{VERSION 2 3 "SUN SPARC SOLARIS" "2.3" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 256 "" 0 1 0 0 0 0 0 1 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 }{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 49 "TP no 4: Racines de polyn omes et alg\350bre lin\351aire" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "with(linalg): # Valider cette ligne." }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 82 "Il faut donner la r\351ponse sur le papier aux question s \351crites avec un mot en gras." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 82 "1. On d\351finit par r\351 currence des polynomes P0(t)=P1(t)=1, Pn+1(t)=Pn(t)-t*Pn-1(t)." }}} {EXCHG {PARA 0 "" 0 "" {TEXT 257 8 "Calculer" }{TEXT -1 27 " \340 la m ain P2, P3,P4 et P5." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 85 "2. Trouver avec Maple leurs racines, tracer leur graphe (en uti lisant solve et plot)." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 59 "3. Avec la procedure jones qui calculent les polynomes \+ Pn, " }{TEXT 258 11 "conjecturer" }{TEXT -1 62 " le degr\351 de Pn et \+ le coefficient du terme de plus haut degr\351." }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 94 "jones:=proc(n) if n=0 then 1 elif n=1 then 1 e lse\njones(n-1)-t*jones(n-2) fi end; # A valider." }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 85 "4. On s'interesse \340 la premi\350re racine positive. Av ec fsolve et la procedure jones, " }{TEXT 259 6 "donner" }{TEXT -1 89 " conjecturalement une limite quand n tend vers l'infini pour la pr emi\350re racine positive." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 35 "On d\351fi nit maintenant les vecteurs " }{XPPEDIT 18 0 "Xn = MATRIX([[Pn+1], [Pn ]])" "/%#XnG-%'MATRIXG6#7$7#,&%#PnG\"\"\"\"\"\"F+7#F*" }{TEXT -1 17 " \+ et la matrice A:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "A:=matr ix(2,2,[1,-t,1,0]); # A valider.\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 3 "5. " }{TEXT 261 8 "Verifier" }{TEXT -1 17 " que Xn+1=AXn et " } {TEXT 269 8 "calculer" }{TEXT -1 29 " Xn+1 en fonction de A et X0." }} }{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 3 "6. " }{TEXT 260 7 "Comment" }{TEXT -1 132 " obtenir Pn avec A et X0? V\351rifier \+ pour n<10 la coincidence des calculs avec la proc\351dure Jones (la \+ puissance ni\350me d'une matrice" }}{PARA 0 "" 0 "" {TEXT -1 19 "se ca lcule par &^)." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 43 "7. Nous faisons le changement de variabl e " }{XPPEDIT 18 0 "t:=1/(2*cos(x))^2" ">%\"tG*&\"\"\"\"\"\"*$*&\"\"#F &-%$cosG6#%\"xGF&\"\"#!\"\"" }{TEXT -1 4 " = " }{XPPEDIT 18 0 "1/(exp (x*I)+exp(-x*I))^2" "*&\"\"\"\"\"\"*$,&-%$expG6#*&%\"xGF$%\"IGF$F$-F(6 #,$*&F+F$F,F$!\"\"F$\"\"#F1" }{TEXT -1 6 " . " }}{PARA 0 "" 0 "" {TEXT -1 1 " " }{TEXT 262 9 "Proposer " }{TEXT -1 69 "une formule conj ecturale pour la premi\350re solution positive de Qn(x)=" }{XPPEDIT 18 0 "Pn(1/(2*cos(x))^2)" "-%#PnG6#*&\"\"\"\"\"\"*$*&\"\"#F'-%$cosG6#% \"xGF'\"\"#!\"\"" }{TEXT -1 2 " " }}{PARA 0 "" 0 "" {TEXT -1 28 "Pour cela vous pouvez, pour " }{XPPEDIT 18 0 "n<=5" "1%\"nG\"\"&" }{TEXT -1 96 ", calculer Qn \340 partir de la matrice B et de l'id\351e de la pr\351c\351dente question et utiliser solve." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "B:=matrix(2,2,[1,-1/(2*cos(x))^2,1,0]);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 3 "8. " }{TEXT 263 7 "Trouver" }{TEXT -1 42 " une colonne C1 non nulle telle que C.C1=" }{XPPEDIT 18 0 "exp(x*I)/( exp(x*I)+exp(-x*I))" "*&-%$expG6#*&%\"xG\"\"\"%\"IGF(F(,&-F$6#*&F'F(F) F(F(-F$6#,$*&F'F(F)F(!\"\"F(F2" }{TEXT -1 64 "C1 avec gausselim et la \+ matrice C (autre forme de la matrice B)." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "C:=matrix(2,2,[1,-1/(exp(x*I)+exp(-x*I))^2,1,0]);" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 3 "9. " }{TEXT 264 8 "Trouver " }{TEXT -1 41 "une colonne C2 non nulle telle que C.C2=" }{XPPEDIT 18 0 "exp(-x*I)/ (exp(x*I)+exp(-x*I))" "*&-%$expG6#,$*&%\"xG\"\"\"%\"IGF)!\"\"F),&-F$6# *&F(F)F*F)F)-F$6#,$*&F(F)F*F)F+F)F+" }{TEXT -1 17 "C2 avec gausselim" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 4 "10. " }{TEXT 265 10 "Quelle est" } {TEXT -1 28 " la relation entre C1 et C2?" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 110 "11. Former la ma trice U=[C1,C2], calculer son inverse. V\351rifier le calcul de Maple \+ en calculant U&*inverse(U)." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 29 "Calc uler D1:=inverse(U)*C*U. " }{TEXT 266 10 "Quelle est" }{TEXT -1 31 " l a propri\351t\351 de la matrice D1?" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 4 "12. " }{TEXT 267 8 "Exprimer" }{TEXT -1 49 " Qn en fonction de n , D1, et U, puis de n,x et U." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 " " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 4 "13. " }{TEXT 268 6 "Donner" }{TEXT -1 110 " les instructions Maple qui vous permettent \+ de conclure que x=Pi/(n+1) est la premi\350re racine positive de Qn. \+ " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 107 "14. V\351rifier que pour n=6, Map le \351tait incapable de simplifier le calcul formellement (essayer au ssi n=10)." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "simplify(subs (t=1/(2*(cos(Pi/7)))^2,jones(6)));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}} {MARK "28 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 }