PC * Les troisième et quatrième parties peuvent

2010 −2011
V
V L(V)f V
fkk
IdVk f k
k
f0=IdV;fk+1 =fk◦f
E n
Enn
E=R[X] ; En=Rn[X].
D E =R[X]Q
Q0Dn
En=Rn[X]Q Q0
λ
λIdE+D g
E λ λIdEn+Dn
g En
V f V.
fk, k = 0,1,2, ...
V
ker f0⊂ker f1⊂ker f2⊂... ⊂ker fk⊂ker fk+1 ⊂...
p
fpfp+1 ker fp= ker fp+1 q
p fqfq+1 ker fq= ker fq+1
k p, ker fk= ker fp.

2010 −2011
V n
fkp
n p ≤nker fnker fn+1
u V
n q 1q≥1
uquq= 0 unun= 0
u
g
g
λ
n n ∈Np
n(0 ≤p≤n)
g En=Rn[X]
g2=λIdEn+Dn
g Dn
g◦Dn=Dn◦g.
Ep=Rp[X] ker (Dn)p+1
Epg Engp
g Ep
(gp)2=λIdEp+Dp.
g E =
R[X]
g2=λIdE+D,
g D
g◦D=D◦g.
n En=
Rn[X]g gn
g En
(gn)2=λIdEn+Dn.

2010 −2011
g E =R[X]
g2=λIdE+D .
F E
n+ 1
D DF
D F
F En=Rn[X]
G E
D
G E
g D
λ < 0
λ
g E0=R0[X]
g2=λIdE0+D0
λ λ < 0
·g E
g2=λIdE+D .
·g En
g2=λIdEn+Dn.
Dn
n1λ
AλAλn+ 1
aij , i = 0,1, ...n, j = 0,1, ....n,
aii =λ, ai i+1 = 1, aij = 0 j6=i j 6=i+ 1.
Aλ=
λ1 0 ··· 0
0λ1··· 0
0 0 λ··· 0
··· ··· ··· ··· ···
000··· λ

2010 −2011
f V n+1
fn+1 fn
fn+1 = 0, fn6= 0 .
y V
B= (fn(y), fn−1(y), ..., y)
f B
BnEn=Rn[X]
Dn
A0λIdEn+Dn
Bn
n2
h E2
D22
D2
h=aIdE2+bD2+c(D2)2.
a, b, c
g E2
g2=λIdE2+D2.
G3
G2=A1.
λ
n1
g g2=Dn
g En=
Rn[X]g2=Dng
g22 dim ker g2≥
2
g
En=Rn[X]g2=Dn

2010 −2011
g
E=R[X]g2=D.
g gk=Dm
m1 (m≥1) k
2 (k≥2) g E =R[X]
gk=Dm.
D g
Eker gp
q k (0 ≤q≤k)
p2k(2 ≤p≤k)
Φ ker gp
Φ : P−→ g(P).
Φ ker gp
ker gp−1.Φ
Φ
ImΦ = ker gp−1
ker gpker gp−1
ker gp
ker g
k m
g E
gk=Dm