Une page de Wikiversité.
En raison de limitations techniques, la typographie souhaitable du titre, «
Exercice : Développements limitésFonctions d'une variable réelle/Exercices/Développements limités », n'a pu être restituée correctement ci-dessus.
Calculer le développement limité de
arctan
x
{\displaystyle \arctan x}
1
{\displaystyle 1}
2
{\displaystyle 2}
Calculer le développement limité de
arsinh
(
1
+
2
x
+
3
x
2
)
{\displaystyle \operatorname {arsinh} \left(1+2x+3x^{2}\right)}
Solution
Commençons par calculer le dl en 0 à l'ordre 3 de
arsinh
(
1
+
u
)
{\displaystyle \operatorname {arsinh} (1+u)}
Trigonométrie hyperbolique/Fonctions hyperboliques réciproques ) :
arsinh
′
(
1
+
u
)
=
1
1
+
(
1
+
u
)
2
=
1
2
1
+
u
+
u
2
/
2
=
1
2
(
1
−
1
2
(
u
+
u
2
/
2
)
+
3
8
u
2
)
+
o
(
u
2
)
=
1
2
(
1
−
u
2
+
u
2
8
)
+
o
(
u
2
)
{\displaystyle \operatorname {arsinh} '(1+u)={\frac {1}{\sqrt {1+(1+u)^{2}}}}={\frac {1}{{\sqrt {2}}{\sqrt {1+u+u^{2}/2}}}}={\frac {1}{\sqrt {2}}}\left(1-{\frac {1}{2}}\left(u+u^{2}/2\right)+{\frac {3}{8}}u^{2}\right)+o\left(u^{2}\right)={\frac {1}{\sqrt {2}}}\left(1-{\frac {u}{2}}+{\frac {u^{2}}{8}}\right)+o\left(u^{2}\right)}
arsinh
(
1
+
u
)
=
arsinh
(
1
)
+
1
2
(
u
−
u
2
4
+
u
3
24
)
+
o
(
u
3
)
{\displaystyle \operatorname {arsinh} (1+u)=\operatorname {arsinh} (1)+{\frac {1}{\sqrt {2}}}\left(u-{\frac {u^{2}}{4}}+{\frac {u^{3}}{24}}\right)+o\left(u^{3}\right)}
arsinh
(
1
+
2
x
+
3
x
2
)
=
arsinh
(
1
)
+
1
2
(
2
x
+
3
x
2
−
4
x
2
+
12
x
3
4
+
8
x
3
24
)
+
o
(
x
3
)
=
ln
(
1
+
2
)
+
1
2
(
2
x
+
2
x
2
−
8
x
3
3
)
+
o
(
x
3
)
=
ln
(
1
+
2
)
+
2
x
+
2
x
2
−
4
2
x
3
3
+
o
(
x
3
)
.
{\displaystyle {\begin{aligned}\operatorname {arsinh} \left(1+2x+3x^{2}\right)&=\operatorname {arsinh} (1)+{\frac {1}{\sqrt {2}}}\left(2x+3x^{2}-{\frac {4x^{2}+12x^{3}}{4}}+{\frac {8x^{3}}{24}}\right)+o\left(x^{3}\right)\\&=\ln \left(1+{\sqrt {2}}\right)+{\frac {1}{\sqrt {2}}}\left(2x+2x^{2}-{\frac {8x^{3}}{3}}\right)+o\left(x^{3}\right)\\&=\ln \left(1+{\sqrt {2}}\right)+{\sqrt {2}}x+{\sqrt {2}}x^{2}-{\frac {4{\sqrt {2}}x^{3}}{3}}+o\left(x^{3}\right).\end{aligned}}}
Calculer le développement limité de
(
sin
2
x
−
sin
(
x
2
)
)
(
e
tan
x
−
e
tanh
x
)
{\displaystyle \left(\sin ^{2}x-\sin(x^{2})\right)\left(\operatorname {e} ^{\tan x}-\operatorname {e} ^{\tanh x}\right)}
Solution
Le premier facteur est équivalent à
−
x
4
/
3
{\displaystyle -x^{4}/3}
2
x
3
/
3
{\displaystyle 2x^{3}/3}
9
−
3
=
6
{\displaystyle 9-3=6}
9
−
4
=
5
{\displaystyle 9-4=5}
sin
2
x
−
sin
(
x
2
)
=
(
x
−
x
3
6
+
x
5
120
)
2
−
(
x
2
−
x
6
6
)
+
o
(
x
6
)
=
−
x
4
3
+
19
x
6
90
+
o
(
x
6
)
{\displaystyle \sin ^{2}x-\sin(x^{2})=\left(x-{\frac {x^{3}}{6}}+{\frac {x^{5}}{120}}\right)^{2}-\left(x^{2}-{\frac {x^{6}}{6}}\right)+o\left(x^{6}\right)=-{\frac {x^{4}}{3}}+{\frac {19x^{6}}{90}}+o\left(x^{6}\right)}
e
tan
x
−
e
tanh
x
=
2
x
3
3
+
(
x
+
x
3
3
)
2
−
(
x
−
x
3
3
)
2
2
+
(
x
+
x
3
3
)
3
−
(
x
−
x
3
3
)
3
6
+
o
(
x
5
)
=
2
x
3
+
2
x
4
+
x
5
3
+
o
(
x
5
)
.
{\displaystyle {\begin{aligned}\operatorname {e} ^{\tan x}-\operatorname {e} ^{\tanh x}&={\frac {2x^{3}}{3}}+{\frac {\left(x+{\frac {x^{3}}{3}}\right)^{2}-\left(x-{\frac {x^{3}}{3}}\right)^{2}}{2}}+{\frac {\left(x+{\frac {x^{3}}{3}}\right)^{3}-\left(x-{\frac {x^{3}}{3}}\right)^{3}}{6}}+o\left(x^{5}\right)\\&={\frac {2x^{3}+2x^{4}+x^{5}}{3}}+o\left(x^{5}\right).\end{aligned}}}
(
sin
2
x
−
sin
(
x
2
)
)
(
e
tan
x
−
e
tanh
x
)
=
x
7
9
(
−
1
+
19
x
2
30
)
(
2
+
2
x
+
x
2
)
+
o
(
x
9
)
=
x
7
9
(
−
2
−
2
x
+
4
x
2
15
)
+
o
(
x
9
)
=
−
2
x
7
9
−
2
x
8
9
+
4
x
9
135
+
o
(
x
9
)
.
{\displaystyle {\begin{aligned}\left(\sin ^{2}x-\sin(x^{2})\right)(\operatorname {e} ^{\tan x}-\operatorname {e} ^{\tanh x})&={\frac {x^{7}}{9}}\left(-1+{\frac {19x^{2}}{30}}\right)\left(2+2x+x^{2}\right)+o\left(x^{9}\right)\\&={\frac {x^{7}}{9}}\left(-2-2x+{\frac {4x^{2}}{15}}\right)+o\left(x^{9}\right)\\&=-{\frac {2x^{7}}{9}}-{\frac {2x^{8}}{9}}+{\frac {4x^{9}}{135}}+o\left(x^{9}\right).\end{aligned}}}
Calculer le développement limité de
1
x
2
−
1
(
arcsin
x
)
2
{\displaystyle {\frac {1}{x^{2}}}-{\frac {1}{(\arcsin x)^{2}}}}
Solution
Puisque
1
x
2
−
1
(
arcsin
x
)
2
=
1
−
(
arcsin
x
x
)
−
2
x
2
{\displaystyle {\frac {1}{x^{2}}}-{\frac {1}{(\arcsin x)^{2}}}={\frac {1-\left({\frac {\arcsin x}{x}}\right)^{-2}}{x^{2}}}}
arcsin
x
x
{\displaystyle {\frac {\arcsin x}{x}}}
arcsin
x
{\displaystyle \arcsin x}
arcsin
′
x
=
1
1
−
x
2
=
1
+
x
2
2
+
3
x
4
8
+
5
x
6
16
+
o
(
x
7
)
{\displaystyle \arcsin 'x={\frac {1}{\sqrt {1-x^{2}}}}=1+{\frac {x^{2}}{2}}+{\frac {3x^{4}}{8}}+{\frac {5x^{6}}{16}}+o\left(x^{7}\right)}
arcsin
x
=
x
+
x
3
6
+
3
x
5
40
+
5
x
7
112
+
o
(
x
8
)
{\displaystyle \arcsin x=x+{\frac {x^{3}}{6}}+{\frac {3x^{5}}{40}}+{\frac {5x^{7}}{112}}+o\left(x^{8}\right)}
1
x
2
−
1
(
arcsin
x
)
2
=
1
−
(
1
+
x
2
6
+
3
x
4
40
+
5
x
6
112
+
o
(
x
7
)
)
−
2
x
2
=
2
(
x
2
6
+
3
x
4
40
+
5
x
6
112
)
−
3
(
x
2
6
+
3
x
4
40
)
2
+
4
(
x
2
6
)
3
x
2
+
o
(
x
5
)
=
1
3
+
x
2
15
+
31
x
4
945
+
o
(
x
5
)
.
{\displaystyle {\begin{aligned}{\frac {1}{x^{2}}}-{\frac {1}{(\arcsin x)^{2}}}&={\frac {1-\left(1+{\frac {x^{2}}{6}}+{\frac {3x^{4}}{40}}+{\frac {5x^{6}}{112}}+o\left(x^{7}\right)\right)^{-2}}{x^{2}}}\\&={\frac {2\left({\frac {x^{2}}{6}}+{\frac {3x^{4}}{40}}+{\frac {5x^{6}}{112}}\right)-3\left({\frac {x^{2}}{6}}+{\frac {3x^{4}}{40}}\right)^{2}+4\left({\frac {x^{2}}{6}}\right)^{3}}{x^{2}}}+o\left(x^{5}\right)\\&={\frac {1}{3}}+{\frac {x^{2}}{15}}+{\frac {31x^{4}}{945}}+o\left(x^{5}\right).\end{aligned}}}
Soit
x
>
0
{\displaystyle x>0}
ln
(
1
+
x
)
>
x
−
x
2
2
{\displaystyle \ln(1+x)>x-{\frac {x^{2}}{2}}}
Redémontrer le théorème de Taylor-Young par application itérée de la première règle de l'Hôpital.
Donner les d.l. à l'ordre 3 :
en 0 de
cos
×
sin
{\displaystyle \cos \times \sin }
en 0 de
sin
∘
k
:=
sin
∘
sin
∘
⋯
∘
sin
{\displaystyle \sin ^{\circ k}:=\sin \circ \sin \circ \dots \circ \sin }
k
{\displaystyle k}
k -ième itérée
sin
{\displaystyle \sin }
k -ième puissance,
sin
k
x
:=
(
sin
x
)
k
=
o
(
x
3
)
{\displaystyle \sin ^{k}x:=(\sin x)^{k}=o(x^{3})}
k
>
3
{\displaystyle k>3}
en 0 de
(
1
+
x
)
α
{\displaystyle \left(1+x\right)^{\alpha }}
α
{\displaystyle \alpha }
en 1 de
e
x
{\displaystyle \operatorname {e} ^{\sqrt {x}}}
Solution
cos
x
sin
x
=
(
1
−
x
2
2
+
o
(
x
2
)
)
(
x
−
x
3
6
+
o
(
x
3
)
)
{\displaystyle \cos x\sin x=\left(1-{\frac {x^{2}}{2}}+o(x^{2})\right)\left(x-{\frac {x^{3}}{6}}+o(x^{3})\right)}
cos
x
sin
x
=
sin
2
x
2
=
2
x
−
(
2
x
)
3
6
+
o
(
x
3
)
2
{\displaystyle \cos x\sin x={\frac {\sin 2x}{2}}={\frac {2x-{\frac {(2x)^{3}}{6}}+o(x^{3})}{2}}}
cos
x
sin
x
=
x
−
2
x
3
3
+
o
(
x
3
)
{\displaystyle \cos x\sin x=x-{\frac {2x^{3}}{3}}+o(x^{3})}
∀
k
∈
N
∗
sin
∘
k
(
x
)
=
x
−
k
6
x
3
+
o
(
x
3
)
{\displaystyle \forall k\in \mathbb {N} ^{*}\quad \sin ^{\circ k}(x)=x-{\frac {k}{6}}x^{3}+o(x^{3})}
k
=
0
{\displaystyle k=0}
sin
∘
0
=
i
d
R
{\displaystyle \sin ^{\circ 0}=\mathrm {id} _{\mathbb {R} }}
Posons
f
(
x
)
=
(
1
+
x
)
α
{\displaystyle f(x)=\left(1+x\right)^{\alpha }}
f
(
0
)
=
1
{\displaystyle f(0)=1}
f
′
(
x
)
=
α
(
1
+
x
)
α
−
1
{\displaystyle f'(x)=\alpha \left(1+x\right)^{\alpha -1}}
f
″
(
x
)
=
α
(
α
−
1
)
(
1
+
x
)
α
−
2
{\displaystyle f''(x)=\alpha \left(\alpha -1\right)\left(1+x\right)^{\alpha -2}}
f
‴
(
x
)
=
α
(
α
−
1
)
(
α
−
2
)
(
1
+
x
)
α
−
3
{\displaystyle f'''(x)=\alpha \left(\alpha -1\right)\left(\alpha -2\right)\left(1+x\right)^{\alpha -3}}
f
′
(
0
)
=
α
{\displaystyle f'(0)=\alpha }
f
″
(
0
)
=
α
(
α
−
1
)
{\displaystyle f''(0)=\alpha \left(\alpha -1\right)}
f
‴
(
0
)
=
α
(
α
−
1
)
(
α
−
2
)
{\displaystyle f'''(0)=\alpha \left(\alpha -1\right)\left(\alpha -2\right)}
f
(
x
)
=
f
(
0
)
+
f
′
(
0
)
x
+
f
″
(
0
)
2
!
x
2
+
f
‴
(
0
)
3
!
x
3
+
o
(
x
3
)
=
1
+
α
x
+
α
(
α
−
1
)
2
x
2
+
α
(
α
−
1
)
(
α
−
2
)
6
x
3
+
o
(
x
3
)
{\displaystyle f(x)=f(0)+f'(0)x+{\frac {f''(0)}{2!}}x^{2}+{\frac {f'''(0)}{3!}}x^{3}+o\left(x^{3}\right)=1+\alpha x+{\frac {\alpha \left(\alpha -1\right)}{2}}x^{2}+{\frac {\alpha \left(\alpha -1\right)\left(\alpha -2\right)}{6}}x^{3}+o\left(x^{3}\right)}
n : cf. développements limités des fonctions usuelles en zéro ).
Le cas particulier
α
=
1
2
{\displaystyle \alpha ={\frac {1}{2}}}
1
+
h
=
1
+
h
2
−
h
2
8
+
h
3
16
+
o
(
h
3
)
=
1
+
u
{\displaystyle {\sqrt {1+h}}=1+{\frac {h}{2}}-{\frac {h^{2}}{8}}+{\frac {h^{3}}{16}}+o\left(h^{3}\right)=1+u}
u
:=
h
2
−
h
2
8
+
h
3
16
+
o
(
h
3
)
→
0
{\displaystyle u:={\frac {h}{2}}-{\frac {h^{2}}{8}}+{\frac {h^{3}}{16}}+o\left(h^{3}\right)\to 0}
h
→
0
{\displaystyle h\to 0}
en 0 de la fonction exponentielle à l'ordre 3 (à nouveau, voir développements limités des fonctions usuelles en zéro ) :
e
1
+
h
=
e
1
+
u
=
e
×
e
u
=
e
×
(
1
+
u
+
u
2
2
+
u
3
6
+
o
(
u
3
)
)
=
e
×
(
1
+
(
h
2
−
h
2
8
+
h
3
16
)
+
1
2
(
(
h
2
)
2
−
2
h
2
h
2
8
)
+
1
6
(
h
2
)
3
+
o
(
h
3
)
)
=
e
×
(
1
+
h
2
+
h
3
48
)
+
o
(
h
3
)
{\displaystyle {\begin{aligned}\operatorname {e} ^{\sqrt {1+h}}&=\operatorname {e} ^{1+u}=\mathrm {e} \times \operatorname {e} ^{u}\\&=\mathrm {e} \times \left(1+u+{\frac {u^{2}}{2}}+{\frac {u^{3}}{6}}+o(u^{3})\right)\\&=\mathrm {e} \times \left(1+\left({\frac {h}{2}}-{\frac {h^{2}}{8}}+{\frac {h^{3}}{16}}\right)+{\frac {1}{2}}\left(\left({\frac {h}{2}}\right)^{2}-2{\frac {h}{2}}{\frac {h^{2}}{8}}\right)+{\frac {1}{6}}\left({\frac {h}{2}}\right)^{3}+o(h^{3})\right)\\&=\mathrm {e} \times \left(1+{\frac {h}{2}}+{\frac {h^{3}}{48}}\right)+o(h^{3})\end{aligned}}}
e
x
=
e
+
e
2
(
x
−
1
)
+
e
48
(
x
−
1
)
3
+
o
(
(
x
−
1
)
3
)
{\displaystyle \operatorname {e} ^{\sqrt {x}}=\mathrm {e} +{\frac {\mathrm {e} }{2}}(x-1)+{\frac {\mathrm {e} }{48}}(x-1)^{3}+o((x-1)^{3})}
![{\displaystyle c\in \left]0,x\right[}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5912057789b1a12aad7064bc6a3f2727e80c8389)
