<a name="ch08c031">　Index begin　Index this file
2010-02-23-20-28 here
Power mean inequality eqn.8.10 
read as
  Ms≦Mt ---eqn.AZ014
where 's' and 't' are inner xk
power, not outer {term} power.
Power mean inequality tell us 
that, 
if s＜t then Ms≦Mt, 
because 1/3＜1/2, we have
  M1/3≦M1/2 ---eqn.AZ015

<a name="ch08c032">
Next exam bottom eqn.8.29.
If we use N to represent terms
in eqn.8.29, we get
  N1/2≦N1/3 ---eqn.AZ016 Ex.8.3
  M1/3≦M1/2 ---eqn.AZ015 PowMean
Here again put two simplified 
equation side by side. 
PowMean is 1/3＜1/2 then M1/3≦M1/2
Ex.8.3  is 1/3＜1/2 then N1/2≦N1/3
<a name="ch08c033">
We found that Exercise 8.3 
eqn.8.29 s,t order is opposite 
to that of power mean inequality !!
Go back to two equation
and exam carefully. We find that
power mean inequality xk has
company probability sequence pk
<a name="ch08c034">
But Ex.8.3 eqn.8.29 ak do not have
probability company ! In this case,
we can scale ak sequence freely. 
Also ak sequence not normalized 
by probability sequence pk. 
This difference is the cause 
of opposite-inequality result.

<a name="ch08c035">
Who said one do not need a wife?
A wife keep everything in good
order.
A wife keep power mean curve in
monotone increase order.
2010-02-23-20-55 stop



<a name="ch08c036">
2010-02-23-20-56 start
■　Exercise 8.3 solution


It is your job for change.
Liu,Hsinhan miss wife Liling.
LiuHH take a break.
2010-02-23-20-57 stop

More related work:　explore　local



<a name="ch08c037">　Index begin　Index this file
2010-02-24-13-52 start
■　Exercise 8.4 problem statement
　　textbook page 132
（Polya's Minimax Characterization）

<a name="ch08c038">
Suppose you must guess the value 
of an unknown number x in the
interval [a,b]⊂(0,∞) and suppose
you will be forced to pay a fine
based on the relative error of 
your guess. How should you guess 
if you want to minimize the worst 
fine that you would have to pay?

<a name="ch08c039">
If you guess is p, then the maximum
fine you would have to pay is

<a name="ch08c041">
One expects p* to be some well-
known mean, but which one is it?
2010-02-24-14-16 stop





<a name="ch08c042">　Index begin　Index this file
2010-02-24-14-19 start
■　Exercise 8.4 hint
　　textbook page 258

We only need to consider p∈[a,b],
and in that case we can write

<a name="ch08c046">
Moreover, we have strict inequality
here, unless (p-a)/a=(b-p)/b, 
so, as Polya (1950) observed,
the unique minimum of F(p) is
attained at p*=2ab/(a+b),
which is the harmonic mean of
a and b.
2010-02-24-14-43 stop

<a name="ch08c047">
2010-02-24-14-52 the exact answer 
x is out of question !! x has no
position in the answer process !?

<a name="ch08c048">
2010-02-24-18-56 start
■　Exercise 8.4 solution


Why start from eqn.AZ017 ?
To understand Exercise 8.4. Above
has a small drawing program.
<a name="ch08c049">
Input is Box11. Output is Box12.
Input need five numbers, they are
described in Box11.
eqn.8.31 is drawn in figure.
eqn.8.31 is rewritten as eqn.AZ020
  F(p)=max(x∈[a,b]){|p-x|/x} ---eqn.AZ020
Actually next equation do the job.
  F(x)=max(x∈[a,b]){|p-x|/x} ---eqn.AZ021
//F(p) means that p is a variable,
//x constant. F(x) is just reverse.

<a name="ch08c050">
There are two run buttons
"Ex.8.4RUN 0" and "Ex.8.4RUN 1"
Program use 't' as variable.
Both p and x are fixed number.
"Ex.8.4RUN 0" draw {|p-t|/t} //x vary
"Ex.8.4RUN 1" draw {|t-x|/x} //p vary

<a name="ch08c051">　Index begin　Index this file
"Ex.8.4RUN 0" draw {|p-t|/t}
x vary. Curve is non-linear.
"If you guess is p" is constant.
"unknown number x" walk through
whole domain. (set x as variable t)
When x meet p, F(x) is zero. //not F(p)
[F(x) is fine, punish money]
Key point to observe in this drawing
is that maximum F(x) occur at two
end points. This is why we start 
from eqn.AZ017 .

<a name="ch08c052">
"Ex.8.4RUN 1" draw {|t-x|/x}
p vary. Curve is linear. Same
observation as above. (two ends
has maximum value)

<a name="ch08c053">
With the knowledge that maximum
occur at end points. We can goto
Exercise 8.4 hint.
Exercise 8.4 key equation is
eqn.AZ017 It tell us to choose
greater one of two end values.
<a name="ch08c054">
If two end values are not equal,
then the greater one may trap us
to pay higher fine. 
If two end values are equal, that
is the best case. Then we require
  (p-a)/a=(b-p)/b ---eqn.AZ022
<a name="ch08c055">　Index begin　Index this file
solve for p get answer
  p*=2ab/(a+b) ---eqn.AZ023
Take reciprocal get
  1/p*=(a+b)/(2ab)
  1/p*=1/(2b) + 1/(2a) ---eqn.AZ024
This is well-known harmonic mean.
Attention: 
"unknown number x" never show 
up in solving procedure!
2010-02-24-19-22 stop

<a name="ch08c056">
Box12 output has
pmin=1.6666666666666667
pmin is harmonic mean of a=1 
and b=5.
If paste 1.6666666666666667
to Box11, change from
2.5 guess value
to
1.6666666666666667 guess value
and click "Ex.8.4RUN 0"
<a name="ch08c057">
output curve two end values
are equal height. 
Maximum fine minimized,
this is our answer.
2010-02-24-19-28 stop

2010-02-25-19-06 done proofread
2010-02-25-19-18 done spelling check



<a name="ch08c058">　Index begin　Index this file
2010-02-26-10-46 start
■　Exercise 8.5 problem statement
　　textbook page 133
（The Geometric Mean as a Minimum）

Prove that the geometric mean has
the representation

<a name="ch08c061">
For practice with this characterization
of the geometric mean, use it to
give another proof that the geometric
mean is superadditive, that is show
that the formula (8.33) implies the 
bound (2.31) on page 34.
2010-02-26-11-03 stop





<a name="ch08c062">　Index begin　Index this file
2010-02-26-11-51 start
■　Exercise 8.5 hint
　　textbook page 259

For all x∈D we have the bound
  (a1*a2*...*an)1/n //*1=x1*x2*...*xn
 =(a1x1*a2x2*...*anxn)1/n //GM≦AM
 ≦{∑[k=1,n](akxk)}/n ---eqn.14.57
by the AM-GM inequality, and we have
equality here if and only if akxk
does not depend on k. 
<a name="ch08c063">
If we take
  xk=ak / (a1*a2*...*an)1/n //by ByronSchmuland
  xk=(a1*a2*...*an)1/n / ak ---eqn.AZ026
then x∈D and the equality holds
in the bound (14.57). This is all
one needs to justify the identity
(8.33)

<a name="ch08c064">　Wonder
Now to prove the bound (2.31) on
page 34, one now just notes

since two choices are better than 
one.
<a name="ch08c065">
Incidentally, this type of argument
is exploited systematically in
Bechenbech and Bellman (1965)
where the formula (8.33) is called
quasilinear representation of the
geometric mean.
2010-02-26-12-26 stop

<a name="ch08c067">　Index begin　Index this file
2010-02-26-12-55 start
■　Exercise 8.5 solution


AM-GM Inequality
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="ch08c068">
Use power mean language, 
AM-GM Inequality is next

<a name="ch08c069">
2010-02-26-13-26 here
eqn.AZ028 has probability coefficients
pk. They are allowed to be unequal.
If we set pk to be equal, then
p1=p2=...=pn. Because eqn.AE20 require
pk sum to one, we find
  p1=p2=...=pn=1/n ---eqn.AZ029
put these pk=1/n to eqn.AZ028,
<a name="ch08c070">
it is almost the same as eqn.8.33.
The difference is that
eqn.8.33 use 1/n which is equal pk
eqn.8.33 use additional ∏xk=1
∏xk=x1*x2*x3=1 multiplied (extra)
∑pk=p1+p2+p3=1 summed (routine)
<a name="ch08c071">
How do we know
  ∏xk＝1, k=1,2,...,n ---eqn.AZ030
This requirement is stated in two
steps.
First step in eqn.8.33 say
  (x1,x2,...,xn)∈D ---eqn.AZ031
<a name="ch08c072">　Index begin　Index this file
Second step in eqn.AZ025 require
set D has the following property
  ∏xk＝1, k=1,2,...,n
In eqn.14.57 equality we used
  1=x1*x2*...*xn ---eqn.AZ032
which is ∏xk＝1.
Any number multiply by one not 
change value.
<a name="ch08c073">
In eqn.14.57 INequality we used
GM≦AM inequality for the sequence
akxk, (consider multiplication of 
two numbers ak, xk as one number)
Insert xk for ak play trick now.
Because
  1=x1*x2*...*xn ---eqn.AZ032
we can define xk to be
  xk=(a1*a2*...*an)1/n / ak ---eqn.AZ026
<a name="ch08c074">
To see whether eqn.AZ026 satisfy
eqn.AZ032, assume n=3 for simpler
example, then
  x1=(a1*a2*a3)1/3 / a1
  x2=(a1*a2*a3)1/3 / a2
  x3=(a1*a2*a3)1/3 / a3
  x1*x2*x3=
  [(a1*a2*a3)1/3]3/[a1a2a3]
  =[a1a2a3]/[a1a2a3] = 1 ---eqn.AZ033
Find eqn.AZ026 do satisfy eqn.AZ032.

<a name="ch08c075">
eqn.AZ026 tell us that
  ak*xk=(a1*a2*...*an)1/n ---eqn.AZ028
which is constant, independent
from index k. All ak*xk have same
value (a1*a2*...*an)1/n
(ak*xk=constant is key point
 for constant sequence GM=AM,
 otherwise GM＜AM)
The section Equality GM=AM is easy
tell us that when all elements 
are equal (same as constant)
then GM≦AM become GM=AM.
This specific GM=AM is minimum
value of AM.
2010-02-26-14-14 stop

<a name="ch08c076">　Index begin　Index this file
2010-02-26-14-22 start
To prove eqn.2.31
We pay attention to that 
eqn.8.33 left side (∏ side) 
do not contain xk
eqn.8.33 right side (∑ side) 
has xk term.
<a name="ch08c077">
eqn.2.31 is same as eqn.8.33
left side.
Three terms in eqn.2.31 change
to three terms in eqn.AZ027 
Give reasoning in eqn.AZ027 is
easy.
<a name="ch08c078">
akxk and bkxk are allowed to 
find minimum independently
the result is less than or 
at most equal to akxk and bkxk 
tied together for a minimum.
eqn.2.31 is proved.
2010-02-26-14-32 stop



<a name="ch08c079">　Index begin　Index this file
2010-02-26-15-50 start
■　Exercise 8.6 problem statement
　　textbook page 133
（More on the Method of Halves）

The method of halves applies to more
than just inequalities, it can
also be used to prove some elegant 
identities. 
<a name="ch08c080">
As an illustration, show
that the familiar half-angle formula
  sin(x)=2*sin(x/2)*cos(x/2) ---eqn.AZ029
implies the infinite product identity

<a name="ch08c083">
Incidentally, the product formula 
(8.34) for sin(x)/x is known as 
Viete's identity, and it has been
known since 1593.
2010-02-26-16-12 stop





<a name="ch08c084">　Index begin　Index this file
2010-02-26-16-17 start
■　Exercise 8.6 hint
　　textbook page 259

The half-angle formula for sine 
gives

<a name="ch08c088">
2010-02-26-16-42 here
and as k→∞ the bracketed term goes
to 1 since 
  sin(t)=t+O(t3) ---eqn.AZ038
as t→0. 
//sin(t)/t=t/t+O(t3)/t=1+O(t2)=1
Upon setting 
  x=PI/2 ---eqn.AZ039
one gets the second formula after
computing the successive values of
cosine with help from its half-
angle formula.  
<a name="ch08c089">
Naor (1998, pp. 139-143) gives a 
full discussion of Viete's formula,
including a fascinating geometric 
proof.
2010-02-26-16-49 stop



<a name="ch08c090">
2010-02-26-16-55 start
■　Exercise 8.6 solution


Exercise 8.6 hint already solved
this problem, and this is an easy
problem. Nothing to discuss here.
2010-02-26-16-56 stop



<a name="ch08c091">　Index begin　Index this file
2010-02-26-17-59 start
■　Exercise 8.7 problem statement
　　textbook page 133
（Differentiation of an Inequality）

In general one can not differentiate 
the two sides of an inequality and
expect any meaningful consequences,
<a name="ch08c092">
but there are special situations where
"differentiation of an inequality"
does make sense. There are even
times when such differentiations
have lead to spectacular new results.
The aspirations of this exercise are 
more modest, but they point the way
to what is possible.

<a name="ch08c093">
(a) Consider a function f that is
    differentiable at t0 and
    that satisfies the bound 
  f(t0)≦f(t) ---eqn.AZ040
    for all t∈[t0,t0+∆)
    and some ∆＞0. Show that one
    then has 
  0≦f'(t0) ---eqn.AZ041

<a name="ch08c094">
(b) Use the preceding observation
    to show that the power mean
    inequality implies that for
    all xk＞0 and all nonnegative
    pk with total
  p1+p2+...+pn=1 ---eqn.AZ042
    one has

2010-02-26-18-22 stop





<a name="ch08c096">　Index begin　Index this file
2010-02-26-18-28 start
■　Exercise 8.7 hint
　　textbook page 259

Our assumptions give us the bound
  [f(t0+h)-f(t0)]/h≧0 ---eqn.AZ043
for all h∈(0,∆], and now we just
let h→0 to prove the first claim. 
To address the second claim, one
first notes by the power mean
inequality, or by Jensen's 
inequality, that one has

<a name="ch08c098">
Since
  0=f(1)≦f(t) for 1≦t ---eqn.AZ045
we also have
  f'(1)≧0 ---eqn.AZ046
and this is precisely the bound
(8.35)
2010-02-26-18-49 stop

<a name="ch08c099">　Index begin　Index this file
2010-03-01-14-55 start
■　Exercise 8.7 solution


part (a) discussion.
Exercise 8.7 find differentiation
of inequality type. Given condition
is
<a name="ch08c100">
[[
a function f that is
    differentiable at t0 and
    that satisfies the bound 
  f(t0)≦f(t) ---eqn.AZ040
    for all t∈[t0,t0+∆)
    and some ∆＞0. 
]]
<a name="ch08c101">
Please see the drawing drawEx.8.7
and click "Ex.8.7RUN" button.
(reading notes use 't' as variable,
 drawing use 'x' as variable, since
 function differentiatef1() for
 d[f(x)]/dx use 'x' as variable.
 Yield to function definition cause
 not consistency. 2010-03-01-15-16)

<a name="ch08c102">
Default curve is
  f(x)=(x-1)*(x-1)+0.5 ---eqn.AZ047
Default point is x=3.2, which is t0
In the problem statement,
given condition
  f(t0)≦f(t) ---eqn.AZ040
tell us that for a positive small
number 0≦h≦∆ we have f(t0)≦f(t)
here
  t=t0+h＞t0 ---eqn.AZ048
//The use of ∆ says that d[f(t)]/dt＞0
//is a local phenomenon. Few pace more
//ahead may not have d[f(t)]/dt＞0;
//2010-03-03-11-34
<a name="ch08c103">
In the drawing, red curve right to
dot vertical line is t＞t0 section.
t0 is where dot line located.
Rewrite given condition eqn.AZ040 as
  [f(t0+h)-f(t0)]≧0 ---eqn.AZ049
divide by positive number h, get
  [f(t0+h)-f(t0)]/h≧0 ---eqn.AZ050
<a name="ch08c104">
We require h approach to zero,
eqn.AZ050 is differentiation at 
point t0 exactly. Then
given condition eqn.AZ040 give us
  f'(t0)= ---eqn.AZ051
  limit[h→0][f(t0+h)-f(t0)]/h≧0
//'=' from derivative definition
//'≧' from given condition.
eqn.AZ051 is same as eqn.AZ041.
Problem part (a) solved easily.
2010-03-01-15-32 here

<a name="ch08c105">　Index begin　Index this file
2010-03-01-15-42 start
For part (b)

2010-03-01-16-15
Next copied from tute0030.htm#ch08a045

<a name="ch08c106">
Please compare next two equations

<a name="ch08c107">
From both equation greater than
side, we see directly that
  f(xk)=xkr ---eqn.AX106
From both equation less than
side, we see
  f(∑px)={∑px}r ---eqn.AX107

<a name="ch08c108"> <a name="ch08a047">
eqn.AX106 and eqn.AX107 suggest
that 
  f(x)=xr ---eqn.AX108

Jensen Inequality for convex 
function require that
  r>1 ---eqn.AX109
Above copied from tute0030.htm#ch08a047
2010-03-01-16-31 here

<a name="ch08c109">
Jensen give us eqn.AZ052, re-write
eqn.AZ052 as same type of eqn.AZ044
next

<a name="ch08c110">　Index begin　Index this file
In eqn.AZ053 if we set r=1, find
  f(r=1)=∑pkxk - ∑pkxk = 0 ---eqn.AZ054
This is just one point r=0 condition.
What if r=2, r=3? Do we have
f(r=2)＞f(r=3) ? (≧0)
f(r=2)＜f(r=3) ? (≧0)
f(r=2)＝f(r=3) ? (≧0)
<a name="ch08c111">
Here we need Power Mean inequality

<a name="ch08c112">
eqn.8.10 short writing is
  Ms≦Mt for s≦t ---eqn.AZ055
Now set s=r and t=r+0.1, we have
  Mr≦Mr+0.1 for r≧1 ---eqn.AZ056
Here Mr has a outer root power 1/r
Mr+0.1 has outer root power 1/(r+0.1)
See eqn.8.10.
eqn.AZ056 whole equation take r th 
power, eqn.AZ056 become
  Mrr≦Mr+0.1r for r≧1 ---eqn.AZ057
<a name="ch08c113">
Write in eqn.AZ053 form for f(r)
and f(r+0.1) get

<a name="ch08c114">
2010-03-01-17-18 here
0.5*pow(0.2,1.1)+0.5*pow(0.3,1.1)
0.21811921483686045
0.5*pow(0.2,2.1)+0.5*pow(0.3,2.1)
0.056922365225850285
in x^p if x<1 and p>1
greater p get smaller x^p

2010-03-01-17-22 stop

<a name="ch08c115">　Index begin　Index this file
2010-03-01-20-01 start
Above calculation
[[
in x^p if x<1 and p>1
greater p get smaller x^p
]]
But we require greater p get greater
function value, that is monotone
increase. Above calculation take
only one term into consideration
(the red or blue) did not consider
the minus_black terms.
<a name="ch08c116">
It is easier to verify numerically.
Here is a program local
http://freeman2.com/powmean2.htm
Added new function, it is marked
"Jensen". There are two buttons
"JensenA" and "JensenB".
"JensenA" modify input boxes value
before call function powMeanf()
modification target at x-sequence
value range in [0,17]. 
<a name="ch08c117">
If you have
x-sequence value range in [0,1]
for example, you need give proper 
setting (input boxes values) then
click "JensenB" button.
"Jensen" button draw three curves.
Red curve is eqn.AZ053 red term
(or blue term same thing)
Blue curve is eqn.AZ053 black term
ignored '-'.
Black curve is whole eqn.AZ053 .

<a name="ch08c118">
First, fill '1' or '1.5' to 10^ box
second click "random5#" button,
third click "JensenA" button
Black curve has zero value at t=1
In Box 3, output, find a line
similar to next line
1; 15.558182090181461; 15.558182090181461; 0
Here t=1, both red term and black
term has value 15.55818, their
difference red-black (eqn.AZ053)
is zero. Zero value is required
by eqn.AZ054. 
<a name="ch08c119">
Key point is
for t＞1, black curve is monotone
increase in limited domain [t0,t0+∆)
Above x-sequence range in [0,17]

<a name="ch08c120">　Index begin　Index this file
Below x-sequence range in [0,1]
First, fill '0' to 10^ box
second click "random5#" button,
third modify the following boxes
(use the following value)
y min: -1 , y max: 1
fourth click "JensenB" button
(alert: do not click "JensenA" )
<a name="ch08c121">
Output curve show that black curve
has negative value for t in [0,1]
black curve has positive value for
 t in ( 1, greater_value )
black curve is monotone increase
within visible domain [1,3]
We need black curve be monotone 
increase.
<a name="ch08c122">
With above observation, (NOT A
 PROOF) we can answer
What if r=2, r=3? Do we have
f(r=2)＞f(r=3) ? (≧0)
f(r=2)＜f(r=3) ? (≧0)
f(r=2)＝f(r=3) ? (≧0)
The answer is monotone increase
we have f(r=2)＜f(r=3)

<a name="ch08c123">
What we have in hand is
eqn.AZ053 is monotone increase,
f(r=1)=0 (eqn.AZ054)
we use part (a) conclusion
get   f'(1)≧0 ---eqn.AZ046
follow textbook
[[
and this is precisely the bound
(8.35)
]]
we solved part (b)
Do you agree?
2010-03-01-20-41 here

<a name="ch08c124">
2010-03-01-20-49 start
Not really done !!
How to go from eqn.AZ044 to
eqn.8.35 It look like remote.
Why log() show up? why not exp()?
why not sin()? Puzzle !!

<a name="ch08c125">　Index begin　Index this file
Function g(t) if variable t is in
power position like E^t, //E=2.71828...
We know if
  g(t)=exp(t) ---eqn.AZ059
then 
  d[g(t)]/dt=exp(t) ---eqn.AZ060
<a name="b2t_power">
We know if
  h(t)=B^t ---eqn.AZ061
then 
  d[h(t)]/dt=h(t)*log(B) ---eqn.AZ062
//if h(t)=B^t were h(t)=E^t, then
//log(B) become log(E) which is ONE.

<a name="ch08c126">
eqn.AZ044 is waiting to be
differentiated. We do as following

<a name="ch08c129">
We need use the condition
  f'(1)≧0 ---eqn.AZ046
then eqn.AZ064≧0
and  eqn.AZ064 change t to one.
After this two steps, eqn.AZ064
is same as eqn.8.35
Now Exercise 8.7 is really solved.
2010-03-01-21-25 stop

2010-03-02-18-56 done proofread
2010-03-03-15-12 done spelling check

<a name="ch08c130">
Update 2010-03-16 made following 
notes about eqn.AZ027

2010-03-16-20-59 copy next from 
tute0034.htm
<a name="ch09b014">
Wonder, textbook page 259, Line 11
whether eqn.AZ027 is better stated as
  OUTER_min{inner_min[x∈D](∑[k=1,n]ak*xk)/n,
            inner_min[x∈D](∑[k=1,n]bk*xk)/n}
  ≦min[x∈D]{∑[k=1,n](ak+bk)*xk}/n ---eqn.BB009

<a name="ch09b015">
eqn.AZ027 two team two min add
together has risk greater than
joint team one min !!
2010-03-16-13-06 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