<a name="ch12b048">
Above is coefficient analysis.
Not involve the root of Q(t).
But Ek of Q(t) is defined on
roots of Q(t). 
See Lowercase ek(x) defined
and Uppercase Ek(x) defined
We take second, root-based 
route to find Ek of Q(t). 
Problem given {x1,x2,...,xn}
as roots of P(t). 
<a name="ch12b049">
After differentiate P(t),
get Q(t), Q(t) has a set of 
roots {y1,y2,...,yn-1}
Which should be different from
{x1,x2,...,xn}, unless there are 
repeated root in {x1,x2,...,xn}.
Repeated root not change analysis.
Build Q(t) from n-1Ek(y) get
<a name="ch12b050">　Index begin　Index this file
  Q(t)=P'(t)/n= ---eqn.BK040
 =(t-y1)(t-y2)...(t-yn-1)

<a name="ch12b051">
eqn.BK041 and eqn.BK036 both
express Q(t), they are the same
equation from two different path
Equate Ek from eqn.BK036
and Ek from eqn.BK041 get
  Ek(x1,x2,...,xn)
 =Ek(y1,y2,...,yn-1) ---eqn.12.6
If use tute0043.htm express 
method, eqn.12.6 is next
  nEk(x)=n-1Ek(y) ---eqn.BK042
eqn.12.6 is our main result.
2010-05-20-21-38 stop

<a name="ch12b052">　Index begin　Index this file
2010-05-20-23-19 start
■　Why is it so remarkable?

What benefit we get from eqn.12.6?
Textbook page 181 explain as 
following.
===textbook copy start===
Why is it so remarkable?
The left-hand side of the identity
(12.6) is a function of the n
vector x=(x1,x2,...,xn) while the 
right side is a function of the 
n-1 vector y=(y1,y2,...,yn-1)
<a name="ch12b053">
Thus if we can prove a relation
such as
  0≦F(E0(y),E1(y),...,En-1(y)) ---eqn.BK043
for all y∈[a,b]n-1.
Then it follows that we also 
have the relation
  0≦F(E0(x),E1(x),...,En-1(x)) ---eqn.BK044
for all x∈[a,b]n.
<a name="ch12b054">
That is, any inequality -- or
identity -- which provides a
relation between the n-1 
quantities E0(y),E1(y),...,En-1(y) 
and which is valid for all 
values of y∈[a,b]n-1 extends 
automatically to a corresponding
relation for the n-1 quantities
E0(x),E1(x),...,En-1(x) which is
valid for all x∈[a,b]n.

<a name="ch12b055">
This presents a rare but valuable
situation where to prove a
relation for function of just n-1
variables. This observation can
be used in an ad hoc way to produce 
many special identities which
<a name="ch12b056">
otherwise would be completely
baffling, and it can also be used
systematically to provide seamless
induction proofs for results such 
as Newton's inequalities.
===textbook copy stop===
2010-05-20-23-39 stop

<a name="ch12b057">　Index begin　Index this file
2010-05-21-10-50 start
We had numerical example
  x=[a,b,c,d,e]=[1,2,3,4,5] ---eqn.BK004
 P5(t)=  ---eqn.BK014
 =(t-a)*(t-b)*(t-c)*(t-d)*(t-e)
 =t*t*t*t*t -t*t*t*t*15 +t*t*t*85
 -t*t*225   +t*274  -120
<a name="ch12b058">
That is
  P5(t)=t5－15*t4+85*t3－225*t2+274*t－120 ---eqn.BK045
  P4(t)=t4－12*t3+51*t2－90*t+54.8 ---eqn.BK046
P4(t) and P5(t) are related by
  P4(t)=d[P5(t)]/dt ---eqn.BK047

<a name="ch12b059">
From one view point that P4(t) is 
created from P5(t), no need go a
redundant loop find P4(t) roots
y and re-assemble P4(t) from y.

<a name="ch12b060">
From second view point that 
n-1Ek(y) should be generated
by P4(t) roots, just like nEk(x) 
are generated by P5(t) roots x.
Textbook equate two E's
  Ek(x1,x2,...,xn)
 =Ek(y1,y2,...,yn-1) ---eqn.12.6
<a name="ch12b061">
or in tute0043.htm notation
  nEk(x)=n-1Ek(y) ---eqn.BK042
where n-1Ek(y) is created by 
'redundant' step, that is n-1Ek(y) 
is created by P4(t) roots y.

<a name="ch12b062">　Index begin　Index this file
The following verify that 
P4(t) roots y create n-1Ek(y) 
Starting point is
  P4(t)=t4－12*t3+51*t2－90*t+54.8 ---eqn.BK046
end point is still eqn.BK046.

<a name="ch12b063">
Please goto (local)
http://freeman2.com/polyroot.htm#begin0
Click [04] button for 4th order
polynomial. Then fill in the
following coefficients
54.8  －90  51  －12  1
where 54.8 goto box marked [C00]
and in [C04] fill 1.
<a name="ch12b064">
Click [fast get roots] button
Box 1, answer has
[[
Below is answer, complex polynomial root ([1,2] same as 1+2i)
1st root is [1.355567131841731,0]
2nd root is [2.456088843227607,0]
3rd root is [3.543910070325765,0]
4th root is [4.644433954604896,0]

Below is answer too, complex number only, no text.
1.355567131841731+0i
2.456088843227607+0i
3.543910070325765+0i
4.644433954604896+0i
]]

<a name="ch12b065">
P4(t) roots y has
1.355567131841731
2.456088843227607
3.543910070325765
4.644433954604896

<a name="ch12b066">　Index begin　Index this file
Next goto(local)
http://freeman2.com/complex2.htm#calculator
in [Box3, input JS command]
fill
[[
f=1.355567131841731
g=2.456088843227607
h=3.543910070325765
i=4.644433954604896
e1=f+g+h+i
E1=e1/bicof(4,1)
e2=f*g+f*h+f*i+g*h+g*i+h*i
E2=e2/bicof(4,2)
e3=f*g*h+f*g*i+f*h*i+g*h*i
E3=e3/bicof(4,3)
e4=f*g*h*i
E4=e4/bicof(4,4)
E1
E2
E3
E4
]]
<a name="ch12b067">
Next click
[Test box3 command, output to box4]
Box 4 show up next
[[
E1
3
E2
8.5
E3
22.500000654128463
E4
54.80000354686018
]]
<a name="ch12b068">
Above is E1,E2,E3,E4 from P4(t)
roots y //P4=4th order polynomial
Please goto P5(t) E's
compare E1,E2,E3,E4 [P4 no E5]
Result show that E1,E2,E3,E4 
from P4 and from P5 are the
same !!
<a name="ch12b069">
They should be identical, 
because we walked a loop.
Above numerical experiments
confirmed textbook eqn.12.6
2010-05-21-11-46 stop

<a name="ch12b070">　Index begin　Index this file
2010-05-21-15-20 start
■　Prove high power end Newton 
　　Inequality 

We have eqn.12.6 in hand, for
example, a data set of ten
numbers 
  x=[x1,x2,...,x10] ---eqn.BK048
and ask to prove
  10E3(x)*10E5(x)≦10E42(x) ---eqn.BK049
is true. 
<a name="ch12b071">
eqn.12.6 tell us that use x to 
build a tenth order polynomial
p10(t) use x as its roots.
Differentiate p10(t) to fifth
order polynomial p5(t). Then
prove next inequality eqn.BK050
<a name="ch12b072">
  5E3(x)* 5E5(x)≦ 5E42(x) ---eqn.BK050
If eqn.BK050 is true, then 
eqn.BK049 is true by eqn.12.6 
Key point is red 5
10E5 is 10th order polynomial
middle term, hard to prove.
5E5 is 5th order polynomial
end term, easy to prove.
<a name="ch12b073">
eqn.BK050 is p5(t)'s high power
end inequality. This point can
be seen from the term 5E5(x)
for uEv(x), we must have u≧v. 
If u=v, that is the end, and 
no more. Because vEv(x) multiply 
by tzero, vEv(x) is a constant. 
Next differentiation v-1Ev(x) 
is gone.  2E3 do NOT exist.

<a name="ch12b074">
For above reason, we only need
to prove high power end Newton 
Inequality.
Newton's Inequality general form
is next
Ej-1(x1,x2,...,xn)*Ej+1(x1,x2,...,xn)≦
Ej2(x1,x2,...,xn) ---eqn.12.7
2010-05-21-15-52 here

<a name="ch12b075">
If we have one number in hand, 
that is 
  x=[x1] ---eqn.BK051
There is nothing to do.

<a name="ch12b076">　Index begin　Index this file
■　Prove Newton Inequality n=2
Textbook page 182

If we have two data in hand, 
that is 
  x=[x1,x2] ---eqn.BK052
Use x1,x2 as two roots, build
a polynomial P2(t).
  P2(t)=(t-x1)*(t-x2) ---eqn.BK053
       =t*t-t*(x1+x2)+x1*x2
       =t*t-2*t*[(x1+x2)/2]+x1*x2
<a name="ch12b077">
On the other hand
  P2(t)=t2*2E0(x1,x2)
     -2*t1*2E1(x1,x2)
     +1*t0*2E2(x1,x2) ---eqn.BK054
<a name="ch12b078">
For the creation of eqn.BK054
please read start from
Chapter 12: Symmetric Sums
to at least 
Uppercase Ek(x) defined.
eqn.BK053 and eqn.BK054 are 
both expression of P2(t).
<a name="ch12b079">
We have
  2E0(x1,x2)=1 ---eqn.BK055
  2E1(x1,x2)=[(x1+x2)/2] ---eqn.BK056
  2E2(x1,x2)=x1*x2 ---eqn.BK057
<a name="ch12b080">
For n=2 simplest case, 
Newton's inequality is
  E0(x1,x2)* E2(x1,x2)
 ≦ E12(x1,x2) ---eqn.12.8
Above is textbook equation.
Below is tute0043.htm equation.
  2E0(x)* 2E2(x)≦ 2E12(x) ---eqn.BK058
Put eqn.BK055, eqn.BK056 and
eqn.BK057 into eqn.BK058, get
  1*[x1*x2]?≦?[(x1+x2)/2]2 ---eqn.BK059
<a name="ch12b081">
We assume that x1＞0 and x2＞0
so that x1*x2＞0. Now
take square root of eqn.BK059
  [x1*x2]1/2?≦?(x1+x2)/2 ---eqn.BK060
eqn.BK060 is certainly true.
<a name="ch12b082">
[x1*x2]1/2 is Geometric Mean
and
(x1+x2)/2 is Arithmetic Mean
GM≦AM is a matter of course.
(one year ago, LiuHH is inequality
 stranger. Now LiuHH is sophomore)
2010-05-21-16-31 here

Above is n=2 case.
<a name="ch12b083">　Index begin　Index this file
■　Prove Newton Inequality n=3
Textbook page 182

Below is n=3 case.
We have three data in hand.
  x=[x1,x2,x3] ---eqn.BK061
We use x build third degree 
polynomial P3(t). To save work
express P3(t) with 3Ek(x) 
directly
<a name="ch12b084">
P3(t)=1*t3*3E0(x1,x2,x3)
     -3*t2*3E1(x1,x2,x3)
     +3*t1*3E2(x1,x2,x3)
     -1*t0*3E3(x1,x2,x3) ---eqn.BK062
<a name="ch12b085">
eqn.BK062 coefficient are
[1,-3,+3,-1]. Neglect +/-
[1,+3,+3,+1] is 
bicof(3,0)=1
bicof(3,1)=3
bicof(3,2)=3
bicof(3,3)=1
<a name="ch12b086">
Why eqn.BK062 involve
binomial coefficients?
Please see from
Lowercase ek(x) defined.
to
Uppercase Ek(x) defined.

<a name="ch12b087">
P3(t) eqn.BK062 contain two
Newton's inequalities. They are
  E0(x1,x2,x3)* E2(x1,x2,x3)≦ E12(x1,x2,x3) ---eqn.12.9
  E1(x1,x2,x3)* E3(x1,x2,x3)≦ E22(x1,x2,x3) ---eqn.12.10
Above is textbook equation.
Below is tute0043.htm equation.
  3E0(x)* 3E2(x)≦ 3E12(x) ---eqn.BK063
  3E1(x)* 3E3(x)≦ 3E22(x) ---eqn.BK064
<a name="ch12b088">　Index begin　Index this file
for n=3 case, eqn.BK063 do not 
have 3E3(x). We can push eqn.BK063
to lower order polynomial. Compare
eqn.BK063 with eqn.BK058 below
  3E0(x)* 3E2(x)≦ 3E12(x) ---eqn.BK063
  2E0(x)* 2E2(x)≦ 2E12(x) ---eqn.BK058
<a name="ch12b089">
Lower right corner 0,2,1 are
the same in both equation.
Lower left  corner 3,2 tell us
that job of P3(t) hand over to
P2(t). Base on eqn.12.6 and 
already proved n=2 case
eqn.BK058, we conclude that 
eqn.BK063 is true.

<a name="ch12b090">
eqn.BK064 is different, because 
it has the term 3E3(x). We can 
not push eqn.BK064 to lower
order polynomial. Because
v-1Ev(x) is undefined.
v-1Ev(x) is non-exist.
<a name="ch12b091">
Lower left corner sub-note 'v-1' 
means polynomial order is v-1.
Lower right corner sub-note 'v' 
means Symmetric Sum has order v.
We have v-1 data in hand, we can
not create v-th multiplication.
Same argument in tute0042.htm
Five numbers can not have power 
six.

<a name="ch12b092">
We can not push eqn.BK064 to lower
order polynomial. We must prove it
right here n=3. This part is done
in textbook page 183

<a name="ch12b093">　Index begin　Index this file
We need to find the definitions
  3E1(x)=(x1+x2+x3)/3 ---eqn.BK065
  3E2(x)=(x1*x2+x1*x3+x2*x3)/3 ---eqn.BK066
  3E3(x)=x1*x2*x3 ---eqn.BK067
Above three equations are 
Symmetric Sum of x1,x2,x3 three
data set.
<a name="ch12b094">
eqn.BK065 and eqn.BK066 divide 
by three are taking average.
Please see one to one
Put eqn.BK065 to eqn.BK067 into
eqn.BK064. The equation to prove
is next

<a name="ch12b096">
To prove high power end Newton 
Inequality, it is always easier 
to prove if use the reciprocal 
of x. In this case we divide 
whole equation by the n-th
multiplied term 
  nEn(x)=(x1*x2*...*xn) ---eqn.BK301
Inserted equation, use number 300+
<a name="ch12b097">
For this reason (divide by x), 
we require that the observed data 
x1,x2,...,xn all be positive. 
Zero is not allowed.
<a name="ch12b098">　Index begin　Index this file
(x1*x2*x3)2 is expressed as 3E32(x) 
Divide eqn.12.11 by (x1*x2*x3)2
get

<a name="ch12b101">
Whether eqn.BK070 is true?
From rearrangement inequality
view point, eqn.BK070 is true.
Because we can assume the order
  0＜x1≦x2≦x3 ---eqn.BK071
then reciprocal give us
  1/x1≧1/x2≧1/x3 ---eqn.BK072
<a name="ch12b102">
Consider 
sequence_A=[1/x1, 1/x2, 1/x3]
sequence_B=[1/x2, 1/x3, 1/x1]
Rearrangement inequality say
sequence_A dot product sequence_A
is greater than or equal to
sequence_A dot product sequence_B

<a name="ch12b103">　Index begin　Index this file
Whether eqn.BK070 is true?
From Cauchy's inequality
view point, eqn.BK070 is true.
Because we can assume the 
sequences
sequence_C=[1/x1, 1/x3, 1/x2]
sequence_D=[1/x2, 1/x1, 1/x3]
<a name="ch12b104">
sequence_C dot product sequence_D
get eqn.BK070 less than side
sequence_C dot product sequence_C
get square root of eqn.BK070 
greater than side.
sequence_D dot product sequence_D
get square root of eqn.BK070 
greater than side.
Net result is eqn.BK070.

<a name="ch12b105">
Whether eqn.BK070 is true?
From AM-GM inequality view point,
eqn.BK070 is true.
Because we can do three times
AM-GM inequality, then sum
them to eqn.BK070. 
<a name="ch12b106">
That is
1/√(x1x1x2x2)≦[1/(x1x1)+1/(x2x2)]/2 ---eqn.BK073
1/√(x1x1x3x3)≦[1/(x1x1)+1/(x3x3)]/2 ---eqn.BK074
1/√(x2x2x3x3)≦[1/(x2x2)+1/(x3x3)]/2 ---eqn.BK075
Sum eqn.BK073 to eqn.BK075 get
eqn.BK070.

<a name="ch12b107">
Conclude:
For n=3 case, Newton Inequality 
   eqn.12.10 
or eqn.12.11 
or eqn.BK064
or eqn.BK070 
is true.
2010-05-21-18-59 stop

<a name="ch12b108">　Index begin　Index this file
2010-05-21-22-30 start
■　Prove Newton Inequality n=any
Textbook page 183

Newton's inequality in general
case, 
  Ek-1(x)*Ek+1(x)≦Ek2(x) ---eqn.12.2
          for 0＜k＜n
<a name="ch12b109">
x contain hidden condition
that is x has n elements,
polynomial has degree=n.
Equation contain Ek+1(x) 
which can not exceed En(x)
k+1≦n, then k＜n .
Equation contain Ek-1(x) 
which can not be smaller than 
E0(x), 0≦k-1 then 0＜k .

<a name="ch12b110">
If k+1＜n, we have
  Ek-1(x)*Ek+1(x)≦Ek2(x) ---eqn.12.12
          for 1≦k＜n-1
eqn.12.12 condition 1≦k＜n-1
eqn.12.2  condition 0＜k＜n
The inequality in the group
k+1＜n can be push down to
lower order polynomial's 
high end inequality. Done
proof in low order polynomial.

<a name="ch12b111">
If k+1=n, Ek+1(x)=En(x), we have
  En-2(x)*En(x)≦En-12(x) ---eqn.12.13
this is a boundary inequality.
We can not push En(x) down to
lower order polynomial. Because
there is no more same E downward.
Must prove eqn.12.13 right here.

<a name="ch12b112">　Index begin　Index this file
We write eqn.12.13 in longhand
and use x^j remind us that this 
xj is a missing term.
Why missing? The only one not
missing any is nEn(x). All
elements multiply together. The
next one nEn-1(x) has one missing 
in symmetric sum.
<a name="ch12b113">
Example, assume x=[a,b,c,d,e]
5E5(x)=a*b*c*d*e no one missing.
5E4(x)=b*c*d*e+a*c*d*e
      +a*b*d*e+a*b*c*e
      +a*b*c*d
In which 
"b*c*d*e" missing an 'a'
"a*c*d*e" missing a  'b'
etc. 
<a name="ch12b114">
If we write "b*c*d*e" as
"a^*b*c*d*e", a^=pure number 1
at the step whole equation 
divide by "a*b*c*d*e" , the 
missing one "x^j" show up at 
denominator. 
Use "a^" or use "x^j" help 
us do bookkeeping right.
2010-05-21-23-22 stop

<a name="ch12b115">
2010-05-22-09-26 start
We begin prove eqn.12.13 which 
is the high power end Newton's 
inequality. eqn.12.13 use the
following symmetric sums

<a name="ch12b118">
2010-05-22-10-28 here
For n elements sequence, we build
n-th order polynomial. There are
n-1 Newton's inequality. They are
<a name="ch12b119">　Index begin　Index this file
  nEn-2(x)*nEn(x)≦nEn-12(x) ---eqn.12.13
  nEn-3(x)*nEn-1(x)≦nEn-22(x) ---eqn.BK079
  nEn-4(x)*nEn-2(x)≦nEn-32(x) ---eqn.BK080
..... //Red are boundary terms
<a name="ch12b120">
  nE2(x)* nE4(x)≦ nE32(x) ---eqn.BK081
  nE1(x)* nE3(x)≦ nE22(x) ---eqn.BK082
  nE0(x)* nE2(x)≦ nE12(x) ---eqn.BK083
<a name="ch12b121">
Boundary Newton's inequality
eqn.12.13 high power end and
eqn.BK083 low  power end are
easy to prove. Because there
are less terms involved.
bicof(10,10)=1, bicof(10,9)=10;
bicof(10,1)=10, bicof(10,0)=1.
<a name="ch12b122">
Middle Newton's inequality has
many terms and hard to sort out
a form for inequality argument.
Please see a hard work example
at here and get trouble at here.
bicof(10,5)=252, bicof(10,4)=210;
after square 252*252=63504 terms
<a name="ch12b123">　Index begin　Index this file
Now we prove boundary Newton's 
inequality eqn.12.14, it is much
easier than the middle Newton's.
2010-05-22-11-09 here

<a name="ch12b124">
We assume all data x1,x2,...,xn
are positive numbers, no zero.
We divide eqn.12.14 by square
of x1*x2*...*xn convert high
power (n-th power x1*x2*...*xn)
to reciprocal of low power 1/x1
or 1/(x2*x3) etc. This division 
let x^j change from invisible 
one to visible 1/xj. 
After division we get

<a name="ch12b127">
2010-05-22-11-43 here
How to prove eqn.12.15?
Textbook page 183 explain as 
following.
===textbook copy start===
<a name="ch12b128">　Index begin　Index this file
We could now stick with the 
pattern that worked for H3, 
but there is a more graceful 
way to finish which is almost 
staring us in the face. 
<a name="ch12b129">
If we adopt the language of
symmetric function, the 
target bound (12.15) may be 
written more systematically as
<a name="ch12b130">

<a name="ch12b131">
and one now sees that this
inequality is covered by the
first bound of the group 
(12.12). Thus the proof of
Newton's inequality is 
complete.
===textbook copy stop===
2010-05-22-11-54 here

<a name="ch12b132">　Index begin　Index this file
First time read here, LiuHH
was very puzzle. 
Why?! 
Newton's inequality is designed 
for x. Now the data become
  y=[1/x1,1/x2,...,1/xn] ---eqn.BK086
How can I use x equation to
prove y property?

<a name="ch12b133">
After more think, find out why.

Newton's inequality is designed 
for x. That is very true.
x and y are different, that is 
also true.
<a name="ch12b134">
We need to see what limitation
applied to data x ?
Review from the beginning,
Newton's inequality require
x has n elements, otherwise
it is another set Newton.
<a name="ch12b135">
Require x has all positive
elements, so that division 
of x1*x2*...*xn is possible.
That is all. 
<a name="ch12b136">
Then y satisfy
these two conditions too. So
x's Newton inequality apply
to y perfect. However,
<a name="ch12b137">
we must differentiate n-th
order polynomial Pn(t) all 
the way down to P2(t).
['2' in E2(y) demand P2(t)]
This is not a problem, as 
long as the proof is DONE.

Problem 12.1 is solved.
2010-05-22-12-13 stop

<a name="ch12b138">　Index begin　Index this file
2010-05-22-16-01 start
■　Prove Maclaurin Inequality

Maclaurin's inequality asserts 
that
  [En(x)]1/n ≦ [En-1(x)]1/(n-1) ≦ ... 
  ≦ [E2(x)]1/2 ≦ E1(x) ---eqn.12.3
<a name="ch12b139">
Here we still use eqn.12.6
With the property
  nEk(x)=n-1Ek(y) ---eqn.BK042
where 1≦k＜n
we only prove high power end 
Maclaurin Inequality. 
<a name="ch12b140">
Middle power Maclaurin Ineq.
go to lower order polynomial's
high power end, like Newton did. 
Please see Push down diagram 
Newton use 3 E, Maclaurin use 2. 
uEv(x) u=v indicate reach end.
Diagram colored for Newton but
same reason apply to Maclaurin.
<a name="ch12b141">
Let us exam high power end 
Maclaurin Inequality
  [En(x)]1/n ?≦? [En-1(x)]1/(n-1) ---eqn.BK087
The requirement to x is that
x has n elements and all 
of x elements are positive.
If we normalize x such that
Geometric Mean of x is one
  x1*x2*...*xn=1 ---eqn.BK088
<a name="ch12b142">　Index begin　Index this file
Normalize x still satisfy
two requirements.
eqn.BK087 raise power to (n-1)
eqn.BK087 become
  1 ?≦? En-1(x) ---eqn.BK089
Divide eqn.BK089 by 
x1*x2*...*xn=1, we have

<a name="ch12b145">
eqn.BK091 is AM-GM inequality.
for the sequence
  y=[1/x1,1/x2,...,1/xn] ---eqn.BK086
which is true.
Proof of Maclaurin Inequality
is done.
2010-05-22-16-44 stop

<a name="ch12b146">　Index begin　Index this file
2010-05-22-17-07 start
■　'high power end'? why it is t0?
The term 'high power end' is 
used frequently in this page.
Sometime it may be confuse.
Please see polynomial in terms 
of E's. In one equation, each 
Eu*tv must have u+v=constant.
Please see related topic.
<a name="ch12b147">
'high power end' for E, u=n is
' low power end' for t, v=0 .
' low power end' for E, u=0 is
'high power end' for t, v=n .
<a name="ch12b148">
In this page, the term
'high power end' is reserved 
for E, not for t. If you see
notes say 'high power end',
look for E, do not look at t!

<a name="ch12b149">
Polynomial  P6(t) differentiate
once become P5(t) 
Eu*tv u+v=constant 6.
change to
Eu*tw u+w=constant 5.
tv reduce power by one.
Eu is constant (no t), 
Eu is unchanged.
2010-05-22-17-25 stop

<a name="ch12b150">　Index begin　Index this file
2010-05-22-18-30 start
■　Equality in Newton and Maclaurin

Newton's inequality eqn.12.2.
Maclaurin's inequality eqn.12.3
equality condition is
  x1=x2=...=xn ---eqn.BK092

<a name="ch12b151">
For n=3 case, Prove Newton's
Inequality, used three methods
From rearrangement inequality
From Cauchy's inequality
From AM-GM inequality.

<a name="ch12b152">
We explain equality condition
from three methods too.
The basic point is that we are
given one data set 
  x=[x1,x2,...,xn] ---eqn.BK093
we can not use xn+1.
<a name="ch12b153">
Use numerical example for easy
illustration. Assume given
  seq_A1=[1,2,3,4,5] ---eqn.BK094

<a name="ch12b154">　Index begin　Index this file
Rearrangement inequality need
just one data set. Second seq.
is re-arranged from first. 
Define
  seq_B1=[3,2,4,1,5] ---eqn.BK095

<a name="ch12b155">
Rearrangement inequality get
  seq_A1*seq_B1 ≦ seq_A1*seq_A1 ---eqn.BK096
  1*3+2*2+3*4+4*1+5*5  //48≦
 ≦1*1+2*2+3*3+4*4+5*5  //55
How can we do re-arrangement
still get equality?
The only answer is that given
  seq_A2=[2,2,2,2,2] ---eqn.BK097
All elements are the same.
Re-arrange please. 
Any seq_B2 is still [2,2,2,2,2]

<a name="ch12b156">
The condition
  x1=x2=...=xn ---eqn.BK092
let us get equality in
Newton's inequality eqn.12.2.
and
Maclaurin's inequality eqn.12.3

<a name="ch12b157">
Next see Cauchy's inequality
Cauchy require two sequences.
If we have
  seq_A1=[1,2,3,4,5] ---eqn.BK094
  seq_C1=[2,4,6,8,10] ---eqn.BK098
Two sequence are proportional.
Cauchy will give us equality!
<a name="ch12b158">
But, we discuss Symmetric Sums
'6,8,10' are not given !
seq_C1 is forbidden. Then 
create second sequence from 
given sequence is same as
re-arrangement. Same result
eqn.BK092 let Cauchy give us 
equality.

<a name="ch12b159">　Index begin　Index this file
Third, see AM-GM inequality.
We applied three times AM-GM 
inequality. From eqn.BK073 to 
eqn.BK075. Each one equality 
condition is 
  xi=xj ---eqn.BK099
Link one by one, all xk are
equal. x1=x2,x2=x3,x3=x4 etc.
Again we get eqn.BK092 Please 
see Equality GM=AM is easy
2010-05-22-19-12 stop

2010-05-23-18-20 done first  proofread
2010-05-23-21-23 done second proofread
2010-05-23-21-46 done spelling check



<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