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{SECT 0 {SECT 0 {PARA 3 "" 0 "" {TEXT -1 43 "CorrigŽ du TD 3 : ArithmŽ
tique et polyn™mes" }}{PARA 256 "" 0 "" {TEXT -1 18 "Jean SŽbastien RO
Y" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 9 "Polyn™mes" }}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 34 "PrŽliminaire utile dans la suite :" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "with(numtheory):" }}{PARA 7 "" 1 "
" {TEXT -1 33 "Warning, new definition for order" }}}{SECT 0 {PARA 4 "
" 0 "" {TEXT -1 27 "Une factorisation difficile" }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 24 "factor(x^2458+x^1229+1);" }}{PARA 7 "" 1 "" 
{TEXT -1 32 "Warning, computation interrupted" }}}{EXCHG {PARA 0 "" 0 
"" {TEXT -1 54 "Manifestement Maple n'y arrive pas. Que dire de 1229 ?
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "isprime(1229);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#%%trueG" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 95 "C'est probablement un nombre premier (pas sžr, car isprim
e est un test probabiliste). De plus :" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 7 "2*1229;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"%eC" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 74 "Factorisons donc quelques polyn™me
s de la forme x^2p+x^p+1 avec p premier." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 53 "seq(factor(x^(2*p)+x^p+1),p=seq(ithprime(i),i=3..5));
" }}{PARA 12 "" 1 "" {XPPMATH 20 "6%*&,(*$%\"xG\"\"#\"\"\"F&F(F(F(F(,0
*$F&\"\")F(*$F&\"\"(!\"\"*$F&\"\"&F(*$F&\"\"%F.*$F&\"\"$F(F&F.F(F(F(*&
F$F(,4*$F&\"#7F(*$F&\"#6F.*$F&\"\"*F(F*F.*$F&\"\"'F(F1F.F3F(F&F.F(F(F(
*&F$F(,@*$F&\"#?F(*$F&\"#>F.*$F&\"#<F(*$F&\"#;F.*$F&\"#9F(*$F&\"#8F.F9
F(*$F&\"#5F.F;F(F,F.F=F(F1F.F3F(F&F.F(F(F(" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 35 "Clairement x^2+x+1 reviens partout." }}{PARA 0 "" 0 "" 
{TEXT -1 38 "Regardons les polyn™mes cyclotomiques." }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 16 "cyclotomic(3,x);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#,(*$%\"xG\"\"#\"\"\"F%F'F'F'" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 11 "Tiens donc." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
49 "seq(cyclotomic(3*p,x),p=seq(ithprime(i),i=3..5));" }}{PARA 12 "" 
1 "" {XPPMATH 20 "6%,0*$%\"xG\"\")\"\"\"*$F%\"\"(!\"\"*$F%\"\"&F'*$F%
\"\"%F**$F%\"\"$F'F%F*F'F',4*$F%\"#7F'*$F%\"#6F**$F%\"\"*F'F$F**$F%\"
\"'F'F-F*F/F'F%F*F'F',@*$F%\"#?F'*$F%\"#>F**$F%\"#<F'*$F%\"#;F**$F%\"#
9F'*$F%\"#8F*F4F'*$F%\"#5F*F6F'F(F*F8F'F-F*F/F'F%F*F'F'" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 42 "Bon ca a l'air clair. Mais pourquoi cela \+
?" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "expand((x^(2*p)+x^p+1)
*(x^p-1));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*$)%\"xG%\"pG\"\"$\"\"
\"!\"\"F)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 44 "Ok : voir la propriŽ
tŽ donnŽe dans l'ŽnoncŽ." }}}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 21 "Un \+
peu d'arithmŽtique" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 60 "VŽrifions la
 propriŽtŽ proposŽe pour de petites valeurs de n" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 29 "seq(n*(n^6-1) mod 7,n=1..10);" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6,\"\"!F#F#F#F#F#F#F#F#F#" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 33 "Tout va bien. On factorise donc :" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 24 "Factor(n*(n^6-1)) mod 7;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#*0%\"nG\"\"\",&F$F%\"\"'F%F%,&F$F%\"\"&F%F%,&F$F%\"\"#F
%F%,&F$F%\"\"%F%F%,&F$F%\"\"$F%F%,&F$F%F%F%F%" }}}{EXCHG {PARA 0 "" 0 
"" {TEXT -1 69 "Ok. Le produit de 7 entiers consŽcutif est forcement d
ivisible par 7." }}{PARA 0 "" 0 "" {TEXT -1 87 "Bien noter l'utilisati
on de Factor et non pas de factor. Ici c'est 'mod' qui travaille." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "factor(n*(n^6-1)) mod 7;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#*,%\"nG\"\"\",&F$F%\"\"'F%F%,&F$F%F%F%
F%,(*$F$\"\"#F%F$F%F%F%F%,(F*F%F$F'F%F%F%" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 43 "Ce n'est pas vraiment le rŽsultat souhaitŽ." }}{PARA 0 "
" 0 "" {TEXT -1 16 "Et avec msolve ?" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 22 "msolve(n*(n^6-1)=0,7);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#<#/%\"nGF%" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 30 "Ok : Tout ent
ier est solution." }}}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 41 "Racines ra
tionnelles de polyn™mes de Z[X]" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 55 
"si p/q irrŽductible est racine de an*x^n+...+a0 alors :" }}{PARA 0 "
" 0 "" {TEXT -1 54 "an*p^n=-q*(a(n-1)*p^(n-1)+...+a0*q^n-1) c'est ˆ di
re :" }}{PARA 0 "" 0 "" {TEXT -1 78 "q divise an*p^n. Or q est premier
 avec p et donc avec p^n et donc q divise an." }}{PARA 0 "" 0 "" 
{TEXT -1 20 "De meme p divise a0." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
128 "Application : Faisons une fonction qui renvoie tous les nombres r
Žpondant ˆ la condition nŽcessaire que l'on vient de dŽmontrer." }}
{PARA 0 "" 0 "" {TEXT -1 12 "Un exemple :" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 25 "poly:=27*(x-4/3)*(x-7/9);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%%polyG,$*&,&%\"xG\"\"\"#!\"%\"\"$F)F),&F(F)#!\"(\"\"*
F)F)\"#F" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "expand(poly);" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6#,(*$%\"xG\"\"#\"#FF%!#d\"#G\"\"\"" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "lcoeff(poly,x);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#\"#F" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 15 "tcoeff(poly,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"#G" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "divisors(28);" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#<(\"\"\"\"\"#\"\"%\"\"(\"#9\"#G" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 18 "D'ou la fonction :" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 101 "possibles:= (p,x) -> \{seq(seq(seq(s*i/j,j=di
visors(lcoeff(p,x))),i=divisors(tcoeff(p,x))),s=\{-1,1\})\}:" }}{PARA 
7 "" 1 "" {TEXT -1 42 "Warning, `s` in call to `seq` is not local" }}
{PARA 7 "" 1 "" {TEXT -1 42 "Warning, `i` in call to `seq` is not loca
l" }}{PARA 7 "" 1 "" {TEXT -1 42 "Warning, `j` in call to `seq` is not
 local" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "possibles(poly,x)
;" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#<R#!\"(\"\"*#!\"#\"\"$#F%F)#F%\"#
F#!#9F)#F.F&#F.F,#!#GF)#F2F&#F2F,#!\"%F,#\"\"\"F&#F8F,#\"\"#F)#F;F&#F;
F,#\"\"%F)#F?F&#F?F,#F8F)!\"\"F8F(F;F6F?F%\"\"(F.\"#9F2\"#G#FEF)#FEF&#
FEF,#FFF)#FFF&#FDF)#FDF&#FDF,#F6F)#FCF&#FCF,#F6F&#FFF,#F(F&#F(F,#FCF)
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 49 "Reste ˆ ne selectionner que l
es 'vraies' racines." }}{PARA 0 "" 0 "" {TEXT -1 103 "On crŽe une fonc
tion qui renvoie true (vrai) si 'v' est une racine du polyn™me 'p' d'i
ndŽterminŽe 'x' :" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "estnul
:=(v,p,x) -> subs(x=v,p)=0:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 14 "Et
 on utilise " }{TEXT 261 6 "select" }{TEXT -1 67 " pour ne prendre dan
s les possibles (poss) que les vraies racines :" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 47 "vraies:= (p,x,poss) -> select(estnul,poss,p,x)
:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 20 "On combine le tout :" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "toutesracines := (p,x) -> vr
aies(p,x,possibles(p,x)):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 10 "RŽsu
ltat :" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "toutesracines(pol
y,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#<$#\"\"%\"\"$#\"\"(\"\"*" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 7 "Great !" }}}}}{SECT 0 {PARA 3 "" 0 
"" {TEXT -1 22 "Fractions rationnelles" }}{SECT 0 {PARA 4 "" 0 "" 
{TEXT -1 33 "DŽcomposition en ŽlŽments simples" }}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 42 "q:=(x^3+x^2-x+1)/(-3*x+7*x^2-3*x^3+7*x^4);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"qG*&,**$%\"xG\"\"$\"\"\"*$F(\"\"#F
*F(!\"\"F*F*F*,*F(!\"$F+\"\"(F'F/*$F(\"\"%F0F-" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 21 "convert(q,parfrac,x);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#,(*$%\"xG!\"\"#F&\"\"$*$,&F%\"\"(!\"$\"\"\"F&#\"$V\"\"#
()*&,&F(F-F%F+F-,&F-F-*$F%\"\"#F-F&#F-\"#H" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 84 "Avouons le : la dŽcomposition en ŽlŽments simple n'est pa
s un sujet tr�s passionant." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
25 "convert(q,fullparfrac,x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*$%
\"xG!\"\"#F&\"\"$-%$SumG6$*&,(#\"$&f\"%BD\"\"\"%'_alphaG#!\"$\"#e*$F2
\"\"##\"$\"e\"%Y]F1,&F%F1F2F&F&/F2-%'RootOfG6#,*F4F1%#_ZG\"\"(*$FAF7F4
*$FAF(FBF1" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "convert(\",ra
dical);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,**$%\"xG!\"\"#F&\"\"$*$,&F
%\"\"\"#!\"$\"\"(F+F&#\"$V\"\"$4'*&,&#F.\"#eF+%\"IG#F-F5F+,&F%F+F6F&F&
F+*&,&F4F+F6#F(F5F+,&F%F+F6F+F&F+" }}}}}{SECT 0 {PARA 3 "" 0 "" {TEXT 
-1 15 "Suites de Sturm" }}{PARA 0 "" 0 "" {TEXT -1 68 "Se rŽfŽrer ˆ l'
annexe une description plus mathŽmatique du probl�me." }}{EXCHG {PARA 
0 "" 0 "" {TEXT 256 12 "Question 1 :" }{TEXT -1 44 " Il suffit de divi
ser par le PGCD de P et P'" }}{PARA 0 "" 0 "" {TEXT -1 20 "Prenons un \+
exemple :" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "poly:=(x-1)^3*
(x-2)^4*(x-3)^5;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%polyG*(,&%\"xG
\"\"\"!\"\"F(\"\"$,&F'F(!\"#F(\"\"%,&F'F(!\"$F(\"\"&" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 12 "La fonction " }{TEXT 262 7 "simples" }{TEXT -1 
44 " divise un polynome P par le PGCD de P et P'" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 45 "simples:=(p,x) -> expand(p/gcd(p,diff(p,x))):
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "factor(simples(poly,x))
;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*(,&%\"xG\"\"\"!\"\"F&F&,&F%F&!\"
#F&F&,&F%F&!\"$F&F&" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 257 12 "Question \+
2 :" }{TEXT -1 48 " Il s'agit de l'agorithme d'Euclide (300 av JC)." }
}}{EXCHG {PARA 0 "" 0 "" {TEXT 258 12 "Question 3 :" }{TEXT -1 72 " Ex
primer la relation de rŽcurence sans modulo et remplacer Pk(x) par 0.
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 259 12 "Question 4 :" }{TEXT -1 77 "
 Sturm(P,x) ne change de valeur que pour un x pour lequel un des Pk s'
annule." }}{PARA 0 "" 0 "" {TEXT -1 219 "Mais comme quand un Pk s'annu
le, pour k>0, Pk-1 et Pk+1 ont un signe different, peu importe comment
 Pk se comporte autour de x, Sturm(P,x) ne change pas de valeur. Ainsi
 le seul changement possible est pour P0(x)=P(x)=0" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT 260 12 "Question 5 :" }{TEXT -1 126 " Quand x cro”t, au \+
moment o� P(x) s'annule, Sturm(P,x) decroit de 1 (quelque soit le sign
e de P'). D'o� la propriŽtŽ cherchŽe." }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 12 "Appliquons :" }}{PARA 0 "" 0 "" {TEXT -1 87 "D'abord, une
 fonction recusive renvoyant la suite de sturm a partir d'un certain r
ang :" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 100 "SuiteDeSturmAPCR:
=proc(p0,p1,x);if p1=0 then p0 else p0,SuiteDeSturmAPCR(p1,-rem(p0,p1,
x),x) fi;end:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 25 "Puis son initial
isation :" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 79 "SuiteDeSturm:=
(p,x) -> [SuiteDeSturmAPCR(simples(p,x),diff(simples(p,x),x),x)]:" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 12 "Un exemple :" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 21 "SuiteDeSturm(poly,x);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#7&,**$%\"xG\"\"$\"\"\"*$F&\"\"#!\"'F&\"#6F+F(,(F)F'F&!#
7F,F(,&#!\"%F'F(F&#F*F'F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 69 "Une \+
version lŽg�rement modifiŽe pour avoir les changements de signe :" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 122 "SignesSturmAPCR:=proc(p0,p1
,x); if p1=0 then 0 else abs(signum(p0)-signum(p1))+SignesSturmAPCR(p1
,-rem(p0,p1,x),x) fi;end:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
77 "SignesSturm:=(p,x) -> SignesSturmAPCR(simples(p,x),diff(simples(p,
x),x),x)/2:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 9 "Exemple :" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "SignesSturm(poly,x);" }}
{PARA 12 "" 1 "" {XPPMATH 20 "6#,(-%$absG6#,&-%'signumG6#,**$%\"xG\"\"
$\"\"\"*$F-\"\"#!\"'F-\"#6F2F/F/-F)6#,(F0F.F-!#7F3F/!\"\"#F/F1-F%6#,&F
4F/-F)6#,&F-F/!\"#F/F8F9-F%6#,&F=F/F8F/F9" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 20 "spoly:=unapply(\",x):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 11 "spoly(-10);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"$
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "spoly(3/2);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#\"\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 11 "spoly(5/2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"\"" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "spoly(10);" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#\"\"!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 40 "Calcul
ons le nombre de racines rŽelles :" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 55 "limit(spoly(x),x=-infinity)-limit(spoly(x),x=infinity
);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"$" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 8 "The End." }}}}}{MARK "3 29 0 0" 8 }{VIEWOPTS 1 1 0 1 1 
1803 }