//img.uscri.be/pth/7682033c7a10b4cb86e71ae517f3ec9047481a73
YouScribe est heureux de vous offrir cette publication
Lire

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

4 pages
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.
Voir plus Voir moins

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