Inequality Study 32nd file   Update 2010-03-16
index   this   program   DocA   Limit  
XYGraph v2.3 - web page graph   ☜☞   donate   get code
The Cauchy-Schwarz Master Class   J. Michael Steele   ★★★★★
This file is personal home work. No one
proofread. Cannot promise correctness.
If you suspect any view point wrong,
please ask a math expert near by.
Freeman 2009-06-19-10-46

Please use MSIE browser to read this file.
Did not test other browser. This file is
written under MSIE 6.0




<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="ch08c001"> Index begin Index this file
2010-02-23-10-27
■ Power Mean attention
In previous work, we discussed
□ Why require p1+p2+...+pn=1 ? ---eqn.AE20
It relate to Power Mean directly. 
Power Mean inequality is next
<a name="ch08c002">
{
k=n
k=1
pkxks }
1/s
 
 
{
k=n
k=1
pkxkt }
1/t
 
 
---page 123 ---Line 26 ---eqn.8.10
width of above equation
<a name="ch08c003">
Here, we consider equation
physics dimension consistency.
pk is probability, which is a pure
number, not carry physics dimension.
Assume xk is measured distance,
eqn.8.10 left side {term} is 
 sum of pure_number*distances
physics dimension is length to s
power. Then {term} rise to 1/s
power again, the net result is 
back to distance first power.
[ s*(1/s)=1 ]
<a name="ch08c004">
eqn.8.10 right side change from s
to t. Same conclusion. Therefore
eqn.8.10 is distance compare with
distance. The math form in right
side of eqn.8.10 is marked as Mt
Left side is Ms. When t approach 
to zero we have M0. 
<a name="ch08c005">
M0 is expressed in eqn.8.2, its 
right side ∏xkpk was discussed
in tute0011.htm#ch02a063
In brief, because SUM{pk}=1, then 
∏xkpk still has the dimension of 
length. M0 can be compared with M1
M-1 etc.
2010-02-23-11-06 stop


<a name="ch08c006"> Index begin Index this file 2010-02-23-16-26 start ■ Exercise 8.2 problem statement   textbook page 131 (Harmonic Means and Recognizable Sums) Suppose x1,x2,...,xn are positive and let S denote their sum. Show that one has the bound
<a name="ch08c007">
 
n2

2n-1
S

2S-x1
S

2S-x2
+...+
S

2S-xn
---page 131 ---line 20 ---eqn.AZ001
width of above equation
<a name="ch08c008">
In this problem (and many like it)
one gets a nice hint from the fact
that there is a simple expression 
for the sum of the denominator on
the right-hand side.
2010-02-23-16-38 stop






<a name="ch08c009"> Index begin Index this file
2010-02-23-16-41 start
■ Exercise 8.2 hint
  textbook page 258

By the upside-down HM-AM inequality
(8.16)
<a name="ch08c010"> //2010-02-23-17-36 change xj to aj
1

p1*a1+p2*a2+...+pn*an
p1

a1
p2

a2
+...+
pn

an
---page 126
---Line 25
---eqn.8.16
width of above equation
<a name="ch08c011">
one has // equal probability p1=p2=...=pn=1/n
n2

a1+a2+...+an
1

a1
1

a2
+...+
1

an
---page 258
---Line 8
---eqn.AZ002
width of above equation
<a name="ch08c012">
2010-02-23-17-00 here
If we set
  ak=2S-xk ---eqn.AZ003
then
  a1+a2+...+an=2nS-S=(2n-1)S ---eqn.AZ004
and the HM-AM bound yields
<a name="ch08c013">
 
n2

(2n-1)S
1

2S-x1
1

2S-x2
+...+
1

2S-xn
---page 258 ---line 11 ---eqn.AZ005
width of above equation
2010-02-23-17-06 stop



<a name="ch08c014"> Index begin Index this file
2010-02-23-17-08 start
■ Exercise 8.2 solution


Exercise 8.2 hint already solved
this problem. 
Few thing to observe.
<a name="ch08c015">
In eqn.8.16 set
  p1=p2=...=pn=1/n ---eqn.AZ006
to get eqn.AZ002.
Then use eqn.AZ003 to relate xk 
and ak and S. Here 
  S=x1+x2+...+xn ---eqn.AZ007
which is given in Exercise 8.2 
problem statement.
<a name="ch08c016">
With
  ak=2S-xk ---eqn.AZ003
relation, we can express eqn.AZ002
less than side denominator
  a1+a2+...+an
  =(2S-x1)+(2S-x2)+...+(2S-xn)
  =2nS - x1-x2-...-xn
  =2nS - S
<a name="ch08c017">
that is
  a1+a2+...+an=(2n-1)S ---eqn.AZ008
Put eqn.AZ008 to eqn.AZ002 left
 side (once)
Put eqn.AZ003 to eqn.AZ002 right
 side (n times)
we get eqn.AZ005.
From eqn.AZ005 to eqn.AZ001 is
just one step. Problem solved.
2010-02-23-17-24 stop


<a name="ch08c018"> Index begin Index this file 2010-02-23-18-37 start ■ Exercise 8.3 problem statement   textbook page 132 (Integral Analogs and Homogeneity in ∑) The importance of weight factor. Weight factor not sum to one? All Mt, t≠0 are still length1.0 BUT GM M0 at t=0 has length1.1 AM:length1.0 compare?! GM:length1.1 Power Mean curve jump to infinity at singular GM. explore local <a name="ch08c019"> (a) Show that for all nonnegative sequence {ak: 1≦k≦n} one has
{
k=n
k=1
ak1/2 }
2
 
 
{
k=n
k=1
ak1/3 }
3
 
 
---page 132
---Line 3
---eqn.8.29
width of above equation
<a name="ch08c020">
and be sure to notice the
differences between this bound and
the power mean inequality (8.10)
with s=1/3 and t=1/2.
//2010-03-02-16-50 key difference !
//eqn.8.29 each term has probability
//coefficient=1, sum to n, NOT one!

<a name="ch08c021">
(b) By analogy with the bound (8.29)
    one might carelessly guess that
    for nonnegative f one has an
    integral bound
<a name="ch08c022">
{
x=1
x=0
f1/2(x)dx }
2
 
 
{
x=1
x=0
f1/3(x)dx }
3
 
 
---page 132
---Line 8
---eqn.8.30
width of above equation
Show by example that the bound
(8.30) does not hold in general.

<a name="ch08c023"> Index begin Index this file
The likehood of an integral analog
can often be explained by a heuristic
principle which Hardy, Littlewood 
and Polya (1952, p4) describes as 
"homogeneity in ∑". The principle
suggests that we consider ∑ in a
bound such as (8.29) as a formal
symbol. 
<a name="ch08c024">
In this case we see that
the left side is "homogeneous of
order two in ∑" while the right
side is "homogeneous of order three
in ∑" The two sides are therefore 
incompatible and one should not
expect any integral analog. 
<a name="ch08c025">
On the other hand, in Cauchy's 
inequality and Holder's inequality
both sides are homogeneous of
order one in ∑. It is therefor
natural -- even inevitable --
that we should have integral
analog for these bounds.
2010-02-23-19-04 stop





<a name="ch08c026"> Index begin Index this file
2010-02-23-19-16 start
■ Exercise 8.3 hint explore
  textbook page 258 local

Both sides of the bound (8.29) are
homogeneous of order one in (a1,a2,
...,an), so we can assume without
loss of generality that
  a11/3+a21/3+...+an1/3=1 ---eqn.AZ009
<a name="ch08c027">
Given this, we only need to show
that
  a11/2+a21/2+...+an1/2≦1 ---eqn.AZ010
and this is remarkable easy. By
the normalization, we have
  ak≦1 for all 1≦k≦n ---eqn.AZ011
so we also have
  ak1/2 ≦ ak1/3 for all 1≦k≦n ---eqn.AZ012
//if ak=0.5, ak1/2=0.70710<0.79370=ak1/3
and we just take the sum to get
our target bound. 
<a name="ch08c028">
One might want
to reflect on what made this 
exercise so much easier than the
proof of the power mean inequality
(8.10). For part (b), if we take
  f(x)=x6 ---eqn.AZ013
to minimize arithmetic, then we
see that the putative bound (8.30)
falsely asserts 1/16 ≦ 1/27.
2010-02-23-19-30 stop





<a name="ch08c029"> explore local
2010-02-23-20-10 start
■ Exercise 8.3 discussion
Put power mean inequality (8.10)
and Exercise 8.3 target eqn.8.29
side by side, easier for comparison.
<a name="ch08c030"> Top line is power mean inequality
{
k=n
k=1
pkxks }
1/s
 
 
{
k=n
k=1
pkxkt }
1/t
 
 
{
k=n
k=1
ak1/2 }
2
 
 
{
k=n
k=1
ak1/3 }
3
 
 
---compare ---eqn.8.10 top, ---eqn.8.29 bottom
width of above equation
<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
F(p) =
max
x∈[a,b]
{
|p-x|

x
}
---page 132
---Line 26
---eqn.8.31
width of above equation
<a name="ch08c040">
So, your analytic challenge is to find the value p*
such that
F(p*) =
min
p
F(p) =
min
p
max
x∈[a,b]
{
|p-x|

x
}
---page 132 ---Line 28 ---eqn.8.32
width of above equation
<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="ch08c043">
F(p) = max {
p-a

a
,
b-p

b
}
---page 258
---Line 24
---eqn.AZ017
width of above equation
<a name="ch08c044">
The identity
a

a+b
{
p-a

a
}
b

a+b
{
b-p

b
}
b-a

a+b
---page 258 ---Line 26 ---eqn.AZ018
width of above equation
<a name="ch08c045">
tell us that (b-a)/(a+b) is a weighted mean
of (p-a)/a and (b-p)/b, so we always have
the bound
F(p) = max {
p-a

a
,
b-p

b
}
b-a

a+b
---page 258
---Line 29
---eqn.AZ019
width of above equation
<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="drawEx.8.4">  Index begin Index this file
2010-02-24-16-40
Box11
Box12



<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="ch08c059">
{
k=n
k=1
ak }
1/n
 
 
= min {
1

n
k=n
k=1
akxk:(x1,x2,...,xn)∈D }
---page 133 ---Line 3 ---eqn.8.33
width of above equation
<a name="ch08c060">
where D is the region of Realn defined by //●●change
D = { (x1,x2,...,xn):
k=n
k=1
xk = 1, xk0, k=1,2,...,n }
---page 133 ---Line 5 ---eqn.AZ025 //change from '≧' to '>'
//because if one xk=0, ∏xk=1 is void. 2010-02-26-11-15

width of above equation
<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
min
x∈D
1

n
k=n
k=1
akxk
min
x∈D
1

n
k=n
k=1
bkxk
min
x∈D
1

n
k=n
k=1
(ak+bk)xk
---page 259 ---Line 11 ---eqn.AZ027
width of above equation
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="ch08c066"> <a name="ch02c097">
 
(
k=n
k=1
ak )
1/n
 
 
(
k=n
k=1
bk )
1/n
 
 
[
k=n
k=1
(ak+bk) ]
1/n
 
 
Superadditivity of the Geometric Mean
---page 34 line 19 ---eqn.2.31
width of above equation



<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
GM = M0 = next line //∑pk=1 , k=1,2,...,n is required
define
k=n
k=1
xkpk {
k=n
k=1
pkxkt=1 }
1/(t=1)
 
 
define
from above = AM = M1 = Mt=1
---GM≦AM power mean style expression ---eqn.AZ028
width of above equation

compare above with eqn.8.33
{
k=n
k=1
ak }
1/n
 
 
= min {
1

n
k=n
k=1
akxk:(x1,x2,...,xn)∈D }
---page 133 ---Line 3 ---eqn.8.33
width of above equation
<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="ch08c081">
sin(x)

x
k=∞
k=1
cos(
x

2k
)
---page 133
---Line 14
---eqn.8.34
width of above equation
<a name="ch08c082">
and verify in turn that this implies the poignant formula
2

PI
√2

2
*
√[2+√2]

2
*
√{2+√[2+√2]}

2
*...
---page 133 ---Line 16 ---eqn.AZ034
width of above equation
<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="ch08c085">
sin(x)

x
2sin(x/2)cos(x/2)

x
=cos(x/2)*
sin(x/2)

x/2
---page 259
---Line 17
---eqn.AZ035
width of above equation <a name="ch08c086">
sin(x)

x
=cos(x/2)*cos(x/4)*
sin(x/4)

x/4
---page 259
---Line 18
---eqn.AZ036
width of above equation <a name="ch08c087">
sin(x)

x
=cos(x/2)*cos(x/4)*...*cos(x/2k)* {
sin(x/2k)

x/2k
}
---page 259 ---Line 19 ---eqn.AZ037
width of above equation
<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
<a name="ch08c095">
{
k=n
k=1
pkxk } log {
k=n
k=1
pkxk } {
k=n
k=1
pkxk*log(xk) }
---page 134 ---Line 4 ---eqn.8.35
width of above equation
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="ch08c097">
f(t)=
k=n
k=1
pkxkt (
k=n
k=1
pkxk )
t
 
 
≧0
for all
t∈[1,∞)
---page 259 ---Line 29 ---eqn.AZ044
width of above equation
<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="drawEx.8.7"> 2010-02-26-19-38
Box22 equation definition must use 'x' as variable.
Box21
Box22







<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
{
k=n
k=1
pkxk }
r
 
 
{
k=n
k=1
pkxkr }
---page 124 ---Line 5 ---eqn.AZ052 (eqn.AX105)
width of above equation
//above Jensen Inequality with base function f(x)=xr, r>1
//below Jensen Inequality with unspecified base function

and
<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
f(r)=
k=n
k=1
pkxkr {
k=n
k=1
pkxk }
r
 
 
≧0
for all r∈[1,∞) , since x^r (r>1) is convex.
---page 259 ---Line 29 ---eqn.AZ053 (eqn.AX105)
width of above equation
<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
{
k=n
k=1
pkxks }
1/s
 
 
{
k=n
k=1
pkxkt }
1/t
 
 
---page 123 ---Line 26 ---eqn.8.10
width of above equation
<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
f(r)=
k=n
k=1
pkxkr {
k=n
k=1
pkxk }
r
 
 
≧0
f(r+0.1)=
k=n
k=1
pkxkr+0.1 {
k=n
k=1
pkxk }
r+0.1
 
 
≧0
for all r∈[1,∞) , since x^r (r>1) is convex.
---page 259 ---Line 29 ---eqn.AZ058 (eqn.AX105)
width of above equation
<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
f(t)=
k=n
k=1
pkxkt (
k=n
k=1
pkxk )
t
 
 
≧0
for all
t∈[1,∞)
---page 259 ---Line 29 ---eqn.AZ044 second
width of above equation <a name="ch08c127">
d[f(t)]

dt
k=n
k=1
pk
d[xkt]

dt
d

dt
(
k=n
k=1
pkxk )
t
 
 
---page 259 ---Line 29 ---eqn.AZ063
red to red, blue to blue, all NOT contain 't' (B here)
width of above equation <a name="ch08c128">
d[f(t)]

dt
k=n
k=1
pkxktlog(xk) (
k=n
k=1
pkxk )
t
 
 
log (
k=n
k=1
pkxk )
---page 134 ---Line 4 ---eqn.AZ064 (eqn.8.35)
width of above equation
<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


Javascript index
http://freeman2.com/jsindex2.htm   local
Save graph code to same folder as htm files.
http://freeman2.com/jsgraph2.js   local


File name tute0032.htm means
TUTor, English, 32nd .htm
Chinese series file name is tutc0001.htm

This page, Inequality file twenty six.
http://freeman2.com/tute0032.htm
First Upload 2010-02-25
(Inequality start from tute0007.htm)

Thank you for visiting Freeman's page. 
Freeman  2010-02-25-19-19

≦ ≠ ≧ <=>±≡≈≌≒∏∑√∛∜∝ →∞ ⊕⊙
〈v,w〉 ∈ ∀∂⊥∃∋∆∇∟∠∫∬∭∮∥○●◎ 
∧∨∩∪∴∵∶∷⊂⊃⊄⊅⊆⊇⊿+-*/
§‰¼½¾ ⅓⅔⅕⅖⅗⅘⅙⅚⅛⅜⅝⅞⅟←↑→↓↔↕↖↗↘↙
■□ ▢▣▤▥▦▧▨▩▪▫ × ÷ ° º¹²³ 
ΑΒΓΔΕΖΗΘΙΚΛΜΝΞΟΠΡ΢ΣΤΥΦΧΨΩ
ΪΫάέήίΰ αβγδεζηθικλμνξοπρςστυφχψω