<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="ch14b001">　Index begin　Index this file
2010-07-29-16-50 start
■　Numerical evaluation square 
　　of complex sequence sum
Below is draft work. Skip.

Textbook problems 14.2 and 14.3
both start from eqn.14.9, both
discuss {∑[n=1,M]cn}2 where cn
is an array of complex numbers.

<a name="ch14b002">
LiuHH hope to get some feeling
about eqn.14.9 and write a short
program to calculate eqn.14.9 
for any complex sequence that
user supplied. Please copy the
following code and paste to
complex calculator page (local)
http://freeman2.com/complex2.htm#calculator
"Box3, input JS command" then
click 
"test box3 command, output to box4"
<a name="ch14b003">
The following code you can change 
the beginning complex definition
part. If you have total five
complex numbers, define them from
cp[0] to cp[4]. Do not jump the
sequence number in [number].

<a name="ch14b004"> goto code
First explain the output form.
Output to box 4 as the following
[[
abscsum //not use bye09(), see many 0000000
9.360000000000001
csum //∑[n=1,M]c_n, eqn.14.9 left side
3,-0.6000000000000001
csq1ERR //(∑[n=1,M]c_n)^2 by cmulf(csum,csum)
8.64,-3.6000000000000005
csq2ERR //(∑[n=1,M]c_n)^2 by cpowf(csum,2)
8.639999999999998,-3.6
<a name="ch14b005">
bye09a(csq3) //complex1 to complex2 power
2.425154105744493,-2.4770598585406622,71.85735237933063,35.65992358084912,-270.9650063361174,1836.4436463333613,-42354.28218568532,7168.738068926585
2-1 //Below is eqn 14.9 right side
1
riteside //eqn.14.9 riteside
9.36
leftside //eqn.14.9 leftside
9.36
<a name="ch14b006">　Index begin　Index this file
dsum  //like index square
35.12
esum  //unlike index multiply
-25.76,0
sumde //dsum + esum
9.36,0
]]

<a name="ch14b007">
abscsum is abs( {∑[n=1,M]cn}2 )
it is eqn.14.9 left side include 
square operation. It is equality's
final answer.

<a name="ch14b008">
csum is ∑[n=1,M]cn, sum of all
complex in one seq. No square.

csq1ERR is {∑[n=1,M]cn}2, result 
by calling cmulf(csum,csum)

csq2ERR is {∑[n=1,M]cn}2, result 
by calling cpowf(csum,2)

<a name="ch14b009">
bye09a(csq3) is complex csum 
raise to power of '1.2-0.5i'
bye09a(csq3) has nothing to do 
with eqn.14.9. Just illustrate
the command
cpowf(csum,'1.2-0.5i',0,3)
 output principle value answer
and non-principle value answer.

<a name="ch14b010">
2-1 //Below is eqn 14.9 right side
1
this is dummy calculation, just 
show up the comment. Output
separator.

<a name="ch14b011">　Index begin　Index this file
riteside //eqn.14.9 riteside
leftside //eqn.14.9 leftside
Above two values should be 
identical. Because eqn.14.9
is equality, not inequality.

<a name="ch14b012">
dsum  //like index square
eqn.14.9 right side has 
∑[n=1,M]{|cn|2}. "like index" 
indicate that cm*cn has m=n.

<a name="ch14b013">
esum  //unlike index multiply
eqn.14.9 right side has 
∑[1≦m＜n≦M]{cm*cnj+cmj*cn}
"unlike index" indicate that 
cm*cn has m≠n.
complex conjugate of c is cj

<a name="ch14b014">
sumde //dsum + esum
This is eqn.14.9 right side
total sum. It should be same 
as left side total sum.
2010-07-29-17-50 stop

Following is test code.

//<a name="ch14b015">　Index begin　Index this file
//2010-07-26-12-02 start code
//2010-07-26-15-20 write function cconj(c1)
//2010-07-26-19-18 write function bye09a(arg1)
//2010-07-27 update complex2.htm complex2.js
//2010-07-28 update tute0049.htm sumpart2.htm
//2010-07-29-11-08  done code
//9907261202 start code
//9907271509 delete debug
//9907282135 delete oType (output type)
//
//copy/paste the following code to
//http://freeman2.com/complex2.htm#calculator
//box 3, click [test box3 command, output to box4]
//following code calculate square
//of {∑[n=1,M]cn} ; cn is complex
//The Cauchy-Schwarz Master Class
//page 212; line 2,3,4; eqn.14.9
//<a name="ch14b016">
var errmsg='Output may contain error. You must exam first.'
errmsg; //9907301438
var cp=[];
cp[0]=' 1.2+2.3i';
cp[1]=' 2.2-3.1i';
cp[2]='-0.2+1.7i';
cp[3]=' 1.8+0.4i';
cp[4]='-2.0-1.9i';
//You can change above value
//You can add more cp[5], cp[6]
//You can delete cp[4],cp[3]
//<a name="ch14b017">
var cLen=cp.length;
var dsum=toCplx('0');
var esum=toCplx('0');
var sumde=caddf(esum,dsum);
var i,j,k,m1,m2,m3;
//<a name="ch14b018">
//do not copy code from source file
//source file less than is '&lt;'
//copy from browser screen.
//screen less than is '<' //9907291043
for(i=0;i<cLen;i++)
cp[i]=toCplx(cp[i]);
//<a name="ch14b019">　Index begin　Index this file
var csum=toCplx('0'); //csum=[0,0]; same
for(i=0;i<cLen;i++) //can not use 'c1+c2'
csum=caddf(csum,cp[i]); //use caddf(c1,c2)
var csq1ERR=cmulf(csum,csum) //csum*csum
var csq2ERR=cpowf(csum,2)    //csum^2
var csq1OK=cmulf(csum,cconj(csum)) //c1*c1_conjugate
var csq3=cpowf(csum,'1.2-0.5i',0,3)//illustration 
var abscsum=cabsf(csq1ERR); //9907261820
//<a name="ch14b020">
abscsum //not use bye09(), see many 0000000
csum //∑[n=1,M]c_n, eqn.14.9 left side
csq1OK  //c1*c1_conjugate
csq1ERR //(∑[n=1,M]c_n)^2 by cmulf(csum,csum)
csq2ERR //(∑[n=1,M]c_n)^2 by cpowf(csum,2)
bye09a(csq3) //complex1 to complex2 power
//9907261213
//<a name="ch14b021">
2-1 //Below is eqn 14.9 right side
dsum=0; //dsum is ∑[n=1,M](c_n*c_n) 
for(i=0;i<cLen;i++) //dsum like index
dsum+=cabsf(cmulf(cp[i],cp[i]));
//<a name="ch14b022">
for(i=1;i<cLen;i++){//9907261226
for(j=0;j<i;j++){ //esum unlike index
conji=cconj(cp[i]);//complex conjugate
conjj=cconj(cp[j]);//cconj(c0) not change c0
//<a name="ch14b023">　Index begin　Index this file
m1=cmulf(cp[j],conji); //9907261413 use conji,conjj
m2=cmulf(cp[i],conjj); //p212 line 2 right end
m3=caddf(m1,m2); //can not use m1+m2
esum=caddf(esum,m3);//9907261239
}}
//<a name="ch14b024">
var realesum=cgetr(esum); //9907261825
sumde=caddf(esum,dsum);
dsum=bye09(dsum)  //bye09(numberOnly)
esum=bye09a(esum) //bye09a(numberArray)
sumde=bye09a(sumde)
//9907261743 add bye09()
//9907261230
//<a name="ch14b025">
var riteside=bye09a(dsum+realesum);
var leftside=bye09a(abscsum);//bye09a(number) OK
riteside //eqn.14.9 riteside
leftside //eqn.14.9 leftside
dsum  //like index square
esum  //unlike index multiply
sumde //dsum + esum
//9907261905 eqn.14.9 left and right match
//copy/paste the above code to
//http://freeman2.com/complex2.htm#calculator
//box 3, click [test box3 command, output to box4]

<a name="ch14b026">
2010-07-29-19-04 start
eqn.14.7B has two bounds
  √{2N[1+log(N)]}  ---eqn.BR001
  √{N+N*[1+log(N/2)]} ---eqn.BR002
and one sum of complex
  ∑[k=1,M]exp(i*2*PI*(kk+bk+c)/N) ---eqn.BR003
What is the relative magnitude 
of these three values? answer
Following is a short code for
two bounds

//<a name="ch14b027">　Index begin　Index this file
//9907291038 separate N,N1,N2 
//to another program
var N=8; //9907282140 use N,N1,N2
var N1=2*N*(1+log(N));
var N2=N+N*(1+log(N/2));
N  //determine your N value at 'v_a_r N'
N1 //2*N*(1+log(N))   CSMC bound
N2 //N+N*(1+log(N/2)) tute0049 bound

<a name="ch14b028">
N1 and N2 are square of the
bounds. Early code has 
riteside and leftside. They
are square of sum of complex
sequence. It is ok to compare 
all square values.
<a name="ch14b029">
Bounds eqn.BR001 and eqn.BR002
have restriction. 
First require complex be exp(iθ)
Can not be any complex c0, since
abs(any complex) may not= 1.
Second require complex sequence 
must be generated by
  ∑[k=1,M]exp(i*2*PI*(kk+bk+c)/N) ---eqn.BR003
Can not be any Euler's formula 
complex. //second cover first.

<a name="ch14b030">
For arbitrary complex, LiuHH
write click button [rand a+ib]
at sumpart2.htm#buildSequence

<a name="ch14b031">
For any Euler's formula complex, 
write click button [exp(ix)]
at sumpart2.htm#buildSequence

<a name="ch14b032">　Index begin　Index this file
For Problem 14.2 complex, LiuHH
write click button [exp(iy)]
at sumpart2.htm#buildSequence

<a name="ch14b033">
To use button [rand a+ib], in
sumpart2.htm#ControlCenter
fill boxes at
[[
10^  ; +/0 , +/0/- ;  digits;  eq.p  
fill  numbers;  integer;   b=a^  
]]
Then click button [rand a+ib]
How to use?

<a name="ch14b034">
To use button [exp(ix)], in
sumpart2.htm#buildSequence
fill in boxes
[[
x bgn  ; x step  ; x end
]]
Then click button [exp(ix)]
How to use?

<a name="ch14b035">
To use button [exp(iy)], in
sumpart2.htm#buildSequence
fill in box
[[
Box 4, debug
]]
Then click button [exp(iy)]
How to use?
// 2010-08-04-17-00 Added [h3] 
// click button in  local
// sumpart2.htm#buildSequence

<a name="ch14b036">
Input four numbers in
Box 4 in the order b,c,M,N
CSMC page 210, equation 14.7
∑[k=1,M]exp(i2π(k*k+b*k+c)/N)
require 0≦M＜N
function expiy() build only
exp(i*2*PI*[(k*k+b*k+c)/N])
(not build [exp(ix)] style)

<a name="ch14b037">　Index begin　Index this file
After finish function expiy()
LiuHH is in a position to test
eqn.BR001,eqn.BR002,eqn.BR003
three values. First data was
2 3 5 8 for sumpart2.htm box 4
<a name="ch14b038">
Output five complex, 
two cancel to zero,
other three are identical.
copy from sumpart2.htm box 2
paste to complex2.htm Box3,
replace the code top at
ch14b016

<a name="ch14b039">
result is eqn.BR003 value
abscsum //not use bye09(), see many 0000000
8.999999999999998

<a name="ch14b040">
Open second browser, run
[[
var N=8; //9907282140 use N,N1,N2
var N1=2*N*(1+log(N));
var N2=N+N*(1+log(N/2));
N  //determine your N value at 'v_a_r N'
N1 //2*N*(1+log(N))   CSMC bound
N2 //N+N*(1+log(N/2)) tute0049 bound
]]
<a name="ch14b041">
output 
[[
N  //determine your N value at 'v_a_r N'
8
N1 //2*N*(1+log(N))   CSMC bound
49.27106466687737
N2 //N+N*(1+log(N/2)) tute0049 bound
27.090354888959126
]]
<a name="ch14b042">　Index begin　Index this file
The magnitude is
  8.9999 ＜ 27.0903 ＜ 49.2710
   BR003 ＜  BR002  ＜  BR001
9 bound by 27 ?? How rich !!
If 9 bound by 9.0 that is sharp.

<a name="ch14b043">
LiuHH start review Problem 14.2
It is too long, build an index
for Problem 14.2. After review
did not find any thing wrong.
In index marked
cause inequality 1
cause inequality 2
one more inequality 3
eqn.14.7 do have three time
inflation !!
2010-07-29-19-53 stop

<a name="ch14b044">
when 
{∑[n=1,M]cn}2
=
∑[n=1,M]{|cn|2}. "like index" 
such that  "unlike index" 
∑[1≦m＜n≦M]{cm*cnj+cmj*cn}
is zero ?
2010-07-29-22-05

<a name="ch14b045">
2010-07-30-10-40
input
[[
var cp=[];
cp[0]='0+1i';
cp[1]='1.+0i';
]]
<a name="ch14b046">
output
[[
riteside //eqn.14.9 riteside
2
leftside //eqn.14.9 leftside
2
dsum  //like index square
2  
esum  //unlike index multiply
0,0   //＜== key, zero
sumde //dsum + esum
2,0
]]

<a name="ch14b047">
[[
var cp=[];
cp[0]='0+1i';
cp[1]='1.+0i';
cp[2]='1.4+0i';
cp[3]='0+1.4i';
]]
<a name="ch14b048">　Index begin　Index this file
2010-07-30-10-43 other similar 
[[
riteside //eqn.14.9 riteside
11.52
leftside //eqn.14.9 leftside
11.52
dsum  //like index square
5.92
esum  //unlike index multiply
5.6,0 //＜== fail to get zero 
sumde //dsum + esum
11.52,0
]]

<a name="ch14b049">
2010-07-30-10-57
Textbook page 212 line 4, 
eqn.14.9 last term is // ---eqn.BR004
+2*Re∑[h=1,M-1]∑[m=1,M-h]cm+h*cmj
complex conjugate of c is cj

<a name="ch14b050">
Textbook page 213 line 11, 
eqn.14.14 last term is // ---eqn.BR005
+2*∑[h=1,N-1]┃∑[m=1,N-h]cm+h*cmj┃

<a name="ch14b051">
At first LiuHH question is that
eqn.BR004 is already real number
why eqn.BR005 take absolute for
real number? After think, know
the difference. 
<a name="ch14b052">
eqn.BR004 is real, not a complex.
but eqn.BR004 can be positive or 
negative. 
eqn.BR005 force unlike index 
multiply term be positive.

<a name="ch14b053">
eqn.BR004 has positive real and
negative real cancellation.
eqn.BR005 inner summation has
cancellation. 
Outer summation no cancellation.
2010-07-30-11-18

//<a name="ch14b054">　Index begin　Index this file
//copy/paste the following code to
//http://freeman2.com/complex2.htm#calculator
//box 3, click [test box3 command, output to box4]
//
//2010-07-30-11-24 start code
//2010-07-30-12-19  done code (but error)
//2010-09-20 & 21 correction
//following code calculate square
//of [∑[n=1,M]cn] ; cn is complex
//The Cauchy-Schwarz Master Class
//page 213; line 11; eqn.14.14
var errmsg='Output may contain error. You must exam first.'
errmsg; //9907301438 Code for CSMC eqn.14.14
//9909202105 start correction
//<a name="ch14b055">
var cp=[];
cp[0]=' 1.2+2.3i';
cp[1]=' 2.2-3.1i';
cp[2]='-0.2+1.7i';
cp[3]=' 1.8+0.4i';
cp[4]='-2.0-1.9i';
//You can change above value
//You can add more cp[5], cp[6]
//You can delete cp[4],cp[3]
//
//<a name="ch14b055b">
//9909210945 use oneValue
var oneValue=' 1+2i'; //example oneValue.
oneValue=''; //null cause multiple value.
//set oneValue to a complex number
//to get all sequence be one value
//
var i,j,k,m1,m2,m3;
//
//<a name="ch14b055c">
m3=toCplx(oneValue);       //9909211710
if(oneValue.length>0 &&1==1
 &&!isNaN(m3[0]) &&2==2 //9909211725
  ) //&&1==1 &&2==2 added for complex2.htm 9909221927
  { //9909210947
for(i=0;i<cp.length;i++)
cp[i]=oneValue
2-1 //use oneValue, leftside=riteside, not leftside < riteside
  }
//<a name="ch14b056">
var N=cp.length;
var csum=toCplx('0'); //csum=SUM[n=1,N](c_n) 
var dsum=toCplx('0'); //dsum=SUM[n=1,N](c_n*c_n) 
var esum=toCplx('0'); //esum=SUM[i,j](c_i*c_j) 
//dsum=sum of two complex of like index 
//esum=sum of two complex of UNlike index 
//do not copy code from source file
//copy from browser screen.
//<a name="ch14b057">
//change string complex to array complex
for(i=0;i<N;i++)    //'1+2i' is string
cp[i]=toCplx(cp[i]);//[1,2]  is array
//only array complex can be manipulated
//next loop CSMC eqn.14.14 1st summation 
//eqn.14.14 less than side sum. no square
for(i=0;i<N;i++) //can not use 'c1+c2'
csum=caddf(csum,cp[i]); //use caddf(c1,c2)
//eqn.14.14 less than side square the sum
//complex number rule: square a complex
//use c1*c1_conjugate ; not use c1*c1
//<a name="ch14b058">
var csq1OK=cmulf(csum,cconj(csum)) //c1*c1_conjugate
var csq1ERR=cmulf(csum,csum) //csum*csum
var abscsum=cabsf(csq1ERR);  //9907261820
csum    //SUM[n=1,N]c_n, plain sum, no square.
csq1OK  //csum*csum_conjugate
csq1ERR //(SUM[n=1,N]c_n)^2 by cmulf(csum,csum)
abscsum //cabsf(csum*csum) NOT use conjugate
//
//9907261213
//<a name="ch14b059">　Index begin　Index this file
3-2 //Below is eqn.14.14 right side
//dsum is page 213, eqn.14.14 
//greater side first term. 9909201753 note
dsum=0; //dsum is SUM[n=1,N](c_n*c_n) 
//next loop CSMC eqn.14.14 2nd summation 
//eqn.14.14 greater than side single sum
for(i=0;i<N;i++) //dsum like index
dsum+=cabsf(cmulf(cp[i],cconj(cp[i])));
bye09(dsum) // SUM[n=1,N](c_n*c_n_conjugate) 
csq1OK      // SUM(c_n)*[SUM(c_n)]_conjugate
//dsum is sum[abs(complex^2)],  sum positive
//csq1OK={abs[sum(complex)]}^2, sum +/0/- cancel
//9909202135 here
//
var rhoN=[]; //CSMC  page 213 eqn.14.15
//Next line code rhoN[0][1] is string, correct,
//rhoN[0][1] is not number. intended. 9908031941
rhoN[0]=[0,'0i // following take abs() and sum = rhoAll // '];
var rhoAll=0; //9907301141
var h,m; //h,m match CSMC eqn.14.14. 9909202256
var conjm; //conjm is complex conjugate of cp[m]
//
//<a name="ch14b060">
//for(h=0;h<N-1;h++) //outer for loop, outer sum
//9909211250 h loop can NOT start from h=0
//if h=0 then {c_(m+h)*c_m_conj} become
//{c_(m)*c_m_conj} which is not UNlike index.
//next loop CSMC eqn.14.14 3rd summation 
//greater than side double sum outer sum
for(h=1;h<N;h++) //9909211253
  {//9907261226
esum=cnewf(0,0)   //9909202311
//esum=SUM[h,m]{c_(m+h)*c_m_conj}
//esum=sum of product of two 
//complex of UNlike index m+h!=m
//
//<a name="ch14b061">
//9909211242 eqn.14.14 sum m from 1 to N-h. But code
//define cp[i] start from i=0, not i=1. Next for loop
//start from m=0 (not m=1) end at m<N-h (not m<N-h-1)
//for(m=0;m<N-h-1;m++)
//next loop CSMC eqn.14.14 4th summation
//greater than side double sum inner sum
for(m=0;m<N-h;m++) //9909211246
    { //inner for loop, inner sum
//<a name="ch14b062">
conjm=cconj(cp[m]);//cconj(c0) not change c0
m2=cmulf(cp[m+h],conjm); //p213 eqn.14.14 right end
//re-write code cut many line here.
//<a name="ch14b063">
//m2 is c_(m+h)*c_m_conj; h > 0
esum=caddf(esum,m2);//9909202306
    } //end inner for(m=0;m<N-h-1;m++)
//<a name="ch14b064">　Index begin　Index this file
rhoN[h]=[0,0];       //9907301138
rhoN[h][0]=bye09(esum[0]);
rhoN[h][1]=bye09(esum[1]);//9907301139
rhoAll=rhoAll+cabsf(rhoN[h]);//9907301142
  } //end outer for(h=0;h<N-1;h++)
//<a name="ch14b065">
//bye09a(bb) change bb=2.19999999 to 2.2
rhoAll=bye09a(rhoAll); //bye09(numberOnly)
esum=bye09a(esum)      //bye09a(numberArray)
//9907261743 add bye09()
//9907261230
//<a name="ch14b066">
var riteside=bye09a(dsum+2*rhoAll);
var leftside=bye09a(abscsum);//bye09a(number) OK
leftside //eqn.14.14 leftside (less than side)
riteside //eqn.14.14 riteside=dsum+2*rhoAll
dsum     // all like index square
2*rhoAll //rhoAll=sum of all |esum|
esum  //LAST unlike index multiply
//<a name="ch14b067">
4-3   //Below is CSMC page 213 line -1
rhoAll //auto correlation replace esum
//dsum+rhoAll //replace riteside 9907301151
dsum+2*rhoAll //replace riteside 9907301151
N*N // N*N<dsum+2*rhoAll page 213 line -1
//<a name="ch14b068">
//9907301426 print rhoN[i] value
var rhoPrint='';
for(i=0;i<rhoN.length;i++)
  { //this for loop change array to string
rhoPrint+=rhoN[i][0]+(rhoN[i][1]<0?'':'+')+rhoN[i][1]+'i\n'
  } //so that output read better.
rhoPrint; //CSMC page 213, eqn.14.15 //9907301431
//9907301609
//copy/paste the above code to
//http://freeman2.com/complex2.htm#calculator
//box 3, click [test box3 command, output to box4]


<a name="ch14b069">　Index begin　Index this file
2010-07-30-18-41 start
■　The role of auto correlations
Rewrite equation 14.9

<a name="ch14b071">
2010-07-30-19-06 here
Above put two equations side 
by side for comparison purpose.
eqn.14.9 is equality.
eqn.14.14 is inequality.
Here has eqn.14.9 calculation
<a name="ch14b072">　Index begin　Index this file
Why take absolute value cause
inequality? 
Use one example to explain.
Assume d sequence=[-2,+5,-8]
|d1+d2+d3| = |-2+5+(-8)|=5
but |d1|+|d2|+|d3| = 15
<a name="ch14b073">
Where to take absolute value?
that is very different.
In the case of take absolute
value first, add all positive
there is no cancellation.
In the case of take absolute
value last, add positive and
negative, it has cancellation.

<a name="ch14b074">
eqn.14.14 has double summation
∑[h=1,N-1]∑[m=1,N-h]c1*c2
From above analysis, we know 
∑[h=1,N-1]∑[m=1,N-h]┃c1*c2┃
is greater or equal to next
∑[h=1,N-1]┃∑[m=1,N-h]c1*c2┃
is greater or equal to next
┃∑[h=1,N-1]∑[m=1,N-h]c1*c2┃

<a name="ch14b075">
eqn.14.9 ≦ eqn.14.14 it has
two causes.
First, eqn.14.9 right side 
not take absolute value at all.
In this case, it is like -5＜+5
Second, if eqn.14.9 right side
take outer absolute value, 
eqn.14.9 is still ≦ eqn.14.14
In this case, it is like +5＜15
<a name="ch14b076">
Not take absolute value,
eqn.14.9 is equality.
Take absolute value any where
eqn.14.9 become inequality.
eqn.14.9 become eqn.14.14
2010-07-30-19-36 here

<a name="ch14b077">　Index begin　Index this file
Textbook define blue term in
eqn.14.14 as auto correlation
sums ρN(h), that is
  ρN(h)=∑[m=1,N-h]cm+h*cmj ---eqn.14.15
for all 1≦h＜N . (p.213 L.14)
complex conjugate of c is cj
If ρN(h) are small on average
then the complex sequence sum 
  ┃c1+c2+...+cN┃ ---eqn.BR007
should be relative small.
<a name="ch14b078">
If complex sequence elements
  |ck|≦1 ---eqn.BR008
and if
  lim[N→∞]{ρN(h)/N}=0 ---eqn.14.16 (eqn.BR009)
for all h=1,2,..., then
<a name="ch14b079">
do we have
  ┃c1+c2+...+cN┃/N→0 ---eqn.BR010
as N→∞
The answer is yes, we will show
it at the following calculation.
2010-07-30-20-05 stop

<a name="ch14b080">
2010-07-31-10-41 start
Textbook say in eqn.14.14 the blue
term evaluate too many times,
the number of evaluation times
is from h=1 to h=N-1, total N-1
times. 
<a name="ch14b081">
Textbook say it is 
possible reduce from N-1 times
to H times, where
  1≦H＜N  ---eqn.BR011
and give Problem 14.3 next.

<a name="ch14b082">　Index begin　Index this file
2010-07-31-10-47 here
■　Problem 14.3 (A Qualitative
van der Corput Inequality)

<a name="ch14b083">
Show that for each complex sequence
  c1,c2,...,cN
and for each integer
  1≦H＜N  ---eqn.BR011
one has the inequality

<a name="ch14b085">
2010-07-31-11-17 here
LiuHH have question about
van der Corput Inequality
Please compare eqn.14.17 with
eqn.14.7
Both eqn.14.17 and eqn.14.7
less than side have about the
same amount work. Sum a whole
complex sequence. eqn.14.17
take extra complex square once.
<a name="ch14b086">
Both equations are bounded by
upper bound. Please compare
two upper bounds.
eqn.14.7 upper bound need just
one input value N and give a
upper bound √{2N[1+log(N)]} 
no complex summation.
<a name="ch14b087">　Index begin　Index this file
eqn.14.7 upper bound is simpler
than its less than side. BUT
eqn.14.17 upper bound is much
more complicate than its less 
than side !! 
Let calculation for less than 
side be 100% work. We do work 
more than 100% to get estimate 
bound for 100% work?
<a name="ch14b088">
eqn.14.17 upper bound calculate
complex absolute value n times,
square it, sum real number n 
times. PLUS sum auto correlations
ρN(h) H times. 
In eqn.14.17, if we ignore ρN(h), 
|∑cn|2 has ≦ ∑|cn|2 see here
<a name="ch14b089">
The evaluation of ρN(h) involve 
complex1 and complex conjugate 
of complex2 multiplication.

Please be alert about this
difference between 
eqn.14.7 //easier bound
and eqn.14.17. //harder bound
2010-07-31-11-35 here



<a name="ch14b090">
//textbook page 213 second half
If we assume that complex 
sequence elements all be
  |ck|≦1 ---eqn.BR008
for k=1,2,...
and if //ρN(h) definition
  lim[N→∞]{ρN(h)/N}=0 ---eqn.14.16 (eqn.BR009)
eqn.14.17 tell us that auto 
correlations in upper bound 
∑[h=1,H]|ρN(h)/N| approach 
to zero and drop out.
<a name="ch14b091">
Since |ck|≦1, then N times 
summation value is less than N.
eqn.14.17 upper bound approach 
to {N+0}*4N/(H+1) = 4N*N/(H+1)
Move N*N to less than side get

<a name="ch14b093">
2010-07-31-12-02 here
In eqn.14.18, H is arbitrary
within 
  1≦H＜N  ---eqn.BR011
eqn.14.18 left side is bounded.
Take square root, we find
  ┃c1+c2+...+cN┃/N→0 ---eqn.BR010
as N→∞ 
A bounded infinite sequence sum, 
its infinity element must be zero,
<a name="ch14b094">
Alert: eqn.14.18 left side is
bounded, finite summation value 
is finite and greater than zero.
eqn.BR010 infinity summation.
summation of complex sequence
is finite. 

<a name="ch14b095">
Summation divide by N 
when N approach to infinity,
finite/N approach to zero.
2010-07-31-12-18 stop

<a name="ch14b096">
2010-07-31-15-29 start
Auto correlation sums ρN(h), is
defined as
  ρN(h)=∑[m=1,N-h]cm+h*cmj ---eqn.14.15
for all 1≦h＜N . (p.213 L.14)
complex conjugate of c is cj
<a name="ch14b097">
Because 1≦h then index m+h and
index m are never equal. cm+h 
and cm are two different complex 
numbers. Textbook present a 
"shift of itself" method.
2010-07-31-15-35 here

<a name="ch14b099">
2010-07-31-17-10 here
eqn.BR012 is a graph of H=2. 
It is one original sequence
and two (H=2) shift sequence.
Boundary is a concern. 
To avoid boundary condition, 
extend sequence to positive 
infinity and negative infinity. 
Extended elements all fill with 
zero. //zero1, zero2
<a name="ch14b100">
Elements in between vertical
bars '┋' are original elements.
Those outside elements are zeros. 
In eqn.14.17, outer summation 
∑[h=1,H] sum auto correlations 
ρN(h) H times. Each ρN(h) has 
inner summation sum N-h times.
  ρN(h)=∑[m=1,N-h]cm+h*cmj ---eqn.14.15
for all 1≦h＜N . (p.213 L.14)
<a name="ch14b101">
cm+h and cm can not be same
complex number, in eqn.BR012
can not pair vertically. We
work "down-left" diagonal. 
In eqn.BR012
Purple c1,c0,c-1 is one diagonal
Blue   c2,c1,c0 is one diagonal
Red    c3,c2,c1 is one diagonal
etc.
<a name="ch14b102">
Why build "shift of itself"?
In red c3,c2,c1 diagonal group
pick any two, multiply together, 
no same index multiplication happen
satisfy auto correlations ρN(h)
definition. Other diagonal group 
are the same.


<a name="ch14b103">　Index begin　Index this file
Above said "two multiply together"
Next eqn.BR013 and eqn.14.19 do
NOT have "two multiply together"
they are just summations.

<a name="ch14b105">
2010-07-31-17-58 here
eqn.BR013 calculate summation 
of three identical complex seq.
in eqn.BR012.

<a name="ch14b106">
eqn.BR013 left side sum three
rows (not sum diagonal) Each
row has n complex numbers. 
Total 3*(one sequence sum)
Total counts = 3*N

<a name="ch14b107">
eqn.BR013 right side sum N+2
diagonal (not rows) elements.
Total counts = (N+2)*3 = 3N+6
6 extra 'complex numbers' are
extended zero. In above H=2
eqn.BR012 case, 
<a name="ch14b108">　Index begin　Index this file
the extra zero elements are
Left to left vertical bar :
purple c0, purple c-1, blue c0
Right to right vertical bar :
red  cN+1,  blue cN+2, blue cN+1

<a name="ch14b109">
Added zero do not change total
sum. Textbook say eqn.BR013 is
an identity.
eqn.BR013 is for H=2 case. For
other H value, in eqn.BR013, 
replace 2 by H to get next 
generalized eqn.14.19.

<a name="ch14b111">
Similarly, eqn.14.19 pad many
zero 'complex number'.
2010-07-31-18-41 stop

<a name="ch14b112">
2010-07-31-20-10 start
Target inequality eqn.14.17 left
side is square of sequence sum.
eqn.14.19 has sequence sum. If
we square eqn.14.19, we reach 
eqn.14.17 less than side.

<a name="ch14b114">
2010-07-31-20-32 here
Equality in eqn.BR014 is an
application of eqn.14.19 .
Inequality in eqn.BR014 is a
result of taking absolute value
first before summation. See
Why take absolute value cause
inequality? Inserted blue one
do not change value.

<a name="ch14b115">
The reason insert blue one in
eqn.BR014 second line is for
applying Cauchy's inequality.
Please see eq.1.7
<a name="ch14b116">
Compare eqn.BR014 second line
with Cauchy's inequality.
eqn.BR014 blue one is eq.1.7
ak sequence. N+H one's
ak={1,1,1, ... , 1} ---eqn.BR302 
N+H come from upper bound of
∑[n=1,N+H] in eqn.BR014 .

<a name="ch14b117">
eqn.BR014 red term is NOT 
eq.1.7 bk whole sequence!
eqn.BR014 red term is just
one element of eq.1.7 bk 
bk whole sequence is
bk={|∑[h=0,H]c1-h|, //---eqn.BR303
    |∑[h=0,H]c2-h|, ...
    |∑[h=0,H]ck-h|, ...
    |∑[h=0,H]cN+H-h|} 
ci has many zeros zero1, zero2
<a name="ch14b118">　Index begin　Index this file
eqn.BR014 red term change n
value from n=1 to n=N+H total
N+H terms that is eq.1.7 bk 
whole sequence.

<a name="ch14b119">
eqn.BR014 line two outer sum
∑[n=1,N+H] control Cauchy's 
inequality. That means, ak 
sequence is {1, 1, ... , 1}
total N+H 1's. 
ak sum to N+H. This is the 
reason that N+H show up at 
next eqn.14.20.

<a name="ch14b121">
2010-07-31-21-02 here
eqn.14.20 is Cauchy's inequality
eq.1.7 squared.
eqn.14.20 left side is square.
When we compare eqn.14.20 with
eq.1.7, we must square eq.1.7 
to get correct comparison.
In eq.1.7 ignore square root.

<a name="ch14b122">
eq.1.7 right side ak sequence
square first then sum.
eqn.BR014 say ak sequence is 
all one sequence. One square 
is still one. Sum N+H times 
get (N+H) in eqn.14.20 .

<a name="ch14b123">　Index begin　Index this file
eq.1.7 right side bk sequence
square first then sum.
eqn.BR014 say bk sequence is 
red term in eqn.BR014.
One red term is one element 
of bk sequence.

<a name="ch14b124">
Cauchy say: square it then 
sum all of them. This is how 
we get eqn.14.20 right side 
double sum.

<a name="ch14b125">
eqn.14.20 has one inequality,
but it is contributed by two
sources. 
First is take absolute value 
before sum cause inequality.
Second is Master Cauchy's magic
work.

Up to here, explained eqn.14.20.
2010-07-31-21-17 stop

<a name="ch14b126">
2010-08-01-12-23 start
In eqn.14.20, expand greater 
than side double summation

<a name="ch14b130">
2010-08-01-13-02 here
eqn.BR015 "AA=" is rewrite square
of a summation to two of same
sequence but different index.
That is change 
from index n-h * n-h
 to  index n-j * n-k
<a name="ch14b131">
Complex number of n-k index
replaced by its complex conjugate.
This is complex number square
rule.

<a name="ch14b132">
eqn.BR015 "BB=" is separate
like index from unlike index.
In cn-a*cn-bj
if a=b then above is product 
   of like index.
if a≠b then above is product 
   of unlike index.
<a name="ch14b133">　Index begin　Index this file
In "BB=", 
|cn-s|2 is product of like index
cn-scn-tj is product of unlike 
index. 
After separation, product of 
unlike index not allow like
index occur in unlike group.
2010-08-01-13-18 stop

<a name="ch14b134">
2010-08-01-13-57 start
eqn.BR015 "CC=" is bring outer
summation ∑[n=1,N+H] to inside
of { two terms } 
First term become 
∑[n=1,N+H]∑[s=0,H]|cn-s|2
which is
<a name="ch14b135">
  ∑[n=1,N+H]∑[s=0,H]|cn-s|2
 = //expand inner sum here
  ∑[n=1,N+H]{|cn-0|2+|cn-1|2+...+|cn-H|2}
 = //expand outer sum here
  {|c1-0|2+|c1-1|2+...+|c1-H|2} //n=1
 +{|c2-0|2+|c2-1|2+...+|c2-H|2} //n=2
 +... ---eqn.BR016
<a name="ch14b136">
 +{|cN-0|2+|cN-1|2+...+|cN-H|2} //n=N
 +{|cN+1-0|2+|cN+1-1|2+...+|cN+1-H|2} //n=N+1
 +{|cN+2-0|2+|cN+2-1|2+...+|cN+2-H|2} //n=N+2
 +...
 +{|cN+H-0|2+|cN+H-1|2+▪▪▪+|cN+H-H|2} //n=N+H
All red terms are zero.

<a name="ch14b137">
Sum all like index elements in 
eqn.BR016 get "CC=" first term
  (H+1)*∑[n=1,N]|cn|2
you can find diagram for H=2 
case at ch14b098
<a name="ch14b138">　Index begin　Index this file
eqn.BR015 "CC=" second term 
put three summation together 
for unlike index elements.

<a name="ch14b139">
eqn.BR015 "DD≦" is easy to 
understand. Please see Why take 
absolute value cause inequality?
2010-08-01-14-56 here

<a name="ch14b140">
eqn.BR015 "EE=" play the same
technique as "CC=" did.
"CC=" first term expanded at 
eqn.BR016, result is drop
one summation and get a factor
of (H+1)
"EE=" do the same thing H times
get factor (H+1-h)
LiuHH still not figure out the
detail step of eqn.BR015 "EE=".
One reason of NOT DONE
<a name="ch14b141">
Above all start from eqn.14.20.
eqn.14.20 less than side has a
factor (H+1)2. Move this
factor to greater than side
denominator. Get the following

<a name="ch14b146">
This is precisely the inequality
given by van der Corput in 1931.
When we introduce the auto 
correlation sums and bound the
coefficients in the simplest way,
we come directly to the inequality
(14.17) which was suggested
by our challenge problem.

<a name="ch14b147">
LiuHH has two blur points.
First. LiuHH still not know
the detail step of eqn.BR015 
"EE=".
<a name="ch14b148">　Index begin　Index this file
Second, LiuHH need more time 
to find out why eqn.BR019 right
end term has summation upper
bound N-h instead of N.

Before know the detail,
Problem 14.3 is NOT DONE
2010-08-01-16-17 stop

<a name="ch14b149">
2010-08-01-18-18 start
Find eqn.BR015 "EE=".

<a name="ch14b152">　Index begin　Index this file
2010-08-01-18-39 here
eqn.BR020 treat unlike index 
product.
Different from "BB=" to "CC=" 
that is like index product.
eqn.BR020 has many zeros.
2010-08-01-18-43 stop

<a name="ch14b153"> skip
2010-08-01-21-25 draft work start
CSMC page 217 line 2
Let φj(x) j∈S be 
a sequence of functions if
∫[x=0,1]φj(x)*φkj(x)dx
=0  if j≠k ---eqn.14.21
=1  if j=k
then φj(x) j∈S is
an orthogonal sequence.
φj(x)=exp(2PI*i*j*x) is
such an orthogonal sequence

<a name="ch14b154">
2010-08-01-21-41
next '-' in '-2PI' is result 
of taking complex conjugate
Next i=√(-1), j,k are index.

<a name="ch14b155">
 ∫[x=0,1]exp(2PI*i*j*x)*exp(-2PI*i*k*x)dx
=∫[x=0,1]exp(2PI*i*j*x-2PI*i*k*x)dx
if j=k
 ∫[x=0,1]exp(2PI*i*j*x-2PI*i*k*x)dx
=∫[x=0,1]exp(2PI*i*j*x-2PI*i*j*x)dx
=∫[x=0,1]exp(0)dx
=∫[x=0,1]1*dx
=1*[1-0]
=1  OK right
<a name="ch14b156">
2010-08-01-21-43
 ∫[x=0,1]exp(2PI*i*j*x)*exp(-2PI*i*k*x)dx
=∫[x=0,1]exp(2PI*i*j*x-2PI*i*k*x)dx
if j≠k
 ∫[x=0,1]exp(2PI*i*j*x-2PI*i*k*x)dx
=∫[x=0,1]exp(2PI*i*x*[j-k])dx
=∫[x=0,1]{cos(2PI*x*[j-k])+i*sin(2PI*x*[j-k])}dx
=∫[x=0,1]{cos(2PI*x*[j-k])+i*sin(2PI*x*[j-k])}d(2PI*x*[j-k])/(2PI*[j-k])
=∫[x=0,1]d{sin(2PI*x*[j-k])-i*cos(2PI*x*[j-k])}/(2PI*[j-k])
={sin(2PI*1*[j-k])-i*cos(2PI*1*[j-k])}/(2PI*[j-k])
-{sin(2PI*0*[j-k])-i*cos(2PI*0*[j-k])}/(2PI*[j-k])
={0-i*1}/(2PI*[j-k])
-{0-i*1}/(2PI*[j-k])
=0  OK right
2010-08-01-21-48 draft work stop

<a name="ch14b157">
2010-08-02-07-37 draft work start
For any finite set A⊂S, the
orthonomorality condition (14.21)
and direct expansion lead one to 
the identity

<a name="ch14b159">
2010-08-02-07-50 here
Next use special case example 
to verify eqn.14.22. Assume
L=1,m=2,n=3 for short example.
Let complex coefficients {ck}
be
  {ck}={cL,cm,cn} ---eqn.BR021
with restriction
  |ck|≦1 for all 1≦k≦n  ---eqn.BR022
<a name="ch14b160">
Let complex function {φk}
be
  {φk}={φL,φm,φn} ---eqn.BR023
  ={exp(2πiLx),exp(2πimx),exp(2πinx)}
Sum has total three elements
  ∑[k∈A]{ck*φk}=cL*exp(2πiLx)
  +cm*exp(2πimx)+cn*exp(2πinx) ---eqn.BR024
2010-08-02-08-10 here

<a name="ch14b161">
Complex square rule c1*c1_conjugate
Product of sum=complex1 can not be
  complex1 * complex1
must be
  complex1 * complex1_conjugate
Because assume total sum has
value a+b*i. Wrong expression
  (a+b*i)*(a+b*i)  //ERROR
 =aa+2abi+bbii 
 =aa-bb+2abi ---eqn.BR025
is still a complex. 
<a name="ch14b162">
The
  complex1 * complex1_conjugate
expression give us
  (a+b*i)*(a-b*i)  //correct
 =aa+abi-abi-bbii 
 =aa+bb ---eqn.BR026
it is a real number

<a name="ch14b163">　Index begin　Index this file
Simplest example is complex2=i
then 
  complex2*complex2=i*i=-1 //error
but
  complex2 * complex2_conjugate
  =i*(-i) =+1 //correct

<a name="ch14b164">
Square of sum is
  sum*sum_conjugate ---eqn.BR027
 =[cL*exp(+2πiLx)+cm*exp(+2πimx)+cn*exp(+2πinx)]
 *[cL*exp(━2πiLx)+cm*exp(━2πimx)+cn*exp(━2πinx)]
Negative ━ is a result of 
         complex conjugate.
<a name="ch14b165">
Use simplified expression
  sum*sum_conjugate ---eqn.BR028
 =[cL*e(+Lx)+cm*e(+mx)+cn*e(+nx)]
 *[cL*e(-Lx)+cm*e(-mx)+cn*e(-nx)]
 =cLe(+Lx)[cLe(-Lx)+cme(-mx)+cne(-nx)]
 +cme(+mx)[cLe(-Lx)+cme(-mx)+cne(-nx)]
 +cne(+nx)[cLe(-Lx)+cme(-mx)+cne(-nx)]

<a name="ch14b166">
  sum*sum_conjugate ---eqn.BR029
 =cLcLe(+Lx)e(-Lx)
+cLcme(+Lx)e(-mx)+cLcne(+Lx)e(-nx)
 +cmcme(+mx)e(-mx)
+cmcne(+mx)e(-nx)+cmcLe(+mx)e(-Lx)
 +cncne(+nx)e(-nx)
+cncLe(+nx)e(-Lx)+cncme(+nx)e(-mx)

<a name="ch14b167">
In eqn.BR029, integrate whole
equation from x=0 to x=1. 
eqn.14.21 tell us that
three red terms are like index 
terms LL,mm,nn. they have 
∫exp()dx value 1 and saved.
<a name="ch14b168">　Index begin　Index this file
six black terms are unlike index 
terms Lm,mn,nL. they have 
∫exp()dx value 0 and dropped.
<a name="ch14b169">
The final result is
  ∫[x=0,1]sum*sum_conjugate*dx
 =cLcL+cmcm+cncn ---eqn.BR030
eqn.BR030 is a special case of 
eqn.14.22 specify n=3. It is
just a check work.
2010-08-02-09-00 stop

<a name="ch14b170">
2010-08-02-10-10 start
eqn.14.22 is integral of SQUARE 
of sum of complex sequence.
instead of SQUARE, now take
integral of sum of complex seq.
We do not square. Schwarz square 
it. Apply Schwarz inequality 
eqn.1.19

<a name="ch14b172">
Integral of 12 is one
Integral of |Sn(x)dx|2 apply
eqn.14.22 get
(|c1|2+|c2|2+...+|cn|2)
Take square root get eqn.BR031
greater than side value.
<a name="ch14b173">
We assumed
  |ck|≦1 for all 1≦k≦n  ---eqn.BR022
eqn.BR031 tell us that the 
integral of |Sn(x)| is bounded
by (|c1|2+|c2|2+...+|cn|2)1/2
<a name="ch14b174">
Each |ck| has maximum value 1
there are total n of them. We 
get ∫[x=0,1]|Sn(x)|dx is bounded 
by √n. Textbook page 217 line 13.
2010-08-02-10-57 draft work stop

<a name="ch14b175">　Index begin　Index this file
2010-08-02-12-01 start
■　Problem 14.4 (Rademacher-
　　Menchoff Inequality)
Textbook page 217 second half page.

<a name="ch14b176">
Given that the functions
  φk:[0,1] to complex 1≦k≦n ---eqn.BR032
are orthonormal. Show that the
partial sum
  Sk(x)=c1φ1(x)+c2φk(x) ---eqn.BR033
   + ... +ckφk(x) ; 1≦k≦n
satisfy the maximal inequality

<a name="ch14b178">
This is known as Rademacher-
Menchoff Inequality, and it is
surely among the most important 
results in the theory of
orthogonal series.
2010-08-02-12-26 stop

2010-08-02-12-37 draw
f(x)=
log(4*x)*log(4*x)/log(2)/log(2)

2010-08-02-15-06 start coding
2010-08-02-17-09 done  coding

<a name="ch14b179">
2010-08-02-18-00 start
If change base 2 to base 3
Then there are gap that base 3 
can not fill. For example
express number 5 in
Sum of Power of 3, find
5<3^2=9 //can not use 3^2
5>3^1=3 //3^1 can be summed
5=3+2=3^1+2
<a name="ch14b180">
here '2' can not be expressed 
by any integer power of 3.

Only 2 can be a reasonable base
for "Integer=Sum of Power of 2"
2010-08-02-18-03 stop

<a name="ch14b181">　Index begin　Index this file
2010-08-03-09-40 start
Above said (residual)
'2' can not be expressed 
by any integer power of 3.
then what do we get if 
express 5(10 base) by 3 base
system?
<a name="ch14b182">
5(10 base) in binary is 101
5(10 base) in 3 base system 
     is 12 = 5(3 base)
we can express 5 as
5 = 1*3^1 + 2*3^0 = 3+2 = 5

<a name="ch14b183">
But the spirit of
"Integer=Sum of Power of 2"
is all coefficient be either
one or zero.
<a name="ch14b184">
If we allow coefficient be 
greater than one, then any
integer can be expressed by
Sum of Power of any other 
integer, for example
<a name="ch14b185">
4 can be sum of 5^0, just 
repeat four times.
  4=5^0+5^0+5^0+5^0
Doing this way, the spirit of
"Integer=Sum of Power of 2"
is gone.
2010-08-03-09-48 stop

<a name="ch14b187">　Index begin　Index this file
2010-08-03-10-19 start
Read textbook Problem 14.4 
several times. LiuHH can not 
figure out why eqn.14.23 
require max[1≦k≦n]|Sk2(x)|
Requirement "max[1≦k≦n]" say
that k=n is not maximum.
<a name="ch14b188">
eqn.14.22 say Sk2(x) value 
is sum of |cj|2. j∈A
|cj|2 can not be negative.
Maximum value should be k=n 
for the whole sequence. Then
no need max() function.

<a name="ch14b189">
Second point LiuHH do not
understand is that for 27
elements complex sequence
Textbook divide 27 into
[1,27]={1,2,3,....,15,16}
      ∪{17,18,...,24}
      ∪{25,26}∪{27}
textbook page 219 line 17.
<a name="ch14b190">
{1,2,3,....,15,16} has 
2^4=16 elements.
{17,18,...,24} has 2^3=8 
elements.
{25,26} has 2^1=2 elements.
{27}    has 2^0=1 element.
<a name="ch14b191">
The maximum sum value is k=n.
Such two base power division 
contribute to what ? 
If division is necessary,
why choose 2^(inetger) power
as unit? If choose every 5
elements as a unit, what 
trouble do we get?

<a name="ch14b192">
LiuHH need more time to think,
before find out the answer,
Problem 14.4 is NOT DONE
2010-08-03-10-37 stop

2010-08-04-13-56 done first proofread
during this proofread, add [SP3]
button and write 
function sumOfPow2Bf()
which is simpler than
function sumOfPow2f()

2010-08-04-20-48 done second proofread
2010-08-04-21-05 done spelling check

<a name="ch14b193">
2010-08-05-17-50 start
LiuHH have question about
van der Corput Inequality.
LiuHH focus on the bound of
van der Corput Inequality.

<a name="ch14b194">
Inequality have two categories
One is bound inequality, the
other is none bound inequality.
Bound inequality has main role
to find a bound for a target 
sequence. The calculation of 
bound should be simpler than 
the calculation of target 
sequence. eqn.14.7 is clearly
a bound inequality. 

<a name="ch14b195">
None bound inequality example
is AM-GM inequality. In AM-GM
which bound which? that is not
important. The importance of
AM-GM inequality is that two
average method give us two
different answer. One (AM) is
certainly greater than or
equal to the other (GM).

<a name="ch14b196">
LiuHH guess 
van der Corput Inequality is 
none bound inequality. 
Focus on complicated upper 
bound, that is not the main 
point (LiuHH guess). 
But what is the key point of 
van der Corput Inequality? 
LiuHH is still thinking.
2010-08-05-18-06 stop



<a name="docB001">　Index begin　Index this file
2010-09-22-19-40 start
Update 2010-09-22 document.
From 2010-09-19 to next few days,
Liu,Hsinhan modify complex2.htm.
One work is add a longer example
(click button 51) This example is
taken from tute0050.htm#ch14b054.
After carefully read this code,
<a name="docB002">
LiuHH found the code was wrong.
2010-09-20 & 21 made correction.
Update 2010-09-22 re-write above
code for CSMC eqn.14.14.
The corrected code LiuHH still 
can NOT promise to be correct. 
Simply no other body proofread, 
no other similar program run for 
comparison. 
<a name="docB003">
The new code added "&&1==1" and 
"&&2==2" to bring in '=' which 
tell old complex2.htm not to echo 
print this line. Next few days 
LiuHH will update complex2.htm. 
New version do not need "&&1==1".
2010-09-22-19-50 stop

<a name="docB004">
2010-09-25-20-18
Update 2010-09-25 delete
var realesum=cgetr(esum);
at ch14b065



<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