ÿþ<HTML> <HEAD> <TITLE>2nd Positive Integer power sum COefficient equation</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" /> <!--jspico_e.htm author --> <meta name="author" content="Liu, HsinHan ‰Rk”"o &#21129;&#37995;&#28450;" /> <script type="text/javascript"> <!-- function initial0(linkNum, plotNum) { tuteLink(0); //9806231221 } 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 //--> </SCRIPT> <script src="http://freeman2.com/tutelink.js" language="javascript"></script><!--9806111210--> <script src="tutelink.js" language="javascript"></script><!--9806111211--> </HEAD> <body link="#FF0000" vlink="#0000FF" alink="#50A000" bgcolor="#ccfcfc" onload="javascript:initial0()" > <span id="gotoFM2msg"></span><!--9712091951--> <a href=jspico_e.htm#docA01> First P.I.CO file </a> 00 <font color=red> This is SECOND page of </font> <br> <a href=jspico2e.htm#docA001> Positive Integer power sum COefficient equation </a> <br> <a href="jspico2e.htm#program0"> program </a> 0 <a href="jspico_e.htm#Arndt01">Arndt</a> 0 <a href=jspico2e.htm#docA001> DocA </a> 0 <a href="jspico_e.htm#getCol5">first9</a> 0 <a href=jspico2e.htm#docA113> Update 2009-09-25 </a> 0 <br> <a href="jspico_e.htm#docB04">Coefficient triangle</a>0 <a href="jspico_e.htm#docB02">Pattern conjectures</a>0 <a href="jspico_e.htm#docB18">Column 5 pattern</a>0 <br> <a href=jspico_e.htm#docB41>Proof</a> coef. column <!--9809051419--> <a href=jspico_e.htm#docB77>01</a>0 <a href=jspico_e.htm#docB82>02</a>0 <a href=jspico_e.htm#docB87>03</a>0 <a href=jspico_e.htm#docB100>04</a>0 <a href=jspico_e.htm#docB109>05</a>0 <a href=jspico_e.htm#docB110>List</a>0 <br> <a href=jspico2e.htm#docB001>Proof</a> coef. <!--9809101009--> <a href=jspico2e.htm#docB006>k+1</a> <a href=jspico2e.htm#docB009>k</a> <a href=jspico2e.htm#docB016>k-1</a> <a href=jspico2e.htm#docB024>k-2</a> <a href=jspico2e.htm#docB043>k-3</a> <a href=jspico2e.htm#docB057>k-4</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> <br> <hr> <pre><font size=+2><a name="docA001">&lt;a name="docA001"&gt;</a><a name="docA01"></a> Section "docA001" prime, goto <a href=http://freeman2.com/jsprime2.htm>freeman2.com</a>/<a href=jsprime2.htm>jsprime2.htm</a> Section "<a href="#docA101">docA101</a>" general introduction Section "<a href="#docB001">docB001</a>" symbolic equation, wondering Section "<a href="#docC001">docC001</a>" program in/output, wondering Section "<a href="#docD001">docD001</a>" program intro., no wondering 2009-09-07-14-55 start To find more Positive Integer power sum COefficient equation We have Coefficient <a href=jspico_e.htm#docB04>triangle</a> and coefficient <a href=jspico_e.htm#docB110>list</a> in hand. But based on these data, we are unable to find next new higher power equation. Second file (this file) http://freeman2.com/jspico2e.htm attempt to find more information. <a name="docA002">&lt;a name="docA002"&gt;</a><a name="docA02"></a> First create <a href=jspico2e.htm#program0>program</a> to build Prime numbers and in huge number numerator/denominator cancel common prime factor and reduce to shorter rational number. 2009-09-07-15-04 stop </font></pre> <center> <table border=2><!--9809172118--> <td> Prime program moved to next URL<a href=http://freeman2.com/jsprime2.htm> <br> http://freeman2.com/jsprime2.htm</a> <br> If you saved a copy and place in the same <br> folder as jspico2e.htm, then please click <br> <a href=jsprime2.htm> jsprime2.htm</a> Freeman 2009-09-17-21-22 </td> </table> </center> <br> <br> <script language="javascript">alert0()</script> <hr> <hr> <pre><font size=+2><a name="docA101">&lt;a name="docA101"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a> 2009-09-17-21-27 start This file jspico2e.htm is second page of Positive Integer power sum COefficient equation. jspico2e.htm has a program help you to build power sum equations. <a name="docA102">&lt;a name="docA102"&gt;</a> The simplest equation is " i to n = +n*n*(1/2)+n*(1/2) Three following equations are " i*i to n = +n*n*n*(1/3)+n*n*(1/2)+n*(1/6) " i^3 to n = +n*n*n*n*(1/4)+n*n*n*(1/2)+n*n*(1/4) " i^4 to n = +n*n*n*n*n*(1/5)+n*n*n*n*(1/2)+n*n*n*(1/3)+n*n*(0)+n*(-1/30) <a name="docA103">&lt;a name="docA103"&gt;</a> Better mathematics equations are</font></pre> <TABLE WIDTH="360" id=T202Liua border=0 ><!--9809172136--> <TR> <TD> &nbsp; </TD> <TD> <TABLE WIDTH="140" border=0 id=T202Liub > <TR> <TD> <CENTER>i=n</CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER>i=1</CENTER> </TD> <TD> i </TD> <TD> = </TD> <TD> <CENTER>n(n+1)</CENTER> <CENTER><hr color=black></CENTER> <CENTER>2</CENTER> </TD> </TR> </TABLE> </TD> <TD> ---integer first power <br> ---summation n-formula </TD> </TR> </TABLE> width of above equation <INPUT name=Button202Liua type="button" value="default" onclick="B202Liua.value=360,B202Liub.value=140,T202Liua.width=B202Liua.value,T202Liub.width=B202Liub.value"> <input id="B202Liua" value=360 size=3 onchange=T202Liua.width=B202Liua.value /> <input id="B202Liub" value=140 size=3 onchange=T202Liub.width=B202Liub.value /> <br> <a name="docA104">&lt;a name="docA104"&gt;</a> <br> <TABLE WIDTH="440" id=T203Liua border=0 ><!--9809172145--> <TR> <TD> &nbsp; </TD> <TD> <TABLE WIDTH="200" border=0 id=T203Liub > <TR> <TD> <CENTER>i=n</CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER>i=1</CENTER> </TD> <TD> i<sup>2</sup> </TD> <TD> = </TD> <TD> <CENTER>n(n+1)(2n+1)</CENTER> <CENTER><hr color=black></CENTER> <CENTER>6</CENTER> </TD> </TR> </TABLE> </TD> <TD> ---integer second power <br> ---summation n-formula </TD> </TR> </TABLE> width of above equation <INPUT name=Button203Liua type="button" value="default" onclick="B203Liua.value=440,B203Liub.value=200,T203Liua.width=B203Liua.value,T203Liub.width=B203Liub.value"> <input id="B203Liua" value=440 size=3 onchange=T203Liua.width=B203Liua.value /> <input id="B203Liub" value=200 size=3 onchange=T203Liub.width=B203Liub.value /> <br> <a name="docA105">&lt;a name="docA105"&gt;</a> <br> <TABLE WIDTH="440" id=T204Liua border=0 ><!--9809172147--> <TR> <TD> &nbsp; </TD> <TD> <TABLE WIDTH="200" border=0 id=T204Liub > <TR> <TD> <CENTER>i=n</CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER>i=1</CENTER> </TD> <TD> i<sup>3</sup> </TD> <TD> = </TD> <TD> <CENTER>n<sup>2</sup>(n+1)<sup>2</sup></CENTER> <CENTER><hr color=black></CENTER> <CENTER>4</CENTER> </TD> </TR> </TABLE> </TD> <TD> ---integer third power <br> ---summation n-formula </TD> </TR> </TABLE> width of above equation <INPUT name=Button204Liua type="button" value="default" onclick="B204Liua.value=440,B204Liub.value=200,T204Liua.width=B204Liua.value,T204Liub.width=B204Liub.value"> <input id="B204Liua" value=440 size=3 onchange=T204Liua.width=B204Liua.value /> <input id="B204Liub" value=200 size=3 onchange=T204Liub.width=B204Liub.value /> <br> <a name="docA106">&lt;a name="docA106"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a> <br> <TABLE WIDTH="520" id=T205Liua border=0 ><!--9809172150--> <TR> <TD> &nbsp; </TD> <TD> <TABLE WIDTH="280" border=0 id=T205Liub > <TR> <TD> <CENTER>i=n</CENTER> <CENTER><FONT SIZE=+3>"</FONT></CENTER> <CENTER>i=1</CENTER> </TD> <TD> i<sup>4</sup> </TD> <TD> = </TD> <TD> <CENTER>n(n+1)(6*n<sup>3</sup>+9*n<sup>2</sup>+n-1)</CENTER> <CENTER><hr color=black></CENTER> <CENTER>30</CENTER> </TD> </TR> </TABLE> </TD> <TD> ---integer fourth power <br> ---summation n-formula </TD> </TR> </TABLE> width of above equation <INPUT name=Button205Liua type="button" value="default" onclick="B205Liua.value=520,B205Liub.value=280,T205Liua.width=B205Liua.value,T205Liub.width=B205Liub.value"> <input id="B205Liua" value=520 size=3 onchange=T205Liua.width=B205Liua.value /> <input id="B205Liub" value=280 size=3 onchange=T205Liub.width=B205Liub.value /> <pre><font size=+2><a name="docA107">&lt;a name="docA107"&gt;</a> 2009-08-28-18-48 LiuHH access http://www.jackpo.org/wp-content/uploads/2006/03/formulas.pdf save as jackpo.org_formulas.pdf page 2 (total 10 pages) has above four equations. <a name="docA108">&lt;a name="docA108"&gt;</a> High power equations are following " i^5 i=1 to n = +n*n*n*n*n*n*(1/6)+n*n*n*n*n*(1/2)+n*n*n*n*(5/12)+n*n*n*(0)+n*n*(-1/12) " i^6 i=1 to n = +n*n*n*n*n*n*n*(1/7)+n*n*n*n*n*n*(1/2)+n*n*n*n*n*(1/2)+n*n*n*n*(0)+n*n*n*(-1/6)+n*n*(0)+n*(1/42) " i^7 i=1 to n = +n*n*n*n*n*n*n*n*(1/8)+n*n*n*n*n*n*n*(1/2)+n*n*n*n*n*n*(7/12)+n*n*n*n*n*(0)+n*n*n*n*(-7/24)+n*n*n*(0)+n*n*(1/12) " i^8 i=1 to n = +n*n*n*n*n*n*n*n*n*(1/9)+n*n*n*n*n*n*n*n*(1/2)+n*n*n*n*n*n*n*(2/3)+n*n*n*n*n*n*(0)+n*n*n*n*n*(-7/15)+n*n*n*n*(0)+n*n*n*(2/9)+n*n*(0)+n*(-1/30) " i^9 i=1 to n = +n*n*n*n*n*n*n*n*n*n*(1/10)+n*n*n*n*n*n*n*n*n*(1/2)+n*n*n*n*n*n*n*n*(3/4)+n*n*n*n*n*n*n*(0)+n*n*n*n*n*n*(-7/10)+n*n*n*n*n*(0)+n*n*n*n*(1/2)+n*n*n*(0)+n*n*(-3/20)+n*(0)+(0) "i^10 i=1 to n = +n*n*n*n*n*n*n*n*n*n*n*(1/11)+n*n*n*n*n*n*n*n*n*n*(1/2)+n*n*n*n*n*n*n*n*n*(10/12)+n*n*n*n*n*n*n*n*(0)+n*n*n*n*n*n*n*(-1)+n*n*n*n*n*n*(0)+n*n*n*n*n*(1)+n*n*n*n*(0)+n*n*n*(-1/2)+n*n*(0)+n*(5/66)+(0) "i^11 i=1 to n = +n*n*n*n*n*n*n*n*n*n*n*n*(1/12)+n*n*n*n*n*n*n*n*n*n*n*(1/2)+n*n*n*n*n*n*n*n*n*n*(11/12)+n*n*n*n*n*n*n*n*n*(0)+n*n*n*n*n*n*n*n*(-165/120)+n*n*n*n*n*n*n*(0)+n*n*n*n*n*n*(462/252)+n*n*n*n*n*(0)+n*n*n*n*(-165/120)+n*n*n*(0)+n*n*(50/120) "i^12 i=1 to n = +n*n*n*n*n*n*n*n*n*n*n*n*n*(1/13)+n*n*n*n*n*n*n*n*n*n*n*n*(1/2)+n*n*n*n*n*n*n*n*n*n*n*(12/12)+n*n*n*n*n*n*n*n*n*n*(0)+n*n*n*n*n*n*n*n*n*(-220/120)+n*n*n*n*n*n*n*n*(0)+n*n*n*n*n*n*n*(792/252)+n*n*n*n*n*n*(0)+n*n*n*n*n*(-396/120)+n*n*n*n*(0)+n*n*n*(2200/1320)+n*n*(0)+n*(-7601/30030) <a name="docA109">&lt;a name="docA109"&gt;</a> " i^5 to " i^8 four equations are program output from <a href=http://freeman2.com/jspico_e.htm#program0>jspico_e.htm</a>#<a href=jspico_e.htm#program0>program0</a> " i^9 to " i^12 are try and error, hand calculated equations. <a name="docA110">&lt;a name="docA110"&gt;</a> Higher power equation is hard to do with hand. LiuHH wrote <a href="jspico2e.htm#program0">program make_km()</a> to do the calculation. Found " i^13 i=1 to n <a href="#docC051">formula</a> is pow(n,14)*1/14+pow(n,13)*1/2+pow(n,12)*13/12-pow(n,10)*286/120+pow(n,8)*1287/252-pow(n,6)*858/120+pow(n,4)*7150/1320-pow(n,2)*98813/60060 " i^14 i=1 to n <a href="#docD027">formula</a> is pow(n,15)*(1/15)+pow(n,14)*(1/2)+pow(n,13)*(14/12)+pow(n,11)*(-364/120)+pow(n,9)*(2002/252)+pow(n,7)*(-1716/120)+pow(n,5)*(20020/1320)+pow(n,3)*(-1383382/180180)+pow(n,1)*(7/6) All of above fourteen formulas are exact. <a name="docA111">&lt;a name="docA111"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a> It is possible to use <a href="jspico2e.htm#program0">program make_km()</a> to get higher " i^15 formulas. Method is described in Section "<a href="#docD001">docD001</a>" program intro., no wondering Possible human errors described in Section "<a href="#docC001">docC001</a>" program in/output, wondering You are welcome to use this program. Freeman 2009-09-17-22-17 <a name="docA112">&lt;a name="docA112"&gt;</a> "Update 2009-09-20" made a correction, put function bicof() back to program. On 2009-09-18 found file end has function bye09(). Because jspico2e.htm file size is large. I delete all Javascript at file end. Include delete function bicof() which cause all "km2" to "m20" button stop work. Without knowing this damage, LiuHH uploaded a not-working page to Internet. Sorry. Now "Update 2009-09-20" should work. Please also visit Prime page at<a href=http://freeman2.com/jsprime2.htm> http://freeman2.com</a>/<a href=jsprime2.htm>jsprime2.htm</a> Add prime gap function. Thank you for visiting Freeman's web site. Freeman 2009-09-20-13-57 <a name="docA113">&lt;a name="docA113"&gt;</a> 2009-09-25-16-45 start 2009-09-25-12-18 found " i^15 i=1 to n formula is +(n^16)/(15+1)+(n^15)/2+(n^14)*15/12 +(n^13)*0-(n^12)*455/120+(n^11)*0 +(n^10)*715/60+(n^9)*0-(n^8)*429/16 +(n^7)*0+(n^6)*455/12+(n^5)*0 -(n^4)*691/24+(n^3)*0+(n^2)*35/4+(n^1)*0 Next line is Javascript readable formula. define n=5 (for example) first n=5; +pow(n,16)/(15+1)+pow(n,15)/2+pow(n,14)*15/12+pow(n,13)*0-pow(n,12)*455/120+pow(n,11)*0+pow(n,10)*715/60+pow(n,9)*0-pow(n,8)*429/16+pow(n,7)*0+pow(n,6)*455/12+pow(n,5)*0-pow(n,4)*691/24+pow(n,3)*0+pow(n,2)*35/4+pow(n,1)*0 " i^15 i=1 to 5 answer is 31605701625 <a name="docA114">&lt;a name="docA114"&gt;</a> 2009-09-25-14-06 found constant in function km15(k) { //9809251332 k=parseInt(k); return '('+eval(k*(k-1)*(k-2)*(k-3)*(k-4)*(k-5)*(k-6)*(k-7)*(k-8)*(k-9)*(k-10)*(k-11)*(k-12)*(k-13)*(k-14))+')/(-22120201153290240000/7498041)'; //9809251406 } is (-22120201153290240000/7498041) 2009-09-25-16-54 stop </font></pre> <hr> <hr> <pre><font size=+2><a name="docB001">&lt;a name="docB001"&gt;</a> Section "docA001" prime, goto <a href=http://freeman2.com/jsprime2.htm>freeman2.com</a>/<a href=jsprime2.htm>jsprime2.htm</a> Section "<a href="#docA101">docA101</a>" general introduction Section "<a href="#docB001">docB001</a>" symbolic equation, wondering Section "<a href="#docC001">docC001</a>" program in/output, wondering Section "<a href="#docD001">docD001</a>" program intro., no wondering 2009-09-08-19-35 start Above write program to find a way for Huge rational reduce to shorter rational. (Prime moved to <a href=http://freeman2.com/jsprime2.htm>freeman2.com</a>/<a href=jsprime2.htm>jsprime2.htm</a>) This function will be used in determine<a href="jspico_e.htm#docB04"> Coefficient triangle</a> bottom new longer coefficients. In first page <a href=http://freeman2.com/jspico_e.htm>freeman2.com</a>/<a href=jspico_e.htm>jspico_e.htm</a> Calculated general equation for "{i}=n*(n+1)/2 for "{i*i}=n*(n+1)*(2*n+1)/6 for "{i^3}=[n*(n+1)/2]^2 ..... for "{i^12}= +n*n*n*n*n*n*n*n*n*n*n*n*n*(1/13)+n*n*n*n*n*n*n*n*n*n*n*n*(1/2)+n*n*n*n*n*n*n*n*n*n*n*(12/12)+n*n*n*n*n*n*n*n*n*(-220/120)+n*n*n*n*n*n*n*(792/252)+n*n*n*n*n*(-396/120)+n*n*n*(2200/1320)+n*(-7601/30030) <a name="docB002">&lt;a name="docB002"&gt;</a> Trouble is that it is not easy to predict future equation from existing equation. The reason is that first page ignore symbolic equation and change to numerical value quickly. From numerical value, the triangle structure disappear. Now, second page, try re-do first page work and reserve symbolic equation as long as possible. Hopefully detect structure from symbolic equation. 2009-09-08-19-58 here <a name="docB003">&lt;a name="docB003"&gt;</a> Re-write a<sub>j</sub> summary j=k+1 to k-7 a<sub>k+1</sub>=1/(k+1) a<sub>k</sub> =1/2 a<sub>k-1</sub>=k/12 a<sub>k-2</sub>=0 a<sub>k-3</sub>=-k*(k-1)*(k-2)/(5*4*3*2*3*2) a<sub>k-4</sub>=0 a<sub>k-5</sub>=k*(k-1)*(k-2)*(k-3)*(k-4)/(7*6*5*4*3*2*3*2) a<sub>k-6</sub>=0 a<sub>k-7</sub>= -k*(k-1)*(k-2)*(k-3)*(k-4)*(k-5)*(k-6)/(8*7*6*5*4*3*2*3*2*5) <a name="docB004">&lt;a name="docB004"&gt;</a> Define b<sub>k+1</sub>=1/(k+1) b<sub>k</sub> =1/2 b<sub>k-1</sub>=k/12 b<sub>k-2</sub>=0 b<sub>k-3</sub>=1/(5*4*3*2*3*2) b<sub>k-4</sub>=0 b<sub>k-5</sub>=1/(7*6*5*4*3*2*3*2) b<sub>k-6</sub>=0 b<sub>k-7</sub>=1/(8*7*6*5*4*3*2*3*2*5) [[ 2009-09-18-08-05 note: define b<sub>k+1</sub> etc. the purpose is hope find a<sub>k+1</sub> etc constant pattern and help to predict future value. Finally, not use b<sub>k+1</sub> and even give up symbolic equation work. Turn to program make_km() for solution. ]] <hr> <a name="general_equation">&lt;a name="general_equation"&gt;</a> From general equation <a href=jspico_e.htm#docB55>eqn.AA209</a> or from expanded whole <a href=jspico_e.htm#docB60>eqn.AA211</a> or from detailed whole <a href=jspico_e.htm#docB71>eqn.AA217</a> Collect n^(k-3) terms. Need its coefficient, set coef to zero to find a<sub>k-2</sub> (set coef to zero? please see <a href="jspico2e.htm#arbitraryN1">n is arbitrary</a>) <a name="docB005">&lt;a name="docB005"&gt;</a> Start symbolic derivation now. 2009-09-08-20-07 here from eqn.AA217 <a href=jspico_e.htm#docB71>start</a> <a href=jspico_e.htm#docB73>end</a> collect coefficients for highest term pow(n,k+1) * [ bino(k+1,k+1)*a<sub>k+1</sub> - a<sub>k+1</sub> ] We know bino(k+1,k+1)=1, so highest term pow(n,k+1) * (zero) Nothing new. <hr> <a name="docB006">&lt;a name="docB006"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a> Next see pow(n,k) coef. to find a<sub>k+1</sub>. From <a href=jspico_e.htm#docB71>eqn.AA217</a> collect coefficients for second highest term. (bold red terms) Set coef to zero to find a<sub>k+1</sub> (set coef to zero? see <a href="jspico2e.htm#arbitraryN1">n is arbitrary</a>) <font size=+2 color=red><b>pow(n,k) * [ +bino(k+1,k)*a<sub>k+1</sub> +bino(k,k) *a<sub>k</sub> <== cancel -bino(k,k) - a<sub>k</sub> <== cancel ] </b></font> <a name="docB007">&lt;a name="docB007"&gt;</a> Where bino(k,k)=1 and a<sub>k</sub> term cancel left pow(n,k)*[bino(k+1,k)*a<sub>k+1</sub>-1] Equation must be true for ANY n We require pow(n,k) coefficients to be zero pow(n,k)* [bino(k+1,k)*a<sub>k+1</sub>-1] = 0 ---eqn.AA218 This requirement result bino(k+1,k)*a<sub>k+1</sub>-1 = 0 or <a name="docB008">&lt;a name="docB008"&gt;</a> a<sub>k+1</sub> = 1/bino(k+1,k) = 1/(k+1) ---eqn.AA219 **new b<sub>k+1</sub>, pow(n,k)** <font color=red> a<sub>k+1</sub> = 1/bino(k+1,k) ---eqn.AB001 define b<sub>k+1</sub> = 1/bino(k+1,k) ---eqn.AB002 </font> 2009-09-08-20-17 here <hr> <a name="docB009">&lt;a name="docB009"&gt;</a> Next see pow(n,k-1) coef. to find a<sub>k</sub>. From <a href=jspico_e.htm#docB71>eqn.AA217</a> collect coefficients for third highest term (bold blue terms) Set coef to zero to find a<sub>k</sub> (set coef to zero? see <a href="jspico2e.htm#arbitraryN1">n is arbitrary</a>) <font size=+2 color=blue><b>pow(n,k-1) * [ +bino(k+1,k-1)*a<sub>k+1</sub> +bino(k,k-1) *a<sub>k</sub> +bino(k-1,k-1)*a<sub>k-1</sub> <== cancel -bino(k,k-1) - a<sub>k-1</sub> <== cancel ] </b></font> <a name="docB010">&lt;a name="docB010"&gt;</a> Where bino(k-1,k-1)=1 and a<sub>k-1</sub> drop left pow(n,k-1) * ---eqn.AA220 [ +bino(k+1,k-1)*a<sub>k+1</sub> +bino(k,k-1) *a<sub>k</sub> -bino(k,k-1) ] eqn.AA220 has three terms. they have the following values bino(k+1,k-1)=(k+1)!/[(k-1)!*2!]=(k+1)*k/2 <a name="docB011">&lt;a name="docB011"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a> from eqn.AA219, a<sub>k+1</sub>=1/(k+1) bino(k+1,k-1)*a<sub>k+1</sub>=(k+1)*k/2/(k+1)=k/2 ---eqn.AA221 bino(k,k-1)*a<sub>k</sub>=k*a<sub>k</sub> ---eqn.AA222 -bino(k,k-1)=-k ---eqn.AA223 eqn.AA220 become pow(n,k-1) * ---eqn.AA224 [ +k/2 +k*a<sub>k</sub> -k ] <a name="docB012">&lt;a name="docB012"&gt;</a> Above is first page work, change to numerical value quickly. Below keep symbolic equation in pow(n,k-1) * ---eqn.AA220 [ +bino(k+1,k-1)*a<sub>k+1</sub> +bino(k,k-1) *a<sub>k</sub> -bino(k,k-1) ] <a name="docB013">&lt;a name="docB013"&gt;</a> set +bino(k+1,k-1)*a<sub>k+1</sub> +bino(k,k-1) *a<sub>k</sub> -bino(k,k-1) to zero, solve for a<sub>k</sub> get a<sub>k</sub> = [bino(k,k-1)-bino(k+1,k-1)*a<sub>k+1</sub>]/bino(k,k-1) a<sub>k</sub> = 1 - a<sub>k+1</sub>*bino(k+1,k-1)/bino(k,k-1) <a name="docB014">&lt;a name="docB014"&gt;</a> from eqn.AA219 a<sub>k+1</sub> = 1/bino(k+1,k) a<sub>k</sub> = 1 - bino(k+1,k-1)/bino(k,k-1)/bino(k+1,k) where bino(k+1,k-1)/bino(k,k-1)/bino(k+1,k)= [(k+1)*k/2!]/k/(k+1) = 1/2! = 1/2 a<sub>k</sub> = 1 - 1/2 = 1/2 <a name="docB015">&lt;a name="docB015"&gt;</a> **new b<sub>k</sub>, pow(n,k-1)** <font color=red> a<sub>k</sub> = 1/2 ---eqn.AB003 define b<sub>k</sub> = 1/2 ---eqn.AB004 </font> 2009-09-08-20-42 here <hr> <a name="docB016">&lt;a name="docB016"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a> Next see pow(n,k-2) coef. to find a<sub>k-1</sub>. From <a href=jspico_e.htm#docB71>eqn.AA217</a> collect coefficients for fourth highest term (bold purple terms) Set coef to zero to find a<sub>k-1</sub> (set coef to zero? see <a href="jspico2e.htm#arbitraryN1">n is arbitrary</a>) <font size=+2 color=purple><b>pow(n,k-2) * [ +bino(k+1,k-2)*a<sub>k+1</sub> +bino(k,k-2)*a<sub>k</sub> +bino(k-1,k-2)*a<sub>k-1</sub> +bino(k-2,k-2)*a<sub>k-2</sub> <== cancel -bino(k,k-2) -a<sub>k-2</sub> <== cancel ] </b></font> <a name="docB017">&lt;a name="docB017"&gt;</a> Evaluate bino(), get pow(n,k-2) * ---eqn.AA226 [ +(k+1)!/[(k-2)!*3!]*1/(k+1) +k!/[(k-2)!*2!]*1/2 +(k-1)*a<sub>k-1</sub> +a<sub>k-2</sub> -k!/[(k-2)!*2!] -a<sub>k-2</sub> ] Here +bino(k-2,k-2)*a<sub>k-2</sub>-a<sub>k-2</sub> cancel Demand coefficient to be zero get ..... <a name="docB018">&lt;a name="docB018"&gt;</a> Above is page one work, change to numerical value quickly. Below is page two work, preserve symbolic equation as long as possible. Start from pow(n,k-2) * [ +bino(k+1,k-2)*a<sub>k+1</sub> +bino(k,k-2)*a<sub>k</sub> +bino(k-1,k-2)*a<sub>k-1</sub> +bino(k-2,k-2)*a<sub>k-2</sub> -bino(k,k-2) -a<sub>k-2</sub> ] <a name="docB019">&lt;a name="docB019"&gt;</a> in which "+bino(k-2,k-2)*a<sub>k-2</sub>-a<sub>k-2</sub>" is zero, since bino(r,r)=1, a<sub>k-2</sub>-a<sub>k-2</sub>=0 <font color=red> Because n is arbitrary, must demand coefficient to be zero, then we have </font> +bino(k+1,k-2)*a<sub>k+1</sub> +bino(k,k-2)*a<sub>k</sub> +bino(k-1,k-2)*a<sub>k-1</sub> -bino(k,k-2) = 0 <a name="docB020">&lt;a name="docB020"&gt;</a> a<sub>k-1</sub> is unknown, a<sub>k+1</sub> and a<sub>k</sub> are solved. +bino(k-1,k-2)*a<sub>k-1</sub> =bino(k,k-2)-bino(k+1,k-2)*a<sub>k+1</sub>-bino(k,k-2)*a<sub>k</sub> a<sub>k-1</sub> =bino(k,k-2)/bino(k-1,k-2) -a<sub>k+1</sub>*bino(k+1,k-2)/bino(k-1,k-2) -a<sub>k</sub>*bino(k,k-2)/bino(k-1,k-2) from eqn.AB001 a<sub>k+1</sub> = 1/bino(k+1,k) from eqn.AB003 a<sub>k</sub> = 1/2 <a name="docB021">&lt;a name="docB021"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a><font color=red> a<sub>k-1</sub> = bino(k,k-2)/bino(k-1,k-2) ---eqn.AB005A -a<sub>k+1</sub>*bino(k+1,k-2)/bino(k-1,k-2) -a<sub>k</sub>*bino(k,k-2)/bino(k-1,k-2) Start from a<sub>k-1</sub>, general equation for a<sub>k-some_value</sub>, show up a<sub>k+1</sub> and a<sub>k</sub> Use eqn.AB005A, ignore eqn.AB005B. (2009-09-09-18-07 note) </font> a<sub>k-1</sub> = bino(k,k-2)/bino(k-1,k-2) -bino(k+1,k-2)/bino(k+1,k)/bino(k-1,k-2) -bino(k,k-2)/bino(k-1,k-2)/2 ---eqn.AB005B Look like must change to numerical value now <a name="docB022">&lt;a name="docB022"&gt;</a> a<sub>k-1</sub> = [k*(k-1)/2!]/[(k-1)/1!] -[(k+1)*k*(k-1)/3!]/[(k+1)/1!]/[(k-1)/1!] -[k*(k-1)/2!]/[(k-1)/1!]/2 = k/2 -k/3! -k/4 = k/2 - k/6 -k/4 = 6k/12 - 2k/12 -3k/12 <a name="docB023">&lt;a name="docB023"&gt;</a> **new b<sub>k-1</sub>, pow(n,k-2)** <font color=red> a<sub>k-1</sub> = k/12 ---eqn.AB006 define b<sub>k-1</sub> = k/12 ---eqn.AB007 </font> 2009-09-08-21-09 stop <hr> <a name="arbitraryN1">&lt;a name="arbitraryN1"&gt;</a> <font color=red>n is arbitrary</font> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a> 2009-09-09-12-46 start If we see an equation a*b = 0 where 'b' is arbitrary and finite. We must conclude that a=0 In the general equation "{i}=n*(n+1)/2 the value of 'n' is arbitrary. (equation is true for any 'n') Then 'n' coefficient must be zero. There are a few explanation, list as below. <a name="arbitraryN2">&lt;a name="arbitraryN2"&gt;</a> <a href="jspico_e.htm#docB69">n=arbitrary_value, coefficient MUST BE ZERO</a> <a href="jspico_e.htm#docB70">move everything to one side. The other side is '=0'</a> <a href="jspico_e.htm#docB76">require pow(n,k) coefficients to be zero</a> <a href="jspico2e.htm#docB019">n is arbitrary, must demand coefficient to be zero</a> 2009-09-09-13-12 stop <hr> <a name="docB024">&lt;a name="docB024"&gt;</a> 2009-09-09-15-20 start Next see pow(n,k-3) coef. to find a<sub>k-2</sub>. From <a href=jspico_e.htm#docB71>eqn.AA217</a> collect n^(k-3) terms. Set coef to zero to find a<sub>k-2</sub> (set coef to zero? see <a href="jspico2e.htm#arbitraryN1">n is arbitrary</a>) <a name="docB025">&lt;a name="docB025"&gt;</a> pow(n,k-3)* ---eqn.AA228 [ +bino(k+1,k-3)*a<sub>k+1</sub> +bino(k,k-3)*a<sub>k</sub> +bino(k-1,k-3)*a<sub>k-1</sub> +bino(k-2,k-3)*a<sub>k-2</sub> +bino(k-3,k-3)*a<sub>k-3</sub> -bino(k,k-3) -a<sub>k-3</sub> ] <a name="docB026">&lt;a name="docB026"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a> All 'bino()' come from (n+1)^k expansion.<a href=jspico_e.htm#docB60> Other side</a> of general equation use n^k, Other side n^k no expansion and no 'bino()' <a href="jspico_e.htm#docB91"> Next is bino(s,t) evaluation.</a> ..... Above is file one, it goto numerical value quickly. Below is file two, preserve symbolic equation as long as possible. 2009-09-09-15-34 here <a name="docB027">&lt;a name="docB027"&gt;</a> pow(n,k-3)* ---eqn.AA228A [ +bino(k+1,k-3)*a<sub>k+1</sub> +bino(k,k-3)*a<sub>k</sub> +bino(k-1,k-3)*a<sub>k-1</sub> +bino(k-2,k-3)*a<sub>k-2</sub> +bino(k-3,k-3)*a<sub>k-3</sub> -bino(k,k-3) -a<sub>k-3</sub> ] <a name="docB028">&lt;a name="docB028"&gt;</a> = (above, below which one is better?) pow(n,k-3)* ---eqn.AA228B [ +bino(k+1,k-3)*1/bino(k+1,k) +bino(k,k-3)*1/2 +bino(k-1,k-3)*{ bino(k,k-2)/bino(k-1,k-2) -bino(k+1,k-2)/bino(k+1,k)/bino(k-1,k-2) -bino(k,k-2)/bino(k-1,k-2)/2 } +bino(k-2,k-3)*a<sub>k-2</sub> <font color=red>+bino(k-3,k-3)*a<sub>k-3</sub> <== cancel</font> -bino(k,k-3) <font color=red>-a<sub>k-3</sub> <== cancel</font> ] where +bino(k-3,k-3)*a<sub>k-3</sub>-a<sub>k-3</sub> = +a<sub>k-3</sub>-a<sub>k-3</sub> = 0 <a name="docB029">&lt;a name="docB029"&gt;</a> In eqn.AA228 <a href="jspico2e.htm#arbitraryN1">n is arbitrary</a>, demand coefficient = 0 set +bino(k+1,k-3)*a<sub>k+1</sub> +bino(k,k-3)*a<sub>k</sub> +bino(k-1,k-3)*a<sub>k-1</sub> +bino(k-2,k-3)*a<sub>k-2</sub> -bino(k,k-3) = 0 <a name="docB030">&lt;a name="docB030"&gt;</a> that is (above, below which one is better?) +bino(k+1,k-3)*1/bino(k+1,k) +bino(k,k-3)*1/2 +bino(k-1,k-3)*{ bino(k,k-2)/bino(k-1,k-2) -bino(k+1,k-2)/bino(k+1,k)/bino(k-1,k-2) -bino(k,k-2)/bino(k-1,k-2)/2 } +bino(k-2,k-3)*a<sub>k-2</sub> -bino(k,k-3) = 0 <a name="docB031">&lt;a name="docB031"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a> Solve for a<sub>k-2</sub> get +bino(k-2,k-3)*a<sub>k-2</sub> = +bino(k,k-3) -bino(k+1,k-3)*a<sub>k+1</sub> -bino(k,k-3)*a<sub>k</sub> -bino(k-1,k-3)*a<sub>k-1</sub> <a name="docB032">&lt;a name="docB032"&gt;</a> that is (above, below which one is better?) +bino(k-2,k-3)*a<sub>k-2</sub> = +bino(k,k-3) // from <a href=jspico_e.htm#docB55>(n+1)^k</a> -bino(k+1,k-3)/bino(k+1,k) // from <a href="jspico2e.htm#docB008">a<sub>k+1</sub></a> -bino(k,k-3)*1/2 // from <a href="jspico2e.htm#docB015">a<sub>k</sub></a> -bino(k-1,k-3)*{ // from <a href="jspico2e.htm#docB021">a<sub>k-1</sub></a> bino(k,k-2)/bino(k-1,k-2) // four lines all a<sub>k-1</sub> -bino(k+1,k-2)/bino(k+1,k)/bino(k-1,k-2) -bino(k,k-2)/bino(k-1,k-2)/2 <a name="docB033">&lt;a name="docB033"&gt;</a> Simpler equation is } a<sub>k-2</sub> = +bino(k,k-3)/bino(k-2,k-3) ---eqn.AB008 -a<sub>k+1</sub>*bino(k+1,k-3)/bino(k-2,k-3) -a<sub>k</sub> *bino(k,k-3) /bino(k-2,k-3) -a<sub>k-1</sub>*bino(k-1,k-3)/bino(k-2,k-3) here a<sub>k+1</sub>, a<sub>k</sub>, a<sub>k-1</sub> are all solved. 2009-09-09-16-07 here <a name="docB034">&lt;a name="docB034"&gt;</a> Let a<sub>k+1</sub>, a<sub>k</sub> be two parameters, express a<sub>k-1</sub> in terms of a<sub>k+1</sub>, a<sub>k</sub> a<sub>k-1</sub> = bino(k,k-2)/bino(k-1,k-2) ---eqn.AB005A -a<sub>k+1</sub>*bino(k+1,k-2)/bino(k-1,k-2) -a<sub>k</sub>*bino(k,k-2)/bino(k-1,k-2) <a name="docB035">&lt;a name="docB035"&gt;</a> then a<sub>k-2</sub> = +bino(k,k-3)/bino(k-2,k-3) ---eqn.AB008 -a<sub>k+1</sub>*bino(k+1,k-3)/bino(k-2,k-3) -a<sub>k</sub> *bino(k,k-3) /bino(k-2,k-3) -bino(k,k-2)*bino(k-1,k-3)/bino(k-1,k-2)/bino(k-2,k-3) +a<sub>k+1</sub>*bino(k+1,k-2)*bino(k-1,k-3)/bino(k-1,k-2)/bino(k-2,k-3) +a<sub>k</sub>*bino(k,k-2)*bino(k-1,k-3)/bino(k-1,k-2)/bino(k-2,k-3) <a name="docB036">&lt;a name="docB036"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a> 2009-09-09-16-20 Dear Reader, this page is not a textbook. This page is thinking record, you will see wondering, puzzle, error, correction etc. for example, "(above, below which one is better?)" Please understand the nature of this page. Thank you. 2009-09-09-16-23 <a name="docB037">&lt;a name="docB037"&gt;</a> Next put similar term a<sub>k+1</sub> a<sub>k</sub> together a<sub>k-2</sub> ---eqn.AB009 = +bino(k,k-3)/bino(k-2,k-3) -bino(k,k-2)*bino(k-1,k-3)/bino(k-1,k-2)/bino(k-2,k-3) +a<sub>k+1</sub>*bino(k+1,k-2)*bino(k-1,k-3)/bino(k-1,k-2)/bino(k-2,k-3) -a<sub>k+1</sub>*bino(k+1,k-3)/bino(k-2,k-3) +a<sub>k</sub>*bino(k,k-2)*bino(k-1,k-3)/bino(k-1,k-2)/bino(k-2,k-3) -a<sub>k</sub> *bino(k,k-3) /bino(k-2,k-3) <a name="docB038">&lt;a name="docB038"&gt;</a> Do I made any mistake? here check if above equation reduce to a<sub>k-2</sub> = 0 a<sub>k-2</sub> = +[k(k-1)(k-2)/3!]/[(k-2)/1!] -[k(k-1)/2!]*[(k-1)(k-2)/2!]/[(k-1)/1!]/[(k-2)/1!] -a<sub>k+1</sub>*[(k+1)k(k-1)(k-2)/4!]/[(k-2)/1!] +a<sub>k+1</sub>*[(k+1)k(k-1)/3!]*[(k-1)(k-2)/2!]/[(k-1)/1!]/[(k-2)/1!] -a<sub>k</sub> *[k(k-1)(k-2)/3!]/[(k-2)/1!] +a<sub>k</sub>*[k(k-1)/2!]*[(k-1)(k-2)/2!]/[(k-1)/1!]/[(k-2)/1!] <a name="docB039">&lt;a name="docB039"&gt;</a> a<sub>k-2</sub> = +[k(k-1)(k-2)/6]/[(k-2)] -[k(k-1)/2]*[(k-1)(k-2)/2]/[(k-1)]/[(k-2)] -[1/(k+1)]*[(k+1)k(k-1)(k-2)/4!]/[(k-2)] +[1/(k+1)]*[(k+1)k(k-1)/6]*[(k-1)(k-2)/2]/[(k-1)]/[(k-2)] -[1/2]*[k(k-1)(k-2)/6]/[(k-2)] +[1/2]*[k(k-1)/2]*[(k-1)(k-2)/2]/[(k-1)]/[(k-2)] <a name="docB040">&lt;a name="docB040"&gt;</a> a<sub>k-2</sub> = +[k(k-1)/6] -[k(k-1)/2]/2 -[k(k-1)/4!] +[k(k-1)/6]/2 -[1/2]*[k(k-1)/6] +[1/2]*[k(k-1)/2]/2 <a name="docB041">&lt;a name="docB041"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a> a<sub>k-2</sub> = +[k(k-1)][1/6-1/4-1/24+1/12-1/12+1/8] = +[k(k-1)][4/24-6/24-1/24+2/24-2/24+3/24] = +[k(k-1)][4-6-1+2-2+3]/24 = 0 2009-09-09-16-44 checked OK, '=0' is right !! <a name="docB042">&lt;a name="docB042"&gt;</a> 2009-09-09-17-47 start Above evaluation for a<sub>k-2</sub> get<font color=red> a<sub>k-2</sub> = 0 ---eqn.AB010</font> this relation is valid for any kg"2 in the equation " i^(k) = general formula In the future a<sub>k-3</sub> , a<sub>k-4</sub> , a<sub>k-5</sub> etc. refer back to a<sub>k-2</sub>, we use eqn.AB010 we do not use <a href="jspico2e.htm#docB037">eqn.AB009</a> For general formula, please see <a href=jspico_e.htm#sum_eqn>sum equations</a> For its coefficient, please see <a href=jspico_e.htm#docB04>triangle</a> a<sub>k-2</sub> is column 04 in Coefficient triangle 2009-09-09-18-02 here <hr> <a name="docB043">&lt;a name="docB043"&gt;</a> 2009-09-09-18-12 start Next see pow(n,k-4) coef. to find a<sub>k-3</sub>. From <a href=jspico_e.htm#docB71>eqn.AA217</a> collect n^(k-4) terms. Set coef to zero to find a<sub>k-3</sub> (set coef to zero? see <a href="jspico2e.htm#arbitraryN1">n is arbitrary</a>) Refer to <a href="#docB024">general equation</a> and <a href=jspico_e.htm#docB101>how to extend to next level</a>. we find <a name="docB044">&lt;a name="docB044"&gt;</a> pow(n,k-4)* ---eqn.AB011 [ +bino(k+1,k-4)*a<sub>k+1</sub> +bino(k ,k-4)*a<sub>k</sub> +bino(k-1,k-4)*a<sub>k-1</sub> +bino(k-2,k-4)*a<sub>k-2</sub> +bino(k-3,k-4)*a<sub>k-3</sub> +bino(k-4,k-4)*a<sub>k-4</sub> -bino(k,k-4) -a<sub>k-4</sub> ] In which bino(k-4,k-4)*a<sub>k-4</sub>-a<sub>k-4</sub>=1*a<sub>k-4</sub>-a<sub>k-4</sub>=0 <a name="docB045">&lt;a name="docB045"&gt;</a> Since <a href="jspico2e.htm#arbitraryN1">n is arbitrary</a>, we demand coefficient to be zero, get +bino(k+1,k-4)*a<sub>k+1</sub> +bino(k ,k-4)*a<sub>k</sub> +bino(k-1,k-4)*a<sub>k-1</sub> +bino(k-2,k-4)*a<sub>k-2</sub> +bino(k-3,k-4)*a<sub>k-3</sub> -bino(k,k-4) = 0 ---eqn.AB012 <a name="docB046">&lt;a name="docB046"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a> Use previous relation <a href="jspico2e.htm#docB042">a<sub>k-2</sub> = 0 </a> and <a href="jspico2e.htm#docB021">a<sub>k-1</sub></a> rewrite eqn.AB012 as following bino(k-3,k-4)*a<sub>k-3</sub> = bino(k,k-4) -bino(k+1,k-4)*a<sub>k+1</sub> -bino(k ,k-4)*a<sub>k</sub> -bino(k-1,k-4)*a<sub>k-1</sub> -bino(k-2,k-4)*0 = bino(k,k-4) -bino(k+1,k-4)*a<sub>k+1</sub> -bino(k ,k-4)*a<sub>k</sub> -bino(k-1,k-4)* [ bino(k,k-2)/bino(k-1,k-2) -a<sub>k+1</sub>*bino(k+1,k-2)/bino(k-1,k-2) -a<sub>k</sub>*bino(k,k-2)/bino(k-1,k-2) ] <a name="docB047">&lt;a name="docB047"&gt;</a> Group a<sub>k+1</sub> and a<sub>k</sub> separately bino(k-3,k-4)*a<sub>k-3</sub> = bino(k,k-4) -bino(k-1,k-4)*bino(k,k-2)/bino(k-1,k-2) +a<sub>k+1</sub>*bino(k-1,k-4)*bino(k+1,k-2)/bino(k-1,k-2) -a<sub>k+1</sub>*bino(k+1,k-4) +a<sub>k</sub>*bino(k-1,k-4)*bino(k,k-2)/bino(k-1,k-2) -a<sub>k</sub>*bino(k ,k-4) <a name="docB048">&lt;a name="docB048"&gt;</a> move bino(k-3,k-4) to other side a<sub>k-3</sub> ---eqn.AB013A = bino(k,k-4)/bino(k-3,k-4) -bino(k,k-2)*bino(k-1,k-4)/bino(k-1,k-2)/bino(k-3,k-4) +a<sub>k+1</sub>*bino(k+1,k-2)*bino(k-1,k-4)/bino(k-1,k-2)/bino(k-3,k-4) -a<sub>k+1</sub>*bino(k+1,k-4)/bino(k-3,k-4) +a<sub>k</sub>*bino(k,k-2)*bino(k-1,k-4)/bino(k-1,k-2)/bino(k-3,k-4) -a<sub>k</sub>*bino(k,k-4)/bino(k-3,k-4) <a name="docB049">&lt;a name="docB049"&gt;</a> Here is a place to observe the pattern. eqn.AB013 is symbolic, reserve pattern structure. 2009-09-09-18-33 here Compare the pattern of a<sub>k-3</sub> eqn.AB013 and a<sub>k-2</sub> <a href="jspico2e.htm#docB037">eqn.AB009</a> 2009-09-09-18-43 here <a name="docB050">&lt;a name="docB050"&gt;</a> 2009-09-09-18-56 start From <a href=jspico_e.htm#docB110>a<sub>k-#</sub> summary</a>, we know a<sub>k-3</sub>=-k*(k-1)*(k-2)/(5*4*3*2*3*2) Now verify eqn.AB013, change to numerical equation, see if it match the know result. <a name="docB051">&lt;a name="docB051"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a> a<sub>k-3</sub> ---eqn.AB013B = bino(k,k-4)/bino(k-3,k-4) -bino(k,k-2)*bino(k-1,k-4)/bino(k-1,k-2)/bino(k-3,k-4) +a<sub>k+1</sub>*bino(k+1,k-2)*bino(k-1,k-4)/bino(k-1,k-2)/bino(k-3,k-4) -a<sub>k+1</sub>*bino(k+1,k-4)/bino(k-3,k-4) +a<sub>k</sub>*bino(k,k-2)*bino(k-1,k-4)/bino(k-1,k-2)/bino(k-3,k-4) -a<sub>k</sub>*bino(k,k-4)/bino(k-3,k-4) <a name="docB052">&lt;a name="docB052"&gt;</a> = [k(k-1)(k-2)(k-3)/4!]/[(k-3)/1!] -[k(k-1)/2!]*[(k-1)(k-2)(k-3)/3!]/[(k-1)/1!]/[(k-3)/1!] +a<sub>k+1</sub>*[(k+1)k(k-1)/3!]*[(k-1)(k-2)(k-3)/3!]/[(k-1)/1!]/[(k-3)/1!] -a<sub>k+1</sub>*[(k+1)k(k-1)(k-2)(k-3)/5!]/[(k-3)/1!] +a<sub>k</sub>*[k(k-1)/2!]*[(k-1)(k-2)(k-3)/3!]/[(k-1)/1!]/[(k-3)/1!] -a<sub>k</sub>*[k(k-1)(k-2)(k-3)/4!]/[(k-3)/1!] <a name="docB053">&lt;a name="docB053"&gt;</a> = [k(k-1)(k-2)/4!] -[k(k-1)/2!]*[(k-2)/3!] +[1/(k+1)]*[(k+1)k(k-1)/3!]*[(k-2)/3!] -[1/(k+1)]*[(k+1)k(k-1)(k-2)/5!] +[1/2]*[k(k-1)/2!]*[(k-2)/3!] -[1/2]*[k(k-1)(k-2)/4!] <a name="docB054">&lt;a name="docB054"&gt;</a> a<sub>k-3</sub> ---eqn.AB013C = [k(k-1)(k-2)]* //from no a<sub>k+1</sub> no a<sub>k</sub> [1/4! -1/(2!*3!)] +[k(k-1)(k-2)]* //from a<sub>k+1</sub> +[1/3! * 1/3! -1/5!] +[k(k-1)(k-2)]* //from a<sub>k</sub> +[1/2 * 1/2! * 1/3! -1/2 * 1/4!] <a name="docB055">&lt;a name="docB055"&gt;</a> = [k(k-1)(k-2)]*[1/24-1/(2*6)+1/36-1/120+1/24-1/48] = [k(k-1)(k-2)]*[(30-60+20-6+30-15)/720] = [k(k-1)(k-2)]*[(-1)/720] <a name="docB056">&lt;a name="docB056"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a> a<sub>k-3</sub> ---eqn.AB013D = [k(k-1)(k-2)]*[(-1)/720] //9809091928 right eqn.AB013 passed necessary condition check 2009-09-09-19-29 stop <hr> <a name="docB057">&lt;a name="docB057"&gt;</a> 2009-09-10-10-00 start Next see pow(n,k-5) coef. to find a<sub>k-4</sub>. From <a href=jspico_e.htm#docB71>eqn.AA217</a> collect n^(k-5) terms. Set coef to zero to find a<sub>k-4</sub> (set coef to zero? see <a href="jspico2e.htm#arbitraryN1">n is arbitrary</a>) Refer to <a href="#docB024">general equation</a> and <a href=jspico_e.htm#docB101>how to extend to next level</a>. we find <a name="docB058">&lt;a name="docB058"&gt;</a> pow(n,k-5)* ---eqn.AB014 [ +bino(k+1,k-5)*a<sub>k+1</sub> +bino(k ,k-5)*a<sub>k</sub> +bino(k-1,k-5)*a<sub>k-1</sub> +bino(k-2,k-5)*a<sub>k-2</sub> +bino(k-3,k-5)*a<sub>k-3</sub> +bino(k-4,k-5)*a<sub>k-4</sub> +bino(k-5,k-5)*a<sub>k-5</sub> <== cancel -bino(k,k-5) -a<sub>k-5</sub> <== cancel ] where bino(k-5,k-5)*a<sub>k-5</sub>-a<sub>k-5</sub>=1*a<sub>k-5</sub>-a<sub>k-5</sub>=0 <a name="docB059">&lt;a name="docB059"&gt;</a> pow(n,k-5)* [ (k+1)*(k+0)*(k-1)*(k-2)*(k-3)*(k-4)*a<sub>k+1</sub>/6! (k+0)*(k-1)*(k-2)*(k-3)*(k-4)*a<sub>k</sub>/5! (k-1)*(k-2)*(k-3)*(k-4)*a<sub>k-1</sub>/4! (k-2)*(k-3)*(k-4)*a<sub>k-2</sub>/3! (k-3)*(k-4)*a<sub>k-3</sub>/2! (k-4)*a<sub>k-4</sub>/1! -bino(k,k-5) ] Above is old calculation, it goto numerical value quickly. <a name="docB060">&lt;a name="docB060"&gt;</a> Below is new calculation, it preserve symbolic equation as long as possible. In eqn.AB014 <a href="jspico2e.htm#arbitraryN1">set coefficient to zero</a> get bino(k+1,k-5)*a<sub>k+1</sub> +bino(k ,k-5)*a<sub>k</sub> +bino(k-1,k-5)*a<sub>k-1</sub> +bino(k-2,k-5)*a<sub>k-2</sub> +bino(k-3,k-5)*a<sub>k-3</sub> +bino(k-4,k-5)*a<sub>k-4</sub> -bino(k,k-5) = 0 ---eqn.AB015 <a name="docB061">&lt;a name="docB061"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a> In eqn.AB015 only unknown is a<sub>k-4</sub> +bino(k-4,k-5)*a<sub>k-4</sub> = ---eqn.AB016 bino(k,k-5) -bino(k+1,k-5)*a<sub>k+1</sub> -bino(k ,k-5)*a<sub>k</sub> -bino(k-1,k-5)*a<sub>k-1</sub> -bino(k-2,k-5)*a<sub>k-2</sub> // <a href=jspico2e.htm#docB042>a<sub>k-2</sub>=0</a> -bino(k-3,k-5)*a<sub>k-3</sub> <a name="docB062">&lt;a name="docB062"&gt;</a> both <a href="jspico2e.htm#docB021">a<sub>k-1</sub></a> and <a href="jspico2e.htm#docB048">a<sub>k-3</sub></a> can be expressed in terms of a<sub>k+1</sub> and a<sub>k</sub> Re-write eqn.AB016 as following +bino(k-4,k-5)*a<sub>k-4</sub> = ---eqn.AB017 bino(k,k-5) -bino(k+1,k-5)*a<sub>k+1</sub> -bino(k ,k-5)*a<sub>k</sub> -bino(k-1,k-5)* { bino(k,k-2)/bino(k-1,k-2) -a<sub>k+1</sub>*bino(k+1,k-2)/bino(k-1,k-2) -a<sub>k</sub>*bino(k,k-2)/bino(k-1,k-2) } -bino(k-3,k-5)* { bino(k,k-4)/bino(k-3,k-4) -bino(k,k-2)*bino(k-1,k-4)/bino(k-1,k-2)/bino(k-3,k-4) +a<sub>k+1</sub>*bino(k+1,k-2)*bino(k-1,k-4)/bino(k-1,k-2)/bino(k-3,k-4) -a<sub>k+1</sub>*bino(k+1,k-4)/bino(k-3,k-4) +a<sub>k</sub>*bino(k,k-2)*bino(k-1,k-4)/bino(k-1,k-2)/bino(k-3,k-4) -a<sub>k</sub>*bino(k,k-4)/bino(k-3,k-4) } <a name="docB063">&lt;a name="docB063"&gt;</a> +bino(k-4,k-5)*a<sub>k-4</sub> // coefficient is NOT one = ---eqn.AB018 bino(k,k-5) -bino(k-1,k-5)*bino(k,k-2)/bino(k-1,k-2) -bino(k-3,k-5)*bino(k,k-4)/bino(k-3,k-4) +bino(k-3,k-5)*bino(k,k-2)*bino(k-1,k-4)/bino(k-1,k-2)/bino(k-3,k-4) +a<sub>k+1</sub>*bino(k+1,k-2)*bino(k-1,k-5)/bino(k-1,k-2) -a<sub>k+1</sub>*bino(k+1,k-5) -a<sub>k+1</sub>*bino(k+1,k-2)*bino(k-1,k-4)*bino(k-3,k-5)/bino(k-1,k-2)/bino(k-3,k-4) +a<sub>k+1</sub>*bino(k+1,k-4)*bino(k-3,k-5)/bino(k-3,k-4) -a<sub>k</sub>*bino(k ,k-5) +a<sub>k</sub>*bino(k,k-2)*bino(k-1,k-5)/bino(k-1,k-2) -a<sub>k</sub>*bino(k,k-2)*bino(k-1,k-4)*bino(k-3,k-5)/bino(k-1,k-2)/bino(k-3,k-4) +a<sub>k</sub>*bino(k,k-4)*bino(k-3,k-5)/bino(k-3,k-4) <a name="docB064">&lt;a name="docB064"&gt;</a> a<sub>k-4</sub> // coefficient is one = ---eqn.AB019 bino(k,k-5)/bino(k-4,k-5) -bino(k-1,k-5)*bino(k,k-2)/bino(k-1,k-2)/bino(k-4,k-5) -bino(k-3,k-5)*bino(k,k-4)/bino(k-3,k-4)/bino(k-4,k-5) +bino(k-3,k-5)*bino(k,k-2)*bino(k-1,k-4)/bino(k-1,k-2)/bino(k-3,k-4)/bino(k-4,k-5) +a<sub>k+1</sub>*bino(k+1,k-2)*bino(k-1,k-5)/bino(k-1,k-2)/bino(k-4,k-5) -a<sub>k+1</sub>*bino(k+1,k-5)/bino(k-4,k-5) -a<sub>k+1</sub>*bino(k+1,k-2)*bino(k-1,k-4)*bino(k-3,k-5)/bino(k-1,k-2)/bino(k-3,k-4)/bino(k-4,k-5) +a<sub>k+1</sub>*bino(k+1,k-4)*bino(k-3,k-5)/bino(k-3,k-4)/bino(k-4,k-5) -a<sub>k</sub>*bino(k ,k-5)/bino(k-4,k-5) +a<sub>k</sub>*bino(k,k-2)*bino(k-1,k-5)/bino(k-1,k-2)/bino(k-4,k-5) -a<sub>k</sub>*bino(k,k-2)*bino(k-1,k-4)*bino(k-3,k-5)/bino(k-1,k-2)/bino(k-3,k-4)/bino(k-4,k-5) +a<sub>k</sub>*bino(k,k-4)*bino(k-3,k-5)/bino(k-3,k-4)/bino(k-4,k-5) 2009-09-10-11-17 here <a name="docB065">&lt;a name="docB065"&gt;</a> Expect eqn.AB018 to be zero. Refer to given result <a href="jspico2e.htm#docB008">a<sub>k+1</sub></a> = 1/(k+1) and <a href="jspico2e.htm#docB015">a<sub>k</sub></a> = 1/2 Verify numerical value as following +bino(k-4,k-5)*a<sub>k-4</sub> // coefficient is NOT one = ---eqn.AB018 [k(k-1)(k-2)(k-3)(k-4)/5!] -[(k-1)(k-2)(k-3)(k-4)/4!]*[k(k-1)/2!]/[k-1] -[(k-3)(k-4)/2!]*[k(k-1)(k-2)(k-3)/4!]/[k-3] +[(k-3)(k-4)/2!]*[k(k-1)/2!]*[(k-1)(k-2)(k-3)/3!]/[k-1]/[k-3] +[1/(k+1)]*[(k+1)k(k-1)/3!]*[(k-1)(k-2)(k-3)(k-4)/4!]/[k-1] -[1/(k+1)]*[(k+1)k(k-1)(k-2)(k-3)(k-4)/6!] -[1/(k+1)]*[(k+1)k(k-1)/3!]*[(k-1)(k-2)(k-3)/3!]*[(k-3)(k-4)/2!]/[k-1]/[k-3] +[1/(k+1)]*[(k+1)k(k-1)(k-2)(k-3)/5!]*[(k-3)(k-4)/2!]/[k-3] -[1/2]*[k(k-1)(k-2)(k-3)(k-4)/5!] +[1/2]*[k(k-1)/2!]*[(k-1)(k-2)(k-3)(k-4)/4!]/[k-1] -[1/2]*[k(k-1)/2!]*[(k-1)(k-2)(k-3)/3!]*[(k-3)(k-4)/2!]/[k-1]/[k-3] +[1/2]*[k(k-1)(k-2)(k-3)/4!]*[(k-3)(k-4)/2!]/[k-3] 2009-09-10-11-37 here <a name="docB066">&lt;a name="docB066"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a> +bino(k-4,k-5)*a<sub>k-4</sub> = ---eqn.AB020 (k*(k-1)*(k-2)*(k-3)*(k-4)/120) -((k-1)*(k-2)*(k-3)*(k-4)/24)*(k*(k-1)/2)/(k-1) -((k-3)*(k-4)/2)*(k*(k-1)*(k-2)*(k-3)/24)/(k-3) +((k-3)*(k-4)/2)*(k*(k-1)/2)*((k-1)*(k-2)*(k-3)/6)/(k-1)/(k-3) +(1/(k+1))*((k+1)*k*(k-1)/6)*((k-1)*(k-2)*(k-3)*(k-4)/24)/(k-1) -(1/(k+1))*((k+1)*k*(k-1)*(k-2)*(k-3)*(k-4)/720) -(1/(k+1))*((k+1)*k*(k-1)/6)*((k-1)*(k-2)*(k-3)/6)*((k-3)*(k-4)/2)/(k-1)/(k-3) +(1/(k+1))*((k+1)*k*(k-1)*(k-2)*(k-3)/120)*((k-3)*(k-4)/2)/(k-3) -(1/2)*(k*(k-1)*(k-2)*(k-3)*(k-4)/120) +(1/2)*(k*(k-1)/2)*((k-1)*(k-2)*(k-3)*(k-4)/24)/(k-1) -(1/2)*(k*(k-1)/2)*((k-1)*(k-2)*(k-3)/6)*((k-3)*(k-4)/2)/(k-1)/(k-3) +(1/2)*(k*(k-1)*(k-2)*(k-3)/24)*((k-3)*(k-4)/2)/(k-3) = (k*(k-1)*(k-2)*(k-3)*(k-4)/120)-((k-1)*(k-2)*(k-3)*(k-4)/24)*(k*(k-1)/2)/(k-1)-((k-3)*(k-4)/2)*(k*(k-1)*(k-2)*(k-3)/24)/(k-3)+((k-3)*(k-4)/2)*(k*(k-1)/2)*((k-1)*(k-2)*(k-3)/6)/(k-1)/(k-3)+(1/(k+1))*((k+1)*k*(k-1)/6)*((k-1)*(k-2)*(k-3)*(k-4)/24)/(k-1)-(1/(k+1))*((k+1)*k*(k-1)*(k-2)*(k-3)*(k-4)/720)-(1/(k+1))*((k+1)*k*(k-1)/6)*((k-1)*(k-2)*(k-3)/6)*((k-3)*(k-4)/2)/(k-1)/(k-3)+(1/(k+1))*((k+1)*k*(k-1)*(k-2)*(k-3)/120)*((k-3)*(k-4)/2)/(k-3)-(1/2)*(k*(k-1)*(k-2)*(k-3)*(k-4)/120)+(1/2)*(k*(k-1)/2)*((k-1)*(k-2)*(k-3)*(k-4)/24)/(k-1)-(1/2)*(k*(k-1)/2)*((k-1)*(k-2)*(k-3)/6)*((k-3)*(k-4)/2)/(k-1)/(k-3)+(1/2)*(k*(k-1)*(k-2)*(k-3)/24)*((k-3)*(k-4)/2)/(k-3) <a name="docB067">&lt;a name="docB067"&gt;</a> 2009-09-10-11-54 numerical verified +bino(k-4,k-5)*a<sub>k-4</sub> = 0 for any k>4 Since bino(k-4,k-5) = k-4 >0 , conclude a<sub>k-4</sub> = 0 ---eqn.AB020 2009-09-10-11-56 stop <a name="docB068">&lt;a name="docB068"&gt;</a> 2009-09-10-13-36 start Above numerical verified a<sub>k-4</sub> = 0 Below symbolic evaluation get "= 0" eqn.AB020 = +bino(k-4,k-5)*a<sub>k-4</sub> = ---eqn.AB021 (k*(k-1)*(k-2)*(k-3)*(k-4)/120) -((k-1)*(k-2)*(k-3)*(k-4)/24)*(k/2) -((k-3)*(k-4)/2)*(k*(k-1)*(k-2)/24) +((k-3)*(k-4)/2)*(k*(k-1)/2)*((k-2)/6) +(k*(k-1)/6)*((k-2)*(k-3)*(k-4)/24) -(k*(k-1)*(k-2)*(k-3)*(k-4)/720) -(k*(k-1)/6)*((k-2)/6)*((k-3)*(k-4)/2) +(k*(k-1)*(k-2)*(k-3)/120)*((k-4)/2) -(1/2)*(k*(k-1)*(k-2)*(k-3)*(k-4)/120) +(1/2)*(k*(k-1)/2)*((k-2)*(k-3)*(k-4)/24) -(1/2)*(k*(k-1)/2)*((k-2)/6)*((k-3)*(k-4)/2) +(1/2)*(k*(k-1)*(k-2)*(k-3)/24)*((k-4)/2) <a name="docB069">&lt;a name="docB069"&gt;</a> +bino(k-4,k-5)*a<sub>k-4</sub> = ---eqn.AB022 (k*(k-1)*(k-2)*(k-3)*(k-4)/120) -(k*(k-1)*(k-2)*(k-3)*(k-4)/(2*24)) -(k*(k-1)*(k-2)*(k-3)*(k-4)/(2*24)) +(k*(k-1)*(k-2)*(k-3)*(k-4)/(2*2*6)) +(k*(k-1)*(k-2)*(k-3)*(k-4)/(6*24)) -(k*(k-1)*(k-2)*(k-3)*(k-4)/(720)) -(k*(k-1)*(k-2)*(k-3)*(k-4)/(6*6*2)) +(k*(k-1)*(k-2)*(k-3)*(k-4)/(120*2)) -(k*(k-1)*(k-2)*(k-3)*(k-4)/(120*2)) +(k*(k-1)*(k-2)*(k-3)*(k-4)/(2*2*24)) -(k*(k-1)*(k-2)*(k-3)*(k-4)/(2*2*2*6)) +(k*(k-1)*(k-2)*(k-3)*(k-4)/(2*2*24)) <a name="docB070">&lt;a name="docB070"&gt;</a> +bino(k-4,k-5)*a<sub>k-4</sub> = ---eqn.AB023 (k*(k-1)*(k-2)*(k-3)*(k-4)* { +1/120 -1/(2*24) -1/(2*24) +1/(2*2*6) +1/(6*24) -1/720 -1/(6*6*2) +1/(120*2) -1/(120*2) +1/(2*2*24) -1/(2*2*2*6) +1/(2*2*24) } = (k*(k-1)*(k-2)*(k-3)*(k-4)* { 6-15-15+30+5-1-10+3-3+7.5-15+7.5 }/720 = 0 <a name="docB071">&lt;a name="docB071"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a> Result is +bino(k-4,k-5)*a<sub>k-4</sub> = 0 Since bino(k-4,k-5) is not zero, must be a<sub>k-4</sub> = 0 ---eqn.AB024 2009-09-10-14-22 here <a name="docB072">&lt;a name="docB072"&gt;</a> <a href="jspico2e.htm#docB064">eqn.AB019</a> will be used as pattern. Now simplify eqn.AB019 as following First see a<sub>k-4</sub> no a<sub>k+1</sub> and no a<sub>k</sub> terms a<sub>k-4</sub> (partial 1/3) // coefficient is one = ---eqn.AB025 (partial 1/3) bino(k,k-5)/bino(k-4,k-5) -bino(k-1,k-5)*bino(k,k-2)/bino(k-1,k-2)/bino(k-4,k-5) +bino(k-3,k-5)* { -bino(k,k-4)/bino(k-3,k-4)/bino(k-4,k-5) +bino(k,k-2)*bino(k-1,k-4)/bino(k-1,k-2)/bino(k-3,k-4)/bino(k-4,k-5) } <a name="docB073">&lt;a name="docB073"&gt;</a> = bino(k,k-5)/(k-4) -bino(k-1,k-5)*bino(k,k-2)/(k-1)/(k-4) +bino(k-3,k-5)* { -bino(k,k-4)/(k-3)/(k-4) +bino(k,k-2)*bino(k-1,k-4)/(k-1)/(k-3)/(k-4) } = [k(k-1)(k-2)(k-3)(k-4)/5!]/(k-4) -[ (k-1)(k-2)(k-3)(k-4)/4!]*[k(k-1)/2!]/(k-1)/(k-4) +[(k-3)(k-4)/2!]* { -[k(k-1)(k-2)(k-3)/4!]/(k-3)/(k-4) +[k(k-1)/2!]*[(k-1)(k-2)(k-3)/3!]/(k-1)/(k-3)/(k-4) } <a name="docB074">&lt;a name="docB074"&gt;</a> = [k(k-1)(k-2)(k-3)/5!] -[(k-2)(k-3)/4!]*[k(k-1)/2!] -[1/2!]*[k(k-1)(k-2)(k-3)/4!] +[1/2!]*[k/2!]*[(k-1)(k-2)(k-3)/3!] } = [k(k-1)(k-2)(k-3)]* [1/5! - 1/(4!*2!) - 1/(4!*2!) + 1/(2!*2!*3!)] <a name="docB075">&lt;a name="docB075"&gt;</a> a<sub>k-4</sub> (partial 1/3) // coefficient is one = ---eqn.AB026 (partial 1/3) [k(k-1)(k-2)(k-3)]* // red term is previous term {-[<font color=red>1/4! -1/(2!*3!)</font>]/2! +1/5! - 1/(4!*2!)} 2009-09-10-14-58 stop <a name="docB076">&lt;a name="docB076"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a> 2009-09-10-17-40 start Second see a<sub>k+1</sub> terms in a<sub>k-4</sub> a<sub>k-4</sub> (partial 2/3) // coefficient is one = ---eqn.AB027 (partial 2/3) -a<sub>k+1</sub>*bino(k+1,k-5)/bino(k-4,k-5) +a<sub>k+1</sub>*[<font color=red>bino(k+1,k-4)/bino(k-3,k-4)</font>]*bino(k-3,k-5)/bino(k-4,k-5) +a<sub>k+1</sub>*bino(k+1,k-2)*bino(k-1,k-5)/bino(k-1,k-2)/bino(k-4,k-5) -a<sub>k+1</sub>*[<font color=red>bino(k+1,k-2)*bino(k-1,k-4)/bino(k-1,k-2)/bino(k-3,k-4)</font>]*bino(k-3,k-5)/bino(k-4,k-5) = // red is previous term -[1/(k+1)]*[(k+1)k(k-1)(k-2)(k-3)(k-4)/6!]/(k-4) +[1/(k+1)]*[(k+1)k(k-1)(k-2)(k-3)/5!]*[(k-3)(k-4)/2!]/[k-3]/[k-4] +[1/(k+1)]*[(k+1)k(k-1)/3!]*[(k-1)(k-2)(k-3)(k-4)/4!]/[k-1]/[k-4] -[1/(k+1)]*[(k+1)k(k-1)/3!]*[(k-1)(k-2)(k-3)/3!]*[(k-3)(k-4)/2!]/[k-1]/[k-3]/[k-4] <a name="docB077">&lt;a name="docB077"&gt;</a> = -[k(k-1)(k-2)(k-3)/6!] +[k(k-1)(k-2)(k-3)/5!]*[1/2!] +[k(k-1)/3!]*[(k-2)(k-3)/4!] -[k(k-1)/3!]*[(k-2)(k-3)/3!]*[1/2!] = [k(k-1)(k-2)(k-3)]* [-1/6! +1/5!*1/2! +1/3!*1/4! -1/3!*1/3!*1/2! ] <a name="docB078">&lt;a name="docB078"&gt;</a> a<sub>k-4</sub> (partial 2/3) // coefficient is one = ---eqn.AB028 (partial 2/3) [k(k-1)(k-2)(k-3)]* // red term is previous term [ -(<font color=red>-1/5! +1/3!*1/3!</font>)*1/2! -1/6! +1/3!*1/4!] 2009-09-10-18-35 here <a name="docB079">&lt;a name="docB079"&gt;</a> Third see a<sub>k</sub> terms in a<sub>k-4</sub> a<sub>k-4</sub> (partial 3/3) // a<sub>k</sub>=1/2 = ---eqn.AB029 (partial 3/3) -a<sub>k</sub>*bino(k,k-5)/bino(k-4,k-5) +a<sub>k</sub>*bino(k,k-2)*bino(k-1,k-5)/bino(k-1,k-2)/bino(k-4,k-5) -a<sub>k</sub>*[<font color=red>bino(k,k-2)*bino(k-1,k-4)/bino(k-1,k-2)/bino(k-3,k-4)</font>]*bino(k-3,k-5)/bino(k-4,k-5) +a<sub>k</sub>*[<font color=red>bino(k,k-4)/bino(k-3,k-4)</font>]*bino(k-3,k-5)/bino(k-4,k-5) <a name="docB080">&lt;a name="docB080"&gt;</a> = // red term is previous term -[1/2]*[k(k-1)(k-2)(k-3)(k-4)/5!]/(k-4) +[1/2]*[k(k-1)/2!]*[(k-1)(k-2)(k-3)(k-4)/4!]/(k-1)/(k-4) -[1/2]*[<font color=red>[k(k-1)/2!]*[(k-1)(k-2)(k-3)/3!]/(k-1)/(k-3)</font>]*[(k-3)(k-4)/2!]/(k-4) +[1/2]*[<font color=red>[k(k-1)(k-2)(k-3)/4!]/(k-3)</font>]*[(k-3)(k-4)/2!]/(k-4) = // red term is previous term k(k-1)(k-2)(k-3)* { -1/2/5! +[1/2]*1/2!*1/4! -[1/2]*[<font color=red>1/2!/3!</font>]/2! +[1/2]*[<font color=red>1/4!</font>]/2! } <a name="docB081">&lt;a name="docB081"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a> = // red term is previous term k(k-1)(k-2)(k-3)* { -1/2/5! +[1/2]*1/2!*1/4! -[1/2]*[<font color=red>1/2!/3!</font>]/2! <== alert !! +[1/2]*[<font color=red>1/4!</font>]/2! <== alert !! } <a name="docB082">&lt;a name="docB082"&gt;</a> 2009-09-10-19-11 here alert !! []/2!] could be previous a<sub>k</sub> if it is true, then should write []/2!] as []/2] and write [1/2]*[1/2] as [1/2]^2 for a<sub>k</sub>*a<sub>k</sub> 2009-09-10-19-14 stop <a name="docB083">&lt;a name="docB083"&gt;</a> 2009-09-11-20-15 start Now put a<sub>k-4</sub> (partial 1/3) a<sub>k-4</sub> (partial 2/3) a<sub>k-4</sub> (partial 3/3) together <a name="docB084">&lt;a name="docB084"&gt;</a> a<sub>k-4</sub> (partial 1/3) // coefficient is one = ---eqn.AB026 (partial 1/3) [k(k-1)(k-2)(k-3)]* // red term is previous term {-[<font color=red>1/4! -1/(2!*3!)</font>]/2! +1/5! - 1/(4!*2!)} + ---eqn.AB028 (partial 2/3) [k(k-1)(k-2)(k-3)]* // red term is previous term [ -(<font color=red>-1/5! +1/3!*1/3!</font>)*1/2! -1/6! +1/3!*1/4!] <a name="docB085">&lt;a name="docB085"&gt;</a> + ---eqn.AB028 (partial 3/3) k(k-1)(k-2)(k-3)* { -1/2/5! +[1/2]*1/2!*1/4! -[1/2]*[<font color=red>1/2!/3!</font>]/2! <== alert !! +[1/2]*[<font color=red>1/4!</font>]/2! <== alert !! } 2009-09-11-20-?? stop <a name="docB086">&lt;a name="docB086"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a> 2009-09-11-20-41 start check if sum to zero {-[1/4! -1/(2!*3!)]/2! +1/5! - 1/(4!*2!)} + [ -(-1/5! +1/3!*1/3!)*1/2! -1/6! +1/3!*1/4!] + { -1/2/5! +[1/2]*1/2!*1/4! -[1/2]*[1/2!/3!]/2! +[1/2]*[1/4!]/2! } <a name="docB087">&lt;a name="docB087"&gt;</a> (-(1/24 -1/(2*6))/2 +1/120 - 1/(24*2)) + ( -(-1/120 +1/6*1/6)*1/2 -1/720 +1/6*1/24) + ( -1/2/120 +(1/2)*1/2*1/24 -(1/2)*(1/2/6)/2 +(1/2)*(1/24)/2 ) <a name="docB088">&lt;a name="docB088"&gt;</a> 2009-09-11-20-47 OK sum to zero !! (-(1/24-1/(2*6))/2+1/120-1/(24*2))+(-(-1/120+1/6*1/6)*1/2-1/720+1/6*1/24)+(-1/2/120+(1/2)*1/2*1/24-(1/2)*(1/2/6)/2+(1/2)*(1/24)/2) =0 <font color=red> 2009-09-11-20-52 change direction not chase symbolic equation now it just like rolling snow ball get more and more bino(.,.) change to numerical solution. </font> <a name="docB089">&lt;a name="docB089"&gt;</a> <a href="jspico2e.htm#arbitraryN1">set coefficient to zero</a> (Please see <a href="jspico2e.htm#arbitraryN1">n is arbitrary</a>) <a href="jspico2e.htm#docA0##"></a> <a name="list_a_k">&lt;a name="list_a_k"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a> from <a href=jspico_e.htm#docB55>(n+1)^k</a> from <a href="jspico2e.htm#docB008">a<sub>k+1</sub></a> from <a href="jspico2e.htm#docB015">a<sub>k</sub></a> from <a href="jspico2e.htm#docB021">a<sub>k-1</sub></a> from <a href="jspico2e.htm#docB042">a<sub>k-2</sub></a> <a name="compare_pattern">&lt;a name="compare_pattern"&gt;</a> Compare the pattern of a<sub>k-2</sub> <a href="jspico2e.htm#docB037">eqn.AB009</a> a<sub>k-3</sub> <a href="jspico2e.htm#docB048">eqn.AB013</a> a<sub>k-4</sub> <a href="jspico2e.htm#docB064">eqn.AB019</a> <a href="jspico2e.htm#general_equation">general equation</a> <a href=jspico_e.htm#sum_eqn>sum equations</a> <a href=jspico_e.htm#docB04>triangle</a> <a href=jspico_e.htm#docB110>a<sub>k-#</sub> summary</a> </font></pre> <a name="program0">&lt;a name="program0"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#coefTriangle">triangle</a> <br> Prime move to <a href=http://freeman2.com/jsprime2.htm>freeman2.com</a>/<a href=jsprime2.htm>jsprime2.htm</a> <br><font color=red> "make_km(k)" button create make_km2(k) etc. functions <br> "km2" etc. button create formula to calculate coefficents <br> Both use k=[ ] box value. k is power in "{i^(k)} i=1 to i=n. <br></font> Box 2 &nbsp; <INPUT onclick='javascript:window.clipboardData.setData("Text",document.getElementById("boxc02").value.toString())' type=button value="copy 2"> <INPUT onclick='document.getElementById("boxc02").value=""' type=button value="del 2"> &nbsp; k=<input id="kValueBox" value="10" size=5 /> <INPUT onclick='javascript:make_km(kValueBox.value)' type=button value="make_km(k)"> <a href="#docD048"> help01 </a> <br><!--9809120925 use box2--> <TEXTAREA id=boxc02 name=boxc02 rows=10 cols=80 > </TEXTAREA> Following button create formula to find a<sub>k-#</sub> coefficients. <br> <INPUT onclick='javascript:boxc03.value=make_km2(kValueBox.value)' type=button value="km2"> <INPUT onclick='javascript:boxc03.value=make_km3(kValueBox.value)' type=button value="km3"> <INPUT onclick='javascript:boxc03.value=make_km4(kValueBox.value)' type=button value="km4"> <INPUT onclick='javascript:boxc03.value=make_km5(kValueBox.value)' type=button value="km5"> <INPUT onclick='javascript:boxc03.value=make_km6(kValueBox.value)' type=button value="km6"> <INPUT onclick='javascript:boxc03.value=make_km7(kValueBox.value)' type=button value="km7"> <INPUT onclick='javascript:boxc03.value=make_km8(kValueBox.value)' type=button value="km8"> <INPUT onclick='javascript:boxc03.value=make_km9(kValueBox.value)' type=button value="km9"> <INPUT onclick='javascript:boxc03.value=make_km10(kValueBox.value)' type=button value="m10"> <INPUT onclick='javascript:boxc03.value=make_km11(kValueBox.value)' type=button value="m11"> <INPUT onclick='javascript:boxc03.value=make_km12(kValueBox.value)' type=button value="m12"> <br> <INPUT onclick='javascript:boxc03.value=make_km13(kValueBox.value)' type=button value="m13"> <INPUT onclick='javascript:boxc03.value=make_km14(kValueBox.value)' type=button value="m14"> <INPUT onclick='javascript:boxc03.value=make_km15(kValueBox.value)' type=button value="m15"> <INPUT onclick='javascript:boxc03.value=make_km16(kValueBox.value)' type=button value="m16"> <INPUT onclick='javascript:boxc03.value=make_km17(kValueBox.value)' type=button value="m17"> <INPUT onclick='javascript:boxc03.value=make_km18(kValueBox.value)' type=button value="m18"> <INPUT onclick='javascript:boxc03.value=make_km19(kValueBox.value)' type=button value="m19"> <INPUT onclick='javascript:boxc03.value=make_km20(kValueBox.value)' type=button value="m20"> <br> Box 3 &nbsp; <INPUT onclick='javascript:window.clipboardData.setData("Text",document.getElementById("boxc03").value.toString())' type=button value="copy 2"> <INPUT onclick='document.getElementById("boxc03").value=""' type=button value="del 2"> <a href="#docD008"> help02 </a> <br><!--9809122041 use box3--> <TEXTAREA id=boxc03 name=boxc03 rows=10 cols=80 > </TEXTAREA> <br> <pre><a name="coefTriangle">&lt;a name="coefTriangle"&gt;</a><a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a><!-- 2009-09-14-14-03 add triangle add kp1 to km12--> 2009-09-01-22-21 <font color=red>Coefficient triangle</font> " i^(k) , (k)=(1) to (k)=(12), <font color=red>left-most column is k</font> coefficients from high power to constant (1) : 1/2, 1/2, 00ÿÿthis is "{i}=n*(n+1)/2=n*n/2+n/2+0 (2) : 1/3, 1/2, 2/12, 00ÿÿthis is "{i*i}=n*(n+1)*(2*n+1)/6 (3) : 1/4, 1/2, 3/12, 0, 00ÿÿthis is "{i^3}=[n*(n+1)/2]^2 (4) : 1/5, 1/2, 4/12, 0, -4/120, 0 <= <font color=red>Each row</font> <a href=jspico_e.htm#docB07><font color=red>sum to one</font></a> (5) : 1/6, 1/2, 5/12, 0, -10/120, 0, 0 (6) : 1/7, 1/2, 6/12, 0, -20/120, 0, 6/252, 0 (7) : 1/8, 1/2, 7/12, 0, -35/120, 0, 21/252, 0, 0 (8) : 1/9, 1/2, 8/12, 0, -56/120, 0, 56/252, 0, -4/120, 0 (9) :1/10, 1/2, 9/12, 0, -84/120, 0,126/252, 0, -18/120, 0, 0 (10):1/11, 1/2,10/12, 0,-120/120, 0,252/252, 0, -60/120, 0, 100/1320, 0 (11):1/12, 1/2,11/12, 0,-165/120, 0,462/252, 0,-165/120, 0, 550/1320, 0, 0 (12):1/13, 1/2,12/12, 0,-220/120, 0,792/252, 0,-396/120, 0,2200/1320, 0, -691/2730, 0<font color=red> (13):1/14, 1/2,13/12, 0,-286/120, 0,1287/252,0,-858/120, 0,7150/1320, 0, -691/420, 0 0 <!-- 98,09,16,15,45 so half factor is here "2*" !!!!! "//2*km11_unknown=" 98,09,16,15,47 change -98813/30030 to -98813/60060 --> (14):1/15, 1/2,14/12, 0,-364/120, 0,2002/252,0,-1716/120,0, 2002/132, 0, -691/90, 0, 7/6 column 01 02 03 04 05 06 07 08 09 10 11 12 13 14 kp1 kp0 km1 km2 km3 km4 km5 km6 km7 km8 km9 km10 km11 km12</font> <a name="coefTriangle2">&lt;a name="coefTriangle2"&gt;</a> <font color=red>In triangle, each row</font> <a href=jspico_e.htm#docB07><font color=red>sum to one</font></a> kp1 means a<sub>k+1</sub> [p]=[plus], km1 means a<sub>k-1</sub> [m]=[minus]. column 01,02,03,04,06,08,10,12,14 see <a href="jspico_e.htm#docB02">Pattern conjectures</a> column 05 <a href="jspico_e.htm#docB18">conjecture</a> and <a href="jspico_e.htm#getCol5">calculator</a>. <font color=red size=+2>Row (13) (14) are solved by program "make_km".</font> <font color=red>" i^(14) n-formula: use pow(n,15)/15, do not use pow(n,14)/15 ! " i^(14) direct sum: use pow(2,14), do not use pow(2,15) ! n-formula start at n^(k+1); direct sum is 1+2^k+...+n^k . ALERT!</font> <a name="docC001">&lt;a name="docC001"&gt;</a> Section "docA001" prime, goto <a href=http://freeman2.com/jsprime2.htm>freeman2.com</a>/<a href=jsprime2.htm>jsprime2.htm</a> Section "<a href="#docA101">docA101</a>" general introduction Section "<a href="#docB001">docB001</a>" symbolic equation, wondering Section "<a href="#docC001">docC001</a>" program in/output, wondering Section "<a href="#docD001">docD001</a>" program intro., no wondering 2009-09-14-15-20 start On top of Box2, there is a button [make_km(k)] This button make_km#() functions, for example [function make_km2(k)] make a<sub>k-2</sub> coefficient. If set k=[16] program build from function make_km2(k) up to function make_km15(k) Two parameter decide one coefficient. First parameter is [k], in Coefficient triangle [k] is left most column, left to [:] it can be labeled as [column 00] (no such label) k is power in " i^(k) , (k)=(1) to (k)=(12) <a name="docC002">&lt;a name="docC002"&gt;</a> Second parameter is [function make_km#(k)] Bottom of Coefficient triangle has a red row [kp1 kp0 km1 km2 ... km12] where kp1=k+1 for a<sub>k+1</sub> coefficient = 1/(k+1) kp0=k+0 for a<sub>k</sub> coefficient = 1/2 km1=k-1 for a<sub>k-1</sub> coefficient = k/12 First few kp1,kp0,km1 program NOT care. No button for them. They are simple, you can hand calculate those values. <a name="docC003">&lt;a name="docC003"&gt;</a> To find " i^(k) k=8 sum from i=1 to i=n (for example) the highest power is n^(k+1) Hand calculate kp1=1/(8+1)=1/9, get n^9/9 Hand calculate kp0=1/2, get n^8/2 Hand calculate km1=k/12=8/12, get (n^7)*8/12 Set k=[box] to k=[8] Click km2 button, get 6*km2_unknown=[newLineHere] -(126*1/9+56*1/2+21*8/12+(-56)) <a name="docC004">&lt;a name="docC004"&gt;</a> Paste "-(126*1/9+56*1/2+21*8/12+(-56))" to http://freeman2.com/<a href=http://freeman2.com/complex2.htm#calculator>complex2.htm</a>#<a href=complex2.htm#calculator>calculator</a> in box3, click [test box3 command output to box4] find answer [0], get (n^6)*0 All [km2],[km4],[km6]... even number button get zero answer. <a name="docC005">&lt;a name="docC005"&gt;</a> Next Click km3 button, get //5*km3_unknown= -(126*1/9+70*1/2+35*8/12+15*0/1+(-70)) Calculate value 5*km3_unknown=-2.333333333333 that is 15*km3_unknown=-7, find km3_unknown=-7/15 In Coefficient triangle table row (8) (k=8) and km3 column, find [-56/120] which is -7/15 get (n^5)*(-7/15) this is one term in a long equation. Rest coef. follow same procedure. <a name="docC006">&lt;a name="docC006"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a> Last two steps, Click km7 button, get //1*km7_unknown= -(1*1/9+1*1/2+1*8/12+1*0/1+1*(-336)/720+1*0/1+1*6720/30240+1*0/1+(-1)) Its value is -1/30, goto Coefficient triangle table row (8) and column km7 find -4/120. -4/120 = -1/30 OK. Get (n^1)*(-1/30) Finally click km8 button, get error message "make_km8(k): need k&gt;8, now k=8" goto Coefficient triangle table row (8) and column km8 find constant=zero.<font color=red> All equation constant must be zero. For make_km8(k), click km2 up to km7.</font> Above procedure should get the following equation " i^8 to n = +n*n*n*n*n*n*n*n*n*(1/9)+n*n*n*n*n*n*n*n*(1/2) +n*n*n*n*n*n*n*(2/3)+n*n*n*n*n*n*(0)+n*n*n*n*n*(-7/15) +n*n*n*n*(0)+n*n*n*(2/9)+n*n*(0)+n*(-1/30) <a name="docC007">&lt;a name="docC007"&gt;</a> Coefficient triangle row (8) has whole coef. Coef. is already here! because worked before. This process help you find NEW EQUATION, find those coefficient not listed. This process has limit. It need function definition for function kp1(k), function kp0(k), function km1(k) <a name="docC008">&lt;a name="docC008"&gt;</a> At writing time defined up to function km8(k) function km9(k) is undefined. Then If set k=15 and click km10, box3 show [15*__km9__unknown_coefficient__]. We can not proceed at this point at writing time. LiuHH tested column km3 and km5 get result match known coefficient value. But can not promise all the way correct. Still working. 2009-09-14-15-58 stop <a name="docC009">&lt;a name="docC009"&gt;</a> 2009-09-17-14-29 start The main engine of this file jspico2e.htm is function make_km() make_km() create Javascript code for calculating Positive Integer power sum COefficients. make_km() started on 2009-09-12-08-09 //9809120809 make_km() done on 2009-09-14-11-26 //9809141126 <a name="docC010">&lt;a name="docC010"&gt;</a> on 2009-09-14-19-24 first time use make_km() to find coefficients for " i^(13) n-formula. There are several error made during first time run. Second time run for " i^(14) n-formula is success. Please comapre the trouble record with the <a href="#docD001">success record</a>. <a name="docC011">&lt;a name="docC011"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a><font color=red> LiuHH present first time trouble run, the main purpose is hope if reader do " i^(15) n-formula avoid the same trouble/error. Save your time. </font> <a name="docC012">&lt;a name="docC012"&gt;</a> //9809141924 start create sum{i^13} equation n^14*1/14 n^13*1/2 n^12*13/12 n^11*0 n^10*(-143/60) n^9*0 n^8*(25740/5040) n^7*0 n^6*(-429/60) n^5*0 n^4*(65/12) n^3*0 n^2*<s>(-98813/30030)</s> // -98813/60060 n^1*0 <a name="docC013">&lt;a name="docC013"&gt;</a> Above is " i^(13) n-formula coefficients. Below calculate coef. put result to above table. 2009-09-17-14-47 here. <a name="docC014">&lt;a name="docC014"&gt;</a> Next " i^(13) km2_unknown=0 please see " i^(14) <a href="#docD011">docD011</a> explanation. //9809141927 //11*km2_unknown= -(1001*1/14+286*1/2+66*13/12+(-286)) =0 <a name="docC015">&lt;a name="docC015"&gt;</a> Next " i^(13) km3_unknown=0 please see " i^(14) <a href="#docD012">docD012</a> explanation. //10*km3_unknown= -(2002*1/14+715*1/2+220*13/12+55*0/1+(-715)) =-23.83333333333337 =-143/6 km3_unknown= -143/60 //9809141929 <a name="docC016">&lt;a name="docC016"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a> Next " i^(13) km4_unknown=0 reason is same as " i^(14) <a href="#docD011">docD011</a> explanation. //9*km4_unknown= -(3003*1/14+1287*1/2+495*13/12+165*0/1+45*(-1716)/720+(-1287)) =0 <a name="docC017">&lt;a name="docC017"&gt;</a> Next " i^(13) km5_unknown=0 reason is same as " i^(14) <a href="#docD012">docD012</a> explanation. //9809141931 //8*km5_unknown= -(3432*1/14+1716*1/2+792*13/12+330*0/1+120*(-1716)/720+36*0/1+(-1716)) = 40.85714285714289 <a name="docC018">&lt;a name="docC018"&gt;</a> //9809141932 -( 3432*1/14 +1716*1/2 +792*13/12 +330*0/1 +120*(-1716)/720 +36*0/1 +(-1716)) //9809141933 -( 3432*1/14 +1716*1/2 +792*13/12 +120*(-1716)/720 +(-1716)) <a name="docC019">&lt;a name="docC019"&gt;</a> there are two major denominator 720 and 14 (12 is covered by 720, 2 is covered by 12) Use <a href=http://freeman2.com/jsprime2.htm#program0>freeman2.com</a>/<a href=jsprime2.htm#program0>jsprime2.htm</a> to find least common multiple for 720 and 14 720/14 2*2*2*2*3*3*5 2*7 720*7 5040 <a name="docC020">&lt;a name="docC020"&gt;</a> //8*km5_unknown= = 40.85714285714289 = -( 3432*360/5040 +1716*2520/5040 +792*13*420/5040 +120*(-1716)*7/5040 -1716*5040/5040 ) = -( 3432*360 +1716*2520 +792*13*420 +120*(-1716)*7 -1716*5040)/5040 = <a name="docC021">&lt;a name="docC021"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a> //8*km5_unknown= 205920/5040 Here, do not forget '8' factor km5_unknown= 205920/5040/8 = 25740/5040 In <a href="#coefTriangle">Coefficient triangle</a> '25740/5040' is recorded as '1287/252' Above done " i^(13) km5_unknown <a name="docC022">&lt;a name="docC022"&gt;</a> Below start " i^(13) km6_unknown //9809141941 //7*km6_unknown= -(3003*1/14+1716*1/2+924*13/12+462*0/1+210*(-1716)/720+84*0/1+28*154440/30240+(-1716)) =0 Zero is quick. <a name="docC023">&lt;a name="docC023"&gt;</a> Below start " i^(13) km7_unknown //9809141942 //6*km7_unknown= -(2002*1/14+1287*1/2+792*13/12+462*0/1+252*(-1716)/720+126*0/1+56*154440/30240+21*0/1+(-1287)) = -42.90000000000009 km7_unknown= -42.9/6 // '6' come from "//6*km7_unknown=" = -429/60 //9809141943 //5*km8_unknown= -(1001*1/14+715*1/2+495*13/12+330*0/1+210*(-1716)/720+126*0/1+70*154440/30240+35*0/1+15*(-8648640)/1209600+(-715)) =0 2009-09-17-15-09 here <a name="docC024">&lt;a name="docC024"&gt;</a> //9809141944 //4*km9_unknown= -(364*1/14+286*1/2+220*13/12+165*0/1+120*(-1716)/720+84*0/1+56*154440/30240+35*0/1+20*(-8648640)/1209600+10*0/1+(-286)) = 21.66666666666663 =65/3 km9_unknown= 65/3/4 = 65/12 //9809141946 //3*km10_unknown= -(91*1/14+78*1/2+66*13/12+55*0/1+45*(-1716)/720+36*0/1+28*154440/30240+21*0/1+15*(-8648640)/1209600+10*0/1+6*(259459200)/47900160+(-78)) =0 <a name="docC025">&lt;a name="docC025"&gt;</a> <font color=red> BELOW LiuHH MADE MISTAKE !! forget '2' in "//2*km11_unknown=" </font> //9809141947 //2*km11_unknown= -(14*1/14+13*1/2+12*13/12+11*0/1+10*(-1716)/720+9*0/1+8*154440/30240+7*0/1+6*(-8648640)/1209600+5*0/1+4*(259459200)/47900160+3*0/1+(-13)) = -3.2904761904761876 = -<s>98813/30030</s> <a name="docC026">&lt;a name="docC026"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a> //9809141950 //1*km12_unknown= -(1*1/14+1*1/2+1*13/12+1*0/1+1*(-1716)/720+1*0/1+1*154440/30240+1*0/1+1*(-8648640)/1209600+1*0/1+1*(259459200)/47900160+1*0/1+1*(3113510400)/(-1892437580.3183791)+(-1)) =4.440892098500626e-16 =0 <a name="docC027">&lt;a name="docC027"&gt;</a> //9809141924 start create sum{i^13} equation //9809141951 done create sum{i^13} equation n^14*1/14 n^13*1/2 n^12*13/12 n^11*0 n^10*(-143/60) n^9*0 n^8*(25740/5040) n^7*0 n^6*(-429/60) n^5*0 n^4*(65/12) n^3*0 n^2*<s>(-98813/30030)</s> // -98813/60060 n^1*0 <a name="docC028">&lt;a name="docC028"&gt;</a> Because made error and not knowing error. then LiuHH fall into big puzzle. //9809141952 1/14+1/2+13/12+(-143/60)+(25740/5040)+(-429/60)+(65/12)+(-98813/30030) =-0.6452380952380956 not sum to one, wrong !! change denominator, compatible with table see which one change abruptly //9809142001 1/14+1/2+13/12 -286/120 +1287/252 -858/120+7150/1320-98813/30030 =-0.6452380952380956 <a name="docC029">&lt;a name="docC029"&gt;</a> On 2009-09-14-20-05 LiuHH goto jspico_e.htm#program0 find numerical solution for " i^(13) //9809142005 //9809142006 Arndt program " i^(13) for i=1 to i=n coefficients from high power to constant 169578329300447/2374096584967660 5658653150261099/11317307882501584 1319612547501560/1218100649278479 -5525172457317795/78508065034983440000 -10583763248713898/4442851227626055 -8608864186755377/680638044958482300 4316292031548731/828854527932351 -3135700589519679/5501623798799788 -24332656978023830/5008515540996507 -48431513811591570/7644314127260959 94430015665479930/5654831203049255 -46688924909745730/4253104636703159 2917122053756420/5283506936000621 41346534510728350/7857465901817549 -4116346999188412/1270823313873759 <a name="docC030">&lt;a name="docC030"&gt;</a> //9809142018 sum to one 0.07142857218791584+0.49999993010796373+1.0833362155115915-0.00007037713201638279-2.3822006874556347-0.01264822654349347+5.207538700809288-0.5699591073827604-4.858257257834633-6.33562580047258+16.69899812651533-10.977610216036717+0.5521185245125374+5.2620698616285795-3.239118258415364 = 1.000000000000008 Arndt program sum is one //9809142020 my sum not one !! 1/14+1/2+13/12+(-143/60)+(25740/5040)+(-429/60)+(65/12)+(-98813/30030) =-0.6452380952380956 <a name="docC031">&lt;a name="docC031"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a> not sum to one, wrong !! " i^(13) for i=1 to i=n coefficients from high power to constant my program wrong !! At this time 2009-09-14-20-20, LiuHH do not know where is wrong. 2009-09-17-15-22 stop <a name="docC032">&lt;a name="docC032"&gt;</a> 2009-09-17-17-07 start Busy two hours, 2009-09-14-22-16 can not find error !! //9809142216 can not find error !! [[ //9809161345 1/14+1/2+13/12-286/120+1287/252-858/120+7150/1320-98813/30030 =-0.6452380952380956 but it should be ONE ]] <a name="docC033">&lt;a name="docC033"&gt;</a> //9809161532 2009-09-16-15-32 suspect last coefficient '-98813/30030' error " i^(13) coefficients 1/14+1/2+13/12-286/120+1287/252 -858/120+7150/1320-98813/30030 not sum to one. (should be one) <a name="docC034">&lt;a name="docC034"&gt;</a> rewrite as 1/14+1/2+13/12-286/120+1287/252 -858/120+7150/1320+x=1 where x is last coefficient. Solve for x get x=1-(1/14+1/2+13/12-286/120+1287/252-858/120+7150/1320) =-1.6452380952380947 //9809161534 1.6452380952380956*30030 = 49406.500000000014 98813-49406.5= 49406.5 <a name="docC035">&lt;a name="docC035"&gt;</a> //9809161535 try 1/14+1/2+13/12-286/120+1287/252 -858/120+7150/1320-49406.5/30030 1/14+1/2+13/12-286/120+1287/252-858/120+7150/1320-49406.5/30030 = 0.9999999999999996 <a name="docC036">&lt;a name="docC036"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a> //9809161536 find suspect point !! last term in 1/14+1/2+13/12-286/120+1287/252 -858/120+7150/1320-<s>98813/30030</s> is -<s>98813/30030</s> it should be half -98813/60060 !! //9809161539 <a name="docC037">&lt;a name="docC037"&gt;</a> set k=13, click km12 get [[ //1*km12_unknown= -(1*1/14+1*1/2 +1*13/12 +1*0/1 +1*(-1716)/720 +1*0/1+1*154440/30240 +1*0/1 +1*(-8648640)/1209600 +1*0/1 +1*(259459200)/47900160 +1*0/1 +1*(3113510400)/<s>(-691/1307674368000)</s> +(-1)) ]] = 5892124087672109000 but it should be zero !! <a name="docC038">&lt;a name="docC038"&gt;</a> //9809161544 set k=13 click km11 //2*km11_unknown= //LiuHH forget "2*" in this line -(14*1/14+13*1/2+12*13/12+11*0/1+10*(-1716)/720+9*0/1+8*154440/30240+7*0/1+6*(-8648640)/1209600+5*0/1+4*(259459200)/47900160+3*0/1+(-13)) = -3.2904761904761876 <a name="docC039">&lt;a name="docC039"&gt;</a> back to -98813/30030 = -3.2904761904761903 //9809161545 so half factor is here "2*" !!!!! "//2*km11_unknown=" <a name="docC040">&lt;a name="docC040"&gt;</a> //9809161547 change -98813/30030 to -98813/60060 //9809161553 -(1*1/14+1*1/2+1*13/12+1*0/1+1*(-1716)/720+1*0/1+1*154440/30240+1*0/1+1*(-8648640)/1209600+1*0/1+1*(259459200)/47900160+1*0/1+1*(3113510400)/(-691/1307674368000)+(-1)) get 5892124087672109000 but it should be zero <a name="docC041">&lt;a name="docC041"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a> 2009-09-17-17-22 Note: Two error show up together then it is really confuse. -(1*1/14+1*1/2+1*13/12+1*0/1+1*(-1716)/720+1*0/1+1*154440/30240+1*0/1+1*(-8648640)/1209600+1*0/1+1*(259459200)/47900160+1*0/1+1*(3113510400)/(-691/1307674368000)+(-1)) //9809161554 suspect (-691/1307674368000) <a name="docC042">&lt;a name="docC042"&gt;</a> [[ //9809141950 //1*km12_unknown= -(1*1/14+1*1/2 +1*13/12 +1*0/1 +1*(-1716)/720 +1*0/1+1*154440/30240 +1*0/1 +1*(-8648640)/1209600 +1*0/1 +1*(259459200)/47900160 +1*0/1 +1*(3113510400)/(-1892437580.3183791)+(-1)) =4.440892098500626e-16 =0 ]] //9809141950 get zero (correct) answer //km11(k) use "k*.../(-1892437580.3183791)" correct <a name="docC043">&lt;a name="docC043"&gt;</a> //9809161607 why changed !!?? why now 2009-09-16-16-07 not get zero? //km11(k) use "k*.../(-691/1307674368000)" error //9809161614 CORRECT to km11(k) use "k*.../(-1307674368000/691)" then <a name="docC044">&lt;a name="docC044"&gt;</a> //9809161615 get zero //1*km12_unknown= -(1*1/14+1*1/2+1*13/12+1*0/1+1*(-1716)/720+1*0/1+1*154440/30240+1*0/1+1*(-8648640)/1209600+1*0/1+1*(259459200)/47900160+1*0/1+1*(3113510400)/(-1307674368000/691)+(-1)) 4.440892098500626e-16 <a name="docC045">&lt;a name="docC045"&gt;</a> 1/14, 1/2,13/12, 0,-286/120, 0,1287/252,0,-858/120, 0,7150/1320, 0,-98813/60060, 0 0 //9809161617 sum to ONE !! 1/14+1/2+13/12-286/120+1287/252-858/120+7150/1320-98813/60060 0.9999999999999996 <a name="docC046">&lt;a name="docC046"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a> <font color=red size=+3>Possible human errors are listed below</font> <font color=red>Below is summary, above is detail.</font> error 1: forget "2" in "<a href="#docC025">//2*km11_unknown=</a>" detect at not <a href=jspico_e.htm#docB07>sum to one</a> (should be one) <a name="docC047">&lt;a name="docC047"&gt;</a> error 2: in equation calculation use k*(k-1)*...*(k-10<font color=red size=+3>)*(</font>-691/1307674368000) in function km11(k) definition use k*(k-1)*...*(k-10<font color=red size=+3>)/(</font>-<s>691/1307674368000</s>) equation change from '*' to '/', but forget to reverse constant. Cause second error. detect at not get zero coefficient All k-[even] constant should be 0. 2009-09-17-17-42 record to here. <a name="docC048">&lt;a name="docC048"&gt;</a> Next is third error. [[ //9809161350 " i^(13) i=1 to i=2 get n=2 pow(n,13)*1/14+pow(n,12)*1/2+pow(n,11)*13/12-pow(n,9)*286/120+pow(n,7)*1287/252-pow(n,5)*858/120+pow(n,3)*7150/1320-pow(n,1)*98813/30030 =4093.209523809522 above wrong answer below correct answer pow(2,13)+ pow(1,13) =8193 ]] <a name="docC049">&lt;a name="docC049"&gt;</a> <font color=red>change from "/30030" to "/60060"</font> //9809161619 n=2 pow(n,13)*1/14+pow(n,12)*1/2+pow(n,11)*13/12-pow(n,9)*286/120+pow(n,7)*1287/252-pow(n,5)*858/120+pow(n,3)*7150/1320-pow(n,1)*98813/60060 get 4096.499999999998 it is half of correct answer !! <a name="docC050">&lt;a name="docC050"&gt;</a> <font color=red>change from pow(n,13)/14 to pow(n,14)/14</font> //9809161624 above error, below right n=2 pow(n,14)*1/14+pow(n,13)*1/2+pow(n,12)*13/12-pow(n,10)*286/120+pow(n,8)*1287/252-pow(n,6)*858/120+pow(n,4)*7150/1320-pow(n,2)*98813/60060 get 8192.999999999996 compare with pow(2,13)+ pow(1,13) =8193 //9809161627 OK k=13 passed !!! 9809161627 <a name="docC051">&lt;a name="docC051"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a> error 3: n=2 pow(n,13)*1/14+pow(n,12)*1/2+pow(n,11)*13/12-pow(n,9)*286/120+pow(n,7)*1287/252-pow(n,5)*858/120+pow(n,3)*7150/1320-pow(n,1)*98813/60060 n=2 above error, below right pow(n,14)*1/14+pow(n,13)*1/2+pow(n,12)*13/12-pow(n,10)*286/120+pow(n,8)*1287/252-pow(n,6)*858/120+pow(n,4)*7150/1320-pow(n,2)*98813/60060 <a name="docC052">&lt;a name="docC052"&gt;</a> After third error, made red line alert below <a href="#coefTriangle">Coefficient triangle</a> <font color=red>" i^(14) n-formula: use pow(n,15)/15, do not use pow(n,14)/15 ! " i^(14) direct sum: use pow(2,14), do not use pow(2,15) ! n-formula start at n^(k+1); direct sum is 1+2^k+...+n^k . ALERT!</font> <a name="docC053">&lt;a name="docC053"&gt;</a> " i^(13) n-formula is a hard journey. After this experience, " i^(14) n-formula is a <a href="#docD001">fast work</a>. If you do " i^(15) n-formula or higher Please read " i^(13) n-formula experience first. Freeman 2009-09-17-17-55 <hr> <a name="docD001">&lt;a name="docD001"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a> Section "docA001" prime, goto <a href=http://freeman2.com/jsprime2.htm>freeman2.com</a>/<a href=jsprime2.htm>jsprime2.htm</a> Section "<a href="#docA101">docA101</a>" general introduction Section "<a href="#docB001">docB001</a>" symbolic equation, wondering Section "<a href="#docC001">docC001</a>" program in/output, wondering Section "<a href="#docD001">docD001</a>" program intro., no wondering 2009-09-16-21-53 <font color=red size=+3>How to build " i^(14) n-formula?</font> (<font color=red>build " i^(14) n-formula is a success story</font>) for example " i^(14) i=1 to i=3 direct sum method is 1^(14) + 2^(14) + 3^(14) = 4799354 or in Javascript code, it is pow(3,14)+pow(2,14)+pow(1,14) = 4799354 here n=3. for large n, direct sum method is slow. for large n, n-formula method is fast. Next is step by step, to build " i^(14) n-formula Next take from working record. <a name="docD002">&lt;a name="docD002"&gt;</a> Please open four web pages. First: <a href=http://freeman2.com/jspico2e.htm#docD001>jspico2e.htm</a>#<a href=jspico2e.htm#docD001>docD001</a> ; discuss how to second: <a href=http://freeman2.com/jspico2e.htm#program0>jspico2e.htm</a>#<a href=jsprime2.htm#program0>program0</a> ; create formula third: <a href=http://freeman2.com/complex2.htm#calculator>complex2.htm</a>#<a href=complex2.htm#calculator>calculator</a>; calculate value fourth: <a href=http://freeman2.com/jsprime2.htm#program0>jsprime2.htm</a>#<a href=jsprime2.htm#program0>program0</a> ; least common multiple <a name="docD003">&lt;a name="docD003"&gt;</a> where<font color=red> second: jspico2e.htm#program0 ; create formula use program "make_km" to get coefficient value. program "make_km" is the heart of this file. </font> <a name="docD004">&lt;a name="docD004"&gt;</a> //9809161927 start (14), " i^(k) = " i^(14) All calculated value put to here, next is general n-formula for " i^(14) pow(n,15)*(1/15) pow(n,14)*(1/2) pow(n,13)*(14/12) pow(n,12)*0 pow(n,11)*(-364/120) pow(n,10)*0 pow(n,9)*(2002/252) pow(n,8)*0 pow(n,7)*(-1716/120) pow(n,6)*0 pow(n,5)*(20020/1320) pow(n,4)*0 pow(n,3)*(-1383382/180180) pow(n,2)*0 pow(n,1)*(7/6) pow(n,0)*0 <a name="docD005">&lt;a name="docD005"&gt;</a> We are working at " i^(14) formula n-formula begin at n^(k+1) (direct sum use 1^k+2^k+...+n^k) here k=14, k is integer power. We begin at n^(k+1) = n^(15) In Javascript code, n^(15) is pow(n,15) here k=14, if start from pow(n,14) that is ERROR!! <a name="docD006">&lt;a name="docD006"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a> Highest three power terms, user calculate by hand. program "make_km" ignore them. Because highest three power terms not fit the general pattern. For (k+1)th term, the derived result is at <a href=jspico_e.htm#docB77>jspico_e.htm#docB77</a>. Equation is pow(n,15)*(1/15) // k+1=15 This is <a href="#docD004">" i^(14) n-formula</a> first line. <a name="docD007">&lt;a name="docD007"&gt;</a> For (k)th term, the derived result is at <a href=jspico_e.htm#docB82>jspico_e.htm#docB82</a>. Equation is pow(n,14)*(1/2) // k=14 This is <a href="#docD004">" i^(14) n-formula</a> second line. For (k-1)th term, the derived result is at <a href=jspico_e.htm#docB87>jspico_e.htm#docB87</a>. Equation is pow(n,13)*(14/12) // k-1=13 This is <a href="#docD004">" i^(14) n-formula</a> third line. 2009-09-16-22-24 here <a name="docD008">&lt;a name="docD008"&gt;</a> Next begin at (k-2)th term. Goto second browser page jspico2e.htm#program0 "Box 2" line has a "k=[ ]" Please fill 14 to the k box, it look like "k=[14 ]" On top of "Box 3", find "km2" "km3" .... "m13" .... "m20" buttons. Click "km2" button, "km2" means "k-2" 'm' is 'minus'. <a name="docD009">&lt;a name="docD009"&gt;</a> After click "km2" button, get next in [[...]] //9809161930 <== this is working time stamp [[ //12*km2_unknown= -(1365*1/15+364*1/2+78*14/12+(-364)) ]] (in "k=[ ]" box fill "14" get above) We are working at " i^(14) formula in "12*km2_unknown" 12+2=14 then it is right result. <a name="docD010">&lt;a name="docD010"&gt;</a> Next goto third: <a href=http://freeman2.com/complex2.htm#calculator>complex2.htm</a>#<a href=complex2.htm#calculator>calculator</a>; calculate value paste next two lines //12*km2_unknown= -(1365*1/15+364*1/2+78*14/12+(-364)) to complex2.htm#calculator box 3 <a name="docD011">&lt;a name="docD011"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a> Click "text box 3 command, output to box 4" button, get next two lines -(1365*1/15+364*1/2+78*14/12+(-364)) 0 We get //12*km2_unknown=0 and km2_unknown=0 is correct result. This is just ONE coefficient. There are total 15 (k+1) coefficients. Go to the <a href="#docD004">" i^(14) n-formula</a> paste '0' to right side of "pow(n,12)*" get "pow(n,12)*0" <a name="docD012">&lt;a name="docD012"&gt;</a> Next find km3_unknown On top of "Box 3", click "km3" button. //9809161931 get next two lines. //11*km3_unknown= -(3003*1/15+1001*1/2+286*14/12+66*0/1+(-1001)) again, in "11*km3_unknown" 11+3=14 is correct sign for " i^(14) formula <a name="docD013">&lt;a name="docD013"&gt;</a> paste //11*km3_unknown= -(3003*1/15+1001*1/2+286*14/12+66*0/1+(-1001)) to complex2.htm#calculator box 3 Click "text box 3 command, output to box 4" button, get next two lines -(3003*1/15+1001*1/2+286*14/12+66*0/1+(-1001)) -33.36666666666679 <a name="docD014">&lt;a name="docD014"&gt;</a> this number is <a href="#coefTriangle">Coefficient triangle</a> row-(14) and column-05 (km3) value "-364/120". Since we are building " i^(14) formula, "-364/120" should be unknown to us at this moment. However, lower power coefficients are known to us. Look at column-05 (km3) earlier coefficients, all use 120 as denominator. <a name="docD015">&lt;a name="docD015"&gt;</a> We try -33.36666666666679*(120/120) -33.36666666666679*120=-4004 So -33.36666666666679=-4004/120 that is //11*km3_unknown=-4004/120 km3_unknown=-(4004/11)/120=-364/120 Go to the <a href="#docD004">" i^(14) n-formula</a> paste '-364/120' to right side of "pow(n,11)*" get "pow(n,11)*(-364/120)" Above, detailed two coefficients. The following is just working record. 2009-09-16-23-12 here <a name="docD016">&lt;a name="docD016"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a> //9809161932 //10*km4_unknown= -(5005*1/15+2002*1/2+715*14/12+220*0/1+55*(-2184)/720+(-2002)) = -2.2737367544323206e-13 //9809161933 //9*km5_unknown= -(6435*1/15+3003*1/2+1287*14/12+495*0/1+165*(-2184)/720+45*0/1+(-3003)) = 71.5 = (18018/252) km5_unknown= (2002/252) <a name="docD017">&lt;a name="docD017"&gt;</a> //9809161935 //8*km6_unknown= -(6435*1/15+3432*1/2+1716*14/12+792*0/1+330*(-2184)/720+120*0/1+36*240240/30240+(-3432)) =0 //9809161936 //7*km7_unknown= -(5005*1/15+3003*1/2+1716*14/12+924*0/1+462*(-2184)/720+210*0/1+84*240240/30240+28*0/1+(-3003)) = -100.10000000000036 = (-12012/120) km7_unknown= (-12012/120)/7 = (-1716/120) <a name="docD018">&lt;a name="docD018"&gt;</a> //9809161937 //6*km8_unknown= -(3003*1/15+2002*1/2+1287*14/12+792*0/1+462*(-2184)/720+252*0/1+126*240240/30240+56*0/1+21*(-17297280)/1209600+(-2002)) = 2.2737367544323206e-13 <a name="docD019">&lt;a name="docD019"&gt;</a> //9809161944 //5*km9_unknown= -(1365*1/15+1001*1/2+715*14/12+495*0/1+330*(-2184)/720+210*0/1+126*240240/30240+70*0/1+35*(-17297280)/1209600+15*0/1+(-1001)) = 75.83333333333348 = 100100/1320 km9_unknown= (20020/1320) <a name="docD020">&lt;a name="docD020"&gt;</a> //9809161946 //4*km10_unknown= -(455*1/15+364*1/2+286*14/12+220*0/1+165*(-2184)/720+120*0/1+84*240240/30240+56*0/1+35*(-17297280)/1209600+20*0/1+10*(726485760)/47900160+(-364)) =0 //9809161947 //3*km11_unknown= -(105*1/15+91*1/2+78*14/12+66*0/1+55*(-2184)/720+45*0/1+36*240240/30240+28*0/1+21*(-17297280)/1209600+15*0/1+10*(726485760)/47900160+6*0/1+(-91)) = -23.033333333333274 = (-1383382/60060) km11_unknown= (-1383382/180180) <a name="docD021">&lt;a name="docD021"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a> //9809161952 //2*km12_unknown= -(15*1/15+14*1/2+13*14/12+12*0/1+11*(-2184)/720+10*0/1+9*240240/30240+8*0/1+7*(-17297280)/1209600+6*0/1+5*(726485760)/47900160+4*0/1+3*(14529715200)/(-1307674368000/691)+(-14)) = 3.552713678800501e-15 //9809161953 //1*km13_unknown= -(1*1/15+1*1/2+1*14/12+1*0/1+1*(-2184)/720+1*0/1+1*240240/30240+1*0/1+1*(-17297280)/1209600+1*0/1+1*(726485760)/47900160+1*0/1+1*(14529715200)/(-1307674368000/691)+1*0/1+(-1)) = 1.1666666666666678 = 7/6 <a name="docD022">&lt;a name="docD022"&gt;</a> 9809161927 start (14), " i^(k) = " i^(14) 9809161954 done (14), " i^(k) = " i^(14) pow(n,15)*(1/15) pow(n,14)*(1/2) pow(n,13)*(14/12) pow(n,12)*0 pow(n,11)*(-364/120) pow(n,10)*0 pow(n,9)*(2002/252) pow(n,8)*0 pow(n,7)*(-1716/120) pow(n,6)*0 pow(n,5)*(20020/1320) pow(n,4)*0 pow(n,3)*(-1383382/180180) pow(n,2)*0 pow(n,1)*(7/6) pow(n,0)*0 <a name="docD023">&lt;a name="docD023"&gt;</a> //9809161955 now check sum (1/15) (1/2) (14/12) (-364/120) (2002/252) (-1716/120) (20020/1320) (-1383382/180180) (7/6) <a name="docD024">&lt;a name="docD024"&gt;</a> //9809161956 (1/15)+(1/2)+(14/12)+(-364/120)+(2002/252)+(-1716/120)+(20020/1320)+(-1383382/180180)+(7/6) = 0.9999999999999989 9809161956 GREAT !! 9809161956 <font color=red>in a row, coefficient sum <a href=jspico_e.htm#docB07>MUST be ONE</a>.</font> <a name="docD025">&lt;a name="docD025"&gt;</a> 9809162006 SUM{i^14} i=1 to i=n coefficients are (1/15) (1/2) (14/12) (-364/120) (2002/252) (-1716/120) (20020/1320) (-1383382/180180) (7/6) <a name="docD026">&lt;a name="docD026"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a> //9809162008 equation is pow(n,15)*(1/15) +pow(n,14)*(1/2) +pow(n,13)*(14/12) +pow(n,11)*(-364/120) +pow(n,9)*(2002/252) +pow(n,7)*(-1716/120) +pow(n,5)*(20020/1320) +pow(n,3)*(-1383382/180180) +pow(n,1)*(7/6) <a name="docD027">&lt;a name="docD027"&gt;</a> //9809162009 n=2 pow(n,15)*(1/15)+pow(n,14)*(1/2)+pow(n,13)*(14/12)+pow(n,11)*(-364/120)+pow(n,9)*(2002/252)+pow(n,7)*(-1716/120)+pow(n,5)*(20020/1320)+pow(n,3)*(-1383382/180180)+pow(n,1)*(7/6) = 16385.000000000003 pow(2,14)+1 16385 //9809162010 <a name="docD028">&lt;a name="docD028"&gt;</a> easy error is that 9809162012 in [[ n=2 pow(n,15)*(1/15)+pow(n,14)*(1/2)+pow(n,13)*(14/12)+pow(n,11)*(-364/120)+pow(n,9)*(2002/252)+pow(n,7)*(-1716/120)+pow(n,5)*(20020/1320)+pow(n,3)*(-1383382/180180)+pow(n,1)*(7/6) ]] use pow(n,15) (CORRECT) and in direct sum, use pow(2,15)+1 WRONG!! next is right pow(2,14)+1 9809162013 stop record 2009-09-16-23-17 stop <a name="docD029">&lt;a name="docD029"&gt;</a> 2009-09-17-09-56 start Above introduce how to build equation for " i^(14) Program worked, because all necessary functions are defined. Function for " i^(14) is function make_km14(k) { ..... } <a name="docD030">&lt;a name="docD030"&gt;</a> It is 30 lines long. Now see the shorter [[ function make_km3(k) { if(k&lt;4)return "make_km"+k+"(k): need k&gt;"+k+", now k="+k; k=parseInt(k);//9809122215 //### pow(n,k-u1) coefficient where u1=4 //### coefficient is valid only if k&gt;= u1=4 //### coefficient has u0 bicof() lines, u0=7 return "//"+bicof(k-3,k-4)+"*km3_unknown=\n-(" +bicof(k+1,k-4)+"*"+kp1(k)+"+" +bicof(k ,k-4)+"*"+kp0(k)+"+" +bicof(k-1,k-4)+"*"+km1(k)+"+" +bicof(k-2,k-4)+"*"+km2(k)+"+" //+bicof(k-4,k-4)+"*"+km4(k)+"+" +"("+(-bicof(k,k-4))+")" //(-km4(k)) +")\n" ; } ]] <a name="docD031">&lt;a name="docD031"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a> To accomplish our goal n-formula for " i^(14) we need function definition function make_km14(k) which is longer than above make_km3(k) we need function definition for kp1(k), kp0(k), km1(k), km2(k) etc. First three functions are kp1(k)=1/(k+1) kp0(k)=1/2 km1(k)=k/12 The following functions are km2(k)=km4(k)=km6(k)= ... km[even](k)=0 <a name="docD032">&lt;a name="docD032"&gt;</a> All km[odd](k) are non-zero, for example function km3(k){ k=parseInt(k); //9809122212 return '('+eval(-k*(k-1)*(k-2))+')/720';//9809122011 } function make_km14(k) need km function up to km13(k). When begin write function make_km(k) only define up to km8(k), km9(k) is undefined. <a name="docD033">&lt;a name="docD033"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a> <font color=red size=+3>km9(k) return k-9 coefficient value</font> <font color=red>Build km9(k) apply to build km[odd number]</font> km9(k) body is next function km9(k) //9809122306 chocked, 9809141831 passed { k=parseInt(k); return '('+eval(k*(k-1)*(k-2)*(k-3)*(k-4)*(k-5)*(k-6)*(k-7)*(k-8))+')/47900160';//9809141831 } The problem is the constant "47900160". How to find this constant? "km9(k)" is "k minus 9 (k)" <a name="docD034">&lt;a name="docD034"&gt;</a> Next equation solve for a<sub>k-9</sub> pow(n,k-10)* [ +bino(k+1,k-10)*a<sub>k+1</sub> +bino(k ,k-10)*a<sub>k</sub> +bino(k-1,k-10)*a<sub>k-1</sub> +bino(k-2,k-10)*a<sub>k-2</sub> +bino(k-3,k-10)*a<sub>k-3</sub> +bino(k-4,k-10)*a<sub>k-4</sub> +bino(k-5,k-10)*a<sub>k-5</sub> +bino(k-6,k-10)*a<sub>k-6</sub> +bino(k-7,k-10)*a<sub>k-7</sub> +bino(k-8,k-10)*a<sub>k-8</sub> <== a<sub>k-8</sub> and early all known +bino(k-9,k-10)*a<sub>k-9</sub> <== a<sub>k-9</sub> unknown +bino(k-10,k-10)*a<sub>k-10</sub> <==cancel -bino(k,k-10) -a<sub>k-10</sub> <==cancel ] <a name="docD035">&lt;a name="docD035"&gt;</a> in which +bino(k-10,k-10)*a<sub>k-10</sub>-a<sub>k-10</sub> =1*a<sub>k-10</sub>-a<sub>k-10</sub> =0 In <a href="#program0">make_km</a> program, set k=[10] then click "km9" button this use "km1" to "km8" value solve for "km9" <a name="docD036">&lt;a name="docD036"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a> Next is working record of how to find constant a<sub>k-9</sub> . [[ return '('+eval(k*(k-1)*(k-2)*(k-3)*(k-4)*(k-5)*(k-6)*(k-7)*(k-8))+')/km9_unknown_constant'; //9809122307 what is denominator? //9809141824 k=10 click km9 get //1*km9_unknown= -(1*1/11+1*1/2+1*10/12+1*0/1+1*(-720)/720+1*0/1+1*30240/30240+1*0/1+1*(-604800)/1209600+1*0/1+(-1)) 0.07575757575757569 <a name="docD037">&lt;a name="docD037"&gt;</a> 1/0.07575757575757569 13.200000000000011 km9_unknown=1/13.200000000000011 km9_unknown=10/132 //9809141827 k=10 (k*(k-1)*(k-2)*(k-3)*(k-4)*(k-5)*(k-6)*(k-7)*(k-8)) =3628800 3628800/constant=10/132 <a name="docD038">&lt;a name="docD038"&gt;</a> //9809141830 constant=132*3628800/10=47900160 function km9(k){ k=parseInt(k); return '('+eval(k*(k-1)*(k-2)*(k-3)*(k-4)*(k-5)*(k-6)*(k-7)*(k-8))+')/47900160';//9809141831 } ]] 2009-09-17-10-34 here <a name="docD039">&lt;a name="docD039"&gt;</a> Above is working record. Below is explanation. (a little bit redundant) In <a href="#program0">make_km</a> program, set k=[10] then click "km9" button, return [[ //1*km9_unknown= -(1*1/11+1*1/2+1*10/12+1*0/1+1*(-720)/720+1*0/1+1*30240/30240+1*0/1+1*(-604800)/1209600+1*0/1+(-1)) ]] Above km9_unknown can be calculated. <a name="docD040">&lt;a name="docD040"&gt;</a> On the other hand, function km9(k) should return the same value. function km9(k) definition is k*(k-1)*(k-2)*(k-3)*(k-4)*(k-5)*(k-6)*(k-7)*(k-8)/km9_unknown_constant'; Now equate k*(k-1)*(k-2)*(k-3)*(k-4)*(k-5)*(k-6)*(k-7)*(k-8)/km9_unknown_constant'; and //1*km9_unknown= -(1*1/11+1*1/2+1*10/12+1*0/1+1*(-720)/720+1*0/1+1*30240/30240+1*0/1+1*(-604800)/1209600+1*0/1+(-1)) Solve for km9_unknown_constant, (we used k=10) get km9_unknown_constant=47900160 <a name="docD041">&lt;a name="docD041"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a> function km9(k) definition is next line k*(k-1)*(k-2)*(k-3)*(k-4)*(k-5)*(k-6)*(k-7)*(k-8)/47900160 Up to here solved function km9(k). To verify function km9(k), need to create function make_km11(k) If " i^(11) equation coefficients sum to one then function km9(k) definition is correct. (km9_unknown_constant=47900160 is correct) <a name="docD042">&lt;a name="docD042"&gt;</a> function make_km##(k) and function km##(k) work together, one can not be two paces ahead of other one! Like our feet, one can only be one pace ahead of other one. <a name="docD043">&lt;a name="docD043"&gt;</a> At writing time, this file jspico2e.htm define function km##(k) up to function km13(k) (non-zero) function km14(k) (zero) define function make_km##(k) up to function make_km20(k) but start from function make_km16(k) output show "_km15_undefined_" function make_km16(k) is not working. <a name="docD044">&lt;a name="docD044"&gt;</a> If you want to explore. Start from function km15(k){ ... } find constant used in function km15(k). To do so, read function km11(k){ ... } // non-zero output function km13(k){ ... } // non-zero output first. <a name="docD045">&lt;a name="docD045"&gt;</a> function km11(k){ ... } use return '('+eval(k*(k-1)*(k-2)*(k-3)*(k-4)*(k-5)*(k-6)*(k-7)*(k-8)*(k-9)*(k-10))+')/(-1307674368000/691)'; //9809161614 CORRECT km11(k) use k* up to *(k-10) function km13(k){ ... } use return '('+eval(k*(k-1)*(k-2)*(k-3)*(k-4)*(k-5)*(k-6)*(k-7)*(k-8)*(k-9)*(k-10)*(k-11)*(k-12))+')/(74724249600)'; //9809161924 km13(k) use k* up to *(k-12) <a name="docD046">&lt;a name="docD046"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a> then function km15(k){ ... } use return '('+eval(k*(k-1)*(k-2)*(k-3)*(k-4)*(k-5)*(k-6)*(k-7)*(k-8)*(k-9)*(k-10)*(k-11)*(k-12)*(k-13)*(k-14))+')/(km15_unknown_constant)'; km15(k) use k* up to *(k-14) find km15_unknown_constant, then put constant in place. (avoid error <a href="#docC047">02</a> !) function make_km16(k) start work. Before return value, please include a code line k=parseInt(k); <a name="docD047">&lt;a name="docD047"&gt;</a> find km15_unknown_constant, you may tripped by careless error. (Error <a href="#docC046">01</a>, <a href="#docC047">02</a>, <a href="#docC048">03</a>) Please read find km13_unknown_constant record first. Avoid those mistakes, you may go as smooth as find km14_unknown_constant. Good Luck !! 2009-09-17-11-15 stop <a name="docD048">&lt;a name="docD048"&gt;</a> 2009-09-17-12-10 start If you have done function km15(k){ ... } function km17(k){ ... } function km19(k){ ... } and want to create function km21(k){ ... } function km23(k){ ... } etc. jspico2e.htm define up to function make_km20(k) You need to create function make_km21(k) function make_km22(k) etc. <a name="docD049">&lt;a name="docD049"&gt;</a> These tedious coding is easiest part in this file jspico2e.htm Please goto <a href="#program0">box 2</a>. Fill 25 in "k=[ ]" box, then click "make_km(k)" button. Output to box 2. Copy from function make_km21(k) to function make_km25(k). Paste function definition code to this web page jspico2e.htm <a name="docD050">&lt;a name="docD050"&gt;</a> Create button for make_km21(k) to make_km25(k) This file has button code up to make_km20(k) See next code &lt;INPUT onclick='javascript:boxc03.value=make_km20(kValueBox.value)' type=button value="m20"&gt; Create &lt;INPUT onclick='javascript:boxc03.value=make_km21(kValueBox.value)' type=button value="m21"&gt; etc. You can go forward as much as you like. 2009-09-17-12-23 stop <a name="docD051">&lt;a name="docD051"&gt;</a> <a href="jspico2e.htm#docD001">k-14</a>, <a href="jspico2e.htm#docD033">km9(k)</a>, <a href="jspico2e.htm#docC046">err.1,2,3</a>, <a href="jspico2e.htm#program0">program</a>, <a href="jspico2e.htm#coefTriangle">triangle</a> 2009-09-18-08-21 Box 2 output only make_km2() etc. Box 2 NOT output km2() etc. function make_km13(k) is much longer than function km13(k) . Program take care longer function. You need some exercise, shorter one is for you. 2009-09-18-08-25 </pre> <SCRIPT language=JavaScript> <!-- //bicof(arg1,arg2) function kp1(k) { //return [1,k+1]; //9809120755 k=parseInt(k); //9809122212 return '1/'+(k+1); //9809122007 } function kp0(k) { //return [1,2]; //9809120756 return '1/2'; //9809122008 } function km1(k) { //return [k,12]; //9809120757 return k+'/12'; //9809122009 } function km2(k) { //return [0,1]; //9809120758 return '0/1'; //9809122010 } function km3(k) { //return [-k*(k-1)*(k-2),(720)]; //9809120759 k=parseInt(k); //9809122212 return '('+eval(-k*(k-1)*(k-2))+')/720';//9809122011 } function km4(k) { //return [0,1]; //9809120800 return '0/1'; //9809122012 } function km5(k) { //return [k*(k-1)*(k-2)*(k-3)*(k-4),(30240)]; //9809120801 k=parseInt(k); //9809122212 return eval(k*(k-1)*(k-2)*(k-3)*(k-4))+'/30240';//9809122013 } function km6(k) { //return [0,1]; //9809120802 return '0/1'; //9809122014 } function km7(k) { //return [-k*(k-1)*(k-2)*(k-3)*(k-4)*(k-5)*(k-6),(1209600)]; //9809120803 k=parseInt(k); //9809122212 return '('+eval(-k*(k-1)*(k-2)*(k-3)*(k-4)*(k-5)*(k-6))+')/1209600';//9809122015 } function km8(k) { //return [0,1]; //9809120804 return '0/1'; //9809122016 } function km9(k) //9809122306 chocked, 9809141831 passed { k=parseInt(k); //return '('+eval(k*(k-1)*(k-2)*(k-3)*(k-4)*(k-5)*(k-6)*(k-7)*(k-8))+')/1209600';//9809122307 what is denominator? //return '__km9__unknown_coefficient__'; //9809130020 return '('+eval(k*(k-1)*(k-2)*(k-3)*(k-4)*(k-5)*(k-6)*(k-7)*(k-8))+')/47900160';//9809141831 } function km10(k) { return '0/1'; //9809122342 } function km11(k) { k=parseInt(k); //return '__km11__unknown_coefficient__'; //9809130020 //return '('+eval(k*(k-1)*(k-2)*(k-3)*(k-4)*(k-5)*(k-6)*(k-7)*(k-8)*(k-9)*(k-10))+')/(-1892437580.3183791)';//9809141920 non integer denominator? //9809161327 hand calculated, above return value is right, same as next exact answer //return '('+eval(k*(k-1)*(k-2)*(k-3)*(k-4)*(k-5)*(k-6)*(k-7)*(k-8)*(k-9)*(k-10))+')/(-691/1307674368000)'; //9809161328 wrong return '('+eval(k*(k-1)*(k-2)*(k-3)*(k-4)*(k-5)*(k-6)*(k-7)*(k-8)*(k-9)*(k-10))+')/(-1307674368000/691)'; //9809161614 CORRECT } function km12(k) { return '0/1'; } function km13(k) { //9809161924 k=parseInt(k); return '('+eval(k*(k-1)*(k-2)*(k-3)*(k-4)*(k-5)*(k-6)*(k-7)*(k-8)*(k-9)*(k-10)*(k-11)*(k-12))+')/(74724249600)'; //9809161924 } function km14(k){return '0/1';} //function km15(k){return '_km15_undefined_';} function km15(k) { //9809251332 k=parseInt(k); return '('+eval(k*(k-1)*(k-2)*(k-3)*(k-4)*(k-5)*(k-6)*(k-7)*(k-8)*(k-9)*(k-10)*(k-11)*(k-12)*(k-13)*(k-14))+')/(-22120201153290240000/7498041)'; //9809251406 } /** 98,09,25,14,09 k=17 //1*km16_unknown= -(1*1/18+1*1/2+1*17/12+1*0/1+1*(-4080)/720+1*0/1+1*742560/30240+1*0/1+1*(-98017920)/1209600+1*0/1+1*(8821612800)/47900160+1*0/1+1*(494010316800)/(-1307674368000/691)+1*0/1+1*(14820309504000)/(74724249600)+1*0/1+1*(177843714048000)/(-22120201153290240000/7498041)+(-1)) -2.1316282072803005e-14 98,09,25,14,10 km16 zero ! CORRECT !! (-22120201153290240000/7498041) is correct !! /**/ function km16(k){return '0/1';} function km17(k){return '_km17_undefined_';} function km19(k){return '_km19_undefined_';} function km21(k){return '_km21_undefined_';} function km23(k){return '_km23_undefined_';} function km25(k){return '_km25_undefined_';} function km27(k){return '_km27_undefined_';} function km29(k){return '_km29_undefined_';} //9809152140 km13(k) to km30(k) function km18(k){return '0/1';} function km20(k){return '0/1';} function km22(k){return '0/1';} function km24(k){return '0/1';} function km26(k){return '0/1';} function km28(k){return '0/1';} function km30(k){return '0/1';} // 9809120809 function make_km(k, ouType) //9809140921 use ouType { if(isNaN(k)) return '[k=]box need an integer.' k=parseInt(k); if(k<2) return 'k>=2 required' boxc02.value=''; //9809141503 /** var bgn0=[ '+bino(k+1,k-q)*kp1(k+1)', '+bino(k ,k-q)*kp0(k)', '+bino(k-1,k-q)*km1(k-1)', '+bino(k-q,k-q)*kmq(k-q)', '-bino(k,k-q)', '-kmq(k-q)' ]; // 9809120817 var bgn0=[ '+bino(k+1,k-q)*kp1(k)', '+bino(k ,k-q)*kp0(k)', '+bino(k-1,k-q)*km1(k)', '+bino(k-q,k-q)*kmq(k)', '-bino(k,k-q)', '-kmq(k)' ]; // 9809121941 /**/ var bgn0=[ '+bicof(k+1,k-q)+"*"+kp1(k)+"+"', '+bicof(k ,k-q)+"*"+kp0(k)+"+"', '+bicof(k-1,k-q)+"*"+km1(k)+"+"', '+bicof(k-q,k-q)+"*"+kmq(k)+"+"', '+"("+(-bicof(k,k-q))+")"', '(-kmq(k))' ]; // 9809122110 var ans0=''; var end0=['']; var u0,u1,u2,u3,u4,u5,u6,u7; var q='q'; //9809120956 for(u0=0;u0<bgn0.length;u0++) {//9809120954 end0[u0]=bgn0[u0].replace(/q/g,'2');//9809120957 all replacement } for(u0=0;u0<end0.length;u0++) {//9809120955 ans0+=end0[u0]+'\n' } var dbg0=''; //9809121040 var end1=['']; //9809120959 var old0=''; var new0=''; var kk=k; var regexp=''; //9809121103 var nextK; //9809121121 var currentK=-1; //9809121147 var kMinus='k-'; //9809121422 /** 9809121719 u1=k is k th equation u0 is k th equation, sub-equation sequence index if change from 'k-5' to 'k-6' use u1 in place for k, NOT use u0 regexp=eval('/'+kMinus+u0+'/g'); //9809121536 above is wrong below is right regexp=eval('/'+kMinus+(u1-2)+'/g'); //9809121620 9809121721 /**/ for(u1=3;u1<=kk;u1++) {//9809121007 end1=['']; //9809120959 old0='k-'+(u1-1) new0='k-'+(u1) for(u0=0;u0<end0.length;u0++) {//9809120959 if(u0<u1) //9809121753 { end1[u0]=end0[u0].replace(old0,new0) continue; //9809121255 } //if(u0<u1) if(u0==u1) //9809121753 { u2=end0[u0].indexOf(kMinus,0); //9809121423 u2=end0[u0].indexOf(kMinus,u2+1); u3=end0[u0].indexOf(')',u2); //9809121424 currentK=end0[u0].substring(u2+2,u3); currentK=parseInt(currentK); //9809121305 nextK=currentK+1; //9809121306 end1[u0]=end0[u0].substring(0,u2+2) //9809121140 ***** +nextK +end0[u0].substring(u3,end0[u0].length) regexp=eval('/'+kMinus+(u1-1)+'/g'); //9809121855 end1[u0+1]=end1[u0].replace(regexp,(kMinus+u1)); //9809121856 end1[u0+1]= //9809121857 end1[u0+1].replace('km'+(u1-1),'km'+(u1)) continue; //9809121255 } //if(u0==u1) end1[u0+1]=end0[u0].replace(old0,new0) end1[u0+1]= //9809121310 end1[u0+1].replace('km'+currentK,'km'+nextK) } //for(u0=0;u0<end0.length;u0++) ans0+='\n' +'function make_km'+(u1-1)+'(k)\n' //9809140928 '-1' +'{\n' //9809121952 add function //+'if(k<'+u1+')return "make_km"+k+"(k): need k>"+k+", now k="+k;\n\n' //9809141753 //+'if(k<'+u1+')return "make_km'+(u1-1)+'(k): need k>"+k+", now k="+k;\n\n' //9809201408 ['+(u1-1)+'] +'if(k<'+u1+')return "make_km'+(u1-1)+'(k): need k>'+(u1-1)+', now k="+k;\n\n' //9809201435 2nd ['+(u1-1)+'] +'k=parseInt(k);//9809122215\n\n' +'//### pow(n,k-u1) coefficient where u1='+u1+'\n' +'//### coefficient is valid only if k>= u1='+u1+'\n' +'//### coefficient has u0 bicof() lines, u0='+u0+'\n' +'return "0="\n' for(u2=0;u2<end1.length;u2++) {//9809121012 if(u2==end1.length-4) { u7=end1[u2].indexOf('(k)'); u6=end1[u2].substring(0,u7) +'_unknown' //9809122132 +end1[u2].substring(u7+3,end1[u2].length) +'\n' u6=u6.replace('"+',''); //9809122141 u6=u6.replace('+"','+@"'); u6=u6.replace('+"',''); u6=u6.replace('+@"','+"'); ans0+=u6; } else //9809122032 use two if() if(u2==end1.length-3)ans0+='//'+end1[u2]+'\n' else //9809122032 use two if() if(u2==end1.length-1)ans0+='//'+end1[u2]+'\n' else ans0+=end1[u2]+'\n' } ans0+='' +';\n}\n\n' for(u0=0;u0<end1.length;u0++) end0[u0]=end1[u0]; //9809121321 update } //for(u1=3;u1<=kk;u1++) var ans1=''; //9809140923 u1=0; u2=0; var returnAt=0; var returnEnd=0; var sectionBgn=0; var sectionEnd=0; var unknownAt=0 var unknownEnd=0 var plusBicof=0; var semiColAt=0; //9809141236 var iter0=0; //9809141016 sectionEnd=ans0.indexOf('function make_',0) do //9809141011 use do while { iter0++; returnAt= returnEnd= sectionBgn= sectionEnd= unknownAt= unknownEnd= plusBicof= sectionEnd; //9809141111 returnAt=ans0.indexOf('return "0="',sectionEnd) returnEnd=ans0.indexOf('\n',returnAt) ans1+=ans0.substring(sectionBgn,returnAt); ans1+='return '; //9809141225 unknownAt=ans0.indexOf('_unknown',returnAt+1) unknownEnd=ans0.indexOf('\n',unknownAt) plusBicof=ans0.lastIndexOf('+bicof(',unknownAt); ans1+='"//"'+ans0.substring(plusBicof,unknownEnd-2)+'=\\n-("'; //9809141248 add ['"//"'+] //begin -(.....) ans1+=ans0.substring(returnEnd,plusBicof-1); //+'\n'; 9809141415 del '\n' semiColAt=ans0.indexOf(';',unknownEnd) //9809141237 ans1+=ans0.substring(unknownEnd,semiColAt)+'+")\\n"\n'; //9809141250 add [\\n] //close -(.....) sectionEnd=ans0.indexOf('function make_',unknownAt) if(sectionEnd<0) { ans1+=ans0.substring(semiColAt,ans0.length)+'\n'; //9809141240 break; } ans1+=ans0.substring(semiColAt,sectionEnd)+'\n'; //9809141241 } while(iter0<1000) //9809141123 ans1.replace('unknown+"','unknown=\\n"') //9809141028 boxc02.value+='' +'//Please paste following function definition to\n' +'//your program page. You need to build a button\n' +'//for each function. File jspico2e.htm already\n' +'//build button km2 to km15. Just paste these code\n' +'//still not enough. You need define functions\n' +'//function km9(k){body} up to function km14(k)\n' +'//Please see document at \n' +'//http://freeman2.com/jspico2e.htm#docC001\n' +'//Freeman 2009-09-14-18-00\n\n' +ans1; //9809141126 } // function make_km(k) /** 9809121057 js bible 863 Chapter 27 &#10022; The String Object var regexp = /be/g; soliloquy.replace(regexp, &#8220;exist&#8221;); /**/ //9809122000 output goto service. // function make_km2(k) { if(k<3)return "make_km2(k): need k>2, now k="+k; k=parseInt(k);//9809122215 //### pow(n,k-u1) coefficient where u1=3 //### coefficient is valid only if k>= u1=3 //### coefficient has u0 bicof() lines, u0=6 return "//"+bicof(k-2,k-3)+"*km2_unknown=\n-(" +bicof(k+1,k-3)+"*"+kp1(k)+"+" +bicof(k ,k-3)+"*"+kp0(k)+"+" +bicof(k-1,k-3)+"*"+km1(k)+"+" //+bicof(k-3,k-3)+"*"+km3(k)+"+" +"("+(-bicof(k,k-3))+")" //(-km3(k)) +")\n" ; } function make_km3(k) { if(k<4)return "make_km3(k): need k>3, now k="+k; k=parseInt(k);//9809122215 //### pow(n,k-u1) coefficient where u1=4 //### coefficient is valid only if k>= u1=4 //### coefficient has u0 bicof() lines, u0=7 return "//"+bicof(k-3,k-4)+"*km3_unknown=\n-(" +bicof(k+1,k-4)+"*"+kp1(k)+"+" +bicof(k ,k-4)+"*"+kp0(k)+"+" +bicof(k-1,k-4)+"*"+km1(k)+"+" +bicof(k-2,k-4)+"*"+km2(k)+"+" //+bicof(k-4,k-4)+"*"+km4(k)+"+" +"("+(-bicof(k,k-4))+")" //(-km4(k)) +")\n" ; } function make_km4(k) { if(k<5)return "make_km4(k): need k>4, now k="+k; k=parseInt(k);//9809122215 //### pow(n,k-u1) coefficient where u1=5 //### coefficient is valid only if k>= u1=5 //### coefficient has u0 bicof() lines, u0=8 return "//"+bicof(k-4,k-5)+"*km4_unknown=\n-(" +bicof(k+1,k-5)+"*"+kp1(k)+"+" +bicof(k ,k-5)+"*"+kp0(k)+"+" +bicof(k-1,k-5)+"*"+km1(k)+"+" +bicof(k-2,k-5)+"*"+km2(k)+"+" +bicof(k-3,k-5)+"*"+km3(k)+"+" //+bicof(k-5,k-5)+"*"+km5(k)+"+" +"("+(-bicof(k,k-5))+")" //(-km5(k)) +")\n" ; } function make_km5(k) { if(k<6)return "make_km5(k): need k>5, now k="+k; k=parseInt(k);//9809122215 //### pow(n,k-u1) coefficient where u1=6 //### coefficient is valid only if k>= u1=6 //### coefficient has u0 bicof() lines, u0=9 return "//"+bicof(k-5,k-6)+"*km5_unknown=\n-(" +bicof(k+1,k-6)+"*"+kp1(k)+"+" +bicof(k ,k-6)+"*"+kp0(k)+"+" +bicof(k-1,k-6)+"*"+km1(k)+"+" +bicof(k-2,k-6)+"*"+km2(k)+"+" +bicof(k-3,k-6)+"*"+km3(k)+"+" +bicof(k-4,k-6)+"*"+km4(k)+"+" //+bicof(k-6,k-6)+"*"+km6(k)+"+" +"("+(-bicof(k,k-6))+")" //(-km6(k)) +")\n" ; } function make_km6(k) { if(k<7)return "make_km6(k): need k>6, now k="+k; k=parseInt(k);//9809122215 //### pow(n,k-u1) coefficient where u1=7 //### coefficient is valid only if k>= u1=7 //### coefficient has u0 bicof() lines, u0=10 return "//"+bicof(k-6,k-7)+"*km6_unknown=\n-(" +bicof(k+1,k-7)+"*"+kp1(k)+"+" +bicof(k ,k-7)+"*"+kp0(k)+"+" +bicof(k-1,k-7)+"*"+km1(k)+"+" +bicof(k-2,k-7)+"*"+km2(k)+"+" +bicof(k-3,k-7)+"*"+km3(k)+"+" +bicof(k-4,k-7)+"*"+km4(k)+"+" +bicof(k-5,k-7)+"*"+km5(k)+"+" //+bicof(k-7,k-7)+"*"+km7(k)+"+" +"("+(-bicof(k,k-7))+")" //(-km7(k)) +")\n" ; } function make_km7(k) { if(k<8)return "make_km7(k): need k>7, now k="+k; k=parseInt(k);//9809122215 //### pow(n,k-u1) coefficient where u1=8 //### coefficient is valid only if k>= u1=8 //### coefficient has u0 bicof() lines, u0=11 return "//"+bicof(k-7,k-8)+"*km7_unknown=\n-(" +bicof(k+1,k-8)+"*"+kp1(k)+"+" +bicof(k ,k-8)+"*"+kp0(k)+"+" +bicof(k-1,k-8)+"*"+km1(k)+"+" +bicof(k-2,k-8)+"*"+km2(k)+"+" +bicof(k-3,k-8)+"*"+km3(k)+"+" +bicof(k-4,k-8)+"*"+km4(k)+"+" +bicof(k-5,k-8)+"*"+km5(k)+"+" +bicof(k-6,k-8)+"*"+km6(k)+"+" //+bicof(k-8,k-8)+"*"+km8(k)+"+" +"("+(-bicof(k,k-8))+")" //(-km8(k)) +")\n" ; } function make_km8(k) { if(k<9)return "make_km8(k): need k>8, now k="+k; k=parseInt(k);//9809122215 //### pow(n,k-u1) coefficient where u1=9 //### coefficient is valid only if k>= u1=9 //### coefficient has u0 bicof() lines, u0=12 return "//"+bicof(k-8,k-9)+"*km8_unknown=\n-(" +bicof(k+1,k-9)+"*"+kp1(k)+"+" +bicof(k ,k-9)+"*"+kp0(k)+"+" +bicof(k-1,k-9)+"*"+km1(k)+"+" +bicof(k-2,k-9)+"*"+km2(k)+"+" +bicof(k-3,k-9)+"*"+km3(k)+"+" +bicof(k-4,k-9)+"*"+km4(k)+"+" +bicof(k-5,k-9)+"*"+km5(k)+"+" +bicof(k-6,k-9)+"*"+km6(k)+"+" +bicof(k-7,k-9)+"*"+km7(k)+"+" //+bicof(k-9,k-9)+"*"+km9(k)+"+" +"("+(-bicof(k,k-9))+")" //(-km9(k)) +")\n" ; } function make_km9(k) { if(k<10)return "make_km9(k): need k>9, now k="+k; k=parseInt(k);//9809122215 //### pow(n,k-u1) coefficient where u1=10 //### coefficient is valid only if k>= u1=10 //### coefficient has u0 bicof() lines, u0=13 return "//"+bicof(k-9,k-10)+"*km9_unknown=\n-(" +bicof(k+1,k-10)+"*"+kp1(k)+"+" +bicof(k ,k-10)+"*"+kp0(k)+"+" +bicof(k-1,k-10)+"*"+km1(k)+"+" +bicof(k-2,k-10)+"*"+km2(k)+"+" +bicof(k-3,k-10)+"*"+km3(k)+"+" +bicof(k-4,k-10)+"*"+km4(k)+"+" +bicof(k-5,k-10)+"*"+km5(k)+"+" +bicof(k-6,k-10)+"*"+km6(k)+"+" +bicof(k-7,k-10)+"*"+km7(k)+"+" +bicof(k-8,k-10)+"*"+km8(k)+"+" //+bicof(k-10,k-10)+"*"+km10(k)+"+" +"("+(-bicof(k,k-10))+")" //(-km10(k)) +")\n" ; } function make_km10(k) { if(k<11)return "make_km10(k): need k>10, now k="+k; k=parseInt(k);//9809122215 //### pow(n,k-u1) coefficient where u1=11 //### coefficient is valid only if k>= u1=11 //### coefficient has u0 bicof() lines, u0=14 return "//"+bicof(k-10,k-11)+"*km10_unknown=\n-(" +bicof(k+1,k-11)+"*"+kp1(k)+"+" +bicof(k ,k-11)+"*"+kp0(k)+"+" +bicof(k-1,k-11)+"*"+km1(k)+"+" +bicof(k-2,k-11)+"*"+km2(k)+"+" +bicof(k-3,k-11)+"*"+km3(k)+"+" +bicof(k-4,k-11)+"*"+km4(k)+"+" +bicof(k-5,k-11)+"*"+km5(k)+"+" +bicof(k-6,k-11)+"*"+km6(k)+"+" +bicof(k-7,k-11)+"*"+km7(k)+"+" +bicof(k-8,k-11)+"*"+km8(k)+"+" +bicof(k-9,k-11)+"*"+km9(k)+"+" //+bicof(k-11,k-11)+"*"+km11(k)+"+" +"("+(-bicof(k,k-11))+")" //(-km11(k)) +")\n" ; } function make_km11(k) { if(k<12)return "make_km11(k): need k>11, now k="+k; k=parseInt(k);//9809122215 //### pow(n,k-u1) coefficient where u1=12 //### coefficient is valid only if k>= u1=12 //### coefficient has u0 bicof() lines, u0=15 return "//"+bicof(k-11,k-12)+"*km11_unknown=\n-(" +bicof(k+1,k-12)+"*"+kp1(k)+"+" +bicof(k ,k-12)+"*"+kp0(k)+"+" +bicof(k-1,k-12)+"*"+km1(k)+"+" +bicof(k-2,k-12)+"*"+km2(k)+"+" +bicof(k-3,k-12)+"*"+km3(k)+"+" +bicof(k-4,k-12)+"*"+km4(k)+"+" +bicof(k-5,k-12)+"*"+km5(k)+"+" +bicof(k-6,k-12)+"*"+km6(k)+"+" +bicof(k-7,k-12)+"*"+km7(k)+"+" +bicof(k-8,k-12)+"*"+km8(k)+"+" +bicof(k-9,k-12)+"*"+km9(k)+"+" +bicof(k-10,k-12)+"*"+km10(k)+"+" //+bicof(k-12,k-12)+"*"+km12(k)+"+" +"("+(-bicof(k,k-12))+")" //(-km12(k)) +")\n" ; } function make_km12(k) { if(k<13)return "make_km12(k): need k>12, now k="+k; k=parseInt(k);//9809122215 //### pow(n,k-u1) coefficient where u1=13 //### coefficient is valid only if k>= u1=13 //### coefficient has u0 bicof() lines, u0=16 return "//"+bicof(k-12,k-13)+"*km12_unknown=\n-(" +bicof(k+1,k-13)+"*"+kp1(k)+"+" +bicof(k ,k-13)+"*"+kp0(k)+"+" +bicof(k-1,k-13)+"*"+km1(k)+"+" +bicof(k-2,k-13)+"*"+km2(k)+"+" +bicof(k-3,k-13)+"*"+km3(k)+"+" +bicof(k-4,k-13)+"*"+km4(k)+"+" +bicof(k-5,k-13)+"*"+km5(k)+"+" +bicof(k-6,k-13)+"*"+km6(k)+"+" +bicof(k-7,k-13)+"*"+km7(k)+"+" +bicof(k-8,k-13)+"*"+km8(k)+"+" +bicof(k-9,k-13)+"*"+km9(k)+"+" +bicof(k-10,k-13)+"*"+km10(k)+"+" +bicof(k-11,k-13)+"*"+km11(k)+"+" //+bicof(k-13,k-13)+"*"+km13(k)+"+" +"("+(-bicof(k,k-13))+")" //(-km13(k)) +")\n" ; } function make_km13(k) { if(k<14)return "make_km13(k): need k>13, now k="+k; k=parseInt(k);//9809122215 //### pow(n,k-u1) coefficient where u1=14 //### coefficient is valid only if k>= u1=14 //### coefficient has u0 bicof() lines, u0=17 return "//"+bicof(k-13,k-14)+"*km13_unknown=\n-(" +bicof(k+1,k-14)+"*"+kp1(k)+"+" +bicof(k ,k-14)+"*"+kp0(k)+"+" +bicof(k-1,k-14)+"*"+km1(k)+"+" +bicof(k-2,k-14)+"*"+km2(k)+"+" +bicof(k-3,k-14)+"*"+km3(k)+"+" +bicof(k-4,k-14)+"*"+km4(k)+"+" +bicof(k-5,k-14)+"*"+km5(k)+"+" +bicof(k-6,k-14)+"*"+km6(k)+"+" +bicof(k-7,k-14)+"*"+km7(k)+"+" +bicof(k-8,k-14)+"*"+km8(k)+"+" +bicof(k-9,k-14)+"*"+km9(k)+"+" +bicof(k-10,k-14)+"*"+km10(k)+"+" +bicof(k-11,k-14)+"*"+km11(k)+"+" +bicof(k-12,k-14)+"*"+km12(k)+"+" //+bicof(k-14,k-14)+"*"+km14(k)+"+" +"("+(-bicof(k,k-14))+")" //(-km14(k)) +")\n" ; } function make_km14(k) { if(k<15)return "make_km14(k): need k>14, now k="+k; k=parseInt(k);//9809122215 //### pow(n,k-u1) coefficient where u1=15 //### coefficient is valid only if k>= u1=15 //### coefficient has u0 bicof() lines, u0=18 return "//"+bicof(k-14,k-15)+"*km14_unknown=\n-(" +bicof(k+1,k-15)+"*"+kp1(k)+"+" +bicof(k ,k-15)+"*"+kp0(k)+"+" +bicof(k-1,k-15)+"*"+km1(k)+"+" +bicof(k-2,k-15)+"*"+km2(k)+"+" +bicof(k-3,k-15)+"*"+km3(k)+"+" +bicof(k-4,k-15)+"*"+km4(k)+"+" +bicof(k-5,k-15)+"*"+km5(k)+"+" +bicof(k-6,k-15)+"*"+km6(k)+"+" +bicof(k-7,k-15)+"*"+km7(k)+"+" +bicof(k-8,k-15)+"*"+km8(k)+"+" +bicof(k-9,k-15)+"*"+km9(k)+"+" +bicof(k-10,k-15)+"*"+km10(k)+"+" +bicof(k-11,k-15)+"*"+km11(k)+"+" +bicof(k-12,k-15)+"*"+km12(k)+"+" +bicof(k-13,k-15)+"*"+km13(k)+"+" //+bicof(k-15,k-15)+"*"+km15(k)+"+" +"("+(-bicof(k,k-15))+")" //(-km15(k)) +")\n" ; } function make_km15(k) { if(k<16)return "make_km15(k): need k>15, now k="+k; k=parseInt(k);//9809122215 //### pow(n,k-u1) coefficient where u1=16 //### coefficient is valid only if k>= u1=16 //### coefficient has u0 bicof() lines, u0=19 return "//"+bicof(k-15,k-16)+"*km15_unknown=\n-(" +bicof(k+1,k-16)+"*"+kp1(k)+"+" +bicof(k ,k-16)+"*"+kp0(k)+"+" +bicof(k-1,k-16)+"*"+km1(k)+"+" +bicof(k-2,k-16)+"*"+km2(k)+"+" +bicof(k-3,k-16)+"*"+km3(k)+"+" +bicof(k-4,k-16)+"*"+km4(k)+"+" +bicof(k-5,k-16)+"*"+km5(k)+"+" +bicof(k-6,k-16)+"*"+km6(k)+"+" +bicof(k-7,k-16)+"*"+km7(k)+"+" +bicof(k-8,k-16)+"*"+km8(k)+"+" +bicof(k-9,k-16)+"*"+km9(k)+"+" +bicof(k-10,k-16)+"*"+km10(k)+"+" +bicof(k-11,k-16)+"*"+km11(k)+"+" +bicof(k-12,k-16)+"*"+km12(k)+"+" +bicof(k-13,k-16)+"*"+km13(k)+"+" +bicof(k-14,k-16)+"*"+km14(k)+"+" //+bicof(k-16,k-16)+"*"+km16(k)+"+" +"("+(-bicof(k,k-16))+")" //(-km16(k)) +")\n" ; } function make_km16(k) { if(k<17)return "make_km16(k): need k>16, now k="+k; k=parseInt(k);//9809122215 //### pow(n,k-u1) coefficient where u1=17 //### coefficient is valid only if k>= u1=17 //### coefficient has u0 bicof() lines, u0=20 return "//"+bicof(k-16,k-17)+"*km16_unknown=\n-(" +bicof(k+1,k-17)+"*"+kp1(k)+"+" +bicof(k ,k-17)+"*"+kp0(k)+"+" +bicof(k-1,k-17)+"*"+km1(k)+"+" +bicof(k-2,k-17)+"*"+km2(k)+"+" +bicof(k-3,k-17)+"*"+km3(k)+"+" +bicof(k-4,k-17)+"*"+km4(k)+"+" +bicof(k-5,k-17)+"*"+km5(k)+"+" +bicof(k-6,k-17)+"*"+km6(k)+"+" +bicof(k-7,k-17)+"*"+km7(k)+"+" +bicof(k-8,k-17)+"*"+km8(k)+"+" +bicof(k-9,k-17)+"*"+km9(k)+"+" +bicof(k-10,k-17)+"*"+km10(k)+"+" +bicof(k-11,k-17)+"*"+km11(k)+"+" +bicof(k-12,k-17)+"*"+km12(k)+"+" +bicof(k-13,k-17)+"*"+km13(k)+"+" +bicof(k-14,k-17)+"*"+km14(k)+"+" +bicof(k-15,k-17)+"*"+km15(k)+"+" //+bicof(k-17,k-17)+"*"+km17(k)+"+" +"("+(-bicof(k,k-17))+")" //(-km17(k)) +")\n" ; } function make_km17(k) { if(k<18)return "make_km17(k): need k>17, now k="+k; k=parseInt(k);//9809122215 //### pow(n,k-u1) coefficient where u1=18 //### coefficient is valid only if k>= u1=18 //### coefficient has u0 bicof() lines, u0=21 return "//"+bicof(k-17,k-18)+"*km17_unknown=\n-(" +bicof(k+1,k-18)+"*"+kp1(k)+"+" +bicof(k ,k-18)+"*"+kp0(k)+"+" +bicof(k-1,k-18)+"*"+km1(k)+"+" +bicof(k-2,k-18)+"*"+km2(k)+"+" +bicof(k-3,k-18)+"*"+km3(k)+"+" +bicof(k-4,k-18)+"*"+km4(k)+"+" +bicof(k-5,k-18)+"*"+km5(k)+"+" +bicof(k-6,k-18)+"*"+km6(k)+"+" +bicof(k-7,k-18)+"*"+km7(k)+"+" +bicof(k-8,k-18)+"*"+km8(k)+"+" +bicof(k-9,k-18)+"*"+km9(k)+"+" +bicof(k-10,k-18)+"*"+km10(k)+"+" +bicof(k-11,k-18)+"*"+km11(k)+"+" +bicof(k-12,k-18)+"*"+km12(k)+"+" +bicof(k-13,k-18)+"*"+km13(k)+"+" +bicof(k-14,k-18)+"*"+km14(k)+"+" +bicof(k-15,k-18)+"*"+km15(k)+"+" +bicof(k-16,k-18)+"*"+km16(k)+"+" //+bicof(k-18,k-18)+"*"+km18(k)+"+" +"("+(-bicof(k,k-18))+")" //(-km18(k)) +")\n" ; } function make_km18(k) { if(k<19)return "make_km18(k): need k>18, now k="+k; k=parseInt(k);//9809122215 //### pow(n,k-u1) coefficient where u1=19 //### coefficient is valid only if k>= u1=19 //### coefficient has u0 bicof() lines, u0=22 return "//"+bicof(k-18,k-19)+"*km18_unknown=\n-(" +bicof(k+1,k-19)+"*"+kp1(k)+"+" +bicof(k ,k-19)+"*"+kp0(k)+"+" +bicof(k-1,k-19)+"*"+km1(k)+"+" +bicof(k-2,k-19)+"*"+km2(k)+"+" +bicof(k-3,k-19)+"*"+km3(k)+"+" +bicof(k-4,k-19)+"*"+km4(k)+"+" +bicof(k-5,k-19)+"*"+km5(k)+"+" +bicof(k-6,k-19)+"*"+km6(k)+"+" +bicof(k-7,k-19)+"*"+km7(k)+"+" +bicof(k-8,k-19)+"*"+km8(k)+"+" +bicof(k-9,k-19)+"*"+km9(k)+"+" +bicof(k-10,k-19)+"*"+km10(k)+"+" +bicof(k-11,k-19)+"*"+km11(k)+"+" +bicof(k-12,k-19)+"*"+km12(k)+"+" +bicof(k-13,k-19)+"*"+km13(k)+"+" +bicof(k-14,k-19)+"*"+km14(k)+"+" +bicof(k-15,k-19)+"*"+km15(k)+"+" +bicof(k-16,k-19)+"*"+km16(k)+"+" +bicof(k-17,k-19)+"*"+km17(k)+"+" //+bicof(k-19,k-19)+"*"+km19(k)+"+" +"("+(-bicof(k,k-19))+")" //(-km19(k)) +")\n" ; } function make_km19(k) { if(k<20)return "make_km19(k): need k>19, now k="+k; k=parseInt(k);//9809122215 //### pow(n,k-u1) coefficient where u1=20 //### coefficient is valid only if k>= u1=20 //### coefficient has u0 bicof() lines, u0=23 return "//"+bicof(k-19,k-20)+"*km19_unknown=\n-(" +bicof(k+1,k-20)+"*"+kp1(k)+"+" +bicof(k ,k-20)+"*"+kp0(k)+"+" +bicof(k-1,k-20)+"*"+km1(k)+"+" +bicof(k-2,k-20)+"*"+km2(k)+"+" +bicof(k-3,k-20)+"*"+km3(k)+"+" +bicof(k-4,k-20)+"*"+km4(k)+"+" +bicof(k-5,k-20)+"*"+km5(k)+"+" +bicof(k-6,k-20)+"*"+km6(k)+"+" +bicof(k-7,k-20)+"*"+km7(k)+"+" +bicof(k-8,k-20)+"*"+km8(k)+"+" +bicof(k-9,k-20)+"*"+km9(k)+"+" +bicof(k-10,k-20)+"*"+km10(k)+"+" +bicof(k-11,k-20)+"*"+km11(k)+"+" +bicof(k-12,k-20)+"*"+km12(k)+"+" +bicof(k-13,k-20)+"*"+km13(k)+"+" +bicof(k-14,k-20)+"*"+km14(k)+"+" +bicof(k-15,k-20)+"*"+km15(k)+"+" +bicof(k-16,k-20)+"*"+km16(k)+"+" +bicof(k-17,k-20)+"*"+km17(k)+"+" +bicof(k-18,k-20)+"*"+km18(k)+"+" //+bicof(k-20,k-20)+"*"+km20(k)+"+" +"("+(-bicof(k,k-20))+")" //(-km20(k)) +")\n" ; } function make_km20(k) { if(k<21)return "make_km20(k): need k>20, now k="+k; k=parseInt(k);//9809122215 //### pow(n,k-u1) coefficient where u1=21 //### coefficient is valid only if k>= u1=21 //### coefficient has u0 bicof() lines, u0=24 return "//"+bicof(k-20,k-21)+"*km20_unknown=\n-(" +bicof(k+1,k-21)+"*"+kp1(k)+"+" +bicof(k ,k-21)+"*"+kp0(k)+"+" +bicof(k-1,k-21)+"*"+km1(k)+"+" +bicof(k-2,k-21)+"*"+km2(k)+"+" +bicof(k-3,k-21)+"*"+km3(k)+"+" +bicof(k-4,k-21)+"*"+km4(k)+"+" +bicof(k-5,k-21)+"*"+km5(k)+"+" +bicof(k-6,k-21)+"*"+km6(k)+"+" +bicof(k-7,k-21)+"*"+km7(k)+"+" +bicof(k-8,k-21)+"*"+km8(k)+"+" +bicof(k-9,k-21)+"*"+km9(k)+"+" +bicof(k-10,k-21)+"*"+km10(k)+"+" +bicof(k-11,k-21)+"*"+km11(k)+"+" +bicof(k-12,k-21)+"*"+km12(k)+"+" +bicof(k-13,k-21)+"*"+km13(k)+"+" +bicof(k-14,k-21)+"*"+km14(k)+"+" +bicof(k-15,k-21)+"*"+km15(k)+"+" +bicof(k-16,k-21)+"*"+km16(k)+"+" +bicof(k-17,k-21)+"*"+km17(k)+"+" +bicof(k-18,k-21)+"*"+km18(k)+"+" +bicof(k-19,k-21)+"*"+km19(k)+"+" //+bicof(k-21,k-21)+"*"+km21(k)+"+" +"("+(-bicof(k,k-21))+")" //(-km21(k)) +")\n" ; } //bicof(arg1,arg2) //--> </script> <br><hr><br> Javascript index <br> <a href=http://freeman2.com/jsindex2.htm> http://freeman2.com/jsindex2.htm </a> <br> Space Curve Projector <br> <a href=http://freeman2.com/curve3d2.htm > http://freeman2.com/curve3d2.htm </a> <br> foot of perpendicular <br> <a href=http://freeman2.com/eyefoot2.htm > http://freeman2.com/eyefoot2.htm </a> <br> Gram-Schmidt Process <br> <a href=http://freeman2.com/gramsch2.htm> http://freeman2.com/gramsch2.htm </a> <br> complex polynomial root <br> <a href=http://freeman2.com/polyroot.htm> http://freeman2.com/polyroot.htm </a> <br> complex variable functions <br> <a href=http://freeman2.com/complex2.htm> http://freeman2.com/complex2.htm </a> <br> Hilbert's Inequality and Schur Constant <br> <a href=http://freeman2.com/tute0009.htm> http://freeman2.com/tute0009.htm </a> <br> 1st Positive Integer power sum COefficient equation. <br> <a href=http://freeman2.com/jspico_e.htm> http://freeman2.com/jspico_e.htm </a> <br> Prime number and prime decomposition <br> <a href=http://freeman2.com/jsprime2.htm> http://freeman2.com/jsprime2.htm </a> <br> <br> 2nd Positive Integer power sum COefficient equation. <br> <a href=http://freeman2.com/jspico2e.htm> http://freeman2.com/jspico2e.htm </a> <br> First upload 2009-09-07 <br> <br> Program name in Chinese: (2009-09-03-11-36) <br> Brief name: cŒTýQxe pico <br> Full name: cktexecxeŒTMzÂOxelQ_ýQxe <br> Brief name: uÂO z_ make_km <br> Full name: "uuÂOxe„v z_ <br> <br> Thank you for visiting Freeman's page.0 <br> Freeman 02009-09-07-16-42 <br> <br> Please visit first page <a href=http://freeman2.com/jspico_e.htm>freeman2.com</a>/<a href=jspico_e.htm>jspico_e.htm</a> <!--9809071643--> <br> <br> <span id="tuteLink2"></span> </BODY> </HTML> <SCRIPT language=JavaScript><!-- // 9507121343 function stepf(t0,bgn0) { if(t0<bgn0) return 0.; return 1.; } // end function // 9507121346 // 9507121356 // stepf(t,3) get t<3 value 0 // stepf(t,3) get t>=3 value 1 // // (1-stepf(t,3)) get t<3 value 1 // (1-stepf(t,3)) get t>=3 value 0 // // (stepf(t,2)-stepf(t,3)) // get 2<=t<3 value 1 // t<2 and t>=3 value 0 // 9507121400 /* 95,07,08,14,55,00 http://www.univie.ac.at/future.media/moe/JavaCalc/jcintro.html c:\$fm\js\math9507\JavaCalc-jcintro.html 95,07,08,15,00 http://www.univie.ac.at/future.media/moe/JavaCalc/parser.js c:\$fm\js\math9507\JavaCalc-jcintro-parser.js Return to MathCollections Ken Kikuchi Comment me: kikuchi@mix.or.jp Last updated: 2/29/2000 9507132210 include function loggamma(x) and function gamma(x) /**/ // Ab hier (17. 3. 2000) von Ken's Script \áernommen: function factorial(n) { /* factorial */ with(Math) { if (n<0) /* if negative */ return gamma(n+1); else if ((n == 0) || (n == 1)) return 1; else if (abs(n)-floor(abs(n))==0 ) /* if positive integer */ return n * factorial(n-1) ; else /* if non-integer */ return gamma(n+1); } } function loggamma(x) { /* log gamma */ var u0; //9507142008 with(Math) { var v=1; var w=0; var z=0; while ( x<8 ) { v*=x; x++ } w=1/(x*x); u0= //9507142009 ((((((((-3617/122400)*w + 7/1092)*w -691/360360)*w + 5/5940)*w -1/1680)*w + 1/1260)*w -1/360)*w + 1/12)/x + 0.5 * log(2*PI)-log(v)-x+(x-0.5)*log(x) ; //status='x=['+x+']; w=['+w+']; u0=['+u0+'] 9507142010'; return u0; } } function gamma(x) { /* gamma */ //9507141940 var i0; var y0=''; var g0,g1,g2; //9507142019 for(i0=0;i0<(x+'').length;i0++) { if(i0==0&&(x+'').charAt(i0)=='(')continue; if((x+'').charAt(i0)==')')break; //9507141944 y0+=(x+'').charAt(i0); } x=parseFloat(y0); with(Math) { if ( x <= 0 ) { if (abs(x)-floor(abs(x))==0 ) return "ComplexInfinity" ; else { g0=loggamma(1-x); g1=PI/( sin(PI*x) * exp( g0 ) ); return g1; } } else { g0=loggamma(x); g1=exp( g0 ) ; return g1 ; } } } //ISBN 0-12-059820-5 page 307 line -4 // LiuHH 200904271706 //function binomialCoef(arg1,arg2) //function binoCoef(arg1,arg2) function bicof(arg1,arg2) { //9804271706 arg1=parseInt(arg1); arg2=parseInt(arg2); if(arg1<arg2)return 0; return factorial(arg1)/(factorial(arg2)*factorial(arg1-arg2)); } //9804271718 //function factorialDouble(arg1) //ISBN 0-12-059820-5 page 545 eqn. 10.33c // LiuHH 200904271746 //function factorialDouble(arg1) //function factorDbl(arg1) // // Double factorial facdb(arg1) // 9!!=9*7*5*3*1 // 8!!=8*6*4*2 function facdb(arg1) { //9804271746 arg1=parseInt(arg1); var i0,i1,i2; var n; if(arg1%2) { n=parseInt(arg1/2); for(i0=0,i1=1;i0<n;i0++) i1=i1*2; return factorial(arg1)/factorial(n)/i1; } //9804271752 n=arg1/2; for(i0=0,i1=1;i0<n;i0++) i1=i1*2; return factorial(n)*i1; } //9804271754 /** 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) //--> </script> <!-- 2009-09-07-11-38 start record 2009-09-01-11-19 start jspico_e.htm copy tute0012.htm jspico_e.htm 2009-09-07-10-21 start jspico2e.htm copy jspico_e.htm jspico2e.htm 2009-09-07-11-41 done record 2009-09-15-10-05 from jspico2e.htm take out prime code section, save as jsprime2.htm -->