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12 0 0 0 1 2 2 2 2 2 2 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Ma
ple Output" -1 11 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 
1 1 1 }3 3 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Maple Output" -1 12 1 
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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 299 17 "Rings and fields." }}
{PARA 0 "" 0 "" {TEXT 321 20 "Differential fields." }}{PARA 0 "" 0 "" 
{TEXT 343 48 "Algebraic extensions and logarithmic extensions." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 300 9 "Notation:
" }}{PARA 0 "" 0 "" {TEXT 301 1 "Q" }{TEXT -1 22 ": the rational numbe
rs" }}{PARA 0 "" 0 "" {TEXT 302 1 "R" }{TEXT -1 18 ": the real numbers
" }}{PARA 0 "" 0 "" {TEXT 303 1 "C" }{TEXT -1 21 ": the complex number
s" }}{PARA 0 "" 0 "" {TEXT 306 1 "C" }{TEXT -1 47 "[x]: All polynomial
s in x with coefficients in " }{TEXT 307 1 "C" }{TEXT -1 1 "." }}
{PARA 0 "" 0 "" {TEXT 304 1 "C" }{TEXT -1 54 "(x): All rational functi
ons in x with coefficients in " }{TEXT 305 2 "C." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 2 "A " }{TEXT 262 5 "field" }
{TEXT -1 15 " is a set like " }{TEXT 256 1 "Q" }{TEXT -1 2 ", " }
{TEXT 257 1 "R" }{TEXT -1 2 ", " }{TEXT 258 1 "C" }{TEXT -1 5 ", or " 
}{TEXT 259 4 "Q(x)" }{TEXT -1 2 ", " }{TEXT 260 4 "R(x)" }{TEXT -1 5 "
, or " }{TEXT 261 4 "C(x)" }{TEXT -1 100 " where the operations +, -, \+
*, / are defined and have the usual properties. If K is a field, e.g. \+
K=" }{TEXT 308 1 "C" }{TEXT -1 109 ", then K[x] is the set of all poly
nomials  a.0*x^0+...+a.n*x^n with a.0..a.n in K. This set K[x] is call
ed a " }{TEXT 263 4 "ring" }{TEXT -1 170 ", meaning that it has the op
erations +, -, * with the usual properties. In K[x] there is not alway
s a division /, because f/g is not always a polynomial when f,g in K[x
]." }}{PARA 0 "" 0 "" {TEXT -1 113 "The set K(x) denotes the set of al
l f/g where f,g in K[x], and g<>0. When K is a field then K(x) is also
 a field." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
100 "Note that when x is not a variable but a number, then the same de
finitions also apply. For example: " }{TEXT 264 1 "Q" }{TEXT -1 99 "[s
qrt(2)] is the ring of all numbers of the form: a.0*sqrt(2)^0+...+a.n*
sqrt(2)^n with a.0..a.n in " }{TEXT 265 1 "Q" }{TEXT -1 102 ". Since w
e don't need powers of sqrt(2) that are higher than 1, i.e. sqrt(2)^7=
2^3*sqrt(2)^1, the set " }{TEXT 266 1 "Q" }{TEXT -1 25 "[sqrt(2)] equa
ls the set " }{TEXT 267 1 "Q" }{TEXT -1 1 "+" }{TEXT 268 1 "Q" }{TEXT 
-1 35 "*sqrt(2)=\{a0 + a1*sqrt(2)|a0,a1 in " }{TEXT 269 1 "Q" }{TEXT 
-1 2 "\}." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "(a0+a1*sqrt(2)
)/(b0+b1*sqrt(2));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&,&%#a0G\"\"\"*
&%#a1GF&-%%sqrtG6#\"\"#F&F&F&,&%#b0GF&*&%#b1GF&F)F&F&!\"\"" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "expand(numer(%)*(b0-b1*sqrt(2))) / \+
expand(denom(%)*(b0-b1*sqrt(2)));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*
&,**&%#a0G\"\"\"%#b0GF'F'*(F&F'%#b1GF'-%%sqrtG6#\"\"#F'!\"\"*(%#a1GF'F
+F'F(F'F'*(F.F'F1F'F*F'F/F',&*$)F(F.F'F'*&F.F')F*F.F'F/F/" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 90 "This is again of the form c0+c1*sqrt(2) f
or some c0 and c1, and hence again an element of " }{TEXT 270 1 "Q" }
{TEXT -1 71 "[sqrt(2)]. So although we could not always divide non-zer
o elements of " }{TEXT 271 1 "Q" }{TEXT -1 20 "[x], in other words " }
{TEXT 272 1 "Q" }{TEXT -1 56 "[x] is not a field,  we can divide non-z
ero elements of " }{TEXT 273 1 "Q" }{TEXT -1 39 "[sqrt(2)] and end up \+
with something in " }{TEXT 274 1 "Q" }{TEXT -1 14 "[sqrt(2)]. So " }
{TEXT 275 1 "Q" }{TEXT -1 10 "[sqrt(2)]=" }{TEXT 276 1 "Q" }{TEXT -1 
15 "(sqrt(2)), but " }{TEXT 277 1 "Q" }{TEXT -1 7 "[x] <> " }{TEXT 
278 1 "Q" }{TEXT -1 4 "(x)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 22 "A field K is called a " }{TEXT 279 18 "differen
tial field" }{TEXT -1 76 " when a differentiation ' is defined on K th
at has the following properties:" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }
{TEXT 350 21 "I)   (a+b)' = a' + b'" }}{PARA 256 "" 0 "" {TEXT -1 26 "
II)   (a*b)' = a'*b + b'*a" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 
"" 0 "" {TEXT -1 14 "Examples are: " }{TEXT 280 4 "Q(x)" }{TEXT -1 45 
" with the usual differentiation a'=diff(a,x)." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 19 "Another example is " }
{TEXT 281 1 "Q" }{TEXT -1 74 "(x,sqrt(x^3+1)), which is the set of all
 functions that can be written as:" }}{PARA 0 "" 0 "" {TEXT -1 73 "(a \+
polynomial in x and t)/(a polynomial in x and t)  where t=sqrt(x^3+1).
" }}{PARA 0 "" 0 "" {TEXT -1 52 "Just like with the sqrt(2) example, a
ll elements of " }{TEXT 282 1 "Q" }{TEXT -1 78 "(x,sqrt(x^3+1)) can be
 written in the form a=a0+a1*sqrt(x^3+1) where a0,a1 in " }{TEXT 283 
1 "Q" }{TEXT -1 41 "(x). The fact that this is a field means:" }}
{PARA 0 "" 0 "" {TEXT -1 75 "1) if you add or substract two things of \+
that form, it's again of that form" }}{PARA 0 "" 0 "" {TEXT -1 67 "2) \+
if you multiply two things of that form, it's again of that form" }}
{PARA 0 "" 0 "" {TEXT -1 74 "3) if you divide two non-zero things of t
hat form, it's again of that form" }}{PARA 0 "" 0 "" {TEXT -1 102 "  N
ote: to get it in that form may require some simplification just like \+
above in the sqrt(2) example." }}{PARA 0 "" 0 "" {TEXT -1 59 "4) these
 operations +, -,*, / satisfy the usual properties." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 18 "The fact that K = " }
{TEXT 284 1 "Q" }{TEXT -1 96 "(x,sqrt(x^3+1)) is a differential field \+
means all of the above 4 properties, plus the following." }}{PARA 0 "
" 0 "" {TEXT -1 145 "5) A differentiation is defined, and if a is an e
lement of  K (in the example that means: if a is of that form), then a
' is also an element of K." }}{PARA 0 "" 0 "" {TEXT -1 59 "6) This dif
ferentiation ' satisfies the two rules I and II." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 174 "Note that if K is any fi
eld, we can always make it into a differential field by defining a'=0 \+
for all a in K. Of course that is not the kind of examples we are inte
rested in." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 48 "The smallest interesting example is the field K=" }{TEXT 285 1 
"Q" }{TEXT -1 309 "(x). We already know two possible ways to define a \+
differentiation on K that satisfies rules I and II, the usual one a'=d
iff(a,x), the trivial one where a'=0 for all a. Can we find more examp
les of differentiations on K? (in other words, can we find another ope
ration ' that satisfies the two rules I and II)." }}{PARA 0 "" 0 "" 
{TEXT -1 431 "Suppose ' is any operation that satisfies rules I and II
. Then by rule II we see that b'=(1*b)'=1'*b + b'*1 = 1'*b + b'  so 1'
 times b is always zero. Hence 1' must be 0. Then by rule I we see tha
t (1+1)'=0+0, and so 2'=0, and if you keep applying rule I you see tha
t n'=0 for all positive integers n. Also, 0'+1'=(0+1)'=1'=0 so 0'=0, a
nd (-1)'+1'=(-1+1)'=0'=0 so (-1)'=0, and so it follows that n'=0 for n
egative integers as well." }}{PARA 0 "" 0 "" {TEXT -1 136 "Now if r=a/
b then by rule II you see that: a'=(r*b)'=r'*b+b'*r  so r'*b=a'-b'*r s
o r'=a'/b-b'*r/b =a'/b-b'*a/b^2=(a'*b-b'*a)/b^2 so the " }{TEXT 309 
34 "quotient rule follows from rule II" }{TEXT -1 27 ":   (a/b)'=(a'*b
-b'*a)/b^2." }}{PARA 0 "" 0 "" {TEXT -1 97 "Since n'=0 for all integer
s n we see by the quotient rule that (n/m)'=0 for all integers n,m. So
 " }{TEXT 345 70 "a'=0 for all rational numbers a, no matter which def
inition of ' we us" }{TEXT -1 43 "e (as long as it satisfies rules I a
nd II)." }}{PARA 0 "" 0 "" {TEXT -1 63 "Let's continue the study of th
e possible differentiations on K=" }{TEXT 286 1 "Q" }{TEXT -1 241 "(x)
. We will see that once we choose a value for x', then the differentia
tion ' is completely defined, there are no more choices. If we define \+
x'=1 we have the usual differentiation and if we define x'=0 then ever
ything is constant (like in " }{TEXT 287 1 "Q" }{TEXT -1 153 "(Pi) whe
re Pi is the constant number 3.1415..). Why does the value of x' deter
mine ' completely? First of all, we've already seen that a'=0 for all \+
a in " }{TEXT 288 1 "Q" }{TEXT -1 413 " no matter which differentiatio
n ' we use. Second, if we have x^2 then by rule II the derivative is 2
*x*x'. Keep on applying rule II and you get that (x^n)'=n*x^(n-1)*x' f
or all integers n. Take sums, and then you see that for every polynomi
al in x we have only 1 way to define its derivative once we've fixed a
 choice for x', and then by the quotient rule the definition is unique
ly defined for every element of " }{TEXT 289 1 "Q" }{TEXT -1 5 "(x). \+
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 34 "So, y
ou can take any element k of " }{TEXT 290 1 "Q" }{TEXT -1 126 "(x), th
en define x' to be that k, and since ' is supposed to satisfy rules I \+
and II, that will then completely determine ' on " }{TEXT 291 1 "Q" }
{TEXT -1 75 "(x). In this case it is easy to see that for this ' we ha
ve a'=k*diff(a,x)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 40 "Of course in practice, whenever we take " }{TEXT 292 1 "Q
" }{TEXT -1 7 "(x) or " }{TEXT 293 1 "C" }{TEXT -1 77 "(x), then the d
ifferentiation will always be defined such that everything in " }
{TEXT 294 1 "Q" }{TEXT -1 4 " or " }{TEXT 295 1 "C" }{TEXT -1 168 " is
 constant (has derivative 0) and that x'=1. But it is good to keep in \+
mind that diff(f,x) is not the only way to define an operation ' that \+
satisfies rules I and II." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 46 "Now consider the following differential field:" }}
{PARA 0 "" 0 "" {TEXT -1 2 "K=" }{TEXT 296 1 "C" }{TEXT -1 41 "(x,t) w
here the field of complex numbers " }{TEXT 297 1 "C" }{TEXT -1 45 " ar
e the constants, x'=1 and t=(x^3+2)^(1/3)." }}{PARA 0 "" 0 "" {TEXT 
-1 12 "What is t' ?" }}{PARA 0 "" 0 "" {TEXT -1 154 "Of course we can \+
apply the rules we know from calcules to find diff(t,x). But are there
 any other choices for the definition of the operation ' such that:" }
}{PARA 0 "" 0 "" {TEXT -1 21 "*) a'=0 for all a in " }{TEXT 298 1 "C" 
}}{PARA 0 "" 0 "" {TEXT -1 45 "*) We choose the value of x' to be equa
l to 1" }}{PARA 0 "" 0 "" {TEXT -1 38 "*) ' satisfies the conditions I
 and II" }}{PARA 0 "" 0 "" {TEXT -1 262 "The answer is NO. If you defi
ne the value of x' (we'll always choose x'=1, but again, that's not th
e only possible choice), then the value of t' where t=(x^3+2)^(1/3) is
 fixed, meaning that there is only possible value for t' such that ' s
atisfies the two rules." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT 310 5 "Note:" }{TEXT -1 184 " You may be surprised to see t
hat the value of diff( (x^3+2)^(1/3) , x) is already fixed without any
 reference to the classic way of introducing differentiation with slop
es of graphs. " }{TEXT 312 112 "All that we need for the theory of dif
ferentiation is just properties I and II, and we need to define what x
' is" }{TEXT -1 95 ". Then, without every taking any limits of (delta \+
y)/(delta x), without any slopes and graphs, " }{TEXT 311 31 "using on
ly algebraic operations" }{TEXT -1 101 " and no analysis, we still see
 that there can be only one correct value for t' where t=(x^3+2)^(1/3)
." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "t:=(x^3+2)^(1/3);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"tG*$),&*$)%\"xG\"\"$\"\"\"\"\"\"\"
\"#F-#F-F+F," }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 132 "Now let's prove \+
the statement that t' has only 1 possible value once x' is fixed. Note
 that t is a root of the following polynomial:" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 20 "T(x)^3 - (x^3+2) =0;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/,(*$)-%\"TG6#%\"xG\"\"$\"\"\"\"\"\"*$)F*F+F,!\"\"!\"#F
-\"\"!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 111 "We have seen that beca
use of rule I, the derivative of 0 equals 0. So if we differentiate on
 both sides we get:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "diff
(lhs(%),x) = diff(rhs(%),x);  # where x'=1" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/,&*&)-%\"TG6#%\"xG\"\"#\"\"\"-%%diffG6$F'F*\"\"\"\"\"$
*$)F*F+F,!\"$\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "solve
(%, diff(T(x),x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&*$)%\"xG\"\"#
\"\"\"F(*$)-%\"TG6#F&\"\"#F(!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 15 "subs(T(x)=t,%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&
*$)%\"xG\"\"#\"\"\"F(*$),&*$)F&\"\"$F(\"\"\"F'F/#\"\"#F.F(!\"\"" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 183 "So we see that no matter how we w
ant to define t', once we fix the value of x' to be 1, then the above \+
expression is the only possible value for t' if we want to obey rules \+
I and II. " }{TEXT 346 99 "No slopes, no limits, no analysis, only rul
es I, II and algebra were used to reach this conclusion." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "diff(t,x);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#*&*$)%\"xG\"\"#\"\"\"F(*$),&*$)F&\"\"$F(\"\"\"F'F/#\"\"
#F.F(!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 313 11 "Definition." }
{TEXT -1 70 " Let K be a subfield of L. Let a be an element of L. Then
 a is called " }{TEXT 314 16 "algebraic over K" }{TEXT -1 71 " when th
ere exists a non-zero polynomial P(x) in K[x] such that P(a)=0." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 18 "Examples:
 If K is " }{TEXT 315 1 "Q" }{TEXT -1 10 " and L is " }{TEXT 316 1 "C
" }{TEXT -1 150 " then sqrt(7+sqrt(-1)) is algebraic over K, but Pi is
 not algebraic over K. To find such a polynomial P(x) one can issue th
e following Maple commands." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
20 "a:=sqrt(7+sqrt(-1));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"aG*$-%
%sqrtG6#,&\"\"(\"\"\"%\"IGF+\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 7 "P:=x-a;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"PG,&%\"
xG\"\"\"*$-%%sqrtG6#,&\"\"(F'%\"IGF'\"\"\"!\"\"" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 231 "Now a is a root of P, but P is not yet in K[x]. Thi
s is remedied by the command evala(Norm( .. )) which produces a polyno
mial in K[x] such that P is a factor of that polynomial. First the squ
are roots have to be in RootOf notation." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 21 "P:=convert(P,RootOf);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%\"PG,&%\"xG\"\"\"-%'RootOfG6#,(*$)%#_ZG\"\"#\"\"\"F'!\"(F'-F)6
#,&F,F'F'F'!\"\"F5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "P:=ev
ala(Norm(P));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"PG,(\"#]\"\"\"*$)
%\"xG\"\"#\"\"\"!#9*$)F*\"\"%F,F'" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 13 "rem(P,x-a,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"
!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 121 "Indeed, the old P, which wa
s x-a, is a factor of the new P, and so a is a root of P. So the numbe
r a is algebraic over K=" }{TEXT 317 1 "Q" }{TEXT -1 1 "." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 15 "Likewise, if K=" }
{TEXT 318 1 "C" }{TEXT -1 281 "(x) and a=sqrt(x^2+x+sqrt(x)) and t=1-x
*a then t is algebraic over K. The function t is an element of K(a), a
nd every element of K(a) is algebraic over K, including the function t
. This means that there exists a polynomial P(T) in K[T] such that P(t
)=0. It can be found as follows:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 23 "a:=sqrt(x^2+x+sqrt(x));" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%\"aG*$-%%sqrtG6#,(*$)%\"xG\"\"#\"\"\"\"\"\"F,F/*$-F'6#F,F.F/F.
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "t:=1-x*a;" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#>%\"tG,&\"\"\"F&*&%\"xGF&-%%sqrtG6#,(*$)F(\"\"#
\"\"\"F&F(F&*$-F*6#F(F0F&F0!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 7 "P:=T-t;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"PG,(%\"
TG\"\"\"!\"\"F'*&%\"xGF'-%%sqrtG6#,(*$)F*\"\"#\"\"\"F'F*F'*$-F,6#F*F2F
'F2F'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "P:=evala(Norm( con
vert(P,RootOf) ));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"PG,@*$)%\"xG
\"\"&\"\"\"!\"\"\"\"\"F,%\"TG!\"%*$)F(\"\"$F*!\"#*$)F(\"\"%F*F2*$)F-\"
\"#F*\"\"'*&F4F*F7F*F2*&F4F*F-F,F5*&F0F*F7F*F2*&F0F*F-F*F5*$)F(\"\")F*
F,*$)F(\"\"(F*F8*$)F(F9F*F,*$)F-F5F*F,*$)F-F1F*F." }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 18 "P:=subs(T=T(x),P);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"PG,@*$)%\"xG\"\"&\"\"\"!\"\"\"\"\"F,-%\"TG6#F(!\"%*
$)F(\"\"$F*!\"#*$)F(\"\"%F*F4*$)F-\"\"#F*\"\"'*&F6F*F9F*F4*&F6F*F-F,F7
*&F2F*F9F*F4*&F2F*F-F*F7*$)F(\"\")F*F,*$)F(\"\"(F*F:*$)F(F;F*F,*$)F-F7
F*F,*$)F-F3F*F0" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 147 "A variable T \+
is always considered constant in Maple. But an undetermined function T
(x) is not constant. Maple will leave diff(T(x),x) un-evaluated." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 27 "Now subs(
T(x)=t,P) is zero." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "evala
(  subs(T(x)=t,P)  );" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 45 "And diff(0,x)=0 so the following i
s zero too:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "diff(P,x);" 
}}{PARA 12 "" 1 "" {XPPMATH 20 "6#,F*$)%\"xG\"\"%\"\"\"!\"&-%%diffG6$-
%\"TG6#F&F&!\"%*$)F&\"\"#F(!\"'*$)F&\"\"$F(!\")*&F-\"\"\"F*F:\"#7*&F6F
()F-F3F(F8*(F%F(F-F(F*F(F0*&F6F(F-F(\"#;*&F%F(F*F(F'*&F2F(F=F(F4*(F6F(
F-F(F*F(F0*&F2F(F-F(F;*&F6F(F*F(F'*$)F&\"\"(F(\"\")*$)F&\"\"'F(\"#9*$)
F&\"\"&F(FL*&)F-F7F(F*F(F'*&F=F(F*F(!#7" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 37 "diff(T(x),x)=solve(%=0,diff(T(x),x));" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#/-%%diffG6$-%\"TG6#%\"xGF*,$*&*&)F*\"\"#\"\"\",6*$
F.F0!\"&!\"'\"\"\"F*!\")*&F*F5)F'F/F0F6*&F*F0F'F5\"#;*$F8F0F4F'\"#7*$)
F*\"\"&F0\"\")*$)F*\"\"%F0\"#9*$)F*\"\"$F0\"\"'F5F0,2F5F5F'!\"$*&FBF0F
'F0F5FA!\"\"*&FFF0F'F0F5FEFL*$)F'FGF0FLF;FG!\"\"#F5FC" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "evala(subs(T(x)=t,rhs(%)));" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&*&-%%sqrtG6#,(*$)%\"xG\"\"#\"\"\"
\"\"\"F,F/*$-F'6#F,F.F/F.,.!\"&F/F0!\"\"*$)F,#\"\"$F-F.!\"$F*\"#9*$)F,
F9F.\"\")F,\"\"'F/F.,*F5F/F*F-F,F/F<F/!\"\"#F5\"\"%" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 94 "Again, we see that the rules I and II leave no \+
choice for t', the above expression must be t'." }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 10 "diff(t,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#
,&*$-%%sqrtG6#,(*$)%\"xG\"\"#\"\"\"\"\"\"F+F.*$-F&6#F+F-F.F-!\"\"*&*&F
+F.,(F+F,F.F.*&F-F-*$-F&6#F+F-!\"\"#F.F,F.F-*$-F&6#F(F-F:#F2F," }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evala(%);" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#,$*&*&-%%sqrtG6#,(*$)%\"xG\"\"#\"\"\"\"\"\"F,F/*$-F'6
#F,F.F/F.,.!\"&F/F0!\"\"*$)F,#\"\"$F-F.!\"$F*\"#9*$)F,F9F.\"\")F,\"\"'
F/F.,*F5F/F*F-F,F/F<F/!\"\"#F5\"\"%" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 32 "The conclusion is the following:" }}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT -1 112 "If K is a differential field, i.e. \+
a field with an operation ' defined on it that satisfies rules I and I
I, and " }{TEXT 319 174 "if an expression a is algebraic over K, then \+
there only one way to define a differentiation ' on the field K(a) tha
t coincides with the differentiation ' we already had on K." }{TEXT 
-1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
101 "This differentiation works as follows. If t in K(a), and P(T) in \+
K[T] such that P(t)=0, then compute:" }}{PARA 0 "" 0 "" {TEXT -1 20 " \+
  diff( P(T(x)) ,x)" }}{PARA 0 "" 0 "" {TEXT -1 26 "which will be of t
he form:" }}{PARA 0 "" 0 "" {TEXT -1 21 "   P1(T) + T' * P2(T)" }}
{PARA 0 "" 0 "" {TEXT -1 33 "and so you find: T'=-P1(T)/P2(T)." }}
{PARA 0 "" 0 "" {TEXT -1 126 "Here P1(T) is the polynomial you get fro
m P(T) if you simply differentiate all the coefficients but you don't \+
differentiate T." }}{PARA 0 "" 0 "" {TEXT -1 113 "P2(T) is the polynom
ial you get if you don't differentiate the coefficients, but you diffe
rentiate w.r.t. T only." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "
P:=subs(T(x)=T,P); # Note: subs(T=t,P) is zero." }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"PG,@*$)%\"xG\"\"&\"\"\"!\"\"\"\"\"F,%\"TG!\"%*$)F(
\"\"$F*!\"#*$)F(\"\"%F*F2*$)F-\"\"#F*\"\"'*&F4F*F7F*F2*&F4F*F-F,F5*&F0
F*F7F*F2*&F0F*F-F*F5*$)F(\"\")F*F,*$)F(\"\"(F*F8*$)F(F9F*F,*$)F-F5F*F,
*$)F-F1F*F." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "P1:=diff(P,x
); # view T as constant, differentiate w.r.t. x" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#P1G,6*$)%\"xG\"\"%\"\"\"!\"&*$)F(\"\"#F*!\"'*$)F(\"
\"$F*!\")*&F1F*)%\"TGF.F*F3*&F1F*F6\"\"\"\"#;*&F-F*F5F*F/*&F-F*F6F*\"#
7*$)F(\"\"(F*\"\")*$)F(\"\"'F*\"#9*$)F(\"\"&F*FC" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 58 "P2:=diff(P,T); # view x as constant, differen
tiate w.r.t T" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#P2G,2!\"%\"\"\"%\"
TG\"#7*&)%\"xG\"\"%\"\"\"F(F'F&*$F+F.F-*&)F,\"\"$F.F(F.F&*$F1F.F-*$)F(
F2F.F-*$)F(\"\"#F.!#7" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "t_
prime:=normal(-P1/P2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%(t_primeG,
$*&*&)%\"xG\"\"#\"\"\",6*$F(F+!\"&!\"'\"\"\"F)!\")*&F)F0)%\"TGF*F+F1*&
F)F+F4F0\"#;*$F3F+F/F4\"#7*$)F)\"\"&F+\"\")*$)F)\"\"%F+\"#9*$)F)\"\"$F
+\"\"'F0F+,2F0F0F4!\"$*&F>F+F4F+F0F=!\"\"*&FBF+F4F+F0FAFH*$)F4FCF+FHF7
FC!\"\"#F0F?" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "t_prime:=ev
ala(subs(T=t,t_prime));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%(t_primeG
,$*&*&-%%sqrtG6#,(*$)%\"xG\"\"#\"\"\"\"\"\"F.F1*$-F)6#F.F0F1F0,.!\"&F1
F2!\"\"*$)F.#\"\"$F/F0!\"$F,\"#9*$)F.F;F0\"\")F.\"\"'F1F0,*F7F1F,F/F.F
1F>F1!\"\"#F7\"\"%" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "evala
(diff(t,x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&*&-%%sqrtG6#,(*$)%
\"xG\"\"#\"\"\"\"\"\"F,F/*$-F'6#F,F.F/F.,.!\"&F/F0!\"\"*$)F,#\"\"$F-F.
!\"$F*\"#9*$)F,F9F.\"\")F,\"\"'F/F.,*F5F/F*F-F,F/F<F/!\"\"#F5\"\"%" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 25 "Let K be a subfield of L." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 20 "Then L is
 called an " }{TEXT 320 24 "algebraic extension of K" }{TEXT -1 181 " \+
if every element of L is algebraic over K. Remember that an element t \+
of L is algebraic over K if there exists a non-zero polynomial P(T) in
 K[T] such that P(t)=0. As we have seen:" }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 177 "If K is a differential field, and
 if L is an algebraic extension of K, then there is only 1 way that th
e differentiation ' can be extended to a differentiation ' on the fiel
d L." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 89 "L
et L be a differential field and K be a subfield. Suppose that theta i
s an element of L." }}{PARA 0 "" 0 "" {TEXT -1 180 "Then K(theta) is t
he subfield of L. The field K(theta) contains all elements of L that c
an be represented as a quotient of polynomials in theta with coefficie
nts in K. For example:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT 324 8 "Example:" }}{PARA 0 "" 0 "" {TEXT -1 5 "*) K=" }
{TEXT 322 1 "C" }{TEXT -1 3 "(x)" }}{PARA 0 "" 0 "" {TEXT -1 27 "*) V \+
= some open subset of " }{TEXT 323 1 "C" }{TEXT -1 76 ", simply connec
ted, that does not contain the point 0. For example: V=\{x in " }
{TEXT 325 4 "C | " }{TEXT -1 15 "abs(x-1) < 0.5\}" }}{PARA 0 "" 0 "" 
{TEXT -1 205 "*) L = the set of all meromorphic functions \{f: V --> C
\}, so these functions are allowed to have poles, but for each of thes
e poles x=c there must exist an integer n such that (x-c)^n*f is analy
tic at x=c." }}{PARA 0 "" 0 "" {TEXT -1 43 "*) theta = log(x) which is
 an element of L." }}{PARA 0 "" 0 "" {TEXT -1 244 "*) Then K(theta) is
 the set of all meromorphic functions \{f: V --> C\} such that f can b
e written as a quotient of polynomials in x and theta. Now K is a diff
erential field, L is a much larger differential field, and K(theta) is
 a subfield of L." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 132 "Now we may wonder, is theta algebraic over K yes or no? \+
Does there exist a polynomial P(T) in K[T] such that P(T)<>0 and P(the
ta)=0?" }}{PARA 0 "" 0 "" {TEXT -1 303 "Suppose such polynomial P(T) e
xists. In that case we will show that there exists an element alpha in
 K such that alpha'=theta'. Since theta'=1/x and since we know that th
ere is no element alpha in K with alpha'=1/x, we then have a contradic
tion and it will follow that theta can not be algebraic over K." }}
{PARA 0 "" 0 "" {TEXT -1 133 "(Intermezzo: how did we know that no alp
ha in K has alpha'=1/x? Remember that we had the following result for \+
any element alpha in K:" }}{PARA 0 "" 0 "" {TEXT -1 66 "  alpha has po
le of order n at x=c  ---> alpha' has pole order n+1" }}{PARA 0 "" 0 "
" {TEXT -1 58 "  alpha has no pole at x=c  ---> alpha' has no pole at \+
x=c" }}{PARA 0 "" 0 "" {TEXT -1 104 "and this implied that at x=c we e
ither have no pole, or we have pole order >=2, but never pole order 1)
." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 374 "Now
 suppose that theta=log(x) and that we have a0..an in K, and that P(T)
=a0*T^0+...+an*T^n, that P(T)<>0 and that P(theta)=0. If P(T)<>0 then \+
we may assume that an<>0 (otherwise take smaller n), and that an=1 (ju
st divide all a.i by an). We may also assume that n is as small as pos
sible (if there is another polynomial with smaller n, then just take t
hat one). So we have:" }}{PARA 0 "" 0 "" {TEXT -1 35 "P(T)=a0*T^0+...+
a.(n-1)*T^(n-1)+T^n" }}{PARA 0 "" 0 "" {TEXT -1 42 "Then for every roo
t theta of P(T) we have:" }}{PARA 0 "" 0 "" {TEXT -1 33 "(**) P1(theta
)+theta'*P2(theta)=0" }}{PARA 0 "" 0 "" {TEXT -1 49 "So, the following
 polynomial vanishes at T=theta:" }}{PARA 0 "" 0 "" {TEXT -1 105 "(**)
 a0'*T^0+...+a.(n-1)'*T^(n-1)+1'*T^n  + theta'*( a0*0*T^(0-1)+...+a.(n
-1)*(n-1)*T^(n-2)+1*n*T^(n-1) )" }}{PARA 0 "" 0 "" {TEXT -1 251 "and s
ince 1'=0 this is a polynomial of degree <n in T. This contradicts min
imality of n unless (**) is the zero polynomial. So (**), which is in \+
K[T], is the zero polynomial. So all coefficients are zero, including \+
the coefficient of T^(n-1) which is:" }}{PARA 0 "" 0 "" {TEXT -1 25 " \+
   a.(n-1)' + theta'*1*n" }}{PARA 0 "" 0 "" {TEXT -1 280 "so theta' = \+
-(1/n) * a.(n-1)'. If you take alpha=-(1/n)*a.(n-1) then alpha is in K
, and alpha'=theta' which we have already argued that that was impossi
ble. Therefore, there can not exist any non-zero polynomial P(T) in K[
T] for which P(theta)=0. So theta is not algebraic over K." }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 327 11 "Definition:" }
{TEXT -1 108 " If K a subfield of L, and if theta in L is not a root o
f any polynomial P(T) in K[T], then theta is called " }{TEXT 326 21 "t
ranscendental over K" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 0 "" 0 "" {TEXT 329 5 "Note:" }{TEXT -1 78 " Transcendental nu
mbers are those complex numbers that are not algebraic over " }{TEXT 
328 1 "Q" }{TEXT -1 78 ". An example is Pi. If you take all roots of a
ll non-zero polynomials P(x) in " }{TEXT 347 1 "Q" }{TEXT -1 53 "[x] t
hen you get the set of all algebraic numbers in " }{TEXT 348 1 "C" }
{TEXT -1 48 ". The remaining numbers are transcendental over " }{TEXT 
349 1 "Q" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT 330 5 "Note:" }{TEXT -1 261 " New variables are also tran
scendental. If K is a field, and A is a new variable that does not sho
w up in K, then A is considered transcendental over K, which means tha
t every polynomial in K[A] that has at least one non-zero coefficient \+
is considered non-zero." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 5 "If K=" }{TEXT 331 1 "Q" }{TEXT -1 339 " and A=sqrt(3)
, which is new in the sense that A does not show up in K, but it is no
t a variable, then not every polynomial in K[A] with non-zero coeffici
ents is non-zero, for example A^2-3 has non-zero coefficients 1 and -3
 but it is still considered to be equal to 0, which implies that A=sqr
t(3) is algebraic over K, not transcendental." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 5 "If K=" }{TEXT 332 1 "C" }
{TEXT -1 292 "(x^2), then x is algebraic over K, because it is a root \+
of T^2-x^2 which is a non-zero element of K[T]. However, if y is an un
assigned, not yet used, variable, then y is transcendental over K beca
use we can't find any non-zero polynomial in K[T] that would become ze
ro if we do subs(T=y,...)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 
"" 0 "" {TEXT -1 35 "Consider again what we have proven:" }}{PARA 0 "
" 0 "" {TEXT -1 5 "If K=" }{TEXT 333 1 "C" }{TEXT -1 44 "(x), where we
 have an operation ' such that:" }}{PARA 0 "" 0 "" {TEXT -1 29 "*) ' s
atisfies rules I and II" }}{PARA 0 "" 0 "" {TEXT -1 7 "*) x'=1" }}
{PARA 0 "" 0 "" {TEXT -1 21 "*) c'=0 for all c in " }{TEXT 334 1 "C" }
}{PARA 0 "" 0 "" {TEXT -1 145 "And if we have any differential field e
xtension L, with an element theta for which theta'=1/x, then theta is \+
automatically transcendental over K." }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 0 "" 0 "" {TEXT -1 61 "In fact, something more general holds, \+
the proof is the same." }}{PARA 0 "" 0 "" {TEXT 335 8 "Theorem:" }}
{PARA 0 "" 0 "" {TEXT -1 169 "Let K be a differential field, and let t
heta be an element of a differential field extension L of K. If theta'
 is an element of K, then there are only two possibilities:" }}{PARA 
0 "" 0 "" {TEXT -1 27 "*) either theta is in K, or" }}{PARA 0 "" 0 "" 
{TEXT -1 34 "*) theta is transcendental over K." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 336 11 "Definition:" }}{PARA 0 "
" 0 "" {TEXT -1 172 "Let K be a differential field, and theta be an el
ement of a differential field extension of K (an element of some diffe
rential field that has K as a differential subfield)." }}{PARA 0 "" 0 
"" {TEXT -1 38 "If there exists an a in K such that   " }{TEXT 337 15 
"theta' = a'/a, " }{TEXT -1 76 "and if there is no element alpha in K \+
with alpha'=a'/a then theta is called " }{TEXT 338 18 "logarithmic ove
r K" }{TEXT -1 46 ". The differential field K(theta) is called a " }
{TEXT 339 26 "logarithmic extension of K" }{TEXT -1 1 "." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 340 4 "Note" }{TEXT -1 
102 ": the theorem says that if theta is logarithmic over K then it is
 automatically transcendental over K." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT 341 5 "Note:" }{TEXT -1 290 " the derivative
 of log(a) is a'/a. So theta can be thought of as log(a) plus perhaps \+
a constant. Note that with this definition of logarithmic, if at any p
oint we replace theta by theta+c where c is a constant, i.e. c'=0, the
n in the above definition that won't make any difference at all: " }
{TEXT 342 263 "If theta is logaritmic over K and we replace theta by t
heta+c for any constant c then nothing changes in the definition (only
 theta' is used in the definition, not theta itself), so we are allowe
d to do that any time we want and it should not change any results. " 
}{TEXT -1 307 "This has serious consequences. If f is some function, a
nd if F is its antiderivative (meaning that F'=f) and if F contains lo
garithmic extensions (i.e. F has a logarithm theta that did not yet oc
cur in f), then we can replace theta by theta+c in F, and the result s
hould again be an antiderivative. Example:" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 21 "f:=1/(x^3-x)+x*ln(x);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%\"fG,&*&\"\"\"F',&*$)%\"xG\"\"$F'\"\"\"F+!\"\"!\"\"F-*&F+F--%#
lnG6#F+F-F-" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "F:=int(f,x);
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"FG,,-%#lnG6#%\"xG!\"\"-F'6#,&F
)\"\"\"F*F.#F.\"\"#-F'6#,&F)F.F.F.F/*&)F)F0\"\"\"F&F.F/*$F5F6#F*\"\"%
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 113 "Now ln(x) already appeared i
n f, so we can't fool around with it. But theta=ln(x+1) is one of the \+
new logarithms." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "theta:=l
n(x+1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&thetaG-%#lnG6#,&%\"xG\"
\"\"F*F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "diff(F,x);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#,**&\"\"\"F%%\"xG!\"\"!\"\"*&F%F%,&F&
\"\"\"F(F+F'#F+\"\"#*&F%F%,&F&F+F+F+F'F,*&F&F+-%#lnG6#F&F+F+" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "c:=17; # some arbitrary cons
tant." }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"cG\"#<" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 26 "FF:=subs(theta=theta+c,F);" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#>%#FFG,.-%#lnG6#%\"xG!\"\"-F'6#,&F)\"\"\"F*F.#F.
\"\"#-F'6#,&F)F.F.F.F/#\"#<F0F.*&)F)F0\"\"\"F&F.F/*$F7F8#F*\"\"%" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "diff(FF,x); # same as diff(F
,x)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,**&\"\"\"F%%\"xG!\"\"!\"\"*&F%
F%,&F&\"\"\"F(F+F'#F+\"\"#*&F%F%,&F&F+F+F+F'F,*&F&F+-%#lnG6#F&F+F+" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 140 "Since F and FF are both antideri
vatives of f, and since antiderivatives are unique up to constants, it
 follows that F-FF must be a constant:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 5 "F-FF;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6##!#<\"\"#" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT 344 15 "An observation:" }}{PARA 0 "" 0 "" {TEXT -1 27 "F - subs
(theta=theta+c, F)," }}{PARA 0 "" 0 "" {TEXT -1 93 "where c is any con
stant you like,  can only turn out to be a constant when: F is of the \+
form:" }}{PARA 0 "" 0 "" {TEXT -1 15 "F = A + c*theta" }}{PARA 0 "" 0 
"" {TEXT -1 53 "where A contains no theta, and where c is a constant.
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 865 "As a
 consequence, in integration, if you get a new logarithm theta in the \+
antiderivative F, then theta never appears in the denominator of F, F \+
has degree 1 in theta, and the coefficient of theta^1 is always a cons
tant. This is one of the things that Liouville's principle says, which
 will be formulated more precisely later. Remember also that the coeff
icient in front of the logarithm theta is a residue. So residues have \+
to be constant, otherwise we can't find an anti-derivative no matter w
hich logarithmic extensions we use. For rational function integration \+
this was always the case (rational functions can always be integrated \+
if you allow logarithmic extensions) but we'll see later that for more
 general functions we do not always get constant residues, and in thos
e cases it is impossible to find an antiderivative if we only allow lo
garithmic extensions." }}}}{MARK "3 5 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 }