ÿþ<HTML> <HEAD> <TITLE>Inequality; Study 8th file</TITLE> <META http-equiv=Content-Type content="text/html; charset=big5"> <style type="text/css"> v\:* { behavior: url(#default#VML); } </style> <!-- The Cauchy-Schwarz Master Class author --> <meta name="author" content="J. Michael Steele" /> <!-- XYGraph v2.3 code author --> <meta name="author" content="J. Gebelein" /> <!--tute0008.htm author --> <meta name="author" content="Liu, HsinHan ‰Rk”"o &#21129;&#37995;&#28450;" /> <script type="text/javascript"> <!-- //If can not load jsgraph2.js next definition is //effective, tell user where to get jsgraph2.js var msg01='document.getElementById("gotoFM2msg").innerHTML+=\''; var msg02=' goto online to get http://freeman2.com/jsgraph2.js<br>\';'; function stepf(t0,bgn0) //9712091955 {eval(msg01+'01'+msg02);} function loggamma(x) {eval(msg01+'02'+msg02);} function gamma(x) {eval(msg01+'03'+msg02);} function factorial(n) {eval(msg01+'04'+msg02);} function wipeCurve(d3) {eval(msg01+'05'+msg02);} function DrawGraf2(argd2) {eval(msg01+'06'+msg02);} function closeStr(klozID) {eval(msg01+'07'+msg02);} function toSyntax(inStr1) {eval(msg01+'08'+msg02);} function x2t(in0) {eval(msg01+'09'+msg02);} function deleteData(workID,probID) {eval(msg01+'10'+msg02);} function delAll(probID) {eval(msg01+'11'+msg02);} function openTable(drawID,table01) {eval(msg01+'12'+msg02);} function open0lxr(xx) {eval(msg01+'13'+msg02);} function closelxr(yy) {eval(msg01+'14'+msg02);} //9712092032 //if receive http://freeman2.com/jsgraph2.js // jsgraph2.js redefine above functions, //otherwise tell where to find jsgraph2.js function XYGraph(yy) {eval(msg01+'101'+msg02);} //9801260855 function XYLine(yy) {eval(msg01+'102'+msg02);} //9801260857 function Label(yy) {eval(msg01+'103'+msg02);} //9801260859 function tutelink() //9806231220 { document.getElementById("gotoFM2msg").innerHTML+= '50 goto online to get http://freeman2.com/tutelink.js<br>'; } function initial0(linkNum, plotNum) { tuteLink(0); //9806231221 //http://freeman2.com/tutelink.js //if use tutelink(0) when offline, get error 9711252125 } function alert0() //9806161720 { document.write('<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>' ); } //9806161724 function ineqIndexEn() //9806170948 { document.write('' +'<a name="index">&lt;a name="index"&gt;</a><a name="index01"></a><br>' +'<font color=red size=+3>Can not load index file</font><br>' +'Please download ineqinde.js Save to same folder as this file.<a href=http://freeman2.com/ineqinde.js><br>' +'http://freeman2.com/ineqinde.js</a><br>' ); } //9806170952 //If receive ineqinde.js //re-define function ineqIndexEn() //print index correctly. /** Cauchy Inequality used frequently create a command to build it any where. 2009-06-26,10,52 here 2009-06-26,11,20 must call HelloCauchy(hcPar) outside of <pre>..</pre> otherwise width wrong. /**/ function HelloCauchy(hcPar1) //9806261053 { strCauchy1='' +'<TABLE WIDTH="520" id=T005L19a' +hcPar1 +' border=0 ><TR><TD> &nbsp; </TD>' +'<TD><TABLE WIDTH="380" border=0 id=T005L19b' +hcPar1 +'>' +'<TR><TD><CENTER><FONT SIZE=-2>k="</FONT></CENTER>\n<CENTER><FONT SIZE=+3>"</FONT></CENTER>\n<CENTER><FONT SIZE=-2>k=1</FONT></CENTER></TD>' +'<TD><CENTER>a<sub><FONT SIZE=-1> k</FONT></sub> b<sub><FONT SIZE=-1> k</FONT></sub></CENTER></TD>' +'<TD>f"</TD>' +'<TD><CENTER><FONT SIZE=+3>[</FONT></CENTER></TD>' +'<TD><CENTER><FONT SIZE=-2>k="</FONT></CENTER>\n<CENTER><FONT SIZE=+3>"</FONT></CENTER>\n<CENTER><FONT SIZE=-2>k=1</FONT></CENTER></TD>' +'<TD><CENTER>a<sub><FONT SIZE=-1> k</FONT></sub><sup>2</sup></CENTER></TD>' +'<TD><CENTER><FONT SIZE=+3>]</FONT></CENTER></TD>' +'<TD><CENTER><FONT SIZE=-2>1/2</FONT></CENTER>\n<CENTER><FONT SIZE=+3>00</FONT></CENTER>\n<CENTER><FONT SIZE=-2>00</FONT></CENTER></TD>' +'<TD><CENTER><FONT SIZE=+3>[</FONT></CENTER></TD>' +'<TD><CENTER><FONT SIZE=-2>k="</FONT></CENTER>\n<CENTER><FONT SIZE=+3>"</FONT></CENTER>\n<CENTER><FONT SIZE=-2>k=1</FONT></CENTER></TD>' +'<TD><CENTER>b<sub><FONT SIZE=-1> k</FONT></sub><sup>2</sup></CENTER></TD>' +'<TD><CENTER><FONT SIZE=+3>]</FONT></CENTER></TD>' +'<TD><CENTER><FONT SIZE=-2>1/2</FONT></CENTER>\n<CENTER><FONT SIZE=+3>00</FONT></CENTER>\n<CENTER><FONT SIZE=-2>00</FONT></CENTER></TD>' +'</TR></TABLE></TD>' +'<TD>Page 5 eqn. 1.7</TD></TR></TABLE>' +'<font color=red><b>Cauchy Inequality</b></font> width<INPUT name=Button004L20a' +hcPar1 +' type="button" value="default" onclick="B005L19a' +hcPar1 +'.value=520,B005L19b' +hcPar1 +'.value=380,T005L19a' +hcPar1 +'.width=B005L19a' +hcPar1 +'.value,T005L19b' +hcPar1 +'.width=B005L19b' +hcPar1 +'.value">' +'<input id=B005L19a' +hcPar1 +' value=520 size=3 onchange=T005L19a' +hcPar1 +'.width=B005L19a' +hcPar1 +'.value />' +'<input id=B005L19b' +hcPar1 +' value=380 size=3 onchange=T005L19b' +hcPar1 +'.width=B005L19b' +hcPar1 +'.value />0calling ID '+hcPar1 ; if(arguments.length==1) document.write(strCauchy1); else { argLen=arguments.length; //9806261457 var w0; //9806261502 var w2; //9806261336 var strCauchy2=strCauchy1; for(w0=1;w0<argLen;w0++) //9806261503 { if(arguments[w0].length==1) status='argument '+w0+' should use two strings.' else { //9806261328 w2=-1; while((w2=strCauchy2.indexOf(arguments[w0][0],(w2+1)))>=0) strCauchy2= //9806261337 strCauchy2.replace(arguments[w0][0],arguments[w0][1]); } //if(arguments[w0].length==1) else } //for(w0=0;w0<argLen;w0++) document.write(strCauchy2); //9806261512 } //if(arguments.length==1) else } //9806261105;9806261516 function HelloCauchy(hcPar1) 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' ); } //function alert1() 9806261748 function alert2() //9806261808 { document.write('' +'0<font color=red size=+3>' +'LiuHH&#39;s sol. has no guarantee!</font>' ); } //function alert2() 9806261809 //--> </SCRIPT> <script src="http://freeman2.com/tutelink.js" language="javascript"></script><!--9806111210--> <script src="tutelink.js" language="javascript"></script><!--9806111211--> <!-- 9807011717 delete, because tute0008.htm no graph. <script type="text/javascript" src=jsgraph1.js> </script><!--9801101328--> </HEAD> <body link="#FF0000" vlink="#0000FF" alink="#50A000" bgcolor="#ccfcfc" onload="javascript:initial0()" > <span id="gotoFM2msg"></span><!--9712091951--> <font size=+3><a href=#index01> Inequality</a> </font> 0<a href=#index06>Study 8th file</a>0 <a href=#20091217> Update 2009-12-17 </a> <br> <a href=tute0008.htm#index> index </a> 0 <a href=tute0008.htm#index06> this </a> 0 <a href=tute0008.htm#Cauchy_6> program </a> 0 <a href=tute0008.htm#docA01> docA </a> 0 <a href=tute0008.htm#docB01> docB </a> <br> <a href=http://www.structura.info/XYGraph/XYGraphDemo.htm> XYGraph v2.3 - web page graph </a> &nbsp; <!--9506231219 add link--> <a href=http://www.structura.info/XYGraph/Links.htm> && <!--##9756;##9758;--> </a> &nbsp; <!--9506231216 add link--> <a href=http://www.structura.info/XYGraph/Purchase.htm> donate<!--9801031950 use donate ##9829;##9835; e&k&--> </a> &nbsp; <a href="http://www.structura.info/XYGraph/XYGraph.zip"> get code</a> <!-- 2006-06-12-18-21 download http://www.structura.info/XYGraph/XYGraph.zip C:\$fm\js\xygraph\XYGraph.zip 2006-06-19-08-21 record --> <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 <br> <span id="tutelink1"></span> <br> <script src="http://freeman2.com/rocsitee.js" language="javascript"></script> <pre><font size=+2> <a name="docA01">&lt;a name="docA01"&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="docA02">&lt;a name="docA02"&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> 0<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="docA03">&lt;a name="docA03"&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="docA04">&lt;a name="docA04"&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="docA05">&lt;a name="docA05"&gt;</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 </font></pre> <br> <!--9806170912 index01 --> <script src="ineqinde.js" language="javascript"></script> <script language="javascript">ineqIndexEn()</script> <!--9806170930--> <br> <pre><font size=+2><a name="ch01b001">&lt;a name="ch01b001"&gt;</a>0<a href="#index">Index begin</a>0<a href="#index06">Index this file</a>  %0Continuous Cauchy Inequality 2009-06-25-15-12 start <a href=tute0007.htm#ch01a026>Cauchy Inequality</a> is discrete type for example, sequence [1, 2, 3, 4, 5] First element 1 and second element 2 nothing in between them and undefined. This is discrete type. On the other hand, Continuous Cauchy Inequality can be defined with integral. If function f(x) is defined between x=1 to x=5. Integration from x=1 to x=5 run over every point in [1,5]. This is continuous type. Cauchy student Bunyakovsky is first one propose Continuous Cauchy Inequality. Equation is next.</font></pre> <a name="ch01b002">&lt;a name="ch01b002"&gt;</a> <TABLE WIDTH="700" id=T010L16a border=0 ><!--9806251528--> <TR> <TD> &nbsp; &nbsp; </TD> <TD> <TABLE WIDTH="500" border=0 id=T010L16b > <TR> <TD> <CENTER><FONT SIZE=-2>x=b</FONT></CENTER> <CENTER><FONT SIZE=+3>+"</FONT></CENTER> <CENTER><FONT SIZE=-2>x=a</FONT></CENTER> </TD> <TD> f(x) g(x) dx </TD> <TD> f" </TD> <TD> <FONT SIZE=+3>[</FONT> </TD> <TD> <CENTER><FONT SIZE=-2>x=b</FONT></CENTER> <CENTER><FONT SIZE=+3>+"</FONT></CENTER> <CENTER><FONT SIZE=-2>x=a</FONT></CENTER> </TD> <TD> f(x) <sup>2</sup> dx </TD> <TD> <FONT SIZE=+3>]</FONT> </TD> <TD> <CENTER><FONT SIZE=-2>1/2</FONT></CENTER> <CENTER>0</CENTER> <CENTER>0</CENTER> </TD> <TD> <FONT SIZE=+3>[</FONT> </TD> <TD> <CENTER><FONT SIZE=-2>x=b</FONT></CENTER> <CENTER><FONT SIZE=+3>+"</FONT></CENTER> <CENTER><FONT SIZE=-2>x=a</FONT></CENTER> </TD> <TD> g(x) <sup>2</sup> dx </TD> <TD> <FONT SIZE=+3>]</FONT> </TD> <TD> <CENTER><FONT SIZE=-2>1/2</FONT></CENTER> <CENTER>0</CENTER> <CENTER>0</CENTER> </TD> </TR> </TABLE> </TD> <TD> -----page 10 eqn.1.19 </TD> </TR> </TABLE> width of above equation <INPUT name=Button010L16a type="button" value="default value" onclick="B010L16a.value=700,B010L16b.value=500,T010L16a.width=B010L16a.value,T010L16b.width=B010L16b.value"> <input id="B010L16a" value=700 size=3 onchange=T010L16a.width=B010L16a.value /> <input id="B010L16b" value=500 size=3 onchange=T010L16b.width=B010L16b.value /> <pre><font size=+2><a name="ch01b003">&lt;a name="ch01b003"&gt;</a>0<a href="#index">Index begin</a>0<a href="#index06">Index this file</a> <script language="javascript">alert0()</script>  %0Schwarz Inequality 2009-06-25-15-45 here Textbook did not introduce Bunyakovsky's proof. However, do talk about Schwarz's proof. Above Bunya. eqn. use one variable x ; f=f(x) and g=g(x) Below Schwarz eqn. use two variables x,y; f=f(x,y) and g=g(x,y) This is the main difference. Let S be x,y domain. Defined by problem statement.</font></pre> <a name="ch01b004">&lt;a name="ch01b004"&gt;</a> <TABLE WIDTH="700" id=T011L04a border=0 ><!--9806251552--> <TR> <TD> &nbsp; </TD> <TD> <TABLE WIDTH="500" border=0 id=T011L04b > <TR> <TD> <FONT SIZE=+3>|</FONT> </TD> <TD> <CENTER><FONT SIZE=-2>0</FONT></CENTER> <CENTER><FONT SIZE=+3>,"</FONT></CENTER> <CENTER><FONT SIZE=-2>x,y"S</FONT></CENTER> </TD> <TD> f g dxdy </TD> <TD> <FONT SIZE=+3>|</FONT> </TD> <TD> f" </TD> <TD> <FONT SIZE=+3>[</FONT> </TD> <TD> <CENTER><FONT SIZE=-2>0</FONT></CENTER> <CENTER><FONT SIZE=+3>,"</FONT></CENTER> <CENTER><FONT SIZE=-2>x,y"S</FONT></CENTER> </TD> <TD> f <sup>2</sup> dxdy </TD> <TD> <FONT SIZE=+3>]</FONT> </TD> <TD> <CENTER><FONT SIZE=-2>1/2</FONT></CENTER> <CENTER>0</CENTER> <CENTER>0</CENTER> </TD> <TD> <FONT SIZE=+3>[</FONT> </TD> <TD> <CENTER><FONT SIZE=-2>0</FONT></CENTER> <CENTER><FONT SIZE=+3>,"</FONT></CENTER> <CENTER><FONT SIZE=-2>x,y"S</FONT></CENTER> </TD> <TD> g <sup>2</sup> dxdy </TD> <TD> <FONT SIZE=+3>]</FONT> </TD> <TD> <CENTER><FONT SIZE=-2>1/2</FONT></CENTER> <CENTER>0</CENTER> <CENTER>0</CENTER> </TD> </TR> </TABLE> </TD> <TD> ---page 11, eqn. 1.20 </TD> </TR> </TABLE> width of above equation <INPUT name=Button011L04a type="button" value="default value" onclick="B011L04a.value=700,B011L04b.value=500,T011L04a.width=B011L04a.value,T011L04b.width=B011L04b.value"> <input id="B011L04a" value=700 size=3 onchange=T011L04a.width=B011L04a.value /> <input id="B011L04b" value=500 size=3 onchange=T011L04b.width=B011L04b.value /> <pre><font size=+2><a name="ch01b005">&lt;a name="ch01b005"&gt;</a> 2009-06-25-16-09 here To prove Schwarz Inequality page 11, eqn. 1.20 As following. To simplify work, define A, B, C as following</font></pre> <TABLE WIDTH="620" id=T011L02a border=0 ><!--9806251612--> <TR> <TD> &nbsp; </TD> <TD> <TABLE WIDTH="480" border=0 id=T011L02b > <TR> <TD> A ÿ </TD> <TD> <CENTER><FONT SIZE=-2>0</FONT></CENTER> <CENTER><FONT SIZE=+3>,"</FONT></CENTER> <CENTER><FONT SIZE=-2>x,y"S</FONT></CENTER> </TD> <TD> f <sup>2</sup> dxdy </TD> <TD> ÿ </TD> <TD> B ÿ </TD> <TD> <CENTER><FONT SIZE=-2>0</FONT></CENTER> <CENTER><FONT SIZE=+3>,"</FONT></CENTER> <CENTER><FONT SIZE=-2>x,y"S</FONT></CENTER> </TD> <TD> f g dxdy </TD> <TD> ÿ </TD> <TD> C ÿ </TD> <TD> <CENTER><FONT SIZE=-2>0</FONT></CENTER> <CENTER><FONT SIZE=+3>,"</FONT></CENTER> <CENTER><FONT SIZE=-2>x,y"S</FONT></CENTER> </TD> <TD> g <sup>2</sup> dxdy </TD> </TR> </TABLE> </TD> <TD> ---page 11 line 2 </TD> </TR> </TABLE> width of above equation <INPUT name=Button011L02a type="button" value="default value" onclick="B011L02a.value=620,B011L02b.value=480,T011L02a.width=B011L02a.value,T011L02b.width=B011L02b.value"> <input id="B011L02a" value=620 size=3 onchange=T011L02a.width=B011L02a.value /> <input id="B011L02b" value=480 size=3 onchange=T011L02b.width=B011L02b.value /> <pre><font size=+2><a name="ch01b006">&lt;a name="ch01b006"&gt;</a> 2009-06-25-16-22 here Now consider a polynomial p(t) of variable t as following</font></pre> <TABLE WIDTH="640" id=T011L14a border=0 ><!--9806251612--> <TR> <TD> &nbsp; </TD> <TD> <TABLE WIDTH="480" border=0 id=T011L14b > <TR> <TD> p(t) ÿ </TD> <TD> <CENTER><FONT SIZE=-2>0</FONT></CENTER> <CENTER><FONT SIZE=+3>,"</FONT></CENTER> <CENTER><FONT SIZE=-2>x,y"S</FONT></CENTER> </TD> <TD> <FONT SIZE=+3>[</FONT> </TD> <TD> t*f(x,y)+g(x,y) </TD> <TD> <FONT SIZE=+3>]</FONT> </TD> <TD> <CENTER><FONT SIZE=-2>2</FONT></CENTER> <CENTER>0</CENTER> <CENTER>0</CENTER> </TD> <TD> dxdy </TD> <TD> = A*t*t+2*B*t+C </TD> </TR> </TABLE> </TD> <TD> ---page 11 line 14 </TD> </TR> </TABLE> width of above equation <INPUT name=Button011L14a type="button" value="default value" onclick="B011L14a.value=640,B011L14b.value=480,T011L14a.width=B011L14a.value,T011L14b.width=B011L14b.value"> <input id="B011L14a" value=640 size=3 onchange=T011L14a.width=B011L14a.value /> <input id="B011L14b" value=480 size=3 onchange=T011L14b.width=B011L14b.value /> <pre><font size=+2><a name="ch01b007">&lt;a name="ch01b007"&gt;</a> 2009-06-25-16-31 here page 11 line 14 equation left hand side expand the square term, take integral with respect to x and y. Because t is not a function of x and y. t can be move out of integration. Use page 11 line 2 symbol A, B, C for integration value, get page 11 line 14 equation right hand side quadratic expression for variable t. Equation left hand side is square of real number t*f(x,y)+g(x,y) then integral in positive x,y direction. Integration result is non-negative. Therefore, <a name="ch01b008">&lt;a name="ch01b008"&gt;</a> Quadratic equation p(t) ÿ A*t*t+2*B*t+C is also non-negative. Quadratic equation's discriminant f" 0, that is (2*B)*(2*B) - 4*A*C f" 0 and then B*B - A*C f" 0 -----eqn.AB01 refer to page 11 line 2 definition of A,B,C, recover integration terms for eqn.AB01 get Schwarz Inequality <a href="#ch01b004">page 11 eq.1.20</a> . 2009-06-25-17-03 here <a name="ch01b009">&lt;a name="ch01b009"&gt;</a>0<a href="#index">Index begin</a>0<a href="#index06">Index this file</a>  %0quadratic discriminant>0 what trouble? f(x)=A*x*x+ B*x+C has discriminant B*B-4*A*C g(x)=A*x*x+2B*x+C has discriminant 4*B*B-4*A*C textbook use g(x) style , so discriminant is B*B-A*C quadratic discriminant>0 means there are two real numbers x1=[-2B+2"(B*B-AC)]/[2A] and x2=[-2B-2"(B*B-AC)]/[2A] let p(x1)=0 and p(x2)=0 ÿ (x1 NOT= x2) when <a name="ch01b010">&lt;a name="ch01b010"&gt;</a> x is in (x1,x2) , p(x)<0 (discriminantÿ0) x=x1 or x=x2 , p(x)=0 (discriminantÿ0) x is out side of (x1,x2) p(x)>0 (discriminantÿ0) But! p(x) is square of real number, p(x)<0 ? impossible ! must exclude p(x)<0 not allow discriminant>0 then Cauchy~Schwarz Inequality is a result. 2009-06-25-17-39 stop <a name="ch01b011">&lt;a name="ch01b011"&gt;</a> 2009-06-25-20-49 start  %0Is it assigned? is it necessary? Number sequence 1: -1, -2, -3 Number sequence 2: -4, -5, -6 vector dot product is (-1)*(-4)+(-2)*(-5)+(-3)*(-6) = 32 Vector 1 length * vector 2 length get 3.74165*8.77496=32.8329 Cauchy Inequality indicate that 32.8329 > 32 <a name="ch01b012">&lt;a name="ch01b012"&gt;</a> Because we made rule : (-3)*(-6) = +18 If we make new rule : (-3)*(-6) = -18 then (-1)*(-1) = -1 and "(-1) = "[(-1)*(-1)] = real number -1 We can discard the headache complex analysis! Because no need imaginary sqrt(-1)! and Negative number sequence 1: -1, -2, -3 Negative number sequence 2: -4, -5, -6 their Cauchy Inequality will be -32.8329 < -32 <a name="ch01b013">&lt;a name="ch01b013"&gt;</a> Mathematician can make it a rule 0! = 1 We ask mathematician make a new rule (-1)*(-1) = -1 is it OK? 2009-06-25-21-07 here This problem ask that the relation (-1)*(-1) = +1 is it assigned? or is it necessary? <font color=red size=+3> Here, Would you please pause for a while and think. Do not read the following discussion immediately. Is it OK? </font> <a name="ch01b014">&lt;a name="ch01b014"&gt;</a> The following is a study notes. Yearning for the Impossible ISBN 1-56881-254-x 2008-03-03 purchase 2008-03-12 receive book Page 26 and 27 algebra distribution law a*(b+c) = a*b + a*c -----eqn.AB02 We must insist this distribution law, because this law describe the facts in our universe. The following is its derived result. <a name="ch01b015">&lt;a name="ch01b015"&gt;</a> ab=a(b+0)00 nothing! just add a 'nothing' 0 =ab + a*00 carry out distribution law 0 = a*0 000 -----eqn.AB03 ab cancel each other from both side. Next see a*(b+c) = a*b + a*c require a=-1, b=+1, c=-1 0 = (-1)*000this is eqn.AB03 rule. = (-1)*(1 +(-1))00because 1 +(-1) is zero = (-1)*1 + (-1)*(-1)0carry out distribution law = (-1) + (-1)*(-1)0"*1" not change value <a name="ch01b016">&lt;a name="ch01b016"&gt;</a> Left and right side add 1 at same time, get<font color=red> +1 = (-1)*(-1)</font>000 -----eqn.AB04 We must insist distribution law, we can NOT assign<font color=silver> -1 = (-1)*(-1)</font> For this reason, Cauchy~Schwarz Inequality is always true, Complex analysis is still hot topic in school and LiuHH must continue study. 2009-06-25-21-32 stop <a name="ch01b017">&lt;a name="ch01b017"&gt;</a>0<a href="#index">Index begin</a>0<a href="#index06">Index this file</a> <script language="javascript">alert0()</script> 2009-06-26-09-55 start The following is chapter one exercise  %0Exercise 1.1 problem statement 00textbook page 12 ÿThe 1-Trick and the Splitting Trick ÿ Given real sequence a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub>, ... a<sub>n</sub> Prove the following two relations.</font></pre> <a name="ch01b018">&lt;a name="ch01b018"&gt;</a> <br> &nbsp; a<sub>1</sub>+a<sub>2</sub>+ ... +a<sub>n</sub> f" ("n)*[ a<sub>1</sub><sup>2</sup>+a<sub>2</sub><sup>2</sup>+ ... +a<sub>n</sub><sup>2</sup> ] <sup>1/2</sup> ---page 12 line 23 <br>and<br> <TABLE WIDTH="540" id=T012L25a border=0 ><!--9806261009--> <TR> <TD> &nbsp; </TD> <TD> <TABLE WIDTH="380" border=0 id=T012L25b > <TR> <TD> <CENTER><FONT SIZE=-2>k=n</FONT></CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER><FONT SIZE=-2>k=1</FONT></CENTER> </TD> <TD> <CENTER>a<sub><FONT SIZE=-1> k</FONT></sub></CENTER> </TD> <TD> f" </TD> <TD> <CENTER><FONT SIZE=+3>[</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>k=n</FONT></CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER><FONT SIZE=-2>k=1</FONT></CENTER> </TD> <TD> <CENTER>|a<sub><FONT SIZE=-1> k</FONT></sub>|<sup>2/3</sup> </CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>]</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>1/2</FONT></CENTER> <CENTER><FONT SIZE=+3>00</FONT></CENTER> <CENTER><FONT SIZE=-2>00</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>[</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>k=n</FONT></CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER><FONT SIZE=-2>k=1</FONT></CENTER> </TD> <TD> <CENTER>|a<sub><FONT SIZE=-1> k</FONT></sub>|<sup>4/3</sup> </CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>]</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>1/2</FONT></CENTER> <CENTER><FONT SIZE=+3>00</FONT></CENTER> <CENTER><FONT SIZE=-2>00</FONT></CENTER> </TD> </TR> </TABLE> </TD> <TD> ---page 12 line 25 </TD> </TR> </TABLE> width of above equation <INPUT name=Button012L25a type="button" value="default value" onclick="B012L25a.value=540,B012L25b.value=380,T012L25a.width=B012L25a.value,T012L25b.width=B012L25b.value"> <input id="B012L25a" value=540 size=3 onchange=T012L25a.width=B012L25a.value /> <input id="B012L25b" value=380 size=3 onchange=T012L25b.width=B012L25b.value /> <pre><font size=+2><a name="ch01b019">&lt;a name="ch01b019"&gt;</a> 2009-06-26-10-17 here Exercise 1.1 use The 1-Trick and the Splitting Trick which is inequality problem-solving important tool. Later discussion will use them again and again. These tricks will show up in different form. In some case these tricks are important step. <script language="javascript">alert1()</script> <a name="ch01b020">&lt;a name="ch01b020"&gt;</a>  %0Exercise 1.1 hint 00textbook page 226 Textbook "The Cauchy-Schwarz Master Class" has hint for every exercise. Help reader to think. Simpler problem, hint is solution. Complicate problem, still need reader to fill in details. <a href="tute0008.htm#ch01b018"> page 12 line 23 eqn.</a> can be done by using 1-Trick. That is in <a href=tute0007.htm#ch01a026>Cauchy Inequality</a> let all b sequence elements b<sub>1</sub>=b<sub>2</sub>= ... =b<sub>n</sub> equal to 1. Although this look so simple. But for a beginner, it is not immediately clear that all single sequence a<sub>1</sub>,a<sub>2</sub>, ... ,a<sub>n</sub> has a all one sequence b<sub>1</sub>=b<sub>2</sub>= ... =b<sub>n</sub>=1 as company. <a name="ch01b021">&lt;a name="ch01b021"&gt;</a><a href="tute0008.htm#ch01b018"> page 12 line 25 eqn.</a> use Splitting Trick. For example a^1 = [a^(1/2)]*[a^(1/2)] a^1 = [a^(1/3)]*[a^(2/3)] Why split sequence a to multiplication of a^(1/3) and a^(2/3) ? Because Cauchy's Inequality require two sequences. For one sequence a<sub>1</sub>,a<sub>2</sub>, ... ,a<sub>n</sub> first method is 1-Trick. Second method is Splitting Trick. Must obey Cauchy's requirement. <a name="ch01b022">&lt;a name="ch01b022"&gt;</a> page 12 line 25 eqn. use a^1 = [a^(1/3)]*[a^(2/3)] Splitting Trick, substitute to Cauchy's Inequality to get answer. Because Cauchy's Inequality show up every where. Create a command to call Cauchy any where need it. 2009-06-26-10-52 here 2009-06-26-11-20 Must call HelloCauchy(hcPar) outside of &lt;pre&gt;..&lt;/pre&gt; othrewise equation width is not designed width.</font></pre> <a name="ch01b023">&lt;a name="ch01b023"&gt;</a> <script lang="javascript"> <!-- HelloCauchy(101) //--> </script> <pre><font size=+2> <a name="ch01b024">&lt;a name="ch01b024"&gt;</a>0<a href="#index">Index begin</a>0<a href="#index06">Index this file</a>  %0Exercise 1.1 solution <script lang="javascript">alert2()</script> 2009-06-26-15-23 here Above is Cauchy's Inequality general form. Change k=" to k=n , because discuss finite terms. Change b<sub>k</sub> to 1<sub>k</sub> because use 1-Trick get following result.</font></pre> <a name="ch01b025">&lt;a name="ch01b025"&gt;</a> <script lang="javascript"> <!-- HelloCauchy(102,['b<','1<'],['="','=n'],['1/2','0.5']); //9806261515 //--> </script> <br> Programmer please pay attention: calling ID 102 made three changes. <br> <pre><font size=+2><a name="ch01b026">&lt;a name="ch01b026"&gt;</a> 2009-06-26-15-27 here a<sub>k</sub> * 1 = a<sub>k</sub> ÿ 1*1 = 1 ÿ add n one get n ÿeqn.1.7 ÿcalling ID 102 ÿ change to</font></pre> &nbsp; a<sub>1</sub>+a<sub>2</sub>+ ... +a<sub>n</sub> f" ("n)*[ a<sub>1</sub><sup>2</sup>+a<sub>2</sub><sup>2</sup>+ ... +a<sub>n</sub><sup>2</sup> ] <sup>1/2</sup> ---page 12 line 23 <pre><font size=+2>2009-06-26-15-36 here Above solved page 12 line 23 equation. <a name="ch01b027">&lt;a name="ch01b027"&gt;</a> Below solve page 12 line 25 equation. Hint give clear instruction, split sequence "(a<sub>k</sub>) to "(a<sub>k</sub><sup>1/3</sup>*a<sub>k</sub><sup>2/3</sup>) where a<sub>k</sub><sup>1/3</sup>*a<sub>k</sub><sup>2/3</sup> IS a<sub>k</sub> take a<sub>k</sub><sup>1/3</sup> as Cauchy's Inequality sequence a<sub>k</sub> take a<sub>k</sub><sup>2/3</sup> as Cauchy's Inequality sequence b<sub>k</sub> then general equation</font></pre> <a name="ch01b028">&lt;a name="ch01b028"&gt;</a> <script lang="javascript"> <!-- HelloCauchy(103) //--> </script> <br> become <TABLE WIDTH="640" id=T012L25c border=0 ><!--9806261544--> <TR> <TD> &nbsp; </TD> <TD> <TABLE WIDTH="480" border=0 id=T012L25d > <TR> <TD> <CENTER><FONT SIZE=-2>k=n</FONT></CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER><FONT SIZE=-2>k=1</FONT></CENTER> </TD> <TD> <CENTER>a<sub>k</sub><sup>1/3</sup>*a<sub>k</sub><sup>2/3</sup></CENTER> </TD> <TD> f" </TD> <TD> <CENTER><FONT SIZE=+3>[</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>k=n</FONT></CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER><FONT SIZE=-2>k=1</FONT></CENTER> </TD> <TD> <CENTER>|a<sub><FONT SIZE=-1> k</FONT></sub>|<sup>2/3</sup> </CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>]</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>1/2</FONT></CENTER> <CENTER><FONT SIZE=+3>00</FONT></CENTER> <CENTER><FONT SIZE=-2>00</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>[</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>k=n</FONT></CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER><FONT SIZE=-2>k=1</FONT></CENTER> </TD> <TD> <CENTER>|a<sub><FONT SIZE=-1> k</FONT></sub>|<sup>4/3</sup> </CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>]</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>1/2</FONT></CENTER> <CENTER><FONT SIZE=+3>00</FONT></CENTER> <CENTER><FONT SIZE=-2>00</FONT></CENTER> </TD> </TR> </TABLE> </TD> <TD> ---page 12 line 25 </TD> </TR> </TABLE> width of above equation <INPUT name=Button012L25a type="button" value="default value" onclick="B012L25c.value=640,B012L25d.value=480,T012L25c.width=B012L25c.value,T012L25d.width=B012L25d.value"> <input id="B012L25c" value=640 size=3 onchange=T012L25c.width=B012L25c.value /> <input id="B012L25d" value=480 size=3 onchange=T012L25d.width=B012L25d.value /> <pre><font size=+2><a name="ch01b029">&lt;a name="ch01b029"&gt;</a> 2009-06-26-15-50 here page 12 line 25 equation say sequence a is real number sequence. We know that real number include negative number. If negative number take power less than one then get complex number. For example (-1)^0.5 is imaginary i When problem involve power less than one, most problem require seq. be non-negative numbers. But exercise 1.1 require seq. be real, allow negative number. <a name="ch01b030">&lt;a name="ch01b030"&gt;</a> page 12 line 25 equation left hand side a<sub>k</sub><sup>1/3</sup>*a<sub>k</sub><sup>2/3</sup> become a<sub>k</sub>. no power less than one. but right hand side has a<sub>k</sub><sup>2/3</sup> and a<sub>k</sub><sup>4/3</sup> ( = a<sub>k</sub>*a<sub>k</sub><sup>1/3</sup> ) Both have power less than one. To avoid show up complex number, page 12 line 25 equation right hand side use absolute value to solve problem. 2009-06-26-16-01 stop <hr> <a name="ch01b031">&lt;a name="ch01b031"&gt;</a>0<a href="#index">Index begin</a>0<a href="#index06">Index this file</a> <script language="javascript">alert0()</script> 2009-06-26-17-20 start  %0Exercise 1.2 problem statement 00textbook page 12 (Products of averages and averages of products) Given non-negative sequence p<sub>1</sub>+p<sub>2</sub>+p<sub>3</sub>+ ... +p<sub>n</sub>=1 ---eqn.AB05 and non-negative seq. a<sub>1</sub>, a<sub>2</sub>, a<sub>3</sub>, ... a<sub>n</sub> and non-negative seq. b<sub>1</sub>, b<sub>2</sub>, b<sub>3</sub>, ... b<sub>n</sub> a and b have the following relation for all j=1 to j=n , 1f"a<sub>j</sub>b<sub>j</sub> ---eqn.AB06 For above given condition, prove the following relation hold.</font></pre> <a name="ch01b032">&lt;a name="ch01b032"&gt;</a> <TABLE WIDTH="500" id=T013L02a border=0 ><!--9806261731--> <TR> <TD> &nbsp; </TD> <TD> <TABLE WIDTH="320" border=0 id=T013L02b > <TR> <TD> 1 f" </TD> <TD> <CENTER><FONT SIZE=+3>[</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>j=n</FONT></CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER><FONT SIZE=-2>j=1</FONT></CENTER> </TD> <TD> <CENTER> p<sub><FONT SIZE=-1> j</FONT></sub> a<sub><FONT SIZE=-1> j</FONT></sub> </CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>]</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>[</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>j=n</FONT></CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER><FONT SIZE=-2>j=1</FONT></CENTER> </TD> <TD> <CENTER> p<sub><FONT SIZE=-1> j</FONT></sub> b<sub><FONT SIZE=-1> j</FONT></sub> </CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>]</FONT></CENTER> </TD> </TR> </TABLE> </TD> <TD> ----page 13 line 2 </TD> </TR> </TABLE> width of above equation <INPUT name=Button013L02a type="button" value="default value" onclick="B013L02a.value=500,B013L02b.value=320,T013L02a.width=B013L02a.value,T013L02b.width=B013L02b.value"> <input id="B013L02a" value=500 size=3 onchange=T013L02a.width=B013L02a.value /> <input id="B013L02b" value=320 size=3 onchange=T013L02b.width=B013L02b.value /> <pre><font size=+2>2009-06-26-17-50 here <script language="javascript">alert1()</script> <a name="ch01b033">&lt;a name="ch01b033"&gt;</a>  %0Exercise 1.2 hint 00textbook page 226 This is another Splitting Trick problem. Consider next relation.</font></pre> <TABLE WIDTH="540" id=T226L15a border=0 ><!--9806261758--> <TR> <TD> &nbsp; </TD> <TD> <TABLE WIDTH="380" border=0 id=T226L15b > <TR> <TD> 1 f" </TD> <TD> <CENTER><FONT SIZE=-2>j=n</FONT></CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER><FONT SIZE=-2>j=1</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>{</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>[</FONT></CENTER> </TD> <TD> <CENTER> p<FONT SIZE=-1><sub> j</sub><sup>1/2</sup></FONT> a<FONT SIZE=-1><sub> j</sub><sup>1/2</sup></FONT> </CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>] [</FONT></CENTER> </TD> <TD> <CENTER> p<FONT SIZE=-1><sub> j</sub><sup>1/2</sup></FONT> b<FONT SIZE=-1><sub> j</sub><sup>1/2</sup></FONT> </CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>]</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>}</FONT></CENTER> </TD> </TR> </TABLE> </TD> <TD> ---page 226 line 15 </TD> </TR> </TABLE> width of above equation <INPUT name=Button226L15a type="button" value="default value" onclick="B226L15a.value=540,B226L15b.value=380,T226L15a.width=B226L15a.value,T226L15b.width=B226L15b.value"> <input id="B226L15a" value=540 size=3 onchange=T226L15a.width=B226L15a.value /> <input id="B226L15b" value=380 size=3 onchange=T226L15b.width=B226L15b.value /> <pre><font size=+2>2009-06-26-18-03 here Take Cauchy Inequality for above equation. <a name="ch01b034">&lt;a name="ch01b034"&gt;</a>0<a href="#index">Index begin</a>0<a href="#index06">Index this file</a> 2009-06-26-18-12 here  %0Exercise 1.2 solution <script lang="javascript">alert2()</script> Given condition is for all j=1 to j=n , 1f"a<sub>j</sub>b<sub>j</sub> -----eqn.AB06 eqn.AB06 whole equation take square root get 1f""(a<sub>j</sub>b<sub>j</sub>) -----eqn.AB07 another given condition is p<sub>1</sub>+p<sub>2</sub>+p<sub>3</sub>+ ... +p<sub>n</sub>=1 -----eqn.AB05 Because p<sub>j</sub> is non-negative, in eqn.AB07 multiply p<sub>j</sub> to both side, not change inequality. Get p<sub>j</sub>f"p<sub>j</sub>"(a<sub>j</sub>b<sub>j</sub>) -----eqn.AB08 <a name="ch01b035">&lt;a name="ch01b035"&gt;</a> In eqn.AB08 sum j from j=1 to j=n get</font></pre> <TABLE WIDTH="500" id=T226M15a border=0 ><!--9806261821--> <TR> <TD> &nbsp; </TD> <TD> <TABLE WIDTH="280" border=0 id=T226M15b > <TR> <TD> <CENTER><FONT SIZE=-2>j=n</FONT></CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER><FONT SIZE=-2>j=1</FONT></CENTER> </TD> <TD> <CENTER> p<FONT SIZE=-1><sub> j</sub></FONT> </CENTER> </TD> <TD> f" </TD> <TD> <CENTER><FONT SIZE=-2>j=n</FONT></CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER><FONT SIZE=-2>j=1</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>[</FONT></CENTER> </TD> <TD> <CENTER> p<FONT SIZE=-1><sub> j</sub></FONT> ( a<FONT SIZE=-1><sub> j</sub></FONT> b<FONT SIZE=-1><sub> j</sub></FONT> )<sup>1/2<sup> </CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>] </FONT></CENTER> </TD> </TR> </TABLE> </TD> <TD> ---page 226 line 15 aux. </TD> </TR> </TABLE> width of above equation <INPUT name=Button226M15a type="button" value="default value" onclick="B226M15a.value=500,B226M15b.value=280,T226M15a.width=B226M15a.value,T226M15b.width=B226M15b.value"> <input id="B226M15a" value=500 size=3 onchange=T226M15a.width=B226M15a.value /> <input id="B226M15b" value=280 size=3 onchange=T226M15b.width=B226M15b.value /> <pre><font size=+2><a name="ch01b036">&lt;a name="ch01b036"&gt;</a> 2009-06-26-18-46 here Left hand side is one ÿsee <a href="#ch01b031">eqn.AB05</a> ÿ ÿ so we get 1f""[p<sub>j</sub>"(a<sub>j</sub>) "(b<sub>j</sub>)] -----eqn.AB09 eqn.AB09 right hand side p<sub>j</sub> split and re-write as "(p<sub>j</sub>)*"(p<sub>j</sub>) re-combine to 1f""{["(p<sub>j</sub>)"(a<sub>j</sub>)]*["(p<sub>j</sub>)"(b<sub>j</sub>)]} -----eqn.AB10 eqn.AB10 is page 226 line 15 equation. <a name="ch01b037">&lt;a name="ch01b037"&gt;</a> take ["(p<sub>j</sub>)"(a<sub>j</sub>)] as (a<sub>k</sub>) in Cauchy Ineq. take ["(p<sub>j</sub>)"(b<sub>j</sub>)] as (b<sub>k</sub>) in Cauchy Ineq. and substitute into Cauchy's Inequality</font></pre> <script lang="javascript"> <!-- HelloCauchy(104) //--> </script> <pre><font size=+2><a name="ch01b038">&lt;a name="ch01b038"&gt;</a> 2009-06-26-18-57 here When substitute, need pay attention to next three points. First. eqn.AB10 right hand side IS page 5 eqn.1.7 left hand side. Page 5 eqn.1.7 can NOT be greater than one directly. It must use eqn.AB10 ÿorigin from <a href="#ch01b034">eqn.AB06</a> ÿ to get greater than or equal to one conclusion. Second. page 5 eqn.1.7 square bracketed terms take square first. This square cancel the square root in eqn.AB10 So <a href="#ch01b032">answer (page 13 line 2)</a> summed term has power one. <a name="ch01b039">&lt;a name="ch01b039"&gt;</a> Third. After get solution, whole equation take square, page 5 eqn.1.7 square root (Cauchy put square root there !) out side of square bracket disappear. After these three attention, answer become page 13 line 2 equation. 2009-06-26-19-05 stop <hr> <a name="ch01b040">&lt;a name="ch01b040"&gt;</a>0<a href="#index">Index begin</a>0<a href="#index06">Index this file</a> <script language="javascript">alert0()</script> 2009-06-27-09-33 start  %0Exercise 1.3 problem statement 00textbook page 13 ÿWhy not three or more? ÿ Cauchy Inequality supply an upper limit for pair product sum of two arrays. ÿfor example a<sub>1</sub>, b<sub>1</sub> is one pair ; and "[a<sub>k</sub> b<sub>k</sub>] is pair product sum ÿ ÿ<a href="#ch01b122">Next equation, greater than side is upper limit of less than side</a> ÿ</font></pre> <script lang="javascript"> <!-- HelloCauchy(105) //--> </script> <pre><font size=+2><a name="ch01b041">&lt;a name="ch01b041"&gt;</a> Why limit to two sequences? If there are three sequences, can we find an upper limit for three terms product sum of three arrays? Exercise 1.3 invite reader to prove two related equations.</font></pre> <TABLE WIDTH="640" id=T013L13a border=0 ><!--9806270954--> <TR> <TD> &nbsp; </TD> <TD> <TABLE WIDTH="460" border=0 id=T013L13b > <TR> <TD> <CENTER><FONT SIZE=+3>[</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>k=n</FONT></CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER><FONT SIZE=-2>k=1</FONT></CENTER> </TD> <TD> <CENTER> a<sub><FONT SIZE=-1> k</FONT></sub> b<sub><FONT SIZE=-1> k</FONT></sub> c<sub><FONT SIZE=-1> k</FONT></sub> </CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>]</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>4</FONT></CENTER> <CENTER><FONT SIZE=+3>00</FONT></CENTER> <CENTER><FONT SIZE=-2>00</FONT></CENTER> </TD> <TD> f" </TD> <TD> <CENTER><FONT SIZE=+3>[</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>k=n</FONT></CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER><FONT SIZE=-2>k=1</FONT></CENTER> </TD> <TD> <CENTER>a<sub><FONT SIZE=-1> k</FONT></sub><sup>2</sup> </CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>]</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>2</FONT></CENTER> <CENTER><FONT SIZE=+3>00</FONT></CENTER> <CENTER><FONT SIZE=-2>00</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>k=n</FONT></CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER><FONT SIZE=-2>k=1</FONT></CENTER> </TD> <TD> <CENTER>b<sub><FONT SIZE=-1> k</FONT></sub><sup>4</sup> </CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>k=n</FONT></CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER><FONT SIZE=-2>k=1</FONT></CENTER> </TD> <TD> <CENTER>c<sub><FONT SIZE=-1> k</FONT></sub><sup>4</sup> </CENTER> </TD> </TR> </TABLE> </TD> <TD> ---page 13 line 13 </TD> </TR> </TABLE> width of above equation <INPUT name=Button013L13a type="button" value="default value" onclick="B013L13a.value=640,B013L13b.value=460,T013L13a.width=B013L13a.value,T013L13b.width=B013L13b.value"> <input id="B013L13a" value=640 size=3 onchange=T013L13a.width=B013L13a.value /> <input id="B013L13b" value=460 size=3 onchange=T013L13b.width=B013L13b.value /> <pre><font size=+2><a name="ch01b042">&lt;a name="ch01b042"&gt;</a> 2009-06-27-10-02 here Here mention an off-topic matter. When build page 13 line 13 equation, copy right hand side "a<sub>k</sub><sup>2</sup> to "b<sub>k</sub><sup>2</sup> and "c<sub>k</sub><sup>2</sup> Did not know there is an error in it. But later calculate power of both side, Left hand side abc is power of 3, and [abc]<sup>4</sup> rise power to 12. Look at right hand side, ["a<sub>k</sub><sup>2</sup>]<sup>2</sup>"b<sub>k</sub><sup>2</sup>"c<sub>k</sub><sup>2</sup>, rise power to 8. But 12 not equal to 8 ! Immediately check what is wrong, found that "b<sub>k</sub><sup>2</sup> and "c<sub>k</sub><sup>2</sup> are wrong. It should be "b<sub>k</sub><sup>4</sup> and "c<sub>k</sub><sup>4</sup> . If a,b,c are all length, then left hand side is length to power 12 right hand side must be to power 12 too it is possible to compare great/small. (Counter example is two acre is greater than one mile? no !! we can not compare area with length!!) <a name="ch01b043">&lt;a name="ch01b043"&gt;</a> 2009-06-27-10-15 here Next is second equation.</font></pre> <TABLE WIDTH="580" id=T013L14a border=0 ><!--9806271018--> <TR> <TD> &nbsp; </TD> <TD> <TABLE WIDTH="420" border=0 id=T013L14b > <TR> <TD> <CENTER><FONT SIZE=+3>[</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>k=n</FONT></CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER><FONT SIZE=-2>k=1</FONT></CENTER> </TD> <TD> <CENTER> a<sub><FONT SIZE=-1> k</FONT></sub> b<sub><FONT SIZE=-1> k</FONT></sub> c<sub><FONT SIZE=-1> k</FONT></sub> </CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>]</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>2</FONT></CENTER> <CENTER><FONT SIZE=+3>00</FONT></CENTER> <CENTER><FONT SIZE=-2>00</FONT></CENTER> </TD> <TD> f" </TD> <TD> <CENTER><FONT SIZE=-2>k=n</FONT></CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER><FONT SIZE=-2>k=1</FONT></CENTER> </TD> <TD> <CENTER>a<sub><FONT SIZE=-1> k</FONT></sub><sup>2</sup> </CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>k=n</FONT></CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER><FONT SIZE=-2>k=1</FONT></CENTER> </TD> <TD> <CENTER>b<sub><FONT SIZE=-1> k</FONT></sub><sup>2</sup> </CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>k=n</FONT></CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER><FONT SIZE=-2>k=1</FONT></CENTER> </TD> <TD> <CENTER>c<sub><FONT SIZE=-1> k</FONT></sub><sup>2</sup> </CENTER> </TD> </TR> </TABLE> </TD> <TD> ---page 13 line 14 </TD> </TR> </TABLE> width of above equation <INPUT name=Button013L14a type="button" value="default value" onclick="B013L14a.value=580,B013L14b.value=420,T013L14a.width=B013L14a.value,T013L14b.width=B013L14b.value"> <input id="B013L14a" value=580 size=3 onchange=T013L14a.width=B013L14a.value /> <input id="B013L14b" value=420 size=3 onchange=T013L14b.width=B013L14b.value /> <br> page 13 line 14 equation answer take a square root, <a href="#Cauchy_6">program available</a>. <br> <pre><font size=+2>2009-06-27-10-22 here <script language="javascript">alert1()</script> <a name="ch01b044">&lt;a name="ch01b044"&gt;</a>  %0Exercise 1.3 hint 00textbook page 226 page 13 line 13 equation need to carry out Cauchy Inequality twice. Group as following a<sub>k</sub>(b<sub>k</sub>c<sub>k</sub>) page 13 line 14 equation if let a<sub>k</sub>=b<sub>k</sub>=c<sub>k</sub>=1, get n<sup>2</sup>f"n<sup>3</sup> (n g" 1) means page 13 line 14 equation is not difficult. To start, try d<sub>k</sub>=c<sub>k</sub>/"["(c<sub>j</sub>*c<sub>j</sub>)] -----eqn.AB11 where "["(c<sub>j</sub>*c<sub>j</sub>)] is length of c sequence. <a name="ch01b045">&lt;a name="ch01b045"&gt;</a> eqn.AB11 is normalized sequence c . Sequence d has length one, its component d<sub>k</sub> must be less than or equal to one. page 13 line 14 equation start from apply Cauchy Inequality to sequence a and b (initially no sequence c) At less than side of Cauchy Inequality, multiply a d<sub>k</sub> which is no greater than one. This will not change inequality sign. Next from d<sub>k</sub> recover c<sub>k</sub> ..... 2009-06-27-10-52 here <a name="ch01b046">&lt;a name="ch01b046"&gt;</a>0<a href="#index">Index begin</a>0<a href="#index06">Index this file</a> 2009-06-27-10-54 here  %0Exercise 1.3 solution <script lang="javascript">alert2()</script> Prove page 13 line 13 equation. Consider a<sub>k</sub>(b<sub>k</sub>c<sub>k</sub>) ÿ Carry out Cauchy Inequality twice. First time group b<sub>k</sub>c<sub>k</sub> as one item.</font></pre> <TABLE WIDTH="660" id=T013L13c border=0 ><!--9806271057--> <TR> <TD> &nbsp; </TD> <TD> <TABLE WIDTH="460" border=0 id=T013L13d > <TR> <TD> <CENTER><FONT SIZE=-2>k=n</FONT></CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER><FONT SIZE=-2>k=1</FONT></CENTER> </TD> <TD> <CENTER> a<sub><FONT SIZE=-1> k</FONT></sub> ( b<sub><FONT SIZE=-1> k</FONT></sub> c<sub><FONT SIZE=-1> k</FONT></sub> ) </CENTER> </TD> <TD> f" </TD> <TD> <CENTER><FONT SIZE=+3>[</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>k=n</FONT></CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER><FONT SIZE=-2>k=1</FONT></CENTER> </TD> <TD> <CENTER>a<sub><FONT SIZE=-1> k</FONT></sub><sup>2</sup> </CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>]</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>1/2</FONT></CENTER> <CENTER><FONT SIZE=+3>00</FONT></CENTER> <CENTER><FONT SIZE=-2>00</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>[</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>k=n</FONT></CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER><FONT SIZE=-2>k=1</FONT></CENTER> </TD> <TD> <CENTER> b<sub><FONT SIZE=-1> k</FONT></sub><sup>2</sup> c<sub><FONT SIZE=-1> k</FONT></sub><sup>2</sup> </CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>]</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>1/2</FONT></CENTER> <CENTER><FONT SIZE=+3>00</FONT></CENTER> <CENTER><FONT SIZE=-2>00</FONT></CENTER> </TD> </TR> </TABLE> </TD> <TD> ---page 13 line 13 step 1 </TD> </TR> </TABLE> width of above equation <INPUT name=Button013L13c type="button" value="default value" onclick="B013L13c.value=660,B013L13d.value=460,T013L13c.width=B013L13c.value,T013L13d.width=B013L13d.value"> <input id="B013L13c" value=660 size=3 onchange=T013L13c.width=B013L13c.value /> <input id="B013L13d" value=460 size=3 onchange=T013L13d.width=B013L13d.value /> <pre><font size=+2><a name="ch01b047">&lt;a name="ch01b047"&gt;</a> 2009-06-27-11-10 here Square whole equation to remove square root.</font></pre> <TABLE WIDTH="660" id=T013L13e border=0 ><!--9806271112--> <TR> <TD> &nbsp; </TD> <TD> <TABLE WIDTH="460" border=0 id=T013L13f > <TR> <TD> <CENTER><FONT SIZE=+3>[</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>k=n</FONT></CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER><FONT SIZE=-2>k=1</FONT></CENTER> </TD> <TD> <CENTER> a<sub><FONT SIZE=-1> k</FONT></sub> ( b<sub><FONT SIZE=-1> k</FONT></sub> c<sub><FONT SIZE=-1> k</FONT></sub> ) </CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>]</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>2</FONT></CENTER> <CENTER><FONT SIZE=+3>00</FONT></CENTER> <CENTER><FONT SIZE=-2>00</FONT></CENTER> </TD> <TD> f" </TD> <TD> <CENTER><FONT SIZE=+3>[</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>k=n</FONT></CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER><FONT SIZE=-2>k=1</FONT></CENTER> </TD> <TD> <CENTER>a<sub><FONT SIZE=-1> k</FONT></sub><sup>2</sup> </CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>]</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>[</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>k=n</FONT></CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER><FONT SIZE=-2>k=1</FONT></CENTER> </TD> <TD> <CENTER> b<sub><FONT SIZE=-1> k</FONT></sub><sup>2</sup> c<sub><FONT SIZE=-1> k</FONT></sub><sup>2</sup> </CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>]</FONT></CENTER> </TD> </TR> </TABLE> </TD> <TD> ---page 13 line 13 step 2 </TD> </TR> </TABLE> width of above equation <INPUT name=Button013L13e type="button" value="default value" onclick="B013L13e.value=660,B013L13f.value=460,T013L13e.width=B013L13e.value,T013L13f.width=B013L13f.value"> <input id="B013L13e" value=660 size=3 onchange=T013L13e.width=B013L13e.value /> <input id="B013L13f" value=460 size=3 onchange=T013L13f.width=B013L13f.value /> <pre><font size=+2><a name="ch01b048">&lt;a name="ch01b048"&gt;</a> 2009-06-27-11-17 here Now take Cauchy Inequality for b<sub>k</sub><sup>2</sup>c<sub>k</sub><sup>2</sup> get</font></pre> <TABLE WIDTH="660" id=T013L13g border=0 ><!--9806271057--> <TR> <TD> &nbsp; </TD> <TD> <TABLE WIDTH="460" border=0 id=T013L13h > <TR> <TD> <CENTER><FONT SIZE=-2>k=n</FONT></CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER><FONT SIZE=-2>k=1</FONT></CENTER> </TD> <TD> <CENTER> b<sub>k</sub><sup>2</sup> c<sub>k</sub><sup>2</sup> </CENTER> </TD> <TD> f" </TD> <TD> <CENTER><FONT SIZE=+3>[</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>k=n</FONT></CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER><FONT SIZE=-2>k=1</FONT></CENTER> </TD> <TD> <CENTER> b<sub><FONT SIZE=-1> k</FONT></sub><sup>4</sup> </CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>]</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>1/2</FONT></CENTER> <CENTER><FONT SIZE=+3>00</FONT></CENTER> <CENTER><FONT SIZE=-2>00</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>[</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>k=n</FONT></CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER><FONT SIZE=-2>k=1</FONT></CENTER> </TD> <TD> <CENTER> c<sub><FONT SIZE=-1> k</FONT></sub><sup>4</sup> </CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>]</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>1/2</FONT></CENTER> <CENTER><FONT SIZE=+3>00</FONT></CENTER> <CENTER><FONT SIZE=-2>00</FONT></CENTER> </TD> </TR> </TABLE> </TD> <TD> ---page 13 line 13 step 3 </TD> </TR> </TABLE> width of above equation <INPUT name=Button013L13g type="button" value="default value" onclick="B013L13g.value=660,B013L13h.value=460,T013L13g.width=B013L13g.value,T013L13h.width=B013L13h.value"> <input id="B013L13g" value=660 size=3 onchange=T013L13g.width=B013L13g.value /> <input id="B013L13h" value=460 size=3 onchange=T013L13h.width=B013L13h.value /> <pre><font size=+2><a name="ch01b049">&lt;a name="ch01b049"&gt;</a> 2009-06-27-11-23 here Step 3 left hand side is step 2 right most square bracket. Replace it with step 3 right hand side. Then whole equation take square, get page 13 line 13 equation. 2009-06-27-11-26 here Next see page 13 line 14 equation. Cauchy Inequality for sequence a and b is</font></pre> <script lang="javascript"> <!-- HelloCauchy(106) //--> </script> <pre><font size=+2><a name="ch01b050">&lt;a name="ch01b050"&gt;</a> 2009-06-27-11-30 here Exercise 1.3 hint said normalize seq. c and let d<sub>k</sub>=c<sub>k</sub>/"["(c<sub>j</sub>*c<sub>j</sub>)] -----eqn.AB11 component c<sub>k</sub> f" whole length "["(c<sub>j</sub>*c<sub>j</sub>)] so d<sub>k</sub> f" 1 Seq. a and seq. b less than side multiply a d<sub>k</sub> which is not greater than one. This will make less than side even smaller. Inequality will not change direction. Finally, recover c<sub>k</sub>/"["(c<sub>j</sub>*c<sub>j</sub>)] from d<sub>k</sub> "["(c<sub>j</sub>*c<sub>j</sub>)] do not change with "[k vary] "["(c<sub>j</sub>*c<sub>j</sub>)] can be moved out of "[k vary] "["(c<sub>j</sub>*c<sub>j</sub>)] is positive term, it can be moved to greater than side without change inequality direction. After all of these steps, we get page 13 line 14 equation. 2009-06-27-11-38 stop <hr> <a name="ch01b051">&lt;a name="ch01b051"&gt;</a> 2009-06-27-22-14 start Above said [[ Seq. a and seq. b less than side multiply a d<sub>k</sub> which is not greater than one. This will make less than side even smaller. Inequality will not change direction. ]] Each time add one more sequence, Cauchy Inequality smaller side will become even smaller. The more sequence involved, the greater gap between greater than side and less than side. LiuHH wrote a program to calculate multiple sequence Cauchy Inequality up to six sequences. Reader can test program and feel the gap. 2009-06-27-22-19 stop <a name="ch01b052">&lt;a name="ch01b052"&gt;</a> 2009-06-27-12-27 start write Program verify page 13 line 14 equation up to six sequences.</font></pre> <form name=form1> <a name="Cauchy_6">&lt;a name="Cauchy_6"&gt;</a>0<a href="#index">Index begin</a>0<a href="#index06">Index this file</a> <br>  %0Program: <font color=red>Six sequences</font> Cauchy's Inequality 2009-06-27-12-30 here <br> <table border=2> <tr> <td> <font color=red> ATTEN <br> TION!</font> </td> <td> tute0008.htm#Cauchy_6 box9 limit matrix rank up to 10 <br><a href=tute0009.htm#Cauchy_6><font color=red> tute0009.htm#Cauchy_6</font></a> <font color=red>has no limit and more function.</font> </td> </tr> </table> Program result <font color=red>square rooted</font>, textbook squared. <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> Program has three functions. <br> First, <a href="tute0009.htm#ch10h008">Regular Cauchy Inequality.</a> <input type="button" value="Regular Cauchy" onclick="javascript:goCauchy6()" /> <br> Second, <a href="tute0009.htm#ch10h004">Generalized Cauchy Inequality</a> <input type="button" value="Generalized Cauchy" onclick="javascript:genCauchy()" /> 0 <br> Third, Matrix eigenvalue and eigenvector <INPUT onclick='eigensMan()' type=button value="eigenvalue eigenvector"> <br> <font color=red> This program accept symmetric matrix. Box9 read lower triangle matrix. </font> <br> Both second and third made note around box9 indicate where is input and output. <br><font color=red> If box1 to box6 all filled data, program handle six array regular Cauchy Inequality. <br> If box4 is empty, program handle three array regular Cauchy Inequality. <br> Please fill array numbers to box1 to box6, for test purpose, you can click 'random number' <br> Regular Cauchy Inequality <b>***** output to box7 *****</b> below.</font> <br> Random range <INPUT name="RadiocA1" type="radio" onclick="boxcA4.value=0" checked >less than one, <INPUT name="RadiocA1" type="radio" onclick="boxcA4.value=1" >less than ten, <br> or <!-- <INPUT name="RadiocA1" type="radio" >10.~^ --> 10^ <INPUT value="0" id=boxcA4 size=5 onchange="RadiocA1[0].checked=RadiocA1[1].checked=0" > ; random number <INPUT name="RadiocA2" type="radio" >+/0 , <INPUT name="RadiocA2" type="radio" checked >+/0/- <br> random number has <INPUT value="3" id=boxcA3 size=5 > digits ; <input type="checkbox" name="chkbxcA1" id="chkbxcA1" /> integer only<!--9806111025--> </form> Each seq. has <INPUT value="5" id=boxcA1 size=5 > numbers <input type="button" value="random number" onclick="javascript:for(var d0=11;d0<17;d0++)eval('fillRand(\'boxc'+d0+'.value\')')" /> <!-- 9806101618 use boxc11.value and boxc12.value --> <input type="button" value="Regular Cauchy" onclick="javascript:goCauchy6()" /> <br> If sequence 1 proportional to sequence 2, inequality become equality. <br> proportional <input type="button" value="ratio" onclick="javascript:eqCauchy(1)" /> = <INPUT value="0" id=boxcA2 size=5 onchange="javascript:eqCauchy(0)" > (box2 change) <br> <table border=0> <a name="ch01tb01"></a> <tr> <td> Box1 &nbsp; <INPUT onclick='javascript:window.clipboardData.setData("Text",document.getElementById("boxc11").value.toString())' type=button value="copy1"> <INPUT onclick='document.getElementById("boxc11").value=""' type=button value="del1"> <br><!--//9806271232 use Box1 --> <TEXTAREA id=boxc11 name=boxc11 rows=6 cols=18 > 1 2 3 4 5 </TEXTAREA> </td> <!-- 1.1824 3.5030 4.6972 2.6731, 0.2926e3;.9243 1.7e2 2.956 8.7575 11.743 6.6827 731.5 2.3107 425 --> <td> Box2 &nbsp; <INPUT onclick='javascript:window.clipboardData.setData("Text",document.getElementById("boxc12").value.toString())' type=button value="copy2"> <INPUT onclick='document.getElementById("boxc12").value=""' type=button value="del2"> <br> <TEXTAREA id=boxc12 name=boxc12 rows=6 cols=18 > 6 7 8 9 10 </TEXTAREA> </td> <td> <a name="ch01a041"></a> Box3 &nbsp; <INPUT onclick='javascript:window.clipboardData.setData("Text",document.getElementById("boxc13").value.toString())' type=button value="copy3"> <INPUT onclick='document.getElementById("boxc13").value=""' type=button value="del3"> <br> <TEXTAREA id=boxc13 name=boxc13 rows=6 cols=18 > 0.556 -0.145 0.308 -0.305 -0.088 </TEXTAREA> </td> </tr> <tr> <td> Box4 &nbsp; <INPUT onclick='javascript:window.clipboardData.setData("Text",document.getElementById("boxc14").value.toString())' type=button value="copy4"> <INPUT onclick='document.getElementById("boxc14").value=""' type=button value="del4"> <br> <TEXTAREA id=boxc14 name=boxc14 rows=6 cols=18 > -4.82 -7.74 1.51 -3.31 1.47 </TEXTAREA> </td> <td> Box5 &nbsp; <INPUT onclick='javascript:window.clipboardData.setData("Text",document.getElementById("boxc15").value.toString())' type=button value="copy5"> <INPUT onclick='document.getElementById("boxc15").value=""' type=button value="del5"> <br> <TEXTAREA id=boxc15 name=boxc15 rows=6 cols=18 > 3.6 7.94 2.94 -7.73 -7.48 </TEXTAREA> </td> <td> Box6 &nbsp; <INPUT onclick='javascript:window.clipboardData.setData("Text",document.getElementById("boxc16").value.toString())' type=button value="copy6"> <INPUT onclick='document.getElementById("boxc16").value=""' type=button value="del6"> <br> <TEXTAREA id=boxc16 name=boxc16 rows=6 cols=18 > 3.49 0.01 9.78 7.58 1.03 </TEXTAREA> </td> </tr> </table> <input type="button" value="random number" onclick="javascript:for(var d0=11;d0<17;d0++)eval('fillRand(\'boxc'+d0+'.value\')')" /> <!--9806280814 change OK onclick="javascript:fillRand('boxc11.value'),fillRand('boxc12.value'),fillRand('boxc13.value'),fillRand('boxc14.value'),fillRand('boxc15.value'),fillRand('boxc16.value')" /> --> &nbsp; <input type="button" value="Regular Cauchy" onclick="javascript:goCauchy6()" /> <br> Example <input type="button" value="1" onclick='javascript:boxc11.value="1 2 3",boxc12.value="4 5 6",boxc13.value="7 8 9",boxc14.value="10 11 12",boxc15.value="13 14 15",boxc16.value="16 17 18",goCauchy6()' /> <input type="button" value="2" onclick="javascript:for(var d0=11;d0<17;d0++)eval('boxc'+d0+'.value=\'2 0 0\''),goCauchy6()" /> <input type="button" value="3" onclick="javascript:for(var d0=11;d0<17;d0++)eval('boxc'+d0+'.value=\'0 0 '+(7-d0%10)+'\''),goCauchy6()" /> <input type="button" value="4" onclick="javascript:for(var d0=11;d0<17;d0++)eval('boxc'+d0+'.value=\'2 1 0\''),goCauchy6()" /> 2 and 3 are equality. <font color=red>output to box7</font> <table border=0> <a name="ch01tb02"></a> <tr> <td> Box 7, answer &nbsp; <INPUT onclick='javascript:window.clipboardData.setData("Text",document.getElementById("boxc17").value.toString())' type=button value="copy7"> <INPUT onclick='document.getElementById("boxc17").value=""' type=button value="del7"> <br> <TEXTAREA id=boxc17 name=boxc17 rows=6 cols=30 > </TEXTAREA> </td> <td> Box 8, debug &nbsp; <INPUT onclick='javascript:window.clipboardData.setData("Text",document.getElementById("boxc18").value.toString())' type=button value="copy8"> <INPUT onclick='document.getElementById("boxc18").value=""' type=button value="del8"> <br> <TEXTAREA id=boxc18 name=boxc18 rows=6 cols=30 > </TEXTAREA> </td> </tr> </table> <a name="gCauchy"></a> <font color=red size=+2> Output may contain error, Please verify first. <br> Program environment is MSIE 6.0, please use MSIE </font> <br> Gen. Cauchy Inequality accept two number sequences. Define them in box1 & 2. <br> Reg. Cauchy Inequality accept six number sequences. Define them in box1 to 6. <br> If delete box3, then Reg. Cauchy accept two number sequences. When on equal <br> foot, it is possible to compare Gen. Cauchy and Reg. Cauchy results. <br> <input type="button" value="Regular Cauchy" onclick="javascript:goCauchy6()" /> <INPUT onclick='genCauchy()' type=button value="Generalized Cauchy"> , <INPUT onclick='testGenCauchy(0)' type=button value="debug1"> , Example<INPUT onclick='testGenCauchy(1)' type=button value="A"> <INPUT onclick='testGenCauchy(2)' type=button value="B"> <INPUT onclick='testGenCauchy(3)' type=button value="C"> <table border=0> <a name="ch01tb03"></a> <tr> <td> Box 9 define lower triangle matrix <br> <font color=red>not full matrix</font> <INPUT onclick='javascript:window.clipboardData.setData("Text",document.getElementById("boxc19").value.toString())' type=button value="copy9"> <INPUT onclick='document.getElementById("boxc19").value=""' type=button value="del9"> <br> <TEXTAREA id=boxc19 name=boxc19 rows=6 cols=30 > 1 0 1 0 0 1 0 0 0 1 0 0 0 0 1 </TEXTAREA><!--9807020947 use box9--> </td> <!-- 1 a b c d 0 1 e f g 0 0 1 h i 0 0 0 1 j 0 0 0 0 1 --> <td> <span id=spantb03> Gen. Cauchy handle two seq.<br> define them in box1 and 2<br> box9 define lower triangle<br> matrix. Output right here.<br> If eigenvalue<0, Inequality<br> reverse direction.9807021018 </span> </td> </tr> </table> <table> <tr> <td> <font color=red> ATTEN <br> TION!</font> </td> <td> tute0008.htm#Cauchy_6 box9 limit matrix rank up to 10 <br><a href=tute0009.htm#Cauchy_6><font color=red> tute0009.htm#Cauchy_6</font></a> <font color=red>has no limit and more function.</font> </td> </tr> </table> <table> <tr> <td> <INPUT onclick='eigensMan()' type=button value="eigenvalue eigenvector"> </td> <td> <font color=red> Box9 input, click left [eigen] button, box8 output;<br> if eigenvalue<0, Gen. Cauchy Inequality reverse. </font> </td> <tr> </table> Regular Cauchy Inequality use identity matrix. Its diagonal elements are all one, other elements are all zero. <br> Cauchy Inequality: <font color=red size=+1><b> [a][M][b]f""{[a][M][a]}*"{[b][M][b]} </b></font> <br><font color=red> Generalized Cauchy Inequality use any symmetrical matrix. </font> Require all three [a][M][b], [a][M][a], [b][M][b] multiplication use SAME matrix. <br><font color=red> If three matrix in one equation are not the same matrix, then it is not Generalized Cauchy Inequality. </font> <br> [debug1] button create a matrix use elements i,j coordinates as element value. <br> Example [A] create <a href=tute0008.htm#docA02>textbook</a> exercise 1.7 matrix. <br> Example [B] create textbook page 63 light-cone inequality matrix. <br> Box 1&2 first element is one, other elements are less than one for light speed is the upper limit. <br> Fill data in box 1&2 then click [Gen. Cauchy] button. 9807021908 <br> Example [C] show that lower triangle matrix definition all numbers can be in just one line. No need to make it triangle shape. 9807022101 <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> <SCRIPT language=JavaScript> <!-- /** 2009-08-03-14-38 start ÿ9808031438 ÿ Other file http://freeman2.com/tutc0009.htm This file http://freeman2.com/tutc0008.htm Other file tutc0009.htm has all code of this file This file tutc0008.htm short many code of other file Other file tutc0009.htm function is better than this file This file tutc0008.htm function is not as complete as other file Other file tutc0009.htm has no limit to matrix size. This file tutc0008.htm limit matrix size up to rank 10 Other file tutc0009.htm reorder eigenvalue from large to small This file tutc0008.htm first eigenvalue may not be largest number Other file tutc0009.htm has code to verify matrix-eigenvalue-eigenvector This file tutc0008.htm no verification code Other file tutc0009.htm will build Hilbert matrix This file tutc0008.htm not build Hilbert matrix Other file tutc0009.htm has lower triangle element number list This file tutc0008.htm no lower triangle element number list Other file tutc0009.htm has reorder number sequence test box This file tutc0008.htm no reorder number sequence test box Please use other file tutc0009.htm program. 2009-08-03-14-47 /**/ /** 9807271639 both regular Cauchy Inequality and Generalized Cauchy Inequality use [v][M][w]f""{[v][M][v]}*"{[w][M][w]} If matrix [M] is identity matrix, it is regular Cauchy Inequality, otherwise it is Generalized Cauchy Inequality function genCauchy() cauculate Generalized Cauchy Inequality find [v][M][w] and "{[v][M][v]}*"{[w][M][w]} vdotw=genDot(mat0,inNum[0],inNum[1]) vector [v] is inNum[0] vector [w] is inNum[1] matrix [M] mat0 is lower triangle matrix [v][M][w] is genDot() main function If [M] is positive definite matrix, for non zero [v] , [v][M][v]>0 If [M] is not positive definite matrix, for non zero [v] , [v][M][v] >0 =0 <0 all possible. if [v][M][v] and [w][M][w] both >0 print regular answer. Otherwise print red alert answer. 9807271654 /**/ function genCauchy() //9807020951 { //Generalized Cauchy Inequality var maxSeq=2; var inNum=new Array(maxSeq) inNum[0]=readdata(boxc11.value); //9807020952 inNum[1]=readdata(boxc12.value); var mat0=readdata(boxc19.value); var v0size=inNum[0].length; var v1size=inNum[1].length; var m0size=mat0.length; //9807020959 boxc18.value='' +'Generalized Cauchy Inequality\n' +'output to box9 righthand side\n' +'Box8 is debug output.\n' +'v0size='+v0size+'\n' +'v1size='+v1size+'\n' +'m0size='+m0size+'\n' +'9807021001\n' var vdotw=genDot(mat0,inNum[0],inNum[1]) if((vdotw+'').charAt(0)=='e') { return; //9807022004 } var vdotv=genDot(mat0,inNum[0],inNum[0]) if((vdotv+'').charAt(0)=='e') { return; } var wdotw=genDot(mat0,inNum[1],inNum[1]) if((wdotw+'').charAt(0)=='e') { return; } var vw=wdotw*vdotv; if(vdotv>=0 && wdotw>=0) { vdotv=Math.sqrt(vdotv); wdotw=Math.sqrt(wdotw); //9807021306 vw=Math.sqrt(vw); spantb03.innerHTML='' +'vector1 length ÿ0'+vdotv+'<br>' +'vector2 length ÿ0'+wdotw+'<br>' +'v1_len * v2_lenÿ '+vw+'<br>' +'v1_dot_v2 ÿ '+vdotw+'<br>' +(vw>=vdotw?'v1_len * v2_leng"dot value': '<font color=red>Alert ! v1_len * v2_lenÿdot value</font>' ) } else { spantb03.innerHTML='' //9807021338 +'vector1 <font color=red>length_square</font>ÿ0'+vdotv+'<br>' +'vector2 <font color=red>length_square</font>ÿ0'+wdotw+'<br>' +'vector1&2 <font color=red>product_square</font>ÿ'+vw+'<br>' +'vector1&2 dot_squareÿ('+vdotw+')^2<br>' +(vw>=vdotw*vdotw?'vector1&2 productg"vector1&2 dot': '<font color=red>alert ! v1&2 productÿv1&2 dot</font>' ) } boxc18.value+='\n' +'If want use second method to\n' +'verify answer, try next page\n' +'real, complex calculator:\n' +'http://freeman2.com/complex2.htm\n' +'ÿcomplex2.htm box3 ÿ' } // function genCauchy() function genDot(mat0,vec0,vec1) //9807021020 { var v0size=vec0.length; var v1size=vec1.length; var m0size=mat0.length; //9807021022 var oStr0=''; var oStr1=''; //9807021618 var oStr2=''; //9807021637 var s0,s1,s2,s3; var diagLoc=[0,0]; //9807021159 for(s0=0,s1=1;s0<60;s1++) { s0+=s1; diagLoc[s1-1]=s0-1; //9807021200 if(m0size==s0)break; //9807021031 oStr0+=s0+'\n' } oStr0+=s0+'\n'; if(m0size!=s0 ||s0<3 ||s0>55 ) { spantb03.innerHTML='' +'Symmetric matrix must use lower<br>' +'triangle matrix. Total number='+m0size+'<br>' +'But program accept only 3,6,10,15,<br>' +'21,28,36,45,55. Suggest delete box3<br>' +'Run with regular Cauchy Ineq.<br>' ; /** boxc13.value=''; goCauchy6(); /**/ return 'err01'; } var sqMatDim=s1; //9807021050 if(v0size<sqMatDim) { //9807021055 spantb03.innerHTML='' +'Vector 1 dimension='+v0size+'<br>' +'less than matrix rank='+sqMatDim+'<br>' +'Can not calculate.<br>' return 'err02'; } if(v1size<sqMatDim) { //9807021057 spantb03.innerHTML='' +'Vector 2 dimension='+v1size+'<br>' +'less than matrix rank='+sqMatDim+'<br>' +'Can not calculate.<br>' return 'err03'; } var dbg0=''; //9807021104 var ans0=0; //9807021059 var big0,sml0; //9807021251 var dbg1=''; //9807021642 for(s0=0;s0<sqMatDim;s0++) { for(s1=0;s1<sqMatDim;s1++) { //9807021100 if(s0==s1) { oStr1+=mat0[diagLoc[s0]]+', '; //9807021620 debug dbg1+='+('+vec0[s0]+')*('+mat0[diagLoc[s0]]+')*('+vec1[s1]+')'; //9807021640 debug ans0+= //9807021254 vec0[s0]*mat0[diagLoc[s0]]*vec1[s1]; continue; } big0=s0; //s0=i sml0=s1; //s1=j if(s1>s0){big0=s1;sml0=s0;} //9807021256 oStr1+=mat0[diagLoc[big0]-big0+sml0]+', '; //9807021621 debug dbg1+='+('+vec0[s0]+')*('+mat0[diagLoc[big0]-big0+sml0]+')*('+vec1[s1]+')'; //9807021641 debug ans0+= //9807021257 vec0[s0]*mat0[diagLoc[big0]-big0+sml0]*vec1[s1]; } oStr1+='\n'; //9807021622 debug oStr2+='\nans0 '+bye09(ans0) //9807022009 bye09() +' include the following new add number\n'+dbg1+'\n'; dbg1=''; //9807021644 debug } boxc18.value+='\n*****\n' +oStr0+'\n' +oStr1+'\n' +oStr2+'\n' +'Above output string to one line\nEasier for paste to box for calculation\n' +'Following output in sequence order\nEasier for visual check.\n' +oStr2.replace(/\+/g,'\n\+') return ans0; //9807021304 } // function genDot(mat0,vec0,vec1) function testGenCauchy(tgArg1) //9807021626 { if(tgArg1==1) { //9807021726 boxc11.value='1 -2'; boxc12.value='6\n7\n'; boxc19.value='5\n1 3\n'; genCauchy(); } else if(tgArg1==2) { //9807021858 boxc11.value='1 0.2 0.3 0.4 0.5'; boxc12.value='1\n0.6\n-0.24\n0.35\n0.2\n'; boxc19.value='1\n0 -1\n0 0 -1\n0 0 0 -1\n0 0 0 0 -1\n'; genCauchy(); } else if(tgArg1==3) { //9807022057 boxc11.value='1.1 2.8 3.6'; boxc12.value='4\n-5\n8.1\n'; boxc19.value='5.35 1.2 3.9 1 -1.5 4.6'; genCauchy(); } else { //9807021628 boxc11.value='1 -2 3 4 5'; boxc12.value='6\n7\n-8\n9\n10\n'; boxc19.value='11\n12 22\n13 23 33\n14 24 34 44\n15 25 35 45 55\n'; genCauchy(); } } //9807021630 function goCauchy6() //9806271256 { var ou7Str=''; var ou8Str=''; var ou9Str=''; var w0=w1=w2=w3=w4=w5=0.; var warn0=''; //9806102207 var w6,w7; //9806271322 var maxSeq=6; var inNum=new Array(maxSeq) inNum[0]=readdata(boxc11.value); //9806101628 inNum[1]=readdata(boxc12.value); inNum[2]=readdata(boxc13.value); //9806271257 inNum[3]=readdata(boxc14.value); inNum[4]=readdata(boxc15.value); inNum[5]=readdata(boxc16.value); var sum6=[0,0,0,0,0,0]; //9806271724 var inLen=new Array(maxSeq); //9806271329 inLen[0]=inNum[0].length; inLen[1]=inNum[1].length; //9806101630 inLen[2]=inNum[2].length; inLen[3]=inNum[3].length; inLen[4]=inNum[4].length; inLen[5]=inNum[5].length; var sum1=sum2=sum3=sum4=sum5=0; //9806271648 w1=0; //9806271631 for(w0=0;w0<inLen.length;w0++) { //decide maxSeq value if(inLen[w0]==0) { maxSeq=w0; if(maxSeq<2) { boxc17.value='' +'Please fill number sequence from box1 to box6' return; } } } // for(w0=0;w0<inLen.length;w0++) sum2=0; // ineq. less than side a1b1c1+a2b2c2+... sum for(w1=0;w1<inNum[0].length;w1++) { sum1=1; // ineq. less than side ak*bk*ck product for(w2=0;w2<maxSeq;w2++) //9806271705 { if(isNaN(inNum[w2][w1])) { inNum[w2][w1]=0; //9806271645 warn0='sequence '+(w2+1)+' too short, fill with zero, output may not be useful.\n' } ou8Str+='' //9806271323 +inNum[w2][w1]+'\t' ou7Str+='' +'sum1*inNum[w2][w1]='+sum1 +'*('+inNum[w2][w1]+')\n' //ineq. less than side ak*bk*ck product //sum1 use '*' in next line sum1=sum1*inNum[w2][w1]; //9806271651 } //for(w2=0;w2<maxSeq;w2++) //ineq. less than side a1b1c1+a2b2c2+... sum //sum2 use '+' in next line sum2+=sum1; ou7Str+='\n' +'ineq. less than side ak*bk*ck product ...\n' +'up to now, product sum1='+sum1+'\n' +'ineq. less than side a1*b1*c1+...+ak*bk*ck\n' +'up to now, summation sum2='+sum2+'\n============\n' ou8Str+='\n' } // for(w1=0;w1<inNum[0].length;w1++) //sum3 is ineq. greater than side //a_len*b_len*c_len product sum3=1; for(w2=0;w2<maxSeq;w2++) { sum6[w2]=0; for(w1=0;w1<inNum[0].length;w1++) sum6[w2]=sum6[w2]+inNum[w2][w1]*inNum[w2][w1] //above sum6[w2] is sequence_a a1*a1+a2*a2+a3*a3 //Below is sequence_a length "(a1*a1+a2*a2+a3*a3) sum6[w2]=Math.sqrt(sum6[w2]); //sum3 is sequence_a length * seq._b len * seq._c len sum3=sum3*sum6[w2]; ou9Str+='' +'w2='+w2+'\n' +'sequence_'+(w2+1)+' length sum6[w2]='+sum6[w2]+'\n' +'sequence length product up to here sum3='+sum3+'\n' +'========\n' } boxc18.value='' +'If input l2 keep 2 drop l(L)\n' +'Handle next '+maxSeq+' sequences, is it right?\n' +ou8Str +'\n\ninequality greater than side' +' sum3 is as following\n' //9806272135 +ou9Str +'\n' +'If sequence #1 has only\n' +'five elements, other seq.\n' +'read only five elements.\n' +'fill with zero if short.\n' boxc17.value='' +warn0 //9806102209 +'sum2='+sum2+'\n' +'sum3='+sum3+'\n' +'Cauchy said: sum2 f" sum3\n' +'\n' +'Inequality less than side\n' +'sum2 is as following\n\n' //9806272130 +ou7Str //9806271657 return; } 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; for(w0=0;w0<inStr0.length;w0++) { numb0=''; //skip those char are NOT [0123456789-+.] // [!(...)] means [is not ...] // [!=] means [not equal] 9807271051 // w1=w0; //9802172117 while(!(inStr0.charAt(w1)>='0' &&inStr0.charAt(w1)<='9') &&inStr0.charAt(w1)!='-' &&inStr0.charAt(w1)!='+' &&inStr0.charAt(w1)!='.' &&w1<inStr0.length ) { //index w1 increase only, //no record what read 9807271058 w1++; } //above while() loop stop at reading //either one in the list [0123456789-+.] //then start next while loop. 9807271055 // while((inStr0.charAt(w1)>='0' &&inStr0.charAt(w1)<='9') ||inStr0.charAt(w1)=='-' ||inStr0.charAt(w1)=='+' ||inStr0.charAt(w1)=='.' ||inStr0.charAt(w1)=='e' ||inStr0.charAt(w1)=='E' &&w1<inStr0.length ) //9802171608 { //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)=='.') { w0=w1; //9802172218 //update index w0=w1 continue; //continue to next iteration. } //skip the following building code. //up to this point, numb0 is useful number //build output string oArray //9807271111 w0=w1; //9806101210 update index //string numb0 to output string oArray //add a newline byte '\n' to separate data // 9807271113 oArray+=numb0+'\n'; //9806101159 } // for(w0=0;w0<inStr0.length;w0++) //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 function fillRand(ouBoxID) //9806101238 { 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 precision0; //9806101257 precision0=parseInt(document.form1.boxcA3.value); //9806101346 if(document.form1.chkbxcA1.checked) precision0=powIndex; //9806111156 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; var dataPerS=parseInt(boxcA1.value); //9806101246 var DecInt=0; //9806111054 if(document.form1.chkbxcA1.checked)DecInt=1; var ouStr1=''; var ouStr2=''; var t0,t1,t2; 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'; } eval(ouBoxID+'=ouStr1'); //9806101542 OK } //9806101352 function fillRand(ouBoxID) /** [[ When sequence 1 proportional to sequence 2, Cauchy inequality become equality. Please give ratio = [0 ] Box 2 change ]] If change ratio number, program call eqCauchy() 9806171917 here /**/ function eqCauchy(EQarg) //9806101738 { //Please give seq2/seq1 value var inStr7=readdata(boxc11.value); //9806101739 var ratio0=parseFloat(boxcA2.value); //9806101747 var w0,w1,w2; w1=''; for(w0=0;w0<inStr7.length;w0++) { w2=''; //9806172109 if(ratio0*inStr7[w0]>=0)w2=' '; //w1+=ratio0*inStr7[w0]+'\n'; //9806101742 if(EQarg!=0) //9806210911 use EQarg show bye09() difference w1+=w2+ //9806172110 use w2 bye09(ratio0*inStr7[w0])+'\n'; //9806171858 bye09() else w1+=w2+ratio0*inStr7[w0]+'\n'; } boxc12.value=w1; //9806101748 //boxc12.value=bye09(w1); //9806171830 //It is an error to call bye09() here, //because here w1 is not a simple number, // w1 is a string of several numbers. //call bye09() at time stamp 9806171858 //that is a correct place. //9806172103 record } /** 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' ÿ var zeroLoc=(in09+'').indexOf('000000',dotLoc); // find '000000' ÿcut start '000000' if(zeroLoc>0) //9806121216 return parseFloat((in09+'').substring(0,zeroLoc)); //9806121209 // if number has '999999' ÿ var nineLoc=(in09+'').indexOf('999999',dotLoc); // find '999999' ÿ 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 eigensMan( A, RR, E, N ) //9807032110 { var mat0=readdata(boxc19.value); var m0size=mat0.length; //9807021022 var s0,s1,s2,s3; var diagLoc=[0,0]; //9807021159 for(s0=0,s1=1;s0<60;s1++) { s0+=s1; diagLoc[s1-1]=s0-1; //9807021200 if(m0size==s0)break; //9807021031 } if(m0size!=s0 ||s0<3 ||s0>55 ) { spantb03.innerHTML='' +'Matrix manager: must use lower<br>' +'triangle matrix. total number='+m0size+'<br>' +'Program accept following total 3,6,<br>' +'10,15,21,28,36,45,55. Can not find<br>' +'eigenvalue & eigenvector. 9807032126' return 'err01'; } var sqMatDim=s1; //9807021050 var RR=new Array(sqMatDim) var E=new Array(sqMatDim) for(s0=0;s0<sqMatDim;s0++) { //9807032129 RR[s0]=new Array(sqMatDim); } // ans0=eigens( A, RR, E, N ); var ans0=eigens( mat0, RR, E, sqMatDim ); var eigVal=ans0[0]; var eigVec=ans0[1]; //9807032116 boxc18.value=''; //boxc18.value+=''+ans0+'\n\n'+eigVal+'\n\n'+eigVec+'\n\n' boxc18.value+='' +'Symmetric matrix has '+sqMatDim+' eigenvalues\n' for(s0=0;s0<sqMatDim;s0++) { //9807032230 boxc18.value+='' +eigVal[s0]+' \n' } boxc18.value+='' +'\n' for(s0=0;s0<sqMatDim;s0++) { //9807032129 boxc18.value+='' +'eigenvalue='+eigVal[s0]+' has eigenvector\n' for(s1=0;s1<sqMatDim;s1++) //9807032141 //boxc18.value+=''+eigVec[s0][s1]+', '; //undefined boxc18.value+='' +eigVec[s0*sqMatDim+s1]+' \n'; //9807032143 boxc18.value+='' +'\n' } } /* eigens.c * * Eigenvalues and eigenvectors of a real symmetric matrix * * * * SYNOPSIS: * * int n; * double A[n*(n+1)/2], EV[n*n], E[n]; * void eigens( A, EV, E, n ); * * * * DESCRIPTION: * * The algorithm is due to J. vonNeumann. * * A[] is a symmetric matrix stored in lower triangular form. * That is, A[ row, column ] = A[ (row*row+row)/2 + column ] * or equivalently with row and column interchanged. The * indices row and column run from 0 through n-1. * * EV[] is the output matrix of eigenvectors stored columnwise. * That is, the elements of each eigenvector appear in sequential * memory order. The jth element of the ith eigenvector is * EV[ n*i+j ] = EV[i][j]. * * E[] is the output matrix of eigenvalues. The ith element * of E corresponds to the ith eigenvector (the ith row of EV). * * On output, the matrix A will have been diagonalized and its * orginal contents are destroyed. * * ACCURACY: * * The error is controlled by an internal parameter called RANGE * which is set to 1e-10. After diagonalization, the * off-diagonal elements of A will have been reduced by * this factor. * * ERROR MESSAGES: * * None. * 9807041118 start Above is document of cephes27\double\eigens.c from cephes27.zip It was C language code. The following change to Javascript code. 9807041120 here function eigens( A, RR, E, N ) work only for symmetric matrix. argument A store lower triangle matrix argument N store matrix rank number Output use next line return [E,RR]; //9807032114 Javascript language must use next line N=parseInt(N); //9807031927 original C language code no need above line. because C language do not allow variable type change. 9807041133 */ function eigens( A, RR, E, N ) //9804141912 { var IND, L, LL, LM, M, MM, MQ, I, J, IA, LQ; var IQ, IM, IL, NLI, NMI; var ANORM, ANORMX, AIA, THR, ALM, ALL, AMM, X, Y; var SINX, SINX2, COSX, COSX2, SINCS, AIL, AIM; var RLI, RMI; var RANGE = 1.0e-10; /*3.0517578e-5;*/ var done01=0; //9804141917 var RR=[0,0]; N=parseInt(N); //9807031927 /* Initialize identity matrix in RR[] */ for( J=0; J<N*N; J++ ) RR[J] = 0.0; MM = 0; for( J=0; J<N; J++ ) { RR[MM + J] = 1.0; MM += N; } ANORM=0.0; for( I=0; I<N; I++ ) { for( J=0; J<N; J++ ) { if( I != J ) { IA = I + (J*J+J)/2; AIA = A[IA]; ANORM += AIA * AIA; } } } if( ANORM <= 0.0 ) // goto done; done01=1; //9804141918 if(!done01) //9804141919 { //ANORM = sqrt( ANORM + ANORM ); ANORM = Math.sqrt( ANORM + ANORM ); ANORMX = ANORM * RANGE / N; THR = ANORM; while( THR > ANORMX ) { THR=THR/N; do { /* while IND != 0 */ IND = 0; for( L=0; L<N-1; L++ ) { for( M=L+1; M<N; M++ ) { MQ=(M*M+M)/2; LM=L+MQ; ALM=A[LM]; // if( fabs(ALM) < THR ) if( Math.abs(ALM) < THR ) continue; //Math. IND=1; LQ=(L*L+L)/2; LL=L+LQ; MM=M+MQ; ALL=A[LL]; AMM=A[MM]; X=(ALL-AMM)/2.0; // Y=-ALM/sqrt(ALM*ALM+X*X); Y=-ALM/Math.sqrt(ALM*ALM+X*X); if(X < 0.0) Y=-Y; // SINX = Y / sqrt( 2.0 * (1.0 + sqrt( 1.0-Y*Y)) ); SINX = Y / Math.sqrt( 2.0 * (1.0 + Math.sqrt( 1.0-Y*Y)) ); SINX2=SINX*SINX; // COSX=sqrt(1.0-SINX2); COSX=Math.sqrt(1.0-SINX2); COSX2=COSX*COSX; SINCS=SINX*COSX; /* ROTATE L AND M COLUMNS */ for( I=0; I<N; I++ ) { IQ=(I*I+I)/2; if( (I != M) && (I != L) ) { if(I > M) IM=M+IQ; else IM=I+MQ; if(I >= L) IL=L+IQ; else IL=I+LQ; AIL=A[IL]; AIM=A[IM]; X=AIL*COSX-AIM*SINX; A[IM]=AIL*SINX+AIM*COSX; A[IL]=X; } NLI = N*L + I; NMI = N*M + I; RLI = RR[ NLI ]; RMI = RR[ NMI ]; RR[NLI]=RLI*COSX-RMI*SINX; RR[NMI]=RLI*SINX+RMI*COSX; } X=2.0*ALM*SINCS; A[LL]=ALL*COSX2+AMM*SINX2-X; A[MM]=ALL*SINX2+AMM*COSX2+X; A[LM]=(ALL-AMM)*SINCS+ALM*(COSX2-SINX2); } /* for M=L+1 to N-1 */ } /* for L=0 to N-2 */ } while( IND != 0 ); } /* while THR > ANORMX */ } // if(!done01) //9804141920 // done: ; //9804141921 /* Extract eigenvalues from the reduced matrix */ L=0; for( J=1; J<=N; J++ ) { L=L+J; E[J-1]=A[L-1]; } return [E,RR]; //9807032114 } //9804141958 // function eigens( A, RR, E, N ) //9804141912 //--> </SCRIPT> <pre><font size=+2><a name="ch01b053">&lt;a name="ch01b053"&gt;</a> 2009-06-27-19-28 start Above is six sequence Cauchy Inequality, allow six sequence at one time. Please pay attention to the following First, If sequence one has five elements other sequence read only five elements. discard the remaining. If other sequence do not have five elements, program fill with zero. Second, if work with two sequences, let box3 be empty. Third, three sequences or more, the condition for Cauchy inequality to become equality is <a href="tute0008.htm#ch01b058"><font color=red>tight</font></a>. <a name="ch01b054">&lt;a name="ch01b054"&gt;</a> Explain third point as following.</font></pre> <script lang="javascript"> <!-- HelloCauchy(107) //--> </script> <br> Above is two sequences Cauchy Inequality. <br> Below is three sequences Cauchy Inequality. <TABLE WIDTH="620" id=T013L14c border=0 ><!--9806271940--> <TR> <TD> &nbsp; </TD> <TD> <TABLE WIDTH="420" border=0 id=T013L14d > <TR> <TD> <CENTER><FONT SIZE=+3>[</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>k=n</FONT></CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER><FONT SIZE=-2>k=1</FONT></CENTER> </TD> <TD> <CENTER> a<sub><FONT SIZE=-1> k</FONT></sub> b<sub><FONT SIZE=-1> k</FONT></sub> c<sub><FONT SIZE=-1> k</FONT></sub> </CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>]</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>2</FONT></CENTER> <CENTER><FONT SIZE=+3>00</FONT></CENTER> <CENTER><FONT SIZE=-2>00</FONT></CENTER> </TD> <TD> f" </TD> <TD> <CENTER><FONT SIZE=-2>k=n</FONT></CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER><FONT SIZE=-2>k=1</FONT></CENTER> </TD> <TD> <CENTER>a<sub><FONT SIZE=-1> k</FONT></sub><sup>2</sup> </CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>k=n</FONT></CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER><FONT SIZE=-2>k=1</FONT></CENTER> </TD> <TD> <CENTER>b<sub><FONT SIZE=-1> k</FONT></sub><sup>2</sup> </CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>k=n</FONT></CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER><FONT SIZE=-2>k=1</FONT></CENTER> </TD> <TD> <CENTER>c<sub><FONT SIZE=-1> k</FONT></sub><sup>2</sup> </CENTER> </TD> </TR> </TABLE> </TD> <TD> ---page 13 line 14 ref.1 </TD> </TR> </TABLE> width of above equation <INPUT name=Button013L14c type="button" value="default value" onclick="B013L14c.value=620,B013L14d.value=420,T013L14c.width=B013L14c.value,T013L14d.width=B013L14d.value"> <input id="B013L14c" value=620 size=3 onchange=T013L14c.width=B013L14c.value /> <input id="B013L14d" value=420 size=3 onchange=T013L14d.width=B013L14d.value /> <pre><font size=+2><a name="ch01b055">&lt;a name="ch01b055"&gt;</a>0<a href="#index">Index begin</a>0<a href="#index06">Index this file</a> 2009-06-27-19-43 here Two sequences Cauchy Inequality ask both sequence a and b to be proportional to get equality. To simplify work, let a=b (proportional constant is one) to get equality.<font color=red> If you test, click [reg. Cauchy] button do not click [Gen. Cauchy] button.<!--9807301048--></font> When come to three sequences Cauchy Inequality, still require a=b=c, Can we get equality from a=b=c? First let us treat ab as one unit, (ab sequence become c*c sequence) is it clear that sequence c*c and sequence c are NOT in proportion. For example, let a=b=c=[1,2,3] Then a*b=[1*1, 2*2, 3*3]=[1,4,9] c*c=a*b=[1,4,9] and c=[1,2,3] are NOT in proportion.<!--9807301057--> One square equal to one. If we use all one to build a, b, c sequence, a=b=c=[1, 1, 1] In this case seq. a*b and seq. c both are [1, 1, 1] , can we get equality? <a name="ch01b056">&lt;a name="ch01b056"&gt;</a> Three sequence Cauchy Inequality less than side is [1, 1, 1] dot [1, 1, 1] = 3 But Cauchy Inequality greater than side is vector length "3 multiply three times=3"3 3"3 not equal to 3 . We still can not get equality for three seq. Cauchy Inequality.<font color=red> Look like</font> <font color=silver>three sequences Cauchy Inequality is absolute inequality.</font> Textbook page 13 line 13 equation and line 14 equation both use "f"". Freeman is still learning, hope find solution. Reader please put question mark here. 2009-06-27-20-16 stop <a name="9807301400">&lt;a name="9807301400"&gt;</a><font color=red> All LiuHH's WRONG GUESS words are in gray (silver) color text.</font> 9807301400 <a name="ch01b057">&lt;a name="ch01b057"&gt;</a> 2009-06-28-09-46 start  %0When multiple sequence become equality? A simple observation: three sequence Cauchy Inequality " a<sub>k</sub> b<sub>k</sub> c<sub>k</sub> f" a_len * b_len * c_length If seq. a has one number A, then a_len = A If seq. b has one number B, then b_len = B If seq. c has one number C, then c_len = C In this case, equation both side are A*B*C. Equality exist. <a name="ch01b058">&lt;a name="ch01b058"&gt;</a> Three and more sequence Cauchy Inequality will be equality, if the following conditions are true. First: all sequence are single value. If sequence are multiple value, only one element is non zero. Second: all non zero element must have same index number in every sequence. For example, all sequence have non zero element as first element. All sequence non-first element are all zero. Six sequence Cauchy Inequality program, <a href="#ch01a041">one line above</a> [Box 7, answer] has Example 1,2,3,4 buttons. Button [2] and [3] both get equality. Please test. 2009-06-28-09-58 stop <hr> <a name="ch01b059">&lt;a name="ch01b059"&gt;</a>0<a href="#index">Index begin</a>0<a href="#index06">Index this file</a> <script language="javascript">alert0()</script> 2009-06-28-16-39 start  %0Exercise 1.4 problem statement 00textbook page 13 ÿSome help from symmetry ÿ Under many conditions, Cauchy Inequality with symmetry create surprise problems. The following are two such problems. Exercise 1.4A: Given x, y, z are all positive, prove</font></pre> <TABLE WIDTH="620" id=T013L20a border=0 ><!--9806281648--> <TR> <TD> &nbsp; </TD> <TD> <TABLE WIDTH="460" border=0 id=T013L20b > <TR> <TD> S= </TD> <TD> <CENTER><FONT SIZE=+3>[</FONT></CENTER> </TD> <TD> <CENTER>x+y</CENTER> <CENTER><hr color=black></CENTER> <CENTER>x+y+z</CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>]</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>1/2</FONT></CENTER> <CENTER><FONT SIZE=+3>00</FONT></CENTER> <CENTER><FONT SIZE=-2>00</FONT></CENTER> </TD> <TD> + </TD> <TD> <CENTER><FONT SIZE=+3>[</FONT></CENTER> </TD> <TD> <CENTER>x+z</CENTER> <CENTER><hr color=black></CENTER> <CENTER>x+y+z</CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>]</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>1/2</FONT></CENTER> <CENTER><FONT SIZE=+3>00</FONT></CENTER> <CENTER><FONT SIZE=-2>00</FONT></CENTER> </TD> <TD> + </TD> <TD> <CENTER><FONT SIZE=+3>[</FONT></CENTER> </TD> <TD> <CENTER>y+z</CENTER> <CENTER><hr color=black></CENTER> <CENTER>x+y+z</CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>]</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>1/2</FONT></CENTER> <CENTER><FONT SIZE=+3>00</FONT></CENTER> <CENTER><FONT SIZE=-2>00</FONT></CENTER> </TD> <TD> f" 6<sup>1/2</sup> </TD> </TR> </TABLE> </TD> <TD> ---page 13 line 20 </TD> </TR> </TABLE> width of above equation <INPUT name=Button013L20a type="button" value="default value" onclick="B013L20a.value=620,B013L20b.value=460,T013L20a.width=B013L20a.value,T013L20b.width=B013L20b.value"> <input id="B013L20a" value=620 size=3 onchange=T013L20a.width=B013L20a.value /> <input id="B013L20b" value=460 size=3 onchange=T013L20b.width=B013L20b.value /> <pre><font size=+2><a name="ch01b060">&lt;a name="ch01b060"&gt;</a> 2009-06-28-16-56 here Exercise 1.4B: Given x, y, z are all positive, prove</font></pre> <TABLE WIDTH="460" id=T013L22a border=0 ><!--9806281657--> <TR> <TD> &nbsp; </TD> <TD> <TABLE WIDTH="280" border=0 id=T013L22b > <TR> <TD> x+y+z f" 2* </TD> <TD> <CENTER><FONT SIZE=+3>[</FONT></CENTER> </TD> <TD> <CENTER>x*x</CENTER> <CENTER><hr color=black></CENTER> <CENTER>y+z</CENTER> </TD> <TD> + </TD> <TD> <CENTER>y*y</CENTER> <CENTER><hr color=black></CENTER> <CENTER>x+z</CENTER> </TD> <TD> + </TD> <TD> <CENTER>z*z</CENTER> <CENTER><hr color=black></CENTER> <CENTER>x+y</CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>]</FONT></CENTER> </TD> </TR> </TABLE> </TD> <TD> ---page 13 line 22 </TD> </TR> </TABLE> width of above equation <INPUT name=Button013L22a type="button" value="default value" onclick="B013L22a.value=460,B013L22b.value=280,T013L22a.width=B013L22a.value,T013L22b.width=B013L22b.value"> <input id="B013L22a" value=460 size=3 onchange=T013L22a.width=B013L22a.value /> <input id="B013L22b" value=420 size=3 onchange=T013L22b.width=B013L22b.value /> <pre><font size=+2> <script language="javascript">alert1()</script> <a name="ch01b061">&lt;a name="ch01b061"&gt;</a> 2009-06-28-17-05 here  %0Exercise 1.4 hint 00textbook page 227 Exercise 1.4A use Cauchy Inequality and the 1-Trick as following</font></pre> <TABLE WIDTH="620" id=T227L09a border=0 ><!--9806281708--> <TR> <TD> &nbsp; </TD> <TD> <TABLE WIDTH="460" border=0 id=T227L09b > <TR> <TD> S*Sf"(1*1+1*1+1*1)* </TD> <TD> <CENTER><FONT SIZE=+3>[</FONT></CENTER> </TD> <TD> <CENTER>x+y</CENTER> <CENTER><hr color=black></CENTER> <CENTER>x+y+z</CENTER> </TD> <TD> + </TD> <TD> <CENTER>x+z</CENTER> <CENTER><hr color=black></CENTER> <CENTER>x+y+z</CENTER> </TD> <TD> + </TD> <TD> <CENTER>y+z</CENTER> <CENTER><hr color=black></CENTER> <CENTER>x+y+z</CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>]</FONT></CENTER> </TD> <TD> = 6 </TD> </TR> </TABLE> </TD> <TD> ---page 227 line 9 </TD> </TR> </TABLE> width of above equation <INPUT name=Button227L09a type="button" value="default value" onclick="B227L09a.value=620,B227L09b.value=460,T227L09a.width=B227L09a.value,T227L09b.width=B227L09b.value"> <input id="B227L09a" value=620 size=3 onchange=T227L09a.width=B227L09a.value /> <input id="B227L09b" value=460 size=3 onchange=T227L09b.width=B227L09b.value /> <pre><font size=+2><a name="ch01b062">&lt;a name="ch01b062"&gt;</a> 2009-06-28-17-24 here Exercise 1.4B use Splitting Trick and Cauchy Inequality.</font></pre> <TABLE WIDTH="700" id=T227L11a border=0 ><!--9806281729--> <TR> <TD> &nbsp; </TD> <TD> <TABLE WIDTH="520" border=0 id=T227L11b > <TR> <TD> x+y+z= </TD> <TD> <CENTER>x</CENTER> <CENTER><hr color=black></CENTER> <CENTER>"(y+z)</CENTER> </TD> <TD> "(y+z) </TD> <TD> + </TD> <TD> <CENTER>y</CENTER> <CENTER><hr color=black></CENTER> <CENTER>"(x+z)</CENTER> </TD> <TD> "(x+z) </TD> <TD> + </TD> <TD> <CENTER>z</CENTER> <CENTER><hr color=black></CENTER> <CENTER>"(x+y)</CENTER> </TD> <TD> "(x+y) </TD> </TR> </TABLE> </TD> <TD> ---page 227 line 11 </TD> </TR> </TABLE> width of above equation <INPUT name=Button227L11a type="button" value="default value" onclick="B227L11a.value=700,B227L11b.value=520,T227L11a.width=B227L11a.value,T227L11b.width=B227L11b.value"> <input id="B227L11a" value=700 size=3 onchange=T227L11a.width=B227L11a.value /> <input id="B227L11b" value=520 size=3 onchange=T227L11b.width=B227L11b.value /> <pre><font size=+2> <a name="ch01b063">&lt;a name="ch01b063"&gt;</a> 2009-06-28-17-36 here  %0Exercise 1.4 solution <script lang="javascript">alert2()</script> Now look at Exercise 1.4A equation at page 13 line 20.</font></pre> <TABLE WIDTH="660" id=T013L20c border=0 ><!--9806281739--> <TR> <TD> &nbsp; </TD> <TD> <TABLE WIDTH="460" border=0 id=T013L20d > <TR> <TD> S= </TD> <TD> <CENTER><FONT SIZE=+3>[</FONT></CENTER> </TD> <TD> <CENTER>x+y</CENTER> <CENTER><hr color=black></CENTER> <CENTER>x+y+z</CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>]</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>1/2</FONT></CENTER> <CENTER><FONT SIZE=+3>00</FONT></CENTER> <CENTER><FONT SIZE=-2>00</FONT></CENTER> </TD> <TD> + </TD> <TD> <CENTER><FONT SIZE=+3>[</FONT></CENTER> </TD> <TD> <CENTER>x+z</CENTER> <CENTER><hr color=black></CENTER> <CENTER>x+y+z</CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>]</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>1/2</FONT></CENTER> <CENTER><FONT SIZE=+3>00</FONT></CENTER> <CENTER><FONT SIZE=-2>00</FONT></CENTER> </TD> <TD> + </TD> <TD> <CENTER><FONT SIZE=+3>[</FONT></CENTER> </TD> <TD> <CENTER>y+z</CENTER> <CENTER><hr color=black></CENTER> <CENTER>x+y+z</CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>]</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>1/2</FONT></CENTER> <CENTER><FONT SIZE=+3>00</FONT></CENTER> <CENTER><FONT SIZE=-2>00</FONT></CENTER> </TD> <TD> f" 6<sup>1/2</sup> </TD> </TR> </TABLE> </TD> <TD> ---page 13 line 20 ref.1 </TD> </TR> </TABLE> width of above equation <INPUT name=Button013L20c type="button" value="default value" onclick="B013L20c.value=660,B013L20d.value=460,T013L20c.width=B013L20c.value,T013L20d.width=B013L20d.value"> <input id="B013L20c" value=660 size=3 onchange=T013L20c.width=B013L20c.value /> <input id="B013L20d" value=460 size=3 onchange=T013L20d.width=B013L20d.value /> <pre><font size=+2><a name="ch01b064">&lt;a name="ch01b064"&gt;</a> 2009-06-28-17-42 here Cauchy Inequality less than side is dot product of two vectors (seq.) Cauchy Inequality greater than side is product of two vector lengths. Length of a vector is the square sum of all of its components. Then take square root to the final sum. Equation page 13 line 20 ref. less than side has three square root terms. If take them as component of a vector. After square operation, they can be added easily. But problem is that Cauchy Inequality require two sequences. Exercise 1.4 give only one. Exercise 1.4 hint tell us that second sequence is invisible [1, 1, 1] vector. With two vectors in hand, it is a easy job. Define vector (sequence) S to be </font></pre> <TABLE WIDTH="660" id=T013L20e border=0 ><!--9807301656--> <TR> <TD> &nbsp; </TD> <TD> <TABLE WIDTH="460" border=0 id=T013L20f > <TR> <TD> Vector S = </TD> <TD> <CENTER><FONT SIZE=+3>[</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>(</FONT></CENTER> </TD> <TD> <CENTER>x+y</CENTER> <CENTER><hr color=black></CENTER> <CENTER>x+y+z</CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>)</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>1/2</FONT></CENTER> <CENTER><FONT SIZE=+3>00</FONT></CENTER> <CENTER><FONT SIZE=-2>00</FONT></CENTER> </TD> <TD> , </TD> <TD> <CENTER><FONT SIZE=+3>(</FONT></CENTER> </TD> <TD> <CENTER>x+z</CENTER> <CENTER><hr color=black></CENTER> <CENTER>x+y+z</CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>)</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>1/2</FONT></CENTER> <CENTER><FONT SIZE=+3>00</FONT></CENTER> <CENTER><FONT SIZE=-2>00</FONT></CENTER> </TD> <TD> , </TD> <TD> <CENTER><FONT SIZE=+3>(</FONT></CENTER> </TD> <TD> <CENTER>y+z</CENTER> <CENTER><hr color=black></CENTER> <CENTER>x+y+z</CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>)</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>1/2</FONT></CENTER> <CENTER><FONT SIZE=+3>00</FONT></CENTER> <CENTER><FONT SIZE=-2>00</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>]</FONT></CENTER> </TD> </TR> </TABLE> </TD> <TD> ----page 13 line 20 ref.2 </TD> </TR> </TABLE> width of above equation <INPUT name=Button013L20e type="button" value="default value" onclick="B013L20e.value=660,B013L20f.value=460,T013L20e.width=B013L20e.value,T013L20f.width=B013L20f.value"> <input id="B013L20e" value=660 size=3 onchange=T013L20e.width=B013L20e.value /> <input id="B013L20f" value=460 size=3 onchange=T013L20f.width=B013L20f.value /><!--9807301711--> <pre><font size=+2><a name="ch01b065">&lt;a name="ch01b065"&gt;</a> Vector [1, 1, 1] dot vector S get page 13 line 20 ref.1 less than side. Cauchy Inequality greater than side is two vector's length product. Find length as following. Vector [1, 1, 1] has length "(1*1+1*1+1*1)="3 Vector S length square is [(x+y)/(x+y+z)]+[(y+z)/(x+y+z)]+[(z+x)/(x+y+z)] =2*[(x+y+z)/(x+y+z)] =2 Vector S length is "2 Cauchy Inequality greater than side is "3*"2="6 Put both less than term and greater than terms into Cauchy Inequality get answer page13 line 20 equation. 2009-06-28-18-01 stop <a name="ch01b066">&lt;a name="ch01b066"&gt;</a> 2009-06-28-18-57 start Exercise 1.4B hint said rewrite x as "(y+z)*[x/"(y+z)] rewrite y as "(z+x)*[y/"(z+x)] rewrite z as "(x+y)*[z/"(x+y)] sequence 1 is [ "(y+z) , "(z+x) , "(x+y) ] sequence 2 is [x/"(y+z), y/"(z+x), z/"(x+y)] Cauchy Inequality less than side is dot product of two vectors, it is x+y+z <a name="ch01b067">&lt;a name="ch01b067"&gt;</a> Cauchy Inequality greater than side is product of two vector's length. sequence 1 length square is (y+z)+(z+x)+(x+y) = [2(x+y+z)] sequence 2 length square is x*x/(y+z) + y*y/(z+x) + z*z/(x+y) This is the value in page 13 line 22 square bracket. Here we talk about SQUARE of inequality squared equation is [v1 dot v2]^2 f" v1_len^2 * v2_len^2 that is (x+y+z)*(x+y+z)f"[2(x+y+z)]* -----eqn.AB12 [x*x/(y+z) + y*y/(z+x) + z*z/(x+y)] Given condition is that x, y, z are all positive, then (x+y+z) is non-zero. Cancel one (x+y+z) from both side of eqn.AB12 We get solution page 13 line 22 equation. 2009-06-28-19-15 stop <hr> <a name="ch01b068">&lt;a name="ch01b068"&gt;</a>0<a href="#index">Index begin</a>0<a href="#index06">Index this file</a> <script language="javascript">alert0()</script> 2009-06-29-08-00 start  %0Exercise 1.5 problem statement 00textbook page 13 ÿA Cyrstallographic Inequality with a Message ÿ Assume f(x) = cos(²x) -----eqn.AB13 Identity f(x)*f(x) = [1+f(2x)]/2 ---eqn.AB14 ÿNoteUþeqn.AB14 is cosine half angle identity ÿ Assume for all 1f"kf"n we have p<sub>k</sub>g"0 ---eqn.AB15 given p<sub>1</sub>+p<sub>2</sub>+ ... p<sub>n</sub>=1 ---eqn.AB16 Let g(x) = "[k=1,n]{p<sub>k</sub>cos(²<sub>k</sub>x)} ---eqn.AB17 Under above given condition, prove that next inequality is true. g(x)*g(x)f"[1+g(2x)]/2 ---eqn.AB18 <a name="ch01b069">&lt;a name="ch01b069"&gt;</a> Inequality eqn.AB18 is called Harker-Kasper Inequality. In Cyrstallography, Harker-Kasper Inequality is important. In mathematics inequality, Harker- Kasper Inequality has one more important factor. Given one identity (here eqn.AB14) We should consider its inequality part. (here eqn.AB18) <script language="javascript">alert1()</script> <a name="ch01b070">&lt;a name="ch01b070"&gt;</a>  %0Exercise 1.5 hint 00textbook page 227 2009-06-29-08-28 here Split p<sub>k</sub><sup>1</sup> as (p<sub>k</sub><sup>1/2</sup>)*(p<sub>k</sub><sup>1/2</sup>) Use identity cos(x)*cos(x) = [1+cos(2x)]/2 ---eqn.AB19 and Cauchy Inequality to get</font></pre> <TABLE WIDTH="480" id=T227L14a border=0 ><!--9806290851--> <TR> <TD> &nbsp; </TD> <TD> <TABLE WIDTH="300" border=0 id=T227L14b > <TR> <TD> g<sup>2</sup>(x) </TD> <TD> f" </TD> <TD> <CENTER><FONT SIZE=-2>k=n</FONT></CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER><FONT SIZE=-2>k=1</FONT></CENTER> </TD> <TD> <CENTER>p<sub><FONT SIZE=-1> k</FONT></sub> </CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>k=n</FONT></CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER><FONT SIZE=-2>k=1</FONT></CENTER> </TD> <TD> <CENTER> p<sub><FONT SIZE=-1> k</FONT></sub> cos<sup>2</sup>(²<sub>k</sub>x) </CENTER> </TD> </TR> </TABLE> </TD> <TD> ---page 227 line 14 </TD> </TR> </TABLE> width of above equation <INPUT name=Button227L14a type="button" value="default value" onclick="B227L14a.value=480,B227L14b.value=300,T227L14a.width=B227L14a.value,T227L14b.width=B227L14b.value"> <input id="B227L14a" value=480 size=3 onchange=T227L14a.width=B227L14a.value /> <input id="B227L14b" value=300 size=3 onchange=T227L14b.width=B227L14b.value /> <TABLE WIDTH="500" id=T227L15a border=0 ><!--9806290901--> <TR> <TD> &nbsp; </TD> <TD> <TABLE WIDTH="320" border=0 id=T227L15b > <TR> <TD> = </TD> <TD> <CENTER><FONT SIZE=-2>k=n</FONT></CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER><FONT SIZE=-2>k=1</FONT></CENTER> </TD> <TD> <CENTER>p<sub><FONT SIZE=-1> k</FONT></sub> </CENTER> </TD> <TD> <CENTER>1</CENTER> <CENTER><hr color=black></CENTER> <CENTER>2</CENTER> </TD> <TD> [1+cos(2²<sub>k</sub>x)] </TD> <TD> =[1+g(2x)]/2 </TD> </TR> </TABLE> </TD> <TD> ---page 227 line 15 </TD> </TR> </TABLE> width of above equation <INPUT name=Button227L15a type="button" value="default value" onclick="B227L15a.value=500,B227L15b.value=320,T227L15a.width=B227L15a.value,T227L15b.width=B227L15b.value"> <input id="B227L15a" value=500 size=3 onchange=T227L15a.width=B227L15a.value /> <input id="B227L15b" value=320 size=3 onchange=T227L15b.width=B227L15b.value /> <pre><font size=+2> <a name="ch01b071">&lt;a name="ch01b071"&gt;</a> 2009-06-29-09-08 here  %0Exercise 1.5 solution <script lang="javascript">alert2()</script> because p<sub>k</sub>g"0 ---eqn.AB15 and p<sub>1</sub>+p<sub>2</sub>+ ... p<sub>n</sub>=1 ---eqn.AB16 We call p<sub>k</sub> as probability. Identity f(x)*f(x) = [1+f(2x)]/2 ---eqn.AB14 Inequality g(x)*g(x) f" [1+g(2x)]/2 ---eqn.AB18 They have same structure, why identity become inequality?! We need to see what is the difference between function f(x) and function g(x) ? <a name="ch01b072">&lt;a name="ch01b072"&gt;</a> Define f(x) = cos(²x) -----eqn.AB13 Define g(x) = "[k=1,n]{p<sub>k</sub> cos(²<sub>k</sub>x)} ---eqn.AB17 Function f(x) and function g(x) difference is Function f(x) equal to cos(²x) directly no probability, no summation. Function g(x) equal to cos(²x) indirectly has probability, has summation. Parameter ²<sub>k</sub> in cos(²<sub>k</sub>x) in function g(x) change with probability. <a name="ch01b073">&lt;a name="ch01b073"&gt;</a> When we consider cosine half angle identity cos(x)*cos(x) = [1+cos(2x)]/2 ---eqn.AB19 If probability enter equation were multiplying whole equation both side with p<sub>k</sub><sup>2</sup> at same time, like the following<!--9807301818--> fabricated equation p<sub>k</sub><sup>2</sup>*cos(x)*cos(x) = p<sub>k</sub><sup>2</sup>*[1+cos(2x)]/2 ---eqn.AB19 fabricated The result is still identity. But, probability enter equation only in front of cosine function, look like the following p<sub>k</sub>cos(x)*p<sub>k</sub>cos(x) ÿÿ [1+p<sub>k</sub>cos(2x)]/2 -----eqn.AB20 Uneven distributed. This cause identity eqn.AB14 become inequality eqn.AB18. 2009-06-29-09-47 here <a name="ch01b074">&lt;a name="ch01b074"&gt;</a> g(x) defined to be "[k=1,n]{p<sub>k</sub> cos(²<sub>k</sub>x)} ---eqn.AB17 probability p<sub>k</sub> multiply with cos(²<sub>k</sub>x) probability p<sub>k</sub> multiply with f<sub>²k</sub>(x) ÿ then take summation Starting point is p<sub>k</sub> multiply with f<sub>²k</sub>(x) -----eqn.AB21 End point require p<sub>k</sub>cos(2x) 0noteUþcos(2x)=2*cos<sup>2</sup>(x)-10 End point require p<sub>k</sub>*[2*f<sub>²k</sub>(x)*f<sub>²k</sub>(x)-1] -----eqn.AB22 f<sub>²k</sub>(x) has square ÿbut p<sub>k</sub> not square ÿ textbook hint is Split p<sub>k</sub><sup>1</sup> as (p<sub>k</sub><sup>1/2</sup>)*(p<sub>k</sub><sup>1/2</sup>) Take p<sub>1</sub><sup>1/2</sup>, p<sub>2</sub><sup>1/2</sup>, ... p<sub>n</sub><sup>1/2</sup> -----eqn.AB23 as sequence 1 Take p<sub>1</sub><sup>1/2</sup>f<sub>²1</sub>(x), p<sub>2</sub><sup>1/2</sup>f<sub>²2</sub>(x), ... p<sub>n</sub><sup>1/2</sup>f<sub>²n</sub>(x) -----eqn.AB24 as sequence 2, then use Cauchy Inequality. <a name="ch01b075">&lt;a name="ch01b075"&gt;</a> Cauchy Inequality less than side is dot product of two vectors, get p<sub>1</sub><sup>1/2</sup>*p<sub>1</sub><sup>1/2</sup>f<sub>²1</sub>(x) + p<sub>2</sub><sup>1/2</sup>*p<sub>2</sub><sup>1/2</sup>f<sub>²2</sub>(x) + ... + p<sub>n</sub><sup>1/2</sup>*p<sub>n</sub><sup>1/2</sup>f<sub>²n</sub>(x) -----eqn.AB25 Less than side is ÿsee eqn.AB17 ÿ p<sub>1</sub>f<sub>²1</sub>(x) + p<sub>2</sub>f<sub>²2</sub>(x) + ... + p<sub>n</sub>f<sub>²n</sub>(x) = g(x) -----eqn.AB26 Cauchy Inequality greater than side is vector 1 length multiply vector 2 length. vector 1 length square is vector 1 elements squared then sum p<sub>1</sub><sup>1/2</sup>*p<sub>1</sub><sup>1/2</sup> + p<sub>2</sub><sup>1/2</sup>*p<sub>2</sub><sup>1/2</sup> + ... + p<sub>n</sub><sup>1/2</sup>*p<sub>n</sub><sup>1/2</sup> that isÿsee eqn.AB16 ÿ p<sub>1</sub> + p<sub>2</sub> + ... + p<sub>n</sub> = 1 -----eqn.AB27 <a name="ch01b076">&lt;a name="ch01b076"&gt;</a> vector 2 length square is vector 2 elements squared then sum [p<sub>1</sub><sup>1/2</sup>f<sub>²1</sub>(x)]*[p<sub>1</sub><sup>1/2</sup>f<sub>²1</sub>(x)] + [p<sub>2</sub><sup>1/2</sup>f<sub>²2</sub>(x)]*[p<sub>2</sub><sup>1/2</sup>f<sub>²2</sub>(x)] + ... + [p<sub>n</sub><sup>1/2</sup>f<sub>²n</sub>(x)]*[p<sub>n</sub><sup>1/2</sup>f<sub>²n</sub>(x)] that is [p<sub>1</sub>f<sub>²1</sub>(x)*f<sub>²1</sub>(x)]+[p<sub>2</sub>f<sub>²2</sub>(x)*f<sub>²2</sub>(x)] + ... + [p<sub>n</sub>f<sub>²n</sub>(x)*f<sub>²n</sub>(x)] -----eqn.AB28 Cauchy Inequality is v1 dot v2 f" v1_len * v2_len Squared version Cauchy Inequality is [v1 dot v2]^2 f" [v1_len]^2 * [v2_len]^2 <a name="ch01b077">&lt;a name="ch01b077"&gt;</a> Squared version Cauchy Inequality become g(x)*g(x) f" 1 * 0000000{[p<sub>1</sub>f<sub>²1</sub>(x)*f<sub>²1</sub>(x)]+[p<sub>2</sub>f<sub>²2</sub>(x)*f<sub>²2</sub>(x)] 0000000+ ... + [p<sub>n</sub>f<sub>²n</sub>(x)*f<sub>²n</sub>(x)]} -----eqn.AB29 <font color=red>Sequence 1 length is one, then [ 1 * ] is invisible. The following will use this [ 1 * ] (eqn.AB27) one more time.</font> eqn.AB29 right hand side has n terms summed. First of them is next. p<sub>1</sub>f<sub>²1</sub>(x)*f<sub>²1</sub>(x) -----eqn.AB30 Use cosine half angle identity cos(x)*cos(x) = [1+cos(2x)]/2 ---eqn.AB19 <a name="ch01b078">&lt;a name="ch01b078"&gt;</a> change eqn.AB30 to the next p<sub>1</sub>*f<sub>²1</sub>(x)*f<sub>²1</sub>(x) = //given f(x)=cos(²x) --eqn.AB13 p<sub>1</sub>*cos(²<sub>1</sub>x)*cos(²<sub>1</sub>x) = //cos half angle id. eqn.AB19 p<sub>1</sub>[1+cos(2*²<sub>1</sub>x)]/2 = //move p<sub>1</sub> into [ ... ] [p<sub>1</sub>+p<sub>1</sub>*cos(2*²<sub>1</sub>x)]/2 -----eqn.AB31 eqn.AB29 right hand side has n terms, eqn.AB31 is just first term. Now sum all of them get eqn.AB29 right hand side = 0.5*"[k=1,n]{p<sub>k</sub>} //<font color=red>use eqn.AB27 get 1</font> 000+0.5*"[k=1,n]{p<sub>k</sub> cos(2*²<sub>k</sub>x)} -----eqn.AB32 <a name="ch01b079">&lt;a name="ch01b079"&gt;</a> g(x) is defined to be "[k=1,n]{p<sub>k</sub> cos(²<sub>k</sub>x)} ---eqn.AB17 eqn.AB32 use 2*²<sub>k</sub>x so, eqn.AB32 has g(2x) eqn.AB29 right hand side = eqn.AB32 = 0.5*{1+g(2x)} Finally Squared version Cauchy Inequality is [v1 dot v2]^2 f" [v1_len]^2 * [v2_len]^2 g(x)*g(x) f" {1+g(2x)}/2 -----eqn.AB33 eqn.AB33 is problem target eqn.AB18 2009-06-29-11-02 stop <hr> <a name="ch01b080">&lt;a name="ch01b080"&gt;</a>0<a href="#index">Index begin</a>0<a href="#index06">Index this file</a> <script language="javascript">alert0()</script> 2009-06-29-14-27 start  %0Exercise 1.6 problem statement 00textbook page 14 ÿA Sum of Inversion Preserving Summands ÿ Assume for all 1f"kf"n we have p<sub>k</sub>g"0 -----eqn.AB34 Given p<sub>1</sub>+p<sub>2</sub>+ ... p<sub>n</sub>=1 -----eqn.AB35 Show the following inequality.</font></pre> <TABLE WIDTH="480" id=T014L10a border=0 ><!--9806291434--> <TR> <TD> &nbsp; </TD> <TD> <TABLE WIDTH="300" border=0 id=T014L10b > <TR> <TD> <CENTER><FONT SIZE=-2>k=n</FONT></CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER><FONT SIZE=-2>k=1</FONT></CENTER> </TD> <TD> <FONT SIZE=+3>(</FONT> </TD> <TD> <CENTER> p<sub><FONT SIZE=-1> k</FONT></sub> + </CENTER> </TD> <TD> <CENTER>1</CENTER> <CENTER><hr color=black></CENTER> <CENTER>p<sub><FONT SIZE=-1> k</FONT></sub> </CENTER> </TD> <TD> <FONT SIZE=+3>)</FONT> </TD> <TD> <CENTER><FONT SIZE=-2>2</FONT></CENTER> <CENTER>0</CENTER> <CENTER>0</CENTER> </TD> <TD> g" </TD> <TD> n<sup>3</sup>+2n+1/n </TD> </TR> </TABLE> </TD> <TD> ---page 14 line 10 </TD> </TR> </TABLE> width of above equation <INPUT name=Button014L10a type="button" value="default value" onclick="B014L10a.value=480,B014L10b.value=300,T014L10a.width=B014L10a.value,T014L10b.width=B014L10b.value"> <input id="B014L10a" value=480 size=3 onchange=T014L10a.width=B014L10a.value /> <input id="B014L10b" value=300 size=3 onchange=T014L10b.width=B014L10b.value /> <pre><font size=+2><a name="ch01b081">&lt;a name="ch01b081"&gt;</a> 2009-06-29-14-43 here Determine the necessary and sufficient condition that equality hold. The future problem ÿpage 206 exercise 13.6 ÿ will see a problem its power is not two. <script language="javascript">alert1()</script> <a name="ch01b082">&lt;a name="ch01b082"&gt;</a>  %0Exercise 1.6 hint 00textbook page 227 In page 14 line 10 equation, expand left hand side square term, get</font></pre> <TABLE WIDTH="580" id=T227L17a border=0 ><!--9806291449--> <TR> <TD> &nbsp; </TD> <TD> <TABLE WIDTH="400" border=0 id=T227L17b > <TR> <TD> <CENTER><FONT SIZE=-2>k=n</FONT></CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER><FONT SIZE=-2>k=1</FONT></CENTER> </TD> <TD> <FONT SIZE=+3>(</FONT> </TD> <TD> <CENTER> p<sub><FONT SIZE=-1> k</FONT></sub> + </CENTER> </TD> <TD> <CENTER>1</CENTER> <CENTER><hr color=black></CENTER> <CENTER>p<sub><FONT SIZE=-1> k</FONT></sub> </CENTER> </TD> <TD> <FONT SIZE=+3>)</FONT> </TD> <TD> <CENTER><FONT SIZE=-2>2</FONT></CENTER> <CENTER>0</CENTER> <CENTER>0</CENTER> </TD> <TD> ÿ </TD> <TD> 2n + </TD> <TD> <CENTER><FONT SIZE=-2>k=n</FONT></CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER><FONT SIZE=-2>k=1</FONT></CENTER> </TD> <TD> p<sub><FONT SIZE=-1> k</FONT></sub><sup>2</sup> + </TD> <TD> <CENTER><FONT SIZE=-2>k=n</FONT></CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER><FONT SIZE=-2>k=1</FONT></CENTER> </TD> <TD> <CENTER>1</CENTER> <CENTER><hr color=black></CENTER> <CENTER>p<sub><FONT SIZE=-1> k</FONT></sub><sup>2</sup> </CENTER> </TD> </TR> </TABLE> </TD> <TD> ---227˜ 14.44_ </TD> </TR> </TABLE> width of above equation <INPUT name=Button227L17a type="button" value="default value" onclick="B227L17a.value=580,B227L17b.value=400,T227L17a.width=B227L17a.value,T227L17b.width=B227L17b.value"> <input id="B227L17a" value=580 size=3 onchange=T227L17a.width=B227L17a.value /> <input id="B227L17b" value=400 size=3 onchange=T227L17b.width=B227L17b.value /> <pre><font size=+2><a name="ch01b083">&lt;a name="ch01b083"&gt;</a> 2009-06-29-14-56 here Next estimate right most two terms. Use 1-Trick and given condition p<sub>1</sub>+p<sub>2</sub>+ ... p<sub>n</sub>=1 -----eqn.AB35 <a href="#ch01b087">"p<sub>k</sub><sup>2</sup> value is at least 1/n </a>-----eqn.AB36 To estimate "[1/p<sub>k</sub><sup>2</sup>] use Cauchy Inequality Sequence 1 is "p<sub>k</sub> ÿ sequence 2 is 1/["p<sub>k</sub>] get n<sup>2</sup> f" "[1/p<sub>k</sub>] -----eqn.AB37 Fianlly use Cauchy Inequality again Sequence 3 is [1, 1, ... 1] ÿ Sequence 4 is 1/p<sub>k</sub> get n<sup>3</sup> f" "[1/p<sub>k</sub><sup>2</sup>] -----eqn.AB38 This exercise has several method to solve. Above hint, use Cauchy Inequality and the 1-Trick. 2009-06-29-15-18 here <a name="ch01b084">&lt;a name="ch01b084"&gt;</a> 2009-06-29-15-23 start  %0Exercise 1.6 solution <script lang="javascript">alert2()</script> Problem ask to show "[p<sub>k</sub> + 1/p<sub>k</sub>]<sup>2</sup> g" n<sup>3</sup>+2n+1/n ----page 14 line 10 Expand left hand side = ["(p<sub>k</sub><sup>2</sup>)]+["(2*p<sub>k</sub>*1/p<sub>k</sub>)]+["(1/p<sub>k</sub><sup>2</sup>)] Middle term p<sub>k</sub>*1/p<sub>k</sub> cancel get one ÿ 2*1 sum n times, get 2n page 14 line 10 left hand side = ["(p<sub>k</sub><sup>2</sup>)]+2n+["(1/p<sub>k</sub><sup>2</sup>)] ---page 227 eqn.14.44 <a name="ch01b085">&lt;a name="ch01b085"&gt;</a> From hint, in order to estimate the value of "[1/p<sub>k</sub><sup>2</sup>] we use Cauchy Inequality. Sequence 1 is "p<sub>k</sub> ÿ Sequence 2 is 1/["p<sub>k</sub>] Sequence 1 dot Sequence 2 get "["p<sub>k</sub>/"p<sub>k</sub>] ÿ Add one n times, get n Above is Cauchy Inequality less than side v1 dot v2 f" [v1_len]^0.5 * [v2_len]^0.5 Below is Cauchy Inequality greater than side Sequence 1 length square is "[("p<sub>k</sub>)<sup>2</sup>] = eqn.AB35 = 1 Sequence 2 length square is "[(1/"p<sub>k</sub>)<sup>2</sup>] = "(1/p<sub>k</sub>) Sequence 1 length is "1 ÿ Sequence 2 length is "["(1/p<sub>k</sub>)] Cauchy Inequality of sequence 1 and 2 is n f" 1*"["(1/p<sub>k</sub>)] Square whole equation, get n*n f" "(1/p<sub>k</sub>) -----eqn.AB37 <a name="ch01b086">&lt;a name="ch01b086"&gt;</a> Based on hint, use Cauchy Inequality again Sequence 3 is [1, 1, ... 1] ÿ Sequence 4 is 1/p<sub>k</sub> k=1,2.... Sequence 3 dot sequence 4 1*1/p<sub>1</sub> + 1*1/p<sub>2</sub> + ... + 1*1/p<sub>n</sub> Sequence 3 dot sequence 4 result is "(1/p<sub>k</sub>) -----eqn.AB39 Above is Cauchy Inequality less than side v1 dot v2 f" [v1_len]^0.5 * [v2_len]^0.5 Below is Cauchy Inequality greater than side Sequence 3 length square is 1*1+1*1+ ... + 1*1 = n ÿadd n times ÿ Sequence 4 length square is "[(1/p<sub>k</sub>)<sup>2</sup>] Sequence 3 length is "n ÿ Sequence 4 length is "{"[(1/p<sub>k</sub>)<sup>2</sup>]} Cauchy Inequality of sequence 3 and 4 is "(1/p<sub>k</sub>) f" "n * "{"[(1/p<sub>k</sub>)<sup>2</sup>]} -----eqn.AB40 eqn.AB40 left hand side and eqn.AB37 right hand side are equal, combine two equations eliminate "(1/p<sub>k</sub>) get n*n f" "n * "{"[(1/p<sub>k</sub>)<sup>2</sup>]} -----eqn.AB41 square eqn.AB41 whole equation, calcel one n, get n<sup>3</sup> f" "[(1/p<sub>k</sub>)<sup>2</sup>] -----eqn.AB42 2009-06-29-16-25 here <a name="ch01b087">&lt;a name="ch01b087"&gt;</a> Next turn our attention to "(p<sub>k</sub><sup>2</sup>) use Sequence 5 p<sub>1</sub>, p<sub>2</sub>, ... p<sub>n</sub> Apply Cauchy Inequality to Sequence 3 and 5 Sequence 3 dot sequence 5 get 1*p<sub>1</sub>+1*p<sub>2</sub>+ ... 1*p<sub>n</sub>=1 Above is Cauchy Inequality less than side v1 dot v2 f" [v1_len]^0.5 * [v2_len]^0.5 Below is Cauchy Inequality greater than side Sequence 3 length square is 1*1+1*1+ ... + 1*1 = n ÿadd n times ÿ Sequence 5 length square is "(p<sub>k</sub><sup>2</sup>) Cauchy Inequality of sequence 3 and 5 is 1 f" "n * "["(p<sub>k</sub><sup>2</sup>)] Square whole equation, and move n to left 1/n f" "(p<sub>k</sub><sup>2</sup>) -----eqn.AB43 <a name="ch01b088">&lt;a name="ch01b088"&gt;</a> Problem ask to show ["(p<sub>k</sub><sup>2</sup>)]+2n+["(1/p<sub>k</sub><sup>2</sup>)] g" n<sup>3</sup>+2n+1/n -----eqn.AB44 ÿthis is combination of page 227 eqn.14.44 and page 14 line 10 equation ÿ eqn.AB42 give eqn.AB44 two inner terms. eqn.AB44 middle terms are both 2n . eqn.AB44 left end term and right end term refer to eqn.AB43 for inequality. Up to here, eqn.AB44 is done 2009-06-29-16-46 stop <a name="ch01b089">&lt;a name="ch01b089"&gt;</a> 2009-06-29-16-59 start There are more to be done. Problem ask [[ Determine the necessary and sufficient condition that equality hold. ]] We use Cauchy Inequality to solve problem. Necessary and sufficient condition for Cauchy Inequality become equality, the same condition apply to Exercise 1.6. <font color=red> Necessary and sufficient condition for Cauchy Inequality become equality is that sequence 1 and 2 are proportional </font> <a name="ch01b090">&lt;a name="ch01b090"&gt;</a> First time apply Cauchy Inequality to Sequence 1: "p<sub>k</sub> ÿ and Sequence 2: 1/["p<sub>k</sub>] The ratio of k th elements from both seq. 1 and 2 is » = "p<sub>k</sub> / 1/["p<sub>k</sub>], that is » = p<sub>k</sub> It is constant for any k, so p<sub>k</sub> is constant. Given "[k=1,n] p<sub>k</sub> = 1 then p<sub>k</sub> = 1/n -----eqn.AB45 <a name="ch01b091">&lt;a name="ch01b091"&gt;</a> Second time apply Cauchy Inequality to Sequence 3: [1, 1, ... 1] ÿ and Sequence 4: 1/p<sub>k</sub> If eqn.AB45 is true, then it apply to sequence 3 and 4 too. Third time apply Cauchy Inequality to Sequence 3: [1, 1, ... 1] ÿ and Sequence 5: p<sub>k</sub> If eqn.AB45 is true, then it apply to sequence 3 and 5 too. eqn.AB45 is the necessary and sufficient condition for Exercise 1.6 equality hold. 2009-06-29-17-14 stop <a name="ch01b092">&lt;a name="ch01b092"&gt;</a> The following is computer program [[ n=5 //Exercise 1.6 equality hold p=1/n //require five probability are the same big0=n*p*p + n*(1/p/p) +2*n sml0= n*n*n+1/n +2*n big0 sml0 ]] Paste above six lines to <a href=http://freeman2.com/complex2.htm> http://freeman2.com/complex2.htm</a> box3 ÿclick [test box3 command, output to box4] get big0 135.2 sml0 135.2 This is equality condition. 2009-06-29-17-28 <hr> <a name="ch01b093">&lt;a name="ch01b093"&gt;</a>0<a href="#index">Index begin</a>0<a href="#index06">Index this file</a> <script language="javascript">alert0()</script> 2009-06-29-18-31 start  %0Exercise 1.7 problem statement 00textbook page 14 ÿFlexibility of Form ÿ Assume x, y, ±, ² are all real numbers, please prove (5±x+±y+²x+3²y)<sup>2</sup>f"(5±<sup>2</sup>+2±²+3²<sup>2</sup>)(5x<sup>2</sup>+2xy+3y<sup>2</sup>) -----page 14 eqn.1.22 Please explain eqn.1.22 is Cauchy~Schwarz Inequality <a href="tute0007.htm#ch01a065">eqn.1.16</a>. For this problem, you need design a special dot product 0.,. 0. <script language="javascript">alert1()</script> <a name="ch01b094">&lt;a name="ch01b094"&gt;</a> 2009-06-29-18-42 here  %0Exercise 1.7 hint 00textbook page 228 The natural dot product 0.,. 0 definition is 0x,y 0 = 5x<sub>1</sub>y<sub>1</sub> + x<sub>1</sub>y<sub>2</sub> + x<sub>2</sub>y<sub>1</sub> + 3x<sub>2</sub>y<sub>2</sub> -----eqn.AB46 Seq. 1: x = (x<sub>1</sub>, x<sub>2</sub>) -----eqn.AB47 Seq. 2: y = (y<sub>1</sub>, y<sub>2</sub>) -----eqn.AB48 <a name="ch01b095">&lt;a name="ch01b095"&gt;</a> eqn.AB46 is a proposed dot product. It must comply with <a href=tute0007.htm#ch01a051>vector dot product rule</a>. Whether satisfy rule t$ÿ00v,v 0g"000 ---eqn.AA27 Whether satisfy rule u$ÿ00v,v 0=000 ---eqn.AA28 Please pay attention to that polynomial 0 5z<sup>2</sup>+2z+3=0 has no real root. (Note, textbook use 5z<sup>2</sup>+3z+3=0 0LiuHH change to 5z<sup>2</sup>+2z+3=0 09807011605) Broadly speak, if a<sub>jk</sub> ÿ 1f"j,kf"n is real elements of a symmetric matrix, that is for all 1f"j,kf"n we have a<sub>jk</sub> = a<sub>kj</sub>, then</font></pre> <TABLE WIDTH="460" id=T228L15a border=0 ><!--9806291906--> <TR> <TD> &nbsp; </TD> <TD> <TABLE WIDTH="260" border=0 id=T228L15b > <TR> <TD> 0x,y 0 </TD> <TD> ÿ </TD> <TD> <CENTER><FONT SIZE=-2>j=n</FONT></CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER><FONT SIZE=-2>j=1</FONT></CENTER> </TD> <TD> <CENTER><FONT SIZE=-2>k=n</FONT></CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER><FONT SIZE=-2>k=1</FONT></CENTER> </TD> <TD> a<sub>jk</sub> x<sub>j</sub> y<sub>k</sub> </TD> </TR> </TABLE> </TD> <TD> ---page 228 eqn.14.45 </TD> </TR> </TABLE> width of above equation <INPUT name=Button228L15a type="button" value="default value" onclick="B228L15a.value=460,B228L15b.value=260,T228L15a.width=B228L15a.value,T228L15b.width=B228L15b.value"> <input id="B228L15a" value=460 size=3 onchange=T228L15a.width=B228L15a.value /> <input id="B228L15b" value=260 size=3 onchange=T228L15b.width=B228L15b.value /> <pre><font size=+2><a name="ch01b096">&lt;a name="ch01b096"&gt;</a> 2009-06-29-19-11 here is a candidate of dot product definition in n dimensional real space. Require the following: First, for all n dimensional real space 000vector (x<sub>1</sub>, x<sub>2</sub>, ... , x<sub>n</sub>) 000A quadratic function 000Q(x<sub>1</sub>, x<sub>2</sub>, ... , x<sub>n</sub>) = "<sub>j</sub> "<sub>k</sub> a<sub>jk</sub> x<sub>j</sub> y<sub>k</sub> 000has only non-negative values. Second, function Q(x<sub>1</sub>, x<sub>2</sub>, ... , x<sub>n</sub>) = 0 000if and only if vector (x<sub>1</sub>, ... , x<sub>n</sub>) 000is a null vector [0,0, ... 0] Polynomial satisfy above two points is called positive definite quadratic form. These kind polynomial give us variety different Cauchy Inequality definitions. 2009-06-29-19-24 here <a name="ch01b097">&lt;a name="ch01b097"&gt;</a> 2009-06-30-10-51 start  %0Exercise 1.7 discussion textbook hint said 00 The natural dot product 0.,. 0 definition is 0x,y 0 = 5x<sub>1</sub>y<sub>1</sub> + x<sub>1</sub>y<sub>2</sub> + x<sub>2</sub>y<sub>1</sub> + 3x<sub>2</sub>y<sub>2</sub> -----eqn.AB46 Seq. 1: x = (x<sub>1</sub>, x<sub>2</sub>) -----eqn.AB47 Seq. 2: y = (y<sub>1</sub>, y<sub>2</sub>) -----eqn.AB48 00 Compare with regular Cauchy Inequality dot product part (less than side) 0x,y 0 = x<sub>1</sub>y<sub>1</sub> + x<sub>2</sub>y<sub>2</sub> -----eqn.AB49 eqn.AB49Uþnon-cross coef. is 1 & 1 ÿ x<sub>1</sub>y<sub>1</sub> + x<sub>2</sub>y<sub>2</sub> ÿ eqn.AB46Uþnon-cross coef. is 5 & 3 ÿ5x<sub>1</sub>y<sub>1</sub> +3x<sub>2</sub>y<sub>2</sub> ÿ eqn.AB49UþcrossTerm coef. is 0 & 0 ÿ0x<sub>1</sub>y<sub>2</sub> +0x<sub>2</sub>y<sub>1</sub> ÿ eqn.AB46UþcrossTerm coef. is 1 & 1 ÿ x<sub>1</sub>y<sub>2</sub> + x<sub>2</sub>y<sub>1</sub> ÿ There are so big difference, is it possible that Cauchy Inequality still be true? <a name="ch01b098">&lt;a name="ch01b098"&gt;</a> eqn.AB46 and eqn.AB49 both use seq. x ÿeqn.AB47 ÿ and seq. y ÿeqn.AB48 ÿ Difference between eqn.AB46 and eqn.AB49 is that dot product coefficient are different. eqn.AB46Uþ 5x<sub>1</sub>y<sub>1</sub> + x<sub>1</sub>y<sub>2</sub> + x<sub>2</sub>y<sub>1</sub> + 3x<sub>2</sub>y<sub>2</sub> eqn.AB49Uþ 1x<sub>1</sub>y<sub>1</sub> +0x<sub>1</sub>y<sub>2</sub> +0x<sub>2</sub>y<sub>1</sub> + 1x<sub>2</sub>y<sub>2</sub> Key point is coefficient. Now rewrite eqn.AB46 and rewrite eqn.AB49 as following <a name="ch01b099">&lt;a name="ch01b099"&gt;</a> rewrite eqn.AB46=</font></pre> <TABLE WIDTH="400" id=TX0107aa border=0 ><!--9806301121 TX0107aa T = Table X0107a = exercise 01-07 ŠÖŠ a --> <TR> <TD> &nbsp; </TD> <TD> <TABLE WIDTH="200" border=0 id=TX0107ab > <TR> <TD> [x<sub>1</sub>, x<sub>2</sub>] </TD> <TD> <CENTER>[ 5 0 1 ]<sub> </sub></CENTER> <CENTER>[ 1 0 3 ]<sub> </sub></CENTER> </TD> <TD> <CENTER>[y<sub>1</sub>]</CENTER> <CENTER>[y<sub>2</sub>]</CENTER> </TD> </TR> </TABLE> </TD> <TD> ---eqn.AB50 </TD> </TR> </TABLE> width of above equation <INPUT name=ButtonX0107aa type="button" value="default value" onclick="BX0107aa.value=400,BX0107ab.value=200,TX0107aa.width=BX0107aa.value,TX0107ab.width=BX0107ab.value"> <input id="BX0107aa" value=400 size=3 onchange=TX0107aa.width=BX0107aa.value /> <input id="BX0107ab" value=200 size=3 onchange=TX0107ab.width=BX0107ab.value /> <pre><font size=+2><a name="ch01b100">&lt;a name="ch01b100"&gt;</a> 2009-06-30-11-32 here rewrite eqn.AB49=</font></pre> <TABLE WIDTH="400" id=TX0107ac border=0 ><!--9806301135 TX0107ac T = Table X0107a = exercise 01-07 ŠÖŠ a --> <TR> <TD> &nbsp; </TD> <TD> <TABLE WIDTH="200" border=0 id=TX0107ad > <TR> <TD> [x<sub>1</sub>, x<sub>2</sub>] </TD> <TD> <CENTER>[ 1 0 0 ]<sub> </sub></CENTER> <CENTER>[ 0 0 1 ]<sub> </sub></CENTER> </TD> <TD> <CENTER>[y<sub>1</sub>]</CENTER> <CENTER>[y<sub>2</sub>]</CENTER> </TD> </TR> </TABLE> </TD> <TD> ---eqn.AB51 </TD> </TR> </TABLE> width of above equation <INPUT name=ButtonX0107ac type="button" value="default value" onclick="BX0107ac.value=400,BX0107ad.value=200,TX0107ac.width=BX0107ac.value,TX0107ad.width=BX0107ad.value"> <input id="BX0107ac" value=400 size=3 onchange=TX0107ac.width=BX0107ac.value /> <input id="BX0107ad" value=200 size=3 onchange=TX0107ad.width=BX0107ad.value /> <pre><font size=+2><a name="ch01b101">&lt;a name="ch01b101"&gt;</a>  %0Matrix and vector multiplication 2009-06-30-11-39 here Above rewrite a polynomial in a matrix and two vector multiplication operation. This method, isolate common term (x<sub>1</sub>, x<sub>2</sub>) and (y<sub>1</sub>, y<sub>2</sub>). Put coefficients in matrix form and stand out clearly. eqn.AB50 rewrite eqn.AB46 and eqn.AB51 rewrite eqn.AB49. How to undo the rewrite? The following explain how a matrix multiply with a vector. Use simpler 3x3 matrix and 1x3 vector as an example. Assume vector 1 is [ a, b, c ], this is a 1x3 row vector. Assume vector 2 is [ x, y, z ], this is a 1x3 row vector. Assume 3x3 matrix is [1,2,3; 4,5,6; 7,8,9] In square form it is<!--9807311025--> [1 2 3] [4 5 6] [7 8 9] <a name="ch01b102">&lt;a name="ch01b102"&gt;</a>  %0Transpose of a Matrix Please pay attention to that row vector [ x, y, z ] must transpose to column vector before matrix and vector multiplication Transpose mark as [ x, y, z ]<sup>T</sup> write it clearly as a column vector [x] [y] [z] Or we can write as [ x; y; z ] . Inter change columns and rows is called transpose of a matrix. Vector1 multiply matrix multiply Vector2 is next.</font></pre> <TABLE WIDTH="420" id=TX0107ae border=0 ><!--9806301155--> <TR> <TD> &nbsp; </TD> <TD> <TABLE WIDTH="200" border=0 id=TX0107af > <TR> <TD> [a, b, c] </TD> <TD> <CENTER>[ 1 0 2 0 3 ]<sub> </sub></CENTER> <CENTER>[ 4 0 5 0 6 ]<sub> </sub></CENTER> <CENTER>[ 7 0 8 0 9 ]<sub> </sub></CENTER> </TD> <TD> <CENTER>[x]<sub> </sub></CENTER> <CENTER>[y]<sub> </sub></CENTER> <CENTER>[z]<sub> </sub></CENTER> </TD> </TR> </TABLE> </TD> <TD> -----eqn.AB52 </TD> </TR> </TABLE> width of above equation <INPUT name=ButtonX0107ae type="button" value="default value" onclick="BX0107ae.value=420,BX0107af.value=200,TX0107ae.width=BX0107ae.value,TX0107af.width=BX0107af.value"> <input id="BX0107ae" value=420 size=3 onchange=TX0107ae.width=BX0107ae.value /> <input id="BX0107af" value=200 size=3 onchange=TX0107af.width=BX0107af.value /> <pre><font size=+2><a name="ch01b103">&lt;a name="ch01b103"&gt;</a> 2009-06-30-12-00 here Vector 1 is a row vector. Put vector 1 left to matrix. Vector 2 is a column vector. Put vector 2 right to matrix. (Vector IS a rectangular matrix)<font color=red> 3 by 3 matrix [1,2,3; 4,5,6; 7,8,9] those number are address number. Easier for visual exam matrix product rule.</font> V1*M1*V2 start from right most two matrix. Next is step one. <a href="#ch01b128">Detail step here</a></font></pre> <TABLE WIDTH="420" id=TX0107ag border=0 ><!--9806301206--> <TR> <TD> &nbsp; </TD> <TD> <TABLE WIDTH="200" border=0 id=TX0107ah > <TR> <TD> [a, b, c] </TD> <TD> <CENTER>[ <font color=red>1*x+2*y+3*z</font> ]<sub> </sub></CENTER> <CENTER>[ <font color=red>4*x+5*y+6*z</font> ]<sub> </sub></CENTER> <CENTER>[ <font color=red>7*x+8*y+9*z</font> ]<sub> </sub></CENTER> </TD> </TR> </TABLE> </TD> <TD> -----eqn.AB53 </TD> </TR> </TABLE> width of above equation <INPUT name=ButtonX0107ag type="button" value="default value" onclick="BX0107ag.value=420,BX0107ah.value=200,TX0107ag.width=BX0107ag.value,TX0107ah.width=BX0107ah.value"> <input id="BX0107ag" value=420 size=3 onchange=TX0107ag.width=BX0107ag.value /> <input id="BX0107ah" value=200 size=3 onchange=TX0107ah.width=BX0107ah.value /> <pre><font size=+2><a name="ch01b104">&lt;a name="ch01b104"&gt;</a> 2009-06-30-12-09 here 3 by 3 square matrix in eqn.AB53 disappear.<font color=red> eqn.AB53 right red strings are new created column vector.</font> eqn.AB53 left black alphabet is original 1 by 3 row vector. From 1*x+2*y+3*z it is easy to see the result is dot product between v2 elements and M1 row 1 elements.<font color=red> v2 first element x multiply with M1 row 1 first element 1 plus v2 second element y multiply with M1 row 1 second element 2 plus v2 third element z multiply with M1 row 1 third element 3 to get 1*x+2*y+3*z </font> Second number 4*x+5*y+6*z and third number 7*x+8*y+9*z use same method. <a name="ch01b105">&lt;a name="ch01b105"&gt;</a> Next step in eqn.AB53 left side V1=[a, b, c] dot product with just created column vector, get a * [1*x+2*y+3*z] plus b * [4*x+5*y+6*z] plus c * [7*x+8*y+9*z] Final answer for eqn.AB52 is a1x+a2y+a3z+b4x+b5y+b6z+c7x+c8y+c9z -----eqn.AB54 ÿ<font color=red> a,b,c,x,y,z,1,...,9 in eqn.AB54 are all address label.</font> ÿ <a name="ch01key1">&lt;a name="ch01key1"&gt;</a><!--9806302036 start add-->0<a href="#index">Index begin</a>0<a href="#index06">Index this file</a>  %0Matrix size (<a href="#ch01b127">detail</a>) All matrix need to specify its row number and column number. Please see eqn.AB52. Left side row vector [a, b, c] is one by three rectangular matrix. Middle [1,2,3; 4,5,6; 7,8,9] is three by three square matrix. Right side V2=[x; y; z] is three by one rectangular matrix. String together, it is [1 by 3]*[3 by 3]*[3 by 1] <a name="ch01key2">&lt;a name="ch01key2"&gt;</a><font color=red size=+3> Final answer eqn.AB54 is 1 by 1 "matrix" (pure number). Matrix multiplication [1 by 3]*[3 by 3]*[3 by 1] their neighbor number must be the same. [1 by 3]*[3 by 3]*[3 by 1] is OK [2 by 4]*[4 by 2]*[2 by 3] is OK [1 by 3]*[4 by 3]*[2 by 1] is WRONG! </font><!--9806302040 stop add--><a href=#ch01b124>WHY WRONGÿ</a> <a name="ch01b106">&lt;a name="ch01b106"&gt;</a> Now go back to <a href="#ch01b099">eqn.AB50</a> matrix multiplication ÿfirst step not touch row vector [x<sub>1</sub>, x<sub>2</sub>] 0first step is [5,1;1,3] multiply [y<sub>1</sub>;y<sub>2</sub>] 0 ÿ First step result is [x<sub>1</sub>, x<sub>2</sub>] multiply [5*y<sub>1</sub>+1*y<sub>2</sub>] [1*y<sub>1</sub>+3*y<sub>2</sub>] Please pay attention to that two lines below 'multiply' is a column vector, 2 by 1 matrix. <a name="ch01b107">&lt;a name="ch01b107"&gt;</a> matrix multiplication second step is x<sub>1</sub>*5*y<sub>1</sub> + x<sub>1</sub>*1*y<sub>2</sub> plus x<sub>2</sub>*1*y<sub>1</sub> + x<sub>2</sub>*3*y<sub>2</sub> -----eqn.AB55 <a href="#ch01b094">eqn.AB46</a>Uþ 5x<sub>1</sub>y<sub>1</sub> + x<sub>1</sub>y<sub>2</sub> + x<sub>2</sub>y<sub>1</sub> + 3x<sub>2</sub>y<sub>2</sub> change to <a href="#ch01b099">eqn.AB50</a> then recover to eqn.AB55 eqn.AB55 and eqn.AB46 are same thing. 2009-06-30-12-30 stop <a name="ch01b108">&lt;a name="ch01b108"&gt;</a> 2009-06-30-13-51 start Problem give the following equation (5±x+±y+²x+3²y)<sup>2</sup>f"(5±<sup>2</sup>+2±²+3²<sup>2</sup>)(5x<sup>2</sup>+2xy+3y<sup>2</sup>) -----page 14 eqn.1.22 Less than side formula and greater than side formula are different. How do I know get which one to convert? <a name="ch01b109">&lt;a name="ch01b109"&gt;</a> Less than side formula is dot product of two different vectors. Greater than side formula is dot product of two same vectors (twice). "two same vectors dot product" is special case "two different vectors dot product" is general case We should not use special case, therefore<font color=red> Study "two different vectors dot product" formula and find matrix from it. If it is regular Cauchy Inequality, study the less than side formula. If it is light cone inequality, study the greater than side formula.</font> <font color=red size=+3>Is this right? Please put a question mark.</font>(9807011625) <a name="ch01b110">&lt;a name="ch01b110"&gt;</a> From less than side formula, how to find matrix? Now mark matrix elements with a name. [m11 m12] [m21 m22] <a href="#ch01b099">eqn.AB50</a> matrix is [5 1] [1 3] so we have m11=5 m12=1 m21=1 m22=3<font color=red><!--9806302054 add--> Positive definite matrix is symmetric square matrix. It has the relation m12=m21 (here=1) Rank three and higher rank have mij=mji .</font> <a name="ch01b111">&lt;a name="ch01b111"&gt;</a> Please see <a href="#ch01b099">eqn.AB50</a> and Seq.1 x = (x<sub>1</sub>, x<sub>2</sub>) -----eqn.AB47 Seq.2 y = (y<sub>1</sub>, y<sub>2</sub>) -----eqn.AB48 Seq.1 element1 x<sub>1</sub> multiply Seq.2 element1 y<sub>1</sub> get coefficient of matrix element m11 x<sub>1</sub>y<sub>1</sub> that is 5x<sub>1</sub>y<sub>1</sub> It is also a1x in eqn.AB54 Seq.1 element1 x<sub>1</sub> multiply Seq.2 element2 y<sub>2</sub> get coefficient of matrix element m12 x<sub>1</sub>y<sub>2</sub> that is 1x<sub>1</sub>y<sub>2</sub> It is also a2y in eqn.AB54 <a name="ch01b112">&lt;a name="ch01b112"&gt;</a> Seq.1 element2 x<sub>2</sub> multiply Seq.2 element1 y<sub>1</sub> get coefficient of matrix element m21 x<sub>2</sub>y<sub>1</sub> that is 1x<sub>2</sub>y<sub>1</sub> It is also b4x in eqn.AB54 Seq.1 element2 x<sub>2</sub> multiply Seq.2 element2 y<sub>2</sub> get coefficient of matrix element m22 x<sub>2</sub>y<sub>2</sub> that is 3x<sub>2</sub>y<sub>2</sub> It is also b5y in eqn.AB54 <font color=red> Based on less than side formula 5±x+±y+²x+3²y Write coefficients to matrix proper element get <a href="#ch01b099">eqn.AB50</a> .</font> 2009-06-30-14-19 here <a name="ch01b113">&lt;a name="ch01b113"&gt;</a>  %0Exercise 1.7 solution <script lang="javascript">alert2()</script> Based on less than side formula 5±x+±y+²x+3²y build square matrix [5 1] -----eqn.AB56 [1 3] <a name="ch01b114">&lt;a name="ch01b114"&gt;</a> Find eigenvalue lambda for eqn.AB56 det{ [5-lambda 1 ] } = 0 { [1 3-lambda ] } (5-lambda)*(3-lambda) - 1*1 = 0 lambda*lambda - 8*lambda +15-1 = 0 lambda*lambda - 8*lambda +14 = 0 -----eqn.AB57 <a name="ch01b115">&lt;a name="ch01b115"&gt;</a> eigenvalue lambda is 2.585786437626905 // =((5+3)-sqrt((5-3)*(5-3)+4*1*1))/2 5.414213562373095 // =((5+3)+sqrt(8))/2 Both are positive number, so matrix [5 1] -----eqn.AB56 [1 3] is positive definite matrix. 2009-06-30-14-37 here <a name="ch01b116">&lt;a name="ch01b116"&gt;</a> [M] is positive definite matrix in the following analysis. positive definite matrix [M] has the property (Please see any matrix textbook) for non-zero vector [X] [X]<sup>T</sup> [M] [X] ÿ 0 -----eqn.AB58 ÿ<font color=red>here [X] is column vector, [X]<sup>T</sup> is row vector.</font> Square0matrix [M] right hand side must be column vector and left hand side must be row vector. 0 ÿ eqn.AB58 satisfy <a href="tute0007.htm#ch01a051">vector product rule t$</a> ÿ <a name="ch01b117">&lt;a name="ch01b117"&gt;</a> If [X] is zero vector [0,0] [X]<sup>T</sup> [M] [X] = [x1, x2] [m11 m12] [x1] [m21 m22] [x2] = [ 0, 0] [m11 m12] [ 0] [m21 m22] [ 0] = 0*m11*0 + 0*m12*0 + 0*m21*0 + 0*m22*0 = 0 satisfy <a href="tute0007.htm#ch01a051">vector product rule u$</a> ÿ vector product rule v$0w$0x$ are all easy part. Exercise 1.7 hint skip v$0w$0x$ ÿ Reader may try to prove them. <a name="ch01b118">&lt;a name="ch01b118"&gt;</a> Positive definite matrix product [X]<sup>T</sup>[M][X] satisfy <a href="tute0007.htm#ch01a051">vector product rules</a> ÿ Next is last step. Reference is next The University of Arizona * Instructor for Math 410: Matrix Analysis (Spring 2009) 2009-06-29-23-23 http://math.arizona.edu/%7Edostert/Math410/ 2009-06-29-23-54 start download http://math.arizona.edu/%7Edostert/Math410/Ch3_123.pdf <a name="ch01b119">&lt;a name="ch01b119"&gt;</a> Ch3_123.pdf page 6<font color=red> Theorem: Every inner product satisfies the Cauchy-Schwarz inequality. |0v,w 0|f"%"v%"%"w%" "v,w"V -----eqn.AB59 </font> If vector w is zero vector [0,0,0 ... 0] ÿ then eqn.AB59 is true. Assume w is not zero vector. Let t be any real number. From <a href="tute0007.htm#ch01a051">vector product rule t$</a> <a name="ch01b120">&lt;a name="ch01b120"&gt;</a> 0f"|0v+tw,v+tw 0| 00=%"v%"<sup>2</sup>+2t0v,w 0+t<sup>2</sup>%"w%"<sup>2</sup> -----eqn.AB60 eqn.AB60 has one variable t other parameters are constant. eqn.AB60 has extreme value if differentiate eqn.AB60 and set result to zero. Mark variable t at this point as t<sub>0</sub> get t<sub>0</sub> = ÿ00v,w 0/%"w%"<sup>2</sup> -----eqn.AB61 ÿAssumed %"w%" not zero ÿ%"w%" can be a denominator ÿ <a name="ch01b121">&lt;a name="ch01b121"&gt;</a> Substitute t<sub>0</sub> in eqn.AB61 to eqn.AB60 get 0f"%"v%"<sup>2</sup> - 0v,w 0<sup>2</sup>/%"w%"<sup>2</sup> rearrange and get |0v,w 0|f"%"v%"%"w%" -----eqn.AB59 Exercise 1.7 is done. 2009-06-30-15-28 stop <hr> <a name="ch01b122">&lt;a name="ch01b122"&gt;</a> 2009-07-01-12-03 Every textbook say <a href="#ch01b040">Cauchy Inequality greater than side is upper limit of less than side</a>. This is relative. I am your upper limit, then you are my lower limit. Can we say Cauchy Inequality less than side is lower limit of greater than side? Please see Cauchy Inequality</font></pre> <script lang="javascript"> <!-- HelloCauchy(108) //--> </script> <pre><font size=+2>2009-07-01-12-08 here Cauchy Inequality less than side is dot product of two different vectors. Answer changes if two vectors relative angle change. Cauchy Inequality greater than side is product of two vectors length. Answer NOT change if two vectors relative angle change. Because vectors length has nothing to do with relative angle. Less than side has infinite many possible values. When relative angle become zero, inequality become equality. Greater than side has just one value. This value is less than side maximum possible value. Under these conditions, we say Cauchy Inequality greater than side is upper limit of less than side. It is a better statement. 2009-07-01-12-14 stop <a name="ch01b124">&lt;a name="ch01b124"&gt;</a> 2009-07-31-10-25 start<!--9807311025-->  %0<a href="#ch01key2">Why</a> matrix multiplication 3x3 * 3x1 00neighbor number must be the same? Write a square matrix [1 2 3] [4 5 6] = M1 [7 8 9] its first row is [1 2 3] second row is [4 5 6] third row is [7 8 9] first col is [1; 4; 7] second col is [2; 5; 8] third col is [3; 6; 9] <a name="ch01b125">&lt;a name="ch01b125"&gt;</a> Column vector [1; 4; 7] is simpler notation. Column vector [1; 4; 7] should be written as following [1] [4] [7] Formal notation take three lines, Simpler notation [1; 4; 7] take just one line. ÿrow vector [1 4 7] transpose get [1 4 7]<sup>T</sup> it is also a column vector written in one line ÿ Column vector [1] [4] [7] write as [1; 4; 7] often. [;] indicate change line. <a name="ch01b126">&lt;a name="ch01b126"&gt;</a> Row vector [1 2 3] is called one by three matrix. Because Row vector [1 2 3] has one row and three columns. Column vector [1] [4] [7] is called three by one matrix. Because column vector [1; 4; 7] has three rows and one column. <a name="ch01b127">&lt;a name="ch01b127"&gt;</a> To count the size of a matrix. Start from upper left corner. Go downward along left side, count how many rows are there. Remember it as start from upper left corner of 'L', goto lower left corner. Next count how many columns. Start from lower left corner, go right along lower side. Count how many columns are there. Remember it as start from lower left corner of 'L', goto lower right corner. Based on above method, count matrix M1 [1 2 3] [4 5 6] = M1 [7 8 9] get three rows and three columns. M1 has size 3 by 3. <a name="ch01b128">&lt;a name="ch01b128"&gt;</a> Suppose V2=[11; 22; 33] Based on above method, count matrix V2 [11] [22] [33] get three rows and one columns. V2 has size 3 by 1. M1 multiply V2 as following. First step, arrange in original form <font color=red>[1 2 3]</font> <font color=blue>[11]</font> <font color=red>[4 5 6]</font>*<font color=blue>[22]</font> <font color=red>[7 8 9]</font> <font color=blue>[33]</font> <a name="ch01b129">&lt;a name="ch01b129"&gt;</a> Second step, push turn column vector<font color=blue> [11 22 33]</font><font color=red> [ 1 2 3] [ 4 5 6] [ 7 8 9]</font> Third step, dot product of two vectors<b><font color=green> [11 22 33] [ 1 2 3]</font></b> [ 4 5 6] [ 7 8 9] <a name="ch01b130">&lt;a name="ch01b130"&gt;</a> Fourth step, finish row one calculation.<b><font color=green> [11*1+22*2+33*3]</font></b> <font color=red>&lt;==one row one number 154</font> [ 4 5 6]0000<font color=red>&lt;==one row three numbers</font> [ 7 8 9]0000<font color=red>&lt;==one row three numbers</font> To carry out row two, follow same step. Push turn column vector as following<b><font color=green> [11*1+22*2+33*3]</font></b><font color=blue> [11 22 33]</font><font color=red> [ 4 5 6] [ 7 8 9]</font> <a name="ch01b131">&lt;a name="ch01b131"&gt;</a> To carry out row three, follow same step. Push turn column vector as following<b><font color=green> [11*1+22*2+33*3]</font></b> <font color=red>&lt;==one row one number 154</font><b><font color=green> [11*4+22*5+33*6]</font></b> <font color=red>&lt;==one row one number 352</font><font color=blue> [11 22 33]0000&lt;==calculation in progress.</font><font color=red> [ 7 8 9]</font>0000<font color=red>&lt;==one row three numbers</font> Finally get three by one matrix. [11*1+22*2+33*3]0 0[154] [11*4+22*5+33*6]0=0[352] [11*7+22*8+33*9]0 0[550] <a name="ch01b132">&lt;a name="ch01b132"&gt;</a> One important point is that at start [1 2 3] [11] [4 5 6]*[22] [7 8 9] [33] Two matrices have size 3 by 3 and 3 by 1. 3 by 3 and 3 by 1, the word 'and' has both 3 at its left and right side. OK. Matrix multiplication is possible. On the other hand, if start at 3 by 3 and 2 by 1 like the following [1 2 3] [11] [4 5 6]*[22] [7 8 9] or if start at 3 by 3 and 4 by 1 like the following [1 2 3] [11] [4 5 6]*[22] [7 8 9] [33] [44] After push turn column vector, it is not possible to do dot product for two different size vectors!! For matrix multiplication, it is necessary that neighbor number must be the same. 2009-07-31-11-08 stop <a name="ch01b133">&lt;a name="ch01b133"&gt;</a> 2009-08-01-10-44 start 4 by 3 and 3 by 5 matrix multiplication get what result? Solve this problem in five steps. 'five' come from '3 by 5' . First solve for 4 by 3 and 3 by 1 matrix multiplication This problem is almost the same as <a href="#ch01b128">above example</a>. The result is 4 by 1 column matrix. After do five times, get five columns. Final answer is 4 by 5 matrix. A simple way to remember is the following<font color=red> "Keep head and tail, forget middle". 4 by 3 and 3 by 5 get 4 by 5 matrix. 1 by 3 and 3 by 3 and 3 by 1 get 1 by 1 2 by 4 and 4 by 2 and 2 by 3 get 2 by 3</font> 2009-08-01-10-52 stop <a name="unicod32">&lt;a name="unicod32"&gt;</a>0<a href=tute0007.htm#unicod01>more</a> 2009-07-01-19-56 Just found that vector dot product 0v,w 0 ÿ is much better than ÿv,wÿ The following is a Unicode section contains '0' and ' 0'. &amp;#12281;ÿù/0&amp;#12282;ÿú/0&amp;#12283;ÿû/0&amp;#12284;ÿü/ &amp;#12285;ÿý/0&amp;#12286;ÿþ/0&amp;#12287;ÿÿ/0&amp;#12288;ÿ0 &amp;#12289;ÿ00&amp;#12290;ÿ00&amp;#12291;ÿ00&amp;#12292;ÿ0 &amp;#12293;ÿ00&amp;#12294;ÿ00&amp;#12295;ÿ00&amp;#12296;ÿ0 &amp;#12297;ÿ 00&amp;#12298;ÿ 00&amp;#12299;ÿ 00&amp;#12300;ÿ 0 &amp;#12301;ÿ 00&amp;#12302;ÿ00&amp;#12303;ÿ00&amp;#12304;ÿ0 &amp;#12305;ÿ00&amp;#12306;ÿ00&amp;#12307;ÿ00&amp;#12308;ÿ0 &amp;#12309;ÿ00&amp;#12310;ÿ00&amp;#12311;ÿ00&amp;#12312;ÿ0 &amp;#12313;ÿ00&amp;#12314;ÿ00&amp;#12315;ÿ00&amp;#12316;ÿ0 &amp;#12317;ÿ00&amp;#12318;ÿ00&amp;#12319;ÿ00&amp;#12320;ÿ 0 &amp;#12321;ÿ!00&amp;#12322;ÿ"00&amp;#12323;ÿ#00&amp;#12324;ÿ$0 &amp;#12325;ÿ%00&amp;#12326;ÿ&00&amp;#12327;ÿ'00&amp;#12328;ÿ(0 &amp;#12329;ÿ)00&amp;#12330;ÿ*00&amp;#12331;ÿ+00&amp;#12332;ÿ,0 &amp;#12333;ÿ-00&amp;#12334;ÿ.00&amp;#12335;ÿ/00&amp;#12336;ÿ00 &amp;#12337;ÿ100&amp;#12338;ÿ200&amp;#12339;ÿ300&amp;#12340;ÿ40 &amp;#12341;ÿ500&amp;#12342;ÿ600&amp;#12343;ÿ700&amp;#12344;ÿ80 &amp;#12345;ÿ900&amp;#12346;ÿ:00&amp;#12347;ÿ;00&amp;#12348;ÿ<0 &amp;#12349;ÿ=00&amp;#12350;ÿ>00 2009-07-01-20-00 stop </font></pre> <pre><font size=+2> <a name=docB01>&lt;a name=docB01&gt;</a>0<a href="#index">Index begin</a>0<a href="#index06">Index this file</a> <script language="javascript">alert0()</script> Working record. 2009-06-10-09-45 Build Cauchy Inequality input box 2009-06-10-18-09 Done Cauchy Inequality<!--9806101809--> 2009-06-11-12-00 Improve Cauchy Inequality program. 2009-06-27-12-27 Start six sequence Cauchy Inequality 2009-06-27-18-30 Done six sequence Cauchy Inequality 2009-07-02-09-51 Start generalized Cauchy Inequality 2009-07-02-13-22 Done generalized Cauchy Inequality 2009-07-01-11-23 start proofread tutc0008.htm (Chinese version) 2009-07-01-16-32 done proofread tutc0008.htm 2009-07-28-10-00 copy tutc0008.htm tute0008.htm start tute0008.htm English version 2009-07-31-18-49 translate to right here. 2009-08-03-18-40 done spelling check <a name=docB02>&lt;a name=docB02&gt;</a> 2009-07-02-21-40 start Update 2009-07-01 add exercise 1.4 to 1.7 Update 2009-07-02 add Generalized Cauchy Inequality Use matrix to define dot product rule. <a href=#gCauchy>Generalized Cauchy Inequality program</a> come from exercise 1.7 Can not promise correct output. Please verify first. Special attention: Generalized Cauchy Inequality can reverse the inequality direction! 2009-07-02-21-47 stop <a name=docB03>&lt;a name=docB03&gt;</a> 2009-07-03-22-40 start Update 2009-07-04 add matrix eigenvalue and eigenvector calculation. To find matrix eigenvalue and eigenvector that is LiuHH's long time hope. But LiuHH capability is limited, can only use other people's code. Change to Javascript. Start from 2009-04-14, LiuHH find online resources Found three C-code for matrix eigenvalue and eigenvector. Convert to Javascript, but failed one after one. Each code has its own difficulty. Now write Generalized Cauchy Inequality, I must know eigenvalue is positive or negative. If get negative answer, Generalized Cauchy Inequality reverse its inequality direction ! Today 2009-07-03 review 2009-04-14 work. Find why fail. In three different codes, eigens() in cephes27.zip has smallest trouble. Finally 2009-07-03-19-27 found what is wrong, add one line code N=parseInt(N); //9807031927 then get solution. Now include eigens() in tutc0008.htm This web page <a href=tute0008.htm#gCauchy>is capable to find matrix eigenvalue</a>. Then I feel relief. Better version is in <a href=tute0009.htm#gCauchy>tute0009.htm</a> Please pay attention to the following function eigens() solve only for symmetry matrix. If you have non-symmetric matrix, eigens() can not help. Input symmetric matrix in box9, output at box8. below box9, there is a button "eigenvalue \n eigenvector" 2009-07-03-23-00 stop <a name=docB04>&lt;a name=docB04&gt;</a> 2009-07-03-20-27 start (9807032027) function eigens( A, RR, E, N ) On 2000-04-18-01-41 start download cephes27.zip from ftp://ftp.simtel.net/pub/simtelnet/msdos/c/cephes27.zip 2000-04-18-02-42 done CEPHES7 ZIP 1,307,024 04-18-00 2:41a cephes7.zip <a name=docB05>&lt;a name=docB05&gt;</a> 2003-07-02-21-21 from ftp://ftp.simtel.net/pub/simtelnet/msdos/c/cephes28.zip You are user #30 of 500 simultaneous users allowed. download cephes28.zip 2003-07-02-21-38 done 07/02/2003 10:38 PM 1,906,887 cephes28.zip <a name=docB06>&lt;a name=docB06&gt;</a> 2009-04-14-19-12 (9804141912) try to convert from C language to Javascript but not success. 2009-07-03-19-27 found key point, add N=parseInt(N); solve problem. Now include eigens() to tutc0008.htm <a name=docB07>&lt;a name=docB07&gt;</a> function eigens( A, RR, E, N ) accept only symmetric matrix. input argument A store lower triangle matrix input argument N store matrix rank output argument RR return eigenvector output argument E return eigenvalue If N =5 A = 15 array of fifteen numbers. 1+2+3+4+5=15 RR = 5*5 matrix ÿeach eigenvector has five components. total five vectors. E = 5 vector for five eigenvalues. 2009-04-14-20-40 (9807032040) stop In fact, this file use next code to get answer return [E,RR]; //9807032114 <a name=docB08>&lt;a name=docB08&gt;</a> 2009-07-04-11-38 start (9807041138) 3 by 3 symmetric matrix [a b c] [b d e] [c e f] store next six values in box9 a b d c e f <a name=docB09>&lt;a name=docB09&gt;</a> On the other hand, if see lower triangle matrix is defined as 11 12 22 13 23 33 14 24 34 44 15 25 35 45 55 then full matrix is 11 12 13 14 15 12 22 23 24 25 13 23 33 34 35 14 24 34 44 45 15 25 35 45 55 store only lower triangle part to save storage memory. 2009-07-04-11-42 stop (9807041142) <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 </font></pre> <pre><font size=+2> <a name="Copyright">&lt;a name="Copyright"&gt;</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> <br><hr><br> Javascript index <br> <a href=http://freeman2.com/jsindex2.htm> http://freeman2.com/jsindex2.htm </a> &nbsp; <a href=jsindex2.htm> local </a> <!--9810201450 add local link--> <br> Save graph code to same folder as htm files. <br> <a href=http://freeman2.com/jsgraph2.js> http://freeman2.com/jsgraph2.js </a> &nbsp; <a href=jsgraph2.js> local </a> <br> <script src="jslist1e.js" language="javascript"></script> <br> <br> URL of this file Inequality file two <br> <a href=http://freeman2.com/tute0008.htm> http://freeman2.com/tute0008.htm </a> <br> first upload 2009-08-04 <br> <br> Thank you for visiting Freeman's page.0 <br> Freeman <br> 2009-07-31-19-39 <br> <br> <span id="tuteLink2"></span> <br> <script src='http://freeman2.com/linksec1.js' language='javascript'></script> <br> <script src='http://freeman2.com/linkall1.js' language='javascript'></script> </BODY> </HTML>