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" }}{PARA 6 "" 1 "" {TEXT -1 123 "\nA PROOF OF A RECURRENCE\n\nBy Shal osh B. Ekhad, Temple University, ekhad@math.temple.edu\n\nTheorem:Let \+ F(n,k) be given by" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%)binomialG 6$%\"nG%\"kG" }}{PARA 6 "" 1 "" {TEXT -1 131 "\nand let SUM(n) be \+ the sum of F(n,k) with\nrespect to k .\n\nSUM(n) satisfies the following linear recurrence equation" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&-%$SUMG6#%\"nG!\"#-F%6#,&F'\"\"\"F,F,F," }}{PARA 11 " " 1 "" {XPPMATH 20 "6#%$=0.G" }}{PARA 6 "" 1 "" {TEXT -1 43 "\nPROOF: \+ We cleverly construct G(n,k) :=" }}{PARA 11 "" 1 "" {XPPMATH 20 "6 #*(%\"kG\"\"\",(%\"nG!\"\"F(F%F$F%F(-%)binomialG6$F'F$F%" }}{PARA 6 " " 1 "" {TEXT -1 20 "with the motive that" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&-%\"FG6$%\"nG%\"kG!\"#-F%6$,&F'\"\"\"F-F-F(F-" }}{PARA 6 "" 1 "" {TEXT -1 96 "= G(n,k+1)-G(n,k) (check!)\n\nand the theorem foll ows upon summing with respect to k .QED." }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 16 "Let's verify it:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 123 "FF := (n,k) -> binomial(n,k);\nGG := (n,k) -> k*binomial(n,k)/(-n-1+k );\nlh := -2*FF(n,k)+FF(n+1,k);\nrh := GG(n,k+1)-GG(n,k);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#FFG%)binomialG" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#GGG:6$%\"nG%\"kG6\"6$%)operatorG%&arrowGF)*(9%\"\"\"-%)binomi alG6$9$F.F/,(F3!\"\"F5F/F.F/F5F)F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6# >%#lhG,&-%)binomialG6$%\"nG%\"kG!\"#-F'6$,&F)\"\"\"F/F/F*F/" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#rhG,&*(,&%\"kG\"\"\"F)F)F)-%)binomialG6$% \"nGF'F),&F-!\"\"F(F)F/F)*(F(F)-F+6$F-F(F),(F-F/F/F)F(F)F/F/" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 82 "Dividing through by F(n,k) and simplifying gives rational functions on both sides." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 75 "lh := \+ sumtools[simpcomb](lh/FF(n,k));\nrh := sumtools[simpcomb](rh/FF(n,k)); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#lhG,$*&,(%\"nG!\"\"F)\"\"\"%\"k G\"\"#F*,(F(F)F)F*F+F*F)F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#rhG,$ *&,(%\"nG!\"\"F)\"\"\"%\"kG\"\"#F*,(F(F)F)F*F+F*F)F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "simplify(lh-rh);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "zeilp ap(binomial(n,k)^2,k,n);" }}{PARA 6 "" 1 "" {TEXT -1 123 "\nA PROOF OF A RECURRENCE\n\nBy Shalosh B. Ekhad, Temple University, ekhad@math.te mple.edu\n\nTheorem:Let F(n,k) be given by" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*$-%)binomialG6$%\"nG%\"kG\"\"#" }}{PARA 6 "" 1 "" {TEXT -1 131 "\nand let SUM(n) be the sum of F(n,k) with\nres pect to k .\n\nSUM(n) satisfies the following linear recurrence equation" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*&,&%\"nG!\"%!\"#\"\"\" F)-%$SUMG6#F&F)F)*&,&F&F)F)F)F)-F+6#F.F)F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%$=0.G" }}{PARA 6 "" 1 "" {TEXT -1 43 "\nPROOF: We clev erly construct G(n,k) :=" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#**,(% \"nG!\"$F&\"\"\"%\"kG\"\"#F'F(F),(F%!\"\"F+F'F(F'!\"#-%)binomialG6$F%F (F)" }}{PARA 6 "" 1 "" {TEXT -1 20 "with the motive that" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#,&*&,&%\"nG!\"%!\"#\"\"\"F)-%\"FG6$F&%\"kGF)F)*& ,&F&F)F)F)F)-F+6$F/F-F)F)" }}{PARA 6 "" 1 "" {TEXT -1 96 "= G(n,k+1) -G(n,k) (check!)\n\nand the theorem follows upon summing with respec t to k .QED." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "zeilpap (binomial(n,k)^3,k,n);" }}{PARA 6 "" 1 "" {TEXT -1 123 "\nA PROOF OF A RECURRENCE\n\nBy Shalosh B. Ekhad, Temple University, ekhad@math.temp le.edu\n\nTheorem:Let F(n,k) be given by" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*$-%)binomialG6$%\"nG%\"kG\"\"$" }}{PARA 6 "" 1 "" {TEXT -1 131 "\nand let SUM(n) be the sum of F(n,k) with\nres pect to k .\n\nSUM(n) satisfies the following linear recurrence equation" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#,(*&,&%\"nG\"\"\"F'F'\"\" #-%$SUMG6#F&F'!\")*&,(*$F&F(!\"(F&!#@!#;F'F'-F*6#F%F'F'*&,&F&F'F(F'F(- F*6#F6F'F'" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%$=0.G" }}{PARA 6 "" 1 " " {TEXT -1 43 "\nPROOF: We cleverly construct G(n,k) :=" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#*.,(*$%\"nG\"\"#\"\"\"F&F'F(F(F(,6*$F&\"\"$! #9F%!#uF&!$G\"!#sF(%\"kG\"#y*&F&F'F0F(\"#F*&F&F(F0F(\"#$**&F&F(F0F'!#= *$F0F'!#I*$F0F+\"\"%F(F0F+,(F&!\"\"F=F(F0F(!\"$,(F&F=!\"#F(F0F(F>-%)bi nomialG6$F&F0F+" }}{PARA 6 "" 1 "" {TEXT -1 20 "with the motive that" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#,(*&,&%\"nG\"\"\"F'F'\"\"#-%\"FG6$F& %\"kGF'!\")*&,(*$F&F(!\"(F&!#@!#;F'F'-F*6$F%F,F'F'*&,&F&F'F(F'F(-F*6$F 7F,F'F'" }}{PARA 6 "" 1 "" {TEXT -1 96 "= G(n,k+1)-G(n,k) (check!) \n\nand the theorem follows upon summing with respect to k .QED." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 20 "Gauss's 2F1 identity" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "zeilpap(GAMMA(k-n)*GAMMA(k+b)/(GAMMA(k+c)*k!),k,n);" }}{PARA 6 "" 1 "" {TEXT -1 123 "\nA PROOF OF A RECURRENCE\n\nBy Shalosh B. Ekhad, T emple University, ekhad@math.temple.edu\n\nTheorem:Let F(n,k) be g iven by" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#**-%&GAMMAG6#,&%\"nG!\"\"% \"kG\"\"\"F+-F%6#,&F*F+%\"bGF+F+-F%6#,&F*F+%\"cGF+F)-%*factorialG6#F*F )" }}{PARA 6 "" 1 "" {TEXT -1 131 "\nand let SUM(n) be the sum of \+ F(n,k) with\nrespect to k .\n\nSUM(n) satisfies the follow ing linear recurrence equation" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*& ,(%\"nG!\"\"%\"bG\"\"\"%\"cGF'F)-%$SUMG6#F&F)F)*(,&F&F)F)F)F),&F&F)F*F )F)-F,6#F/F)F'" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%$=0.G" }}{PARA 6 " " 1 "" {TEXT -1 43 "\nPROOF: We cleverly construct G(n,k) :=" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#*0,(%\"nG!\"\"F&\"\"\"%\"kGF'F&,(F(F'F &F'%\"cGF'F'F(F'-%&GAMMAG6#,&F%F&F(F'F'-F,6#,&F(F'%\"bGF'F'-F,6#,&F(F' F*F'F&-%*factorialG6#F(F&" }}{PARA 6 "" 1 "" {TEXT -1 20 "with the mot ive that" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*&,(%\"nG!\"\"%\"bG\"\" \"%\"cGF'F)-%\"FG6$F&%\"kGF)F)*(,&F&F)F)F)F),&F&F)F*F)F)-F,6$F0F.F)F' " }}{PARA 6 "" 1 "" {TEXT -1 96 "= G(n,k+1)-G(n,k) (check!)\n\nand the theorem follows upon summing with respect to k .QED." }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 44 "Bailey's 4F3. Last identity on Ko epf p. 84." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 106 "zeilpap(GAMMA(k+a)*G AMMA(k+1+a/2)*GAMMA(k+b)*GAMMA(k-n)/(GAMMA(k+a/2)*GAMMA(k+1+a-b)*GAMMA (2+2*b-n)),k,n);" }}{PARA 6 "" 1 "" {TEXT -1 123 "\nA PROOF OF A RECUR RENCE\n\nBy Shalosh B. Ekhad, Temple University, ekhad@math.temple.edu \n\nTheorem:Let F(n,k) be given by" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*0-%&GAMMAG6#,&%\"kG\"\"\"%\"aGF)F)-F%6#,(F(F)F)F)F*#F)\"\"#F)-F %6#,&F(F)%\"bGF)F)-F%6#,&%\"nG!\"\"F(F)F)-F%6#,&F(F)F*F.F8-F%6#,*F(F)F )F)F*F)F3F8F8-F%6#,(F/F)F3F/F7F8F8" }}{PARA 6 "" 1 "" {TEXT -1 131 "\n and let SUM(n) be the sum of F(n,k) with\nrespect to k . \n\nSUM(n) satisfies the following linear recurrence equation" }} {PARA 12 "" 1 "" {XPPMATH 20 "6#,**,,(%\"nG\"\"\"%\"bG!\"#F'F'F',&F&F' F(F)F',(F&F'F(F)!\"\"F'F',**&F&F'%\"aGF'\"\"#*$F/F0F'F/\"\"'!\"%F'F'-% $SUMG6#F&F'F'**F%F'F*F',:*&F&F0F/F'\"\"%*&F&F'F/F0F:*(F&F'F/F'F(F'F0*$ F/\"\"$F'*&F/F0F(F'F'F.\"#=F1\"\"**&F/F'F(F'F2F&!\")F/\"#9F(F3!#;F'F'- F56#,&F&F'F'F'F'F,*(F%F',J*&F&F>F/F'F0*&F&F0F/F0F>*(F&F0F/F'F(F'F0*&F& F'F/F>F'*(F&F'F/F0F(F'F>*&F/F>F(F'F'F9FDF;FDF<\"#5F=F0F?\"\")*$F&F0F3F .\"#E*&F&F'F(F'F3F1\"#:FBFRF&!#?F/FRF(!#7FWF'F'-F56#,&F&F'F0F'F'F'*(,* F.F0F1F'F/F:F3F'F',*F&F'F/F'F(F,F>F'F'-F56#,&F&F'F>F'F'F'" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%$=0.G" }}{PARA 6 "" 1 "" {TEXT -1 43 "\nPROOF: We cleverly construct G(n,k) :=" }}{PARA 12 "" 1 "" {XPPMATH 20 " 6#*@,(%\"nG\"\"\"%\"bG!\"#!\"\"F&F&,,*$F%\"\"#F&*&F%F&F'F&!\"%F%F&*$F' F,\"\"%F'F(F&,4*&F%F&%\"aGF,F,*$F3F,F0*$F3\"\"$F&F3!#;*&%\"kGF&F3F,F,* &F9F&F3F&\"#7*(F%F&F9F&F3F&F0F9!\")*&F%F&F3F&F.F&,(F9F&F3F&F'F)F&,(F%F )F)F&F9F&F),&F9F,F3F&F),(F%F)F(F&F9F&F),(F%F)!\"$F&F9F&F)-%&GAMMAG6#,& F9F&F3F&F&-FF6#,(F9F&F&F&F3#F&F,F&-FF6#,&F9F&F'F&F&-FF6#,&F%F)F9F&F&-F F6#,&F9F&F3FLF)-FF6#,*F9F&F&F&F3F&F'F)F)-FF6#,(F,F&F'F,F%F)F)" }} {PARA 6 "" 1 "" {TEXT -1 20 "with the motive that" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#,**,,(%\"nG\"\"\"%\"bG!\"#F'F'F',&F&F'F(F)F',(F&F'F(F)! \"\"F'F',**&F&F'%\"aGF'\"\"#*$F/F0F'F/\"\"'!\"%F'F'-%\"FG6$F&%\"kGF'F' **F%F'F*F',:*&F&F0F/F'\"\"%*&F&F'F/F0F;*(F&F'F/F'F(F'F0*$F/\"\"$F'*&F/ F0F(F'F'F.\"#=F1\"\"**&F/F'F(F'F2F&!\")F/\"#9F(F3!#;F'F'-F56$,&F&F'F'F 'F7F'F,*(F%F',J*&F&F?F/F'F0*&F&F0F/F0F?*(F&F0F/F'F(F'F0*&F&F'F/F?F'*(F &F'F/F0F(F'F?*&F/F?F(F'F'F:FEFF0F@\"\")*$F&F0F3F.\"#E*&F&F' F(F'F3F1\"#:FCFSF&!#?F/FSF(!#7FXF'F'-F56$,&F&F'F0F'F7F'F'*(,*F.F0F1F'F /F;F3F'F',*F&F'F/F'F(F,F?F'F'-F56$,&F&F'F?F'F7F'F'" }}{PARA 6 "" 1 "" {TEXT -1 96 "= G(n,k+1)-G(n,k) (check!)\n\nand the theorem follows upon summing with respect to k .QED." }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 28 "AZpapd(1/(1-x)/x^(n+1),x,n);" }}{PARA 6 "" 1 "" {TEXT -1 207 "\nA PROOF OF A LINEAR RECURRENCE SATISFIED BY AN INTEGRA L\n\nBy Shalosh B. Ekhad, Temple University, ekhad@math.temple.edu\n\n I will give a short proof of the following result.\n\nTheorem:Let F( n,x) be given by" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&,&\"\"\"F%%\"x G!\"\"F')F&,&%\"nGF%F%F%F'" }}{PARA 6 "" 1 "" {TEXT -1 146 "\nand let \+ INTEGRAL(n) be the integral of F(n,x) with\nrespect to x \+ .\n\nINTEGRAL(n) satisfies the following linear recurrence equation " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*&,&%\"nG!\"\"F'\"\"\"F(-%)INTEG RALG6#F&F(F(*&,&F&F(F(F(F(-F*6#F-F(F(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%$=0.G" }}{PARA 6 "" 1 "" {TEXT -1 43 "\nPROOF: We cleverly constru ct G(n,x) :=" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*(,&!\"\"\"\"\"%\" xGF&F&,&F&F&F'F%F%)F',&%\"nGF&F&F&F%" }}{PARA 6 "" 1 "" {TEXT -1 20 "w ith the motive that" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*&,&%\"nG!\" \"F'\"\"\"F(-%\"FG6$F&%\"xGF(F(*&,&F&F(F(F(F(-F*6$F.F,F(F(" }}{PARA 6 "" 1 "" {TEXT -1 88 "= diff(G(n,x),x)\n\nand the theorem follows upo n integrating with respect to x .QED." }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}}{MARK "0 0 0" 13 }{VIEWOPTS 1 1 0 1 1 1803 }