Exercices - Page Personnelle de Jérôme Von Buhren

(P, Q)∈K[X]2P∧Q= 1
F=P
Q⇔P Q .
F∈C(X)F(X+ 1) = F(X)
F∈C(X)F0=1
X
F∈K(X) deg F0<deg F−1 deg F= 0
(p, q)∈(N∗)2p∧q= 1
Xp−1
Xq−1.
n∈N∗ω= exp(2iπ/n)
F∈C(X)
F(ωX) = F(X)⇔ ∃G∈C(X), F (X) = G(Xn).
F=
n−1
X
k=0
X+ωk
X−ωk
(F, G0, . . . , Gn−1)∈C(X)n+1
Fn+Gn−1Fn−1+. . . +G0= 0.
F Gi
F∈C(X)A F (C\A)
F∈K(X)a∈Kn∈N
Gn∈K(X)a
F(X) =
n
X
k=0
F(k)(a)
k!(X−a)k+ (X−a)n+1Gn(X).
(i)X3
(X−1)(X−2)(X−3),(ii)X3+ 1 (X−1)3,(iii)X4−X2 (X+ 1)2(X2+ 1),
(iv)2X2+ 1
(X−1)2(X+ 1)2,(v)4 (X2+ 1)2,(vi)1 (X+ 1)2(X3−1).
n∈N∗
(i)n!
X(X−1) . . . (X−n),(ii)Xn−1 Xn−1,(ii)1 (X−1)(Xn−1).

n∈N∗p∈J1, n −1Kωk= exp(2ikπ/n)
(i)
n−1
X
k=0
1
X−ωk
,(ii)
n−1
X
k=0
ωp
k
X−ωk
.
P∈C[X]P0/P
P∈C[X]P0|P
P∈C[X]z1, . . . , zn
P00/P
n
X
k=1
P00(zk)
P0(zk).
P(0) 6= 0 1/P
n
X
k=1
1
zkP0(zk).
p∈J0, n −1KXp/P
n
X
k=1
zp
k
P0(zk).
P=anXn+. . . +a0∈R[X]R
P0/P
∀x∈R,(P02−P P 00)(x)>0.
∀k∈J1, n −1K, ak−1ak+1 6a2
k.
(i) lim
n→+∞
n
X
k=1
1
k(k+ 1)(k+ 2),(ii) lim
n→+∞
n
X
k=1
k
k4+k2+ 1.
F= (X2+ 1)−1∈C(X)
Pn∈Rn[X]
F(n)=Pn
(X2+ 1)n+1 .
Pn
n∈N∗n
1
X(X2+ 1).
F∈R(X)
F(X+ 1) −F(X) = X+ 3
X(X−1)(X+ 1).

F=P/Q P ∧Q= 1 Q2=X(P0Q−Q0P)
0Q α P P 0Q−Q0P0
α−1−1 = (α−1) −2α
α= 0
F=P/Q
deg(P0Q−QP 0)<deg(P0Q) deg(P) =
deg(Q) deg(F)=0
F
F= (aXn+b)/(Xn−1)
X= 0 F=n(Xn+ 1)/(Xn−1)
F=P/Q P ∧Q= 1 λ∈C
P(z)−λQ(z) = 0 λ∈CP−λQ = 0
F(C\A) = {λ}λ∈CP−λQ =µ∈C∗
F(C\A) = C\ {λ}P(z)−λQ(z)=0
z∈CP∧Q= 1 z /∈A F (C\A) = C
(i)X3
(X−1)(X−2)(X−3) = 1 + 1 2(X−1) −8 X−2+27 2(X−3),
(ii)X3+ 1
(X−1)3= 1 + 2 (X−1)3+3 (X−1)2+3 X−1,
(iii)X4−X2
(X+ 1)2(X2+ 1) =X2(X−1) (X+ 1)(X2+ 1) = 1 + −1 X+ 1 +−1/2 X−i+−1/2 X+i,
(iv)2X2+ 1
(X−1)2(X+ 1)2=−1/4 X+ 1 +3/4 (X+ 1)2+1/4 X−1+3/4 (X−1)2,
(v)4
(X2+ 1)2=−1 (X−i)2+−i X−i+−1 (X+i)2+i X+i,
(vi)1
(X+ 1)2(X3−1) =−1/2 (X+ 1)2+−3/4 X+ 1 +1/12 (X−1) +1/3 X−j+1/3 X−j2.
ω1, . . . , ωnn
(i)n!
X(X−1) . . . (X−n)=
n
X
k=0
(−1)n−kn
k1 X−k,
(ii)Xn−1
Xn−1=1 n
n−1
X
k=0
1
X−ωk
,
(iii)1
(X−1)(Xn−1) =1−n 2n(X−1) +1 n(X−1)2+
n−1
X
k=1
ωk
ωk−1
1
X−ωk
.
ω1, . . . , ωnn
(i)
n−1
X
k=0
1
X−ωk
=nXn−1
Xn−1,(ii)
n−1
X
k=0
ωp
k
X−ωk
=nXp−1
Xn−1.
0
−1/P (0)
p=n−1 1 0
1/4 1/2
F∈R(X)
F(X+ 1) −F(X) = 2
X−1−3 X+1 X+ 1.
F1
F0 ]0,1[
F
F(X) = 1
X−2 X−1+C C ∈R.