En raison de limitations techniques, la typographie souhaitable du titre, «
Exercice : Inversion locale, fonctions implicitesCalcul différentiel/Exercices/Inversion locale, fonctions implicites », n'a pu être restituée correctement ci-dessus.
Montrer que l'application
φ
:
(
r
,
θ
)
↦
(
x
,
y
)
=
(
r
cos
θ
,
r
sin
θ
)
{\displaystyle \varphi :(r,\theta )\mapsto (x,y)=(r\cos \theta ,r\sin \theta )}
1 -difféomorphisme de l'ouvert
]
0
,
∞
[
×
]
−
π
,
π
[
{\displaystyle \left]0,\infty \right[\times \left]-\pi ,\pi \right[}
R
−
{\displaystyle \mathbb {R} _{-}}
f
(
x
,
y
)
=
g
(
r
,
θ
)
{\displaystyle f(x,y)=g(r,\theta )}
f
{\displaystyle f}
g
{\displaystyle g}
Soit
U
{\displaystyle U}
f
(
x
,
y
)
=
(
x
2
−
y
2
,
2
x
y
)
{\displaystyle f(x,y)=(x^{2}-y^{2},2xy)}
f
{\displaystyle f}
U
{\displaystyle U}
Soit
g
{\displaystyle g}
R
2
{\displaystyle \mathbb {R} ^{2}}
R
2
{\displaystyle \mathbb {R} ^{2}}
g
(
x
,
y
)
=
(
x
+
y
,
x
y
)
{\displaystyle g(x,y)=(x+y,xy)}
U
⊂
R
2
{\displaystyle U\subset \mathbb {R} ^{2}}
g
{\displaystyle g}
U
{\displaystyle U}
g
(
U
)
{\displaystyle g(U)}
Soit
h
{\displaystyle h}
R
2
{\displaystyle \mathbb {R} ^{2}}
R
2
{\displaystyle \mathbb {R} ^{2}}
h
(
x
,
y
)
=
(
e
x
cos
y
,
e
x
sin
y
)
{\displaystyle h(x,y)=(\operatorname {e} ^{x}\cos y,\operatorname {e} ^{x}\sin y)}
h
{\displaystyle h}
1 et que
D
(
x
,
y
)
h
{\displaystyle D_{(x,y)}h}
(
x
,
y
)
∈
R
2
{\displaystyle (x,y)\in \mathbb {R} ^{2}}
h
{\displaystyle h}
R
2
{\displaystyle \mathbb {R} ^{2}}
h
(
R
2
)
{\displaystyle h(\mathbb {R} ^{2})}
Solution
L'application
φ
{\displaystyle \varphi }
∞ car ses coordonnées le sont. Pour montrer que c'est un difféormorphisme global, il suffit de montrer que c'est un difféo local (théorème d'inversion locale) et qu'elle est bijective. Calculons sa matrice jacobienne :
J
φ
(
r
,
θ
)
=
(
cos
θ
−
r
sin
θ
sin
θ
r
cos
θ
)
{\displaystyle J\varphi (r,\theta )={\begin{pmatrix}\cos \theta &-r\sin \theta \\\sin \theta &r\cos \theta \end{pmatrix}}}
φ
{\displaystyle \varphi }
det
(
J
φ
(
r
,
θ
)
)
=
r
(
cos
2
θ
+
sin
2
θ
)
=
r
>
0
{\displaystyle \det(J\varphi (r,\theta ))=r(\cos ^{2}\theta +\sin ^{2}\theta )=r>0}
(
x
,
y
)
≠
(
0
,
0
)
{\displaystyle (x,y)\neq (0,0)}
φ
{\displaystyle \varphi }
]
0
,
∞
[
×
]
−
π
,
π
[
{\displaystyle \left]0,\infty \right[\times \left]-\pi ,\pi \right[}
φ
{\displaystyle \varphi }
(
r
,
θ
)
{\displaystyle (r,\theta )}
(
x
,
y
)
{\displaystyle (x,y)}
φ
{\displaystyle \varphi }
x
2
+
y
2
{\displaystyle x^{2}+y^{2}}
y
/
x
{\displaystyle y/x}
J
f
(
x
,
y
)
=
2
(
x
−
y
y
x
)
{\displaystyle Jf(x,y)=2{\begin{pmatrix}x&-y\\y&x\end{pmatrix}}}
(
x
,
y
)
≠
(
0
,
0
)
{\displaystyle (x,y)\neq (0,0)}
f
{\displaystyle f}
z
↦
z
2
{\displaystyle z\mapsto z^{2}}
(
S
,
P
)
{\displaystyle (S,P)}
g
{\displaystyle g}
S
2
−
4
P
≥
0
{\displaystyle S^{2}-4P\geq 0}
(
x
1
,
x
2
)
{\displaystyle (x_{1},x_{2})}
(
x
2
,
x
1
)
{\displaystyle (x_{2},x_{1})}
x
1
{\displaystyle x_{1}}
x
2
{\displaystyle x_{2}}
x
2
−
S
x
+
p
=
0
{\displaystyle x^{2}-Sx+p=0}
g
{\displaystyle g}
U
:=
{
(
x
,
y
)
∈
R
2
∣
x
<
y
}
{\displaystyle U:=\{(x,y)\in \mathbb {R} ^{2}\mid x<y\}}
g
(
U
)
=
{
(
S
,
P
)
∈
R
2
∣
S
2
−
4
P
>
0
}
{\displaystyle g(U)=\{(S,P)\in \mathbb {R} ^{2}\mid S^{2}-4P>0\}}
U
{\displaystyle U}
U
{\displaystyle U}
g
{\displaystyle g}
∞ sur
U
{\displaystyle U}
∀
(
x
,
y
)
∈
U
det
J
g
(
x
,
y
)
=
|
1
y
1
x
|
=
x
−
y
≠
0
{\displaystyle \forall (x,y)\in U\quad \det J_{g}(x,y)={\begin{vmatrix}1&y\\1&x\end{vmatrix}}=x-y\neq 0}
h
{\displaystyle h}
∞ car ses deux composantes (de
R
2
{\displaystyle \mathbb {R} ^{2}}
R
{\displaystyle \mathbb {R} }
R
{\displaystyle \mathbb {R} }
R
2
{\displaystyle \mathbb {R} ^{2}}
∞ de
R
{\displaystyle \mathbb {R} }
R
{\displaystyle \mathbb {R} }
det
J
h
(
x
,
y
)
=
e
2
x
|
cos
y
−
sin
y
sin
y
cos
y
|
=
e
2
x
≠
0
{\displaystyle \det J_{h}(x,y)=\operatorname {e} ^{2x}{\begin{vmatrix}\cos y&-\sin y\\\sin y&\cos y\end{vmatrix}}=\operatorname {e} ^{2x}\neq 0}
C
{\displaystyle \mathbb {C} }
C
{\displaystyle \mathbb {C} }
h
{\displaystyle h}
0
≤
y
<
2
π
{\displaystyle 0\leq y<2\pi }
Soit
φ
:
R
2
→
R
2
,
(
x
,
y
)
↦
(
x
−
sin
(
y
/
2
)
,
y
−
sin
(
x
/
2
)
)
{\displaystyle \varphi :\mathbb {R} ^{2}\to \mathbb {R} ^{2},\ (x,y)\mapsto (x-\sin(y/2),y-\sin(x/2))}
Justifier que
φ
{\displaystyle \varphi }
1 , calculer sa différentielle et voir que
D
φ
(
x
,
y
)
{\displaystyle \mathrm {D} \varphi (x,y)}
(
x
,
y
)
∈
R
2
{\displaystyle (x,y)\in \mathbb {R} ^{2}}
Montrer que
φ
{\displaystyle \varphi }
1 -difféomorphisme de
R
2
{\displaystyle \mathbb {R} ^{2}}
φ
(
R
2
)
{\displaystyle \varphi (\mathbb {R} ^{2})}
φ
(
R
2
)
{\displaystyle \varphi (\mathbb {R} ^{2})}
Montrer que
φ
−
1
{\displaystyle \varphi ^{-1}}
φ
(
R
2
)
{\displaystyle \varphi (\mathbb {R} ^{2})}
R
2
{\displaystyle \mathbb {R} ^{2}}
En déduire que
φ
{\displaystyle \varphi }
R
2
{\displaystyle \mathbb {R} ^{2}}
R
2
{\displaystyle \mathbb {R} ^{2}}
Soit
p
=
(
π
/
2
−
1
,
π
−
1
/
2
)
{\displaystyle p=(\pi /2-1,\pi -1/{\sqrt {2}})}
D
φ
−
1
(
p
)
{\displaystyle \mathrm {D} \varphi ^{-1}(p)}
Solution
φ
{\displaystyle \varphi }
1 ; elle l'est donc aussi.
J
φ
(
x
,
y
)
=
(
1
−
1
/
2
cos
(
y
/
2
)
−
1
/
2
cos
(
x
/
2
)
1
)
{\displaystyle J\varphi (x,y)={\begin{pmatrix}1&-1/2\cos(y/2)\\-1/2\cos(x/2)&1\end{pmatrix}}}
det
(
J
φ
(
x
,
y
)
)
=
1
−
1
/
4
cos
(
x
/
2
)
cos
(
y
/
2
)
≥
3
/
4
>
0
{\displaystyle \det(J\varphi (x,y))=1-1/4\cos(x/2)\cos(y/2)\geq 3/4>0}
D
φ
(
x
,
y
)
{\displaystyle \mathrm {D} \varphi (x,y)}
Supposons
φ
(
x
1
,
y
1
)
=
φ
(
x
2
,
y
2
)
{\displaystyle \varphi (x_{1},y_{1})=\varphi (x_{2},y_{2})}
x
1
−
sin
(
y
1
/
2
)
=
x
2
−
sin
(
y
2
/
2
)
{\displaystyle x_{1}-\sin(y_{1}/2)=x_{2}-\sin(y_{2}/2)}
y
1
−
sin
(
x
1
/
2
)
=
y
2
−
sin
(
x
2
/
2
)
{\displaystyle y_{1}-\sin(x_{1}/2)=y_{2}-\sin(x_{2}/2)}
sin
(
y
1
/
2
)
−
sin
(
y
2
/
2
)
=
x
1
−
x
2
{\displaystyle \sin(y_{1}/2)-\sin(y_{2}/2)=x_{1}-x_{2}}
sin
(
x
1
/
2
)
−
sin
(
x
2
/
2
)
=
y
1
−
y
2
{\displaystyle \sin(x_{1}/2)-\sin(x_{2}/2)=y_{1}-y_{2}}
∀
a
,
b
∈
R
|
sin
a
−
sin
b
|
≤
|
a
−
b
|
{\displaystyle \forall a,b\in \mathbb {R} \quad |\sin a-\sin b|\leq |a-b|}
inégalité des accroissements finis ). Donc
|
x
1
−
x
2
|
≤
|
y
1
/
2
−
y
2
/
2
|
{\displaystyle |x_{1}-x_{2}|\leq |y_{1}/2-y_{2}/2|}
|
y
1
−
y
2
|
≤
|
x
1
/
2
−
x
2
/
2
|
{\displaystyle |y_{1}-y_{2}|\leq |x_{1}/2-x_{2}/2|}
|
x
1
−
x
2
|
≤
1
/
4
|
x
1
−
x
2
|
{\displaystyle |x_{1}-x_{2}|\leq 1/4|x_{1}-x_{2}|}
x
1
=
x
2
{\displaystyle x_{1}=x_{2}}
y
1
=
y
2
{\displaystyle y_{1}=y_{2}}
φ
{\displaystyle \varphi }
1 -difféomorphisme de
R
2
{\displaystyle \mathbb {R} ^{2}}
φ
(
R
2
)
{\displaystyle \varphi (\mathbb {R} ^{2})}
φ
(
R
2
)
{\displaystyle \varphi (\mathbb {R} ^{2})}
Soient
(
X
1
,
Y
1
)
,
(
X
2
,
Y
2
)
∈
φ
(
R
2
)
{\displaystyle (X_{1},Y_{1}),(X_{2},Y_{2})\in \varphi (\mathbb {R} ^{2})}
φ
(
x
1
,
y
1
)
=
(
X
1
,
Y
1
)
{\displaystyle \varphi (x_{1},y_{1})=(X_{1},Y_{1})}
φ
(
x
2
,
y
2
)
=
(
X
2
,
Y
2
)
{\displaystyle \varphi (x_{2},y_{2})=(X_{2},Y_{2})}
φ
−
1
(
X
1
,
Y
1
)
=
(
x
1
,
y
1
)
{\displaystyle \varphi ^{-1}(X_{1},Y_{1})=(x_{1},y_{1})}
φ
−
1
(
X
2
,
Y
2
)
=
(
x
2
,
y
2
)
{\displaystyle \varphi ^{-1}(X_{2},Y_{2})=(x_{2},y_{2})}
R
2
{\displaystyle \mathbb {R} ^{2}}
‖
(
x
,
y
)
‖
=
|
x
|
+
|
y
|
{\displaystyle \|(x,y)\|=|x|+|y|}
‖
φ
−
1
(
X
1
,
Y
1
)
−
φ
−
1
(
X
2
,
Y
2
)
‖
=
‖
(
x
1
,
y
1
)
−
(
x
2
,
y
2
)
‖
=
‖
(
x
1
−
x
2
,
y
1
,
y
2
)
‖
=
|
x
1
−
x
2
|
+
|
y
1
−
y
2
|
{\displaystyle \|\varphi ^{-1}(X_{1},Y_{1})-\varphi ^{-1}(X_{2},Y_{2})\|=\|(x_{1},y_{1})-(x_{2},y_{2})\|=\|(x_{1}-x_{2},y_{1},y_{2})\|=|x_{1}-x_{2}|+|y_{1}-y_{2}|}
x
1
−
sin
(
y
1
/
2
)
=
X
1
{\displaystyle x_{1}-\sin(y_{1}/2)=X_{1}}
x
2
−
sin
(
y
2
/
2
)
=
X
2
{\displaystyle x_{2}-\sin(y_{2}/2)=X_{2}}
y
1
−
sin
(
x
1
/
2
)
=
Y
1
{\displaystyle y_{1}-\sin(x_{1}/2)=Y_{1}}
y
2
−
sin
(
x
2
/
2
)
=
Y
2
{\displaystyle y_{2}-\sin(x_{2}/2)=Y_{2}}
x
1
−
x
2
=
sin
(
y
1
/
2
)
+
X
1
−
sin
(
y
2
/
2
)
−
X
2
{\displaystyle x_{1}-x_{2}=\sin(y_{1}/2)+X_{1}-\sin(y_{2}/2)-X_{2}}
y
1
−
y
2
=
sin
(
x
1
/
2
)
+
Y
1
−
sin
(
x
2
/
2
)
−
Y
2
{\displaystyle y_{1}-y_{2}=\sin(x_{1}/2)+Y_{1}-\sin(x_{2}/2)-Y_{2}}
|
x
1
−
x
2
|
+
|
y
1
−
y
2
|
≤
|
X
1
−
X
2
|
+
|
sin
(
y
1
/
2
)
−
sin
(
y
2
/
2
)
|
+
|
Y
1
−
Y
2
|
+
|
sin
(
x
1
/
2
)
−
sin
(
x
2
/
2
)
|
≤
|
X
1
−
X
2
|
+
1
/
2
|
y
1
−
y
2
|
+
|
Y
1
−
Y
2
|
+
1
/
2
|
x
1
−
x
2
|
{\displaystyle {\begin{aligned}|x_{1}-x_{2}|+|y_{1}-y_{2}|&\leq |X_{1}-X_{2}|+|\sin(y_{1}/2)-\sin(y_{2}/2)|+|Y_{1}-Y_{2}|+|\sin(x_{1}/2)-\sin(x_{2}/2)|\\&\leq |X_{1}-X_{2}|+1/2|y_{1}-y_{2}|+|Y_{1}-Y_{2}|+1/2|x_{1}-x_{2}|\end{aligned}}}
|
x
1
−
x
2
|
+
|
y
1
−
y
2
|
≤
2
(
|
X
1
−
X
2
|
+
|
Y
1
−
Y
2
|
)
=
2
‖
(
X
1
,
Y
1
)
−
(
X
2
,
Y
2
)
‖
{\displaystyle |x_{1}-x_{2}|+|y_{1}-y_{2}|\leq 2(|X_{1}-X_{2}|+|Y_{1}-Y_{2}|)=2\|(X_{1},Y_{1})-(X_{2},Y_{2})\|}
φ
−
1
{\displaystyle \varphi ^{-1}}
φ
−
1
{\displaystyle \varphi ^{-1}}
(
X
1
,
Y
1
)
=
(
X
2
,
Y
2
)
{\displaystyle (X_{1},Y_{1})=(X_{2},Y_{2})}
φ
{\displaystyle \varphi }
|
|
|
J
φ
−
1
(
φ
(
x
,
y
)
)
|
|
|
=
|
|
|
1
1
−
1
/
4
cos
(
x
/
2
)
cos
(
y
/
2
)
(
1
1
/
2
cos
(
y
/
2
)
1
/
2
cos
(
x
/
2
)
1
)
|
|
|
≤
2
{\displaystyle \left|\left|\left|J_{\varphi ^{-1}}(\varphi (x,y))\right|\right|\right|=\left|\left|\left|{\frac {1}{1-1/4\cos(x/2)\cos(y/2)}}{\begin{pmatrix}1&1/2\cos(y/2)\\1/2\cos(x/2)&1\end{pmatrix}}\right|\right|\right|\leq 2}
0
<
1
1
−
1
/
4
cos
(
x
/
2
)
cos
(
y
/
2
)
≤
4
3
et
|
u
+
v
/
2
cos
(
y
/
2
)
|
+
|
u
/
2
cos
(
x
/
2
)
+
v
|
≤
3
2
(
|
u
|
+
|
v
|
)
{\displaystyle 0<{\frac {1}{1-1/4\cos(x/2)\cos(y/2)}}\leq {\frac {4}{3}}\quad {\text{et}}\quad |u+v/2\cos(y/2)|+|u/2\cos(x/2)+v|\leq {\frac {3}{2}}(|u|+|v|)}
φ
−
1
{\displaystyle \varphi ^{-1}}
R
2
{\displaystyle \mathbb {R} ^{2}}
M
2
(
R
)
{\displaystyle \operatorname {M} _{2}(\mathbb {R} )}
M
2
(
R
)
{\displaystyle \operatorname {M} _{2}(\mathbb {R} )}
(
x
,
y
)
↦
|
|
|
J
φ
−
1
(
φ
(
x
,
y
)
)
|
|
|
{\displaystyle (x,y)\mapsto \left|\left|\left|J_{\varphi ^{-1}}(\varphi (x,y))\right|\right|\right|}
J
φ
−
1
:
φ
(
R
2
)
→
M
2
(
R
)
{\displaystyle J_{\varphi ^{-1}}:\varphi (\mathbb {R} ^{2})\to \operatorname {M} _{2}(\mathbb {R} )}
φ
(
R
2
)
{\displaystyle \varphi (\mathbb {R} ^{2})}
R
{\displaystyle \mathbb {R} }
Il reste à montrer que l'ensemble
φ
(
R
2
)
{\displaystyle \varphi (\mathbb {R} ^{2})}
R
2
{\displaystyle \mathbb {R} ^{2}}
connexité de
R
2
{\displaystyle \mathbb {R} ^{2}}
(
X
n
,
Y
n
)
{\displaystyle (X_{n},Y_{n})}
φ
(
R
2
)
{\displaystyle \varphi (\mathbb {R} ^{2})}
(
X
,
Y
)
{\displaystyle (X,Y)}
R
2
{\displaystyle \mathbb {R} ^{2}}
(
X
n
,
Y
n
)
{\displaystyle (X_{n},Y_{n})}
φ
−
1
{\displaystyle \varphi ^{-1}}
(
x
n
,
y
n
)
:=
φ
−
1
(
X
n
,
Y
n
)
{\displaystyle (x_{n},y_{n}):=\varphi ^{-1}(X_{n},Y_{n})}
(
x
,
y
)
{\displaystyle (x,y)}
R
2
{\displaystyle \mathbb {R} ^{2}}
(
X
n
,
Y
n
)
=
φ
(
x
n
,
y
n
)
→
φ
(
x
,
y
)
{\displaystyle (X_{n},Y_{n})=\varphi (x_{n},y_{n})\to \varphi (x,y)}
(
X
,
Y
)
=
φ
(
x
,
y
)
∈
φ
(
R
2
)
{\displaystyle (X,Y)=\varphi (x,y)\in \varphi (\mathbb {R} ^{2})}
p
=
φ
(
π
/
2
,
π
)
{\displaystyle p=\varphi (\pi /2,\pi )}
J
φ
−
1
(
p
)
=
(
J
φ
(
π
/
2
,
π
)
)
−
1
=
(
1
0
−
1
2
2
1
)
−
1
=
(
1
0
1
2
2
1
)
{\displaystyle J\varphi ^{-1}(p)=(J\varphi (\pi /2,\pi ))^{-1}={\begin{pmatrix}1&0\\-{\frac {1}{2{\sqrt {2}}}}&1\end{pmatrix}}^{-1}={\begin{pmatrix}1&0\\{\frac {1}{2{\sqrt {2}}}}&1\end{pmatrix}}}
Généralisation : soit
F
:
R
n
→
R
n
{\displaystyle F:\mathbb {R} ^{n}\to \mathbb {R} ^{n}}
C
>
0
{\displaystyle C>0}
x
,
y
∈
R
n
{\displaystyle x,y\in \mathbb {R} ^{n}}
(*)
‖
F
(
x
)
−
F
(
y
)
‖
≥
C
‖
x
−
y
‖
{\displaystyle \|F(x)-F(y)\|\geq C\|x-y\|}
Justifier que
F
{\displaystyle F}
Montrer que pour tous
y
,
e
∈
R
n
{\displaystyle y,e\in \mathbb {R} ^{n}}
‖
D
F
(
y
)
[
e
]
‖
≥
C
‖
e
‖
{\displaystyle \|DF(y)[e]\|\geq C\|e\|}
x
=
y
+
t
e
{\displaystyle x=y+te}
t
{\displaystyle t}
0
{\displaystyle 0}
Soit
a
∈
R
n
{\displaystyle a\in \mathbb {R} ^{n}}
lim
‖
x
‖
→
+
∞
‖
F
(
x
)
−
a
‖
=
+
∞
{\displaystyle \lim _{\|x\|\to +\infty }\|F(x)-a\|=+\infty }
En déduire que la fonction
g
:
x
↦
‖
F
(
x
)
−
a
‖
2
2
{\displaystyle g:x\mapsto \|F(x)-a\|_{2}^{2}}
R
n
{\displaystyle \mathbb {R} ^{n}}
x
a
∈
R
n
{\displaystyle x_{a}\in \mathbb {R} ^{n}}
F
(
x
a
)
=
a
{\displaystyle F(x_{a})=a}
Conclure que
F
{\displaystyle F}
R
n
{\displaystyle \mathbb {R} ^{n}}
R
n
{\displaystyle \mathbb {R} ^{n}}
Application : soit
f
:
R
→
R
{\displaystyle f:\mathbb {R} \to \mathbb {R} }
κ
∈
]
0
,
1
[
{\displaystyle \kappa \in \left]0,1\right[}
∀
t
∈
R
|
f
′
(
t
)
|
≤
κ
{\displaystyle \forall t\in \mathbb {R} \quad |f'(t)|\leq \kappa }
φ
:
R
2
→
R
2
,
(
x
,
y
)
↦
(
x
+
f
(
y
)
,
y
+
f
(
x
)
)
{\displaystyle \varphi :\mathbb {R} ^{2}\to \mathbb {R} ^{2},\;(x,y)\mapsto (x+f(y),y+f(x))}
|
f
(
t
)
−
f
(
s
)
|
≤
κ
|
t
−
s
|
{\displaystyle |f(t)-f(s)|\leq \kappa |t-s|}
⟨
φ
(
x
,
y
)
−
φ
(
u
,
v
)
,
(
x
,
y
)
−
(
u
,
v
)
⟩
≥
(
1
−
κ
)
‖
(
x
,
y
)
−
(
u
,
v
)
‖
2
2
{\displaystyle \langle \varphi (x,y)-\varphi (u,v),(x,y)-(u,v)\rangle \geq (1-\kappa )\|(x,y)-(u,v)\|_{2}^{2}}
‖
φ
(
x
,
y
)
−
φ
(
u
,
v
)
‖
2
≥
(
1
−
κ
)
‖
(
x
,
y
)
−
(
u
,
v
)
‖
2
{\displaystyle \|\varphi (x,y)-\varphi (u,v)\|_{2}\geq (1-\kappa )\|(x,y)-(u,v)\|_{2}}
φ
:
R
2
→
R
2
{\displaystyle \varphi :\mathbb {R} ^{2}\to \mathbb {R} ^{2}}
Solution
F
{\displaystyle F}
F
(
x
)
=
F
(
y
)
{\displaystyle F(x)=F(y)}
C
‖
x
−
y
‖
≤
0
{\displaystyle C\|x-y\|\leq 0}
x
=
y
{\displaystyle x=y}
En suivant l'indication, on obtient :
‖
F
(
y
+
t
e
)
−
F
(
y
)
‖
≥
C
|
t
|
‖
e
‖
{\displaystyle \|F(y+te)-F(y)\|\geq C|t|\|e\|}
|
t
|
{\displaystyle |t|}
t
≠
0
{\displaystyle t\neq 0}
t
→
0
{\displaystyle t\to 0}
‖
D
F
(
y
)
[
e
]
‖
≥
C
‖
e
‖
{\displaystyle \|DF(y)[e]\|\geq C\|e\|}
‖
F
(
x
)
−
a
‖
≥
‖
F
(
x
)
−
F
(
0
)
‖
−
‖
a
−
F
(
0
)
‖
≥
C
‖
x
‖
−
‖
a
−
F
(
0
)
‖
{\displaystyle \|F(x)-a\|\geq \|F(x)-F(0)\|-\|a-F(0)\|\geq C\|x\|-\|a-F(0)\|}
lim
‖
x
‖
→
+
∞
‖
F
(
x
)
−
a
‖
=
+
∞
{\displaystyle \lim _{\|x\|\to +\infty }\|F(x)-a\|=+\infty }
La fonction
g
:
x
↦
‖
F
(
x
)
−
a
‖
2
2
{\displaystyle g:x\mapsto \|F(x)-a\|_{2}^{2}}
exercice « Extrema d'une fonction continue » de la leçon sur les espaces vectoriels normés). En un tel point
x
a
{\displaystyle x_{a}}
x
{\displaystyle x}
e
{\displaystyle e}
⟨
∇
g
(
x
)
,
e
⟩
=
D
g
(
x
)
(
e
)
=
2
⟨
F
(
x
)
−
a
,
D
F
(
x
)
[
e
]
⟩
{\displaystyle \langle \nabla _{g}(x),e\rangle =D_{g}(x)(e)=2\langle F(x)-a,DF(x)[e]\rangle }
∇
g
(
x
a
)
{\displaystyle \nabla _{g}(x_{a})}
F
(
x
a
)
−
a
{\displaystyle F(x_{a})-a}
D
F
(
x
a
)
[
e
]
{\displaystyle DF(x_{a})[e]}
e
{\displaystyle e}
x
∈
R
n
{\displaystyle x\in \mathbb {R} ^{n}}
D
F
(
x
)
:
R
n
→
R
n
{\displaystyle DF(x):\mathbb {R} ^{n}\to \mathbb {R} ^{n}}
F
(
x
a
)
−
a
{\displaystyle F(x_{a})-a}
R
n
{\displaystyle \mathbb {R} ^{n}}
F
(
x
a
)
=
a
{\displaystyle F(x_{a})=a}
Pour tout
a
∈
R
n
{\displaystyle a\in \mathbb {R} ^{n}}
x
a
∈
R
n
{\displaystyle x_{a}\in \mathbb {R} ^{n}}
F
(
x
a
)
=
a
{\displaystyle F(x_{a})=a}
F
{\displaystyle F}
R
n
{\displaystyle \mathbb {R} ^{n}}
R
n
{\displaystyle \mathbb {R} ^{n}}
|
f
(
t
)
−
f
(
s
)
|
≤
κ
|
t
−
s
|
{\displaystyle |f(t)-f(s)|\leq \kappa |t-s|}
|
f
′
|
{\displaystyle |f'|}
⟨
φ
(
x
,
y
)
−
φ
(
u
,
v
)
,
(
x
,
y
)
−
(
u
,
v
)
⟩
=
(
x
−
u
)
2
+
(
y
−
v
)
2
+
(
f
(
y
)
−
f
(
v
)
)
(
x
−
u
)
+
(
f
(
x
)
−
f
(
u
)
)
(
y
−
v
)
≥
(
x
−
u
)
2
+
(
y
−
v
)
2
−
2
κ
|
x
−
u
|
|
y
−
v
|
≥
(
1
−
κ
)
‖
(
x
,
y
)
−
(
u
,
v
)
‖
2
2
.
{\displaystyle {\begin{aligned}\langle \varphi (x,y)-\varphi (u,v),(x,y)-(u,v)\rangle &=(x-u)^{2}+(y-v)^{2}+(f(y)-f(v))(x-u)+(f(x)-f(u))(y-v)\\&\geq (x-u)^{2}+(y-v)^{2}-2\kappa |x-u||y-v|\\&\geq (1-\kappa )\|(x,y)-(u,v)\|_{2}^{2}.\end{aligned}}}
Montrer que si
a
{\displaystyle a}
b
{\displaystyle b}
1
{\displaystyle 1}
x
,
y
∈
R
{\displaystyle x,y\in \mathbb {R} }
y
+
e
x
y
=
a
{\displaystyle y+\mathrm {e} ^{xy}=a}
x
+
e
−
x
y
=
b
{\displaystyle x+\mathrm {e} ^{-xy}=b}
Soit
f
:
R
2
→
R
2
,
(
x
,
y
)
↦
(
x
sin
(
x
y
)
+
y
,
y
cos
(
x
y
)
+
x
)
{\displaystyle f:\mathbb {R} ^{2}\to \mathbb {R} ^{2},\ \left(x,y\right)\mapsto \left(x\sin \left(xy\right)+y,y\cos \left(xy\right)+x\right)}
(
(
a
n
,
b
n
)
)
{\displaystyle \left(\left(a_{n},b_{n}\right)\right)}
(
0
,
0
)
{\displaystyle \left(0,0\right)}
f
(
a
n
,
b
n
)
=
(
0
,
0
)
{\displaystyle f\left(a_{n},b_{n}\right)=(0,0)}
n
{\displaystyle n}
(
(
a
n
,
b
n
)
)
{\displaystyle \left(\left(a_{n},b_{n}\right)\right)}
Solution
Considérons l'application
f
:
R
2
→
R
2
,
(
x
,
y
)
↦
(
y
+
e
x
y
,
x
+
e
−
x
y
)
{\displaystyle f:\mathbb {R} ^{2}\to \mathbb {R} ^{2},\ \left(x,y\right)\mapsto \left(y+\mathrm {e} ^{xy},x+\mathrm {e} ^{-xy}\right)}
∞ ). On a
f
(
0
,
0
)
=
(
1
,
1
)
{\displaystyle f\left(0,0\right)=\left(1,1\right)}
J
f
(
0
,
0
)
=
(
y
e
x
y
1
+
x
e
x
y
1
−
y
e
−
x
y
−
x
e
−
x
y
)
(
0
,
0
)
=
(
0
1
1
0
)
{\displaystyle Jf\left(0,0\right)={\begin{pmatrix}y\mathrm {e} ^{xy}&1+x\mathrm {e} ^{xy}\\1-y\mathrm {e} ^{-xy}&-x\mathrm {e} ^{-xy}\end{pmatrix}}\left(0,0\right)={\begin{pmatrix}0&1\\1&0\end{pmatrix}}}
Si
f
(
x
,
y
)
=
(
0
,
0
)
{\displaystyle f\left(x,y\right)=\left(0,0\right)}
|
y
|
=
|
x
sin
(
x
y
)
|
≤
|
x
2
y
|
{\displaystyle \left|y\right|=\left|x\sin \left(xy\right)\right|\leq \left|x^{2}y\right|}
|
x
|
=
|
y
cos
(
x
y
)
|
≤
|
y
|
{\displaystyle \left|x\right|=\left|y\cos \left(xy\right)\right|\leq \left|y\right|}
|
y
|
≤
|
y
3
|
{\displaystyle \left|y\right|\leq \left|y^{3}\right|}
|
y
|
<
1
{\displaystyle \left|y\right|<1}
y
=
0
{\displaystyle y=0}
x
=
0
{\displaystyle x=0}
f
(
0
,
0
)
=
(
0
,
0
)
{\displaystyle f\left(0,0\right)=\left(0,0\right)}
J
f
(
0
,
0
)
=
(
sin
(
x
y
)
+
x
y
cos
(
x
y
)
x
2
cos
(
x
y
)
+
1
−
y
2
sin
(
x
y
)
+
1
cos
(
x
y
)
−
x
y
sin
(
x
y
)
)
(
0
,
0
)
=
(
0
1
1
1
)
{\displaystyle Jf\left(0,0\right)={\begin{pmatrix}\sin(xy)+xy\cos(xy)&x^{2}\cos(xy)+1\\-y^{2}\sin(xy)+1&\cos(xy)-xy\sin(xy)\end{pmatrix}}\left(0,0\right)={\begin{pmatrix}0&1\\1&1\end{pmatrix}}}
(
x
,
y
)
{\displaystyle \left(x,y\right)}
(
0
,
0
)
{\displaystyle \left(0,0\right)}
(
0
,
0
)
{\displaystyle \left(0,0\right)}
(
0
,
0
)
{\displaystyle \left(0,0\right)}
f
(
x
,
y
)
{\displaystyle f\left(x,y\right)}
Donner l'allure de la courbe d'équation
x
4
+
y
3
−
x
2
−
y
2
+
x
−
y
=
0
{\displaystyle x^{4}+y^{3}-x^{2}-y^{2}+x-y=0}
(
0
,
0
)
{\displaystyle \left(0,0\right)}
(
1
,
1
)
{\displaystyle \left(1,1\right)}
Soit
f
(
x
,
y
)
=
x
3
+
y
3
−
3
x
y
−
1
{\displaystyle f\left(x,y\right)=x^{3}+y^{3}-3xy-1}
f
(
x
,
y
)
=
0
{\displaystyle f\left(x,y\right)=0}
(
0
,
1
)
{\displaystyle \left(0,1\right)}
y
=
ϕ
(
x
)
{\displaystyle y=\phi \left(x\right)}
∞ , et donner le développement limité à l'ordre
3
{\displaystyle 3}
ϕ
{\displaystyle \phi }
0
{\displaystyle 0}
Soit
g
(
x
,
y
)
=
2
e
x
+
y
−
1
+
ln
(
x
−
y
)
−
2
x
+
y
3
{\displaystyle g\left(x,y\right)=2\operatorname {e} ^{x+y-1}+\ln(x-y)-2x+y^{3}}
g
(
x
,
y
)
=
0
{\displaystyle g\left(x,y\right)=0}
(
1
,
0
)
{\displaystyle \left(1,0\right)}
y
=
ϕ
(
x
)
{\displaystyle y=\phi \left(x\right)}
∞ , et donner le développement limité à l'ordre
3
{\displaystyle 3}
ϕ
{\displaystyle \phi }
1
{\displaystyle 1}
Soit
h
(
x
,
y
,
z
)
=
x
3
+
y
3
+
z
3
−
2
z
(
x
+
y
)
−
2
x
+
y
−
2
z
−
1
{\displaystyle h\left(x,y,z\right)=x^{3}+y^{3}+z^{3}-2z\left(x+y\right)-2x+y-2z-1}
h
(
x
,
y
,
z
)
=
0
{\displaystyle h\left(x,y,z\right)=0}
(
0
,
0
,
−
1
)
{\displaystyle \left(0,0,-1\right)}
z
=
ϕ
(
x
,
y
)
{\displaystyle z=\phi \left(x,y\right)}
∞ , et donner le développement limité à l'ordre
2
{\displaystyle 2}
ϕ
{\displaystyle \phi }
(
0
,
0
)
{\displaystyle \left(0,0\right)}
Solution
Soit
f
(
x
,
y
)
:=
x
4
+
y
3
−
x
2
−
y
2
+
x
−
y
{\displaystyle f\left(x,y\right):=x^{4}+y^{3}-x^{2}-y^{2}+x-y}
∂
f
∂
x
(
x
,
y
)
=
4
x
3
−
2
x
+
1
{\displaystyle {\frac {\partial f}{\partial x}}\left(x,y\right)=4x^{3}-2x+1}
∂
f
∂
y
(
x
,
y
)
=
3
y
2
−
2
y
−
1
{\displaystyle {\frac {\partial f}{\partial y}}\left(x,y\right)=3y^{2}-2y-1}
Les deux points sont réguliers :
∂
f
∂
x
(
0
,
0
)
=
1
{\displaystyle {\frac {\partial f}{\partial x}}\left(0,0\right)=1}
∂
f
∂
y
(
0
,
0
)
=
−
1
{\displaystyle {\frac {\partial f}{\partial y}}\left(0,0\right)=-1}
y
=
x
+
a
x
2
+
o
(
x
2
)
{\displaystyle y=x+ax^{2}+o\left(x^{2}\right)}
0
=
−
x
2
−
x
2
+
x
−
(
x
+
a
x
2
)
+
o
(
x
2
)
{\displaystyle 0=-x^{2}-x^{2}+{\cancel {x}}-\left({\cancel {x}}+ax^{2}\right)+o\left(x^{2}\right)}
donc
a
=
−
2
{\displaystyle a=-2}
1
{\displaystyle 1}
∂
f
∂
x
(
1
,
1
)
=
3
{\displaystyle {\frac {\partial f}{\partial x}}\left(1,1\right)=3}
∂
f
∂
y
(
1
,
1
)
=
0
{\displaystyle {\frac {\partial f}{\partial y}}\left(1,1\right)=0}
y
=
1
+
h
{\displaystyle y=1+h}
x
=
1
+
a
h
2
+
o
(
h
2
)
{\displaystyle x=1+ah^{2}+o\left(h^{2}\right)}
0
=
(
1
+
a
h
2
)
4
+
(
1
+
h
)
3
−
(
1
+
a
h
2
)
2
−
(
1
+
h
)
2
+
(
1
+
a
h
2
)
−
(
1
+
h
)
+
o
(
h
2
)
=
1
+
4
a
h
2
+
1
+
3
h
+
3
h
2
−
1
−
2
a
h
2
−
1
−
2
h
−
h
2
+
1
+
a
h
2
−
1
−
h
+
o
(
h
2
)
=
h
2
(
4
a
+
3
−
2
a
−
1
+
a
)
+
o
(
h
2
)
{\displaystyle {\begin{aligned}0&=\left(1+ah^{2}\right)^{4}+\left(1+h\right)^{3}-\left(1+ah^{2}\right)^{2}-\left(1+h\right)^{2}+\left(1+ah^{2}\right)-\left(1+h\right)+o\left(h^{2}\right)\\&={\cancel {1}}+4ah^{2}+{\cancel {1}}+3h+3h^{2}-{\cancel {1}}-2ah^{2}-{\cancel {1}}-2h-h^{2}+{\cancel {1}}+ah^{2}-{\cancel {1}}-h+o\left(h^{2}\right)\\&=h^{2}\left(4a+3-2a-1+a\right)+o\left(h^{2}\right)\end{aligned}}}
donc
a
=
−
2
/
3
{\displaystyle a=-2/3}
f
{\displaystyle f}
∞ et
∂
f
∂
y
(
x
,
y
)
=
3
y
2
−
3
x
{\displaystyle {\frac {\partial f}{\partial y}}\left(x,y\right)=3y^{2}-3x}
f
(
0
,
1
)
=
0
{\displaystyle f\left(0,1\right)=0}
∂
f
∂
y
(
0
,
1
)
=
3
≠
0
{\displaystyle {\frac {\partial f}{\partial y}}\left(0,1\right)=3\neq 0}
ϕ
{\displaystyle \phi }
Soit
ϕ
(
x
)
=
1
+
a
x
+
b
x
2
+
c
x
3
+
o
(
x
3
)
{\displaystyle \phi \left(x\right)=1+ax+bx^{2}+cx^{3}+o\left(x^{3}\right)}
3
{\displaystyle 3}
0
{\displaystyle 0}
0
=
x
3
+
(
1
+
a
x
+
b
x
2
+
c
x
3
)
3
−
3
x
(
1
+
a
x
+
b
x
2
+
c
x
3
)
−
1
+
o
(
x
3
)
=
x
3
+
1
+
3
a
x
+
3
b
x
2
+
3
c
x
3
+
3
a
2
x
2
+
6
a
b
x
3
+
a
3
x
3
−
3
x
(
1
+
a
x
+
b
x
2
)
−
1
+
o
(
x
3
)
=
x
(
3
a
−
3
)
+
x
2
(
3
b
+
3
a
2
−
3
a
)
+
x
3
(
1
+
3
c
+
6
a
b
+
a
3
−
3
b
)
+
o
(
x
3
)
{\displaystyle {\begin{aligned}0&=x^{3}+\left(1+ax+bx^{2}+cx^{3}\right)^{3}-3x\left(1+ax+bx^{2}+cx^{3}\right)-1+o\left(x^{3}\right)\\&=x^{3}+{\cancel {1}}+3ax+3bx^{2}+3cx^{3}+3a^{2}x^{2}+6abx^{3}+a^{3}x^{3}-3x\left(1+ax+bx^{2}\right)-{\cancel {1}}+o\left(x^{3}\right)\\&=x\left(3a-3\right)+x^{2}\left(3b+3a^{2}-3a\right)+x^{3}\left(1+3c+6ab+a^{3}-3b\right)+o\left(x^{3}\right)\end{aligned}}}
donc
a
=
1
{\displaystyle a=1}
b
=
0
{\displaystyle b=0}
c
=
−
2
/
3
{\displaystyle c=-2/3}
g
{\displaystyle g}
∞ et
∂
g
∂
y
(
x
,
y
)
=
2
e
x
+
y
−
1
−
1
x
−
y
+
3
y
2
{\displaystyle {\frac {\partial g}{\partial y}}\left(x,y\right)=2\operatorname {e} ^{x+y-1}-{\frac {1}{x-y}}+3y^{2}}
g
(
1
,
0
)
=
0
{\displaystyle g\left(1,0\right)=0}
∂
g
∂
y
(
1
,
0
)
=
1
≠
0
{\displaystyle {\frac {\partial g}{\partial y}}\left(1,0\right)=1\neq 0}
ϕ
{\displaystyle \phi }
Soit
ϕ
(
1
+
u
)
=
a
u
+
b
u
2
+
c
u
3
+
o
(
u
3
)
{\displaystyle \phi \left(1+u\right)=au+bu^{2}+cu^{3}+o\left(u^{3}\right)}
3
{\displaystyle 3}
1
{\displaystyle 1}
0
=
2
e
(
1
+
a
)
u
+
b
u
2
+
c
u
3
+
ln
(
1
+
(
1
−
a
)
u
−
b
u
2
−
c
u
3
)
−
2
−
2
u
+
(
a
u
)
3
+
o
(
u
3
)
=
2
+
2
(
1
+
a
)
u
+
2
b
u
2
+
2
c
u
3
+
(
1
+
a
)
2
u
2
+
2
(
1
+
a
)
b
u
3
+
(
1
+
a
)
3
u
3
/
3
+
(
1
−
a
)
u
−
b
u
2
−
c
u
3
−
(
1
−
a
)
2
u
2
/
2
+
(
1
−
a
)
b
u
3
+
(
1
−
a
)
3
u
3
/
3
−
2
−
2
u
+
a
3
u
3
+
o
(
u
3
)
=
(
1
+
a
)
u
+
u
2
[
b
+
1
/
2
+
3
a
+
a
2
/
2
]
+
u
3
[
c
+
3
b
+
a
b
+
2
/
3
+
2
a
2
+
a
3
]
+
o
(
u
3
)
{\displaystyle {\begin{aligned}0&=2\operatorname {e} ^{(1+a)u+bu^{2}+cu^{3}}+\ln \left(1+(1-a)u-bu^{2}-cu^{3}\right)-2-2u+\left(au\right)^{3}+o\left(u^{3}\right)\\&={\cancel {2}}+2(1+a)u+2bu^{2}+2cu^{3}+(1+a)^{2}u^{2}+2(1+a)bu^{3}+(1+a)^{3}u^{3}/3\\&\quad +(1-a)u-bu^{2}-cu^{3}-(1-a)^{2}u^{2}/2+(1-a)bu^{3}+(1-a)^{3}u^{3}/3-{\cancel {2}}-2u+a^{3}u^{3}+o\left(u^{3}\right)\\&=(1+a)u+u^{2}\left[b+1/2+3a+a^{2}/2\right]+u^{3}\left[c+3b+ab+2/3+2a^{2}+a^{3}\right]+o\left(u^{3}\right)\end{aligned}}}
donc
a
=
−
1
{\displaystyle a=-1}
b
=
2
{\displaystyle b=2}
c
=
−
17
/
3
{\displaystyle c=-17/3}
h
{\displaystyle h}
∞ et
∂
h
∂
z
(
x
,
y
,
z
)
=
3
z
2
−
2
(
x
+
y
)
−
2
{\displaystyle {\frac {\partial h}{\partial z}}\left(x,y,z\right)=3z^{2}-2\left(x+y\right)-2}
h
(
0
,
0
,
−
1
)
=
0
{\displaystyle h\left(0,0,-1\right)=0}
∂
h
∂
z
(
0
,
0
,
−
1
)
=
1
≠
0
{\displaystyle {\frac {\partial h}{\partial z}}\left(0,0,-1\right)=1\neq 0}
ϕ
{\displaystyle \phi }
Soit
ϕ
(
x
,
y
)
=
−
1
+
a
x
+
b
y
+
c
x
2
+
d
x
y
+
e
y
2
+
o
(
x
2
+
y
2
)
{\displaystyle \phi \left(x,y\right)=-1+ax+by+cx^{2}+dxy+ey^{2}+o\left(x^{2}+y^{2}\right)}
2
{\displaystyle 2}
(
0
,
0
)
{\displaystyle \left(0,0\right)}
0
=
(
−
1
+
a
x
+
b
y
+
c
x
2
+
d
x
y
+
e
y
2
)
3
−
2
(
−
1
+
a
x
+
b
y
)
(
x
+
y
)
−
2
x
+
y
−
2
(
−
1
+
a
x
+
b
y
+
c
x
2
+
d
x
y
+
e
y
2
)
−
1
+
o
(
x
2
+
y
2
)
=
−
1
+
3
a
x
+
3
b
y
+
3
c
x
2
+
3
d
x
y
+
3
e
y
2
−
3
(
a
2
x
2
+
2
a
b
x
y
+
b
2
y
2
)
−
2
(
−
x
+
a
x
2
+
b
x
y
−
y
+
a
x
y
+
b
y
2
)
−
2
x
+
y
−
2
(
−
1
+
a
x
+
b
y
+
c
x
2
+
d
x
y
+
e
y
2
)
−
1
+
o
(
x
2
+
y
2
)
=
x
(
3
a
+
2
−
2
−
2
a
)
+
y
(
3
b
+
2
+
1
−
2
b
)
+
x
2
(
3
c
−
3
a
2
−
2
a
−
2
c
)
+
x
y
(
3
d
−
6
a
b
−
2
b
+
2
a
−
2
d
)
+
y
2
(
3
e
−
3
b
2
−
2
b
−
2
e
)
+
o
(
x
2
+
y
2
)
{\displaystyle {\begin{aligned}0&=\left(-1+ax+by+cx^{2}+dxy+ey^{2}\right)^{3}-2\left(-1+ax+by\right)\left(x+y\right)-2x+y-2\left(-1+ax+by+cx^{2}+dxy+ey^{2}\right)-1+o\left(x^{2}+y^{2}\right)\\&=-{\cancel {1}}+3ax+3by+3cx^{2}+3dxy+3ey^{2}-3\left(a^{2}x^{2}+2abxy+b^{2}y^{2}\right)-2\left(-x+ax^{2}+bxy-y+axy+by^{2}\right)-2x+y-2\left(-{\cancel {1}}+ax+by+cx^{2}+dxy+ey^{2}\right)-{\cancel {1}}+o\left(x^{2}+y^{2}\right)\\&=x\left(3a+2-2-2a\right)+y\left(3b+2+1-2b\right)+x^{2}\left(3c-3a^{2}-2a-2c\right)+xy\left(3d-6ab-2b+2a-2d\right)+y^{2}\left(3e-3b^{2}-2b-2e\right)+o\left(x^{2}+y^{2}\right)\end{aligned}}}
donc
a
=
0
{\displaystyle a=0}
b
=
−
3
{\displaystyle b=-3}
c
=
0
{\displaystyle c=0}
d
=
−
6
{\displaystyle d=-6}
e
=
21
{\displaystyle e=21}
Soient
P
0
{\displaystyle P_{0}}
d
{\displaystyle d}
α
0
{\displaystyle \alpha _{0}}
P
0
{\displaystyle P_{0}}
R
d
[
X
]
×
R
→
R
,
(
P
,
α
)
↦
P
(
α
)
{\displaystyle \mathbb {R} _{d}\left[X\right]\times \mathbb {R} \to \mathbb {R} ,\ \left(P,\alpha \right)\mapsto P\left(\alpha \right)}
U
×
V
{\displaystyle U\times V}
(
P
0
,
α
0
)
{\displaystyle \left(P_{0},\alpha _{0}\right)}
P
∈
U
{\displaystyle P\in U}
α
{\displaystyle \alpha }
V
{\displaystyle V}
P
↦
α
{\displaystyle P\mapsto \alpha }
∞ .
Montrer que si
P
0
{\displaystyle P_{0}}
k
{\displaystyle k}
U
{\displaystyle U}
P
0
{\displaystyle P_{0}}
k
{\displaystyle k}
∞
ϕ
1
,
…
,
ϕ
k
:
U
→
R
{\displaystyle \phi _{1},\dots ,\phi _{k}:U\to \mathbb {R} }
P
∈
U
{\displaystyle P\in U}
ϕ
1
(
P
)
,
…
,
ϕ
k
(
P
)
{\displaystyle \phi _{1}\left(P\right),\dots ,\phi _{k}\left(P\right)}
P
{\displaystyle P}
Que se passe-t-il pour les racines multiples ?
Solution
R
d
[
X
]
{\displaystyle \mathbb {R} _{d}\left[X\right]}
R
d
+
1
{\displaystyle \mathbb {R} ^{d+1}}
∂
P
(
α
)
∂
α
=
P
′
(
α
)
{\displaystyle {\frac {\partial {P\left(\alpha \right)}}{\partial \alpha }}=P'\left(\alpha \right)}
(
P
,
α
)
{\displaystyle \left(P,\alpha \right)}
P
′
(
α
)
≠
0
{\displaystyle P'\left(\alpha \right)\neq 0}
(
P
0
,
α
0
)
{\displaystyle \left(P_{0},\alpha _{0}\right)}
∞ car polynomiale).On considère la fonction
R
d
[
X
]
×
R
k
→
R
k
,
(
P
,
α
1
,
…
,
α
k
)
↦
(
P
(
α
1
)
,
…
,
P
(
α
k
)
)
{\displaystyle \mathbb {R} _{d}\left[X\right]\times \mathbb {R} ^{k}\to \mathbb {R} ^{k},\ \left(P,\alpha _{1},\dots ,\alpha _{k}\right)\mapsto \left(P\left(\alpha _{1}\right),\dots ,P\left(\alpha _{k}\right)\right)}
∂
(
P
(
α
1
)
,
…
,
P
(
α
k
)
)
∂
(
α
1
,
…
,
α
k
)
=
diag
(
P
′
(
α
1
)
,
…
,
P
′
(
α
k
)
)
{\displaystyle {\frac {\partial {\left(P\left(\alpha _{1}\right),\dots ,P\left(\alpha _{k}\right)\right)}}{\partial {\left(\alpha _{1},\dots ,\alpha _{k}\right)}}}=\operatorname {diag} {\left(P'\left(\alpha _{1}\right),\dots ,P'\left(\alpha _{k}\right)\right)}}
(
P
,
α
1
,
…
,
α
k
)
{\displaystyle \left(P,\alpha _{1},\dots ,\alpha _{k}\right)}
α
1
,
…
,
α
k
{\displaystyle \alpha _{1},\dots ,\alpha _{k}}
P
′
(
α
i
)
{\displaystyle P'\left(\alpha _{i}\right)}
Une racine mutiple
α
0
{\displaystyle \alpha _{0}}
P
0
{\displaystyle P_{0}}
P
{\displaystyle P}
P
0
{\displaystyle P_{0}}
α
0
{\displaystyle \alpha _{0}}
P
0
(
X
)
=
X
2
{\displaystyle P_{0}(X)=X^{2}}
Soient
a
,
b
∈
R
{\displaystyle a,b\in \mathbb {R} }
0
<
|
a
b
|
≤
1
{\displaystyle 0<|ab|\leq 1}
l'application
f
:
(
x
,
y
)
↦
(
x
+
a
sin
y
,
y
+
b
sin
x
)
{\displaystyle f:(x,y)\mapsto (x+a\sin y,y+b\sin x)}
toute suite dans
R
2
{\displaystyle \mathbb {R} ^{2}}
f
{\displaystyle f}
l'ensemble
F
:=
f
(
R
2
)
{\displaystyle F:=f(\mathbb {R} ^{2})}
f
−
1
:
F
→
R
2
{\displaystyle f^{-1}:F\to \mathbb {R} ^{2}}
f
{\displaystyle f}
R
2
{\displaystyle \mathbb {R} ^{2}}
|
a
b
|
<
1
{\displaystyle |ab|<1}
si
|
a
b
|
=
1
{\displaystyle |ab|=1}
f
{\displaystyle f}
R
2
{\displaystyle \mathbb {R} ^{2}}
Solution
Supposons que
f
(
x
1
,
y
1
)
=
f
(
x
2
,
y
2
)
{\displaystyle f\left(x_{1},y_{1}\right)=f\left(x_{2},y_{2}\right)}
|
x
1
−
x
2
|
=
|
a
|
|
sin
y
2
−
sin
y
1
|
≤
|
a
|
|
y
2
−
y
1
|
{\displaystyle |x_{1}-x_{2}|=|a||\sin y_{2}-\sin y_{1}|\leq |a||y_{2}-y_{1}|}
|
y
2
−
y
1
|
≤
|
b
|
|
x
1
−
x
2
|
{\displaystyle |y_{2}-y_{1}|\leq |b||x_{1}-x_{2}|}
|
a
b
|
{\displaystyle |ab|}
|
y
2
−
y
1
|
=
|
sin
y
2
−
sin
y
1
|
{\displaystyle |y_{2}-y_{1}|=|\sin y_{2}-\sin y_{1}|}
|
∫
y
1
y
2
d
t
|
=
|
∫
y
1
y
2
cos
t
d
t
|
≤
∫
y
1
y
2
|
cos
t
|
d
t
{\displaystyle \left|\int _{y_{1}}^{y_{2}}\mathrm {d} t\right|=\left|\int _{y_{1}}^{y_{2}}\cos t\,\mathrm {d} t\right|\leq \int _{y_{1}}^{y_{2}}|\cos t|\,\mathrm {d} t}
y
1
=
y
2
{\displaystyle y_{1}=y_{2}}
x
1
=
x
2
{\displaystyle x_{1}=x_{2}}
Soit
X
n
=
(
x
n
,
y
n
)
=
X
n
{\displaystyle X_{n}=(x_{n},y_{n})=X_{n}}
(
x
n
′
,
y
n
′
)
=
X
n
′
:=
f
(
X
n
)
→
X
′
{\displaystyle (x'_{n},y'_{n})=X'_{n}:=f(X_{n})\to X'}
(
X
n
)
{\displaystyle (X_{n})}
|
x
n
|
≤
|
x
n
′
|
+
|
a
|
{\displaystyle |x_{n}|\leq |x'_{n}|+|a|}
|
y
n
|
≤
|
y
n
′
|
+
|
b
|
{\displaystyle |y_{n}|\leq |y'_{n}|+|b|}
X
′
{\displaystyle X'}
f
{\displaystyle f}
(
X
n
)
{\displaystyle (X_{n})}
Résulte immédiatement des deux questions précédentes.
det
J
f
(
x
,
y
)
=
|
1
a
cos
y
b
cos
x
1
|
=
1
−
a
b
cos
x
cos
y
{\displaystyle \det Jf\left(x,y\right)={\begin{vmatrix}1&a\cos y\\b\cos x&1\end{vmatrix}}=1-ab\cos x\cos y}
(
x
,
y
)
∈
R
2
{\displaystyle (x,y)\in \mathbb {R} ^{2}}
|
a
b
|
<
1
{\displaystyle |ab|<1}
f
{\displaystyle f}
R
2
{\displaystyle \mathbb {R} ^{2}}
det
J
f
(
x
,
y
,
z
)
{\displaystyle \det Jf\left(x,y,z\right)}
f
(
R
2
)
{\displaystyle f(\mathbb {R} ^{2})}
R
2
{\displaystyle \mathbb {R} ^{2}}
R
2
{\displaystyle \mathbb {R} ^{2}}
f
:
R
2
→
R
2
{\displaystyle f:\mathbb {R} ^{2}\to \mathbb {R} ^{2}}
si
|
a
b
|
=
1
{\displaystyle |ab|=1}
f
{\displaystyle f}
(
x
′
,
y
′
)
∈
R
2
{\displaystyle (x',y')\in \mathbb {R} ^{2}}
(
a
n
,
b
n
)
{\displaystyle (a_{n},b_{n})}
(
a
,
b
)
{\displaystyle (a,b)}
|
a
n
b
n
|
<
1
{\displaystyle |a_{n}b_{n}|<1}
(
x
n
,
y
n
)
∈
R
2
{\displaystyle (x_{n},y_{n})\in \mathbb {R} ^{2}}
(
x
′
,
y
′
)
=
(
x
n
+
a
n
sin
y
n
,
y
n
+
b
n
sin
x
n
)
{\displaystyle (x',y')=(x_{n}+a_{n}\sin y_{n},y_{n}+b_{n}\sin x_{n})}
(
x
n
,
y
n
)
{\displaystyle (x_{n},y_{n})}
(
x
,
y
)
{\displaystyle (x,y)}
f
(
x
,
y
)
=
(
x
′
,
y
′
)
{\displaystyle f(x,y)=(x',y')}
Soit
f
:
R
3
→
R
2
{\displaystyle f:\mathbb {R} ^{3}\to \mathbb {R} ^{2}}
f
(
x
,
y
,
z
)
=
(
x
2
−
y
2
+
z
2
−
1
,
x
y
z
−
1
)
{\displaystyle f(x,y,z)=(x^{2}-y^{2}+z^{2}-1,xyz-1)}
Soit
(
x
0
,
y
0
,
z
0
)
∈
R
3
{\displaystyle (x_{0},y_{0},z_{0})\in \mathbb {R} ^{3}}
f
(
x
0
,
y
0
,
z
0
)
=
(
0
,
0
)
{\displaystyle f(x_{0},y_{0},z_{0})=(0,0)}
Montrez que dans un voisinage de
(
x
0
,
y
0
,
z
0
)
{\displaystyle \left(x_{0},y_{0},z_{0}\right)}
f
(
x
,
y
,
z
)
=
(
0
,
0
)
{\displaystyle f\left(x,y,z\right)=(0,0)}
I
{\displaystyle I}
x
0
{\displaystyle x_{0}}
φ
:
I
→
R
2
{\displaystyle \varphi :I\to \mathbb {R} ^{2}}
∞ .
Solution
J
f
(
x
,
y
,
z
)
=
(
2
x
−
2
y
2
z
y
z
x
z
x
y
)
{\displaystyle Jf\left(x,y,z\right)={\begin{pmatrix}2x&-2y&2z\\yz&xz&xy\end{pmatrix}}}
|
2
x
−
2
y
y
z
x
z
|
=
2
z
(
x
2
+
y
2
)
≠
0
{\displaystyle {\begin{vmatrix}2x&-2y\\yz&xz\end{vmatrix}}=2z\left(x^{2}+y^{2}\right)\neq 0}
(
x
,
y
,
z
)
{\displaystyle \left(x,y,z\right)}
(
x
0
,
y
0
,
z
0
)
{\displaystyle \left(x_{0},y_{0},z_{0}\right)}
x
0
y
0
z
0
=
1
≠
0
{\displaystyle x_{0}y_{0}z_{0}=1\neq 0}
(
x
,
y
,
z
)
{\displaystyle \left(x,y,z\right)}
f
(
x
,
y
,
z
)
=
(
0
,
0
)
{\displaystyle f\left(x,y,z\right)=(0,0)}
(
x
0
,
y
0
,
z
0
)
{\displaystyle \left(x_{0},y_{0},z_{0}\right)}
(
x
,
y
)
{\displaystyle \left(x,y\right)}
z
{\displaystyle z}
φ
{\displaystyle \varphi }
f
{\displaystyle f}
∞ .
Wikipedia-logo-v2.svg Soient
f
:
R
2
→
R
{\displaystyle f:\mathbb {R} ^{2}\to \mathbb {R} }
f
(
x
,
y
)
=
x
3
+
y
3
−
3
x
y
{\displaystyle f(x,y)=x^{3}+y^{3}-3xy}
C
{\displaystyle C}
(
x
,
y
)
∈
R
2
{\displaystyle (x,y)\in \mathbb {R} ^{2}}
f
(
x
,
y
)
=
0
{\displaystyle f(x,y)=0}
Au voisinage de quels points la relation
f
(
x
,
y
)
=
0
{\displaystyle f(x,y)=0}
y
{\displaystyle y}
x
{\displaystyle x}
x
{\displaystyle x}
y
{\displaystyle y}
Calculer la dérivée de la fonction implicite lorsqu'elle existe et écrire l'équation de la tangente à
C
{\displaystyle C}
Montrer que l'équation
e
x
+
e
y
+
x
+
y
−
2
=
0
{\displaystyle \operatorname {e} ^{x}+\operatorname {e} ^{y}+x+y-2=0}
0
{\displaystyle 0}
φ
{\displaystyle \varphi }
x
{\displaystyle x}
3
{\displaystyle 3}
0
{\displaystyle 0}
Montrer que les équations
x
+
y
−
z
t
=
0
,
x
y
−
z
+
t
=
0
{\displaystyle x+y-zt=0,\ xy-z+t=0}
(
0
,
1
)
{\displaystyle (0,1)}
x
=
φ
1
(
z
,
t
)
,
y
=
φ
2
(
z
,
t
)
{\displaystyle x=\varphi _{1}(z,t),\ y=\varphi _{2}(z,t)}
φ
1
(
0
,
1
)
=
1
{\displaystyle \varphi _{1}(0,1)=1}
Solution
y
{\displaystyle y}
x
{\displaystyle x}
0
≠
∂
f
∂
y
(
x
,
y
)
=
3
(
y
2
−
x
)
{\displaystyle 0\neq {\frac {\partial f}{\partial y}}(x,y)=3(y^{2}-x)}
x
≠
y
2
{\displaystyle x\neq y^{2}}
x
{\displaystyle x}
y
{\displaystyle y}
y
≠
x
2
{\displaystyle y\neq x^{2}}
Au voisinage d'un couple
(
a
,
b
)
{\displaystyle (a,b)}
a
≠
b
2
{\displaystyle a\neq b^{2}}
f
(
x
,
y
)
=
0
⇔
y
=
φ
(
x
)
{\displaystyle f(x,y)=0\Leftrightarrow y=\varphi (x)}
φ
′
(
x
)
=
−
x
2
−
φ
(
x
)
φ
(
x
)
2
−
x
{\displaystyle \varphi '(x)=-{\frac {x^{2}-\varphi (x)}{\varphi (x)^{2}-x}}}
x
{\displaystyle x}
y
{\displaystyle y}
C
{\displaystyle C}
(
a
,
b
)
≠
(
0
,
0
)
{\displaystyle (a,b)\neq (0,0)}
(
a
2
−
b
)
(
x
−
a
)
+
(
b
2
−
a
)
(
y
−
b
)
=
0
{\displaystyle (a^{2}-b)(x-a)+(b^{2}-a)(y-b)=0}
(
a
2
−
b
)
x
+
(
b
2
−
a
)
y
+
a
b
=
0
{\displaystyle (a^{2}-b)x+(b^{2}-a)y+ab=0}
Remarques : Toute équation cubique se ramène (par translation de la variable) à la forme réduite
x
3
+
p
x
+
q
=
0
{\displaystyle x^{3}+px+q=0}
p
,
q
∈
R
{\displaystyle p,q\in \mathbb {R} }
Δ
=
−
4
p
3
−
27
q
2
{\displaystyle \Delta =-4p^{3}-27q^{2}}
Δ
>
0
{\displaystyle \Delta >0}
Δ
<
0
{\displaystyle \Delta <0}
Δ
=
0
{\displaystyle \Delta =0}
p
,
q
≠
0
{\displaystyle p,q\neq 0}
0
{\displaystyle 0}
p
=
q
=
0
{\displaystyle p=q=0}
p
=
−
3
y
{\displaystyle p=-3y}
q
=
y
3
{\displaystyle q=y^{3}}
Δ
=
27
y
3
(
4
−
y
3
)
{\displaystyle \Delta =27y^{3}(4-y^{3})}
y
<
0
{\displaystyle y<0}
y
>
4
3
{\displaystyle y>{\sqrt[{3}]{4}}}
x
{\displaystyle x}
f
(
x
,
y
)
=
0
{\displaystyle f(x,y)=0}
0
<
y
<
4
3
{\displaystyle 0<y<{\sqrt[{3}]{4}}}
∞ de
y
{\displaystyle y}
x
{\displaystyle x}
y
{\displaystyle y}
(
a
,
b
)
{\displaystyle (a,b)}
b
≠
a
2
{\displaystyle b\neq a^{2}}
b
=
a
2
{\displaystyle b=a^{2}}
a
3
+
b
3
−
3
a
b
=
a
3
(
a
3
−
2
)
{\displaystyle a^{3}+b^{3}-3ab=a^{3}(a^{3}-2)}
(
a
,
b
)
=
(
0
,
0
)
{\displaystyle (a,b)=(0,0)}
(
2
3
,
4
3
)
{\displaystyle ({\sqrt[{3}]{2}},{\sqrt[{3}]{4}})}
Au point
(
0
,
0
)
{\displaystyle (0,0)}
x
(
y
)
{\displaystyle x(y)}
y
>
0
{\displaystyle y>0}
y
=
0
{\displaystyle y=0}
x
=
0
{\displaystyle x=0}
Au point
(
2
3
,
4
3
)
{\displaystyle ({\sqrt[{3}]{2}},{\sqrt[{3}]{4}})}
x
(
y
)
{\displaystyle x(y)}
y
<
4
3
{\displaystyle y<{\sqrt[{3}]{4}}}
y
=
4
3
{\displaystyle y={\sqrt[{3}]{4}}}
a
≠
b
2
{\displaystyle a\neq b^{2}}
y
=
φ
(
x
)
{\displaystyle y=\varphi (x)}
b
=
a
2
{\displaystyle b=a^{2}}
φ
′
(
a
)
=
0
{\displaystyle \varphi '(a)=0}
∂
(
e
x
+
e
y
+
x
+
y
−
2
)
∂
y
=
e
y
+
1
≠
0
{\displaystyle {\frac {\partial (\operatorname {e} ^{x}+\operatorname {e} ^{y}+x+y-2)}{\partial y}}=\operatorname {e} ^{y}+1\neq 0}
φ
{\displaystyle \varphi }
∞ ). Soit
φ
(
x
)
=
a
+
b
x
+
c
x
2
+
d
x
3
+
o
(
x
3
)
{\displaystyle \varphi (x)=a+bx+cx^{2}+dx^{3}+o(x^{3})}
3
{\displaystyle 3}
0
{\displaystyle 0}
0
=
1
+
x
+
x
2
2
+
x
3
6
+
e
a
(
1
+
b
x
+
c
x
2
+
d
x
3
+
1
2
(
b
2
x
2
+
2
b
c
x
3
)
+
b
3
6
x
3
)
+
x
+
a
+
b
x
+
c
x
2
+
d
x
3
−
2
{\displaystyle 0=1+x+{\frac {x^{2}}{2}}+{\frac {x^{3}}{6}}+\operatorname {e} ^{a}\left(1+bx+cx^{2}+dx^{3}+{\frac {1}{2}}(b^{2}x^{2}+2bcx^{3})+{\frac {b^{3}}{6}}x^{3}\right)+x+a+bx+cx^{2}+dx^{3}-2}
a
=
0
,
b
=
−
1
,
c
=
−
1
/
2
,
d
=
−
1
/
4
{\displaystyle a=0,\quad b=-1,\quad c=-1/2,\quad d=-1/4}
Remarque : alternativement, le d.l. de la fonction implicite
φ
{\displaystyle \varphi }
k
=
1
,
2
,
…
{\displaystyle k=1,2,\dots }
φ
{\displaystyle \varphi }
soit en dérivant
k
{\displaystyle k}
f
(
x
,
φ
(
x
)
)
=
0
{\displaystyle f(x,\varphi (x))=0}
k
{\displaystyle k}
φ
{\displaystyle \varphi }
f
{\displaystyle f}
φ
′
(
x
)
=
−
∂
f
∂
x
(
x
,
φ
(
x
)
)
∂
f
∂
y
(
x
,
φ
(
x
)
)
,
{\displaystyle \varphi '(x)=-{\frac {{\frac {\partial f}{\partial x}}(x,\varphi (x))}{{\frac {\partial f}{\partial y}}(x,\varphi (x))}},}
φ
″
(
x
)
=
−
∂
2
f
∂
x
2
(
x
,
φ
(
x
)
)
+
2
∂
2
f
∂
x
∂
y
(
x
,
φ
(
x
)
)
φ
′
(
x
)
+
∂
2
f
∂
y
2
(
x
,
φ
(
x
)
)
φ
′
2
(
x
)
∂
f
∂
y
(
x
,
φ
(
x
)
)
{\displaystyle \varphi ''(x)=-{\frac {{\frac {\partial ^{2}f}{\partial x^{2}}}\left(x,\varphi (x)\right)+2{\frac {\partial ^{2}f}{\partial x\partial y}}\left(x,\varphi (x)\right)\varphi '(x)+{\frac {\partial ^{2}f}{\partial y^{2}}}\left(x,\varphi (x)\right){\varphi '}^{2}(x)}{{\frac {\partial f}{\partial y}}\left(x,\varphi (x)\right)}}}
et (en notation abrégée) :
φ
‴
=
−
∂
3
f
∂
x
3
+
3
∂
3
f
∂
x
2
∂
y
φ
′
+
3
∂
3
f
∂
x
∂
y
2
φ
′
2
+
∂
3
f
∂
y
3
φ
′
3
+
3
∂
2
f
∂
x
∂
y
φ
″
+
3
∂
2
f
∂
y
2
φ
′
φ
″
∂
f
∂
y
{\displaystyle \varphi '''=-{\frac {{\frac {\partial ^{3}f}{\partial x^{3}}}+3{\frac {\partial ^{3}f}{\partial x^{2}\partial y}}\varphi '+3{\frac {\partial ^{3}f}{\partial x\partial y^{2}}}{\varphi '}^{2}+{\frac {\partial ^{3}f}{\partial y^{3}}}{\varphi '}^{3}+3{\frac {\partial ^{2}f}{\partial x\partial y}}\varphi ''+3{\frac {\partial ^{2}f}{\partial y^{2}}}\varphi '\varphi ''}{\frac {\partial f}{\partial y}}}}
ce qui donne ici :
φ
′
=
−
1
+
e
x
1
+
e
φ
,
φ
″
=
−
e
x
+
e
φ
φ
′
2
1
+
e
φ
,
φ
‴
=
−
e
x
+
e
φ
φ
′
3
+
3
e
φ
φ
′
φ
″
1
+
e
φ
,
{\displaystyle \varphi '=-{\frac {1+\operatorname {e} ^{x}}{1+\operatorname {e} ^{\varphi }}},\quad \varphi ''=-{\frac {\operatorname {e} ^{x}+\operatorname {e} ^{\varphi }{\varphi '}^{2}}{1+\operatorname {e} ^{\varphi }}},\quad \varphi '''=-{\frac {\operatorname {e} ^{x}+\operatorname {e} ^{\varphi }{\varphi '}^{3}+3\operatorname {e} ^{\varphi }\varphi '\varphi ''}{1+\operatorname {e} ^{\varphi }}},}
donc en
0
{\displaystyle 0}
φ
(
0
)
=
0
{\displaystyle \varphi (0)=0}
φ
′
=
−
2
2
=
−
1
,
φ
″
=
−
1
+
φ
′
2
2
=
−
1
,
φ
‴
=
−
1
+
φ
′
3
+
3
φ
′
φ
″
2
=
−
3
2
{\displaystyle \varphi '=-{\frac {2}{2}}=-1,\quad \varphi ''=-{\frac {1+{\varphi '}^{2}}{2}}=-1,\quad \varphi '''=-{\frac {1+{\varphi '}^{3}+3\varphi '\varphi ''}{2}}=-{\frac {3}{2}}}
soit en n'appliquant cette méthode que pour exprimer
φ
′
{\displaystyle \varphi '}
k
−
1
{\displaystyle k-1}
φ
″
=
−
[
∂
2
f
∂
x
2
+
∂
2
f
∂
y
∂
x
φ
′
]
∂
f
∂
y
−
∂
f
∂
x
[
∂
2
f
∂
x
∂
y
+
∂
2
f
∂
y
2
φ
′
]
(
∂
f
∂
y
)
2
{\displaystyle \varphi ''=-{\frac {\left[{\frac {\partial ^{2}f}{\partial x^{2}}}+{\frac {\partial ^{2}f}{\partial y\partial x}}\varphi '\right]{\frac {\partial f}{\partial y}}-{\frac {\partial f}{\partial x}}\left[{\frac {\partial ^{2}f}{\partial x\partial y}}+{\frac {\partial ^{2}f}{\partial y^{2}}}\,\varphi '\right]}{\left({\frac {\partial f}{\partial y}}\right)^{2}}}}
φ
‴
=
−
∂
3
f
∂
x
3
+
2
∂
3
f
∂
y
∂
x
2
φ
′
+
∂
3
f
∂
y
2
∂
x
φ
′
2
+
∂
2
f
∂
y
∂
x
φ
″
∂
f
∂
y
+
∂
f
∂
x
[
∂
3
f
∂
x
2
∂
y
+
2
∂
3
f
∂
x
∂
y
2
φ
′
+
∂
3
f
∂
y
3
φ
′
2
+
∂
2
f
∂
y
2
φ
″
]
(
∂
f
∂
y
)
2
{\displaystyle \varphi '''=-{\frac {{\frac {\partial ^{3}f}{\partial x^{3}}}+2{\frac {\partial ^{3}f}{\partial y\partial x^{2}}}\varphi '+{\frac {\partial ^{3}f}{\partial y^{2}\partial x}}\varphi '^{2}+{\frac {\partial ^{2}f}{\partial y\partial x}}\varphi ''}{\frac {\partial f}{\partial y}}}+{\frac {{\frac {\partial f}{\partial x}}\left[{\frac {\partial ^{3}f}{\partial x^{2}\partial y}}+2{\frac {\partial ^{3}f}{\partial x\partial y^{2}}}\varphi '+{\frac {\partial ^{3}f}{\partial y^{3}}}\,\varphi '^{2}+{\frac {\partial ^{2}f}{\partial y^{2}}}\,\varphi ''\right]}{\left({\frac {\partial f}{\partial y}}\right)^{2}}}}
+
2
(
[
∂
2
f
∂
x
2
+
∂
2
f
∂
y
∂
x
φ
′
]
∂
f
∂
y
−
∂
f
∂
x
[
∂
2
f
∂
x
∂
y
+
∂
2
f
∂
y
2
φ
′
]
)
(
∂
2
f
∂
x
∂
y
+
∂
2
f
∂
y
2
φ
′
)
(
∂
f
∂
y
)
3
{\displaystyle +2\,{\frac {{\Big (}\left[{\frac {\partial ^{2}f}{\partial x^{2}}}+{\frac {\partial ^{2}f}{\partial y\partial x}}\varphi '\right]{\frac {\partial f}{\partial y}}-{\frac {\partial f}{\partial x}}\left[{\frac {\partial ^{2}f}{\partial x\partial y}}+{\frac {\partial ^{2}f}{\partial y^{2}}}\,\varphi '\right]{\Big )}\left({\frac {\partial ^{2}f}{\partial x\partial y}}+{\frac {\partial ^{2}f}{\partial y^{2}}}\varphi '\right)}{\left({\frac {\partial f}{\partial y}}\right)^{3}}}}
ce qui donne ici :
φ
″
=
−
e
x
(
1
+
e
φ
)
−
(
1
+
e
x
)
e
φ
φ
′
(
1
+
e
φ
)
2
{\displaystyle \varphi ''=-{\frac {\operatorname {e} ^{x}\left(1+\operatorname {e} ^{\varphi }\right)-\left(1+\operatorname {e} ^{x}\right)\operatorname {e} ^{\varphi }\,\varphi '}{\left(1+\operatorname {e} ^{\varphi }\right)^{2}}}}
φ
‴
=
−
e
x
(
1
+
e
φ
)
+
(
1
+
e
x
)
e
φ
[
φ
′
2
+
φ
″
]
(
1
+
e
φ
)
2
+
2
(
e
x
(
1
+
e
φ
)
−
(
1
+
e
x
)
e
φ
φ
′
)
e
φ
φ
′
(
1
+
e
φ
)
3
{\displaystyle \varphi '''=-{\frac {\operatorname {e} ^{x}}{\left(1+\operatorname {e} ^{\varphi }\right)}}+{\frac {\left(1+\operatorname {e} ^{x}\right)\operatorname {e} ^{\varphi }\left[\varphi '^{2}+\varphi ''\right]}{\left(1+\operatorname {e} ^{\varphi }\right)^{2}}}+2\,{\frac {{\Big (}\operatorname {e} ^{x}\left(1+\operatorname {e} ^{\varphi }\right)-\left(1+\operatorname {e} ^{x}\right)\operatorname {e} ^{\varphi }\,\varphi '{\Big )}\operatorname {e} ^{\varphi }\varphi '}{\left(1+\operatorname {e} ^{\varphi }\right)^{3}}}}
et l'on retrouve, en
0
{\displaystyle 0}
φ
(
0
)
=
0
{\displaystyle \varphi (0)=0}
φ
′
(
0
)
=
−
1
{\displaystyle \varphi '(0)=-1}
φ
″
=
−
2
+
2
2
2
=
−
1
,
φ
‴
=
−
1
2
+
2
[
1
+
φ
″
]
2
2
+
2
−
4
2
3
=
−
3
2
{\displaystyle \varphi ''=-{\frac {2+2}{2^{2}}}=-1,\quad \varphi '''=-{\frac {1}{2}}+{\frac {2\left[1+\varphi ''\right]}{2^{2}}}+2\,{\frac {-4}{2^{3}}}=-{\frac {3}{2}}}
On retrouve ainsi le d.l.
φ
(
x
)
=
a
+
b
x
+
c
x
2
+
d
x
3
+
o
(
x
3
)
{\displaystyle \varphi (x)=a+bx+cx^{2}+dx^{3}+o(x^{3})}
a
=
φ
(
0
)
=
0
,
b
=
φ
′
(
0
)
=
−
1
,
c
=
φ
′
(
0
)
2
=
−
1
2
,
d
=
φ
″
(
0
)
6
=
−
1
4
{\displaystyle a=\varphi (0)=0,\quad b=\varphi '(0)=-1,\quad c={\frac {\varphi '(0)}{2}}=-{\frac {1}{2}},\quad d={\frac {\varphi ''(0)}{6}}=-{\frac {1}{4}}}
J
[
(
x
,
y
)
↦
(
x
+
y
−
z
t
,
x
y
−
z
+
t
)
]
(
x
0
,
y
0
)
=
(
1
1
y
0
x
0
)
{\displaystyle J\left[(x,y)\mapsto (x+y-zt,xy-z+t)\right]_{(x_{0},y_{0})}={\begin{pmatrix}1&1\\y_{0}&x_{0}\end{pmatrix}}}
x
0
≠
y
0
{\displaystyle x_{0}\neq y_{0}}
(
x
0
,
y
0
,
z
0
,
t
0
)
∈
S
{\displaystyle (x_{0},y_{0},z_{0},t_{0})\in S}
(
z
0
,
t
0
)
{\displaystyle (z_{0},t_{0})}
(
x
,
y
)
=
φ
(
z
,
t
)
{\displaystyle (x,y)=\varphi (z,t)}
φ
(
z
0
,
t
0
)
=
(
x
0
,
y
0
)
{\displaystyle \varphi (z_{0},t_{0})=(x_{0},y_{0})}
J
φ
(
z
,
t
)
=
−
(
1
1
y
x
)
−
1
(
−
t
−
z
−
1
1
)
=
1
y
−
x
(
x
−
1
−
y
1
)
(
−
t
−
z
−
1
1
)
=
1
y
−
x
(
1
−
x
t
−
1
−
x
z
y
t
−
1
1
+
y
z
)
{\displaystyle J\varphi (z,t)=-{\begin{pmatrix}1&1\\y&x\end{pmatrix}}^{-1}{\begin{pmatrix}-t&-z\\-1&1\end{pmatrix}}={\frac {1}{y-x}}{\begin{pmatrix}x&-1\\-y&1\end{pmatrix}}{\begin{pmatrix}-t&-z\\-1&1\end{pmatrix}}={\frac {1}{y-x}}{\begin{pmatrix}1-xt&-1-xz\\yt-1&1+yz\end{pmatrix}}}
(
x
0
,
y
0
,
z
0
,
t
0
)
=
(
1
,
−
1
,
0
,
1
)
{\displaystyle (x_{0},y_{0},z_{0},t_{0})=(1,-1,0,1)}
J
φ
(
z
0
,
t
0
)
=
(
0
1
2
−
1
)
{\displaystyle J\varphi (z_{0},t_{0})={\begin{pmatrix}0&1\\2&-1\end{pmatrix}}}
On considère l'application
Ψ
:
M
n
(
R
)
×
M
n
(
R
)
→
M
n
(
R
)
{\displaystyle \Psi :M_{n}(\mathbb {R} )\times M_{n}(\mathbb {R} )\to M_{n}(\mathbb {R} )}
Ψ
(
A
,
B
)
=
A
B
−
I
{\displaystyle \Psi (A,B)=AB-I}
φ
:
G
L
(
n
,
R
)
→
G
L
(
n
,
R
)
,
A
↦
A
−
1
{\displaystyle \varphi :GL(n,\mathbb {R} )\to GL(n,\mathbb {R} ),\,A\mapsto A^{-1}}
Solution
Ψ
{\displaystyle \Psi }
∞ . Pour tout
(
A
,
B
)
=
(
A
,
A
−
1
)
∈
Ψ
−
1
(
{
0
}
)
{\displaystyle (A,B)=(A,A^{-1})\in \Psi ^{-1}(\{0\})}
∂
Ψ
∂
B
(
A
,
B
)
:
K
↦
A
K
{\displaystyle {\frac {\partial \Psi }{\partial B}}_{(A,B)}:K\mapsto AK}
K
↦
A
−
1
K
{\displaystyle K\mapsto A^{-1}K}
φ
{\displaystyle \varphi }
∞ .
d
φ
A
(
H
)
=
[
−
(
∂
Ψ
∂
B
(
A
,
A
−
1
)
)
−
1
∘
∂
Ψ
∂
A
(
A
,
A
−
1
)
]
(
H
)
=
−
(
∂
Ψ
∂
B
(
A
,
A
−
1
)
)
−
1
(
H
A
−
1
)
=
−
A
−
1
H
A
−
1
{\displaystyle \mathrm {d} \varphi _{A}(H)=\left[-\left({\frac {\partial \Psi }{\partial B}}_{(A,A^{-1})}\right)^{-1}\circ {\frac {\partial \Psi }{\partial A}}_{(A,A^{-1})}\right](H)=-\left({\frac {\partial \Psi }{\partial B}}_{(A,A^{-1})}\right)^{-1}(HA^{-1})=-A^{-1}HA^{-1}}
Soit l'application
F
:
R
3
→
R
,
(
x
,
y
,
z
)
↦
x
2
+
y
2
+
2
z
2
−
cos
z
{\displaystyle F:\mathbb {R} ^{3}\to \mathbb {R} ,\;(x,y,z)\mapsto {\sqrt {x^{2}+y^{2}+2z^{2}}}-\cos z}
Démontrer que la relation
F
(
x
,
y
,
z
)
=
0
{\displaystyle F(x,y,z)=0}
a
=
(
0
,
1
,
0
)
∈
R
3
{\displaystyle a=(0,1,0)\in \mathbb {R} ^{3}}
φ
(
x
,
z
)
=
y
{\displaystyle \varphi (x,z)=y}
1 .
Calculer les dérivées partielles d'ordre 1 de
φ
{\displaystyle \varphi }
(
0
,
0
)
{\displaystyle (0,0)}
Solution
F
{\displaystyle F}
1 et pour
(
x
,
y
,
z
)
≠
(
0
,
0
,
0
)
{\displaystyle (x,y,z)\neq (0,0,0)}
∂
F
∂
y
=
2
y
2
x
2
+
y
2
+
2
z
2
{\displaystyle {\frac {\partial F}{\partial y}}={\frac {2y}{2{\sqrt {x^{2}+y^{2}+2z^{2}}}}}}
∂
F
∂
y
(
a
)
=
1
{\displaystyle {\frac {\partial F}{\partial y}}(a)=1}
U
{\displaystyle U}
(
0
,
0
)
{\displaystyle (0,0)}
V
{\displaystyle V}
0
{\displaystyle 0}
ϕ
{\displaystyle \phi }
1 de
U
{\displaystyle U}
V
{\displaystyle V}
∀
(
x
,
y
,
z
)
∈
U
×
V
F
(
x
,
y
,
z
)
=
0
⇔
y
=
ϕ
(
x
,
z
)
{\displaystyle \forall (x,y,z)\in U\times V\quad F(x,y,z)=0\Leftrightarrow y=\phi (x,z)}
On a
ϕ
(
0
,
0
)
=
1
{\displaystyle \phi (0,0)=1}
x
2
+
ϕ
(
x
,
z
)
2
+
2
z
2
−
cos
z
=
0
{\displaystyle {\sqrt {x^{2}+\phi (x,z)^{2}+2z^{2}}}-\cos z=0}
x
{\displaystyle x}
z
{\displaystyle z}
∂
F
∂
x
(
x
,
ϕ
(
x
,
z
)
,
z
)
+
∂
F
∂
y
(
x
,
ϕ
(
x
,
z
)
,
z
)
∂
ϕ
∂
x
=
0
{\displaystyle {\frac {\partial F}{\partial x}}(x,\phi (x,z),z)+{\frac {\partial F}{\partial y}}(x,\phi (x,z),z){\frac {\partial \phi }{\partial x}}=0}
(
0
,
0
)
{\displaystyle (0,0)}
∂
ϕ
∂
x
(
0
,
0
)
=
0
{\displaystyle {\frac {\partial \phi }{\partial x}}(0,0)=0}
∂
ϕ
∂
z
(
0
,
0
)
=
0
{\displaystyle {\frac {\partial \phi }{\partial z}}(0,0)=0}
Soient
E
{\displaystyle E}
f
:
E
→
E
{\displaystyle f:E\to E}
1 . On suppose que
f
{\displaystyle f}
‖
d
f
x
(
h
)
‖
=
‖
h
‖
{\displaystyle \|\mathrm {d} f_{x}(h)\|=\|h\|}
x
,
h
∈
E
{\displaystyle x,h\in E}
Montrer que tout
a
∈
E
{\displaystyle a\in E}
Ω
{\displaystyle \Omega }
f
{\displaystyle f}
‖
f
(
x
)
−
f
(
y
)
‖
=
‖
x
−
y
‖
{\displaystyle \|f(x)-f(y)\|=\|x-y\|}
x
,
y
∈
Ω
{\displaystyle x,y\in \Omega }
En déduire que
f
{\displaystyle f}
a
{\displaystyle a}
En déduire que
f
{\displaystyle f}
Solution
Wikipedia-logo-v2.svg Wikipedia-logo-v2.svg
d
f
a
{\displaystyle \mathrm {d} f_{a}}
|
|
|
d
f
x
|
|
|
=
|
|
|
(
d
f
x
)
−
1
|
|
|
=
1
{\displaystyle |\!|\!|\mathrm {d} f_{x}|\!|\!|=|\!|\!|(\mathrm {d} f_{x})^{-1}|\!|\!|=1}
Sur une boule ouverte de centre
a
{\displaystyle a}
r
{\displaystyle r}
f
{\displaystyle f}
h
↦
f
(
a
+
h
)
−
f
(
a
)
{\displaystyle h\mapsto f(a+h)-f(a)}
B
(
0
,
r
)
{\displaystyle B(0,r)}
Z
[
1
/
2
]
{\displaystyle \mathbb {Z} [1/2]}
Z
[
1
/
2
]
{\displaystyle \mathbb {Z} [1/2]}
nombres dyadiques puis, par continuité,
R
{\displaystyle \mathbb {R} }
D'après la question précédente,
d
f
{\displaystyle \mathrm {d} f}
E
{\displaystyle E}
connexe ,
d
f
{\displaystyle \mathrm {d} f}
f
{\displaystyle f}
![{\displaystyle \left]0,\infty \right[\times \left]-\pi ,\pi \right[}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a75377be37b6f3d795f90f904fbb1b7a92952b10)
![{\displaystyle \|DF(y)[e]\|\geq C\|e\|}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aa4f5a0c7913448eee2b3a496e271660225bc509)
![{\displaystyle \kappa \in \left]0,1\right[}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4bf9de80df0461fb67c04da1b2589aff3cc64dbc)
![{\displaystyle \langle \nabla _{g}(x),e\rangle =D_{g}(x)(e)=2\langle F(x)-a,DF(x)[e]\rangle }](https://wikimedia.org/api/rest_v1/media/math/render/svg/5711af2b123f84a2dad66cfa0f16daa79724c826)
![{\displaystyle DF(x_{a})[e]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a25f69092bcd904892b441275105e59cd48c50dd)
![{\displaystyle {\begin{aligned}0&=2\operatorname {e} ^{(1+a)u+bu^{2}+cu^{3}}+\ln \left(1+(1-a)u-bu^{2}-cu^{3}\right)-2-2u+\left(au\right)^{3}+o\left(u^{3}\right)\\&={\cancel {2}}+2(1+a)u+2bu^{2}+2cu^{3}+(1+a)^{2}u^{2}+2(1+a)bu^{3}+(1+a)^{3}u^{3}/3\\&\quad +(1-a)u-bu^{2}-cu^{3}-(1-a)^{2}u^{2}/2+(1-a)bu^{3}+(1-a)^{3}u^{3}/3-{\cancel {2}}-2u+a^{3}u^{3}+o\left(u^{3}\right)\\&=(1+a)u+u^{2}\left[b+1/2+3a+a^{2}/2\right]+u^{3}\left[c+3b+ab+2/3+2a^{2}+a^{3}\right]+o\left(u^{3}\right)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4c707bacecbc3be8c228749ddb9be7a4f7602757)
![{\displaystyle \mathbb {R} _{d}\left[X\right]\times \mathbb {R} \to \mathbb {R} ,\ \left(P,\alpha \right)\mapsto P\left(\alpha \right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1bf8d9a2dd62917e99fdae095c2aafcef95f3725)
![{\displaystyle \mathbb {R} _{d}\left[X\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5098cd8bb56153c3cf06a58331f2257f7e24cd86)
![{\displaystyle \mathbb {R} _{d}\left[X\right]\times \mathbb {R} ^{k}\to \mathbb {R} ^{k},\ \left(P,\alpha _{1},\dots ,\alpha _{k}\right)\mapsto \left(P\left(\alpha _{1}\right),\dots ,P\left(\alpha _{k}\right)\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/80e9b4a2d4e73de615a356ea1d219efcb65d46ff)
![{\displaystyle y>{\sqrt[{3}]{4}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5d11a8973ab67a1eca1e896f23e52f71be48e440)
![{\displaystyle 0<y<{\sqrt[{3}]{4}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/94c5c2f090e6bdd83a33a6df6371fa48dc62be7b)
![{\displaystyle ({\sqrt[{3}]{2}},{\sqrt[{3}]{4}})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2a1dde25062596dcb542d379224346208041cbf9)
![{\displaystyle y<{\sqrt[{3}]{4}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e1d0ca478ec7437658228e95ed067c643f24f9de)
![{\displaystyle y={\sqrt[{3}]{4}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f080f19fd611cb9af78812e06779a3d5ef0915e1)
![{\displaystyle \varphi ''=-{\frac {\left[{\frac {\partial ^{2}f}{\partial x^{2}}}+{\frac {\partial ^{2}f}{\partial y\partial x}}\varphi '\right]{\frac {\partial f}{\partial y}}-{\frac {\partial f}{\partial x}}\left[{\frac {\partial ^{2}f}{\partial x\partial y}}+{\frac {\partial ^{2}f}{\partial y^{2}}}\,\varphi '\right]}{\left({\frac {\partial f}{\partial y}}\right)^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a6a6eab19acd38ce54d5c0da0db70504056dd26f)
![{\displaystyle \varphi '''=-{\frac {{\frac {\partial ^{3}f}{\partial x^{3}}}+2{\frac {\partial ^{3}f}{\partial y\partial x^{2}}}\varphi '+{\frac {\partial ^{3}f}{\partial y^{2}\partial x}}\varphi '^{2}+{\frac {\partial ^{2}f}{\partial y\partial x}}\varphi ''}{\frac {\partial f}{\partial y}}}+{\frac {{\frac {\partial f}{\partial x}}\left[{\frac {\partial ^{3}f}{\partial x^{2}\partial y}}+2{\frac {\partial ^{3}f}{\partial x\partial y^{2}}}\varphi '+{\frac {\partial ^{3}f}{\partial y^{3}}}\,\varphi '^{2}+{\frac {\partial ^{2}f}{\partial y^{2}}}\,\varphi ''\right]}{\left({\frac {\partial f}{\partial y}}\right)^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c5bc0a11cae946e2161b3c1ce8a3b76a7eeae5b6)
![{\displaystyle +2\,{\frac {{\Big (}\left[{\frac {\partial ^{2}f}{\partial x^{2}}}+{\frac {\partial ^{2}f}{\partial y\partial x}}\varphi '\right]{\frac {\partial f}{\partial y}}-{\frac {\partial f}{\partial x}}\left[{\frac {\partial ^{2}f}{\partial x\partial y}}+{\frac {\partial ^{2}f}{\partial y^{2}}}\,\varphi '\right]{\Big )}\left({\frac {\partial ^{2}f}{\partial x\partial y}}+{\frac {\partial ^{2}f}{\partial y^{2}}}\varphi '\right)}{\left({\frac {\partial f}{\partial y}}\right)^{3}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/762c14fc171b03ba80963484e6149844a80aa984)
![{\displaystyle \varphi '''=-{\frac {\operatorname {e} ^{x}}{\left(1+\operatorname {e} ^{\varphi }\right)}}+{\frac {\left(1+\operatorname {e} ^{x}\right)\operatorname {e} ^{\varphi }\left[\varphi '^{2}+\varphi ''\right]}{\left(1+\operatorname {e} ^{\varphi }\right)^{2}}}+2\,{\frac {{\Big (}\operatorname {e} ^{x}\left(1+\operatorname {e} ^{\varphi }\right)-\left(1+\operatorname {e} ^{x}\right)\operatorname {e} ^{\varphi }\,\varphi '{\Big )}\operatorname {e} ^{\varphi }\varphi '}{\left(1+\operatorname {e} ^{\varphi }\right)^{3}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5d1a8a2b220924ac38e18bea8538b0a94ccc2d9e)
![{\displaystyle \varphi ''=-{\frac {2+2}{2^{2}}}=-1,\quad \varphi '''=-{\frac {1}{2}}+{\frac {2\left[1+\varphi ''\right]}{2^{2}}}+2\,{\frac {-4}{2^{3}}}=-{\frac {3}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4c70023ba8b47b988203f72f4ae3b309ff0d51d3)
![{\displaystyle J\left[(x,y)\mapsto (x+y-zt,xy-z+t)\right]_{(x_{0},y_{0})}={\begin{pmatrix}1&1\\y_{0}&x_{0}\end{pmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bfac0f0e762c6f13b0332af403d919a7cf7e5088)
=-\left({\frac {\partial \Psi }{\partial B}}_{(A,A^{-1})}\right)^{-1}(HA^{-1})=-A^{-1}HA^{-1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a1a08f85ba3b04af7e9da1d344c08a732463548e)
![{\displaystyle \mathbb {Z} [1/2]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ac6560a29f54d1403d584259891b958b24453fa1)
