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10 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 }{PSTYLE "" 0 318 1 {CSTYLE "" -1 -1 "" 1 10 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 }{PSTYLE "" 0 319 1 {CSTYLE "" -1 -1 "" 1 10 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 }{PSTYLE "" 5 320 1 {CSTYLE "" -1 -1 "" 1 10 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 }{PSTYLE "" 0 321 1 {CSTYLE "" -1 -1 "" 1 10 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 20 "Calcul formel Mass1 " } {TEXT -1 9 "2003-2004" }}{PARA 0 "" 0 "" {TEXT 257 9 "feuille 5" } {TEXT 258 64 " Alg\350bre Lin\351aire (suite): puissances d'une matr ice sym\351trique" }}{PARA 0 "" 0 "" {TEXT 259 24 "calcul_mass1_2003_f 5.mws" }}}{SECT 0 {PARA 256 "" 0 "" {TEXT -1 7 "Outils " }}{SECT 0 {PARA 257 "" 0 "" {TEXT -1 48 "Des fonctions ( hors biblioth\350que) q u'on utilise" }}{EXCHG {PARA 258 "> " 0 "" {MPLTEXT 1 0 274 "with(lina lg):\nracine:=proc(n)\nlocal k;\nk:=1;\nwhile k*k <= n do k:=k+1; od; \nRETURN(k-1);\nend:\n\nmatrice:=proc(L)# nops(L) ( nombre d'\351l \351ments de L) est un carr\351 \nlocal j;\nj:=racine(nops(L));\nRETUR N(evalm(matrix(j,j,L)));\nend:\n\nvecteur:=proc(L)\nRETURN(evalm(vecto r(L)));\nend:" }}{PARA 7 "" 1 "" {TEXT -1 80 "Warning, the protected n ames norm and trace have been redefined and unprotected\n" }}}}{SECT 0 {PARA 259 "" 0 "" {TEXT -1 37 "Les fonctions \"solve\" et \" linsol ve\"" }}{EXCHG {PARA 260 "> " 0 "" {MPLTEXT 1 0 7 "? solve" }}}{EXCHG {PARA 261 "" 0 "" {TEXT -1 74 "Si on se donne un syst\350me d'\351quat ions lin\351aires ( en Maple, un ensemble : " }{TEXT 272 1 "\{" } {TEXT -1 3 "..." }{TEXT 273 1 "\}" }{TEXT -1 3 "), " }{TEXT 275 5 "sol ve" }{TEXT -1 71 " permet d'en calculer les solutions (quand elles exi stent!...)\nExemple:" }}}{EXCHG {PARA 262 "> " 0 "" {MPLTEXT 1 0 44 "e quations := \{u+v+w=1, 3*u+v=3, u-2*v-w=0\}:\n" }{TEXT -1 28 "Noter le s accolades \{ et \}" }{MPLTEXT 1 0 33 "\nsolutions := solve( equati ons ):" }}}{EXCHG {PARA 263 "" 0 "" {TEXT -1 38 "Pour v\351rifier, on \+ utilise la fonction " }{TEXT 276 4 "subs" }{TEXT -1 2 " :" }}}{EXCHG {PARA 264 "> " 0 "" {MPLTEXT 1 0 29 "subs( solutions, equations ):" }} }{EXCHG {PARA 265 "" 0 "" {TEXT -1 97 "Cela semble tout \340 fait rais onnable..\nSi, maintenant, on se donne le syst\350me sous la forme Ax= b :" }}}{EXCHG {PARA 266 "> " 0 "" {MPLTEXT 1 0 42 "A:=matrice([1,2,3, 4]):\nb:= vecteur([1,0]):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 285 54 "On \+ peut r\351soudrele syst\350me avec la fonction linsolve :" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "u:=linsolve(A,b):" }}}{EXCHG {PARA 267 "" 0 "" {TEXT -1 53 "Si on veut \"r\351cup\351rer\" les composante s de u ou de A:" }}}{EXCHG {PARA 268 "> " 0 "" {MPLTEXT 1 0 17 "u[1], u[2],A[1,2]:" }}}{EXCHG {PARA 269 "" 0 "" {TEXT -1 45 "On peut aussi a voir un second membre matrice:" }}}{EXCHG {PARA 270 "> " 0 "" {MPLTEXT 1 0 22 "B:=matrice([1,0,0,1]):" }}}{EXCHG {PARA 271 "> " 0 " " {MPLTEXT 1 0 17 "F:=linsolve(A,B):" }}}{EXCHG {PARA 272 "" 0 "" {TEXT -1 94 "Comme B est la matrice de l'identit\351, et que AF = B, F est la matrice inverse de A, v\351rifions" }}}{EXCHG {PARA 273 "> " 0 "" {MPLTEXT 1 0 18 "test:=evalm(A&*F):" }}}{EXCHG {PARA 274 "" 0 "" {TEXT -1 38 "On peut aussi faire cela formellement:" }}}{EXCHG {PARA 275 "> " 0 "" {MPLTEXT 1 0 22 "G:=matrice([a,b,c,d]):" }}}{EXCHG {PARA 276 "> " 0 "" {MPLTEXT 1 0 17 "H:=linsolve(G,B):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 282 12 "La fonction " }{TEXT 283 3 "det" }{TEXT 284 21 " est d\351finie en Maple" }}}{EXCHG {PARA 277 " > " 0 "" {MPLTEXT 1 0 14 "det_G:=det(G):" }}}{EXCHG {PARA 278 "> " 0 " " {MPLTEXT 1 0 66 "K:=matrice([det_G*H[1,1],det_G*H[1,2],det_G*H[2,1], det_G*H[2,2]]):" }}}}}{SECT 0 {PARA 279 "" 0 "" {TEXT -1 25 "Et mainte nant, au boulot!" }}{EXCHG {PARA 280 "" 0 "" {TEXT 260 10 "D\351finiti on" }{TEXT -1 61 " : On dira que G :=matrice([a,b,c,d]) est sym\351t rique si b=c" }}}{SECT 0 {PARA 281 "" 0 "" {TEXT -1 14 "Exercice 1 : \+ " }}{EXCHG {PARA 282 "" 0 "" {TEXT -1 18 "Soit G la matrice " }}} {EXCHG {PARA 283 "> " 0 "" {MPLTEXT 1 0 56 "G:=matrice([r,s,s,t]):\nal ias(Id2 = matrice([1,0,0,1])):\n" }{TEXT -1 17 "Cette \"fonction\" " } {TEXT 280 6 "alias " }{TEXT -1 61 "permet de d\351finir des \"constant es \" et de leur donner un nom." }{MPLTEXT 1 0 1 "\n" }{TEXT -1 46 "Ma intenant la matrice de l'identit\351 s'appelle " }{TEXT 278 3 "Id2" }} }{EXCHG {PARA 284 "> " 0 "" {MPLTEXT 1 0 26 "evalm(Id2): evalm(G&*Id2) :" }}}{EXCHG {PARA 285 "" 0 "" {TEXT 261 3 "Q1 " }{TEXT -1 49 ": Calcu ler le d\351terminant de G -x*Id2 (fonction " }{TEXT 274 3 "det" } {TEXT -1 2 " )" }}}{EXCHG {PARA 286 "" 0 "" {TEXT -1 28 "On note PG ce d\351terminant. \n" }{TEXT 262 2 "Q2" }{TEXT -1 134 " : V\351rifier q ue PG est un polynome en x de degr\351 2 qui a deux racines r\351elle s; pour quelles valeurs de r, s et t sont-elles \351gales ?\n" }{TEXT 281 5 "Apr\350s" }{TEXT -1 68 " avoir fait ce calcul \"\340 la main\", on pourra essayer l'instruction " }{TEXT 277 27 "solutions:= [solve( PG,x) ] " }{TEXT -1 37 "pour voir la r\351ponse donn\351e par Maple" } }}{SECT 0 {PARA 287 "" 0 "" {TEXT -1 7 "R\351ponse" }}{EXCHG {PARA 288 "> " 0 "" {MPLTEXT 1 0 18 "PG:=det(G -x*Id2):" }}{PARA 289 "" 1 " " {XPPMATH 20 "6#>%\"PG,,*&%\"rG\"\"\"%\"tGF(F(*&F'F(%\"xGF(!\"\"*&F+F (F)F(F,*$)F+\"\"#F(F(*$)%\"sGF0F(F," }}}{EXCHG {PARA 290 "> " 0 "" {MPLTEXT 1 0 25 "solutions:=[solve(PG,x)]:" }}{PARA 291 "" 1 "" {XPPMATH 20 "6#>%*solutionsG7$,(*&\"\"#!\"\"%\"rG\"\"\"F+*&F(F)%\"tGF+ F+*&F(F),**$)F*F(F+F+*(F(F+F*F+F-F+F)*$)F-F(F+F+*&\"\"%F+)%\"sGF(F+F+# F+F(F+,(*&F(F)F*F+F+*&F(F)F-F+F+*&F(F)F/F9F)" }}}}}{SECT 0 {PARA 292 " " 0 "" {TEXT -1 13 "Exercice 2 : " }}{EXCHG {PARA 293 "" 0 "" {TEXT -1 38 "Soit G la matrice sym\351trique suivante " }}}{EXCHG {PARA 294 "> " 0 "" {MPLTEXT 1 0 26 "G:=matrice([1.,2.,2.,3.]):" }}}{EXCHG {PARA 295 "" 0 "" {TEXT 263 3 "Q1 " }{TEXT -1 65 "Calculer les racines de PG ( faire le calcul en r\351el , fonction " }{TEXT 279 5 "evalf " }{TEXT -1 70 " ) ; on note h la plus grande en valeur absolue et lv \+ la plus petite.\n" }{TEXT 264 6 "Q1 bis" }{TEXT -1 55 " : Calculer les d\351terminants de G -h*Id2et de G -lv*Id2" }}{PARA 296 "" 0 "" {TEXT 265 2 "Q2" }{TEXT -1 67 " On note K := (1/h)*G. Calculer M1:=K^ 101 puis K^102 puis K^103 \n" }{TEXT 266 2 "Q3" }{TEXT -1 61 " Calcul er det(M1)\nSoit vp1 le premier vecteur colonne de M1:" }}}{EXCHG {PARA 297 "> " 0 "" {MPLTEXT 1 0 33 "vp1:= vecteur([M1[1,1],M1[2,1]]): " }{TEXT -1 0 "" }}}{EXCHG {PARA 298 "" 0 "" {TEXT 267 2 "Q4" }{TEXT -1 48 " Calculer G&*vp1 - h*vp1. Que peut-on conclure?" }}}{SECT 0 {PARA 299 "" 0 "" {TEXT -1 7 "R\351ponse" }}{EXCHG {PARA 300 "> " 0 " " {MPLTEXT 1 0 50 "PG:=det(G -x*Id):\nsolutions:=evalf([solve(PG,x)]): " }}}{EXCHG {PARA 301 "> " 0 "" {MPLTEXT 1 0 16 "h:=solutions[1]:" }}} {EXCHG {PARA 302 "> " 0 "" {MPLTEXT 1 0 89 "K:= evalm((1/h)*G):\nN:=10 1:\nM1:=evalm(Id):\nfor k from 1 to N do \n M1:=evalm(M1&*K):\nod:" }}}{EXCHG {PARA 303 "> " 0 "" {MPLTEXT 1 0 10 "evalm(M1):" }}}{EXCHG {PARA 304 "> " 0 "" {MPLTEXT 1 0 13 "evalm(K&*M1):" }}}{EXCHG {PARA 305 "> " 0 "" {MPLTEXT 1 0 13 "evalm(K&*M1):" }}}{EXCHG {PARA 306 "> \+ " 0 "" {MPLTEXT 1 0 8 "det(M1):" }}}{EXCHG {PARA 307 "> " 0 "" {MPLTEXT 1 0 22 "evalm(G&*vp1 - h*vp1):" }}}}}{SECT 0 {PARA 308 "" 0 " " {TEXT -1 13 "Exercice 3 : " }}{EXCHG {PARA 309 "" 0 "" {TEXT 268 2 " Q1" }{TEXT -1 93 " : Calculer M2:=G^101, M3:=G^102, calculer q:= eva lf( M3[1,1]/M2[1,1] ) puis det(G -q*Id2)." }}{PARA 310 "" 0 "" {TEXT -1 101 "En d\351duire que ce calcul donne \351galement une valeur num \351rique de h .\nSoit L la matrice inverse de G\n" }{TEXT 269 2 "Q2" }{TEXT -1 114 " Faire le meme calcul que pr\351c\351demment en rempla \347ant G par L et en d\351duire une valeur num\351rique de lv (Exerc ice2)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 271 150 "Si 101 est une puissance trop petite, essayer avec des puissances plus grandes, 150 \+ , 200, 400 par exemple mais attention \340 la dur\351e de votre calcul !" }}}{SECT 0 {PARA 311 "" 0 "" {TEXT -1 7 "R\351ponse" }}{EXCHG {PARA 312 "> " 0 "" {MPLTEXT 1 0 117 "N:=101:\nM2:=evalm(Id):\nfor k f rom 1 to N do \n M2:= evalm(M1&*G):\nod:\nM3:=evalm(M2&*G):\nq:= eva lf(M3[1,1]/M2[1,1]):\n" }}}{EXCHG {PARA 313 "> " 0 "" {MPLTEXT 1 0 13 "det(G -q*Id):" }}}}}{SECT 0 {PARA 314 "" 0 "" {TEXT -1 10 "Exercice 4 " }{TEXT 270 27 ": de plus en plus fort!...." }}{EXCHG {PARA 315 "" 0 "" {TEXT -1 32 "Soit M la matrice sym\351trique 3x3" }}}{EXCHG {PARA 316 "> " 0 "" {MPLTEXT 1 0 36 "M:=matrice([4,-1,0,-1,4,-1,0,-1,4]):" } }}{EXCHG {PARA 317 "" 0 "" {TEXT -1 66 "et , attention , en dimension \+ 3 la matrice de l'identit\351 s'\351crit :" }}}{EXCHG {PARA 318 "> " 0 "" {MPLTEXT 1 0 34 "Id3:=matrice([1,0,0,0,1,0,0,0,1]):" }}}{EXCHG {PARA 319 "" 0 "" {TEXT -1 53 "Refaire les calculs de l'exercice 3 ave c la matrice M" }}}{SECT 0 {PARA 320 "" 0 "" {TEXT -1 7 "R\351ponse" } }{EXCHG {PARA 321 "> " 0 "" {MPLTEXT 1 0 132 "N:=150:\nR1:=evalm(Id3): \nfor k from 1 to N do \n R1:= evalm(R1&*M):\nod:\nR2:=evalm(R1&*M): \nq:= evalf(R2[1,1]/R1[1,1]):\ndet(M-q*Id3):\n" }}}}}}}{MARK "2 5 6" 0 }{VIEWOPTS 1 1 0 1 1 1803 }