<a name="docA001">　Index begin　Index this file
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="ch06e001">　Index begin　Index this file
2010-01-06-09-43 start
■　Exercise 6.9 problem statement
　　textbook page 102
（Hadwiger-Finsler Inequality）

Foa any triangle with the traditional 
labeling of Figure 6.2, the law of
cosines tells us that
  a2 = b2 + c2 - 2bc*cos(α) ---eqn.AT001
<a name="ch06e002">
Show that this law implies the area
formula
  a2 = (b-c)2 + 4A*tan(α/2) ---eqn.AT002
then show how Jensen's inequality
implies that in any triangle one
has
  a2+b2+c2 ≧  ---eqn.AT003
  (a-b)2+(b-c)2+(c-a)2+4√3A
This bound is known as the Hadwiger-
Finsler Inequality and it provides
one of the nicest refinement of
Weitzenbock's inequality.
2010-01-06-09-55 stop





<a name="ch06e003">
2010-01-06-11-44 start
■　Exercise 6.9 hint
　　textbook page 250

To prove the first formula, we note
<a name="ch06e004">
  a2 = b2 + c2 - 2bc*cos(α)
// [//] lines are LiuHH added notes.
// = b2+c2 - 2bc + 2bc - 2bc*cos(α)
   = (b-c)2 + 2bc*(1-cos(α))
// = (b-c)2 + 2bc*(1-cos(α))*[2sin(α)]/[2sin(α)]
// = (b-c)2 + 2*{bc*sin(α)/2}*(1-cos(α))*[2]/[sin(α)]
   = (b-c)2 + 4{A}*(1-cos(α))/sin(α)
   = (b-c)2 + 4A*tan(α/2) ---eqn.AT004
<a name="ch06e005">
So by symmetry and summing, we see
that
  a2+b2+c2 = (a-b)2 + (b-c)2 + (c-a)2 
  + 4A[tan(α/2)+tan(β/2)+tan(γ/2)] ---eqn.AT005
<a name="ch06e006">　Index begin　Index this file
Since x→tan(x) is convex on [0,PI/2]
Jensen's inequality gives us
  [tan(α/2)+tan(β/2)+tan(γ/2)]/3
  ≧ tan((α+β+γ)/2/3) ＝ tan(PI/6) ---eqn.AT006
and 
  tan(PI/6)＝√3   ---eqn.AT007 error //textbook
  tan(PI/6)＝1/√3 ---eqn.AT007 //Byron Schmuland

// tan() is convex '︶' on [0,PI/2]
// cos() is concave '︵' on [0,PI/2]

<a name="ch06e007">
//eqn.AT006 and eqn.AT007 together get
// [tan(α/2)+tan(β/2)+tan(γ/2)]/3 ≧ 1/√3
// [tan(α/2)+tan(β/2)+tan(γ/2)]   ≧ 3/√3 =√3
// [tan(α/2)+tan(β/2)+tan(γ/2)]   ≧ √3 ---eqn.AT008
so, this completes the proof.
<a name="ch06e008">
Engel (1998, p.173) gives this as 
the eighth among his eleven amusing
proofs of Weitzenbock's inequality
and its refinements.
2010-01-06-12-08 stop




<a name="ch06e009">
2010-01-06-14-20 start
■　Exercise 6.9 solution


Exercise 6.9 hint already solved this
problem. Here present Jensen's inequality
and compare with eqn.AT006

<a name="ch06e010">
Jensen's inequality is next

<a name="ch06e011">　Index begin　Index this file
In eqn.6.2, let n=3 and let
  p1=p2=p3=1/3  ---eqn.AT009
set f(x)=tan(x). 
Variable x is α/2, β/2, γ/2 (not α,β,γ)
apply generalized AM to DOMAIN get
  f(x_AM)=tan((α/2 + β/2 + γ/2)/3)
that is
  f(x_AM)=tan((α + β + γ)/6)
that is // α+β+γ=PI=180 degree
  f(x_AM)=tan(PI/6)  ---eqn.AT010

<a name="ch06e012">
Next apply generalized AM to RANGE
get
  f_AM=[tan(α/2)+tan(β/2)+tan(γ/2)]/3 ---eqn.AT011

<a name="ch06e013">
Jensen's inequality say:
for convex function f(x) we have
  f(x_AM) ≦ f_AM ---eqn.AT012
that is
  f_AM ≧ f(x_AM) ---eqn.AT013
<a name="ch06e014">
Then we conclude
  [tan(α/2)+tan(β/2)+tan(γ/2)]/3
  ≧  ---eqn.AT006
  tan((α+β+γ)/2/3)＝tan(PI/6)＝1/√3
that is
  [tan(α/2)+tan(β/2)+tan(γ/2)] ≧ √3 ---eqn.AT008

<a name="ch06e015">
Put eqn.AT008 into eqn.AT005 get
  a2+b2+c2 = (a-b)2 + (b-c)2 + (c-a)2 
  + 4A[tan(α/2)+tan(β/2)+tan(γ/2)] ---eqn.AT005
  ≧ (a-b)2 + (b-c)2 + (c-a)2 + 4A*√3 ---eqn.AT003

Here complete the proof.

<a name="ch06e016">
LiuHH only start from Jensen's inequality.
Hint first few steps are basic and easy.
LiuHH added red terms for easier reading.
2010-01-06-15-03 stop



<a name="ch06e017">　Index begin　Index this file
2010-01-06-18-03 start
■　Rolle's Theorem
The following is a copy from the book
Calculus and Analytic Geometry
by George B. Thomas, Jr. and
by Ross L. Finney. Sixth edition
ISBN 0-201-16290-3
Reprinted with correction, May 1994
Page 222 and 223 and 224

<a name="ch06e018">
Rolle's Theorem
There is strong geometric evidence 
that between two points where a
smooth curve y=f(x) crosses the
x-axis there is at least one point 
on the curve where the tangent is 
horizontal. (Fig. 3.57a) //Click [11]
This would
say that if a differentiable function
is zero at two values, then its
derivative is zero somewhere in
between. But this need not be the 
case if the graph has a corner where
f' fails to exist (Fig. 3.57b).  //Click [12]
More precisely, we have the following
theorem.

<a name="ch06e019">　Index begin　Index this file
//following is page 223
Rolle's Theorem
[[
Suppose that y=f(x) is continuous over
the closed interval a≦x≦b and 
differentiable  on the open interval
a＜x＜b . If
  f(a)=f(b)=0  ---eqn.AT014
then there is at least one number c 
between a and b where f' is zero.
That is
  f'(c)=0 for some c.  a＜c＜b ---eqn.AT015
]]

<a name="ch06e020">
Proof. If f(x)=0 for all x between a 
and b, then f(x)=0 for all x between
a and b (the derivative of a constant 
function is zero) and the theorem is
true.

<a name="ch06e021">
But if f(x) is not zero everywhere 
between a and b, then either it is
positive some place, or negative
some place, or both. In any case,
the function will then have a maximum
positive value or a minimum negative
value, or both (by theorem 9, Article
1.11) That is, it has an extreme value
at a point c where f(c) is negative
(in the case of minimum) or f(c) is
positive (in the case of maximum).
<a name="ch06e022">
In either case, c is neither a nor b,
since
  f(a)=f(b)=0 , f(c)≠0 ---eqn.AT016
Therefore c lies between a and b and
Theorem 1 of Article 3.4 applies,
showing that the derivative must be
zero at x=c
  f'(c)=0 for some c.  a＜c＜b ---eqn.AT015
■ (proof done symbol)

<a name="ch06e023">　Index begin　Index this file
Example 1
The polynomial
  y=x3-4x = f(x) ---eqn.AT017
is continuous and differentiable for
all x, -∞＜x＜+∞. If we take
  a=-2, b=+2
the hypotheses of Rolle's Theorem are
satisfied, since
  f(-2)=f(+2)=0  ---eqn.AT018
<a name="ch06e024">
Thus
  f'(x)=3x2-4  ---eqn.AT019
must be zero at least once between
-2 and +2. In fact, we find
  3x2-4=0  ---eqn.AT020
at
  x=c1=-2√3/3 ---eqn.AT021
and
  x=c1=+2√3/3 ---eqn.AT022
□ (example done symbol)

<a name="ch06e025">
//following is page 224
Rolle's Theorem can be combined with
the Intermediate Value Theorem of
Article 1.11 to obtain a criterion 
for isolating the real solution of
an equation f(x)=0. Suppose that a 
and b are two real numbers such that
<a name="ch06e026">
a) f(x) and its first derivative 
   f'(x) are continuous for a＜x＜b
b) f(a) and f(b) have opposite signs.
c) f'(x) is different from zero for
   all values of x between a and b.
Then there is one and only one solution
of the equation f(x)=0 between a and b.

<a name="ch06e027">　Index begin　Index this file
Example 2
Show that the equation
  x3+3x+1=0  ---eqn.AT023
has exactly one real root.

<a name="ch06e028">
Solution
Let
  f(x)=x3+3x+1  ---eqn.AT024
Then the derivative
  f'(x)=3x2+3  ---eqn.AT025
is never zero (because it is always
positive). Now, if there were even
two points x=a and x=b where f(x) was
zero. 
<a name="ch06e029">
Rolle's Theorem would guarantee
the existence of a point x=c where f' 
was zero, because the Intermediate 
Value Theorem tells us that the graph 
of y=f(x) crosses the x-axis somewhere
between x=-1 (where y=-3) and x=0 
(where y=1) (see Fig. 3.58).  //Click [13]
□ (example done symbol)

The Above is a copy from the book
Calculus and Analytic Geometry

2010-01-06-18-57 stop

<a name="ch06e030">　Index begin　Index this file
2010-01-07-16-53 start
■　Exercise 6.10 problem statement
　　textbook page 102
（The f'' Criterion and Rolle's theorem）

<a name="ch06e031">
We saw earlier (page 90) that the
fundamental theorem of calculus
implies that if one has
  f''(x)≧0 for all x∈[a,b] ---eqn.AT026
then f is convex on [a,b]. This
exercise sketches how one can also
prove this important fact by 
estimating the difference
  f(px1+qx2)-pf(x1)-qf(x2)
by comparison with an appropriate 
polynomial.

<a name="ch06e032">
(a) Take  0＜p＜1, q=1-p ---eqn.AT027
    and set μ=px1+qx2 ---eqn.AT028
    where x1＜x2 ---eqn.AT029
// LiuHH: more precisely x1＜μ＜x2 
    Find the unique quadratic polynomial
    Q(x) such that
    Q(x1)=f(x1) ---eqn.AT030
    Q(x2)=f(x2) ---eqn.AT031
    Q(μ)=f(μ) ---eqn.AT032

<a name="ch06e033">
(b) Use the fact that
    ∆(x)=f(x)-Q(x) ---eqn.AT033
    has three distinct zeros in [a,b]
    to show that there is an x*, such
    that  ∆''(x*)=0 ---eqn.AT034

<a name="ch06e034">
(c) Finally, explain how
    f''(x)≧0 for all x∈[a,b] ---eqn.AT026
    and  ∆''(x*)=0 ---eqn.AT034
    imply that
    f(px1+qx2)-pf(x1)-qf(x2)≧0 ---eqn.AT035
    f(px1+qx2)-pf(x1)-qf(x2)≦0 ---eqn.AT035 correction 2009-01-12-18-43
why change ≧ to ≦ ?
2010-01-07-17-10 stop



<a name="ch06e035">　Index begin　Index this file
2010-01-07-17-16 start
■　Exercise 6.10 hint
　　textbook page 250

The polynomial Q(x) can be written as 
a sum of three simple quadratics.

<a name="ch06e037">
2010-01-07-17-31 here
By two application of Rolle's theorem
we see that Q'(x)-f'(x) has a zero in 
(x1,μ) and a zero in (μ,x2), so a third 
application of Rolle's theorem shows
there is an x* between these zeros
<a name="ch06e038">
for which we have
  0 = Q''(x*) - f''(x*) ---eqn.AT037
We therefore have 
  Q''(x*) = f''(x*) ≧ 0 ---eqn.AT038
But

<a name="ch06e040">
2010-01-07-17-50 here
so, by setting
  p=(x2－μ)/(x2－x1) ---eqn.AT040
and
  q=(μ－x1)/(x2－x1) ---eqn.AT041
and simplifying, one finds that the 
last inequality reduces to the 
definition of the convexity of f.
2010-01-07-17-54 stop

<a name="ch06e041">　Index begin　Index this file
2010-01-09-08-09 start
■　About Interactive f(x)_OR_x(t)&y(t) drawing
The graph program included in this
tute0026.htm file has three parts.
(1) RolleDraw [11], [12], [13] three 
    buttons.
(2) x(t),y(t) parametric drawing 
    [51], [52] two buttons.
(3) Exercise 6.10 [14], [15], [16]
    and [17] four buttons.

<a name="ch06e042">
RolleDraw is the initial motivation
of this graph program.
RolleDraw three figures are from
Calculus and Analytic Geometry
At first, this program draw only x=x
and y=f(x). This method can not draw
a circle. Because y=f(x) not allow 
two function values. But a circle
has two points at a given x value.

<a name="ch06e043">
Write a few more code lines, added 
program function allow parametric 
drawing x=x(t) and y=y(t).

RolleDraw talk about Rolle's Theorem
it is part of textbook exercise 6.10
RolleDraw related to textbook. But 
parametric drawing has no relation
to inequality. parametric drawing is
added for general interesting.

<a name="ch06e044">
Third function of this drawing program
is for Exercise 6.10. It has parameter
boxes as following
[[
x1=[  ]<x2=[  ]; Dot □
0<p=[  ]<1 ; Ufunction[   ] "U"=convex 
d[Uf]/dx[   ]  dd[Uf]/dx/dx[   ] 
]]
<a name="ch06e045">
Exercise 6.10 use x1,x2 and p
program control panel design input 
boxes for x1,x2 and p.
Exercise 6.10 use convex function
f(x) and quadratic function Q(x). 
Build 
    ∆(x)=f(x)-Q(x) ---eqn.AT033
<a name="ch06e046">
Find ∆'(x) and ∆''(x). For this
application, program control panel
provided three boxes Ufunction[   ] 
d[Uf]/dx[   ]  dd[Uf]/dx/dx[   ] 
let user define f(x),f'(x),f''(x)

<a name="ch06e047">　Index begin　Index this file
Ufunction IS f(x). 'U' is shape of 
a convex curve. Although concave
function is also allowed.

<a name="ch06e048">
Click buttons [A] ... [E] express
fill function definition.
[A], [B], [C] are convex.
[D] is concave.
[E] is concave AND convex. Then [E]
is neither. Just let user compare
the difference.

<a name="ch06e049">
Drawing buttons are [14], [15], [16]
and [17].

[14] main point is quadratic Q(x).
     Red curve is Q(x), black,blue
     and purple are three components
     of Q(x)
<a name="ch06e050">　Index begin　Index this file
[15] main point is Blue curve two legs
     pass [x1,mu] and [mu,x2]. Please
     see textbook
     [[
     By two application of Rolle's theorem
     we see that Q'(x)-f'(x) has a zero in 
     (x1,μ) and a zero in (μ,x2), 
     ]]
<a name="ch06e051">
     Red straight line is Q'(x)
     Black curve is d[Ufunction]/dx
     Blue curve is d[Q(x)-Uf]/dx.
     Blue = Red - Black
     Purple horizontal line is Q''
     which is a constant. (quadratic
     function differentiate twice
     must be a constant)
2010-01-09-09-03 here
<a name="ch06e052">
2010-01-09-09-16 start
[16] main point is blue -∆(x)=-[f(x)-Q(x)]
     has three zeros. -∆(x) pass x-axis
     three times at x1, mu, x2.
     Red curve is quadratic Q(x)=eqn.AT036
     Black curve is Ufunction f(x)
     Blue curve is Q(x)-Ufunction. 
     Main curve of button [16].
     Purple curve is d[Q(x)-f(x)]/dx 
     Purple must one leg pass [x1,mu],
     other leg pass [mu,x2]
<a name="ch06e053">
[17] main point is -∆(x)=[Q(x)-f(x)]
     and its derivatives.
     red curve Q(x)-Ufunction must 
     pass x1,mu,x2
     black curve d[Q-Uf]/dx, 
     blue curve dd[Q-Uf]/dx/dx
     Purple curve is dd[Uf]/dx/dx. 
<a name="ch06e054">
     Program has four equation definition
     boxes. Button [17] draw these four
     curves plus one more Q''=constant.
     Blue curve and Purple curve are
     negative to each other. up/down
     parallel shift Q'' amount.
     Delete f0, delete f1, click
     [Draw f0(x) to f3(x)] button
     see Blue and Purple clearly.
2010-01-09-09-32 stop

<a name="ch06e055">　Index begin　Index this file
2010-01-09-12-22 start
■　why choose a quadratic equation?
We need to discuss two points
(1) why choose a quadratic equation?
(2) why choose a quadratic equation
    this way?

<a name="ch06e056">
Exercise 6.10 problem statement suggest
build a quadratic equation. But did
not give a direct explanation why.
Exercise 6.10 hint mentioned Rolle's 
theorem. This is an indirect reasoning.
Rolle's theorem need f(a)=f(b)=0
hint said
[[
By two application of Rolle's theorem
]]
<a name="ch06e057">
then we need f(b)=f(c)=0 for second
application. All together, we need
f(a)=f(b)=f(c)=0 at three points. We 
create a polynomial (degree unknown
at this time) P(x), such that
  P(a)=f(a) ---eqn.AT042
  P(b)=f(b) ---eqn.AT043
  P(c)=f(c) ---eqn.AT044
<a name="ch06e058">
then the defined composite equation
  δ(x) = P(x)-f(x) ---eqn.AT045
has three roots at x=a,b,c
First Rolle's theorem application
to δ(x) at x=a and x=b.
Second Rolle's theorem application
to δ(x) at x=b and x=c.
Satisfy our needs. 

<a name="ch06e059">　Index begin　Index this file
For three given points, what degree
polynomial we need? 
Second degree polynomial has three
coefficients c2,c1,c0. 
  P2(x) = c2*x*x + c1*x + c0 ---eqn.AT046
Three given condition eqn.AT042
eqn.AT043 and eqn.AT044 determine
three coefficients. Then we need
second degree polynomial, which is
called quadratic equation.
<a name="ch06e060">
Third degree? need fourth point 'd',
but we do not have fourth point.
First degree? where to use third
point 'c'? For our problem in hand
three given conditions, quadratic
equation is the only choice.

Here answered
(1) why choose a quadratic equation?
2010-01-09-12-49 here

<a name="ch06e061">
2010-01-09-12-56 start
Next see
(2) why choose a quadratic equation
    this way?
Exercise 6.10 hint suggest eqn.AT036
Why so complicate?! Everybody do
simple work! 
<a name="ch06e062">
We can choose the following
  Q1(x) ＝ (x－x2)(x－μ) + (x－x1)(x－μ)
         + (x－x1)(x－x2) ---eqn.AT047
here 'x' is variable, x1 and x2 and μ
are constants. 
Q1(x) IS a quadratic equation and
Q1(x) is simpler than Q(x)=eqn.AT036

<a name="ch06e063">
Dear reader, 
   do we choose Q1(x)?
or do we choose Q(x)? 
or do we choose P2(x)?
Please think.
2010-01-09-13-07 stop






Would you please hold for ten minutes
here?
Please think for a while. Thank you.









<a name="ch06e064">　Index begin　Index this file
2010-01-09-15-09 start
To answer
[[
   do we choose Q1(x)?
or do we choose Q(x)? 
or do we choose P2(x)?
]]
as the following.

<a name="ch06e065">
Choose Q(x) is correct for two reasons
(A) Q(x) will cancel 
    δ(x) = P(x)-f(x) ---eqn.AT045
    at x=a, x=b, x=c. In Exercise 6.10
    case, at x=x1, x=mu, x=x2
    On the other hand, Q1(x) can not
    get zero value at x=x1, x=mu, x=x2
<a name="ch06e066">
(B) Q(x) meet f(x) OK, 
    δ(x) = Q(x)-f(x) OK. But
    δ(x) =Q1(x)-f(x) WRONG !!
    Q1(x)-f(x) is chicken talk to duck,
    totally wrong.

<a name="ch06e067">
Why Q(x) can get zero value at x=x1, 
x=mu, x=x2 ?
Please see Q(x) definition
For Q(x1) eqn.AT036 tell us that
if x=x1, first term 
  f(x)*(x-x2)*(x-μ)/[(x1-x2)*(x1-μ)]
become 
  f(x1)*(x1-x2)*(x1-μ)/[(x1-x2)*(x1-μ)]
become 
  f(x1)
<a name="ch06e068">
eqn.AT036 second term and third term 
both have (x-x1), when x=x1, (x-x1)=0
Second term and third term auto cancel
at x=x1, we have
  Q(x1) = f(x1) ---eqn.AT048
Similarly
at x=x2, Q(x2) = f(x2) ---eqn.AT049
at x=x3, Q(x3) = f(x3) ---eqn.AT050

<a name="ch06e069">　Index begin　Index this file
So 
    ∆(x)=f(x)-Q(x) ---eqn.AT033
at x=x1 become
    ∆(x1)=f(x1)-Q(x1) ---eqn.AT051
become
    ∆(x1)=f(x1)-f(x1) = 0
2010-01-09-15-42 stop

<a name="ch06e070">
2010-01-09-16-55 start
Similarly
    ∆(x2)=f(x2)-Q(x2)=0 ---eqn.AT052
    ∆(μ)=f(μ)-Q(μ)=0 ---eqn.AT053

But f(x)-Q1(x) is not zero at x=x1
Reader can do simple substitution
let x=x1, f(x1)-Q1(x1) is not zero.

Above is (A) δ(x1)=P(x1)-f(x1)=0

<a name="ch06e071">
Below is (B) chicken talk to duck.
First term of Q(x) is
   Qa(x)=f(x1)*(x-x2)*(x-μ)/[(x1-x2)*(x1-μ)] ---eqn.AT054
First term of Q1(x) is
   Q1a(x)=(x－x2)(x－μ) ---eqn.AT055
What is the difference?

<a name="ch06e072">
All mathematics equation ultimately
serve for our real world needs.
We do dimension analysis. Assume x 
is time. Assume f(x) is distance.
f'(x) is velocity.
f''(x) is acceleration.
Qa(x) is distance*time*time/time/time
Qa(x) has distance dimension.

<a name="ch06e073">
How about Q1a(x)?
Q1a(x) is time*time. But 
    ∆(x)=f(x)-Q(x) ---eqn.AT033
require distance f(x) add/subtract 
with Q(x).
distance f(x) subtract distance Qa(x)
is OK

<a name="ch06e074">　Index begin　Index this file
distance f(x) subtract time_squared 
Q1a(x) ? That is chicken talk to duck!
WRONG !! The question
[[
   do we choose Q1(x)?
or do we choose Q(x)? 
or do we choose P2(x)?
]]
We must choose Q(x), 
   must not choose Q1(x).

<a name="ch06e075">
How about P2(x)?
  P2(x) = c2*x*x + c1*x + c0 ---eqn.AT046
Compare P2(x) with distance formula
  s(t)=0.5*g*t*t + v0*t + s0 ---eqn.AT056
'g' is gravity
v0 is initial velocity
s0 is initial distance

<a name="ch06e076">
If we require P2(x) be distance, require
x be time. Then
 c2 has acceleration dimension,
 c1 has velocity dimension,
 c0 has distance dimension.
P2(x) is compatible with f(x) in
dimensional analysis.
But to find out c2,c1,c0 in terms 
of x1, μ, x2 and f(x1), f(μ), f(x2)
This is a complicated process.
Q(x) is much easier than P2(x).

<a name="ch06e077">
Exercise 6.10 problem statement say
[[
Find the unique quadratic polynomial
Q(x)
]]
This word 'unique' is justified.
Q(x) is easier and correct.
2010-01-09-17-30 stop

<a name="ch06e078">
2010-01-09-22-47 start
(2) why choose a quadratic equation
    this way?
Start from the quadratic equation
  Q1(x) ＝ (x－x2)(x－μ) + (x－x1)(x－μ)
         + (x－x1)(x－x2) ---eqn.AT047

<a name="ch06e079">　Index begin　Index this file
This is a equation of x2. From 
dimension analysis assumption, x2 is
time2. But f(x) is distance. We need
to convert from time2 to distance.
First thing to do with Q1(x) is to
cancel time2. Divide time2 by time2
that is (x-x2)*(x-μ)/[(x1-x2)*(x1-μ)]
which is a pure number. Next multiply
pure number by distance f(x1), we get
   Qa(x)=f(x1)*(x-x2)*(x-μ)/[(x1-x2)*(x1-μ)] ---eqn.AT054
which is first term of Q(x). Other two 
terms treated equally.

<a name="ch06e080">
Please pay attention to the fact that 
the dimension convert terms are all
constants. They do not contain variable
'x'. Not multiply 'x' but multiply f(x1)
Not divide by 'x' but divide by constant
[(x1-x2)*(x1-μ)]. Let quadratic equation
continue quadratic.

<a name="ch06e081">
We choose a quadratic equation this way
let us turn to chicken talk to chicken
and duck talk to duck. Distance meet
distance. Teacher happy, student get
100% score.
2010-01-09-23-12 stop


<a name="ch06e082">　Index begin　Index this file
2010-01-12-11-35 start
■　Exercise 6.10 solution

Exercise 6.10 hint suggest use a
quadratic equation Previous study
reviewed
Why choose quadratic equation?
Why choose quadratic equation this way?

<a name="ch06e083">
Now back to hint
Rolle's theorem say
[[
If
  f(a)=f(b)=0  ---eqn.AT014
then there is at least one number c 
between a and b where f' is zero.
]]
<a name="ch06e084">
We created quadratic equation by 
requiring three zeros at
  x=x1 ＜ x=mu (μ) ＜ x=x2
for the composite equation
    ∆(x)=f(x)-Q(x) ---eqn.AT033
We can apply Rolle's theorem to ∆(x)
twice. Once for the interval [x1, mu]
Second time for the interval [mu, x2]

<a name="ch06e085">
Rolle's theorem tell us 
there is one point x=m for which
    ∆'(x)=f'(x)-Q'(x) ---eqn.AT057
equal zero at x=m. AND x=n too.
m in [x1, mu] and n in [mu, x2]

<a name="ch06e086">
That is ∆'(x)=∆'(m)=0 ---eqn.AT058
 and    ∆'(x)=∆'(n)=0 ---eqn.AT059
We can go one level higher !!
because ∆'(m)=0 and ∆'(n)=0 
<a name="ch06e087">　Index begin　Index this file
Rolle's theorem tell us 
there is one point x=x* for which
    ∆''(x)=f''(x)-Q''(x) ---eqn.AT060
equal zero at x=x*.
(textbook use x*,
 drawing button [17] use Q''
 for the point x=x*, y=Q''(x) . )

(Why Q''(x) ≧ 0 ?)
2010-01-12-12-12 here (think)

<a name="ch06e088">
2010-01-12-12-14 start
We have ∆''(x*)=0, that is
  f''(x*)-Q''(x*)=0 ---eqn.AT061
that is 
  Q''(x*)=f''(x*)

<a name="ch06e089">
For a convex function f(x), it has the
property f''(x*)≧0, then we find
  Q''(x*)=f''(x*)≧0 ---eqn.AT062

<a name="ch06e090">
Go back to eqn.AT036. Differentiate
eqn.AT036 twice get eqn.AT039
From eqn.AT036 to eqn.AT039 is a
simple differentiation for a quadratic
equation. Result is a constant.
convex function f(x) property tell us
that this constant is a positive value.
(see eqn.AT062)

<a name="ch06e091">
The remaining work is simplifying.
Start from eqn.AT039 Q''(x*)≧0
and setting
  p=(x2－μ)/(x2－x1) ---eqn.AT040
and
  q=(μ－x1)/(x2－x1) ---eqn.AT041

<a name="ch06e093">
2010-01-12-12-38 here
Above is starting point
Below is end equation
    f(px1+qx2)-pf(x1)-qf(x2))≧0 ---eqn.AT035
where
    μ=px1+qx2 ---eqn.AT028
2010-01-12-12-40 stop

<a name="ch06e094">
2010-01-12-13-42 start
Refer to eqn.AT035, write eqn.AT063 
as following

2010-01-12-18-00 correction start

<a name="ch06e095">
Let us review three number's magnitude
order. It is
  x1 ＜ μ ＜ x2 ---eqn.AT067
Because Exercise 6.10 problem statement
give
[[
    Take  0＜p＜1, q=1-p ---eqn.AT027
    and set μ=px1+qx2 ---eqn.AT028
    where x1＜x2 ---eqn.AT029
]]

<a name="ch06e096">
With the order x1 ＜ μ ＜ x2 ---eqn.AT067
we know negative*negative＞0
  (x1－x2)*(x1－μ)＞0 ---eqn.AT068
and positive*positive＞0
  (x2－x1)*(x2－μ)＞0 ---eqn.AT069
and positive*negative＜0
  (μ－x1)*(μ－x2)＜0 ---eqn.AT070

in eqn.AT066, we cancel '2'.
<a name="ch06e097">　Index begin　Index this file
Move NEGATIVE number (μ－x1)*(μ－x2)
from left denominator to right 
numerator, 
WE MUST REVERSE THE INEQUALITY
and find next equation.
2010-01-12-18-18 correction stop

<a name="ch06e100">
2010-01-12-14-04 here
Note: μ=px1+qx2 ---eqn.AT028
eqn.AT065 is the end equation
eqn.6.1, NOT eqn.AT035
Exercise 6.10 is done.
2010-01-12-14-06 stop

<a name="ch06e101">
2010-01-12-18-22 correction note start
When solve Exercise 6.10, LiuHH 
follow Exercise 6.10 problem statement
and Exercise 6.10 hint
But 2010-01-12-17-32 paid attention
to that
Exercise 6.10 requested result 
eqn.AT035 inequality is opposite
to convex definition eqn.6.1
<a name="ch06e102">
After few minutes think. Find out
why. The key point is at the term
(μ－x1)*(μ－x2) is negative.
Made correction.
2010-01-12-18-38 correction note stop





<a name="ch06e103">　Index begin　Index this file
2010-01-12-19-55 start
■　Exercise 6.11 problem statement
　　textbook page 102
（Transformation to Achieve Convexity）

Show that for positive a, b and c
such that
  a + b + c = abc ---eqn.AT071
one has

<a name="ch06e105">
2010-01-12-20-05 here
This problem from the 1998 Korean
National Olympiad is not easy, even
with the hint provided by the exercise's
title. Someone who is lucky may draw 
a link between the hypothesis a+b+c=abc
and the reasonably well-known fact that
in a triangle labeled as in Figure 6.2
<a name="ch06e106">
one has
  tan(α)+tan(β)+tan(γ)
  =tan(α)tan(β)tan(γ) ---eqn.AT073
This identity is easily checked by
applying the addition formula for the
tangent to the sum 
  γ = π -(α+β) ---eqn.AT074
But it is surely easier to remember
than to discover on the spot.
2010-01-12-20-12 here






<a name="ch06e107">　Index begin　Index this file
2010-01-12-20-15 start
■　Exercise 6.11 hint
　　textbook page 250

Given the hint, we obviously want
to consider the change of variables,
  α = arctan(a) ---eqn.AT075
  β = arctan(b) ---eqn.AT076
  γ = arctan(c) ---eqn.AT077
<a name="ch06e108">
The condition
  a＞0, b＞0, c＞0  ---eqn.AT078
  a+b+c=abc ---eqn.AT079
now tell us that
  α＞0, β＞0, γ＞0  ---eqn.AT080
  α+β+γ=π ---eqn.AT081

<a name="ch06e109">
The tangent inequality also becomes
  cos(α)+cos(β)+cos(γ) ≦ 3/2 ---eqn.AT082
and this follow directly from Jensen's
inequality in view of the concavity of
cosine on [0,π] and the evaluation
  cos(π/3)=1/2 ---eqn.AT083
This solution follows Andreescu and
Feng (2000, p.86). Hojoo Lee has
given another solution which exploits
the homogenization trick which we 
discuss in Chapter 12 (page 189)
2010-01-12-20-26 stop






<a name="ch06e110">　Index begin　Index this file
2010-01-13-18-37 start
■　Exercise 6.11 discussion

First verify
[[
in a triangle labeled as in Figure 6.2
one has
  tan(α)+tan(β)+tan(γ)
  =tan(α)tan(β)tan(γ) ---eqn.AT073
]]

<a name="ch06e111">
From B. O. Peirce integration table 
page 75, item 593 find

<a name="ch06e114">
2010-01-13-19-09 here
Move right denominator to left side
find
  tan(γ) － tan(γ)tan(α)tan(β)
  ＝ －tan(α)－tan(β)
move all negative terms to the other 
side, get
  tan(α)＋tan(β)＋tan(γ)
  ＝tan(γ)tan(α)tan(β) ---eqn.AT073
2010-01-13-19-13 here

<a name="ch06e115">　Index begin　Index this file
2010-01-13-19-17 here
why
[[
The tangent inequality also becomes
  cos(α)+cos(β)+cos(γ) ≦ 3/2 ---eqn.AT082
]]
eqn.AT073 is tangent equality

<a name="ch06e116">
2010-01-13-19-23 here
when reached
[[
tell us that
  α＞0, β＞0, γ＞0  ---eqn.AT080
  α+β+γ=π ---eqn.AT081
]]

<a name="ch06e117">
Tangent formula is off duty. Forget
eqn.AT073

Start fresh new from triangle ABC
and Jensen's inequality base function
f(x)=cos(x)
to draw conclusion.
2010-01-13-19-26 stop




<a name="ch06e118">　Index begin　Index this file
2010-01-13-21-38 start
■　Exercise 6.11 solution


Change of variables, give us
  α = arctan(a) ---eqn.AT075
  β = arctan(b) ---eqn.AT076
  γ = arctan(c) ---eqn.AT077
<a name="ch06e119">
equivalently, we have
  a = tan(α) ---eqn.AT087
  b = tan(β) ---eqn.AT088
  c = tan(γ) ---eqn.AT089
<a name="ch06e120">
Given condition
  a + b + c = abc ---eqn.AT071
is same as
  tan(α)+tan(β)+tan(γ)
  =tan(α)tan(β)tan(γ) ---eqn.AT073
It is the result of change of variable
eqn.AT087, eqn.AT088, eqn.AT089.

<a name="ch06e121">
In Exercise 6.11 discussion we 
found that in an triangle, we have
relation of eqn.AT073.

<a name="ch06e122">
The given condition eqn.AT071 and
tangent identity eqn.AT073 conclude
change of variable to a plane triangle.
Once plane triangle is in our hand.
Both eqn.AT071 and eqn.AT073 finish
their duty and both retire from the
front stage.

<a name="ch06e123">　Index begin　Index this file
In a plane triangle, we consider the
function 
  f(x)=cos(x) ---eqn.AT090

BUT cos(x) has convexity on [π/2, π]
and cos(x) has concavity on [0, π/2]
2010-01-13-21-58 stop

<a name="ch06e124">
2010-01-13-22-37 start
cos() function is CONCAVE on [0, PI/2]
(not on [0,π] )
LiuHH add the next extra limit
  α<π/2, β<π/2, γ<π/2 ---eqn.AT091
Since when a→∞ , arctangent(a)→π/2
α, β, γ can NEVER touch π/2. Triangle
ABC MUST be acute-angled triangle, 
can never be obtuse-angled triangle.




<a name="ch06e125">
We use Jensen's inequality for CONCAVE 
cos() get
  AM in range ≦ AM in domain
that is
  [cos(α) + cos(β) + cos(γ)]/3
  ≦ cos((α+β+γ)/3)  ---eqn.AT092

  cos((α+β+γ)/3)=cos(PI/3)
  =cos(60 degree)=0.5 ---eqn.AT093

tan() is convex '︶' on [0,PI/2]
cos() is concave '︵' on [0,PI/2]

<a name="ch06e126">
eqn.AT092 become
  cos(α) + cos(β) + cos(γ)
  ≦ 3*0.5 =3/2 ---eqn.AT094

<a name="ch06e127">
Last step
sqrt(1+a*a) = sqrt(1+tan(α)*tan(α))
 = sqrt(1+sin(α)*sin(α)/(cos(α)*cos(α)))
 = sqrt((cos(α)*cos(α)+sin(α)*sin(α))/(cos(α)*cos(α)))
 = sqrt(1/(cos(α)*cos(α)))
 = 1/cos(α) ---eqn.AT095

<a name="ch06e128">
  1/sqrt(1+a*a) = cos(α) ---eqn.AT096
Similarly
  1/sqrt(1+b*b) = cos(β) ---eqn.AT097
  1/sqrt(1+c*c) = cos(γ) ---eqn.AT098

Put eqn.AT096, eqn.AT097, eqn.AT098 in
eqn.AT094 get our answer eqn.AT072
2010-01-13-22-48 stop

2010-01-14-13-07 done proofread 
2010-01-14-14-44 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

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ISBN 978-0-521-54677-5
2009-06-19-10-56