{VERSION 3 0 "SGI MIPS UNIX" "3.0" } {USTYLETAB {CSTYLE "" -1 256 "" 1 18 0 0 0 0 0 0 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 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 262 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 263 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 264 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 } {CSTYLE "" -1 265 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 266 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 267 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 268 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 } {CSTYLE "" -1 269 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 270 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 271 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 272 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 } {CSTYLE "" -1 273 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 274 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 275 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 276 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 } {CSTYLE "" -1 277 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 278 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 279 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 280 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 } {CSTYLE "" -1 281 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 282 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 283 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 284 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 } {CSTYLE "" -1 285 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 286 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 287 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 288 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 } {CSTYLE "" -1 289 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 290 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 291 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 292 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 } {CSTYLE "" -1 293 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 294 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 295 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT 256 21 "Liouvilles principle." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 422 "Let K be a differential field. The set of constants of K is always a field, le t's call this field Const. We will always assume in the algorithm for \+ integration that the field of constants is algebraically closed, meani ng that every non-zero polynomial in Const[T] has a root in Const (by \+ repeatedly applying this it then follows that all roots are in Const). This assumption will also be necessary in Liouvilles principle." }} {PARA 0 "" 0 "" {TEXT -1 104 "In all examples we will always assume th at the field of constants Const is the field of complex numbers " } {TEXT 257 2 "C." }{TEXT -1 46 " The fundamental theorem of algebra say s that " }{TEXT 258 1 "C" }{TEXT -1 37 " is algebraically closed, ever y f in " }{TEXT 259 1 "C" }{TEXT -1 29 "[T] has degree(f,T) roots in \+ " }{TEXT 260 1 "C" }{TEXT -1 29 ", counting with multiplicity." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 263 12 "Definiti ons:" }}{PARA 0 "" 0 "" {TEXT -1 9 "theta is " }{TEXT 264 9 "algebraic " }{TEXT -1 74 " over K when there is a non-zero polynomial P in K[T] \+ such that P(theta)=0" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 9 "theta is " }{TEXT 265 14 "transcendental" }{TEXT -1 43 " over K when theta is not algebraic over K." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 9 "theta is " }{TEXT 266 11 "logari thmic" }{TEXT -1 13 " over K when:" }}{PARA 0 "" 0 "" {TEXT -1 3 " \+ " }{TEXT 276 4 "L1) " }{TEXT -1 30 "theta' = a'/a for some a in K" }} {PARA 0 "" 0 "" {TEXT -1 3 " " }{TEXT 277 3 "L2)" }{TEXT -1 207 " th ere is no b in K with b'=a'/a, which implies that theta is transcenden tal over K (if theta was algebraic then we could find such b from the \+ minimum polynomial P, as was explained in a previous worksheet)." }} {PARA 0 "" 0 "" {TEXT -1 3 " " }{TEXT 278 3 "L3)" }{TEXT -1 98 " (no new constants) The field of constants of K(theta) is the same as the \+ field of constants of K." }}{PARA 0 "" 0 "" {TEXT -1 27 "Note that L1+ L2 implies L3." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 9 "theta is " }{TEXT 267 11 "exponential" }{TEXT -1 13 " over K when:" }}{PARA 0 "" 0 "" {TEXT -1 3 " " }{TEXT 279 3 "E1)" } {TEXT -1 31 " theta'=a*theta for some a in K" }}{PARA 0 "" 0 "" {TEXT -1 3 " " }{TEXT 280 3 "E2)" }{TEXT -1 31 " theta is transcendental o ver K" }}{PARA 0 "" 0 "" {TEXT -1 3 " " }{TEXT 281 3 "E3)" }{TEXT -1 98 " (no new constants) The field of constants of K(theta) is the s ame as the field of constants of K." }}{PARA 0 "" 0 "" {TEXT -1 3 " \+ " }{TEXT 282 3 "E4)" }{TEXT -1 82 " for every positive integer n, ther e exists no b in K for which b<>0 and b'=n*a*b." }}{PARA 0 "" 0 "" {TEXT -1 210 "Note that E1+E2+E3 is equivalent to E1+E4. We will show \+ that if E2 does not hold, i.e. when theta is algebraic, then E4 does n ot hold, i.e., there is a non-zero b in K with b'=n*a*b for some posit ive integer n." }}{PARA 0 "" 0 "" {TEXT -1 732 "If theta is algebraic, then let P in K[T] be its minimum polynomial, i.e. the monic (lcoeff( P,T)=1) polynomial of minimal degree for which P(theta)=0. Let n be th e degree of P. The minimality of n implies that P is an irreducible po lynomial. Let P=T^n+a.(n-1)*T^(n-1)+...+a.0*T^0. Then a.0 is (-1)^n ti mes the product of the roots of P. Since the root theta of P satisfies theta'=a*theta, and the polynomial P determines how all roots of P ar e differentiated, and P is irreducible, it follows that (alpha.i)'=a*a lpha.i for all roots alpha.i of P. Let b=(-1)^n*a.0 be the product of \+ all these alpha.i, so b=alpha.1*..*alpha.n. Now (alpha.1*alpha.2)'=(al pha.1)'*alpha.2 + (alpha.2)'*alpha.1=2*a*alpha.1*alpha.2, and likewise b'=n*a*b." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 9 "theta is " }{TEXT 268 10 "elementary" }{TEXT -1 46 " over K when one of the following three holds:" }}{PARA 0 "" 0 "" {TEXT -1 27 " * ) theta is algebraic, or" }}{PARA 0 "" 0 "" {TEXT -1 29 " *) theta is logarithmic, or" }}{PARA 0 "" 0 "" {TEXT -1 25 " *) theta is exponen tial" }}{PARA 0 "" 0 "" {TEXT -1 223 "over K. If theta1 is elementary \+ over K, and theta2 is elementary over K(theta1), and theta3 is element ary over K(theta1,theta2), etc, and theta.n is elementary over K(theta 1,..,theta.(n-1)) then K(theta1,..,theta.n) is an e" }{TEXT 269 20 "le mentary extension " }{TEXT -1 522 "of K. Note that if K has an algebra ically closed field, then K(theta) has the same field of constants. Wh y, well, well theta is algebraic, then the field of constants of K(the ta) must be an algebraic extension of the field of constants of K, and if the field of constants of K is algebraically closed then any algeb raic extension of that field must be that field itself. If theta is lo garithmic, it follows from assumption L3, which in turn follows from a ssumption L1+L2. If theta is exponential, then it is assumption E3." } }{PARA 0 "" 0 "" {TEXT -1 62 "So, if K has an algebraically closed fie ld of constants (like " }{TEXT 272 1 "C" }{TEXT -1 72 "), then every e lementary extension of K has the same field of constants." }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 3 "An " }{TEXT 270 19 "elementary function" }{TEXT -1 45 " is an element of an elementary ex tension of " }{TEXT 271 1 "C" }{TEXT -1 159 "(x). Every function you c an build up from rational functions with exp, log, and algebraic exten sions, nested in any way you want, is an elementary function. A " } {TEXT 273 34 "transcendental elementary function" }{TEXT -1 114 " is a n elementary function where you only use exponential and logarithmic e xtensions, but no algebraic extensions." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 274 43 "General problem of elementary integ ration: " }{TEXT -1 116 " Given an elementary function f, decide if th ere exists an elementary function F for which F'=f. If so, find such F ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 275 56 "Pro blem of elementary integration, transcendental case: " }{TEXT -1 130 " Given a transcendental elementary function f, decide if there exists \+ an elementary function F for which F'=f. If so, find such F." }}{PARA 0 "" 0 "" {TEXT -1 511 "This is what we will do in this class. So the \+ input function f should be built up with just exp's and log's, and fur thermore, you're not allowed to construct algebraic extensions with th ese exp's and log's, so any function that contains something like exp( 1/2 * log(x^5+x-3)) is not allowed in the transcendental case. In the \+ general case, such things are allowed, which makes it harder, in the g eneral case, if one has an algebraic extension, one has to study algeb raic curves in order to study the valuations." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 261 9 "Theorem. \+ " }{TEXT -1 12 "(Liouville)." }}{PARA 0 "" 0 "" {TEXT -1 90 "Let K be \+ a differential field with an algebraically closed field of constants ( we'll take " }{TEXT 262 1 "C" }{TEXT -1 212 "). Let f be in K and supp ose that there is an elementary extension L=K(theta1,..,theta.n) of K \+ such that L contains an element F for which F'=f. So theta.i is elemen tary over K(theta.1,...,theta.(i-1)) for all i." }}{PARA 0 "" 0 "" {TEXT -1 5 "Then:" }}{PARA 0 "" 0 "" {TEXT -1 43 " (1) f = u' + sum(c. i * (v.i)'/v.i, i=1..m)" }}{PARA 0 "" 0 "" {TEXT -1 82 "for some u,v.1 ,..v.n in K, some non-negative integer n, and some constantc c.i in " }{TEXT 283 1 "C" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 25 "This c an be rewritten as:" }}{PARA 0 "" 0 "" {TEXT -1 40 " (2) F = u + sum(c .i * log(v.i), i=1..m)" }}{PARA 0 "" 0 "" {TEXT -1 33 "for some u,v.1. .v.n in K, c.i in " }{TEXT 284 1 "C" }{TEXT -1 1 "." }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 47 "We can look at the prob lem in two ways, A or B:" }}{PARA 0 "" 0 "" {TEXT -1 276 " (A) Startin g with f, we may wonder what kind of elementary field extensions would be useful for finding an anti-derivative F and what kind of elementar y field extensions would be useless (in other words: which extensions \+ may help to find an antiderivative and which don't?).:" }}{PARA 0 "" 0 "" {TEXT -1 271 " (B) Starting with F in L, we may wonder what struc ture should F have in order for it to be possible that F' ends up in a smaller differential field K. In other words: which kind of elementar y extensions theta can disappear from F when you differentiate, and wh ich can't?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 671 "Liouville's theorem says that the only extensions you're ever \+ going to need are logarithmic extensions. So the only exp's you're goi ng to find in F are exp's that already appear in f. However, F might h ave more log's than f has, although that can only happen in a very spe cial way, namely the only way you can ever need a logarithmic extensio n is when that log appears with degree 1, and with a constant lcoeff(. .., theta), i.e. with a residue that is constant. Furthermore, Liouvil les theorem says (look at it in way (A)) that algebraic extensions are useless for finding F, you don't need any algebraic extensions to fin d F other than the ones that already appear in f." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 490 "For example, when we did integration in K[theta] where theta was logarithmic over K, if we had f of degree d>0, and we did int(lcoeff(f,Theta),x), if we would get a ny extension other than Theta in that integral, then the integral we'r e attempting to compute can't have the form (2), and it won't be too h ard to prove that in such case f does not have form (1), and so no ele mentary integral F will exist. The same holds when we calculated resid ues and one or more of them were non-constant." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 34 "If v in K then v'/v is ca lled the " }{TEXT 285 22 "logarithmic derivative" }{TEXT -1 160 " of v . The name explains itself, because the derivative of log(v) is v'/v. The logarithmic derivative shares some properties with the logarithm, in particular:" }}{PARA 0 "" 0 "" {TEXT -1 81 "LD1) The logarithmic d erivative of v^n is n times the logarithmic derivative of v" }}{PARA 0 "" 0 "" {TEXT -1 21 "(v^n)'/(v^n) = n*v'/v" }}{PARA 0 "" 0 "" {TEXT -1 91 "LD2) The logarithmic derivative of v*w is the sum of the logari thmic derivatives of v and w" }}{PARA 0 "" 0 "" {TEXT -1 26 "(v*w)'/(v *w) = v'/v + w'/w" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 832 "With these properties of the logarithmic derivative, if \+ we have two residues c1 and c2 such that say c2=k*c1 for some integer \+ k then we can combine the terms c1*(v.1)'/(v.1) + c2*(v.2)'/(v.2) as c 1*V'/V where V=v.1*v.2^k. This can be used to reduce the number of new log's we use for F. This trick will not play a role in the case that \+ we consider in this course, the transcendental case, except for the fa ct that it could be used to make answers a little nicer (i.e. fewer ne w log's). However, for the general case (K has algebraic extensions) t hen it will be important to get as few as possible new log's, in such \+ case one calculates all residues r.1...r.M, then one calculates a basi s for the residues, which is a minimal set of complex numbers c.1..c.m such that each residue r.j is of the form sum(k.ij*c.i) with k.ij int egers." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 286 18 "Proof of theorem: " }{TEXT -1 246 "To prove the theorem we will us e induction w.r.t. n where n is the number of extensions theta.i. So w e will assume that the theorem is true whenever we have two differenti al field extensions K' subset L' that consists of infinity in the logarithmic case, and in the exponential case P<>0 and P<>infinity) and consider the valuation v_P , which measures how often the factor theta-P appears in an expression (negative v_P means theta-P is a factor of the denominator, positive \+ v_P means theta-P is a factor of the numerator). Since P is not a spec ial point, we have:" }}{PARA 0 "" 0 "" {TEXT -1 24 "v_P(a')>=0 when v_ P(a)=0" }}{PARA 0 "" 0 "" {TEXT -1 32 "v_P(a')=v_P(a)-1 when v_P(a)<>0 ." }}{PARA 0 "" 0 "" {TEXT -1 59 "Since v_P(a/b)=v_P(a)-v_P(b) we see \+ that when a<>0 we have:" }}{PARA 0 "" 0 "" {TEXT -1 39 "v_P(a'/a) = -1 if and only if v_P(a)<>0" }}{PARA 0 "" 0 "" {TEXT -1 38 "v_P(a'/a) >= 0 if and only if v_P(a)=0" }}{PARA 0 "" 0 "" {TEXT -1 98 "In particul ar, v_P(S) is never smaller than -1, in other words S can not have \"p oles\" of order >1." }}{PARA 0 "" 0 "" {TEXT -1 420 "Now v_P(f)=0 beca use f contains no theta. And u'=f-S, so v_P(u') >= min(v_P(f), v_P(-S) ) = min(0, v_P(S)) >= -1. However, v_P(u') can never be -1, the pole o rder at a regular point P of a derivative can never be 1. So v_P(u')>= 0. So u' can not have a pole at P. Then S=f-u' can not have a pole at \+ P either. Note that at a regular point, if v_P(u)<0 then v_P(u')<0, so we must have v_P(u)>=0 for all non-special points P." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 211 "So u can only have po les at the special points, and nowhere else. In the logarithmic case, \+ this means that u must be an element of K[theta], and in the exponenti al case it means that u is in K[theta,theta^(-1)]." }}{PARA 0 "" 0 "" {TEXT -1 554 "Case2). theta is exponential. Now let P be a special poi nt (0 or infinity). Then v_P(a')=v_P(a) when v_P(a)<>0 and v_P(a')>=0 \+ when v_P(a)=0. So v_P(a'/a) is always >=0. So v_P(S)>=0, so v_P(u')=v_ P(f-S)>=min(v_P(f),v_P(-S))=0. So v_P(u') is >=0 for special points as well. Note that in the exponential case, if v_P(u)<0 then v_P(u') is \+ also <0, so it follows that v_P(u)>=0 for all P. If v_P(u)>=0 for all \+ finite P, then u is a polynomial in theta, but if v_P(u)>=0 for P=infi nity then the degree must be zero, so u does not contain any theta, so u in K." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 646 "Case1). In the logarithmic case, we already have u in K[theta], s o we'll have to look at v_P(u) where P=infinity. When P=infinity we ha ve v_P(a')>=v_P(a) so v_P(a'/a)<=0 so v_P(S)>=0, so v_P(u')=v_P(f-S)>= min(0,0)=0. So v_P(u')>=0, but since u' was a polynomial in theta, we \+ see that the degree is -v_P(u')<=0, so u' in K. So for P=infinity, we \+ have v_P(u')=0, which leaves only two possible values for v_P(u), name ly 0 or -1, and -1 is only possible if the lcoeff is constant. This me ans that either u in K, or u=c*theta+u1 with u1 in K. If u=c*theta+u1, then replace u by u1, and add c*v'/v to S to make up for the differen ce where v'/v=theta'." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 910 "We have now shown that we can indeed get u in the rig ht form (1new). Now we need to handle S. There are a couple of cases, \+ for example in the exponential case we may have something like v.i=the ta^5 and then have c.i * (v.i)'/v.i inside S. However, theta=a' for so me a in K, and then c.i*(v.i)'/v.i equals c.i*5*a' which we can get ri d of by putting it under the u' term. If theta is exponential, then in this way we can handle the case where v_P(v.i)<>0 for special points \+ P and v_P(v.i)=0 for the non-special points. We then need to show that we can also get rid of the v_P(v.i)<>0 for non-special points P by sh owing that such a thing is only possible if you have terms in S that c ancel each other (at least partially cancel each other). We need to as sume that we have a minimal number of terms in S, m minimal, and then \+ we can prove in the logarithmic and exponential case that the v.i do n ot contain theta." }}}}{MARK "0 105 0" 910 }{VIEWOPTS 1 1 0 1 1 1803 }