Décomposition en éléments simples des fractions rationnelles

K R C
P
QP Q
K[X]Q6= 0 QK(X)
K
RP
QP
Q
F=P
Q(P)−(Q)F(F)
(X2−2X+4)(X−2)
X3(X−2)2
(X2−2X+4)
X3(X−2) 3−5 = −2
−∞
K(X) + ×
P
Q+R S=P S +QR QS
P
Q×R S=P R QS ,
XK
F=P
Qx7→ F(x) = P(x) Q(x)F
KQ F
F=P
Q∈K(X)α∈K
α F α P
α F α Q
R(X)(X2+X+ 1)(X−1)2X
(X−2)(X2+ 1)(X+ 1)4
4
R(X)
F=P
Q∈K(X)E G
F=E+G(G)<0.
E F P
Q

F=P
Q
•(F)<0
•(F)≥0P Q
P=QE +R(R)<(Q)
P
Q=QE +R Q=QE Q+R Q=E
|{z} +R
Q
|{z}
<0
.
F0=X
X2−40
F1=X5+ 1
X(X−1)2X2+ 2X+ 3
F2=1
(X2−1)(X2+ 1)20
F3=4X3
(X2−1)20
R
F=P
Q∈R(X)E Q
R[X]
Q=λ
r
Y
k=1
(X−αk)mk
s
Y
l=1
(X2+βlX+γl)nl.
(Ak,i)1≤k≤r
1≤i≤mk
(Bl,j )1≤l≤s
1≤j≤nl
(Cl,j )1≤l≤s
1≤j≤nl
F=E
|{z} +
r
X
k=1
mk
X
i=1
Ak,i
(X−αk)i
|{z }
αk
+
s
X
l=1
nl
X
j=1
Bl,j X+Cl,j
(X2+βlX+γl)j.
FR
F=P
Q∈R(X)m α
Q= (X−α)mQ1Q1∈R[X]Q1(α)6= 0 F
F=A1
(X−α)+A2 (X−α)2+· · · +Am (X−α)m+F0=
m
X
i=1
Ai
(X−α)i
| {z }α
+F0,
F0α Aii= 1, . . . , m
A, B, C, D...
F0=X
X2−4F0=A (X−2)
| {z } 2
+B
(X+ 2)
| {z } −2
.
F1=X5+ 1
X(X−1)2F1=X2+ 2X+ 3
| {z }+A
X
|{z} 0
+B
X−1+C (X−1)2
| {z }
−1
.

F2=1
(X2−1)(X2+ 1)2F2=A X−1+B X+ 1 +CX +D X2+ 1 +EX +F (X2+ 1)2.
F3=4X3
(X2−1)2F3=A X−1+B (X−1)2+C X+ 1 +D (X+ 1)2
R(X)
α m F
1
(X−α)mF F (X−α)m
X α
F−A
(X−α)mA
α m −1
F0=X
X2−4F0=A
(X−2) +B (X+2)
•A α = 2 m= 1
(X−2)
X
X2−4×(X−2)X=2
=A
(X−2) ×(X−2) + B (X+ 2) ×(X−2)X=2
⇐⇒ X
X+ 2X=2
=1
2=A.
•B(X+ 2) −2B=1
2
F0F0=1
2(X−2) +1 2(X+2)
F1=X5+ 1
X(X−1)2F1=X2+ 2X+3+ A
X+B X−1+C (X−1)2
•A C
X0A= 1
(X−1)21C= 2
F1=X2+ 2X+3+ 1
X+B X−1+2 (X−1)2
•B
F1−2
(X−1)2=X5+ 1 X(X−1)2−2 (X−1)2=X5−2X+ 1 (X−1)2X.
X5−2X+ 1 X−1X5−2X+ 1 = (X−1)(X4+X3+X2+X−1)
F1−2
(X−1)2=X4+X3+X2+X−1 X(X−1) .
1
B= 3 F1F1=X2+ 2X+3+ 1
X+3 X−1+2 (X−1)2
X
F1=X5+ 1
X(X−1)2F1=X2+ 2X+ 3 + 1
X+B X−1+2 (X−1)2
X
F1(−1) = (−1)5+ 1
(−1)(−1−1)2= (−1)2+ 2(−1) + 3 + 1 −1+B −1−1+2 (−1−1)2⇐⇒ 0 = −B 2+3 2,
B= 3

F α m F
−α m F F (X)F(−X) = ±F(X)
F
F2=1
(X2−1)(X2+ 1)2F2(X) = F2(−X)
F2(X) = A
X−1+B X+ 1 +CX +D X2+ 1 +EX +F (X2+ 1)2=−A X+ 1 +−B X−1+−CX +D X2+ 1 +−EX +F (X2+ 1)2=F2(−X).
A=−B C =E= 0
F2(X) = A
X−1−A X+ 1 +D X2+ 1 +F (X2+ 1)2.
•A(X−1) X= 1 A= 1/8
•F(X2+ 1)2X=i F =−1/2
•D0X−1 = −1/8−1/8 + D−1/2D=−1/4
F2(X) = 1
8(X−1) −1 8(X+1) −1 4(X2+1) −1 2(X2+1)2.
F x 7→
xF (x)F
F3=4X3
(X2−1)2F3=A X−1+B (X−1)2+C X+ 1 +D (X+ 1)2
•F3
−F3(X) = −A
X−1+−B (X−1)2+−C X+ 1 +−D (X+ 1)2=−A X+ 1 +B (X+ 1)2+−C X−1+D (X−1)2=F3(−X).
A=C B =−D F3=A
X−1+B (X−1)2+A X+1 +B (X+1)2
•B(X−1)21B= 1
•Alimx→∞ xF3(x) = limx→∞ 4x4
(x2−1)2= 4
lim
x→∞ xF3(x) = lim x→∞
Ax
x−1+x (x−1)2+Ax x+ 1 +x (x+ 1)2= 2A.
2A= 4 A= 2
F3=2
X−1+1 (X−1)2+2 X+ 1 +1 (X+ 1)2.
F=X2+ 3X+ 1
(X−1)2(X−2), G =X5 X4−1H=4 (X2−1)2
F=1
X(X+1)
G=1
X3(X3+1)
(a3+b3) = (a+b)(a2−ab +b2)a, b ∈R.