. λ* carry the responsibility to
change complex to real. Third term complex
is ∑akbk . If we define λ = ∑akbk， then
<a name="ch01a099">
third term λ* multiply with ∑akbk, get
real number. Real can be compared in
inequality. For this reason, we need
define λ = ∑akbk carry out λ Ｘ λ* and
get |∑akbk|2 this is smaller value side
of the inequality.
Greater value side of the inequality is
∑|ak|2 Ｘ ∑|bk|2 But eq.AA42 do not
have the structure for ∑|ak|2 Ｘ ∑|bk|2<a name="ch01a100">
Can we ask λ to carry out the second
mission? If denominator of λ is not one.
If it is ∑|bk|2 instead, after common-
denominator process, first term of eq.AA42
become ∑|ak|2 Ｘ ∑|bk|2 ! For this reason,
λ definition need to have ∑|bk|2 as
denominator. eq.AA43 definition require
λ0 = ∑akbk / ∑|bk|2 see-again eq.AA43
Thank God ! eq.AA42 second term has ∑|bk|2
which cancel |λ|2 created denominator
extra ∑|bk|2 . Finally success.
Whether above λ definition reasoning can
be applied to future problem? Hope so.
Every one use his/her own thinking to
solve the problem in hand.
2009-06-16-10-40 stop
<a name="ch01a101">
■ Real reason for Complex Inequality
2009-06-16-16-36 start
Below is another method to prove (CBS)−
inequality. Not use complex reasoning,
that is not use complex conjugate. Use
real number concept to prove. Easier to
understand.
Real sequence one: 1, -2
Real sequence two: 3, 4
Take absolute value first, then sum get
Seq. Sum 1: 1*3 + |-2|*4 = 3+8 = 11
Sum first, then take absolute value get
Seq. Sum 2: |1*3 + (-2)*4| = |3-8| = 5
We can see
Take absolute value first, then sum get
greater value.
Sum first, then take absolute value get
smaller value.
Above is real number.
<a name="ch01a102">
Below is complex number.
Complex m+ni take absolute value get
√(m*m+n*n) this is a positive number.
Positive add positive no cancellation.
But complex number can be viewed as
vector in 2D space. Addition of two
complex number is same as addition of
two vectors. If two complex number have
different phase angle, then they are
not in same direction. Like two sides
of a triangle -- not collinear. Sum of
two complex number has shorter norm
then the norm sum of two complex numbers.
Although no negative value and no
cancellation, still cause inequality.
<a name="ch01a102B">
Complex sequence 1, take absolute value
first, then sum get greater value answer.
Complex sequence 2, product sum first,
then take absolute value, get smaller
value answer. Therefore
|∑ akbk| is smaller (has cancellation)
∑(|ak|Ｘ|bk|) is greater (no cancel)
|∑ akbk| ≦ ∑(|ak|Ｘ|bk|) -----eq.AA51
If eq.AA51 elements all be positive value,
no cancellation, then triangle inequality
dominant.
<a name="ch01a103">
ak is complex, bk is complex.
|ak| is real，|bk| is real，
Next is REAL valued Cauchy inequality
[∑(|ak|Ｘ|bk|)]2≦∑|ak|2Ｘ∑|bk|2 -----eq.AA52
eq.AA52 left hand side is eq.AA51 right
hand side. Put two equations together, get
(CBS)−inequality emis.de theorem 2.3
|∑akbk|2≦∑|ak|2 ∑|bk|2 -----eq.AA53
Above reasoning did not use conjugate.
2009-06-16-17-09 stop
<a name="ch01a104">
■ Two inequalities, two reasons
2009-06-16-18-55 start
eq.AA51 reason for inequality and
eq.AA52 reason for inequality are different.
eq.AA52 reason for inequality is two
sequences (vectors) angle is not zero.
Sequences not in proportion,
vectors not collinear.
[∑|ak|Ｘ|bk|]2≦∑|ak|2 ∑|bk|2 -----eq.AA52
eq.AA52 right hand side is a product of
two vector norms. Ignore their relative
angle.
eq.AA52 left hand side is a dot product
of two vectors. Their relative angle is
important.
<a name="ch01a105">
eq.AA51 reason for inequality is
cancellation or triangle inequality.
|∑ akbk| ≦ ∑(|ak|Ｘ|bk|) -----eq.AA51
eq.AA51 right hand side is product sum
of two positive real sequences. All
positive has no cancellation.
eq.AA51 left hand side is two complex
sequence product sum. Not take absolute
value first, allow cancellation.
If sequence elements are all positive,
no cancellation. then
eq.AA51 right hand side is travel between
two points along two sides of a triangle.
(longer distance)
eq.AA51 left hand side is travel between
two points along a straight line between
two points. (shorter distance)
Complex multiplication a1b1 = complex c1
and a2b2 = complex c2
Although |a1b1+a2b2|≦...
do not look like triangle inequality, but
|c1 + c2| ≦ |c1| + |c2|
IS triangle inequality.
2009-06-16-19-10 here
<a name="ch01a106">
■ Complex calculator, sample code
This file tute0007.htm do not have complex
number calculator. Another web page
http://freeman2.com/complex2.htm
has complex number calculator. Please copy
next 19 line code paste to complex2.htm
box 3, then click
"Test box3 command, output to box4"
[[
a1='1+2i'
a2='3+4i'
b1='-1-1i'
b2='-1-2i'
aa1=cabsf(a1)*cabsf(a1)
aa2=cabsf(a2)*cabsf(a2)
bb1=cabsf(b1)*cabsf(b1)
bb2=cabsf(b2)*cabsf(b2)
big0=sqrt(aa1+aa2)*sqrt(bb1+bb2)
s1=cabsf(a1)*cabsf(b1)
s2=cabsf(a2)*cabsf(b2)
mid0=s1+s2
c1=cmulf(a1,b1)
c2=cmulf(a2,b2)
c3=caddf(c1,c2)
sml0=cabsf(c3)
big0
mid0
sml0
]]
Above simple code allow one sequence two
complex number only.
<a name="ch01a107">
Box 4 output the following
[[
big0
14.49137674618944
mid0
14.342617547667329
sml0
14.317821063276353
]]
<a name="ch01a108">
then change the value of complex definition
a1='1+2i'
a2='3+4i'
b1='-1-1i'
b2='-1-2i'
click
"Test box3 command, output to box4"
again.
<a name="ch01a109">
big0 calculate eq.AA52 right hand side
mid0 calculate eq.AA52 left hand side, also eq.AA51 right hand side
sml0 calculate eq.AA51 left hand side.
Answer order is big0 ≧ mid0 ≧ sml0
We can not compare complex numbers.
Only complex number property in real number
can be compared.
2009-06-16-19-19 stop
<a name="Copyright">
2009-06-19-10-48
If you are interested in inequality, suggest
you buy Professor J. Michael Steele's book
The Cauchy-Schwarz Master Class.
Above book is the textbook of this file.
To buy textbook that is to show thank to
Professor J. Michael Steele's wonderful
work.
To buy textbook that is also a sign of
respect copyright law.
Thank you. Freeman 2009-06-19
The Cauchy-Schwarz Master Class
J. MichaelSteele★★★★★
2009-06-19-10-56
<a name=docB01>Index
Working record
2008-11-03-20-15 click 'purchase' button
2008-11-06-13-11 receive
An introduction to the art of mathematical inequalities
http://www-stat.wharton.upenn.edu/~steele/Publications/Books/CSMC/CSMC_index.html
The Cauchy-Schwarz Master Class.
ISBN 978-0-521-54677-5
J. Michael Steele
http://www-stat.wharton.upenn.edu/~steele/index.html
2009-01-26-08-41 create inequal0.htm
2009-06-08-18-29
copy inequal0.htm ineqc001.htm
2009-06-11-11-20
ren ineqc001.htm tute0007.htm
<a name=docB02>
2009-06-11-19-03 start
tutc0005.htm (Chinese page) record calculus
reading notes. About four or five years ago
thought I should study calculus. Freeman
studied mechanical engineering. In school
learned how to apply calculus and how to
solve engineering problem. Did not learn
how to prove calculus theorems. During
work, found many formula are effective.
Apply the calculus rule then get correct
answer. For example, chain rule, implicit
differentiation, L'Hopistal rule, partial
integration, Leibniz's differentiation for
integral upper/lower limit etc. I know how
but do not know why. During write tutc0005
my main purpose is to learn how to prove
these theorems. I should continue work on
calculus. But
<a name=docB03>
later time, my attention shift to
Space Curve Projector
http://freeman2.com/curve3d2.htm
foot of perpendicular
http://freeman2.com/eyefoot2.htm
Gram-Schmidt Process
http://freeman2.com/gramsch2.htm
complex polynomial root
http://freeman2.com/polyroot.htm
complex variable functions
http://freeman2.com/complex2.htm
then start read
The Cauchy-Schwarz Master Class
About one month ago, I think
Should I put my time to calculus?
Should I put my time to inequality?
Finally, decide put my time to inequality.
Because there are many calculus material
online and very few inequality pages.
In the future, I will return to calculus,
because how to prove calculus theorems?
that is always my concern. In the near
future, my main work is inequality.
2009-06-11-19-38 here
<a name=docB04>
draft name inequal0.htm （not upload）
initial name ineqc001.htm （not upload）
current name tute0007.htm （will upload）
All programming pages do not use tut*.htm
name.
All programming pages have English version
name.
Initially, inequality file name use ineqc001.htm
it is an intention to create English ineqe001.htm
But program in inequality file is secondary
work, the main work is study and write notes.
File name tutc*.htm fit better. So rename
to tutc0007.htm and English version is
tute0007.htm
Chinese tutc0001.htm ... tutc0006.htm uploaded.
English tute0001.htm ... tute0006.htm not exist.
Because my time is limited, and the meterial
in tutc0001.htm ... tutc0006.htm is too
elementry. There are many English elementry
math pages online. In the future, if I have
time, and if I find some one proofread for
me. I will upload tute0001.htm ... tute0006.htm
2009-06-11-19-50 stop
<a name=docB05>
2009-06-16-11-08 start
Freeman web page use big file. Range from
200k to 300k bytes. Now tutc0007.htm
(Chinese version) file size reached
200k. I shall change file. In the future
if add more material to tutc0007.htm
its size should be within 300k bytes.
2009-06-16-11-11 stop
<a name=docB06>
2009-06-17-11-12 start proofread
2009-06-17-13-18 done tutc0007.htm
initially
"width of above equation default value"
has only value, no button.
2009-06-17-16-40 start
2009-06-17-17-53 change "width of above equation"
to button.
<a name=docB07>
2009-06-18-15-31 start English version
copy tutc0007.htm tute0007.htm
copy ineqindc.js ineqinde.js
2009-06-18-16-35 paid attention that
[[
<a name="ch01a083">
■ Complex number Cauchy Inequality
]]
is in textbook page 65, example 4.6
2009-06-23-15-37 start proofread tute0007.htm
2009-06-23-20-51 done proofread
<a name=docB08>
2009-08-04-10-08 start
[update 2009-08-04] copy function readdata()
from tute0009.htm to tute0007.htm
no major change.
2009-08-04 first time upload tute0008.htm
tute0007.htm and tute0009.htm made minor
change.
tute0007.htm first time upload 2009-06-24
tute0009.htm first time upload 2009-07-26
tute0008.htm first time upload 2009-08-04
upload tute0009.htm earlier than tute0008.htm
because program in tute0009.htm is better than
program in tute0008.htm.
2009-08-04-10-13 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