ÿþ<HTML> <HEAD> <TITLE>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;" /> <meta name="author" content="Arndt Brünner"> <meta name="date" content="2007-04-06T12:58:16+02:00"> <meta name="keywords" content="gleichungssysteme,lineare,gleichungssystem,lösen,gauss,gauß,lösungsverfahren,eliminationsverfahren,algorithmus,rechner,javascript,programm,arndt brünner,mathematik,beispiele,erklärung"><title>Lineare Gleichungssysteme lösen mit dem Gaußschen Eliminationsverfahren</title> <script type="text/javascript"> <!-- function initial0(linkNum, plotNum) { tuteLink(0); //9806231221 } //9809011320 include //gleichungssysteme.js // Javascript zur Lösung linearer Gleichungssysteme // (c) Arndt Brünner, Gelnhausen, 27.8.2001 // Version: 18.4.2003 var m=new Array(),L=new Array(),B=new Array(); var vvv=new Array("a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"); var lambda='µ'; var t1lmerker=0; var v=new Array(); var n,maxgrad=5,mingrad=2; function Analyse() { t=new String; t=document.f.t1.value; ErklTyp=(document.f.erkltyp[0].checked)?0:1; if(document.f.immererkl.checked)Erkl=true; ganzzahlig=document.f.ganzzahlig.checked; document.f.gauss.value=""; if(t.indexOf("=")==-1){Analyse2(t);return;} if (t.indexOf("(")>-1){alert("unzulässiges Zeichen: ("); return;} if (t.indexOf(")")>-1){alert("unzulässiges Zeichen: )"); return;} if (t.indexOf("^")>-1){alert("unzulässiges Zeichen: ^"); return;} if (t.indexOf("²")>-1){alert("unzulässiges Zeichen: ²"); return;} if (t.indexOf("³")>-1){alert("unzulässiges Zeichen: ³"); return;} if ((t.search(/\/\D/)>-1)||(t.search(/\D\//)>-1)){alert("Eine Variable als Zähler oder Nenner ist unzulässig");return;} t=t.replace(/\*/g,"").replace(/·/g,"").toLowerCase().replace(/\,/g,".").replace(/ +/g,""); t=t.replace(/ /g,"-").replace(/ /g,"-").replace(/\t/g,"").replace(/ /g,"").replace(/=\+/g,"="); while(t.indexOf("+-")>-1)t=t.replace(/\+-/g,"-"); while(t.indexOf("--")>-1)t=t.replace(/--/g,"+"); while(t.indexOf("++")>-1)t=t.replace(/\++/g,"+"); homogen=1; n=0;li=-1; for(i=0 ; i<26;i++) if (t.indexOf(vvv[i])>-1){v[n]=vvv[i];n++;} g=t.split("\n"); var Z=new Array(),N=new Array(); for(i=0;i<ng;i++){if (g[i].indexOf("=")==-1){ng=i;break;}} for(i=0;i<(n+1)*n;i++){ m[i]=0;N[i]=1;Z[i]=0;} for(i=0;i<g.length;i++)g[i]=g[i].replace(/\n/g,"").replace(/\r/g,""); var ig=0;var ng=0; for(var iig=0;iig<g.length;iig++) //Schleife für Gleichungen { if((g[iig].indexOf("=")==-1)||(g[iig].length<3))continue; var GG=g[iig].split("="); if(GG[1].substr(0,1)!="-")GG[1]="+"+GG[1]; if(GG[0].substr(0,1)!="-")GG[0]="+"+GG[0]; var G=GG[0]+GG[1].replace(/-/g,"#").replace(/\+/g,"-").replace(/#/g,"+"); G=G.replace(/-/g,"#-").replace(/\+/g,"#+"); var T=G.split("#"); for(i=0;i<T.length;i++) { status="analysiere Eingabe und erzeuge Matrix"; var t=T[i],tt=t.charAt(t.length-1).toLowerCase(); var iDD=t.search(/\D\D/) if((iDD>-1)&&(t.length>2)&&(iDD<t.length-1)) {alert("Variablennamen aus zwei Buchstaben \noder Produkte aus Variablen sind unzulässig:\n"+t.replace(/-/,"").replace(/\+/,""));return;} for(j=0;j<n;j++)if(tt==v[j])break; var Bruch=t.substr(0,t.length-((j==n)?0:1)).split("/"); if(Bruch.length==1)Bruch[1]=1; var I=(j==n)?(ig+1)*(n+1)-1:ig*(n+1)+j; if(I>=Z.length)for(var ii=Z.length;ii<=I;ii++){Z[ii]=0;N[ii]=1;m[ii]=0;} var vz=(j==n)?-1:1; if(Bruch[0]=="-")Bruch[0]=-1; if(Bruch[0]=="+")Bruch[0]=1; Z[I]=Z[I]*Number(Bruch[1])+vz*Number(Bruch[0])*N[I]; N[I]*=Number(Bruch[1]); if((vz==-1)&&(Z[I]!=0))homogen=0; if((isNaN(Z[I]))||(isNaN(N[I]))){alert("Fehler in der Eingabe bei "+t+"\n"+g[iig]);return;} } g[iig]=g[iig].replace(/=/," = ").replace(/\+/g," + ").replace(/-/g," - ").replace(/= - /,"= -"); if(g[iig].substr(0,3)==" - ")g[iig]="-"+g[iig].substring(3,g[iig].length); g[iig]=g[iig].replace(/\./g,",").replace(/ ,/g," 0,").replace(/-,/g,"-0,"); if(g[iig].charAt(0)==",")g[iig]="0"+g[iig]; ig++;ng++; } // ig-Schleife t=""; for(i=0;i<n*(n+1);i++)m[i]=Z[i]+"/"+N[i]; document.f.t1.value=g.join("\n").replace(/\n+/g,"\n"); if(document.f.t1.value.charAt(document.f.t1.value.length-1)!="\n")document.f.t1.value+="\n"; v[n]=" "; if ((homogen==0)) { if (n>ng){ alert("Mehr Variablen als Gleichungen"); return;} else if (n<ng){ alert("Mehr Gleichungen als Variablen"); return;} //GLSL(n,n+1); GLSL_(m,n); if(Erkl)document.f.gauss.value=ET; Erkl=false; for(i=0;i<n;i++) { s=0; for(j=0;j<n;j++){if(m[(n+1)*i+j]!=0)s=1;} //alert(s + "\n"+ m[(n+1)*(i+1)-1] +"\n"+((s==0)&&(m[(n+1)*(i+1)-1]!=0))); if((s==0)&&(m[(n+1)*(i+1)-1]!=0)){t=" \n Gleichungssystem\n hat keine Lösung";break;} if(m[(n+1)*i+i]==0){t=" \n keine eindeutige\n Lösung gefunden";break;} t+=" "+ v[i]+" = "+String(Runden(m[(n+1)*(i+1)-1],10)).replace(/\./g,",")+"\n"; } t=t.replace(/ ,/g," 0,").replace(/-,/g,"-0,"); document.f.t2.value=t; } else { j=0; for(i=0;i<n*(n+1);i++)m[i]=Z[i]/N[i]; for(i=0;i<n*ng;i++){B[i]=m[i+j];if ((i+1)%n==0)j++;} //while(ng<n){for(i=0;i<n;i++)B[n*ng+i]=B[i];ng++;} homogenGLSL(ng,n,0); for(i=0;i<n;i++) t+=" "+ v[i]+" = "+String(Runden(L[i],10)).replace(/\./,",")+"·"+lambda+"\n"; t=t.replace(/ \-1·/g," -").replace(/ 1·/g," ").replace(/ 0·µ/g," 0").replace(/·/g,""); t+="\n µ aus R"; for (i=0;i<n;i++)if(L[i]==0){t="";for(i=0;i<n;i++)t+=" "+v[i]+" = 0\n";t+="\n (sonst keine Lösung\n mit einem Parameter)";break;} t=t.replace(/ ,/g," 0,").replace(/-,/g,"-0,"); document.f.t2.value=t; document.f.gauss.value=" \n Erklärungen zu homogenen Gleichungssystemen sind noch nicht verfügbar."; } status=""; } function Analyse2(t) { if(t.length<3){if(Erkl)document.f.gauss.value=" \n Geben Sie oben zunächst ein Gleichungssystem ein\n\n oder lassen Sie ein Zufallsbeispiel erzeugen!";return;} document.f.erkltyp[0].checked=true; t=t.replace(/\n/g," ").replace(/\r/g," ").replace(/\t/g," ").replace(/,/g,"."); t=t.replace(/ +/g," "); while(t.indexOf(" ")==0){t=t.substr(1,t.length);}; while(t.lastIndexOf(" ")==t.length-1){t=t.substr(0,t.length-1);}; m=t.split(" "); mm=m.length; //for(i=0;i<mm;i++)m[i]=Number(m[i]); nn=(Math.sqrt(4*mm+1)-1)/2; if(nn!=Math.floor(nn)){alert("Matrix hat nicht die Form n x (n+1)");return;} //GLSL(nn,nn+1); GLSL_(m,nn); if(Erkl)document.f.gauss.value=ET; Erkl=false; t=""; for(i=0;i<mm;i++){t+=Runden(m[i],10)+" ";if(i%(nn+1)==nn)t+="\n";} t=t.replace(/\./g,",").replace(/ ,/g," 0,").replace(/-,/g,"-0,"); document.f.t2.value=t; } function Runden(t,i) { var T=String(t); if(T.indexOf("/")>0)if(document.f.alsBruch.selectedIndex==0)return t;else t=eval(t); var d=Math.pow(10,i); return Math.round(t*d)/d; } function GLSL(nz, ns) { var i, j, k ; var q; for (j = 0;j<ns-1;j++) { // Diagonalenfeld normalisieren q = m[j * ns + j]; if (q == 0){ //Gewährleisten, daß keine 0 in der Diagonale steht for (i = j+1 ;i< nz;i++) { // Suche Reihe mit Feld <> 0 und addiere dazu if (m[i * ns + j] != 0) { for (k = 0 ; k<ns; k++) { m[j * ns + k] += m[i * ns + k]; } q = m[j * ns + j]; break; } } } if (q != 0) { // Diagonalen auf 1 bringen for (k = 0;k< ns;k++) { m[j * ns + k] = m[j * ns + k] / q; } } // Spalten außerhalb der Diagonalen auf 0 bringen for (i = 0 ; i< nz ; i++) { if (i != j ) { q = m[i * ns + j]; for (k = 0; k< ns;k++) { m[i * ns + k] -= q * m[j * ns + k]; } } } } } function homogenGLSL(nz, ns) { var i=0, j=0, k=0 ; var q=0.0; for (j = 0;j<nz;j++) { // Diagonalenfeld normalisieren q = B[j * ns + j]; if (q == 0.0){ //Gewährleisten, daß keine 0 in der Diagonale steht for (i = j+1;i< nz;i++) { // Suche Reihe mit Feld <> 0 und addiere dazu if (B[i * ns + j] !=0.0 ) { for (k = 0 ; k<ns; k++) { B[j * ns + k] += B[i * ns + k]; } q = B[j * ns + j]; break; } } } if (Math.abs(q)<0.000001){B[j*ns+j]=0;q=0;} if (q != 0.0) { // Diagonalen auf 1 bringen for (k = 0;k< ns;k++) { B[j * ns + k] = B[j * ns + k] / q; } } // Spalten außerhalb der Diagonalen auf 0 bringen for (i = 0 ; i< nz ; i++) { if (i!=j) { q = B[i * ns + j]; for (k = 0; k< ns;k++) { B[i * ns + k] -= q * B[j * ns + k]; } } } } // freie Parameter bestimmen j=ns-1; for(i=0;i<ns;i++) L[i]=0.0; for (i=nz-1;i>=0;i--) { if (Math.abs(B[i*ns+j])<0.0000001) { L[i]=1.0; for (k=i-1;k>=0;k--) L[k]-=B[k*ns+j]; // Abhängigkeit vom freien Parameter im Lösungsvektor eintragen j--; } else j--; } min=10000000000; GGT=1; q=1.0; for(i=0;i<ns;i++){if(L[i]==0)return; q*=L[i];} Nenner=Runden(GetNenner(q,0.00001,1000),8); for(i=0;i<ns;i++) L[i]=Runden(L[i]*Nenner,8); GGT=L[ns-1]; for(i=0;i<ns-1;i++) { if((min>Math.abs(L[i]))&&(L[i]!=0))min=Math.abs(L[i]); GGT=ggT(L[i],GGT); } if (Math.abs(GGT)>=1) for(i=0;i<ns;i++) L[i]=L[i]/GGT; else for(i=0;i<ns;i++) L[i]=L[i]/min; } function ggT(a,b) { var r=-1; while(r!=0) {r=a%b;a=b;b=r;} return a; } function GetNenner(x,d,max) { var y; for(var i=1;i<=max;i++) { y=x*i; if(Math.abs(y-Math.floor(y+0.5))<=d) return i; } return 1; } function erkl() { Erkl=true;Analyse(); } function bsp() { var x=new Array(),vv=new Array(),i,j,g,gl="",gr,t="@\n",k=new Array(),s,ik,jj,f,z0,z1; mingrad=Number(document.f.mingrad.value); if(mingrad<2)mingrad=2;if(isNaN(mingrad))mingrad=2; maxgrad=Number(document.f.maxgrad.value); if(maxgrad>9)maxgrad=9; if((maxgrad<2)||isNaN(maxgrad))maxgrad=mingrad; if(mingrad>maxgrad)mingrad=maxgrad; document.f.mingrad.value=mingrad; document.f.maxgrad.value=maxgrad; n=Math.floor(Math.random()*(maxgrad-mingrad)+mingrad); if(isNaN(n))n=2; switch(Math.floor(Math.random()*4)) { case 0: vv[0]="a";vv[1]="b";vv[2]="c";vv[3]="d";vv[4]="e";break; case 1: vv[0]="x";vv[1]="y";vv[2]="z";vv[3]="w";vv[4]="u";break; case 2: vv[0]="m";vv[1]="n";vv[2]="p";vv[3]="q";vv[4]="r";break; default:vv[0]=vvv[j=Math.floor(Math.random()*vvv.length)]; for(i=1;i<n;i++) { j=(j+Math.floor(Math.random()*5)+1)%vvv.length; if(vvv[j]=="o")j++;if(vvv[j]=="l")j++; vv[i]=vvv[j]; } } j=Math.floor(Math.random()*vvv.length); for(i=vv.length;i<n;i++) { do{ j=(j+Math.floor(Math.random()*5)+1)%vvv.length; if(vvv[j]=="o")j++;if(vvv[j]=="l")j++; for(jj=0;jj<i;jj++)if(vv[jj]==vvv[j])break; }while(jj<i); vv[i]=vvv[j]; } vv[n]=""; for(i=0;i<n*(n+1);i++)m[i]=0; for(i=0;i<n;i++) { x[i]=Math.round(Math.random()*40-20);if(Math.random()>.8)x[i]+=.5; //if(Math.random()>.9)x[i]-=.25;if(Math.random()>.9)x[i]-=Math.floor(Math.random()*10)/10; do{f=Math.floor(Math.random()*10-5);}while(f==0); m[(i+1)*(n+1)-1]=-x[i]*f;m[i*(n+1)+i]=f; } for(i=0;i<n*n;i++) { var fmin=10000000000,ff,fi=-0.5; z0=Math.floor(Math.random()*n); do{z1=Math.floor(Math.random()*n)}while(z1==z0); for(f=-10;f<10;f++) { ff=0; for(jj=0;jj<=n;jj++)ff+=Math.abs(m[z0*(n+1)+jj]+f*m[z1*(n+1)+jj]); if((ff<fmin)&&(ff>4*n)){fmin=ff;fi=f;} } for(jj=0;jj<n;jj++)if(m[z0*(n+1)+jj]+Math.floor(fi+.1)*m[z1*(n+1)+jj]!=0)break; if(jj==n)fi=1; for(jj=0;jj<n+1;jj++)m[z0*(n+1)+jj]+=Math.floor(fi+.1)*m[z1*(n+1)+jj]; } for(i=0;i<n;i++) { z0=Math.floor(Math.random()*n); do{z1=Math.floor(Math.random()*n)}while(z1==z0); for(j=0;j<n+1;j++){jj=m[z0*(n+1)+j];m[z0*(n+1)+j]=m[z1*(n+1)+j];m[z1*(n+1)+j]=jj;} } for(i=0;i<n*(n+1);i++) { if((m[i]==1)&&((i%(n+1))<n))m[i]="+"+vv[i%(n+1)]; else if((m[i]==-1)&&((i%(n+1))<n))m[i]="-"+vv[i%(n+1)]; else if(m[i]!=0) m[i]=((m[i]>0)?"+":"")+m[i]+vv[i%(n+1)]; else m[i]=""; } //alert(m); for(i=0;i<n;i++) { gl="";gr=""; for(j=0;j<n+1;j++) { if(m[i*(n+1)+j]!="") { jj=Math.floor(Math.random()*4) switch(jj) { case 0: gl+=m[i*(n+1)+j];break; case 1: gr+="-"+m[i*(n+1)+j];break; case 2: gl=m[i*(n+1)+j]+gl;break; default: gr="-"+m[i*(n+1)+j]+gr;break; } } } if(gl=="")gl="0";if(gr=="")gr="0"; //if(gl.length<gr.length){var gg=gl;gl=gr;gr=gg;} t+=gl+"="+gr+"\n"; } t=t.replace(/\./g,",").replace(/-,/g,"-0,").replace(/\+,/g,"+0,"); t=t.replace(/--/g,"+").replace(/-\+/g,"-").replace(/\+-/g,"-").replace(/\n\+/g,"\n"); t=t.replace(/=\+/g,"=").replace(/-/g," - ").replace(/\+/g," + "); t=t.replace(/= - /g,"=-").replace(/\n - /g,"\n-").replace(/@\n/,""); t=t.replace(/=/g," = "); document.f.t1.value=t; document.f.t2.value=""; } function bsp_() { bsp();Analyse(); } function bsp__() { Erkl=true; bsp(); //var t=(ErklTyp==0)?" Umformen und sortieren (Variablen alphabetisch links, Konstanten rechts):\n\n"+getGS()+"\n\n":""; Analyse(); //document.f.gauss.value=" \n Gleichungssystem:\n\n "+document.f.t1.value.replace(/\n/g,"\n\n ")+"\n"+t+document.f.gauss.value; } function getGS(vz) { var i,j,jj,t="@\n "; if(vz==null)vz=""; for(i=0;i<n;i++) { for(j=0;j<n;j++) if(m[i*(n+1)+j]!="") { if(vz=="") t+=m[i*(n+1)+j]; else { var mm=String(m[i*(n+1)+j]).replace(/,/,"."); if(mm=="")mm="0";var emm=eval(mm); if(emm==1)t+="+"+v[j]; else if(emm==-1)t+="-"+v[j]; else if(emm!=0)t+="+"+emm+v[j]; } } if(vz=="") t+="=-"+vz+m[i*(n+1)+n]+"\n\n "; else { var mm=String(m[i*(n+1)+n]).replace(/,/,"."); if(mm=="")mm="0";var emm=eval(mm); t+="="+emm+"\n\n "; } } t=t.replace(/\./g,",").replace(/-,/g,"-0,").replace(/\+,/g,"+0,"); t=t.replace(/--/g,"+").replace(/\+-/g,"-").replace(/\++/g,"+").replace(/\n +\+/g,"\n ").replace(/-\+/g,"-"); t=t.replace(/=-\n/g,"=0\n").replace(/=\+/g,"=").replace(/-/g," - ").replace(/\+/g," + "); t=t.replace(/= - /g,"=-").replace(/\n +- /g,"\n -").replace(/@\n/,""); t=t.replace(/=/g," = "); return t; } function uebungen() { var t="<html>\n<head>\n<meta http-equiv=\"content-type\" content=\"text/html; charset=iso-8859-1\">\n"; t+="<meta name=\"author\" content=\"Arndt Brünner\">\n"; t+="<title>Übungen</title>\n<link rel=stylesheet type=\"text/css\" href=\"standard1.css\">\n"; t+="</head>\n<body bgcolor=\"#FFFAEE\" link=\"#000066\" alink=\"#000066\" vlink=\"#000066\">\n"; t+='<h3>Übungsaufgaben</h3>\n<p><small>erzeugt auf: <a href="http://www.arndt-bruenner.de/mathe/scripts/gleichungssysteme.htm">http://home.t-online.de/home/arndt.bruenner/mathe/scripts/gleichungssysteme.htm</a></small></p>\n'; var l="<p>&nbsp;</p>\n<hr><p>&nbsp;</p>\n<h5>Lösungen</h5>\n<p>&nbsp;<br><small>"; var i; t+="<p>&nbsp;</p>\n<h5>Lineare Gleichungssysteme lösen</h5>\n"; t+='<table cellpadding=10>\n'; for(i=0;i<10;i++) { bsp_(); if((i%2)==0)t+=((i>0)?'</tr>':'')+'<tr>\n'; else t+='<td>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</td>\n'; t+='<td style="text-align:right">('+(i+1)+')</td><td>\n'; t+='<table style="margin-left:50px"><tr>\n'; t+='<td rowspan=2><img src="vstr.gif" width="2" height="'+(20*n)+'">&nbsp;</td>\n'; t+='<td align=center >\n'; t+=document.f.t1.value.replace(/\n/g,"<br>\n"); t+='</td>\n'; t+='<td rowspan=2>&nbsp;<img src="vstr.gif" width="2" height="'+(20*n)+'"></td>\n'; t+='</tr><tr><td align=center>\n'; t+='</td></tr></table>\n</td>\n'; l+='('+(i+1)+') &nbsp;&nbsp;&nbsp;'+document.f.t2.value.replace(/\n/g,"&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;")+'<br>\n'; } t+='</tr></table>\n'; t+=l+"<p>&nbsp;</p><hr><p><small>&copy; Arndt Brünner&nbsp;&nbsp;<br>eMail: arndt.bruenner@t-online.de\n"; t+='<br>Homepage: <a href="http://www.arndt-bruenner.de/mathe/mathekurse.htm">http://www.arndt-bruenner.de/mathe</a></small>\n</p>\n'; t+='<p><small><a href="javascript:history.back()">zur&uuml;ck</a></small></p>\n</body>\n</html>\n'; document.open(); document.write(t); document.close(); } //lgsb.js // Javascript (c) Arndt Brünner // Alle Rechte vorbehalten! // Version: 18.4.2003 var Erkl=(1==0), ErklTyp=0; var ET=""; var ganzzahlig=(1==9); function test() { var x=23;y=-13 var z=new Array( 2,-3,2*x-3*y , -1,2, -1*x+2*y ); var n=new Array(1,1,1,1,1,1); var lz=new Array(),ln=new Array(); GLSLB(z,n,lz,ln,2); alert("Zähler: "+ lz.join(" ")+"\nNenner: "+ln.join(" ")); } function GLSL_(m,nz) { var n=new Array(),z=new Array(); var ln=new Array(),lz=new Array(); var i,nn=nz*(nz+1); status="Wandle in Brüche um"; for(i=0;i<nn;i++) { z[i]=m[i]; var Z=String(m[i]).replace(/,/g,".");n[i]=1;var ZZ=Z.split("/"); if(Z.indexOf("/")>-1){z[i]=ZZ[0];n[i]=ZZ[1];kuerzen(z,n,i);} if(ZZ[0].indexOf(".")>-1){n[i]=Number("1"+"0000000000000000000".substr(0,ZZ[0].length-ZZ[0].indexOf(".")-1));z[i]=Number(ZZ[0].replace(/\./g,""));kuerzen(z,n,i);} //while(z[i]!=Math.floor(z[i])){n[i]*=10;z[i]*=10;} } GLSLB(z,n,lz,ln,nz); for(i=0;i<nn;i++)m[i]=String(z[i])+(((n[i]!=1)&&(z[i]!=0))?"/"+String(n[i]):""); for(i=0;i<nz;i++)z[i*(nz+1)+nz]=lz[i]+(((ln[i]!=1)&&(lz[i]!=0))?"/"+ln[i]:""); } // z: Zähler, n: Nenner // lz,ln: Lösungen // nz: Anzahl Zeilen function GLSLB(z,n,lz,ln,nz) { var i, j, k, d, ns=nz+1, g=ErklTyp==1; var qz,qn,abbruch=(1==0); if(ganzzahlig)return GLSLBgz(z,n,lz,ln,nz); ET=""; v[nz]=""; if(Erkl) { ET+=" \n Gleichungssystem:\n\n "+document.f.t1.value.replace(/\n/g,"\n\n ")+"\n\n"; ET+=(!g)?" Umformen und sortieren (Variablen alphabetisch links, Konstanten rechts):\n\n"+getGS("+")+"\n\n":""; } if(g) { ET+=" \n Die Gleichungen werden so umgeformt und untereinander geschrieben, daß alle gleichen Variablen\n\n auf der linken Seite"; ET+=" der Gleichung untereinander stehen und die konstanten Zahlen\n\n auf der rechten Seite.\n\n"; } else ET+=" \n Stelle die Koeffizientenmatrix auf. Reihenfolge der Variablen: "+v.join(", ")+"Konstante\n\n"; if(Erkl)ET+=toStr(z,n,nz)+"\n\n\n"; for (j = 0;j<ns-1;j++) { status="Löse das Gleichungssystem - zu "+Math.floor(j*100/(ns-1))+"% fertig"; // Diagonalenfeld normalisieren d=j*ns+j; // Index Diagonalenfeld if (z[d]==0) { //Gewährleisten, daß keine 0 in der Diagonale steht for(i=j+1;i<nz;i++) { // Suche Reihe mit Feld <> 0 und addiere dazu if (z[i*ns+j]!=0) { var fi=i*ns+j; if(Erkl) { if(g) ET+=" Da in der "+(j+1)+". Gleichung kein "+v[j]+" vorkommt"; else ET+=" Da das Diagonalenfeld in der "+(j+1)+". Zeile 0 ist"; ET+=", tausche die "+(j+1)+". und die "+(i+1)+". Zeile:\n\n"; } // for(k=0;k<ns;k++)plus(z,n,j*ns+k,i*ns+k,j*ns+k,1,1,n[fi],z[fi],false); for(k=0;k<ns;k++){var X=n[j*ns+k];n[j*ns+k]=n[i*ns+k];n[i*ns+k]=X;X=z[j*ns+k];z[j*ns+k]=z[i*ns+k];z[i*ns+k]=X;} if(Erkl) { ET+=toStr(z,n,nz)+"\n\n\n"; } break; } } } if((z[d]!=0)&&(z[d]!=n[d])) { // Diagonalen auf 1 bringen if(Erkl) { if((/*(Math.abs(n[d])<Math.abs(z[d]))||*/(Math.abs(n[d])==1))&&(Math.abs(n[d])!=Math.abs(z[d]))) ET+=" Durch Division der "+(j+1)+((g)?". Gleichung":". Zeile")+" durch "+BrStr(z[d],n[d]); else ET+=" Durch Multiplikation der "+(j+1)+((g)?". Gleichung":". Zeile")+" mit "+BrStr(n[d],z[d]); if(g)ET+=" wird der Faktor vor "+v[j]+" eliminiert:\n\n"; else ET+=" wird das Diagonalelement zu 1 gemacht:\n\n"; } for(k=0;k<ns;k++)if(k!=j)geteilt(z,n,j*ns+k,d,j*ns+k,true); z[d]=1;n[d]=1; if(Erkl)ET+=toStr(z,n,nz)+"\n\n\n"; } else if((Erkl)&&(z[d]!=0)) { if(g)ET+=" In der "+(j+1)+". Gleichung steht "+v[j]+" bereits ohne Faktor.\n\n\n"; else ET+=" Das Diagonalenfeld der "+(j+1)+". Zeile ist bereits 1.\n\n\n"; } // Spalten außerhalb der Diagonalen auf 0 bringen if(Erkl) { for(i=0;i<nz;i++) { if((i!=j)&&(z[i*ns+j]!=0)) { if(Erkl) { if(g)ET+=" Mit der "+(j+1)+". Gleichung wird in allen anderen Gleichung der Summand mit "+v[j]+" eliminiert.\n\n\n"; else ET+=" Mit der "+(j+1)+". Zeile werden alle anderen Zeilen in der "+(j+1)+". Spalte auf 0 gebracht.\n\n\n"; } break; } } if(i==nz) { if(g)ET+=" In allen übrigen Gleichungen ist kein "+v[j]+" mehr enthalten.\n\n\n"; else ET+=" Alle übrigen Zeilen haben in der "+(j+1)+". Spalte bereits eine 0.\n\n\n"; } } if(!Erkl)for(i=0;i<nz*(nz+1);i++)kuerzen(z,n,i); for (i = 0 ; i< nz ; i++) { if (i != j ) { var qz=z[i*ns+j],qn=n[i*ns+j]; if(qz!=0) { for(k=0;k<ns;k++)minus(z,n,i*ns+k,j*ns+k,i*ns+k,1,1,qz,qn,true); //alert("Zeile "+j+"\nZähler: "+ z.join(" ")+"\nNenner: "+n.join(" ")); if(Erkl) { if(g)ET+=" Zur "+(i+1)+". Gleichung wird "+((-qz!=qn)?"das "+BrStr(-qz,qn)+"fache der ":"die ")+(j+1)+". Gleichung addiert:\n\n"; else ET+=" Zur "+(i+1)+". Zeile wird "+((-qz!=qn)?"das "+BrStr(-qz,qn)+"fache der ":"die ")+(j+1)+". Zeile addiert:\n\n"; ET+=toStr(z,n,nz)+"\n\n\n"; } } for(k=0;k<nz;k++)if(z[i*ns+k]!=0)break; if(k==nz) { abbruch=true; if(Erkl) { if(g)ET+=" In der "+(i+1)+". Gleichung sind alle Variablen verschwunden."; else ET+=" In der "+(i+1)+". Zeile sind alle Koeffizienten 0."; ET+=" \n\n   > Das Gleichungssystem hat keine "+((z[(i+1)*ns-1]!=0)?"":"eindeutige ")+"Lösung.\n\n\n"; } break; } } } if(abbruch)break; } // Lösungsvektor füllen //alert("Zähler: "+ z.join(" ")+"\nNenner: "+n.join(" ")); for(i=0;i<nz;i++) { lz[i]=z[(i+1)*ns-1];ln[i]=n[(i+1)*ns-1]; if(ln[i]<0){lz[i]=-lz[i];ln[i]=-ln[i];} kuerzen(lz,ln,i); } if((Erkl)&&(ErklTyp==0))if(!abbruch)ET+=" In der letzten Spalte stehen die Lösungen.\n\n"; if((Erkl)&&(ErklTyp==1)) { } } // z: Zähler, n: Nenner // lz,ln: Lösungen // nz: Anzahl Zeilen function GLSLBgz(z,n,lz,ln, nz) { var i, j, k, d, ns=nz+1, g=ErklTyp==1,gg,e1,e2,nnorm=1==0; var qz,qn,abbruch=(1==0); ET=""; v[nz]=""; if(Erkl) { ET+=" \n Gleichungssystem:\n\n "+document.f.t1.value.replace(/\n/g,"\n\n ")+"\n\n"; ET+=(g)?" Umformen und sortieren (Variablen alphabetisch links, Konstanten rechts):\n\n"+getGS("+")+"\n\n\n":""; } if(g) { ET+=" \n Die Gleichungen werden so umgeformt und untereinander geschrieben, daß alle gleichen Variablen\n\n auf der linken Seite"; ET+=" der Gleichung untereinander stehen und die konstanten Zahlen\n\n auf der rechten Seite.\n\n"; } else ET+=" \n Stelle die Koeffizientenmatrix auf. "+((v[0]==null)?"":"Reihenfolge der Variablen: "+v.join(", ")+"Konstante")+"\n\n"; if(Erkl)ET+=toStr(z,n,nz)+"\n\n\n"; status="mache ganzzahlig und teile zeilenweise durch den ggT"; e1=e2=false; for(j=0;j<nz;j++) { gg=n[j*ns];for(i=1;i<ns;i++){gg*=n[j*ns+i]/ggT(gg,n[j*ns+i]);} for(i=0;i<ns;i++){z[j*ns+i]*=gg/n[j*ns+i];n[j*ns+i]=1;} if(gg!=1)e2=true; gg=z[j*ns];for(i=1;i<ns;i++)gg=ggT(gg,z[j*ns+i]); for(i=0;i<ns;i++)z[j*ns+i]/=gg; if(gg!=1)e1=true; } if((Erkl)&&(e1||e2)) { ET+=" \n Die Koeffizienten werden zeilenweise durch "+(e2?"Multiplikation mit dem kgV der Nenner\n\n und anschließende ":"")+"Division durch den ggT ganzzahlig gemacht.\n\n"; ET+=toStr(z,n,nz)+"\n\n\n"; } for (j = 0;j<ns-1;j++) { status="Löse das Gleichungssystem - zu "+Math.floor(j*100/(ns-1))+"% fertig"; d=j*ns+j; // Index Diagonalenfeld if (z[d]==0) { //Gewährleisten, daß keine 0 in der Diagonale steht for(i=j+1;i<nz;i++) { // Suche Reihe mit Feld <> 0 und addiere dazu if (z[i*ns+j]!=0) { var fi=i*ns+j; if(Erkl) { if(g) ET+=" Da in der "+(j+1)+". Gleichung kein "+v[j]+" vorkommt"; else ET+=" Da das Diagonalenfeld in der "+(j+1)+". Zeile 0 ist"; ET+=", tausche die "+(j+1)+". und die "+(i+1)+". Zeile:\n\n"; } // for(k=0;k<ns;k++)plus(z,n,j*ns+k,i*ns+k,j*ns+k,1,1,n[fi],z[fi],false); for(k=0;k<ns;k++){var X=n[j*ns+k];n[j*ns+k]=n[i*ns+k];n[i*ns+k]=X;X=z[j*ns+k];z[j*ns+k]=z[i*ns+k];z[i*ns+k]=X;} if(Erkl) { ET+=toStr(z,n,nz)+"\n\n\n"; } break; } } } // Spalten außerhalb der Diagonalen auf 0 bringen if(Erkl) { for(i=0;i<nz;i++) { if((i!=j)&&(z[i*ns+j]!=0)) { if(Erkl) { if(g)ET+=" Mit der "+(j+1)+". Gleichung wird in allen anderen Gleichung der Summand mit "+v[j]+" eliminiert.\n\n\n"; else ET+=" Mit der "+(j+1)+". Zeile werden alle anderen Zeilen in der "+(j+1)+". Spalte auf 0 gebracht.\n\n\n"; } break; } } if(i==nz) { if(g)ET+=" In allen übrigen Gleichungen ist kein "+v[j]+" mehr enthalten.\n\n\n"; else ET+=" Alle übrigen Zeilen haben in der "+(j+1)+". Spalte bereits eine 0.\n\n\n"; } } //if(!Erkl)for(i=0;i<nz*(nz+1);i++)kuerzen(z,n,i); for(i=0;i<nz;i++) { if(i!=j) { var qz=z[i*ns+j],qn=n[i*ns+j],qzz=z[j*ns+j],gg=ggT(qzz,qz); qz/=gg;qzz/=gg;if(qzz<0){qzz=-qzz;qz=-qz;} if(qz!=0) { for(k=0;k<ns;k++)plus(z,n,i*ns+k,j*ns+k,i*ns+k,qzz,1,-qz,1,true); //alert("Zeile "+j+"\nZähler: "+ z.join(" ")+"\nNenner: "+n.join(" ")); if(Erkl) { if(qz!=1) { if(g)ET+=((qzz==1)?" Zur ":" Zum "+qzz+"fachen der ")+(i+1)+". Gleichung wird "+((-qz!=qzz)?"das "+(-qz)+"fache der ":"die ")+(j+1)+". Gleichung addiert:\n\n"; else ET+=((qzz==1)?" Zur ":" Zum "+qzz+"fachen der ")+(i+1)+". Zeile wird "+((-qz!=qzz)?"das "+(-qz)+"fache der ":"die ")+(j+1)+". Zeile addiert:\n\n"; } else { if(g)ET+=((qzz==1)?" Von der ":" Vom "+qzz+"fachen der ")+(i+1)+". Gleichung wird die "+(j+1)+". Gleichung subtrahiert:\n\n"; else ET+=((qzz==1)?" Von der ":" Vom "+qzz+"fachen der ")+(i+1)+". Zeile wird die "+(j+1)+". Zeile subtrahiert:\n\n"; } ET+=toStr(z,n,nz)+"\n\n\n"; } } for(k=0;k<nz;k++)if(z[i*ns+k]!=0)break; if(k==nz) { abbruch=true; if(Erkl) { if(g)ET+=" In der "+(i+1)+". Gleichung sind alle Variablen verschwunden."; else ET+=" In der "+(i+1)+". Zeile sind alle Koeffizienten 0."; ET+=" \n\n   > Das Gleichungssystem hat keine "+((z[(i+1)*ns-1]!=0)?"":"eindeutige ")+"Lösung.\n\n\n"; } break; } } } if(abbruch)break; if(j<nz-1) { e1=false; var p="",pp; for(i=0;i<nz;i++) { gg=z[i*ns];for(k=1;k<ns;k++)gg=ggT(gg,z[i*ns+k]); if(gg!=1){e1=true;for(k=0;k<ns;k++)z[i*ns+k]/=gg;if(p!="")p+=",";p+=(i+1);} } pp=p.split(",");p=pp[0];for(i=1;i<pp.length-1;i++)p+=", "+pp[i];if(pp.length>1)p+=" und "+pp[i]; if(e1&&Erkl)ET+=" \n Die Zeile"+((pp.length==1)?" "+p+" kann":"n "+p+" können")+" per Division durch ihren ggT vereinfacht werden:\n\n"+toStr(z,n,nz)+"\n\n\n"; } } e1=false; if(!abbruch) { for(i=0;i<nz;i++) { if(z[i*ns+i]!=1){n[i*ns+ns-1]=z[i*ns+i];z[i*ns+i]=1;e1=true;} } if(Erkl&&e1)ET+=" \n Nun wird zeilenweise durch die "+(g?"Koeffizienten der ":"")+"Diagonalelemente dividiert:\n\n"+toStr(z,n,nz)+"\n\n\n"; } // Lösungsvektor füllen //alert("Zähler: "+ z.join(" ")+"\nNenner: "+n.join(" ")); for(i=0;i<nz;i++) { kuerzen(n,z,(i+1)*ns-1); if(n[(i+1)*ns-1]<0){z[(i+1)*ns-1]=-z[(i+1)*ns-1];n[(i+1)*ns-1]=-n[(i+1)*ns-1];} lz[i]=z[(i+1)*ns-1];ln[i]=n[(i+1)*ns-1]; //kuerzen(lz,ln,i); } if((Erkl)&&(ErklTyp==0))if(!abbruch)ET+=" In der letzten Spalte stehen die Lösungen.\n\n"; if((Erkl)&&(ErklTyp==1)) { } ET=ET.replace(/ 2fach/g," Doppelt").replace(/ 3fach/g," Dreifach").replace(/ 4fach/g," Vierfach"); ET=ET.replace(/das 1fache der/g,"die"); } function BrStr(z,n) { if(n<0){n*=-1;z*=-1;} if(n==1)return z; if(z==0)return 0; return z+"/"+n; } function toStrV(z,n,nz) { var i,j,max=new Array(nz+1),t="",bz; for(i=0;i<=nz;i++) { max[i]=Math.max(Math.abs(z[i]),Math.abs(n[i])); for(j=0;j<nz;j++)max[i]=Math.max(Math.max(max[i],Math.abs(z[j*(nz+1)+i])),Math.abs(n[j*(nz+1)+i])); } for(i=0;i<nz*(nz+1);i++)kuerzen(z,n,i); for(i=0;i<=nz;i++){max[i]=String(max[i]).length;} for(i=0;i<nz;i++) { bz=(1==0); for(j=0;j<nz+1;j++){if(z[i*(nz+1)+j]==0)n[i*(nz+1)+j]=1;} var b=(1==0); for(j=0;j<nz+1;j++){if(n[i*(nz+1)+j]!=1){b=true;break;}} if(b) { t+="\n " for(j=0;j<nz+1;j++) { if(j==nz)t+=" "; t+=" "+((n[i*(nz+1)+j]!=1)?cStr(Math.abs(z[i*(nz+1)+j]),max[j]):" ".substr(0,max[j]))+" "; } } t+="\n "; for(j=0;j<nz+1;j++) { if(j==nz)t+=" = "; if((!bz)&&(j==nz)){t=t.substr(0,t.length-8)+" 0 = "; } var maxx=Math.max(Math.abs(n[i*(nz+1)+j]),Math.abs(z[i*(nz+1)+j])); maxx=String(maxx).length; var zz=z[i*(nz+1)+j]; var tt=((zz<0)?" - ":((zz>0)&&(bz))?" + ":" "); if(j<nz)tt+=((n[i*(nz+1)+j]==1)?rStr(Math.abs(zz)+"·"+v[j],max[j]+2):cStr("                       ".substr(0,maxx),max[j])+"·"+v[j])+" "; else tt+=((n[i*(nz+1)+j]==1)?cStr(Math.abs(zz)+"·"+v[j],max[j]):cStr("                       ".substr(0,maxx),max[j])+"·"+v[j])+" "; if((j==0)||(j==nz))tt=tt.replace(/\+/g," "); if((j==nz)&&(zz==0)){;} t+=tt; if(zz!=0)bz=true; } if(b) { t+="\n "; for(j=0;j<nz+1;j++) { if(j==nz)t+=" "; t+=" "+((n[i*(nz+1)+j]!=1)?cStr(Math.abs(n[i*(nz+1)+j]),max[j]):" ".substr(0,max[j])) +" "; } } t+="\n"; } // t=t.replace(/- /g," - ").replace(/- /g," - ").replace(/- /g," - ").replace(/- /g," - ").replace(/\+/g," "); t=t.replace(/· /g,""); t=t.replace(/\n \+/g,"\n ");//.replace(/\n - /g,"\n - ").replace(/\n - /g,"\n - "); jj=0; // t=t.replace(/ 0·\D/g," ").replace(/- 0·/g," "); t=t.replace(/= \+/g,"= ").replace(/ 0·\D/g," ").replace(/ 1·/g," ").replace(/- 1·/g," -"); while((i=t.search(/\n +- /))>-1) { j=t.substring(i,t.length).indexOf("-"); t=t.substr(0,i+j)+" -"+t.substring(i+j+2,t.length); jj++;if(jj>n)break; } t=t.replace(/= - /g,"= - ").replace(/= - /g,"= - ").replace(/= - /g,"= - ").replace(/= - /g,"= - "); return t; } function toStr(z,n,nz) { var i,j,max=new Array(nz+1),t="",vv=ErklTyp==1,ns=(nz==Math.sqrt(z.length))?nz:nz+1; if(vv)return toStrV(z,n,nz); for(i=0;i<=nz;i++) { max[i]=Math.max(Math.abs(z[i]),Math.abs(n[i])); for(j=0;j<nz;j++)max[i]=Math.max(Math.max(max[i],Math.abs(z[j*(ns)+i])),Math.abs(n[j*(ns)+i])); } for(i=0;i<nz*(ns);i++)kuerzen(z,n,i); for(i=0;i<ns;i++){max[i]=String(max[i]).length;} for(i=0;i<nz;i++) { for(j=0;j<ns;j++){if(z[i*(ns)+j]==0)n[i*(ns)+j]=1;} var b=(1==0); for(j=0;j<ns;j++){if(n[i*(ns)+j]!=1){b=true;break;}} if(b) { t+="\n " for(j=0;j<ns;j++)t+=" "+((n[i*(ns)+j]!=1)?cStr(Math.abs(z[i*(ns)+j]),max[j]):" ".substr(0,max[j]))+(vv?" ":"")+" "; } t+="\n "; for(j=0;j<ns;j++) { var maxx=Math.max(Math.abs(n[i*(ns)+j]),Math.abs(z[i*(ns)+j])); maxx=String(maxx).length; t+=((z[i*(ns)+j]<0)?"- ":"+ ")+((n[i*(ns)+j]==1)?cStr(Math.abs(z[i*(ns)+j]),max[j]):cStr("                       ".substr(0,maxx),max[j]))+" "; } if(b) { t+="\n "; for(j=0;j<ns;j++)t+=" "+((n[i*(ns)+j]!=1)?cStr(Math.abs(n[i*(ns)+j]),max[j]):" ".substr(0,max[j])) +(vv?" ":"")+" "; } t+="\n"; } t=t.replace(/- /g," - ").replace(/- /g," - ").replace(/- /g," - ").replace(/- /g," - ").replace(/\+/g," "); return t; } function cStr(t,max) { var L0=String(t).length,L1=Math.max(Math.floor((max-L0)/2),0); L0=max-L1-L0; if(L1>L0){var LL=L0;L0=L1;L1=LL;} //alert(t+"\n"+max+"\n"+" ".substr(0,L0)+t+" ".substr(0,L1)); return " ".substr(0,L0)+t+" ".substr(0,L1); } function rStr(t,max) { var L0=max-String(t).length; return " ".substr(0,L0)+t; } function kuerzen(z,n,i) { if(z[i]==0){n[i]=1;return;} var g=ggT(z[i],n[i]); z[i]/=g;n[i]/=g; if(n[i]<0){z[i]=-z[i];n[i]=-n[i];} return g; } function plus(z,n,i,j,k,fiz,fin,fjz,fjn,kuerzn) { var nn=n[i]*n[j]*fin*fjn,zz=fiz*z[i]*n[j]*fjn+fjz*z[j]*n[i]*fin; z[k]=zz;n[k]=nn; if(kuerzn)kuerzen(z,n,k); } function minus(z,n,i,j,k,fiz,fin,fjz,fjn,kuerzn) { return plus(z,n,i,j,k,fiz,fin,-fjz,fjn,kuerzn); } function mal(z,n,i,j,k,fz,fn,kuerzn) { n[k]=n[i]*n[j]*fn;z[k]=f*z[i]*z[j]; if(kuerzn)kuerzen(z,n,k); } function geteilt(z,n,i,j,k,kuerzn) { // alert("vor geteilt"+i+":"+j+"->"+k+"\nZähler: "+ z.join(" ")+"\nNenner: "+n.join(" ")+"\n"); var nn=n[i]*z[j]; z[k]=z[i]*n[j]; n[k]=nn; // alert("nach geteilt\nZähler: "+ z.join(" ")+"\nNenner: "+n.join(" ")+"\n"); if(kuerzn)kuerzen(z,n,k); } function ggT(a,b) { if(isNaN(a)||isNaN(b))return 1; if((a==0)&&(b==0))return 1; if(a*b==0)return Math.max(Math.abs(a),Math.abs(b));//if(!confirm(a+" "+b))return 1; a=Math.abs(a);b=Math.abs(b); var c;do{c=a%b;a=b;b=c;}while(c!=0);return a; } function kgV(a,b){return a*b/ggT(a,b);} function toBruch(m) { var Z=String(m),n=1,ZZ=Z.split("/"),z=0; if(Z.indexOf("/")>-1){z=ZZ[0];n=ZZ[1];}else z=Number(m); if(ZZ[0].indexOf(".")>-1){n=Number("1"+"0000000000000000000".substr(0,ZZ[0].length-ZZ[0].indexOf(".")-1));z=Number(ZZ[0].replace(/\./g,""));} var s=(z*n<0)?-1:1;z=Math.abs(z);n=Math.abs(n); var g=ggT(z,n);z/=g;n/=g; return new Array(s*z,n); } //showrechner.js var rechnerwnd; function showRechner(pfad) { if((!rechnerwnd)||(rechnerwnd.closed)) rechnerwnd=window.open(pfad+'rechner.htm','Rechner','width=384,height=170'); rechnerwnd.focus(); } function alert0() //9806161720 { document.write('<font color=red>' +'This file is personal home work. No one<br>' +'proofread. Cannot promise correctness.<br>' +'If you suspect any view point wrong,<br>' +'please ask a math expert near by. <br>' +'Freeman 2009-06-19-10-46</font><br>' ); } //9806161724 function alert2() //9806261808 { document.write('' +'0<font color=red size=+3>' +'LiuHH&#39;s sol. has no guarantee!</font>' ); } //function alert2() 9806261809 //--> </SCRIPT> <script src="http://freeman2.com/tutelink.js" language="javascript"></script><!--9806111210--> <script src="tutelink.js" language="javascript"></script><!--9806111211--> <script type="text/javascript" src=jsgraph2.js> </script><!--9808261915--> </HEAD> <body link="#FF0000" vlink="#0000FF" alink="#50A000" bgcolor="#ccfcfc" onload="javascript:initial0()" > <span id="gotoFM2msg"></span><!--9712091951--> <font size=+1> <a href=jspico_e.htm#docA01> Positive Integer power sum COefficient equation </a> </font> <br> Please visit second page <a href=http://freeman2.com/jspico2e.htm>freeman2.com</a>/<a href=jspico2e.htm>jspico2e.htm</a> <!--9809071445--> <br> <a href="jspico_e.htm#program0"> program </a> 0 <a href="#Arndt01">Arndt</a> 0 <a href=jspico_e.htm#docA01> DocA </a> 0 <a href="#getCol5">first9</a> 0 <a href=jspico_e.htm#docB111> Update 2009-09-07 </a> 0 <br> <a href="#docB04">Coefficient triangle</a>0 <a href="#docB02">Pattern conjectures</a>0 <a href="#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=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="docA01">&lt;a name="docA01"&gt;</a> <a href="#docB41">proof</a>, <a href="#program0">program</a>, <a href="#docB02">Conjecture</a>, <a href="#docB04">triangle</a> 2009-09-01-20-08 start This file <a href=http://freeman2.com/jspico_e.htm> http://freeman2.com/jspico_e.htm</a> find out Positive Integer power sum COefficient equation. Brief as PICO program. If we want to sum 1+2+3+4+5+...+n to find numerical answer, write loop code to add 1, 2, 3, .... up to n. The other quicker method is to use equation to get answer. The equation is "[i=1 to i=n]{i} = n*(n+1)/2 ---eqn.AA01 <a name="docA02">&lt;a name="docA02"&gt;</a> Next problem, how to calculate "[i=1 to i=n]{i<sup>2</sup>} ? numerical method is easy. Next is the sum equation. (omit '[i=1 to i=n]') "{i<sup>2</sup>} = n*(n+1)*(2*n+1)/6 ---eqn.AA02 Similarly "{i<sup>3</sup>} = n*n*(n+1)*(n+1)/4 ---eqn.AA03 "{i<sup>4</sup>} = n*(n+1)*(6*n*n*n+9*n*n+n-1)/30 ---eqn.AA04 <a name="docA03">&lt;a name="docA03"&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. The immediate question is what equation for "{i<sup>5</sup>} ? Guess work just waste time, no use. Spend two days find out how to get "{i<sup>5</sup>} equation. This method can be extended to higher "{i<sup>HI</sup>} equations. But in realty, "{i<sup>9</sup>} break down due to overflow/underflow/truncation error etc. <a name="docA04">&lt;a name="docA04"&gt;</a> How to find out equation for "{i<sup>5</sup>}? = n*n*n*n*n*n/6+n*n*n*n*n/2 +5*n*n*n*n/12-n*n/12 ---eqn.AA05 Explain as following. Look at "[i=1 to i=n]{i} = n*(n+1)/2 ---eqn.AA01 Here i to first power i^1 get highest polynomial n*n in second power. <a name="docA05">&lt;a name="docA05"&gt;</a> Observation: The general equation for "{i<sup>power</sup>} has highest power n<sup>power+1</sup> in 'n' Set up a general equation for "{i<sup>5</sup>} It is polynomial in 5+1 power (n*n*n*n*n*n) 5+1 power polynomial has 5+2 coefficients. Let "{i<sup>5</sup>}= a*n*n*n*n*n*n +b*n*n*n*n*n +c*n*n*n*n +d*n*n*n +e*n*n +f*n +g ---eqn.AA06 Here a,b,c,d,e,f,g are seven unknowns. To solve for seven unknowns need seven linearly independent equation. To get this much information, let n=1 to n=7. <a name="docA06">&lt;a name="docA06"&gt;</a> <a href="#docB41">proof</a>, <a href="#program0">program</a>, <a href="#docB02">Conjecture</a>, <a href="#docB04">triangle</a> For n=1, eqn.AA06 become a*1 +b*1 +c*1 +d*1 +e*1 +f*n +g = numerical result = 1 (n=1, 1<sup>5</sup>) (Above is a rule: all coef. sum to one) For n=2, eqn.AA06 become a*2<sup>6</sup> +b*2<sup>5</sup> +c*2<sup>4</sup> +d*2<sup>3</sup> +e*2<sup>2</sup> +f*2<sup>1</sup> +g*2<sup>0</sup> = numerical result = 33 (n=2, 1<sup>5</sup>+2<sup>5</sup>) <a name="docA07">&lt;a name="docA07"&gt;</a> do the same process up to n=7 (n=7, 1<sup>5</sup>+2<sup>5</sup>+3<sup>5</sup>+4<sup>5</sup>+5<sup>5</sup>+6<sup>5</sup>+7<sup>5</sup>=29008) Finally get the following equation [ 1 1 1 1 1 1 1][a] [ 1] [ 2^6 2^5 2^4 2^3 2*2 2 1][b]=[ 33] ---eqn.AA07 [ 3^6 3^5 3^4 3^3 3*3 3 1][c] [ 276] [ 4^6 4^5 4^4 4^3 4*4 4 1][d] [ 1300] [ 5^6 5^5 5^4 5^3 5*5 5 1][e] [ 4425] [ 6^6 6^5 6^4 6^3 6*6 6 1][f] [12201] [ 7^6 7^5 7^4 7^3 7*7 7 1][g] [29008] <a name="docA08">&lt;a name="docA08"&gt;</a> eqn.AA07 right hand side column vector is numerical result. eqn.AA07 left hand side 7*7 matrix is coefficient matrix. a,b,c,d,e,f,g are seven unknowns. <a name="docA09">&lt;a name="docA09"&gt;</a> Here we ask <a href="#Arndt04">Arndt Br&#252;nner</a> for help Arndt Br&#252;nner's program solve for a,b,c,d,e,f,g These number output to box1 in two styles. First is coefficients only [[ " i^(5) for i=1 to i=7 coefficients from high power to constant 1/6, 1/2, 5/12, 0, -1/12, 0, 0 ]] Second is mathematics equation [[ Math equation is +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)+n*(0)+(0) ]] <a name="docA10">&lt;a name="docA10"&gt;</a> Nine equations coefficients are listed at <a href="#docB01">docB01</a> What guide line can we find? 1: Highest power coefficients are 1/(k+1) k is power in " i^(k) for i=1 to i=n 2: Second high power coefficients are 1/2 3: Constant is always zero. What else can we get from <a href="#docB04">this triangle</a>? 2009-09-01-21-12 stop <a name="docA11">&lt;a name="docA11"&gt;</a> <a href="#docB41">proof</a>, <a href="#program0">program</a>, <a href="#docB02">Conjecture</a>, <a href="#docB04">triangle</a> 2009-09-01-21-24 start One more observation 4: each coefficient set sum to 1. 1=1/2+1/2+0 1=1/3+1/2+1/6+0 1=1/4+1/2+1/4+0+0 1=1/5+1/2+1/3+0+(-1/30)+0 1=1/6+1/2+5/12+0+(-1/12)+0+0 1=1/7+1/2+1/2+0+(-1/6)+0+1/42+0 1=1/8+1/2+7/12+0+(-7/24)+0+1/12+0+0 1=1/9+1/2+2/3+0+(-7/15)+0+2/9+0+(-1/30)+0 1=1/10+1/2+3/4+0+(-7/10)+0+1/2+0+(-3/20)+0+0 2009-09-01-21-26 stop <a name="docA12">&lt;a name="docA12"&gt;</a> 2009-09-02-19-26 start Consider next equation for " i^(3) from i=1 to i=n [ 1 1 1 1 1][a] [1^3 ] [2^4 2^3 2*2 2 1][b]=[1^3+2^3 ] ---eqn.AA08 [3^4 3^3 3*3 3 1][c] [1^3+2^3+3^3 ] [4^4 4^3 4*4 4 1][d] [1^3+2^3+3^3+4^3 ] [5^4 5^3 5*5 5 1][e] [1^3+2^3+3^3+4^3+5^3] <a name="docA13">&lt;a name="docA13"&gt;</a> It has the following value [ 1 1 1 1 1][a] [ 1] [ 16 8 4 2 1][b]=[ 9] ---eqn.AA09 [ 81 27 9 3 1][c] [ 36] [256 64 16 4 1][d] [ 100] [625 125 25 5 1][e] [ 225] <a name="docA14">&lt;a name="docA14"&gt;</a> Please pay attention to eqn.AA08 Left hand side is matrix evaluate equation Right hand side is vector evaluate summation See eqn.AA08 last row [5^4 5^3 5*5 5 1][Î%]=[1^3+2^3+3^3+4^3+5^3] ('Î%' indicate not important) <a name="docA15">&lt;a name="docA15"&gt;</a> Matrix row [5^4 5^3 5*5 5 1 ] know n only Last row is [n^4 n^3 n^2 n^1 n^0] ---eqn.AA10 Compare above left side with right side below Vector elem.[1^3+2^3+3^3+4^3+5^3] ---eqn.AA11 Vector element know k=3rd power only.<font color=red> eqn.AA10: n=constant, power=0,1,2,3,4 eqn.AA11: Sum from 1 to n power=3=const. Reader should pay attention to this difference.</font> <a name="docA16">&lt;a name="docA16"&gt;</a> <a href="#docB41">proof</a>, <a href="#program0">program</a>, <a href="#docB02">Conjecture</a>, <a href="#docB04">triangle</a> To solve for coefficient a,b,c,d,e we must know right hand side value. Right hand side is numerical data. 2009-09-02-19-52 stop </font></pre> <a name="lowrank">&lt;a name="lowrank"&gt;</a> <br> 2009-09-01-15-25 <br> Lower rank formulas are here <br> " i^(k) from i=1 to i=n, k=1,2,3,4. k=positive integer only <br> " i^(1) = n*(n+1)/2 <br> " i^(2) = n*(n+1)*(2*n+1)/6 <br> " i^(3) = n*n*(n+1)*(n+1)/4 <br> " i^(4) = n*(n+1)*(6*n*n*n+9*n*n+n-1)/30 <br> " i^(5)= n*n*n*n*n*n/6+n*n*n*n*n/2+5*n*n*n*n/12-n*n/12 <br> <a name="program0">&lt;a name="program0"&gt;</a> 0<a href="#Arndt04">Arndt program</a> 0<a href="#getCol5">first9</a> <br> To solve for Positive Integer power sum COefficient <br> need to use Arndt Br&#252;nner&#39;s work below. Thank you Arndt ! <!--9809031148 thank--> <br> Program take care Arndt codes, user just click <INPUT onclick='javascript:picoMan(matrixRankN.value)' type=button value="Calculate Equation"> <br> Output to Box1. &nbsp; for " i^( <INPUT value="5" id="matrixRankN" size=5 onchange='javascript:if(isNaN(matrixRankN.value))matrixRankN.value=0;if(matrixRankN.value<0)matrixRankN.value=0' ><!--9809011535--> ) from i=1 to i=n <br> <font color=red> <span id=span_msg01> Try " i^(0) to " i^(9), program break down at " i^(9). </span> </font> <br> Box 1 Output &nbsp; <!--9809011541--> <INPUT onclick='javascript:window.clipboardData.setData("Text",document.getElementById("boxc01").value.toString())' type=button value="copy1"> <INPUT onclick='document.getElementById("boxc01").value=""' type=button value="del_1"> <br> <TEXTAREA id=boxc01 name=boxc01 rows=8 cols=60 > </TEXTAREA> <br> Goto <a href=http://freeman2.com/complex2.htm#calculator>freeman2.com</a>/<a href=complex2.htm#calculator>complex2.htm</a> box3, define line 1 "n=8" Paste <br><!--9809031155--> math equation at line 2, click [test box3 command, output to box4] <br> <a name="sum_i_from_1_to_n">&lt;a name="sum_i_from_1_to_n"&gt;</a> <a href="#Arndt04">Arndt program</a> 0<a href="#getCol5">first9</a> <!--9809021133 --> <br> Power series summation. 2009-09-02-12-30 <br> upper bound n <INPUT value="8" id=upper_bound_n size=5 ><!--9808301845--> <INPUT onclick='javascript:span_eqn_to_n.innerHTML="[From eqn. get] and [From sum. get] should be same value.<br>",span_sum_to_n.innerHTML="If different value, possibly eqn is wrong. (sum right)<br>",upper_bound_n.value="0"' type=button value="Clear"> Lower bound is 1. <br> <INPUT onclick='javascript:span_eqn_to_n.innerHTML="From eqn. get: "+sumi2n(upper_bound_n.value,1)+"<br>",summation0(upper_bound_n.value,1)' type=button value="" i^1 to n"> &nbsp; <INPUT onclick='javascript:span_eqn_to_n.innerHTML="From eqn. get: "+sumi2n(upper_bound_n.value,2)+"<br>",summation0(upper_bound_n.value,2)' type=button value="" i^2 to n"> &nbsp; <INPUT onclick='javascript:span_eqn_to_n.innerHTML="From eqn. get: "+sumi2n(upper_bound_n.value,3)+"<br>",summation0(upper_bound_n.value,3)' type=button value="" i^3 to n"> &nbsp; <INPUT onclick='javascript:span_eqn_to_n.innerHTML="From eqn. get: "+sumi2n(upper_bound_n.value,4)+"<br>",summation0(upper_bound_n.value,4)' type=button value="" i^4 to n"> &nbsp; <INPUT onclick='javascript:span_eqn_to_n.innerHTML="From eqn. get: "+sumi2n(upper_bound_n.value,5)+"<br>",summation0(upper_bound_n.value,5)' type=button value="" i^5 to n"> <br> <INPUT onclick='javascript:span_eqn_to_n.innerHTML="From eqn. get: "+sumi2n(upper_bound_n.value,6)+"<br>",summation0(upper_bound_n.value,6)' type=button value="" i^6 to n"> &nbsp; <INPUT onclick='javascript:span_eqn_to_n.innerHTML="From eqn. get: "+sumi2n(upper_bound_n.value,7)+"<br>",summation0(upper_bound_n.value,7)' type=button value="" i^7 to n"> &nbsp; <INPUT onclick='javascript:span_eqn_to_n.innerHTML="From eqn. get: "+sumi2n(upper_bound_n.value,8)+"<br>",summation0(upper_bound_n.value,8)' type=button value="" i^8 to n"> &nbsp; <INPUT onclick='javascript:span_eqn_to_n.innerHTML="From eqn. get: "+sumi2n(upper_bound_n.value,9)+"<br>",summation0(upper_bound_n.value,9)' type=button value="" i^9 to n"> &nbsp; <INPUT onclick='javascript:span_eqn_to_n.innerHTML="From eqn. get: "+sumi2n(upper_bound_n.value,10)+"<br>",summation0(upper_bound_n.value,10)' type=button value=""i^10 to n"> <br> <INPUT onclick='javascript:span_eqn_to_n.innerHTML="From eqn. get: "+sumi2n(upper_bound_n.value,11)+"<br>",summation0(upper_bound_n.value,11)' type=button value=""i^11 to n"> <INPUT onclick='javascript:span_eqn_to_n.innerHTML="From eqn. get: "+sumi2n(upper_bound_n.value,12)+"<br>",summation0(upper_bound_n.value,12)' type=button value=""i^12 to n"> &nbsp; <INPUT onclick='javascript:span_eqn_to_n.innerHTML="From eqn. get: "+sumi2nPower(upper_bound_n.value)+"<br>",summation0(upper_bound_n.value,sum_i_power.value)' type=button value="" i^"> <!--"" i^5 to n"--> <INPUT value="6" id=sum_i_power size=5 ><!--9808310832--> i^1.2 fraction power OK<!--9809032150--> <!--9808302318 = unknown equation--> <br> <span id=span_eqn_to_n> <br> </span> <span id=span_sum_to_n> Find answer here<br> </span> <!-- 2009-08-31-09-18 done Power series summation--> <br> <span id=span_range> <br> </span> <pre><a name="sum_eqn">&lt;a name="sum_eqn"&gt;</a> 2009-09-02-11-40 sum equations " i to n = +n*n*(1/2)+n*(1/2) " 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) " i^5 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 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 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 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 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 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 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 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) 2009-09-02-11-48 done sum equations "i to "i^10 2009-09-03-19-58 found "i^11 2009-09-03-20-11 add "i^11 2009-09-04-10-17 found "i^12 2009-09-04-10-22 add "i^12 </pre> <script lang="javascript"> <!-- var msg='Please input positive integer at [upper bound n] box.'; // 9809021150 // function sumi2n(nn,dd) { if(isNaN(nn)) return 0; nn=parseInt(nn); if(nn<=0) return msg; var n=nn; var eqn0= [ '', /* " i to n = */ '+n*n*(1/2)+n*(1/2)', /* " ii 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)', /* " i^5 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 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 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 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 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 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 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 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)' ]; /* 2009-09-03-21-40 add "i^11 equation 2009-09-04-10-43 add "i^12 equation */ var limit0=eqn0.length-1; //9809021158 if(dd>limit0) { //span_range.innerHTML='<br><br>'; //9809021226 span_range.innerHTML='' //9809021228 +'" i^('+dd+') from i=1 to i=' +upper_bound_n.value +'. Sorry, no equation.' ; return 'dimemsion='+dd+'>'+limit0; } span_range.innerHTML='' //9809021220 +'" i^('+dd+') from i=1 to i=' +upper_bound_n.value +'. Please see equation below.' ; return bye09(eval(eqn0[dd])); } //9809021202 // 9808310856 function sumi2nPower(nn) { span_range.innerHTML='<br><br>'; //9809021223 return '" i^('+sum_i_power.value+') from i=1 to i='+upper_bound_n.value+', unknown equation.'; } // 9808310858 //9808302257 function summation0(nn,id0) { if(isNaN(nn)) { span_sum_to_n.innerHTML='From sum. get: Upper bound is-Not-a-Number'; return 0; } nn=parseInt(nn); if(nn<=0) { span_sum_to_n.innerHTML='From sum. get: Upper bound is-Non-Positive-Number'; return 0; } if(isNaN(id0)) { span_sum_to_n.innerHTML='From sum. get: i-rise-to-power is-Not-a-Number'; return 0; //9808310851 } id0=parseFloat(id0); //9808310853 var sum0=0; for(var p0=1;p0<=nn;p0++) { sum0+=Math.pow(p0,id0); } var type0=''; //9808310900 if((id0+'')=='1' ||(id0+'')=='2' ||(id0+'')=='3' ||(id0+'')=='4' ||(id0+'')=='5' //9809010050 ||(id0+'')=='6' ||(id0+'')=='7' ||(id0+'')=='8' ||(id0+'')=='9' ||(id0+'')=='10' ||(id0+'')=='11' //9809032142 ||(id0+'')=='12' //9809041044 ) type0=' ; Please verify results!' else type0=' ; Numerical solution only.'; //9808310900 span_sum_to_n.innerHTML='From sum. get: '+sum0+type0; return 0; } //9808302308 /** 2009-09-01-00-55 add function bye09(in09) to tute0011.htm 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> <hr> <pre><font size=+2><a name="docB01">&lt;a name="docB01"&gt;</a> <a href="#docB41">proof</a>, <a href="#program0">program</a>, <a href="#docB02">Conjecture</a>, <a href="#docB04">triangle</a> 2009-09-01-18-30 " i^(1) coefficients from high power to constant 1/2, 1/2, 0 " i^(2) 1/3, 1/2, 1/6, 0 " i^(3) 1/4, 1/2, 1/4, 0, 0 " i^(4) 1/5, 1/2, 1/3, 0, -1/30, 0 " i^(5) 1/6, 1/2, 5/12, 0, -1/12, 0, 0 " i^(6) 1/7, 1/2, 1/2, 0, -1/6, 0, 1/42, 0 " i^(7) 1/8, 1/2, 7/12, 0, -7/24, 0, 1/12, 0, 0 " i^(8) 1/9, 1/2, 2/3, 0, -7/15, 0, 2/9, 0, -1/30, 0 " i^(9) coefficients from high power to constant 1/10, 1/2, 3/4, 0, -7/10, 0, 1/2, 0, -3/20, 0, 0 " i^(9) Math equation is +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) Program break down at " i^(9) " i^(9) coefficients are repaired 2009-09-01-18-52 <a name="docB02">&lt;a name="docB02"&gt;</a> 2009-09-02-00-15 Conjectures for <a href="#docB04">coef. triangle</a> 1: Highest power coefficients are 1/(k+1) k is power in " i^(k) for i=1 to i=n 2: Second high power coefficients are 1/2 3: Constant is always zero. 4: each coefficient set sum to 1. 5: Third high power coefficients are #/12 '#' increase one when power increase one. <a name="docB03">&lt;a name="docB03"&gt;</a> 6: Fourth high power coefficients are zero 7: Fifth high power coefficients are negative. 8: Start from third high power coefficients, next power has zero coef. Then negative, then zero, then positive, then zero etc. 9: <a href="#docB18">Column five</a> 2009-09-03-23-58 These conjectures have no proof. We can not predict the unknown (11)th coefficient set. 2009-09-02-00-25 <a name="docB04">&lt;a name="docB04"&gt;</a> <a href="#docB02">Pattern conjectures</a>, <a href="#docB41">proof</a> 2009-09-01-22-21 <font color=red>Coefficient triangle</font> " i^(k) , (k)=(1) to (k)=(12) 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 (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,-7601/30030, 0<font color=red> column 01 02 03 04 05 06 07 08 09 10 11 12 13 14</font> column 01,02,03,04,06,08,10,12,14 see <a href="#docB02">Pattern conjectures</a> column 05 <a href="#docB18">conjecture</a> and <a href="#getCol5">calculator</a>. <a name="docB05">&lt;a name="docB05"&gt;</a> (10):1/11, 1/2,10/12, 0,-120/120,0, 1/1, 0,-30/60, 0, 5/66, 0 2009-09-01-22-48 guess " i^(10) From coefficient sum to 1 find '5/66' " i^(10) Math equation is +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) 2009-09-01-22-51 <a name="docB06">&lt;a name="docB06"&gt;</a> <a href="#docB41">proof</a>, <a href="#program0">program</a>, <a href="#docB02">Conjecture</a>, <a href="#docB04">triangle</a> +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) <a name="docB07">&lt;a name="docB07"&gt;</a> 2009-09-02-06-15 4: each coefficient set sum to 1. is easy to explain. Any summation equation must be true for n=1 and 1^any_power=1, this condition force all coefficient sum to 1. 2009-09-02-06-18 <a name="docB08">&lt;a name="docB08"&gt;</a> 2009-09-02-06-42 3: Constant is always zero. any equation is valid for n=0 right hand side is 0 then constant must be = 0 2009-09-02-06-43 <a name="docB09">&lt;a name="docB09"&gt;</a> 2009-09-03-10-59 'update 2009-09-03' add the following code [[ ouStr1=ouStr1.replace(/, /g,'; ');//9809031050 ouStr1=ouStr1.replace(/,/g,'.'); ouStr1=ouStr1.replace(/; /g,', '); ]] <a name="docB10">&lt;a name="docB10"&gt;</a> Before add '9809031050' code, Math equation is (too long, omit) ***** big error equation ***** " i^(9) from i=1 to i=10. get 8.115895624042362e+44 but correct answer is From eqn. get: 1574304985 <a name="docB11">&lt;a name="docB11"&gt;</a> <a href="#docB41">proof</a>, <a href="#program0">program</a>, <a href="#docB02">Conjecture</a>, <a href="#docB04">triangle</a> After add '9809031050' code, Math equation is (too long, omit) ***** correct equation ***** " i^(9) from i=1 to i=10. get 1574304985.0000004 correct answer is 1574304985 <a name="docB12">&lt;a name="docB12"&gt;</a> The reason that before 'update 2009-09-03' generate big error answer 8.115895624042362e+44 is that when request " i^(9) Arndt Brünner's work output to the following [[ coefficients from high power to constant 6164081965611661/61640819656134050, 5155004252058715/10310008504085338, 4908405681047445/6544540908389189, 2471137929879305/4,811589347427001e+24, -6252256789683410/8931795356868823, 766295064850303/3,0021419880618152e+22, 4994368678553267/9988739288667978, 1612987475556151/6,778429079461882e+21, -4555028353434601/30366782630979680, 6036212294932357/2,0070355172700415e+22, -8816619631634209/8,586325069295574e+22 ]] where comma in [...3, 7...] and comma in [8,586325069295574e+22] have different meaning. LiuHH did not pay attention to this difference. Use both comma as separator and cause big error. <a name="docB13">&lt;a name="docB13"&gt;</a> 2009-09-03-14-13 repeat words deleted, 2009-09-03-11-12 <a name="docB14">&lt;a name="docB14"&gt;</a> "Update 2009-09-04" add " i^(11) equation "Update 2009-09-04" add <a href="#huge_to_sml">huge to small</a> code. But no help. 2009-09-03-22-01 record "Update 2009-09-04" add " i^(12) equation. 2009-09-04-11-18 record "Update 2009-09-04" add <a href="#getCol5">column 5 calculator</a>. 2009-09-04-15-43 record [[ " i^(12) for i=1 to i=n coefficients from high power to constant 6291476941929729/81789200323615760, 2024007280027114/4048014431611459, 4889929392869138/4889934481241627, 2820674523245057/80225921439448660000, -12676584503961494/6911755417179109, 698226521061421/69727488742124080, 3033039204646851/995193537484100, 3919984334779285/6162049301525286, -44938138956937790/7138933101912657, 26108525088125380/2665864212641073, -8541165113894874/429708028862003, 156590788714843520/5221488374424287, -172590431458016930/7285082543775481, 41899526063808296/5482558770388439 Math equation is +n*n*n*n*n*n*n*n*n*n*n*n*n*(6291476941929729/81789200323615760)+n*n*n*n*n*n*n*n*n*n*n*n*(2024007280027114/4048014431611459)+n*n*n*n*n*n*n*n*n*n*n*(4889929392869138/4889934481241627)+n*n*n*n*n*n*n*n*n*n*(2820674523245057/80225921439448660000)+n*n*n*n*n*n*n*n*n*(-12676584503961494/6911755417179109)+n*n*n*n*n*n*n*n*(698226521061421/69727488742124080)+n*n*n*n*n*n*n*(3033039204646851/995193537484100)+n*n*n*n*n*n*(3919984334779285/6162049301525286)+n*n*n*n*n*(-44938138956937790/7138933101912657)+n*n*n*n*(26108525088125380/2665864212641073)+n*n*n*(-8541165113894874/429708028862003)+n*n*(156590788714843520/5221488374424287)+n*(-172590431458016930/7285082543775481)+(41899526063808296/5482558770388439) constant term 41899526063808296/5482558770388439 =7.642330491760459 ]] 2009-09-03-13-55 " i^(12) from i=1 to i=n Why constant is not zero ??!! dig !! <a name="docB15">&lt;a name="docB15"&gt;</a> 2009-09-03-20-00 LiuHH target at " i^(12) Why " i^(12) constant is not zero? Find " i^(11) first half try and error, half follow pattern start fron 2009-09-03-13-55 on 2009-09-03-19-58 found <a name="docB16">&lt;a name="docB16"&gt;</a> <a href="#docB41">proof</a>, <a href="#program0">program</a>, <a href="#docB02">Conjecture</a>, <a href="#docB04">triangle</a> " i^(11) Math equation is +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) +n*(0) +(0) " i^11 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) 2009-09-03-20-03 <a name="docB17">&lt;a name="docB17"&gt;</a> 2009-09-03-20-57 " i^(12) for i=1 to i=n is out of reach !!! Too much deviation from the pattern !! 2009-09-03-20-59 <a name="docB18">&lt;a name="docB18"&gt;</a> 2009-09-03-23-22 <a href="#docB04">Coefficient triangle</a> column 1,2,3,4 pattern are easy to observe. Observed pattern for column five is next. copy column five to below at left column -4/120 , 4+ 6=10 , 6<font color=red><b>+4</b></font>=10 -10/120 , 10+10=20 , 10<font color=red><b>+5</b></font>=15 -20/120 , 20+15=35 , 15<font color=red><b>+6</b></font>=21 -35/120 , 35+21=56 , 21<font color=red><b>+7</b></font>=28 -56/120 , 56+28=84 , 28<font color=red><b>+8</b></font>=36 -84/120 , 84+36=120 , 36<font color=red><b>+9</b></font>=45 !!! RULE 2009-09-03-23-27 -120/120 ,120+45=165 , <font color=red><b>+10</b></font> <==future '+10' -165/120 , <a name="docB19">&lt;a name="docB19"&gt;</a> 2009-09-03-23-38 for " i^(12) column five predict_step_1 45+10=55 predict_step_2 165+55=220 predict_step_3 -220/120 Observed rule for column five is CORRECT In LiuHH's search for " i^(12) work record show (-220/120) used +n*n*n*n*n*n*n*n*n*(-220/120) 2009-09-03-23-40 </font></pre> <a href="#program0">program</a> <a name="getCol5">&lt;a name="getCol5"&gt;</a> <font color=red>Do not confuse i with n</font> <!--200909061500--> <br> "i^5 [for i=1 to i=3] = 1^5+2^5+3^5 = 276 (n=3, k=5) <br> Above is numerical calculation use 1,2,3 fifth power. <br> Below is same problem formula calculation for n=3, k=5 <br> n^6*(1/6)+n^5*(1/2)+n^4*(5/12)+n*n*(-1/12) = 276 <br> Above is power=k=5 equation. It is true for any upper <br> bound n. If let n=3, above equation give answer 276. <br> [col 1-9] button give you <font color=red>first nine</font> <a href="#docB110">coefficients</a>. <br> 2009-09-04-15-19 A utility help decide <a href="#docB04">column 5</a> <script type="text/javascript"> <!-- //98,09,04,14,26 //write a program for column five //9809041426 function colFive() { rank0=boxCol5a.value; rank0=parseInt(rank0); if(isNaN(rank0)) return 'isNaN('+rank0+')'; if(rank0<4) return 'rank0='+rank0+', but >=4 required'; if(rank0==4) return '"i^4 column five coef. suggest: -4/120'; //if(rank0==5) return '-'+ (4+sumi2n(rank0-2,1)) +'/120'; var bgn0=4; //9809041442 //var ans1=0; var ans1=4; //9809041504 for(var i=4;i<rank0;i++) ans1+=sumi2n(i-1,1); span_eqn_to_n.innerHTML='<br>'; //9809041512 summation0(upper_bound_n.value,rank0) var k=rank0; //9809061132 return '"i^'+rank0+' column five coef. suggest: -'+ans1+'/120' + ' ; 2nd method : -'+(120*k*(k-1)*(k-2)/(3*3*4*4*5))+'/120' ; } //9809041506 success //9809052211 exact formula is // '-'+(120*k*(k-1)*(k-2)/(3*3*4*4*5))+'/120' /** ak+1=1/(k+1) ak =1/2 ak-1=k/12 ak-2=0 ak-3=-k*(k-1)*(k-2)/(5*4*3*2*3*2) ak-4=0 ak-5=k*(k-1)*(k-2)*(k-3)*(k-4)/(7*6*5*4*3*2*3*2) ak-6=0 ak-7= -k*(k-1)*(k-2)*(k-3)*(k-4)*(k-5)*(k-6)/(8*7*6*5*4*3*2*3*2*5) /**/ function column1to9(arg0) //9809061330 { var eqns=['', //9809061337 '1/(k+1)', '1/2', 'k/12', '0', '-k*(k-1)*(k-2)/(5*4*3*2*3*2)', '0', 'k*(k-1)*(k-2)*(k-3)*(k-4)/(7*6*5*4*3*2*3*2)', '0', '-k*(k-1)*(k-2)*(k-3)*(k-4)*(k-5)*(k-6)/(8*7*6*5*4*3*2*3*2*5)' ] var numerator0=['', //9809061400 '1', '1', 'k', '0', '-k*(k-1)*(k-2)', '0', 'k*(k-1)*(k-2)*(k-3)*(k-4)', '0', '-k*(k-1)*(k-2)*(k-3)*(k-4)*(k-5)*(k-6)' ]; var denominator0=['', //9809061405 '(k+1)', '2', '12', '', '720', '', '30240', '', '1209600' ]; var divider0=['', //9809061407 '/', '/', '/', '', '/', '', '/', '', '/' ]; power0=boxCol5a.value; power0=parseInt(power0); if(isNaN(power0)) return 'isNaN('+boxCol5a.value+')'; if(power0<=0) return 'power0='+power0+', but >=1 required'; var ans1=''; //9809061332 var limit0=eqns.length-1; //9809061341 var limit1=limit0; //9809061422 if(power0<limit1)limit1=power0; //9809061348 var k=power0; //9809061132 if(arg0==0) for(var i=1;i<=limit1+1;i++) { ans1+='n^'+(power0+2-i)+'*('+eval(eqns[i])+')<br>\n'; if(i==limit0)break; //9809061424 } else for(var i=1;i<=limit1+1;i++) { ans1+='n^'+(power0+2-i)+'*(' +eval(numerator0[i]) +divider0[i] +(i==1?k+1:denominator0[i]) +')<br>\n'; if(i==limit0)break; } if(power0>limit0) ans1+='<font color=red>More coefficient not included. Above equation not complete.</font><br>'; span_eqn_to_n.innerHTML='<br>'; //98090613 summation0(upper_bound_n.value,power0) //ans1+='i,limit0,limit1,power0='+i+', '+limit0+', '+limit1+', '+power0+'<br>' return ans1; } //9809061355 success , but decimal output. //9809061447 success , rational number output /**/ //--> </SCRIPT> <br> "i^( <INPUT value="7 k value" id="boxCol5a" size=5 ><!--9809041445--> ) &nbsp; <INPUT onclick='javascript:spanColAns.innerHTML=colFive()' type=button value="get Col 5"> &nbsp; <INPUT onclick='javascript:spanColAns.innerHTML=column1to9(0)' type=button value="Col 1-9 A"> &nbsp; <INPUT onclick='javascript:spanColAns.innerHTML=column1to9(1)' type=button value="Col 1-9 B"> &nbsp; <INPUT onclick='javascript:spanColAns.innerHTML="<br>",boxCol5a.value=""' type=button value="clear"> <br> <span id=spanColAns> <br> </span> <br> <pre><font size=+2><a name="docB20">&lt;a name="docB20"&gt;</a> <a href="#docB41">proof</a>, <a href="#program0">program</a>, <a href="#docB02">Conjecture</a>, <a href="#docB04">triangle</a> 2009-09-04-10-46 Motivation for finding " i^(12) equation is that Arndt Brünner's output for " i^(12) equation has non-zero constant. It get reasonable answer. So the conjecture that <a href="#docB02">constant be zero</a> break down? LiuHH hope to find the answer. Dug for " i^(11) equation then <a href="#docB15">" i^(12) equation</a>. Final answer is that " i^(12) equation constant is zero. <a name="docB21">&lt;a name="docB21"&gt;</a> Find " i^(12) equation, first follow the pattern and guess coefficient constant. Failed. Twice decide give up. The success path is the following. Build full matrix equation. <a name="docB22">&lt;a name="docB22"&gt;</a> 2009-09-04-08-06 " i^(12) a,b,c,d,e,f,g,h,i,j,k,l,m,n 14 variables [ 1 1 1 1 1 1 1 1 1 1 1 1 1 1][a] [ 1] [ 2^13 2^12 2^11 2^10 2^9 2^8 2^7 2^6 2^5 2^4 2^3 2*2 2 1][b]=[ 4097] ---eqn.AA101 [ 3^13 3^12 3^11 3^10 3^9 3^8 3^7 3^6 3^5 3^4 3^3 3*3 3 1][c] [ 535538] [ 4^13 4^12 4^11 4^10 4^9 4^8 4^7 4^6 4^5 4^4 4^3 4*4 4 1][d] [ 17312754] [ 5^13 5^12 5^11 5^10 5^9 5^8 5^7 5^6 5^5 5^4 5^3 5*5 5 1][e] [ 261453379] [ 6^13 6^12 6^11 6^10 6^9 6^8 6^7 6^6 6^5 6^4 6^3 6*6 6 1][f] [2438235715] [ 7^13 7^12 7^11 7^10 7^9 7^8 7^7 7^6 7^5 7^4 7^3 7*7 7 1][g] [16279522916] [ 8^13 8^12 8^11 8^10 8^9 8^8 8^7 8^6 8^5 8^4 8^3 8*8 8 1][h] [84998999652] [ 9^13 9^12 9^11 9^10 9^9 9^8 9^7 9^6 9^5 9^4 9^3 9*9 9 1][i] [367428536133] [ 10^13 10^12 10^11 10^10 10^9 10^8 10^7 10^6 10^5 10^4 10^3 10*10 10 1][j] [1367428536133] [ 11^13 11^12 11^11 11^10 11^9 11^8 11^7 11^6 11^5 11^4 11^3 11*11 11 1][k] [4505856912854] [ 12^13 12^12 12^11 12^10 12^9 12^8 12^7 12^6 12^5 12^4 12^3 12*12 12 1][l] [13421957361110] [ 13^13 13^12 13^11 13^10 13^9 13^8 13^7 13^6 13^5 13^4 13^3 13*13 13 1][m] [36720042483591] [ 14^13 14^12 14^11 14^10 14^9 14^8 14^7 14^6 14^5 14^4 14^3 14*14 14 1][n] [93413954858887] 2009-09-04-08-29 <a name="docB23">&lt;a name="docB23"&gt;</a> No need to solve whole equation, because [[ a=1/13 b=1/2 c=12/12 d=0 e=-220/120 f=0 g=w=792/252 h=0 i=x=-396/120 j=0 k=y=2200/1320 l=0 m=z=-7601/30030 n=0 2009-09-04-10-15 ]] a=1/13 b=1/2 c=12/12 d=0 f=0 h=0 j=0 l=0 n=0 are described in <a href="#docB02">Conjectures</a> <a href="#docB19">e=-220/120</a> is described <a href="#docB18">here</a>. only w,x,y,z four coefficients are uncertain. <a name="docB24">&lt;a name="docB24"&gt;</a> Build four equations [[ 2009-09-04-09-40 1/13 +1/2 +1 -1*(220/120) 1*w 1*x 1*y 1*z = 1 2^13/13 +2^12/2 +2^11 -2^9*(220/120) 2^7*w 2^5*x 2^3*y 2*z = 4097 (=1^12+2^12) 3^13/13 +3^12/2 +3^11 -3^9*(220/120) 3^7*w 3^5*x 3^3*y 3*z = 535538 (=1^12+2^12+3^12) 4^13/13 +4^12/2 +4^11 -4^9*(220/120) 4^7*w 4^5*x 4^3*y 4*z = 17312754 (=1^12+2^12+3^12+4^12) [=1+pow(2,12)+pow(3,12)+pow(4,12)] 2009-09-04-09-58 1*w+ 1*x+ 1*y+ 1*z = 1.2564102564102564 128*w+ 32*x+ 8*y+ 2*z = 309.5128205128208 2187*w+ 243*x+ 27*y+ 3*z = 6115.769230769249 16384*w+1024*x+ 64*y+ 4*z = 48219.02564102411 ]] <a name="docB25">&lt;a name="docB25"&gt;</a> Ask Arndt Brünner for help, get solution. [[ 2009-09-04-09-59 w = 9104177589384788/2896783778440913 x = -17220659735585978/5218381738063757 y = 9187617697792860/5512570618730461 z = -768920039355504/3037846175892421 ]] <a name="docB26">&lt;a name="docB26"&gt;</a> <a href="#docB41">proof</a>, <a href="#program0">program</a>, <a href="#docB02">Conjecture</a>, <a href="#docB04">triangle</a> w is column 7. Previous work, lower rank equation column 7 denominator is 252. then w = 9104177589384788/2896783778440913 become g=w=792/252 Similarly refer to previous denominator find i=x=-396/120 k=y=2200/1320 <a name="docB27">&lt;a name="docB27"&gt;</a> Only m=z=-7601/30030 no previous denominator. How to find z ? (coef. m=z unknown) Working record is next [[ 2009-09-04-10-05 1/13+1/2+1-220/120+792/252-396/120+2200/1320+z=1 z=1-1/13-1/2-1+220/120-792/252+396/120-2200/1320 z=-0.2531135531135533 <a name="docB28">&lt;a name="docB28"&gt;</a> 1320=10*11*12 120=10*12 252=3*12*7 10*11*12*13*7=120120 z=-30404/120120 2009-09-04-10-12 z=-7601/30030 ]] Up to here, general equation for " i^(12) is solved. 2009-09-04-11-12 </font></pre> <a name="huge_to_sml">&lt;a name="huge_to_sml"&gt;</a> <a href="#docB41">proof</a>, <a href="#program0">program</a>, <a href="#docB02">Conjecture</a>, <a href="#docB04">triangle</a> <br> 2009-09-03-14-16 <br> " i^(k) when k>8, program do not output small integer rational number. But output huge integer rational number. Box below try to find equivalent small integer rational number (polynomial coefficients) <br> huge rational: <INPUT value="5174582439943824/2819723675734019" id="boxDig11a" size=50 ><!--9809031420--> <br> Iterations: &nbsp; &nbsp;&nbsp; <INPUT value="1000" id="boxDig11b" size=5 ><!--9809031446--> &nbsp; <INPUT onclick='javascript:huge2sml()' type=button value="huge to small"> <br> Box 2 Output &nbsp; <!--9809031422--> <INPUT onclick='javascript:window.clipboardData.setData("Text",document.getElementById("boxc02").value.toString())' type=button value="copy2"> <INPUT onclick='document.getElementById("boxc02").value=""' type=button value="del_2"> <br> <TEXTAREA id=boxc02 name=boxc02 rows=10 cols=60 > </TEXTAREA> <br> Box2 not really help, if denominator blur a little bit, <br> Program will not return to the right denominator. 2009-09-03-20-32 <br> <pre><font size=+2><a name="docB29">&lt;a name="docB29"&gt;</a> 2009-09-04-21-50 Following is a small calculation for " i^3 to n = +n*n*n*n*(1/4)+n*n*n*(1/2)+n*n*(1/4) ---eqn.AA102 case, Find out coefficient values. <font color=red>The key point to get answer is that n=arbitrary.</font> Following is working record. 2009-09-04-21-53 <a name="docB30">&lt;a name="docB30"&gt;</a> 2009-09-04-21-12 for power=3 " i^3 to n = +n*n*n*n*(1/4)+n*n*n*(1/2)+n*n*(1/4) SUM i*i = n*n*n*n*a + n*n*n*b + n*n*c + n*d + e =1^3 + 2^3 + 3^3 + 4^3 + .... + n^3 ---eqn.AA103 why a=1/4 ? why b=1/2 ? big puzzle. <a name="docB31">&lt;a name="docB31"&gt;</a> <a href="#docB41">proof</a>, <a href="#program0">program</a>, <a href="#docB02">Conjecture</a>, <a href="#docB04">triangle</a> <font color=red> KEY observation 1^3 +...+ n^3 use n*n*n*n*a + n*n*n*b + n*n*c + n*d + e ---eqn.AA104 1^3 +...+ (n-1)^3 use (n-1)*(n-1)*(n-1)*(n-1)*a + (n-1)*(n-1)*(n-1)*b + (n-1)*(n-1)*c + (n-1)*d + e ---eqn.AA105 </font> <a name="docB32">&lt;a name="docB32"&gt;</a> move n^3 from right hand side to left hand side. n*n*n*n*a + n*n*n*(b-1) + n*n*c + n*d + e ---eqn.AA106 =1^3 + 2^3 + 3^3 + 4^3 + .... = use KEY observation at this step = (n-1)*(n-1)*(n-1)*(n-1)*a ---eqn.AA107 + (n-1)*(n-1)*(n-1)*b + (n-1)*(n-1)*c + (n-1)*d + e = n*n*n*n*a + n*n*n*(b-1) + n*n*c + n*d + e ---eqn.AA106 e cancel here , e is undetermined here <a name="docB33">&lt;a name="docB33"&gt;</a> 2009-09-04-21-18 here (n-1)*(n-1)*(n-1)*(n-1)*a + (n-1)*(n-1)*(n-1)*b + (n-1)*(n-1)*c + (n-1)*d = ---eqn.AA108 n*n*n*n*a + n*n*n*(b-1) + n*n*c + n*d <a name="docB34">&lt;a name="docB34"&gt;</a> 2009-09-04-21-20 [[ n=18 a=1/4 b=1/2 c=3/12 d=0 (n-1)*(n-1)*(n-1)*(n-1)*a + (n-1)*(n-1)*(n-1)*b + (n-1)*(n-1)*c + (n-1)*d -n*n*n*n*a - n*n*n*(b-1) - n*n*c - n*d //---eqn.AA109 ]] Goto <a href=http://freeman2.com/complex2.htm#calculator>freeman2.com</a>/<a href=complex2.htm#calculator>complex2.htm</a> paste above six lines to box3, change value of n, answer is always zero. <a name="docB35">&lt;a name="docB35"&gt;</a> eqn.AA109 is zero for ANY n (n*n-2*n+1)*(n*n-2*n+1)*a + (n*n-2*n+1)*(n-1)*b + (n*n-2*n+1)*c + (n-1)*d -n*n*n*n*a - n*n*n*(b-1) - n*n*c - n*d ---eqn.AA110 =0 = <a name="docB36">&lt;a name="docB36"&gt;</a> <a href="#docB41">proof</a>, <a href="#program0">program</a>, <a href="#docB02">Conjecture</a>, <a href="#docB04">triangle</a> (n*n*n*n-2*n*n*n+n*n -2*n*n*n+2*2*n*n-2*n*1 n*n-2*n+1 )*a + ( n*n*n-2*n*n+1*n -n*n+2*n-1 )*b + (n*n-2*n+1)*c + (n-1)*d -n*n*n*n*a - n*n*n*(b-1) - n*n*c - n*d ---eqn.AA111 <a name="docB37">&lt;a name="docB37"&gt;</a> =0 = n*n*n*n*(a-a) +n*n*n*(-2a-2a+b-b+1) +n*n*(6a-3b+c-c) +n*(-4a+3b-2c+d-d) +a-b+c-d ---eqn.AA112 <a name="docB38">&lt;a name="docB38"&gt;</a> 2009-09-04-21-34 since n is arbitrary must its coef be zero, get n*n*n*n*(a-a) ==> a-a=0 no help ---eqn.AA113 +n*n*n*(-2a-2a+b-b+1) ==> -2a-2a+b-b+1=0 ---eqn.AA114 +n*n*(6a-3b+c-c) ==> 6a-3b+c-c=0 ---eqn.AA115 +n*(-4a+3b-2c+d-d) ==> -4a+3b-2c+d-d=0 ---eqn.AA116 +a-b+c-d =0 ---eqn.AA117 <a name="docB39">&lt;a name="docB39"&gt;</a> 2009-09-04-21-36 here eqn.AA114: -2a-2a+1=0 ==> a=1/4 ---eqn.AA118 eqn.AA115: 6a-3b =0 ==> b=2a=2/4=1/2 ---eqn.AA119 eqn.AA116: -4a+3b-2c=0 ==> 2c=-4a+3b ==> 2c=-4/4+3/2=1/2 ==> c=1/4 ---eqn.AA120 eqn.AA117: +a-b+c-d =0 ==> d=a-b+c ==> d=1/4-1/2+1/4=0 ---eqn.AA121 <a name="docB40">&lt;a name="docB40"&gt;</a> Now find e SUM i*i = n*n*n*n*a + n*n*n*b + n*n*c + n*d + e since n is arbitrary, and solution is unique it must be e=0 ---eqn.AA122 Put all coefficient together get " i^3 to n = +n*n*n*n*(1/4)+n*n*n*(1/2)+n*n*(1/4) Above reasoning is special case for power=3 not general for power=any power=any derivation is similar. 2009-09-04-21-47 <hr> <a name="docB41">&lt;a name="docB41"&gt;</a> <font color=red size=+3><b>The following is proof section.</b></font> Update 2009-09-05 add the following proof. Proof coef. column <a href=jspico_e.htm#docB77>01</a> <a href=jspico_e.htm#docB82>02</a> <a href=jspico_e.htm#docB87>03</a> <a href=jspico_e.htm#docB100>04</a> <a href=jspico_e.htm#docB109>05</a> <a href=jspico_e.htm#docB110>List</a> <a href="#docB41">proof</a>, <a href="#program0">program</a>, <a href="#docB02">Conjecture</a>, <a href="#docB04">triangle</a> 2009-09-05-08-32 Please view <a href="#docB04">coefficient triangle</a>. Column 01 coefficient is 1/(power+1) Column 02 coefficient is 1/2 Column 03 coefficient is power/12 etc. <a name="docB42">&lt;a name="docB42"&gt;</a> Is this pattern a coincidence? Is this pattern a must? If it is a must result, how to prove? The following is an explanation. <a name="docB43">&lt;a name="docB43"&gt;</a> First observe a special case " i^3 from i=1 to n Answer from equation, use formula " i^3 = ... = n*n*n*n*(1/4)+n*n*n*(1/2)+n*n*(1/4) ---eqn.AA201 calculate only i=n th term. from n_power_0 to n_power_k+1 (here k=3) Formula contain coefficients (1/4), (1/2), (1/4). For large n, formula method is quick. <a name="docB44">&lt;a name="docB44"&gt;</a> In contrast, Numerical calculation, use next equation 1^3 + 2^3 + 3^3 + 4^3 + ..... + n^3 ---eqn.AA202 Answer from summation calculate only 3 rd power. from i=1 to i=n This summation do not use coefficients. For large n, summation method is slow. <a name="docB45">&lt;a name="docB45"&gt;</a> Calculation use only n th term and calculation use only 3 rd power are very different. They get same answer! For " i^3 we write eqn.AA201=eqn.AA202 as below +n*n*n*n*(1/4)+n*n*n*(1/2)+n*n*(1/4) = ---eqn.AA203 1^3 + 2^3 + 3^3 + 4^3 + ..... + n^3 2009-09-05-08-59 here <a name="docB46">&lt;a name="docB46"&gt;</a> <a href="#docB41">proof</a>, <a href="#program0">program</a>, <a href="#docB02">Conjecture</a>, <a href="#docB04">triangle</a> Above equation is for third power " i^3. For general case " i^k , write as next Answer from equation, use formula " i^k = ... "[i=1 to n] i^k = "[q=0,k+1]pow(n,q)*a<sub>q</sub> ---eqn.AA204 Answer from summation use loop sum "[i=1 to n] i^k = "[i=1,n]i^k ---eqn.AA205 <a name="docB47">&lt;a name="docB47"&gt;</a> Since eqn.AA204 is general formula, it cover the case k=3. Check if this is true. In eqn.AA204 set k=3 get "[i=1 to n] i^3 = "[q=0,3+1]pow(n,q)*a<sub>q</sub> expand "[q=0,3+1] get "[i=1 to n] i^3 = pow(n,0)*a<sub>0</sub> + pow(n,1)*a<sub>1</sub> + pow(n,2)*a<sub>2</sub> + pow(n,3)*a<sub>3</sub> + pow(n,4)*a<sub>4</sub> <a name="docB48">&lt;a name="docB48"&gt;</a> It is same as next more familiar form "[i=1 to n] i^3 = n^0*a<sub>0</sub> + n^1*a<sub>1</sub> + n^2*a<sub>2</sub> + n^3*a<sub>3</sub> + n^4*a<sub>4</sub> ---eqn.AA206 for ""[i=1 to n] i^3" coefficients are a<sub>4</sub>=1/4, a<sub>3</sub>=1/2, a<sub>2</sub>=1/4, a<sub>1</sub>=0, a<sub>0</sub>=0 Then eqn.AA206 is same as <a href="#docB43">eqn.AA201</a>. <a name="docB49">&lt;a name="docB49"&gt;</a> How to find a<sub>4</sub>=1/4 etc. ? Please goto <a href="#program0">program</a> input '3' for third power summation, then click [calculate equation] button. Output to box1. Above checked eqn.AA204 it cover answer from equation power=3 case <a name="docB50">&lt;a name="docB50"&gt;</a> Below check eqn.AA205 see whether it cover answer from summation power=3 case "[i=1 to n] i^k = "[i=1,n]i^k ---eqn.AA205 In eqn.AA205, set k=3 for power=3 case, get "[i=1 to n] i^3 = "[i=1,n]i^3 = 1^3 + 2^3 + 3^3 + ..... + n^3 It is same as <a href="#docB44">eqn.AA202</a> Now "eqn.AA204 and eqn.AA205 is general" is true for k=3 case. For k=other_value, exam procedure is the same. 2009-09-05-09-33 here <a name="docB51">&lt;a name="docB51"&gt;</a> <a href="#docB41">proof</a>, <a href="#program0">program</a>, <a href="#docB02">Conjecture</a>, <a href="#docB04">triangle</a> eqn.AA204 and eqn.AA205 get same answer for power k. We can write eqn.AA204 = eqn.AA205 General equation for positive integer power sum. "[i=1 to n] i^k = "[q=0,k+1]pow(n,q)*a<sub>q</sub> = "[i=1,n]i^k ---eqn.AA207 <a name="docB52">&lt;a name="docB52"&gt;</a> eqn.AA207 is general, eqn.AA207 is true for sum to n eqn.AA207 is true for sum to n+1 too We have next equation for sum to n+1 "[i=1 to n+1] i^k = "[q=0,k+1]pow(n+1,q)*a<sub>q</sub> = "[i=1,n+1]i^k ---eqn.AA208 <a name="docB53">&lt;a name="docB53"&gt;</a><font color=red> To prove Column 01 coefficient is 1/(power+1) Column 02 coefficient is 1/2 Column 03 coefficient is power/12 etc. key step is to relate eqn.AA207 and eqn.AA208 </font> Clearly eqn.AA207 NOT EQUAL eqn.AA208 eqn.AA207 sum from 1^k up to n^k eqn.AA208 sum from 1^k up to (n+1)^k Both equation sum to power k. (200909061142) <a name="docB54">&lt;a name="docB54"&gt;</a> Two answer from equations are hard to relate, because expansion of "[q=0,k+1]pow(n+1,q) is complicate Two answer from summations are next "[i=1,n]i^k ---eqn.AA207 right hand side "[i=1,n+1]i^k ---eqn.AA208 right hand side What a simple job ?! <a name="docB55">&lt;a name="docB55"&gt;</a> Sum to n+1 is "[i=1,n+1]i^k = "[i=1,n]i^k + (n+1)^k from equation right hand side we get eqn.AA208 = eqn.AA207 + (n+1)^k then goto equation left hand side, write <font color=red> "[q=0,k+1]pow(n+1,q)*a<sub>q</sub> = ... ---eqn.AA209 = "[q=0,k+1]pow(n,q)*a<sub>q</sub> + (n+1)^k</font> 2009-09-05-10-12 here <a name="docB56">&lt;a name="docB56"&gt;</a> <a href="#docB41">proof</a>, <a href="#program0">program</a>, <a href="#docB02">Conjecture</a>, <a href="#docB04">triangle</a> Above walk long way, not start yet. eqn.AA209 is our start equation. Slow part in eqn.AA209 is expansion of pow(n+1,q). Assume q=3 then pow(n+1,3) is next (n+1)^3 = (n+1)*(n+1)*(n+1) = 1*n*n*n+3*n*n+3*n+1 <a name="docB57">&lt;a name="docB57"&gt;</a> Define bino(s,t) as following.</font></pre> <TABLE WIDTH="460" id=T201Liua border=0 ><!--9809051021--> <TR> <TD> &nbsp; </TD> <TD> <TABLE WIDTH="260" border=0 id=T201Liub > <TR> <TD> bino(s,t) := </TD> <TD> <CENTER><FONT SIZE=+3>(</FONT></CENTER> </TD> <TD> <CENTER>s</CENTER> <CENTER>0</CENTER> <CENTER>t</CENTER> </TD> <TD> <CENTER><FONT SIZE=+3>)</FONT></CENTER> </TD> <TD> ÿ </TD> <TD> <CENTER>s!</CENTER> <CENTER><hr color=black></CENTER> <CENTER>t!(s-t)!</CENTER> </TD> </TR> </TABLE> </TD> <TD> ---binomial definition <br> ---eqn.AA210 </TD> </TR> </TABLE> width of above equation <INPUT name=Button201Liua type="button" value="default" onclick="B201Liua.value=460,B201Liub.value=260,T201Liua.width=B201Liua.value,T201Liub.width=B201Liub.value"> <input id="B201Liua" value=460 size=3 onchange=T201Liua.width=B201Liua.value /> <input id="B201Liub" value=260 size=3 onchange=T201Liub.width=B201Liub.value /> <pre><font size=+2><a name="docB58">&lt;a name="docB58"&gt;</a> 2009-09-05-10-30 here Then for q=3 pow(n+1,q) = (n+1)^3 = (n+1)*(n+1)*(n+1) can be written as pow(n+1,3) = bino(3,3)*n^3 + bino(3,2)*n^2 + bino(3,1)*n^1 + bino(3,0)*n^0 Please goto <a href=http://freeman2.com/tute0010.htm#factorial>freeman2.com</a>/<a href=tute0010.htm#factorial>tute0010.htm</a> in "binomial m[ ], n[ ]" box, fill '3' for m and '2' for n click [get binomial] button for answer. <a name="docB59">&lt;a name="docB59"&gt;</a> Numerical value is pow(n+1,3) = 1*n^3 + 3*n^2 + 3*n^1 + 1*n^0 Numerical value is good for understanding. Symbol equation is good for proof for general case. <a name="docB60">&lt;a name="docB60"&gt;</a> Now back to <a href="#docB55">eqn.AA209</a> <font color=red> "[q=0,k+1]pow(n+1,q)*a<sub>q</sub> = ... ---eqn.AA209 = "[q=0,k+1]pow(n,q)*a<sub>q</sub> + (n+1)^k</font> Expand get pow(n+1,k+1)*a<sub>k+1</sub> +pow(n+1,k)*a<sub>k</sub> +pow(n+1,k-1)*a<sub>k-1</sub> +pow(n+1,k-2)*a<sub>k-2</sub> + "[q=0,k-3]pow(n+1,q)*a<sub>q</sub> = ---eqn.AA211 (whole eqn.) (n+1)^k + pow(n,k+1)*a<sub>k+1</sub> + pow(n,k)*a<sub>k</sub> + pow(n,k-1)*a<sub>k-1</sub> + pow(n,k-2)*a<sub>k-2</sub> + "[q=0,k-3]pow(n,q)*a<sub>q</sub> 2009-09-05-10-48 here <a name="docB61">&lt;a name="docB61"&gt;</a> <a href="#docB41">proof</a>, <a href="#program0">program</a>, <a href="#docB02">Conjecture</a>, <a href="#docB04">triangle</a> eqn.AA211 both side let highest four terms stand out of summation sign. eqn.AA211 left hand side is (n+1)^power terms need to use binomial coef. to expand. eqn.AA211 right hand side is (n)^power terms no need expansion. <a name="docB62">&lt;a name="docB62"&gt;</a> Do expansion one by one pow(n+1,k+1)*a<sub>k+1</sub> = ---eqn.AA212 (one term) [ bino(k+1,k+1)*pow(n,k+1) +bino(k+1,k) *pow(n,k) +bino(k+1,k-1)*pow(n,k-1) +bino(k+1,k-2)*pow(n,k-2) ] *a<sub>k+1</sub> + "[q=0,k-3]bino(k+1,q)*pow(n,q)*a<sub>k+1</sub> <a name="docB63">&lt;a name="docB63"&gt;</a> Next +pow(n+1,k)*a<sub>k</sub> = ---eqn.AA213 (one term) [ bino(k,k) *pow(n,k) +bino(k,k-1)*pow(n,k-1) +bino(k,k-2)*pow(n,k-2) ] *a<sub>k</sub> + "[q=0,k-3]bino(k,q)*pow(n,q)*a<sub>k</sub> <a name="docB64">&lt;a name="docB64"&gt;</a> Next +pow(n+1,k-1)*a<sub>k-1</sub> = ---eqn.AA214 (one term) [ bino(k-1,k-1)*pow(n,k-1) +bino(k-1,k-2)*pow(n,k-2) ] *a<sub>k-1</sub> + "[q=0,k-3]bino(k,q)*pow(n,q)*a<sub>k-1</sub> <a name="docB65">&lt;a name="docB65"&gt;</a> Next is the special term which link "[i=1,n]i^k and "[i=1,n+1]i^k expand (n+1)^k = ---eqn.AA215 (one term) bino(k,k)*pow(n,k) +bino(k,k-1)*pow(n,k-1) +bino(k,k-2)*pow(n,k-2) + "[q=0,k-3]bino(k,q)*pow(n,q) this term (n+1)^k no a<sub>k</sub> coef. <a name="docB66">&lt;a name="docB66"&gt;</a> <a href="#docB41">proof</a>, <a href="#program0">program</a>, <a href="#docB02">Conjecture</a>, <a href="#docB04">triangle</a> Now put eqn.AA212 (one term) eqn.AA213 (one term) eqn.AA214 (one term) eqn.AA215 (one term) back to eqn.AA211 (whole eqn.) get <a name="docB67">&lt;a name="docB67"&gt;</a> [ bino(k+1,k+1)*pow(n,k+1) +bino(k+1,k) *pow(n,k) +bino(k+1,k-1)*pow(n,k-1) +bino(k+1,k-2)*pow(n,k-2) ] *a<sub>k+1</sub> + "[q=0,k-3]bino(k+1,q)*pow(n,q)*a<sub>k+1</sub> + [ bino(k,k) *pow(n,k) +bino(k,k-1)*pow(n,k-1) +bino(k,k-2)*pow(n,k-2) ] *a<sub>k</sub> + "[q=0,k-3]bino(k,q)*pow(n,q)*a<sub>k</sub> + [ bino(k-1,k-1)*pow(n,k) +bino(k-1,k-2)*pow(n,k-1) ] *a<sub>k-1</sub> + "[q=0,k-3]bino(k,q)*pow(n,q)*a<sub>k-1</sub> <!-- + "[q=0,k-2]pow(n+1,q)*a<sub>q</sub> 9809051546 replace above to next two lines--> + bino(k-2,k-2)*pow(n,k-2)*a<sub>k-2</sub> + "[q=0,k-3]bino(k-2,q)*pow(n,q)*a<sub>k-2</sub> <a name="docB68">&lt;a name="docB68"&gt;</a> = ---eqn.AA216 (whole eqn.) bino(k,k)*pow(n,k) +bino(k,k-1)*pow(n,k-1) +bino(k,k-2)*pow(n,k-2) + "[q=0,k-3]bino(k,q)*pow(n,q) + pow(n,k+1)*a<sub>k+1</sub> + pow(n,k)*a<sub>k</sub> + pow(n,k-1)*a<sub>k-1</sub> + pow(n,k-2)*a<sub>k-2</sub> + "[q=0,k-3]pow(n,q)*a<sub>q</sub> <a name="docB69">&lt;a name="docB69"&gt;</a> 2009-09-05-11-08 here <font color=red> The important point for eqn.AA216 (whole eqn.) is that eqn.AA216 satisfy n=arbitrary_value for this reason n^(k+1) coefficient MUST BE ZERO n^(k) coefficient MUST BE ZERO n^(k-1) coefficient MUST BE ZERO etc.</font> <a name="docB70">&lt;a name="docB70"&gt;</a> We need group coefficients together rewrite eqn.AA216 (whole eqn.) move everything to one side. The other side is '=0', get the following <a name="docB71">&lt;a name="docB71"&gt;</a> <a href="#docB41">proof</a>, <a href="#program0">program</a>, <a href="#docB02">Conjecture</a>, <a href="#docB04">triangle</a> [ ---eqn.AA217 (whole eqn.) START (<a href="#docB73">END</a>) bino(k+1,k+1)*pow(n,k+1) <font size=+2 color=red><b>+bino(k+1,k) *pow(n,k)</b></font> <font size=+2 color=blue><b>+bino(k+1,k-1)*pow(n,k-1)</b></font> <font size=+2 color=purple><b>+bino(k+1,k-2)*pow(n,k-2)</b></font> ] *a<sub>k+1</sub> + "[q=0,k-3]bino(k+1,q)*pow(n,q)*a<sub>k+1</sub> + [ <font size=+2 color=red><b> bino(k,k) *pow(n,k)</b></font> <font size=+2 color=blue><b>+bino(k,k-1)*pow(n,k-1)</b></font> <font size=+2 color=purple><b>+bino(k,k-2)*pow(n,k-2)</b></font> ] *a<sub>k</sub> + "[q=0,k-3]bino(k,q)*pow(n,q)*a<sub>k</sub> <a name="docB72">&lt;a name="docB72"&gt;</a> + [ <font size=+2 color=blue><b> bino(k-1,k-1)*pow(n,k-1)</b></font> <font size=+2 color=purple><b>+bino(k-1,k-2)*pow(n,k-2)</b></font> ] *a<sub>k-1</sub> + "[q=0,k-3]bino(k,q)*pow(n,q)*a<sub>k-1</sub> <font size=+2 color=purple><b>+ bino(k-2,k-2)*pow(n,k-2)*a<sub>k-2</sub></b></font> + "[q=0,k-3]bino(k-2,q)*pow(n,q)*a<sub>k-1</sub> <a name="docB73">&lt;a name="docB73"&gt;</a> <font size=+2 color=red><b>- bino(k,k)*pow(n,k)</b></font> <font size=+2 color=blue><b>-bino(k,k-1)*pow(n,k-1)</b></font> <font size=+2 color=purple><b>-bino(k,k-2)*pow(n,k-2)</b></font> - "[q=0,k-3]bino(k,q)*pow(n,q) - pow(n,k+1)*a<sub>k+1</sub> <font size=+2 color=red><b>- pow(n,k)*a<sub>k</sub></b></font> <font size=+2 color=blue><b>- pow(n,k-1)*a<sub>k-1</sub></b></font> <font size=+2 color=purple><b>- pow(n,k-2)*a<sub>k-2</sub></b></font> - "[q=0,k-3]pow(n,q)*a<sub>q</sub> = ZERO ---eqn.AA217 (whole eqn.) END (<a href="#docB71">START</a>) <a name="docB74">&lt;a name="docB74"&gt;</a> from eqn.AA217 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. <a name="docB75">&lt;a name="docB75"&gt;</a> Next from eqn.AA217 collect coefficients for second highest term (bold red terms) <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> -bino(k,k) - a<sub>k</sub> ] </b></font> 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] <a name="docB76">&lt;a name="docB76"&gt;</a> <a href="#docB41">proof</a>, <a href="#program0">program</a>, <a href="#docB02">Conjecture</a>, <a href="#docB04">triangle</a> 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<sub>k+1</sub> = 1/bino(k+1,k) = 1/(k+1) ---eqn.AA219 <a name="docB77">&lt;a name="docB77"&gt;</a> <font color=red size=+3>Up to here Proved Coefficient <a href="#docB04">triangle</a> column 01 coefficient = 1/bino(k+1,k) column 01 coefficient = 1/(k+1)</font> 2009-09-05-11-39 stop <a name="docB78">&lt;a name="docB78"&gt;</a> 2009-09-05-13-00 start Above proved column 01 coefficient = 1/(k+1) Next prove column 02 coefficient = 1/2 From eqn.AA217 collect coefficients for third highest term (bold blue terms) <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> -bino(k,k-1) - a<sub>k-1</sub> ] </b></font> Where bino(k-1,k-1)=1 and a<sub>k-1</sub> drop left <a name="docB79">&lt;a name="docB79"&gt;</a> 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 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 <a name="docB80">&lt;a name="docB80"&gt;</a> 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="docB81">&lt;a name="docB81"&gt;</a> <a href="#docB41">proof</a>, <a href="#program0">program</a>, <a href="#docB02">Conjecture</a>, <a href="#docB04">triangle</a> <a name="docB71">eqn.AA217</a> demand equal to zero for ANY n. It is possible only if coefficients are all zero. We require eqn.AA224 equal to zero, get +k/2 +k*a<sub>k</sub> -k = 0 then a<sub>k</sub>=1/2 ---eqn.AA225 <a name="docB82">&lt;a name="docB82"&gt;</a> <font color=red size=+3>Up to here Proved Coefficient <a href="#docB04">triangle</a> column 02 coefficient = 1/2</font> 2009-09-05-13-23 here Above proved column 01 coefficient = 1/(k+1) Above prove column 02 coefficient = 1/2 <a name="docB83">&lt;a name="docB83"&gt;</a> Next prove column 03 coefficient = k/12 From eqn.AA217 collect coefficients for fourth highest term (bold purple terms) <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> -bino(k,k-2) -a<sub>k-2</sub> ] </b></font> <a name="docB84">&lt;a name="docB84"&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> ] <a name="docB85">&lt;a name="docB85"&gt;</a> Here +bino(k-2,k-2)*a<sub>k-2</sub>-a<sub>k-2</sub> cancel Demand coefficient to be zero get +(k+1)!/[(k-2)!*3!]*1/(k+1) +k!/[(k-2)!*2!]*1/2 +(k-1)*a<sub>k-1</sub> -k!/[(k-2)!*2!] =0 <a name="docB86">&lt;a name="docB86"&gt;</a> <a href="#docB41">proof</a>, <a href="#program0">program</a>, <a href="#docB02">Conjecture</a>, <a href="#docB04">triangle</a> +(k-1)*a<sub>k-1</sub> = k!/[(k-2)!*2!] -(k+1)!/[(k-2)!*3!]*1/(k+1) -k!/[(k-2)!*2!]*1/2 = k(k-1)/2 -k(k-1)/6 -k(k-1)/4 = k(k-1)/12 (k-1)*a<sub>k-1</sub> = k(k-1)/12 a<sub>k-1</sub> = k/12 ---eqn.AA227 <a name="docB87">&lt;a name="docB87"&gt;</a> <font color=red size=+3>Up to here Proved Coefficient <a href="#docB04">triangle</a> column 03 coefficient = k/12</font> Other coefficients can be build with same method. 2009-09-05-13-59 stop <a name="docB88">&lt;a name="docB88"&gt;</a> "Update 2009-09-06" add following few more calculations and add <a href="#getCol5">first9</a> coefficients. 2009-09-06-12-20 document start Few more calculations. Next is Coefficient <a href="#docB04">triangle</a> column 04 expect get all zero. Equations below is 2009-09-05 draft work. On 2009-09-06 include draft work to web page and put explanation in web page. <a name="docB89">&lt;a name="docB89"&gt;</a> 2009-09-05-21-04 From general equation <a href=#docB55>eqn.AA209</a> or from expanded whole <a href=#docB60>eqn.AA211</a> or from detailed whole <a href=#docB71>eqn.AA217</a> Collect n^(k-3) terms. Need its coefficient, set coef to zero to find a<sub>k-2</sub> <a name="docB90">&lt;a name="docB90"&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="docB91">&lt;a name="docB91"&gt;</a> <a href="#docB41">proof</a>, <a href="#program0">program</a>, <a href="#docB02">Conjecture</a>, <a href="#docB04">triangle</a> All 'bino()' come from (n+1)^k expansion.<a href=#docB60> Other side</a> of general equation use n^k, Other side n^k no expansion and no 'bino()' Next is bino(s,t) evaluation. <a name="docB92">&lt;a name="docB92"&gt;</a> 2009-09-05-21-11 pow(n,k-3)* ---eqn.AA229 [ (k+1)*(k)*(k-1)*(k-2)*a<sub>k+1</sub>/4! (k)*(k-1)*(k-2)*a<sub>k</sub>/3! (k-1)*(k-2)*a<sub>k-1</sub>/2! (k-2)*a<sub>k-2</sub>/1! 1*a<sub>k-3</sub>/0! -(k)*(k-1)*(k-2)/3! ] <a name="docB93">&lt;a name="docB93"&gt;</a> bino(k-3,k-3)*a<sub>k-3</sub> -a<sub>k-3</sub> = zero Use <a href=#docB77>a<sub>k+1</sub></a> value Use <a href=#docB82>a<sub>k</sub></a> value Use <a href=#docB87>a<sub>k-1</sub></a> value Above equation change to next. <a name="docB94">&lt;a name="docB94"&gt;</a> 2009-09-05-21-16 pow(n,k-3)* ---eqn.AA230 [ (k+1)*(k)*(k-1)*(k-2)*1/(k+1)/4! +(k)*(k-1)*(k-2)/2/3! +(k-1)*(k-2)*k/12/2! +(k-2)*a<sub>k-2</sub> -(k)*(k-1)*(k-2)/3! ] <a name="docB95">&lt;a name="docB95"&gt;</a> Write clearly factor '(k)*(k-1)*(k-2)' as next 2009-09-05-21-18 pow(n,k-3)* ---eqn.AA231 [ (k)*(k-1)*(k-2)/4! +(k)*(k-1)*(k-2)/2/3! + k *(k-1)*(k-2)/12/2! +(k-2)*a<sub>k-2</sub> -(k)*(k-1)*(k-2)/3! ] <a name="docB96">&lt;a name="docB96"&gt;</a> <a href="#docB41">proof</a>, <a href="#program0">program</a>, <a href="#docB02">Conjecture</a>, <a href="#docB04">triangle</a> Collect coefficient for '(k)*(k-1)*(k-2)' 2009-09-05-21-20 pow(n,k-3)* ---eqn.AA232 [ (k)*(k-1)*(k-2) *[1/4! + 1/2/3! + 1/12/2! -1/3!] +(k-2)*a<sub>k-2</sub> ] <a name="docB97">&lt;a name="docB97"&gt;</a> Change factorial to numbers 2009-09-05-21-21 pow(n,k-3)* ---eqn.AA233 [ (k)*(k-1)*(k-2) *[1/24 + 1/12 + 1/24 -1/6] +(k-2)*a<sub>k-2</sub> ] <a name="docB98">&lt;a name="docB98"&gt;</a> Calculate 2009-09-05-21-22 pow(n,k-3)* ---eqn.AA234 [ (k)*(k-1)*(k-2) *[0] +(k-2)*a<sub>k-2</sub> ] <a name="docB99">&lt;a name="docB99"&gt;</a> Terms in square bracket is pow(n,k-3) coefficient. Since equation is valid for ANY n. MUST n's coefficient be zero demand (k)*(k-1)*(k-2)*[0] ---eqn.AA235 +(k-2)*a<sub>k-2</sub> =0 get a<sub>k-2</sub>=0 ---eqn.AA236 <a name="docB100">&lt;a name="docB100"&gt;</a> <font color=red size=+3>Up to here Proved Coefficient <a href="#docB04">triangle</a> column 04 coefficient = 0</font> 2009-09-05-21-23 <hr> <a name="docB101">&lt;a name="docB101"&gt;</a> <a href="#docB41">proof</a>, <a href="#program0">program</a>, <a href="#docB02">Conjecture</a>, <a href="#docB04">triangle</a> 2009-09-06-13-05 document to here Next is Coefficient <a href="#docB04">triangle</a> column 05 When repeat same work few times, it is easy to find out next new coefficient equation. 1. change all ",k-3)" to ",k-4)" 2. add one new line "+bino(k-4,k-4)*a<sub>k-4</sub>" 3. change "-a<sub>k-3</sub>" to "-a<sub>k-4</sub>" then start new calculation. Must carry out point 2. add new line. In current calculation, point 2. and 3. two terms cancel each other. In next iteration, point 2. added new line provide next iteration with a chance to change from "+bino(k-4,k-4)*a<sub>k-4</sub>" to "+bino(k-4,k-5)*a<sub>k-4</sub>". so next equation is complete. <a name="pownk">&lt;a name="pownk"&gt;</a> 2009-09-06-18-04 What is "pow(n,k-4)" ? If k=7, "pow(n,k-4)" = "pow(n,3)" = n*n*n <a name="docB102">&lt;a name="docB102"&gt;</a> 2009-09-05-21-24 pow(n,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)*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> ] <a name="docB103">&lt;a name="docB103"&gt;</a> Above equation has cancellation. +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 This type cancellation occurs in each calculation. 2009-09-05-21-29 pow(n,k-4)* [ +(k+1)*(k+0)*(k-1)*(k-2)*(k-3)*a<sub>k+1</sub>/5! +(k+0)*(k-1)*(k-2)*(k-3)*a<sub>k</sub>/4! +(k-1)*(k-2)*(k-3)*a<sub>k-1</sub>/3! +(k-2)*(k-3)*a<sub>k-2</sub>/2! +(k-3)*a<sub>k-3</sub> -k*(k-1)*(k-2)*(k-3)/4! ] <a name="docB104">&lt;a name="docB104"&gt;</a> 2009-09-05-21-37 Use previous calculated values a<sub>k+1</sub>, a<sub>k</sub> a<sub>k-1</sub>, a<sub>k-2</sub>. Please see <a href="#docB110">List</a>. Change above equation to next equation. pow(n,k-4)* [ +(k+1)*(k+0)*(k-1)*(k-2)*(k-3)/(k+1)/5! +(k+0)*(k-1)*(k-2)*(k-3)/2/4! +(k-1)*(k-2)*(k-3)*k/12/3! +(k-2)*(k-3)*0/2! +(k-3)*a<sub>k-3</sub> -k*(k-1)*(k-2)*(k-3)/4! ] <a name="docB105">&lt;a name="docB105"&gt;</a> 2009-09-05-21-39 pow(n,k-4)* [ (k+0)*(k-1)*(k-2)*(k-3)/5! +(k+0)*(k-1)*(k-2)*(k-3)/2/4! + k *(k-1)*(k-2)*(k-3)/12/3! +(k-3)*a<sub>k-3</sub> -k*(k-1)*(k-2)*(k-3)/4! ] <a name="docB106">&lt;a name="docB106"&gt;</a> <a href="#docB41">proof</a>, <a href="#program0">program</a>, <a href="#docB02">Conjecture</a>, <a href="#docB04">triangle</a> 2009-09-05-21-41 pow(n,k-4)* [ (k+0)*(k-1)*(k-2)*(k-3)* [1/5! + 1/2/4! + 1/12/3! -1/4!] +(k-3)*a<sub>k-3</sub> ] 2009-09-05-21-42 pow(n,k-4)* [ (k+0)*(k-1)*(k-2)*(k-3)* [ 1/120 + 1/48 + 1/72 -1/24 ] +(k-3)*a<sub>k-3</sub> ] <a name="docB107">&lt;a name="docB107"&gt;</a> 2009-09-05-21-44 pow(n,k-4)* [ (k+0)*(k-1)*(k-2)*(k-3)* [ 1/(3*4*2*5) + 1/(3*4*4) + 1/(2*2*2*3*3) -1/(2*2*2*3) ] +(k-3)*a<sub>k-3</sub> ] 2009-09-05-21-48 pow(n,k-4)* [ (k+0)*(k-1)*(k-2)*(k-3)* (6+15+10-30)/(3*3*4*4*5) +(k-3)*a<sub>k-3</sub> ] <a name="docB108">&lt;a name="docB108"&gt;</a> 2009-09-05-21-50 pow(n,k-4)* [ (k+0)*(k-1)*(k-2)*(k-3)* (+1)/(3*3*4*4*5) +(k-3)*a<sub>k-3</sub> ] require (k+0)*(k-1)*(k-2)*(k-3)* (+1)/(3*3*4*4*5) +(k-3)*a<sub>k-3</sub> =0 <a name="docB109">&lt;a name="docB109"&gt;</a> a<sub>k-3</sub>= -k*(k-1)*(k-2)/(3*3*4*4*5) <font color=red size=+3>Up to here Proved Coefficient <a href="#docB04">triangle</a> column 05 coefficient = -k*(k-1)*(k-2)/(3*3*4*4*5)</font> 2009-09-05-21-52 The following work is routine and longer. Omit here. Few more results are listed here. Below is a summary of coefficients up to a<sub>k-7</sub>. <a name="docB110">&lt;a name="docB110"&gt;</a> 2009-09-05-21-58 summary 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) 2009-09-06-13-21 document stop <a href="#docB41">proof</a>, <a href="#program0">program</a>, <a href="#docB02">Conjecture</a>, <a href="#docB04">triangle</a> <a name="docB111">&lt;a name="docB111"&gt;</a> 2009-09-07-14-48 "Update 2009-09-07" add link to second page Please visit second page <a href=http://freeman2.com/jspico2e.htm> http://freeman2.com/jspico2e.htm</a> If you saved a copy, click <a href=jspico2e.htm>jspico2e.htm</a><!--9809071445--> 2009-09-07-14-51 </font></pre> <script type="text/javascript"> <!-- //9809031426 function huge2sml() { var inStr=boxDig11a.value; inStr=inStr.replace(/, /g,' ');//9809031438 inStr=inStr.replace(/,/g,'.'); // 766295064850303/3,0021419880618152e+22, 4994368678553267 var iter0=boxDig11b.value; //9809031448 if(isNaN(iter0))iter0=boxDig11b.value=1000; else { iter0=parseInt(iter0); if(iter0<0)iter0=boxDig11b.value=1000; } if(iter0==0) { boxc02.value='' +'iter0==0\n' return; } var inp0=inStr.split('/'); if(inp0.length!=2 ||isNaN(inp0[0]) ||isNaN(inp0[1]) ) { boxc02.value='' +'Please input rational numbers to [huge rational] box.\n' +'Example: "5174582439943824/2819723675734019".\n' return; } inp0[0]=parseFloat(inp0[0]); inp0[1]=parseFloat(inp0[1]); var sign0=1; //9809031520 if(inp0[0]<0) { inp0[0]=-inp0[0]; sign0=-1; } if(parseInt(inp0[1])==0) { boxc02.value='' +'divide by zero\n' return; } var k0,k1,k2; var inNumb=inp0[0]/inp0[1] var minErr=9999; //9809031453 var err0=0; var better0=0; for(k0=2;k0<iter0;k0++) { k1=k0*inNumb; k2=parseInt(k1); err0=k1-k2; if(err0<minErr) { minErr=err0; better0=k0; } } var butter1=better0*inNumb; if((butter1-parseInt(butter1))>=0.5) butter1=parseInt(butter1)+1; //9809031516 else butter1=parseInt(butter1); boxc02.value='' +'inp0[0]/inp0[1]='+ ((sign0==-1)?'-':'') +inp0[0]+'/'+inp0[1]+'\n' +'inNumb='+inNumb+'\n' +'iter0='+iter0+'\n' +'minErr='+minErr+'\n' +'better0='+better0+'\n' +'better0*inNumb='+(better0*inNumb)+'\n' +'Suggest: '+(sign0*butter1)+'/'+better0+'\n' +'9809031504\n' } //9809011559 function picoMan(pmArg1) { var i0,i1,i2,i3; if(isNaN(pmArg1)) { span_msg01.innerHTML= 'Please fill a positive integer in above power box.' return ; } var sum0=0; pmArg1=parseInt(pmArg1); var rank0=pmArg1+2; //9809011628 var ouStr0=''; for(i0=0;i0<=rank0;i0++)ouStr0+=' 1'; ouStr0+='\n'; for(i0=2;i0<=rank0;i0++) for(i1=rank0-1;i1>=1;i1--) { if(i1>1) ouStr0+=Math.pow(i0,i1)+' '; else //i1==1 { sum0=0; for(var p0=1;p0<=i0;p0++) { sum0+=Math.pow(p0,pmArg1); } ouStr0+=i0+' 1 ' + sum0 +'\n' } //if(i1>1) else } //for(i0=...;i0++) for(i1=rank0-1;i1>=1;i1--) document.f.t1.value=ouStr0; //9809011722 Analyse(); //9809011723 var ouStr1=document.f.t2.value ouStr1=ouStr1.replace(/, /g,'; ');//9809031050 ouStr1=ouStr1.replace(/,/g,'.'); ouStr1=ouStr1.replace(/; /g,', '); ouStr1=ouStr1.split('\n');//9809011727 var ouStr2='';//9809011730 for(i0=0;i0<ouStr1.length;i0++) { i2=ouStr1[i0].length-1;//9809011751 add '-1' while(ouStr1[i0].charCodeAt(i2)<32&&i2>0) i2--;//9809011749 while(ouStr1[i0].charAt(i2)==' '&&i2>0) i2--; i1=i2; while(ouStr1[i0].charAt(i1)!=' '&&i1>0) i1--; ouStr2+=ouStr1[i0].substring(i1+1,i2+1)+', ';//9809011752 } var ans0=''; //9809011757 i2=ouStr2.length-1;//9809011754 while(i2>0&&(ouStr2.charAt(i2)==' ' ||ouStr2.charAt(i2)==',')) i2--; ans0+='" i^('+pmArg1+') for i=1 to i=n\n' +'coefficients from high power to constant\n' +ouStr2.substring(0,i2+1) //9809011756 +'\n\n'; ouStr2=ouStr2.replace(/ /g,''); //9809011815 ouStr2=ouStr2.split(','); //9809011805 ans0+='Math equation is\n' for(i0=rank0-1;i0>=0;i0--) { if(i0>0) { ans0+='+n' for(i1=1;i1<i0;i1++) ans0+='*n'; ans0+='*('+ouStr2[rank0-1-i0]+')' } else //9809011822 { ans0+='+('+ouStr2[rank0-1-i0]+')' } } ans0+='' +'\n\n'; boxc01.value=ans0 } /** 1 0 0 0 0 0 0 1/6 0 1 0 0 0 0 0 1/2 0 0 1 0 0 0 0 5/12 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 -1/12 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 9809011654 get OK right !! 1 1 1 1 1 1 1 1 64 32 16 8 4 2 1 33 729 243 81 27 9 3 1 276 4096 1024 256 64 16 4 1 1300 15625 3125 625 125 25 5 1 4425 46656 7776 1296 216 36 6 1 12201 117649 16807 2401 343 49 7 1 29008 9809011655 get rank 6 answer QUICK !! 1 0 0 0 0 0 0 0 1/7 0 1 0 0 0 0 0 0 1/2 0 0 1 0 0 0 0 0 1/2 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 -1/6 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 1/42 0 0 0 0 0 0 0 1 0 /**/ //--> </SCRIPT> <br> <hr> <br> <a name="Arndt01">&lt;a name="Arndt01"&gt;</a><br> linear sets of equations compute <br> 2009-05-28-23-33 LiuHH access <br><a href=http://www.arndt-bruenner.de/mathe/scripts/gleichungssysteme.htm> http://www.arndt-bruenner.de/mathe/scripts/gleichungssysteme.htm</a> <br> <br><font color=red size=+3> Below is Arndt Br&#252;nner&#39;s work</font> <br> <br> <a name="Arndt02">&lt;a name="Arndt02"&gt;</a><br> <h2>Computer for the release of linear sets of equations</h2> <p><big>With interactive explanations to the Gauss solution procedure</big></p> <p>This Javascript solves linear sets of equations up to 26 variables and homogeneous sets of equations, whose solutions depend all on exactly a free parameter. The solution method is represented when desired in detail on the basis the entered set of equations. The Script counts recently on breaks, i.e. the results are exactly, so far counters and denominators of inputs, intermediate steps or results the border of 10^15 do not exceed. <br><a name="Arndt03">&lt;a name="Arndt03"&gt;</a> <p>In the left text window the equations are line by line entered, so many, as altogether variables are contained. It is not necessary that in each equation all variable emerge, also is all the same the sequence. Necessary it is only that possibly on the left and on the right from the equals sign a linear sum of variables (with signs and/or factors) stands and absolute member (number without variable). <br><u><b>Again:</b> Also break numbers can be entered (e.g.: <tt>2/3x</tt>).</u> <br>Clasped terms, break terms or powers cannot be processed.<br> The variable names must be individual letters. The entire alphabet (without umlauts) is available, large and lower case is not differentiated. Multiplication symbol (* or &#183;) are not necessary. </p> <p><a href="javascript:bsp()">Random example produce</a> <a href="javascript:bsp_()">&#8230; solve and</a></p> <a name="Arndt04">&lt;a name="Arndt04"&gt;</a> &nbsp; <a href="#program0">program0</a> &nbsp; <a href="#sum_i_from_1_to_n">sum_i_from_1_to_n</a> <form name="f"> <table><tr><td>Here <a name="rechner">equations</a> <small>(or coefficient matrix)</small> enter</td> <td>Solutions (computed)</td></tr> <tr><td> <span> <textarea name="t1" cols="45" rows="10" onkeyup="document.f.t2.value=''" wrap="off"> </textarea></span></td> <td><span><textarea name="t2" cols="25" rows="10" readonly="readonly" wrap="off"></textarea></span></td></tr> <tr><td><input value="Solve linear equations" onclick="Analyse()" type="button"> <!--9809011331 "Lösen des Gleichungssystems" => "Solve linear equations" "Eingaben löschen" => "clear input" --> <input value="clear input" onclick="document.f.t1.focus()" type="reset"> <input value="try5" type="button" onclick="document.f.t1.value=' 1 1 1 1 1 1 1 1\n 64 32 16 8 4 2 1 33\n 729 243 81 27 9 3 1 276\n 4096 1024 256 64 16 4 1 1300\n 15625 3125 625 125 25 5 1 4425\n 46656 7776 1296 216 36 6 1 12201\n117649 16807 2401 343 49 7 1 29008\n'"> </td> <td> <center>Expenditure <select name="alsBruch" onchange="Analyse()"><option>as break</option><option>as decimal number</option></select> </center> </td> </tr> <tr><td>&nbsp;&nbsp;&nbsp;<input name="immererkl" type="checkbox"> produce explanations always immediately </td></tr> </tbody></table> <a name="Arndt05">&lt;a name="Arndt05"&gt;</a><br> [[ <br> LiuHH add "try5" to generate <font color=red> " i^(5) from i=1 to i=7 input data. </font> get answer <br> " i^(5)= n*n*n*n*n*n/6+n*n*n*n*n/2+5*n*n*n*n/12-n*n/12 <br> 2009-09-01-14-32 LiuHH made above notes. <br> ]] <br> <a name="Arndt06">&lt;a name="Arndt06"&gt;</a><br> <p>The solutions of homogeneous sets of equations are indicated in dependence of a free parameter. If a not-homogeneous set of equations has infinitely many solutions, the message &#8220;no clear solution is indicated found&#8221;.<br> <i>A homogeneous</i> set of equations does not possess (after simplification) absolute members. </p> <p>Javascript uses the Gauss algorithm, which is called also Gauss elimination procedure, since successively in the equations systematically variables are eliminated. In addition only two methods are used: Multiply an equation by a number, an adding (a multiples) of an equation to another.</p> <a name="Arndt07">&lt;a name="Arndt07"&gt;</a><br> <p>Over the following switching surface an explanation of the solution method can be produced for the sets of equations or coefficient stencils entered above:</p> <table> <tbody><tr><td><a href="javascript:bsp__()">Random example produce</a></td></tr> <tr><td style="vertical-align: middle;"><input value="Detailed minutes of the solution method generate" onclick="erkl()" type="button"> </td><td>&nbsp;&nbsp;&nbsp;</td><td> <nobr><input name="erkltyp" value="k" checked="checked" type="radio"> Coefficient matrix</nobr><br> <input name="erkltyp" value="g" type="radio"> Equations<br>&nbsp;</td><td>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</td> <td style="vertical-align: middle;"><nobr><input name="ganzzahlig" type="checkbox"> integral count</nobr></td> </tr></tbody></table> <font style="font-size: 9pt; line-height: 8pt;"> <textarea name="gauss" rows="30" cols="100" style="font-size: 9pt; line-height: 6pt;" wrap="off"></textarea> </font> <p><a name="uebungen">&nbsp;</a></p> <a name="Arndt08">&lt;a name="Arndt08"&gt;</a><br> <table align="center" border="1" bgcolor="#E4DAB7" cellpadding="20"><tr><td> <center> <h4><a name="ueben">To</a> practicing:<hr width="80%"></h4> <input type="button" class="button" onclick="javascript:uebungen()" style="font-size:12pt" value="Side with ten sets of equations produce"> <p>The solutions are smallprinted provided.</p> <p>Minimum dimension: <input name="mingrad" size="2" maxlength="1" value="2"> Maximum dimension: <input name="maxgrad" maxlength="1" size="2" value="5"> </center></td></tr></table> <a name="Arndt09">&lt;a name="Arndt09"&gt;</a><br> <p><small><b>Notes:</b></small><br><small>The attitudes for minimum and maximum dimension are effective also for the example production on this side.</small><br><small> All examples are solved to the sample by the algorithm, although the solutions are first produced with the production of the tasks and for the program to that extent already admit are. If solutions are thus indicated, then that means that the task is also clearly solvable.</small></p> <p>Possibly helpfully: <a href="http://66.163.168.225/babelfish/translate_url_content?.intl=us&lp=de_en&trurl=http%3a%2f%2fwww.arndt-bruenner.de%2fmathe%2fscripts%2fbruchrechnung2.htm"><font face="Symbol">&#174;</font></a><a href="http://66.163.168.225/babelfish/translate_url_content?.intl=us&lp=de_en&trurl=http%3a%2f%2fwww.arndt-bruenner.de%2fmathe%2fscripts%2fbruchrechnung2.htm"> basic operations of arithmetic of the Bruchrechnung</a></p> <p> </p> <p>Sample applications for the algorithm are on these three sides:<br> <a href="http://66.163.168.225/babelfish/translate_url_content?.intl=us&lp=de_en&trurl=http%3a%2f%2fwww.arndt-bruenner.de%2fmathe%2fscripts%2fregr.htm">Linear involution</a> &#149; <a href="http://66.163.168.225/babelfish/translate_url_content?.intl=us&lp=de_en&trurl=http%3a%2f%2fwww.arndt-bruenner.de%2fmathe%2fscripts%2fkreis3p.htm">Circle by 3 points</a> &#149; <a href="http://66.163.168.225/babelfish/translate_url_content?.intl=us&lp=de_en&trurl=http%3a%2f%2fwww.arndt-bruenner.de%2fmathe%2fscripts%2fkugel4p.htm">Ball by 4 points</a><br></p> </form> <br><font color=red size=+3> Above is Arndt Br&#252;nner&#39;s work</font> <br> <a href="#docB41">proof</a>, <a href="#program0">program</a>, <a href="#docB02">Conjecture</a>, <a href="#docB04">triangle</a> <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> <br> This page, Positive Integer power sum COefficient equation. <br> <a href=http://freeman2.com/jspico_e.htm> http://freeman2.com/jspico_e.htm </a> <br> First upload 2009-09-02 <br> <br> Program name in Chinese: (2009-09-03-11-36) <br> Brief name: cŒTýQxe <br> Full name: cktexecxeŒTMzÂOxelQ_ýQxe <br> <br> Thank you for visiting Freeman's page.0 <br> Freeman 02009-09-01-23-36 <br> <br> Please visit second page <a href=http://freeman2.com/jspico2e.htm>freeman2.com</a>/<a href=jspico2e.htm>jspico2e.htm</a> <!--9809071445--> <br> <br> <span id="tuteLink2"></span> </BODY> </HTML> <!-- 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 -->