Solution du TP Maple n°3

O O O

O

O

O O O

O O O

Solution du TP Maple n°3

MG2 - automne 2010

Exercice 1

restart:

A:=X^n-n*X^7+1; B:=X^5+2*X^4-6*X^3-4*X^2+13*X-6;

A:=XⁿKnX⁷C1

B:=X⁵C2 X⁴K6 X³K4 X²C13 XK6 eq:=A=B*Q(X)+R:

R:=add(a[i]*X^i,i=0..4);

R:=a₀Ca₁XCa₂X²Ca₃X³Ca₄X⁴ racines:=solve(B);

racines:=K3,K2, 1, 1, 1

eqs:=subs(X=-3,eq),subs(X=-2,eq),subs(X=1,eq),subs(X=1,diff (eq,X)),subs(X=1,diff(eq,X$2));

eqs:= K3ⁿC2187 nC1 =a₀K3 a₁C9 a₂K27 a₃C81 a₄, K2 ⁿC128 nC1 =a₀K2 a₁ C4 a₂K8 a₃C16 a₄, 2Kn=a₀Ca₁Ca₂Ca₃Ca₄,K6 n=a₁C2 a₂C3 a₃C4 a₄,n² K43 n= 2 a₂C6 a₃C12 a₄

vars:=coeffs(R,X);

vars:=a₀,a₁,a₄,a₃,a₂ sols:=solve({eqs},{vars});

sols:= a₀=1

9 K1 ⁿ 2ⁿK 1465 24 nC 553

288 C 1

32 K1^1Cn 3ⁿC1

4 n²,a₁= 20551 144 n K 7

24 n²C 377 1728C 5

64 K1 ⁿ 3ⁿC 8

27 K1 ^1Cn 2ⁿ,a₂=K1

8 n²K 3427 48 nK 101

576 C 1

64 K1 ^1Cn 3^1CnC 1

9 K1 ⁿ 2^1Cn,a₃= 1 64 K627

16 nC 1

64 K1 ^1Cn 3ⁿ C 1

8 n²,a₄= 1

64 K1 ⁿ 3ⁿC4019 144 nC 37

1728 C 1 24 n²C 1

27 K1 ^1Cn 2ⁿ Rsol:=subs(sols,R);

Rsol:= 1

9 K1 ⁿ 2ⁿK1465 24 nC553

288 C 1

32 K1 ^1Cn 3ⁿC 1

4 n²C 20551 144 nK 7

24 n² C 377

1728C 5

64 K1 ⁿ 3ⁿC 8

27 K1 ^1Cn 2ⁿ XC K1

8 n²K3427 48 nK101

576 C 1

64 K1 ^1Cn 3^1CnC 1

9 K1 ⁿ 2^1Cn X²C 1 64K 627

16 nC 1

64 K1 ^1Cn 3ⁿ C 1

8 n² X³C 1

64 K1 ⁿ 3ⁿC 4019 144 nC 37

1728C 1 24 n²C 1

27 K1^1Cn 2ⁿ X⁴

# vérification pour n=100 :

R1:=rem(subs(n=100,A),B,X); R2:=subs(n=100,Rsol);

O O

O O

O O

R1:=K16105547522875353954039343613150175716751618623 C8052773761437677000494682921912669367574018225 X⁴ K8052773761437677047444705152587832385970432700 X³ K24158321284313030860633982073712519047532808400 X² C40263868807188384861623347917537857782680841400 X R2:=K16105547522875353954039343613150175716751618623

C8052773761437677000494682921912669367574018225 X⁴ K8052773761437677047444705152587832385970432700 X³ K24158321284313030860633982073712519047532808400 X² C40263868807188384861623347917537857782680841400 X R1-R2;

0

Exercice 2

restart:

product(X-r,r=RootOf(X^7-2,X));

X⁷K2

H:=product(product(X-r-s,r=RootOf(X^7-2,X)),s=RootOf(X^17 -3)):

H:=sort(H);

H:=X¹¹⁹K34 X¹¹²C544 X¹⁰⁵K21 X¹⁰²K5440 X⁹⁸K10295880 X⁹⁵C38080 X⁹¹ K11865431928 X⁸⁸C189 X⁸⁵K198016 X⁸⁴K1753404217656 X⁸¹K958322232 X⁷⁸ C792064 X⁷⁷K73058971757184 X⁷⁴C10251916815168 X⁷¹K2489344 X⁷⁰ K945 X⁶⁸K1171409550359328 X⁶⁷K5650506012497976 X⁶⁴C6223360 X⁶³ K10593209592 X⁶¹K8407568031198336 X⁶⁰C441668319396677856 X⁵⁷ K12446720 X⁵⁶K176998340024352 X⁵⁴K29134921153548672 X⁵³C2835 X⁵¹ K7338824689392054432 X⁵⁰C19914752 X⁴⁹K87673540330647936 X⁴⁷ K50194522347443712 X⁴⁶K22874426568 X⁴⁴C31059905300189006976 X⁴³ K25346048 X⁴²K3346477961073716448 X⁴⁰K42660269259761664 X³⁹

C156757268001600 X³⁷K35439911729774568576 X³⁶C25346048 X³⁵K5103 X³⁴ K14193305989102745568 X³³K17061635654784000 X³²K15692214152659968 X³⁰ C10451514240106845696 X²⁹K19496960 X²⁸K8687592144 X²⁷

K7277809809405502080 X²⁶K2897877384880128 X²⁵C49170413590869312 X²³ K679128586744172544 X²²C11141120 X²¹K4271574360816 X²⁰

K388388753217360000 X¹⁹K170567675228160 X¹⁸C5103 X¹⁷ K5655170032317792 X¹⁶C6775977758957568 X¹⁵K4456448 X¹⁴ K5054275195200 X¹³K1308242376709632 X¹²K2228765736960 X¹¹

O

O O O O O

O O O

O

O

O

K198486288 X¹⁰C13829136531840 X⁹K4436523012096 X⁸C1114112 X⁷ K25744920768 X⁶K56418840576 X⁵K1666990080 X⁴C13880160 X³ K30844800 X²C4386816 XK133259

# vérification : irreduc(H);

true simplify(subs(X=2^(1/7)+3^(1/17),H));

0

# autre méthode :

e:=eliminate({X^7-2,(Y-X)^17-3},{X}):

Hbis:=sort(op(e[2]));

Hbis:=Y¹¹⁹K34 Y¹¹²C544 Y¹⁰⁵K21 Y¹⁰²K5440 Y⁹⁸K10295880 Y⁹⁵C38080 Y⁹¹ K11865431928 Y⁸⁸C189 Y⁸⁵K198016 Y⁸⁴K1753404217656 Y⁸¹K958322232 Y⁷⁸ C792064 Y⁷⁷K73058971757184 Y⁷⁴C10251916815168 Y⁷¹K2489344 Y⁷⁰ K945 Y⁶⁸K1171409550359328 Y⁶⁷K5650506012497976 Y⁶⁴C6223360 Y⁶³ K10593209592 Y⁶¹K8407568031198336 Y⁶⁰C441668319396677856 Y⁵⁷ K12446720 Y⁵⁶K176998340024352 Y⁵⁴K29134921153548672 Y⁵³C2835 Y⁵¹ K7338824689392054432 Y⁵⁰C19914752 Y⁴⁹K87673540330647936 Y⁴⁷ K50194522347443712 Y⁴⁶K22874426568 Y⁴⁴C31059905300189006976 Y⁴³ K25346048 Y⁴²K3346477961073716448 Y⁴⁰K42660269259761664 Y³⁹

C156757268001600 Y³⁷K35439911729774568576 Y³⁶C25346048 Y³⁵K5103 Y³⁴ K14193305989102745568 Y³³K17061635654784000 Y³²K15692214152659968 Y³⁰ C10451514240106845696 Y²⁹K19496960 Y²⁸K8687592144 Y²⁷

K7277809809405502080 Y²⁶K2897877384880128 Y²⁵C49170413590869312 Y²³ K679128586744172544 Y²²C11141120 Y²¹K4271574360816 Y²⁰

K388388753217360000 Y¹⁹K170567675228160 Y¹⁸C5103 Y¹⁷ K5655170032317792 Y¹⁶C6775977758957568 Y¹⁵K4456448 Y¹⁴ K5054275195200 Y¹³K1308242376709632 Y¹²K2228765736960 Y¹¹ K198486288 Y¹⁰C13829136531840 Y⁹K4436523012096 Y⁸C1114112 Y⁷ K25744920768 Y⁶K56418840576 Y⁵K1666990080 Y⁴C13880160 Y³ K30844800 Y²C4386816 YK133259

# vérification : Hbis-subs(X=Y,H);

0

Exercice 3

restart:

S:=series(1/((1-X)*(1-X^2)*(1-X^4)),X=0,401):

coeff(S,X,400);

O

O O

O O

O O O

10201 L:=[seq(coeff(S,X,i),i=0..40)];

L:= 1, 1, 2, 2, 4, 4, 6, 6, 9, 9, 12, 12, 16, 16, 20, 20, 25, 25, 30, 30, 36, 36, 42, 42, 49, 49, 56, 56, 64, 64, 72, 72, 81, 81, 90, 90, 100, 100, 110, 110, 121

with(genfunc):

eq:=rgf_findrecur(20,L,u,n);

eq:=u n =u nK1 Cu nK2 Ku nK3 Cu nK4 Ku nK5 Ku nK6 Cu n K7

y:=rsolve({eq,seq(u(i)=L[i+1],i=0..6)},u);

y:= 5

32 K1 ⁿC 1 16 nC 1

16 K1ⁿC 1 16 K 1

16 I IⁿC 1 16 C 1

16 I KIⁿ C1

8 nC1 1

2 nC1 C 1 4 nC 17

32 assume(n,integer): z:=evalc(Re(y));

z:= 5

32 K1 ^n~C 1 16 n~C 1

16 K1 ^n~C 1

16 n~²C 7

16 n~C 21 32 C1

8 cos 1 2 n~ ! C1

8 sin 1 2 n~ !

# vérification pour n=400 : simplify(subs(n=400,z));

10201