<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="ch03b001">　Index begin　Index this file
2009-10-17-17-28 start
■　Positive polynomials in one variable

The following is reading from two sources
First is textbook. Second is next paper
2009-06-24-23-33 LiuHH access
http://folk.uio.no/torkelah/Prosjekt.pdf
saveas folk.uio.no_torkelah_Prosjekt.pdf

<a name="ch03b002">
Next is Lagrange's Identity
It is derived in tute0014.htm#ch03a024
and concluded in tute0014.htm#ch03a037

<a name="ch03b003">
2009-10-17-17-40 here
Lagrange's Identity can be expressed 
by sum of two squared polynomials.
Its value is always non-negative. Then
we have a question:
If a polynomial is positive (or non-
negative), is it possible to express
positive polynomial as sum of squared
polynomials?
<a name="ch03b004">
Positive and non-negative are different.
Positive exclude zero, non-negative 
include zero. "positive polynomial" is 
a mathematics term and zero is allowed.
The following use "positive" loosely to
mean non-negative.
<a name="ch03b005">
If we find a polynomial it is positive
but its sum contains a negative of 
squared polynomial, then the answer is
negative. Since minus a square term do
not satisfy the positive polynomial
definition -- all addition.
2009-10-17-17-49 stop

<a name="ch03b006">　Index begin　Index this file
2009-10-17-18-25 start
■　second-degree positive polynomial
Let us see example first.

Next example come from
textbook page 41 and 42, also
folk.uio.no_torkelah_Prosjekt.pdf
page 12/20
Define second-degree positive polynomial
in one variable p2(x) to be
<a name="ch03b007">
  p2(x) = a*x*x + b*x + c ---eqn.AI001
Re-arrange, we get

<a name="ch03b008">
2009-10-17-18-40 here
We started at p2(x) is positive polynomial.
This assumption require the coefficient a
to be positive, and for ALL x value, p2(x)
be positive. Include x=-b/(2*a). Now let
<a name="ch03b009">
x=-b/(2*a), eqn.3.5 become
p2(x) = 0 + (4ac-b2)/(4*a) ---eqn.AI002
Because p2(x) is positive polynomial,
then (4ac-b2)/(4*a) must be positive.
<a name="ch03b010">
Let us define

<a name="ch03b011">
2009-10-17-18-53 here
Both Q1(x) and Q0(x) are positive
polynomial. we can write p2(x) as
  p2(x) = [Q1(x)]2 + [Q0(x)]2 ---eqn.AI005
eqn.AI005 is true under the constraint
  a ＞ 0 ---eqn.AI006
  4ac-b2 ≧ 0 ---eqn.AI007
2009-10-17-18-58 stop

<a name="ch03b012">　Index begin　Index this file
2009-10-18-12-15 start
Please pay attention to p2(x), Q1(x),
Q0(x) relation.
 p2(x) = a*x*x + b*x + c ---eqn.AI001
also
  p2(x) = [Q1(x)]2 + [Q0(x)]2 ---eqn.AI005
then
<a name="ch03b013">
  [Q1(x)]2 + [Q0(x)]2 = a*x*x + b*x + c ---eqn.AI008
eqn.AI008 is x to second power. 
[Q1(x)]2 can not exceed x second power, 
so does [Q0(x)]2. Then Q1(x) and Q0(x) 
can not exceed x first power.
Q1(x) is defined in eqn.AI003 a straight
         line.
Q0(x) is defined in eqn.AI004 a constant.
<a name="ch03b014">
Square of a straight line become 
quadratic curve. Let us see the 
data. Assume straight line is
  L(x) = 2*x-1  ---eqn.AI009
x increment, L(x) value and L(x)*L(x)
(short as LL(x)) values are next
<a name="ch03b015">
  x   : -1  -0.5  0  0.5  1  1.5  etc
 L(x) : -3  -2   -1  0    1  2    etc
LL(x) : +9  +4   +1  0    1  4    etc

<a name="ch03b016">
Third row for LL(x) is a quadratic
curve. LL(x) ALWAYS tangent to x-axis
at L(x)=0 point.
Above is eqn.3.5 left two terms.
both contains x. eqn.3.5 right end 
is not in above story.
<a name="ch03b017">
eqn.3.5 right end is Q0(x) 
  Q0(x) = √[(4ac-b*b)/(4a)] ---eqn.AI004
Q0(x) do not contain x, when x
varies, Q0(x) not change value! so
Q0(x) is a constant. With constraint
eqn.AI006 and eqn.AI007, Q0(x) is
a non-negative constant. If Q0(x)＞0
whole quadratic curve L(x)*L(x) parallel
move upward and L(x)*L(x) is always
positive. If Q0(x)=0, L(x)*L(x) stay
as is, not change.
2009-10-18-13-00 stop

<a name="ch03b018">　Index begin　Index this file
2009-10-18-14-19 start
■　third-degree polynomial
Above we discussed
 p2(x) = a*x*x + b*x + c ---eqn.AI001
Now let us increase power by one,
from second power of x to third power
  p3(x) = a*x*x*x + b*x*x + c*x + d ---eqn.AI010
New problem is to find next expression
  p3(x) = [Q0(x)]2 + [Q1(x)]2 + [Q2(x)]2 ---eqn.AI011
<a name="ch03b019">
Do you have any comment at this point?
Please click [draw 607] button below,
and pay attention to red curve.

<a name="ch03b020">
2009-10-18-15-10 here
Above curve show clearly that 
ODD power polynomial can NEVER be positive.
Because x→＋∞ and x→－∞ are of opposite
sign, ODD power polynomial must be
half positive and half negative. Our
topic is positive polynomial, p(x) is 
necessary even power polynomial.
<a name="ch03b021">
Since Qi(x) squared contribute to
peven(x), so Qi(x) can be even or odd.
(odd Qi(x) will be squared, become even)
peven(x) include p2(x), p4(x), p6(x) etc.
Each is composed by Q0(x), Q1(x), Q2(x),
Q3(x) etc. For general case
consider x be a variable vector with 
elements x1,x2,...,xn
<a name="ch03b022">　Index begin　Index this file
■　general variables general power
Now our question is
For positive polynomial P(x1,x2,...,xn)
is it possible to write next expression?
  P(x1,x2,...,xn)=Q12+Q22+...+Qs2 ---eqn.AI012
where
  Qk(x1,x2,...,xn) 1≦k≦s ---eqn.AI013
are real polynomials.
This is textbook page 41 problem 3.2

<a name="ch03b023">
The answer to this question shall be
done in two categories, one variable
and many variables.

one variable answer is positive.
two and more variable answer is negative.
2009-10-18-15-53 stop

<a name="ch03b023a">
2009-10-20-10-48 start
For question in ch03b022, 
the answer is:
if Qk(x1,x2,...,xn) allow to be
   polynomial divide by one, then answer
   is negative.
if Qk(x1,x2,...,xn) allow to be
   polynomial divide by another polynomial,
   then answer is positive.
Negative answer is Motzkin's example.
Positive answer is Hilbert's 17th problem.
Positive answer is Emil Artin solution.
Positive answer is Motzkin solved.
2009-10-20-10-57 stop

<a name="ch03b024">
2009-10-18-17-00 start
■　one variable general power
eqn.3.5 showed that if we define
as in eqn.AI003 and eqn.AI004, then
second power polynomial p2(x) can be
expressed as square sum of two lower
degree polynomials Q1(x) and Q0(x) 
Third power polynomial p3(x) can not
be positive polynomial.
<a name="ch03b025">
Fourth power polynomial p4(x) can be
positive polynomial. We use induction
method to prove p4(x). p4(x) equal to 
product of two second power polynomial
Q(x) and R(x) Following notation p4(x)
implies it is fourth and higher degree
polynomial.
The following analysis is valid for
any even power polynomial. Footnote 4
just remind us it is even power.

<a name="ch03b026">
  p4(x) = Q(x)*R(x) ---eqn.AI014
  p4(x) = (Q12+Q02)*(R12+R02) ---eqn.AI015
eqn.AI015 is not the form we want.
Look at eqn.AI003 eqn.AI004 eqn.AI005
Can we transform p4(x) to sum of two 
squared polynomial? Just like eqn.AI005
did. The following show eqn.AI015 right
hand side can be written as sum of two
squares. 
<a name="ch03b027">　Index begin　Index this file
■　Lagrange's Identity help
The tool we use is Lagrange's Identity
Lagrange's Identity left hand side is
product of two sum of squares. Right
hand side is sum of two squares. Exact
what we need !!
<a name="ch03b028">
Compare eqn.AI015 with eqn.3.4 aux2
we write the following
  p4(x) =  ---eqn.AI016
  [Q1(x)R1(x)+Q0(x)R0(x)]2
 +[Q1(x)R0(x)-Q0(x)R1(x)]2
eqn.AI016 is what we want, sum of two
squared polynomials.
2009-10-18-17-38 here

<a name="ch03b029">
■　factor to product of two lower power
Lagrange's Identity is handy, transform
to the format we want quickly. But the
starting point eqn.AI014 need discussion.
For a given polynomial p4(x), how to 
factor it to product of two lower power
terms? Textbook discuss this point in
<a name="twocase">
two different cases.
First  p4(x) has real roots.
Second p4(x) has NO real roots.

<a name="ch03b030">
■　p(x) has real roots
First, study p4(x) has real roots.
We can conclude positive polynomial
real roots must be repeated one root
only. If polynomial has two or more
distinct roots, then there must be 
some section in domain for which its
function value be negative. Please 
see draw607 black curve.
Negative value function can not be 
qualified as positive polynomial.
So positive polynomial can have only
repeated ONE real root.

<a name="ch03b031">
Assume positive polynomial can be
factored as
  P(x) = (x-r)m*R(x) ---eqn.AI017
r is repeated one root,
m is r repeat times.
Can m be even multiplicity? or odd 
multiplicity? If variable x stand 
exactly on top of r. It is hard to 
tell. 
<a name="ch03b032">　Index begin　Index this file
■　perturb x a little bit
Now let us perturb x a little bit. 
Let
  x = r + ε  ---eqn.AI018
where ε is a tiny number, ε can be 
positive or negative.
Put eqn.AI018 into eqn.AI017 get
  P(x) = (r+ε-r)m*R(r+ε)
  P(x) = (ε)m*R(r+ε)  ---eqn.AI019

<a name="ch03b033">
Since x=r is NOT a root of R(x), so
R(r) is not zero. If r were root of
R(x), then r should be factored as
(x-r) one more time, and R(x) reduce
x-power by one.
We assume R(x) is continuous,
We assume  ε   is small +/- value.
<a name="ch03b034">
R(r) is not zero, then there is a
δ＞0 with |ε|＜δ, in this section
eqn.AI019 tell us no matter how ε
vary positive or negative, the sign
of (ε)m not change, because P(x) not
change sign, R(x) not change sign.
how can tiny (ε)m change sign?
<a name="ch03b035">
For +ε or -ε, (ε)m keep same sign,
then m is even number. We can set
  m = 2*k  ---eqn.AI020
write
  P(x) = Q2(x)*R(x) ---eqn.AI021
where Q(x) = (x-r)k  ---eqn.AI022
<a name="ch03b036">
Please note that in eqn.AI022, power
is k, not 2k, not m. Because Q(x) 
will be squared! In eqn.AI021, 
P(x) and Q2(x) are positive, the
remaining R(x) must be positive.

Up to here, we started, P(x) factored
to product of two lower power terms.
2009-10-18-19-05 stop

<a name="ch03b037">　Index begin　Index this file
2009-10-18-19-52 start
■　p(x) has NO real roots
How to factor p(x) to product of two 
lower power terms? Above discussed
First  p(x) has real roots.
The following discuss
Second p(x) has NO real roots.
p(x) not tangent with x-axis.
<a name="ch03b038">
By the fundamental theorem of algebra,
there is a complex root r for which
p(r) = 0
We work with positive polynomial, all
of its coefficient are real number.
then p(r) = 0 implies 
complex_conjugate(p(r)) = 0
and
p(complex_conjugate(r)) = 0
<a name="ch03b039">
Let rc denote complex_conjugate(r)
We can write p(x) as
  p(x) = (x-r)(x-rc)*R(x) ---eqn.AI023
define Q(x) = (x-r)(x-rc) ---eqn.AI024
then
  p(x) = Q(x)*R(x) ---eqn.AI025
Polynomial Q(x) is positive for large
x and Q(x) has no real root, Q(x) not
cross x-axis. Q(x) is always positive.
<a name="ch03b040">
Simplest example is
  Q(x) = x*x + 1 ---eqn.AI026
Q(x) has roots +sqrt(-1) and -sqrt(-1)
Q(x) is positive when x is large.
for any real x, Q(x) always ≧ 1 ＞ 0

<a name="ch03b041">
NO-real-roots p(x) can be factored 
to Q(x)*R(x) and established start
point. Next Lagrange's Identity take
over the proof process.

By induction, we find that any non-
negative polynomial in one variable
can be written as the sum of squares
of two real polynomials.
2009-10-18-20-23 stop

<a name="ch03b042">　Index begin　Index this file
2009-10-19-10-25 start
■　multi-variables positive polynomial
This include two variables to many 
variables. Whether two-variables 
positive polynomial can be expressed
as sum of several lower power squares?
Answer to this question is either yes
or no.
If yes, then we should find a way to
prove it. 
If no, we should find an example.
Textbook page 44, 45 give one example
http://folk.uio.no/torkelah/Prosjekt.pdf
page 14/20 give same example.

Let us state Minkowski's conjecture first.
2009-10-19-10-52 stop

<a name="ch03b043">
2009-10-19-13-05 start
■　Minkowski's conjecture
Textbook page 44, line 1 to 6.
There exist nonnegative polynomial of
two variables that cannot be written as
the sum of squares of real polynomials.
This is precisely what the great
mathematician Hermann Minkowski first
suggested.

<a name="ch03b044">
The following is more about Minkowski's
conjecture.
2009-10-17-13-29 LiuHH accessed next page
http://www.isibang.ac.in/~sury/hilbert17.pdf
save as
positive_polynomial_isibang.ac.in_~sury_hilbert17.pdf
[[
<a name="ch03b045">
The 21-year old Minkowski presenting his
Inaugural Dissertation in July 1885 on 
quadratic forms made the bold conjecture
that there must exist homogeneous, real,
positive semi-definite polynomials of any
degree > 2 in n > 2 variables which are 
not sums of squares of homogeneous real 
polynomials. At the public defense of 
this Dissertation, it was the task of 
Hilbert to attack it but the defense 
ended with Hilbert declaring that he 
<a name="ch03b046">
"was convinced by Minkowski's exposition
 that already for n = 3 there may well 
 be such remarkable forms, which are so
 stubborn as to remain positive without
 allowing themselves to submit to a
 representation as sums of squares of
 forms." In 1888, Hilbert proved
Minkowski's conjecture giving examples 
where a positive semi-definite real 
polynomial f cannot be written as a 
sum of squares of polynomials.
]]

<a name="ch03b047">
In 1967 T. S. Motzkin give an example that
is a positive polynomial but can not be
expressed as sum of square of lower power
polynomials. (textbook page 46 middle and
folk.uio.no_torkelah_Prosjekt.pdf 
page 11/20)

<a name="ch03b048">　Index begin　Index this file
■　Motzkin's example start at AM-GM inequality
Motzkin's example is two variable positive
polynomial. Start from AM-GM inequality.

<a name="ch03b049">
2009-10-19-13-36 here
Let us set
  a1=x2 ---eqn.AI027
  a2=y2 ---eqn.AI028
  a3=1/(x2*y2) ---eqn.AI029
<a name="ch03b050">
Put a1, a2, a3 into eqn.2.3
we find

<a name="ch03b051">
2009-10-19-13-50 here
eqn.AI030 left hand side is one. 
Simplify to next equation.

<a name="ch03b052">
2009-10-19-14-00 here
re-arrange eqn.AI031 get
 P(x,y) = x4y2+x2y4-3x2y2+1≧0 ---eqn.AI032
eqn.AI032 is already in the form
  [x2y1] square +[x1y2] square
 -3*[x1y1] square +1 square ≧ 0
Clearly eqn.AI032 is not sum (addition)
of squared polynomials. Because '-3'
present in eqn.AI032. 
<a name="ch03b053">　Index begin　Index this file
But whether there is other combination 
for some term to absorb '-3' ? and show
up as sum of positive square only? Let 
us do further analysis. Our goal is
  P(x,y)=Q12(x,y)+Q22(x,y)+...+Qs2(x,y) ---eq.3.8
Target eqn.AI032 has highest power of 6
(in x4y2, 4+2=6)
Qi2(x,y) can not exceed power 6
Qi(x,y) can not exceed power 3
<a name="ch03b054">
Third power terms are x3y0, x2y1, x1y2
and x0y3. Exam eqn.AI032 again
when x=0, P(0,y) = 0 + 1 = 1 ---eqn.AI033
when y=0, P(x,0) = 0 + 1 = 1 ---eqn.AI034
eqn.AI033 not allow  x0y3 be used in 
Qi(x,y), because if y3 present in Qi(x,y)
then Qi(0,y)=y3+1 which can change value 
and not be a constant 1.
<a name="ch03b055">
Similarly, eqn.AI034 not allow x3y0 be
used in Qi(x,y). For third power 
term, we can use only x2y1, x1y2 .
Now define Qk(x,y) as following
  Qk(x,y)=ak+bkxy+ckx2y+dkxy2 ---eqn.3.9
where ak, bk, ck, dk are constants.
x and y are two variables. 

<a name="ch03b056">
eq.3.8 tell us:
sum of Qk(x,y) squares equal to P(x,y)
eqn.AI032 tell us:
this sum P(x,y) is x4y2+x2y4-3x2y2+1
then -3x2y2 must come from x1y1 in
Qk(x,y)
<a name="ch03b057">
Exam eqn.3.9, square of "+bkxy" is
b12+b22+...+bs2
Sum of square of real numbers can never
be negative (-3 in this case).
Up to here, we proved that NO combination 
for some term to absorb '-3' .
Positive polynomial eqn.AI032 can not
be expressed as sum of square of lower
power of polynomials.
Minkowski's conjecture is true.
2009-10-19-14-53 stop

<a name="ch03b058">　Index begin　Index this file
2009-10-19-17-32 start
■　Hilbert's 17th problem

2006-12-09-19-15 LiuHH access
http://aleph0.clarku.edu/~djoyce/hilbert/
save as hilbert-problem.htm This is
a short file with Hilbert picture.

2006-12-09-19-18 LiuHH access
http://aleph0.clarku.edu/~djoyce/hilbert/problems.html
save as hilbert-problems.html This is
longer file include all 23 problems.

The following is a copy from
http://folk.uio.no/torkelah/Prosjekt.pdf
page 19 of 20
[[
<a name="ch03b059">
In 1900 David Hilbert held a speech
 outlining 23 problems he considered
 to be fruitful questions for mathe-
maticians of that time. The 17th 
problem is the one that has been under
discussion here, and essentially it's
a conjecture by Hilbert -- namely that
<a name="ch03b060">
any positive polynomial can be written as
a sum of squares of rational functions.
As it turns out, this is the correct 
defining property of positive polynomials.
Hilbert's conjecture was resolved 
affirmatively in 1928 by Emil Artin, and
the proof is considered one of the 
greatest triumphs of modern algebra. 
It uses a number of results beyond what
 we have room for here, though, and so
 we shall not attempt to reproduce it --
it's given for instance in [3].

<a name="ch03b061">
For instance the problematic polynomial 
 s(x,y) = 1 -3x2y2 + x2y4 + x4y2 ---eqn.AI035
we considered earlier can indeed be
 written as a sum of squares of rational
functions, as required by Artin's result.
 It is done in the following way, due to
[3]:
<a name="ch03b062">

<a name="ch03b063">　Index begin　Index this file
Note also that in 1967 it was proven by
 Pfister (see [3]) that if p is a positive
polynomial in n variables then 2n squares
 will always be enough. It will often be
 possible, however, to write p as a sum
 of more squares than that. Note that
the representation of Motzkin's polynomial
 above illustrates this, using 22 = 4
squares.

<a name="ch03b064">
[3] Murray Marshall. Positive Polynomials
    and Sums of Squares. American
    Mathematical Society, 2008.
]]

Above is a copy from
http://folk.uio.no/torkelah/Prosjekt.pdf
page 19 of 20

2009-10-19-18-00 stop

2009-10-20-12-17 done proofread
2009-10-20-14-23 done spelling check



<a name="ch03b065">　Index begin　Index this file
2009-10-21-09-53 start
■　Exercise 3.1 problem statement
　　textbook page 47
（A Trigonometric Path to Discovery）

One only needs multiplication to verify 
the identity of Diophantus.
  (a1b1+a2b2)2 =  ---eqn.3.10
  (a12+a22)(b12+b22) - (a1b2-a2b1)2
yet multiplication does not suggest 
how such an identity might have been
discovered. Take the more inventive 
path suggested by figure 3.1 and show
that the identity of Diophantus is a 
consequence of the most famous theorem
of all, the one universally attributed
to Pythagoras (circa 497 B.C.)
2009-10-21-10-03 stop

<a name="ch03b066">
2009-10-21-15-12 start
Textbook fig 3.1 notes:
[[
The classic identity
  1=cos2(α+β)+sin2(α+β) ---eqn.AI038
permit one to deduce that
  (a12+a22)*(b12+b22) ---eqn.AI039
 =(a1b1+a2b2)2+(a1b2-a2b1)2

<a name="ch03b067">
Fig.3.1 In the right light, the identity
(3.10) of Diophantus and the theorem of
Pythagoras can be seen to be fraternal 
twins, though one is algebraic and the
other geometric.
]]
2009-10-21-15-21 stop



<a name="ch03b068">
2009-10-21-15-51 start
■　Exercise 3.1 hint
　　textbook page 236
From the four geometric tautologies.

2009-10-21-16-37 here






<a name="ch03b074">
■　Exercise 3.1 solution

Exercise 3.1 hint already solved 
problem. LiuHH skip Exercise 3.1.
2009-10-21-16-40 stop



<a name="ch03b075">　Index begin　Index this file
2009-10-21-17-41 start
■　Exercise 3.2 problem statement
　　textbook page 47
（Brahmagupta's Identity）

Brahmagupta (cirea 600 A.D.) established
an identity which shows that for any
integer D the product of two numbers
which can be written in the form 
a2-Db2 with a,b ∈ Z must be an integer
of the same form. More precisely, 
<a name="ch03b076">
Brahmagupta's identity says
 (a2-Db2)(α2-Dβ2)=(aα+Dbβ)2-D(aβ+αb)2 ---eqn.AI047

(a) Prove Brahmagupta's Identity by
evaluating the product
 (a+b√D)(a-b√D)(α+β√D)(α-β√D) ---eqn.AI048
in two different ways. Incidentally, the 
computation is probably more interesting
than you might guess.

<a name="ch03b077">
(b) Can you modify the pattern used to 
prove Brahmagupta's Identity to give
another proof of the identity (3.10)
of Diophantus?
2009-10-21-17-56 stop




<a name="ch03b078">
2009-10-21-18-05 start
■　Exercise 3.2 hint
　　textbook page 237
Here we just prove the identity of
Diophantus since Brahmagupta's Identity
is analogous. As expected, one first
factors. What is amusing is how one 
then recombines twice.
  (x12+x22)*(y12+y22) ---eqn.AI049
 =(x1-ix2)(x1+ix2)(y1-iy2)(y1+iy2)
 =[(x1-ix2)(y1+iy2)][(x1+ix2)(y1-iy2)]
<a name="ch03b079">
The first factor is 
 {(x1y1+x2y2)+i(x1y2-x2y1)} ---eqn.AI050
and the second factor is its conjugate
 {(x1y1+x2y2)-i(x1y2-x2y1)} ---eqn.AI051
so these have product
 (x1y1+x2y2)2+(x1y2-x2y1)2 ---eqn.AI052
a computation which reveals the power
of the factorization 
  a2+b2=(a+ib)(a-ib) ---eqn.AI053
in a most remarkable way.
2009-10-21-18-19 stop





<a name="ch03b080">
2009-10-21-18-28 start
■　Exercise 3.2 solution


The following is Brahmagupta's Identity
red/black group separately.

<a name="ch03b081">
 (a+b√D)(a-b√D)(α+β√D)(α-β√D) ---eqn.AI048
 =
 [(a+b√D)(α-β√D)]*[(a-b√D)(α+β√D)]
 =
 [aα-aβ√D+bα√D-bβ√D√D]*[aα-bα√D+aβ√D-bβ√D√D]
 =
 [aα-bβD-(aβ√D-bα√D)]*[aα-bβD+(aβ√D-bα√D)]
 =
 [aα-bβD]2-[aβ√D-bα√D]2
 =
 [aα-bβD]2-D[aβ-bα]2 ---eqn.AI049
2009-10-21-18-36

<a name="ch03b082">
 (a+b√D)(a-b√D)(α+β√D)(α-β√D)
Above is first  combination, red/black.
Below is second combination, red/black.
 (a+b√D)(a-b√D)(α+β√D)(α-β√D) ---eqn.AI048
 =
 [(a+b√D)(α+β√D)]*[(a-b√D)(α-β√D)]
 =
 [aα+aβ√D+bα√D+bβ√D√D]*[aα-bα√D-aβ√D+bβ√D√D]
 =
 [aα+bβD+(aβ√D+bα√D)]*[aα+bβD-(aβ√D+bα√D)]
 =
 [aα+bβD]2-[aβ√D+bα√D]2
 =
 [aα+bβD]2-D[aβ+bα]2 ---eqn.AI050
2009-10-21-18-50

<a name="ch03b083">
eqn.AI050 is same as eqn.AI047 in
Exercise 3.2 problem statement.
2009-10-21-18-52 stop

<a name="ch03b084">
2009-10-21-18-55
Next use
http://freeman2.com/complex2.htm#calculator
to find out the numerical result.

Paste next five lines in complex2.htm box3
[[
a1=1.23; b1=2.36; dd=7; a2=3.6; b2=4.12;
//next is eqn.AI048
(a1*a1-dd*b1*b1)*(a2*a2-dd*b2*b2)
//next is eqn.AI047
(a1*a2+dd*b1*b2)*(a1*a2+dd*b1*b2)-dd*(a1*b2+a2*b1)*(a1*b2+a2*b1)
]]

<a name="ch03b085">
both equation same answer
3967.0593774399995
3967.0593774399985
2009-10-21-19-01

a1=1.23; b1=2.36; dd=7; a2=3.6; b2=4.12;
//next is eqn.AI048
(a1*a1-dd*b1*b1)*(a2*a2-dd*b2*b2)
//next is eqn.AI049
(a1*a2-dd*b1*b2)*(a1*a2-dd*b1*b2)-dd*(a1*b2-a2*b1)*(a1*b2-a2*b1)

both equation same answer
3967.0593774399995
3967.05937744
2009-10-21-19-05



<a name="ch03b086">　Index begin　Index this file
2009-10-21-19-52 start
■　Exercise 3.3 problem statement
　　textbook page 48
（A Continuous Analogue of Lagrange's
 Identity）

Formulate and prove a continuous 
analogue of Lagrange's  Identity
Next, show that your identity implies
Schwarz's inequality and finally use
your identity to derive a necessary 
and sufficient condition for equality
to hold.



<a name="ch03b087">
2009-10-21-19-56 here
■　Exercise 3.3 hint
　　textbook page 237

One can pass from the discrete identity
to a continuous version by appealing
to the definition of the Riemann 
integral as a limit of sums, but it is
both easier and more informative to
consider the anti-symmetric form
 s(x,y)=f(x)g(y)-g(x)f(y) ---eqn.AI054
and to integrate s2(x,y) over the
square [a,b]2. In this way, one
finds

<a name="ch03b089">
2009-10-21-20-23 here
provided that all of the indicated 
integrals are well defined. 
Incidentally, anti-symmetric forms
often merit exploration. Surprisingly
often, they lead us to useful algebraic
relations.
2009-10-21-20-27 stop





<a name="ch03b090">
2009-10-21-20-30
■　Exercise 3.3 solution


  0.5*[f(x)g(y)－f(y)g(x)]*[f(x)g(y)－f(y)g(x)]
= 0.5*[f(x)g(y)*f(x)g(y) + f(y)g(x)*f(y)g(x)
     －f(y)g(x)*f(x)g(y) － f(y)g(x)*f(x)g(y)]
= [f(x)g(y)]^2 －[f(y)g(x)*f(x)g(y)]
=  f(x)f(x)*g(y)g(y)  －[f(y)g(x)*f(x)g(y)] ---eqn.AI055

<a name="ch03b091">
x and y are dummy variables
so         f(x)f(x)*g(y)g(y)
change to  f(x)f(x)*g(x)g(x) ---eqn.AI056
and        f(y)g(x)*f(x)g(y)
change to  f(x)g(x)*f(x)g(x) ---eqn.AI057

put integral sign back to eqn.AI056
and eqn.AI057 get eqn.14.50 second line.

<a name="ch03b092">
eqn.14.50 first line ≧ 0 and it is
Schwarz's inequality error part.

If drop eqn.14.50 first line, then
eqn.14.50 second line ≧ 0, it is
Schwarz's inequality.

eqn.14.50 first line and second line
together is Continuous Analogue of 
Lagrange's Identity.
2009-10-21-20-50 stop



<a name="ch03b093">　Index begin　Index this file
2009-10-22-09-47 start
■　Exercise 3.4 problem statement
　　textbook page 48
（A Cauchy Interpolation）

Show for 0≦x≦1 and for any pair of
real vectors (a1,a2,...,an) and
(b1,b2,...,bn) that the quantity

<a name="ch03b096">
2009-10-22-10-10 here
The charm of this bound is that for x=0
it reduce to Cauchy's inequality and for
x=1 it reduce to the algebraic identity
  {(a1+a2+...+an)(b1+b2+...+bn)}2
= (a1+a2+...+an)2(b1+b2+...+bn)2 ---eqn.AI059
Thus we have an inequality that 
interpolates between two known results.
2009-10-22-10-16 stop




<a name="ch03b097">　Index begin　Index this file
2009-10-22-10-39 start
■　Exercise 3.4 hint
　　textbook page 237

The two sides of the proposed inequality
can be written respectively as

<a name="ch03b102">
2009-10-22-11-22 here
The first term is a sum of squares 
and the second term is nonnegative
term, and the solution is complete.
The inequality of the problem is
from Wagner (1965) and the solution
is from Flor (1965)
2009-10-22-11-24 stop



<a name="ch03b103">　Index begin　Index this file
2009-10-22-14-22 start
■　Exercise 3.4 solution


in eqn.AI058 let n=3, get
{a1b1+a2b2+a3b3 //---eqn.AI064
+x*[ a1b2+a2b1+a1b3+a3b1+a2b3+a3b2]}2
= // below ---eqn.AI065
{a1b1+a2b2+a3b3
+x*[ a1b2+a2b1+a1b3+a3b1+a2b3+a3b2]
+a1b2+a1b3+a2b1+a2b3+a3b2+a3b1
-a1b2-a1b3-a2b1-a2b3-a3b2-a3b1
}2 // red terms are added-zero
<a name="ch03b104">
= // below ---eqn.AI066
{a1b1+a2b2+a3b3
+a1b2+a1b3+a2b1+a2b3+a3b2+a3b1
+x*[ a1b2+a2b1+a1b3+a3b1+a2b3+a3b2]
    -a1b2-a2b1-a1b3-a3b1-a2b3-a3b2
}2
= // below ---eqn.AI067
{[a1+a2+a3]*[b1+b2+b3]
+x*[ a1b2+a2b1+a1b3+a3b1+a2b3+a3b2]
-1*[+a1b2+a2b1+a1b3+a3b1+a2b3+a3b2
}2
<a name="ch03b105">
= // below ---eqn.AI068
{SUM[j=1,3]aj*SUM[k=1,3]bk
+(x-1)*[ a1b2+a2b1+a1b3+a3b1+a2b3+a3b2]
}2
= // below ---eqn.AI069
{SUM[j=1,3]aj*SUM[k=1,3]bk
+(x-1)*[ a1b2+a2b1+a1b3+a3b1+a2b3+a3b2]
+(1-x)*[a1b1+a2b2+a3b3]
-(1-x)*[a1b1+a2b2+a3b3]
}2  // red terms are added-zero again
<a name="ch03b106">
= // below ---eqn.AI070
{SUM[j=1,3]aj*SUM[k=1,3]bk
+(1-x)*[a1b1+a2b2+a3b3]
// next two lines contribute to SUM[]SUM[]
+(x-1)*[ a1b2+a2b1+a1b3+a3b1+a2b3+a3b2]
+(x-1)*[a1b1+a2b2+a3b3]
}2
// 2009-10-22-14-55 here
= // below ---eqn.AI071
{SUM[j=1,3]aj*SUM[k=1,3]bk
+(1-x)*[a1b1+a2b2+a3b3]
+(x-1)*SUM[j=1,3]aj*SUM[k=1,3]bk
}2
<a name="ch03b107">
= // below ---eqn.AI072
{ // (+1-1)SUM[]SUM[] cancel
+(1-x)*[a1b1+a2b2+a3b3]
+x*SUM[j=1,3]aj*SUM[k=1,3]bk
}2

Start from in eqn.AI058, end at eqn.AI072
eqn.AI072 is same as eqn.AI060
// 2009-10-22-15-00 here

<a name="ch03b108">
Use numerical value check equation
validity.
a1=1;a2=2;a3=3;b1=4;b2=5;b3=6;xx=0.6;
//next in eqn.AI058
a1*b1+a2*b2+a3*b3+xx*(a1*b2+a2*b1+a1*b3+a3*b1+a2*b3+a3*b2)
//next eqn.AI072
(1-xx)*(a1*b1+a2*b2+a3*b3)+xx*(a1+a2+a3)*(b1+b2+b3)
// 2009-10-22-15-12 here

<a name="ch03b109">
http://freeman2.com/complex2.htm#calculator
answer
a1*b1+a2*b2+a3*b3+xx*(a1*b2+a2*b1+a1*b3+a3*b1+a2*b3+a3*b2)
66.8
(1-xx)*(a1*b1+a2*b2+a3*b3)+xx*(a1+a2+a3)*(b1+b2+b3)
66.8

start equation and end equation same value.
2009-10-22-15-13 stop

<a name="ch03b110">
2009-10-22-16-38 start
Above is a simple method to verify
eqn.AI058 and eqn.AI060 are the same.
Above process use n=3, it is easy to
extend to n=n general case. Reader may 
wish to try. 
Up to here verified eqn.AI060 

<a name="ch03b111">
Next, verify eqn.3.11 and eqn.AI061 
aj side and bj are same thing. The 
following do aj side.

<a name="ch03b116">
2009-10-22-17-33 here
Above start from textbook page48 line 15 
and stop at textbook page 238 line 2. It
is aj only. bj side same story.
Up to here verified eqn.AI061 

<a name="ch03b117">
Exercise 3.4 hint say
[[
From which one finds that B－A can be written 
as the sum of the term .....
]]
Now continue expand B term
2009-10-22-17-45 here

<a name="ch03b122">
2009-10-22-18-25  here
B-A is the sum of eqn.AI077, eqn.AI078
eqn.AI079 and eqn.AI080.

x2 terms cancel.
(1-x)2 terms form eqn.AI063
x(1－x) terms form eqn.AI062

<a name="ch03b123">
Exercise 3.4 hint conclude
[[
The first term is a sum of squares 
and the second term is nonnegative
term, and the solution is complete.
]]
2009-10-22-18-31 stop

2009-10-23-11-28 done proofread
2009-10-23-15-00 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