DM de Mathématiques, Suites récurrentes linéaires

u0= 1
u1= 1
un+2 = 2un+1 −2un
(un)n∈
Hn:un.
H0H1u0= 1 u1= 1
HnHn+1 un+2 = 2un+1 −2un
unun+1
(,+,∗)un+2
un+2 −2un+1 + 2un= 0 P(t) = t2−2t+ 2
(un)n∈(x1=2+i√4
2= 1 + i
x2=2−i√4
2= 1 −i
x1=x2
(un)n∈(xn
1)n∈(xn 2)n∈
λ µ n ∈
un=λ(1 + i)n+µ(1 −i)n.
λ µ
n= 0 ⇒u0= 1 = λ+µ
n= 1 ⇒u1= 1 = λ(1 + i) + µ(1 −i)
⇔1 = λ+µ
1 = λ+µ+i(λ−µ)
1 = λ+µ
0 = λ−µ
λ=µ=1
2(un)n∈
un=1
2(1 + i)n+1 2(1 −i)n.
(un)n∈

♠
•(1 + i)n(1 −i)n
•(1 + i) (1 −i)
(a+b)n=
n
X
k=0 n
iakbn−k,n k=n! k!(n−k)! ∈.
un
un=1
2 n
X
k=0 n
kik1n−k!+1 2 n
X
k=0 n
k(−i)k1n−k!
un=1
2
n
X
k=0 n
kik(1 + (−1)k)
k(1 + (−1)k)=0
k
un=1
2
bn
2c
X
k=0 n
2k2i2k
i2 = −1
un=bn
2c
X
k=0 n
k(−1)k.
un
n
kun∈
(1 + i) (1 −i) 1 + i=ρeiθ 1 + i=ρ0eiθ0
√2
(1 + i=√2( 1
√2+i √2)
1−i=√2( 1
√2−i √2)
√2
(1 + i=√2(√2
2+i√2 2)
1−i=√2(√2
2−i√2 2)
cos(π
4) = sin(π 4) = √2 2cos(−π 4) = √2 2sin(−π 4) = −√2 2
1 + i=√2cos(π
4) + isin(π 4)
1−i=√2cos(−π
4)−isin(−π 4)
1 + i=√2eiπ
4
1−i=√2e−iπ
4
un

un=1
2(√2eiπ
4)n+1
2(√2e−iπ
4)n,
2eiθ +e−iθ θ=nπ
4
un=1
2(√2)neinπ
4+e−inπ 4,
eiθ +e−iθ = 2 cos θ
un= (√2)ncos(nπ
4).
♠un
n∈
(√2)ncos(nπ
4)∈.
(√2)ncos(nπ
4)
•n= 2knπ
40,π 2, π 3π 22πcos(nπ 4)∈ {0,−1,1}
un=∈ {0,−2n,2n}un
•n= 2k+ 1 nπ
4
π
4,3π 4,−π 4−3π 42πcos(nπ 4)∈
{√2
2,−√2 2}n= 2k+ 1 (√2)n= 2k√2un∈ {2k√2√2 2,−2k√2√2 2}
un∈ {2k,−2k}un
n un
D1=α=α
D1=
α β
β α =α2−β2
Dn
Dn=
αβ
βαβ
βα
;
;
;
;
;
;
;
;β
;
;
β
;
;
β
βα
|{z }
n
=α×
αβ
βα
;
;
;
;
;
;
;
;β
;
;
β
;
;
β
βα
| {z }
n−1
−β×
β0
βα
;
;
;
;
;
;
;
;β
;
;
β
;
;
β
βα
| {z }
n−1
.
Dn=α×
αβ
βα
;
;
;
;
;
;
;
;β
;
;
β
;
;
β
βα
| {z }
n−1
−β2×
α
;
;
;
;
;
;
;
;β
;
;
β
;
;
β
βα
| {z }
n−2
.

∀n >= 3 Dn=α×Dn−1−β2×Dn−2
Dn=1
√6.
 4 + √6
2!n+1
− 4−√6
2!n+1
.
♠Dn
D1=α,
D2=α2−β2,
Dn=α.Dn−1−β2.Dn−2.
D0
n= 2 D2=α.D1−β2.D0D0= 1
Dn
D0= 1,
D1=α,
Dn=α.Dn−1−β2.Dn−2.
P(t) = t2−αt+β2∆ = α2−4β4
∆6= 0 P r±Dn
(rn
+)n∈(rn −)n∈Dn=θ.rn
++γ.rn −.
θ γ n = 0 n= 1
D0= 1 = θ+γ
D1=a=r++r−=θ.r++γ.r−
θ=r+
r+−r−
γ=r−
r−−r+
Dn=1
r+−r−
(rn+1
+−rn+1
−)r±=α±c(∆)√∆
2c(∆) = 1 ∆ ≥0
i∆<0.
∆=0 P r Dn
(rn)n∈(n×rn)n∈Dn=θ.rn+γ.n.rn.
θ γ n = 0 n= 1
D0= 1 = θ
D1=a= 2.r =θ.r +γ.r
θ= 1 γ= 1
Dn= (1 + n).rnr=α
2.

S(t) = Pk≥0DktkDn=αDn−1−β2Dn−2
1×S(t) = 1 + αt +D2.t +. . .
−α×t.S(t) = 1.t +D1.t2+. . .
+β2×t2.S(t) = D0.t2+. . .
s(t)−α.t.S(t) + β2.t2.S(t) = 1 + 0 + 0 + . . .
.
S(t) = 1
1−αt +β2t2.
β= 0 S(t) = PkαktkDn=αn
β6= 0
Q(t)=1−αt +β2t2Q∆ = α2−4β4
∆6= 0 Q s+s−S(t)
S(t) = 1
β2(t−s+)(t−s−)
(t−s+)(t−s−) = t2−(s++s−)t+s+s−=t2−α
β2.t +1 β2
s+.s−=1
β2.
S(t) = s+s−
(t−s+)(t−s−).
S(t)
S(t) = s+.s−(A
t−s+
+B
t−s−
)A= lim
t→s+
1
t−s−
B= lim
t→s−
1
t−s+
.
S(t) =
s+.s−
s+−s−
t−s+
+
s+.s−
s−−s+
t−s−
=s+.s−
s+−s−
.1
t−s+
+−1
t−s−.
1
t−a=−1 aX
k≥0
(t
a)k.
S(t) = −s+.s−
s+−s−X
k
(( t
s+
)k+1 −(t
s−
)k+1).
Dn=−s+.s−
s+−s− 1
sn+1
+−1
sn+1
−!.