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function alert1() //9806261745
{
document.write(''
+'<br><font color=red size=+3>'
+'Following is exercise hint and solution.<br>'
+'Suggest reader solve problem first, then<br>'
+'use your solution compare with LiuHH&#39;s<br>'
+'solution. LiuHH&#39;s sol. has no guarantee.<br>'
+'Your solution could be better and correct.<br>'
+'2009-07-28-10-16<br>'
+'</font><br>\n<br>\n<br>\n'
);
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//2009-09-23-15-01 from tute0009.htm
//find function readdata(inStr0)
//
function readdata(inStr0) //9806101151
{
//9807270955 read other's no-comment code
//very headache, now back to my code, add
//more comments.
//
//if input inStr0 is Not a Number (isNaN=true)
//then inStr0 is an array or string of char.
if(isNaN(inStr0) //9807161944
//and if inStr0[0] is a Number
//(isNaN=false) (!isNaN=true)
//inStr0 is a number array, return oroginal form
&&!isNaN(inStr0[0])) return inStr0;

//if input inStr0 is empty, return oroginal form
if(inStr0.length==0)return inStr0; //9806101153

//if input inStr0 is a number, return oroginal form
if(!isNaN(inStr0[0]))return inStr0; //9807271000

//up to here input inStr0 is one string of character
//below change string of char '1.23 4.56 7.89 ...'
//to number array aa[0]=1.23, aa[1]=4.56, aa[2]=7.89  ...
//9807271010 here

//assume input string is '1.23 4.56 7.89 ...'
//oArray[0] store '1.23' this is a string
//oArray[1] store '4.56' not a number yet
//oArray[2] store '7.89' convert to number later
//9807271043
var oArray=''; //9806101158
var w0,w1,numb0;

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  {
  numb0='';

//skip those char are NOT [0123456789-+.] 
// [!(...)] means [is not ...]
// [!=] means [not equal] 9807271051
//
w1=w0; //9802172117
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//above while() loop stop at reading 
//either one in the list [0123456789-+.] 
//then start next while loop. 9807271055
//
while((inStr0.charAt(w1)>='0'
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//If it is either one of [0123456789-+.eE]
//continue store data into numb0 . 9807271102
numb0+=inStr0.charAt(w1);
  w1++;
    }

//if all I get is just '.' for example
// 'aa.txt'  it is not a number,
//no need to build output oArray
//9807271106
if(numb0.length==1&&numb0.charAt(0)=='.')
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w0=w1; //9802172218 //update index w0=w1
continue; //continue to next iteration.
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//up to this point,  numb0 is useful number
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w0=w1; //9806101210 update index

//string numb0 to output string oArray
//add a newline byte '\n' to separate data
// 9807271113
oArray+=numb0+'\n'; //9806101159

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//done for() loop, done output string oArray
//use newline byte '\n' to separate numbers.
// oArray.split('\n') in oArray cut data at
//newline byte '\n' and change string
// '1.23\n4.56\n7.89\n...'
//to array oArray[0]='1.23', oArray[1]='4.56', 
//oArray[2]='7.89'  ...
//no matter input data use ',' or tab or
//newline to separate data, here all done
//9807271125
oArray=oArray.split('\n'); //9806101643

//string '1' + string '2' = '12'
//number  1  + number  2  =  3
//above is string array oArray[0]='1.23',
//below is number array  w2[0]=1.23 
//9807271130
var w2=[0,0];

  for(w0=0;w0<oArray.length;w0++)
  { //9806101704
//parseFloat() convert string number
//             to number number.
w1=parseFloat(oArray[w0]);
//no more number? done
if(isNaN(w1))break; //9806101714
w2[w0]=w1; //create number array.
  }

return w2;

} //   function readdata()  9806101214
//from function update0() //9802171511





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<font size=+2>
<a href=tute0012.htm#index01>
Inequality</a>
</font>

<a href=tute0012.htm#index13>
Study 12th file
</a>
 &nbsp;
<a href=#20091217>
Update 2009-12-17
</a>

<br>
<a href=tute0012.htm#index>index</a>
　
<a href=tute0012.htm#index13>this</a>
　
<a href="tute0012.htm#program0">
program
</a>
　
<a href=tute0012.htm#docA001>
DocA
</a>
　
<a href=tute0012.htm#docA006>
Limit
</a>
　
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XYGraph v2.3 - web page graph
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<br>

<a href=http://www-stat.wharton.upenn.edu/~steele/Publications/Books/CSMC/CSMC_index.html>
The Cauchy-Schwarz Master Class
</a>
 &nbsp;
<a href=http://www-stat.wharton.upenn.edu/~steele/index.html>
J. Michael</a>
 
<a href=http://www.wharton.upenn.edu/faculty/steele.html>
Steele</a>
 &nbsp;
<a href=http://www.amazon.com/review/product/052154677X/ref=cm_cr_dp_all_helpful?%5Fencoding=UTF8&coliid=&showViewpoints=1&colid=&sortBy=bySubmissionDateDescending>
★★★★★
</a><!--9806081839 add link-->


<br>
<font color=red>
This file is personal home work. No one
<br>
proofread. Cannot promise correctness.
<br>
If you suspect any view point wrong, 
<br>
please ask a math expert near by.
<br>
Freeman 2009-06-19-10-46</font>
<br>
Please use MSIE browser to read this file.
<br>
Did not test other browser. This file is
<br> <!--9806081846-->
written under MSIE 6.0

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<pre><font size=+2>
<a name="docA001">&lt;a name="docA001"&gt;</a>
2009-06-08-19-10 start
This file <a href=http://freeman2.com/tute0008.htm>http://freeman2.com/tute0008.htm</a>
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">&lt;a name="docA002"&gt;</a>,<a name="textbook">&lt;a name="textbook"&gt;</a>
This file is a note for reading inequality
book written by <a href=http://www.wharton.upenn.edu/faculty/steele_michael.jpg>Professor</a> <a href=http://www-stat.wharton.upenn.edu/~steele/index.html>J. Michael</a> <a href=http://www.wharton.upenn.edu/faculty/steele.html>Steele</a>
<a href=http://www-stat.wharton.upenn.edu/~steele/Publications/Books/CSMC/CSMC_index.html>The Cauchy-Schwarz Master Class</a>  　<a href=http://www.amazon.com/review/product/052154677X/ref=cm_cr_dp_all_helpful?%5Fencoding=UTF8&coliid=&showViewpoints=1&colid=&sortBy=bySubmissionDateDescending>★★★★★</a>
Below use 'textbook' as abbreviation.
Freeman also read web pages online, and will 
indicate the source URL at discussion point.

<a name="docA003">&lt;a name="docA003"&gt;</a>
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">&lt;a name="docA004"&gt;</a>
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">&lt;a name="docA005"&gt;</a>　<a href="#index">Index begin</a>　<a href="#index13">Index this file</a>
On 2009-01-27-10-08 Freeman accessed
the next page<a href=http://www.sftw.umac.mo/~fstitl/10mmo/inequality.html>
http://www.sftw.umac.mo/~fstitl/10mmo/inequality.html</a>
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 <a href="tute0007.htm#unicod01">math symbol and internal code</a>.

This file third function is to display how
to draw curves in web page. The main engine
is <a href=http://www.structura.info/XYGraph/XYGraphDemo.htm>XYGraph v2.3 - Technical Figures</a>

Thank you for read Freeman's inequality page.
2009-06-08-19-47 stop

<a name="docA006">&lt;a name="docA006"&gt;</a>
2009-08-23-15-00 start
□　Exercise 1.14 solution
<script lang="javascript">alert2()</script>
Exercise 1.14 <a href="tute0011.htm#ch01d002">problem (a)</a> is solved by <a href="tute0011.htm#ch01d004">hint</a>
Exercise 1.14 <a href="tute0011.htm#ch01d003">problem (b)</a> is out of LiuHH's
reach. What is "cardinality of a set B⊂Z<sup>3</sup>" ?
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">&lt;a name="docA007"&gt;</a>
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 here
</font></pre>

<hr>



<a name="program0"></a>

Arithmetic Mean and Geometric Mean of a number sequence

<br>

<font color=red size=+2>
Output may contain error, Please verify first.
<br>
Program environment is MSIE 6.0, please use MSIE
</font>

<br>

Box 1 input
&nbsp;
<INPUT onclick='javascript:window.clipboardData.setData("Text",document.getElementById("boxc01").value.toString())'
 type=button value="copy 1">
<INPUT onclick='document.getElementById("boxc01").value=""'
 type=button value="del 1">

<a href=http://freeman2.com/complex2.htm#calculator>
Calculator</a>
 <a href=complex2.htm#calculator>local</a>
<!--9809092259 add link to calculator-->

<br>
<TEXTAREA
 id=boxc01
 name=boxc01 rows=10
 cols=70
 >
</TEXTAREA>
<!--9809261017 use box 1-->

<br>
<font color=red><!--980910011850-->
Carleman's inequality require input sequence infinite sum be bounded.
</font>
<br>

<table border=1>
<tr>
<td>
<input onclick="javascript:boxc02.value=amgmf1(boxc01.value,1)"
 type="button" value="　AM
 GM 1" />
</td>
<td>

<input onclick="javascript:boxc02.value=amgmf1(boxc01.value,2);if(spanAlert1.innerHTML=='')spanAlert1.innerHTML='Carleman1 inequality GG<=E*IN, general purpose. Use E=2.7...'"
 type="button" value="　Carle
 man1" />

<!--9809281718-->
<input type="button" value="Carle
 man2"
 onclick="javascript:carlemanf2();if(spanAlert1.innerHTML=='')spanAlert1.innerHTML='<font color=red>Carleman2 inequality, monotone in/decrease data. read [powData]</font>'" />

<!--9809291117del; 9809300028 use-->
<input type="button" value="Carle
 man3"
 onclick="javascript:carlemanf3();if(spanAlert1.innerHTML=='')spanAlert1.innerHTML='<font color=red>Carleman3 inequality with Gradual AM seq. & Gradual GM seq.</font>'" />


<input type="button" value="dbg"
 onclick="chkbxcD1.checked=chkbxcD2.checked=chkbxcD3.checked=chkbxcD4.checked=0" />
<!--
 onclick="chkbxcD1.checked='1'" />
-->
a<input
 type="checkbox"
 name="chkbxcD1" id="chkbxcD1"
 />,b<input
 type="checkbox"
 name="chkbxcD2" id="chkbxcD2"
 />,c<input
 type="checkbox"
 name="chkbxcD3" id="chkbxcD3"
 />;d<input
 type="checkbox"
 name="chkbxcD4" id="chkbxcD4" />


</td>
</tr>



<tr>
<td>

<a href="#docA011">
help_1</a><!--9809291319-->

</td>

<td>


<input type="button" value="α"
 onclick="boxAlpha.value=''" />

<INPUT value="" id=boxAlpha size=2
 ><!--9809271021-->
,
<input type="button" value="ω"
 onclick="boxOmega.value=''" />

<INPUT value="" id=boxOmega size=2
 ><!--9809271022-->

<INPUT id="chkbxcB0" type="checkbox"
 onclick="if(chkbxcB0.checked)boxOmega.value=0.36787944117144233;else boxOmega.value=''"
>1/E
 ; 
<INPUT id="chkbxcB1" type="checkbox"
 onclick="chkbxcB2.checked=0"
>123 , 
<INPUT id="chkbxcB2" type="checkbox"
 onclick="chkbxcB1.checked=0"
>321


</td>

</tr>



<tr>
<td>

<!--9809261456-->
<input type="button" value="equal #"
 onclick="javascript:var eq0=boxcA4.value; var ou0=''; for(var d0=0;d0<parseInt(boxcA1.value);d0++)ou0+=eq0+','; boxc01.value=ou0" />
</td>


<td>
10^
<INPUT value="2" id=boxcA4 size=5 >
 ; 

<INPUT id="chkbxcA2" type="checkbox"
 onclick="chkbxcA3.checked=!chkbxcA2.checked"
 checked
>+/0 , 
<INPUT id="chkbxcA3" type="checkbox"
 onclick="chkbxcA2.checked=!chkbxcA3.checked"
>+/0/- ; 

<INPUT value="3" id=boxcA3 size=5
 > digits

</td>

</tr>


<tr>

<td>

<input type="button" value="random#"
 onclick="javascript:for(var d0=1;d0<2;d0++)eval('fillRand3(\'boxc0'+d0+'.value\')')" />

</td>
<td>

fill
<INPUT value="5" id=boxcA1 size=5
 > numbers;

<input type="checkbox"
 name="chkbxcA1" id="chkbxcA1" />
integer; power
<INPUT value="-2" id=boxcE1 size=4
 > <!--9809291148-->


</td>

</tr>


<tr>
<td>

<input type="button" value="powData"
 onclick="javascript:boxc01.value=powdata1()" />

</td>

<td>

Bgn
<INPUT value="1" id=boxcC1 size=4
 > 
,# Step
<INPUT value="99" id=boxcC2 size=4
 > 
,End
<INPUT value="100" id=boxcC3 size=4
 > 

</td>

</tr>

</table>

<span id=spanAlert1>
<font color=red>
Set power value < -1, click [powData] get bounded sequence.
</font>
</span><!--9809291516-->

<br>

Box 2 output
&nbsp;
<INPUT onclick='javascript:window.clipboardData.setData("Text",document.getElementById("boxc02").value.toString())'
 type=button value="copy 1">
<INPUT onclick='document.getElementById("boxc02").value=""'
 type=button value="del 1">

<a href=http://freeman2.com/complex2.htm#calculator>
Calculator</a>
 <a href=complex2.htm#calculator>local</a>
<!--9809092259 add link to calculator-->

<br>
<TEXTAREA
 id=boxc02
 name=boxc02 rows=10
 cols=70
 >
Carleman's inequality require that input sequence
all elements be positive, infinite many elements,
sum of infinite many elements be finite.
Only monotone decrease infinite many elements sequence
can participate Carleman's inequality. 
In Bgn box fill '1', in End box fill '100'
in power box fill a value < -1 for Carleman sequence.
If power=-1, integral of 1/x = log(x) is divergent !!
Click "powData" then click "Carle man3" for answer.
</TEXTAREA><!--9810012001 to 9810012009 add box2 comment-->

<br>
You can click "Carleman" button, test any input sequence,
<br>
observe its Gradual GMed sum and Gradual AMed sum.
<br>
But only infinite and convergent sequence can be called 
<br>
Carleman's inequality.  LiuHH 2009-10-02-22-56
<br>
<script language="javascript">alert0()</script>


<!--9809261024 use box 2-->

<script language="javascript">
<!--

//both "AM/GM 1" button and 
//"Carleman1" button use function amgmf1()
//but
//"Carleman2" button use function carlemanf2()
//"AM/GM 1" button do not use debug checkbox
//"Carleman1" button use debug checkbox
//
function amgmf1(agArg1,agArg2) //9809261150
{
var inp0=readdata(agArg1)
var j0,j1,j2,j3,j4,j5;
var len0=inp0.length
var am0=0;
var gm0=1;
//j1=1/len0;
var gmPow=1/len0; //9810051143

spanAlert1.innerHTML=''; //9809291518
j1=0;

for(j0=0;j0<len0;j0++)
  {
if(inp0[j0]<0)j1=1; //9809291521
am0+=inp0[j0]/len0;
//gm0*=Math.pow(inp0[j0],j1); //9809261046
gm0*=Math.pow(inp0[j0],gmPow); //9810051143
//gm0*=Math.pow(inp0[j0],1/len0); //9810051155
//9810051600   pow($,gmPow) is 
//better than  pow($,1/len0)
//if sequence has 2000 elements
//gmPow=1/len0; carry out division once
//pow($,1/len0) carry out division 2000 times
//9810051602
  }

if(j1==1)
spanAlert1.innerHTML=''  //9809291518
+'<font color=red>ALERT ! box1 has negative number! AMGM do not treat negative!</font>'
;

var sum0=bye09(am0*len0);

if(agArg2==1) //"AM/GM 1" button return here
return 'AM='+bye09(am0)+'\n' //9809261039
+'GM='+bye09(gm0)+'\n'
+'SUM='+sum0+'\n'
+'Product='+bye09(gm0)+'^'+len0+'='+bye09(Math.pow(bye09(gm0),len0))+'\n'
+'Number of elements='+len0+'\n9809261039\n'

//9809261700
if(agArg2==2) //Carleman's inequality, 
  {      //textbook page 27, eqn.2.15
var carlrite=Math.E*sum0; //9809261705
var carlleft=0;
var carleman=0;
var carlpart=''; //9809270755
var inp0sum=0;   //9809270813
var dbg1=chkbxcD1.checked; //9809281920 twoseq
var dbg2=chkbxcD2.checked; //9809291252 carlrite
var dbg3=chkbxcD3.checked; //9809291253 carlleft

var crite=""     //9809271047
+"Weight alpha (has theory base) and\n"
+"weight omega (NO  theory base) are\n"
+"both not set. The result is true\n"
+"Carleman's inequality."

var twoseq=''; //9809272121

if(dbg1) twoseq=
 "Carleman's inequality work with the following two sequences\n"
+"input sequence and gradual GMed sequence\n";

var alpha0=parseFloat(boxAlpha.value); //9809271025
var omega0=parseFloat(boxOmega.value);
var weight0=0;
var weight1=1;
var wchoose=0;
if(!isNaN(alpha0))
{weight0=alpha0;wchoose=1;}
else
if(!isNaN(omega0)) //9809271029
    {
wchoose=2;
crite=''
+'if set omega<1, Carleman inequality greater\n'
+'than side value reduce. There is no theory\n'
+'base for this setting. Purely try and error.\n'
+'Now omega setting is '+omega0+'\n'
+'***  if set omega=1/E=0.36787944117144233\n'
+'***  cancel E in E*SUM[k=1,n]{weight*ak}\n'
+'***  inequality become uncertainty.\n'
+'Without omega setting, carlrite='+carlrite+'\n'
+'With    omega setting, carlrite='
+(carlrite*omega0)+'\n'; //9809271059

//if(dbg2) //9809291258use,9809291304del not in loop, let go
carlrite=carlrite*omega0; //9809271100
    } //if(!isNaN(omega0))

if(wchoose==1) //9809271035
    {
sum0=0;
for(j0=0;j0<len0;j0++) //9809261729
      { //modify carlrite
weight1=(1-(1-Math.pow(2,1+alpha0)/Math.E)/(j0+1))/Math.pow(1+1/(j0+1),alpha0);
sum0+=weight1*inp0[j0];
      }

crite='' //9809271054
+'2009-09-02-22-06 LiuHH access\n'
+'http://www.emis.de/journals/JIPAM/images/029_00_JIPAM/029_00_www.pdf\n'
+'if set 0<=alpha<=1/log(2)=1.4426950408889633\n'
+'Carleman inequality two side difference will\n'
+'reduce. Now alpha setting is '+alpha0+'\n'
+'Without alpha setting, carlrite='+carlrite+'\n'
+'With    alpha setting, carlrite='+(Math.E*sum0)+'\n'
+'alpha change weight in E*SUM[k=1,n]{weight*ak}\n'
;
carlrite=Math.E*sum0; //9809271045

    } //if(wchoose==1)
/** JIPAM-029_00 page 9/10 theorem 3.2 add weight
2009-09-29-11-00 add next working record, show
 JIPAM-029_00 page 9/10 theorem 3.2 used here

2009-09-02-22-06 LiuHH access
http://www.emis.de/journals/JIPAM/images/029_00_JIPAM/029_00_www.pdf
save as carleman-029_00_JIPAM-029_00_www.pdf

[[
Carleman’s inequality without weight
SUM[k=1,n]{a1*a2*...*ak}^(1/k) <=
<= E*SUM[k=1,n]{ak} and n goto inf.
]]
9809270849
carleman-029_00_JIPAM-029_00_www.pdf
page 9/10 theorem 3.2 add weight
[[
Carleman’s inequality with weight
SUM[k=1,n]{a1*a2*...*ak}^(1/k) <=
<= E*SUM[k=1,n]{weight*ak} and n goto inf.
]]
k in weight match k in ak (a1,a2,a3...ak)
weight=(1-beta/k)/[(1+1/k)^alpha]
0<=alpha<=1/log(2)=1.4426950408889633
0<=beta <=1-2/E=0.26424111765711533
and
E*bata+2^(1+alpha)=E
9809270854
if alpha=0, beta=1-2/E
if beta=0,  alpha=-1+1/log(2)

bata=1-2^(1+alpha)/E
weight=(1-beta/k)/[(1+1/k)^alpha]
weight=(1-(1-2^(1+alpha)/E)/k)/[(1+1/k)^alpha]

weight=(1-(1-pow(2,1+alpha)/E)/k)/pow(1+1/k,alpha)
9809270906

above is comment for if(wchoose==1)
when user fill number in alpha box, and click
Carleman1 button, program use alpha value.
2009-09-29-11-11
/**/



for(j0=1;j0<=len0;j0++) //9809261729
    {
inp0sum+=inp0[j0-1]; //9809270814
carleman=1;
j2=1/j0; //9809261710
j3=1;
for(j1=0;j1<j0;j1++)
      {
carleman*=Math.pow(inp0[j1],j2); //9809261711
j3=j3*inp0[j1]; //9809270803
      }
carlleft+=carleman;  //9809261712

if(dbg3) //this is heavy output
      {
carlpart+='' //9809270805
+j0+stndrdth(j0)+' term: product='+j3
+', power=1/'+j0+'='+j2
+', value='+bye09(carleman)
+', sum='+bye09(carlleft)
+'\n'
      }

if(dbg1) 
      {
twoseq+=inp0[j0-1]+'\t\t'
+bye09(carleman)+'\n'; //9809272119
      }
    }

inp0sum=bye09(inp0sum);

carlrite=bye09(carlrite);
carlleft=bye09(carlleft); //9809262216

//9809281252
//Carleman's inequality GG<=E*IN
//Gradual GMed sum <= E*INput sum
// SUM[k=1,n]{a1*a2*...*ak}^(1/k)<= E*SUM[k=1,n]{ak} 
// next ggin means GG/IN, drop E=2.71828...
// because if data is monotone increase
// GG/IN always <1, no need multiply E
var ggin=carlleft/inp0sum; //9809281257
// on the other hand if data is monotone
// decrease, GG/IN always >1, and
// worst case input data 9,8,7,6,5,4,3,2,1
// GG/IN=1.3493016626997747
// 9809281307

if(dbg1) 
twoseq+=''
+'E*(sum of input) is carlrite, gradual GMed sum is carlleft.\n'
+'Sum of gradual GMed <= E*(sum of input).\n'

return "Carleman's inequality GG<=E*IN\n"
+"Gradual GMed sum <= E*INput sum\n"
+"Drop E, GG/IN="+ggin+"\n" //9809281249
+"SUM[k=1,n]{a1*a2*...*ak}^(1/k) <=\n"
+"<= E*SUM[k=1,n]{ak} and n goto inf.\n"
+"where E=2.718281828459045...\n"
+"carlleft <= carlrite answer is next\n"
+carlleft+"<="+carlrite+"\n"
+((wchoose==1)?"You fill data in alpha box, weights added\n":"")
+((wchoose==2)?"You fill data in omega box, E is not used\n":"")
+((wchoose==2&&chkbxcB0.checked)?"You checked 1/E box, E replaced by 1\n":"")
+"Equality occurs if and only if any\n"
+"one input sequence number is zero.\n"
+"Answer is not confirmed, may be wrong!!\n\n"
+crite+"\n"
+((dbg2)?""
+"===== carlrite is next ===== 9809261719\n"
+"inp0="+inp0+"\n"
+"inp0sum="+inp0sum+"\n"
+"E=2.718281828459045...\n"
+"carlrite="+carlrite+"\n"
 :"")
+((dbg3)?""
+"===== carlleft is next =====\n"
+carlpart
+"carlleft=last sum="+carlleft+"\n\n"
 :"")
+((dbg1)?""
+"=====  Two  sequences  =====\n"
+twoseq //9809272059
 :"")

;
  } //if(agArg2==2) //Carleman's inequality

return 'working';

} //function amgmf1(agArg1,agArg2)  9809261202



/**
2009-09-28-17-23
Carleman's inequality
SUM[k=1,n]{a1*a2*...*ak}^(1/k) <=
<= E*SUM[k=1,n]{ak} and n goto inf.
where E=2.718281828459045...

right hand side {ak} is input data
left  hand side {a1*a2*...*ak}^(1/k)
is gradually GMed term

if {ak} is monotone increase, no need E
each term {a1*a2*...*ak}^(1/k) < {ak}

if {ak} is monotone decrease, if no E
each term {a1*a2*...*ak}^(1/k) > {ak}
in this case E*{ak} reverse inequality

if {ak} is monotone increase and equal 
step size, result is best case
{a1*a2*...*ak}^(1/k)/{ak} < 1 and maximum

if {ak} is monotone decrease and equal 
step size, result is worst case
{a1*a2*...*ak}^(1/k)/{ak} > 1 and minimum

In worst case SUM E*{ak} will reverse the
inequality.

function carlemanf2() drop E, try evaluate
{a1*a2*...*ak}^(1/k)/{ak} see if this value
approach E = 2.718281828459045
2009-09-28-17-35
/**/
function carlemanf2() //9809281736
{
var inStr0=boxc01.value=powdata1()
var ouStr0='';
var d0,d1,d2,d3,d4,d5;
var sumin=0; //sum if input data
var sumgm=0; //sum of gradually GMed terms
var progm=1; //product of gradually GMed terms
//var onegm=0; //sumgm^(1/d0)
var onegm=1; //sumgm^(1/d0); //9809281830
var dbg4=chkbxcD4.checked; //9809291150

for(d0=inStr0.length-1;d0>0;d0--)
    {
if(inStr0.charCodeAt(d0)<32) continue;
else break; //9809281203
    }
d1=inStr0.substring(0,d0+1);
d1=d1.split('\n');
d2=[''];
for(d0=0;d0<d1.length;d0++)
    {
d2[d0]=parseFloat(d1[d0]);
    }
inStr0=d2;
d1=d2='';

ouStr0='' //9809281858
+'If input data monotone decreasing, sumgm/sumin>1, need E\n'
+'   to reverse the inequality to sumgm < E*sumin .\n'
+'If input data monotone increasing, sumgm/sumin<1, no need E\n'
+'If data increase/decrease/in/de etc., sumgm ?<?>? sumin\n'
+'The following output, no E involve. Hope find sumgm/sumin>1\n'
+'case, see sumgm/sumin ratio approach E=2.718281828459045...\n'
//+'input data sum, GMed data sum, sumgm/sumin, GM 1 term\n'
+(dbg4?'Debug box_d is checked.\n'
+'\nsumgm/sumin, GMed data sum, input data sum, GM 1 term'
 :'Debug box_d is not checked.\n\nsumgm/sumin'
 )
+'\n'
;

d2=0; //9809291532

//for(d0=0;d0<inStr0.length;d0++)
for(d0=1;d0<=inStr0.length;d0++)
  {
if(inStr0[d0-1]<0)d2=1; //9809291533

sumin+=inStr0[d0-1];

//progm*=inStr0[d0]; //this line cause infinite
//must do it in smaller scale. 9809281807
//onegm =Math.pow(progm,1/(d0+1)); //9809281746
//onegm*=Math.pow(inStr0[d0],1/(d0+1)); //9809281828

//if(d0==0)onegm=inStr0[d0];
//if(d0==1)onegm=inStr0[d0-1];
onegm=1; //9809281847

//for(d1=1;d1<=d0;d1++)
for(d1=0;d1<d0;d1++)
      { //9809281836
//onegm*=Math.pow(inStr0[d1],1/(d0+1));
onegm*=Math.pow(inStr0[d1],1/(d0));
      }
sumgm+=onegm;

ouStr0+=(sumgm/sumin)
+(dbg4?''
+'\t'+sumgm
+'\t'+sumin
+'\t'+(onegm)
 :''
 )
+'\n'

/**
+'\t'+progm
+'\t'+(1/(d0+1))
+'\t'+(Math.pow(progm,1/(d0+1)))
+'\n';
/**/
  } //for(d0=1;d0<=inStr0.length;d0++)

spanAlert1.innerHTML='';
if(d2==1)
spanAlert1.innerHTML=''  //9809291534
+'<font color=red>ALERT ! box1 has negative number! Carleman do not treat negative!</font>'
;

//boxc02.value=inStr0;
boxc02.value=ouStr0
+'\n'
+'Only last line is real Carlemans inequality answer.\n'
+'Because C.ineq. say input data TOTAL sum eqn. is\n'
+'Gradual GMed total sum <= E*INput total sum\n'
+'9809281906\n'

} //function carlemanf2() //9809281736








//9809300030
//input raw data sequence
//gradual  GMed  sequence
//gradual  AMed  sequence
function carlemanf3() //9809300030
{
var inp0=readdata(boxc01.value)
var j0,j1,j2,j3,j4,j5;
var len0=inp0.length
var am0=0;
var gm0=1;
//j1=1/len0;
var gmPow=1/len0; //9810051150

spanAlert1.innerHTML=''; //9809291518
j1=0;
for(j0=0;j0<len0;j0++)
  {
if(inp0[j0]<0)j1=1; //9809291521
am0+=inp0[j0]/len0;
//gm0*=Math.pow(inp0[j0],j1); //9809261046
gm0*=Math.pow(inp0[j0],gmPow); //9810051151
//gm0*=Math.pow(inp0[j0],1/len0); //9810051158
  }

if(j1==1)
spanAlert1.innerHTML=''  //9809291518
+'<font color=red>ALERT ! box1 has negative number! AMGM do not treat negative!</font>'
;

var sum0=bye09(am0*len0);
var threeseq=''; //9809300035
var gradualAM='';//9809300040
//var gradualSum=0;//9809300041use,9809300043del
var inp0sum=0;
var grAMsum=0;
var grGMsum=0;

for(j0=1;j0<=len0;j0++) //9809261729
    {
inp0sum+=inp0[j0-1]; //9809270814
gradualAM=inp0sum/j0;
grAMsum+=gradualAM; //9809300107

carleman=1;
j2=1/j0; //9809261710
j3=1;
for(j1=0;j1<j0;j1++)
      {
carleman*=Math.pow(inp0[j1],j2); //9809261711
//j3=j3*inp0[j1]; //9809270803
//gradualSum+=inp0[j1]; //9809300042
//inp0sum IS gradualSum   9809300043
      }
//carlleft+=carleman;  //9809261712
grGMsum+=carleman; //9809300108

threeseq+=inp0[j0-1]+'\t\t'
+bye09(gradualAM)+'\t\t'  //9809300045
+bye09(carleman)+'\n'     //9809272119
;
    } //for(j0=1;j0<=len0;j0++) //9809261729

threeseq+='\t\t=====\n' //9809300110
+inp0sum+'\t\t'
+bye09(grAMsum)+'\t\t'
+bye09(grGMsum)+'\n'
+'GA/IN='+(grAMsum/inp0sum)+'\n'
+'GG/IN='+(grGMsum/inp0sum)+'\n'
+'IN = sum of INput sequence\n'
+'GA = sum of Gradual AMed sequence\n'
+'GG = sum of Gradual GMed sequence\n'
+'Many test show GG/IN  far  below E=2.718281828459045...\n'
+'This test show GA/IN still below E=2.718281828459045...\n'
+'Only monotone decrease sequence approach E\n'
+'If   monotone increase sequence we can relax.\n'
+'(Carleman inequality: GG<=E*IN or max(GG/IN)=E .\n'
+' if monotone increase GG/IN<1<E=2.718281...\n'
+' for a GG/IN<1 case, why do we bother GG/IN=E ?\n'
+' so, relax!  2009-09-30-07-58 LiuHH add note.)\n'


inp0sum=bye09(inp0sum);

boxc02.value=""
+"GA/IN="+(grAMsum/inp0sum)+"\n"
+"GG/IN="+(grGMsum/inp0sum)+"\n"
+"Carleman's inequality forget gradual AMed sequence\n"
+"Carleman's inequality say the following (GG<=E*IN)\n"
+"Gradual GMed sum <= E*INput sum\n"
+"SUM[k=1,n]{a1*a2*...*ak}^(1/k) <=\n"
+"<= E*SUM[k=1,n]{ak} and n goto inf.\n"
+"This output not multiply E=2.718281828459045...\n" //9809300048
+"This output add gradual AMed sequence\n" //9809300054
+"Output the following three sequences\n"
+"INput sequence; Gradual AMed sequence; Gradual GMed sequence\n"
+threeseq //9809300050
;
} //function carlemanf3()  9809300051
//




//function lindata1(ldArg1) //9809281403
function powdata1(ldArg1) //9809291129
{
var bgn0=parseFloat(boxcC1.value)
var nos0=parseInt(boxcC2.value); //number of steps
var end0=parseFloat(boxcC3.value)
var pow0=parseFloat(boxcE1.value);//9809291209
if(isNaN(bgn0)) return 'Bgn box value error'
if(isNaN(nos0)) return '[# Step] box value error'
if(isNaN(end0)) return 'End box value error'
if(isNaN(pow0)) return 'Power box value error'

if(nos0<=0) return '[# Step] box value <=0'
//if(pow0< 0) pow0=-pow0; //9809291212

var iter0=0;
var i0,i1,i2;
var step0=(end0-bgn0)/nos0
var out0='';
i1=bgn0;

//if(step0<0)if(pow0>1)pow0=-pow0; //9809291217add/del
//because iter0 control 

while(iter0<=nos0)
  {
iter0++;
//out0+=i1+'\n'
out0+=Math.pow(i1,pow0)+'\n'; //9809291219
i1=i1+step0;
  }

return out0
} //9809281421







//9809261416 
//fillRand(ouBoxID)   use   ".form1."
//fillRand2(ouBoxID) delete ".form1."
//fillRand3(ouBoxID) delete ".form1."
//use chkbxcB1, chkbxcB2 for sorting
//
function fillRand3(ouBoxID) //9806101238
{

//alert('ouBoxID=['+ouBoxID+'] 9809261423')

if(arguments.length==0    //9806101558
 ||arguments[0].length==0 //9806101602
  )
  {
alert("Call function fillRand(ouBoxID) need parameter.\n"
+"It tell program where output go. Example next\n"
+" onclick=\"javascript:fillRand('boxc12.value')\"\n"
+"must have a box ID=boxc12, Now no parameter.")
;
return; //9806101551
  }

//var powIndex=parseInt(document.form1.boxcA4.value); //9806111044
//var powIndex=parseInt(document.boxcA4.value); //9809261410
var powIndex=parseInt(boxcA4.value); //9809261424

var precision0;  //9806101257
//precision0=parseInt(document.form1.boxcA3.value); //9806101346
//precision0=parseInt(document.boxcA3.value); //9809261411
precision0=parseInt(boxcA3.value); //9809261425

//if(document.form1.chkbxcA1.checked)
//precision0=powIndex; //9806111156
//if(document.chkbxcA1.checked)precision0=powIndex; //9809261412
if(chkbxcA1.checked)precision0=powIndex; //9809261426

powIndex=Math.pow(10,powIndex) //9806111050

var precision1=Math.pow(10,precision0); //9806101312
var randRang=precision1; //9806101240

if(powIndex) //9806111045
 randRang=randRang/powIndex;

var randPoNe=1; //9806101243
var randPN=1; //9806101337
//if(document.form1.RadiocA2[1].checked)randPN=-1;
//if(document.chkbxcA3.checked)randPN=-1;
if(chkbxcA3.checked)randPN=-1; //9809261427

var dataPerS=parseInt(boxcA1.value); //9806101246

var DecInt=0; //9806111054
//if(document.form1.chkbxcA1.checked)DecInt=1;
//if(document.chkbxcA1.checked)DecInt=1; //9809261413
if(chkbxcA1.checked)DecInt=1; //9809261428

var ouStr1='';
var ouStr2='';
var t0,t1,t2,t3;

for(t0=0;t0<dataPerS;t0++)
  {
t1=Math.random()*precision1; //error ?! 9806101327

if(randPN<0)
    {
t2=(t1+'').charAt((t1+'').length-1); //9806101332
if(t2%2)randPoNe=-1
else randPoNe=+1; //9806101322
    }
t1=Math.floor(t1)*randPoNe;
t1=t1/randRang;
if(DecInt)t1=parseInt(t1);  //9806111057
if(randPoNe==1)ouStr1+=' '; //9806101352
ouStr1+=t1+'\n';
  }

if(chkbxcB1.checked
 ||chkbxcB2.checked //9809281210
  ) 
  { //9809281140
//ouStr1=sortNumb(ouStr1).replace(/,/g,'\n')
//alert('9809282243');
//alert('ouStr1.length='+ouStr1.length+'; 9809282245');
//alert('ouStr1='+ouStr1+'; 9809282246');

for(t0=ouStr1.length-1;t0>0;t0--)
    {
if(ouStr1.charCodeAt(t0)<32) continue;
else break; //9809281203
    }
t1=ouStr1.substring(0,t0+1);
t1=t1.split('\n');
t2=[''];
for(t0=0;t0<t1.length;t0++)
    {
//t1[t0]=parseFloat(t1[t0]);
//if(!isNaN(t1[t0])) //9809281159
t2[t0]=parseFloat(t1[t0]);
//else break;
    }
//t3=bubble_1(t2,t2.length);

bubble_1(t2,t2.length);

if(chkbxcB1.checked) //1,2,3 increase
    {
t3=[''];
for(t0=0;t0<t2.length;t0++)
      {
t3[t0]=t2[t2.length-1-t0];
      }
t2=t3; //9809281213
    }

ouStr1=(''+t2).replace(/,/g,'\n')

//alert('t1='+t1+'\n'+'t2='+t2+'\n'+'t3='+t3+'\n9809281156');
  }

eval(ouBoxID+'=ouStr1'); //9806101542 OK

}
//9806101352 function fillRand(ouBoxID)
//9809261417 function fillRand2(ouBoxID)
//fillRand3(ouBoxID) delete ".form1."
//use chkbxcB1, chkbxcB2 for sorting


//-->
</script>

<br>
<a name="sortnumb">&lt;a name="sortnumb"&gt;</a>　<a href="#index">Index begin</a>　<a href="#index13">Index this file</a>
<br>
<font color=red size=+2>Sort number</font> not text string

<br>
<font color=red size=+2>
Output may contain error, Please verify first.
<br>
Program environment is MSIE 6.0, please use MSIE
</font>

<br>

<INPUT onclick='javascript:sortNumbMan(0),spanSort1.innerHTML="Output new index for both eigenvalue and vector. Run by sortNumb()"'
 type=button value="Output
New Index"><!--9807151815-->
 
 &nbsp;

<INPUT onclick='javascript:bubbleMan(0),spanSort1.innerHTML="Output new eigenvalue sequence. Ignore eigenvector. Run by bubble_1()"'
 type=button value="Output
New Seq."><!--9807151816-->
 

<br>
<span id=spanSort1><font color=red>
Box11 accept only numbers, not name list, not non-number string.
</font></span>
<br>

<br>

<a name="ch01tb04"></a>
<table border=0>
<tr>

<td>
Box 11, input number sequence
<br>
<INPUT onclick='javascript:window.clipboardData.setData("Text",document.getElementById("boxc21").value.toString())'
 type=button value="copy11">
<INPUT onclick='document.getElementById("boxc21").value=""'
 type=button value="del11">

<INPUT onclick='upsidedown()'
 type=button value="up side down"><!--9809291951-->

<br>
<TEXTAREA
 id=boxc21
 name=boxc21 rows=6
 cols=30
 >
 0.556
-0.145
 0.302e3
-0.305e2
-0.088
</TEXTAREA>
</td>

<td>
Box 12, output answer
<br>
<INPUT onclick='javascript:window.clipboardData.setData("Text",document.getElementById("boxc22").value.toString())'
 type=button value="copy12">
<INPUT onclick='document.getElementById("boxc22").value=""'
 type=button value="del12">

<br>
<TEXTAREA
 id=boxc22
 name=boxc22 rows=6
 cols=30
 >
</TEXTAREA>
</td>

</tr>
</table>
This file sort numbers, if it were sort string, then e2, e3 go wrong.
<br>
2009-07-15-11-18 start sortNumb()
<br>
2009-07-15-18-40  done sortNumb()
<br>
2009-09-29-20-50  done upsidedown()
<br>
<br>


<pre><font size=+2>
<a name="docA011">&lt;a name="docA011"&gt;</a>
2009-09-29-15-39 help_1
<a href="#program0">Program</a>
Arithmetic Mean and Geometric Mean of a number sequence 
usage is as following.
Box 1 is input. Paste numbers into box 1.
Box 2 is output. There are two programs.
First  AM/GM inequality 
Second Carleman's inequality 
<a name="docA012">&lt;a name="docA012"&gt;</a>
There are three buttons for them
"AM GM 1"    it run amgmf1(boxc01.value,1)
"Carle man1" it run amgmf1(boxc01.value,2)
Above two buttons run function amgmf1()
"Carle man2" it run carlemanf2()

"Carle man1" read user supplied arbitrary data
"Carle man2" create monotone increase/decrease
data

<a name="docA013">&lt;a name="docA013"&gt;</a>
"Carle man1" use Gradual GMed sum <= E*INput sum
"Carle man2" drop E=2.718281828459045...
for monotone increase data change E to a value
less than one, for example 0.9, still perfect.
       Gradual GMed sum <= 0.9*INput sum
for monotone decrease data need E to rise input
data sum over Gradual GMed sum.
"Carle man2" main purpose is to see the ratio of
Gradual GMed sum / INput sum (no E here)

<script language="javascript">alert0()</script>
<a name="docA014">&lt;a name="docA014"&gt;</a>　<a href="#index">Index begin</a>　<a href="#index13">Index this file</a>
There are four debug checkbox. dbg a,b,c;d.
"AM GM 1" do not use debug checkbox
"Carle man1" use dbg a,b,c three checkbox
"Carle man2" use dbg d one checkbox
If check box are unchecked. Output less material
and run faster for shorter information.
If check box are checked. Output more material
and run slower for detail information.
"dbg" button is all-clear button, not all-check.
<a name="docA015">&lt;a name="docA015"&gt;</a>
dbg a checked, "Carle man1" output input sequence
      and gradual GMed number sequence.
dbg b checked, "Carle man1" output carlrite
      that is input sequence related output.
      include echo print input sequence, make
      sure program read right data.
dbg c checked, "Carle man1" output carlleft
      include carlpart, carlpart is long string
      debug data.
dbg d checked, "Carle man2" output longer output.

<a name="docA016">&lt;a name="docA016"&gt;</a>
α box and ω box are add for "Carle man1" use.
If α box fill a number 
     0<=alpha<=1/log(2)=1.4426950408889633
"Carle man1" use alpha to lower input sum
reference is next
[[
2009-09-02-22-06 LiuHH access
http://www.emis.de/journals/JIPAM/images/029_00_JIPAM/029_00_www.pdf
page 9/10 theorem 3.2 
]]

<a name="docA017">&lt;a name="docA017"&gt;</a>
ω box has no theory base. ω simply add to right
hand side of
 Gradual GMed sum <= E*INput sum
as
 Gradual GMed sum <= ω*E*INput sum
If set ω=1/E=0.36787944117144233 then equation
become
 Gradual GMed sum <= INput sum
Check box 1/E fill 1/E=0.36787944117144233 for 
you.

<a name="docA018">&lt;a name="docA018"&gt;</a>
The remaining boxes/buttons create data for you.
"equal #" create equal numbers. It take number 
         from "10^" box and repeat fill-box times.
         output to box 1 as input data.
"random#" create random numbers.
"10^" box determine random number order of magnitude.
"+/0" box checked, output positive/zero random numbers.
"+/0/-" box checked, output +/0/- random numbers.
"digits" box determine the length of each random number.
"fill" box determine build how many random number.
"integer" checkbox force output integer.
"123" checkbox sort random number to increase order.
"321" checkbox sort random number to decrease order.

<a name="docA019">&lt;a name="docA019"&gt;</a>
"powData" button create x^power data.
"bgn" box is begin value.
"# Step" box determine how many step to divide.
"End" box determine stop value.
"power" box determine power in x^power.
If set same value to "bgn" box "End" box 
output constant data.

"Carle man2" button automatically click "powData"
button and read data.

This is designed for you to build random data
or build power data, and not necessary to use
data here. You can build data and copy box1 
value, use data in other application.

<a name="docA020">&lt;a name="docA020"&gt;</a>
Box11 and box12 has nothing to do with AMGM or
Carleman inequality. Program ("123", "321")
use function bubble_1(). Add few more line, 
build box11,12, let you to sort number. 
A handy tool.
2009-09-29-16-32 stop

<a name="docA021">&lt;a name="docA021"&gt;</a>　<a href="#index">Index begin</a>　<a href="#index13">Index this file</a>
2009-09-29-20-58 start
2009-09-29 add "upside down" function. 
Purpose is to reverse increase/decrease
numbers. Text (non-number) can be reversed
too.
"123", "321" checkbox care only this program
generated data. If you have your own
increase/decrease data and want to reverse
them, "123", "321" checkbox can not help.
"up side down" button will help you with
your data.
2009-09-29-21-02 stop

<a name="docA022">&lt;a name="docA022"&gt;</a>
2009-09-29-21-10 start
Alert !!
Both Arithmetic Mean and Geometric Mean
inequality and Carleman inequality work
with positive or zero number only. This
page allow you to create sequence contain
negative number. You can create number 
sequence here and use them in other place.
If you use number sequence right here for
AM/GM or Carleman inequalities, <font color=red>it is your
responsibility not to generate negative
number!</font>
2009-09-29-21-15 stop

<a name="docA023">&lt;a name="docA023"&gt;</a>
2009-09-30-07-33 start
On 2009-09-30-00-51 write function carlemanf3()
The button is "Carle man3". Main point is to
involve sum of Gradual AMed sequence.
Define
IN = sum of INput sequence
GA = sum of Gradual AMed sequence
GG = sum of Gradual GMed sequence
Carleman's inequality in brief is GG<=E*IN
Many test show GG/IN  far  below E=2.718281828459045...
Carleman3 show GA/IN still below E=2.718281828459045...
(AM >= GM, therefore  GA/IN >= GG/IN )

carlemanf3() output 
INput sequence
Gradual AMed sequence
Gradual GMed sequence
and sum each, find GG/IN GA/IN two ratio.

function carlemanf3() help thinking.
2009-09-30-07-41 stop

<a name="docA024">&lt;a name="docA024"&gt;</a>
2009-09-30-08-03 start
If you have a set test data pasted to box 1.
Click "Carle man1", program read your data.
Click "Carle man3", program read your data.
But
Click "Carle man2", your input data changed
you may be surprised and wondering what is 
going on?

"Carle man2" button read data from "powData"
related data, Bgn box, # Step box, End box, 
power box all contribute to "Carle man2" 
button. "powData" generated data send to 
box 1 and erase previous value. 
"Carle man2" button main point is to test
monotone increase or monotone decrease data.
<font color=red>If your data is not generated from "powData"
button do not touch "Carle man2" button.</font>
2009-09-30-08-12 stop

<a name="docA025">&lt;a name="docA025"&gt;</a>
2009-10-05-12-36 start
Update 2009-10-05 correct GM error.
nineth draft and before
tute0012a09.htm output GM correct.
tenth draft and after
tute0012a10.htm output GM wrong.

<a name="docA026">&lt;a name="docA026"&gt;</a>
correct code is
[[
j1=1/len0;

for(j0=0;j0&lt;len0;j0++)
  {
am0+=inp0[j0]/len0;
gm0*=Math.pow(inp0[j0],j1); //9809261046
  }
]]

<a name="docA027">&lt;a name="docA027"&gt;</a>
wrong code is
[[
j1=1/len0;

spanAlert1.innerHTML=''; //9809291518
j1=0;
for(j0=0;j0&lt;len0;j0++)
  {
if(inp0[j0]&lt;0)j1=1; //9809291521
am0+=inp0[j0]/len0;
gm0*=Math.pow(inp0[j0],j1); //9809261046
  }

if(j1==1)
spanAlert1.innerHTML=''  //9809291518
+'&lt;font color=red&gt;ALERT ! box1 has negative number! AMGM do not treat negative!&lt;/font&gt;'
;
]]

<a name="docA028">&lt;a name="docA028"&gt;</a>
wrong code re-use j1 for different purpose.
for three element sequence q3=[1,3,9] its 
GM = (1*3*9)^(1/3) = 27^(1/3) = 3
correct code assign
j1=1/len0;  that is  j1=1/3
use j1 in
gm0*=Math.pow(inp0[j0],j1); //9809261046
get correct answer.
Wrong code change j1 definition to j1=0
then
gm0*=Math.pow(inp0[j0],j1);
is
gm0*=Math.pow(inp0[j0],0);
always get 1, wrong.

<a name="docA029">&lt;a name="docA029"&gt;</a>
On 2009-09-29-15-?? LiuHH add code to warn
negative value element. The flag used is 
again j1 (here wrong!!), j1 is redefined.
Because GM answer is already correct, LiuHH
pay attention to correct warning, and forget
to check GM answer.

<a name="docA030">&lt;a name="docA030"&gt;</a>
2009-10-05-11-18 found GM answer is wrong!
read code, find out the redefinition error.

change
j1=1/len0; 
gm0*=Math.pow(inp0[j0],j1); //9809261046
to
gmPow=1/len0; 
gm0*=Math.pow(inp0[j0],gmPow); //9810051143
solve problem
2009-10-05-12-52 stop

</font></pre>

<hr>

<pre><font size=+2><a name="ch02b001">&lt;a name="ch02b001"&gt;</a>　<a href="#index">Index begin</a>　<a href="#index13">Index this file</a>
2009-10-01-11-49 start
■　Carleman warm up

<script language="javascript">alert0()</script>
<!--2009-10-01-19-36 start--><font color=red>
Carleman's Inequality require input sequence
has infinite many elements and sum to finite
value. Only monotone decreasing sequence can
be a candidate of Carleman's Inequality seq.
<!--2009-10-01-19-39 stop--></font>
<a href="#textbook">Text book</a> page 27 discuss Carleman's Inequality.
For each positive real sequence
a<sub>1</sub>, a<sub>2</sub>, ... a<sub>k</sub>, one has next inequality.
</font></pre>

<a name="ch02b002">&lt;a name="ch02b002"&gt;</a>
<br>
<TABLE WIDTH="380"
 id=T027L14a
  border=0
 ><!--9810011154-->
<TR>
<TD>
 &nbsp;
</TD>
<TD>
<TABLE WIDTH="260" border=0
 id=T027L14b
>
<TR>


<TD>
<CENTER><FONT SIZE=-2>k=∞</FONT></CENTER>
<CENTER><FONT SIZE=+3>∑</FONT></CENTER>
<CENTER><FONT SIZE=-2>k=1</FONT></CENTER>
</TD>


<TD>
(a<sub>1</sub>a<sub>2</sub>...a<sub>k</sub>)<sup>1/k</sup>
</TD>

<TD>
≦ e 
</TD>


<TD>
<CENTER><FONT SIZE=-2>k=∞</FONT></CENTER>
<CENTER><FONT SIZE=+3>∑</FONT></CENTER>
<CENTER><FONT SIZE=-2>k=1</FONT></CENTER>
</TD>


<TD>
a<sub>k</sub>
</TD>

</TR>
</TABLE>
</TD>

<TD>

 ---page 27
<br>
 ---eqn.2.15

</TD>
</TR>
</TABLE>

where e=2.718281828459045...
<br>
width of above equation
<INPUT name=Button027L14a type="button"
 value="default"
 onclick="B027L14a.value=380,B027L14b.value=260,T027L14a.width=B027L14a.value,T027L14b.width=B027L14b.value">

<input id="B027L14a"
 value=380
 size=3
 onchange=T027L14a.width=B027L14a.value
 />
<input id="B027L14b"
 value=260
 size=3
 onchange=T027L14b.width=B027L14b.value
 />


<pre><font size=+2><a name="ch02b003">&lt;a name="ch02b003"&gt;</a>
2009-10-01-12-03 here
Problem statement exclude a<sub>k</sub>=0
and not allow a<sub>k</sub><0 .
One online page
[[
2009-09-02-22-06 LiuHH access
http://www.emis.de/journals/JIPAM/images/029_00_JIPAM/029_00_www.pdf
carleman-029_00_JIPAM-029_00_www.pdf
page 3/10 introduction
]]
say equality in <a href="#ch02b002">eqn.2.15</a> hold if and only if
one element of sequence is zero.

<a name="ch02b004">&lt;a name="ch02b004"&gt;</a>
First time see <a href="#ch02b002">eqn.2.15</a> , LiuHH is wondering
what is the meaning of this equation? After
re-read <a href="#textbook">textbook</a> several times, gradually
understand that 
eqn.2.15 right hand side is sum of original 
         input sequence
eqn.2.15 left hand side is sum of gradual GMed
         new sequence from original input seq.

<a name="ch02b005">&lt;a name="ch02b005"&gt;</a>
To explain clearly, let k=1 to k=5 (was k=inf)

original input sequence is a<sub>1</sub>,a<sub>2</sub>,a<sub>3</sub>,a<sub>4</sub>,a<sub>5</sub>
When first term goto front stage, we have
first  original term  a<sub>1</sub>  ---eqn.AF001
first  gr. GMed term (a<sub>1</sub>)^(1/1) ---eqn.AF002

When first two terms goto front stage, we have
second original term a<sub>2</sub> ---eqn.AF003
second gr. GMed term (a<sub>1</sub>*a<sub>2</sub>)^(1/2) ---eqn.AF004
<font color=red>eqn.AF003 no a<sub>1</sub> shadow,
eqn.AF004 a<sub>1</sub> has influence.
</font>
<a name="ch02b006">&lt;a name="ch02b006"&gt;</a>　<a href="#index">Index begin</a>　<a href="#index13">Index this file</a>
When first three terms goto front stage, we have
third  original term a<sub>3</sub> ---eqn.AF005
third  gr. GMed term (a<sub>1</sub>*a<sub>2</sub>*a<sub>3</sub>)^(1/3) ---eqn.AF006
<font color=red>eqn.AF005 no a<sub>1</sub>,a<sub>2</sub> shadow,
eqn.AF006 a<sub>1</sub>,a<sub>2</sub> have influence etc.
</font>
When first four terms goto front stage, we have
fourth original term a<sub>4</sub> ---eqn.AF007
fourth gr. GMed term (a<sub>1</sub>*a<sub>2</sub>*a<sub>3</sub>*a<sub>4</sub>)^(1/4) ---eqn.AF008

When all five terms goto front stage, we have
fifth  original term a<sub>5</sub> ---eqn.AF009
fifth  gr. GMed term (a<sub>1</sub>*a<sub>2</sub>*a<sub>3</sub>*a<sub>4</sub>*a<sub>5</sub>)^(1/5) ---eqn.AF010

<a name="ch02b007">&lt;a name="ch02b007"&gt;</a>
Carleman's Inequality sum both original sequence
and gradual GMed new sequence, result is

Sum of original seq. a<sub>1</sub>+a<sub>2</sub>+a<sub>3</sub>+a<sub>4</sub>+a<sub>5</sub> ---eqn.AF011
(extend to infinite many elements for Carleman)

Sum of gr. GMed seq.  ---eqn.AF012
 (a<sub>1</sub>)^(1/1) + (a<sub>1</sub>*a<sub>2</sub>)^(1/2)+(a<sub>1</sub>*a<sub>2</sub>*a<sub>3</sub>)^(1/3)
+(a<sub>1</sub>*a<sub>2</sub>*a<sub>3</sub>*a<sub>4</sub>)^(1/4)+(a<sub>1</sub>*a<sub>2</sub>*a<sub>3</sub>*a<sub>4</sub>*a<sub>5</sub>)^(1/5)
(extend to infinite many elements for Carleman)

<a name="ch02b008">&lt;a name="ch02b008"&gt;</a>
re-write Sum of original seq. eqn.AF011 and 
sum of gr. GMed seq. eqn.AF012 in compact form
and Carleman demand that the ratio of
 GMed sum / original sum 
not exceed E=2.718281828459045... 
then we get <a href="#ch02b002">eqn.2.15</a> 

Carleman's Inequality push this sequence to
infinity.

2009-10-01-12-36 here

<a name="ch02b009">&lt;a name="ch02b009"&gt;</a>
■　Gradual GMed seq. sum

Follow the same pattern, we can create
gradual AMed sequence as following
<font color=red>
Define IN</font> = sum of INput sequence
       IN = eqn.AF011
<font color=red>
Define GG</font> = Gradual GMed seq. sum
       GG = eqn.AF012

<a name="ch02b010">&lt;a name="ch02b010"&gt;</a>
■　Gradual AMed seq. sum

<font color=red>Define GA</font> = eqn.AF013
Gradual AMed seq. sum =   ---eqn.AF013
 (a<sub>1</sub>)/1 + (a<sub>1</sub>+a<sub>2</sub>)/2 + (a<sub>1</sub>+a<sub>2</sub>+a<sub>3</sub>)/3
+(a<sub>1</sub>+a<sub>2</sub>+a<sub>3</sub>+a<sub>4</sub>)/4 + (a<sub>1</sub>+a<sub>2</sub>+a<sub>3</sub>+a<sub>4</sub>+a<sub>5</sub>)/5

Because AM >= GM , 
each individual term of gr. AMed seq. >=
each individual term of gr. GMed seq. 

then we know  GA >= GG  ---eqn.AF014

<a name="ch02b011">&lt;a name="ch02b011"&gt;</a>
Carleman proved if multiply <a href="#ch02b009">IN</a> by E=2.718281828459045... 
then GG <= E*IN  ---eqn.AF015
 is guaranteed.

Above introduce background information.
Hope you will get better picture now.
2009-10-01-12-51 here

<script language="javascript">alert0()</script>

<a name="ch02b012">&lt;a name="ch02b012"&gt;</a>　<a href="#index">Index begin</a>　<a href="#index13">Index this file</a>
■　GG and GA but not Carleman

2009-10-01-13-58 start
Carleman's inequality is  GG <= E*IN 
Input sequence must be infinite many elements
and elements summation is finite value.
<font color=red><!--9810011950-->
The following is NOT Carleman's inequality.
Because discuss finite many elements
Because discuss monotone increase sequence.
</font>
Is it always need an 'E*' to rise '<a href="#ch02b009">IN</a>'.
Whether there is case that GG <= IN ?
without 'E*' ! (E=2.718281828459045...)

Let us find answer for the simplest case
two number sequence S2 = [a<sub>1</sub>, a<sub>2</sub>]

<a name="ch02b013">&lt;a name="ch02b013"&gt;</a>
If a<sub>1</sub>=a<sub>2</sub> ---eqn.AF016
 then
IN = sum of INput sequence = a<sub>1</sub>+a<sub>2</sub> = 2*a<sub>1</sub>
GG = Gradual GMed seq. sum
   = (a<sub>1</sub>)^(1/1) + (a<sub>1</sub>*a<sub>2</sub>)^(1/2)
   = (a<sub>1</sub>) + (a<sub>1</sub>*a<sub>1</sub>)^(1/2)
   = 2*a<sub>1</sub> ---eqn.AF017
If a<sub>1</sub>=a<sub>2</sub> IN = GG
we do not need 'E*' between <a href="#ch02b009">IN</a> and GG

<a name="ch02b014">&lt;a name="ch02b014"&gt;</a>
If a<sub>1</sub> < a<sub>2</sub>  ---eqn.AF018
that is 
if sequence is increasing.
IN = sum of INput sequence = a<sub>1</sub>+a<sub>2</sub>  ---eqn.AF019
GG = Gradual GMed seq. sum
   = (a<sub>1</sub>)^(1/1) + (a<sub>1</sub>*a<sub>2</sub>)^(1/2) ---eqn.AF020

<a name="ch02b015">&lt;a name="ch02b015"&gt;</a>
Now consider their difference 
IN =  a<sub>1</sub> +     <font color=red>a<sub>2</sub> </font>       ---eqn.AF019
GG =  a<sub>1</sub> + <font color=red>(a<sub>1</sub>*a<sub>2</sub>)^(1/2)</font> ---eqn.AF020
Because (a<sub>1</sub>*a<sub>2</sub>)^(1/2) is geometric mean
between a<sub>1</sub> and a<sub>2</sub>

<a name="ch02b016">&lt;a name="ch02b016"&gt;</a>
For an increasing sequence a<sub>1</sub> < a<sub>2</sub>
    (a<sub>1</sub>*a<sub>2</sub>)^(1/2) < a<sub>2</sub> then GG < <a href="#ch02b009">IN</a>

For a  decreasing sequence a<sub>1</sub> > a<sub>2</sub>
    (a<sub>1</sub>*a<sub>2</sub>)^(1/2) > a<sub>2</sub> then GG > IN

<a name="ch02b017">&lt;a name="ch02b017"&gt;</a>
Consider a simple numerical example
Increase sequence [1, 2, 3]
IN = 1+2+3
GG = 1 + (1*2)^(1/2) + (1*2*3)^(1/3)
   = 1 + 1.414213562373095 + 1.8171205928321396

IN = 6
GG = 4.231334155205235
GG < IN and
GG << E*IN  (E=2.718281828459045)

<a name="ch02b018">&lt;a name="ch02b018"&gt;</a>　<a href="#index">Index begin</a>　<a href="#index13">Index this file</a>
■　monotone increase, GG < IN

For a monotone increase sequence, we conclude
that we do not need 'E*' and
Gradual GMed sequence sum < INput sequence sum

For a monotone decrease sequence, the condition
is reversed, in general
Gradual GMed sequence sum > INput sequence sum

<a name="ch02b019">&lt;a name="ch02b019"&gt;</a>
Consider a simple numerical example
Decrease sequence [3, 2, 1]
<a href="#ch02b009">IN</a> = 3+2+1
GG = 3 + (3*2)^(1/2) + (3*2*1)^(1/3)
   = 3 + 2.4494897427831783 + 1.8171205928321396

IN = 6
GG = 7.266610335615319
GG > IN  and
GG < E*IN  (E=2.718281828459045)
<font color=red><!--9810011956-->
Above is NOT Carleman's inequality.
Because discuss finite many elements
Because discuss monotone increase sequence.
</font>
2009-10-01-14-48 stop

<a name="ch02b020">&lt;a name="ch02b020"&gt;</a>
■　Order critical for GA and GG

2009-10-02-14-28 start
Restate above mentioned
monotone increase sequence and
monotone decrease sequence
in a slightly different way.
Assume we have a sequence 
Q1=[1,4,3,5,2] ---eqn.AF021
Put Q1 in monotone increase order get
Q2=[1,2,3,4,5] ---eqn.AF022
Put Q1 in monotone decrease order get
Q3=[5,4,3,2,1] ---eqn.AF023
<a name="ch02b021">&lt;a name="ch02b021"&gt;</a>
If order is not important, then Q1, Q2
and Q3 are the same. For example: sum
SUM Q1<sub>i</sub>=1+4+3+5+2=15 ---eqn.AF024A
SUM Q2<sub>i</sub>=1+2+3+4+5=15 ---eqn.AF024B
SUM Q3<sub>i</sub>=5+4+3+2+1=15 ---eqn.AF024C
Above is sequence element sum
<a name="ch02b022">&lt;a name="ch02b022"&gt;</a>
Below is Gradual Geometric_Mean_ed sum 
first  gr. GMed term (a<sub>1</sub>)^(1/1) ---eqn.AF002
second gr. GMed term (a<sub>1</sub>*a<sub>2</sub>)^(1/2) ---eqn.AF004
third  gr. GMed term (a<sub>1</sub>*a<sub>2</sub>*a<sub>3</sub>)^(1/3) ---eqn.AF006
.....
 k th  gr. GMed term (a<sub>1</sub>*a<sub>2</sub>*...*a<sub>k</sub>)^(1/k) ---eqn.AF025
Can you see if Q1, Q2, Q3 produce same result
or different result for Gradual GMed sum ?
<a name="ch02b023">&lt;a name="ch02b023"&gt;</a>
Sum of input sequence eqn.AF024 each term do
not have memory about earlier term value.
BUT, Gradual GMed sum has memory from earlier
term. For example if sequence is monotone
decrease, then a<sub>1</sub> has largest value, this
value influence the following terms all the 
way to the end. Create a high value Gradual
GMed sum. On the other hand, if sequence is
monotone increase, everything reverse.
<a name="ch02b024">&lt;a name="ch02b024"&gt;</a><font color=red>
For Gradual Geometric_Mean_ed sum, order of
input sequence is very important. For
Gradual Arithmetic_Mean_ed sum, same thing.
</font>
2009-10-02-15-04 stop

<a name="ch02b025">&lt;a name="ch02b025"&gt;</a>　<a href="#index">Index begin</a>　<a href="#index13">Index this file</a>
■　Begin textbook chapter two, Carleman

2009-10-02-15-10 start
Above is general reading understanding.
Below is <a href="#textbook">textbook</a> page 27, chapter two
Carleman's inequality.
For each positive real sequence
a<sub>1</sub>, a<sub>2</sub>, ... a<sub>k</sub>, one has next inequality.
</font></pre>

<a name="ch02b026">&lt;a name="ch02b026"&gt;</a>

<br>
<TABLE WIDTH="380"
 id=T027L14c
  border=0
 ><!--9810021513-->
<TR>
<TD>
 &nbsp;
</TD>
<TD>
<TABLE WIDTH="260" border=0
 id=T027L14d
>
<TR>


<TD>
<CENTER><FONT SIZE=-2>k=∞</FONT></CENTER>
<CENTER><FONT SIZE=+3>∑</FONT></CENTER>
<CENTER><FONT SIZE=-2>k=1</FONT></CENTER>
</TD>


<TD>
(a<sub>1</sub>a<sub>2</sub>...a<sub>k</sub>)<sup>1/k</sup>
</TD>

<TD>
≦ e 
</TD>


<TD>
<CENTER><FONT SIZE=-2>k=∞</FONT></CENTER>
<CENTER><FONT SIZE=+3>∑</FONT></CENTER>
<CENTER><FONT SIZE=-2>k=1</FONT></CENTER>
</TD>


<TD>
a<sub>k</sub>
</TD>

</TR>
</TABLE>
</TD>

<TD>

 ---page 27
<br>
 ---eqn.2.15

</TD>
</TR>
</TABLE>

where e=2.718281828459045...
<br>
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<pre><font size=+2><a name="ch02b027">&lt;a name="ch02b027"&gt;</a>
2009-10-02-15-21 here
eqn.2.15 right hand side is the summation
of input sequence
a<sub>1</sub>+a<sub>2</sub>+a<sub>3</sub>+...+a<sub>k</sub> ---eqn.AF027
multiply by e=2.718281828459045... 

eqn.2.15  left hand side is the summation
of Gradual GMed terms of input sequence.
Explicit form is
(a<sub>1</sub>)^(1/1)+(a<sub>1</sub>*a<sub>2</sub>)^(1/2)+(a<sub>1</sub>*a<sub>2</sub>*a<sub>3</sub>)^(1/3)+
 ... +(a<sub>1</sub>*a<sub>2</sub>*...*a<sub>k</sub>)^(1/k) ---eqn.AF028

<a name="ch02b028">&lt;a name="ch02b028"&gt;</a><font color=red>
Carleman's inequality treat infinite long
sequence, number of terms in a<sub>k</sub> sequence
is infinity many. Require SUM{a<sub>k</sub>} be finite.</font>
Then for large k, a<sub>k</sub> must be monotone
decrease. When
 k approach to infinity,
 a<sub>k</sub> approach to zero.

<a href="#ch02b002">eqn.2.15</a> suggest if SUM{a<sub>k</sub>} be finite, then 
SUM{(a<sub>1</sub>*a<sub>2</sub>*...*a<sub>k</sub>)^(1/k)} be finite too.

<a name="ch02b029">&lt;a name="ch02b029"&gt;</a>　<a href="#index">Index begin</a>　<a href="#index13">Index this file</a>
■　First try and error (failure)

One possible way to solve this problem is
to apply AM-GM inequality to eqn.2.15 

Look at eqn.AF027, apply AM-GM inequality
to each term individually

for first term
(a<sub>1</sub>)^(1/1) ---eqn.AF002
AM-GM inequality say
(a<sub>1</sub>)^(1/1) = (a<sub>1</sub>) ---eqn.AF029

<a name="ch02b030">&lt;a name="ch02b030"&gt;</a>
for second term
(a<sub>1</sub>*a<sub>2</sub>)^(1/2) ---eqn.AF004
AM-GM inequality say
(a<sub>1</sub>*a<sub>2</sub>)^(1/2) <= (a<sub>1</sub>+a<sub>2</sub>)/2 ---eqn.AF030

for third term
(a<sub>1</sub>*a<sub>2</sub>*a<sub>3</sub>)^(1/3) ---eqn.AF006
AM-GM inequality say
(a<sub>1</sub>*a<sub>2</sub>*a<sub>3</sub>)^(1/3) <= (a<sub>1</sub>+a<sub>2</sub>+a<sub>3</sub>)/3 ---eqn.AF031
.....
<a name="ch02b031">&lt;a name="ch02b031"&gt;</a>
for k th term
(a<sub>1</sub>*a<sub>2</sub>*a<sub>3</sub>*...*a<sub>k</sub>)^(1/k) ---eqn.AF025
AM-GM inequality say
(a<sub>1</sub>*a<sub>2</sub>*a<sub>3</sub>*...*a<sub>k</sub>)^(1/k) <= (a<sub>1</sub>+a<sub>2</sub>+a<sub>3</sub>+...+a<sub>k</sub>)/k ---eqn.AF032
continue up to k=n terms 
(k is general index, n is end index)

<a name="ch02b032">&lt;a name="ch02b032"&gt;</a>
Add eqn.AF029, eqn.AF030, eqn.AF031, eqn.AF032
get
(a<sub>1</sub>)^(1/1)+(a<sub>1</sub>*a<sub>2</sub>)^(1/2)+(a<sub>1</sub>*a<sub>2</sub>*a<sub>3</sub>)^(1/3)+
 ... +(a<sub>1</sub>*a<sub>2</sub>*...*a<sub>k</sub>)^(1/k)
 ... +(a<sub>1</sub>*a<sub>2</sub>*...*a<sub>k</sub>*...*a<sub>n</sub>)^(1/n)
<=                          ---eqn.AF033
(a<sub>1</sub>) + (a<sub>1</sub>+a<sub>2</sub>)/2 + (a<sub>1</sub>+a<sub>2</sub>+a<sub>3</sub>)/3
+ ... + (a<sub>1</sub>+a<sub>2</sub>+a<sub>3</sub>+...+a<sub>k</sub>)/k
+ ... + (a<sub>1</sub>+a<sub>2</sub>+a<sub>3</sub>+...+a<sub>k</sub>+...+a<sub>n</sub>)/n

<a name="ch02b033">&lt;a name="ch02b033"&gt;</a>
eqn.AF033 left hand side is <a href="#ch02b002">eqn.2.15</a> left
hand side (we start from eqn.2.15 left side)

eqn.AF033 right hand side collect terms, get
 (a<sub>1</sub>)*[1+1/2+1/3+ ... +1/k+ ... +1/n]
+(a<sub>2</sub>)*[0+1/2+1/3+ ... +1/k+ ... +1/n]
+(a<sub>3</sub>)*[0+0/2+1/3+ ... +1/k+ ... +1/n]
+.....
+(a<sub>k</sub>)*[0+0/2+0/3+ .0. +1/k+ ... +1/n]
+.....                       ---eqn.AF034
+(a<sub>n</sub>)*[0+0/2+0/3+ .0. +0/k+ .0. +1/n]

<a name="ch02b034">&lt;a name="ch02b034"&gt;</a>　<a href="#index">Index begin</a>　<a href="#index13">Index this file</a>
■　Two summation explained

eqn.AF033 right hand side use two summation
first  ∑ from k=1 to k=n
second ∑ from j=k to j=n
Why it is so? Let use look at a<sub>3</sub>
+(a<sub>3</sub>)*[0+0/2+1/3+ ... +1/k+ ... +1/n]
first  ∑ where k=3 is a<sub>3</sub>, its coefficient
is [0+0/2+1/3+ ... +1/k+ ... +1/n]

<a name="ch02b035">&lt;a name="ch02b035"&gt;</a>
a<sub>3</sub> start after a<sub>1</sub>
eqn.AF029 say a<sub>1</sub> has 1/1, but
a<sub>3</sub> do not (or a<sub>3</sub> has 0/1)

a<sub>3</sub> start after a<sub>2</sub>
eqn.AF030 say a<sub>1</sub>,a<sub>2</sub> have 1/2, but
a<sub>3</sub> do not (or a<sub>3</sub> has 0/2)

a<sub>3</sub> get non-zero coefficient start
from <a href="#ch02b030">eqn.AF031</a>
second ∑ from j=k=3 to j=n just right for k=3
a<sub>3</sub> coefficient is [0+0/2+1/3+ ... +1/k+ ... +1/n]
or a<sub>3</sub>*∑[j=k=3 to j=n]{1/j}

<a name="ch02b036">&lt;a name="ch02b036"&gt;</a>
Above is example of k=3. 
In general, equation is same for every k
a<sub>k</sub>*<font color=red>∑[j=k to j=n](1/j)</font> ---eqn.AF035
//red term approach to <a href="#ch02b038">infinity</a> if n => inf.
in eqn.AF035, k is constant, j is variable.
eqn.AF035 is just one term. 
Use two summation write eqn.AF034 in compact
form as following
eqn.AF034 = //red term approach to <a href="#ch02b038">infinity</a>.
∑[k=1 to k=n]{a<sub>k</sub>*<font color=red>∑[j=k to j=n](1/j)</font>} ---eqn.AF036
In eqn.AF036 
left  ∑, k is variable, not constant.
right ∑, k is constant, j is variable.
2009-10-02-16-30 here

<a name="ch02b037">&lt;a name="ch02b037"&gt;</a>
eqn.AF034 or eqn.AF035 is right hand side
of eqn.AF033. Put eqn.AF035 to eqn.AF033 get
eqn.AF033 left side <= eqn.AF035 as following
∑[k=1 to k=n]{(a<sub>1</sub>*a<sub>2</sub>*...*a<sub>k</sub>)^(1/k)} ---eqn.AF037
<= ∑[k=1 to k=n]{a<sub>k</sub>*<font color=red>∑[j=k to j=n](1/j)</font>}

<a name="ch02b038">&lt;a name="ch02b038"&gt;</a>
■　Sum to infinity, get failure

eqn.AF037 start from <a href="#ch02b002">eqn.2.15</a> left side, but
failed to reach eqn.2.15 right hand side.
Because Carleman's inequality require n 
approach to infinity, in this case
<font color=red>∑[j=k to j=n](1/j)</font> ---eqn.AF038
is infinity. 
eqn.AF038 is integral of f(x)=1/x, get
log(x) [x=k to x=>infinity] is infinity

Above plan get no conclusion.
2009-10-02-16-47 stop

<a name="ch02b039">&lt;a name="ch02b039"&gt;</a>　<a href="#index">Index begin</a>　<a href="#index13">Index this file</a>
■　Change direction, use fudge factor

2009-10-02-18-09 start
<a href="#textbook">textbook</a> page 28 suggest to use fudge factor.
start from eqn.2.15 left side write as below

∑[k=1 to k=∞]{(a<sub>1</sub>*a<sub>2</sub>*...*a<sub>k</sub>)^(1/k)}
=
∑[k=1 to k=∞]{(a<sub>1</sub>*a<sub>2</sub>*...*a<sub>k</sub>)^(1/k) * 1}
=
∑[k=1 to k=∞]{(a<sub>1</sub>*a<sub>2</sub>*...*a<sub>k</sub>)^(1/k)
 * (c<sub>1</sub>*c<sub>2</sub>*...*c<sub>k</sub>)^(1/k)/[(c<sub>1</sub>*c<sub>2</sub>*...*c<sub>k</sub>)^(1/k)]}
=
∑[k=1 to k=∞]{[(a<sub>1</sub>c<sub>1</sub>*a<sub>2</sub>c<sub>2</sub>*...*a<sub>k</sub>c<sub>k</sub>)^(1/k)]
/[(c<sub>1</sub>*c<sub>2</sub>*...*c<sub>k</sub>)^(1/k)]} ---eqn.AF039

Apply AM-GM inequality to eqn.AF039 get
</font></pre>

<a name="ch02b040">&lt;a name="ch02b040"&gt;</a>

<br>

<TABLE WIDTH="540"
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<TR>
<TD>
 &nbsp;
</TD>
<TD>
<TABLE WIDTH="420" border=0
 id=T028L23b
>
<TR>


<TD>
<CENTER><FONT SIZE=-2>k=∞</FONT></CENTER>
<CENTER><FONT SIZE=+3>∑</FONT></CENTER>
<CENTER><FONT SIZE=-2>k=1</FONT></CENTER>
</TD>


<TD>

(a<sub>1</sub>a<sub>2</sub>...a<sub>k</sub>)<sup>1/k</sup>

</TD>

<TD>
 ＝ 
</TD>


<TD>
<CENTER><FONT SIZE=-2>k=∞</FONT></CENTER>
<CENTER><FONT SIZE=+3>∑</FONT></CENTER>
<CENTER><FONT SIZE=-2>k=1</FONT></CENTER>
</TD>


<TD>
<font color=red>
<CENTER>(a<sub>1</sub>c<sub>1</sub>*a<sub>2</sub>c<sub>2</sub>*...*a<sub>k</sub>c<sub>k</sub>)<sup>1/k</sup></CENTER>
</font>
<CENTER><hr color=black></CENTER>
<CENTER>(c<sub>1</sub>*c<sub>2</sub>*...*c<sub>k</sub>)<sup>1/k</sup></CENTER>

</TD>

</TR>
</TABLE>
</TD>

<TD>

 ---page 28
<br>
 ---eqn.2.17a

</TD>
</TR>
</TABLE>

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<br>


<TABLE WIDTH="540"
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<TR>
<TD>
 &nbsp;
</TD>
<TD>
<TABLE WIDTH="420" border=0
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>
<TR>


<TD>
<CENTER><FONT SIZE=-2>k=∞</FONT></CENTER>
<CENTER><FONT SIZE=+3>∑</FONT></CENTER>
<CENTER><FONT SIZE=-2>k=1</FONT></CENTER>
</TD>


<TD>

(a<sub>1</sub>a<sub>2</sub>...a<sub>k</sub>)<sup>1/k</sup>

</TD>

<TD>
≦ 
</TD>


<TD>
<CENTER><FONT SIZE=-2>k=∞</FONT></CENTER>
<CENTER><FONT SIZE=+3>∑</FONT></CENTER>
<CENTER><FONT SIZE=-2>k=1</FONT></CENTER>
</TD>


<TD>
<font color=red>
<CENTER>a<sub>1</sub>c<sub>1</sub>+a<sub>2</sub>c<sub>2</sub>+...+a<sub>k</sub>c<sub>k</sub></CENTER>
</font>
<CENTER><hr color=black></CENTER>
<CENTER><font color=red>k</font>*(c<sub>1</sub>*c<sub>2</sub>*...*c<sub>k</sub>)<sup>1/k</sup></CENTER>

</TD>

</TR>
</TABLE>
</TD>

<TD>

 ---page 28
<br>
 ---eqn.2.17b

</TD>
</TR>
</TABLE>

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<pre><font size=+2><a name="ch02b041">&lt;a name="ch02b041"&gt;</a>
2009-10-02-18-44 here
Red text in eqn.2.17 is AM-GM inequality 
result. 
Red text at eqn.2.17a is Geometric Mean. 
Red text at eqn.2.17b is Arithmetic Mean.
Arithmetic Mean need to divide by number of
terms k, eqn.2.17b has a k at denominator.
2009-10-02-18-46 here

<a name="ch02b042">&lt;a name="ch02b042"&gt;</a>
■　Study coefficient one by one

Just for understanding,
Let us expand eqn.2.17b, for k=3 case
</font></pre>

<TABLE WIDTH="540"
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<TR>
<TD>
 &nbsp;
</TD>
<TD>
<TABLE WIDTH="420" border=0
 id=T028L23f
>
<TR>


<TD>
<CENTER>a<sub>1</sub>c<sub>1</sub></CENTER>
<CENTER><hr color=black></CENTER>
<CENTER>1*(c<sub>1</sub>)<sup>1/1</sup></CENTER>
</TD>


<TD>
＋
</TD>


<TD>
<CENTER>a<sub>1</sub>c<sub>1</sub>+a<sub>2</sub>c<sub>2</sub></CENTER>
<CENTER><hr color=black></CENTER>
<CENTER>2*(c<sub>1</sub>*c<sub>2</sub>)<sup>1/2</sup></CENTER>
</TD>


<TD>
＋
</TD>


<TD>
<CENTER>a<sub>1</sub>c<sub>1</sub>+a<sub>2</sub>c<sub>2</sub>+a<sub>3</sub>c<sub>3</sub></CENTER>
<CENTER><hr color=black></CENTER>
<CENTER>3*(c<sub>1</sub>*c<sub>2</sub>*c<sub>3</sub>)<sup>1/3</sup></CENTER>
</TD>

</TR>
</TABLE>
</TD>

<TD>

 more terms
<br>

 ---page 28
<br>
 ---eqn.2.17c

</TD>
</TR>
</TABLE>

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<br>
<br>
<a name="ch02b043">&lt;a name="ch02b043"&gt;</a>
<br>
a<sub>1</sub>c<sub>1</sub> has coefficients
<br>

<TABLE WIDTH="540"
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<TD>
 &nbsp;
</TD>
<TD>
<TABLE WIDTH="420" border=0
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>
<TR>


<TD>
<CENTER>1</CENTER>
<CENTER><hr color=black></CENTER>
<CENTER>1*(c<sub>1</sub>)<sup>1/1</sup></CENTER>
</TD>


<TD>
＋
</TD>


<TD>
<CENTER>1</CENTER>
<CENTER><hr color=black></CENTER>
<CENTER>2*(c<sub>1</sub>*c<sub>2</sub>)<sup>1/2</sup></CENTER>
</TD>


<TD>
＋
</TD>


<TD>
<CENTER>1</CENTER>
<CENTER><hr color=black></CENTER>
<CENTER>3*(c<sub>1</sub>*c<sub>2</sub>*c<sub>3</sub>)<sup>1/3</sup></CENTER>
</TD>

</TR>
</TABLE>
</TD>

<TD>

 more terms
<br>

 ---page 28
<br>
 ---eqn.2.17d

</TD>
</TR>
</TABLE>

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<br>
<br>
<a name="ch02b044">&lt;a name="ch02b044"&gt;</a>
<br>
a<sub>2</sub>c<sub>2</sub> has coefficients
<br>
<br>

<TABLE WIDTH="540"
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<TD>
 &nbsp;
</TD>
<TD>
<TABLE WIDTH="420" border=0
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>
<TR>

<TD>
<CENTER>1</CENTER>
<CENTER><hr color=black></CENTER>
<CENTER>2*(c<sub>1</sub>*c<sub>2</sub>)<sup>1/2</sup></CENTER>
</TD>


<TD>
＋
</TD>


<TD>
<CENTER>1</CENTER>
<CENTER><hr color=black></CENTER>
<CENTER>3*(c<sub>1</sub>*c<sub>2</sub>*c<sub>3</sub>)<sup>1/3</sup></CENTER>
</TD>


<TD>
＋ more terms
</TD>


</TR>
</TABLE>
</TD>

<TD>

 ---page 28
<br>
 ---eqn.2.17e

</TD>
</TR>
</TABLE>

width of above equation
<INPUT name=Button028L23i type="button"
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<input id="B028L23i"
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<input id="B028L23j"
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 size=3
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<br>
<br>
<a name="ch02b045">&lt;a name="ch02b045"&gt;</a>
<br>
a<sub>3</sub>c<sub>3</sub> has coefficients
<br>
<br>

<TABLE WIDTH="440"
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<TR>
<TD>
 &nbsp;
</TD>
<TD>
<TABLE WIDTH="320" border=0
 id=T028L23l
>
<TR>


<TD>
<CENTER>1</CENTER>
<CENTER><hr color=black></CENTER>
<CENTER>3*(c<sub>1</sub>*c<sub>2</sub>*c<sub>3</sub>)<sup>1/3</sup></CENTER>
</TD>


<TD>
＋ more terms
</TD>


</TR>
</TABLE>
</TD>

<TD>

 ---page 28
<br>
 ---eqn.2.17f

</TD>
</TR>
</TABLE>

width of above equation
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<input id="B028L23k"
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<pre><font size=+2><a name="ch02b046">&lt;a name="ch02b046"&gt;</a>　<a href="#index">Index begin</a>　<a href="#index13">Index this file</a>
■　Write double sum

2009-10-02-19-12 here
After above observations, we write eqn.2.17b 
in double sum as following</font></pre>


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<TR>
<TD>
 &nbsp;
</TD>
<TD>
<TABLE WIDTH="460" border=0
 id=T028L23n
>
<TR>


<TD>
<CENTER><FONT SIZE=-2>k=∞</FONT></CENTER>
<CENTER><FONT SIZE=+3>∑</FONT></CENTER>
<CENTER><FONT SIZE=-2>k=1</FONT></CENTER>
</TD>


<TD>

(a<sub>1</sub>a<sub>2</sub>...a<sub>k</sub>)<sup>1/k</sup>

</TD>

<TD>
≦ 
</TD>


<TD>
<CENTER><FONT SIZE=-2>k=∞</FONT></CENTER>
<CENTER><FONT SIZE=+3>∑</FONT></CENTER>
<CENTER><FONT SIZE=-2>k=1</FONT></CENTER>
</TD>


<TD>
<CENTER><FONT SIZE=+3>[</FONT></CENTER>
</TD>


<TD>
a<sub>k</sub>c<sub>k</sub>
</TD>


<TD>
<CENTER><FONT SIZE=-2>j=∞</FONT></CENTER>
<CENTER><FONT SIZE=+3>∑</FONT></CENTER>
<CENTER><FONT SIZE=-2>j=k</FONT></CENTER>
</TD>


<TD>
<CENTER>1</CENTER>
<CENTER><hr color=black></CENTER>
<CENTER>j*(c<sub>1</sub>*c<sub>2</sub>*...*c<sub>j</sub>)<sup>1/j</sup></CENTER>

</TD>


<TD>
<CENTER><FONT SIZE=+3>]</FONT></CENTER>
</TD>

</TR>
</TABLE>
</TD>

<TD>

 ---page 28
<br>
 ---eqn.2.17g

</TD>
</TR>
</TABLE>

width of above equation
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 value="default"
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<input id="B028L23m"
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<pre><font size=+2><a name="ch02b047">&lt;a name="ch02b047"&gt;</a>
2009-10-02-19-22 here
In eqn.2.17g square bracket, set k=1,
we have eqn.2.17d (plus more terms)

In eqn.2.17g square bracket, set k=2,
we have eqn.2.17e (plus more terms)

In eqn.2.17g square bracket, set k=3,
we have eqn.2.17f (plus more terms)
etc.

Hope this explanation help you understand
<a href="#textbook">textbook</a> page 28 equation 2.17

<a name="ch02b048">&lt;a name="ch02b048"&gt;</a>
■　Double sum compared with Carleman

Compare eqn.2.17g with <a href="#ch02b002">eqn.2.15</a></font></pre>


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<TR>
<TD>
 &nbsp;
</TD>
<TD>
<TABLE WIDTH="500" border=0
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>
<TR>

<TD>
<font color=red>
<CENTER><FONT SIZE=-2>k=∞</FONT></CENTER>
<CENTER><FONT SIZE=+3>∑</FONT></CENTER>
<CENTER><FONT SIZE=-2>k=1</FONT></CENTER>
</font>
</TD>


<TD>

<font color=red>
(a<sub>1</sub>a<sub>2</sub>...a<sub>k</sub>)<sup>1/k</sup>
</font>

</TD>

<TD>
≦ 
</TD>


<TD>
<font color=red>
<CENTER><FONT SIZE=-2>k=∞</FONT></CENTER>
<CENTER><FONT SIZE=+3>∑</FONT></CENTER>
<CENTER><FONT SIZE=-2>k=1</FONT></CENTER>
</font>
</TD>


<TD>
<font color=red>
a<sub>k</sub>
</font>
</TD>

</font>

<TD>
<CENTER><FONT SIZE=+3>[</FONT></CENTER>
</TD>


<TD>
<font color=blue>
c<sub>k</sub>
</font>
</TD>


<TD>
<font color=blue>
<CENTER><FONT SIZE=-2>j=∞</FONT></CENTER>
<CENTER><FONT SIZE=+3>∑</FONT></CENTER>
<CENTER><FONT SIZE=-2>j=k</FONT></CENTER>
</font>
</TD>


<TD>
<font color=blue>
<CENTER>1</CENTER>
<CENTER><hr color=black></CENTER>
<CENTER>j*(c<sub>1</sub>*c<sub>2</sub>*...*c<sub>j</sub>)<sup>1/j</sup></CENTER>
</font>
</TD>


<TD>
<CENTER><FONT SIZE=+3>]</FONT></CENTER>
</TD>

</TR>
</TABLE>
</TD>

<TD>

 ---page 28
<br>
 ---eqn.2.17h

</TD>
</TR>
</TABLE>

width of above equation
<INPUT name=Button028L23o type="button"
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<input id="B028L23o"
 value=620
 size=3
 onchange=T028L23o.width=B028L23o.value
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<input id="B028L23p"
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<pre><font size=+2><a name="ch02b049">&lt;a name="ch02b049"&gt;</a>
2009-10-02-19-42 here
In eqn.2.17h, 
left  side red  part is gradual GMed sum
right side red  part is input sequence sum
right side blue part is 
      coefficient  E=2.718281828459045...
The following work is to calculate blue
part and relate to E=2.718281828459045...

Define eqn.2.17h blue part as s<sub>k</sub>
</font></pre>

<a name="ch02b050">&lt;a name="ch02b050"&gt;</a>

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<TR>
<TD>
 &nbsp;
</TD>
<TD>
<TABLE WIDTH="280" border=0
 id=T028L29b
>
<TR>

<TD>
s<sub>k</sub>
</TD>

<TD>
＝
</TD>

<TD>
c<sub>k</sub>
</TD>


<TD>
<CENTER><FONT SIZE=-2>j=∞</FONT></CENTER>
<CENTER><FONT SIZE=+3>∑</FONT></CENTER>
<CENTER><FONT SIZE=-2>j=k</FONT></CENTER>
</TD>


<TD>
<CENTER>1</CENTER>
<CENTER><hr color=black></CENTER>
<CENTER>j*(c<sub>1</sub>*c<sub>2</sub>*...*c<sub>j</sub>)<sup>1/j</sup></CENTER>

</TD>

</TR>
</TABLE>
</TD>

<TD>

 ---page 28
<br>
 ---eqn.2.18

</TD>
</TR>
</TABLE>

width of above equation
<INPUT name=Button028L29a type="button"
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<input id="B028L29a"
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<pre><font size=+2><a name="ch02b051">&lt;a name="ch02b051"&gt;</a>　<a href="#index">Index begin</a>　<a href="#index13">Index this file</a>
■　Telescope look far side

2009-10-02-19-55 here
In eqn.2.18, the difficult part is j go
to infinity. How to make an infinity sum
a simple expression? the method to use
"In fact, almost all of these come from
the telescoping identity" (<a href="#textbook">textbook</a> words)
</font></pre>


<TABLE WIDTH="420"
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<TR>
<TD>
 &nbsp;
</TD>
<TD>
<TABLE WIDTH="280" border=0
 id=T029L08b
>
<TR>


<TD>
<CENTER><FONT SIZE=-2>j=∞</FONT></CENTER>
<CENTER><FONT SIZE=+3>∑</FONT></CENTER>
<CENTER><FONT SIZE=-2>j=k</FONT></CENTER>
</TD>

<TD>
<FONT SIZE=+3>{</FONT>
</TD>

<TD>
<CENTER>1</CENTER>
<CENTER><hr color=black></CENTER>
<CENTER>b<sub>j</sub></CENTER>
</TD>

<TD>
－
</TD>


<TD>
<CENTER>1</CENTER>
<CENTER><hr color=black></CENTER>
<CENTER>b<sub>j+1</sub></CENTER>
</TD>

<TD>
<FONT SIZE=+3>}</FONT>
</TD>

<TD>
＝
</TD>


<TD>
<CENTER>1</CENTER>
<CENTER><hr color=black></CENTER>
<CENTER>b<sub>k</sub></CENTER>
</TD>


</TR>
</TABLE>
</TD>

<TD>

 ---page 29
<br>
 ---line 8
<br>
 ---eqn.AF040

</TD>
</TR>
</TABLE>

width of above equation
<INPUT name=Button029L08a type="button"
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<input id="B029L08a"
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<pre><font size=+2><a name="ch02b052">&lt;a name="ch02b052"&gt;</a>
2009-10-02-20-05 here
In eqn.AF040 b<sub>j</sub> approach to infinity
and 1/b<sub>j</sub> approach to zero.

The best choice of telescoping identity 
is next.</font></pre>

<a name="ch02b053">&lt;a name="ch02b053"&gt;</a>

<TABLE WIDTH="460"
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<TR>
<TD>
 &nbsp;
</TD>
<TD>
<TABLE WIDTH="340" border=0
 id=T029L12b
>
<TR>


<TD>
<CENTER><FONT SIZE=-2>j=∞</FONT></CENTER>
<CENTER><FONT SIZE=+3>∑</FONT></CENTER>
<CENTER><FONT SIZE=-2>j=k</FONT></CENTER>
</TD>


<TD>
<CENTER>1</CENTER>
<CENTER><hr color=black></CENTER>
<CENTER>j(j+1)</CENTER>
</TD>


<TD>
＝
</TD>


<TD>
<CENTER><FONT SIZE=-2>j=∞</FONT></CENTER>
<CENTER><FONT SIZE=+3>∑</FONT></CENTER>
<CENTER><FONT SIZE=-2>j=k</FONT></CENTER>
</TD>


<TD>
<FONT SIZE=+3>{</FONT>
</TD>

<TD>
<CENTER>1</CENTER>
<CENTER><hr color=black></CENTER>
<CENTER>j</CENTER>
</TD>

<TD>
－
</TD>

<TD>
<CENTER>1</CENTER>
<CENTER><hr color=black></CENTER>
<CENTER>j+1</CENTER>
</TD>

<TD>
<FONT SIZE=+3>}</FONT>
</TD>


<TD>
＝
</TD>


<TD>
<CENTER>1</CENTER>
<CENTER><hr color=black></CENTER>
<CENTER>k</CENTER>
</TD>


</TR>
</TABLE>
</TD>

<TD>

 ---page 29
<br>
 ---eqn.2.19

</TD>
</TR>
</TABLE>

width of above equation
<INPUT name=Button029L12a type="button"
 value="default"
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<input id="B029L12a"
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<br>
<br>
<a name="ch02b054">&lt;a name="ch02b054"&gt;</a>
<br>
Expand eqn.2.19 , first three terms are

<br>


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<TR>
<TD>
 &nbsp;
</TD>
<TD>
<TABLE WIDTH="500" border=0
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>
<TR>

<TD>
<FONT SIZE=+3>(</FONT>
</TD>

<TD>
<CENTER>1</CENTER>
<CENTER><hr color=black></CENTER>
<CENTER>k</CENTER>
</TD>

<TD>
－
</TD>

<TD>
<CENTER>1</CENTER>
<CENTER><hr color=black></CENTER>
<CENTER>k+1</CENTER>
</TD>

<TD>
<FONT SIZE=+3>)</FONT>
</TD>

<TD>
＋
</TD>


<TD>
<FONT SIZE=+3>(</FONT>
</TD>

<TD>
<CENTER>1</CENTER>
<CENTER><hr color=black></CENTER>
<CENTER>k+1</CENTER>
</TD>

<TD>
－
</TD>

<TD>
<CENTER>1</CENTER>
<CENTER><hr color=black></CENTER>
<CENTER>k+2</CENTER>
</TD>

<TD>
<FONT SIZE=+3>)</FONT>
</TD>

<TD>
＋
</TD>


<TD>
<FONT SIZE=+3>(</FONT>
</TD>

<TD>
<CENTER>1</CENTER>
<CENTER><hr color=black></CENTER>
<CENTER>k+2</CENTER>
</TD>

<TD>
－
</TD>

<TD>
<CENTER>1</CENTER>
<CENTER><hr color=black></CENTER>
<CENTER>k+3</CENTER>
</TD>

<TD>
<FONT SIZE=+3>)</FONT>
</TD>

<TD>
＋.....
</TD>

<TD>
＝
</TD>

<TD>
<CENTER>1</CENTER>
<CENTER><hr color=black></CENTER>
<CENTER>k</CENTER>
</TD>


</TR>
</TABLE>
</TD>

<TD>

 ---page 29
<br>
 ---eqn.2.19
<br>
 ---aux

</TD>
</TR>
</TABLE>

width of above equation
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<input id="B029L12c"
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<pre><font size=+2><a name="ch02b055">&lt;a name="ch02b055"&gt;</a>
2009-10-02-20-27 here
In eqn.2.19 aux, we can see
-1/(k+1) + 1/(k+1) = 0
-1/(k+2) + 1/(k+2) = 0
-1/(k+3) + 1/(k+3) = 0
all the way to infinity. Only term left 
is first term 1/k

Compare eqn.2.18 right hand side denominator
 with   eqn.2.19  left hand side denominator
That is to compare
 j*(c<sub>1</sub>*c<sub>2</sub>*...*c<sub>j</sub>)<sup>1/j</sup> with j(j+1)
If we define
   (c<sub>1</sub>*c<sub>2</sub>*...*c<sub>j</sub>)<sup>1/j</sup> = (j+1) ---eqn.2.20
for j=1,2,3 ...
then s<sub>k</sub> in eqn.2.18 become the simplest
telescoping identity !
</font></pre>

<a name="ch02b056">&lt;a name="ch02b056"&gt;</a>　<a href="#index">Index begin</a>　<a href="#index13">Index this file</a>
<br>
■　Goodbye infinity, first fruit
<br>

<TABLE WIDTH="580"
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<TR>
<TD>
 &nbsp;
</TD>
<TD>
<TABLE WIDTH="460" border=0
 id=T029L18b
>
<TR>

<TD>
s<sub>k</sub>
</TD>

<TD>
＝
</TD>

<TD>
c<sub>k</sub>
</TD>


<TD>
<CENTER><FONT SIZE=-2>j=∞</FONT></CENTER>
<CENTER><FONT SIZE=+3>∑</FONT></CENTER>
<CENTER><FONT SIZE=-2>j=k</FONT></CENTER>
</TD>


<TD>
<CENTER>1</CENTER>
<CENTER><hr color=black></CENTER>
<CENTER>j*(c<sub>1</sub>*c<sub>2</sub>*...*c<sub>j</sub>)<sup>1/j</sup></CENTER>
</TD>


<TD>
＝
</TD>

<TD>
c<sub>k</sub>
</TD>


<TD>
<CENTER><FONT SIZE=-2>j=∞</FONT></CENTER>
<CENTER><FONT SIZE=+3>∑</FONT></CENTER>
<CENTER><FONT SIZE=-2>j=k</FONT></CENTER>
</TD>



<TD>
<CENTER>1</CENTER>
<CENTER><hr color=black></CENTER>
<CENTER>j*(j+1)</CENTER>
</TD>


<TD>
＝
</TD>


<TD>
<CENTER>c<sub>k</sub></CENTER>
<CENTER><hr color=black></CENTER>
<CENTER>k</CENTER>
</TD>


</TR>
</TABLE>
</TD>

<TD>

 ---page 29
<br>
 ---eqn.2.21

</TD>
</TR>
</TABLE>

width of above equation
<INPUT name=Button029L18a type="button"
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<input id="B029L18a"
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<pre><font size=+2><a name="ch02b057">&lt;a name="ch02b057"&gt;</a>
2009-10-02-20-42 here
The remaining work is to find out c<sub>k</sub> .
Look at
 (c<sub>1</sub>*c<sub>2</sub>*...*c<sub>j</sub>)<sup>1/j</sup> = (j+1) ---eqn.2.20
one term before above term is
 (c<sub>1</sub>*c<sub>2</sub>*...*c<sub>j-1</sub>)<sup>1/(j-1)</sup> = (j-1+1) ---eqn.AF041

Rise eqn.AF041 to j-1 th power get
 (c<sub>1</sub>*c<sub>2</sub>*...*c<sub>j-1</sub>) = j<sup>(j-1)</sup> ---eqn.AF042
Rise eqn.2.20 to j th power get
 (c<sub>1</sub>*c<sub>2</sub>*...*c<sub>j</sub>) = (j+1)<sup>j</sup> ---eqn.AF043

Divide two equations
eqn.AF043 / eqn.AF042 to solve for c<sub>j</sub> get
</font></pre>

<a name="ch02b058">&lt;a name="ch02b058"&gt;</a>


<TABLE WIDTH="520"
 id=T029L25a
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<TR>
<TD>
 &nbsp;
</TD>
<TD>
<TABLE WIDTH="360" border=0
 id=T029L25b
>
<TR>

<TD>
c<sub>j</sub>
</TD>

<TD>
＝
</TD>


<TD>
<CENTER>c<sub>1</sub>*c<sub>2</sub>*...*c<sub>j</sub></CENTER>
<CENTER><hr color=black></CENTER>
<CENTER>c<sub>1</sub>*c<sub>2</sub>*...*c<sub>j-1</sub></CENTER>
</TD>


<TD>
＝
</TD>


<TD>
<CENTER>(j+1)<sup>j</sup></CENTER>
<CENTER><hr color=black></CENTER>
<CENTER>j<sup>j-1</sup></CENTER>
</TD>


<TD>
＝ j
</TD>

<TD>
<CENTER><FONT SIZE=+3>(</FONT></CENTER>
</TD>

<TD>
1 + 
</TD>

<TD>
<CENTER>1</CENTER>
<CENTER><hr color=black></CENTER>
<CENTER>j</CENTER>
</TD>

<TD>
<CENTER><FONT SIZE=+3>)</FONT></CENTER>
</TD>

<TD>
<CENTER>j</CENTER>
<CENTER>　</CENTER>
<CENTER>　</CENTER>
</TD>

</TR>
</TABLE>
</TD>

<TD>

 ---page 29
<br>
 ---line 25
<br>
 ---eqn.AF044

</TD>
</TR>
</TABLE>

width of above equation
<INPUT name=Button029L25a type="button"
 value="default"
 onclick="B029L25a.value=520,B029L25b.value=360,T029L25a.width=B029L25a.value,T029L25b.width=B029L25b.value">

<input id="B029L25a"
 value=520
 size=3
 onchange=T029L25a.width=B029L25a.value
 />
<input id="B029L25b"
 value=360
 size=3
 onchange=T029L25b.width=B029L25b.value
 />


<pre><font size=+2><a name="ch02b059">&lt;a name="ch02b059"&gt;</a>
2009-10-02-21-03 here
j in eqn.AF044 is dummy variable. Change
from j to k, get
c<sub>k</sub>/k = (1+1/k)^k ---eqn.AF045

<a href="#ch02b048">eqn.2.17h</a> blue part is defined to be s<sub>k</sub>
<a href="#ch02b056">eqn.2.21</a>  found s<sub>k</sub> is c<sub>k</sub>/k
eqn.AF045 found c<sub>k</sub>/k expression (1+1/k)^k
How to link (1+1/k)^k with
 E=2.718281828459045...
?

<a name="ch02b060">&lt;a name="ch02b060"&gt;</a>
■　Taylor series and Peirce Table made Carleman smile

From Taylor series expansion
exp(x) = 1 + x + higher_order_term
</font></pre>


<TABLE WIDTH="480"
 id=T631Liua
  border=0
 ><!--9810031150-->
<TR>
<TD>
 &nbsp;
</TD>
<TD>
<TABLE WIDTH="320" border=0
 id=T631Liub
>
<TR>

<TD>
e<sup>x</sup>
</TD>

<TD>
 ＝ 1 + x +
</TD>


<TD>
<CENTER>x<sup>2</sup></CENTER>
<CENTER><hr color=black></CENTER>
<CENTER>2 !</CENTER>
</TD>

<TD>
 + 
</TD>

<TD>
<CENTER>x<sup>3</sup></CENTER>
<CENTER><hr color=black></CENTER>
<CENTER>3 !</CENTER>
</TD>

<TD>
 + ... 　　　 [x<sup>2</sup>＜∞]
</TD>



</TR>
</TABLE>
</TD>

<TD>

 ---Peirce Table
<br>
 ---Item 759

</TD>
</TR>
</TABLE>

width of above equation
<INPUT name=Button631Liua type="button"
 value="default"
 onclick="B631Liua.value=480,B631Liub.value=320,T631Liua.width=B631Liua.value,T631Liub.width=B631Liub.value">

<input id="B631Liua"
 value=480
 size=3
 onchange=T631Liua.width=B631Liua.value
 />
<input id="B631Liub"
 value=320
 size=3
 onchange=T631Liub.width=B631Liub.value
 />

<pre><font size=+2>
or for non-zero positive x
exp(x) > 1 + x  ---eqn.AF046
let y=1/x, that is
x=1/y
exp(1/y) > 1 + 1/y
it is all positive term, 
rise to y th power get
exp(1) > (1 + 1/y)^y
that is

 e > (1+1/k)^k ---eqn.AF047
(y and k are both dummy variable)

<a name="ch02b061">&lt;a name="ch02b061"&gt;</a>　<a href="#index">Index begin</a>　<a href="#index13">Index this file</a>
Substitute eqn.AF047 to blue term in
<a href="#ch02b048">eqn.2.17h</a> we conclude 
Carleman's Inequality <a href="#ch02b002">eqn.2.15</a> 
is true.

Above proof is Professor Polya's work.
See <a href="#textbook">text book</a> page 27 upper one third 
page.
2009-10-02-21-26 stop

<a name="ch02b062">&lt;a name="ch02b062"&gt;</a>
2009-10-02-22-22 start
on
2009-10-01-22-41 LiuHH accessed
http://emis.u-strasbg.fr/journals/JIPAM/images/135_02_JIPAM/135_02_www.pdf
carleman_history_emis.u-strasbg.fr_981001.pdf
Above page has more than ten proofs for
Carleman's Inequality
emis.u-strasbg.fr page 8/42 and page 9/42 
has proof same as this page tute0012.htm
proof. emis.u-strasbg.fr page proof is very
compact.
2009-10-02-22-27 stop

<script language="javascript">alert0()</script>

</font></pre>


<!--[=][][]
a<sub>1</sub>+a<sub>2</sub>+a<sub>3</sub>
a<sub>1</sub>,a<sub>2</sub>,...,a<sub>n</sub>
a<sub>1</sub><sup><sup>p<sub>1</sub></sup></sup>
AM-GM Inequality 
Arithmetic mean and geometric mean
■　
a<sub>1</sub>+a<sub>2</sub>+a<sub>3</sub>+a<sub>4</sub>
-->

<hr>

<pre><font size=+2>
<a name=20091217>&lt;a name=20091217&gt;</a>
2009-12-17-11-09 start
Update 2009-12-17 change all tute*.htm
(from tute0007.htm to tute0023.htm)
first:
Correct 'Limit' link from '#docA06'
to '#docA006'
second:
Change Javascript index to read from
jslist1e.js so that update jslist1e.js
then update ALL tute*.htm.
2009-12-17-11-23 stop

<a name="Copyright">&lt;a name="Copyright"&gt;</a>　<a href="#index">Index begin</a>　<a href="#index13">Index this file</a>
2009-06-19-10-48
If you are interested in inequality,
suggest you buy the book<a href=http://www.amazon.com/review/product/052154677X/ref=cm_cr_dp_all_helpful?%5Fencoding=UTF8&coliid=&showViewpoints=1&colid=&sortBy=bySubmissionDateDescending>
The Cauchy-Schwarz Master Class</a>
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
<a href=http://www-stat.wharton.upenn.edu/~steele/Publications/Books/CSMC/CSMC_index.html>
The Cauchy-Schwarz Master Class</a><a href=http://www-stat.wharton.upenn.edu/~steele/index.html>
J. Michael</a> <a href=http://www.wharton.upenn.edu/faculty/steele.html>Steele</a> <a href=http://www.amazon.com/review/product/052154677X/ref=cm_cr_dp_all_helpful?%5Fencoding=UTF8&coliid=&showViewpoints=1&colid=&sortBy=bySubmissionDateDescending>★★★★★</a>
ISBN 978-0-521-54677-5<!--9807061815 add ISBN-->
2009-06-19-10-56
</font></pre>


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Save graph code to same folder as htm files.
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First upload 2009-10-03
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<script language="javascript" type="text/javascript">
<!--



/**
2009-06-12-15-31 start
Javascript use float number for calculation.
Answer may contain long string of '000000'
or '999999', for example
2.00000000000001 
-25.999999999999996 
On 2009-06-12-11-58 write a function
function bye09(in09) //9806121158
To use this function, call as following
coef2=bye09(coef2);
where coef2 is a number. After call
2.00000000000001 change to 2
-25.999999999999996 change to -26
It is handy, you can use it too.
If number is 1.23111111111 there is no change.

function bye09(in09) handle output number
for better looking. Do not call bye09()
during calculation, that is just slow down
process. If expect answer to be irrational
Do not call bye09().
If expect answer to be integer or short
decimal number like 1.2. In these case
you can call bye09().
2009-06-12-15-40 stop

2009-06-17-19-05
bye09() process only one number at a time.
If you have a string of several numbers.
Do not put number string as input argument
bye09() will send string back immediately.
This file has sample code at time stamp
'9806171858'
2009-06-17-19-09 stop

/**/
function bye09(in09) //9806121158
{ // in09 is input number,
// for example 2.0000000001
// If input is not a number, return here
if(isNaN(in09)) return in09;

// where is decimal point '.' ?
var dotLoc=(in09+'').indexOf('.',0);
// no decimal point? integer? return.
if(dotLoc==-1) return in09;

// change input string to number
in09=parseFloat(in09);

// if number has '000000' ？
var zeroLoc=(in09+'').indexOf('000000',dotLoc);
// find '000000' ，cut start '000000' 
if(zeroLoc>0) //9806121216
return parseFloat((in09+'').substring(0,zeroLoc)); //9806121209

// if number has '999999' ？
var nineLoc=(in09+'').indexOf('999999',dotLoc);
// find '999999' ， cut start '999999'
if(nineLoc>0) //9806121223 here
  { // copy number left to '999999'
//but immediately neighbor digit not copy
// nineLoc-1 because '999999' need increase
// digit by one.
//If input is 1.23999999999 copy only 1.2
//not copy 1.23 , because 3 will change to 4.
var nine0=(in09+'').substring(0,nineLoc-1);
//between '.' and '999999' has other digit.
if(nineLoc-dotLoc>1) //9806121220
    { //neighbor digit add one, paste to end
nine0+=(parseInt((in09+'').charAt(nineLoc-1))+1);
    }
else //between '.' and '999999' no other digit.
    {// modified number is an integer
nine0=parseInt(nine0); //9806121313
if(nine0<0)nine0--; // negative integer minus one
else nine0++; //9806121314 positive integer add one
    }
return parseFloat(nine0);
//9806121213 return new number
  }
// no '000000' no '999999' return original form
return parseFloat(in09); //9806121224
} //function bye09(in09) 





function stndrdth(i0) //9802141617 begin js code
{
  var sth = ["  ", "st", "nd", "rd", "th"];
  if(i0<0) return sth[0];
  if(i0%100>3&&i0%100<21) return sth[4];
  if(i0%10==1) return sth[1];
  if(i0%10==2) return sth[2];
  if(i0%10==3) return sth[3];
  return sth[4];

} //function stndrdth(i0) 
//9802141626 done js code




//sortNumbMan() all ...Man() are MANager
//function. Handle input/output. Other
//programmer may not need this manager
//function. 
//This manager function call main engine
// ouArray=sortNumb(inArray)
//Other programmer may want to use sortNumb()
// 9807161228
function sortNumbMan(smArg1) //9807151818
{
var inArray=readdata(boxc21.value); //9807151819
boxc22.value='' //9807151824
+(''+sortNumb(inArray)).replace(/,/g,'\n');
} //9807151826 function sortNumbMan(smArg1) 


/**
2009-07-22-11-05 found old record
8902WK01.DLY
[[
A tutorial on pointers and arrays in C
2000-02-16-01-57
http://www.netcom.com/~tjensen/ptr/pointers.htm
save as  pointer_01  to  pointer_14
2000-02-16-02-05
http://home.netcom.com/~tjensen/
]]
02/16/2000  02:04 AM  19,015 ch10x.htm
/**/
//bubbleMan() all ...Man() are MANager
//function. Handle input/output. Other
//programmer may not need this manager
//function. 
//This manager function call main engine
// bubble_1(inArray,
//          inArray.length
//         )
//Other programmer may want to use bubble_1()
// 9807161229
function bubbleMan(bmArg1) //9807151827
{
var inArray=readdata(boxc21.value); //9807151819
 bubble_1(inArray,
          inArray.length
         )
boxc22.value='' //9807151830
+(''+inArray).replace(/,/g,'\n');
}  //9807151832 function bubbleMan(bmArg1)


/**
9807161850 
Assume input sequence is
 inNumb[]=[
 0.556
-0.145
 0.302e3
-0.305e2
-0.088
]
then
inNumb[0] store  0.556
inNumb[1] store -0.145
inNumb[2] store  302.0
inNumb[3] store -30.5
inNumb[4] store -0.088

then call
newSeq=sortNumb(inNumb)

get newSeq=[2,0,4,1,3]
it means
inNumb[2] store  302.0
greater than
inNumb[0] store  0.556
greater than
inNumb[4] store -0.088
greater than
inNumb[1] store -0.145
greater than
inNumb[3] store -30.5

newSeq=[2,0,4,1,3] 
this new sequence can be used
 to re-order eigenvalue and
 to re-order eigenvector
9807161856 stop

9807271355 start
above
inNumb[0]
...
inNumb[4]

old index is  bb = [0,1,2,3,4]
function sortNumb() main job is
to change index to [2,0,4,1,3]

If sort number directly, use
if (aa[j] > aa[i])

sortNumb() sort number IN-directly
sort number index bb[i] directly, use
if (aa[bb[j]] > aa[bb[i]])
9807271401 stop

/**/
//function bubble(aa, N //9807151528
function sortNumb(      //9807151528
  aa //aa is input number sequence. 
                 )
{
// 9807151612 and 9807151616 are two key points
  var N=aa.length; //9807151625

var bb=new Array(N);
for(var m0=0;m0<N;m0++)
bb[m0]=m0; //9807250842
//above line, create bb=[0,1,2,3,4]

  var i, j, k;
  for (i = 0; i < N-1; i++)
  {
    for (j = i+1; j <= N-1; j++)
    {
//    if (aa[j] > aa[i]) //from greatest to smallest
      if (aa[bb[j]] > aa[bb[i]]) //9807151616
      {
k=bb[i];      //9807151612
bb[i]=bb[j];  //9807151611
bb[j]= k   ;  //9807151613
      } //if (aa[bb[j]] > aa[bb[i]])
    }   //for (j = i+1; j <= N-1; j++)
  }     //for (i = 0; i < N-1; i++)
return bb; // 9807161630
} //9807151617 function sortNumb()

/**
9807271416 start
loop command
  for (i = 0; i < N-1; i++)
    for (j = i+1; j <= N-1; j++)
work as following
assume input data is
aa[0] =  0.556
aa[1] = -0.145
aa[2] =  302.0
aa[3] = -30.5
aa[4] = -0.088
original index is bb=[0,1,2,3,4]

Goal is to find new order as below
aa[2] =  302.0
aa[0] =  0.556
aa[4] = -0.088
aa[1] = -0.145
aa[3] = -30.5
right hand side of '=' is in great 
to small order
new index is bb=[2,0,4,1,3]

First enter outer loop
  for (i = 0; i < N-1; i++)
Index [0,1,2,3,4] has five number, so
N=5
i start from 0, 1, 2 to 3 .
i not touch 4 （i < N-1）
If i=4 then last number aa[4] no one 
to compete with aa[4] . Need at least 
reserve one number to compete with aa[i]
Therefore,
Five number sequence, i not reach 4 
  N  number sequence, i not reach N-1

If it is just begin, i=0 .
aa[0] is main role.
Next see inner loop
    for (j = i+1; j <= N-1; j++)
index j start from i+1=1 
j goto 4 and stop. （j <= N-1）
In inner loop, outer loop index i
is constant, never change.
Take aa[1],aa[2],aa[3],aa[4] 
One after one, compare with aa[0].

If it is just begin,
old order is bb=[0,1,2,3,4]
command
 if (aa[bb[j]] > aa[bb[i]])
is
 if (aa[bb[1]] > aa[bb[0]])
is
 if (aa[1] > aa[0])
is
 if (-0.145 > 0.556)
Not true, not run if(true) commands
Inner loop index j increase get j=2
At this time, 
outer loop index i still =0 

command
 if (aa[bb[j]] > aa[bb[i]])
is
 if (aa[2] > aa[0])
is
 if (302.0 > 0.556)
It is TRUE !!
run if(true) commands
k=bb[i];     bb[i] value store at k 
bb[i]=bb[j]; bb[i] receive bb[j] value
bb[j]= k;    bb[j] receive bb[i] value
that is bb[i] and bb[j] exchange value.
Before first exchange bb=[0,1,2,3,4]
After  first exchange bb=[2,1,0,3,4]

bb[j] compare with bb[i] continue as
above fashion. Until all done, get
new index sequence
bb=[2,0,4,1,3]

command
return bb; // 9807161630
send new index back to calling function.
9807271441 stop
/**/






/* Program bubble_1.c from PTRTUT10.HTM   6/13/97
Ted Jensen
Redwood City, California
tjensen@ix.netcom.com
July 1998

//9807151221 open
C:\lc0\c_online\pointer\ch10x.htm
void bubble(int a[], int N)
 */
function bubble_1(a, N
// , bb //9807151249 bb store new sequence, but
)       //failed, then have to write sortNumb()
{

//alert('bubble_1, a='+a+'\n'+'N='+N+', 9809281207');

//  int i, j, t;
    var i, j, t;
    for (i = N-1; i >= 0; i--)
    {
        for (j = 1; j <= i; j++)
        {
//          if (a[j-1] > a[j]) //9807151238 change to next
            if (a[j-1] < a[j]) //from greatest to smallest
            {
                t = a[j-1];
                a[j-1] = a[j];
                a[j] = t;
            }
        }
    }
}
/*---------------------- end bubble_1.c -----------------------*/





/**
//9809291954
if box11 has
[[ 0.556
-0.145
 0.302e3
-0.305e2
-0.088]]
last line no newline byte,
next line
inStr0=boxc21.value+'\n';
do not help
.split() know box11 last line no newline byte,
even if I attach '\n', .split() still drop '\n'
from last line. then after upside down
first line (was list line and no newline byte)
and second line string to one line.
98,09,29,20,36 record

//9809292038 version two
in output string add '\n', success
ouStr0+='\n'; //9809292040

/**/
function upsidedown()
{
var inStr0=boxc21.value
var ouStr0='';
var p0,p1,p2;

inStr0=inStr0.split('\n');
ouStr0=inStr0[inStr0.length-1];
if(ouStr0.charCodeAt(ouStr0.length-1)>=32)
ouStr0+='\n'; //9809292040

for(p0=inStr0.length-2;p0>=0;p0--)
  {
 ouStr0+=inStr0[p0];
  }

boxc22.value=ouStr0;
}//9809292050




//-->
</script>

�