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Exercice : Orthonormalisation de Gram-SchmidtEspace euclidien/Exercices/Orthonormalisation de Gram-Schmidt », n'a pu être restituée correctement ci-dessus.
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E
{\displaystyle E}
φ
{\displaystyle \varphi }
appliquer la méthode de Gram-Schmidt à la famille libre
F
{\displaystyle F}
Vect
(
F
)
{\displaystyle \operatorname {Vect} (F)}
calculer la projection orthogonale de
v
∈
E
{\displaystyle v\in E}
Vect
(
F
)
{\displaystyle \operatorname {Vect} (F)}
donner les équations de
Vect
(
F
)
{\displaystyle \operatorname {Vect} (F)}
E
=
R
3
{\displaystyle E=\mathbb {R} ^{3}}
φ
{\displaystyle \varphi }
F
=
(
(
1
,
0
,
−
1
)
,
(
1
,
−
1
,
0
)
)
{\displaystyle F=((1,0,-1),(1,-1,0))}
v
=
(
1
,
1
,
1
)
{\displaystyle v=(1,1,1)}
E
=
R
4
{\displaystyle E=\mathbb {R} ^{4}}
φ
{\displaystyle \varphi }
F
=
(
(
1
,
1
,
0
,
0
)
,
(
1
,
0
,
−
1
,
1
)
,
(
0
,
1
,
1
,
1
)
)
{\displaystyle F=((1,1,0,0),(1,0,-1,1),(0,1,1,1))}
v
=
(
1
,
1
,
1
,
1
)
{\displaystyle v=(1,1,1,1)}
E
=
R
3
{\displaystyle E=\mathbb {R} ^{3}}
φ
(
x
,
y
)
=
3
x
1
y
1
−
x
1
y
2
−
x
2
y
1
+
3
x
2
y
2
+
3
x
3
y
3
{\displaystyle \varphi (x,y)=3x_{1}y_{1}-x_{1}y_{2}-x_{2}y_{1}+3x_{2}y_{2}+3x_{3}y_{3}}
F
=
(
(
1
,
0
,
0
)
,
(
0
,
1
,
0
)
)
{\displaystyle F=((1,0,0),(0,1,0))}
v
=
(
0
,
0
,
1
)
{\displaystyle v=(0,0,1)}
E
=
R
3
[
X
]
{\displaystyle E=\mathbb {R} _{3}[X]}
φ
(
P
,
Q
)
=
∫
−
1
1
P
(
X
)
Q
(
X
)
d
X
{\displaystyle \varphi (P,Q)=\int _{-1}^{1}P(X)Q(X)\,\mathrm {d} X}
F
=
(
1
,
X
,
X
2
)
{\displaystyle F=(1,X,X^{2})}
v
=
X
3
{\displaystyle v=X^{3}}
E
=
R
3
[
X
]
{\displaystyle E=\mathbb {R} _{3}[X]}
φ
(
P
,
Q
)
=
∫
0
1
P
(
X
)
Q
(
X
)
d
X
{\displaystyle \varphi (P,Q)=\int _{0}^{1}P(X)Q(X)\,\mathrm {d} X}
F
=
(
1
,
X
,
X
2
)
{\displaystyle F=(1,X,X^{2})}
v
=
X
3
{\displaystyle v=X^{3}}
Solution
e
1
=
(
1
,
0
,
−
1
)
/
2
{\displaystyle e_{1}=(1,0,-1)/{\sqrt {2}}}
e
2
=
u
‖
u
‖
{\displaystyle e_{2}={\frac {u}{\|u\|}}}
u
=
(
1
,
−
1
,
0
)
−
1
2
φ
(
(
1
,
−
1
,
0
)
,
(
1
,
0
,
−
1
)
)
(
1
,
0
,
−
1
)
=
(
1
,
−
1
,
0
)
−
(
1
/
2
,
0
,
−
1
/
2
)
=
(
1
/
2
,
−
1
,
1
/
2
)
{\displaystyle u=(1,-1,0)-{\frac {1}{2}}\varphi ((1,-1,0),(1,0,-1))(1,0,-1)=(1,-1,0)-(1/2,0,-1/2)=(1/2,-1,1/2)}
‖
u
‖
=
3
/
2
{\displaystyle \|u\|={\sqrt {3/2}}}
e
2
=
2
/
3
(
1
/
2
,
−
1
,
1
/
2
)
=
1
6
(
1
,
−
2
,
1
)
{\displaystyle e_{2}={\sqrt {2/3}}(1/2,-1,1/2)={\frac {1}{\sqrt {6}}}(1,-2,1)}
p
(
v
)
=
0
{\displaystyle p(v)=0}
v
⊥
F
{\displaystyle v\perp F}
Vect
(
F
)
=
v
⊥
{\displaystyle \operatorname {Vect} (F)=v^{\bot }}
x
+
y
+
z
=
0
{\displaystyle x+y+z=0}
e
1
=
(
1
,
1
,
0
,
0
)
/
2
{\displaystyle e_{1}=(1,1,0,0)/{\sqrt {2}}}
e
2
=
u
‖
u
‖
{\displaystyle e_{2}={\frac {u}{\|u\|}}}
u
=
(
1
,
0
,
−
1
,
1
)
−
1
2
φ
(
(
1
,
0
,
−
1
,
1
)
,
(
1
,
1
,
0
,
0
)
)
(
1
,
1
,
0
,
0
)
=
(
1
,
0
,
−
1
,
1
)
−
1
2
(
1
,
1
,
0
,
0
)
=
(
1
/
2
,
−
1
/
2
,
−
1
,
1
)
{\displaystyle u=(1,0,-1,1)-{\frac {1}{2}}\varphi ((1,0,-1,1),(1,1,0,0))(1,1,0,0)=(1,0,-1,1)-{\frac {1}{2}}(1,1,0,0)=(1/2,-1/2,-1,1)}
‖
u
‖
=
5
/
2
{\displaystyle \|u\|={\sqrt {5/2}}}
e
2
=
2
/
5
(
1
/
2
,
−
1
/
2
,
−
1
,
1
)
=
1
10
(
1
,
−
1
,
−
2
,
2
)
{\displaystyle e_{2}={\sqrt {2/5}}(1/2,-1/2,-1,1)={\frac {1}{\sqrt {10}}}(1,-1,-2,2)}
e
3
=
w
‖
w
‖
{\displaystyle e_{3}={\frac {w}{\|w\|}}}
w
=
{\displaystyle w=}
(
0
,
1
,
1
,
1
)
−
1
2
φ
(
(
0
,
1
,
1
,
1
)
,
(
1
,
1
,
0
,
0
)
)
(
1
,
1
,
0
,
0
)
−
1
10
φ
(
(
0
,
1
,
1
,
1
)
,
(
1
,
−
1
,
−
2
,
2
)
)
(
1
,
−
1
,
−
2
,
2
)
=
{\displaystyle (0,1,1,1)-{\frac {1}{2}}\varphi ((0,1,1,1),(1,1,0,0))(1,1,0,0)-{\frac {1}{10}}\varphi ((0,1,1,1),(1,-1,-2,2))(1,-1,-2,2)=}
(
−
2
/
5
,
2
/
5
,
4
/
5
,
6
/
5
)
=
2
5
(
−
1
,
1
,
2
,
3
)
{\displaystyle (-2/5,2/5,4/5,6/5)={\frac {2}{5}}(-1,1,2,3)}
‖
w
‖
=
2
5
15
{\displaystyle \|w\|={\frac {2}{5}}{\sqrt {15}}}
e
3
=
1
15
(
−
1
,
1
,
2
,
3
)
{\displaystyle e_{3}={\frac {1}{\sqrt {15}}}(-1,1,2,3)}
p
(
v
)
=
φ
(
v
,
e
1
)
e
1
+
φ
(
v
,
e
2
)
e
2
+
φ
(
v
,
e
3
)
e
3
=
(
1
,
1
,
0
,
0
)
+
1
3
(
−
1
,
1
,
2
,
3
)
=
(
2
/
3
,
4
/
3
,
2
/
3
,
1
)
{\displaystyle p(v)=\varphi (v,e_{1})e_{1}+\varphi (v,e_{2})e_{2}+\varphi (v,e_{3})e_{3}=(1,1,0,0)+{\frac {1}{3}}(-1,1,2,3)=(2/3,4/3,2/3,1)}
Vect
(
F
)
{\displaystyle \operatorname {Vect} (F)}
0
=
|
1
1
0
x
1
0
1
y
0
−
1
1
z
0
1
1
t
|
=
2
(
x
−
y
+
z
)
{\displaystyle 0={\begin{vmatrix}1&1&0&x\\1&0&1&y\\0&-1&1&z\\0&1&1&t\end{vmatrix}}=2(x-y+z)}
y
=
x
+
z
{\displaystyle y=x+z}
F
{\displaystyle F}
Vérifions d'abord que la forme bilinéaire symétrique
φ
{\displaystyle \varphi }
x
∈
R
3
{\displaystyle x\in \mathbb {R} ^{3}}
φ
(
x
,
x
)
=
3
x
1
2
−
2
x
1
x
2
+
3
x
2
2
+
3
x
3
2
=
(
x
1
−
x
2
)
2
+
2
x
1
2
+
2
x
2
2
+
3
x
3
2
>
0
{\displaystyle \varphi (x,x)=3x_{1}^{2}-2x_{1}x_{2}+3x_{2}^{2}+3x_{3}^{2}=(x_{1}-x_{2})^{2}+2x_{1}^{2}+2x_{2}^{2}+3x_{3}^{2}>0}
e
1
=
(
1
,
0
,
0
)
/
φ
(
(
1
,
0
,
0
)
,
(
1
,
0
,
0
)
=
(
1
/
3
,
0
,
0
)
{\displaystyle e_{1}=(1,0,0)/{\sqrt {\varphi ((1,0,0),(1,0,0)}}=(1/{\sqrt {3}},0,0)}
e
2
=
u
φ
(
u
,
u
)
{\displaystyle e_{2}={\frac {u}{\sqrt {\varphi (u,u)}}}}
=
(
0
,
1
,
0
)
−
1
3
φ
(
(
0
,
1
,
0
)
,
(
1
,
0
,
0
)
)
(
1
,
0
,
0
)
=
(
1
/
3
,
1
,
0
)
{\displaystyle =(0,1,0)-{\frac {1}{3}}\varphi ((0,1,0),(1,0,0))(1,0,0)=(1/3,1,0)}
φ
(
u
,
u
)
=
1
/
3
−
2
/
3
+
3
=
8
/
3
{\displaystyle \varphi (u,u)=1/3-2/3+3=8/3}
e
2
=
3
8
(
1
/
3
,
1
,
0
)
=
1
2
6
(
1
,
3
,
0
)
{\displaystyle e_{2}={\sqrt {\frac {3}{8}}}(1/3,1,0)={\frac {1}{2{\sqrt {6}}}}(1,3,0)}
p
(
v
)
=
0
{\displaystyle p(v)=0}
v
⊥
F
{\displaystyle v\perp F}
φ
{\displaystyle \varphi }
x
1
y
3
{\displaystyle x_{1}y_{3}}
x
2
y
3
{\displaystyle x_{2}y_{3}}
Vect
(
F
)
=
v
⊥
{\displaystyle \operatorname {Vect} (F)=v^{\bot }}
=
φ
(
(
x
1
,
x
2
,
x
3
)
,
v
)
=
3
x
3
{\displaystyle =\varphi ((x_{1},x_{2},x_{3}),v)=3x_{3}}
x
3
=
0
{\displaystyle x_{3}=0}
F
{\displaystyle F}
e
1
=
1
/
2
{\displaystyle e_{1}=1/{\sqrt {2}}}
e
2
=
u
‖
u
‖
{\displaystyle e_{2}={\frac {u}{\|u\|}}}
u
=
X
−
1
2
∫
−
1
1
x
d
x
=
X
{\displaystyle u=X-{\frac {1}{2}}\int _{-1}^{1}x\,\mathrm {d} x=X}
‖
u
‖
=
2
/
3
{\displaystyle \|u\|={\sqrt {2/3}}}
e
2
=
3
/
2
X
{\displaystyle e_{2}={\sqrt {3/2}}X}
e
3
=
w
‖
w
‖
{\displaystyle e_{3}={\frac {w}{\|w\|}}}
w
=
X
2
−
1
2
∫
−
1
1
x
2
d
x
−
X
3
2
∫
−
1
1
x
3
d
x
=
X
2
−
1
3
{\displaystyle w=X^{2}-{\frac {1}{2}}\int _{-1}^{1}x^{2}\,\mathrm {d} x-X{\frac {3}{2}}\int _{-1}^{1}x^{3}\,\mathrm {d} x=X^{2}-{\frac {1}{3}}}
‖
w
‖
=
∫
−
1
1
(
x
4
−
(
2
/
3
)
x
2
+
1
/
9
)
d
x
=
∫
−
1
1
(
x
4
−
(
2
/
3
)
x
2
+
1
/
9
)
d
x
=
8
45
{\displaystyle \|w\|={\sqrt {\int _{-1}^{1}(x^{4}-(2/3)x^{2}+1/9)\,\mathrm {d} x}}={\sqrt {\int _{-1}^{1}(x^{4}-(2/3)x^{2}+1/9)\,\mathrm {d} x}}={\sqrt {\frac {8}{45}}}}
e
3
=
45
8
(
X
2
−
1
3
)
=
5
8
(
3
X
2
−
1
)
{\displaystyle e_{3}={\sqrt {\frac {45}{8}}}(X^{2}-{\frac {1}{3}})={\sqrt {\frac {5}{8}}}(3X^{2}-1)}
p
(
v
)
=
φ
(
v
,
e
1
)
e
1
+
φ
(
v
,
e
2
)
e
2
+
φ
(
v
,
e
3
)
e
3
{\displaystyle p(v)=\varphi (v,e_{1})e_{1}+\varphi (v,e_{2})e_{2}+\varphi (v,e_{3})e_{3}}
=
1
2
∫
−
1
1
x
3
d
x
+
3
2
X
∫
−
1
1
x
4
d
x
+
5
8
(
3
X
2
−
1
)
∫
−
1
1
x
3
(
3
x
2
−
1
)
d
x
=
3
5
X
{\displaystyle ={\frac {1}{2}}\int _{-1}^{1}x^{3}\,\mathrm {d} x+{\frac {3}{2}}X\int _{-1}^{1}x^{4}\,\mathrm {d} x+{\frac {5}{8}}(3X^{2}-1)\int _{-1}^{1}x^{3}(3x^{2}-1)\,\mathrm {d} x={\frac {3}{5}}X}
Vect
(
F
)
{\displaystyle \operatorname {Vect} (F)}
t
=
0
{\displaystyle t=0}
(
x
,
y
,
z
,
t
)
{\displaystyle (x,y,z,t)}
R
3
[
X
]
{\displaystyle \mathbb {R} _{3}[X]}
(
1
,
X
,
X
2
,
X
3
)
{\displaystyle (1,X,X^{2},X^{3})}
e
1
=
1
{\displaystyle e_{1}=1}
e
2
=
u
‖
u
‖
{\displaystyle e_{2}={\frac {u}{\|u\|}}}
u
=
X
−
∫
0
1
x
d
x
=
X
−
1
/
2
{\displaystyle u=X-\int _{0}^{1}x\,\mathrm {d} x=X-1/2}
‖
u
‖
2
=
∫
0
1
(
x
2
−
x
+
1
/
4
)
d
x
=
1
/
3
−
1
/
2
+
1
/
4
=
1
/
12
{\displaystyle \|u\|^{2}=\int _{0}^{1}(x^{2}-x+1/4)\,\mathrm {d} x=1/3-1/2+1/4=1/12}
e
2
=
(
X
−
1
/
2
)
12
=
(
2
X
−
1
)
3
{\displaystyle e_{2}=(X-1/2){\sqrt {12}}=(2X-1){\sqrt {3}}}
e
3
=
w
‖
w
‖
{\displaystyle e_{3}={\frac {w}{\|w\|}}}
w
=
X
2
−
1
3
−
3
(
2
X
−
1
)
∫
0
1
(
2
x
3
−
x
2
)
d
x
=
X
2
−
X
+
1
6
{\displaystyle w=X^{2}-{\frac {1}{3}}-3(2X-1)\int _{0}^{1}(2x^{3}-x^{2})\,\mathrm {d} x=X^{2}-X+{\frac {1}{6}}}
‖
w
‖
2
=
∫
0
1
(
x
4
−
2
x
3
+
4
x
2
/
3
−
x
/
3
+
1
/
36
)
d
x
=
1
/
5
−
1
/
2
+
4
/
9
−
1
/
6
+
1
/
36
=
1
/
180
{\displaystyle \|w\|^{2}=\int _{0}^{1}(x^{4}-2x^{3}+4x^{2}/3-x/3+1/36)\,\mathrm {d} x=1/5-1/2+4/9-1/6+1/36=1/180}
e
3
=
180
(
X
2
−
X
+
1
/
6
)
=
5
(
6
X
2
−
6
X
+
1
)
{\displaystyle e_{3}={\sqrt {180}}(X^{2}-X+1/6)={\sqrt {5}}(6X^{2}-6X+1)}
p
(
v
)
=
φ
(
v
,
e
1
)
e
1
+
φ
(
v
,
e
2
)
e
2
+
φ
(
v
,
e
3
)
e
3
{\displaystyle p(v)=\varphi (v,e_{1})e_{1}+\varphi (v,e_{2})e_{2}+\varphi (v,e_{3})e_{3}}
=
∫
0
1
x
3
d
x
+
3
(
2
X
−
1
)
∫
0
1
(
2
x
4
−
x
3
)
d
x
+
5
(
6
X
2
−
6
X
+
1
)
∫
0
1
x
3
(
6
x
2
−
x
+
1
)
d
x
{\displaystyle =\int _{0}^{1}x^{3}\,\mathrm {d} x+3(2X-1)\int _{0}^{1}(2x^{4}-x^{3})\,\mathrm {d} x+5(6X^{2}-6X+1)\int _{0}^{1}x^{3}(6x^{2}-x+1)\,\mathrm {d} x}
=
1
/
4
+
(
6
X
−
3
)
3
/
20
+
(
6
X
2
−
6
X
+
1
)
/
4
=
(
30
X
2
−
12
X
+
1
)
/
20
{\displaystyle =1/4+(6X-3)3/20+(6X^{2}-6X+1)/4=(30X^{2}-12X+1)/20}
Trouver une base orthonormée de
E
=
R
3
[
X
]
{\displaystyle E=\mathbb {R} _{3}[X]}
φ
(
P
,
Q
)
=
∫
−
1
1
P
(
x
)
Q
(
x
)
d
x
{\displaystyle \varphi (P,Q)=\int _{-1}^{1}P(x)Q(x)\,\mathrm {d} x}
Pour tout réel
t
{\displaystyle t}
Q
t
{\displaystyle Q_{t}}
∀
P
∈
E
P
(
t
)
=
φ
(
P
,
Q
t
)
{\displaystyle \forall P\in E\quad P(t)=\varphi (P,Q_{t})}
Q
t
{\displaystyle Q_{t}}
t
{\displaystyle t}
Solution
D'après l'exercice précédent (question 4),
e
1
=
1
/
2
{\displaystyle e_{1}=1/{\sqrt {2}}}
e
2
=
3
/
2
X
{\displaystyle e_{2}={\sqrt {3/2}}X}
e
3
=
5
8
(
3
X
2
−
1
)
{\displaystyle e_{3}={\sqrt {\frac {5}{8}}}(3X^{2}-1)}
X
3
{\displaystyle X^{3}}
Vect
(
e
1
,
e
2
,
e
3
)
{\displaystyle \operatorname {Vect} (e_{1},e_{2},e_{3})}
3
X
/
5
{\displaystyle 3X/5}
e
4
=
z
‖
z
‖
{\displaystyle e_{4}={\frac {z}{\|z\|}}}
z
=
X
3
−
3
X
/
5
{\displaystyle z=X^{3}-3X/5}
‖
z
‖
2
=
∫
−
1
1
(
x
6
−
6
x
4
/
5
+
9
x
2
/
25
)
d
x
=
2
(
1
/
7
−
6
/
25
+
3
/
25
)
=
8
/
(
7
×
25
)
{\displaystyle \|z\|^{2}=\int _{-1}^{1}(x^{6}-6x^{4}/5+9x^{2}/25)\,\mathrm {d} x=2(1/7-6/25+3/25)=8/(7\times 25)}
e
4
=
7
8
(
5
X
3
−
3
X
)
{\displaystyle e_{4}={\sqrt {\frac {7}{8}}}(5X^{3}-3X)}
polynômes de Legendre .
Pour tout
t
∈
R
{\displaystyle t\in \mathbb {R} }
P
↦
P
(
t
)
{\displaystyle P\mapsto P(t)}
E
{\displaystyle E}
théorème de représentation de Riesz , elle est de la forme
P
↦
φ
(
P
,
Q
t
)
{\displaystyle P\mapsto \varphi (P,Q_{t})}
Q
t
∈
E
{\displaystyle Q_{t}\in E}
Q
t
=
∑
i
=
1
4
φ
(
e
i
,
Q
t
)
e
i
=
∑
i
=
1
4
e
i
(
t
)
e
i
=
1
2
+
3
2
t
X
+
5
8
(
3
t
2
−
1
)
(
3
X
2
−
1
)
+
7
8
(
5
t
3
−
3
t
)
(
5
X
3
−
3
X
)
{\displaystyle Q_{t}=\sum _{i=1}^{4}\varphi (e_{i},Q_{t})e_{i}=\sum _{i=1}^{4}e_{i}(t)e_{i}={\frac {1}{2}}+{\frac {3}{2}}tX+{\frac {5}{8}}(3t^{2}-1)(3X^{2}-1)+{\frac {7}{8}}(5t^{3}-3t)(5X^{3}-3X)}
=
35
8
(
5
t
3
−
3
t
)
X
3
+
15
8
(
3
t
2
−
1
)
X
2
+
15
8
(
−
7
t
3
+
5
t
)
X
+
3
8
(
−
5
t
2
+
3
)
{\displaystyle ={\frac {35}{8}}(5t^{3}-3t)X^{3}+{\frac {15}{8}}(3t^{2}-1)X^{2}+{\frac {15}{8}}(-7t^{3}+5t)X+{\frac {3}{8}}(-5t^{2}+3)}
Trouver une base orthonormée de
E
=
R
3
[
X
]
{\displaystyle E=\mathbb {R} _{3}[X]}
φ
(
P
,
Q
)
=
∑
i
=
0
3
P
(
i
)
Q
(
i
)
{\displaystyle \varphi (P,Q)=\sum _{i=0}^{3}P(i)Q(i)}
Solution
e
1
=
1
‖
1
‖
=
1
2
{\displaystyle e_{1}={\frac {1}{\|1\|}}={\frac {1}{2}}}
e
2
=
u
‖
u
‖
{\displaystyle e_{2}={\frac {u}{\|u\|}}}
u
=
X
−
φ
(
e
1
,
X
)
e
1
=
X
−
3
/
2
{\displaystyle u=X-\varphi (e_{1},X)e_{1}=X-3/2}
‖
u
‖
2
=
(
−
3
/
2
)
2
+
(
−
1
/
2
)
2
+
(
1
/
2
)
2
+
(
3
/
2
)
2
=
5
{\displaystyle \|u\|^{2}=(-3/2)^{2}+(-1/2)^{2}+(1/2)^{2}+(3/2)^{2}=5}
e
2
=
X
−
3
/
2
5
{\displaystyle e_{2}={\frac {X-3/2}{\sqrt {5}}}}
e
3
=
v
‖
v
‖
{\displaystyle e_{3}={\frac {v}{\|v\|}}}
v
=
X
2
−
φ
(
e
1
,
X
2
)
e
1
−
φ
(
e
2
,
X
2
)
e
2
{\displaystyle v=X^{2}-\varphi (e_{1},X^{2})e_{1}-\varphi (e_{2},X^{2})e_{2}}
=
X
2
−
1
4
(
1
+
4
+
9
)
−
1
5
(
−
1
/
2
+
2
+
27
/
2
)
(
X
−
3
/
2
)
=
X
2
−
3
X
+
1
{\displaystyle =X^{2}-{\frac {1}{4}}(1+4+9)-{\frac {1}{5}}(-1/2+2+27/2)(X-3/2)=X^{2}-3X+1}
‖
v
‖
2
=
1
2
+
(
−
1
)
2
+
(
−
1
)
2
+
1
2
=
4
{\displaystyle \|v\|^{2}=1^{2}+(-1)^{2}+(-1)^{2}+1^{2}=4}
e
3
=
1
2
(
X
2
−
3
X
+
1
)
{\displaystyle e_{3}={\frac {1}{2}}(X^{2}-3X+1)}
e
4
=
w
‖
w
‖
{\displaystyle e_{4}={\frac {w}{\|w\|}}}
w
=
X
3
−
φ
(
e
1
,
X
3
)
e
1
−
φ
(
e
2
,
X
3
)
e
2
−
φ
(
e
3
,
X
3
)
e
3
=
{\displaystyle w=X^{3}-\varphi (e_{1},X^{3})e_{1}-\varphi (e_{2},X^{3})e_{2}-\varphi (e_{3},X^{3})e_{3}=}
X
3
−
1
4
(
1
+
8
+
27
)
−
1
5
(
−
1
/
2
+
4
+
81
/
2
)
(
X
−
3
/
2
)
−
1
4
(
−
1
−
8
+
27
)
(
X
2
−
3
X
+
1
)
=
{\displaystyle X^{3}-{\frac {1}{4}}(1+8+27)-{\frac {1}{5}}(-1/2+4+81/2)(X-3/2)-{\frac {1}{4}}(-1-8+27)(X^{2}-3X+1)=}
X
3
−
9
−
44
5
(
X
−
3
/
2
)
−
9
2
(
X
2
−
3
X
+
1
)
=
X
3
−
9
X
2
/
2
+
47
X
/
10
−
3
/
10
{\displaystyle X^{3}-9-{\frac {44}{5}}(X-3/2)-{\frac {9}{2}}(X^{2}-3X+1)=X^{3}-9X^{2}/2+47X/10-3/10}
‖
w
‖
2
=
(
−
3
/
10
)
2
+
(
9
/
10
)
2
+
(
−
9
/
10
)
2
+
(
3
/
10
)
2
=
9
/
5
{\displaystyle \|w\|^{2}=(-3/10)^{2}+(9/10)^{2}+(-9/10)^{2}+(3/10)^{2}=9/5}
e
4
=
5
3
(
X
3
−
9
X
2
/
2
+
47
X
/
10
−
3
/
10
)
{\displaystyle e_{4}={\frac {\sqrt {5}}{3}}(X^{3}-9X^{2}/2+47X/10-3/10)}
On définit un produit scalaire sur
E
:=
R
n
[
X
]
{\displaystyle E:=\mathbb {R} _{n}[X]}
∀
P
,
Q
∈
E
⟨
P
,
Q
⟩
=
∫
0
1
P
(
t
)
Q
(
t
)
d
t
{\displaystyle \forall P,Q\in E\quad \langle P,Q\rangle =\int _{0}^{1}P(t)Q(t)\,\mathrm {d} t}
Soit
f
{\displaystyle f}
E
{\displaystyle E}
f
(
P
)
=
P
(
1
/
3
)
{\displaystyle f(P)=P(1/3)}
On sait (théorème de représentation de Riesz ) qu'il existe un unique
Q
∈
E
{\displaystyle Q\in E}
∀
P
∈
E
f
(
P
)
=
⟨
P
,
Q
⟩
{\displaystyle \forall P\in E\quad f(P)=\langle P,Q\rangle }
Pour
n
=
1
,
2
,
3
{\displaystyle n=1,2,3}
Q
{\displaystyle Q}
Solution
Pour
P
=
∑
a
i
X
i
{\displaystyle P=\sum a_{i}X^{i}}
Q
=
∑
b
j
X
j
{\displaystyle Q=\sum b_{j}X^{j}}
⟨
P
,
Q
⟩
=
∑
0
≤
i
,
j
≤
n
a
i
b
j
i
+
j
+
1
{\displaystyle \langle P,Q\rangle =\sum _{0\leq i,j\leq n}{a_{i}b_{j} \over i+j+1}}
n
=
1
{\displaystyle n=1}
a
,
b
∈
R
{\displaystyle a,b\in \mathbb {R} }
∀
a
′
,
b
′
∈
R
b
b
′
+
a
b
′
+
b
a
′
2
+
a
a
′
3
=
a
′
/
3
+
b
′
{\displaystyle \forall a',b'\in \mathbb {R} \quad bb'+{ab'+ba' \over 2}+{aa' \over 3}=a'/3+b'}
b
/
2
+
a
/
3
=
1
/
3
{\displaystyle b/2+a/3=1/3}
b
+
a
/
2
=
1
{\displaystyle b+a/2=1}
b
=
2
,
a
=
−
2
{\displaystyle b=2,a=-2}
Q
=
−
2
X
+
2
{\displaystyle Q=-2X+2}
n
=
2
{\displaystyle n=2}
a
,
b
,
c
∈
R
{\displaystyle a,b,c\in \mathbb {R} }
∀
a
′
,
b
′
,
c
′
∈
R
c
c
′
+
b
c
′
+
c
b
′
2
+
a
c
′
+
b
b
′
+
c
a
′
3
+
a
b
′
+
b
a
′
4
+
a
a
′
5
=
a
′
/
9
+
b
′
/
3
+
c
′
{\displaystyle \forall a',b',c'\in \mathbb {R} \quad cc'+{bc'+cb' \over 2}+{ac'+bb'+ca' \over 3}+{ab'+ba' \over 4}+{aa' \over 5}=a'/9+b'/3+c'}
c
/
3
+
b
/
4
+
a
/
5
=
1
/
9
{\displaystyle c/3+b/4+a/5=1/9}
c
/
2
+
b
/
3
+
a
/
4
=
1
/
3
{\displaystyle c/2+b/3+a/4=1/3}
c
+
b
/
2
+
a
/
3
=
1
{\displaystyle c+b/2+a/3=1}
c
=
1
/
3
,
b
=
8
,
a
=
−
10
{\displaystyle c=1/3,b=8,a=-10}
Q
=
−
10
X
2
+
8
X
+
1
/
3
{\displaystyle Q=-10X^{2}+8X+1/3}
n
=
3
{\displaystyle n=3}
a
,
b
,
c
,
d
∈
R
{\displaystyle a,b,c,d\in \mathbb {R} }
∀
a
′
,
b
′
,
c
′
,
d
′
∈
R
d
d
′
+
c
d
′
+
d
c
′
2
+
b
d
′
+
c
c
′
+
d
b
′
3
+
a
d
′
+
b
c
′
+
c
b
′
+
d
a
′
4
+
a
c
′
+
b
b
′
+
c
a
′
5
+
a
b
′
+
b
a
′
6
+
a
a
′
7
=
a
′
/
27
+
b
′
/
9
+
c
′
/
3
+
d
′
{\displaystyle \forall a',b',c',d'\in \mathbb {R} \quad dd'+{cd'+dc' \over 2}+{bd'+cc'+db' \over 3}+{ad'+bc'+cb'+da' \over 4}+{ac'+bb'+ca' \over 5}+{ab'+ba' \over 6}+{aa' \over 7}=a'/27+b'/9+c'/3+d'}
d
/
4
+
c
/
5
+
b
/
6
+
a
/
7
=
1
/
27
{\displaystyle d/4+c/5+b/6+a/7=1/27}
d
/
3
+
c
/
4
+
b
/
5
+
a
/
6
=
1
/
9
{\displaystyle d/3+c/4+b/5+a/6=1/9}
d
/
2
+
c
/
3
+
b
/
4
+
a
/
5
=
1
/
3
{\displaystyle d/2+c/3+b/4+a/5=1/3}
d
+
c
/
2
+
b
/
3
+
a
/
4
=
1
{\displaystyle d+c/2+b/3+a/4=1}
a
=
4.5.7.11
/
27
=
1540
/
27
{\displaystyle a=4.5.7.11/27=1540/27}
b
=
−
4.5.43
/
9
=
−
860
/
9
{\displaystyle b=-4.5.43/9=-860/9}
c
=
4.5.19
/
9
=
380
/
9
{\displaystyle c=4.5.19/9=380/9}
d
=
−
4.17
/
27
=
−
68
/
27
{\displaystyle d=-4.17/27=-68/27}
Dans
R
n
+
1
{\displaystyle \mathbb {R} ^{n+1}}
E
{\displaystyle E}
x
0
+
…
+
x
n
=
0
{\displaystyle x_{0}+\ldots +x_{n}=0}
Trouver l'unique
y
∈
E
{\displaystyle y\in E}
∀
x
=
(
x
0
,
…
,
x
n
)
∈
E
⟨
x
,
y
⟩
=
x
0
{\displaystyle \forall x=(x_{0},\ldots ,x_{n})\in E\quad \langle x,y\rangle =x_{0}}
Même question pour
x
1
{\displaystyle x_{1}}
x
0
−
2
x
1
{\displaystyle x_{0}-2x_{1}}
Solution
∀
(
x
1
,
…
,
x
n
)
∈
R
n
(
−
x
1
−
…
−
x
n
)
y
0
+
x
1
y
1
+
…
x
n
y
n
=
−
x
1
−
…
−
x
n
{\displaystyle \forall (x_{1},\ldots ,x_{n})\in \mathbb {R} ^{n}\quad (-x_{1}-\ldots -x_{n})y_{0}+x_{1}y_{1}+\ldots x_{n}y_{n}=-x_{1}-\ldots -x_{n}}
−
y
0
+
y
1
=
…
=
−
y
0
+
y
n
=
−
1
{\displaystyle -y_{0}+y_{1}=\ldots =-y_{0}+y_{n}=-1}
y
1
=
…
=
y
n
=
y
0
−
1
{\displaystyle y_{1}=\ldots =y_{n}=y_{0}-1}
0
=
∑
y
i
=
y
0
+
n
(
y
0
−
1
)
{\displaystyle 0=\sum y_{i}=y_{0}+n(y_{0}-1)}
y
=
(
n
,
−
1
,
−
1
,
…
,
−
1
)
/
(
n
+
1
)
{\displaystyle y=(n,-1,-1,\ldots ,-1)/(n+1)}
De même pour
x
1
{\displaystyle x_{1}}
y
′
=
(
−
1
,
n
,
−
1
,
…
,
−
1
)
/
(
n
+
1
)
{\displaystyle y'=(-1,n,-1,\ldots ,-1)/(n+1)}
et pour
x
0
−
2
x
1
{\displaystyle x_{0}-2x_{1}}
y
″
=
y
−
2
y
′
=
(
n
+
2
,
−
1
−
2
n
,
1
,
…
,
1
)
/
(
n
+
1
)
{\displaystyle y''=y-2y'=(n+2,-1-2n,1,\ldots ,1)/(n+1)}
![{\displaystyle E=\mathbb {R} _{3}[X]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ccb1eab9b5163c8bd91a2e0cd7467bf3cf561506)
![{\displaystyle \mathbb {R} _{3}[X]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/84382a2e05fcb9ef1ab49949eb5a0c7e0e753819)
![{\displaystyle E:=\mathbb {R} _{n}[X]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/68e7203f7b4465fc47b31103632bb50ea8e3bca4)
