{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 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 257 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 } {CSTYLE "" -1 260 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 261 "terminal" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 262 "" 1 24 0 0 0 0 0 0 0 0 0 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Tim es" 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 "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 "R3 Font 0" -1 256 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" -1 257 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 "" 3 258 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 }} {SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 261 46 "/home/m262f99/CHYZAK/work sheetsV.4/hw7ansm.mws" }{MPLTEXT 1 0 0 "" }}{PARA 258 "" 0 "" {TEXT 262 54 "Math 262a, Fall 1999, Glenn Tesler\nHomework 7\n11/19/99" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 85 "with(Ore_algebra):\ncleanpol := (f,Sn) -> sort(collect(expand(f),Sn,factor),[Sn,x,n]):" }}}{EXCHG {PARA 3 "" 0 "" {TEXT -1 10 "Problem 1." }}{PARA 0 "" 0 "" {TEXT -1 18 "Set up an algebra." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "An := shi ft_algebra([Sn,n]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#AnG%,Ore_alg ebraG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 45 "Express the recurrences \+ in operator notation." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "recF := Sn ^2 - (n+3)*Sn + 2*n;\nrecG := Sn^2 - (2*n+1)*Sn + n^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%recFG,(*$%#SnG\"\"#\"\"\"*&,&%\"nGF)\"\"$F)F)F' F)!\"\"F,F(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%recGG,(*$%#SnG\"\"# \"\"\"*&,&%\"nGF(F)F)F)F'F)!\"\"*$F,F(F)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 38 "Find the gcd and lcm of the operators." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "G := skew_gcdex(recF,recG,Sn,An);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"GG7',*%\"nG\"\"#*&%#SnG\"\"\"F'F+F+F*!\"#*$F'F (!\"\"F+F.,*F-F+F'F.F)F.F*F(,*F'F,F(F+F)F+F*F," }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 29 "Decode this into gcd and lcm:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 77 "gcdFG := cleanpol(G[1],Sn);\nlcmFG := cleanpol(skew_p roduct(G[4],recF,An),Sn);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&gcdFGG ,&*&,&%\"nG\"\"\"!\"#F)F)%#SnGF)F)*&F'F)F(F)!\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&lcmFGG,**&,&%\"nG!\"\"\"\"#\"\"\"F+%#SnG\"\"$F+*&,(* $F(F*F*F(F+!\")F+F+F,F*F+*&,&F(F+F)F+F+F(F*F**&,**$F(F-F)F0!\"%F(\"\"& \"\"%F+F+F,F+F+" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 256 3 "(a)" }{TEXT -1 118 " The lcm is the operator notation for the minimal annihilator \+ satisfied by both functions. In function notation it is" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "applyopr(lcmFG,a(n),An) = 0;" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#/,**(,&%\"nG\"\"\"!\"\"F(F(F'\"\"#-%\"aG6#F'F(F**& ,**$F'\"\"$F)*$F'F*!\"%F'\"\"&\"\"%F(F(-F,6#,&F'F(F(F(F(F(*&,(F2F*F'F( !\")F(F(-F,6#,&F'F(F*F(F(F(*&,&F'F)F*F(F(-F,6#,&F'F(F1F(F(F(\"\"!" }}} {EXCHG {PARA 0 "" 0 "" {TEXT 257 3 "(b)" }{TEXT -1 286 " If indeed f(n )=g(n), then the gcd of the two operators is an operator annihilating \+ f(n), and rational multiples of it are too, except their poles/zeros m ay give exceptional values of n. Here, the gcd is (n-2)(Sn-n), so the function satisfies (Sn-n)f(n)=0 except possibly at n=2. Thus" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "f(n)=C*(n-1)!;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"fG6#%\"nG*&%\"CG\"\"\"-%*factorialG6#,&F'F*!\"\"F* F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 128 "for n>=1, for some constan t C, with a possible exception at n=2. However, setting n=1 in the or iginal recurrence for f(n) gives" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "subs(n=1,applyopr(recF,f(n),An))=0;" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6#/,(-%\"fG6#\"\"\"\"\"#-F&6#F)!\"%-F&6#\"\"$F(\"\"!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 91 "and the values f(1)=C*0!=C, f(3)=C*2!=2C, force f(2)=C=C*(2-1)!, so it's still true at n=2." }{MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT 258 4 "(c) " }{TEXT -1 721 " The Euclidea n algorithm produced lcm = u*recF = v*recG. The way we did it in cla ss, u and v could have denominators (functions of n), but this would b e bad if we divided by 0. Chyzak's implementation clears denominators , resulting in extra factors in the lcm. A complete procedure would b e to find the lcm of some order D, s.t. u,v have no denominators. Her e, D=3. Verify the first D initial conditions, f(1)=g(1), f(2)=g(2), \+ f(3)=g(3). Then, all further values of f(n), g(n) are equal by iterat ing the recursion, EXCEPT we cannot deduce f(N+D)=g(N+D) when the lead ing term of the lcm has a nonnegative integer root N, so we must check this separately. Here this happens for N=2, so we also must check f( 5)=g(5)." }}}{EXCHG {PARA 0 "" 0 "" {TEXT 259 3 "(d)" }{TEXT -1 41 " U sing the common right factor Sn-n gives" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "divF:=skew_pdiv(recF,Sn-n,Sn,An); divG:=skew_pdiv(recG,Sn-n,Sn ,An);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%divFG7%\"\"\",&%#SnGF&!\"# F&\"\"!" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%divGG7%\"\"\",&%#SnGF&% \"nG!\"\"\"\"!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 236 "meaning 1*recF - (Sn-2)*(Sn-n) = 0 (the reason for the multiple of recF is again to clear denominators; instead of 1 it could be a function of n, but is \+ still n-free). So this says recF = (Sn-2)(Sn-n) and recG=(Sn-n)(Sn-n) . Check it:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "skew_product(Sn-2,S n-n,A); simplify(\"-recF);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,(%\"nG \"\"#*&,&F$!\"\"!\"$\"\"\"F*%#SnGF*F**$F+F%F*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "ske w_product(Sn-n,Sn-n,A); simplify(\"-recG);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,(*$%\"nG\"\"#\"\"\"*&,&F%!\"#!\"\"F'F'%#SnGF'F'*$F,F&F '" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 3 "" 0 "" {TEXT -1 43 "Problem 2. \+ The old way is with Wronskians:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 605 "wronsk := proc(f,D,A)\n local n,m,i,j;\n n := nops(f);\n \+ m := linalg[matrix](n,n);\n for i from 1 to n do\n for j from 0 to n-1 do\n m[i,j+1] := applyopr(skew_power(D,j,A),f[i],A);\n \+ od od;\n RETURN(evalm(m))\nend:\nwronskeq := proc(f,D,A)\n lo cal W, yy,n,eqn;\n W := wronsk(f,D,A);\n n := nops(f);\n yy : = ['W[1,n+1-j]'$j=1..n];\n eqn := linalg[det](W);\n\n # clear th e leading coefficient & simplify\n eqn := eqn / coeff(eqn,yy[1]);\n eqn := simplify(eqn); eqn := numer(eqn);\n\n # collect it and c lean it up\n eqn := sort(collect(eqn,\{op(yy)\},factor),yy) = 0;\ne nd: " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "Ax := diff_algebra ([Dx,x]): An := shift_algebra([Sn,n]):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "W := wronsk([y(x),sin(x),x],Dx,Ax);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"WG-%'MATRIXG6#7%7%-%\"yG6#%\"xG-%%diffG6$F*F--F/ 6$F.F-7%-%$sinGF,-%$cosGF,,$F4!\"\"7%F-\"\"\"\"\"!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 29 "Minimal order diffeq possible" }}{PARA 0 "> " 0 " " {MPLTEXT 1 0 32 "wronskeq([y(x),sin(x),x],Dx,Ax);" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#/,(*&,&-%$sinG6#%\"xG!\"\"*&F*\"\"\"-%$cosGF)F-F-F--% %diffG6$-F16$-%\"yGF)F*F*F-F-*(F*F-F'F-F3F-F-*&F'F-F5F-F+\"\"!" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 212 "Or, we know the minimal diffeq wi th rational coeffs having sin(x) as a solution is y''+y=0, and cos(x) \+ is another solution of it. So the minimal equation with only rational functions of x as coefficients will be" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "wronskeq([y(x),sin(x),cos(x),x],Dx,Ax);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/,**&%\"xG\"\"\"-%%diffG6$-F)6$-F)6$-%\"yG6#F&F&F&F&F'F 'F+!\"\"*&F&F'F-F'F'F/F2\"\"!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 57 " Similarly in (b) the minimal order recurrence equation is" }}{PARA 0 " > " 0 "" {MPLTEXT 1 0 42 "fibeq := wronskeq([f(n),n!,Fib(n)],Sn,An);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%&fibeqG/,(*&,(-%$FibG6#,&%\"nG\"\" \"F.F.!\"\"*&-F*6#F-F.F-F.F.F1F.F.-%\"fG6#,&F-F.\"\"#F.F.F.*&,*-F*F5F. F0!\"$F1!\"#*&F1F.F-F7F/F.-F4F+F.F.*(F,F.,(*&F)F.F-F.F.F)F7F:F/F.-F4F2 F.F.\"\"!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 440 "and we could plug i n the explicit formula for Fibonacci numbers\n Fib(n) = ( w1 ^ (n +1) - w2 ^ (n+1) ) / sqrt(5)\nwith w1 = (1+sqrt(5))/2, w2 = -1/w1 = (1-sqrt(5))/2\nto make this explicit, but that will make it uglier.\n Let Sr be the algebra of shift operators with rational functions of n \+ as coefficients.\nTo get a recurrence in Sr with n! and Fib(n) as solu tions, we must take all the solutions of the minimal equation Fib(n) i n Sr.\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "fibeq2 := wronsk eq([f(n),n!,w1^n,(-1/w1)^n],Sn,An);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6 #>%'fibeq2G/,**(%#w1G\"\"\",0*&F(\"\"#%\"nGF)!\"\"*$F(F,F.F(F)*&F(F)F- F)\"\"$*&F(F)F-F,F)F-F)F)F)F)-%\"fG6#,&F-F)F1F)F)F)*&,6F)F)F-F)F(F.*$F (\"\"%F)*&F(F:F-F)F)F/!\"(F+!#7*&F(F,F-F1F.*&F(F,F-F,!\"'*$F(F1F)F)-F4 6#,&F-F)F,F)F)F)*&,D*&F(F1F-F,\"\"'!\"#F)F-!\"$F(F@F9FI*$F-F,F.FAFH*&F -F1F(F)F.F0!#6F/F1F2F@F;FJ*&F(F:F-F,F.F+F1*&F(F1F-F1F)*&F(F1F-F)\"#6F? F)F)-F46#,&F-F)F)F)F)F)**F(F)FTF),0F2F)F+F.F-F)F0\"\"&F/FIF,F)F(FWF)-F 46#F-F)F)\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 195 "ff := [' f(n+k)'$k=0..3]:\nfibeq3 := subs(w1=(1+sqrt(5))/2,lhs(fibeq2)):\nfibeq 3 := fibeq3/coeff(fibeq3,f(n+3)):\nfibeq3 := collect(fibeq3,ff,factor) :\nfibeq3 := collect(numer(fibeq3),ff,factor) = 0;\n" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%'fibeq3G/,**(,&%\"nG\"\"\"\"\"$F*F*,&F)F*F*F*\"\"# -%\"fG6#F)F*F**&,*F)\"\"&*$F)F+F*F+F**$F)F-\"\"%F*-F/6#F,F*F**&,*F5!\" '!\"$F*F)!\"*F4!\"\"F*-F/6#,&F)F*F-F*F*F**(F)F*FAF*-F/6#F(F*F*\"\"!" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 " " {TEXT 260 26 "The new way is with LCM's." }{TEXT -1 258 " The minim al operator in the Weyl algebra annihilating sin(x) is from the first \+ order equation\n (sin x)*y'(x) - (sin x)'*y(x) = 0, or (Dx - tan(x)) *y(x)=0.\nThe minimal operator annihilating x is from the equation\n \+ x*y' - x'*y = 0 or (x Dx - 1) y = 0" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "Isinx := Dx-tan(x); Ix := x*Dx-1; G := skew_gcdex(Isi nx,Ix,Dx,Ax);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&IsinxG,&%#DxG\"\" \"-%$tanG6#%\"xG!\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#IxG,&*&%\" xG\"\"\"%#DxGF(F(!\"\"F(" }}{PARA 8 "" 1 "" {TEXT -1 92 "Error, skew_g cdex expects its 1st argument, P, to be of type polynom, but received \+ Dx-tan(x)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 325 "The package won't d o this computation because it expects polynomials. If we did it by ha nd we would have the results given previously.\nThe minimal operator i n the rational Weyl algebra Wr is obtained by using the minimal annihi lators of the two functions in Wr. The one for sin(x) is changed to\n y''+1=0 or (Dx^2+1)y=0" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "Isinx := Dx^2+1;\nG := skew_gcdex(Isinx,Ix,Dx,Ax);" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#>%&IsinxG,&*$%#DxG\"\"#\"\"\"F)F)" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#>%\"GG7'%\"xGF&,$%#DxG!\"\",&*&F&\"\"#F(\"\"\"F) F&F-,(F&F-*&F&F-F(F,F-F(!\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "lc := cleanpol(skew_product(G[4],Isinx,Ax),Dx);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#lcG,**&%\"xG\"\"#%#DxG\"\"$!\"\"*&F'\"\"\"F)F(F -*&F'F(F)F-F+F'F-" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "lc := \+ cleanpol(\"/x,Dx);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#lcG,**&%\"xG \"\"\"%#DxG\"\"$!\"\"*$F)\"\"#F(*&F'F(F)F(F+F(F(" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 34 "The minimal order equation is then" }}{PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 24 "applyopr(\",y(x),Ax) = 0;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/,*-%\"yG6#%\"xG\"\"\"*&F(F)-%%diffG6$F%F(F)!\"\"-F,6$F +F(F)*&F(F)-F,6$F/F(F)F.\"\"!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 223 "For (b) we'll have the same problem with the software for the minimal operator in the shift algebra. In the rational shift algebra, we hav e\n (Sn-(n+1)) y(n) = 0 for y(n)=n!\n (Sn^2 - Sn - 1) y(n) = 0 f or y(n) = Fib(n)" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "I1 := Sn-(n+1): I2 := Sn^2-Sn-1: G := skew_gcdex(I1,I2,Sn,An);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%\"GG7',&%\"nG\"\"#*$F'F(\"\"\",(!\"\"F*%#SnGF,F'F,F*, 2F'!\"%F)F,!\"$F**&F-F*F'F*F0F-F0*&F'F(F-F*F,*&F-F(F'F*F(*&F'F(F-F(F*, .\"\"$F**$F'F6F*F'\"\"(F)\"\"&F1!\"#F2F," }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 56 "The minimal operator in Sr annihilating n! and Fib(n) is " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "lc := cleanpol(skew_product(G[4 ],I1,An),Sn);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%#lcG,**(,&%\"nG\"\" \"\"\"#F)F)%#SnG\"\"$F(F)F)*&,**$F(F,!\"\"*$F(F*!\"'F(!\"*!\"$F)F)F+F* F)*&,*F/F)F1\"\"%F(\"\"&F,F)F)F+F)F)*&,&F(F)F,F)F),&F(F)F)F)F*F)" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 32 "and in function notation this is" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "applyopr(lc,a(n),An) = 0;" }} {PARA 12 "" 1 "" {XPPMATH 20 "6#/,**(,&%\"nG\"\"\"\"\"$F(F(,&F'F(F(F( \"\"#-%\"aG6#F'F(F(*&,**$F'F)F(*$F'F+\"\"%F'\"\"&F)F(F(-F-6#F*F(F(*&,* F1!\"\"F2!\"'F'!\"*!\"$F(F(-F-6#,&F'F(F+F(F(F(*(F'F(F?F(-F-6#F&F(F(\" \"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 20 "fn := 5*n^3 + 4*n^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#fnG,&*$%\"nG\"\"$\"\"&*$F'\"\"#\"\"%" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "expand(applyopr(Sn-1,fn,An));" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#,(*$%\"nG\"\"#\"#:F%\"#B\"\"*\"\"\"" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "expand(applyopr(Sn-1,\",An ));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&\"#Q\"\"\"%\"nG\"#I" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "expand(applyopr(Sn-1,\",An)) ;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"#I" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 49 "FF := 9/2*ff(n,2) + 38/6*ff(n,3) + 30/24*ff(n,4);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#FFG,(*&%\"nG\"\"\",&F'F(!\"\"F(F( #\"\"*\"\"#*(F'F(F)F(,&F'F(!\"#F(F(#\"#>\"\"$**F'F(F)F(F/F(,&F'F(!\"$F (F(#\"\"&\"\"%" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "expand(FF );" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,**$%\"nG\"\"##!\"$\"\"%F%#F&\" \"$*$F%F+#!\"(\"\"'*$F%F)#\"\"&F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "expand(subs(n=m+1,\"));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,**$%\"mG\"\"##\"#8\"\"%F%#F&\"\"$*$F%F+#\"#B\"\"'*$F%F)#\"\"&F) " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "0 0 0" 42 } {VIEWOPTS 1 1 0 1 1 1803 }