{VERSION 2 3 "SUN SPARC SOLARIS" "2.3" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 }{CSTYLE "2D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 256 "terminal" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE " " -1 257 "" 1 24 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "T imes" 1 14 0 0 0 0 2 2 2 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Text Output" -1 2 1 {CSTYLE "" -1 -1 "Courier" 1 10 0 0 255 1 0 0 0 0 0 1 3 0 0 }1 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Heading 1" 0 3 1 {CSTYLE "" -1 -1 "" 1 18 0 0 0 0 0 1 0 0 0 0 0 0 0 }1 0 0 0 6 6 0 0 0 0 0 0 -1 0 }{PSTYLE "Heading 2" 3 4 1 {CSTYLE "" -1 -1 "" 1 14 0 0 0 0 0 0 0 0 0 0 0 0 0 }0 0 0 -1 4 4 0 0 0 0 0 0 -1 0 }{PSTYLE " Warning" 2 7 1 {CSTYLE "" -1 -1 "" 0 1 0 0 255 1 0 0 0 0 0 0 1 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Error" 7 8 1 {CSTYLE "" -1 -1 "" 0 1 255 0 255 1 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Maple Output" 0 11 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 3 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 11 12 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }1 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 3 256 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "R3 Font 0" -1 257 1 {CSTYLE "" -1 -1 "Helvetica" 1 14 0 0 0 0 2 1 2 0 0 0 0 0 0 } 0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "R3 Font 2" -1 258 1 {CSTYLE "" -1 -1 "Courier" 1 14 0 0 0 0 2 2 2 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 259 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 260 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 11 261 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 262 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 11 263 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 264 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 11 265 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 266 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 11 267 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 268 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 11 269 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 270 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 271 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 11 272 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 256 44 "/home/m262f99/KOEPF/works heetsV.4/WZdemo.mws" }{MPLTEXT 1 0 0 "" }}{PARA 256 "" 0 "" {TEXT 257 58 "Math 262a, Fall 1999, Glenn Tesler\nWZ method demo\n10/29/99" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "read `hsum.mpl`;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%ZCopyright~1998~~Wolfram~Koepf,~Konrad-Zuse -Zentrum~BerlinG" }}}{EXCHG {PARA 3 "" 0 "" {TEXT -1 16 "Example 1. Pr ove" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "Sum(binomial(n,k),k)=2^n;" } }{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$SumG6$-%)binomialG6$%\"nG%\"kGF+) \"\"#F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 28 "Divide summand by righ t side" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "F1 := (n,k) -> binomial(n ,k) / 2^n;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#F1G:6$%\"nG%\"kG6\"6$ %)operatorG%&arrowGF)*&-%)binomialG6$9$9%\"\"\")\"\"#F1!\"\"F)F)" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 34 "Set up term for Gosper's algorithm " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "a1 := F1(n+1,k) - F1(n,k);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%#a1G,&*&-%)binomialG6$,&%\"nG\"\"\"F ,F,%\"kGF,)\"\"#F*!\"\"F,*&-F(6$F+F-F,)F/F+F0F0" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 19 "G1 := gosper(a1,k);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#G1G,$*(%\"kG\"\"\",(%\"nG!\"\"F+F(F'\"\"#F+,&*&-%)bi nomialG6$,&F*F(F(F(F'F()F,F2F+F(*&-F06$F*F'F()F,F*F+F+F(F+" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 224 "Check F(n+1,k) - F(n,k) = G(n,k+1) - G(n ,k).\nTo do this, divide both sides by F(n,k); that makes both sides r ational functions of n,k.\nSimplify them till they are in rational for m. Then it's easy to compare them.\nleft side:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "lh1 := (F1(n+1,k) - F1(n,k))/F1(n,k);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$lh1G*(,&*&-%)binomialG6$,&%\"nG\"\"\"F-F-%\"kGF -)\"\"#F+!\"\"F-*&-F)6$F,F.F-)F0F,F1F1F-F3F1F5F-" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 32 "Maple's built-in simplification:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "simplify(lh1);" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6#,$*&,(%\"nG!\"\"F'\"\"\"%\"kG\"\"#F(,(F)F(F'F(F&F'F'#F'F*" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 88 "And if that had failed, Koepf has \+ a fancier one designed for hypergeometric expressions:" }}{PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 14 "simpcomb(lh1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&,(%\"nG!\"\"F'\"\"\"%\"kG\"\"#F(,(F)F(F'F(F&F'F'#F'F*" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 243 "right side:\n(This is ugly... F1 \+ was defined in Maple as a function of two variables, while G1 is just \+ an algebraic expression, so parameters must be given to F1, and cannot be given to G1; but we want k replaced by k+1 in G1, so we use subs.) " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "rh1 := (subs(k=k+1,G1) - G1)/F 1(n,k);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%$rh1G*(,&*(,&%\"kG\"\"\"F *F*F*,(F)\"\"#F*F*%\"nG!\"\"F.,&*&-%)binomialG6$,&F-F*F*F*F(F*)F,F4F.F **&-F26$F-F(F*)F,F-F.F.F*F.*(F)F*,(F-F.F.F*F)F,F.,&*&-F26$F4F)F*F5F.F* *&-F26$F-F)F*F9F.F.F*F*F*FAF.F9F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "simplify(rh1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$* &,(%\"nG!\"\"F'\"\"\"%\"kG\"\"#F(,(F)F(F'F(F&F'F'#F'F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "simplify(lh1-rh1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" }}}{EXCHG {PARA 259 "" 0 "" {TEXT -1 260 "N ote: the comparison function doesn't work because it asks whether the \+ expressions are coded identically, not whether they are algebraically \+ equal. The specific formulas given for lh1, rh1 above, are different \+ expressions, though algebraically they are equal." }}{PARA 260 "> " 0 "" {MPLTEXT 1 0 15 "evalb(lh1=rh1);" }}{PARA 261 "" 1 "" {XPPMATH 20 " 6#%&falseG" }}}{EXCHG {PARA 262 "> " 0 "" {MPLTEXT 1 0 25 "evalb((x+1) ^2=x^2+2*x+1);" }}{PARA 263 "" 1 "" {XPPMATH 20 "6#%&falseG" }}} {EXCHG {PARA 264 "> " 0 "" {MPLTEXT 1 0 8 "(x+1)^2;" }}{PARA 265 "" 1 "" {XPPMATH 20 "6#*$,&%\"xG\"\"\"F&F&\"\"#" }}}{EXCHG {PARA 266 "> " 0 "" {MPLTEXT 1 0 10 "x^2+2*x+1;" }}{PARA 267 "" 1 "" {XPPMATH 20 "6#, (*$%\"xG\"\"#\"\"\"F%F&F'F'" }}}{EXCHG {PARA 268 "> " 0 "" {MPLTEXT 1 0 10 "2*(x^2+1);" }}{PARA 269 "" 1 "" {XPPMATH 20 "6#,&*$%\"xG\"\"#F&F &\"\"\"" }}}{EXCHG {PARA 270 "" 0 "" {TEXT -1 87 "Maple does some simp lifications on its own, but only simple ones, so this test is true:" } }{PARA 271 "> " 0 "" {MPLTEXT 1 0 25 "evalb(2*(x^2+1)=2*x^2+2);" }} {PARA 272 "" 1 "" {XPPMATH 20 "6#%%trueG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 37 "Now, back to WZ... The certificate is" }}{PARA 0 "> " 0 " " {MPLTEXT 1 0 19 "R1 := G1 / F1(n,k);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#R1G,$*,%\"kG\"\"\",(%\"nG!\"\"F+F(F'\"\"#F+,&*&-%)binomialG6$ ,&F*F(F(F(F'F()F,F2F+F(*&-F06$F*F'F()F,F*F+F+F(F5F+F7F(F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "R1 := simplify(R1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#R1G,$*&%\"kG\"\"\",(F'F(!\"\"F(%\"nGF*F*#F(\"\" #" }}}{EXCHG {PARA 3 "" 0 "" {TEXT -1 92 "Now do the same problem over again with Koepf's single function that does much of the above:" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "R1b := WZcertificate(F1(n,k),k,n); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$R1bG,$*&%\"kG\"\"\",(F'F(!\"\"F (%\"nGF*F*#F(\"\"#" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 84 "The fact th at it succeeded implies f1(n) = sum_k F1(n,k) is a constant. By contra st," }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "WZcertificate(binomial(n,k), k,n);" }}{PARA 8 "" 1 "" {TEXT -1 50 "Error, (in WZcertificate) extend ed WZ method fails" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 57 "Now, to pro ve our identity from the certificate, do this:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "G1b := F1(n,k) * R1b; " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$G1bG,$**-%)binomialG6$%\"nG%\"kG\"\"\")\"\"#F*!\"\"F+F,,(F+F, F/F,F*F/F/#F,F." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "simplify ((F1(n+1,k) - F1(n,k))/F1(n,k));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$ *&,(%\"nG!\"\"F'\"\"\"%\"kG\"\"#F(,(F)F(F'F(F&F'F'#F'F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "simplify((subs(k=k+1,G1b) - G1b)/F1 (n,k));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&,(%\"nG!\"\"F'\"\"\"%\" kG\"\"#F(,(F)F(F'F(F&F'F'#F'F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "simplify(\"\"-\");" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" } }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 73 "The equation F(n+1,k)-F(n,k) = G (n,k+1)-G(n,k) holds (where F=F1, G=G1b)." }}{PARA 0 "" 0 "" {TEXT -1 40 "Thus f1(n) is constant. To determine it," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "sum(F1(0,k),k=-infinity..infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$sumG6$-%)binomialG6$\"\"!%\"kG/F*;,$%)infini tyG!\"\"F." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 71 "It can't figure tha t out, so figure the support by hand and plug it in." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "sum(F1(0,k),k=0..0);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 4 "" 0 " " {TEXT -1 19 "Companion identity:" }}{PARA 0 "" 0 "" {TEXT -1 146 "De fine a new function g1(k) = sum_\{n>=0\} G1(n,k):\n(Using = instead of := makes it print nicely, but it doesn't actually assign a value to \+ \"g1(k)\")" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "g1(k) = Sum(G1,n=0..i nfinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%#g1G6#%\"kG-%$SumG6$,$ *(F'\"\"\",(%\"nG!\"\"F0F-F'\"\"#F0,&*&-%)binomialG6$,&F/F-F-F-F'F-)F1 F7F0F-*&-F56$F/F'F-)F1F/F0F0F-F0/F/;\"\"!%)infinityG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 2 "or" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "simpG1 \+ := simpcomb(G1):\ng1 = Sum(simpG1,n=0..infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%#g1G-%$SumG6$,$**-%&GAMMAG6#,&%\"nG\"\"\"F/F/F/-F+6#% \"kG!\"\"-F+6#,(F.F/\"\"#F/F2F3F3)F7F.F3#F3F7/F.;\"\"!%)infinityG" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 38 "Note this is -binomial(n,k-1)/2^(n +1):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "bG1 := -binomial(n,k-1)/2^( n+1):\nconvert(bG1,GAMMA);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$**-%&G AMMAG6#,&%\"nG\"\"\"F*F*F*-F&6#%\"kG!\"\"-F&6#,(F)F*\"\"#F*F-F.F.)F2F( F.F." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 131 "From F1(n+1,k)-F1(n,k) = G1(n,k+1) - G1(n,k), we can obtain a telescoping relation in n on the left side. Sum it over n=0,1,2,... " }}{PARA 0 "" 0 "" {TEXT -1 57 " So F1(infinity,k)-F1(0,k) = g1(k+1)-g1(k). In this case:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "F1(infinity,k);" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#*&-%)binomialG6$%)infinityG%\"kG\"\"\")\"\"#F'!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "limit(F1(n,k),n=infinity) ;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 8 "F1(0,k);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%)binom ialG6$\"\"!%\"kG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 214 "So for integ ers k<>0, we get 0-0 = g1(k+1)-g1(k),\nand for k=0 we get 0-1 = g1(1)- g1(0).\nThus, for k=1,2,3,4,... we have g1(k)=C for some constant C,\n and for k=0,-1,-2,-3,-4,... we have g1(k)=different constant=C+1:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "Sum(bG1,n=0..infinity) = pie cewise(k>=1,C,C+1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$SumG6$,$*&- %)binomialG6$%\"nG,&%\"kG\"\"\"!\"\"F/F/)\"\"#,&F,F/F/F/F0F0/F,;\"\"!% )infinityG-%*PIECEWISEG6$7$%\"CG1F/F.7$,&F " 0 "" {MPLTEXT 1 0 63 "Sum(binomial(n,k)/2^n,n=0. .infinity) = piecewise(k>=0,2,k<0,0);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$SumG6$*&-%)binomialG6$%\"nG%\"kG\"\"\")\"\"#F+!\"\"/F+;\"\"!%)i nfinityG-%*PIECEWISEG6$7$F/1F3F,7$F32F,F3" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 40 "Let's find \+ C and check this empirically." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "g1 := sum(G1,n=0..infinity);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%#g1G,$ **%\"kG\"\"\",&!\"\"F(F'\"\"#F*,&-%)binomialG6$F(F'#F(F+-F.6$\"\"!F'F* F(-%*hypergeomG6%7$F(F(7#,&F+F(F'F*F0F(F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 9 "Check it:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 87 "for kk fro m -5 to 5 do\n val := subs(k=kk,g1):\n print(` g1`(kk)=evalf(v al));\nod:" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$~g1G6#!\"&\"\"!" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$~g1G6#!\"%\"\"!" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#/-%$~g1G6#!\"$\"\"!" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6#/-%$~g1G6#!\"#\"\"!" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$~g1G6#!\" \"\"\"!" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$~g1G6#\"\"!F'" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$~g1G6#\"\"\"$!\"\"\"\"!" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$~g1G6#\"\"#\"\"!" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$~g1G6#\"\"$\"\"!" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$~g1G 6#\"\"%\"\"!" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$~g1G6#\"\"&\"\"!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 57 "Something is wrong... Maple is \+ confused by singularities." }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "subs(k=1,g1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6 #,$*&,&-%)binomialG6$\"\"\"F)#F)\"\"#-F'6$\"\"!F)!\"\"F)-%*hypergeomG6 %7$F)F)7#F)F*F)F/" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf( \");" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$!\"\"\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "subs(k=2,g1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&,&-%)binomialG6$\"\"\"\"\"##F)F*-F'6$\"\"!F*!\"\"F) -%*hypergeomG6%7$F)F)7#F.F+F)#!\"#\"\"$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(\");" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" } }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 133 "It evaluated binomial(1,2)=bino mial(0,2)=0, and then never evaluated hypergeom(...) because it though t 0*something=0. Let's redo it." }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "g1;" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#,$* *%\"kG\"\"\",&!\"\"F&F%\"\"#F(,&-%)binomialG6$F&F%#F&F)-F,6$\"\"!F%F(F &-%*hypergeomG6%7$F&F&7#,&F)F&F%F(F.F&F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 65 "Rewrite the hypergeom as an infinite sum over a new varia ble, r.." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 74 "g1r := subs(hypergeom([ 1,1],[2-k],1/2) = hyperterm([1,1],[2-k],1/2,r),g1);" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#>%$g1rG,$*.%\"kG\"\"\",&!\"\"F(F'\"\"#F*,&-%)binomial G6$F(F'#F(F+-F.6$\"\"!F'F*F(-%*factorialG6#%\"rGF(-%+pochhammerG6$,&F+ F(F'F*F7F*)F0F7F(F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "g1r \+ := simpcomb(g1r);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$g1rG,$**-%&GAM MAG6#,&%\"rG\"\"\"F,F,F,-F(6#,(\"\"#F,%\"kG!\"\"F+F,F2)#F,F0F+F,-F(6#F 1F2#F2F0" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 58 "Check that summing ov er r would give what we started with." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "g1r2:=sumtools[sumtohyper](g1r,r);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%g1r2G,$**-%$sinG6#*&%#PiG\"\"\"%\"kGF,F,F+!\"\",&F-F,F.F,F.-% *hypergeomG6%7$F,F,7#,&\"\"#F,F-F.#F,F6F,F7" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 60 "Using the reflection formula for the Gamma function con verts" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "binomial(0,k) = convert(bi nomial(0,k),GAMMA);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%)binomialG6$ \"\"!%\"kG*&-%&GAMMAG6#,&F(\"\"\"F.F.!\"\"-F+6#,&F.F.F(F/F/" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "simplify(\");" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#/-%)binomialG6$\"\"!%\"kG,$*(-%$sinG6#*&%#PiG\" \"\",&F(F0F0F0F0F0F(!\"\"F/F2F2" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 45 "explaining what just happened. Now continue." }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 101 "for kk from -5 to 5 do\n g1rk := limit(g1r,k=kk):\n print(` g1`(kk)=sum(g1rk,r=0..infini ty));\nod:" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$~g1G6#!\"&\"\"!" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$~g1G6#!\"%\"\"!" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#/-%$~g1G6#!\"$\"\"!" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6#/-%$~g1G6#!\"#\"\"!" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$~g1G6#!\" \"\"\"!" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$~g1G6#\"\"!F'" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$~g1G6#\"\"\"!\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$~g1G6#\"\"#!\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6 #/-%$~g1G6#\"\"$!\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$~g1G6#\" \"%!\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$~g1G6#\"\"&!\"\"" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 87 "Note: We had to do it term by term , maple doesn't know how to take the limit otherwise:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "limit(g1,k=2);" }}{PARA 12 "" 1 "" {XPPMATH 20 " 6#-%&limitG6$,$**%\"kG\"\"\",&!\"\"F)F(\"\"#F+,&-%)binomialG6$F)F(#F)F ,-F/6$\"\"!F(F+F)-%*hypergeomG6%7$F)F)7#,&F,F)F(F+F1F)F+/F(F," }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "limit(g1r2,k=2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%&limitG6$,$**-%$sinG6#*&%#PiG\"\"\"%\"kGF- F-F,!\"\",&F.F-F/F-F/-%*hypergeomG6%7$F-F-7#,&\"\"#F-F.F/#F-F7F-F8/F.F 7" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "limit(g1r2,k=-2);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#-%&limitG6$,$**-%$sinG6#*&%#PiG\"\"\"% \"kGF-F-F,!\"\",&F.F-F/F-F/-%*hypergeomG6%7$F-F-7#,&\"\"#F-F.F/#F-F7F- F8/F.!\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 3 "" 0 "" {TEXT -1 23 "Example 2. Gauss's 2F1:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "H ypergeom([a,b],[c],1) = pochhammer(c-b,-a)/pochhammer(c,-a);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%*HypergeomG6%7$%\"aG%\"bG7#%\"cG\"\"\"*&- %+pochhammerG6$,&F+F,F)!\"\",$F(F2F,-F/6$F+F3F2" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 25 "which also can be written" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "convert(rhs(\"),GAMMA);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#**-%&GAMMAG6#,(%\"cG\"\"\"%\"bG!\"\"%\"aGF+F)-F%6#,&F(F)F*F+F+-F %6#,&F(F)F,F+F+-F%6#F(F)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 118 "Ther e's no summation symbol in the above, but remember the hypergeometric \+ series notation abbreviates an infinite sum:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "F2a := hyperterm([a,b],[c],1,k);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$F2aG**-%+pochhammerG6$%\"aG%\"kG\"\"\"-F'6$%\"bGF*F+ -F'6$%\"cGF*!\"\"-%*factorialG6#F*F2" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "r2 := pochhammer(c-b,-a)/pochhammer(c,-a);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#r2G*&-%+pochhammerG6$,&%\"cG\"\"\"%\"bG! \"\",$%\"aGF-F+-F'6$F*F.F-" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "F2b := F2a/r2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$F2bG*.-%+poch hammerG6$%\"aG%\"kG\"\"\"-F'6$%\"bGF*F+-F'6$%\"cGF*!\"\"-%*factorialG6 #F*F2-F'6$,&F1F+F.F2,$F)F2F2-F'6$F1F9F+" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 45 "Also, there's no \"n\" in the above. Let a=-n." }}{PARA 0 "" 0 "" {TEXT -1 53 "We want to prove f2(n) = sum_k F2(n,k) equ als 1." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "F2c := subs(a=-n,F2b);" } }{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$F2cG*.-%+pochhammerG6$,$%\"nG!\"\" %\"kG\"\"\"-F'6$%\"bGF,F--F'6$%\"cGF,F+-%*factorialG6#F,F+-F'6$,&F3F-F 0F+F*F+-F'6$F3F*F-" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "R2 := WZcertificate(F2c,k,n);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#R2G**,( %\"cG\"\"\"%\"kGF(!\"\"F(F(F)F(,(F)F(F*F(%\"nGF*F*,(F'F(%\"bGF*F,F(F* " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 157 "This proves the original sum is true for integer n, i.e., integer a. Gauss didn't have the restri ction that it must be an integer. Here's a way around it.." } {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "F2d := \+ subs(a=-n-eps,F2b);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%$F2dG*.-%+poc hhammerG6$,&%\"nG!\"\"%$epsGF+%\"kG\"\"\"-F'6$%\"bGF-F.-F'6$%\"cGF-F+- %*factorialG6#F-F+-F'6$,&F4F.F1F+,&F*F.F,F.F+-F'6$F4F;F." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "R2d := WZcertificate(F2d,k,n);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%$R2dG**,(%\"cG\"\"\"%\"kGF(!\"\"F(F( F)F(,*%\"nGF*F*F(%$epsGF*F)F(F*,*F'F(%\"bGF*F,F(F-F(F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 195 "eps, b, c are constants that were just c arried along for the ride. For all integer n, Gauss's 2F1 sum holds f or arbitrary b,c and for a=-n-eps, where eps is also arbitrary; thus, \+ a is arbitrary." }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 36 "If we want, we can verify the proof:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "G2d := F2d * R2d;" }}{PARA 12 "" 1 "" {XPPMATH 20 "6# >%$G2dG*6-%+pochhammerG6$,&%\"nG!\"\"%$epsGF+%\"kG\"\"\"-F'6$%\"bGF-F. -F'6$%\"cGF-F+-%*factorialG6#F-F+-F'6$,&F4F.F1F+,&F*F.F,F.F+-F'6$F4F;F .,(F4F.F-F.F+F.F.F-F.,*F*F+F+F.F,F+F-F.F+,*F4F.F1F+F*F.F,F.F+" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "(subs(n=n+1,F2d)-F2d)/F2d = \+ (subs(k=k+1,G2d)-G2d)/F2d;" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#/*0,&*.- %+pochhammerG6$,(%\"nG!\"\"F,\"\"\"%$epsGF,%\"kGF--F(6$%\"bGF/F--F(6$% \"cGF/F,-%*factorialG6#F/F,-F(6$,&F5F-F2F,,(F+F-F-F-F.F-F,-F(6$F5F " 0 "" {MPLTEXT 1 0 12 "simplify(\");" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#/,$*(,0*&%\"nG \"\"\"%\"bGF)F)F*F)*&%$epsGF)F*F)F)*&%\"kGF)%\"cGF)F)*&F.F)F*F)!\"\"*& F(F)F.F)F)*&F.F)F,F)F)F),*F(F1F1F)F,F1F.F)F1,*F/F)F*F1F(F)F,F)F1F1F$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalb(\");" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%%trueG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 102 "S ee the tables on Koepf pp. 84, 87 for a lot of other identities that a re proved in this same fashion." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 " " }}}{EXCHG {PARA 3 "" 0 "" {TEXT -1 29 "Example 3. A=B Example 7.3.1 " }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "F3 : = binomial(n,k)^2/binomial(2*n,n);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6# >%#F3G*&-%)binomialG6$%\"nG%\"kG\"\"#-F'6$,$F)F+F)!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "R3 := WZcertificate(F3,k,n);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%#R3G,$**,(%\"nG!\"$F)\"\"\"%\"kG\"\" #F*F+F,,&F(F,F*F*!\"\",(F+F*F.F*F(F.!\"##F*F," }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 37 "So this implies that for n=0,1,2,...," }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "Sum(F3,k)=constant;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$SumG6$*&-%)binomialG6$%\"nG%\"kG\"\"#-F)6$,$F+F-F+! \"\"F,%)constantG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 22 "Evaluate the constant." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "sum(subs(n=0,F3),k=0. .0);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 44 "Thus, the above sum equals 1, so this proves" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "Sum(binomial(n,k)^2,k)=binomial(2*n ,n);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$SumG6$*$-%)binomialG6$%\"n G%\"kG\"\"#F,-F)6$,$F+F-F+" }}}{EXCHG {PARA 4 "" 0 "" {TEXT -1 18 "Com panion identity" }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "G3 := R3*F3;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#G3 G,$*.,(%\"nG!\"$F)\"\"\"%\"kG\"\"#F*F+F,,&F(F,F*F*!\"\",(F+F*F.F*F(F.! \"#-%)binomialG6$F(F+F,-F26$,$F(F,F(F.#F*F," }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 6 "Define" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "g3(k) = Sum (G3,n=0..infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%#g3G6#%\"kG- %$SumG6$,$*.,(%\"nG!\"$F/\"\"\"F'\"\"#F0F'F1,&F.F1F0F0!\"\",(F'F0F3F0F .F3!\"#-%)binomialG6$F.F'F1-F76$,$F.F1F.F3#F0F1/F.;\"\"!%)infinityG" } }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 65 "Summing F(n+1,k)-F(n,k) = G(n,k+ 1)-G(n,k) for n=0..infinity gives" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "'F3(infinity,k)'-'F3(0,k)' = g3(k+1)-g3(k);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/,&-%#F3G6$%)infinityG%\"kG\"\"\"-F&6$\"\"!F)!\"\",&-%# g3G6#,&F)F*F*F*F*-F16#F)F." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "subs(n=infinity,F3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&-%)binom ialG6$%)infinityG%\"kG\"\"#-F%6$F'F'!\"\"" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 21 "limit(F3,n=infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "subs(n=0,F3) ;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&-%)binomialG6$\"\"!%\"kG\"\"#-F %6$F'F'!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 86 "which is 1 if k=0 and is 0 otherwise, i.e., delta(0,k). So this proves for integer k, " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "g3(k+1)-g3(k) = piecewise(k<>0, 0-0,k=0,0-1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/,&-%#g3G6#,&%\"kG\" \"\"F*F*F*-F&6#F)!\"\"-%*PIECEWISEG6$7$\"\"!0F)F27$F-/F)F2" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 23 "Now evaluate at any k. " }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "sum(subs(k=0,G3),n=0.. infinity);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 38 "This proves g3(0)=0 so g3(1)=-1. Thus" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "g3(k) = piecewise(k>=1,-1,k<=0,0); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%#g3G6#%\"kG-%*PIECEWISEG6$7$!\" \"1\"\"\"F'7$\"\"!1F'F0" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 5 "i.e.," }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "Sum(G3,n=0..infinity)=rhs(\");" } }{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$SumG6$,$*.,(%\"nG!\"$F+\"\"\"%\"k G\"\"#F,F-F.,&F*F.F,F,!\"\",(F-F,F0F,F*F0!\"#-%)binomialG6$F*F-F.-F46$ ,$F*F.F*F0#F,F./F*;\"\"!%)infinityG-%*PIECEWISEG6$7$F01F,F-7$F<1F-F<" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 28 "which can be rearranged into" } }{PARA 0 "> " 0 "" {MPLTEXT 1 0 93 "Sum(convert(simpcomb(-2*subs(k=k+1 ,G3)),binomial),n=0..infinity) = piecewise(k>=0,2,k<=-1,0);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%$SumG6$**,(%\"nG\"\"$\"\"\"F+%\"kG!\"#F+- %)binomialG6$F)F,\"\"#,&F)F+F+F+!\"\"-F/6$,&F)F1F+F+F)F3/F);\"\"!%)inf inityG-%*PIECEWISEG6$7$F11F9F,7$F91F,F3" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 3 "" 0 "" {TEXT -1 47 "Example 4. Id entities where the WZ method fails" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 130 "F4 := (n,k) -> (-1)^k / (-3)^n * binomial(n,k) * binomial(3*k,n); \nf4 := n -> sum(F4(n,k),k=0..n); # interval 0..n contains support" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%#F4G:6$%\"nG%\"kG6\"6$%)operatorG%&a rrowGF)**)!\"\"9%\"\"\")!\"$9$F/-%)binomialG6$F4F0F1-F66$,$F0\"\"$F4F1 F)F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#f4G:6#%\"nG6\"6$%)operatorG %&arrowGF(-%$sumG6$-%#F4G6$9$%\"kG/F3;\"\"!F2F(F(" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 27 "WZcertificate(F4(n,k),k,n);" }}{PARA 8 "" 1 "" {TEXT -1 50 "Error, (in WZcertificate) extended WZ method fails" }} }{EXCHG {PARA 0 "" 0 "" {TEXT -1 36 "But the function really is consta nt:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "'f4(nn)'$nn=0..15;" }}{PARA 11 "" 1 "" {XPPMATH 20 "62\"\"\"F#F#F#F#F#F#F#F#F#F#F#F#F#F#F#" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 33 "Find a recursion it does satisfy: " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "sumrecursion(F4(n,k),k,s(n));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/,(*&,&%\"nG\"\"#\"\"$\"\"\"F*-%\"sG 6#,&F'F*F(F*F*F(*&,&F'\"\"&\"\"(F*F*-F,6#,&F'F*F*F*F*!\"\"*&F5F*-F,6#F 'F*F*\"\"!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 64 "Thus, L F = 0 where L=L4=2(2n+3)E^2 - (5n+7)E + (n+1) and F=F4." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "L4 := 2*(2*n+3)*E^2 - (5*n+7)*E + (n+1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#L4G,**&,&%\"nG\"\"#\"\"$\"\"\"F+%\"EGF)F)*&, &F(\"\"&\"\"(F+F+F,F+!\"\"F(F+F+F+" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 434 "The software we're using now doesn't handle this noncommutativ e algebra, but we can \"fake\" it in this case: if we systematically k eep n's on the left and E's on the right, then any factorization of th e form\n(*) p(n) * q(n,E) * r(E)\n(where within q, the n's a re on the left and E's are on the right) results in the same computati on whether it's done commutatively or noncommutatively. Do the normal \+ commutative factorization:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "facto r(L4);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&,&%\"EG\"\"\"!\"\"F&F&,**& F%F&%\"nGF&\"\"%F%\"\"'F*F'F'F&F&" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 303 "This expression is literally false. As far as maple is concerned , n*E=E*n, although in reality, (n+1)*E = E*n. But we can make sense \+ of it: since this was treated as a commutative polynomial, Maple think s it's the same as\n L4 = ((4n-6)E - (n+1)) * (E-1)\nand this h as the form (*) given above. " }{TEXT 258 4 "This" }{TEXT -1 31 " is \+ a valid factorization. So\n" }{MPLTEXT 1 0 42 "((4*n-6)*E - (n+1)) * \+ (E-1) * 'f4(n)' = 0;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/*(,(*&,&%\"nG \"\"%!\"'\"\"\"F+%\"EGF+F+F(!\"\"F-F+F+,&F,F+F-F+F+-%#f4G6#F(F+\"\"!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 130 "is true. Now we want to prove that in fact (E-1) f4(n) = 0, which is a right factor of this whole m ess. Let h4(n) = (E-1) f4(n)." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "h4 := n -> f4(n+1) - f4(n);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6 #>%#h4G:6#%\"nG6\"6$%)operatorG%&arrowGF(,&-%#f4G6#,&9$\"\"\"F2F2F2-F. 6#F1!\"\"F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 165 "Then ((4n-6)E - (n+1)) h4(n) = 0. This has order 1, so it has a solution A * h4a(n) \+ for some function h4a(n), and the constant A is determined by one init ial value:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "h4(0); 'h4(nn)'$'nn'= 0..10; " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" }}{PARA 11 "" 1 "" {XPPMATH 20 "6-\"\"!F#F#F#F#F#F#F#F#F#F#" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 185 "Observe that h4(n)=0 satisfies the recursion equation an d the initial value, and therefore the sole solution is h4(n) = 0 iden tically. So (E-1) f4(n)=0, so f4(n+1)=f4(n), as we wanted." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "0 0 0" 13 }{VIEWOPTS 1 1 0 1 1 1803 }