<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="ch04c001">　Index begin　Index this file

2009-11-11-12-31 start
■　Exercise 4.1 problem statement
　　textbook page 67
（Triangle Inequality "Pot Shots"）

The triangle inequality in Rd may seem 
obvious, but some of its consequences 
can be puzzling when they are presented
out of context. Here the next three
exercises are not at all hard, but you
might ask yourself "Would these have
been so easy yesterday?"

<a name="ch04c002">
(a) Show that the nonnegative x,y,z that
    (√2)*(x+y+z)≦  ---eqn.AL001
    √(x2+y2)+√(z2+y2)+√(z2+x2)
(b) Show for 0＜x≦y≦z that
    √(y2+z2)≦  ---eqn.AL002
    x√2+√[(y-x)2+(z-x)2]
(c) Show for positive x,y,z that
    2√3≦  ---eqn.AL003
    √(x2+y2+z2)+√(x-2+y-2+z-2)

<a name="ch04c003">
The list can be continued almost without
limit, yet there is really only one theme:
any time you see a sum of square roots in
an inequality, you should give a moment's
thought to the possibility that the triangle
inequality may help.
2009-11-11-12-50 here




<a name="ch04c004">　Index begin　Index this file
2009-11-11-13-00 start
■　Exercise 4.1 hint
　　textbook page 239
Each case follows by an application of the
triangle inequality to an appropriate sum.
Those sums are:
(a) (x+y+z,x+y+z)=(x,y)+(y,z)+(z,x) ---eqn.AL004
(b) (y,z)=(x,x)+(y-x,z-x)           ---eqn.AL005
(c) (2,2,2)≦(x+1/x,y+1/y,z+1/z)    ---eqn.AL006
           =(x,y,z)+(1/x,1/y,1/z)
2009-11-11-13-05 here






<a name="ch04c005">
■　Exercise 4.1 (a) solution


Exercise 4.1 (a) use graphic solution 
as following. Please change parameter
and click [Plot10] button.
Graph use "a,b,c" extend to "d,e"
Graph did not use "x,y,z".
2009-11-11-13-09 stop

<a name="ch04c005a">　Index begin　Index this file
2009-11-11-13-17 start
■　Exercise 4.1 (b) solution
Hint give the identity
(b) (y,z)=(x,x)+(y-x,z-x)           ---eqn.AL005
(x,x) vector add with (y-x,z-x) vector
become (y,z) vector.
(y,z) vector is straight line from (0,0)
to (y,z). (No figure for this problem)
Above is a vector equation, each
vector has two components.
Following consider vector length, no
component, only pure number length.
Please see caution

<a name="ch04c005b">
Apply Pythagoras theorem get
   (y,z) vector length = √(y2+z2)
   (x,x) vector length = x√2
   (y-x,z-x) vector length = √[(y-x)2+(z-x)2]
(x,x) vector length add (y-x,z-x) vector
length is longer "two side" sum.
Triangle inequality tell us that
"two side" length sum is greater than 
or equal to "straight line" side, then
we have the answer
(b) √(y2+z2)≦  ---eqn.AL002
    x√2+√[(y-x)2+(z-x)2]

<a name="ch04c005c">
The requested condition 0＜x≦y≦z is 
used at 
(b) (y,z)=(x,x)+(y-x,z-x)           ---eqn.AL005
with 0＜x≦y≦z, we get positive component
for the vector (y-x,z-x) [y-x≧0 ,z-x≧0]
when we work with distance, there is no
negative distance concept.
y≦z or y≧z does not matter.
2009-11-11-13-37 here

<a name="ch04c005d">　Index begin　Index this file
■　Exercise 4.1 (c) solution
Problem statement require "positive x,y,z"
Hint give the next relation
(c) (2,2,2)≦(x+1/x,y+1/y,z+1/z)    ---eqn.AL006
           =(x,y,z)+(1/x,1/y,1/z)

<a name="ch04c005e">
For positive x, we have 2≦x+1/x   ---eqn.AL007
Because x is positive, we can 
rewrite eqn.AL007 as 2x≦x*x+1
rewrite   again   as 0≦x*x-2x+1
We come to  0≦(x-1)*(x-1)  ---eqn.AL008
eqn.AL008 is true, then 
eqn.AL007 is true. 
Alert! 
eqn.AL007 is NOT triangle inequality !

So one side length (x+1/x,y+1/y,z+1/z)
is greater than or equal to the length
of vector (2,2,2). Which is
√[2*2+2*2+2*2] = 2√3 ≦ ---eqn.AL009
√[(x+1/x)2+(y+1/y)2+(z+1/z)2]
When x=y=z=1, equality exist.

<a name="ch04c005f">
In eqn.AL006 we have
  (x+1/x,y+1/y,z+1/z)=
  (x,y,z)+(1/x,1/y,1/z) ---eqn.AL006B
This is triangle, left vector is one side
from (0,0,0) to (x+1/x,y+1/y,z+1/z)
Right two vectors are two sides. (x,y,z) 
and (1/x,1/y,1/z) go longer distance to
same point as one side does.

Caution:
eqn.AL006B is a vector equation and 
an equality equation. The following 
eqn.AL010 is a scalar and inequality
equation.

<a name="ch04c005g">
Triangle inequality tell us that
length of  (x+1/x,y+1/y,z+1/z) is less 
than or at most equal to the sum of two
sides length. That is
√[(x+1/x)2+(y+1/y)2+(z+1/z)2]≦  ---eqn.AL010
√(x2+y2+z2)+√(x-2+y-2+z-2)
Combine eqn.AL009 and eqn.AL010 we 
proved eqn.AL003.
2009-11-11-14-01 stop

<a name="ch04c006">　Index begin　Index this file
2009-11-11-17-33 start
■　Exercise 4.2 problem statement
　　textbook page 67
（The Geometry of "Steepest Ascent"）

If f:Rn→R is a differentiable 
function, then one often hears that the
gradient

<a name="ch04c008">
2009-11-11-17-48 here
points in the direction of steepest 
ascent of f provided that ∇f≠0. In 
longhand, this says for any unit 
vector u one has the bound

<a name="ch04c009">
2009-11-11-17-56 here
where v=∇f(x)/∥∇f(x)∥ ---eqn.AL012
Prove the inequality and show that it is
strict unless u=v.



<a name="ch04c010">　Index begin　Index this file
2009-11-11-18-00 here
■　Exercise 4.2 hint
　　textbook page 239
The derivative on the left is equal to
〈∇f(x),u〉 which is bounded by
  ∥∇f(x)∥∥u∥=∥∇f(x)∥ ---eqn.AL013
by the Cauchy-Schwarz inequality. 
(problem given u is unit vector, then
 ∥u∥=1; LiuHH 2009-11-15-15-32)

On the other hand, the derivative on the
right is equal to
  〈∇f(x),v〉=∥∇f(x)∥ ---eqn.AL014
by direct calculation and the definition
of v. These observations yield the
inequality (4.21).

<a name="ch04c011">
We have equality in the application of
the Cauchy-Schwarz inequality only if
u and ∇f(x) are proportional, so the
bound (4.21) reduces to an equality if
and only if    u=λ∇f(x) ---eqn.AL015
Since u is a unit vector, this implies
  λ=±1/∥∇f(x)∥  ---eqn.AL016
Only the positive sign can give equality
in the bound (4.21), and in that case we
have u=v.
2009-11-11-18-15




<a name="ch04c012">
2009-11-11-18-22
■　Exercise 4.2 solution


First we need to know the difference
between vector u and vector v.
vector u is any vector
vector u is a variable vector.
vector u is unit vector, length=1

vector v is ∇f(x)/∥∇f(x)∥ ---eqn.AL012
vector v is not any vector.
vector v is a constant vector.
vector v is unit vector, length=1

<a name="ch04c013">
Although both u vector and v vector are
unit vector, length=1. Both vector's
length are varied (changed, played with)
because the pure number t in tu and tv
in eqn.4.21 modify the vector length.

<a name="ch04c014">
eqn.4.21 left side 
  d[f(x+tu)]/dt evaluate at t=0
is inner product of gradient f(x) and u
vector that is 〈∇f(x),u〉. The term
〈∇f(x),u〉 is bounded by Cauchy inequality
  〈∇f(x),u〉≦∥∇f(x)∥∥u∥ ---eqn.AL019
because u is unit vector, ∥u∥=1, then
eqn.AL019 reduce to
  〈∇f(x),u〉≦∥∇f(x)∥     ---eqn.AL020

<a name="ch04c015">　Index begin　Index this file
On the other hand, the derivative on the
right side of eqn.4.21 is 〈∇f(x),v〉
  〈∇f(x),v〉=∥∇f(x)∥∥v∥ ---eqn.AL021

<a name="ch04c016">
Why eqn.AL019 use ≦ link left and right?
why eqn.AL021 use ＝ link left and right?
because eqn.AL019 include u vector and u
is arbitrary unit vector, u can change to 
any direction. Only if u parallel to ∇f
then eqn.AL019 has equality, otherwise
eqn.AL019 is inequality.
But v vector is given as 
  v=∇f(x)/∥∇f(x)∥ ---eqn.AL012
v is ∇f(x) vector and length=1, v can not
vary. So eqn.AL021 has equality.

Compare eqn.AL013 and eqn.AL015, target
equation eqn.4.21 is proved.

<a name="ch04c017">
Cauchy-Schwarz inequality become equality
only if two sequences are in proportion
or two vectors are in same direction.
Now we have inner product 〈∇f(x),u〉,
two vectors are ∇f(x) and u. When they
are in same direction, we can write
  u=λ∇f(x) ---eqn.AL015
eqn.AL015 say ∇f(x) and u are in same 
direction, but may have different length.
   λ=∥u∥/∥∇f(x)∥  ---eqn.AL017
is two vector length ratio.

<a name="ch04c018">
If two vectors are not in same direction
we can NOT write equation like eqn.AL015.
u is unit vector, so ∥u∥=1, eqn.AL017
change to
   λ=1/∥∇f(x)∥  ---eqn.AL022
Put eqn.AL022 into eqn.AL015, we find
  u=∇f(x)/∥∇f(x)∥ ---eqn.AL018
but
  v=∇f(x)/∥∇f(x)∥ ---eqn.AL012
<a name="ch04c019">
We conclude that 
when variable unit vector u coincide with
constant non-variable vector v, at this 
moment, the inequality
  〈∇f(x),u〉≦∥∇f(x)∥∥u∥ ---eqn.AL019
become equality.
2009-11-11-19-10 stop

<a name="ch04c020">　Index begin　Index this file
2009-11-12-17-47 start
■　Exercise 4.3 problem statement
　　textbook page 68
（Cauchy via another identity）

Lagrange's identity is not the only 
formula that gives an instant proof 
of Cauchy's inequality. Check that 
in any real inner product space the
difference 
  〈v,v〉〈w,w〉-〈v,w〉2 ---eqn.AL023
can be written as

<a name="ch04c022">
2009-11-12-18-00 here
and explain why this also implies the
general Cauchy-Schwarz inequality.

Incidentally, one does not need a flash
of algebraic insight to discover the
representation (4.22). As Figure 4.6 
suggests, this formula cannot remain
hidden for long once we ask ourselves 
about minimization of the polynomial
  P(t)=〈v-tw,v-tw〉 ---eqn.AL024
2009-11-12-18-05 here






<a name="ch04c023">
2009-11-12-18-10 start
■　Exercise 4.3 hint
　　textbook page 240
Direct expansion proves the representation
(4.22). To minimize P(t) we solve 
   P'(t)=2t〈w,w〉-2〈v,w〉=0 ---eqn.AL025
and find P(t)≧P(t0) where 
   t0=〈v,w〉/〈w,w〉 ---eqn.AL026
The evaluation of P(t0) then leads one to 
the expression (4.22)
2009-11-12-18-14 stop





<a name="ch04c024">　Index begin　Index this file
2009-11-12-19-16 start
■　Exercise 4.3 solution


Start from eqn.AL023, 
  〈v,v〉〈w,w〉-〈v,w〉2 ---eqn.AL023
move toward to eqn.4.22
We assume 〈w,w〉 is not zero,
eqn.AL023 is same as
  〈w,w〉[〈v,v〉-〈v,w〉2/〈w,w〉]
<a name="ch04c025">
Let us add a zero
  〈w,w〉[〈v,v〉-〈v,w〉2/〈w,w〉
      +〈v,w〉2/〈w,w〉-〈v,w〉2/〈w,w〉
      ]
It become
  〈w,w〉[〈v,v〉-2*〈v,w〉2/〈w,w〉
       +〈v,w〉2/〈w,w〉 ]
<a name="ch04c026">
Next let us multiply by one
  〈w,w〉[〈v,v〉-2*〈v,w〉2/〈w,w〉
       +(〈w,w〉/〈w,w〉)〈v,w〉2/〈w,w〉 ]
It is same as
  〈w,w〉[〈v,v〉-2*〈v,w〉2/〈w,w〉
       +〈w,w〉*〈v,w〉2/〈w,w〉2 ] ---eqn.AL027
Let us define 
  s = 〈v,w〉/〈w,w〉 ---eqn.AL028
then eqn.AL027 simplify to
  〈w,w〉[〈v,v〉-2s*〈v,w〉+〈w,w〉*s2 ] ---eqn.AL029
<a name="ch04c027">
Let s goto w side, equation become
  〈w,w〉[〈v,v〉-2〈v,sw〉+〈sw,sw〉] ---eqn.AL030
Here s is a pure number, v and w are
sequence (same as vector).
Now we can see square bracket in eqn.AL030 
is a vector dot product, it is
  〈w,w〉[〈v-sw,v-sw〉] ---eqn.AL031
Now put s (eqn.AL028) back to eqn.AL031
we get eqn.4.22.

<a name="ch04c028">
Above is from eqn.AL023 to eqn.4.22.
Need add zero, need multiply by one.
Actually LiuHH draft work start from
eqn.4.22 move toward eqn.AL023, drop
one and drop zero is much easier.
2009-11-12-19-39 here

<a name="ch04c029">　Index begin　Index this file
2009-11-12-19-49 
eqn.AL028 define s. s is same as t0
s or t0 are constant.
Because v-sequence and w-sequence are
both given, we can not change it.
eqn.AL028 define s with v and w,
result is a constant. This constant s
coincide with constant t0.
<a name="ch04c030">
On the other hand, t is a variable in
eqn.AL024. Because eqn.AL024 use P(t)
t is in P(t), t is a parameter, t MUST
be a variable. Problem suggest us 
consider
  P(t)=〈v-tw,v-tw〉 ---eqn.AL024
The expansion of P(t) is
  P(t)=〈v,v〉-2t*〈v,w〉+〈w,w〉*t2  ---eqn.AL032
t is a variable, 
<a name="ch04c031">
we can differentiate P(t) to get
  P'(t)=2t〈w,w〉-2〈v,w〉 ---eqn.AL033
this moment, eqn.AL033 can be ＜＝＞0
because t is free to vary.
Calculus theory tell us that if we set
 P'(t)=0  ---eqn.AL034
we get the t value where minimum P(t)
or maximum P(t) (or stationary) occur.
<a name="ch04c032">
Then t is not a variable any more,
because eqn.AL034 lock t to a specific
value. Let us set P'(t) to zero
  P'(t)=2t〈w,w〉-2〈v,w〉=0 ---eqn.AL035
solve for t we find
  t=〈v,w〉/〈w,w〉 ---eqn.AL036
This t is a constant, not a variable.
Let us use t0 to represent
this constant t. We write
  t0=〈v,w〉/〈w,w〉 ---eqn.AL037
<a name="ch04c033">
Fig4.6 suggest at t0 we find minimum
P(t) value. Let us put t0=〈v,w〉/〈w,w〉
into eqn.AL024, we will find the minumum
of P(t) is
  minimum of P(t) = h = ---eqn.AL038
  [〈v,v〉〈w,w〉-〈v,w〉2]/〈w,w〉2
LiuHH skip the detail process of
"find the minimum of P(t)". Result is
eqn.AL038. ALERT

<a name="ch04c034">　Index begin　Index this file
In eqn.AL038, denominator 〈w,w〉2 is
positive 
(assume w-sequence is not all zero)
eqn.AL038 numerator 
〈v,v〉〈w,w〉-〈v,w〉2 = 〈v-sw,v-sw〉 
(from eqn.AL023 to eqn.AL031)
〈v-sw,v-sw〉 is one vector v-sw self 
dot product, it must be positive or 
zero.
eqn.AL038 denominator ＞ 0
eqn.AL038 numerator ≧ 0

<a name="ch04c035">
from eqn.AL038 numerator ≧ 0 we get
"Cauchy's inequality for the nth time"

Note 1: Cauchy's inequality says
for v-sequence and w-sequence
two sequences, we have
〈v,v〉〈w,w〉≧〈v,w〉2 ---eqn.AL039
or same expression
〈v,w〉2≦〈v,v〉〈w,w〉 ---eqn.AL040

<a name="ch04c036">
Note 2: Cauchy's inequality work with
        TWO sequences.
        AM-GM inequality work with
        ONE sequence.
2009-11-12-20-36 stop

<a name="ch04c037">
2009-11-13-10-27 start
If you use numerical value to check the
minimum value point. Please be careful

  minimum of P(t) = h1 = ---eqn.AL038
  [〈v,v〉〈w,w〉-〈v,w〉2]/〈w,w〉2
and
  minimum of P(t) = h2 = ---eqn.AL041
  [〈v,v〉〈w,w〉-〈v,w〉2]/〈w,w〉

both are answer, but scale differently.
Please 
see eqn.AL038 no pre-multiply 〈w,w〉
and eqn.4.22 has pre-multiply 〈w,w〉
2009-11-13-10-31 stop



<a name="ch04c038">　Index begin　Index this file

2009-11-13-10-36 start
■　Exercise 4.4 problem statement
　　textbook page 68
（Lagrange's Identity for Complex Numbers）

Prove that for complex ak and bk
1≦k≦n one has

<a name="ch04c040">
2009-11-13-11-00 here
and show that this identity yields the
complex Cauchy inequality as well as the
necessary and sufficient conditions for
equality. Here one should note that this
identity does not follow from direct
substitution of complex numbers into the
Lagrange's identity for real numbers;
those pesky absolute values get in the
way. A slight more sophisticated approach
is required.
2009-11-13-11-05 here






<a name="ch04c041">
2009-11-13-11-07 here
■　Exercise 4.4 hint
　　textbook page 240
This exercise provides a remainder that
one sometimes needs a more elaborate
algebraic identity to deal with the
absolute values of complex numbers
than to deal with the absolute values 
of real numbers. Here the key is to use
Cauchy-Binet four letter identity 
Exercise 3.7 on page 49. The proof of 
that identity was purely algebraic (no
absolute value or complex conjugates 
were used) so the identity is also valid
for complex numbers. 
<a name="ch04c042">
One then just makes
the replacement
 ak to akbar
 bk to bk
 sk to ak and
 tk to bkbar
2009-11-13-11-15 here




<a name="ch04c043">
■　Exercise 4.4 solution


Not familiar with polarization 
transformation.
LiuHH skip Exercise 3.7 solution
 and  skip Exercise 4.4 solution
2009-11-13-11-19 stop



<a name="ch04c044">　Index begin　Index this file
2009-11-13-13-01 start
■　Exercise 4.5 problem statement
　　textbook page 69
（A Vector-Scalar Melange）

Consider real weights pj＞0 j=1,2,...,n
arbitrary real numbers αj j=1,2,...,n
and an inner product space (V,〈.,.〉).
Find an analog of Lagrange's identity 
which suffices to prove that one has
the inequality

<a name="ch04c046">
for all xk 1≦k≦n in V. Also,
check that your identity implies that
equality holds if and only if we have
  αjxk=αkxj ---eqn.AL042
for all 1≦j,k≦n 
2009-11-13-13-21 here







<a name="ch04c047">
2009-11-13-13-24
■　Exercise 4.5 hint
　　textbook page 240
The observation of S.S. Dragomir (2000)
shows how the principles behind Lagrange's
identity continue to bear fruit. Here one
 just takes the natural double sum and
expands:

<a name="ch04c052">
2009-11-13-13-56 here
This identity gives us our target bound
(4.24) and shows that the inequality is
strict unless 
  αjxk=αkxj ---eqn.AL042
for all j and k. Finally one should also
note that a corresponding inequality for
a complex inner product spaces can 
obtained by a similar calculation.
2009-11-13-13-59 stop



<a name="ch04c053">　Index begin　Index this file
2009-11-13-15-20 start
■　Exercise 4.5 solution


Exercise 4.5 hint already solved this
problem. Here make some observation.
Please pay attention to
real weights pj＞0 j=1,2,...,n
arbitrary real numbers αj j=1,2,...,n
and all xk 1≦k≦n in V.
<a name="ch04c054">
The difference between three sequences
is
 pj＞0 and pj*pk≠0 if j≠k
 αj＜＝＞0  αj*αk≠0 if j≠k
 xj＜＝＞0  xj*xk*cos(90)＝0 if j≠k
Given pj＞0 and not mention pj is a 
vector component sequence, then we
can not use pj*pk=0 , we must use
the general condition pj*pk≠0 if j≠k
Similar condition for αj＜＝＞0 .
<a name="ch04c055">
 xj＜＝＞0 is different, problem give
an inner product space (V,〈.,.〉) and
for all xk 1≦k≦n in V.
Vector has inner product. Inner product
use the rule
 xj*xk*cos(90)＝0 if j≠k
2009-11-13-15-40 here

<a name="ch04c056">　Index begin　Index this file
2009-11-13-17-42 start
■　Egg first ? or chicken first ?
(dot product first? or five rules first?)

When LiuHH write to "2009-11-13-15-40"
refer back to 
dot product rule. (Text page 7)
and wondering why dot product rule do 
not say a word about 
 〈v,w〉=v1*w1+v2*w2+v3*w3+.....+vn*wn ---eq.AA26
?
<a name="ch04c057">
How can I start from five rules
and draw conclusion that eq.AA26 is 
necessary true? If no special reason
two sequence products use eqn.AK029.
If sequence is vector components
then use eqn.AK032
<a name="ch04c058">
After think, LiuHH find that 
[[
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]
]]
two rules force calculation 
<a name="ch04c059">
from regular sequence products
 [-10, 10]*[-10, 10] =
  (-10)*(-10) + (-10)*(+10)
 +(+10)*(-10) + (+10)*(+10) = 0
change to vector sequence products
 [-10, 10]*[-10, 10] =
  (-10)*(-10) + 0
 +0 + (+10)*(+10) = 200
Because vector [-10, 10] has non-zero
length, [-10, 10] dot itself can not 
be zero.

<a name="ch04c060">
But many reference do not say this way.
"Mathematical Methods for Physicists"
third edition by George Arfken, 
ISBN 0-12-059820-5 page 14 line 5 say
[[
we define
 A dot B =Ax*Bx+Ay*By+Az*Bz ---eqn.AL047
]]

<a name="ch04c061">
Another book: Calculus Volume I 
second edition by Tom M. Apostol
page 451 bottom say
[[
If A=(a1,...,an) and B=(b1,...,bn)
are two vectors in Vn their dot product 
is defined by
  A dot B = ∑[k=1,n]{akbk} ---eqn.AL048
.....
This multiplication has the following 
algebraic properties. 
.....(same as fice rules)
]]

<a name="ch04c062">
2009-11-13-18-27 here
The key point of Exercise 4.5 is that 
when problem statement say
 "all xk 1≦k≦n in V."
then when j≠k, in eqn.AL045
the term pjpkαjαk〈xk,xj〉
pjpk is not zero,
αjαk is not zero, but
〈xk,xj〉  IS zero !
So vector dot product rule dominant
in eqn.AL045.

Exercise 4.5 is solved in 4.5 hint
LiuHH just point out the key 
difference between p-sequence,
α-sequence and x-sequence.
2009-11-13-18-45 stop

<a name="ch04c063">　Index begin　Index this file

2009-11-14-10-57 start
■　Exercise 4.6 problem statement
　　textbook page 69
（Ptolemy's Inequality）

Ptolemy may be best known for founding
a theory of planetary motion which was
overturned by Copernicus, but parts of
Ptolemy's legacy have stood the test
of time. Among these, Ptolemy has a
namesake inequality which even today is 
a workhorse of the theory of geometric
inequalities. 
<a name="ch04c064">
Ptolemy's inequality 
asserts that in a convex quadrilateral 
"the product of diagonals is bounded by
the sum of the products of the opposite
sides" or in the notation of Figure 4.7
  pq≦ac+bd ---eqn.4.25
Prove this inequality and show that
equality holds if and only if the four
vertices A,B,C,D are all on the 
circumference of a circle.
2009-11-14-11-07 here



<a name="ch04c065">
2009-11-14-11-09
■　Exercise 4.6 hint
　　textbook page 240
There are proofs of this inequality 
that use only the tools of plane 
geometry, but there is also an 
exceptionally interesting proof that
uses the transformation x→1/x for
complex number. There is no loss of
generality in setting A=0, B=z1, C=z2
and D=z3 and the triangle inequality
then gives us

<a name="ch04c067">
2009-11-14-11-22 here
which may be re-written as
  |z2||z1－z3|≦  ---eqn.AL050
  |z3||z1－z2|＋|z1||z2－z3|
After identifying these terms with help 
from Figure 4.7 we see that it is precisely
Ptolemy's Inequality!

<a name="ch04c068">
To prove the converse, we first note
that one has equality in this 
application of the triangle inequality
if and only if the points z1-1, z2-1,
z3-1 are on line. One then obtains the 
required characterization by appealing
to the fact that z→1/z takes a circle 
through the origin to a line and vise 
versa.

<a name="ch04c069">
The transformation z→1/z is perhaps
the leading example of a Möbius
transformation, which more generally
are the maps of the form 
  z→(az+b)/(cz+d) ---eqn.AL051
Every book on complex variables
examine these transformation, but the
treatment of Needham (1997), page 122-
188 is especially attractive. Needham 
also discusses Ptolemy's Inequality
with the help of inversion, but the 
quick treatment given here is closer
to that of Treibergs (2002).
2009-11-14-11-40 stop




<a name="ch04c070">　Index begin　Index this file
2009-11-14-15-56 start
■　Exercise 4.6 solution


Exercise 4.6 hint suggest use complex
number concept to solve problem.
Figure 4.7 has four points, A,B,C,D.
Their absolute location is not important,
only their relative location is important.
<a name="ch04c071">
We can put point A to (0,0) to simplify 
our work and still keep generality, also 
set
point B=z1 (a complex number)
point C=z2 (a complex number)
point D=z3 (a complex number)
Hint also suggest use the transformation
 x→1/x  ---eqn.AL052
for complex number. Then we need to
consider 1/z1 and 1/z2 and 1/z3 three 
numbers. Let us arrange it in the 
following form

<a name="ch04c073">
This is an identity. The red terms are
inserted zero.
eqn.AL053 left side say:
we start from 1/z3 move toward 1/z1
eqn.AL053 right side say:
we start from 1/z3 move toward 1/z2
then start from 1/z2 move toward 1/z1.
Both route have same start and end
points, but different path. 
left  side is  direct  path,
right side is indirect path.
<a name="ch04c074">
This is triangle relation. Let us
apply triangle inequality to three
sides.
side one is 1/z1 － 1/z3 ---eqn.AL054
side two is 1/z1 － 1/z2 ---eqn.AL055
side three: 1/z2 － 1/z3 ---eqn.AL056
<a name="ch04c075">
Triangle inequality says
In a triangle, the length of one side
is not greater than the sum of length
of other two sides.
The length of side one is its absolute
value |1/z1 － 1/z3|
Same condition apply to other two sides.
Based on triangle inequality we have
eqn.AL049
In eqn.AL049 we make common denominator
get

<a name="ch04c077">
eqn.AL057 multiply by |z1|*|z2|*|z3|
get
|z2|*|z3－z1|≦ ---eqn.AL058
|z3|*|z2－z1|＋|z1|*|z3－z2|

Now eqn.AL058 contains non-reciprocal
complex number and easy to identify as
following, 
<a name="ch04c078">
please see Page 69 fig.4.7
|z2| = AC = p
|z3－z1| = BD = q
|z3| = AD = d
|z2－z1| = BC = b
|z1| = AB = a
|z3－z2| = CD = c
eqn.AL058 is
  pq≦ac+bd ---eqn.4.25
exactly !
Problem solved.
2009-11-14-17-36 stop

<a name="ch04c079">
2009-11-14-19-19 start
The proof of Ptolemy's Inequality is a
smart concept. LiuHH first time see the
problem, think from direct approach, 
then how to create square areas for pq
and ac and bd three rectangles? It is
nearly impossible. After read the hint
section, still puzzle 
<a name="ch04c080">
why use complex number?
why take reciprocal of complex number?
the answer show up until the last step.
After make common denominator and 
multiply by |z1|*|z2|*|z3|
Suddenly beautiful picture show up.
Treat complex number as length, then
eqn.AL058 is exactly an area equation.
From 1/length to length squared !!
Genius !!
<a name="ch04c081">
LiuHH goto Internet search for 
"Ptolemy's Inequality", most page use
complex number to prove. Did not find
one use geometry to prove. Only one
page give geometry proof hint 

2009-11-13-22-06 LiuHH access
http://www.ideamath.org/samplegeo6.pdf
save as Ptolemy_ideamath.org_samplegeo6.pdf
page 1/3 has
<a name="ch04c082">
[[
Problem 3. Let A;B;C;D be four points in the
 plane. Let B*;C*;D* be the images of B;C;D
respectively under the inversion with center
 A and radius 1: Find the distances between
 B*;C*;D*
in terms of the distances between A;B;C;D:
 Deduce Ptolemy's Inequality.
]]
This is the only geometry proof related
hint LiuHH found online. This geometry
method still use inversion same as 
reciprocal.
2009-11-14-19-38 stop



<a name="ch04c083">
2009-11-15-13-14 start
■　Caution about triangle inequality

<a name="ch04c084">
2009-11-15-13-44 here
The Cauchy-Schwarz Master Class.
Vector equation oz + zy = oy is an equality.
Above vectors oz,zy,oy form a triangle OZY.
Vectors oz,zy,oy each has a x/y component.
When we apply triangle inequality to above
vector equation, the result is (CAUTION !)
(1)  a scalar equation |oz| + |zy| ≧ |oy|
(2)  a inequality eqn. |oz| + |zy| ≧ |oy|
Apply triangle inequality, changes are
vector equation become scalar equation,
equality equation become inequality equation.
2009-11-15-13-53 stop

2009-11-15-16-30 done proofread
2009-11-15-16-50 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