<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="ch14a001">
2010-07-17-18-08 start
■■Chapter 14: Cancellation 
　　and Aggregation

<a name="ch14a002">　Index begin　Index this file
■　Problem 14.1 
　　(Abel's Inequality)
Textbook page 208

<a name="ch14a003">
Let z1,z2,...,zn denote a 
sequence of complex numbers
with partial sums
  Sk=z1+z2+...+zk ---eqn.BQ001
1≦k≦n. 
<a name="ch14a004">
For each sequence of 
real numbers such that 
  a1≧a2≧...≧an≧0 ---eqn.BQ002
one has 
  |a1z1+a2z2+...+anzn|
 ≦a1*max[1≦k≦n]|Sk| ---eqn.14.1
//Abel's Inequality key point 
//partial sum of data sequence.
2010-07-17-18-20 stop

<a name="ch14a005">　Index begin　Index this file
2010-07-17-18-54 start
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="ch14a006">
//201007171835
//program start here
//define real number coefficient
//require a1≧a2≧a3≧a4≧a5≧0
a1=5
a2=4
a3=3
a4=2
a5=1
//<a name="ch14a007">
//define arbitrary five complex 
//numbers in string form.
c1='-0.7260+5.7410i'
c2=' 1.6399-15.759i'
c3=' 11.229-11.630i'
c4=' 9.9490+16.750i'
c5='-5.9902+0.1433i'
//<a name="ch14a008">
//you can change above complex 
//numbers and a1 to a5 values.
//
//<a name="ch14a009">
//toCplx() convert string complex
//to array complex [-0.7, 5.7]
z1=toCplx(c1)
z2=toCplx(c2)
z3=toCplx(c3)
z4=toCplx(c4)
z5=toCplx(c5)
//
//<a name="ch14a010">　Index begin　Index this file
//start build eqn.14.1 less than side
//cmulr() is complex multiply real
//result save to y1 
y1=cmulr(z1,a1); 
//caddf(): complex add complex
//easy error:    '1+2i' + '3+4i'  NO
//must use caddf('1+2i' , '3+4i') OK
//<a name="ch14a011">
y2=caddf(y1,cmulr(z2,a2))
y3=caddf(y2,cmulr(z3,a3))
y4=caddf(y3,cmulr(z4,a4))
y5=caddf(y4,cmulr(z5,a5))
//above y5 is still complex. 
//cabsf() take absolute value
t5=cabsf(y5)
//done build eqn.14.1 less than side
//
//<a name="ch14a012">
//start eqn.14.1 greater than side
s1=z1;
s2=caddf(z2,s1)
s3=caddf(z3,s2)
s4=caddf(z4,s3)
s5=caddf(z5,s4)
//above five partial sum
//<a name="ch14a013">
//below find their absolute value
u1=cabsf(s1)
u2=cabsf(s2)
u3=cabsf(s3)
u4=cabsf(s4)
u5=cabsf(s5)
//<a name="ch14a014">
//five partial sum, five absolute
//value. Choose the maximum value
max0=max(u1,u2,u3,u4,u5)
//textbook page 208, eqn.14.1
//greater than side
maxa=a1*max0
//done eqn.14.1 greater than side
//
//<a name="ch14a015">
//t5 is eqn.14.1 less than side
t5   //smaller |a1z1+...+a5z5|
maxa //greater a1*max|Sk|
//201007171922


//<a name="ch14a016">
//Use next data at beginning
//to get equality output.
//all same coefficient value
a1=5
a2=5
a3=5
a4=5
a5=5
//<a name="ch14a017">
//all positive, no cancellation
c1=' 0.7260+5.7410i'
c2=' 1.6399+15.759i'
c3=' 11.229+11.630i'
c4=' 9.9490+16.750i'
c5=' 5.9902+0.1433i'
//program end here
//2010-07-17-19-31


<a name="ch14a018">　Index begin　Index this file
2010-07-17-19-35
above program last two lines 
do not have '=', program echo 
print variable value. This 
function can be found ONLY in
http://freeman2.com/complex2.htm
<a name="ch14a019">
If you code one line one 'else'
get trouble. (in complex2.htm)
Because 'else' is reserved word, 
'else' can not be variable name. 
Next line 'else' OK, it has '='.
 else { dummy=0; ...etc.

<a name="ch14a020">
max() is standard to javascript.
any browser support max(). But
cmulr(), caddf(), cabsf() etc
are complex handling functions.
You can use above commands in
http://freeman2.com/complex2.htm
(and complex1.htm), not in other 
page, because other page may use
different complex function name
or even not define complex
function.
2010-07-17-20-02

<a name="ch14a021">
2010-07-17-21-09
Other complex calculator web page
give you principle value only.
But
http://freeman2.com/complex2.htm
give you principle value AND
non-principle value.
2010-07-17-21-10

<a name="ch14a022">
2010-07-17-21-35 start
eqn.14.1 inequality less than
side is
|a1z1+a2z2+...+anzn| ---eqn.BQ003
it is easy to evaluate.
Greater than side is
a1*max[1≦k≦n]|Sk| ---eqn.BQ004
We need choose a maximum value 
from |S1|, |S2|, ... , |Sn|.

<a name="ch14a023">　Index begin　Index this file
Above example use five complex
numbers, they 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="ch14a024">
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="ch14a025">
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="ch14a026">
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|
<a name="ch14a027">
We can compare these five real
numbers and find maximum value.
Absolute value of partial sums
do not have monotone increase
property. For example,
S4 is sum of four complex numbers
S5 is sum of five complex numbers
We can NOT say |S5|≧|S4|
because opposite sign cancel.
<a name="ch14a028">　Index begin　Index this file
If S4=5+6i
If z5= -1-2i then
S5=S4+z5=5+6i+(-1-2i)
S5=4+4i
|S5|=5.6568≦7.8102=|S4|
One of |S1|,|S2|,|S3|,|S4|,|S5|
have maximum value. 
Not necessary the last one.
<a name="ch14a029">
The following is calculation
start from eqn.14.1 less than
side end up at greater than 
side. Start from
  a1z1+a2z2+...+anzn ---eqn.BQ020
eqn.BQ020 is eqn.BQ003 but dropped 
absolute sign.
<a name="ch14a030">
Find z's and S's relation
eqn.BQ010 say z1=S1, easy
eqn.BQ011 say z2=S2-S1 ---eqn.BQ021
eqn.BQ012 say z3=S3-S2 ---eqn.BQ022
eqn.BQ013 say z4=S4-S3 ---eqn.BQ023
eqn.BQ014 say z5=S5-S4 ---eqn.BQ024
<a name="ch14a031">
In eqn.BQ020, change z's to S's
get //set n=5
  a1z1+a2z2+a3z3+a4z4+a5z5
 = ---eqn.BQ025
  a1(S1)+a2(S2-S1)+a3(S3-S2)
        +a4(S4-S3)+a5(S5-S4)
 = //regroup
  S1(a1-a2)+S2(a2-a3)
 +S3(a3-a4)+S4(a4-a5)+S5(a5)
<a name="ch14a032">
Here is key step. There are
five different S's. We replace 
all of S's by max[1≦k≦5]|Sk|
This replacement generate
inequality. After replace by
max[1≦k≦5]|Sk| eqn.BQ025 right
side value increased. Also 
<a name="ch14a033">　Index begin　Index this file
max[1≦k≦5]|Sk| is common to all
terms, we move max[1≦k≦5]|Sk|
out to front. Find
  |a1z1+a2z2+...+anzn| ---eqn.BQ026
 ≦ max[1≦k≦5]|Sk|*
   |(a1-a2) + (a2-a3) + (a3-a4)
  + (a4-a5) + (a5)|
//2010-08-05-16-35 add '|' for abs()
<a name="ch14a034">
Here we find cancellation
 red   cancel red
 blue  cancel blue
purple cancel purple
black  cancel black
Only term left is a1
<a name="ch14a035">
The final result is
  |a1z1+a2z2+...+a5z5| ---eqn.BQ027
 ≦ max[1≦k≦5]|Sk|*a1
//2010-08-05-16-36 add '|' for abs()

<a name="ch14a036">
Above is for case n=5.
Change '5' to 'n' above analysis
is still valid for general 'n'.
The proof of Abel's inequality 
is complete.
2010-07-17-22-36 stop

<a name="ch14a037">　Index begin　Index this file
2010-07-18-12-30 start
■　Abel’s Summation by Parts 
　　Formula
Textbook page 208 line -2,-1
mention Summation by Parts.

<a name="ch14a038">
Start from integration by parts.
Assume uppercase functions
F(x) and G(x) are both analytic 
equations. For example,
assume
  F(x)=log(x) ---eqn.BQ028
  G(x)=x*x ---eqn.BQ029
<a name="ch14a039">
Assume their differentiation are
lowercase functions f(x) and g(x)
That is
  f(x)=d[F(x)]/dx ---eqn.BQ030
  g(x)=d[G(x)]/dx ---eqn.BQ031
For above examples, find
  f(x)=d[log(x)]/dx=1/x ---eqn.BQ032
  g(x)=d[x*x]/dx=2*x ---eqn.BQ033

<a name="ch14a040">
Integration by parts formula is
  ∫d[F(x)G(x)] = ∫d[F(x)]*G(x) ---eqn.BQ034
                +∫F(x)*d[G(x)]
The equality is product rule
of differentiation.
From eqn.BQ030 find
  d[F(x)]=f(x)dx ---eqn.BQ035
From eqn.BQ031 find
  d[G(x)]=g(x)dx ---eqn.BQ036
<a name="ch14a041">
re-write eqn.BQ034 as next
  ∫d[F(x)G(x)] = ∫G(x)*f(x)dx ---eqn.BQ037
                +∫F(x)*g(x)dx
eqn.BQ037 left side is a total
differentiation. Assume 
lower bound is x=a
upper bound is x=b
<a name="ch14a042">
Total differentiation is same
as next
  ∫d[F(x)G(x)]=F(b)G(b)-F(a)G(a) ---eqn.BQ038
Put eqn.BQ037 and eqn.BQ038
together, get
  F(b)G(b)-F(a)G(a)  ---eqn.BQ039
 =∫G(x)*f(x)dx + ∫F(x)*g(x)dx
It is better write eqn.BQ039 as
<a name="ch14a043">　Index begin　Index this file
integration by parts formula
  ∫F(x)*g(x)dx=  ---eqn.BQ040
  F(b)G(b)-F(a)G(a)-∫G(x)*f(x)dx
Beginning of this derivation
Previous similar work
<a name="ch14a044">
The term "F(b)G(b)-F(a)G(a)"
has nothing to do with the
interior of domain [a,b].
"F(b)G(b)-F(a)G(a)" is a 
function of boundary x=a and 
x=b only. In some case (for
example, calculus of variation)
the boundary terms become zero
which simplify the right side 
result.

<a name="ch14a045">
Now put eqn.BQ028 to eqn.BQ033
into eqn.BQ040 as a check. Get
  ∫log(x)*[2*x]dx=
  b*b*log(b)-a*a*log(a)
 -∫x*x*(1/x)dx ---eqn.BQ041

<a name="ch14a046">
Simplify eqn.BQ041 as following
  ∫x*log(x)dx=  ---eqn.BQ042
 (b*b*log(b)-a*a*log(a)-b*b/2+a*a/2)/2
Because there is '1/x', it is 
necessary that integration
domain be positive real. If 
x=a=0, log(a)=log(0) undefined
2010-07-18-13-06 here

<a name="ch14a047">　Index begin　Index this file
Please goto integral.htm#NumIntegral
assume a=1, b=3. Integration
equation is ∫  x*log(x)  *dx
<a name="ch14a048">
fill in the following data
[[
g(x)   x*log(x)
steps  5
x-bgn  1
x-end  3
Newton-Cotes formula n=6
]]
<a name="ch14a049">
output next
[[
From x=1 to x=3
integration value is 2.9437552990474165
]]

<a name="ch14a050">
Go to complex2.htm#calculator
fill in
[[
a=1
b=3
(b*b*log(b)-a*a*log(a)-b*b/2+a*a/2)/2
]]
to Box3, then 
<a name="ch14a051">
click
[Test box3 command, output to box4]
output to box 4
[[ //red analytic, blue numerical
2.9437552990064937 //eqn.BQ042 right side
]]
2.9437552990474165 //eqn.BQ042  left side

<a name="ch14a052">
Numerically confirmed eqn.BQ041
is correct for one domain [1,3]
2010-07-18-13-14 stop

The following is reading of
next two web pages
2010-04-25-15-24 LiuHH access
http://www.math.uconn.edu/~kconrad/math121/121sumbypartsproj.pdf
2010-04-25-16-04 LiuHH access
http://fractal.math.unr.edu/~ejolson/311-06/handout/sumpart.pdf

<a name="ch14a053"> derive:1 2 3 
2010-07-18-19-40 start
Above is integration by parts.
Next is 
Abel’s Summation by Parts Formula
Given two real number sequences
each have total n elements.
a.seq: a1,a2,...,an and 
b.seq: b1,b2,...,bn
<a name="ch14a054">　Index begin　Index this file
Define uppercase A and B as the
partial sum of a.seq and b.seq
  Ak=∑[i=1,k]ai ---eqn.BQ043
  Bk=∑[i=1,k]bi ---eqn.BQ044
where 1≦k≦n. Define
  A0=0 ---eqn.BQ045
  B0=0 ---eqn.BQ046
<a name="ch14a055">
Assume k=5, from eqn.BQ043 find
  A5=∑[i=1,5]ai ---eqn.BQ047
  A5=a1+a2+a3+a4+a5 ---eqn.BQ048
Similarly, Assume k=6, find
  A6=∑[i=1,6]ai ---eqn.BQ049
  A6=a1+a2+a3+a4+a5+a6 ---eqn.BQ050
<a name="ch14a056">
From eqn.BQ048 and eqn.BQ050
it is easy to see that
  a6=A6-A5 ---eqn.BQ051
In general
  aj=Aj-Aj-1 ---eqn.BQ052
Similarly
  bj=Bj-Bj-1 ---eqn.BQ053

<a name="ch14a057">
Uppercase Aj, Bk are partial sum,
they are similar to integral in 
a domain.

<a name="ch14a058">　Index begin　Index this file
eqn.BQ039 is equation for
integration by parts, it has
  F(b)G(b)-F(a)G(a)  ---eqn.BQ054
//∫[x=a,b]f(x)dx ; a=begin,b=end
//∑[k=m,n]ak ; m=begin,n=end
Similar term in summation by 
parts, first guess is
  An*Bn-Am*Bm ---eqn.BQ055 //error
But actually we will get
  An*Bn-Am-1*Bm-1 ---eqn.BQ056
We need define A0=B0=0 for m=1
<a name="ch14a059">
In order to reach eqn.BQ056,
we start from the following
  Ai*Bi-Ai-1*Bi-1 ---eqn.BQ057
Here insert a zero
  Ai*Bi-Ai-1*Bi-1
 =  ---eqn.BQ058
  Ai*Bi-Ai-1*Bi-1
 +Ai*Bi-1-Ai*Bi-1
<a name="ch14a060">
 =  ---eqn.BQ059
  Ai*Bi-Ai*Bi-1
 +Ai*Bi-1-Ai-1*Bi-1
 =  ---eqn.BQ060
  Ai*(Bi-Bi-1)
 +(Ai-Ai-1)*Bi-1
<a name="ch14a061">
eqn.BQ053 tell us that 
blue term Bi-Bi-1=bi ---eqn.BQ061
eqn.BQ052 tell us that 
purple term Ai-Ai-1=ai ---eqn.BQ062
<a name="ch14a062">
Now eqn.BQ057 to eqn.BQ062
give us
  Ai*Bi-Ai-1*Bi-1 ---eqn.BQ063
 =Ai*bi+ai*Bi-1
Be alert that 
uppercase A and B are 
          partial sum
lowercase a and b are 
          sequence elements.

<a name="ch14a063">　Index begin　Index this file
Next sum eqn.BQ063 from i=m 
to i=n, where m＜n, we get

<a name="ch14a065">
Left side of equality has
cancellation,
   red cancel red
  blue cancel blue
purple cancel purple
all the way to the end.
<a name="ch14a066">
Summation of left side is
  +An+0*Bn+0－Am+0-1*Bm+0-1
or
  +An*Bn－Am-1*Bm-1 ---eqn.BQ066
left side = right side, get
  +An*Bn－Am-1*Bm-1 =

<a name="ch14a068">
2010-07-18-20-56 here
Both red and blue are eqn.BQ064
right side.
eqn.BQ067 red terms are summation
in next expression
∑[k=m,n]Ak*bk ---eqn.BQ068
eqn.BQ067 blue terms are summation
in next expression
∑[k=m,n]ak*Bk-1 ---eqn.BQ069

<a name="ch14a069">　Index begin　Index this file
eqn.BQ064 is next combination
eqn.BQ066=eqn.BQ068+eqn.BQ069
finally we get 

<a name="ch14a071">
2010-07-18-21-23 here
Again, remember:
uppercase A and B are partial sum
defined at eqn.BQ043 to eqn.BQ046
lowercase a and b are sequence 
elements.
<a name="ch14a072">　Index begin　Index this file
Integration by parts formula
eqn.BQ040 does not look like
summation by parts formula
eqn.BQ070 
Rewrite summation by parts formula 
as following and compare with 
integration by parts.

<a name="ch14a074">
2010-07-18-21-57 here
Please pay attention to 
Summation by parts eqn.BQ072
index range.
Left side index k=m to k=n
Right side, boundary term has
－Am-1*Bm-1, instead of －Am*Bm
<a name="ch14a075">
Left side sum Ak*bk have same
index k, but
right side sum ak*Bk-1 has 
shifted index k and k-1
2010-07-18-22-06 stop

Above is reading of next two pages
http://www.math.uconn.edu/~kconrad/math121/121sumbypartsproj.pdf
http://fractal.math.unr.edu/~ejolson/311-06/handout/sumpart.pdf

<a name="ch14a076">　Index begin　Index this file
2010-07-20-09-45 start
■　Application of Abel's
　　inequality

<a name="ch14a077">
Let us find the upper bound of
Q, which is defined below.

<a name="ch14a078">
Abel's inequality need two
sequences. One is data z sequence
 z1,z2,...,zn 
contribute to partial sum
Second is coefficient a sequence
  a1≧a2≧...≧an≧0 ---eqn.BQ002
they are all nonnegative and
in descending order.

<a name="ch14a079">
eqn.BQ074 has two sequences
-1,+1,-1,+1,...,-1,+1...
1/√1,1/√2,1/√3,...,,1/√n,...
which is z sequence?
which is a sequence?
<a name="ch14a080">
Since coefficient a sequence 
must be nonnegative and
must be in descending order
therefore 
coefficient 'a' sequence is
1/√1,1/√2,1/√3,...,,1/√n,...
data z sequence is
-1,+1,-1,+1,...,-1,+1...

<a name="ch14a081">　Index begin　Index this file
Abel's inequality say two
sequence dot product is less
or equal to maximum coefficient
multiply by absolute value of
data sequence maximum partial 
sum.
<a name="ch14a082">
coefficient sequence 
1/√1,1/√2,1/√3,...,,1/√n,...
has maximum value at its first 
element 1/√1.
About maximum partial sum of
-1,+1,-1,+1,...,-1,+1...
<a name="ch14a083">
it is easy to see the partial 
sum sequence is
-1, 0,-1, 0,-1, 0,-1, 0,...
absolute value of maximum 
partial sum is 1.
<a name="ch14a084">
Then eqn.BQ074 Q value is
less than or equal to
maximum coefficient 1/√1
multiply by
maximum data partial sum 1
That is
  Q≦1 ---eqn.BQ075
<a name="ch14a085">
If change problem a little
bit. Sum lower limit change 
from k=1 to k=M.
Sum upper limit change 
from k=∞ to k=N with
  1≦M≦N＜∞
then similar analysis get

<a name="ch14a087">
2010-07-20-10-39 here
Next see second application
of Abel's inequality. Define
R as following

<a name="ch14a089">
Sequence R also contain
two sequences
cos(kπ/6)   k=1,2,...,∞
and
1/log(k+1)  k=1,2,...,∞

<a name="ch14a090">
We know coefficient sequence 
must be nonnegative and
must be in descending order
cos(kπ/6) can NOT be coef. seq.
because cos(kπ/6) oscillate in
magnitude and change sign.
log(k+1) can be coefficient seq.
since 1/log(k+1) has descending 
magnitude and always positive.

<a name="ch14a091">　Index begin　Index this file
coefficient seq. is 1/log(k+1)
data seq. is cos(kπ/6)
Maximum coefficient occur at
k=1, that is 1/log(2)
<a name="ch14a092">
Maximum partial sum of data
sequence ?
Textbook say "by brute force"
(page 209, line -7) find
  max[M,N]|∑[k=M,N]cos(kπ/6)|
  =2+√3=3.73205080 ---eqn.14.3
2010-07-20-10-59 here

//<a name="ch14a093">
//2010-07-20-11-00
var m=0
var n=5555
var outp=''; //default short output
//outp='long'; //drop left end '//' for long output
var k;
var sum0=[];
k=m;
sum0[0]=cos(k*PI/6); //k=m=0
//<a name="ch14a094">
for(k=m+1;k<=n;k++)
sum0[k-m]=sum0[k-m-1]+cos(k*PI/6);
var ans0='';
var max0=[0,0];
for(k=0;k<sum0.length;k++)
{ if(outp)ans0+=sum0[k]+'\n';
//if(abs(sum0[k])>max0[1]){dummy=0; //9907232007 change
  if(abs(sum0[k])>abs(max0[1])){dummy=0;
  max0[0]=k;max0[1]=sum0[k];}
}
ans0=max0[0]+' has '+max0[1]+'\n'+ans0;
ans0 // #th partial sum has maximum
//2010-07-20-11-13

//<a name="ch14a095">
//2010-07-20-11-15 get
1970 has 2.3660254037848442
but 2+√3=3.73205080 
Above calculation may be wrong.

<a name="ch14a096">　Index begin　Index this file
2010-07-20-13-16 start
Above program code is a "brute 
force" to find maximum value of
partial sum of data sequence.
From m=0 to n=5555 get partial 
sum of first 1970 elements to 
be 2.3660254037848442 which is
less than 2+√3=3.73205080 
<a name="ch14a097">
This result can not prove that
2+√3 is too loose. Because sum
to infinity. Future sum may be
bound by 2+√3. Now accept 2+√3
as maximum partial sum then
<a name="ch14a098">
Abel's inequality tell us that
R in eqn.BQ076 be bounded by
maximum coefficient 1/log(1+1)
multiply by
maximum data partial sum 2+√3
That is // next  ---eqn.BQ077
  R≦(2+√3)/log(1+1)=5.3842111

<a name="ch14a099">
If R summation start from k=M
end at k=N, then the bound is

<a name="ch14a100">
Textbook say above equation
the bound coefficient 2+√3
is sharp and can not be 
replaced by a smaller value.
Textbook page 109, line -1.
2010-07-20-13-40 here

<a name="ch14a101">　Index begin　Index this file
Before goto Problem 14.2,
define two symbols first.
exp(2πit) is simplified to e(t)
that is
  e(t)=exp(2πit) ---eqn.14.5A
here π=3.141592653589793...  ---eqn.BQ078
     i=√(-1) ---eqn.BQ079
and t is independent variable.
If t is real, 2πit is imaginary,
in this case |e(t)|=1.
If t is complex, then |e(t)|≠1

<a name="ch14a102">
Next define ∥t∥ to be // ---eqn.14.5B
  ∥t∥=min{|t-k|: k in integer Z}
t is any real number, t is in
between two integers, k is the
integer closer to t. 
<a name="ch14a103">
For example;
if t=-1.1 -2＜t＜-1 
∥t∥=min{|-1.1-(-1)|, |-1.1-(-2)|}
    =min{0.1, 0.9}=0.1 ---eqn.BQ080
Z include negative integer -1,-2.

<a name="ch14a104">
if t=+2.7  2＜t＜3
∥t∥=min{|2.7-(2)|, |2.7-(3)|}
    =min{0.7, 0.3}=0.3 ---eqn.BQ081

Here t is a pure number,
     t is not a vector
<a name="ch14a105">
Pure number ∥t∥ and
vector norm ∥v∥ both use 
double bar. Be careful about
the difference. They can not
be mixed. Because a pure
number is different from a
vector.
2010-07-20-13-57 stop

<a name="ch14a106">　Index begin　Index this file
2010-07-20-15-52 start
■　Problem 14.2 (Linear and
Quadratic Exponential Sums)

First as a useful warm-up, show 
that for all t in Real and all
integers M and N one has the
bounds

<a name="ch14a108">
Then, for a more engaging 
challenge, show that for
b,c in Real and all integers
0≦M＜N one also has a uniform
bound for the quadratic 
exponential sums

2010-07-20-16-17 stop

<a name="ch14a110">　Index begin　Index this file
2010-07-20-17-26 start
■　Euler's formula exp(iθ)

  e(t)=exp(2πit) ---eqn.14.5A
and eqn.14.6 both use exp(iθ)
Next is Euler's formula exp(iθ)
using Taylor's series expansion.

2010-07-20-17-54 here
<a name="ch14a113">
The question is:
Whether Taylor's series can be
used for complex number iθ ?
There is one view point
consider eqn.BQ082 as the
definition equation for eiθ
We define eiθ to be eqn.BQ082
No need to prove eqn.BQ082, 
no need worry whether Taylor's 
series can be used for complex.
2010-07-20-18-00 stop

<a name="ch14a114">　Index begin　Index this file
2010-07-20-19-28 start
Next show
(xn-1)/(x-1)=xn-1+xn-2+...+1 ---eqn.BQ083
We show next equivalent equation
(xn-1+xn-2+...+1)*(x-1)=xn-1 ---eqn.BQ084
For convenience, set n=5

<a name="ch14a116">　Index of Problem 14.2
Above is same as next
(x4+x3+x2+x+1)*(x-1)=x5-1 ---eqn.BQ086
If change 5 to n, we get general
result eqn.BQ083 or eqn.BQ084.
2010-07-20-19-59 here

<a name="ch14a117">
Now return to problem 14.2
for Linear Exponential Sums
Start from eqn.14.6 less than 
side. //Quadratic

<a name="ch14a121">　Index of Problem 14.2
2010-07-20-20-30 here
Factor "e(Nt)－1" as following
  e(Nt)=e((Nt)/2)*e((Nt)/2) ---eqn.BQ089
this is like 2=(√2)*(√2)
    1  =e((Nt)/2)*e((-Nt)/2) ---eqn.BQ090
this is like 1=3*1/3. 
<a name="ch14a122">
Then
  e(Nt)－1=e((Nt)/2)*e((Nt)/2)
         -e((Nt)/2)*e((-Nt)/2) ---eqn.BQ091
purple part is common factor.
  e(Nt)－1=e((Nt)/2)
  *[e((Nt)/2)-e((-Nt)/2)] ---eqn.BQ092
<a name="ch14a123">
Similarly for N=1 we have
  e(t)－1=e((t)/2)
  *[e((t)/2)-e((-t)/2)] ---eqn.BQ093
Now put eqn.BQ092 and eqn.BQ093
into eqn.BQ088 get textbook 
page 211 line 10 equation next.

<a name="ch14a125">
2010-07-20-20-50 here
eqn.BQ094 left end term come 
from eqn.14.6. It has absolute 
value. Now in eqn.BQ094, we take
absolute value for whole equation
<a name="ch14a126">　Index of Problem 14.2
Red terms in eqn.BQ094 and black
term left to red all have absolute 
value one. 
see Euler's formula eqn.BQ082

<a name="ch14a128">
2010-07-20-21-13 here
Explain equality in eqn.BQ095,
result is eqn.BQ098, BQ099 below.
eqn.BQ094 blue part cancel 2i.
e(t) is defined as next
  e(t)=exp(2πit) ---eqn.14.5A
then
  e(Nt/2)－e(－Nt/2) ---eqn.BQ096
 =exp(2πit*N/2)-exp(-2πit*N/2)
 =exp(πit*N)-exp(-πit*N)
<a name="ch14a129">　Index begin　Index this file
apply Euler's formula eqn.BQ082 
to eqn.BQ096
  e(Nt/2)－e(－Nt/2) ---eqn.BQ097
 =[cos(+πNt)+i*sin(+πNt)]
 -[cos(-πNt)+i*sin(-πNt)]
 =[cos(πNt)+i*sin(+πNt)
 - cos(πNt)+i*sin( πNt)]
<a name="ch14a130">
  e(Nt/2)－e(－Nt/2) ---eqn.BQ098
 =2i*sin( πNt)
Similarly, for N=1 case
  e( t/2)－e(－ t/2) ---eqn.BQ099
 =2i*sin( πt)

<a name="ch14a131">　Index of Problem 14.2
eqn.BQ094 blue part become
  |sin( πNt)/sin( πt)| ---eqn.BQ100
eqn.BQ100 is eqn.BQ095 middle 
term.
Because |sin( πNt)|≦1
drop |sin( πNt)| cause inequality.
drop |sin( πNt)| get eqn.BQ095
right side which is free from
the bound N.
<a name="ch14a132">
We get eqn.14.6 left inequality.
If sin( πt) approach to zero
 1/sin( πt) approach to infinity.
min{N, 1/sin( πt)} cut off the
infinity and bound by N.
2010-07-20-21-33 stop

<a name="ch14a133">　Index begin　Index this file
2010-07-21-11-08 start
About eqn.14.6 right side 
inequality, textbook page 211
line 13 say
[[
<a name="ch14a134">
Finally, to get the second part 
of the bound (14.6), one only
needs to notice that the graph
of t→sin(πt) makes it obvious 
that
  2∥t∥≦|sin(πt)| ---eqn.BQ101
]]
2010-07-21-11-14 stop

<a name="ch14a135">
2010-07-21-12-38 start
The second part of bound (14.6)
is 
  1/|sin(πt)| ≦ 1/(2∥t∥) ---eqn.BQ102
which is same as eqn.BQ101.
Textbook say "notice the graph"
<a name="ch14a136">　Index of Problem 14.2
LiuHH write a function 
x_lt_sin() to find 2∥t∥ value 
and |sin(πt)| value.
Please goto (local)
http://freeman2.com/sumpart2.htm
Control panel look like next
<a name="ch14a137">
[[

]]
<a name="ch14a138">　Index begin　Index this file
You can fill in values for
x bgn  ; x step  ; x end 
for example
x bgn 0 ; x step 0.25 ; x end 5
then click [Build Seq.] 
<a name="ch14a139">
Program build x seq. at box 4. 
It is next
0,0.25,0.5,0.75,1,1.25,1.5,1.75,
2,2.25,2.5,2.75,3,3.25,3.5,3.75,
4,4.25,4.5,4.75,5
<a name="ch14a140">
If click [2∥x∥≦|sin(πx)|] button
above x seq. is internal data
not show up, box 4 is other
information (Not x seq.)

<a name="ch14a141">　Index of Problem 14.2
click [2∥x∥≦|sin(πx)|] //has π
Box 1,2,3,4 all have output.
Main output is box 3, if you
see a "***" at line end, 
that line data violate
2*∥x∥≦|sin(πx)| 
"***" line has the reverse 
2*∥x∥＞|sin(πx)| 

<a name="ch14a142">
Box 1 output sin(πx) value
Box 2 output 
2*∥x∥, double box4 "*min" value
Box 4 output 
ceil,floor,x,xc,xf; *=min
They are for debug purpose.

<a name="ch14a143">　Index begin　Index this file
Output confirmed that 
textbook page 211 line 13 
  2∥t∥≦|sin(πt)| ---eqn.BQ101
is true.

<a name="ch14a144">
During coding, at first time
LiuHH code  2∥t∥≦|sin(t)| //no π
instead of  2∥t∥≦|sin(πt)|
LiuHH forget π ! then output
violate textbook equation.
 Second click button
[2∥x∥?≦?sin(x)] 
has no π, if you click this 
one, box 3 will show "***"

<a name="ch14a145">
There are two set default 
input. 
First if click [dx] delete 
x bgn  ; x step  ; x end 
three boxes, then click 
[2∥t∥≦|sin(πt)|] button
<a name="ch14a146">　Index of Problem 14.2
program use
x bgn 0.1 ; x step sqrt(PI) ; x end E*E*PI
that is
xbgn=0.1
xstp=1.7724538509055158
xend=23.213404357363384

<a name="ch14a147">
Second if click [11] button
program fill
x bgn 0.5 ; x step 1 ; x end 8
x sequence is
0.5,1.5,2.5,3.5,4.5,5.5,6.5,7.5
Click [2∥x∥≦|sin(πx)|]  all equal
click [2∥x∥?≦?sin(x)] all violate

<a name="ch14a148">　Index begin　Index this file
If you fill in
x bgn 0. ; x step 1 ; x end 8
then 2∥x∥ are all zero, because
x sequence 0,1,2,3,4,5,6,7,8
each one has zero distance from
an integer. This is a surprise
LiuHH saw at initial coding time.

<a name="ch14a149">
Finally,
above is just SHOW 2∥x∥≦|sin(πx)|
above is not a proof.
2010-07-21-13-20 stop

<a name="ch14a150">
2010-07-21-14-36 start
Next prove 
  2∥t∥≦|sin(πt)| ---eqn.BQ101
variable t or x it does not matter.
If use x, same thing
  2∥x∥≦|sin(πx)|

<a name="ch14a151">　Index of Problem 14.2
Define real number t = n.f
where n is integer≦t
      f is fraction f=t-n
∥t∥ definition let ∥t∥ be a
slope 45 degree monotone increase 
straight line from n to n.5, 
For example from 7 to 7.5
<a name="ch14a152">
After pass n.5, the min{|t-k|} 
function assign value n+1-n.f
to ∥t∥ value. 
For example t=7.8 , n.f=7.8
where n=7, fraction f=0.8
7.8 ＞ n.5=7.5
∥t∥ get value n+1-n.f which is
7+1-7.8 = 0.2
<a name="ch14a153">　Index begin　Index this file
Above definition let ∥t∥ be a
hill-like, upside down v shape
broken line.
∥t∥ period is one
∥t∥ amplitude is 0.5

<a name="ch14a154">
Next see sin(t), it is easy
sin(t) period is 2*PI
sin(t) amplitude is one
modify to
|sin(t)| period is PI
|sin(t)| amplitude is one
modify to
|sin(PI*t)| period is one
|sin(PI*t)| amplitude is one

<a name="ch14a155">
Here
|sin(PI*t)| period is one
|sin(PI*t)| amplitude is one
and
∥t∥ period is one
∥t∥ amplitude is 0.5
are close in terms of period
and amplitude.
<a name="ch14a156">　Index of Problem 14.2
Modify ∥t∥ to next
2∥t∥ period is one
2∥t∥ amplitude is one

<a name="ch14a157">
We find
2∥t∥ period is one
2∥t∥ amplitude is one
and
|sin(PI*t)| period is one
|sin(PI*t)| amplitude is one
have same period and same
amplitude plus one more
<a name="ch14a158">　Index begin　Index this file
same function value at 
.f=0.0 and at .f=0.5
When t=0
2∥t∥=2∥0∥=0
|sin(PI*t)|=|sin(PI*0)|=0
When t=0.5
2∥t∥=2∥0.5∥=1
|sin(PI*t)|=|sin(PI*0.5)|=1
<a name="ch14a159">
Any integer multiple of 0 
and 0.5 be the same, because
both 2∥t∥ and |sin(PI*t)| 
are periodic function with
same period one.

<a name="ch14a160">
With three same properties,
next see convexity/concavity
property.
2∥t∥ is linear.
|sin(PI*t)| is concave ╭╮ .
A concave curve always have
curve value ≧ chord value.

<a name="ch14a161">　Index of Problem 14.2
Here we proved that
  2∥t∥≦|sin(πt)| ---eqn.BQ101
is true.
2010-07-21-15-12 stop

<a name="ch14a162"> Problem 14.2 
■　Quadratic exponential sum.
2010-07-21-18-40 start //Linear

eqn.14.7 has exponential of
  (k2+bk+c)/N ---eqn.BQ103
this is a quadratic equation 
of k and k is integer vary 
from k=1 to k=M.
<a name="ch14a163">　Index begin　Index this file
Define P(k) as next
  P(k)=αk2+βk+γ ---eqn.BQ104
α,β,γ are all real numbers.
k is -/0/+ integer (Z).
<a name="ch14a164">
If set
  α=1/N ---eqn.BQ105
  β=b/N ---eqn.BQ106
  γ=c/N ---eqn.BQ107
P(k) become eqn.14.7 exponential
part. 
<a name="ch14a165">
Follow eqn.14.7, define SM(P) 
as next
  SM(P)=∑[k=1,M]e(P(k)) ---eqn.BQ108
SM(P) is a short writing
SM(P) contains previous short 
writing e(t). Two short writing?
Complete expression for SM(P) 
in terms of exp() is next

<a name="ch14a167">
2010-07-21-19-15 here
where "2πi" is absorbed by e(t)
definition, not show up in e(t)
expression. Here i is imaginary
  i=√(-1) ---eqn.BQ079
If we adopt eqn.BQ105 to BQ107
definition for α,β,γ, then 
<a name="ch14a168">　Index begin　Index this file
SM(P) is equation 14.7 left side 
sum term. It is inside of 
absolute sign. If we start from 
SQUARE of SM(P), it is same
as taking absolute sign. At the
end we take square root, the
result is same as eqn.14.7 left
side (with absolute sign)
Next start from [SM(P)]2

<a name="ch14a169">
Textbook write eqn.14.8 exp() 
term as cn, that is define
  cn=exp(2πi[αn2+βn+γ]) ---eqn.BQ109
Alert: eqn.BQ109 use n in place 
of k in eqn.14.8.
(Wonder ! where is divide by N?)
eqn.BQ109 do not have divide by N.
After setting α=1/N, β=b/N, γ=c/N 
"/N" enter eqn.14.13 left side.
Sum cn and square the sum

<a name="ch14a171">　Index of Problem 14.2
2010-07-21-20-03 here
eqn.14.9 first line is expansion 
of square of a complex summation.
Those product with like index n 
contribute to sum of |cn|2
Those product with unlike index 
m＜n, contribute to sum of 
  {cm*cnj+cmj*cn} ---eqn.BQ110
complex conjugate of c is cj
<a name="ch14a172">
To understand eqn.14.9 line 2
set cm=(1+2i) ---eqn.BQ111
set cn=(3+4i) ---eqn.BQ112
cm*cnj=(1+2i)*(3-4i) ---eqn.BQ113
cmj*cn=(1-2i)*(3+4i) ---eqn.BQ114
Sum above two terms get
(1+2i)*(3-4i)+(1-2i)*(3+4i) ---eqn.BQ115
=1*3-2*4ii+2*3i-1*4i //imaginary cancelled
+1*3-2*4ii-2*3i+1*4i //imaginary cancelled
=1*3+2*4 //-ii=+1
+1*3+2*4
=2*(1*3+2*4)
=2*Re[(1+2i)*(3+4i)] ---eqn.BQ116
<a name="ch14a173">　Index begin　Index this file
imaginary cancelled, need just
real part and double.
2010-07-21-20-24 here
Above is the reason that eqn.14.9
change from line 1 to line 2.

<a name="ch14a174">
eqn.14.9 line 2 right summation
do not allow m=n. Because all
m=n goto line 2 left summation.
line 2 right summation require
1≦m＜n≦M
Gap   h=n-m ---eqn.BQ316
 is an integer. 
Gap h can be 1, for example
 m=5,n=6,M=8 ; h=1
Gap h can be M-1, for example
 m=1,n=8,M=8 ; h=7
<a name="ch14a175">
Write n as m plus gap h, 
  n=m+h ---eqn.BQ317
that is write cn*cmj as cm+h*cmj
complex conjugate of c is cj
m,n double summation change
to m,h double summation.
This is the reason of eqn.14.9
line 2 to line 3.
2010-07-21-20-37 here
<a name="ch14a176">　Index of Problem 14.2
If cn in eqn.14.9 is arbitrary
complex, |cn| do not have value
one. For example cn=1+2i
|cn|=√(1*1+2*2) not= 1

<a name="ch14a177">
If we specialize eqn.14.9 by 
setting //textbook p.212 L.5
  cn=e(P(n)) ---eqn.BQ117
The definition
  e(t)=exp(2πit) ---eqn.14.5A
exp(2πit) has everything "2πt"
multiply by i, this allow us 
apply Euler's formula
<a name="ch14a178">　Index begin　Index this file
// on the other hand, one drop 
// real 0.01 in exp(2πit+0.01)
// void this analysis
//cabsf(cexpf(caddf('2i','.01')))
//1.010050167084168
//cabsf(cexpf(caddf('2i','.0')))
//1
//|exp(ai+b)|=|exp(ai)|*|exp(b)|
//if real b=0, |exp(ai+0)|=1
//|exp(ai)| apply Euler formula
//|exp(b)| do NOT apply.
<a name="ch14a179">
Absolute value of Euler's formula
is sin(θ)*sin(θ)+cos(θ)*cos(θ)
that is ONE !
Textbook page 212 line 5 say 
setting cn=e(P(n)) then
next eqn.14.10 has only M
(sum 1 M times ∑[n=1,M]1 get M)
|cn|2 is gone, actually |cn|2 is 
hidden, value one never show up.

<a name="ch14a180">
After we require cn in eqn.14.9 
be 
  cn=e(P(n)) ---eqn.BQ117
equation 14.9 become next
  |SM(P)|2=M*12 //Hello cn ! you are one
 +2Re∑[h=1,M-1]∑[m=1,M-h]
  {e(P(m+h)━P(m))} ---eqn.14.10
2010-07-21-21-05 here

<a name="ch14a181">　Index of Problem 14.2
Why eqn.14.10 has P(m+h)-P(m)?
exp() of two multiplied terms
should be adding, not subtracting
Let us goto
  e(t)=exp(2πit) ---eqn.14.5A
<a name="ch14a182">
cn=e(P(n)) become exp(2πi*P(n))
cm+h=e(P(m+h)) become exp(2πi*P(m+h))
cm=e(P(m)) become exp(2πi*P(m))
Super j is complex conjugate
cmj=e(-P(m)) become exp(-2πi*P(m))

<a name="ch14a183">　Index begin　Index this file
P(k) is defined as next
  P(k)=αk2+βk+γ ---eqn.BQ104
P(k) is all real.
complex conjugate //θ is real
of exp(+i*θ) //cos(θ)+i*sin(θ) 
is exp(-i*θ) //cos(θ)-i*sin(θ) 
eqn.14.9 third line right end
term cmj complex conjugate cause 
negative sign '━' in eqn.14.10
2010-07-21-21-29 stop

<a name="ch14a184">
2010-07-22-13-15 start
eqn.14.10 look complicate. Is 
there a chance to simplify it?
Last term in eqn.14.10 is e()
of the difference of P(m+h) 
and P(m) {e(P(m+h)-P(m))}

<a name="ch14a185">
P(k) is defined as next
  P(k)=αk2+βk+γ ---eqn.BQ104
The term P(m+h)-P(m) cancel
quadratic term m*m !!
Calculate as following
<a name="ch14a186">　Index of Problem 14.2
  P(m+h)-P(m) //cancel quadratic mm
 =α(m+h)2+β(m+h)+γ ---eqn.BQ118
 -[α(m)2+β(m)+γ] //now, linear in m
 =α(mm+2mh+hh -mm)+β(m+h -m)+γ-γ
 =α(2mh+hh)+β(h)+0
<a name="ch14a187">
get textbook page 212 line 13
equation
  P(m+h)-P(m)=2αmh+αhh+βh ---eqn.BQ119
quadratic term m*m cancelled.
e() operation to P(m+h)-P(m)
get
<a name="ch14a188">　Index begin　Index this file
  e(P(m+h)-P(m))
 =e(2αmh+αhh+βh)
 =e(αhh+βh)*e(2αmh) ---eqn.BQ120
e(t) is defined as next
  e(t)=exp(2πit) ---eqn.14.5A
<a name="ch14a189">
Addition 2αmh+αhh+βh become
multiplication e(αhh+βh)*e(2αmh)
this is a property of exp()
function.
<a name="ch14a190">
eqn.14.9 third line right most
summation sum over m . 
eqn.BQ120 separate has_m e(2αmh)
and no_m e(αhh+βh) two terms.
no_m e(αhh+βh) has absolute value
equal to one.
has_m e(2αmh) is linear in m
(m*m cancelled)
2010-07-22-13-55 here
//why e(2αmh) become sin(π1αh) ?
//why '2' become '1' ?

<a name="ch14a191">　Index of Problem 14.2
2010-07-22-14-15 open
2009-02-01-15-27 received 
http://www-stat.wharton.upenn.edu/~steele/Publications/Books/CSMC/LangeListCSMCTypos.pdf
find
[[
<a name="ch14a192">
9. Page 212, equations (14.11) 
and (14.12). The denominator
 sin(πhα) is missing a factor
of 2 in its argument. This 
propagates to a missing 2 in 
the denominator of ∥hα∥ and
possibly further in the problem.
]]
2010-07-22-14-20 here
//why '2αmh' become sin(π1αh) ?
//why '2' become '1' ?
has answer: typo.

<a name="ch14a193">　Index begin　Index this file
2010-07-22-14-35 start
Now compare ∑[m=1,M-h]e(2αmh)
with eqn.BQ095
∑[m=1,M-h]e(2αmh) is eqn.14.10
inner summation. Variable is m
eqn.BQ095 use k as variable.
Sum of e(kt) ≦ 1/|sin(πt)|
k drop off from greater than side.
Apply eqn.BQ095 to eqn.14.10 get

<a name="ch14a195">
2010-07-22-14-47 here
Left side summation variable m
dropped at right side.
"A=" come from eqn.BQ120
"B=" come from |e(αhh+βh)|=1
<a name="ch14a196">　Index of Problem 14.2
eqn.14.11 inequality come from
eqn.BQ095
eqn.14.11 allow us replace the
inner summation ∑[m=1,M-h] of 
eqn.14.9 by an upper bound
1/|sin(π*2hα)|
Quadratic term eqn.14.9 reduce
to

<a name="ch14a198">　Index begin　Index this file
2010-07-22-15-34 here
eqn.14.12 first inequality come
from eqn.14.11
Big red 2 come from eqn.14.9 "2Re"
Small red 2 come from reciprocal 
of eqn.BQ101. Two red 2 cancel out.
Blue 2 from LangeListCSMCTypos.pdf
eqn.14.12 second inequality come
from Problem 14.2 given condition
0≦M＜N and come from eqn.BQ101
eqn.BQ101 result take reciprocal
Reverse inequality for eqn.14.12

<a name="ch14a199">
Observe 
  2∥t∥≦|sin(πt)| ---eqn.BQ101
In eqn.14.12, equivalent to
  2∥2hα∥≦|sin(π*2hα)| ---eqn.BQ121
eqn.BQ121 small red 2 cancel 
eqn.14.12 big red 2
eqn.BQ121 blue 2 is eqn.14.12 
blue 2
Above explained eqn.14.12.
<a name="ch14a200">
Next in eqn.14.12 set
  α=1/N ---eqn.BQ105
  β=b/N ---eqn.BQ106
  γ=c/N ---eqn.BQ107
find  //Gap h=n-m

<a name="ch14a202">
First line of eqn.14.13 is in 
eqn.14.12 replace α by 1/N.
eqn.14.12 red 2 cancelled not
show up in eqn.14.13 first line.
<a name="ch14a203">　Index begin　Index this file
Second line of eqn.14.13 is turn 
denominator denominator N to 
numerator N and write symmetry 
∥2h/N∥ function as double of 
half period summation. Purple 2 
in eqn.14.13 is this double.
After take double, h upper bound
change from N-1 to N/2.
2010-07-22-16-25 here

<a name="ch14a204">
2010-07-22-16-42 start
eqn.14.13 greater than side 
gap h=n-m vary from 1 to N/2. 
Its summation is
  ∑1/h=1/1+1/2+1/3+...+1/m ---eqn.BQ122
here m=N/2. This sum is less than
or equal to 1+log(m). //prove
2010-07-22-16-47 here

<a name="ch14a205">
n=12
sum0=0;
for(var i=1;i<n;i++)
sum0+=1/i
sum0 //smaller
1+log(n) //greater
//2010-07-22-16-49
answer
sum0 //smaller
3.0198773448773446
1+log(n) //greater
3.4849066497880003

<a name="ch14a206">　Index of Problem 14.2
2010-07-22-19-31 start
Next prove
  1/1+1/2+1/3+...+1/m ≦ 1+log(m) ---eqn.BQ123
(Textbook page 213 line 5)
We need next relation
  log(x)-log(x-1)-1/x ＞ 0 ---eqn.BQ124
for x ＞1
<a name="ch14a207">
Define
  f(x)=log(x)-log(x-1)-1/x ---eqn.BQ125
Let x=1.1, calculation give us
log(x)-log(x-1)-1/x = 1.4888＞ 0
When x approach to infinity,
the limit value of f(x) is
  lim[x→∞]{log(x/(x-1))-1/x}
 = log(1) -1/∞ = 0 ---eqn.BQ126
Our goal is to show f(x)＞0
for x ＞1. Now x=1.1 and x→∞
two points value found.
<a name="ch14a208">　Index begin　Index this file
If f(x) ever become negative
for x in [1.1,∞) then 
f'(x) must be partial negative 
and partial positive. 
f'(x) partial negative go from 
positive f(x) to negative f(x).
f'(x) partial positive go from 
negative f(x) to zero f(∞).
If f'(x) is always negative,
then f(x) is always positive.
<a name="ch14a209">
Next show that f'(x) is always 
negative for x in [1.1,∞).
  f(x)=log(x)-log(x-1)-1/x ---eqn.BQ125
then
  f'(x)=d[f(x)]/dx ---eqn.BQ127
 =d[log(x)-log(x-1)-1/x]/dx
 =1/x -1/(x-1)-(-1)/x/x
 =1/x -1/(x-1)+1/x/x
 =((x-1)*x -x*x +(x-1))/(x-1)/x/x
<a name="ch14a210">
we find
  f'(x)=(-1)/(x-1)/x/x ---eqn.BQ128
for x ＞1, -1/(x-1)/x/x ＜ 0
<a name="ch14a211">　Index of Problem 14.2
The slope f'(x) is always negative
for x in [1.1,∞) 
  lim[x→∞]f(x)=0
  f(x=1.1)=1.4888 ＞ 0
therefore, for x ＞1 
  f(x)= ---eqn.BQ124
  log(x)-log(x-1)-1/x ＞ 0
is true.

<a name="ch14a212">　Index begin　Index this file
Return to
  1/1+1/2+1/3+...+1/m ≦ 1+log(m) ---eqn.BQ123
write eqn.BQ123 as
  g(m)=1+log(m)   ---eqn.BQ129
      -(1/1+1/2+1/3+...+1/m)
expect g(m)≧0 for all m≧1

<a name="ch14a213">
for m=1 //  ---eqn.BQ130
  g(1)=1+log(1) -1 = 1+0 -1 =0
for m=2 //  ---eqn.BQ131
  g(2)=1+log(2) -(1+1/2)
      =log(2)-1/2 =0.19 ＞ 0
<a name="ch14a214">
for m=3 //  ---eqn.BQ132
  g(3)=1+log(3) -(1+1/2+1/3)
      =log(3)-1/2-1/3
      =log(3)-1/2-1/3
      +log(2)-log(2)  //insert zero
      =log(3)-log(2)-1/3 //eqn.BQ124
      +log(2)-1/2        //eqn.BQ131
       ＞ 0
<a name="ch14a215">　Index of Problem 14.2
for m=4 //  ---eqn.BQ133
  g(4)=1+log(4) -(1+1/2+1/3+1/4)
      =log(4)-1/2-1/3-1/4
      =log(4)-1/2-1/3-1/4
      +log(2)-log(2)   //insert zero
      +log(3)-log(3)   //insert zero
      =log(4)-log(3)-1/4 //eqn.BQ124
      +log(3)-log(2)-1/3 //eqn.BQ124
      +log(2)-1/2        //eqn.BQ131
       ＞ 0
<a name="ch14a216">　Index begin　Index this file
Above method is induction, apply
to m to infinity.
g(1) is zero, g(2) and higher 
are always positive. Since
  g(m)=1+log(m)   ---eqn.BQ129
      -(1/1+1/2+1/3+...+1/m)≧0
<a name="ch14a217">
then we have 
  1+log(m) ---eqn.BQ134
  ≧ 1/1+1/2+1/3+...+1/m
(Textbook page 213 line 5)
LiuHH believe there are more
compact method to prove. But
for now, this is it.
2010-07-22-20-22 here

<a name="ch14a218">
Now return to eqn.14.13
Cancel purple 2 and blue 2.
Greater than side have 
∑[1≦h≦N/2]1/h
eqn.BQ134 tell us that 
  1+log(N/2) ---eqn.BQ135
  ≧ 1/1+1/2+1/3+...+1/(N/2)
eqn.BQ135 let us push eqn.14.13
greater than side even greater
to N+N*(1+log(N/2))
<a name="ch14a219">
In eqn.14.13, replace greater 
than side with N+N*(1+log(N/2))
and eqn.14.13 whole equation
take square root, get the
target equation eqn.14.7.

<a name="ch14a221">　Index of Problem 14.2
Problem 14.2 is done.
The final result is different
from textbook eqn.14.7. Base 
on LangeListCSMCTypos.pdf
possibly there is typo in either
textbook or in this study notes.
Reader please find the correct
one.
2010-07-22-20-42 stop



2010-07-23-17-50 done first proofread.
2010-07-23-21-28 done second proofread.
2010-07-23-22-29 done spelling check

<a name="ch14a223"> derive:1 2 3 
2010-07-24-15-38 start
Summation by Parts Formula has
several different version.
eqn.BQ072 is a version from
http://fractal.math.unr.edu/~ejolson/311-06/handout/sumpart.pdf
When read a problem at
http://www.math.uconn.edu/~kconrad/math121/121sumbypartsproj.pdf
LiuHH wondered for a while. The
math.uconn.edu/~kconrad/
problem is
[[
<a name="ch14a224">　Index begin　Index this file
Equation 3.1 ∑[n=1,N]un(vn-vn-1)=
uNvN-u1v0-∑[n=1,N-1]vn(un+1-un)
Exercise 4. You know how to 
calculate
  ∫[x=a,b]{x*ex}dx ---eqn.BQ136
using integration by parts
with u = x and dv = exdx
Compute an explicit closed 
formula for the analogous sum 
  ∑[n=1,N]n*2n  ---eqn.BQ137
using summation by parts. 
Check your formula works 
numerically for N =1,2,3,4,5.
]] Can you try first?
   Here is answer.
<a name="ch14a225">
math.uconn.edu/~kconrad/
give next instruction
[[
The sums which can be analyzed with
 summation by parts are those whose
terms can be written as a product 
where one of the factors is the discrete
difference of a_sequence that we can 
say something about (analogous to
<a name="ch14a226">
choosing dv as a function we can 
actually integrate). That is, the sum
should look like ∑[n=1,N]an*bn
 where one of the sequences, say bn, 
is a discrete difference sequence
 cn+1 − cn where we have useful
knowledge of cn.
]]

<a name="ch14a227">
This problem ask to find
  ∑[n=1,N]n*2n ---eqn.BQ137
LiuHH goto eqn.BQ072 and find
a method to solve
math.uconn.edu/~kconrad/
problem. eqn.BQ137 is in the 
form of 
  ∑[k=m,n]ak*bk ---eqn.BQ138
<a name="ch14a228">
both a,b are lowercase. But
eqn.BQ072 has next two sum
 ∑[k=m,n]Ak*bk ---eqn.BQ139
 ∑[k=m,n]ak*Bk ---eqn.BQ140
A uppercase and B uppercase
Uppercase is partial sum.
I can not use eqn.BQ139,BQ140
for eqn.BQ138. 
2010-07-24-16-06 here

<a name="ch14a229"> derive:1 2 3 　Index begin　Index this file
The following is second version
of Abel’s Summation by Parts 
Formula.

2010-04-25-15-27 LiuHH access 
http://www.math.ust.hk/excalibur/v11_n3.pdf
The following is reading of 
math.ust.hk page

<a name="ch14a232">
2010-07-24-16-33 here
Red color eqn.BQ373 is 
Summation by parts formula.
It sum lowercase ak*bk
This is more useful version.

<a name="ch14a233">　Index begin　Index this file
Now try to solve
math.uconn.edu/~kconrad/ 
Exercise 4
[[
 You know how to 
calculate
  ∫[x=a,b]{x*ex}dx ---eqn.BQ136
using integration by parts
with u = x and dv = exdx
Compute an explicit closed 
formula for the analogous sum 
  ∑[n=1,N]n*2n  ---eqn.BQ137
using summation by parts. 
Check your formula works 
numerically for N =1,2,3,4,5.
]]

<a name="ch14a234">
a_sequence is 1,2,3,4,5
b sequence is 21,22,23,24,25
Variable k start from m=1
Variable k  end  at   n=5
Refer to eqn.BQ373 
  ∑[k=1,n]k*2k
 =An*bn
 -∑[k=1,n-1]Ak*(bk+1-bk)
2010-07-24-16-46 here
still have trouble with
uppercase Ak

<a name="ch14a235"> derive:1 2 3 

<a name="ch14a239">　Index begin　Index this file
2010-07-24-19-59 here
eqn.BQ375 is summation by parts
equation.
eqn.BQ375 do not have partial sum
eqn.BQ375 do not have uppercase A,B

Now try to solve
math.uconn.edu/~kconrad/ 
Exercise 4
2010-07-24-20-14 here

<a name="ch14a240">
Start from // another solution
  ∑[k=m,n]k*2k ---eqn.BQ141
a_sequence is 1,2,3...,N ---eqn.BQ142
b sequence is 22,23,...,2k+1,...,2n+1 ---eqn.BQ143
m = 1 in Exercise 4
First
  2k+1=2*2k ---eqn.BQ144
  2k+1-2k=2k ---eqn.BQ145
<a name="ch14a241">
Rewrite target as next
  ∑[k=1,n]k*2k  ---eqn.BQ146
 =∑[k=1,n]k*(2k+1-2k)
From eqn.BQ375, find
  ∑[k=1,n]k*2k ---eqn.BQ147
 =n*2n+1 - 0*20+1
 -∑[k=1,n][k-(k-1)]*2(k-1)+1
 =n*2n+1 -∑[k=1,n]1*2k

<a name="ch14a242">
The term ∑[k=1,n]1*2k apply
eqn.BQ375 one more time
here a_sequence is 1,1,1,1 ... 1
  ∑[k=1,n]1*2k
 =1*2n+1 - 1*20+1
 -∑[k=1,n][1-1]*2(k-1)+1
 =2n+1 - 2 ---eqn.BQ148

<a name="ch14a243">
Substitute eqn.BQ148 to eqn.BQ147
get
  ∑[k=1,n]k*2k ---eqn.BQ147
 =n*2n+1 -(2n+1 - 2)
 =(n-1)*2n+1+2

<a name="ch14a244">
Above is study note from reading
math.uic.edu/~leon/
2010-04-25-15-29 LiuHH access
http://www.math.uic.edu/~leon/cs-mcs401-s08/handouts/summation_by_parts.pdf

2010-07-24-20-58 stop



<a name="ch14a245">　Index begin　Index this file
2010-07-28-14-32 start
On 2010-07-25-16-51 LiuHH access
http://www.math.uconn.edu/~kconrad/math121/121sumbypartsprojsoln.pdf
found solution for exercise 4
It is as following.
<a name="ch14a246"> another solution
For the summation 
  ∑[n=1,N]n*2n  ---eqn.BQ137
set
  un=n ---eqn.BQ149
  vn=2n+1 ---eqn.BQ150
This vn setting is for the
difference consideration.
Equation 3.1 left side is
∑[n=1,N]un(vn-vn-1)
it take difference of vk sequence.
<a name="ch14a247">
If we define
  vn=2n+1 ---eqn.BQ150
then 
  vn-vn-1=2n+1-2n+1-1=2n ---eqn.BQ151
this 2n satisfy eqn.BQ137
When n=0
  v0=20+1=2 ---eqn.BQ152
<a name="ch14a248">
Now continue as next 
  ∑[n=1,N]n*2n = ∑[n=1,N]un(vn-vn-1)
  //apply Equation 3.1 // ---eqn.BQ153
 =uNvN-u1v0-∑[n=1,N-1]vn(un+1-un)
<a name="ch14a249">
From eqn.BQ149,
  u1=1  ---eqn.BQ155
  un=n  ---eqn.BQ149
  un+1=n+1  ---eqn.BQ156
  uN=N  ---eqn.BQ157
<a name="ch14a250">　Index begin　Index this file
From eqn.BQ150,
  v0=2  ---eqn.BQ152
  vn=2n+1 ---eqn.BQ150
  vN=2N+1 ---eqn.BQ158
Alert lowercase n is variable.
Alert uppercase N is upper bound.
<a name="ch14a251">
eqn.BQ153 become
  ∑[n=1,N]n*2n = ∑[n=1,N]un(vn-vn-1)
 =N*2N+1-1*2-∑[n=1,N-1]2n+1(n+1-n) ---eqn.BQ159
Here use one handy difference
  (un+1-un)=(n+1-n)=1 ---eqn.BQ160
<a name="ch14a252">
eqn.BQ159 become
  ∑[n=1,N]n*2n 
 =N*2N+1-2-∑[n=1,N-1]2n+1
 =N*2N+1-2-(4+8+16+...+2N)
 =N*2N+1-(2+4+8+16+...+2N) ---eqn.BQ161
Red terms refer to eqn.BQ148
<a name="ch14a253">
  ∑[n=1,N]n*2n=N*2N+1-(2N+1-2)
 =(N-1)*2N+1-2 ---eqn.BQ162
is final answer. No summation
Need just one 'N' get answer.
another solution
2010-07-28-15-12 stop

<a name="ch14a254">　Index begin　Index this file
■　Index of Problem 14.2
2010-07-28-16-28 start
LiuHH review eqn.14.7B found it 
is too long. Start from
[a name="ch14a106"], end at
[a name="ch14a220"].
The following is an index for 
this long work.

<a name="ch14a255">
2010-07-28-10-15 start index work
to prove eqn.14.7
start at ch14a109
step 1 Euler's formula 
exp(iθ)=cos(θ)+i*sin(θ) ---eqn.BQ082
step 2
(xn-1)/(x-1)=xn-1+xn-2+...+1 ---eqn.BQ083
<a name="ch14a256">
step 3 linear eqn.BQ087
Factor "e(Nt)－1"
purple part is common factor.
  e(Nt)－1=e((Nt)/2)*
  [e((Nt)/2)-e((-Nt)/2)] ---eqn.BQ092
Linear sum of e(kt) eqn.BQ094
step 4 take abs() eqn.BQ095 use 
|sin(πNt)|≦1 cause inequality 1
<a name="ch14a257">
step 5  2∥t∥≦|sin(πt)| ---eqn.BQ101
cause inequality 2.
prove 2∥t∥≦|sin(πt)|, done proof
step 6 Quadratic exponential sum
define SM(P) eqn.14.8. take square 
of |∑[n=1,M]cn| create quadratic.
α,β,γ bring 'N' into equation.
Expand quadratic equality eqn.14.9
<a name="ch14a258">
step 7 gap h=n-m ---eqn.BQ316
defined.  m,n double summation 
change to m,h double summation.
step 8 set cn=e(P(n)). This is 
an important step, all |cn|=1
Before step 8, |cn| not= 1
<a name="ch14a259">
step 9 apply Euler's formula
Hello cn ! you are one
step 10 expand P(m+h)-P(m)
cancel quadratic at eqn.BQ119
why e(2αmh) become sin(π1αh) ?
<a name="ch14a260">　Index begin　Index this file
step 11 apply linear result to 
quadratic
inequality show up (from step 4)
step 12 set α=1/N, β=b/N, γ=c/N
create sum ∑1/h=1/1+1/2+...+1/m 
<a name="ch14a261">
step 13 ch14a206 to ch14a217 prove
  1/1+1/2+1/3+...+1/m ≦ 1+log(m) ---eqn.BQ123
step 14 apply eqn.BQ123 to eqn.14.13
introduce one more inequality 3. 
push greater than side even greater.
done proof at ch14a220
2010-07-28-12-06 done index work
2010-07-28-17-25 done include index

<a name="ch14a262">
Update 2010-08-05 is next

2010-08-05-16-30
when read Exercise 14.1
Abel's Second Inequality
found
Abel's first Inequality
study notes
tute0049.htm#ch14a033
---eqn.BQ026
---eqn.BQ027
are error
forget take abs() for complex

<a name="ch14a263">
2010-08-05-16-40 in
tute0049.htm
change from
max[1≦k≦5]{Sk}
to
max[1≦k≦5]|Sk|
because complex Sk
no max, no min
only abs(complex) |Sk|
has max, has min

<a name="ch14a264">
＜ a name="ch14a032" ＞
all of S's by max[1≦k≦5]{Sk}
max[1≦k≦5]{Sk} eqn.BQ025 right
＜ a name="ch14a033" ＞
max[1≦k≦5]{Sk} is common to all
terms, we move max[1≦k≦5]{Sk}
 ≦ max[1≦k≦5]{Sk}*
＜ a name="ch14a035" ＞
 ≦ max[1≦k≦5]{Sk}*a1

<a name="ch14a265">
2010-08-05-16-52 in
tute0049.htm
done change six occurance of
max[1≦k≦5]{Sk}
 to
max[1≦k≦5]|Sk|
2010-08-05-16-58 record




<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