Bases d'algèbre et d'analyse 2007 Tronc Commun Université de Technologie de Belfort Montbéliard

Description

Examen du Supérieur Université de Technologie de Belfort Montbéliard. Sujet de Bases d'algèbre et d'analyse 2007. Retrouvez le corrigé Bases d'algèbre et d'analyse 2007 sur Bankexam.fr.

Sujets

• 2007

Informations

• Publié par: bankexam
• Publié le: 27 janvier 2008
• Nombre de lectures: 26
• Langue: Français

Extrait

16 2007 2 4 (un
) +∞ (un
) +∞ (un
) +∞ C C z
2
+(3+i)z+
7
4
+i=0.
H ={(x,y)∈R
2
,x
2
+y
2
=1} R
2
(R
2
,+) x y∈Q xy∈Q x∈R−Q y∈Q xy∈R−Q xy∈Q x∈R−Q y∈R−Q xy∈R−Q 7 ab=ba (a+b)
n
=
P
n
k=0
C
k
n
a
k
b
n−k
C
k
n
=
n!
k!(n−k)!
A =
0
@
1 1 1
0 1 1
0 0 1
1
A
n∈N A
n
A=
0
@
1 0 0
0 1 0
0 0 1
1
A
+
0
@
0 1 1
0 0 1
0 0 0
1
A
B =
0
@
2 0 1
0 3 0
0 0 2
1
A
B
n
n∈N 7 F F
11 2007 16 2007 2 C C z
1 =a
1 +ib
1
z
2 =a
2 +ib
2
z
1 „z
2
a
1 <a
2
(a
1 =a
2
b
16b
2
).
2+5i„ 3+i 2 < 3 2+i„ 2+5i 2 = 2 36 5 3+7i6„2+8i „ C „ z
1
,z
2
,z
3 ∈C z
1 „z
2 ⇒z
1 +z
3 „z
2 +z
3
.
0„i „ (C,„) „ 7 11 2007 16 2007 2 (un
) un = −
1
n
n ∈ N
∗
limn→+∞un =0 (un
) u
2k = 2k u
2k+1 = 2k−1 limn→+∞un =+∞ +∞ (un
) M > 0 N ∈N uN >M n>N un>uN >M M >0 N ∈N n>N ⇒un >M limn→∞un =+∞ z
1 =
−3+
q
√
5+1
2
+i(−1+
q
√
5−1
2
)
2
z
1 =
−3−
q
√
5+1
2
+i(−1−
q
√
5−1
2
)
2
H R
2
(0,0) R
2
H x,y ∈ Q p
1
,p
2 ∈ Z q
1
,q
2 ∈ N
∗
x =
p1
q1
y =
p2
q2
xy =
p1p2
q1q2
∈Q 1 y 6= 0 x ∈ R−Q y ∈ Q
∗
xy ∈ R−Q y∈Q
∗
xy∈Q p∈Z
∗
r∈Z q,s∈N y =
p
q
xy =
r
s
p6=0 x=
q
p
×
r
s
∈Q 2 y =0 y =0 xy =0∈Q x=
√
2∈R−Q y =
√
2∈R−Q xy =2∈Q A = C +D C =
0
@
1 0 0
0 1 0
0 0 1
1
A
D =
0
@
0 1 1
0 0 1
0 0 0
1
A
CD = DC
C =I
3
A
n
=(C+D)
n
=C
n
+nC
n−1
D+
n(n−1)
2
C
n−2
D
2
+
n(n−1)(n−2)
6
C
n−3
D
3
+···+D
n
.
D
2
=
0
@
0 0 1
0 0 0
0 0 0
1
A
D
n
=(0) n>3 C
k
=C A
n
=
0
@
1 0 0
0 1 0
0 0 1
1
A
+n
0
@
0 1 1
0 0 1
0 0 0
1
A
+
n(n−1)
2
0
@
0 0 1
0 0 0
0 0 0
1
A
=
0
@
1 n
n(n+1)
2
0 1 n
0 0 1
1
A
11 2007 16 2007 2 B =
0
@
2 0 1
0 3 0
0 0 2
1
A
B =E+F E =
0
@
2 0 0
0 3 0
0 0 2
1
A
F =
0
@
0 0 1
0 0 0
0 0 0
1
A
EF =FE E
n
=
0
@
2
n
0 0
0 3
n
0
0 0 2
n
1
A
n=0 E
0
= I
3
n E
n
=
0
@
2
n
0 0
0 3
n
0
0 0 2
n
1
A
E
n+1
= E
n
×E =
0
@
2
n
0 0
0 3
n
0
0 0 2
n
1
A
×
0
@
2 0 0
0 3 0
0 0 2
1
A
=
0
@
2
n
×2 0 0
0 3
n
×3 0
0 0 2
n
×2
1
A
=
0
@
2
n+1
0 0
0 3
n+1
0
0 0 2
n+1
1
A
n+1 n∈N F
n
=(0) n>2 B
n
=E
n
+nE
n−1
F
=
0
@
2
n
0 0
0 3
n
0
0 0 2
n
1
A
+n
0
@
2
n−1
0 0
0 3
n−1
0
0 0 2
n−1
1
A
0
@
0 0 1
0 0 0
0 0 0
1
A
=
0
@
2
n
0 n2
n−1
0 3
n
0
0 0 2
n
1
A
„ z =a+ib∈C a=a b6b z„z z
1 = a
1 +ib
1
z
2 = a
2 +ib
2
z
1 „ z
2
z
2 „ z
1
a
1 < a
2
a
1 = a
2
b
1 6 b
2
a
1 < a
2
z
2 „ z
1
a
1 = a
2
‰
z
1 „z
2 ⇒ b
16b
2
z
2 „z
1 ⇒ b
26b
1
b
1 =b
2
z
1 =z
2
z
1 = a
1 +ib
1
z
2 = a
2 +ib
2
z
3 = a
3 +ib
3
z
1 „ z
2 ⇔ a
1 < a
2
(a
1 = a
2
b
1 6 b
2
) z
2 „ z
3 ⇔ a
2 < a
3
(a
2 = a
3
b
2 6 b
3
) a
1 <a
2
a
1 <a
3
z
2 „z
3 ⇒a
26a
3
z
1 „z
3
a
1 = a
2
b
1 6 b
2
a
2 < a
3 ⇒ a
1 < a
3 ⇒ z
1 „ z
3
a
2 = a
3
b
26b
3 ⇒a
1 =a
3
b
16b
3 ⇒z
1 „z
3
z
1 =a
1 +ib
1
z
2 =a
2 +ib
2
a
1 <a
2
z
1 „z
2
a
1 =a
2
b
16b
2 ⇒z
1 „z
2
b
1 >b
2 ⇒z
2 „z
1
a
1 >a
2
z
2 „z
1
z
1
z
2
z
1 = a
1 +ib
1
z
2 = a
2 +ib
2
z
3 = a
3 +ib
3
z
1 „ z
2
a
1 < a
2
(a
1 = a
2
b
1 6 b
2
) R a
1 < a
2
(a
1 = a
2
b
1 6 b
2
) ⇔
a
1 +a
3 <a
2 +a
3
(a
1 +a
3 =a
2 +a
3
b
1 +b
36b
2 +b
3
)⇔z
1 +z
3 „z
2 +z
3
0 = 0+i0 i = 0+i 0 „ i 06 1 0×i = 0 i×i = i
2
= −1
i
2
„ 0 11 2007