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{SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT 256 33 "Correction des colles M
aple n\2601&2" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 
"" {TEXT 257 248 "Les solutions propos\351es sont loin d'\352tre les s
eules. Vous avez vu par exemple sept m\351thodes pour calculer 7!, et \+
il en existe d'autres. Et puis, les solutions propos\351es ne sont pas
 forc\351ment les meilleures : n'h\351sitez pas \340 faire part de vos
 id\351es." }}}{EXCHG {PARA 0 "" 0 "" {TEXT 258 10 "Exercice 1" }}
{PARA 0 "" 0 "" {TEXT 259 69 "La partie enti\350re d'un r\351el  x est
 le plus grand entier inf\351rieur \340 x" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 60 "E:=proc(x)\nlocal n;\nn:=0;\nwhile n<x do n:=n+1 od
: n-1;\nend:\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "E(32.32),
E(1.7),E(101.23);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6%\"#K\"\"\"\"$,\"
" }}}{EXCHG {PARA 257 "" 0 "" {TEXT 260 10 "Exercice 2" }}{PARA 0 "" 
0 "" {TEXT 261 63 "La valeur absolue d'un r\351el x vaut x si x est po
sitif, -x sinon" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "ab:=proc
(x)\nif x>=0 then x else -x fi;\nend:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 15 "ab(32),ab(-32);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$\"
#KF#" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 262 10 "Exercice 3" }}{PARA 0 "
" 0 "" {TEXT 263 74 "La moyenne, c'est la somme des op\351randes divis
\351e par le nombre d'op\351randes" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 86 "moy:=proc(l)\nlocal s,k;\ns:=0;\nfor k from 1 to nops
(l) do s:=s+l[k] od: s/nops(l);\nend:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 28 "moy([12,15,14,12,12,18,20]);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6##\"$.\"\"\"(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 264 120 "
Pour rem\351dier \340 ce probl\351me, on peut utiliser evalf qui donne
 par d\351faut un r\351sultat arrondi \340 10 chiffres significatifs" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 92 "moy:=proc(l)\nlocal s,k;
\ns:=0;\nfor k from 1 to nops(l) do s:=s+l[k] od:evalf(s/nops(l));\nen
d:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "moy([12,15,14,12,12,18,20]);
" }}{PARA 11 "" 0 "" {TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$
\"+r&G9Z\"!\")" }}}{EXCHG {PARA 258 "" 0 "" {TEXT 265 10 "Exercice 4" 
}}{PARA 0 "" 0 "" {TEXT -1 2 "1)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 74 "bn:=proc(n)\nlocal a,b,k;\na
:=1+1/n;\nb:=1;\nfor k to n do b:=a**b od:\nb\nend:\n" }}{PARA 0 "> " 
0 "" {MPLTEXT 1 0 6 "bn(5);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#)#\"\"'
\"\"&)F$)F$,$*&F%#\"\"\"F&F&#\"\"%F&#F%\"#D" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 12 "simplify(\");" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*
&)\"\"'*&)F%*&)F%,$*&F%#\"\"\"\"\"&F.#\"\"%F.#F%\"#DF-)F.,$F+#!\"'F2F-
F-)F.,$F(!\"\"F-F-)F.,$F&F9F-" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 266 47 
"Bof...donc, comme d'habitude, on utilise evalf." }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 81 "bn:=proc(n)\nlocal a,b,k;\na:=1+1/n;\nb:=1;\n
for k to n do b:=a**b od:\nevalf(b)\nend:\n" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 3 "2) " }{TEXT 267 73 "\" ...le plus petit entier tel que..
.\" nous fait penser \340 une boucle while" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 40 "m:=1:\nwhile bn(m)>1.001 do m:=m+1 od:\nm;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#\"%-5" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT 268 59 "Votre premier calcul o\371 maple semble ramer autant que
 vous." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 269 
10 "Exercice 5" }}{PARA 259 "" 0 "" {TEXT 270 99 "1) La ruse est d'int
roduire deux variables en plus de l'indice : t pour le terme et s pour
 la somme" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 175 "S:=proc(n)\nl
ocal s,t,k:\ns:=1: t:=1:\nfor k to n do t:=t/k: s:=s+t # il faut bien \+
s\373r d\351finir t avant s #\n                             # s utilis
e donc le nouveau t #\nod:\ns\nend:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 6 "S(32);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6##\"DdgkW<\")
>y$G<j<!4b5'G\"D++S'[?()o17uZtZL__5" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 9 "evalf(\");" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+G=G=
F!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "\"-exp(1);# par cu
riosit\351 #" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&$\"+G=G=F!\"*\"\"\"-
%$expG6#F'!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(\")
;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT 271 64 "fichtre...aurait-on trouv\351 une valeur d\351cimale  ex
acte de e ?!!?" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "evalf(S(3
2)-exp(1),100);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$!jnH*)*Gb]%4C[5P)Q
>*)el#=*e\\/c.#R*4D0]'=\"!#**" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 272 96 
"Ouf....Nous pouvons passer au 2) avec bien s\373r une boucle while co
mme dans l'exercice pr\351c\351dent." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 273 "seuil:=proc(s)\nlocal n:\nn:=0:\nwhile evalf(exp(1)-
S(n),1000)>10^(-s) do n:=n+1 od: #pourquoi n'a-t-on pas#              \+
                                                                      \+
                                               # besoin de valeur abso
lue ? #\nn\nend:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "seuil(3
6);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"#K" }}}{EXCHG {PARA 260 "" 0 
"" {TEXT 273 10 "Exercice 6" }}{PARA 0 "" 0 "" {TEXT 274 167 "Rien de \+
tr\350s difficile math\351matiquement parlant, mais cela nous permet d
e d\351couvrir la fonction elif, contraction de else if qui permet d'a
cc\351der \340 une sous-boucle if." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 312 "sol:=proc(a,b,c)\nlocal d,x1,x2:\nd:=b**2-4*a*c:\nif
 d>0 then  x1:=(-b-sqrt(d))/(2*a): x2:=(-b+sqrt(d))/(2*a):\nelif d=0 t
hen  x1:=-b/(2*a): x2:=x1 :\nelse  x1:=(-b-I*sqrt(-d))/(2*a): x2:=(-b+
I*sqrt(-d))/(2*a) :\nfi;\nif d<>0 then print(`Les solutions sont `.x1.
 `et ` .x2):\nelse print (`La solution est `.x1):\nfi;\nend:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "sol(1,-3,2);" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#%9Les~solutions~sont~1et~2G" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 11 "sol(1,2,1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#%3La~solution~est~-1G" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "s
ol(1,2,3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#(((%4Les~solutions~sont~
G,&!\"\"\"\"\"*&%\"IGF)\"\"##F)F,F(%$et~G,&F(F)F*F)" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT 275 26 "Il y a bien s\373r plus court" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "solv:=proc(a,b,c)\nsolve(a*x**2+b*x
+c=0,x);\nend:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "solv(1,2,
3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$,&!\"\"\"\"\"*&%\"IGF%\"\"##F%F
(F%,&F$F%F&F$" }}}{EXCHG {PARA 261 "" 0 "" {TEXT -1 29 "mais c'est moi
ns constructif." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT 276 10 "Exercice 7" }}{PARA 0 "" 0 "" {TEXT 277 327 "Voici un ex
ercice plus abstrait, mais qui permet de travailler sur les listes et \+
imbrique une boucle if dans une boucle for. Le principe est de passer \+
en revue les couples de termes successifs de la suite. Si au moins l'u
n d'entre eux n'est pas \"dans le bon ordre\", alors la suite des \351
l\351ments de la liste n'est pas croissante." }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 392 "test:=proc(l)\nlocal i,T:\nT:= croissante: #la \+
suite est a priori croissante#\nfor i from 1 to nops(l)-1 do\n        \+
if l[i]>l[i+1] then T:= `pas croissante` fi: #on peut avoir if sans   \+
                                                                      \+
                                                                      \+
                                                     else#\nod:\nT\nen
d:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "test([1,2,3,4,5,6]);
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%+croissanteG" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 16 "test([2,3,4,1]);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#%/pas~croissanteG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 278 
10 "Exercice 8" }}{PARA 0 "" 0 "" {TEXT 279 158 "Pour m\351moire, p pa
rmi n = p parmi n-1 + p-1 parmi n-1. Voici un autre exemple simple de \+
proc\351dure r\351cursive. N'oubliez pas que si p>n, alors p parmi n v
aut 0." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 75 "c:=proc(n,p)\nif \+
p=0 then 1\nelif p>n then 0\nelse c(n-1,p)+c(n-1,p-1)\nfi\nend:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "c(32,3);\nc(3,32);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#\"%g\\" }}{PARA 11 "" 0 "" {TEXT -1 0 "" }}
{PARA 11 "" 0 "" {TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"!
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 511 "branche:=proc(f)\n loc
al a,b:\n a:=limit(f(x)/x,x=infinity):\n if type(a,`..`)=true then lpr
int(`pas d'asymptote`);\n elif type(a,infinity)=true then lprint(`pas \+
d'asymptote`);\n elif a=0 then lprint(`direction asymptotique horizont
ale`);\n else b:=limit(f(x)-a*x,x=infinity):\n if type(b,`..`)=true th
en printf(`direction asymptotique d'equation y=%f x`,a);\n elif type(b
,infinity)=true then printf(`branche parabolique dans la direction y=%
f x`,a);\n else printf(`asymptote d'equation y=%f x+%f`,a,b);\n fi;\n \+
fi;\n end:\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "branche(x->
2*x+cos(x));" }}{PARA 6 "" 1 "" {TEXT -1 39 "direction asymptotique d'
equation y=2 x" }}}}{MARK "47 1 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 }