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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 256 62 "Differenti
ation of solutions of linear differential operators." }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 171 "Let L be a differential operator. Consider the
 set of solutions of L which is a vector space of dimension n=degree(L
,Dx). Now consider the following map (differentiation):" }}{PARA 0 "" 
0 "" {TEXT -1 4 "    " }{TEXT 263 1 "D" }{TEXT -1 19 ": V(L)  ---> ?  \+
(*)" }}{PARA 0 "" 0 "" {TEXT -1 62 "(note in the Maple commands we use
 the notation Dx instead of " }{TEXT 266 1 "D" }{TEXT -1 56 "). What i
s the image of this map? We can write it as Im(" }{TEXT 264 1 "D" }
{TEXT -1 4 ") = " }{TEXT 265 1 "D" }{TEXT -1 120 "( V(L) ). Can we als
o write Im as V(M) for some differential operator M? First we have to \+
know the order of M, which is:" }}{PARA 0 "" 0 "" {TEXT -1 6 "   1) " 
}{TEXT 269 23 "order(M) = dim( Im(D) )" }}{PARA 0 "" 0 "" {TEXT -1 60 
"(assuming that such M exists, but we will see that it does)." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 56 "So what i
s this dimension? What is the dimension of Ker(" }{TEXT 267 1 "D" }
{TEXT -1 9 ") and Im(" }{TEXT 268 1 "D" }{TEXT -1 73 ") in (*)? Recall
 that dim(V(L))=n. Then we know from linear algebra that:" }}{PARA 0 "
" 0 "" {TEXT -1 6 "  2)  " }{TEXT 262 33 "dim( Im(D) ) = n - dim( Ker(
D) )." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 12 "
If we apply " }{TEXT 271 1 "D" }{TEXT -1 37 " on all functions then th
e kernel of " }{TEXT 272 1 "D" }{TEXT -1 27 " is the field of constant
s " }{TEXT 273 3 "C. " }{TEXT -1 39 "So then the dimension of the kern
el of " }{TEXT 274 1 "D" }{TEXT -1 20 " is a 1-dimensional " }{TEXT 
275 1 "C" }{TEXT -1 39 "-vector space. However, we don't apply " }
{TEXT 276 1 "D" }{TEXT -1 64 " on all possible functions, but only on \+
the set V(L). Therefore:" }}{PARA 0 "" 0 "" {TEXT -1 5 "  3) " }{TEXT 
277 59 "For the map (*) we have:  Ker(D) = C intersected with V(L)." }
}{PARA 0 "" 0 "" {TEXT -1 39 "and as an immediate consequence we get:
" }}{PARA 0 "" 0 "" {TEXT -1 10 "  4)      " }{TEXT 278 65 "If C is a \+
subset of V(L) then dim(Ker(D)) = 1 and  dim(Im(D))=n-1" }}{PARA 256 "
" 0 "" {TEXT -1 71 "    If C is not a subset of V(L) then dim(Ker(D)) \+
= 0  and dim(Im(D))=n" }}{PARA 0 "" 0 "" {TEXT -1 8 "When is " }{TEXT 
279 1 "C" }{TEXT -1 25 " a subset of V(L)? Well, " }{TEXT 280 1 "C" }
{TEXT -1 3 "=V(" }{TEXT 281 1 "D" }{TEXT -1 107 "). Now for every K, w
e have that V(K) is a subset of V(L) if and only if K is a right-hand \+
factor of L. So:" }}{PARA 0 "" 0 "" {TEXT -1 6 "  5)  " }{TEXT 282 67 
"C is a subset of V(L) if and only if D is a right-hand factor of L." 
}}{PARA 0 "" 0 "" {TEXT -1 8 "Whether " }{TEXT 283 1 "D" }{TEXT -1 
212 " is a right-hand factor of L is easy to detect, it is a right-han
d factor if and only if: coeff(L,Dx,0)=0. Note that in Maple commands \+
we use Dx instead of D because D is already used in Maple for somethin
g else." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
47 "Without loss of generality, we may assume that " }{TEXT 284 1 "C" 
}{TEXT -1 36 " is a subset of V(L). How? Well, if " }{TEXT 285 2 "C " 
}{TEXT -1 73 "is not a subset of L then we can replace L by: new_L := \+
LCLM(L, Dx), the " }{TEXT 289 26 "Least Common Left Multiple" }{TEXT 
-1 10 " of L and " }{TEXT 288 1 "D" }{TEXT -1 15 ". The operator " }
{TEXT 290 28 "LCLM(L1,L2) has the property" }{TEXT -1 1 ":" }}{PARA 0 
"" 0 "" {TEXT -1 7 "   6)  " }{TEXT 286 31 "V(LCLM(L1,L2)) = V(L1) + V
(L2)." }}{PARA 0 "" 0 "" {TEXT -1 17 "So in particular:" }}{PARA 0 "" 
0 "" {TEXT -1 7 "   7)  " }{TEXT 287 24 "V(LCLM(L, D)) = V(L) + C" }}
{PARA 0 "" 0 "" {TEXT -1 6 "So if " }{TEXT 291 1 "C" }{TEXT -1 317 " i
s not a subset of V(L), then we can replace L by LCLM(L,Dx). The new L
 has all the solutions that the old L has, but in addition it also has
 the constants as solutions. Note that this will not affect the image \+
of V(L) under the map D, because we only added constants to V(L), and \+
constants have image 0 under D. So:" }}{PARA 0 "" 0 "" {TEXT -1 6 "   \+
8) " }{TEXT 292 67 "Image under D of V(L) is the same as image under D
 of V(LCLM(L,D))." }}{PARA 0 "" 0 "" {TEXT -1 134 "So after replacing \+
L by LCLM(L,Dx)  (note that if L was already a left multiple of Dx the
n this does not change L) we may assume that " }{TEXT 293 1 "C" }
{TEXT -1 63 " is a subset of V(L), which means that L is a left-multip
le of " }{TEXT 294 1 "D" }{TEXT -1 17 ", in other words " }{TEXT 295 
1 "D" }{TEXT -1 51 " is a right-hand factor of L. So then we may write
:" }}{PARA 0 "" 0 "" {TEXT -1 5 "  9) " }{TEXT 296 122 "If L=LCLM(L,D)
, in other words C is a subset of V(L), then we can write L = M * D, w
here M is obtained by dividing L by D." }}{PARA 0 "" 0 "" {TEXT -1 
200 "Recall that the multiplication * here represents composition of o
perators. Now take a solution y of L, so y in V(L). Then: L(y)=0. But \+
L=M*D so M(D(y))=0, so M(y')=0. So y' is in V(M). We see now that" }}
{PARA 0 "" 0 "" {TEXT -1 7 "   10) " }{TEXT 297 66 "V(M) is exactly th
e set of derivatives of elements of V(L)=V(M*D)." }{TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 8 "Example:" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "with(DEtools): _Envdiffop
domain:=[Dx,x];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%1_Envdiffopdomain
G7$%#DxG%\"xG" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "L:=Dx^2-x;
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"LG,&*$)%#DxG\"\"#\"\"\"\"\"\"%
\"xG!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "new_L := LCLM(
L,Dx);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&new_LG,(*&%#DxG\"\"\"%\"x
GF(!\"\"*&*$)F'\"\"#\"\"\"F/F)!\"\"F**$)F'\"\"$F/F(" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 77 "Note: doing this step a second time makes no di
fference, as explained before:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 37 "new_L := LCLM(new_L, Dx); # no change" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%&new_LG,(*&%#DxG\"\"\"%\"xGF(!\"\"*&*$)F'\"\"#\"\"\"F
/F)!\"\"F**$)F'\"\"$F/F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 
"rightdivision(new_L, Dx);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7$,(*$)%
#DxG\"\"#\"\"\"\"\"\"*&F'F)%\"xG!\"\"!\"\"F,F.\"\"!" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 8 "M:=%[1];" }}{PARA 11 "" 1 "" {XPPMATH 20 "
6#>%\"MG,(*$)%#DxG\"\"#\"\"\"\"\"\"*&F(F*%\"xG!\"\"!\"\"F-F/" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 91 "Now lets check. Take a solution of
 L, differentiate it, and check if it is a solution of M." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "eqL:=diffop2de(L, y(x));" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%$eqLG,&*&%\"xG\"\"\"-%\"yG6#F'F(!\"\"-%%di
ffG6$F)-%\"$G6$F'\"\"#F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 
"dsolve(eqL,y(x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"yG6#%\"xG,&
*&%$_C1G\"\"\"-%'AiryAiGF&F+F+*&%$_C2GF+-%'AiryBiGF&F+F+" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "sol_L:=rhs(%);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%&sol_LG,&*&%$_C1G\"\"\"-%'AiryAiG6#%\"xGF(F(*&%$_C2GF
(-%'AiryBiGF+F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "der_so
l_L := diff(sol_L,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%*der_sol_LG
,&*&%$_C1G\"\"\"-%'AiryAiG6$F(%\"xGF(F(*&%$_C2GF(-%'AiryBiGF+F(F(" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "eqM:=diffop2de(M, y(x));" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eqMG,(*&%\"xG\"\"\"-%\"yG6#F'F(!\"
\"*&-%%diffG6$F)F'\"\"\"F'!\"\"F,-F/6$F)-%\"$G6$F'\"\"#F(" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "subs(y(x) = der_sol_L, eqM);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#,(*&%\"xG\"\"\",&*&%$_C1GF&-%'AiryAiG6
$F&F%F&F&*&%$_C2GF&-%'AiryBiGF,F&F&F&!\"\"*&-%%diffG6$F'F%\"\"\"F%!\"
\"F1-F46$F'-%\"$G6$F%\"\"#F&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 12 "simplify(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 29 "Indeed, der_sol_L is in V(M)." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 298 29 "Can we a
lso do the opposite? " }{TEXT -1 41 "That is, if we start with an oper
ator M, " }{TEXT 299 36 "can we integrate the solutions of M?" }{TEXT 
-1 46 " That is, can we find an operator L such that:" }}{PARA 0 "" 0 
"" {TEXT -1 25 "  (*)  D: V(L)  ---> V(M)" }}{PARA 0 "" 0 "" {TEXT -1 
348 "is onto? The answer is yes, as we've seen before we may simply ta
ke L=M*D. Then the image is D(V(L))=D(V(M*D)) and we've seen that D(V(
M*D)) always equals V(M). So finding an operator L for which (*) is on
to is trivial. But can we also find an operator L for which (*) is not
 only onto but also 1-1? Sometimes this is possible, like in the examp
le:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "M;" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#,(*$)%#DxG\"\"#\"\"\"\"\"\"*&F&F(%\"xG!\"\"!\"\"F+F-
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "L;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#,&*$)%#DxG\"\"#\"\"\"\"\"\"%\"xG!\"\"" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 199 "in this example we have that (*) is onto, but \+
since the orders of L and M are the same, we have an onto map D: V(L) \+
--> V(M) between two vector spaces of the same dimension. Such a map i
s always 1-1." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 30 "Lets consider another example:" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 10 "M:=Dx+1/x;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"
MG,&%#DxG\"\"\"*&\"\"\"F)%\"xG!\"\"F'" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 45 "sol_M := rhs(dsolve(diffop2de(M,y(x)),y(x)));" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&sol_MG*&%$_C1G\"\"\"%\"xG!\"\"" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "sol_L := int(sol_M,x);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&sol_LG*&%$_C1G\"\"\"-%#lnG6#%\"xGF'
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 155 "If we want (*) to be 1-1, th
en order(L) must be order(M), which is 1 in the example. Then we have \+
no other choice for L than to take (assuming L is monic):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "L1:=Dx - diff(sol_L,x)/sol_L;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#L1G,&%#DxG\"\"\"*&\"\"\"F)*&%\"xG\"
\"\"-%#lnG6#F+\"\"\"!\"\"!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 68 
"If we don't insist on (*) to be 1-1 but just onto, then we can take:
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "L2:=mult(M,Dx);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#L2G,&*&%#DxG\"\"\"%\"xG!\"\"\"\"\"*
$)F'\"\"#F(F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "rightdivis
ion(L2,L1); remainder=%[2];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7$,&%#D
xG\"\"\"*&,&-%#lnG6#%\"xGF&F&F&\"\"\"*&F,\"\"\"F)\"\"\"!\"\"F&\"\"!" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%*REMAINDERG\"\"!" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 74 "We see that L1 is a right-hand factor of L2, wh
ich is clear because V(L1)=" }{TEXT 300 1 "C" }{TEXT -1 35 "*sol_L, wh
ich is a subset of V(L2)." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 20 "So i
n the example M=" }{TEXT 301 1 "D" }{TEXT -1 39 "+1/x we have the foll
owing two choices:" }}{PARA 0 "" 0 "" {TEXT -1 137 "*) we insist that \+
(*) is 1-1, so that order(L)=order(M). Then we get L=L1, but we no lon
ger have rational functions as coefficients of L." }}{PARA 0 "" 0 "" 
{TEXT -1 3 "*) " }{TEXT 304 2 "OR" }{TEXT -1 134 ": we do not insist t
hat (*) is 1-1, then we get L=M*D, so order(L)=order(M)+1, but we do g
et rational functions as coefficients for L." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 25 "whereas in the example M=" }
{TEXT 302 1 "D" }{TEXT -1 7 "^2-1/x*" }{TEXT 303 1 "D" }{TEXT -1 60 "-
x we got the best of both worlds, there was an L such that:" }}{PARA 
0 "" 0 "" {TEXT -1 44 "*) the map (*) is 1-1, so order(L)=order(M)." }
}{PARA 0 "" 0 "" {TEXT -1 3 "*) " }{TEXT 305 4 "AND:" }{TEXT -1 42 " L
 has rational functions as coefficients." }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 233 "The question now is: starting wit
h an operator M with rational functions as coefficients, when can we g
et the best of both worlds, in other words: when does there exist an o
perator L with rational functions as coefficients such that:" }}{PARA 
0 "" 0 "" {TEXT -1 22 "  (*) D: V(L) --> V(M)" }}{PARA 0 "" 0 "" 
{TEXT -1 122 "is onto and 1-1, so order(L)=order(M). We've seen one ex
ample where such L exists and one example where it does not exist." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 169 "Why is i
t so important that (*) is 1-1? Well, if (*) is 1-1 then we will show \+
that we can compute an inverse map of (*). That map will then integrat
e the solutions of M." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}}{MARK "28 0
 0" 31 }{VIEWOPTS 1 1 0 1 1 1803 }