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0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 263 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 264 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 1 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 18 "" 0 "" {TEXT -1 47 "Musterloesung zur Aufgabe 6 aus der 18. Sitzung" }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 256 "" 0 " " {TEXT -1 68 "Eine abstrakte mathematische Struktur: Lie -Algebren un d Lie-Gruppen" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 8 "Es sei " }{TEXT 318 1 "g" }{TEXT -1 42 " eine Algebra ue ber dem Skalarenkoerper " }{TEXT 321 1 "K" }{TEXT -1 3 " (=" }{TEXT 324 1 "R" }{TEXT -1 8 " bzw. = " }{TEXT 325 1 "C" }{TEXT -1 56 "), d. h., es existiert ein Produkt (Multiplikation) " }}{PARA 0 "" 0 "" {TEXT 319 24 " &m : g x g -> g " }{TEXT -1 3 ", " }}{PARA 0 " " 0 "" {TEXT -1 25 "das dem Distibutivgesetz" }}{PARA 0 "" 0 "" {TEXT -1 8 " i) " }{TEXT 315 59 "(x+y) &m (a+b) = x &m a + x \+ &m b + y &m a + y &m b" }{TEXT -1 9 " , " }{TEXT 316 8 "x,y, a,b " }{TEXT -1 5 " in " }{TEXT 317 1 "g" }}{PARA 0 "" 0 "" {TEXT -1 8 " ii) " }{TEXT 320 64 "(alpha x ) &m (beta y ) = alpha beta ( x &m y ) , x,y " }{TEXT -1 2 "in" }{TEXT 322 21 " g , alpha, beta " }{TEXT -1 2 "in" }{TEXT 323 3 " K" }}{PARA 0 "" 0 "" {TEXT -1 37 "genuege. Das Produkt sei assoziativ" }}{PARA 0 "" 0 "" {TEXT -1 7 "iii) " }{TEXT 326 36 "( x &m y ) &m z = x &m ( y &m z)" }} {PARA 0 "" 0 "" {TEXT -1 45 "aber nicht notwendigerweise kommmutativ: " }{TEXT 327 20 "x &m y <> y &m x" }{TEXT -1 42 " . Es existi ere ein Eins-Element namens " }{TEXT 256 6 "Eins " }{TEXT -1 2 "in" }{TEXT 258 3 " g" }{TEXT -1 21 " mit der Eigenschaft" }}{PARA 0 "" 0 "" {TEXT -1 7 "iv) " }{TEXT 328 34 "x &m Eins = Eins &m x = x " }{TEXT -1 15 " fuer jedes " }{TEXT 329 2 "x " }{TEXT -1 3 "i n " }{TEXT 330 1 "g" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }}{EXCHG {PARA 264 "" 0 "" {TEXT -1 0 "" }}{PARA 257 "" 0 "" {TEXT -1 15 "Aufgabe 6.a): " }{TEXT 257 30 "implementiere eine Funktion " } {TEXT 302 1 "`" }{TEXT 259 24 "&m`:=proc(a,b) ... end ," }{TEXT 303 98 " die dieses Produkt repraesentiert. Sie soll die Eigenschaften \+ i) ... iv) realisieren, d.h., " }}{PARA 263 "" 0 "" {TEXT 331 108 "i )+ ii) Summen in den Argumenten sollen expandiert werden, skalare Fak toren sollen vorgezogen werden, z.B.:" }}{PARA 0 "" 0 "" {TEXT -1 15 " " }{TEXT 332 56 "(3*x + y ) &m (2* b ) --> 6 (x &m b) + 2 (y &m b)" }}{PARA 0 "" 0 "" {TEXT -1 15 " " } {TEXT 333 37 "(3*x + y ) &m Eins --> 3 x + y" }}{PARA 0 "" 0 " " {TEXT -1 101 "iii) Die Assoziativitaet kann dadurch implementiert \+ werden, dass man in vernesteten Aufrufen von " }{TEXT 334 2 "&m" } {TEXT -1 52 " Klammerungen systematisch ``nach rechts schiebt'':" }} {PARA 0 "" 0 "" {TEXT -1 15 " " }{TEXT 335 40 "(x &m y) \+ &m z --> x &m ( y &m z )" }}{PARA 0 "" 0 "" {TEXT -1 15 "iv) D er Name " }{TEXT 336 5 "Eins " }{TEXT -1 12 " soll von " }{TEXT 337 2 "&m" }{TEXT -1 43 " als neutrales Element verarbeitet werden." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 312 10 "Weiter hin:" }}{PARA 0 "" 0 "" {TEXT -1 131 "Es soll die Moeglichkeit vorgese hen sein, beliebige Namen als Skalare zu deklarieren, die dann mit i i) aus dem formalen Produkt " }{TEXT 304 2 "&m" }{TEXT -1 53 " heraus gezogen werden. Lege dazu eine globale Menge " }{TEXT 260 9 "scalars \+ " }{TEXT -1 48 "an, welche diejenigen Namen enthaelt, die von " } {TEXT 305 2 "&m" }{TEXT -1 93 " als Skalare aufgefasst werden sollen ( alle anderen symbolischen Namen sind als Elemente von " }{TEXT 338 1 " g" }{TEXT -1 69 " aufzufassen). Die Skalare sind damit die MAPLE-Konst anten (vom Typ " }{TEXT 261 12 "constant ), " }{TEXT -1 82 "sowie alg ebraische Ausdruecke, welche aus Konstanten und (Potenzen von) Namen \+ in " }{TEXT 262 8 " scalars" }{TEXT -1 17 " aufgebaut sind:" }}{PARA 0 "" 0 "" {TEXT -1 4 " " }{TEXT 277 24 "scalars:= \{eps , alpha\}; " }}{PARA 0 "" 0 "" {TEXT -1 4 " " }{TEXT 306 87 "(3*x + eps* y) &m (alpha^2*beta) --> 3 alpha^2 (x &m x) + eps alpha^2 (y &m beta) " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 313 10 "Weit erhin:" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 46 "fuer die spaete r folgenden Aufgabenteile soll " }{TEXT 307 2 "&m" }{TEXT -1 40 " so i mplementiert werden, dass Produkte " }{TEXT 280 16 " x &m y &m z ..." }{TEXT -1 2 " " }}{PARA 0 "" 0 "" {TEXT -1 14 "mit mehr als " } {TEXT 279 5 "Tiefe" }{TEXT -1 23 " Faktoren als Symbol " }{TEXT 281 14 "O(g^(Tiefe+1))" }{TEXT -1 39 " zurueckgeliefert werden. Hierbei s ei " }{TEXT 282 5 "Tiefe" }{TEXT -1 44 " ein global vorzugebender ganz zahliger Wert:" }}{PARA 0 "" 0 "" {TEXT -1 2 " " }{TEXT 283 43 "Tiefe :=3; ( 2*x+ y&m z) &m a &m (Eins+b) " }}{PARA 258 "" 0 "" {TEXT -1 79 " --> 2 ( x &m a) + 2 (x &m a &m b) + y &m z \+ &m a + O(g^4)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 339 9 "Anleitung" }{TEXT -1 27 ": beachte Kapitel 8.10 (``" }{TEXT 314 20 "Operatorschreibweise" }{TEXT -1 32 "'') der Vorlesung/d es Skriptes. " }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 284 7 "Achtung" }{TEXT -1 67 " beim Austesten: es bietet sich an, die beno etigten Prozeduren mit " }{TEXT 287 15 "option remember" }{TEXT -1 46 " zu implementieren. Aendert man den Wert von " }{TEXT 286 5 "Tiefe" }{TEXT -1 39 " , so ist vor weiteren Rechnungen die " }{TEXT 285 14 " remember table" }{TEXT -1 13 " zu loeschen!" }}}{SECT 1 {PARA 3 "" 0 " " {TEXT -1 7 "Loesung" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 2425 "r estart:\n\n# folgende Prozedur ueberprueft, ob das Argument eine Zahl \n# oder ein Skalar aus der Menge scalars oder eine Potenz \n# eines s olchen Skalars ist:\n \nis_scalar:=proc(expr)\nglobal scalars;\noption remember:\n if not assigned(scalars) then\n ERROR(`die Menge der Skalaren-Namen ist nicht definiert`);\n fi;\n if type(expr,constant ) or member(expr,scalars) then RETURN(true);fi;\n if type(expr,`^`) a nd member(op(1,expr),scalars) then RETURN(true);fi;\n false;\nend:\n \n# folgende Prozedur stellt fest, aus wievielen Multiplikationen\n# e in Ausdruck der Form x &m y &m .. besteht. Sollte als Faktor\n# das \+ Symbol O(g^..) auftauchen, so wird diesem Ausdruck Tiefe+1\n# Multipli kationen zugeschrieben\n\nMultiplikationen:=proc(expr) global Tiefe; o ption remember;\n if type(expr,name) then RETURN(1);fi;\n if op(0, expr)=`&m` then RETURN(Multiplikationen(op(1,expr))\n \+ +Multiplikationen(op(2,expr)));\n fi;\n# ein Ausdruc k O(g^..) repraesentiert mindestens \n# Tiefe+1 Multiplikationen:\n \+ if op(0,expr)=O then RETURN(Tiefe+1);fi;\n# andere Faelle sollten nic ht auftreten:\n ERROR(`hier sollte Multiplikationen nie ankommen`); \nend:\n\n# der Multiplikationsoperator, der die Eigenschaften des\n# \+ des Algebraprodukts besitzt:\n\n`&m` :=proc(a,b) local aa;global Tiefe ;option remember;\n if a=0 or b=0 then RETURN(0);fi;\n# Implementier ung des Eins-Elements:\n if a=Eins then RETURN(b);fi;\n if b=Eins \+ then RETURN(a);fi;\n# Implementierung der Linearitaet:\n if type(exp and(a),`+`) then RETURN(map( `&m`,expand(a),b));fi;\n if type(expand (b),`+`) then RETURN(map2(`&m`,a,expand(b)));fi;\n# Vorziehen von skal aren Faktoren:\n if type(a,`*`) then\n for aa in a do\n \+ if is_scalar(aa) then RETURN( aa * ( (a/aa) &m b ) );fi;\n od;\n fi;\n if type(b,`*`) then\n for aa in b do\n if is_s calar(aa) then RETURN( aa * ( a&m (b/aa) ) );fi;\n od;\n fi;\n# Implementierung der Assoziativitaet: (x &m y) &m b --> x &m (y &m b) \n if type(a,function) and op(0,a)=`&m`\n then RETURN(op(1,a) & m (op(2,a) &m b));\n fi; \n# Berechne die Multiplikationstiefen von \+ a und b. Wenn der globale Wert\n# Tiefe ueberschritten wird, dann lief ere als Ergebnis O(g^(Tiefe+1))\n if not assigned(Tiefe) then\n \+ ERROR(`der globale Wert Tiefe ist nicht gesetzt!`);\n fi;\n if Mu ltiplikationen(a)+Multiplikationen(b)>Tiefe\n then RETURN(O(g^(Ti efe+1))); \n fi;\n RETURN('a &m b' );\nend:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 6 "Tests:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 90 "readlib( forget):\ninit:=proc();forget(`&m`);forget(Multiplikationen);forget(is _scalar);end:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "scalars:= \{\};" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "Tiefe:=3: (x+2*y&m z) &m (z &m c) &m (a+b);init();" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "Tiefe:=4: (x+2*y&m z) &m (z &m c) &m (a+b);init();" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "Tiefe:=5: (x+2*y&m z) &m ( z &m c) &m (a+b);init();" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT 266 13 "Aufgabe 6.b):" }{TEXT -1 20 " Implementi ere auf " }{TEXT 340 1 "g" }{TEXT -1 7 " den " }{TEXT 288 14 "``Kom mutator``" }{TEXT -1 24 " , dies ist das Produkt " }}{PARA 0 "" 0 "" {TEXT -1 12 " " }{TEXT 341 55 "&c: g x g -> g , x &c \+ y := x &m y - y &m x ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 120 "Dieses ``Lie-Produkt`` macht g \+ zu einer sogenannten Lie-Algebra. Der Kommutator is offensichtlich \+ antisymmetrisch " }}{PARA 0 "" 0 "" {TEXT -1 12 " " } {TEXT 342 22 "x &c y = - y &c x " }{TEXT -1 3 " , " }}{PARA 0 "" 0 "" {TEXT -1 65 " aber nicht assoziativ. Es gilt die sogenannte Jaco bi-Identitaet" }}{PARA 0 "" 0 "" {TEXT -1 6 " " }{TEXT 343 65 "x \+ &c ( y &c z ) + y &c ( z &c x ) + z &c ( x &c y ) = 0" } {TEXT -1 18 " fuer alle " }{TEXT 344 5 "x,y,z" }{TEXT -1 4 " in " }{TEXT 345 1 "g" }{TEXT -1 1 "." }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 7 "Loesung" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 1 " " }{MPLTEXT 1 0 34 "`&c` := (a,b) -> a &m b - b &m a:" }}{PARA 0 "" 0 "" {TEXT -1 27 "Test der Jacobi-Identitaet:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "x & c (y &c z) + y &c (z &c x) + z &c (x &c y) ;" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 4 "Die " }{TEXT 265 19 "Exponentialfunktion" }{TEXT -1 7 " auf " }{TEXT 346 1 "g" }{TEXT -1 27 " ist die (formale) Reihe \+ " }}{PARA 0 "" 0 "" {TEXT -1 4 " " }{TEXT 270 70 "Exp(x) = Eins \+ + x + (x &m x) / 2! + (x &m x &m x) / 3! + ..." }}{PARA 0 "" 0 "" {TEXT -1 24 "mit der Eigenschaft " }{TEXT 271 54 "Exp(x) &m E xp (-x) = Exp (-x) &m Exp (x) = Eins" }{TEXT -1 3 " . " }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 263 13 "A ufgabe 6.c):" }{TEXT -1 43 " Implementiere die Exponentialfunktion \+ " }{TEXT 264 24 "Exp:=proc(x) ... end !" }{TEXT -1 5 " Da " } {TEXT 308 2 "&m" }{TEXT -1 16 " nicht mehr als " }{TEXT 278 5 "Tiefe" }{TEXT -1 101 " Multiplikationen zulaesst, braucht die Reihe nur bis \+ zur entsprechenden Potenz berechnet zu werden." }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 6 "Teste " }{TEXT 289 3 "Exp" }{TEXT -1 22 ": es muss z.B. gelten:" }}{PARA 0 "" 0 "" {TEXT -1 4 " " }{TEXT 347 47 "Exp(x) &m Exp(-x) = Eins + O(g^ (Tiefe+1))" }}{PARA 0 "" 0 "" {TEXT -1 4 " " }{TEXT 348 46 "Exp(x) \+ &m Exp( x) = Exp(2x) + O(g^(Tiefe+1))" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 7 "Loesung" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 124 "Exp: =proc(a) local result,aa,i;\n result:=Eins;aa:=Eins;\n for i from 1 \+ to Tiefe do\n aa:=a &m aa; result:=result+aa/i!;" }}{PARA 0 "" 0 "" {TEXT -1 9 " " }{MPLTEXT 1 0 37 "od;\n result+O(g^(Tiefe +1));\n end:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 5 "Test:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 90 "readlib(forget):\ninit:=proc();forget(`&m`) ;forget(Multiplikationen);forget(is_scalar);end:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "Tiefe:=4:init(); Exp(x);" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 35 "Tiefe:=5:init(): Exp(x) &m Exp(-x);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "Tiefe:=9:init(); Exp(x) &m E xp(x) - Exp(2*x);" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 70 "Die Theorie der Lie-Algebren und Gruppen liefert das Resultat, dass " }{TEXT 274 16 "Exp(x) &m Exp(y)" }{TEXT -1 7 " mit " }{TEXT 273 8 "Exp(x+y) " }{TEXT -1 23 " uebereinstimmt, wenn " }{TEXT 276 1 "x" }{TEXT -1 5 " und " }{TEXT 275 1 "y" }{TEXT -1 24 " vertauschen, d.h., wenn" }} {PARA 0 "" 0 "" {TEXT -1 2 " " }{TEXT 309 30 "x &c y = x &m y - y &m \+ x = 0" }}{PARA 0 "" 0 "" {TEXT -1 139 "gilt. Allgemein ist der Unter schied zwischen den Exponentialtermen nur durch Kommutatoren bestimmt. So gilt z.B. die folgende Entwicklung (" }{TEXT 290 31 "Baker-Campbel l-Hausdorff-Formel" }{TEXT -1 2 "):" }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{PARA 0 "" 0 "" {TEXT -1 1 " " }{TEXT 310 62 "Exp(-x) &m Exp(-y) &m E xp(x+y) \n = Exp( alpha1 (x &c y)" }}{PARA 259 "" 0 "" {TEXT -1 42 " + alpha2 (x &c ( x &c y) )" }}{PARA 260 "" 0 " " {TEXT -1 93 " + alpha3 (y &c ( x &c y) )\n \+ + alpha4 (x &c ( x &c (x &c y) ) )" }}{PARA 261 "" 0 "" {TEXT -1 473 " + alpha5 (x &c ( y &c (x &c y) ) )\n \+ + alpha6 (y &c ( y &c (y &c x) ) )\n + alpha7 (x &c \+ ( x &c (x &c (x &c y)) ) )\n + alpha8 (x &c ( x &c (y & c (x &c y)) ) )\n + alpha9 (y &c ( x &c (x &c (x &c y)) ) )\n + alpha10 (x &c ( y &c (y &c (x &c y)) ) )\n \+ + alpha11 (y &c ( y &c (x &c (x &c y)) ) )\n + alpha12 (y &c ( y &c (y &c (x &c y)) ) )\n + O(g^6)" }} {PARA 262 "" 0 "" {TEXT -1 14 " )" }}{PARA 0 "" 0 "" {TEXT -1 13 "mit gewissen " }{TEXT 291 12 "universellen" }{TEXT -1 53 " (von der Algebra unabhaengigen) skalaren Konstanten " }{TEXT 269 14 "alphal, alpha2" }{TEXT -1 7 ", ... " }}{PARA 0 "" 0 "" {TEXT -1 74 " Anmerkung: Diese Entwicklung ist folgendermassen zu verstehen: setzt m an " }{TEXT 272 17 "x=eps*a , y=eps*b" }{TEXT -1 31 " mit einem skal aren Paremeter " }{TEXT 268 3 "eps" }{TEXT -1 38 " , so gilt die Forme l bis zur Ordnung " }{TEXT 311 8 "O(eps^6)" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 267 13 "Aufgabe \+ 6.d):" }{TEXT -1 99 " Bestimme die universellen Konstanten alpha1,.. .,alpha12 der Baker-Campbell-Hausdorff-Formel ! " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 7 "Loesung" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 748 "Tiefe:=5;readlib(forget):\n forget(`&m`):forget(Multiplikationen):forget(is_scalar): \nscalars:=\{ seq(alpha.i,i=1..12) \};\ntmp1:=Exp(-x) &m Exp(-y) &m Exp(x+y):\ntmp2 :=Exp( alpha1 *(x &c y)\n + alpha2 *(x &c ( x &c y) )\n \+ + alpha3 *(y &c ( x &c y) )\n + alpha4 *(x &c ( x &c (x \+ &c y) ) )\n + alpha5 *(x &c ( y &c (x &c y) ) )\n + \+ alpha6 *(y &c ( y &c (y &c x) ) )\n + alpha7 *(x &c ( x &c (x &c (x &c y)) ) )\n + alpha8 *(x &c ( x &c (y &c (x &c y)) ) \+ )\n + alpha9 *(y &c ( x &c (x &c (x &c y)) ) )\n + a lpha10*(x &c ( y &c (y &c (x &c y)) ) )\n + alpha11*(y &c ( y &c (x &c (x &c y)) ) )\n + alpha12*(y &c ( y &c (y &c (x &c \+ y)) ) )\n + O(g^6)\n ):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 23 "Zerlege die Differenz " }{TEXT 292 9 "tmp1-tmp2" }{TEXT -1 8 " nach " }{TEXT 295 3 "x,y" }{TEXT -1 33 ", unterschiedlichen A ufrufen von " }{TEXT 294 2 "&m" }{TEXT -1 27 " und nach dem Ordnungste rm " }{TEXT 293 6 "O(g^6)" }{TEXT -1 1 ":" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "tmp:=collect(tmp1-tmp2,[x,y,`&m`,O],distributed):" }} }{EXCHG {PARA 0 "" 0 "" {TEXT -1 31 "Die Koeffizienten (Polynome in " }{TEXT 296 19 "alpha1, .. ,alpha12" }{TEXT -1 75 ") muessen verschwind en. Dies sind die Gleichungen, aus denen die Parameter " }{TEXT 297 6 "alpha1" }{TEXT -1 2 ", " }{TEXT 298 6 "alpha2" }{TEXT -1 25 ", ... z u bestimmen sind:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "eqs:=\{op(tmp) \};" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 37 "Diese Ausdruecke sind von der Form " }{TEXT 299 32 "Koeff(alpha1,..)*(x &m y &m ...)" }{TEXT -1 64 ". Die Koeffizienten werden isoliert, indem alle Terme der Form \+ " }{TEXT 300 13 "x &m y &m ..." }{TEXT -1 42 " durch 1 ersetzt werden . Der Ordnungsterm " }{TEXT 301 6 "O(g^6)" }{TEXT -1 22 " wird durch 0 ersetzt:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "subs(`&m`= proc(a,b); \+ 1 end, O=proc() 0 end,eqs):eqs:=\";" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 16 "Nun loese diese " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "nops(eqs );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 36 "Gleichungen fuer die 12 Unb ekannten:" }{MPLTEXT 1 0 45 "\nErgebnis:=solve(eqs,\{seq(alpha.i,i=1.. 12)\});" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 17 "Uebersichtlicher:" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "subs(Ergebnis,[seq(alpha[i]=alpha.i ,i=1..12)]);" }}}}}{MARK "15" 0 }{VIEWOPTS 1 1 0 3 2 1804 }