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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 256 48 "Multiplica
tion of Linear Differential Operators." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
1412 "mult := proc(L1, L2, Dx, x)\n   local answer,c0,the_rest,DxL2,i;
\n   # options trace;\n   if nargs<4 then\n      # The Dx and x are no
t specified.\n      if assigned(_Envdiffopdomain) then\n          # Th
e names of Dx and x are in _Envdiffopdomain:\n          RETURN(procnam
e(L1,L2, op(_Envdiffopdomain)))\n      else\n          # The two entri
es Dx,x are not specified, and\n          # these names are also not s
pecified in\n          # _Envdiffopdomain so we have no way to figure \+
out\n          # which variable (here denoted by x) is the\n          \+
# independent variable, and which one (here denoted\n          # by Dx
) stands for differentiation. So all we can\n          # do is:\n     \+
     ERROR(\"Variables were not specified\")\n      fi\n   fi;\n   if \+
not has(L1,Dx) then\n      # Dx does not appear in L1, in other words:
\n      # degree(L1,Dx)<1. So the product is commutative:\n      answe
r := L1*L2\n   else\n      c0:=coeff(L1,Dx,0);\n      the_rest := add(
coeff(L1,Dx,i)*Dx^(i-1),i=1..degree(L1,Dx));\n      # Now L1 = c0 + th
e_rest * Dx\n      # Now compute Dx*L2 with the formula Dx*a=a'+a*Dx\n
      DxL2 := diff(L2,x) + L2*Dx;\n      answer := c0*L2 + procname(th
e_rest,DxL2,Dx,x)\n  fi;\n  # Now we have the answer, lets apply the f
ollowing\n  # two commands to make it look nicer:\n  answer := collect
(answer,Dx,normal);\n  # Make sure Maple puts the x on the left and\n \+
 # the Dx on the right side:\n  sort(answer,[x,Dx])\nend:" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "mult(Dx,x);" }}{PARA 8 "" 1 "" 
{TEXT -1 45 "Error, (in mult) Variables were not specified" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "mult(Dx,x,Dx,x);" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#,&*&%\"xG\"\"\"%#DxGF&F&F&F&" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 27 "_Envdiffopdomain := [Dx,x];" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%1_EnvdiffopdomainG7$%#DxG%\"xG" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 11 "mult(Dx,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#,&*&%\"xG\"\"\"%#DxGF&F&F&F&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 14 "mult(Dx^10,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*&%\"xG\"\"
\")%#DxG\"#5\"\"\"F&*$)F(\"\"*F*F)" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 14 "mult(Dx,x^10);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*
&)%\"xG\"#5\"\"\"%#DxG\"\"\"F**$)F&\"\"*F(F'" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 20 "mult(Dx^2+x,Dx^2-x);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#,(*$)%#DxG\"\"%\"\"\"\"\"\"*$)%\"xG\"\"#F(!\"\"F&!\"#" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "mult(Dx^2-x,Dx^2+x);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#,(*$)%#DxG\"\"%\"\"\"\"\"\"*$)%\"xG\"
\"#F(!\"\"F&F-" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 257 42 
"Division of Linear Differential Operators." }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 29 "L1 := x*Dx^5+3*x^2*Dx^3+Dx-1;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#L1G,**&%\"xG\"\"\")%#DxG\"\"&\"\"\"F(*&)F'\"\"#F,)F*
\"\"$F,F1F*F(!\"\"F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "L2 \+
:= 2*Dx^3-x^3*Dx+x;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#L2G,(*$)%#Dx
G\"\"$\"\"\"\"\"#*&)%\"xGF)F*F(\"\"\"!\"\"F.F/" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 34 "Do a Right-Division, meaning that:" }}{PARA 0 "" 0 "
" {TEXT -1 80 "Find an operator Q (right-hand quotient) and R (right-h
and remainder) such that:" }}{PARA 0 "" 0 "" {TEXT -1 20 "L1 = Q*L2 + \+
R    (*)" }}{PARA 0 "" 0 "" {TEXT -1 4 "AND:" }}{PARA 0 "" 0 "" {TEXT 
-1 37 "order(R) < order(L2), in other words:" }}{PARA 0 "" 0 "" {TEXT 
-1 34 "degree(R,Dx) < degree(L2,Dx)  (**)" }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 54 "Note: A left-hand division would b
e: L1 = L2 * Ql + Rl" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "Q:=0
;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"QG\"\"!" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 6 "R:=L1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"
RG,**&%\"xG\"\"\")%#DxG\"\"&\"\"\"F(*&)F'\"\"#F,)F*\"\"$F,F1F*F(!\"\"F
(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "L1 = mult(Q,L2) + R;" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6#/,**&%\"xG\"\"\")%#DxG\"\"&\"\"\"F'*
&)F&\"\"#F+)F)\"\"$F+F0F)F'!\"\"F'F$" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 28 "degree(R,Dx), degree(L2,Dx);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6$\"\"&\"\"$" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 203 "As \+
we can see, with our choice of Q,R, we have that (*) holds, and (**) d
oes not hold. Now we want to change Q,R, in such a way that (*) stays \+
true, and that order(R) drops so that we get closer to (**)." }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 77 "To make the ord
er of R go down, we need to eliminate its highest coefficient:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "lcoeff(R,Dx);" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#%\"xG" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
76 "s := normal(lcoeff(R,Dx)/lcoeff(L2,Dx)) * Dx^(degree(R,Dx) - degre
e(L2,Dx));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"sG,$*&%\"xG\"\"\")%#
DxG\"\"#\"\"\"#F(F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "Q:=Q
+s;   R:=R-mult(s,L2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"QG,$*&%
\"xG\"\"\")%#DxG\"\"#\"\"\"#F(F+" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%
\"RG,.*&)%\"xG\"\"#\"\"\")%#DxG\"\"$F*F-F,\"\"\"!\"\"F.*&)F(\"\"%F*F+F
*#F.F)*(,&F(\"\"'F/F.F.F'F*)F,F)F*F3*(,&F(F-F/F.F.F(F.F,F.F." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "R:=collect(R,Dx,normal);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"RG,**&,&*$)%\"xG\"\"%\"\"\"#\"\"\"
\"\"#*$)F*F/F,\"\"$F.)%#DxGF2F,F.*(,&F*\"\"'!\"\"F.F.F1F,)F4F/F,F-*&,&
F.F.*&F*F.,&F*F2F8F.F.F.F.F4F.F.F8F." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 32 "normal( L1 - (mult(Q,L2) + R) );" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#\"\"!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 27 "0, so (*)
 still holds, but:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "degre
e(R,Dx), degree(L2,Dx);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$\"\"$F#" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 69 "(**) does not yet hold. However, \+
we're getting closer. One more step:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 76 "s := normal(lcoeff(R,Dx)/lcoeff(L2,Dx)) * Dx^(degree(
R,Dx) - degree(L2,Dx));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"sG,&*$)
%\"xG\"\"%\"\"\"#\"\"\"F)*$)F(\"\"#F*#\"\"$F/" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 26 "Q:=Q+s;   R:=R-mult(s,L2);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"QG,(*&%\"xG\"\"\")%#DxG\"\"#\"\"\"#F(F+*$)F'\"\"%F,
#F(F0*$)F'F+F,#\"\"$F+" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"RG,,*(,&
%\"xG\"\"'!\"\"\"\"\"F+)F(\"\"#\"\"\")%#DxGF-F.#F+F-*&,&F+F+*&F(F+,&F(
\"\"$F*F+F+F+F+F0F+F+F*F+*(,&*$F,F.F+F)F+F+)F(\"\"&F.F0F.#F+\"\"%*&F8F
.)F(F6F.#F*F=" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "R:=collect
(R,Dx,normal);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"RG,**(,&%\"xG\"
\"'!\"\"\"\"\"F+)F(\"\"#\"\"\")%#DxGF-F.#F+F-*&,(F+F+*&F(F+,&F(\"\"$F*
F+F+F+*&)F(\"\"&F.,&*$F,F.F+F)F+F+#F+\"\"%F+F0F+F+*&F:F.)F(F6F.#F*F=F*
F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "normal( L1 - (mult(Q,
L2) + R) );" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 28 "degree(R,Dx), degree(L2,Dx);" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6$\"\"#\"\"$" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
52 "Now (*) and (**) both hold. So we have written L1 as" }}{PARA 0 "
" 0 "" {TEXT -1 45 "L1 = Q*L2 + R   where   order(R) < order(L2)." }}
{PARA 0 "" 0 "" {TEXT -1 59 "We've computed the (right-hand) quotient \+
Q and remainder R." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }{TEXT 258 31 "Assignment: Write a procedure:\n" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 259 109 "rightdivision := proc(L1
, L2, Dx, x)\n   ... compute Q, R as above\n   ....\n  [Q, R]  # Outpu
t is Q and R.\nend:" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 80 "You can get hints on how to write this pr
ogram from the usual remainder program:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 220 "myREM:=proc(a,b,x)\n  local R,Q,s;\n  # options trac
e;\n  R:=a;\n  Q:=0;\n  while degree(R,x) >= degree(b,x) do\n    s:=lc
oeff(R,x)/lcoeff(b,x) * x^(degree(R,x)-degree(b,x));\n    Q:=Q+s;\n   \+
 R:=expand(R-s*b)\n  od;\n  [Q,R]\nend:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 35 "a:=2*x^4-3*x^2; b:=2*x^3-3*x^2+x-1;" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#>%\"aG,&*$)%\"xG\"\"%\"\"\"\"\"#*$)F(F+F*!\"$" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"bG,**$)%\"xG\"\"$\"\"\"\"\"#*$)F(F
+F*!\"$F(\"\"\"!\"\"F/" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "m
yREM(a,b,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7$,&%\"xG\"\"\"#\"\"$
\"\"#F&,(*$)F%F)\"\"\"#F&F)F%#!\"\"F)F'F&" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 13 "q,r := op(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>6
$%\"qG%\"rG6$,&%\"xG\"\"\"#\"\"$\"\"#F*,(*$)F)F-\"\"\"#F*F-F)#!\"\"F-F
+F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "a = q*b+r;" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#/,&*$)%\"xG\"\"%\"\"\"\"\"#*$)F'F*F)!\"$,**&
,&F'\"\"\"#\"\"$F*F1F1,**$)F'F3F)F*F+F-F'F1!\"\"F1F1F1F+#F1F*F'#F7F*F2
F1" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "expand(lhs(%) - rhs(%
));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 3 "OK." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }{TEXT 260 30 "Greatest Common Right Divisor." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 250 "myGCD:=proc(a,b,x)\n   loca
l r;\n   if b=0 then\n       # Make a monic, leads to smaller numbers:
\n       collect(a/lcoeff(a,x), x, normal)\n   else\n       r:=myREM(a
,b,x);\n       r:=r[2]; # 1'st=quotient, 2'nd=remainder\n       procna
me(b,r,x)\n   fi:\nend:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "
myGCD(x^6-1,x^8-1,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&!\"\"\"\"\"
*$)%\"xG\"\"#\"\"\"F%" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 
261 245 "Assignment:\nWrite a procedure:\nGCRD := proc(L1,L2, Dx,x)\n \+
  ....\n   ... use your procedure rightdivision\n   ...\nend:\n\nThat \+
computes the Greatest Common Right Divisor of two operators L1, L2. If
 R=GCRD(L1,L2) then:\n  V(R) = V(L1) intersect V(L2)." }}{PARA 0 "" 0 
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 133 "Why? Well, during the
 computation L1 and L2 are constantly changing. L2 is replaced by its \+
remainder modulo L1. So L2 is replaced by:" }}{PARA 0 "" 0 "" {TEXT 
-1 38 " new_L2 = remainder = L2 - quotient*L1" }}{PARA 0 "" 0 "" 
{TEXT -1 81 "Now it is not hard to see that V(L1) intersect V(L2) = V(
L1) intersect V(new_L2)." }}{PARA 0 "" 0 "" {TEXT -1 359 "So this stay
s truth throughtout the algorithm. However, at the end we have that L1
, L2 have become R, 0  (one of the operators becomes 0, that's where t
he algorithm ends, and then the other one is the GCRD). So we have V(L
1) intersect V(L2) = V(R) intersect V(0), which equals V(R) because ev
ery function in V(R) is automatically a solution of the operator 0." }
}}}{MARK "43 5 0" 359 }{VIEWOPTS 1 1 0 1 1 1803 }