problème - al9ahira

n∈Nk∈[[0, n]] pk,n(X)n
kXk(1 −X)n−k
pk,n = (t) = n
ktk(1 −t)n−k=n! k!(n−k)!tk(1 −t)n−k
φnBnRn[X]
Rn[X]
BnRn[X]
Bn(f)f[0,1] Bn
Bn(f)f[0,1] fC1[0,1]
A0A1A2R2A0= (0,1) A1= (1,1) A2= (1,0)
T T ={(x, y)∈[0,1]2|x+y≥1}
t∈[0,1] p0,1(t)=1−t p1,1(t) = t
A(t) = p0,1(t)A0+p1,1(t)A1B(t) = p0,1(t)A1+p1,1(t)A2C(t) = p0,1(t)A(t) + p1,1(t)B(t)
t∈[0,1]
p0,2(t)p1,2(t)p2,2(t)t
A(t)B(t)C(t) = (2t−t2,1−t2)
C(t) =
2
X
k=0
pk,2(t)Ak
TR2
Cf:t7→ C(t)
[0; 1] →R2
.
C T
t∈[0,1] DtCC(t)
t∈[0,1] [A(t), B(t)] Dt
C T [A(t), B(t)]
t= 0 t= 1/2t= 1

n∈Nn≥2Rn[X]n
P(X)P0(X)
FRn[X] (p0,n(X), p1,n(X), ..., pn,n(X))
P∈Rn[X]φn(P)Bn(P)
φn(P) = nXP (X) + X(1 −X)P0(X)
Bn(P)(X) =
n
X
k=0
Pk
npk,n(X).
φnBnRn[X]
k∈[[0, n]] φn(pk,n)(X) = kpk,n(X)
FRn[X]φn
φnBn
r∈N∗t∈[0,1] (Ω,A,P)Tr
(Ω,A,P)B(r, t)Tr=Tr/r
Y(Ω,A,P)E(Y)Y
V(Y)Y
Y(Ω) ⊂[[0, r]] h[[0, r]] h(Y)
E(h(Y)) =
r
X
k=0
h(k)P(Y=k)
B(r, t)
Tt(Ω) k∈[[0, r]] P(Tr=k) = pk,r(t)
E(Tr), E(Tr), V (Tr), V (Tr), E(T2
r)E((Tr)2);
E((Tr)2) = t
r+t2 r(r−1)
t∈[0,1]
r
X
k=0
pk,r (t)=1,
r
X
k=0
k
rpk,r (t) = 1
r
X
k=0 k
r2
pk,r (t) = 1−1
rt2+1 rt.
t∈R
R2[X]Rn[X]Bn
∼
BnR2[X]BnP∈R2[X]
∼
Bn(P) = Bn(P)An
∼
BnR2[X]
M3(R) 3
I3=
100
010
001
H=
100
011
001
D=
100
010
000
Dn=
1 0 0
0 1 0
001−1
n
An=
1 0 0
0 1 1
n
0 0 1 −1
n
=1−1
nI3+1 nH
H
a b Q =
1 0 a
0 1 b
0 0 1
Q
Q−1a b H =QDQ−1

a b H =QDQ−1
Q=
1 0 a
0 1 b
0 0 1
M3(R)M3(R) (M`)
Mlim
`→+∞
(M`) = Mlim
`→+∞
(M`) = M
(i, j)∈[[1,3]]2lim
`→+∞
(M`)i,j =Mi,j
lim
n→+∞(An) = I3
ΨM3(R) Ψ(M) = QMQ−1
lim
`→+∞
(M`) = Mlim
`→+∞
(QM`Q−1) = QMQ−1
An=QDnQ−1
n≥2 lim
`→+∞
(A`
n)
lim
n→+∞(An
n)
n∈N∗f[0,1] Rx∈R
Bn(f)(x) =
n
X
k=0
fk
npk,n(x).
r=n t ∈[0,1]
f(t)−Bn(f)(t) = E(f(t)−f(Tn)) =
n
X
k=0
(f(t)−f(k/n))pk,n(t).
Y
E(Y)≤pE(Y2)
t∈[0,1] E|t−Tn|≤rt(1 −t)
n.
fC1[0,1]
Mf∀(a, b)∈[0,1]2|f(a)−f(b)| ≤ Mf|a−b|
Mf
∀(a, b)∈[0,1]2,|f(a)−f(b)| ≤ Mf|a−b|;
t∈[0,1] E(|f(t)−f(Tn)|)≤Mfrt(1 −t)
n
tß[0,1] |f(t)−Bn(f)(t)| ≤ Mf
2√n
(Bn(f))n∈N∗f[0,1]
fC1[0,1]
Bn(f)
lim
n→+∞Z1
0
Bn(f)(x)x=Z1
0
f(x)x
Sn(f) = 1
n+ 1
n
X
k=0
fk
n

a∈N∗b∈NZ1
0
xa(1 −x)bx=a
b+ 1 Z1
0
xa−1(1 −x)b+1 x
n∈Nk∈[[0, n]] Z1
0
pk,n(x)x k
Z1
0
pk,n(x)x=1
(n+ 1)
lim
n→+∞Sn(f) = Z1
0
f(x)x
f[0,1]
(a, b, c)∈N3a+b≤c−2
x∈[0,1] Z+∞
0
ua(1 + xu)b
(1 + u)cu
b≥1F:x7→ Z+∞
0
ua(1 + xu)b
(1 + u)cuC1[0,1]
h:t7→ t
1−t
[0,1[ →[0,+∞[C1
u=t
1−tF(0) F(1)
k∈N∗n∈Nt∈[0,1]
fn(t) = pk,n(t)n≥k
0n < k, fn(t) = n
ktk(1 −t)kn≥k
0n < k.
n
k∼nk k!n+∞t∈]0,1[ fn(t)
n+∞
Pfn[0,1]
t∈[0,1] S(t) =
+∞
X
n=0
fn(t)
S(t)t= 0 t= 1
0u7→ 1
1−u
u∈[0,1[
+∞
X
n=k
n(n−1)...(n−k+ 1)un−k=k!
(1 −u)k+1
u∈]0,1] S(t) = 1
t
Pfn[0,1]