<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="ch10c001">　Index begin　Index this file
2010-04-16-11-38 start
■　Exercise 10.4 problem statement
　　textbook page 163
（Integral Version）

Prove the integral form of Hilbert's
inequality. That is, show that for
any f,g:[0,∞)→Real, one has

<a name="ch10c003">
The discrete Hilbert's inequality 
(10.1) can be used to prove a 
continuous version, but the strict 
inequality would be lost in the 
process. Typically, it is better 
to mimic the earlier argument 
rather than to apply the earlier 
result.
2010-04-16-11-54 stop




<a name="ch10c004">
2010-04-16-11-55 start
■　Exercise 10.4 hint
　　textbook page 268

One can repeat the proof of the
discrete case line-by-line, and
to do so is worth one's time.
The parallel between the discrete
and continuous problems is really
quite striking.
2010-04-16-11-57 stop



<a name="ch10c005">　Index begin　Index this file
2010-04-16-12-13 start
■　Exercise 10.4 solution


Exercise 10.4 hint said
[[
One can repeat the proof of the
discrete case line-by-line
]]
LiuHH need copy line-by-line.
<a name="ch10c006">
But Cauchy's Inequality for 
countable set is really a puzzle.

Textbook page 14 eqn.1.22, 
greater side 2αβ is 2aman and 
greater side 2xy is 2bmbn and
less side αy+βx is ambn+anbm
This mean that Cauchy use same
matrix at both side of inequality.

<a name="ch10c007">
To prove Hilbert's inequality 
start from generalized Cauchy's 
Inequality, this explain 
Hilbert's less than side. But
generalized Cauchy use same
matrix (see eqn.1.22) at greater
than side, which shows aman and 
bmbn.

<a name="ch10c008">
To prove Hilbert from Cauchy, 
we should start from greater 
than side aman and convert 
greater than side Hilbert 
matrix (show aman) to identity 
matrix (set aman m≠n to zero).
But Problem 10.1 use Cauchy's 
Inequality for countable set 
eqn.10.2 and did not say how
to convert.

<a name="ch10c009">
Textbook page 157 line 16
eqn.BE120 has double sum for
1/(m+n) and for (m/n), But not 
use aman. equation use amam.

Liu,Hsinhan still think why.
Exercise 10.4 is  NOT DONE
2010-04-16-12-45 stop



<a name="ch10c010">　Index begin　Index this file
2010-04-16-14-20 start
■　Exercise 10.5 problem statement
　　textbook page 164
（Homogeneous Kernel Version）

If the function K:[0,∞)×[0,∞)→[0,∞)
has the homogeneity property
<a name="ch10c011">
  K(λx,λy)=λ-1K(x,y) ---eqn.BG001
for all λ＞0, then for any pair
of function f,g:[0,∞)→Real one
has

2010-04-16-14-45 stop





<a name="ch10c015">　Index begin　Index this file
2010-04-16-17-30 start
■　Exercise 10.5 hint
　　textbook page 269

The first step exploits the
homogeneity condition of K(x,y)
by a change of variables
<a name="ch10c016">
  y=ux ---eqn.BG005

<a name="ch10c021">　Index begin　Index this file
2010-04-16-18-16 here
Now, once K has been pulled outside,
we can apply Schwarz's inequality to
the inside integral to find

<a name="ch10c025">　Index begin　Index this file
2010-04-16-18-46 here
This completes the solution of the
exercise with c given by the first
of the three indicated integrals,
and we can make a simple change of
variables to check that all three
of the integrals are equal.

<a name="ch10c026">
This argument is yet another of the
gems from Schur's remarkable 1911
paper. Actually, Schur proved the 
tricker finite range result.

<a name="ch10c028">
where 0≦a≦b≦∞.
In this case, the domain of 
integration changes with the
change of variables, but the 
original plan still works.
2010-04-16-19-00 stop



<a name="ch10c029">　Index begin　Index this file
2010-04-16-19-28 start
■　Exercise 10.5 solution


The requirement
  K(λx,λy)=λ-1K(x,y) ---eqn.BG001
satisfy Hilbert inequality
matrix element mij function.
<a name="ch10c030">
For element located at (i,j)
place, its element value is
  mij=1/(i+j) ---eqn.BG016
See a 3×3 Hilbert matrix eqn.BE108
How to build matrix element value.

<a name="ch10c031">
If change i to 3*i and change
j to 3*j, then
  m3i,3j=1/(3i+3j)=1/[3(i+j)] ---eqn.BG017
here '3' is 'λ' in eqn.BG001
'3' does factor out with a 
power of -1. (we get 1/3)

<a name="ch10c032">
Start from eqn.BG006, which
can be Hilbert's Inequality
eqn.10.1 less than side
integral version with
  K(i,j)=mij=1/(i+j) ---eqn.BG018
<a name="ch10c033">
From eqn.BG006 to eqn.BG007 is
separation of variable partially.
Since K(x,y) is in y integration
and K(x,y) still contain x.
<a name="ch10c034">　Index begin　Index this file
From eqn.BG007 to eqn.BG008 use
the condition 
  K(λx,λy)=λ-1K(x,y) ---eqn.BG001
The magic part is that K(λx,λy)
factor out a 1/λ and y=ux let dy
factor out a λ. λ multiply 1/λ
get one. λ disappear !!
<a name="ch10c035">
From eqn.BG008 to eqn.BG009 is
the cancellation of λ/λ=1 or 
x/x=1 
K(x,y) change to K(1,u) where '1'
is constant. K(1,u) is a function 
of u only. 
From eqn.BG009 to eqn.BG010 is
separation of variable.

<a name="ch10c036">
Integration in {...} of eqn.BG010
is a function of x and u is a
constant to x integration. We can
apply Schwarz's inequality to x 
integration. Get eqn.BG011 and 
eqn.BG012.
eqn.BG012 has g(ux), we change 
variable again, let v=ux, to 
eliminate 'u' from g(ux). That 
is the process from eqn.BG012 
to eqn.BG013.

<a name="ch10c037">
Put partial calculation back to
whole equation of eqn.BG010, we
have eqn.BG014. Two square root 
of integration are Hilbert's
greater than side two-norms. 
u integration in eqn.BG014 is 
Hilbert's greater than side
coefficient.

Exercise 10.5 is done
2010-04-16-20-06 stop



<a name="ch10c038">　Index begin　Index this file
2010-04-16-21-42 start
■　Exercise 10.6 problem statement
　　textbook page 164
（The Method of "Parameterized
 Parameters"）

For any positive weights wk, 
k=1,2,...,n, Cauchy's inequality 
can be restated as a bound on
the square of a general sum

<a name="ch10c040">
and given such a bound it is
sometimes useful to note the
values wk, k=1,2,...,n, can be
regarded as free parameters.
The natural question then
becomes, "What can be done with
this freedom?" Oddly enough,
<a name="ch10c041">
one may then benefit from
introducing yet another real
parameter t so that we can 
write each weight wk as wk(t).
<a name="ch10c042">
This purely psychological step
hopes to simplify our search
for a wise choice of the wk by
refocusing our attention on 
desirable properties of the
functions wk(t), k=1,2,...,n.

<a name="ch10c043">　Index begin　Index this file
Here we want to squeeze 
information out of the bound 
(10.23), and one concrete idea
is to look for choices where
<a name="ch10c044">
(1) the first factor of the
    product (10.23) is bounded
    uniformly in t and where
(2) one can calculate the 
    minimum value over all t
    of the second factor.
<a name="ch10c045">
These may seem like tall orders,
but they can be filled and the
next three steps show how this
plan leads to some marvelous
inferences.

<a name="ch10c046">
(a) Show that if one takes
    wk(t)=t+k2/t ---eqn.BG019
    for k=1,2,...,n. then the
    first factor of the
    inequality (10.23) is 
    bounded by π/2 for all t≧0
    and all n=1,2,...
<a name="ch10c047">
(b) Show that for this choice
    we also have the identity

<a name="ch10c048">　Index begin　Index this file
2010-04-16-22-22 here
(c) Combine the proceeding
    observations to conclude
    that

<a name="ch10c050">
This curious bound is known as
Carlson's inequality, and it
has been known since 1934. 
Despite several almost arbitrary
steps on the path to the 
inequality (10.24), 
<a name="ch10c051">
the value π2
cannot be replaced by a smaller
one, as one can prove by the
stress testing method (page 159)
though not without thought.
2010-04-16-22-33 stop





<a name="ch10c052">　Index begin　Index this file
2010-04-16-23-24 start
■　Exercise 10.6 hint
　　textbook page 269

Integral comparison gives part (a)
by

2010-04-16-23-41 here
<a name="ch10c055">
and for part (b) we note that

is minimized by taking
  t=(B/A)1/2 ---eqn.BG024
<a name="ch10c056">
Part (c) just assembles the pieces.
2010-04-16-23-50 stop


<a name="ch10c057">　Index begin　Index this file
2010-04-17-00-05
Chapter 10 title is 
"Hilbert's Inequality and
 Compensating Difficulties"

what is "compensating difficulties"?

2009-10-01-22-28 LiuHH access
http://www.carma.newcastle.edu.au/~jb616/hw.pdf
hilbert_carma.newcastle.edu.au_981001.pdf
this file page 3/15 has
[[
Proof of Theorem 1. We may assume the 
right-hand side is finite. We apply
H¨older’s inequality with what Hardy 
calls “compensating difficulties” 
(inserting a term and its reciprocal) 
.....
]]

<a name="ch10c058">
For example, please see eqn.BF044
red term.
First appearance is eqn.10.4
it is black print, not red.
2010-04-17-00-17 stop


<a name="ch10c059">　Index begin　Index this file
2010-04-17-12-12 start
■　One data, two data, one data+prob.
Index use "UQ" link to here. "UQ" 
is arbitrary assigned two bytes 
string not used elsewhere.

Exercise 10.6 Problem statement 
start from eqn.10.23 and describe 
using "Cauchy's inequality"

<a name="ch10c060">
If you read textbook chapter by 
chapter and come to this point,
you should know Cauchy's rule
very well. If you did not read
previous chapters and just stop
by here. Following is a brief
for you.

<a name="ch10c061">
If we have observed data in hand,
for example, pressure data,
temperature data, distance data
etc. Different calculation give
us different answer. Inequality
math. compare several answers.

<a name="ch10c062">
Cauchy's inequality need two data
sequences. Chapter 01
Rearrangement inequality need 
two data sequences. Chapter 05
Holder Inequality is broader than
Cauchy. Chapter 09
Hilbert's inequality need two data
sequences. Chapter 10

<a name="ch10c063">
Simple AM-GM inequality need one 
data sequence. Chapter 02
Generalized AM-GM ineq. is under
power mean next.
Power mean inequality need one 
data sequence, one probability 
sequence. Chapter 08 
Jensen's inequality need one 
data sequence, one probability 
sequence. Chapter 06 

<a name="ch10c064">　Index begin　Index this file
Data sequence indicate numbers
with physics dimension, for 
example 'meter' for length.
Probability sequence can not
have physics dimension. The 
word "mean" ask for averaging 
coefficients. Probability is
good candidate for "mean".
Prob. seq. must be pure numbers.
Prob. seq. must be 0≦pk≦1
Prob. seq. must be ∑[k]pk=1

<a name="ch10c065">
Please read each chapter for 
detail. Here just tell you for
Cauchy, we need two sequences.
Problem statement give us one
sequence a1,a2,...,an
Can Cauchy work with just one
sequence? The answer is no.
<a name="ch10c066">
Cauchy require two sequences.
For sequences a1,a2,...,an and
b1,b2,...,bn
Cauchy's inequality less than 
side is dot product of two 
sequences a1b1+a2b2+...+anbn
Cauchy's inequality greater 
than side is product of 2-norm 
of each sequence. ∥a∥2∥b∥2

<a name="ch10c067">
Textbook tell us two tricks for
Cauchy to meet ONE sequence.
First is 1-trick
Second is splitting trick

<a name="ch10c068">
If given one sequence {a} and 
not given second sequence {b}
1-trick build {b} as [1,1,...,1]
Given a1,a2,...,an dot product 
with [1,1,...,1] get a1+a2+...+an 
satisfy Cauchy.

<a name="ch10c069">　Index begin　Index this file
If given one sequence {a} and 
not given second sequence {b}
splitting-trick split {a} as 
first seq.=[√a1,√a2,...,√an]
second seq=[√a1,√a2,...,√an]
<a name="ch10c070">
Dot product two sequences get
a1+a2+...+an 
Square root is equal power split. 
It is possible do other split. 
For example
 first seq. use power 0.3,
second seq. use power 0.7 etc.
Must sum power to ONE 0.3+0.7=1
<a name="ch10c071">
If not use power splitting, 
Cauchy accept negative seq.
elements.
If use power splitting, 
Cauchy must use non-negative 
seq. elements. Since square 
root of negative is complex
which we want to avoid.

<a name="ch10c072">
Exercise 10.6 use third method
to satisfy Cauchy.
Third method close to 1-trick
1-trick use honest 1 as one.
Third method use wk/wk as one.

<a name="ch10c073">
Hope above brief introduction
help you understand the back-
ground of Exercise 10.6.
2010-04-17-13-19 stop



<a name="ch10c074">　Index begin　Index this file
2010-04-17-14-48 start
■　Exercise 10.6 solution


We have one data sequence in 
hand a1,a2,...,an. We build 
second sequence [1,1,...1]
which is 

<a name="ch10c076">
Problem statement suggest
    wk(t)=t+k2/t ---eqn.BG019
one_sequence eqn.BG025 become

<a name="ch10c078">
Why use square root of wk?
Compare eqn.10.23 with Cauchy's
inequality eqn.1.7. eqn.1.7 
greater than side ak2 show up at
eqn.10.23 greater than side.
<a name="ch10c079">　Index begin　Index this file
ak2 get a company wk (that is ak2wk)
ak squared get ak2, then
wk original face must be √(wk)
Since follow ak, square operation 
for √(wk) get wk.
That is why we use √(wk) to 
build two sequences as following.

<a name="ch10c082">
2010-04-17-15-28 here
Dot product two sequences get
a1+a2+...+an 

Cauchy's ineq. greater than side 
use 2-norm for both sequences 
eqn.BG027 and eqn.BG028.

<a name="ch10c083">
2-norm for eqn.BG027 is take
square for each element as
first step. Second step is
sum these squares. We start 
from 1/√wk, now 2-norm 
square 1/√wk first, recover
original face 1/wk Second
step is sum 1/wk then we
get eqn.10.23 {∑1/wk} term

<a name="ch10c086">
Each term in (division) in 
eqn.BG030 is 1/wk(t)
Red power 1/2 in eqn.BG030 is
build by us at eqn.BG027
Blue power 2 in eqn.BG030 is
Cauchy's operation.
eqn.BG030 become eqn.BG021
2010-04-17-16-21 here

<a name="ch10c087">
Now solve problem (a)
Exercise 10.6 hint guide us
  eqn.BG021 ＜ eqn.BG022
this step is an application of
eqn.BE123

<a name="ch10c088">
What is t domain? Problem (a) 
tell us that "for all t≧0"
so eqn.BG021 are all positive 
terms and gradual decrease term 
value. Satisfy "nonnegative 
decreasing function" requirement.
Application of eqn.BE123 is
reasonable.

<a name="ch10c089">　Index begin　Index this file
Next in eqn.BG022 left integral
is less/equal to right integral.
First is change variable
  x=t*y ---eqn.BG031 
This change of variable require
that t≠0, otherwise we can not
cancel t from numerator and
denominator. "for all t≧0"
become "for all t＞0"

<a name="ch10c090">
Second is extend integration
upper bound from x=n to y=∞
This change bound cause '≦'
in eqn.BG022.
(if n initial value is ∞, 
 equality in '≦' active.)

<a name="ch10c091">
Right end equality in eqn.BG022
is a simple formula from integral
table. (Peirce, page 62, item 480)
2010-04-17-16-54 here

<a name="ch10c092">
Now solve problem (b)
Problem (b) ask for minimum value.
Minimum or maximum problem involve
define f(t) and set d[f(t)]/dt to
zero.
Problem (b) specified f(t) be

<a name="ch10c094">　Index begin　Index this file
Put wk(t) into f(x) we get 
eqn.BG023. Here t is variable. 
Lump all non-t term to constants
'A' and 'B' as shown in eqn.BG023
right side. f(t) look like
  f(t)=A*t+B/t  ---eqn.BG032

<a name="ch10c095">
If t is small (0＜t＜1),
A*t drop out and B/t dominate.
Curve similar to g(t)=1/t

If t is great (t＞10),
B/t drop out and A*t dominate.
Curve similar to h(t)=t

<a name="ch10c096">
g(t) shoot to infinity as t→0
h(t) go to infinity as t→∞
In between we have minimum f(t)
value. Mark this special t as t0
(t is variable, t0 is constant)
Find t0 as following. Set
  d[f(t)]/dt=0  ---eqn.BG033
that is set
  d[A*t+B/t]/dt=0  ---eqn.BG034
get  A+(-1)B/t/t=0
or   A*t*t=B 
<a name="ch10c097">
The result is
  t0=(B/A)1/2 ---eqn.BG024
Constants A and B are defined 
in eqn.BG023
Put t0 to eqn.BG032
  min_f(t)=A*(B/A)1/2+B*(A/B)1/2 
          =2*A1/2*B1/2=eqn.BG020
problem (b) solved.
2010-04-17-17-27 here

<a name="ch10c098">
Now solve problem (c)
Problem (c) ask for proving
eqn.10.24

We start at eqn.10.23
with given
    wk(t)=t+k2/t ---eqn.BG019

<a name="ch10c102">
In eqn.BG037, cancel 2.
Whole equation take square, the
result is our goal eqn.10.24.

Exercise 10.6 is done.
2010-04-17-18-05 stop

<a name="ch10c103">
We start at
    wk(t)=t+k2/t ---eqn.BG019
with 't' as variable. The final
answer eqn.10.24 do not have t !
Can you find out which step 
eliminated t? How to remove t?
2010-04-17-18-10 stop

<a name="ch10c103a">
2010-04-19-16-03 start
eqn.BG033 is next
  d[f(t)]/dt=0  ---eqn.BG033
If we define
  g(t)=d[f(t)]/dt  ---eqn.BG077
and not set d[f(t)]/dt to zero
then t vary, g(t) value change.
<a name="ch10c103b">
g(t) is a curve which has one
degree of freedom.
Now we set d[f(t)]/dt to zero.
(or set to any non-zero, same)
One degree of freedom evaporated.
1 dof - 1 constraint = 0 dof
Variable t become constant t0.
2010-04-19-16-11 stop



<a name="ch10c104">　Index begin　Index this file
2010-04-17-19-54 start
■　Exercise 10.7 problem statement
　　textbook page 165
（Hilbert's Inequality via the
 Toeplitz Method）

Show that the elementary integral

<a name="ch10c106">
for n≠0, implies that for
real ak, bk, 1≦k≦N, one has 
the integral representation

<a name="ch10c109">
then show that this representation
and Schwarz's inequality yield
a quick and easy proof of
Hilbert's inequality.
2010-04-17-20-22 stop





<a name="ch10c110">　Index begin　Index this file
2010-04-17-20-24 start
■　Exercise 10.7 hint
　　textbook page 270

Since |t－π|≦π ---eqn.BG041
for t in [0,2π], we have

<a name="ch10c114">
2010-04-17-20-56 here
This remarkably quick way of
obtaining Hilbert's inequality
is known as Toeplitz's method.
Hilbert's original proof also
used trigonometric integrals,
<a name="ch10c115">　Index begin　Index this file
but those used by Hilbert were
not quite as efficient.
Toeplitz's argument tells us 
more generally that if φ is any
bounded function on [0,2π] with
Fourier coefficients cn, 
－∞＜n＜∞, then one has the bound

<a name="ch10c117">
Integral representation can also
be used to prove more distinctive
generalization of Hilbert's 
inequality. For example, if 
α is not an integer, one finds

2010-04-17-21-26 stop


<a name="ch10c120">　Index begin　Index this file
2010-04-18-09-52 start
■　Exercise 10.7 solution


First verify eqn.BG038
The following 
i = sqrt(-1)
n = positive integer, 1/n OK.
t = variable, 0≦t≦PI
e = 2.718281828459045

<a name="ch10c121">
2010-04-18-09-55
   ∫[t=0,2PI]t*e^(int)*dt ---eqn.BG048
//insert 1=(in)/(in)
 = ∫[t=0,2PI]t*e^(int)*d(int)/(in)
//d[e^x]/dx=e^x, here x=int
 = ∫[t=0,2PI]t*d(e^(int))/(in) ---eqn.BG049
<a name="ch10c122">
//next line integration by parts
 = [t=0 & 2PI]t*(e^(int))/(in) -  ---eqn.BG050
 - ∫[t=0,2PI](dt)*(e^(int))/(in)
//insert 1=(in)/(in)
 = {2PI*[e^(in2PI)]-0*e^0}/(in) -  ---eqn.BG051
 - ∫[t=0,2PI](d(int))*(e^(int))/(in)/(in)
<a name="ch10c123">
//Euler formula e^(iθ)=cos(θ)+i*sin(θ)
 = {2PI*[cos(n2PI)+i*sin(n2PI)]}/(in) -  ---eqn.BG052
 - ∫[t=0,2PI]d(e^(int))/(in)/(in)
 = [2PI*(1+0)]/(in) -  ---eqn.BG053
 - [(e^(in2PI))-(e^(in*0))]/(in)/(in)
<a name="ch10c124">
 = 2PI/(in) -  ---eqn.BG054
 - {[cos(n2PI)+i*sin(n2PI)]-1}/(in)/(in)
 = 2PI/(in) -  ---eqn.BG055
 - [(1+i*0)-1]/(in)/(in)
 = 2PI/(in)  ---eqn.BG056
2010-04-18-10-06

<a name="ch10c125">　Index begin　Index this file
2010-04-18-10-07
   ∫[t=0,2PI]e^(int)*dt ---eqn.BG057
 = ∫[t=0,2PI]e^(int)*d(int)/(in) ---eqn.BG058
 = ∫[t=0,2PI]d[e^(int)]/(in) ---eqn.BG059
 = [e^(in2PI)-e^(in*0)]/(in) ---eqn.BG060
 = [cos(n2PI)+i*sin(n2PI)-1]/(in) ---eqn.BG061
 = [1+i*0-1]/(in) ---eqn.BG062
 = 0
<a name="ch10c126">
2010-04-18-10-12
    ∫[t=0,2PI]t*e^(int)*dt ---eqn.BG063
-PI*∫[t=0,2PI]e^(int)*dt
 =  2PI/(in) -PI*0
 =  2PI/(in)

<a name="ch10c127">
2010-04-18-10-14
Next is eqn.BG038

 { ∫[t=0,2PI]t*e^(int)*dt ---eqn.BG064
 -PI*∫[t=0,2PI]e^(int)*dt
 }/(2PI)
 = 2PI/(in)/(2PI)
 = 1/(in)
correct !! same as eqn.BG038 
2010-04-18-10-15 stop

<a name="ch10c128">
2010-04-18-12-20 start
Above is to verify eqn.BG038
How to change from eqn.BG038
to eqn.BG039?
The difference between eqn.BG038
and eqn.BG039 is that e^(int) in
eqn.BG038 change to product of
two summations. 
<a name="ch10c129">
It is better write eqn.BG039 as
following
change ak to am and change relatives
change bk to bn and change relatives

<a name="ch10c136">
Why i=sqrt(-1) show up in eqn.BG068?
2010-04-18-13-00 here

<a name="ch10c137">　Index begin　Index this file
2010-04-18-13-22 continue
Above is Hilbert's Inequality
less than side.
Below is Hilbert's Inequality
greater than side.

<a name="ch10c138">
Greater than side also start 
from eqn.BG039
eqn.BG039 integrand has a 
factor (t－π) We know the
integration 
start from t=0 , end at t=2PI
all shift －π get
<a name="ch10c139">
start: t－π=0－π , end: t－π=2π－π
that is
start: t－π=－π , end: t－π=π
Last line tell us that
－π≦t－π≦π, here t is variable.
－π≦t－π≦π again tell us that
  |t－π|≦π ---eqn.BG069
We write equality eqn.BG039 to
next inequality

<a name="ch10c141">
2010-04-18-13-43 here
eqn.BG039 is equality equation.
The condition
  |t－π|≦π ---eqn.BG069
change eqn.BG039 to eqn.BG070.
eqn.BG070 has same bound as
eqn.BG069. We can add absolute
sign to eqn.BG070, which is
eqn.BG042

<a name="ch10c142">　Index begin　Index this file
From eqn.BG042 to eqn.BG043, it
is application of Schwarz's 
inequality eqn.1.19

how eqn.BG043 change to 
eqn.BG044?
2010-04-18-14-44 stop

<a name="ch10c143">
2010-04-18-14-55 start
eqn.BG043 use absolute value.
For the term akeikt we need 
only its absolute value. 
We know
  |eikt|=1 ---eqn.BG071
<a name="ch10c144">
We simply drop eikt from ak
and drop eikt from bk. Left
only ∑ak and ∑bk they are
not function of t. Separate 
k summation from t integration.
k summation goto eqn.BG044 
t integration get value 2π
and cancel denominator 2π to
one. 
<a name="ch10c145">
That is how we get eqn.BG044 .
eqn.BG044 is Hilbert's Inequality
greater than side. 
eqn.BG068 is Hilbert's Inequality
less than side. Let
eqn.BG068 ≦ eqn.BG044
It is NOT Hilbert's Inequality
because the factor 1/i in
eqn.BG068

Exercise 10.7 is NOT DONE.
2010-04-18-15-08 stop



<a name="ch10c146">　Index begin　Index this file
2010-04-18-16-33 start
■　Exercise 10.8 problem statement
　　textbook page 165
（Functional Equation for the
 Gamma Function）

Recall that the gamma function is
defined by the integral

<a name="ch10c149">
As a consequence, one finds that
the evaluation of the integral
(10.8) yields the famous functional
equation for the Gamma function,

2010-04-18-16-58 stop




<a name="ch10c151">　Index begin　Index this file
2010-04-18-17-36 start
■　Exercise 10.8 hint
　　textbook page 271

We substitute and change orders:

2010-04-18-18-02 stop


<a name="ch10c153">　Index begin　Index this file
2010-04-18-18-22 start
■　Exercise 10.8 solution


Exercise 10.8 problem statement
say
"use an integral representation
 for 1/(1+y)"
<a name="ch10c154">
eqn.BG074 first line red term
is an integral representation
for 1/(1+y). Now verify first.

<a name="ch10c156">
2010-04-18-18-46 here
In eqn.BG075 
"aa＝" is insert a "1=r/r"
"bb＝" is change variable
       from 'tr' to 's'
"cc＝" insert "1=(-1)/(-1)"
"dd＝" is exdx=d[ex], x=-s

<a name="ch10c157">
In eqn.BG075, replace 'r' with
'1+y', indeed, eqn.BG074 first 
line red term is an integral 
representation for 1/(1+y).

<a name="ch10c158">　Index begin　Index this file
From eqn.BG074 line 1 to line 2
group all y-term into {...} and
calculate as following

<a name="ch10c160">
2010-04-18-19-35 here
In eqn.BG076, start term has
e-ty*dy and y2λ
where e-ty is master, we unify
variable from 'y' to '-ty'
<a name="ch10c161">
"ee＝" insert a red 1=t2λ*t-2λ
"ee＝" insert a blue 1=d(ty)/t
"ff＝" merge to (ty)2λ
"gg＝" change variable from ty to x
"gg＝" right side numerator has
form match eqn.BG072
<a name="ch10c162">　Index begin　Index this file
eqn.BG072 x power (λ－1) plus 1
become Γ(λ) parameter λ.
"gg＝" numerator x power (-2λ) 
      plus 1 become Γ(α) parameter
      1-2λ, that is Γ(1-2λ)
"gg＝" denominator t1-2λ is constant
      not influenced by x or y.
eqn.BG076 answer is Γ(1-2λ)/t1-2λ

Above is from eqn.BG074 line 1 to
line 2.
<a name="ch10c163">
From eqn.BG074 line 2 to line 3
is easy. In line 3, write 1/t1-2λ 
as t-1+2λ. Γ(1-2λ) is not a 
function of t. it is a constant.
In line 3, all 't' term is in
Γ function definition form.
<a name="ch10c164">
Variable t power '-1+2λ' plus one
is 2λ, 
Variable t power '-1+2λ' plus one
is Γ function variable. We get Γ(2λ)
Γ(2λ) together with first Γ(1-2λ) 
become final answer Γ(1-2λ)Γ(2λ).

<a name="ch10c165">　Index begin　Index this file
Gamma function parameter must be
greater than zero, we have
for Γ(1-2λ): 1-2λ＞0 ⇒ 1/2＞λ
for Γ(2λ)  :  2λ＞0  ⇒ λ＞0
That is 1/2＞λ＞0

2010-04-18-20-00 stop

2010-04-19-17-01 done proofread
2010-04-19-19-16 done spelling check

========= Chapter ten end here =========




<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