<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="ch06d001">　Index begin　Index this file
2009-12-29-18-16 start
■　Exercise 6.5 problem statement
　　textbook page 100
（Superadditivity of the Geometric Mean）
We have seen before in Exercise 2.11
that for nonnegative aj and bj,
j=1,2,...,n one has superadditivity
of the geometric mean:
<a name="ch06d002">
  (a1*a2*...*an)1/n+(b1*b2*...*bn)1/n
 ≦ [(a1+b1)*(a2+b2)*...*(an+bn)]1/n ---eqn.AS001
Does this also follow from Jensen's 
inequality?
2009-12-29-18-26 stop






<a name="ch06d003">
2009-12-29-18-27 start
■　Exercise 6.5 hint
　　textbook page 248
To build a proof with Jensen's inequality,
we first divide by (a1*a2*...*an)1/n and
write ck for bk/ak, so the target
inequality takes the form
  1 + (c1*c2*...*cn)1/n  ---eqn.AS002
 ≦ [(1+c1)*(1+c2)*...*(1+cn)]1/n
<a name="ch06d004">
Now if we take logs and write cj as 
exp(dj), we find it takes the form
  log(exp(d_bar))≦  ---eqn.AS003
  ∑[j=1,n]{log(1+exp(dj))}/n
where
  d_bar=(d1+d2+...+dn)/n  ---eqn.AS004
<a name="ch06d005">
Finally the last inequality is simply
Jensen's inequality for the convex
function x → log(1+exp(x)), so the
solution is complete. One feature of
this solution worth noting is that
progress came quickly after division
reduced the number of variables
from 2n to n. This phenomenon is
actually rather common, and such
reductions are almost always worth 
a try.

<a name="ch06d006">
Here it is perhaps worth noting that
Minkowski's proof used yet another
idea. Specifically, he built his proof
on analysis of the polynomial
  p(t)=∏(aj+t*bj) ---eqn.AS005
Can you recover his proof?
2009-12-29-18-46 stop




<a name="ch06d007">　Index begin　Index this file
2009-12-29-20-07 start
■　Exercise 6.5 solution

Read the given equation eqn.AS001
with Prof. Steele's guidance, task
become easy. Divide eqn.AS001 by
(a1*a2*...*an)1/n, and define
  ck = bk/ak ---eqn.AS006
<a name="ch06d008">
so the target inequality takes the 
form
  1 + (c1*c2*...*cn)1/n  ---eqn.AS002
 ≦ [(1+c1)*(1+c2)*...*(1+cn)]1/n
Take nth power to eqn.AS002, find
  [1 + (c1*c2*...*cn)1/n]n  ---eqn.AS002 aux1
 ≦ [(1+c1)*(1+c2)*...*(1+cn)]1
Compare eqn.AS002 aux1 with Jensen's
Inequality

<a name="ch06d009">
■　from f1(x)=1+x to f3(x)=log(1+exp(x))
Jensen's Inequality need a base 
function. eqn.AS002 aux1 right side
strongly suggest a base function of 
the form
  f1(x)=1+x ---eqn.AS007
But f1(x)=1+x is not the right choice.
f1(x) is linear, convex and concave.
<a name="ch06d010">
Jensen's Inequality says
for convex function (see)
AM in DOMAIN ≦ AM in RANGE
for concave function (see)
AM in DOMAIN ≧ AM in RANGE

<a name="ch06d011">
f1(x)=1+x first thing wrong is that
we expact f(x_AM)=1+x_AM,
x_AM is Arithmetic Mean of function
parameters. 
Arithmetic Mean use addition divide
           by n.
 Geometric Mean use multiplication
           then take n th root.
<a name="ch06d012">　Index begin　Index this file
eqn.6.2 less than side function
parameter (DOMAIN) use AM ∑pjxj
But 
eqn.AS002 less than side function
parameter use GM (c1*c2*...*cn)1/n
AM and GM are not compatible.
<a name="ch06d013">
Instead of
  f1(x)=1+x ---eqn.AS007
if we use
  f2(x)=1+exp(x) ---eqn.AS008
Assume we have x1,x2,x3
then 
<a name="ch06d014">
  f2(x_AM)=1+exp(x_AM)
          =1+exp((x1+x2+x3)/3)
          =1+[exp(x1)*exp(x2)*exp(x3)]1/3
if define yk=exp(xk), then
  f2(x_AM)=
  f2(y_GM)=1+[y1*y2*y3]1/3
Above transformation achieve our goal: 
express x_AM in y_GM

<a name="ch06d015">
Jensen's Inequality says
for convex function (see)
AM in DOMAIN ≦ AM in RANGE
Next we need see AM in RANGE.
Our improved base function choice is
  f2(x)=1+exp(x) ---eqn.AS008
Assume we have x1,x2,x3
<a name="ch06d016">
AM in RANGE is
  f2_AM={[1+exp(x1)]+[1+exp(x2)]
        +[1+exp(x3)]}/3 ---eqn.AS009
eqn.AS002 greater than side is
  [(1+c1)*(1+c2)*...*(1+cn)]1/n ---eqn.AS010
It is again AM/GM trouble.
Jensen  ask for AM in range,
Problem ask for GM in range.
<a name="ch06d017">　Index begin　Index this file
We know that 
  [log(a)+log(b)+log(c)]/3
  = log[(a*b*c)1/3] ---eqn.AS011
For example, if a=2, b=3, c=4
  (log(2)+log(3)+log(4))/3
  =  ---eqn.AS012
  log(pow((2*3*4),1/3))
  =1.0593512767826485
<a name="ch06d018">
In Javascript code, it is
  (Math.log(2)+Math.log(3)+Math.log(4))/3
  = Math.log(Math.pow(2*3*4,1/3)) ---eqn.AS013
Please paste next two lines 
[[
(Math.log(2)+Math.log(3)+Math.log(4))/3
Math.log(Math.pow(2*3*4,1/3))
]]
to complex2.htm#box03
Box3, and click
"test box3 command, output to box4"
answer are both
1.0593512767826485

<a name="ch06d019">
With the observation
  [log(a)+log(b)+log(c)]/3
  = log[(a*b*c)1/3] ---eqn.AS011
we change from
  f2(x)=1+exp(x) ---eqn.AS008
to
  f3(x)=log(1+exp(x)) ---eqn.AS014
because f3(x) is the final function
that help us solve problem. Drop '3'
write as
  f(x)=log(1+exp(x)) ---eqn.AS015

<a name="ch06d020">
This is how we get Prof. Steele 
suggested 
function x → log(1+exp(x))
We already meet log(1+exp(x)) before
Its convecity is derived at eqn.AR103
f(x)=log(1+exp(x)) ---eqn.AR103

Please goto tute0022.htm#clickJensenA
and click button [14] for log(1+exp(x)).

<a name="ch06d021">
Now let us start from
  f(x)=log(1+exp(x)) ---eqn.AS015
and use
  ck = bk/ak ---eqn.AS006
write ck as exp(dk)

In log(1+exp(x)) 'x' is replaced by dk
x is NOT replaced by ck

<a name="ch06d022">　Index begin　Index this file
AM in DOMAIN is
  f(x_AM)=log(1+exp(x_AM))
  f(x_AM)=log(1+exp([∑dk]/n))
  f(x_AM)=log(1 + (c1*c2*...*cn)1/n) ---eqn.AS016
note: exp(∑(dk)/n)=[∏exp(dk)]1/n ---eqn.AR102
note: define ck = exp(dk)
<a name="ch06d023">
AM in RANGE is
  f_AM=[log(1+exp(d1))+log(1+exp(d2))
       +...+log(1+exp(dn))]/n
  f_AM=[log(1+c1)+log(1+c2)
       +...+log(1+cn)]/n
  f_AM=log{[(1+c1)*(1+c2)*...*(1+cn)]1/n} ---eqn.AS017
note: ∑log(1+ck)=log{∏(1+ck)) ---eqn.AR059&60
<a name="ch06d024">
Jensen's Inequality says
for convex function (see)
AM in DOMAIN ≦ AM in RANGE
So
  f(x_AM)=log(1 + (c1*c2*...*cn)1/n)
   ≦   ---eqn.AS018
  f_AM=log{[(1+c1)*(1+c2)*...*(1+cn)]1/n}
<a name="ch06d025">
From eqn.AS018, take both side
parameter out of log(), we find
  1 + (c1*c2*...*cn)1/n
   ≦   ---eqn.AS019
  [(1+c1)*(1+c2)*...*(1+cn)]1/n
next is simple step, recover ak and bk
from
  ck = bk/ak ---eqn.AS006
<a name="ch06d026">
multiply a1*a2*...*an to the
result and get eqn.AS001.
Problem solved.
Last few steps, let you fill the detail.
2009-12-29-21-57 stop

2009-12-29-23-47 start
<a name="ch06d027">　Index begin　Index this file
■　if log(t1)＞log(t2) then t1＞t2
log(t) is a monotone increase function
if log(t1)＞log(t2) then their parameter
t1 and t2 MUST have same inequality
that is  t1＞t2
Monotone decrease function reverse.

<a name="ch06d028">
For example 
f(x)=log(x) is a monotone increase 
function. Require x > 0
log(8)=2.0794 ＞ 1.7917=log(6)
Their parameter has SAME inequality
   8 ＞ 6

<a name="ch06d029">
f(x)=exp(x) is a monotone increase 
function. x is any real number.
exp(-2)=0.13533 ＞ 0.04978=exp(-3)
Their parameter has SAME inequality
   -2 ＞ -3
Please goto convexMaxPt and
click [02] button to see monotone 
increase function.

Above is monotone increase function
<a name="ch06d030">
Below is Monotone decrease function

f(x)=1/x is a monotone decrease 
function. require x NOT= 0
f(-2)=-1/2 ＜ -1/3=f(-3)
Their parameter has REVERSE inequality
   -2 ＞ -3
Please goto convexMaxPt and
click [03] button to see monotone 
decrease function.

<a name="ch06d031">
f(x)=cos(x) is a monotone decrease 
function on x in (0, PI/2)=(0, 1.57079)
cos(0.1)=0.9950 ＞ 0.36235=cos(1.2)
Their parameter has REVERSE inequality
   0.1 ＜ 1.2
Please goto convexMaxPt and
click [04] button to see cos(x) for
x in [0, PI]

2009-12-30-00-10 stop





<a name="ch06d032">　Index begin　Index this file
2009-12-30-10-34 start
■　Exercise 6.6 problem statement
　　textbook page 101
（Cauchy's Technique and Jensen's
 Inequality）

In 1906, J.L.W.V. Jensen wrote an
article that was inspired by the
proof given by Cauchy's for the
AM-GM inequality, and, in an effort to
get to the heart of Cauchy's argument,
Jensen introduced the class of functions
that satisfy the inequality

<a name="ch06d034">
2009-12-30-10-46 here
Such functions are now called J-convex
function, and, as we note below in 
Exercise 6.7, they are just slightly
more general than the convex functions
defined by condition (6.1)

<a name="ch06d035">
For a moment, step into Jensen's shoe
and show how one can modify Cauchy's
leap-forward fall back induction
(page 20) to prove that for all 
J-convex functions one has

<a name="ch06d037">
2009-12-30-10-59 here
Here we might note that near the end
of his 1906 article, Jensen expressed
the bold view that perhaps someday
the class of convex function might
seen to be as fundamental as the class
of positive function or the class of 
increasing functions. If one allows 
for the mild shift from the specific
notion of J-convexity to the more
modern interpretation of convexity
(6.1), then Jensen's view turned out
to be quite prescient.
2009-12-30-11-04 stop







<a name="ch06d038">　Index begin　Index this file
2009-12-30-11-06 start
■　Exercise 6.6 hint
　　textbook page 249
Essentially no change is needed in 
Cauchy's argument (page 20). First, 
for the cases n=2k, k=1,2,... one 
just applies the defining relation 
(6.25) to successive halves. 
<a name="ch06d039">
For the
fall-back step, one chooses k such
that n≦2k and apply the 2k result 
to the padded sequence yj 1≦j≦2k 
which one defines by taking yj=xj 
for 1≦j≦n and by taking
yj=(x1+x2+...+xn)/n for n＜j≦2k.
2009-12-30-11-15 stop





<a name="ch06d040">
2009-12-30-11-27 start
■　Exercise 6.6 solution


Cauchy's leap-forward fall back 
induction is at tute0011.htm#ch02a021 

<a name="ch06d041">
Cauchy leap forward method is to pad dummy
sequence elements from a6, a7 to a8
This section is at tute0011.htm#ch02a029

Cauchy fall back section, cut dummy
is at tute0011.htm#ch02a033

Please review previous work.
LiuHH skip Exercise 6.6 solution.
2009-12-30-11-31 stop



<a name="ch06d042">　Index begin　Index this file
2009-12-30-12-12 start
■　Exercise 6.7 problem statement
　　textbook page 101
（Convexity and J-Convexity）

Show that if f:[a,b]→Real is continuous
and J-convex, then f must be convex in
the modern sense expressed by the
condition (6.1). As a curiosity, we
<a name="ch06d043">
should note that there do exist
J-Convex function that are not
convex in the modern sense. Never-
theless, such functions are wildly
discontinuous, and they are quite
unlikely to turn up unless they 
are explicitly invited.
2009-12-30-12-18 stop








<a name="ch06d044">
2009-12-30-12-20 start
■　Exercise 6.7 hint
　　textbook page 249
As we noted in the proceeding solution,
iteration of the defining condition 
(6.24) gives us for all k=1,2,... that

<a name="ch06d046">
2009-12-30-12-29 here
so setting xj=x for 1≦j≦m and
xj=y for m＜j≦2k, we also have

<a name="ch06d047">
2009-12-30-12-41 here
If we now choose mt and kt such that
mt/2kt→p as t→∞, then continuity of
f and the proceeding bound give us
convexity of the kind required by
the modern definition (6.1).
2009-12-30-12-45 stop







<a name="ch06d048">　Index begin　Index this file
2009-12-30-14-40 start
■　Exercise 6.7 discussion
Some problem is easier to understand
if use numerical example to explain.
The following is a numerical example.

First compare convex function definition
eqn.6.1 with J-convex definition eqn.6.24

<a name="ch06d049">
A function f:[a,b]→Real is said to be
convex function provided that for all
x,y∈[a,b] and all
  0≦p≦1  ---eqn.AQ003
one has　　Do we prove eqn.6.1?
  f(px +(1-p)y)≦
  pf(x)+(1-p)f(y)    ---eqn.6.1
Above convex, below J-convex
if f:[a,b]→Real is continuous

<a name="ch06d051">
In eqn.6.1, if we set p=1/2, then 
convex eqn.6.1 is same as J-convex
eqn.6.24.

Exercise 6.7 given eqn.6.24 and 
ask to prove eqn.6.24 satisfy
convex eqn.6.1.

<a name="ch06d052">　Index begin　Index this file
Exercise 6.7 hint suggest repeat
apply J-convex eqn.6.24 many times
and in the limit, J-convex approach 
to convex eqn.6.1.

<a name="ch06d053">
For 2k, take k=1 get 21=2
J-convex first iteration is
  f([x1+x2]/2)≦[f(x1)+f(x2)]/2 ---eqn.AS022
[x1+x2]/2 is a new point in domain.
To start next iteration, need one
more such point. Create next equation
  f([x3+x4]/2)≦[f(x3)+f(x4)]/2 ---eqn.AS023
<a name="ch06d054">
Now apply J-convex to two points
[x1+x2]/2 and [x3+x4]/2
we find
  f([x1+x2+x3+x4]/2/2)
  ≦ ---eqn.AS024
  {f([x1+x2]/2)+f([x3+x4]/2)}/2
<a name="ch06d055">
Put eqn.AS022 and eqn.AS023 to right
side of eqn.AS024, get
  f([x1+x2+x3+x4]/2/2)
  ≦ ---eqn.AS025
  [f(x1)+f(x2)+f(x3)+f(x4)]/2/2
We started from 2k, k=1 get 21=2
arrive to 2k, k=2 get 22=4
<a name="ch06d056">
In eqn.AS025, "/2/2" is 4=22
eqn.AS025 created one point 
[x1+x2+x3+x4]/2/2
We need another such point, 
[x5+x6+x7+x8]/2/2

<a name="ch06d057">　Index begin　Index this file
next iteration is k=3, 2k=23=8
next iteration we shall see "/2/2/2"
Above is numerical example, explain
how each iteration grow. 
Exercise 6.7 hint eqn.AS020 use
symbolic equation for same thing.
symbolic equation is general.

<a name="ch06d058">
After grow big enough, need smooth.
Exercise 6.7 hint say
[[
setting xj=x for 1≦j≦m and
xj=y for m＜j≦2k,
]]
That is to set
  x1=x2=...=xm=x ---eqn.AS026
and set
  xm+1=xm+2=...=x2k=y ---eqn.AS027
<a name="ch06d059">
we get eqn.AS021.
Let m and k be function of t. We continue
J-convex divide by 2 process. In the limit
[[
mt/2kt→p as t→∞, then continuity of
f and the proceeding bound give us
convexity of the kind required by
the modern definition (6.1).
]]
The condition t→∞ is to break rational
number limitation and take care 
irrational number.

Possibly you can do better explanation
than LiuHH did.
2009-12-30-15-46 stop






<a name="ch06d060">
2009-12-30-15-47 start
■　Exercise 6.7 solution


Please read 
Exercise 6.7 hint
and
Exercise 6.7 discussion.
2009-12-30-15-48 stop



<a name="ch06d061">　Index begin　Index this file
2009-12-30-18-25 start
■　Exercise 6.8 problem statement
　　textbook page 101
（A "One-Liner" That Could Have
 Taken All Day）

Show that for all 0≦x,y,z≦1, one
has the bound

<a name="ch06d063">
2009-12-30-18-39 here
Placed suggestively in a chapter on
convexity, this problem is not much
more than a one-liner, but in a less
informative location, it might send
one down a long trail of fruitless 
algebra.
2009-12-30-18-42 stop






<a name="ch06d064">
2009-12-30-18-53 start
■　Exercise 6.8 hint
　　textbook page 249
The function L(x,y,z) is convex in each
of its three variables separately and, 
by the argument detailed below, this
implies that L must attain its maximum
at one of the vertices of the cube.
(LiuHH note: Please see Theorem 2)
After eight easy evaluation we find
that L(1,0,0)=2 and that no other
corner has a larger value, so the
solution is complete.

<a name="ch06d065">
It is also easy to show that if a 
function on the cube is convex in
each variable separately, then the
function must attain its maximum on
one of the corner points. In essence 
one argues by induction but, for the
cube in R3, one may as 
well give all of the steps.

<a name="ch06d066">
First, one notes that a convex 
function on [0,1] must take its 
maximum at one of the end points 
of the interval, so, for any fixed
values of y and z, we have the bound
  L(x,y,z)≦max{L(0,y,z),L(1,y,z)} ---eqn.AS029
Similarly, by convexity of y→L(0,y,z)
and y→L(1,y,z), so 
L(0,y,z) is bounded by max{L(0,0,z),L(0,1,z)} 
and 
L(1,y,z) is bounded by max{L(1,0,z),L(1,1,z)}
<a name="ch06d067">
All together, we have for each value
of z that L(x,y,z) is bounded by 
max{L(0,0,z),L(0,1,z),L(1,0,z),L(1,1,z)}

Convexity of z→L(x,y,z) applied four
times then gives us the final bound
  L(x,y,z)≦max{L(e1,e2,e3): ---eqn.AS030
           ek=0 or ek=1 for k=1,2,3}

<a name="ch06d068">
One should note that this argument
does not show that one can find the
maximum by the "greedy algorithm"
that performs three successive 
maximums. In fact the greedy algorithm
can fail miserably here, as easy 
example show.
2009-12-30-19-21 stop


<a name="ch06d069">　Index begin　Index this file
2009-12-30-19-22 start
■　Exercise 6.8 CSMC_Errata.pdf notes

2009-02-01-15-22 LiuHH accessed
http://www-stat.wharton.upenn.edu/~steele/Publications/Books/CSMC/CSMC_Errata.pdf
find

Pages 101/249－250, Exercise 6.8: 
Robin Chapman points out that the 
solution asserts that L(x; y; z) 
is convex in all variables, but

<a name="ch06d071">
2009-12-30-19-37 here
and so
  ∂2L(1,0,z)/∂z2 = -1 ---eqn.AS032

Thus, L(x, y, z) is not convex in 
z. This remains on my "to fix list"
which I expect to get to in early 
2007.

(here "my" is Prof. J.M. Steele
 Liu,Hsinhan note)
2009-12-30-19-40 stop


<a name="ch06d072">　Index begin　Index this file
2009-12-30-20-18 start
■　Exercise 6.8 LangeListCSMCTypos.pdf
　　Study LangeListCSMCTypos.pdf
2009-02-01-15-27 LiuHH accessed
http://www-stat.wharton.upenn.edu/~steele/Publications/Books/CSMC/LangeListCSMCTypos.pdf
find

<a name="ch06d073">
[ Next "me" is the author of
  LangeListCSMCTypos.pdf
  Liu,Hsinhan note.]

14. It took me "all day" to come up 
with the following solution:
It is easy to show by convexity or 
algebra that
  1/(1+u) ≦ 1 - u/2 ---eqn.AS033
for u∈[0, 1] and
  1/(1+u) ≦ 1 - u/3 ---eqn.AS034
for u∈[0, 2]. Hence

<a name="ch06d076">　Index begin　Index this file
Define eqn.AS036 = g(x, y, z)

on the stated domain x, y, z ∈[0, 1]. 
The second derivative test demonstrates 
that g(x, y, z) is convex in x for y and
 z fixed. Hence, g(x, y, z) attains its 
maximum at x = 0 or x = 1. Now
<a name="ch06d077">
  g(1,y,z)=(1-y/2)+y2(1-z/2)+z2(2/3-y/3)
            +(y2-1)(z2-1) ---eqn.AS037
  g(0,y,z)=y2(1-z/2)+z2(1-y/3) ---eqn.AS038
and so
  g(1,y,z)-g(0,y,z)=(1-y/2)-z2/3
          +(y2-1)(z2-1) ≧ 0 ---eqn.AS039

<a name="ch06d078">
Since 1≧y/2+z2/3 and the product 
(y2-1)(z2-1) is nonnegative. 
Finally, g(1,y,z)≦2
if and only if
  (1-y/2)+y2(1-z/2)+z2(2/3-y/3)
   +y2z2-y2-z2+1≦2 ---eqn.AS040

<a name="ch06d079">
if and only if
  -y/2 -y*y*z/2 -z*z*y/3
  +y*y*z*z -z*z/3 ≦0 ---eqn.AS041

The last inequality is true because
  y*y*z*z ≦ y/2 + y*y*z/2 ---eqn.AS042
　　Study LangeListCSMCTypos.pdf
2009-12-30-20-57 stop

<a name="ch06d080">　Index begin　Index this file
2009-12-31-14-17 start
■　Maximum value point on convex function

in Exercise 6.8 hint we read
[[
The function L(x,y,z) is convex in each
of its three variables separately and, 
by the argument detailed below, this
implies that L must attain its maximum
at one of the vertices of the cube.
]]

<a name="ch06d081">
1-D convex function shape like 'U'. Its
minimum value point is an interior point.
Now the question is that where is the
maximum value point on convex function?
Hint suggest the maximum value point is
at the end of convex domain. Is this 
right? If it is right, how to prove?
2009-12-31-14-22 here

<a name="ch06d082">
2009-12-31-15-15 start
From 14:25 to 15:12 LiuHH modified page
tute0022.htm put example button to three
groups, convex/concave/both. Added
three button 10/11/12.

<a name="ch06d083">
Next continue
Maximum value point on convex function

On 2009-12-01-20-46 LiuHH access next page
http://www.nzamt.org.nz/nzimo/wp-content/uploads/2009/01/convex-functions.pdf
page 4/7 has a related theory as following

<a name="ch06d084">　Index begin　Index this file
■　www.nzamt.org.nz/.../convex-functions.pdf
(LiuHH explanation)
Next Theorem 2 is copied from 
www.nzamt.org.nz/.../convex-functions.pdf

Theorem 2. A strongly convex function 
on a closed bounded interval attains 
its maximum at one of its end-points.

Proof. Suppose f : [a, b]→R is strongly
 convex. Then for any t∈[a, b], taking
 x, y, μ to be a, b, (b-t)/(b-a)
respectively, we find

<a name="ch06d086">
2009-12-31-15-45 here
That is, f(t) is at most some weighted
 average of f(a) and f(b). This means 
that at least one of f(a) and f(b)
is greater than or equal to f(t):

It follows that
  max{t∈[a,b]}f(t)=max[f(a),f(b)] ---eqn.AS045

<a name="ch06d088">
The idea here is:
If the average of a set of numbers is
 m, then at least one of the numbers 
is at least m.
The Pigeonhole Principle is another 
version of this same trick.

Above is copied from
http://www.nzamt.org.nz/nzimo/wp-content/uploads/2009/01/convex-functions.pdf
page 4/7 Theorem 2.
2009-12-31-15-58 stop

<a name="ch06d089">　Index begin　Index this file

■　When convex function attains maximum
　　at interior point? answer
2009-12-31-19-32 start
Explain convex-functions.pdf Theorem 2
as following.
[[
Theorem 2. A strongly convex function 
on a closed bounded interval attains 
its maximum at one of its end-points.
]]

<a name="ch06d090">
The if conditions are
first  if "A strongly convex function"
second if "on a closed bounded interval"
the conclusion is
attains its maximum at one of its end-points.

<a name="ch06d091">
first  if 
"A strongly convex function" relative to
"A  weakly  convex function"
Please click Draw Convex [01] button
Red curve is  f(x)=-sin(x)
              for x in [0,PI].
Blue curve is f(x)=-sin(x)
              for x in [0,PI/2].
Plus          f(x)=-1 for x in [PI/2,PI].
(Avoid overlap, blue shift -0.035, ignore it)
Both red and blue are convex.

<a name="ch06d092">
Red  curve is "strongly convex function"
Blue curve is "weakly   convex function"
Blue curve is not strongly convex, because
blue curve has one linear section.
Both attains its maximum at left end.

<a name="ch06d093">
Since blue curve still attains its maximum
at end point. Not attain maximum at interior
point. The linear section of blue curve
provide only minimum value point, not give
max. value.
Can we relax 
first  if "A strongly convex function"
to
first  if "A convex function"
?
<a name="ch06d094">
LiuHH think
"if a strongly convex function"
is better than
"if a convex function"
Because a whole horizontal straight line
can also be called as "A convex function"
in this case, "max. value" point is any
interior point.

above is
first  if "A strongly convex function"
<a name="ch06d095">　Index begin　Index this file
■　"closed bounded" vs. "opened unbounded"
below is
second if "on a closed bounded interval"
relatively is
"on a opened bounded interval" or
"on a (opened) unbounded interval"

<a name="ch06d096">
The condition "closed bounded" is important.
Instead of
f(x)=-sin(x) x in [0, PI]
Now we define
g(x)=-sin(x) x in (0, PI)
In g(x), x is NOT allow to be 0 or PI

<a name="ch06d097">
The interval [0, PI] is closed,
x=0 is allowed, x=PI is allowed

The interval (0, PI) is opened
x=0 disallowed, x=PI disallowed

For an open interval do we have a 
maximum value point?

<a name="ch06d098">
Just think a same question:
Given real number x<1 find maximum x.
The answer is "no solution". Because
0.999 > 0.99
non-stop pile up '9'. Impossible to get
a solution. If we say the answer is 
0.9... infinity '9'
But  "0.9... infinity '9'" IS ONE, which
is excluded !!

<a name="ch06d099">
With the aid of 0.999 > 0.99 example,
It is easy to understand that
g(x)=-sin(x) x in open interval (0, PI)
has no maximum value solution.
Since x=0 is not allowed. for the open 
end '0', start from 0.1 approach zero. 
In 0.1 pile up '0' between '.' and '1'
non-stop, forever, that is no solution !

<a name="ch06d100">　Index begin　Index this file
If domain defined to be (0, infinity)
Left end '0' is open, because used '('
Right end 'infinity' is unbounded
also cause trouble. No trouble? then
what is the value of sin(infinity)?

The condition "closed bounded" in
second if "on a closed bounded interval"
is necessary for a reasonable problem.
2009-12-31-20-30 stop

<a name="ch06d101">
2010-01-01-10-59 start
Above discussed
When convex function attains maximum
　　at interior point?
and
"closed bounded" vs. "opened unbounded"
They are discussion, not mathematical
proving process. Next discuss Theorem 2

<a name="ch06d102">
Theorem 2. A strongly convex function 
on a closed bounded interval attains 
its maximum at one of its end-points.

<a name="ch06d103">
Let us compare
  f(px +(1-p)y)≦
  pf(x)+(1-p)f(y)    ---eqn.6.1
with

<a name="ch06d105">　Index begin　Index this file
2010-01-01-11-08 here
Do not be scared by the complicated
structure of eqn.AS043.
'x' in eqn.6.1 is 'a' in eqn.AS043 
'y' in eqn.6.1 is 'b' in eqn.AS043 
'p' in eqn.6.1 is '(b-t)/(b-a)' in eqn.AS043 
x,y;a,b are constant. p and t are variable.
<a name="ch06d106">
p vary in between [0,1]
t vary in between [a,b]
when t=a it is same as p=1,
when t=b it is same as p=0,
eqn.6.1 and eqn.AS043 are IDENTICAL !!
Both are convexity definition equation.
We can write blue ≦ in eqn.AS044
(blue ≦ is part of definition !
 f(t) is a short write for f(x_AM) 
 see left end of eqn.AS043
 )

<a name="ch06d107">
Also pay attention to that
in eqn.6.1    p+(1-p)=1 ---eqn.AS046
in eqn.AS043  [(b-t)/(b-a)]+[1-(b-t)/(b-a)]=1 ---eqn.AS047
eqn.AS043 tell us that f(t) is an weighted
average of f(a) and f(b).
Whatever weight it is, an average can not
be greater than the maximum value of its
elements. Similarly, an average can not
be smaller than the minimum value of its
elements. Please see Exercise 5.1 titled
Baseball and Cauchy's Third Inequality

<a name="ch06d108">
Because an average can not be greater than
the maximum value of its elements. 
therefore, in eqn.AS043 case, at least one
of f(a) and f(b) is greater than or equal 
to the average value f(t). Replace the 
smaller of f(a) f(b) by the greater, then 
both f(a) and f(b) are same value and both 
greater than or equal to f(t). This 
replacement allow us write red ≦ in 
eqn.AS044

<a name="ch06d109">
in eqn.AS044, take left end and
right end. Apply eqn.AS047 to right end
We find 
  f(t)≦max[f(a),f(b)]  ---eqn.AS048
eqn.AS048 left side has variable 't'
Let us lock left side at maximum value
max{t∈[a,b]}f(t), then we get
  max{t∈[a,b]}f(t)=max[f(a),f(b)] ---eqn.AS045
Here the equality in eqn.AS045 is a
result of applying the baseball team 
analysis.

<a name="ch06d110">　Index begin　Index this file
eqn.AS045 tell us that 
the maximum value point of a convex
function occurs at one of its boundary
point. Either f(a) or f(b).
Remind: function domain is t∈[a,b]
'a' and 'b' are boundary points.
Theorem 2 is proved.
2010-01-01-12-04 stop

<a name="ch06d111">
2010-01-01-14-55 start
Above are preparation for Exercise 6.8
The following is still preparation for 
Exercise 6.8. But not build from other's 
work. Following is LiuHH's own calculation.
(then you must keep your eye wide open and
 bend your finger verify LiuHH's calculation
 Good news is that answer range from 0 to 2
 so, five fingers is more than enough. )

Exercise 6.8 give L(x,y,z) definition.
We need to know ∂L/∂x, ∂L/∂y, ∂L/∂z 
and ∂2L/∂x2, ∂2L/∂y2, ∂2L/∂z2 

<a name="ch06d112">
We start from L(x,y,z) eqn.AS028

<a name="ch06d113">　Index begin　Index this file
2010-01-01-15-05 here
■　∂L/∂x,∂2L/∂x2;∂L/∂y,∂2L/∂y2;∂L/∂z,∂2L/∂z2
L(x,y,z) definition
∂L/∂x =
  ∂[x*x/(1+y)]/∂x
 +∂[y*y/(1+z)]/∂x
 +∂[z*z/(1+x+y)]/∂x
 +∂[x*x(y*y-1)(z*z-1)]/∂x
 =
  2*x/(1+y)
 +0
 -z*z/(1+x+y)2
 +2*x(y*y-1)(z*z-1)

<a name="ch06d114">
∂L/∂x =2*x/(1+y)-z*z/(1+x+y)2
       +2*x(y*y-1)(z*z-1) ---eqn.AS049

∂2L/∂x2 =
  ∂[2*x/(1+y)]/∂x
 +∂[-z*z/(1+x+y)2]/∂x
 +∂[2*x(y*y-1)(z*z-1)]/∂x
 =
  2/(1+y)
 +(-1)(-2)z*z/(1+x+y)3
 +2*(y*y-1)(z*z-1)

<a name="ch06d115">
∂2L/∂x2 = 2/(1+y)+2*z*z/(1+x+y)3
          +2*(1-y*y)(1-z*z) ---eqn.AS050

Above is ∂L/∂x and ∂2L/∂x2
<a name="ch06d116">
Below is ∂L/∂y and ∂2L/∂y2

∂L/∂y =
  ∂[x*x/(1+y)]/∂y
 +∂[y*y/(1+z)]/∂y
 +∂[z*z/(1+x+y)]/∂y
 +∂[x*x(y*y-1)(z*z-1)]/∂y
 =
  -x*x/(1+y)2
 +2*y/(1+z)
 -z*z/(1+x+y)2
 +2*x*x*y(z*z-1)

<a name="ch06d117">
∂L/∂y =-x*x/(1+y)2 +2*y/(1+z)
 -z*z/(1+x+y)2 +2*x*x*y(z*z-1) ---eqn.AS051

∂2L/∂y2 =
  ∂[-x*x/(1+y)2]/∂y
  ∂[+2*y/(1+z)]/∂y
  ∂[-z*z/(1+x+y)2]/∂y
  ∂[+2*x*x*y(z*z-1)]/∂y
 =
  -(-2)*x*x/(1+y)3
  +2/(1+z)
  -(-2)*z*z/(1+x+y)3
  +2*x*x(z*z-1)

<a name="ch06d118">　Index begin　Index this file
∂2L/∂y2 =2*x*x/(1+y)3+2/(1+z)
    +2*z*z/(1+x+y)3-2*x*x(1-z*z) ---eqn.AS052

Above is ∂L/∂y and ∂2L/∂y2
<a name="ch06d119">
Below is ∂L/∂z and ∂2L/∂z2

∂L/∂z = 
  ∂[x*x/(1+y)]/∂z
 +∂[y*y/(1+z)]/∂z
 +∂[z*z/(1+x+y)]/∂z
 +∂[x*x(y*y-1)(z*z-1)]/∂z
 =
  0
 -y*y/(1+z)2
 +2*z/(1+x+y)
 +x*x(y*y-1)*2*z

<a name="ch06d120">
∂L/∂z = -z*z/(1+z)2 +2*z/(1+x+y)
         +2*x*x(y*y-1)*z ---eqn.AS053

∂2L/∂z2 =
  ∂[-y*y/(1+z)2]/∂z
 +∂[2*z/(1+x+y)]/∂z
 +∂[2*x*x(y*y-1)*z]/∂z
  =
  -(-2)*y*y/(1+z)3
 +2/(1+x+y)
 +2*x*x(y*y-1)

<a name="ch06d121">
∂2L/∂z2 =  ---eqn.AS054
  2*y*y/(1+z)3+2/(1+x+y) -2*x*x(1-y*y)
2010-01-01-15-46 here

<a name="ch06d122">
For x=0 or 1, y=0 or 1, z=0 or 1, there
are 2*2*2=8 cases to consider for three
second partial derivatives
∂2L/∂x2 = 2/(1+y)+2*z*z/(1+x+y)3
          +2*(1-y*y)(1-z*z) ---eqn.AS050
∂2L/∂y2 =2*x*x/(1+y)3+2/(1+z)
    +2*z*z/(1+x+y)3-2*x*x(1-z*z) ---eqn.AS052
∂2L/∂z2 =  ---eqn.AS054
  2*y*y/(1+z)3+2/(1+x+y) -2*x*x(1-y*y)
as following

<a name="ch06d123">　Index begin　Index this file
■　Eight cases values.
Case 1, 2, 3ALERT, 4ALERT, 5, 6, 7, 8.
maximum value 2.0

Case 1: x=1,y=1,z=1 L(x,y,z) definition
L(x,y,z)=L(1,1,1)=1.3333333333333332

∂2L/∂x2 = 
2/(1+y)+2*z*z/(1+x+y)/(1+x+y)/(1+x+y)+2*(1-y*y)*(1-z*z)
=1.074074074074074
∂2L/∂y2 =
2*x*x/(1+y)/(1+y)/(1+y)+2/(1+z)+2*z*z/(1+x+y)/(1+x+y)/(1+x+y)-2*x*x*(1-z*z)
=1.324074074074074
∂2L/∂z2 =
2*y*y/(1+z)/(1+z)/(1+z)+2/(1+x+y)-2*x*x*(1-y*y)
=0.9166666666666666
<a name="ch06d124">
Case 2: x=1,y=1,z=0
L(x,y,z)=L(1,1,0)=1.5
∂2L/∂x2 = 
2/(1+y)+2*z*z/(1+x+y)/(1+x+y)/(1+x+y)+2*(1-y*y)*(1-z*z)
=1
∂2L/∂y2 =
2*x*x/(1+y)/(1+y)/(1+y)+2/(1+z)+2*z*z/(1+x+y)/(1+x+y)/(1+x+y)-2*x*x*(1-z*z)
=0.25
∂2L/∂z2 =
2*y*y/(1+z)/(1+z)/(1+z)+2/(1+x+y)-2*x*x*(1-y*y)
=2.6666666666666665
<a name="ch06d125">
Case 3: x=1,y=0,z=1
L(x,y,z)=L(1,0,1)=1.5
∂2L/∂x2 = 
2/(1+y)+2*z*z/(1+x+y)/(1+x+y)/(1+x+y)+2*(1-y*y)*(1-z*z)
=2.25
∂2L/∂y2 =
2*x*x/(1+y)/(1+y)/(1+y)+2/(1+z)+2*z*z/(1+x+y)/(1+x+y)/(1+x+y)-2*x*x*(1-z*z)
=3.25
∂2L/∂z2 =
2*y*y/(1+z)/(1+z)/(1+z)+2/(1+x+y)-2*x*x*(1-y*y)
=-1 ＜＝ALERT ! negative !
<a name="ch06d126">
Case 4: x=1,y=0,z=0 L(x,y,z) definition
L(x,y,z)=L(1,0,0)=2 ＜＝ALERT ! MAXIMUM !
∂2L/∂x2 = //second derivative is not L max.
2/(1+y)+2*z*z/(1+x+y)/(1+x+y)/(1+x+y)+2*(1-y*y)*(1-z*z)
=4  //∂2L/∂x2=4 is not L maximum, L=2 is.
∂2L/∂y2 =
2*x*x/(1+y)/(1+y)/(1+y)+2/(1+z)+2*z*z/(1+x+y)/(1+x+y)/(1+x+y)-2*x*x*(1-z*z)
=2
∂2L/∂z2 =
2*y*y/(1+z)/(1+z)/(1+z)+2/(1+x+y)-2*x*x*(1-y*y)
=-1 ＜＝ALERT ! negative !
<a name="ch06d127">
Case 5: x=0,y=1,z=1
L(x,y,z)=L(0,1,1)=1
∂2L/∂x2 = 
2/(1+y)+2*z*z/(1+x+y)/(1+x+y)/(1+x+y)+2*(1-y*y)*(1-z*z)
=1.25
∂2L/∂y2 =
2*x*x/(1+y)/(1+y)/(1+y)+2/(1+z)+2*z*z/(1+x+y)/(1+x+y)/(1+x+y)-2*x*x*(1-z*z)
=1.25
∂2L/∂z2 =
2*y*y/(1+z)/(1+z)/(1+z)+2/(1+x+y)-2*x*x*(1-y*y)
=1.25
<a name="ch06d128">　Index begin　Index this file
Case 6: x=0,y=1,z=0
L(x,y,z)=L(0,1,0)=1
∂2L/∂x2 = 
2/(1+y)+2*z*z/(1+x+y)/(1+x+y)/(1+x+y)+2*(1-y*y)*(1-z*z)
=1
∂2L/∂y2 =
2*x*x/(1+y)/(1+y)/(1+y)+2/(1+z)+2*z*z/(1+x+y)/(1+x+y)/(1+x+y)-2*x*x*(1-z*z)
=2
∂2L/∂z2 =
2*y*y/(1+z)/(1+z)/(1+z)+2/(1+x+y)-2*x*x*(1-y*y)
=3
<a name="ch06d129">
Case 7: x=0,y=0,z=1 L(x,y,z) definition
L(x,y,z)=L(0,0,1)=1
∂2L/∂x2 = 
2/(1+y)+2*z*z/(1+x+y)/(1+x+y)/(1+x+y)+2*(1-y*y)*(1-z*z)
=4
∂2L/∂y2 =
2*x*x/(1+y)/(1+y)/(1+y)+2/(1+z)+2*z*z/(1+x+y)/(1+x+y)/(1+x+y)-2*x*x*(1-z*z)
=3
∂2L/∂z2 =
2*y*y/(1+z)/(1+z)/(1+z)+2/(1+x+y)-2*x*x*(1-y*y)
=2
<a name="ch06d130">
Case 8: x=0,y=0,z=0
L(x,y,z)=L(0,0,0)=0
∂2L/∂x2 = 
2/(1+y)+2*z*z/(1+x+y)/(1+x+y)/(1+x+y)+2*(1-y*y)*(1-z*z)
=4
∂2L/∂y2 =
2*x*x/(1+y)/(1+y)/(1+y)+2/(1+z)+2*z*z/(1+x+y)/(1+x+y)/(1+x+y)-2*x*x*(1-z*z)
=2
∂2L/∂z2 =
2*y*y/(1+z)/(1+z)/(1+z)+2/(1+x+y)-2*x*x*(1-y*y)
=2
Above calculation ignored ∂2L/∂x/∂y, 
∂2L/∂y/∂z, ∂2L/∂z/∂x.
2010-01-01-16-07 stop L(x,y,z) definition

<a name="ch06d131">
2010-01-01-18-55 numerical confirmed that
L(x,y,z) has maximum value 2.0 in cube [0,1]3

<a name="ch06d132">　Index begin　Index this file
2010-01-02-11-32 start
■　Study LangeListCSMCTypos.pdf

The following is a record of study 
LangeListCSMCTypos.pdf. It begin with
[[
It is easy to show by convexity or 
algebra that
  1/(1+u) ≦ 1 - u/2 ---eqn.AS033
for u∈[0, 1] 

and
  1/(1+u) ≦ 1 - u/3 ---eqn.AS034
for u∈[0, 2].
]]

<a name="ch06d133">
Let us see whether
  1/(1+u) ?≦? 1 - u/2 ---eqn.AS033
  for u∈[0, 1]
is true. Because u∈[0, 1], 1+u＞0
eqn.AS033 can be written as
  1 ?≦? (1 - u/2)(1+u)
and
  1 ?≦? 1+u +1*(-u/2)+u*(-u/2)
and
  0 ?≦? 0+u -u/2 -u*u/2
and
  0 ?≦? u/2 -u*u/2
and
  0 ?≦? u/2*(1-u) ---eqn.AS055
given u∈[0, 1], then 
u/2 ≧0 and (1-u)≧0, therefore eqn.AS055
is true. Its original eqn.AS033 is true.

<a name="ch06d134">
Next see whether
  1/(1+u) ≦ 1 - u/3 ---eqn.AS034
for u∈[0, 2] is true.
for u∈[0, 2], (1+u)＞0, write eqn.AS034 as
  1 ?≦? (1 - u/3)*(1+u)
and
  1 ?≦? 1+u +1*(-u/3)+u*(-u/3)
and
  0 ?≦? 0+u -u/3 -u*u/3
and
  0 ?≦? 2*u/3 -u*u/3
and
  0 ?≦? (2 -u)*u/3 ---eqn.AS056
given u∈[0, 2], then
(2 -u)≧0 and u/3≧0. Therefore eqn.AS056
is true. Its original eqn.AS034 is true.

<a name="ch06d135">
In eqn.AS035, apply eqn.AS033 to x2/(1+y) 
find
  x2/(1+y) ≦ x2*(1-y/2) ---eqn.AS057
Similarly
  y2/(1+z) ≦ y2*(1-z/2) ---eqn.AS058

<a name="ch06d136">
In eqn.AS035, apply eqn.AS034 to z2/(1+x+y) 
(0≦x,y≦1 then 0≦x+y≦2, need eqn.AS034)
find
  z2/(1+x+y) ≦ z2*[1-(x+y)/3] ---eqn.AS059
Apply eqn.AS057,eqn.AS058,eqn.AS059 to
eqn.AS035 get eqn.AS036

eqn.AS036 is defined to be g(x,y,z)

<a name="ch06d137">　Index begin　Index this file
∂g(x,y,z)/∂x =
   ∂[x*x*(1-y/2)]/∂x
  +∂[y*y*(1-z/2)]/∂x
  +∂[z*z*(1-(x+y)/3)]/∂x
  +∂[x*x*(y*y-1)*(z*z-1)]/∂x
=
   2*x*(1-y/2)
  +0
  +[z*z*(0-1/3)]
  +2*x*(y*y-1)*(z*z-1)

∂g(x,y,z)/∂x = 2*x*(1-y/2) -z*z/3
    +2*x*(y*y-1)*(z*z-1) ---eqn.AS060

<a name="ch06d138">
Next see ∂g2(x,y,z)/∂x2

∂g2(x,y,z)/∂x2= g,x,x =
  ∂[2*x*(1-y/2)]/∂x
 +∂[-z*z/3]/∂x
 +∂[+2*x*(y*y-1)*(z*z-1)]/∂x
=
  2*(1-y/2)
 +0
 +2*(y*y-1)*(z*z-1)

∂g2(x,y,z)/∂x2 g,x,x = ---eqn.AS061
   =2*(1-y/2)+2*(1-y*y)*(1-z*z)

<a name="ch06d139">
Next see ∂g2(x,y,z)/∂x/∂y

∂g2(x,y,z)/∂x/∂y= g,x,y =
  ∂[2*x*(1-y/2)]/∂y
 +∂[-z*z/3]/∂y
 +∂[+2*x*(y*y-1)*(z*z-1)]/∂y
=
  2*x*(-1/2)
 +0
 +2*x*(2*y)*(z*z-1)

∂g2(x,y,z)/∂x/∂y= g,x,y = 
   2*x*(-1/2)+2*x*(2*y)*(z*z-1)

∂g2(x,y,z)/∂x/∂y= g,x,y = ---eqn.AS062
   -x-4*x*y*(1-z*z)

<a name="ch06d140">
Next see ∂g2(x,y,z)/∂x/∂z

∂g2(x,y,z)/∂x/∂z= g,x,z =
  ∂[2*x*(1-y/2)]/∂z
 +∂[-z*z/3]/∂z
 +∂[+2*x*(y*y-1)*(z*z-1)]/∂z
=
  0
 -2*z/3
 +2*x*(y*y-1)*(2*z)

∂g2(x,y,z)/∂x/∂z= g,x,z =
   -2*z/3 +2*x*(y*y-1)*(2*z)

∂g2(x,y,z)/∂x/∂z= g,x,z = ---eqn.AS063
   -2*z/3 -4*x*(1-y*y)*z

<a name="ch06d141">
LangeListCSMCTypos.pdf said
[[
on the stated domain x, y, z ∈[0, 1]. 
The second derivative test demonstrates 
that g(x, y, z) is convex in x for y and
 z fixed. Hence, g(x, y, z) attains its 
maximum at x = 0 or x = 1.
]]
We have
g,x,x = 2*(1-y/2)+2*(1-y*y)*(1-z*z) ---eqn.AS061
g,x,y =-x-4*x*y*(1-z*z)             ---eqn.AS062
g,x,z =-2*z/3 -4*x*(1-y*y)*z        ---eqn.AS063
2010-01-02-12-30 here

<a name="ch06d142">　Index begin　Index this file
Because x, y, z ∈[0, 1]. then
x≧0, y≧0, z≧0, (1-y*y)≧0, (1-z*z)≧0.
(1-y/2)＞0. Therefore
g,x,x＞0
g,x,y≦0
g,x,z≦0

<a name="ch06d143">
LiuHH is not sure how g,x,y≦0 and
g,x,z≦0 influence the answer.
Forget this kink, and continue study
<a name="ch06d144">
LangeListCSMCTypos.pdf
[[
Now
  g(1,y,z)=(1-y/2)+y2(1-z/2)+z2(2/3-y/3)
            +(y2-1)(z2-1) ---eqn.AS037
  g(0,y,z)=y2(1-z/2)+z2(1-y/3) ---eqn.AS038
and so
  g(1,y,z)-g(0,y,z)=(1-y/2)-z2/3
          +(y2-1)(z2-1) ≧ 0 ---eqn.AS039

Since 1≧y/2+z2/3 and the product 
(y2-1)(z2-1) is nonnegative. 
]]
g(1,y,z)-g(0,y,z) is simple mathematics.

<a name="ch06d145">
(1-y/2)-z2/3 is 1-(y/2+z2/3)
the maximum value of (y/2+z2/3) is
at y=z=1, which is (1/2+12/3)=5/6
then the minimum value of 1-(y/2+z2/3)
is 1-5/6 = 1/6 ＞ 0
We have 1-(y/2+z2/3) always  ＞ 0
Also (y2-1)(z2-1)=(1-y2)(1-z2)
is nonnegative for y, z ∈[0, 1]. 
That is g(1,y,z)-g(0,y,z) ＞ 0
 g(1,y,z) ＞ g(0,y,z) ---eqn.AS064

<a name="ch06d146">
What is the minimum value of g(0,y,z)?
Go to g(x,y,z), set x=0, find
  g(0,y,z)=y2(1-z/2)+z2(1-y/3) ---eqn.AS038
for y,z∈[0, 1] (1-z/2)＞0 and (1-y/3)＞0
g(0,y,z)= attain its minimum at y=z=0
g(0,y,z)=g(0,0,0)=0
Since g(1,y,z) ＞ g(0,y,z) , g(1,y,z)＞ 0

<a name="ch06d147">　Index begin　Index this file
LangeListCSMCTypos.pdf said
[[
Finally, g(1,y,z)≦2
if and only if
  (1-y/2)+y2(1-z/2)+z2(2/3-y/3)
   +y2z2-y2-z2+1≦2 ---eqn.AS040
]]
<a name="ch06d148">
In g(x,y,z), set x=1 get eqn.AS040
At this point eqn.AS040≦2 is uncertain.
Continue
[[
if and only if
  -y/2 -y*y*z/2 -z*z*y/3
  +y*y*z*z -z*z/3 ≦0 ---eqn.AS041

The last inequality is true because
  y*y*z*z ≦ y/2 + y*y*z/2 ---eqn.AS042
]]
<a name="ch06d149">
Expand last term in eqn.AS039
+(y2-1)(z2-1)
to
+y2z2-y2-z2+1
get eqn.AS040

<a name="ch06d150">
From eqn.AS040 to eqn.AS041 is simply 
cancel integer 1+1 (eqn.AS040 left)
with +2 (eqn.AS040 right), and simplify
the remain of eqn.AS040.

<a name="ch06d151">
Move all negative term in eqn.AS041 to
right, we get
  +y*y*z*z ?≦? +y/2 +y*y*z/2
                +z*z*y/3 +z*z/3 ---eqn.AS065
Because y, z ∈[0, 1], all terms in 
eqn.AS065 are positive/zero.

<a name="ch06d152">　Index begin　Index this file
eqn.AS065 right side has four terms.
Now take two of them, +y/2 +y*y*z/2
to beat left side +y*y*z*z
How to show
  y*y*z*z ?≦? y/2 + y*y*z/2 ---eqn.AS042
is correct? Move left side to right side
and multiply whole equation by 2, we see
  0 ?≦? (y/2 + y*y*z/2 -y*y*z*z)*2 ---eqn.AS066

<a name="ch06d153">
Define
  h(y,z)=y + y*y*z -2*y*y*z*z ---eqn.AS067
  h(y,z)= y + y*y*z*(1-2*z)   ---eqn.AS068
if h(y,z) has negative value at a point 
then dis-prove LangeListCSMCTypos.pdf 
analysis.
if h(y,z) ≧0 at all y, z ∈[0, 1], then
proved LangeListCSMCTypos.pdf analysis.

<a name="ch06d154">
For  y, z ∈[0, 1]
we are interested at z*(1-2*z)≦0
not z*(1-2*z)≧0,
function z*(1-2*z) has extreme values at
z=0, z=1 and z0, where z0 is a point for
  d[z*(1-2*z)]/dz = 0 ---eqn.AS069
<a name="ch06d155">
Find z0 as following
set  d[z*(1-2*z)]/dz = 0 
     d[z-2*z*z]/dz = 1-2*2*z=0
z0= 1/4 ---eqn.AS070
Now evaluate z*(1-2*z) for z=1, z=1/4
and z=0

<a name="ch06d156">
for z=1,   z*(1-2*z)=-1
for z=1/4  z*(1-2*z)=1/8
for z=0    z*(1-2*z)=0
when z=1 expression z*(1-2*z)=-1 give 
us the critical condition negative 1.
Back to
  h(y,z)= y + y*y*z*(1-2*z)   ---eqn.AS068
set z=1, find
  h(y,1)= y + y*y*1*(1-2*1)
  h(y,1)= y - y*y
  h(y,1)= y*(1-y)   ---eqn.AS071
For  y ∈[0, 1], h(y,1)=y*(1-y)≧0

<a name="ch06d157">　Index begin　Index this file
Back to
  +y*y*z*z ≦+y/2 +y*y*z/2 +z*z*y/3 +z*z/3 ---eqn.AS065
without "+z*z*y/3 +z*z/3", the inequality
  +y*y*z*z ≦+y/2 +y*y*z/2
is true, h(y,1)≧0.
Additional term "+z*z*y/3 +z*z/3" is positive
So eqn.AS065 is true for sure.

<a name="ch06d158">
LangeListCSMCTypos.pdf stopped at
[[
The last inequality is true because
  y*y*z*z ≦ y/2 + y*y*z/2 ---eqn.AS042
]]

LangeListCSMCTypos.pdf use ∂g2(x,y,z)/∂x2
to draw conclusion
<a name="ch06d159">
Textbook use ∂g2(x,y,z)/∂x2,
 ∂g2(x,y,z)/∂y2, ∂g2(x,y,z)/∂z2
to get the answer and stopped at
 ∂g2(x,y,z)/∂z2＜0
LiuHH is unable to say who is what.
LiuHH wait for Prof. J. Michael Steele
second edition book.
2010-01-02-14-10 stop




<a name="ch06d160">
■　Exercise 6.8 solution


LiuHH is unable to say who is what.
LiuHH wait for Prof. J. Michael Steele
second edition book.
2010-01-02-14-10 stop



<a name="ch06d161">　Index begin　Index this file
2010-01-02-9-29 start
■　How to use program "fMaxMinCube"
program fMaxMinCube or cubeMaxf()
read a function definition
read x,y,z upper/lower bounds
read x,y,z grid numbers
create a 3-D grid array and calculate
function value at each node.
Find the over-all maximum and over-all
minimum. Report result just above the
label [<a name="box31">]. Report same
result at first two lines in Box 33.

<a name="ch06d162">
Run button is [Find cube Max/Min].

This [Find cube Max/Min] is an utility for
Exercise 6.8. Find numerical answer.
Later add few code line allow program to
record those points whose function value
is in user defined range.

<a name="ch06d163">
The added control boxes are
[[
output x,y,z □  x,y,z,f □  f,x,y,z □  ; Coord □ 
equal potential □  value [  ]  tolerance [  ]
]]

"□" are check boxes, "[  ]" are data boxes

<a name="ch06d164">
If check "equal potential □ " check box
both  value [  ]  tolerance [  ]
become active. If you fill 0.8 and 0.05 
      value [ 0.8 ]  tolerance [ 0.05 ]
Program assign 
eqPotMin=0.8-0.05 =0.75
eqPotMax=0.8+0.05 =0.85
All points with function value in [0.75,0.85]
are recorded. Report to box 32, box 33.

<a name="ch06d165">
If fill the following data
[[
x domain, left 0. , right 1. , grid 11
y domain, left 0. , right 1. , grid 11
z domain, left 0. , right 1. , grid 11
]]
<a name="ch06d166">　Index begin　Index this file
Program assign
xstep=(xrite-xleft)/(xgrid-1);
xstep=( 1.0 - 0.0 )/( 11 -1) = 0.1
Grid  0 has x-coordinate 0.0
Grid  1 has x-coordinate 0.1
Grid  2 has x-coordinate 0.2
.....
Grid 10 has x-coordinate 1.0

<a name="ch06d167">
Grid number is always integer,
coordinate  is  float number.

Similar case in y and z coordinates.
Box 32 output Equal potential coordinates
Box 33 output Equal potential node number

<a name="ch06d168">
Only if check "equal potential □ " check box
then Box 32 and 33 has equal potential output

<a name="ch06d169">
In
[[
output x,y,z □  x,y,z,f □  f,x,y,z □  ; Coord □ 
]]
If check "x,y,z □" Box 32 and 33 has node/coord
output. No function value.

<a name="ch06d170">
If check "x,y,z,f □" Box 32 and 33 has node/coord
output with function value at right end.

If check "f,x,y,z □" Box 32 and 33 has node/coord
output with function value at left end.

"Coord □" has nothing to do with Box 32 and 33.

<a name="ch06d171">　Index begin　Index this file
if NOT check "equal potential □ " check box.
and if check " x,y,z □ " or " x,y,z,f □" 
or  check "f,x,y,z □"
Box 31 output answer as "x,y,z" "x,y,z,f" 
"f,x,y,z" suggested.

<a name="ch06d172">
If check "Coord □", 
Box 31 print x,y,z coordinates (Not nodes)

Too many user, too many cases, write
a general program for each possible 
application.
If you know Javascript language. You 
can change source code to your specific 
need.

Thank you for visiting Freeman's web site.
Liu,Hsinhan 2010-01-02-20-10

<a name="ch06d173">
if equal potential unchecked, then box32 □ and
box33 □ disabled. reverse if, reverse result.
2010-01-04-21-38 start
"Update 2010-01-05" add a little more
control for user.
If user need just max/min function value
and not need all points value,
"Upload 2010-01-04" version user has no
choice, must output all points value.
Program run very slow.
"Update 2010-01-05" version allow user 
turn off box31, not output long string.

<a name="ch06d174">
If all box 31,32,33 turned off and run
51*51*51 grid cost time 7 seconds for
just min/max value output.
If open box 31 for 132654 output lines
51*51*51 grid cost time 8 minutes 
without function value. If include
function value for 51*51*51 grid,
guess cost 20 to 30 minutes. Much
longer than seven seconds simple 
answer time.

<a name="ch06d175">
One line above 
"You can define any f(x,y,z) and any cube"
has
"GO! box31 □ ; box32 □ ; box33 □ help5 "
If box □ is checked, the correspond output
box will store data. box31 record ALL points
function value. box31 'kill time' !!

<a name="ch06d176">
box32 □ ; box33 □ depend on the definition 
of "equal potential  value  tolerance "
If it is thin shell (tolerance nearly zero),
output very few data, not really 'kill time'.
2010-01-04-22-03 stop



2010-01-03-21-11 done proofread
2010-01-04-09-20 done spelling check

<a name="ch06d177">
2010-01-05-17-02 start
On 2010-01-05 add code to improve code
efficiency. Control panel has a line
[[
core code type: A  ; B 
]]
<a name="ch06d178">
If click [type: A ], program use next 
simplyfied sample code
[[
  for(iz=0;iz<zgrid;iz++)
        {
  z=zleft+iz*zstep;
  fval=eval('with(Math){'+fxyz+'}'); //9901012147
  fv1=fv2=''; //9901030704
  if(eqPotOutput==2)//this if can be dropped
  fv2=', '+fval;
  else//this else-if can be dropped
  if(eqPotOutput==3)
  fv1=fval+', ';
        }
]]
<a name="ch06d179">
If click [type: B ], if user selected
checkbox for (eqPotOutput==2), program
use next improved code
[[
  for(ix=0;ix<xgrid;ix++)
  for(iy=0;iy<ygrid;iy++)
  for(iz=0;iz<zgrid;iz++)//heavy loop
        {
  z=zleft+iz*zstep;
  fval=eval('with(Math){'+fxyz+'}'); //9901012147
  fv1=fv2=''; //9901030704
  fv2=', '+fval; //no if in heavy loop
        }
]]
<a name="ch06d180">
[type: B ] try to remove all if-code
that can be removed.
[type: B ] create simplified string,
save in variable 'coreCode', then use
eval(coreCode); //9901051303
to carry out calculation.
<a name="ch06d181">
You can try either one. LiuHH is not
sure whether type B really improve.
Because "Javascript Bible" suggest
user not to use eval() function.
2010-01-05-17-12 stop

<a name="ch06d182">
2010-01-05-17-54 start
At Box 34, there is a [test01] button.
It test piece code section copied from
Box 34 after running
[Find cube Max/Min] & core code type: B 
Paste code to 
function test01()
make sure core code work. 
[test01] button is debug tool.
User choose different setting,
program generate different code.
2010-01-05-17-58 stop



<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