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{SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT -1 37 "Automorphisms and anti-a
utomorphisms." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 3 "An " }{TEXT 262 12 "automorphism" }{TEXT -1 13 " of a ring
 R." }}{PARA 0 "" 0 "" {TEXT -1 60 "A map:   sigma: R ----> R is calle
d an automorphism of R if:" }}{PARA 0 "" 0 "" {TEXT -1 24 "*) sigma is
 1-1 and onto" }}{PARA 0 "" 0 "" {TEXT -1 43 "*) sigma(a+b)=sigma(a)+s
igma(b) for all a,b" }}{PARA 0 "" 0 "" {TEXT -1 44 "*) sigma(a*b)=sigm
a(a)*sigma(b) for all a,b." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 
"" 0 "" {TEXT 263 10 "Example 1)" }{TEXT -1 5 "   R=" }{TEXT 256 1 "C
" }{TEXT -1 35 "[x]  and sigma(a) = subs(x=x+17, a)" }}{PARA 0 "" 0 "
" {TEXT 264 10 "Example 2)" }{TEXT -1 5 "   R=" }{TEXT 257 1 "C" }
{TEXT -1 14 "(x)[Dx], k in " }{TEXT 258 1 "C" }{TEXT -1 55 "(x), and s
igma( sum(a.i * Dx^i) ) = sum(a.i * (Dx-k)^i)" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 122 "Example 2 is interesting
 automorphism of the ring R of differential operators. Suppose L=sum(a
.i*Dx^i,i=0..n) with a.i in " }{TEXT 259 1 "C" }{TEXT -1 145 "(x). Sup
pose that y is a solution of L. So: L(y) = 0, so sum(a.i * diff(y,[x$i
], i=0..n) = 0. Now notice that (you can prove this with induction)" }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 
260 11 "Property 1:" }{TEXT -1 89 "  If z = exp(int(k,x))*y, and if L1
=(Dx-k)^i and L2=Dx^i, then L1(z) = exp(int(k,x)) * y." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 236 "So we see that if z
=exp(int(k,x))*y then sum(a.i * (Dx-k)^i,i=0..n)(z) = exp(int(k,x))*su
m(a.i * Dx^i,i=0..n)(y) = exp(int(k,x))*L(y) = exp(int(k,x))*0 = 0. So
 then z is a solution of sigma(L)=sum(a.i * (Dx-k)^i,i=0..n). So we se
e that:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 261 
11 "Property 2:" }{TEXT -1 146 " For every solution y of L, we have th
at exp(int(k,x))*y is a solution of sigma(L) and vice versa, where sig
ma is as in example 2. In other words:" }}{PARA 0 "" 0 "" {TEXT 265 
38 " V( sigma(L) ) = exp(int(k,x)) * V(L)." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 8 "Example:" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 49 "restart; with(DEtools): _Envdiffopdomain:=[Dx,
x]:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "n:=2; a[2]:=1; a[1]:
=1/x; a[0]:=x;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"nG\"\"#" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>&%\"aG6#\"\"#\"\"\"" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#>&%\"aG6#\"\"\"*&F'F'%\"xG!\"\"" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>&%\"aG6#\"\"!%\"xG" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 33 "L := add(a[i]*mult(Dx$i),i=0..n);" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>%\"LG,(%\"xG\"\"\"*&%#DxGF'F&!\"\"F'*$)F)\"\"#F'F'" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "k:=x^2;" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>%\"kG*$)%\"xG\"\"#\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 44 "sigmaL := add(a[i]*mult( (Dx-k)$i ),i=0..n);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'sigmaLG,,%\"xG!\"\"*&,&%#DxG\"\"\"*
$)F&\"\"#F+F'F+F&F'F+*$)F*F.F+F+*(F.F+F-F+F*F+F'*$)F&\"\"%F+F+" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "dsolve( diffop2de(L,y(x)) );
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"yG6#%\"xG,&*&%$_C1G\"\"\"-%(B
esselYG6$\"\"!,$*$)F'#\"\"$\"\"#F+#F5F4F+F+*&%$_C2GF+-%(BesselJGF.F+F+
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "dsolve( diffop2de(sigma
L,y(x)) );" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"yG6#%\"xG,&*(%$_C1G
\"\"\"-%(BesselYG6$\"\"!,$*$)F'#\"\"$\"\"#F+#F5F4F+-%$expG6#,$*$)F'F4F
+#F+F4F+F+*(%$_C2GF+-%(BesselJGF.F+F7F+F+" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 14 "exp(int(k,x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-
%$expG6#,$*$)%\"xG\"\"$\"\"\"#F+F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 227 "The command L := symmetric_product(L1, L2)  produces a differe
ntial operator such that for every solution y1 of L1 and solution y2 o
f L2, the product y1*y2 is a solution of L. So we can obtain sigma(L) \+
by the following command:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
27 "symmetric_product(L, Dx-k);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,**
$)%#DxG\"\"#\"\"\"F(*&*&,&!\"\"F(*&F'F()%\"xG\"\"$F(F(F(F&F(F(F/F,F,*&
F'F(F/F(F,*$)F/\"\"%F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 
"expand(%-sigmaL); # check if this equals sigmaL" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#\"\"!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 3 "An " }
{TEXT 266 17 "anti-automorphism" }{TEXT -1 13 " of a ring R." }}{PARA 
0 "" 0 "" {TEXT -1 65 "A map:   sigma: R ----> R is called an anti-aut
omorphism of R if:" }}{PARA 0 "" 0 "" {TEXT -1 24 "*) sigma is 1-1 and
 onto" }}{PARA 0 "" 0 "" {TEXT -1 43 "*) sigma(a+b)=sigma(a)+sigma(b) \+
for all a,b" }}{PARA 0 "" 0 "" {TEXT -1 96 "*) sigma(a*b)=sigma(b)*sig
ma(a) for all a,b. So sigma interchanges the order of multiplication, \+
" }{TEXT 272 44 "it interchanges the order of multiplication." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 267 10 "Example \+
3)" }{TEXT -1 270 "  R is the set of all 3 by 3 matrices. This set is \+
a ring because we have +, - and *. Now when A is a matrix let sigma(A)
 be its transpose. Then, as you know from linear algebra, sigma is an \+
anti-automorphism of R. The transpose interchanges the order of multip
lication." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
268 10 "Example 4)" }{TEXT -1 107 " Let R be a commutative ring, so a*
b=b*a for all a,b. Then every automorphism is also an anti-automorphis
m." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 269 51 "Ex
ample 5), the adjoint of a differential operator." }{TEXT -1 7 " Let R
=" }{TEXT 270 1 "C" }{TEXT -1 49 "(x)[Dx]. Define the map sigma: R--->
R as follows:" }}{PARA 0 "" 0 "" {TEXT -1 29 "*) sigma(a)=a for every \+
a in " }{TEXT 271 1 "C" }{TEXT -1 3 "(x)" }}{PARA 0 "" 0 "" {TEXT -1 
18 "*) sigma(Dx) = -Dx" }}{PARA 0 "" 0 "" {TEXT -1 135 "*) Whenever si
gma(a) and sigma(b) have already been defined, then define sigma(a+b)=
sigma(a)+sigma(b) and sigma(a*b)=sigma(b)*sigma(a)." }}{PARA 0 "" 0 "
" {TEXT -1 67 "The above three *)'s define sigma(a) for every a in R. \+
For example:" }}{PARA 0 "" 0 "" {TEXT -1 54 "  sigma(x*Dx) = sigma(Dx)
*sigma(x)=(-Dx)*x = -x*Dx - 1" }}{PARA 0 "" 0 "" {TEXT -1 52 "  sigma(
Dx*x) = sigma(x)*sigma(Dx) = x*(-Dx) = -x*Dx" }}{PARA 0 "" 0 "" {TEXT 
-1 42 "  sigma(Dx-x) = sigma(Dx)+sigma(-x)= -Dx-x" }}{PARA 0 "" 0 "" 
{TEXT -1 180 "It is not hard to prove that this sigma is indeed an ant
i-automorphism and that sigma(sigma(L))=L for every L in R. Now if L i
s a differential operator then sigma(L) is called the " }{TEXT 273 7 "
adjoint" }{TEXT -1 5 " of L" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 133 "The Maple command for the adjoint is: adjoint(
L). Note that this command does not compute sigma(L) as defined above,
 but it computes:" }}{PARA 0 "" 0 "" {TEXT -1 3 "   " }{TEXT 274 30 "a
djoint(L) = (-1)^n * sigma(L)" }}{PARA 0 "" 0 "" {TEXT -1 215 "where n
=degree(L,Dx) is the order of L. It does this so that if the leading c
oefficient of L is 1, then so is the leading coefficient of adjoint(L)
. Without the (-1)^n the leading coefficient would have been (-1)^n." 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 78 "Lets ch
eck that adjoint(L1 * L2) is indeed equal to adjoint(L2) * adjoint(L1)
." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "L1 := Dx^2+Dx+x; L2:=D
x^3+Dx^2-5*x*Dx+3; L:=mult(L1,L2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#
>%#L1G,(*$)%#DxG\"\"#\"\"\"F*F(F*%\"xGF*" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%#L2G,**$)%#DxG\"\"$\"\"\"F**$)F(\"\"#F*F**(\"\"&F*%\"xGF*F(F*!
\"\"F)F*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"LG,4*$)%#DxG\"\"&\"\"
\"F**&\"\"#F*)F(\"\"%F*F**$)F(\"\"$F*F**(F.F*%\"xGF*F0F*!\"\"*&\"\"(F*
)F(F,F*F4*(F.F*F3F*F7F*F4*(F)F*)F3F,F*F(F*F4*&F,F*F(F*F4*&F1F*F3F*F*" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "mult(adjoint(L2), adjoint
(L1));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,4*$)%#DxG\"\"&\"\"\"F(*&\"
\"#F()F&\"\"%F(!\"\"*(F,F(%\"xGF()F&\"\"$F(F-*$F0F(F(*(F,F(F/F()F&F*F(
F(*&F'F(F4F(F-*&\"\"'F(F&F(F(*(F'F()F/F*F(F&F(F-*&\"#8F(F/F(F-" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "adjoint(L);" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#,4*$)%#DxG\"\"&\"\"\"F(*&\"\"#F()F&\"\"%F(!\"\"*(F
,F(%\"xGF()F&\"\"$F(F-*$F0F(F(*(F,F(F/F()F&F*F(F(*&F'F(F4F(F-*&\"\"'F(
F&F(F(*(F'F()F/F*F(F&F(F-*&\"#8F(F/F(F-" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 9 "OK, same." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 
"" {TEXT -1 353 "The adjoint can be used to interchange the role of le
ft and right-hand factors. Suppose that y is a non-zero solution of L.
 Now y is also a non-zero solution of Dx-y'/y because (Dx-y'/y)(y) = D
x(y) - (y'/y)*y = y' - y' = 0. Now C*y is the solution space V(Dx-y'/y
) of Dx-y'/y and this is a subset of V(L) because y in V(L). That mean
s that Dx-y'/y is a " }{TEXT 276 17 "right-hand factor" }{TEXT -1 6 " \+
of L." }}{PARA 0 "" 0 "" {TEXT -1 86 "Then by applying the adjoint, yo
u get that adjoint(Dx-y'/y) = (-1)^n * (Dx+y'/y) is a " }{TEXT 275 16 
"left-hand factor" }{TEXT -1 24 " of adjoint(L). Example:" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "y:=exp(x^2);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"yG-%$expG6#*$)%\"xG\"\"#\"\"\"" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 54 "L1 := Dx^2-Dx+x^2; L2:=Dx-diff(y,x)/y; L:=mu
lt(L1,L2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#L1G,(*$)%#DxG\"\"#\"
\"\"F*F(!\"\"*$)%\"xGF)F*F*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#L2G,
&%#DxG\"\"\"*&\"\"#F'%\"xGF'!\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>
%\"LG,2*$)%#DxG\"\"$\"\"\"F**$)F(\"\"#F*!\"\"*(F-F*%\"xGF*F,F*F.*&)F0F
-F*F(F*F**&\"\"%F*F(F*F.*(F-F*F0F*F(F*F**&F-F*)F0F)F*F.F-F*" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "adjoint(L2);" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#,&%#DxG\"\"\"*&\"\"#F%%\"xGF%F%" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 11 "adjoint(L);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#,0*$)%#DxG\"\"$\"\"\"F(*$)F&\"\"#F(F(*(F+F(%\"xGF(F*F(F(*(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 "" {MPLTEXT 1 0 11 "DFactor(%);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#7$,&%#DxG\"\"\"*&\"\"#F&%\"xGF&F&,(*$)F%F(F&F&F%F&*$)F)F(F&F&" }
}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 111 "Now suppose we have a different
ial operator M of order n and we search for a differential operator L \+
such that:" }}{PARA 0 "" 0 "" {TEXT -1 40 "  D: V(L) ----> V(M)  is a \+
1-1 onto map." }}{PARA 0 "" 0 "" {TEXT -1 162 "As we've seen in the pr
evious worksheet, if we insist on L having rational functions as coeff
icients (which is what we will do) then such L does not always exist.
" }}{PARA 0 "" 0 "" {TEXT -1 286 "The elements of V(L) are anti-deriva
tives of the elements of V(M). Since D is supposed to be 1-1, we have \+
that the constants C=V(D) is not a subset of V(L), in other words: the
 intersection of V(D) and V(L) is V(1)=\{0\}. That means: GCRD(L,D)=1,
 which is equivalent to: coeff(L,Dx,0)<>0." }}{PARA 0 "" 0 "" {TEXT 
-1 121 "By the extended Euclidean algorithm there exist two operators \+
s,r with order(s)<order(D) and order(s)<order(L) such that:" }}{PARA 
0 "" 0 "" {TEXT -1 31 "  (***) s*L+r*D = GCRD(L,D) = 1" }}{PARA 0 "" 
0 "" {TEXT -1 72 "Now order(s)<order(D)=1, so order(s)=0, so the opera
tor s is in fact in " }{TEXT 277 1 "C" }{TEXT -1 8 "(x), so " }{TEXT 
278 10 "sigma(s)=s" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 18 "Now
 r*D=1-s*L. So;" }}{PARA 0 "" 0 "" {TEXT -1 104 "order(r*D)=order(1-s*
L)=order(s*L)=order(s)+order(L)=0+order(L)=order(L)=dim(V(L))=dim(V(M)
)=order(M)=n." }}{PARA 0 "" 0 "" {TEXT -1 48 "And order(r*D)=order(r)+
order(D)=order(r)+1. So " }{TEXT 279 13 "order(r)=n-1." }}{PARA 0 "" 
0 "" {TEXT -1 106 "Now take a solution y of L. So D(y)=y' is a solutio
n of M. Now apply both sides of (***) on y and you get:" }}{PARA 0 "" 
0 "" {TEXT -1 19 "(s*L+r*D)(y) = 1(y)" }}{PARA 0 "" 0 "" {TEXT -1 22 "
s(L(y)) + r(D(y)) = y." }}{PARA 0 "" 0 "" {TEXT -1 45 "Now y is a solu
tion of L so L(y)=0 so we get:" }}{PARA 0 "" 0 "" {TEXT -1 19 "r(D(y))
= r(y') = y." }}{PARA 0 "" 0 "" {TEXT -1 314 "We see now that D sends \+
the solution y of L to the solution y' of M. And the operator r sends \+
y' right back to y. So if D is 1-1 from V(L) to V(M), then the action \+
of the operator r must be the inverse map from V(M) to V(L). In other \+
words: if y'=z is a solution of M, then r(z)=r(y')=y is an anti-deriva
tive of z." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 134 "If we knew what L was, then we could find r by the extended Eu
clidean algorithm, all we would have to do is to write 1 as 1 = s*L+r*
D." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 144 "We
 now note that if the inverse of the map D:V(L)--->V(M) is the map r, \+
then the inverse of the map r:V(M)--->V(L) must then automatically be \+
D." }}{PARA 0 "" 0 "" {TEXT -1 82 "So r*D must be the identity map on \+
V(L), and D*r must be the identity map on V(M)." }}{PARA 0 "" 0 "" 
{TEXT -1 162 "So 1-D*r must be the zero map on V(M). So V(M) is a subs
et of V(1-D*r) which is only true when M is a right-hand factor of 1-D
*r. So there must exist an operator " }{TEXT 280 1 "l" }{TEXT -1 11 " \+
such that:" }}{PARA 0 "" 0 "" {TEXT -1 2 "  " }{TEXT 281 1 "l" }{TEXT 
-1 10 "*M = 1-D*r" }}{PARA 0 "" 0 "" {TEXT -1 15 "in other words:" }}
{PARA 0 "" 0 "" {TEXT -1 2 "  " }{TEXT 282 1 "l" }{TEXT -1 15 "*M+D*r \+
= 1 (**)" }}{PARA 0 "" 0 "" {TEXT -1 64 "Since order(D*r)=1+n-1=n and \+
order(M)=n we must have that order(" }{TEXT 283 1 "l" }{TEXT -1 131 ")
=0 otherwise the order of the left-hand side would be >n and therefore
 not equal to the order of the right-hand side. So we have: " }{TEXT 
284 25 "l in C(x), so sigma(l)=l." }}{PARA 0 "" 0 "" {TEXT -1 202 "Now
 apply the adjoint sigma on both sides of (**). Remember that sigma di
ffers from Maple's adjoint by a factor (-1)^n, but we take here the si
gma defined in example 5. Denote: sM as short for sigma(M)." }}{PARA 
0 "" 0 "" {TEXT -1 29 "  sigma(l*M + D*r) = sigma(1)" }}{PARA 0 "" 0 "
" {TEXT -1 11 "  sM*sigma(" }{TEXT 285 1 "l" }{TEXT -1 25 ") + sigma(r
)*sigma(D) = 1" }}{PARA 0 "" 0 "" {TEXT -1 5 "  sM*" }{TEXT 286 1 "l" 
}{TEXT -1 20 " + sigma(r)*(-D) = 1" }}{PARA 0 "" 0 "" {TEXT -1 52 "Now
 apply both sides on the function 1, and you get:" }}{PARA 0 "" 0 "" 
{TEXT -1 33 "  sM*l(1) + sigma(r)*(-1') = 1(1)" }}{PARA 0 "" 0 "" 
{TEXT -1 5 "  sM(" }{TEXT 287 1 "l" }{TEXT -1 9 ") + 0 = 1" }}{PARA 0 
"" 0 "" {TEXT -1 3 "So " }{TEXT 288 1 "l" }{TEXT -1 55 " is a rational
 solution of the equation sM( y(x) ) = 1." }}{PARA 0 "" 0 "" {TEXT -1 
22 "Once we've calculated " }{TEXT 289 1 "l" }{TEXT -1 17 " (note that
 such " }{TEXT 290 1 "l" }{TEXT -1 308 " need not exist as the operato
rs L and r we are looking for need not exist) then we can find r easil
y from equation (**). And then we can find s*L easily from equation (*
**). We may just as well take s=1 because s is an operator of order 0 \+
(i.e. a rational function) so this will make no difference for V(L)." 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 48 "Here is
 the example from the previous worksheet:" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 19 "M := Dx^2-1/x*Dx-x;" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%\"MG,(*$)%#DxG\"\"#\"\"\"F**&F(F*%\"xG!\"\"F-F,F-" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "n:=degree(M,Dx);" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>%\"nG\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 24 "sM:=(-1)^n * adjoint(M);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#s
MG,(*$)%#DxG\"\"#\"\"\"F**&F(F*%\"xG!\"\"F**&,&*$)F,\"\"$F*F*F*F*F**$)
F,F)F*F-F-" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 32 "Note: sigma(M) is a
lso equal to:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "add(mult( \+
(-Dx)^i, coeff(M,Dx,i) ),i=0..n); normal(%-sM);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#,*%\"xG!\"\"*&%#DxG\"\"\"F$F%F(*&F(F(*$)F$\"\"#F(F%F%*$
)F'F,F(F(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 32 "y:='y'; diffop2de(sM, y(x)) = 1;" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%\"yGF$" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#
/,(*&*&,&*$)%\"xG\"\"$\"\"\"F,F,F,F,-%\"yG6#F*F,F,*$)F*\"\"#F,!\"\"F3*
&-%%diffG6$F-F*F,F*F3F,-F66$F--%\"$G6$F*F2F,F," }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 17 "ratsols(%, y(x));" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#7$7\",$*&\"\"\"F'%\"xG!\"\"F)" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 8 "l:=%[2];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"l
G,$*&\"\"\"F'%\"xG!\"\"F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
23 "leftdivision(1-l*M,Dx);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7$*&%#D
xG\"\"\"%\"xG!\"\"\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "r
:=%[1];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"rG*&%#DxG\"\"\"%\"xG!\"
\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "L:=1-mult(r,Dx);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"LG,&\"\"\"F&*&*$)%#DxG\"\"#F&F&%\"
xG!\"\"F-" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 38 "This L has the same \+
solution space as:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "colle
ct( L/lcoeff(L,Dx), Dx, normal);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&
*$)%#DxG\"\"#\"\"\"F(%\"xG!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
51 "which is the same L as from the previous worksheet." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 15 "Now lets check:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "sol_M := rhs( dsolve(diffop2
de(M,y(x)),y(x)) );" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&sol_MG,&*(%$
_C1G\"\"\"%\"xGF(-%(BesselIG6$#\"\"#\"\"$,$*$)F)#F/F.F(F-F(F(*(%$_C2GF
(F)F(-%(BesselKGF,F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 83 "a
pplyL := proc(L,Y) normal(add(coeff(L,Dx,i)*diff(Y,[x$i]),i=0..degree(
L,Dx))) end;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'applyLGR6$%\"LG%\"Y
G6\"F)F)-%'normalG6#-%$addG6$*&-%&coeffG6%9$%#DxG%\"iG\"\"\"-%%diffG6$
9%7#-%\"$G6$%\"xGF6F7/F6;\"\"!-%'degreeG6$F4F5F)F)F)" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 18 "applyL( r, sol_M);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#*&-%%sqrtG6#%\"xG\"\"\",&*&%$_C1GF(-%(BesselIG6$#!\"\"
\"\"$,$*$)F'#F1\"\"#F(#F6F1F(F(*&%$_C2GF(-%(BesselKG6$#F(F1F2F(F0F(" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "int(sol_M,x);" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#,&*(%$_C1G\"\"\")*$)%\"xG#\"\"$\"\"#F&#F&F,F&-%
(BesselIG6$#!\"\"F,,$F(#F-F,F&F&*(%$_C2GF&F'F&-%(BesselKG6$F.F4F&F3" }
}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 78 "These two expressions are the sa
me if you replace  (x^(3/2))^(1/3) by sqrt(x)." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 17 "One more example." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "restart; with(DEtools): _Env
diffopdomain:=[Dx,x]:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "f:
=hypergeom([b, a],[c],x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fG-%*
hypergeomG6%7$%\"bG%\"aG7#%\"cG%\"xG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 96 "How can we obtain an antiderivative of the hypergeometric funct
ion with the methods given above?" }}{PARA 0 "" 0 "" {TEXT -1 87 "Well
, f is one of the solutions of the following linear ordinary different
ial equation:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 72 "ode := x*(
1-x)*diff(y(x),x$2) + (c - (a+b+1)*x)*diff(y(x),x) - a*b*y(x);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$odeG,(*(%\"xG\"\"\",&F(F(F'!\"\"F(-
%%diffG6$-%\"yG6#F'-%\"$G6$F'\"\"#F(F(*&,&%\"cGF(*&,(%\"aGF(%\"bGF(F(F
(F(F'F(F*F(-F,6$F.F'F(F(*(F:F(F;F(F.F(F*" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 27 "simplify(subs(y(x)=f,ode));" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "M:=
de2diffop(ode,y(x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"MG,(*&,&%
\"xG\"\"\"*$)F(\"\"#F)!\"\"F))%#DxGF,F)F)*&,*%\"cGF)*&F(F)%\"aGF)F-*&F
(F)%\"bGF)F-F(F-F)F/F)F)*&F6F)F4F)F-" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 35 "sM:=(-1)^degree(M,Dx) * adjoint(M);" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#>%#sMG,8*&%\"xG\"\"\")%#DxG\"\"#F(F(*&)F'F+F(F)F(!\"
\"*&%\"cGF(F*F(F.*(\"\"$F(F'F(F*F(F.*&F+F(F*F(F(*(F'F(%\"aGF(F*F(F(*(F
'F(%\"bGF(F*F(F(F(F.*&F7F(F5F(F.F5F(F7F(" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 24 "diffop2de(sM, y(x)) = 1;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/,(*&,**&%\"bG\"\"\"%\"aGF)!\"\"F*F)F(F)F)F+F)-%\"yG6#%
\"xGF)F)*&,,%\"cGF+*&F/F)F*F)F)*&F/F)F(F)F)*&\"\"$F)F/F)F+\"\"#F)F)-%%
diffG6$F,F/F)F)*&,&F/F)*$)F/F7F)F+F)-F96$F,-%\"$G6$F/F7F)F)F)" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "ratsols(%,y(x));" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#7$7\",$*&\"\"\"F',**&%\"bGF'%\"aGF'F'F+!\"\"
F*F,F'F'F,F," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "l:=%[2];" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"lG,$*&\"\"\"F',**&%\"bGF'%\"aGF'F'
F+!\"\"F*F,F'F'F,F," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "left
division(1-l*M,Dx);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7$,&*&*&,&%\"xG
\"\"\"*$)F(\"\"#F)!\"\"F)%#DxGF)F),**&%\"bGF)%\"aGF)F)F2F-F1F-F)F)F-F)
*&,,%\"cGF-*&F(F)F2F)F)*&F(F)F1F)F)F(F-F)F)F)F/F-F-\"\"!" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "r:=%[1];" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"rG,&*&*&,&%\"xG\"\"\"*$)F)\"\"#F*!\"\"F*%#DxGF*F*,*
*&%\"bGF*%\"aGF*F*F3F.F2F.F*F*F.F**&,,%\"cGF.*&F)F*F3F*F**&F)F*F2F*F*F
)F.F*F*F*F0F.F." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "antideri
vative_f := add(coeff(r,Dx,i)*diff(f,[x$i]),i=0..degree(r,Dx));" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%1antiderivative_fG,&*&*&,,%\"cG!\"\"
*&%\"xG\"\"\"%\"aGF-F-*&F,F-%\"bGF-F-F,F*F-F-F--%*hypergeomG6%7$F0F.7#
F)F,F-F-,**&F0F-F.F-F-F.F*F0F*F-F-F*F**&**,&F,F-*$)F,\"\"#F-F*F-F0F-F.
F--F26%7$,&F0F-F-F-,&F.F-F-F-7#,&F)F-F-F-F,F-F-*&F6F-F)F-F*F-" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "simplify(diff(%,x)-f);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 51 "Note that the antiderivative of f is a solution of:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "L:=collect(  1-mult(r,Dx)  ,
Dx,normal);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"LG,(*&*(%\"xG\"\"\"
,&!\"\"F)F(F)F))%#DxG\"\"#F)F),**&%\"bGF)%\"aGF)F)F2F+F1F+F)F)F+F)*&*&
,,%\"cGF+*&F(F)F2F)F)*&F(F)F1F)F)F(F+F)F)F)F-F)F)F/F+F)F)F)" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "diffop2de(L,y(x));" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#,(-%\"yG6#%\"xG\"\"\"*&*&,,%\"cG!\"\"*&F'F(%
\"aGF(F(*&F'F(%\"bGF(F(F'F-F(F(F(-%%diffG6$F$F'F(F(,**&F1F(F/F(F(F/F-F
1F-F(F(F-F(*&*(F'F(,&F-F(F'F(F(-F36$F$-%\"$G6$F'\"\"#F(F(F5F-F(" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "simplify(subs(y(x)=antideriv
ative_f, %));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 410 "Both L and r are computed by the procedu
re integrate_sols, which gives as output [L, r] when there exists L fo
r which D:V(L)--->V(M) is 1-1. If such L (with rational functions as c
oefficients) does not exist (i.e. when ratsols(sigma(M)(y)=1) has no s
olutions) then integrate_sols returns L=M*Dx so then the map is onto b
ut not 1-1, and the inverse map r (the map that integrates) can not be
 found in that case." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "int
egrate_sols(M);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7$,(*&*&,&%\"xG\"\"
\"*$)F(\"\"#F)!\"\"F))%#DxGF,F)F),**&%\"bGF)%\"aGF)F)F3F-F2F-F)F)F-F-*
&*&,,%\"cGF-*&F(F)F3F)F)*&F(F)F2F)F)F(F-F)F)F)F/F)F)F0F-F)F)F),&*&*&F'
F)F/F)F)F0F-F)*&F6F)F0F-F-" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 17 "One
 more example:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "M:=Dx^2+1
/x*Dx+1/x^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"MG,(*$)%#DxG\"\"#
\"\"\"F**&F(F*%\"xG!\"\"F**&F*F**$)F,F)F*F-F*" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 18 "diffop2de(M,y(x));" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#,(*&-%\"yG6#%\"xG\"\"\"*$)F(\"\"#F)!\"\"F)*&-%%diffG6$F%F(F)F(F-
F)-F06$F%-%\"$G6$F(F,F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "
sols:=rhs(dsolve(%));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%solsG,&*&%
$_C1G\"\"\"-%$sinG6#-%#lnG6#%\"xGF(F(*&%$_C2GF(-%$cosGF+F(F(" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "sol:=subs(_C1=1,_C2=0,sols);
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solG-%$sinG6#-%#lnG6#%\"xG" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "int(sol,x);" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#,&*&-%$cosG6#-%#lnG6#%\"xG\"\"\"F+F,#!\"\"\"\"#*(#
F,F/F,-%$sinGF'F,F+F,F," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "
integrate_sols(M);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7$,(*&)%\"xG\"\"
#\"\"\")%#DxGF(F)#F)F(*&#F)F(F)*&F'F)F+F)F)!\"\"F)F),&*&F&F)F+F)#F0F(*
&F,F)F'F)F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "r:=%[2];" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"rG,&*&)%\"xG\"\"#\"\"\"%#DxGF*#!\"
\"F)*&#F*F)F*F(F*F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 75 "anti
derivative_sol := add(coeff(r,Dx,i)*diff(sol,[x$i]),i=0..degree(r,Dx))
;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%3antiderivative_solG,&*&-%$cosG
6#-%#lnG6#%\"xG\"\"\"F-F.#!\"\"\"\"#*(#F.F1F.-%$sinGF)F.F-F.F." }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 137 "However, whenever integrate_sols \+
returns M*Dx instead of [L,r] then we have no operator r to integrate \+
sol. This happens for example for:" }}{PARA 0 "" 0 "" {TEXT -1 260 "M:
=D+1/x which has as solution y=1/x. The integral is ln(x), but no matt
er what the operator r is, as long as r in C(x)[D] you never have that
 r(1/x)=ln(x). So such r will not exist, therefore the equation you fe
ed to ratsols will not have a rational solution." }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 10 "M:=Dx+1/x;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#>%\"MG,&%#DxG\"\"\"*&F'F'%\"xG!\"\"F'" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 18 "integrate_sols(M);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#*&,&%#DxG\"\"\"*&F&F&%\"xG!\"\"F&F&F%F&" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 60 "sM:=(-1)^degree(M,Dx) * adjoint(M);  diffop2de(sM,y(x
)) = 1;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#sMG,&%#DxG!\"\"*&\"\"\"F
)%\"xGF'F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/,&*&-%\"yG6#%\"xG\"\"\"
F)!\"\"F*-%%diffG6$F&F)F+F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
16 "ratsols(%,y(x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7#7#%\"xG" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 276 "The homogeneous equation (replace
 =1 by =0) has a rational solution x, but the inhomogeneous equation h
as no rational solutions (if it had a rational solution the answer wou
ld have looked like [ [solutions of homogeneous equation], one solutio
n of the inhomogeneous equation]." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}}{MARK "76 0 0" 276 }{VIEWOPTS 1 1 0 1 1 1803 1 
1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }