<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="ch06c001">　Index begin　Index this file
2009-12-21-14-50 start
■　Jensen Inequality use AM, not GM
Given condition
  f(x) = 1/√(1+x)  ---eqn.AR001
is convex function. Textbook 
page 96, Problem 6.6, eqn.6.17 is next

<a name="ch06c002">
Now we say:
[[
Based on Jensen's Inequality and given
condition, eqn.AR001 is convex function
so, eqn.6.17 is Jensen's Inequality and
is true.
]]
Do you agree? Please think a while 
before move forward.

<a name="ch06c003">
Let us review two points
first: 
AM-GM Inequality is general for any
 irrational weights (page 23, eqn.2.9)
a1p1a2p2...anpn ≦ p1a1+p2a2+...+pnan ---eqn.2.9
ALERT: p1+p2+...+pn=1 ---eqn.AE20

<a name="ch06c004">
Second:
definition of Jensen's Inequality

<a name="ch06c005">
2009-12-21-15-00 here
eqn.2.9 left side a1p1a2p2...anpn
is generalized GM (Geometric Mean)
 ≦ 
eqn.2.9 right side p1a1+p2a2+...+pnan
is generalized AM (Arithmetic Mean)

Now look at definition of Jensen's 
Inequality eqn.6.2
Jensen Inequality use generalized AM
Jensen Ineq.  NOT use generalized GM
Please pay attention to the difference
and to the key points.

<a name="ch06c006">　Index begin　Index this file
■　AM,GM mixed Jensen-like Inequality
After above review, Let us goto text
book page 96, Problem 6.6, eqn.6.17

<a name="ch06c007">
2009-12-21-15-19 here
If we define 
  f(x) = 1/√(1+x)  ---eqn.AR001
In eqn.6.17 right side, correctly
apply generalized AM to RANGE
But f(x) = 1/√(1+x)
In eqn.6.17 left side, wrongly
apply generalized GM to DOMAIN (NO)
remind comparison
apply generalized AM to DOMAIN (OK)

<a name="ch06c008">
We need modify this problem.
First, change eqn.6.17 left side
from 1/√[1+(abc)1/3]   //GM in domain
 to  1/√[1+(a+b+c)/3] //AM in domain
so that we can use 
  f(x) = 1/√(1+x)  ---eqn.AR001

<a name="ch06c009">
Second, change eqn.6.17 both side
from  1/√[1+a]
 to   1/√[1+exp(x)]
then we can use
  g(x) = 1/√(1+exp(x))  ---eqn.AR002

Textbook page 96 to 99 use second 
method.
2009-12-21-15-40 stop


<a name="ch06c010">　Index begin　Index this file
2009-12-21-17-05 start
■　Jensen-like first modification
The following study first modification
change eqn.6.17 left side
from 1/√[1+(abc)1/3]   //GM in domain
 to  1/√[1+(a+b+c)/3] //AM in domain
so that we can use 
  f(x) = 1/√(1+x)  ---eqn.AR001
Now the whole inequality read as next

<a name="ch06c012">
2009-12-21-17-12 here
For eqn.AR003 , use
  f(x) = 1/√(1+x)  ---eqn.AR001
We have three points a,b,c in domain.
eqn.AR003 left side is AM of a,b,c
in function domain.
eqn.AR003 right side is AM of f(a),f(b)
and f(c) in function range. Both satisfy
Jensen Inequality definition.
We find
  f(x) = 1/√(1+x) = (1+x)^(-0.5)  ---eqn.AR001
  f'(x)= (-0.5)*(1+x)^(-1.5)  ---eqn.AR004
 f''(x)= (-0.5)*(-1.5)*(1+x)^(-2.5)  ---eqn.AR005
<a name="ch06c013">　Index begin　Index this file
■　Jensen Example 7
Please goto Jensen Example and 
click [7] button.
 Red  curve is   f(x) 
Black curve is  f'(x)
Blue  curve is f''(x)
their domain are x greater than -1
x can not be equal to -1 since
1/(1-1) = 1/0 = infinity.
Blue curve f''(x) is positive for
x in (-1, infinity), then red curve
is convex for x in (-1, infinity).

<a name="ch06c014">
Above of Jensen Example
click [random#2], 
program fill x sequence in box11
program fill p sequence in box12
then click [Draw JensenA] button,
goto Box 24, 
Top 5 lines give Jensen inequality answer 
If you see
*** Convex: fsumx ＜ sumpf
then program correct. otherwise
program wrong.

<a name="ch06c015">
fsumx is function of (a+b+c)/3
that is  fsumx = 1/√[1+(a+b+c)/3]
Here p1=p2=p3=1/3, but 
Box 12 may give you different value.

sumpf is sum of p*function value
that is eqn.6.17 right side.

<a name="ch06c016">
Box 11 and Box 12 may give you 
different value. But you must always 
see (in tute0022.htm Box 24, ex. 7)
*** Convex: fsumx ＜ sumpf
because 
  f(x) = 1/√(1+x) = (1+x)^(-0.5)  ---eqn.AR001
is convex for x in (-1, infinity).
2009-12-21-17-38 stop

<a name="ch06c017">　Index begin　Index this file
2009-12-22-15-06 start
■　Jensen-like second modification
Now back to eqn.6.17 again.
If we use
  f(x) = 1/√(1+x)  ---eqn.AR001
as convex function, clearly, we can
NOT use Jensen's inequality. Because
Jensen inequality for convex function
eqn.6.2 use Arithmetic Mean, not use 
Geometric Mean. (abc)1/3 in eqn.6.17 
use GM, not use AM (a+b+c)/3.
<a name="ch06c018">
Second method to modify eqn.AR001 and
eqn.6.17 set is to change eqn.AR001
from
  f(x) = 1/√(1+x)  ---eqn.AR001
to
  g(x) = 1/√(1+exp(x))  ---eqn.AR002
For short writing, define
  d=AM=(x+y+z)/3 ---eqn.AR006
<a name="ch06c019">
Apply AM d=(x+y+z)/3 to eqn.AR002, 
we find
  g(d) = 1/√(1+exp((x+y+z)/3))
       = 1/√(1+[exp(x+y+z)]^(1/3))
       = 1/√{1+[exp(x)*exp(y)*exp(z)]^(1/3)}  ---eqn.AR007
<a name="ch06c020">
Define new expression
  a=exp(x);  ---eqn.AR008
  b=exp(y);  ---eqn.AR009
  c=exp(z);  ---eqn.AR010
Then, eqn.AR007 become
  g(d) = 1/√{1+[exp(x)*exp(y)*exp(z)]^(1/3)}
  g(d) = 1/√{1+[a*b*c]^(1/3)} ---eqn.AR011
[remind: d=(x+y+z)/3]
<a name="ch06c021">
Compare eqn.AR011 with eqn.6.17 left 
side. eqn.AR011 is GM Geometric Mean
The important contribution come from
eqn.AR002
  g(x) = 1/√(1+exp(x))  ---eqn.AR002
Allow us write x as AM, change to GM
AM   exp((x+y+z)/3) ==>
GM  [exp(x)*exp(y)*exp(z)]^(1/3)
<a name="ch06c022">
This modification is effective, we
can not use
  f(x) = 1/√(1+x)  ---eqn.AR001
must use
  g(x) = 1/√(1+exp(x))  ---eqn.AR002
f(x) and g(x) are two different 
functions. Although f(x) is convex in
whole domain, x in (-1, +infinity)
<a name="ch06c023">
But g(x) may not be so. Let us check
  g(x) = 1/√(1+exp(x))    ---eqn.AR002
  g(x) = (1+exp(x))^(-0.5) ---eqn.AR012
 g'(x) = (-0.5)*[(1+exp(x))^(-1.5)]*exp(x) ---eqn.AR013
 g''(x)= (-0.5)*[(1+exp(x))^(-1.5)]*exp(x)
        +(-0.5)*(-1.5)*[(1+exp(x))^(-2.5)]*exp(x)*exp(x) ---eqn.AR014

<a name="ch06c024">　Index begin　Index this file
■　Jensen Example 8
Computer program code in tute0022.htm
function exampleJensen(ceArg1)
if(ceArg1==8) //9812191506
use next lines for g(x) g'(x) g''(x)
g(x)  ="pow((1+exp(x)),(-0.5))"; ---eqn.AR015
g'(x) ="-0.5*exp(x)*pow((1+exp(x)),(-1.5))"; ---eqn.AR016
g''(x)="-0.5*exp(x)*pow((1+exp(x)),(-1.5))-0.5*(-1.5)*exp(2*x)*pow((1+exp(x)),(-2.5))"; ---eqn.AR017
<a name="ch06c025">
Please goto Jensen Example
click [8] button. Program draw 
 g(x) g'(x) g''(x)
Red   curve is g(x)
black curve is g'(x)
blue  curve is g''(x)
Key curve is g''(x), 
within g''(x)≧0, g(x) is convex
within g''(x)≦0, g(x) is concave
The border point is g''(x)=0
2009-12-22-15-55 here

<a name="ch06c026">
Set g''(x) to 0, find x0
g''(x)=0= 
 (-0.5)*[(1+exp(x))^(-1.5)]*exp(x)
+(-0.5)*[(1+exp(x))^(-1.5)]*exp(x)*exp(x)*(-1.5)*[(1+exp(x))^(-1)]
=[(-0.5)*[(1+exp(x))^(-1.5)]*exp(x)]*
 {1+exp(x)*(-1.5)*[(1+exp(x))^(-1)]}=0 ---eqn.AR018

<a name="ch06c027">
 {1+exp(x)*(-1.5)*[(1+exp(x))^(-1)]}=0
 {1+exp(x)*(-1.5)/(1+exp(x))}=0
 {1+exp(x)-1.5*exp(x)}=0
 {1-0.5*exp(x)}=0 ---eqn.AR019
 exp(x)=2
 x0=log(2)=0.6931471805599453  ---eqn.AR020
If all x≦log(2), we have concave function
  g(x) = 1/√(1+exp(x))  ---eqn.AR002
If all x≧log(2), we have convex function
  g(x) = 1/√(1+exp(x))
If some x≧log(2) some x≦log(2), we have
non-convex problem.
2009-12-22-16-01 stop

<a name="ch06c028">　Index begin　Index this file
2009-12-22-16-15 start Example 7
■　Small code for Jensen Example 8
Jensen-like second modification
Next is a small code for Jensen Example [8]
Please open complex2.htm#box03
paste
[[
x=0.1;  //concave. 200912221612 
y=0.2;  //you can change x,y,z value (-inf, +inf)
z=0.6;  //log(2)=0.6931471805599453
ga=1/sqrt(1+pow(exp(x)*exp(y)*exp(z),1/3))
gb=(1/sqrt(1+exp(x))+1/sqrt(1+exp(y))+1/sqrt(1+exp(z)))/3
ga //ga≦gb convex  if x,y,z all ≧ log(2)
gb //ga≧gb concave if x,y,z all ≦ log(2)
gb-ga //+ convex, - concave, must be all
]]
to [Box3, input], then 
<a name="ch06c029">
click [test box3 command, output to box4]
button.
If x,y,z all >= log(2) get ga<=gb
because g''(x)>=0, convex problem

If x,y,z all <= log(2) get ga>=gb
because g''(x)<=0, concave problem

<a name="ch06c030">
If x,y,z partial < log(2) partial > log(2) 
get ga >= or <= gb Jensen's inequality
NOT apply. because domain is not convex.

If x=y=z, get ga = gb 
Please see 
When Jensen's Inequality become equality?
2009-12-22-16-28 stop

<a name="ch06c031">　Index begin　Index this file
2009-12-22-21-36 start
■　Small code for Jensen Example 7
Jensen-like first modification
Please see eqn.AR003. Next is a 
small code for Jensen Example [7]
Please open complex2.htm#box03
paste next
[[
x=-0.56; // 200912222140 
y=0.25;  //you can change x,y,z value
z=-0.6;  //x,y,z all in (-1, infinity)
fa=1/sqrt(1+(x+y+z)/3) //AM in domain for Jensen
fb=(1/sqrt(1+x)+1/sqrt(1+y)+1/sqrt(1+z))/3
fa //fa≦fb convex only, never concave
fb //fa≧fb must be program error
fb-fa //must be positive/zero else program error
]]
to [Box3, input], then 
<a name="ch06c032">
click [test box3 command, output to box4]
button.
Jensen-like first modification domain is
(-1, infinity) it is all convex domain.
2009-12-22-21-48 stop

<a name="ch06c033">
2009-12-23-10-06 start
one more code, 
find out why require a*b*c≧512
[[
//textbook page 96, line 16, abc≧2^9
//why require a*b*c≧512 ? 200912231008
k0=511; //2^9-1=512-1
a=2; //you can change a,b,k0 value
b=3; //Let c value =k0/a/b
c=k0/a/b; //a,b,c all in (0, infinity)
//next line use GM, not Jensen's AM
f1=1/sqrt(1+pow(a*b*c,1/3)) //GM in domain
//next line is Jensen's AM in range
f2=(1/sqrt(1+a)+1/sqrt(1+b)+1/sqrt(1+c))/3
f1 //=1/sqrt(1+pow(a*b*c,1/3))
f2 //=(1/sqrt(1+a)+1/sqrt(1+b)+1/sqrt(1+c))/3
f2-f1 //expect positive (f2≧f1)
]]
Above
find out why require a*b*c≧2^9=512
not get conclusion, need your effort.
2009-12-23-10-50 stop

<a name="ch06c034">　Index begin　Index this file
2009-12-23-18-30 start
■　Problem 6.6 (AMM 2002 proposed by M. Mazur)
Show that if a,b and c are positive real
numbers for which one has the lower bound
  abc≧29=512 ---eqn.AR021
then

<a name="ch06c036">
2009-12-23-18-37 here
Study file tute0024.htm from beginning
to here, always discuss Problem 6.6.

Jensen Inequality use AM, not GM, but
eqn.6.17 use GM at left side for domain
and use AM at right side for range.
use GM at left side is not compatible
with Jensen Inequality.
Second modification use
  g(x) = 1/√(1+exp(x))  ---eqn.AR002
to change GM to AM. 
<a name="ch06c037">
Because
AM in exp((x+y+z)/3) can be written as
GM [exp(x)*exp(y)*exp(z)]^(1/3)
The function we need to consider is
  g(x) = 1/√(1+exp(x))  ---eqn.AR002
and it is not
  f(x) = 1/√(1+x)  ---eqn.AR001
<a name="ch06c038">
Curve for g(x), g'(x), g''(x) are at
Jensen Example 8
Blue curve is g''(x),
where g''(x)≦0, g(x) is concave
where g''(x)≧0, g(x) is convex
Here has calculation for g''(x)=0
 x0=log(2)=0.6931471805599453  ---eqn.AR020
divide the x-axis to concave x≦log(2)
and to convex x≧log(2).
<a name="ch06c039">　Index begin　Index this file
If all test points are ≦log(2),
   we have concave problem
If all test points are ≧log(2),
   we have convex problem.
If some points ≦log(2) and other
points ＞log(2), this is not a 
convexity problem.

<a name="ch06c040">
Problem 6.6 (AMM 2002) allow one or 
two of a,b,c three points be ≦log(2)
require only abc≧512. 
We are proving not-a-convexity-problem
for its convexity result !! MAGIC !!
2009-12-23-19-16 stop


<a name="ch06c041">
2009-12-23-20-27 start
Textbook use extrapolation, draw a 
tangent line shield partial of 
concave g(x)=1/√(1+exp(x)) x＜log(2)
Make sure the tangent line is below
concave g(x) part. Because straight 
line (tangent) can be convex (can be
concave too). 
<a name="ch06c042">
We define h(x) as following
for x≧3*log(2)=2.0794
  h(x)=g(x)=1/√(1+exp(x))  ---eqn.AR022
for x in [0, 3*log(2)] = [0, 2.0794]
  h(x)=g(3*log(2))+(x-3*log(2))*g'(3*log(2))
  h(x)= 1/3 + (3*log(2)-x)*4/27 ---eqn.AR023
h(x) is composite function,
eqn.AR023 is a straight tangent line.
Tangent point is
  (x,y)=(3*log(2), g(3*log(2)))
tangent slope is g'(3*log(2))
[[
<a name="ch06c042a">
2009-12-28-13-08 add start
  h(x)= 1/3 + (3*log(2)-x)*4/27 ---eqn.AR023
how to get 1/3?
how to get 4/27?
in
  g(x) = (1+exp(x))^(-0.5) ---eqn.AR012
 g'(x) = (-0.5)*[(1+exp(x))^(-1.5)]*exp(x) ---eqn.AR013
<a name="ch06c042b">
set x=3*log(2)
  g(3*log(2)) = (1+exp(3*log(2)))^(-0.5)
  = (1+[exp(log(2))]^3)^(-0.5)
  = (1+[2]^3)^(-0.5)
  = (1+8)^(-0.5)
  = 1/sqrt(9) = 1/3
<a name="ch06c042c">
Next re-use : exp(3*log(2))=8
 g'(x) = (-0.5)*[(1+exp(x))^(-1.5)]*exp(x) 
 g'(3*log(2)) = (-0.5)*[(1+ 8 )^(-1.5)]* 8  
  = (-0.5)*[1/(9)^(3/2)]* 8  
  = (-0.5)*[1/(3)^(3)]* 8  
  = (-1/2)*(1/27)* 8  
  = (-4/27)
<a name="ch06c042d">
The '-' in (-4/27) change
from (x-3*log(2))
 to  (3*log(2)-x)
in eqn.AR023
2009-12-28-13-19 add stop
]]
<a name="ch06c043">
Note:   textbook use     |  this file use
Note: f(x)=1/√(1+exp(x)) |  g(x)=1/√(1+exp(x))
Note: g(x)=composite f.  |  h(x)=composite func.
Note:  textbook not use  |  f(x)=1/√(1+x)
Note:  2009-12-23-20-44 here

<a name="ch06c044">　Index begin　Index this file
■　One equality two less than surfaced
Tangent point is at x=3*log(2)=2.0794
given condition is a*b*c≧29=512
that is exp(x+y+z)≧29=512=exp(log(512))
because exp() is monotone increase,
we can remove exp() from both side,
still keep SAME inequality (black),
then  x+y+z≧log(512)=6.2383＞2.0794
this is in h(x)=g(x) section, therefore
we can write equality in next equation.

<a name="ch06c046">
2009-12-23-21-15 here
In eqn.AR024, the left end ≦ right end
this is exact what we need.
Jensen's inequality
AM in domain g((x+y+z)/3)  ≦ 
AM in range [g(x)+g(y)+g(z)]/3
2009-12-23-21-22 stop

<a name="ch06c047">
2009-12-24-15-25 start
eqn.AR024 has one equality and two '≦'.
equality is explained in bold red at
ch06c044
The following need to explain
 purple  ≦
and blue ≦

<a name="ch06c048">
purple ≦ is due to Jensen inequality
for function h(x), h(x) is defined at
eqn.AR022 and eqn.AR023.
h(x) is convex part of g(x) eqn.AR022
plus a linear tangent line eqn.AR023.
linear tangent line will not change
convexity. So convex h(x) is an easy
conclusion.

<a name="ch06c049">　Index begin　Index this file
Next blue ≦ is due to h(x)≦g(x) in
the domain x in [0, 3*log(2)]
Please see Jensen Example 8 for x in
the section [0, 3*log(2)]
Tangent line (red, '\' direction)
start at x=3*log(2)=2.0794415
Between x in [log(2), 3*log(2)],
red curve is convex, red curve go
higher and higher than red line.

<a name="ch06c050">
Start from x=log(2), red curve change
to concave and approach to red line
when x move in reduce value direction.
Next consider the section [0, log(2)]
At x=0, red line
h(x)= 1/3 + (3*log(2)-x)*4/27
h(0)= 1/3 + (3*log(2)-0)*4/27=0.641398
At x=0, red curve
g(x)=1/√(1+exp(x))
g(0)=1/sqrt(1+exp(0))=0.707106
At x=0, 
red curve is higher than red line.

<a name="ch06c051">
At x=log(2), red line
h(x)= 1/3 + (3*log(2)-x)*4/27
h(log(2))= 1/3 + (3*log(2)-log(2))*4/27=0.53871
At x=log(2), red curve
g(x)=1/√(1+exp(x))
g(log(2))=1/sqrt(1+exp(log(2)))=0.577350
At x=log(2), 
red curve is higher than red line.

<a name="ch06c052">
Then in [0, log(2)], red curve g(x)
is greater than red line h(x). 
x in [0, log(2)] g(x)＞h(x)

<a name="ch06c053">
The whole picture is
x in [3*log(2), infinity]  g(x)=h(x)
x in [log(2), 3*log(2)]    g(x)＞h(x)
x in [0, log(2)]           g(x)＞h(x)
So for x in [0, infinity]  g(x)＞h(x)
Here conclude blue ≦ in eqn.AR024 is
true.
Up to here, textbook problem 6.6 is
solved. But incomplete, why?
2009-12-24-16-19 stop

<a name="ch06c054">　Index begin　Index this file
■　fix this gap for negative real
2009-02-01-15-22 LiuHH access
http://www-stat.wharton.upenn.edu/~steele/Publications/Books/CSMC/CSMC_Errata.pdf
find
[[
Pages 96 to 99: Li Zhou has pointed out
 a big gap in this discussion. As the
<a name="ch06c055">
text states, the bound (6.18) needs to
 be proved for all real x,y,z such that
exp(x,y,z)≧2^9=512, but my argument only
 does this for nonnegative x,y,z. I hope
to find a cheap way to fix this gap, or 
in the next edition the convex minorant
trick will be illustrated with a different
 example. Li Zhou has provided a nice
alternative.
]]
Here "my argument", "I hope" are
Prof. J.M. Steele words.
2009-12-24-16-24 Liu,Hsinhan notes.

<a name="ch06c056">
2009-12-24-16-35 start
Why 512?
Please see eqn.6.17
if set
  a=b=c  ---eqn.AR025
eqn.6.17 always become equality.
Please go to drawing program
In 'a' and 'b' and 'c' box fill same
number, then click [One data] get
[[
For given a,b,c, eqn.6.17 right－left=0
]]
click [One data] not draw curve.
click ['c'] not draw curve.

<a name="ch06c057">
In (n*n-1)*(n*n-1)*(n*n-1)
set n=3, get 512
if set n=2, get 27
(2*2-1)*(2*2-1)*(2*2-1)=27
27 is in similar position as 512
and a*b*c≧27 can also support the
equality in eqn.AR024
log(27)=3.2958 > 2.0794 =3*log(2)
512 is not special.
2009-12-24-16-50 stop

<a name="ch06c058">
2009-12-24-19-46 start
Original problem is
[[
if a,b and c are positive real numbers
]]
That is a＞0, b＞0, c＞0.

<a name="ch06c059">　Index begin　Index this file
After a,b,c to x,y,z transformation
a=exp(x)＞0 then x ∈ (-inf, +inf)
Similarly, y and z ∈ (-inf, +inf)
Textbook solved only x ∈ [0, +inf)
Lack of x ∈ (-inf, 0]
So Prof. J.M. Steele will re-write
problem 6.6 in second edition.
2009-12-24-19-53 stop

2009-12-24-21-08 done proofread
2009-12-24-21-23 done spelling check



<a name="ch06c060">　Index begin　Index this file
2009-12-25-15-52 start
■　Exercise 6.1 problem statement
　　textbook page 99
（A Renaissance Inequality）
The Renaissance mathematician Pietro 
Mengoli (1625-1686) only needed simple
algebra to prove the pleasing symmetric
inequality

<a name="ch06c062">
2009-12-25-16-02 here
for  x＞1  ---eqn.AR026

yet he achieved a modest claim on
intellectual immortality when he used
it to give one of the earliest proofs
of the divergence of the harmonic 
series

<a name="ch06c064">
2009-12-25-16-13 here
Rediscover Mengoli's algebraic proof 
of the inequality (6.21) and check
that it also follows from Jensen's 
inequality. Further, show, as Mengoli
did, that the inequality (6.21) 
implies the divergence of Hn.
2009-12-25-16-16 stop





<a name="ch06c065">　Index begin　Index this file
2009-12-25-17-45 start
■　Exercise 6.1 hint
　　textbook page 247
cancelling 1/x from both sides and
adding the fractions, one sees that 
Mengoli's inequality is equivalent to
the trivial bound
  x2＞x2-1 ---eqn.AR027
<a name="ch06c066">
For a proof using Jensen's inequality,
just note that x→1/x is convex. Finally
for a modern version of Mengoli's proof
that Hn diverges, we assume Hn＜∞ and
write H∞ as
  1+(1/2+1/3+1/4)+(1/5+1/6+1/7)
   +(1/8+1/9+1/10)+ ...  ---eqn.AR028
<a name="ch06c067">
Now by applying Mengoli's inequality 
within the indicated groups we find
the lower bound
  1 + 3/3 + 3/6 + 3/9 + ...
  = 1 + H∞ ---eqn.AR029
which yield the contradiction
  H∞ ＞ 1+H∞ ---eqn.AR030
<a name="ch06c068">
By the way, according to Havil (2003,
p.38) it was Mengoli who in 1650 first
posed the corresponding problem of
determining the value of the sum
  1 + 1/22 + 1/32 + ... ---eqn.AR031
The problem resisted the efforts of
Europe's finest mathematicians until
1731 when L. Euler determined the
value to be π2/6
2009-12-25-18-03 stop





<a name="ch06c069">　Index begin　Index this file
2009-12-25-18-10 start
■　Exercise 6.1 solution


Start from eqn.6.21
  1/(x-1) + 1/x + 1/(x+1)
  ?＞? 3/x  ---eqn.AR032
cancel 1/x from both side, get
  1/(x-1) + 1/(x+1) ?＞? 2/x ---eqn.AR033
<a name="ch06c070">
Carry out reduction of fractions 
to a common denominator, get
  [(x+1)+(x-1)]/[(x-1)*(x+1)]
  ?＞? 2/x ---eqn.AR034
or
  [2*x]/[x*x-1] ?＞? 2/x ---eqn.AR035
<a name="ch06c071">
The given condition is x＞1,
rewrite eqn.AR035 as next
  x*x ?＞? x*x-1 ---eqn.AR036
eqn.AR036 is definitely true, then 
the start equation eqn.6.21 is
true. That is
  1/(x-1)+1/x+1/(x+1) ＞ 3/x ---eqn.AR037
eqn.AR032 is uncertain equation
eqn.AR037 is proved equation
2009-12-25-18-24 here

<a name="ch06c072">
Next use Jensen's inequality to 
prove eqn.6.21 (eqn.AR032)
  1/(x-1)+1/x+1/(x+1) ?＞? 3/x  ---eqn.AR032

Next line is Jensen's inequality

<a name="ch06c073">
Important thing is to select a convex
base function f(x). Read eqn.AR032
it is naturally to consider
  f(x) = 1/x ---eqn.AR038
Find f''(x) to see whether f(x)=1/x
is convex.
  f'(x) = -1*x-2 ---eqn.AR039
<a name="ch06c074">
 f''(x) = -1*(-2)*x-3 ---eqn.AR040
that is
 f''(x) = 2/x/x/x ---eqn.AR041
Given condition is x＞1, then f''(x)
is always positive and f(x)=1/x is
convex.

<a name="ch06c075">　Index begin　Index this file
Refer to
  1/(x-1)+1/x+1/(x+1) ?＞? 3/x  ---eqn.AR032
Let us select three points
 x1=m+1  ---eqn.AR042 (three lines)
 x2=m
 x3=m-1

<a name="ch06c076">
AM to function DOMAIN is
 (x1+x2+x3)/3
or 
 AM_DOMAIN = (m+1 +m +m-1)/3
 AM_DOMAIN = m
function(AM) = f(m) = 1/m ---eqn.AR043

<a name="ch06c077">
Next
AM to function RANGE is
  AM_RANGE=[f(x1)+f(x2)+f(x3)]/3
  AM_RANGE=[f(m+1)+f(m)+f(m-1)]/3
  AM_RANGE=[1/(m+1)+1/(m)+1/(m-1)]/3 ---eqn.AR044

<a name="ch06c078">
Jensen's inequality is 
from a convex base function f(x)
apply AM to function DOMAIN
its value is less than or equal to
apply AM to function RANGE

<a name="ch06c079">
We must have
function(AM) = 1/m
≦
[1/(m+1)+1/(m)+1/(m-1)]/3
Move 3 to other side, get
  3/m ≦ 1/(m+1) +1/(m) +1/(m-1) ---eqn.AR045
What is eqn.AR045 ??!!
<a name="ch06c080">
Change dummy variable m to x
get
  3/x ≦ 1/(x+1) +1/(x) +1/(x-1) ---eqn.AR046
eqn.AR046 is Mengoli's inequality
eqn.6.21 !!

Prove Mengoli's inequality from
Jensen's inequality is DONE !!

<a name="ch06c081">　Index begin　Index this file
LiuHH used the expression
  AM in function DOMAIN ≦
  AM in function RANGE
Hope it will help you understand 
that AM in function DOMAIN is 
simply x (or m) !!
2009-12-25-18-58 stop

<a name="ch06c082">
2009-12-25-19-16 start
To prove eqn.6.22, write
  H∞ = //from eqn.AR028
  1+(1/2+1/3+1/4)
   +(1/5+1/6+1/7)
   +(1/8+1/9+1/10)+ ...  
  // apply Mengoli's inequality
  ＞ 1+ 3/3
      + 3/6 
      + 3/9 + ...
  = 1+ [1 +1/2 +1/3 + ... ]
  = 1+ H∞ ---eqn.AR047
<a name="ch06c083">
Take the two ends, get
  H∞ ＞ 1+ H∞ ---eqn.AR048
If H∞ were any finite number
eqn.AR048 can not be true. Now we
have eqn.AR048, only infinity has
such strange property
  infinity ＞ 1+ infinity
We conclude that H∞ is infinity.
The proof is done.
2009-12-25-19-27 stop



<a name="ch06c084">　Index begin　Index this file
2009-12-26-15-25 start
■　Exercise 6.2 problem statement
　　textbook page 99
（A Perfect Cube and a Triple Product）

Show that if
  x,y,z ＞ 0 ---eqn.AR049
and
  x+y+z=1 ---eqn.AR050
the one has

<a name="ch06c086">
2009-12-26-15-36 here
■　Exercise 6.2 hint
　　textbook page 247
The bound follows by applying Jensen's
inequality to the function
  f(t)=log(1+1/t)=log(1+t)-log(t) ---eqn.AR052
which is convex because

2009-12-26-15-45 here




<a name="ch06c088">　Index begin　Index this file
2009-12-26-18-04 start
■　Exercise 6.2 solution

From given condition
  x,y,z ＞ 0 ---eqn.AR049
  x+y+z=1 ---eqn.AR050
and the target equation eqn.AR051
how to get the base function eqn.AR052

<a name="ch06c089">
Let us review Jensen's inequality first

<a name="ch06c091">
2009-12-26-18-21 here
Assume Jensen inequality base function
is convex, from eqn.6.2 to eqn.AR051
If use base function
  f(x)=(1 + 1/x) ---eqn.AR054
will suggest us to have
 (1 + 1/x)/3 + (1 + 1/y)/3 + (1 + 1/z)/3  ---eqn.AR055
How to change adding three terms to
multiplying three terms? 

<a name="ch06c092">
Log function
  log(a)+log(b)+log(c) = log(a*b*c) ---eqn.AR056
will help us achieve this goal.
Instead of f(x)=(1 + 1/x) 
Let us try new base function
  f(x)=log(1 + 1/x) ---eqn.AR057

<a name="ch06c093">
Assume p1=p2=p3=1/3 ---eqn.AR058
∑pj*f(xj) in Jensen's inequality
become
  [log(1 + 1/x)+log(1 + 1/y)
  +log(1 + 1/z)]/3  ---eqn.AR059
this can be written as
  log[(1+1/x)*(1+1/y)*(1+1/z)]/3 ---eqn.AR060
In eqn.AR060 we have 
  (1+1/x)*(1+1/y)*(1+1/z) ---eqn.AR061
which is used in eqn.AR051. This is
a good sign.

<a name="ch06c094">　Index begin　Index this file
■　domain and range are different
For a function f(x), its variable x
defined section is called domain of
f(x). x can be considered as input.
After evaluation, f(x) has a value,
this value has a possible section,
This is output and is called range.
<a name="ch06c095">
For example, sine function sin(x),
its domain is (-infinity,+infinity)
its  range is [-1,+1], we evaluate
sin(12), 12 is within (-inf,+inf)
sin(12)=-0.5365 is in [-1,+1]

eqn.AR059 is Arithmetic Mean in range.
Jensen inequality also work in domain,
review Jensen as next

<a name="ch06c096">
eqn.AR060 is Jensen Inequality right
side. We need calculate left side.
We determined to use new base function
  f(x)=log(1 + 1/x) ---eqn.AR057
Now do AM in DOMAIN, that is
  f(x_AM)=log(1 + 1/x_AM) ---eqn.AR062
See eqn.AR051, here in eqn.AR062
x will split to x,y,z (two different x)
<a name="ch06c097">
Now AM in DOMAIN is
  x_AM=(x+y+z)/3  ---eqn.AR063
Put eqn.AR063 x_AM to eqn.AR062 find
  f(x_AM)=f((x+y+z)/3) ---eqn.AR064
         = log(1 + 1/[(x+y+z)/3])
We have a given condition
  x+y+z=1 ---eqn.AR050
<a name="ch06c098">
then eqn.AR064 become
  f(x_AM)=log(1 + 1/[1/3])
or
  f(x_AM)=log(4) ---eqn.AR065
Jensen Inequality eqn.6.2 tell us
AM in DOMAIN ≦ AM in RANGE
AM in DOMAIN is eqn.AR065
AM in RANGE  is eqn.AR060, 
base on Jensen Inequality, we have
  log(4) ≦ log[(1+1/x)*(1+1/y)*(1+1/z)]/3
<a name="ch06c099">　Index begin　Index this file
Move 3 to less than side, get
  3*log(4) ≦ log[(1+1/x)*(1+1/y)*(1+1/z)]
that is 
  log(43) ≦ ---eqn.AR066
  log[(1+1/x)*(1+1/y)*(1+1/z)]
Now <a name="ch06c100">
because log() is monotone increase,
we can remove log() from both side,
still keep SAME inequality.
After remove log(), keep only log()
arguments, get
  43 ≦ (1+1/x)*(1+1/y)*(1+1/z) ---eqn.AR067
eqn.AR067 is exactly eqn.AR051
same thing !!

<a name="ch06c101">
Are we done ?! NO !!
Jensen Inequality says
for convex function (see)
AM in DOMAIN ≦ AM in RANGE
for concave function (see)
AM in DOMAIN ≧ AM in RANGE

<a name="ch06c102">
We choose base function log(1+1/x)
which is convex.
If we choose base function log(x)
which is concave!!

We must show that base function 
  f(x)=log(1+1/x) ---eqn.AR057
is convex before finish our work.

<a name="ch06c103">
f(x) is convex if f''(x) ≧ 0 in
its whole domain.
f(x) is concave if f''(x) ≦ 0 in
its whole domain.
If within domain, f''(x) ≧ 0 and
f''(x) ≦ 0 both exist, this f(x)
is not a convexity problem.

<a name="ch06c104">　Index begin　Index this file
For  f(x)=log(1+1/x) ---eqn.AR057
it is easy to calculate
f'(x)= (-1/(1+x)/x)
f''(x)= (1+2*x)/(x+x*x)/(x+x*x)
for x＞0, f''(x)＞0, then
  f(x)=log(1+1/x) ---eqn.AR057
is convex. We apply Jensen Inequality
for convex function is correct.
Exercise 6.2 problem solved.

<a name="ch06c105">
Please goto Jensen Example and 
click [9] button for Exercise 6.2
curves.
2009-12-26-19-39 stop

<a name="ch06c106">　Index begin　Index this file
2009-12-27-13-43 start
■　Exercise 6.3 problem statement
　　textbook page 100
（Area Inequality for n-gons）

Figure 6.4 suggests that among all 
convex n-sided convex polygons
that one can inscribed in a circle,
only the regular n-gon has maximal 
area. Can Jensen's inequality be 
used to confirm this suggestion?
2009-12-27-13-46 stop







<a name="ch06c107">
2009-12-27-14-00 start
■　Exercise 6.3 hint
　　textbook page 248
From the geometry of figure 6.4, the 
area A of an inscribed polygon with 
n sides can be written as
  Area A=∑[k=1,n]{sin(θk)}/2 ---eqn.AR068
where 0＜θk＜PI ---eqn.AR069
and  ∑[k=1,n]{θk}=2*PI ---eqn.AR070
Since sin(.) is strictly concave 
on [0,PI], we have

<a name="ch06c109">
2009-12-27-14-32 here
and we have equality if and only if
  θk=2*PI/n for all 1≦k≦n ---eqn.AR072
Since A' is the area of a regular 
inscribed n-gon, the conjecture
optimality is confirmed.
2009-12-27-14-36 stop






<a name="ch06c110">　Index begin　Index this file
2009-12-27-15-45 start
■　Exercise 6.3 solution

Problem ask to use Jensen inequality
to confirm the equal-angle n-gon has
maximal area conjecture. Before we do
anything, we must know the equation
of n-gon total area.

<a name="ch06c111">
For a triangle ABC three vertices 
angles A, B and C and three sides
a, b and c, the triangle area Atr
is
  Atr=b*c*sin(A)/2 ---eqn.AR073
  Atr=c*a*sin(B)/2 ---eqn.AR074
  Atr=a*b*sin(C)/2 ---eqn.AR075
<a name="ch06c112">
Please see figure 6.4 and 
click [random #2] button then 
click [Draw figure 6.4] button.
Any triangle is an isosceles triangle.
Two equal sides are circle radius r. 
We can write k-th triangle area as
  Atr=r*r*sin(θk)/2 ---eqn.AR076
<a name="ch06c113">
Total area Ato is the sum of all
isosceles triangles
  Ato=SUM[k=1,n]r*r*sin(θk)/2 ---eqn.AR077
Radius r is common, assume r*r/2=1,
Total area Ato simplify to
  Ato=SUM[k=1,n]sin(θk) ---eqn.AR078
From eqn.AR078, we know 
Jensen's inequality base function
should be 
  f(x)=sin(x) ---eqn.AR079
<a name="ch06c114">
For a convex polygon, we must have
  0＜x＜PI ---eqn.AR080
eqn.AR080 and eqn.AR069 are the same.

A Jensen's inequality problem need
a base function, need independent
variable lower bound and upper bound.
Equivalently, need domain definition.

<a name="ch06c115">
Domain definition let us decide if
the domain is a convex set. For one
dimensional problem, continuous 
domain is a convex set. Now we have
eqn.AR080 and eqn.AR069, we know
Exercise 6.3 has a convex set.

Domain definition also let us decide
if the base function is convex or
concave or neither.

<a name="ch06c116">　Index begin　Index this file
From tute0023.htm#ch06b049
  to tute0023.htm#ch06b071
textbook proved
Twice differentiation ≧0 get convexity
Equivalent statement
Twice differentiation ≦0 get concavity
although not proved, but it is same
work just do the other way.

<a name="ch06c117">
From
  f(x)=sin(x) ---eqn.AR079
and
  0＜x＜PI ---eqn.AR080
the result is
  f''(x)=-sin(x)≦0  ---eqn.AR081
we know 
  f(x)=sin(x) ---eqn.AR079
is concave on 0＜x＜PI (NOT convex)
<a name="ch06c118">
Jensen Inequality for concave function
is next. ALERT not use '≦' but use '≧'

<a name="ch06c119">
Jensen Inequality for concave function
we have
AM in DOMAIN ≧ AM in RANGE
for
  f(x)=sin(x) ---eqn.AR079
AM in DOMAIN is
  x_AM=(SUM[k=1,n]{θk})/n
for a whole circle
  x_AM=(2*PI)/n ---eqn.AR082
function of AM in DOMAIN is
  f(x_AM)=sin(2*PI/n) ---eqn.AR083

<a name="ch06c120">　Index begin　Index this file
Next we need AM in RANGE 
  f_AM=[f(θ1)+f(θ2)+...+f(θn)]/n
  f_AM=[sin(θ1)+sin(θ2)
       +...+sin(θn)]/n ---eqn.AR084

Jensen Inequality for concave function
say
AM in DOMAIN ≧ AM in RANGE
we have
  sin(2*PI/n) ≧ [sin(θ1)+sin(θ2)
       +...+sin(θn)]/n
<a name="ch06c121">
Move n to other side, get
  n*sin(2*PI/n) ≧ [sin(θ1)+sin(θ2)+...
                  +sin(θn)] ---eqn.AR085
eqn.AR085 left side is area-sum of
n pieces equal angle isosceles triangle
which is greater than or equal to
area-sum of
n pieces UN-equal angle isosceles triangle

<a name="ch06c122">
We are done !!
Wonderful logic !!
Thank you, Prof. Jensen !
Thank you, Prof. Steele !
2009-12-27-17-28 stop



<a name="ch06c123">　Index begin　Index this file
2009-12-27-19-00 start
■　Exercise 6.4 problem statement
　　textbook page 100
（Investment Inequality）
If 0＜rk＜∞ ---eqn.AR086
and if our investment of one dollar
in year k grows to 1+rk dollars at
the end of the year, we call rk the
return on our investment in year k. 
<a name="ch06c124">
Show that the value
  V=(1+r1)(1+r2)...(1+rn) ---eqn.AR087
of out investment after n years must
satisfy the bounds

<a name="ch06c126">
2009-12-27-19-18 here
where
  rG=(r1r2...rn)1/n ---eqn.AR088
and
  rA=(r1+r2+...+rn)/n ---eqn.AR089
Also explain why this bound might be
viewed as a refinement of the AM-GM
inequality.
2009-12-27-19-22 here






<a name="ch06c127">
■　Exercise 6.4 hint
　　textbook page 248
2009-12-27-19-35 start
The second bound is the AM-GM inequality
for
  ak=1+rk, k=1,2,...,n ---eqn.AR090
The first bound follows from Jensen's
inequality applied to the convex
function x → log(1+exp(x)). 
<a name="ch06c128">
Finally, by taking n th roots and 
subtracting 1, we see that the
investment inequality (6.23) refines
the AM-GM bound
  rG≦rA ---eqn.AR091
by slipping V1/n-1 between the
two means.
2009-12-27-19-43 here





<a name="ch06c129">　Index begin　Index this file
2009-12-27-19-46 start
■　Exercise 6.4 solution


Special reasoning, caution !!

<a name="ch06c130">
Define
  ak=1+rk ---eqn.AR092
that is
  a1=1+r1 ---eqn.AR093
  a2=1+r2 ---eqn.AR094
  ...
  an=1+rn ---eqn.AR095

<a name="ch06c131">
GM of sequence ak is
  GMa=(a1a2...an)1/n ---eqn.AR096

AM of sequence ak is
  AMa=(a1+a2+...+an)/n ---eqn.AR097
<a name="ch06c132">
Substitute eqn.AR093,eqn.AR094,eqn.AR095
to eqn.AR097, get
  AMa=(1+r1 + 1+r2 +...+ 1+rn)/n
  AMa=(n +r1+r2+...+rn)/n
  AMa= 1+(r1+r2+...+rn)/n
  AMa= 1+rA ---eqn.AR098

<a name="ch06c133">
We have GMa ≦ AMa ---eqn.AR099
Put eqn.AR096 and eqn.AR098 to eqn.AR099
whole equation rise to n th power
get
  (a1a2...an)n/n ≦ (1+rA)n ---eqn.AR100
Use eqn.AR093,eqn.AR094,eqn.AR095
to recover r from a in eqn.AR100
we get eqn.6.23 middle ≦ right
answer.
  (a1a2...an) ≦ (1+rA)n ---eqn.AR100 aux1
  ∏[k=1,n]{1+rk} ≦ (1+rA)n ---eqn.AR100 aux2
2009-12-27-20-04 here

<a name="ch06c134">　Index begin　Index this file
Second half: eqn.6.23 left ≦ middle
  (1+rG)n≦∏[k=1,n]{1+rk} ---eqn.AR101

Exercise 6.4 hint suggest use 
Jensen's inequality applied to the
convex function x → log(1+exp(x)). 

<a name="ch06c135">
Jensen Inequality says
for convex function
AM in DOMAIN ≦ AM in RANGE

AM (Arithmetic Mean) use summation ∑
but eqn.AR101 use multiplication ∏
Summation of log()s become log() of
multiplication.
2009-12-27-20-23 here

<a name="ch06c136">
This explain the log() part of
[[
Exercise 6.4 hint suggest use 
convex function x → log(1+exp(x))
]]
How about exp(x)?
If instead of log(1+exp(x))
we use base function log(1+x)
<a name="ch06c137">
When we take AM in DOMAIN, we get
log(1+x_AM), again AM is addition
AM is not multiplication, problem 
ask to prove
  (1+rG)n≦∏[k=1,n]{1+rk} ---eqn.AR101
rG is GM
  rG=(r1r2...rn)1/n ---eqn.AR088
<a name="ch06c138">
log(1+x_AM) can not reach rG, but
log(1+exp(x)) can reach rG. Since

log(1+exp((r1+r2+...+rn)/n))  ---eqn.AR102
=
log(1+[exp(r1)*exp(r2)*...*exp(rn)]1/n) 
2009-12-27-20-37 here

<a name="ch06c139">　Index begin　Index this file
2009-12-27-20-46 
Take
  f(x)=log(1+exp(x)) ---eqn.AR103
as Jensen's inequality function.
Prove convexity later. First see
[[
Jensen Inequality says
for convex function
AM in DOMAIN ≦ AM in RANGE
]]
<a name="ch06c140">
  x_AM=(r1+r2+...+rn)/n ---eqn.AR104
  f(x_AM)=log(1+exp(x_AM))
  =log(1+exp((r1+r2+...+rn)/n))

  f(x_AM)  ---eqn.AR105
  =log(1+[exp(r1)*exp(r2)*...*exp(rn)]1/n) 

<a name="ch06c141">
Next see AM in RANGE
  f_AM=[log(1+exp(r1))+log(1+exp(r2))
       +...+log(1+exp(rn))]/n ---eqn.AR106

  f_AM=log[(1+exp(r1))*(1+exp(r2))
       *...*(1+exp(rn))]/n ---eqn.AR107

AM in DOMAIN ≦ AM in RANGE give us
  f(x_AM) ≦ f_AM
so
  log(1+[exp(r1)*exp(r2)*...*exp(rn)]1/n) 
  ≦  ---eqn.AR108
  log[(1+exp(r1))*(1+exp(r2))*...*(1+exp(rn))]/n
<a name="ch06c142">
move 1/n from f_AM to f(x_AM), then
remove log() from both side, get
  {1+[exp(r1)*exp(r2)*...*exp(rn)]1/n)}n
  ≦  ---eqn.AR109
  (1+exp(r1))*(1+exp(r2))*...*(1+exp(rn))
<a name="ch06c143"> why !!
We must replace rk with exp(rk) to
meet the answer eqn.AR101.
Is this right? or
where is wrong !?
2009-12-27-21-04 stop

Special reasoning, caution !!

<a name="ch06c144">
2009-12-27-21-16 start
for non-financial problem, we can set
  a=exp(x);  ---eqn.AR008
But for financial problem, can we can
set interest rate from rk to exp(rk) ?
2009-12-27-21-18 stop

<a name="ch06c145">　Index begin　Index this file
2009-12-27-22-32 start
compare
  {1+[exp(r1)*exp(r2)*...*exp(rn)]1/n)}n
  ≦  ---eqn.AR109
  (1+exp(r1))*(1+exp(r2))*...*(1+exp(rn))
with
  (1+rG)n≦∏[k=1,n]{1+rk} ---eqn.AR101
where
  rG=(r1r2...rn)1/n ---eqn.AR088
<a name="ch06c146"> caution !!
If we interpret
  rj is exp(rj) ---eqn.AR110
then eqn.AR109 is same as eqn.AR101
eqn.AR110 require us let exp(rj)
to represent rj. This can be
done at first step.
<a name="ch06c147">
Since  exp(log(y))=y ---eqn.AR111
At first step, we transform all rj
to log(rj) still write log(rj) as rj,
then at the final step exp(rj) is
original rj.
<a name="ch06c148">
If we do this transformation process
at first step and last step, then the
proof is done.
2009-12-27-22-40 stop

<a name="ch06c149">
2009-12-27-22-47 start
Now need to show that
  f(x)=log(1+exp(x)) ---eqn.AR103
is convex.
  f'(x)=[1/(1+exp(x))]*(1+exp(x))'
  f'(x)=[1/(1+exp(x))]*exp(x)
  f'(x)=exp(x)/(1+exp(x)) ---eqn.AR104

<a name="ch06c150">　Index begin　Index this file
  f''(x)=[exp(x)]' *[(1+exp(x))^(-1)]
    +[exp(x)] *[(1+exp(x))^(-1)]'
  f''(x)=[exp(x)]*[(1+exp(x))^(-1)]
    +[exp(x)] *[(-1)*(1+exp(x))^(-2)]*(1+exp(x))'
  f''(x)=[exp(x)]*[(1+exp(x))^(-1)]
    +[exp(x)] *[(-1)*(1+exp(x))^(-2)]*exp(x)
  f''(x)=[exp(x)/(1+exp(x))]
    -[exp(x)*exp(x)]/[(1+exp(x))^2]
  f''(x)=[exp(x)/(1+exp(x))]
        -[exp(x)/(1+exp(x))]^2 ---eqn.AR105

<a name="ch06c151">
here exp(x)/(1+exp(x))＜1
then f''(x)＞0
for example, y=0.7 ＜1
  y-y*y = 0.7*(1-0.7) = 0.21 ＞0

 f''(x) always ＞0
then 
  f(x)=log(1+exp(x)) ---eqn.AR103
is convex.

<a name="ch06c152">
See
Twice differentiation ≧0 get convexity
2009-12-27-23-00 stop

2009-12-28-15-20 done proofread
2009-12-28-15-38 done spelling check



<a name="docB001">
2009-12-21-11-52 start
On 2009-12-15 Liu,Hsinhan's physical
condition suddenly show up fatigue 
symptom. LiuHH have to stop work for 
few days and slow down pace in future
work. Liu,Hsinhan guess the cause of 
fatigue is desk, chair, monitor, key 
board improper arrangement. 
<a name="docB002">
My desk is small, put two computer 
and one monitor and few books. No 
room for key board, key board sit 
on drawer, drawer open about two 
third. Not all four corner supported.
Back fatigue, eye fatigue and finger 
fatigue. Plus constant radiation from
bulk monitor Sylvania F96 19 inch. I 
have no other choice. Either turn on 
bulk monitor absorb radiation or turn
off computer.
2009-12-21-12-08 stop

<a name="docB003">
2009-12-28-15-40 start
fatigue update.
On 2009-12-15-13-15 Liu,Hsinhan's body
start fatigue symptom. Left eye-white
become all red. No eye-pain, vision 
normal, but has sour eye muscle.
Back pain, can not stand, can not sit. 
Lie on bed, it is pain to move body.
Finger no trouble this time.
<a name="docB004">
Three days later, 90% back pain is gone.
Five days later, 90% red-eye become white.
One week later, both fully recovered.
Now LiuHH take more rest than before.
Hope not occur again. 
<a name="docB005">
Last time red-eye (right eye) is 2003-04-29
possible cause NOT fatigue, but external
chemical smoke. (LiuHH guess; doctor denied
that medication procedure cause red eye.)
Last time back pain, about 6 month to
8 month ago. It is caused by fatigue.
Every day sit in front of computer, 
minimum exercise (walk dog). That is 
LiuHH's life style for long time.
2009-12-28-15-49 stop

<a name="docB006">
2009-12-28-16-51 start
On 2009-11-02 LiuHH get one month 
rent deposit $400 to bank account.
On 2009-12-02 LiuHH get one month 
rent deposit $420 to bank account.

On 2009-12-26 LiuHH get one month 
food supply. Learned that next month
rent ($390) will be deposit to my
bank account.

<a name="docB007">
Before 2009-March/April every spending
paid from my bank account.
Between 2009-April to 2009-July my 
bank account show up red ink.
freeman2.com is about to shut down.
2009-July/Aug. is financial clean up
period.
<a name="docB008">
Start from 2009-July/Aug.
Web site monthly fee and phone bill
are both paid by other means. 
Not paid from my bank account.
2009-12-28-17-28 stop

<a name="docB009">
2009-12-29-11-17 start
Update 2009-12-29 change 
Draw Problem 6.6 code to alert missing
'abc' initial value. See source code 
at time stamp '9812282232'
Second change is in Box 31 fill in
data to match the polygon in textbook
page 100, figure 6.4
Next to Box 31 add [default] button.
2009-12-29-11-20 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