<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="ch14c001">　Index begin　Index this file
2010-08-05-10-14 start
■　Exercise 14.1 problem statement
　　textbook page 221
（Abel's Second Inequality）

Show that for each nondecreasing
sequence of nonnegative real
numbers 
  0≦b1≦b2≦...≦bn ---eqn.BS001
<a name="ch14c002">
one has a bound which differ
slightly from Abel's first
Inequality.
  |b1z1+b2z2+...+bnzn|
 ≦2*bn*max[1≦k≦n]|Sk| ---eqn.14.29
2010-08-05-10-24 stop

2010-08-10-21-06 add Abel's first
Inequality for comparison purpose.
  |a1z1+a2z2+...+anzn|
 ≦a1*max[1≦k≦n]|Sk| ---eqn.14.1
and
  a1≧a2≧...≧an≧0 ---eqn.BQ002





<a name="ch14c003">
2010-08-05-15-02 start
■　Exercise 14.1 hint
　　textbook page 279

To prove the second bound
(14.29), we again sum
  b1z1+b2z2+...+bnzn ---eqn.BS002
<a name="ch14c004">
by parts to get
  S1(b1-b2)+S2(b2-b3)+...
 +Sn-1(bn-1-bn)+Snbn ---eqn.BS003
but this time we bound the sum
 |b1z1+b2z2+...+bnzn| ---eqn.BS004
<a name="ch14c005">
by noting
  |S1||b1-b2|+|S2||b2-b3|+...
 +|Sn-1||bn-1-bn|+|Sn|bn ---eqn.BS005
 ≦max[1≦k≦n]|Sk|{(b2-b1)+(b3-b2)
  +...+(bn-bn-1)+bn} ---eqn.BS006
 ={(bn-b1)+bn}max[1≦k≦n]|Sk| ---eqn.BS007
 ≦2*bn*max[1≦k≦n]|Sk| ---eqn.BS008
2010-08-05-15-19 stop

<a name="ch14c006">　Index begin　Index this file
2010-08-05-15-26
Abel's first Inequality
Problem 14.1 (Abel's Inequality)　eqn.14.1
calculation start at
tute0049.htm#ch14a023
tute0049.htm#ch14a029
<a name="ch14c007">
for special case n=5
real calculation end at
tute0049.htm#ch14a035
 ---eqn.BQ027
2010-08-05-15-30 record

<a name="ch14c008">
2010-08-05-20-35 start
■　Exercise 14.1 discussion

Abel's first Inequality and
Abel's second Inequality are
parallel.
<a name="ch14c009">
from tute0049.htm#ch14a023
 to  tute0049.htm#ch14a035
has special case analysis for
n=5. Five complex numbers and
five real coefficients. Here
do same thing, 
special case analysis for n=5
copy Abel's first Inequality 
study notes as following

<a name="ch14c010">
Assume five complex numbers are
z1='-0.7260+5.7410i' ---eqn.BQ005
z2=' 1.6399-15.759i' ---eqn.BQ006
z3=' 11.229-11.630i' ---eqn.BQ007
z4=' 9.9490+16.750i' ---eqn.BQ008
z5='-5.9902+0.1433i' ---eqn.BQ009

<a name="ch14c011">　Index begin　Index this file
Partial summation values are
S1=z1=-0.7260+5.7410i ---eqn.BQ010
S2=z2+S1=0.9139-10.018i ---eqn.BQ011
S3=z3+S2=12.1429-21.648i ---eqn.BQ012
S4=z4+S3=22.0919-4.898i ---eqn.BQ013
S5=z5+S4=16.1017-4.7547i ---eqn.BQ014

<a name="ch14c012">
Write it symbolic longhand
S1=z1 ---eqn.BQ015
S2=z1+z2 ---eqn.BQ016
S3=z1+z2+z3 ---eqn.BQ017
S4=z1+z2+z3+z4 ---eqn.BQ018
S5=z1+z2+z3+z4+z5 ---eqn.BQ019

<a name="ch14c013">
We can not compare complex number
however, we can compare the
absolute value of a complex 
number. Because absolute value
is a real number. Next step,
take absolute value of partial 
sums |S1|,|S2|,|S3|,|S4|,|S5|
2010-08-05-20-48 here

//<a name="ch14c014">
//2010-08-05-20-49 start code
//Abel's 1st Ineq eqn.BQ027
//Abel's 2nd Ineq eqn.14.29
//If both 1st and 2nd use same
//coefficient sequence, change
//descend to ascend. Do they 
//have same upper bound?
//why Abel's 2nd inequality
//upper bound double?
//run test code and see why.
//
//<a name="ch14c015">　Index begin　Index this file
var errmsg='Output may contain error. You must exam first.'
errmsg; //9908052050
var cp=[];
cp[0]='-0.7260+5.7410i';
cp[1]=' 1.6399-15.759i';
cp[2]=' 11.229-11.630i';
cp[3]=' 9.9490+16.750i';
cp[4]='-5.9902+0.1433i';
var reverseComplex=0; //9908061542
//If you set reverseComplex=1
//program will reverse cp[] order
//<a name="ch14c016">
//Abel's first Inequality
//coef. 5,4,3,2,1
var coef1st=[];
coef1st[0]=5;
coef1st[1]=4;
coef1st[2]=3;
coef1st[3]=2;
coef1st[4]=1;
//<a name="ch14c017">
//Abel's second Inequality
//coef. 1,2,3,4,5
var coef2nd=[];
coef2nd[0]=1;
coef2nd[1]=2;
coef2nd[2]=3;
coef2nd[3]=4;
coef2nd[4]=5;
//You can change above value
//You can add more cp[5], cp[6]
//coef1st[], coef2nd[]
//You can delete cp[4],cp[3]
//coef1st[], coef2nd[]
//
//<a name="ch14c018">
var equalCoef=0;  //9908061140
//if set equalCoef=1, program
//change all coefficient to one
//attempt to reach equality for
//Abel's first Inequality.
//*** To get equality, must have
//*** both equalCoef=1
//*** and  equalFirst=1
//
//<a name="ch14c019">
var equalFirst=0; //9908061157
//if set equalFirst=1, program
//change all complex number to 
//zero, except first complex
//be '3+4i'. Attempt to reach 
//equality for Abel's first 
//Inequality
//
//<a name="ch14c020">　Index begin　Index this file
var cLen=cp.length;
var aLen=coef1st.length; //9908061545
var bLen=coef2nd.length;
var errorMsg=''
//<a name="ch14c021">
if(aLen!=cLen) //9908061546
  {
errorMsg='aLen!=cLen, '+aLen+'!='+cLen;
errorMsg
  }
//<a name="ch14c022">
if(bLen!=cLen) //9908061547
  {
errorMsg='bLen!=cLen, '+bLen+'!='+cLen;
errorMsg
  }
//<a name="ch14c023">
if(reverseComplex==1)
  { //9908061548
var temp=[];
for(i=0;i<cLen;i++)
temp[i]=cp[cLen-i-1]
for(i=0;i<cLen;i++)
cp[i]=temp[i];
temp=''; //9908061551
  }
//<a name="ch14c024">
if(equalCoef==1) //9908061141
  {
for(i=0;i<cLen;i++)
coef1st[i]=coef2nd[i]=1; //9908061142
  }
//<a name="ch14c025">　Index begin　Index this file
var partialSum=[];
partialSum[0]=[0,0]; //9908052127
var i,j,k,m1,m2,m3;
//
//<a name="ch14c026">
if(equalFirst==1) //9908061159
  {
cp[0]=[3,4];
for(i=1;i<cLen;i++)
cp[i]=[0,0]; //try get equality
  } //for Abel 1st ineq. 9908061201
//<a name="ch14c027">
//do not copy code from source file
//source file less than is '&lt;'
//copy from browser screen.
//screen less than is '<'
else for(i=0;i<cLen;i++)//change string complex
cp[i]=toCplx(cp[i]);  //to array complex
//<a name="ch14c028">
partialSum[0][0]=cp[0][0]
partialSum[0][1]=cp[0][1]
for(i=1;i<cLen;i++) //9908052054
partialSum[i]=caddf(cp[i],partialSum[i-1]);
var absPSum=[]; //absolute value of partial sum
var maxPS; //maximum partial sum
var maxPi; //index of max p. sum
maxPi=0;
//<a name="ch14c029">
for(i=0;i<cLen;i++){ //9908052100
absPSum[i]=cabsf(partialSum[i])
if(i==0)maxPS=absPSum[i];
else if(maxPS<absPSum[i]){dummy=0;
 maxPS=absPSum[i];maxPi=i}//[else] button
  }//for(i=0;i<cLen;i++)
//<a name="ch14c030">　Index begin　Index this file
var Abel1stleft=0;
var Abel1strite=0;
var Abel2ndleft=0; //Abel's 2nd Ineq left side
var Abel2ndrite=0; //Abel's 2nd Ineq rite side
var leftSum1st=[0,0]
//<a name="ch14c031">
for(i=0;i<cLen;i++) //9908052115
leftSum1st=caddf(leftSum1st,cmulr(cp[i],coef1st[i]));
Abel1stleft=cabsf(leftSum1st);
Abel1strite=maxPS*coef1st[0];
Abel1stleft //Abel's 1st Ineq less than side
Abel1strite //Abel's 1st Ineq greater than side
//2010-08-05-21-34 stop
//2010-08-05-22-26 start
2-1 //above Abel's 1st below Abel's 2nd
//<a name="ch14c032">
var leftSum2nd=[0,0]
for(i=0;i<cLen;i++) //9908052228
leftSum2nd=caddf(leftSum2nd,cmulr(cp[i],coef2nd[i]));
Abel2ndleft=cabsf(leftSum2nd);
Abel2ndrite=2*maxPS*coef2nd[cLen-1];
Abel2ndleft //Abel's 2nd Ineq less than side
Abel2ndrite //Abel's 2nd Ineq greater than side
3-2 //ALERT Abel2ndrite=2*Abel1strite; drop [2*] OK?
if(reverseComplex==1) //9908061619
reverseComplex='Complex sequence reversed. Different answer.'
reverseComplex
//<a name="ch14c033">
maxPS //maximum partial sum
maxPi //index of maxPS
var absPSumNewLine=''; //9908052234
absPSumNewLine=(absPSum+'').replace(/,/g,'\n'); //9908052237
absPSumNewLine   //abs(partial sum)
//2010-08-05-22-38 stop
//copy/paste the above code to
//http://freeman2.com/complex2.htm#calculator
//box 3, click [test box3 command, output to box4]


<a name="ch14c034">
2010-08-06-11-18 start
In above numerical example,
absolute value of partial sums 
are
[[
absPSumNewLine   //abs(partial sum)
5.786722474769289
10.05959925692868
24.82107822819146
22.628354991249363
16.789041514630906
]]
Maximum partial sum is 24.821

<a name="ch14c035">　Index begin　Index this file
Abel's first Inequality bound is
a1*max[1≦k≦n]|Sk| ---eqn.BQ004
Above numerical example has
  5*24.821=124.105

<a name="ch14c036">
Abel's second Inequality bound is
  |b1z1+b2z2+...+bnzn|
 ≦2*bn*max[1≦k≦n]|Sk| ---eqn.14.29
Above numerical example gives
  2*5*24.821=248.210
(LiuHH guess "2*" is not necessary)

<a name="ch14c037">
Abel's first Inequality less than 
side value is
|a1z1+a2z2+...+anzn| ---eqn.BQ003
It has value
Abel1stleft //Abel's 1st Ineq less than side
61.793913394848204

<a name="ch14c038">
Abel's second Inequality less 
than side value is
  |b1z1+b2z2+...+bnzn| ---eqn.BS004
It has value
Abel2ndleft //Abel's 2nd Ineq less than side
46.62184479286507

<a name="ch14c039">　Index begin　Index this file
Both first and second Inequality
less than side value are less 
than 124.105
LiuHH guess that 
Abel's second Inequality eqn.14.29
upper bound factor '2' is too loose.
2010-08-06-11-32 here

<a name="ch14c040">
2010-08-06-12-05 start
Added code to build equality
for Abel's first Inequality
If in the code section, set
var equalCoef=1;  //9908061140
var equalFirst=1; //9908061157
both to one. Output
<a name="ch14c041">
Abel's first Inequality  5=5
Abel's second Inequality 5＜10
this indicate that
Abel's first Inequality has
chance to become equality.
but
<a name="ch14c042">
Abel's second Inequality is
always lessThan＜2*greaterThan
Factor '2' in Abel's second 
Inequality is unnecessary.
2010-08-06-12-11 here

<a name="ch14c043">　Index begin　Index this file
2010-08-06-12-32 start
■　Exercise 14.1 solution


Exercise 14.1 hint described 
the derivation method.
<a name="ch14c044">
Abel's second Inequality and
Abel's first Inequality have
parallel step 
from tute0049.htm#ch14a023
 to  tute0049.htm#ch14a031
<a name="ch14c045">
first ineq. eqn.BQ025 is same as 
second ineq. eqn.BS003
  S1(b1-b2)+S2(b2-b3)+...
 +Sn-1(bn-1-bn)+Snbn ---eqn.BS003
Same data sequence z1,z2,...,zn
Same partial sum   S1,S2,...,Sn
<a name="ch14c046">
Abel's second Inequality (red)
and
Abel's first Inequality (blue)
difference is next
2nd 0≦b1≦b2≦...≦bn ---eqn.BS001
1st a1≧a2≧...≧an≧0 ---eqn.BQ002
Coef. a_seq. and b_seq are real
numbers. They can compare and 
no need take absolute value.

<a name="ch14c047">
Abel's 2nd, bound
  |b1z1+b2z2+...+bnzn| ---eqn.BS004
 ≦|S1||b1-b2|+|S2||b2-b3|+...
 +|Sn-1||bn-1-bn|+|Sn|bn ---eqn.BS005
<a name="ch14c048">　Index begin　Index this file
Abel's 1st, bound
  |a1z1+a2z2+...+anzn|
 ≦a1*max[1≦k≦n]|Sk| ---eqn.14.1
because b1-b2≦0, b2-b3≦0 etc.
b's are real number coefficients.
<a name="ch14c049">
In eqn.BS006, Exercise 14.1 hint
reverse from |b1-b2| to b2-b1 and
remove absolute sign '|'. At the 
same time replace |Si| 1≦i≦n by
max[1≦k≦n]|Sk|. 
<a name="ch14c050">
Abel's second Inequality become
  |b1z1+b2z2+...+bnzn| ---eqn.BS004
 ≦max[1≦k≦n]|Sk|{(b2-b1)+(b3-b2)
  +...+(bn-bn-1)+bn} ---eqn.BS006
Greater than side cancel +b2-b2
and +b3-b3 ... +bn-1-bn-1 
Left +bn+bn-b1 which is bounded 
by 2bn.

<a name="ch14c051">
The following need your judge.
LiuHH's opinion could be wrong.

We can do Abel's second Ineq.
like Abel's first Ineq. as 
following

<a name="ch14c052">　Index begin　Index this file
we start at beginning 
change complex
from z1 to yn
from z2 to yn-1
...
from zn to y1
<a name="ch14c053">
and change nonnegative real
from b1 to cn
from b2 to cn-1
...
from bn to c1
<a name="ch14c054">
After this change
  |b1z1+b2z2+...+bnzn| ---eqn.BS004
is same as
  |c1y1+c2y2+...+cnyn| ---eqn.BS009
<a name="ch14c055">
We have 
  c1≧c2≧...≧cn≧0 ---eqn.BS010
which is same as
  a1≧a2≧...≧an≧0 ---eqn.BQ002
ci and ai sequences satisfy 
Abel's first Inequality 
requirement eqn.BQ002
<a name="ch14c056">
We apply Abel's first Inequality
for Abel's second Inequality and
get better upper bound. 
Dropped "2*" in eqn.14.29

Is this right?
Please think, please suspect.
2010-08-06-13-28 stop





<a name="ch14c057">　Index begin　Index this file
2010-08-06-16-25 start
■　Exercise 14.2 problem statement
　　textbook page 221
（The Integral Mean Value Formulas）

<a name="ch14c058">
The first integral mean value
formula (IMVF) asserts that for
each continuous function
  f:[a,b]→Real ---eqn.BS011
and each integrable 
  g:[a,b]→[0,∞) ---eqn.BS012
there is a ξ∈[a,b] such that

<a name="ch14c060">
while the second IMVF is the
slightly tricker assertion 
that for each differentiable 
nonincreasing function
  ψ:[a,b]→(0,∞) ---eqn.BS013
and each integrable function
  φ:[a,b]→Real ---eqn.BS014
there is a ξ0∈[a,b] ---eqn.BS015
such that

<a name="ch14c062">
Prove these formulas. They are 
both quite handy, and the 
second one may be tricker 
than you might guess.
2010-08-06-16-46 stop




<a name="ch14c063">　Index begin　Index this file
2010-08-06-18-27 start
■　Exercise 14.2 hint
　　textbook page 279

From the nonnegativity of g
one has the bounds

<a name="ch14c065">
2010-08-06-18-44 here
and by continuity of f it takes
on all values between its
minimum and its maximum. These
observations give us the first
IMVF (14.30)

<a name="ch14c066">
To prove the second, choose Φ
with
  Φ(a)=0 ---eqn.BS017
such that 
  Φ'(x)=φ(x) ---eqn.BS018
then integrate by parts and apply
the first IMVF with
  f(x)=Φ(x) ---eqn.BS019
and
  g(x)=-ψ'(x)≧0 ---eqn.BS020
to find

<a name="ch14c069">
2010-08-06-19-22 here
Since 0＜ψ(b)≦ψ(a) the bracketed
quantity is an average of Φ(b) 
and Φ(ξ) so it must equal to 
Φ(ξ0) for some ξ0∈[ξ,b]⊂[a,b]
by the continuity of Φ.
2010-08-06-19-26 stop


<a name="ch14c070">
■　Exercise 14.2 solution


Start from

<a name="ch14c071">
we have a continuous function
  f:[a,b]→Real ---eqn.BS011
and each integrable 
  g:[a,b]→[0,∞) ---eqn.BS012
"→Real" tell us that f(x) can 
be negative/zero/positive.
"→[0,∞)" tell us that g(x) can 
be zero/positive not negative.
a,b not specified, x in [a,b]
can be negative/zero/positive.
x and f(x) are of different
physics quantity in general.
For example x be distance,
f(x) be work/energy.

<a name="ch14c072">　Index begin　Index this file
In domain [a,b] continuous 
function f(x) has a minimum 
value at point c0. That is 
f(c0)≦f(x) for x in [a,b].
Same reason,
In domain [a,b] continuous 
function f(x) has a maximum 
value at point c1. That is 
f(c1)≧f(x) for x in [a,b].

<a name="ch14c073">
Integrand of eqn.BS022 is 
f(x)g(x).
Change f(x)g(x) to f(c0)g(x)
get a integration lower bound
f(c0) is constant, move f(c0) 
to outside of integration sign.
Write f(c0) as min[a≦y≦b]f(y)
get eqn.BS016 first inequality. 

<a name="ch14c073a">
The inequality direction in
eqn.BS016 can be true only if
g(x) is nonnegative in domain
[a,b]. Explain this point from
the opposite direction. Assume
g(x) is negative in [a,b] then
∫[x=a,b]g(x)dx is negative.
Suppose its value is -5. 
<a name="ch14c073b">
The value of f(x) assume to be 
in [1,3]. Then min f(x) is 1
    min f(y)*∫g(x)dx= 1*(-5)
non-min f(y)*∫g(x)dx= 2*(-5)
we can not conclude -5＜-10
The nonnegativity of g(x)
  g:[a,b]→[0,∞) ---eqn.BS012
is important in this exercise.

<a name="ch14c074">
Similarly,
Change f(x)g(x) to f(c1)g(x)
get a integration upper bound
f(c1) is constant, move f(c1) 
to outside of integration sign.
Write f(c1) as max[a≦y≦b]f(y)
get eqn.BS016 second inequality.

<a name="ch14c075">
Given f(x) is continuous in [a,b]
we can find a point x=ξ in [a,b]
such that eqn.14.30 is true.
This is a consequence of given
continuous condition.

<a name="ch14c076">
First integral mean value formula 
promise there is a point ξ in [a,b]
such that eqn.14.30 holds. But did 
not tell us what is the value of ξ.
2010-08-06-21-59 here

<a name="ch14c077">　Index begin　Index this file
Next is second integral mean value 
formula eqn.14.31
Both eqn.14.30 and eqn.14.31 their
left side are the same. Integration 
of product of two functions. But,
right side are different.

<a name="ch14c078">
In eqn.14.30 f(ξ) get mean point ξ
and f(ξ) stay out side of integral.
ξ has nothing to do with integral.
g(x)*dx integrate over whole domain
[a,b].

<a name="ch14c079">
eqn.14.31 ψ(a) evaluate at boundary
point x=a, not at mean value point
ξ0. φ(x)*dx integrate over partial 
domain [a,ξ0] a≦ξ0≦b. 
Mean value point ξ0 is integration
upper bound. Smart idea is to use
integration by parts, and let
interpolation relation stand out.
<a name="ch14c080">
To use integration by parts, need
define ∫φ(x)dx as Φ(x) , 
  Φ(x)=∫φ(x)dx  ---eqn.BS318
if take derivative, we find
  Φ'(x)=φ(x) ---eqn.BS018
<a name="ch14c081">
We have one freedom to choose
boundary condition of Φ(x). For
this concern, require Φ(x) has
the property
  Φ(a)=0 ---eqn.BS017
one boundary term of integration 
by parts drop out. 

<a name="ch14c082">　Index begin　Index this file
In eqn.BS021 
"AA=" replace φ(x) by Φ'(x)
"BB=" is integration by parts
 －0 caused by Φ(a)=0 .
 ψ(b)Φ(b) is none zero boundary term
 ψ(x)Φ'(x)dx change to -ψ'(x)Φ(x)dx 
<a name="ch14c083">
"CC=" apply First Integral Mean 
  Value Formula. Φ(ξ) be constant.
"DD=" rewrite "CC="
  "CC=" blue goto "DD=" blue
  "CC=" purple goto "DD=" purple
  purple part change from
  -[ψ(b)－ψ(a)] to ψ(a)－ψ(b) 

<a name="ch14c084">
Given condition is differentiable 
nonincreasing function
  ψ:[a,b]→(0,∞) ---eqn.BS013
In other words, ψ(x) is decreasing
or constant in [a,b]
<a name="ch14c085">
for a＜b we have
  ψ(a)≧ψ(b) ---eqn.BS023
f(x)=1/x is monotone decrease
for x in [0.1, infinity)
2＜3, then 1/2＞1/3
monotone increase example
monotone decrease is opposite

<a name="ch14c086">
Because ψ(a)≧ψ(b)＞0, 
//in ψ:[a,b]→(0,∞) ---eqn.BS013
//"→(0,∞)" say ψ(b)＞0
then
  0＜ψ(b)/ψ(a)≦1 ---eqn.BS024
and
  0≦[ψ(a)－ψ(b)]/ψ(a)≦1 ---eqn.BS025
and  // next line ---eqn.BS026
  ψ(b)/ψ(a) + [ψ(a)－ψ(b)]/ψ(a) =1
eqn.BS024,eqn.BS025,eqn.BS026 say
that "DD=" line bracketed quantity 
is an interpolation (average).
Especially not extrapolation.
<a name="ch14c087">
In "DD=" line, the bracketed
quantity is an average of Φ(b) 
and Φ(ξ) so it must equal to 
Φ(ξ0) for some ξ0∈[ξ,b]⊂[a,b]
by the continuity of Φ.
2010-08-06-23-06 stop



<a name="ch14c088">　Index begin　Index this file
2010-08-07-11-24 start
■　Exercise 14.3 problem statement
　　textbook page 222
（A Integral Analog to Abel's
 Inequality）

<a name="ch14c089">
If f:[a,b]→(0,∞) ---eqn.BS027
is a nonincreasing function, then
for each integrable function 
  g:[a,b]→Real ---eqn.BS028
one has the bound

2010-08-07-11-41 stop




<a name="ch14c092">　Index begin　Index this file
2010-08-07-11-43 start
■　Exercise 14.3 hint
　　textbook page 280

The bound (14.32) is immediately
from the second IMVF (14.31). 

<a name="ch14c093">
The sine bound (14.33) then follows
by taking 
  f(x)=1/x    ---eqn.BS029
  g(x)=sin(x) ---eqn.BS030
and by noting that the integral
of g over [a,b] is cos(b)-cos(a)
which is bounded by 2 in absolute
value.
2010-08-07-11-46 stop



<a name="ch14c094">
2010-08-08-10-21 start
■　Exercise 14.3 solution


Alert Abel's Inequality apply
to complex sequence. But
g:[a,b]→Real ---eqn.BS028 
is real.
Compare eqn.14.32 with eqn.14.31

<a name="ch14c095">
Exercise 14.3 eqn.14.32 given
 f:[a,b]→(0,∞) ---eqn.BS027
is a nonincreasing function
and integrable function 
  g:[a,b]→Real ---eqn.BS028

<a name="ch14c096">
Exercise 14.2 eqn.14.31 given
for each differentiable 
nonincreasing function
  ψ:[a,b]→(0,∞) ---eqn.BS013
and each integrable function
  φ:[a,b]→Real ---eqn.BS014

<a name="ch14c097">　Index begin　Index this file
Two given conditions are the
same. why
Exercise 14.3 eqn.14.32 is 
inequality
Exercise 14.2 eqn.14.31 is
equality ?

<a name="ch14c098">
The difference is that mean 
value point x=ξ0 is unspecified
in eqn.14.31 which is the source
of equality. 

<a name="ch14c099">
Exercise 14.3 eqn.14.32 is 
inequality. That is caused 
by specify mean value point 
x=ξ0=y to be 
sup[a≦y≦b]|∫[x=a,y]g(x)dx|
<a name="ch14c100">
here y is not arbitrary,
y is a point get maximum
function value. Because
maximum function value ≧ 
arbitrary function value
that is f(ξ0)≦f(y) This
property create inequality.

<a name="ch14c101">
For equality no need absolute 
sign.
For inequality do need absolute 
sign.
If not use absolute sign, then
we do not know the bound is
upper bound only no lower bound.
 -infinity＜f(x)≦f(ξ0)=3
or both upper/lower bound
 -5=-f(y)≦f(x)≦f(y)=5
Use absolute sign to emphasize
the bound is both upper and below.

<a name="ch14c102">　Index begin　Index this file
Apply eqn.14.31 to Exercise 14.3 
we have equality.

<a name="ch14c104">
2010-08-08-11-15 here
The second part of Exercise 14.3
is easy. Set
  f(x)=1/x    ---eqn.BS029
  g(x)=sin(x) ---eqn.BS030
and use
  sin(x)=-d[cos(x)]/dx 
or
  sin(x)dx=-d[cos(x)]  ---eqn.BS032
eqn.14.32 become

<a name="ch14c106">
2010-08-08-11-37 here
eqn.BS033 greater than side
integration evaluate to
[-cos(y)] - [-cos(a)]
Exercise 14.3 ask sup[a≦y≦b].
where a,b,y are all unspecified.
<a name="ch14c107">
We are free to choose y and a.
The maximum value of
[-cos(y)] - [-cos(a)]
is +1 - (-1) =2
for the case y=-PI and a=0
eqn.BS033 upper bound is 2/a
we reached the target eqn.14.33.
Exercise 14.3 is done.
2010-08-08-11-41 stop



<a name="ch14c108">　Index begin　Index this file
2010-08-08-12-48 start
■　Exercise 14.4 problem statement
　　textbook page 222
（van der Corput on Oscillatory
 Integrals）

<a name="ch14c109">
(a) Given a differentiable function
  θ:[a,b]→Real ---eqn.BS034
for which the derivative θ'(.) is 
monotonic and satisfy
  θ'(x)≧ν＞0 ---eqn.BS035
for all x in [a,b]
Show that one has the bound

<a name="ch14c111">
2010-08-08-13-08 here
(b) Use the bound (14.34) to
    show that if
  θ:[a,b]→Real ---eqn.BS036
    is a twice differentiable
    function with 
  θ''(x)≧ρ＞0 ---eqn.BS037
    for all x in [a,b], then

<a name="ch14c113">　Index begin　Index this file
These work horses lie behind
many basic cancellation arguments
for integrals and sums. They also 
come to us from the same J.G. van
der Corput who gave us our third 
challenge problem. 
<a name="ch14c114">
In fact, these
may be the best known of van der 
Corput many inequalities, even
though they are notably less 
subtle than the bound (14.17).
2010-08-08-13-19 stop





<a name="ch14c115">
2010-08-08-15-26 start
■　Exercise 14.4 hint
　　textbook page 280

We are given that θ'(.) is 
monotonic and without loss of
generality, we assume it is 
nondecreasing. From the second
IMVP of Exercise 14.2, we find
that

<a name="ch14c117">
2010-08-08-15-50 here
The last ratio has modulus bounded
by 2/ν , //see θ'(x)≧ν＞0 
so to complete the proof,
one only needs to check that an
exactly analogous argument applies
to the imaginary part of the 
integral in our target inequality
(14.34)

<a name="ch14c118">　Index begin　Index this file
Since θ'(x) is strictly monotone,
it vanishes at most once in the
interval [a,b], and, for the
moment, suppose it vanishes at c.
To prove the second bound (14.35)
we write the integral I over [a,b]
as the sum I1+I2+I3 of integrals
over [a,c-δ],[c-δ,c+δ], and [c+δ,b]
<a name="ch14c119">
In the interval [c+δ,b], one has 
  θ'(x)≧ρδ ---eqn.BS039
so by the bound (14.34), we have
  |I3|≦4/(ρδ)
An analogous bound applies to I1,
while for the integral I2 we have
the trivial bound
  |I2|≦2δ ---eqn.BS040
<a name="ch14c120">
In sum we have
  |I|≦|I1|+|I2|+|I3|≦8/(ρδ)+2δ ---eqn.BS041
which we can minimize by setting
  δ=2/√ρ ---eqn.BS042
to obtain the target bound (14.35).
<a name="ch14c121">
To be 100% complete, one finally
needs to note that the target
bound continues to hold if 
  c±2/√ρ NOT∈ [a,b] ---eqn.BS043
or indeed, if θ'(x) does not 
vanish in [a,b]
2010-08-08-16-10 stop


<a name="ch14c122">　Index begin　Index this file
2010-08-08-16-44 start
■　Exercise 14.4 solution


Second IMVP of Exercise 14.2 require
differentiable nonincreasing function
  ψ:[a,b]→(0,∞) ---eqn.BS013

<a name="ch14c123">
Exercise 14.4 hint say
assume θ'(x) is nondecreasing. 
and apply the second IMVP of 
Exercise 14.2

How to explain the nonincreasing
and nondecreasing difference?
If read the equation carefully,
we find that two are consistent.

<a name="ch14c124">
Exercise 14.2 nonincreasing ψ(x)
is at numerator (denominator=1)
See eqn.14.31 "ψ(a)*"

Exercise 14.4 nondecreasing θ'(x)
is at denominator.
See eqn.bs038 "1/θ'(a)"

<a name="ch14c125">
Integrand of eqn.14.34 is
exp(iθ(x)). Euler's formula say
  eiθ = cos(θ)+i*sin(θ) ---eqn.BQ082
Then Integrand of eqn.14.34 is
  cos(θ(x))+i*sin(θ(x)) ---eqn.BS044
eqn.bs038 is REAL part of exp(iθ(x))
because eqn.bs038 use cos(θ(x)).
<a name="ch14c126">
"EE=" in eqn.bs038 is insert a
  one = θ'(x)/θ'(x)  ---eqn.BS045
θ'(x) is not zero because problem
given this non zero condition at
  θ'(x)≧ν＞0 ---eqn.BS035

<a name="ch14c127">　Index begin　Index this file
"FF=" is an application of 
second IMVF 
1/θ'(a) has maximum value in [a,b]
there is a mean point x=ξ in [a,b]
for which EQUALITY of "FF=" is true.

<a name="ch14c128">
After move denominator θ'(x) out
of integration become denominator 
θ'(a). The integrand simplified to
  {cos(θ(x))}θ'(x)dx   ---eqn.BS046
which is // next line  ---eqn.BS047
  cos(θ(x))d[θ(x)] = d[sin(θ(x))]

<a name="ch14c129">
"GG=" is to carry out the simple
integration to get  // ---eqn.BS048
  [sin(θ(ξ))－sin(θ(a))]/θ'(a)
numerator of above expression
[sin(θ(ξ))－sin(θ(a))] is bounded 
by 2, same reason as ch14c107
Denominator of eqn.BS047 θ'(a) is 
bounded by ν. see θ'(x)≧ν＞0
The bound of real part is 2/ν

<a name="ch14c130">
Above is REAL part of exp(iθ(x))
Do same thing for IMAGINARY part
of exp(iθ(x)) get a bound 2/ν
Very rough upper bound is 
real bound 2/ν plus imaginary 
bound 2/ν get 4/ν.
2010-08-08-18-06 here

<a name="ch14c131">
2010-08-08-18-18 start
LiuHH wondered for a while.
If '≦' sign in eqn.14.34 is valid
when '=' become effective?

<a name="ch14c132">　Index begin　Index this file
4/ν is eqn.14.34 upper bound.
Is it possible to reach this
upper bound?
cos(θ(x)) and sin(θ(x)) are linked 
by Euler's Formula
cos(θ(x)) and sin(θ(x)) can not 
reach maximum 2/ν independently.
2010-08-08-18-28 stop

<a name="ch14c133">
2010-08-08-19-37 start
In the interval [c+δ,b], one has 
  θ'(x)≧ρδ ---eqn.BS039

Whether "θ'(x)≧ρδ" is correct
physics dimensionally? //answer

<a name="ch14c134">
given
  θ''(x)≧ρ＞0 ---eqn.BS037
so ρ and θ''(x) have same physics 
dimension.
given
over [a,c-δ],[c-δ,c+δ], and [c+δ,b]
so δ and a,b,c have same physics 
dimension, which is x.

<a name="ch14c135">
check dimension of ρδ
ρ*δ is θ''(x)*x and is
  {d[θ'(x)]/dx}*x 
dx and x cancel dimensionally
left physics dimension θ'(x)

<a name="ch14c136">　Index begin　Index this file
therefore
  θ'(x)≧ρδ ---eqn.BS039
is reasonable.
2010-08-08-19-41 here

<a name="ch14c137">
then why
  θ'(x)≧ρδ ---eqn.BS039
??
change
  θ'(x)≧ρδ ---eqn.BS039
to
  θ'(x)/δ ≧ ρ ＞0
chord slope ≧ tangent slope
then what? //LiuHH wondering

<a name="ch14c138">
Problem given
  θ'(x)≧ν＞0 ---eqn.BS035
Exercise 14.4 hint say 
differently
[[
Since θ'(x) is strictly monotone,
it vanishes at most once in the
interval [a,b], 
]]
Given θ'(x)＞0 
 why  θ'(x)=0? why vanish?
If θ'(x) vanish, then "EE=" in
eqn.bs038 insert a 0/0 =θ'(x)/θ'(x)
which is not allowed.

<a name="ch14c139">
LiuHH need more time to think.
Exercise 14.4 is NOT DONE
2010-08-08-19-57 stop



<a name="ch14c140">　Index begin　Index this file
2010-08-08-20-50 start
■　Exercise 14.5 problem statement
　　textbook page 222
（The 'Extend and Conquer' Paradigm）

<a name="ch14c141">
First show that for integers m
and j one has the formula

<a name="ch14c143">
This formula tells us that for 
such a complete sum one either 
has total cancellation or no
cancellation at all. There are 
many remarkable consequences 
of this elementary observation.

<a name="ch14c144">
For example, use it to show 
that for each prime p≧3 and //why?
each pair A and B of subset of
  Fp={0,1,2,3,...,p-1}
one has //i=√(-1); j,k=index

2010-08-08-21-12 stop





<a name="ch14c146">
2010-08-08-21-17 start
■　Exercise 14.5 hint
　　textbook page 281

To begin let W denote the target 
sum and note that

<a name="ch14c149">
Now we come to a devilish trick:
we extend the outside sum to all
of Fp={0,1,2,3,...,p-1}. This is
feasible because we are just
adding positive terms, and it
is sensible because it sets up
the application of the
cancellation identity (14.36).
<a name="ch14c150">　Index begin　Index this file
To put the algebra neatly, we
first define the function δ(x)
by setting 
  δ(0)=1 ---eqn.BS051
and
  δ(x)=0 ---eqn.BS052
for x≠0. then we note

<a name="ch14c153">
This problem and the description
"extend and conquer" are from
the informative exposition of 
Shparlinski (2002) where one
finds several further examples 
of the ways to exploit complete
sums. 
<a name="ch14c154">
Shparlinski links bound 
of the type (14.37) back to the
work of I.M. Vinogradov; in
particular, Exercise 14 of 
Vinogradov (1954, p.128) is of
this kind.
2010-08-08-22-03 stop

2010-08-09-10-06 start coding
2010-08-09-10-32 done  coding
2010-08-09-11-18 done add javascript code


<a name="ch14c155">　Index begin　Index this file
2010-08-09-12-39 start
■　Exercise 14.5 solution


First observe eqn.14.36
e(t) is defined at eqn.14.5A
  e(t)=exp(2πit) ---eqn.14.5A
e(jk/m) is e(t) with t=jk/m
therefore
  e(jk/m)=exp(2πi*jk/m) ---eqn.BS054
<a name="ch14c156">
parameters i,j,k,m have the 
following meaning
  i=sqrt(-1) ---eqn.BS055
k is variable integer index.
j,m are given integers.
2,π,j,k,m are all real numbers
m require to be a prime. then 
  θ=2π*jk/m ---eqn.BS056
is a real number.
<a name="ch14c157">
Euler's formula say
  eiθ = cos(θ)+i*sin(θ) ---eqn.BQ082
we find
  e(jk/m)=exp(2πi*jk/m)
    =  cos(2π*jk/m)
    +i*sin(2π*jk/m) ---eqn.BS057
<a name="ch14c158">
If jk/m is an integer, then 
  cos(2π*integer)=1 ---eqn.BS058
  sin(2π*integer)=0 ---eqn.BS059
and
  e(integer)=exp(2πi*integer)
   = 1+i*0 = 1 ---eqn.BS060
k in jk/m is variable integer
start from k=0 to k=m-1
j,m in jk/m are two given integers.

<a name="ch14c159">
If j/m is an integer, then jk/m is
an integer, we have
  e(integer)=1 ---eqn.BS061
for k=0 to k=m-1.
We have total m one's. Sum to m.
Here answer
[[
sumK = ∑[k=1,m-1]exp( 2πijk/m ) 
sumK =m 　if m does divide j 
]]

<a name="ch14c160">　Index begin　Index this file
If j/m is not an integer, that
is "if m does not divide j"
then jk/m is not an integer, 
we have
  e(quotient)=
      cos(2π*quotient)
   +i*sin(2π*quotient) ---eqn.BS062
which is not 1.
We sum from k=0 to k=m-1 a cycle
of 2PI. What do we get?

<a name="ch14c161">
If j＞1, jk/m is a leap cycle. 
Sum one cycle m complex numbers 
in j*2PI period.
If j=1 then jk/m=k/m this is a 
basic cycle. Sum one cycle m 
complex numbers in 2PI period.

<a name="ch14c162">
Why require m be prime?
In jk/m if m is not a prime,
for example m=10, if j=5. then
variable integer k range from
k=0 to k=m-1=9, at k=2, jk/m=1
2πijk/m is an integer multiple 
of 2πi, k go a "whole cycle" 
pre-maturely. If ask m be prime 
number, no pre-mature happen. 
Goto Full Cycle Euler, fill in
j:5 , m:10 (m=10 is not a prime)
Click [FCEular] button, get ten
complex numbers, five are 1+0i
five are -1+0i. A no fun case.

<a name="ch14c163">
Now set j=1, e(jk/m) become
  e(k/m)=exp(2πi*k/m)
    =  cos(2π*k/m)
    +i*sin(2π*k/m) ---eqn.BS063
For k=0
  e(0/m)=cos(2π*0/m)+i*sin(2π*0/m)
  e(0)=1+0i ---eqn.BS064

<a name="ch14c164">
k=1 and k=m-1 has cancellation.
For k=1
  e(1/m)=cos(2π*1/m)+i*sin(2π*1/m) ---eqn.BS065
For k=m-1 
  e((m-1)/m)=cos(2π*(m-1)/m)
          +i*sin(2π*(m-1)/m) ---eqn.BS066
//2010-08-09-13-20 here
//cos(x)+cos(y)= 2cos((x+y)/2)cos((x-y)/2)
//sin(x)+sin(y)= 2sin((x+y)/2)cos((x-y)/2)
<a name="ch14c165">　Index begin　Index this file
Add two real 
  R1=cos(2*π*1/m)+cos(2*π*(m-1)/m)
    =2*cos((2*π*1/m+2*π*(m-1)/m)/2)
      *cos((2*π*1/m-2*π*(m-1)/m)/2)
    =2*cos(π)*cos(2*π/m-π)
    =-2*cos(2*π/m-π)
<a name="ch14c166">
Add two imaginary
  I1=sin(2*π*1/m)+sin(2*π*(m-1)/m)
    =2*sin((2*π*1/m+2*π*(m-1)/m)/2)
      *cos((2*π*1/m-2*π*(m-1)/m)/2)
    =2*sin(π)*cos(2*π/m-π)
    =0
Pair k=1 and k=m-1 has cancellation.
imaginary cancel to zero.
2010-08-09-13-37 here

<a name="ch14c167">
In general, for k＞0, 
k=q＞0 and k=m-q＞0
has cancellation
For k=q
  e(q/m)=cos(2π*q/m)+i*sin(2π*q/m) ---eqn.BS067
For k=m-q
  e((m-q)/m)=cos(2π*(m-q)/m)
          +i*sin(2π*(m-q)/m) ---eqn.BS068
<a name="ch14c168">
Add two real 
//cos(x)+cos(y)= 2cos((x+y)/2)cos((x-y)/2)
  Rq=cos(2*π*q/m)+cos(2*π*(m-q)/m)
    =2*cos((2*π*q/m+2*π*(m-q)/m)/2)
      *cos((2*π*q/m-2*π*(m-q)/m)/2)
    =2*cos(π)*cos(2*π*q/m-π)
    =-2*cos(2*π*q/m-π) ---eqn.BS069
<a name="ch14c169">
Add two imaginary
//sin(x)+sin(y)= 2sin((x+y)/2)cos((x-y)/2)
  Iq=sin(2*π*q/m)+sin(2*π*(m-q)/m)
    =2*sin((2*π*q/m+2*π*(m-q)/m)/2)
      *cos((2*π*q/m-2*π*(m-q)/m)/2)
    =2*sin(π)*cos(2*π*q/m-π)
    =0 ---eqn.BS070
Pair k=q and k=m-q has cancellation.
imaginary cancel to zero.
2010-08-09-13-42 here

<a name="ch14c170">　Index begin　Index this file
2010-08-09-15-10 start
To see the over all calculation
it is less efficient if add all
complex real part to zero. It is
possible to add all complex real
and imaginary at same time.
Sum e(jk/m) for k=0,1,2,..., m-1
as following. 
<a name="ch14c171">
Consider complex number as two
dimensional vector. Each vector
has modulus one. Vectors phase 
angle increment are 2*PI/m.
There are total m vectors. They
form a m-polygon, equal angle,
equal side length. m-polygon is
closed. 
<a name="ch14c172">
That is head and tail two points 
coincide. 
m-polygon, m vectors net sum is 
zero. So we get 
sumK = ∑[k=1,m-1]exp( 2πijk/m ) 
sumK =0 　if m does not divide j
2010-08-09-15-27 stop

<a name="ch14c173">
2010-08-09-16-07 start
Exercise 14.5 hint start from
W definition which is eqn.BS049
left half and drop absolute
sign. Inequality in eqn.BS049
is a result of taking absolute
value. Please see Why take 
absolute value cause inequality?

<a name="ch14c174">
eqn.BS049 greater than side is
ready for Cauchy's inequality
eqn.BS049 inserted a blue 1*
like eqn.BR014 did. Blue 1 is 
Cauchy inequality's 1-trick

<a name="ch14c175">　Index begin　Index this file
In eqn.BS049, greater than side
outer summation ∑[j∈A] sum two 
sequences. 
First sequence is
  {1,1,1,...,1} ---eqn.BS071
there are |A| of them.
Each element is simply one.

<a name="ch14c176">
Second sequence is 
{∑[k∈B]exp(2πij1k/p),
 ∑[k∈B]exp(2πij2k/p),
 ... ---eqn.BS072
 ∑[k∈B]exp(2πij|A|k/p)}
there are still |A| of them.
Each element is a summation
of another sequence.

<a name="ch14c177">
What is |A| ?
A and B are subset of
  Fp={0,1,2,3,...,p-1} ---eqn.BS073
For example assume p=11=prime
  F11={0,1,2,3,4,5,6,7,8,9,10} ---eqn.BS074
Assume A, B are subset of F11
  A={0,1,2,3,6} ---eqn.BS075
  B={1,2,5,6,8,10} ---eqn.BS076
|A| is number of elements in 
set A . Above example |A|=5 
and |B|=6 (6 elements)

<a name="ch14c178">
summation ∑[j∈A] 
j vary in between {0,1,2,3,6}
summation  ∑[k∈B]
k vary in between {1,2,5,6,8,10}
LiuHH GUESSED What is |A| ?
Could be wrong.



<a name="ch14c179">
Back to eqn.BS071, eqn.BS072.

Cauchy's Inequality equation
in verbal description is
Seq.A dot product with seq.B
this value is less or equal to
the square root of
[[
Seq.A dot product with itself
and multiply with
Seq.B dot product with itself
]]

<a name="ch14c180">　Index begin　Index this file
In eqn.BS050, we use square 
version of Cauchy's inequality.
Square above words in your mind
please.

<a name="ch14c181">
From eqn.BS049 to eqn.BS050 is
a result of applying Cauchy's 
inequality. First sequence
  {1,1,1,...,1} ---eqn.BS071
It dot product with itself get
 1*1 + 1*1 + ... + 1*1 = |A| ---eqn.BS077
|A| is number of elements in 
set A . Above example |A|=5 

<a name="ch14c182">
Master Cauchy take square root 
for |A|. Prof. Steele square it. 
Net result is no '√', no power 2,
just |A|. Wonderful.
Reader should be able to explain
the |A| in eqn.BS050 how do we
get |A|?

<a name="ch14c183">
From eqn.BS049 to eqn.BS050 ,
second sequence eqn.BS072 
experience same process as first
did. But second sequence is not
simple one. Result is a little
bit complicate.
<a name="ch14c184">
Second sequence dot product with 
itself get absolute square in
eqn.BS050. Sum them just like
eqn.BS077 did for first sequence.
Again, Master Cauchy and Prof. 
Steele balance each other. We
have eqn.BS050 j∈A summation.
<a name="ch14c185">　Index begin　Index this file
eqn.BS050 has two square terms.
|W|2 is done by Prof. Steele.
eqn.BS050 right side longer 
square term is Master Cauchy's
signature.

<a name="ch14c186">
From eqn.BS050 to eqn.BS053
"II≦" is to enlarge outer sum
from ∑[j∈A] to ∑[j∈Fp] 
Remember A is a subset of Fp
Above example
  F11={0,1,2,3,4,5,6,7,8,9,10} ---eqn.BS074
  A={0,1,2,3,6} ---eqn.BS075
<a name="ch14c187">
enlarge outer sum, add more
positive term, push greater 
than side even greater. This
push is "devilish trick" in
hint description.

<a name="ch14c188">
In eqn.BS053
"JJ=" is square of complex number
Please see Complex square rule 
"k1" in "JJ=" line is complex
"-k2" in "JJ=" line is conjugate
"k1-k2" absorbed square sign.
k1, k2 like index and unlike
index are mixed together. All
in "JJ=" line

<a name="ch14c189">
In eqn.BS053
"KK=" is switch summation sign.
"LL≦" is to apply eqn.14.36
Possible after "devilish trick"
If k1=k2 eqn.14.36 give value one
δ(k1-k2) record value one.
If k1 NOT= k2 eqn.14.36 give value 
zero. δ(k1-k2) record value zero.

<a name="ch14c190">　Index begin　Index this file
eqn.14.36 cause number p in 
"LL≦" line.
eqn.14.36 use denominator m
and give value m.
eqn.BS053 "KK=" line use 
denominator p, receive value p.

<a name="ch14c191">
How many p do we get?
Summation ∑[k1,k2∈B] there are |B|
cases to apply eqn.14.36. 
This sum contribute to |B|.

<a name="ch14c192">
Over all, greater than side is
p*|A|*|B| 
eqn.BS053 "LL≦" line take square
root, end up with target eqn.14.37

Exercise 14.5 is done.
2010-08-09-18-00 stop

<a name="ch14c193">
2010-08-09-19-10 start
The program Full Cycle Euler
help you calculate eqn.14.36
User input j and m two integers
and click [FCEuler] button.
Box 2 has output .

<a name="ch14c194">　Index begin　Index this file
You can draw complex numbers at
http://freeman2.com/tute0014.htm#fig2.5
If you saved a copy, click local
Copy from tute0051.htm Box 2,
paste to  tute0014.htm Box 1
check "auto scale", 
click [Plot3] get error message
Because tute0014.htm default complex
number separator to be ';', but
tute0051.htm Box 2 data use newline
as number separator. 

<a name="ch14c195">
In tute0014.htm find next line
Box 1 complex separator: newline ';' ',' tab 

check 'newline', auto uncheck ';'
paste data to tute0014.htm Box 1
check "auto scale", 
click [Plot3] get drawing output.

<a name="ch14c196">
In tute0051.htm#fullCycleEuler, 
click [add js code], output to 
Box 2. It has javascript code for
complex number seq. definition.
This js code can be used to
replace tute0050.htm#ch14b016
complex definition.
<a name="ch14c197">
Copy tute0050.htm#ch14b016 long
string code and paste to (local)
http://freeman2.com/complex2.htm#calculator
box 3, then click 
[test box3 command, output to box4]
It is calculation, not drawing.

<a name="ch14c198">
Page complex2.htm has example
buttons from 0 to 13.
Click button 7 and 8 are 
Exercise 14.5 problem
Both button 7 and 8 are for 
eqn.14.36, not for eqn.14.37.
Both are work in 2009-05-18
2010-08-09-19-39 stop



<a name="ch14c199">　Index begin　Index this file
2010-08-09-20-20 start
■　Exercise 14.6 problem statement
　　textbook page 223
（Another Dyadic Passage）

<a name="ch14c200">
Sometimes we have an estimate for
f(x) and we would like an estimate
of g(x), but we cannot show
  g(x)≦f(x) ---eqn.BS078
We may still be able to get a useful
bound on g(x) if we only know that
<a name="ch14c201">
f(x) dominated "half" of g in the
sense that
  g(x)-g(x/2)≦f(x) ---eqn.BS079
for all x≧0
To be specific, assume such a 
function is continuous and 
<a name="ch14c202">
show that if
  f(x)≦Ax+B for x≧0 ---eqn.BS080
then g satisfies the (only slightly
worse) bound 
  g(x)≦A'x+B'log2(x)+C' ---eqn.BS081
for x≧1
for appropriate constants A', B' 
and C'.
2010-08-09-20-28 stop





<a name="ch14c203">　Index begin　Index this file
2010-08-09-20-29 start
■　Exercise 14.6 hint
　　textbook page 281

<a name="ch14c207">　Index begin　Index this file
2010-08-09-21-58 start
eqn.BS083 ●●change to eqn.BS383 
it has two reasons
<a name="ch14c208">
(1) in eqn.BS079
    g(x)-g(x/2)≦f(x)  //given
    g(x) and f(x) have same
    variable x.
    -g() have half x/2.
<a name="ch14c209">
    eqn.BS083 make -g() and  //error
    f() be the same x/2k     //error
    eqn.BS383 make +g() and  //right
    f() be the same x/2k-1   //right
<a name="ch14c210">
(2) k start from k=1. Put this
    initial value k=1 into 
    eqn.BS083 get Ax(1/2 + ...)
    eqn.BS383 get Ax(1 + ...)
    eqn.BS083 get 1/21=1/2
    eqn.BS383 get 1/21-1=1/1
    eqn.BS084 has Ax(1+1/2+...)
    eqn.BS383 provide a '1' for BS084
    eqn.BS083 can not provide a '1'
2010-08-09-22-10 stop




<a name="ch14c211">　Index begin　Index this file
2010-08-10-10-54 start
■　Exercise 14.6 solution


<a name="ch14c212">
The key point of Exercise 14.6
is that given f(x) be bounded
and NOT given
  g(x)≦f(x) ---eqn.BS078
instead, we have
  g(x)-g(x/2)≦f(x) ---eqn.BS079

<a name="ch14c213">
Please pay attention to this
given eqn.BS079.
Three terms g(x) and g(x/2) and
f(x). we know that
f(x), +g(x) have same variable x
f(x), -g(x/2) have different 
variable x vs. x/2 For this
observation, LiuHH made change
from eqn.BS083 to eqn.BS383 

<a name="ch14c214">
Next given condition is
  f(x)≦Ax+B for x≧0 ---eqn.BS080
A and B are constants.
eqn.BS080 say that 
f(x) is proportional to xp
where power
  p≦1 ---eqn.BS089

<a name="ch14c215">
Assume variable x is bounded by
  0≦x≦m ---eqn.BS090
Define ceiling integer for 
real number y be //eqn.BS082
  Y=ceiling(y) ---eqn.BS091
For example
  ceiling(1.23)=2 ---eqn.BS092
  ceiling(log(0.001))
 =ceiling(-6.90)=-6 ---eqn.BS093
<a name="ch14c216">　Index begin　Index this file
etc. Then, For each
  1≦k≦ceiling(m)=M ---eqn.BS094 
we have the bound
  g(x/2k－1)-g(x/2k)≦Ax/2k－1+B ---eqn.BS383
  g(x)     -g(x/2) ≦f(x) ---eqn.BS079
eqn.BS383 is an application of
given condition eqn.BS079 and 
eqn.BS080:f(x)≦Ax+B for x≧0

<a name="ch14c217">
eqn.BS383 has one restriction,
that is in g(x/2k－1)-g(x/2k)
two variables x/2k－1 and x/2k
have half relation.
  (x/2k－1)/2 = x/2k ---eqn.BS095
But target equation eqn.BS081
no such restriction.
<a name="ch14c218">
Apply eqn.BS383 successively
allow x approach to zero as
following. In eqn.BS383
k=1: g(x/21－1)-g(x/21)≦Ax/21－1+B ---eqn.BS096
k=2: g(x/22－1)-g(x/22)≦Ax/22－1+B ---eqn.BS097
k=3: g(x/23－1)-g(x/23)≦Ax/23－1+B ---eqn.BS098
k= ...
k=M: g(x/2M－1)-g(x/2M)≦Ax/2M－1+B ---eqn.BS099
<a name="ch14c219">
Sum above k=1 to k=M equations
Less than side, red cancel red
blue cancel blue, purple cancel
purple. Only begin g(x/21－1) and
end -g(x/2M) two terms left.

<a name="ch14c220">
Sum above k=1 to k=M equations
greater than side has 
Ax*(1/21－1+1/22－1+...+1/2M－1)
and +B+B+B+...+B M times.
Above use upper limit M, 
textbook use K, change M to K
Sum result is
  g(x)-g(x/2K)  ---eqn.BS084
 ≦Ax(1+1/2+1/22+...+1/2k－1)+KB
2010-08-10-11-55 here

<a name="ch14c221">　Index begin　Index this file
2010-08-10-12-10
Please copy next code, paste to
[Box3, input JS command] of
calculator page, then click 
[Test box3 command, output to box4]
Box 4 has output result. (local)

//<a name="ch14c222">
//2010-08-10-12-12
//sum 1/2^0+1/2^1+1/2^2+...+1/2^(k-1)
var i,j,k,s;
j=2; //sum 1/2^power
k=100;
s=0
for(i=0;i<k;i++)
s+=1/pow(j,i);
s
//2010-08-10-12-16
//answer s=2
//sum 1/2^0+1/2^1+1/2^2+...+1/2^99 = 2

<a name="ch14c223">
2010-08-10-12-23 start
In eqn.BS084, move blue g(x/2K)
from less than side to greater 
than side and reverse its sign.
Rewrite eqn.BS084 as
  g(x)  ---eqn.BS100
 ≦Ax(1+1/2+1/22+...+1/2k－1)+KB
  +g(x/2K)
<a name="ch14c224">
eqn.BS100 is a bound to g(x).
Computer program output show
that (1+1/2+1/22+...+1/2k－1)
has limit value 2, accept this
bound, then we have 2Ax.
<a name="ch14c225">
In eqn.BS100 change KB to 
ceiling(m), then we have 
B*ceiling(m) where m is 
upper bound of x 
  0≦x≦m ---eqn.BS090
<a name="ch14c226">　Index begin　Index this file
Change +g(x/2K) to bound too
replace it with max[0≦t≦1]g(t) 
We get the bound equation
eqn.BS085
eqn.BS085 tell us that g(x)
has one constraint that is
  0≦x≦m≦1 ---eqn.BS101

<a name="ch14c227">
Follow textbook hint
so we can take 
    A'=2A ---eqn.BS086 
    B'=B ---eqn.BS087 
and 
    C'=B+max[0≦t≦1]g(t) ---eqn.BS088 
2010-08-10-12-38 here

<a name="ch14c228">
2010-08-10-12-44
study to the end, learned that
g(x) has domain limitation. 
The word "max[0≦t≦1]g(t)"
limit 0≦x≦1 for g(x). See
eqn.BS088. But in eqn.BS081 
g(x) has constraint x≧1.

<a name="ch14c229">
Next two equations are contradict 
to each other
  ceiling(log(0.001))
 =ceiling(-6.90)=-6 ---eqn.BS093
and //eqn.BS094=eqn.BS082
  1≦k≦ceiling(m)=M ---eqn.BS094 

<a name="ch14c230">
eqn.BS093 say integer variable k
is negative integer.
eqn.BS094 say integer variable k
is positive integer.
It can not be true at the same 
time.

<a name="ch14c231">　Index begin　Index this file
LiuHH need more time to think
where I made mistake.
Exercise 14.6 is NOT DONE
2010-08-10-12-50 stop

2010-08-11-17-13 done first proofread
2010-08-11-21-22 done second proofread
2010-08-11-21-41 done spelling check




<a name="Copyright">　Index begin　Index this file
2009-06-19-10-48
If you are interested in inequality,
suggest you buy the book
The Cauchy-Schwarz Master Class
written by Prof. J. Michael Steele
The Cauchy-Schwarz Master Class is
this web page's textbook.

To buy textbook, that is to show thanks
to Prof. J.M. Steele's great work.
and it is also respect copyright law.
Thank you. Freeman 2009-06-19

The Cauchy-Schwarz Master Class
J. Michael Steele ★★★★★
ISBN 978-0-521-54677-5
2009-06-19-10-56