énoncé

\
f
f(x) = 1−e−x
xx > 0
f(0) = 1
f(x) 0 f
[0,+∞[.
f0f0(0)
f]0,+∞[ϕ
∀x > 0f0(x) = ϕ(x)
x2
ϕ f
f+∞
(un)n∈N∗
∀n∈N∗un=Zn
0
e−u
n
1 + udu
n
un>1
eln (n+ 1)
(un)n∈N∗
R1
0f(x)dx
n
06Zn
0
1
1 + udu −un6Z1
0
f(x)dx
unn+∞.

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M3(R)I
M3(R),0M3(R)
AM3(R)E1(A)E2(A)
E1(A) = {M∈ M3(R) ; A M =M}
E2(A) = M∈ M3(R) ; A2M=AM
E1(A)M3(R)
E2(A)M3(R)
E1(A)⊂E2(A)
A E1(A) = E2(A)
A−I E1(A) = {0}
B=
−1 1 0
0−1 1
0 0 2
. E1(B)E2(B)
C=
3−2−1
1 0 −1
2−2 0
λ(C−λI)
λKer(C−λI)
D
P
•P
λ D
•P
C=P DP −1
M∈ M3(R)N=P−1M.
M∈E1(C)⇐⇒ N∈E1(D)
N∈E1(D)a, b, c N =
000
a b c
000
E1(C)
E1(C)
E2(C)
E2(C)
E1(C) = E2(C)

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A B
A60% B40%
A0.1
B0.2
A
Y λ = 20.
X
A
Y Y
k n P [X=k/Y =n]
k6n k > n
(Y=i)i∈N, X
Y X
fR
f(t) = 2
(1 + t)3t>0
f(t) = 0 t < 0
f Z
FZZ
+∞
Z
0
2t
(1 + t)3dt
u=t+ 1
Z
Z
A B Z1Z2Z1Z2
Z
C=B
D=B
P(C), P (D), P (D/C).

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T= max(Z1, Z2)GTT
(T6x) (Z16x) (Z26x)
∀x∈R, GT(x)=[FZ(x)]2
T
A
B
A M
B N
[0,1]
M N
v M w N
S
S M N
M N S