Déterminer les polynômes
P
∈
C
[
X
]
{\displaystyle P\in \mathbb {C} [X]}
Soit
P
∈
C
[
X
]
{\displaystyle P\in \mathbb {C} [X]}
P
(
P
(
X
)
)
−
X
{\displaystyle P(P(X))-X}
P
(
X
)
−
X
{\displaystyle P(X)-X}
Pour tout
n
∈
N
∗
{\displaystyle n\in \mathbb {N} ^{*}}
T
n
{\displaystyle T_{n}}
∀
θ
∈
R
cos
(
n
θ
)
=
T
n
(
cos
θ
)
{\displaystyle \forall \theta \in \mathbb {R} \quad \cos(n\theta )=T_{n}(\cos \theta )}
Préciser la parité, le degré et les racines de
T
n
{\displaystyle T_{n}}
Décomposer
1
T
n
{\displaystyle {\frac {1}{T_{n}}}}
Déterminer les coefficients du polynôme
U
n
{\displaystyle U_{n}}
∀
θ
∈
R
sin
(
(
n
+
1
)
θ
)
=
U
n
(
cos
θ
)
sin
θ
{\displaystyle \forall \theta \in \mathbb {R} \quad \sin((n+1)\theta )=U_{n}(\cos \theta )\sin \theta }
U
n
{\displaystyle U_{n}}
Solution
Solution des questions 1, 2 et 4 : voir Sommation/Exercices/Formule du binôme#Exercice 5-11 et Wikipédia : Polynôme de Tchebychev .
Solution de la question 3 : les racines de
T
n
{\displaystyle T_{n}}
x
k
=
cos
(
2
k
−
1
)
π
2
n
{\displaystyle x_{k}=\cos {\frac {(2k-1)\pi }{2n}}}
k
∈
{
1
,
2
,
…
,
n
}
{\displaystyle k\in \{1,2,\dots ,n\}}
1
T
n
=
∑
k
=
1
n
a
k
X
−
x
k
{\displaystyle {\frac {1}{T_{n}}}=\sum _{k=1}^{n}{\frac {a_{k}}{X-x_{k}}}}
a
k
=
1
T
n
′
(
x
k
)
{\displaystyle a_{k}={\frac {1}{T'_{n}(x_{k})}}}
Or
∀
θ
∈
R
−
n
sin
(
n
θ
)
=
−
sin
θ
T
n
′
(
cos
θ
)
{\displaystyle \forall \theta \in \mathbb {R} \quad -n\sin(n\theta )=-\sin \theta \,T'_{n}(\cos \theta )}
a
k
=
sin
(
2
k
−
1
)
π
2
n
n
sin
(
2
k
−
1
)
π
2
=
(
−
1
)
k
+
1
n
sin
(
2
k
−
1
)
π
2
n
{\displaystyle a_{k}={\frac {\sin {\frac {(2k-1)\pi }{2n}}}{n\sin {\frac {(2k-1)\pi }{2}}}}={\frac {(-1)^{k+1}}{n}}\sin {\frac {(2k-1)\pi }{2n}}}
Soit
M
(
X
)
=
X
3
+
2
X
2
+
X
+
1
{\displaystyle M(X)=X^{3}+2X^{2}+X+1}
Soit a une racine complexe de M,
K
a
=
{
P
(
a
)
∣
P
∈
Q
[
X
]
}
{\displaystyle K_{a}=\{P(a)\mid P\in \mathbb {Q} [X]\}}
I
a
=
{
P
∈
Q
[
X
]
∣
P
(
a
)
=
0
}
{\displaystyle I_{a}=\{P\in \mathbb {Q} [X]\mid P(a)=0\}}
I
a
{\displaystyle I_{a}}
Q
[
X
]
{\displaystyle \mathbb {Q} [X]}
K
a
{\displaystyle K_{a}}
Q
{\displaystyle \mathbb {Q} }
K
a
{\displaystyle K_{a}}
Soient
b
∈
K
a
{\displaystyle b\in K_{a}}
u
:
z
↦
b
z
{\displaystyle u:z\mapsto bz}
Q
{\displaystyle \mathbb {Q} }
K
a
{\displaystyle K_{a}}
On note
τ
i
,
j
{\displaystyle \tau _{i,j}}
S
n
{\displaystyle S_{n}}
i
,
j
∈
[
[
1
,
n
]
]
,
i
≠
j
{\displaystyle i,j\in [\![1,n]\!],i\neq j}
∀
σ
∈
S
n
,
σ
∘
τ
i
,
j
∘
σ
−
1
=
τ
σ
(
i
)
,
σ
(
j
)
{\displaystyle \forall \sigma \in S_{n},\sigma \circ \tau _{i,j}\circ \sigma ^{-1}=\tau _{\sigma (i),\sigma (j)}}
En déduire que pour
n
≥
3
{\displaystyle n\geq 3}
C
=
{
u
∈
S
n
∣
∀
v
∈
S
n
u
∘
v
=
v
∘
u
}
{\displaystyle C=\{u\in S_{n}\mid \forall v\in S_{n}\quad u\circ v=v\circ u\}}
S
n
{\displaystyle S_{n}}
i
d
[
[
1
,
n
]
]
{\displaystyle id_{[[1,n]]}}
Soit E un ensemble. On note
Δ
{\displaystyle \Delta }
Mq
(
P
(
E
)
,
Δ
,
∩
)
{\displaystyle ({\mathcal {P}}(E),\Delta ,\cap )}
Solution
(
P
(
E
)
,
Δ
,
∩
)
{\displaystyle ({\mathcal {P}}(E),\Delta ,\cap )}
(
Z
/
2
Z
,
+
,
×
)
{\displaystyle (\mathbb {Z} /2\mathbb {Z} ,+,\times )}
Résoudre
x
3
=
1
{\displaystyle x^{3}=1}
Z
/
19
Z
{\displaystyle \mathbb {Z} /19\mathbb {Z} }
Trouver les entiers
x
{\displaystyle x}
x
≡
5
mod
8
et
x
≡
1
mod
27
{\displaystyle x\equiv 5{\bmod {8}}{\text{ et }}x\equiv 1{\bmod {27}}}
Résoudre (dans
Z
2
{\displaystyle \mathbb {Z} ^{2}}
88
u
+
27
v
=
1
{\displaystyle 88u+27v=1}
Z
{\displaystyle \mathbb {Z} }
x
≡
2
mod
88
et
x
≡
1
mod
27
{\displaystyle x\equiv 2{\bmod {88}}{\text{ et }}x\equiv 1{\bmod {27}}}
Solution
Un générateur du groupe multiplicatif de
Z
/
19
Z
{\displaystyle \mathbb {Z} /19\mathbb {Z} }
3
6
≡
7
mod
19
{\displaystyle 3^{6}\equiv 7{\bmod {19}}}
7
mod
19
{\displaystyle 7{\bmod {19}}}
1
,
7
,
−
8
{\displaystyle 1,7,-8}
Voir Arithmétique/Exercices/Congruences#Exercice 9-10 .
88
=
27
×
3
+
7
{\displaystyle 88=27\times 3+7}
27
=
7
×
4
−
1
{\displaystyle 27=7\times 4-1}
1
=
7
×
4
−
27
=
(
88
−
27
×
3
)
×
4
−
27
=
88
×
4
−
27
×
13
{\displaystyle 1=7\times 4-27=(88-27\times 3)\times 4-27=88\times 4-27\times 13}
88
u
+
27
v
=
1
{\displaystyle 88u+27v=1}
(
u
,
v
)
=
(
u
0
−
27
k
,
v
0
+
88
k
)
{\displaystyle (u,v)=(u_{0}-27k,v_{0}+88k)}
u
0
=
4
,
v
0
=
−
13
,
k
∈
Z
{\displaystyle u_{0}=4,v_{0}=-13,k\in \mathbb {Z} }
x
≡
2
mod
88
et
x
≡
1
mod
27
{\displaystyle x\equiv 2{\bmod {88}}{\text{ et }}x\equiv 1{\bmod {27}}}
x
0
=
88
u
0
+
27
v
0
×
2
=
−
350
{\displaystyle x_{0}=88u_{0}+27v_{0}\times 2=-350}
x
0
+
27
×
88
k
{\displaystyle x_{0}+27\times 88k}
k
∈
Z
{\displaystyle k\in \mathbb {Z} }
x
0
{\displaystyle x_{0}}
Wikipedia-logo-v2.svg
Soit
U
=
{
z
∈
C
∣
|
z
|
=
1
}
{\displaystyle \mathbb {U} =\{z\in \mathbb {C} \mid |z|=1\}}
n
∈
N
∗
,
U
n
=
{
z
∈
C
∣
z
n
=
1
}
{\displaystyle n\in \mathbb {N} ^{*},\,U_{n}=\{z\in \mathbb {C} \mid z^{n}=1\}}
Mq
ϕ
:
z
↦
z
z
¯
{\displaystyle \phi :z\mapsto {\frac {z}{\bar {z}}}}
(
C
∗
,
.
)
{\displaystyle (\mathbb {C} ^{*},.)}
(
U
,
.
)
{\displaystyle (\mathbb {U} ,.)}
Soient
n
,
m
∈
N
∗
{\displaystyle n,m\in \mathbb {N} ^{*}}
U
n
∩
U
m
=
U
d
{\displaystyle U_{n}\cap U_{m}=U_{d}}
d
=
pgcd
(
n
,
m
)
{\displaystyle d=\operatorname {pgcd} (n,m)}
X
n
−
1
{\displaystyle X^{n}-1}
X
m
−
1
{\displaystyle X^{m}-1}
Déterminer tous les endomorphismes de
(
U
n
,
.
)
{\displaystyle (U_{n},.)}
Déterminer tous les automorphismes de
(
U
n
,
.
)
{\displaystyle (U_{n},.)}
Soit
V
=
⋃
n
∈
N
∗
U
n
{\displaystyle V=\bigcup _{n\in \mathbb {N} ^{*}}U_{n}}
U
{\displaystyle \mathbb {U} }
U
{\displaystyle \mathbb {U} }
Quels sont les sous-groupes finis de
(
C
∗
,
.
)
{\displaystyle (\mathbb {C} ^{*},.)}
Solution
ϕ
(
r
e
i
θ
)
=
e
2
i
θ
{\displaystyle \phi (r\mathrm {e} ^{\mathrm {i} \theta })=e^{2\mathrm {i} \theta }}
ϕ
{\displaystyle \phi }
R
∗
{\displaystyle \mathbb {R} ^{*}}
Soient
u
,
v
∈
Z
{\displaystyle u,v\in \mathbb {Z} }
d
=
u
m
+
v
n
{\displaystyle d=um+vn}
z
d
=
(
z
m
)
u
(
z
n
)
v
{\displaystyle z^{d}=(z^{m})^{u}(z^{n})^{v}}
U
n
∩
U
m
⊂
U
d
{\displaystyle U_{n}\cap U_{m}\subset U_{d}}
pgcd
(
X
n
−
1
,
X
m
−
1
)
=
X
d
−
1
{\displaystyle \operatorname {pgcd} (X^{n}-1,X^{m}-1)=X^{d}-1}
Pour cette question de cours, voir Théorie des groupes/Homomorphismes d'un groupe fini dans le groupe multiplicatif du corps des nombres complexes .
Idem.
V sous-groupe : voir Théorie des groupes/Exercices/Groupes, premières notions#Problème 4 (Sous-groupe réunion de deux sous-groupes ?) , question 2.
e
i
π
2
U
∖
V
{\displaystyle \mathrm {e} ^{\mathrm {i} \pi {\sqrt {2}}}\mathbb {U} \setminus V}
Les
U
n
{\displaystyle U_{n}}
On note classiquement
Z
[
i
]
=
{
a
+
i
b
∣
a
,
b
∈
Z
}
{\displaystyle \mathbb {Z} [\mathrm {i} ]=\{a+\mathrm {i} b\mid a,b\in \mathbb {Z} \}}
Q
(
i
)
=
{
a
+
i
b
∣
a
,
b
∈
Q
}
{\displaystyle \mathbb {Q} (\mathrm {i} )=\{a+\mathrm {i} b\mid a,b\in \mathbb {Q} \}}
Mq
Z
[
i
]
{\displaystyle \mathbb {Z} [\mathrm {i} ]}
Q
(
i
)
{\displaystyle \mathbb {Q} (\mathrm {i} )}
Mq
z
∈
Z
[
i
]
{\displaystyle z\in \mathbb {Z} [\mathrm {i} ]}
Z
[
i
]
{\displaystyle \mathbb {Z} [\mathrm {i} ]}
|
z
|
2
=
1
{\displaystyle |z|^{2}=1}
Z
[
i
]
{\displaystyle \mathbb {Z} [\mathrm {i} ]}
Déterminer les automorphismes du corps
Q
(
i
)
{\displaystyle \mathbb {Q} (\mathrm {i} )}
Q
(
i
)
{\displaystyle \mathbb {Q} (\mathrm {i} )}
Z
[
i
]
{\displaystyle \mathbb {Z} [\mathrm {i} ]}
s
(
i
)
{\displaystyle s(\mathrm {i} )}
Mq si
q
∈
Q
{\displaystyle q\in \mathbb {Q} }
u
∈
Z
{\displaystyle u\in \mathbb {Z} }
|
q
−
u
|
≤
1
2
{\displaystyle |q-u|\leq {\frac {1}{2}}}
∀
(
z
,
z
′
)
∈
Z
[
i
]
2
,
z
′
≠
0
∃
(
u
,
v
)
∈
Z
[
i
]
2
z
=
u
z
′
+
v
,
|
v
|
2
<
|
z
′
|
2
{\displaystyle \forall (z,z')\in \mathbb {Z} [\mathrm {i} ]^{2},z'\neq 0\quad \exists (u,v)\in \mathbb {Z} [\mathrm {i} ]^{2}\quad z=uz'+v,|v|^{2}<|z'|^{2}}
Z
=
z
z
′
∈
Q
(
i
)
{\displaystyle Z={\frac {z}{z'}}\in \mathbb {Q} (\mathrm {i} )}
Z
[
i
]
{\displaystyle \mathbb {Z} [\mathrm {i} ]}
principal .
Faire une étude analogue avec
Z
[
j
]
=
{
a
+
j
b
∣
a
,
b
∈
Z
}
{\displaystyle \mathbb {Z} [\mathrm {j} ]=\{a+\mathrm {j} b\mid a,b\in \mathbb {Z} \}}
Q
(
j
)
=
{
a
+
j
b
∣
a
,
b
∈
Q
}
{\displaystyle \mathbb {Q} (\mathrm {j} )=\{a+\mathrm {j} b\mid a,b\in \mathbb {Q} \}}
Trouver les polynômes
P
∈
R
[
X
]
{\displaystyle P\in \mathbb {R} [X]}
X
P
(
X
+
1
)
=
(
X
+
4
)
P
(
X
)
{\displaystyle XP(X+1)=(X+4)P(X)}
Mq
(
X
2
+
X
+
1
)
2
{\displaystyle (X^{2}+X+1)^{2}}
(
X
+
1
)
6
n
+
1
−
X
6
n
+
1
−
1
{\displaystyle (X+1)^{6n+1}-X^{6n+1}-1}
Pour tout
n
∈
N
∗
{\displaystyle n\in \mathbb {N} ^{*}}
P
n
{\displaystyle P_{n}}
∀
θ
∈
R
∖
π
Z
sin
(
(
2
n
+
1
)
θ
)
=
P
n
(
cot
2
θ
)
sin
2
n
+
1
θ
{\displaystyle \forall \theta \in \mathbb {R} \setminus \pi \mathbb {Z} \quad \sin \left((2n+1)\theta \right)=P_{n}(\cot ^{2}\theta )\sin ^{2n+1}\theta }
Préciser le degré, les racines de
P
n
{\displaystyle P_{n}}
Montrer que
∀
θ
∈
]
0
,
π
/
2
[
cot
2
θ
≤
1
θ
2
≤
1
+
cot
2
θ
{\displaystyle \forall \theta \in \left]0,\pi /2\right[\quad \cot ^{2}\theta \leq {\frac {1}{\theta ^{2}}}\leq 1+\cot ^{2}\theta }
Calculer
∑
k
≥
1
1
k
2
{\displaystyle \sum _{k\geq 1}{\frac {1}{k^{2}}}}
∑
k
≥
1
1
k
4
{\displaystyle \sum _{k\geq 1}{\frac {1}{k^{4}}}}
Montrer qu'il existe une infinité de nombres premiers congrus à –1 modulo 4.
![{\displaystyle P\in \mathbb {C} [X]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d8e0f3045d57f5926108bade04c243a473b2cdf1)
![{\displaystyle K_{a}=\{P(a)\mid P\in \mathbb {Q} [X]\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7ec21371f5c48ad73a1087b2c877e58929cae826)
![{\displaystyle I_{a}=\{P\in \mathbb {Q} [X]\mid P(a)=0\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7303a80c23aa61b6ce6270eedccad78389a1ec44)
![{\displaystyle \mathbb {Q} [X]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/59f611a32bc84287517e9f4b283ac998c0394479)
![{\displaystyle \mathbb {Q} [X]\to \mathbb {C} ,\;P\mapsto P(a)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5bab6945e3ba973fa6bf9bbae24c4b709cd0b9c4)
![{\displaystyle K_{a}\simeq \mathbb {Q} [X]/(M)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/08d9bcc9b37185ba87fc240b89025fc96219241d)
![{\displaystyle i,j\in [\![1,n]\!],i\neq j}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3a1d3847c9568ae3a2edd120c2d595a5a7f10678)
![{\displaystyle id_{[[1,n]]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f38a797c358799310c0dd725b3e3cc8e1a7ded81)
![{\displaystyle \mathbb {Z} [\mathrm {i} ]=\{a+\mathrm {i} b\mid a,b\in \mathbb {Z} \}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bbd70b87e5f2fbebd75076a198fba966e956d258)
![{\displaystyle \mathbb {Z} [\mathrm {i} ]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b2f7e30b2083c5a1ffb637bd8c9ec111a486cd06)
![{\displaystyle z\in \mathbb {Z} [\mathrm {i} ]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/742dffdffdf2f55225dd8330a9e03186ed1eba49)
![{\displaystyle \forall (z,z')\in \mathbb {Z} [\mathrm {i} ]^{2},z'\neq 0\quad \exists (u,v)\in \mathbb {Z} [\mathrm {i} ]^{2}\quad z=uz'+v,|v|^{2}<|z'|^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0c9d69f553bb1983cffa33e691d48b8d4a4961ab)
![{\displaystyle \mathbb {Z} [\mathrm {j} ]=\{a+\mathrm {j} b\mid a,b\in \mathbb {Z} \}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1f35eac3da0a53893689b47ad5adaa077b529825)
![{\displaystyle P\in \mathbb {R} [X]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e73c3091b0cfaa94eecbec01be48bef276cd6f84)
![{\displaystyle A,B\in \mathbb {R} [X]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b7a874f708a05ec65be5c68e8a8fa52658efac79)
![{\displaystyle \mathbb {C} [X]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b6c0e47242946827f33aad1803d66106d2f0307e)
![{\displaystyle \forall \theta \in \left]0,\pi /2\right[\quad \cot ^{2}\theta \leq {\frac {1}{\theta ^{2}}}\leq 1+\cot ^{2}\theta }](https://wikimedia.org/api/rest_v1/media/math/render/svg/eb07a8a23814df82a9596a59415f31fecbbb3ad5)
![{\displaystyle \mathbb {Z} [X]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5a538d203a057d4c604f799c28e9a7be410fdcac)
