CCP 2004 - COMPOSITION de MATHÉMATIQUES II

k MX
n
P
j=1
akj xjk∈[[1, n]],
|
n
P
j=1
akj xj|6
n
P
j=1 |akj | kXk∞6kMkkXk∞,kM X k∞6kMkkXk∞
M=
d
P
k=1
xkBkM0=
d
P
k=1
x0
kBkMλ
N(λM) = max
16k6d|λxk|=|λ|max 16k6d|xk|=|λ|N(M)
∀k∈[[1, d]],|xk+x0
k|6|xk|+|x0 k|6N(M)+N(M0)N(M+M0)6N(M)+N(M0)
N(M) = 0 ∀k∈[[1, d]], xk= 0 M= 0 βM
N M
a b
¡kMpk¢p∈N0¡N(Mp)¢p∈N0¡xp(k)¢p∈N
0
f(x) =
`−1
P
k=0
f(k)(λ)
k!(x−λ)k+Zx
λ
(x−u)`−1
(`−1)! f(`)(u)u
` f λ
u=x+t(λ−x),
f(x) = −(x−λ)`Z1
0
t`−1
(`−1)!f(`)(x+t(x−λ)) t.
h h(x) = −Z1
0
t`−1
(`−1)!f(`)(x+t(x−λ)) tZ1
0
g(x, t)t, g
C1I×[0,1] R
h
C1I∀x∈I, h0(x) = Z1
0
∂g
∂x(x, t)t
gC∞I×[0,1]
hC∞I
j∈[[1, r]] k∈[[0, mj−1]],
f(k)(λj) = g(k)(λj) +
k
X
p=0 µk
p¶h(k−p)(λj)Π(p)
A(λj).
∀p6mj−1,Π(p)
A(λj) = 0 ∀j∈[[1, r]],∀k∈[[0, mj−1]], f(k)(λj) = g(k)(λj)f≡
Ag
j∈[[1, r]], Pj(X)=(X−λ1)m1. . . (X−λj)mj
j
(Hj) : hj∈ C∞
I∀x∈I, f (x)−g(x) = Pj(x)hj(x)

(H1)
(Hj−1) 2 6j6r(Hj)
hj−1= (f−g)×1
Pj−1
(f−g)(k)(λj)=0 k∈[0, mj−1],
∀k∈[0, mj−1], h(k)
j−1(λj) = 0 hj∈ C∞
I
∀x∈I, hj−1(x) = (x−λj)mjhj(x),(Hj)
(2) =⇒(1)
(1) =⇒(2) P≡
AQ j ∈[[1, r]], λj
P−Q mjP−Q(X−λj)mj.
(X−λj)mjλjP−Q
ΠA
ϕ ϕ(P) = ϕ(Q)
P≡
AQ P −QΠAP=Qdeg P−Q6m−1
deg ΠA=m
Rm−1[X]Rmϕ
PfϕRm−1[X]
ϕ(Pf) = ³¡P(k1)(λ1)¢06k16m1−1,...,¡P(kr)(λr)¢06kr6mr−1´
Q R f =QΠA+Rdeg R6m−1
f≡
AR, Pf=R A X,
N
P
k=0
akAk= 0 + Pf(A)f(A) =
N
P
k=0
akAk
A X2−2X+1 = (X−1)2
A(X−1)2,(X−1) A,
ΠA(X) = (X−1)2
f(x) = ax+b, f(A) = aA +bI2
f(x) = sin πx, Pf1Pf(1) = 0 P0
f(1) = −π
Pf(X) = π(1−X)f(A) = π(I2−A)
f(x) (x−1)2g(x), f(A)=0 R1[X]Pf(1) =
P0
f(1) = 0
ϕ, Qj,k Q(k)
j,k (λj) = 1
`∈[[0, mj−1]] `6=k, Q(`)
j,k(λi) = 0.
f∈ C∞
I
r
P
j=1
mj−1
P
k=0
f(k)(λj)Qj,k m f
A, Pf
b
Qj,k f Qj,k,
f=Pf= 1.b Qj,k + 0,
Zj,k αj,k
r
P
j=1
mj−1
P
k=0
αj,kZj,k = 0.
ϕ, f, m,

∀j∈[[1, r]],∀k∈[0, mj−1], f(k)(λj) = αj,k f=Pf=
r
P
j=1
mj−1
P
k=0
αj,kQj,k
f(A) =
r
P
j=1
mj−1
P
k=0
αj,kZj,k = 0,deg f < m f 6= 0
f∈ C∞
If(A) =
r
P
j=1
mj−1
P
k=0
f(k)(λj)Zj,k
f(A) = Pf(A)Zj,k =Qj,k(A)
AΠA(X) = (X−1)2, r = 1, m1= 2 λ1= 1
∀f∈ C∞
I, f(A) = f(1)Z1+f0(1)Z2Z1=Z1,1Z2=Z1,2
f(x) = 1 x∈R∗
+, I2=Z1.
∀x∈R∗
+, f(x) = x, Z1+Z2=A
Z1=I2Z2=A−I2
α > 0, f :x7−→ xαC∞R∗
+
f(A) = I2+α(A−I2)
Aα=αA + (1−α)I2A2004 = 2004A−2003I2√A=1
2(A+I2)
A−X(X2+X) = −X2(X+1) ΠA,
X(X+1)
X2(X+1) A(A+I3)6= 0 ΠA(X) = X2(X+1)
AMn(R)
λ1= 0 λ2=−1Z1,1, Z1,2, Z2,1
∀f∈ C∞
R, f(A) = f(0)Z1,1+f0(0)Z1,2+f(−1)Z2,1.
f(x) = x+1, x x2A+I3=Z1,1, A =Z1,2−Z2,1
A2=Z2,1
Z1,1=A+I3=
2−1 1
2−1 1
2−1 1
, Z1,2=A+A2=
1−1 1
1−1 1
0 0 0
, Z2,1=A2=
0 0 0
−110
−110
.
f7−→ Pf
Pαf =αPfPf+g=Pf+Pg
PfPg, h, k ∈ C∞
If=Pf+hΠAg=Pg+kΠA
fg =PfPg+h1ΠAh1∈ C∞
Ifg ≡
APfPg.
≡
APfPg≡ APfg
H∈R[X]Pfg =PfPg+HΠA
f, g ∈ C∞
Iα∈R, S(αf) = Pαf (A) = αPf(A) = αS(A)S(f+g) = Pf+g(A) =
Pf(A) + Pg(A)
f, g ∈ C∞
IS(fg) = Pfg (A) = Pf(A)Pg(A) +H(A)ΠA(A) = Pf(A)Pg(A) = S(f)S(g)
S(1) = In1 1
S(C∞
I,+,×,·) (Mn(R),+,×,·)
f∈S f(A) = 0 Pf(A) = 0 Pf= 0 f≡
A0
fC∞
I∀j∈[[1, r]],∀k∈[0, mj−1], f(k)(λj) = 0
hΠA, h ∈ C∞
I
(cos A)2+ (sin A)2= (S(cos))2+ (S(sin))2=S(cos2) + S(sin2) =
S(cos2+sin2) = S(1) = In

λjf1C∞
II=R∗
+(√A)2=
(S(f1))2=S(f2
1) = S( ) = A
A×1
A=S( )S(f1) = S(1) = In
1
A
A, 1
A=A−1
SRMA
Mn(R)C∞
IMAf, g ∈ C∞ I,
f(A)g(A) = (fg)(A) = (gf)(A) = g(A)f(A)
f(A) = Pf(A)f∈Rm−1[X],MA={P(A), P ∈Rm−1[X]}=
S(Rm−1[X]),dim MA6m S Rm−1[X]
dim MA6m
A, λjA
−λ1. . . λr,ΠA(A) = 0 α1A+α2A2+···+αmAm=In,
A×(α1In+α2A+··· +αmAm−1) = InA
MA, A−1∈ MA
(1) =⇒(2) f(A)Mn(R),MA
g∈ C∞
If(A)g(A) = In, Pf(A)Pg(A) = In. Pfg(A) = Pf(A)Pg(A) = In,
Pfg (X) = 1
∀j∈[[1, r]], Pf(λj)Pg(λj) = 1 Pf(λj)6= 0
(2) =⇒(1) ∀j∈[[1, r]], Pf(λj)6= 0 PfΠA
U V UPf+VΠA= 1 U(A)Pf(A) = In,
U(A)f(A) = Inf(A)Mn(R)
λ f(A)f(A)−λIn= (f−λ)(A)
j∈[[1, r]] f(λj) = λ
Λf(A)=f(ΛA)
β Zj,k MA
¡fp(A)−f(A)¢p∈N0∀j∈[[1, r]] k∈[[0, mj−1]],
¡f(k)
p(λj)−f(k)(λj)¢p∈Nβ
f:x7−→ etx fp:x7−→
p
P
`=0
(tx)`
`!
f(k)
p=tkfp−kj∈[[1, r]] k∈[[0, mj−1]] k6p¡f(k) p(λj)¢p∈N
¡f(k)(λj)¢p∈N
ft(A) =
+∞
P
`=0
t`
`!A`
X(t) = 
x(t)
y(t)
z(t)
I=R
X(t) = exp(tA)X0, X(t) = ft(A)X0, X0=X(0)
ft(A) = ft(0)Z1,1+f0
t(0)Z1,2+ft(−1)Z2,1=
2+t−1−t1+t
2+t−e−t−1−t+e−t1+t
2−e−t−1+e−t1
X(t)X(0)