/**/
function alert4() //9812060000
{
document.write(''
+'Output may contain error, Please verify first. '
+'Program environment is MSIE 6.0, please use MSIE '
+''
);
} //function alert4() 9812060002
/**
textbook page 87 line 19 eqn.6.2
convex minorant used three times
create a command to build it any
where.
must call HelloJensen(hbPar)
outside of
..
otherwise width
wrong.
200912251340 here
calling code is next line
or
or
9902141245 add w1 to HelloJensen()
in tute0030.htm
99,02,07,21,45 the use of w1
from 9902072131
to 9902072143
solve infinite loop problem
and no need change from uppercase
to lowercase.
/**/
function HelloJensen(hjPar1) //9812251342
{
strJensen1=''
+'<a name="Jensen'
+hjPar1
+'"> '
+'Jensen Inequality for convex function'
+'
'
+'
'
+'
f
(
j=n
∑
j=1
pjxj
)
≦
j=n
∑
j=1
pjf(xj)
'
+'
page 87 line 19 eqn.6.2
'
+'Left side apply generalized AM to DOMAIN. Right side apply generalized AM to RANGE. '
+'width of above equation'
+''
+' calling ID '+hjPar1
;
var iter0=0; //9902161139
var iterMax=5; //[=][][]
if(arguments.length==1)
document.write(strJensen1);
else
{
argLen=arguments.length; //9806261457
var w0; //9806261502
var w2; //9806261336
var w1; //9902072131
var strJensen2=strJensen1;
for(w0=1;w0=0)
while((w2=strJensen2.indexOf(arguments[w0][0],(w2+w1)))>=0)
{ //9902161158 add this '{', '}'
strJensen2= //9806261337
strJensen2.replace(arguments[w0][0],arguments[w0][1]);
w1=arguments[w0][1].length; //9902072133
if(w1==0)w1=1; //9902072143
iter0++;
//if(iter0>iterMax)
//alert('9902161140 ##iter0='+iter0+'\n'+strJensen2);
//return; //9902161142
} //9902161158 add this '{', '}'
} //if(arguments[w0].length==1) else
} //for(w0=0;w0..
otherwise width wrong.
201001310932 here
calling code is next line
or
or
or eqn.7.14
99,02,07,21,45 the use of w1
from 9902072131
to 9902072143
solve infinite loop problem
and no need change from uppercase
to lowercase.
/**/
function HelloSchwarz(hsPar1) //9901310935
{
strSchwarz1=''
+'<a name="Schwarz'
+hsPar1
+'"> '
+'Schwarz Inequality for integration'
+'
'
+'
'
// next line is equation 9901310942
+'
x=b
∫
x=a
f(x)g(x)dx
≦
(
x=b
∫
x=a
f2(x)dx
)
1/2
(
x=b
∫
x=a
g2(x)dx
)
1/2
'
//above line is equation
+'
'
+' page 10 line 16 eqn.1.19 '
+'width of above equation'
+''
+' calling ID '+hsPar1
;
if(arguments.length==1)
document.write(strSchwarz1);
else
{
argLen=arguments.length; //9806261457
var w0; //9806261502
var w2; //9806261336
var w1; //9902072131
var strSchwarz2=strSchwarz1;
var iter0=0;//9902071928
for(w0=1;w0=0)
{
strSchwarz2= //9806261337
strSchwarz2.replace(arguments[w0][0],arguments[w0][1]);
w1=arguments[w0][1].length; //9902072133
if(w1==0)w1=1; //9902072143
}
} //if(arguments[w0].length==1) else
} //for(w0=0;w0
Inequality
Study 30th file
Upload 20100216
indexthis
program
DocA
Limit
XYGraph v2.3  web page graph
☜☞
donate
get code
The CauchySchwarz Master Class
J. Michael
Steele
★★★★★
This file is personal home work. No one
proofread. Cannot promise correctness.
If you suspect any view point wrong,
please ask a math expert near by.
Freeman 200906191046
Please use MSIE browser to read this file.
Did not test other browser. This file is
written under MSIE 6.0
<a name="docA001">Index beginIndex this file
200906081910 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 ProfessorJ. MichaelSteeleThe CauchySchwarz 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 beginIndex this file
On 200901271008 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.
200906081947 stop
<a name="docA006">
200908231500 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⊂Z^{3}" ?
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 CauchySchwarz 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.
200908231512 stop
<a name="ch07c001">Index beginIndex this file
201002130959 start
■ Exercise 7.9 problem statement
textbook page 118
（Littlewood's Middle Derivative Squeeze）
<a name="ch07c002">
Show that if f:[0,∞)→Real is twice
differentiable and if f''(x) is
bounded then
lim[x→∞]f(x)=0 implies
lim[x→∞]f'(x)=0 eqn.AX001
In his Miscellany, J.E. Littlewood
suggests this "pictorial arguments,
while not so purely conventional
can be quite legitimate" The result
of this exercise is his leading
example, and the picture he offered
is essentially that of Figure 7.2.
201002131009 stop
<a name="ch07c003">
201002131518 start
■ Exercise 7.9 hint
textbook page 255
In order to argue by contradiction,
we assume without loss of generality
that there is a sequence x_{n}→∞ such
that
f'(x_{n})≧ε＞0 eqn.AX002
//eqn.AX002 is wrong on purpose
Now by Littlewood's figure 7.2 (or by
triangle lower bound of Exercise 7.7)
we note that
<a name="ch07c005">
201002131536 here
where
B=supf''(x)＜∞ eqn.AX004
and
δ=ε/B eqn.AX005
This bound implies that
f(x)≠o(1) eqn.AX006
so we have our desired
contradiction.
201002131540 stop
<a name="ch07c006">Index beginIndex this file
201002131542 start
■ Exercise 7.9 solution
First clarify the dimension of
δ, ε, B in δ=ε/B eqn.AX005
Because integration upper bound
is x_{n}+δ, then
dim(δ)=dim(x) eqn.AX007
x is f(x) independent variable.
<a name="ch07c007">
Because they can compare each other
f'(x_{n})≧ε＞0 eqn.AX002
then
dim(ε)=dim(f'(x))=dim(f/x) eqn.AX008
Because the equality in
B=supf''(x)＜∞ eqn.AX004
then
dim(B)=dim(f''(x))=dim(f/x/x) eqn.AX009
<a name="ch07c008">
Exam the relation involve
δ, ε, B in δ=ε/B eqn.AX005
dim(δ) ?=? dim(ε)/dim(B) eqn.AX010
that is
dim(x) ?=? dim(f/x)/dim(f/x/x)
that is
dim(x) ?=? dim(x)
<a name="ch07c009">
Yes! Confirmed consistency in
dimension analysis,
δ is same as x
ε is same as f/x
B is same as f/x/x
201002131553 here
<a name="ch07c010">
201002131622 start
If reader review Graph0707
That figure may give better
explanation than Graph0709
Exercise 7.9 problem is that
given f:[0,∞)→Real is twice
differentiable and
given f''(x) is bounded
given lim[x→∞]f(x)=0
imply lim[x→∞]f'(x)=0
<a name="ch07c011">Index beginIndex this file
With our direct sense, we feel
given lim[x→∞]f(x)=0
then lim[x→∞]f'(x)=0
then lim[x→∞]f''(x)=0
otherwise it is hard to imagine
that a f(x)=0_at_x=infinity line
has up/down tendency like wave
at infinity. Express direct sense
in mathematics, it is
<a name="ch07c013">
201002131646 here
Because it is hard to argue that
eqn.AX011 integration part = 0
imply f'(x=infinity)=0, textbook
use method of contradiction.
Assume there is a sequence x_{n}→∞
such that
f'(x_{n})≧ε＞0 eqn.AX002
See what ill result we get.
<a name="ch07c014">
We can apply Exercise 7.7 result
to eqn.AX003 inequality right
half.
Exercise 7.7 say
domain left end function value
domain left end d[f(t)]/dt value
(two boundary information only)
determine a lower bound for the
whole domain of integration.
It is true even if x approach
to infinity.<a name="ch07c015">
Here Exercise 7.9, we
replace f(x) with f'(x)
replace f'(x) with f''(x)
Here Exercise 7.9, it is
domain RIGHT end d[f(t)]/dt value
Maximum dd[f(t)]/dt/dt in domain
(NOT two boundary information)
determine a lower bound for the
whole domain of f'(x) integration.
(eqn.AX003 use f'(x) integration)
<a name="ch07c016">Index beginIndex this file
Triangle area value here in
Exercise 7.9 is (1/2)*ε*ε/B
Exercise 7.9 hint made a
contradiction assumption
if x→∞, f'(x→∞)=ε＞0
and //above red wrong on purpose
if x→∞, f''(x→∞)=supf''(x)＜∞
then ε and B information
created lower bound triangle
(1/2)*ε*ε/B > 0
Nonzero ε in eqn.AX003 say
f(x_{n}+δ)－f(x_{n}) should be > 0
<a name="ch07c017">
BUT, when x_{n} approach to infinity
f(x_{n}+δ) and f(x_{n}) should both
approach to zero and
f(x_{n}+δ)－f(x_{n}) should be ＝ 0
The assumption
limit[f'(x_{n})] ≧ ε＞0 eqn.AX002
must be wrong.
<a name="ch07c018">
Contradiction result tell us
the other side story is true,
that is
given lim[x→∞]f(x)=0
then lim[x→∞]f'(x)=0
Proof is done by method of
contradiction.
201002131721 stop
<a name="ch07c019">
201002131727 start
Exercise 7.7 require that curve
be convex for positive function
value.
For a curve function value be
positive and approach to zero
when x approach to infinity,
This curve must be convex ╰╯
<a name="ch07c020">
For a curve function value be
negative and approach to zero
when x approach to infinity,
This curve must be concave ╭╮
We can apply Exercise 7.7 result
by turn positive triangle to
negative triangle.
<a name="ch07c021">
For a curve function value be
wavelike and approach to zero
when x approach to infinity,
This curve is not concave and
is not convex. But it is
enveloped by a positive convex
and a negative concave and
squeeze f'(x=infinity) to zero
for example
f(x)=exp(x)*sin(x) eqn.AX012
f'(x)=exp(x)*(cos(x)sin(x)) eqn.AX013
201002131747 stop
<a name="ch07c022">Index beginIndex this file
201002131916 start
■ Exercise 7.10 problem statement
textbook page 118
（Monotonicity and Integral Estimates）
Although the point was not stressed
in this chapter, many of the most
useful daytoday estimates of
integrals are found with help from
monotonicity. Gain some practical
experience by proving that
<a name="ch07c024">
201002131928 here
and by showing that 2*log(2)
cannot be replaced by a smaller
constant.
Incidentally, this particular
inequality is one we will see
again when it helps us with
Exercise 11.6.
201002131930 stop
<a name="ch07c025">
201002131931 start
■ Exercise 7.10 hint
textbook page 255/256
Differentiation suffices to confirm
that on (0,1) the map
t→t^{1}log(1+t) eqn.AX015
is decreasing and
t→(1+t^{1})log(1+t) eqn.AX016
is increasing so we have the bounds
<a name="ch07c030">
201002132102 start
■ Exercise 7.10 solution
eqn.AX017 inequality is because
change integrand log(1+t)/t to
its maximum value.
<a name="ch07c031">
Integration domain is t in [x,1]
and require "for all 0＜x＜1"
From exGraph0710 we see red
curve is monotone decrease. The
maximum integrand value occur
at left end where t=x. eqn.AX017
less than side change integrand
[log(1+t)]/t to maximum value and
<a name="ch07c032">Index beginIndex this file
move out of integration, what left
is integral of t from t=x to t=1.
This simple integration has value
1x. (1x) reunion with maximized
body [log(1+x)]/x become eqn.AX017
greater than side
(1x)*[log(1+x)]/x eqn.AX022
<a name="ch07c033">
From eqn.AX017 greater than side
to eqn.AX018 less than side is a
equality. That is because
(1x)*[log(1+x)]/x multiplied by
one. This one is (1+x)/(1+x)
Numerator (1+x) absorbed 1/x and
become (1+1/x). Result is
[(1x)/(1+x)] * (1+1/x)*log(1+x) eqn.AX023
<a name="ch07c034">
In eqn.AX023, do not bother
[(1x)/(1+x)],
because [(1x)/(1+x)] is part
of output. Just look at
(1+1/x)*log(1+x) eqn.AX024
eqn.AX024 is g(x) in exGraph0710
(1+1/x)*log(1+x) is a monotone
increase function. In eqn.AX017
and eqn.AX018 chain, g(x) is at
greater than side. To keep same
inequality, we replace g(x) with
its maximum value in domain
<a name="ch07c035">
which is domain right end t=1
In
(1+1/x)*log(1+x) eqn.AX024
replace x by 1 get constant
2*log(2).
Because 2*log(2) come from g(x)
evaluated at x=1. We can not
replace 2*log(2) by smaller value,
<a name="ch07c036">
If we do, then at the high end
(t=1) greater than condition is
not true any more. Violate exercise
given condition.
201002132128 stop
<a name="ch07c037">Index beginIndex this file
201002140930 start
■ Exercise 7.11 problem statement
textbook page 118
（A Continuous CarlemanType Inequality）
Given an integrable f:[a,b]→[0,∞)
and an integrable weight function
w:[a,b]→[0,∞) with integral 1 on
[a,b], show that one has
201002140941 stop
<a name="ch07c039">
201002140942 start
■ Exercise 7.11 hint
textbook page 257
If W(x) is the integral of w on
[a,x], then
W(a)=0 eqn.AX025
W(b)=1 eqn.AX026
and
W'(x)=w(x) eqn.AX027
so we have
page 257
Line 6
eqn.AX030
width of above equation
<a name="ch07c043">
201002141002 here
where we used first the fact that
0≦W(x)≦1 for all x in [a,b] and
then we applied Jensen's inequality.
201002141004 stop
<a name="ch07c044">
201002141101 start
■ Exercise 7.11 solution
Problem given
[[
an integrable weight function
w:[a,b]→[0,∞) with integral 1 on
[a,b]
]]
that is given
∫[x=a,b]w(x)dx=1 eqn.AX031
<a name="ch07c045">
Exercise 7.11 hint suggest to define
W(x)=∫[t=a,x]w(t)dt eqn.AX032
where a≦x≦b.
From eqn.AX032, we see
W(a)=∫[t=a,a]w(t)dt=0 eqn.AX033
(The area from t=a to t=a is zero)
and
W(b)=∫[t=a,b]w(t)dt=1 eqn.AX034
eqn.AX034 come from the given
relation eqn.AX031
<a name="ch07c046">
To see the relation between
uppercase W(x) and lowercase w(x)
we need refer to Leibniz's rule
Leibniz's rule
eqn.AX035
width of above equation
<a name="ch07c048">
Refer to eqn.AX032, find
W'(x)＝
d
dx
B(x)=x
∫
t=a
w(t)dt
＝
x
∫
a
∂w(t)
∂x
dt
＋ w(x)
d(x)
dx
－ w(a)
d(a)
dx
Apply Leibniz's rule to W(x) eqn.AX036
Blue terms are zero, red terms are one.
width of above equation
<a name="ch07c049">
201002141143 here
In eqn.AX036
∂f(x,t)/∂x = ∂w(t)/∂x = 0 eqn.AX037
d(x)/dx = 1 eqn.AX038
d(a)/dx = 0 eqn.AX039
eqn.AX037=0 because variable t
and variable x are independent
eqn.AX038=1 because identical
terms in deno. and numerator
eqn.AX039=0 because 'a' is
constant.
<a name="ch07c050">
Then eqn.AX036 reduce to
W'(x)＝w(x) eqn.AX040 (eqn.AX027)
which is
d[W(x)]/dx＝w(x)
or
d[W(x)]＝w(x)*dx eqn.AX041
We need one more preparation
equation, that is eqn.AX028
page 257
Line 3 aux
eqn.AX042
width of above equation
<a name="ch07c052">Index beginIndex this file
201002141200 here
In eqn.AX042, red term come from
eqn.AX041.
In eqn.AX042, integration lower
limit change from x=a to W(x)=0
come from eqn.AX033.
In eqn.AX042, integration upper
limit change from x=b to W(x)=1
come from eqn.AX034.
<a name="ch07c053">
In eqn.AX042 right side, change
variable from 'W(x)' to 'v'
then we have Exercise 7.11 hint
eqn.AX028 right end
∫[v=0,1]{log[v]}dv ＝ －1 eqn.AX043
How to prove eqn.AX043?
<a name="ch07c054">
Take an integration table, find
∫{log(x)}dx = x*log(x)  x eqn.AX044
eqn.AX043 has lower and upper
bound, then eqn.AX044 give us
∫[x=0,1]{log[x]}dx ＝
= [x*log(x)  x][x=1]
[x*log(x)  x][x=0]
= [1*log(1)  1][0*log(0)  0]
= 1 eqn.AX045
This is how we get 1 in eqn.AX043<a name="ch07c055">
Above are all preparation work..
Next begin exercise equation.
Start from eqn.7.27 left side
exp{∫[x=a,b]log[f(x)]*w(x)*dx} eqn.AX046
=exp{∫[x=a,b]log[f(x)]*w(x)*dx +11}
Here insert a zero=+11
+1 become exp(1) which is 'e'
1 become ∫[x=0,1]{log[x]}dx
it is preparation work of eqn.AX043
1 become ∫[x=a,b]{log[W(x)]}w(x)dx
it is preparation work of eqn.AX028<a name="ch07c056">
Continue //blue e from exp(+1)
exp{∫[x=a,b]log[f(x)]*w(x)*dx} eqn.AX047
=e*exp{∫[x=a,b]log[f(x)]*w(x)*dx
+∫[x=a,b]{log[W(x)]}w(x)dx
} //red line is 1
<a name="ch07c057">Index beginIndex this file
Now use log(a)+log(b)=log(a*b) eqn.AX048
eqn.AX047 become eqn.AX029
right side integral
exp{∫[x=a,b]log[f(x)]*w(x)*dx} eqn.AX049
=e*exp{∫[x=a,b]log[f(x)*W(x)]*w(x)*dx
We know
W(a)=0 eqn.AX025
W(b)=1 eqn.AX026
<a name="ch07c058">
W(x) is in the range 0≦W(x)≦1
In eqn.AX049, if drop W(x), create
an inequality which is greater
than beforedrop.
(5*0.6 = 3, drop 0.6 get 5 > 3)
Inequality from eqn.AX029 right
term to eqn.AX030 left term is
the result of drop W(x).
<a name="ch07c059">
Finally come to eqn.AX030 right
inequality.
201002141237 here
201002141248 start
We need apply Jensen Inequality
for convex function.
Jensen Inequality is next
<a name="ch07c060">
compare with
e*exp{∫[log[f(x)]*w(x)dx}≦
e*∫exp{log[f(x)]}*w(x)dx eqn.AX050
g(x) in eqn.6.2 is exp() in
eqn.AX050.
Function exp(x) is convex. We
can apply Jensen Inequality.
<a name="ch07c061">
The greater than side of eqn.AX050
has exp{log(y)) which is y,
because exp() and log() cancel
each other. Then greater than side
of eqn.AX050 become
e*∫f(x)*w(x)dx eqn.AX051
This is eqn.7.27 greater than side
term. Problem solved.
201002141308 stop
<a name="ch07c062">
To solve Exercise 7.11, Liu,Hsinhan
rely upon previous reading notes
at book margin on 20090214 and
20090607. Otherwise, today's
work should spend longer time.
201002141311 stop
<a name="ch07c063">Index beginIndex this file
201002141525 start
■ Exercise 7.12 problem statement
textbook page 119
（Gruss's InequalityIntegral
of Products）
<a name="ch07c064">
Suppose that ∞＜α≦A＜∞ and
∞＜β≦B＜∞ and suppose that
function f and g satisfy the
bounds
α≦f(x)≦A eqn.AX052
and
β≦g(x)≦B eqn.AX053
for all 0≦x≦1
Show that one has the bound
≦ (Aα)(Bβ)/4 //This is part of eqn.AX054
page 119
Line 6
eqn.AX054
width of above equation
and show by example that the
factor of 1/4 cannot be replaced
by a smaller number.
201002141542 stop
<a name="ch07c066">
201002141547 start
■ Exercise 7.12 hint
textbook page 257
Setting I_{f} to the integral of f
we have
≦(A－I_{f})(I_{f}α) //This is part of eqn.AX055
page 257
Line 10
eqn.AX055
width of above equation
<a name="ch07c068">Index beginIndex this file
and an analogous inequality holds for g.
Schwarz's inequality then gives

x=1
∫
x=0
f(x)g(x)dx
－I_{f}*I_{g}

2
＝

x=1
∫
x=0
(f(x)I_{f})(g(x)I_{g})dx

2
page 257
Line 13
eqn.AX056
continue next line
width of above equation
<a name="ch07c069">
≦
x=1
∫
x=0
(f(x)－I_{f})^{2}dx
x=1
∫
x=0
(g(x)－I_{g})^{2}dx
page 257
Line 14
eqn.AX057
continue next line
width of above equation
≦ (AI_{f})(I_{f}α)(BI_{g})(I_{g}β) eqn.AX058
≦ [(Aα)^{2}/4]*[(Bβ)^{2}/4] eqn.AX059
<a name="ch07c070">
Where in the last step we used
the fact that
(Ux)(xL)≦(UL)^{2}/4 eqn.AX060
for all L≦x≦U
Finally to see that Gruss's inequality
is sharp,
set f(x)=1 for 0≦x≦1/2 eqn.AX061
set f(x)=0 for 1/2＜x≦1 eqn.AX062
and
set g(x)=1f(x) for all 0≦x≦1. eqn.AX063
201002141731 stop
<a name="ch07c071">Index beginIndex this file
201002141751 start
■ Exercise 7.12 solution
LiuHH do not know how to show
equality in eqn.AX055 is true.
LiuHH skip Exercise 7.12 solution
<a name="ch07c072">
If view from dimension analysis
eqn.AX055 left end integrand is
special. Since
[[
Setting I_{f} to the integral of f
]]
<a name="ch07c073">
I_{f} has dimension of
f*dx which is f*x (dimension)
eqn.AX055 left end integrand is
(f(x)  I_{f}) that is
(f  f*x) it can be true only
if independent variable x is
pure number. Even if this is
true, LiuHH still can not find
why equality in eqn.AX055 is true.
201002141758 stop
========= Chapter seven end here =========
<a name="ch08a001">Index beginIndex this file
201002151416 start
■■Chapter 08: The Ladder of
Power Means
Please read textbook, textbook
has better explanation and more
material. This file is just
study notes. May contain error
ans has less material.
<a name="ch08a002">Index beginIndex this file
■ Problem 8.1 (The Geometric
Mean as a Limit)
For nonnegative real numbers x_{k}
k=1,2,...,n and nonnegative
weights p_{k} k=1,2,...,n with
total mass
p_{1}+p_{2}+...+p_{n}=1 eqn.AX064
one has the limit
<a name="ch08a003">
Attention! eqn.8.2 left summation become
right side multiplication. Coefficient p_{k}
become power p_{k}, and t disappear.
lim
t→0
{
k=n
∑
k=1
p_{k}x_{k}^{t}
}
1/t
＝
k=n
∏
k=1
x_{k}^{pk}
define
＝
M_{0}
page 120
Line 18
eqn.8.2
width of above equation
Alert! eqn.8.2 is valid for t→0
If t＞0, ∑ side ＞ ∏ side, eqn.8.6
If t＜0, ∑ side ＜ ∏ side, eqn.8.12
<a name="ch08a004">
201002151438 here
If two functions f(t) and g(t)
ratio f(t)/g(t) has a limit value
P as variable t approach to zero.
If limit value P is zero
<a name="ch08a006">
We use lowercase symbol o()
f(t)=o(g(t)) as t→0 eqn.AX066
to represent this zero limit
case.
Similarly, if limit value P is
finite (or bounded)
<a name="ch08a008">
We use uppercase symbol O()
f(t)=O(g(t)) as t→0 eqn.AX068
to represent this finite limit
case.
This little o() and big O() method
is called Landau's Notation.
<a name="ch08a009">
Big O help us at the following
solution. Big O relate to two
functions.
First function, exam
h(x)=log(1+x) eqn.AX069
From integration table, find
log(1+x)=x x^2/2 +x^3/3
x^4/4 ++ ... eqn.AX070
require x^2＜1
(Peirce table page 91, item 768)
<a name="ch08a010">
Our focus is x approach to zero,
requirement x^2＜1 match our goal.
In eqn.AX070, when x approach to
zero, x^1 become dominant, higher
power term (x^2 or higher) are
smaller than x^1. We write
h(x)=log(1+x)=x + f(x) eqn.AX071
f(x)=x^2/2 +x^3/3x^4/4 ++ ... eqn.AX072
g(x)= x^2 eqn.AX073
<a name="ch08a011">
Because
lim[x→0]{f(x)/g(x)}=1/2 eqn.AX074
it is finite (qualified for o,O)
and nonzero (qualified for O)
we write
h(x)=x + O(x^2) as x→0 eqn.AX075
or write
log(1+x)=x + O(x^2) as x→0 eqn.8.3
<a name="ch08a012">Index beginIndex this file
Second function, exam
h(x)=exp(x) eqn.AX076
Peirce table page 89, item 759
give us
exp(x)=1 +x +x^2/2! +x^3/3!
+x^4/4! +... eqn.AX077
<a name="ch08a013">
Again,
Our focus is x approach to zero,
In eqn.AX077, when x approach to
zero, x^1 become dominant, higher
power term (x^2 or higher) are
smaller than x^1.
We write
h(x)=exp(x)=1 +x + f(x) eqn.AX078
f(x)=+x^2/2! +x^3/3!+x^4/4! ... eqn.AX079
g(x)= x^2 eqn.AX080
<a name="ch08a014">
Because
lim[x→0]{f(x)/g(x)}=+1/2! eqn.AX081
it is finite (qualified for o,O)
and nonzero (qualified for O)
we write
h(x)=1+x + O(x^2) as x→0 eqn.AX082
or write
exp(x)=1+x+O(x^2) as x→0 eqn.8.4
<a name="ch08a015">
We will use eqn.8.3 (log())
and eqn.8.4 (exp()) in the
following discussion.
201002151536 here
<a name="ch08a016">
Now back to our target eqn.8.2
To take down the power (1/t)
in eqn.8.2, we can take log
for the whole eqn.8.2. To
simplify the following math
equation, we omit "lim t→0"
"lim t→0" is not present in
the following equation, but
"lim t→0" is still working!
<a name="ch08a017">Index beginIndex this file
Take log and hide "lim t→0"
eqn.8.2 become the following
log
{(
k=n
∑
k=1
p_{k}x_{k}^{t}
)
1/t
}
＝
1
t
log
{
k=n
∑
k=1
p_{k}e^{t*log(xk)}
}
page 121
Line 20
eqn.AX083
width of above equation
<a name="ch08a018">
We used twice
log(a^b)=b*log(a) eqn.AX084
We used once
exp(log(c))=c eqn.AX085
eqn.AX083 left side whole term
apply eqn.AX084 once and move
power (1/t) to right side leading
term.
<a name="ch08a019">
eqn.AX083 right side treat x_{k}^{t}
as 'c' in eqn.AX085. Push x_{k}^{t}
to power above 'e'. Then second
time apply eqn.AX084 to x_{k}^{t} and
move 't' to become 't*log(x_{k})'
All operation result equality.
<a name="ch08a020">
eqn.AX083 right side exp() is
exp(t*log(x_{k})) and big O() is
exp(x)=1+x+O(x^2) as x→0 eqn.8.4
we have "t*log(x_{k})" as x
then 1+x become 1+t*log(x_{k})
eqn.8.4 has x approach to zero,
"t*log(x_{k})" has t approach to zero.
<a name="ch08a021">
Now use eqn.8.4 to simplify
exp(). This simplification is
valid only if t approach to zero.
eqn.AX083
right side
＝
1
t
log
{
k=n
∑
k=1
p_{k}
(
1 + t*log(x_{k}) +O(t^{2})
)}
red term applied eqn.8.4
page 121
Line 21
eqn.AX086
width of above equation
<a name="ch08a022">Index beginIndex this file
201002151626 stop
201002151707 start
In eqn.AX086 change {term} as
following
∑[k=1,n]p_{k}[1 + t*log(x_{k}) +O(t^{2})]
=∑[k=1,n]p_{k}*1 //blue term = 1
+∑[k=1,n]{p_{k}*t*log(x_{k})} eqn.AX087
+∑[k=1,n]p_{k}*O(t^{2}) //blue term = 1
<a name="ch08a023">
We have
p_{1}+p_{2}+...+p_{n}=1 eqn.AX064
so two blue terms in eqn.AX087
are one. eqn.AX087 become
1+∑[k=1,n]{p_{k}*t*log(x_{k})}+O(t^{2}) eqn.AX088
t is independent from k summation
move t out of k summation, get
1+t*∑[k=1,n]{p_{k}*log(x_{k})}+O(t^{2}) eqn.AX089
<a name="ch08a024">
Put eqn.AX089 back to {term} in
eqn.AX086
eqn.AX086+
eqn.AX089
＝
1
t
log
{
1 + t*
k=n
∑
k=1
p_{k}*log(x_{k}) +O(t^{2})
}
page 121
Line 22
eqn.AX090
width of above equation
<a name="ch08a025">
In eqn.AX090, we have
log(1 + something), eqn.8.3 is
called here. Log{} in eqn.AX090
become
eqn.AX090
+ eqn.8.3
＝
1
t
{
t*
k=n
∑
k=1
p_{k}*log(x_{k}) +O(t^{2})
}
red term applied eqn.8.3
page 121
Line 22
eqn.AX091
width of above equation
<a name="ch08a026">
In eqn.AX091, denominator 't'
cancel numerator 't' in summation
term and cancel one 't' in big O
We find
eqn.AX091=∑[k=1,n]p_{k}*log(x_{k}) +O(t^{one})
Complete equation start from
left side of eqn.AX083 and end
at eqn.AX091. Put two end into
one equation as following
page 121
Line 20/23
eqn.AX092
width of above equation
<a name="ch08a028">
201002151748 here
We always have t approach to
zero, eqn.AX092 right side
O(t) drop out when t→0.
eqn.AX092 right side, let p_{k}
regain power, change
from p_{k}*log(x_{k})
to log(x_{k}^{pk})
eqn.AX092 become
page 121
Line 20/23
eqn.AX093
width of above equation
<a name="ch08a030">
201002151756 here
Take exp() for eqn.AX093 whole
equation.
Left side exp(log(y)) become y
right side exp(summation) become
product of exp(single), and
exp(single) has one more time
exp(log(z)) become z.
The result is our target eqn.8.2
Is it an easy problem ?
Yes, only if thorough understand !!
201002151800 stop
<a name="ch08a031">Index beginIndex this file
201002151937 start
■ Problem 8.2 (Power Mean Bound
for the Geometric Mean)
Follow in Siegel's footsteps and
prove that for any nonnegative
weights p_{k} k=1,2,...,n with total
mass
p_{1}+p_{2}+...+p_{n}=1 eqn.AX094
and for any nonnegative real
numbers x_{k} k=1,2,...,n, one has
the bound
page 122
Line 24
eqn.8.6
width of above equation
Alert: eqn.8.2, t→0, ∏ term ＝ ∑ term
Alert: eqn.8.6, t＞0, ∏ term ≦ ∑ term
Alert: eqn.8.12, t＜0, ∏ term ≧ ∑ term
<a name="ch08a033">
201002151958 here
Textbook defined M_{t} as following
M_{t} = M_{t}[x;p] ≡
{
k=n
∑
k=1
p_{k}x_{k}^{t}
}
1/t
page 120
Line 3
eqn.8.1
width of above equation
<a name="ch08a034">
With this definition, we have the following variation.
M_{t}^{t} =
k=n
∑
k=1
p_{k}x_{k}^{t}
M_{2t} =
{
k=n
∑
k=1
p_{k}x_{k}^{2t}
}
1/(2t)
M_{2t}^{t} =
{
k=n
∑
k=1
p_{k}x_{k}^{2t}
}
t/(2t)
eqn.AX095
eqn.AX096
eqn.AX097
width of above equation
<a name="ch08a035">
201002152026 here
Read eqn.8.6 and think this
way. Assume t=2＞0, what happen
if t is halved?
M_{t}=M_{2}, t=2 is far away from t=0,
eqn.8.6 two side gap is bigger.
M_{t}=M_{1}, t=1 is closer to t=0,
eqn.8.6 two side gap is smaller.
M_{t}=M_{0}, t=0 is same as t→0,
eqn.8.6 two side gap is zero.
<a name="ch08a036">
With above observation, we can
see
M_{0}≦M_{1}≦M_{2} eqn.AX098
We hope to prove
M_{t}≦M_{2t} eqn.8.7
How can we create a 1t and 2t
relation? Cauchy ! Cauchy !
<a name="ch08a037">Index beginIndex this file
Cauchy's splitting trick will
help us. Write
p_{k}x_{k}^{t}=p_{k}^{1/2}*p_{k}^{1/2}x_{k}^{t} eqn.AX099
Apply Cauchy's inequality for
two sequences p_{k}^{1/2} and p_{k}^{1/2}x_{k}^{t}
Use eqn.AX095 for M_{t}^{t}, we find
M_{t}^{t}=∑[k=1,n]p_{k}x_{k}^{t} eqn.AX100 begin
M_{t}^{t}=∑[k=1,n]p_{k}^{1/2}*p_{k}^{1/2}x_{k}^{t}
Blue 2,t before Cauchy,
Red 2, 1/2 from Cauchy
page 123
Line 4
eqn.AX100 end
width of above equation
<a name="ch08a039">
201002152107 here
Refer to eqn.AX094
"eqn.AX100 end" (left term)=1
Refer to eqn.AX097
"eqn.AX100 end" (right term)=M_{2t}^{t}
From "eqn.AX100 begin"
to "eqn.AX100 end" the whole
equation is
M_{t}^{t}≦M_{2t}^{t} eqn.AX101
eqn.AX101 take tth root get
M_{t}≦M_{2t} eqn.8.7 (M_{t} definition)
<a name="ch08a040">
Up to now, proved M_{t}≦M_{2t}, we can
iterate prove, for all real t
we have the following halves
inequality.
First present a numerical example,
then present a symbolic equation.
Assume t=16, then
M_{0.5}≦M_{1}≦M_{2}≦M_{4}≦M_{8}≦M_{16} eqn.AX102
in symbol, eqn.AX102 become
M_{t/2j}≦M_{t/2j1}≦...≦M_{t/2}≦M_{t} eqn.AX103
<a name="ch08a041">
In eqn.AX103, increase j to infinity
Push M_{t/2j} to M_{0}
In the limit, we have
lim[j→∞]M_{t/2j}=M_{0}
From eqn.8.2, we know
M_{0} = ∏[k=1,n]x_{k}^{pk}eqn.8.6 is proved. (M_{t} definition)
201002152133 stop
<a name="ch08a042">Index beginIndex this file
201002161033 start
■ Problem 8.3 (Power Mean
Inequality)
Consider positive weights p_{k}
k=1,2,...,n which have total mass
p_{1}+p_{2}+...+p_{n}=1 eqn.AX104
and show that for nonnegative
real numbers x_{k} k=1,2,...,n the
mapping t→M_{t} is a nondecreasing
function on all of Real.
<a name="ch08a043">
That is show that for all
∞＜s＜t＜∞
one has
{
k=n
∑
k=1
p_{k}x_{k}^{s}
}
1/s
≦
{
k=n
∑
k=1
p_{k}x_{k}^{t}
}
1/t
page 123
Line 26
eqn.8.10
width of above equation
<a name="ch08a044">
Finally, show that then one has
equality in the bound (8.10) if
and only if x_{1}=x_{2}=...=x_{n}
201002161051 stop
<a name="ch08a045">
201002161058 start
Please compare next two equations
{
k=n
∑
k=1
p_{k}x_{k}
}
p
≦
{
k=n
∑
k=1
p_{k}x_{k}^{p}
}
page 124
Line 5
eqn.AX105
width of above equation
and //Alert: power index p is NOT p_{k}
<a name="ch08a046">
Jensen Inequality need a base
function and this function must
be convex on domain. How to find
the base function from given?
From both equation greater than
side, we see directly that
f(x_{k})=x_{k}^{p} eqn.AX106
From both equation less than
side, we see
f(∑px)={∑px}^{p} eqn.AX107
<a name="ch08a047">Index beginIndex this file
eqn.AX106 and eqn.AX107 suggest
that
f(x)=x^{p} eqn.AX108
Jensen Inequality for convex
function require that
p>1 eqn.AX109
Please pay attention to p and p_{k}
they are different, because
P＞1 but 0≦p_{k}≦1.
201002161128 stop
//debug HelloJensen() correction at 9902161158
<a name="ch08a048">
201002161214 start
f1(x)=x^{p} and p＞1 eqn.AX110
f2(x)=x^{p} and p＝1 eqn.AX111
f3(x)=x^{p} and p＜1 eqn.AX112
f1(x) is convex; shape like "╰╯"
f2(x) convex and concave; shape "/"
f3(x) is concave; shape like "╭╮"
<a name="ch08a049">
We will solve Problem 8.3 use
Jensen Inequality and convex
function
f(x)=x^{p} eqn.AX108
Let us assume that
0＜s＜t eqn.AX113
Define
p=t/s＞1 eqn.AX114
This definition make sure that p＞1
and f1(x)=x^{p} is convex.
<a name="ch08a050">
In eqn.AX105, replace
p → t/s＞1 eqn.AX115
replace nonnegative real numbers
x_{k} → y_{k}^{s} eqn.AX116
We find