Démonstration de la formule de Taylor-Young

## Développement limité à l'ordre 1

Soit

f

{\displaystyle f}

 une fonction dérivable en

x
0

{\displaystyle x\_0}

Donc

lim

x
⟶

x

0

(

f
(
x
)
−
f
(

x

0

)

x
−

x

0

)

=

f
′

(

x

0

)

{\displaystyle \lim\limits\_{x\longrightarrow x\_{0}}\left(\frac{f(x) - f(x\_{0})}{x-x\_{0}}\right) = f'(x\_{0})}

Donc

lim

x
⟶

x

0

(

f
(
x
)
−
f
(

x

0

)

x
−

x

0

−

f
′

(

x

0

)

)

=
0

{\displaystyle \lim\limits\_{x\longrightarrow x\_{0}}\left(\frac{f(x) - f(x\_{0})}{x-x\_{0}} - f'(x\_{0})\right) = 0}

Donc

lim

x
⟶

x

0

(

f
(
x
)
−
f
(

x

0

)

x
−

x

0

−

f
′

(

x

0

)
×
(
x
−

x

0

)

x
−

x

0

)

=
0

{\displaystyle \lim\limits\_{x\longrightarrow x\_{0}}\left(\frac{f(x) - f(x\_{0})}{x-x\_{0}} - \frac{f'(x\_{0})\times(x-x\_{0})}{x-x\_{0}}\right) = 0}

Donc

lim

x
⟶

x

0

(

f
(
x
)
−
f
(

x

0

)
−

f
′

(

x

0

)
×
(
x
−

x

0

)

x
−

x

0

)

=
0

{\displaystyle \lim\limits\_{x\longrightarrow x\_{0}}\left(\frac{f(x) - f(x\_{0}) - f'(x\_{0})\times(x-x\_{0})}{x-x\_{0}} \right) = 0}

Donc

f
(
x
)
−
f
(

x

0

)
−

f
′

(

x

0

)
×
(
x
−

x

0

# )
o
(
x
−

x

0

)

{\displaystyle f(x) - f(x\_{0}) - f'(x\_{0})\times(x-x\_{0}) = o(x-x\_{0})}

 (quand

x
⟶

x

0

{\displaystyle x\longrightarrow x\_{0}}

)

Donc

f
(
x
# )
f
(

x

0

)

f
′

(

x

0

)
×
(
x
−

x

0

)
o
(
x
−

x

0

)

{\displaystyle f(x) = f(x\_{0}) + f'(x\_{0})\times(x-x\_{0}) + o(x-x\_{0})}

 (quand

x
⟶

x

0

{\displaystyle x\longrightarrow x\_{0}}

) DL à l'ordre 1 de

f
(
x
)

{\displaystyle f(x)}

## Développement limité à l'ordre 2

Soit

f

{\displaystyle f}

 est 2 fois dérivable en

x
0

{\displaystyle x\_0}

On calcule

lim

x
⟶

x

0

(

f
(
x
)
−
f
(

x

0

)
−

f
′

(

x

0

)
(
x
−

x

0

)

(
x
−

x

0

)

2

)

{\displaystyle \lim\limits\_{x\longrightarrow x\_{0}} \left(\frac{f(x) - f(x\_{0}) - f'(x\_{0})(x-x\_{0})}{(x-x\_{0})^{2}}\right)}

On pose

g
(
x
# )
f
(
x
)
−
f
(

x

0

)
−

f
′

(

x

0

)
(
x
−

x

0

)

{\displaystyle g(x) = f(x) - f(x\_{0}) - f'(x\_{0})(x-x\_{0})}

et

h
(
x
# )
(
x
−

x

0

)

2

{\displaystyle h(x) = (x-x\_{0})^{2}}

Donc

g
′

(
x
# )

f
′

(
x
)
−

f
′

(

x

0

)

{\displaystyle g'(x) = f'(x) - f'(x\_{0})}

et

h
′

(
x
# )
2
(
x
−

x

0

)

{\displaystyle h'(x) = 2(x-x\_{0})}

Donc

g
′

(
x
)

h
′

(
x
)

=

f
′

(
x
)
−

f
′

(

x

0

)

2
(
x
−

x

0

)

{\displaystyle \frac{g'(x)}{h'(x)} = \frac{f'(x) - f'(x\_{0})}{2(x-x\_{0})}}

Donc

lim

x
⟶

x

0

(

g
′

(
x
)

h
′

(
x
)

)

=

lim

x
⟶

x

0

(

f
′

(
x
)
−

f
′

(

x

0

)

2
(
x
−

x

0

)

)

=

f
″

(

x

0

)

2

{\displaystyle \lim\limits\_{x\longrightarrow x\_{0}}\left(\frac{g'(x)}{h'(x)}\right) = \lim\limits\_{x\longrightarrow x\_{0}}\left(\frac{f'(x) - f'(x\_{0})}{2(x-x\_{0})}\right) = \frac{f''(x\_{0})}{2}}

D'après la [Règle de l'Hôpital],

lim

x
⟶

x

0

(

g
(
x
)

h
(
x
)

)

=

lim

x
⟶

x

0

(

g
′

(
x
)

h
′

(
x
)

)

=

f
″

(

x

0

)

2

{\displaystyle \lim\limits\_{x\longrightarrow x\_{0}}\left(\frac{g(x)}{h(x)}\right) = \lim\limits\_{x\longrightarrow x\_{0}}\left(\frac{g'(x)}{h'(x)}\right) = \frac{f''(x\_{0})}{2}}

Donc

lim

x
⟶

x

0

(

f
(
x
)
−
f
(

x

0

)
−

f
′

(

x

0

)
(
x
−

x

0

)

(
x
−

x

0

)

2

)

=

f
″

(

x

0

)

2

{\displaystyle \lim\limits\_{x\longrightarrow x\_{0}} \left(\frac{f(x) - f(x\_{0}) - f'(x\_{0})(x-x\_{0})}{(x-x\_{0})^{2}}\right) = \frac{f''(x\_{0})}{2}}

Donc

lim

x
⟶

x

0

(

f
(
x
)
−
f
(

x

0

)
−

f
′

(

x

0

)
(
x
−

x

0

)

(
x
−

x

0

)

2

−

f
″

(

x

0

)

2

)

=
0

{\displaystyle \lim\limits\_{x\longrightarrow x\_{0}} \left(\frac{f(x) - f(x\_{0}) - f'(x\_{0})(x-x\_{0})}{(x-x\_{0})^{2}} - \frac{f''(x\_{0})}{2} \right) = 0}

Donc

lim

x
⟶

x

0

(

f
(
x
)
−
f
(

x

0

)
−

f
′

(

x

0

)
(
x
−

x

0

)

(
x
−

x

0

)

2

−

f
″

(

x

0

)

2

(
x
−

x

0

)

2

(
x
−

x

0

)

2

)

=
0

{\displaystyle \lim\limits\_{x\longrightarrow x\_{0}} \left(\frac{f(x) - f(x\_{0}) - f'(x\_{0})(x-x\_{0})}{(x-x\_{0})^{2}} - \frac{\frac{f''(x\_{0})}{2}(x-x\_{0})^{2}}{(x-x\_{0})^{2}} \right) = 0}

Donc

lim

x
⟶

x

0

(

f
(
x
)
−
f
(

x

0

)
−

f
′

(

x

0

)
(
x
−

x

0

)
−

(
x
−

x

0

)

2

2

f
″

(

x

0

)

(
x
−

x

0

)

2

)

=
0

{\displaystyle \lim\limits\_{x\longrightarrow x\_{0}} \left(\frac{f(x) - f(x\_{0}) - f'(x\_{0})(x-x\_{0}) - \frac{(x-x\_{0})^{2}}{2}f''(x\_{0})}{(x-x\_{0})^{2}} \right) = 0}

Donc

f
(
x
)
−
f
(

x

0

)
−

f
′

(

x

0

)
(
x
−

x

0

)
−

(
x
−

x

0

)

2

2

f
″

(

x

0

# )
o

[

(
x
−

x

0

)

2

]

{\displaystyle f(x) - f(x\_{0}) - f'(x\_{0})(x-x\_{0}) - \frac{(x-x\_{0})^{2}}{2}f''(x\_{0}) = o\left[(x-x\_{0})^{2}\right]}

 quand

x
⟶

x

0

{\displaystyle x\longrightarrow x\_{0}}

Donc

f
(
x
# )
f
(

x

0

)

f
′

(

x

0

)
(
x
−

x

0

)

(
x
−

x

0

)

2

2

f
″

(

x

0

)
o

[

(
x
−

x

0

)

2

]

{\displaystyle f(x) = f(x\_{0}) + f'(x\_{0})(x-x\_{0}) + \frac{(x-x\_{0})^{2}}{2}f''(x\_{0}) + o\left[(x-x\_{0})^{2}\right]}

 quand

x
⟶

x

0

{\displaystyle x\longrightarrow x\_{0}}

 DL à l'ordre 2 de

f

{\displaystyle f}

## Développement limité l'ordre 3

Soit

f

{\displaystyle f}

 3 fois dérivable en

x
0

{\displaystyle x\_0}

On calcule

lim

x
⟶

x

0

(

f
(
x
)
−
f
(

x

0

)
−

f
′

(

x

0

)
(
x
−

x

0

)
−

(
x
−

x

0

)

2

2

f
″

(

x

0

)

(
x
−

x

0

)

3

)

{\displaystyle \lim\limits\_{x\longrightarrow x\_{0}}\left(\frac{f(x) - f(x\_{0}) - f'(x\_{0})(x-x\_{0}) - \frac{(x-x\_{0})^{2}}{2}f''(x\_{0})}{(x-x\_{0})^{3}}\right)}

On pose

g
(
x
# )
f
(
x
)
−
f
(

x

0

)
−

f
′

(

x

0

)
(
x
−

x

0

)
−

(
x
−

x

0

)

2

2

f
″

(

x

0

)

{\displaystyle g(x) = f(x) - f(x\_{0}) - f'(x\_{0})(x-x\_{0}) - \frac{(x-x\_{0})^{2}}{2}f''(x\_{0})}

et

h
(
x
# )
(
x
−

x

0

)

3

{\displaystyle h(x) = (x-x\_{0})^{3}}

Donc

g
′

(
x
# )

f
′

(
x
)
−

f
′

(

x

0

)
−

f
″

(

x

0

)
(
x
−

x

0

)

{\displaystyle g'(x) = f'(x) - f'(x\_{0}) - f''(x\_{0})(x-x\_{0})}

et

h
′

(
x
# )
3
(
x
−

x

0

)

2

{\displaystyle h'(x) = 3(x-x\_{0})^{2}}

Donc

g
″

(
x
# )

f
″

(
x
)
−

f
″

(

x

0

)

{\displaystyle g''(x) = f''(x) - f''(x\_{0})}

et

h
″

(
x
# )
6
(
x
−

x

0

)

{\displaystyle h''(x) = 6(x-x\_{0})}

Donc

lim

x
⟶

x

0

(

g
″

(
x
)

h
″

(
x
)

)

=

f

(
3
)

(

x

0

)

6

{\displaystyle \lim\limits\_{x\longrightarrow x\_{0}}\left(\frac{g''(x)}{h''(x)}\right) = \frac{f^{(3)}(x\_{0})}{6}}

D'après la [Règle de l'Hôpital],

lim

x
⟶

x

0

(

g
(
x
)

h
(
x
)

)

=

lim

x
⟶

x

0

(

g
′

(
x
)

h
′

(
x
)

)

=

lim

x
⟶

x

0

(

g
″

(
x
)

h
″

(
x
)

)

=

f

(
3
)

(

x

0

)

6

{\displaystyle \lim\limits\_{x\longrightarrow x\_{0}}\left(\frac{g(x)}{h(x)}\right) = \lim\limits\_{x\longrightarrow x\_{0}}\left(\frac{g'(x)}{h'(x)}\right) = \lim\limits\_{x\longrightarrow x\_{0}}\left(\frac{g''(x)}{h''(x)}\right) = \frac{f^{(3)}(x\_{0})}{6}}

Donc

lim

x
⟶

x

0

(

f
(
x
)
−
f
(

x

0

)
−

f
′

(

x

0

)
(
x
−

x

0

)
−

(
x
−

x

0

)

2

2

f
″

(

x

0

)

(
x
−

x

0

)

3

)

=

f

(
3
)

(

x

0

)

6

{\displaystyle \lim\limits\_{x\longrightarrow x\_{0}}\left(\frac{f(x) - f(x\_{0}) - f'(x\_{0})(x-x\_{0}) - \frac{(x-x\_{0})^{2}}{2}f''(x\_{0})}{(x-x\_{0})^{3}}\right) = \frac{f^{(3)}(x\_{0})}{6}}

Donc

lim

x
⟶

x

0

(

f
(
x
)
−
f
(

x

0

)
−

f
′

(

x

0

)
(
x
−

x

0

)
−

(
x
−

x

0

)

2

2

f
″

(

x

0

)

(
x
−

x

0

)

3

−

f

(
3
)

(

x

0

)

6

)

=
0

{\displaystyle \lim\limits\_{x\longrightarrow x\_{0}}\left(\frac{f(x) - f(x\_{0}) - f'(x\_{0})(x-x\_{0}) - \frac{(x-x\_{0})^{2}}{2}f''(x\_{0})}{(x-x\_{0})^{3}} - \frac{f^{(3)}(x\_{0})}{6}\right) = 0}

Donc

lim

x
⟶

x

0

(

f
(
x
)
−
f
(

x

0

)
−

f
′

(

x

0

)
(
x
−

x

0

)
−

(
x
−

x

0

)

2

2

f
″

(

x

0

)

(
x
−

x

0

)

3

−

f

(
3
)

(

x

0

)

6

(
x
−

x

0

)

3

(
x
−

x

0

)

3

)

=
0

{\displaystyle \lim\limits\_{x\longrightarrow x\_{0}}\left(\frac{f(x) - f(x\_{0}) - f'(x\_{0})(x-x\_{0}) - \frac{(x-x\_{0})^{2}}{2}f''(x\_{0})}{(x-x\_{0})^{3}} - \frac{\frac{f^{(3)}(x\_{0})}{6}(x-x\_{0})^{3}}{(x-x\_{0})^{3}}\right) = 0}

Donc

lim

x
⟶

x

0

(

f
(
x
)
−
f
(

x

0

)
−

f
′

(

x

0

)
(
x
−

x

0

)
−

(
x
−

x

0

)

2

2

f
″

(

x

0

)
−

(
x
−

x

0

)

3

6

f

(
3
)

(

x

0

)

(
x
−

x

0

)

3

)

=
0

{\displaystyle \lim\limits\_{x\longrightarrow x\_{0}}\left(\frac{f(x) - f(x\_{0}) - f'(x\_{0})(x-x\_{0}) - \frac{(x-x\_{0})^{2}}{2}f''(x\_{0}) - \frac{(x-x\_{0})^{3}}{6}f^{(3)}(x\_{0})}{(x-x\_{0})^{3}}\right) = 0}

Donc

f
(
x
)
−
f
(

x

0

)
−

f
′

(

x

0

)
(
x
−

x

0

)
−

(
x
−

x

0

)

2

2

f
″

(

x

0

)
−

(
x
−

x

0

)

3

6

f

(
3
)

(

x

0

# )
o

[

(
x
−

x

0

)

3

]

{\displaystyle f(x) - f(x\_{0}) - f'(x\_{0})(x-x\_{0}) - \frac{(x-x\_{0})^{2}}{2}f''(x\_{0}) - \frac{(x-x\_{0})^{3}}{6}f^{(3)}(x\_{0}) = o\left[(x-x\_{0})^{3}\right]}

Donc

f
(
x
# )
f
(

x

0

)

f
′

(

x

0

)
(
x
−

x

0

)

(
x
−

x

0

)

2

2

f
″

(

x

0

)

(
x
−

x

0

)

3

6

f

(
3
)

(

x

0

)
o

[

(
x
−

x

0

)

3

]

{\displaystyle f(x) = f(x\_{0}) + f'(x\_{0})(x-x\_{0}) + \frac{(x-x\_{0})^{2}}{2}f''(x\_{0}) + \frac{(x-x\_{0})^{3}}{6}f^{(3)}(x\_{0}) + o\left[(x-x\_{0})^{3}\right]}

 DL à l'ordre 3 de

f

{\displaystyle f}

## Démonstration

On veut démontrer que

f
(
x

# k
0

n

(

f

(
k
)

(

x

0

)
×

(
x
−

x

0

)

k

k
!

)

o

[

(
x
−

x

0

)

n

]

{\displaystyle f(x) = \sum\limits\_{k=0}^{n}\left( f^{(k)}(x\_{0}) \times \frac{(x-x\_{0})^{k}}{k!}\right) + o\left[(x-x\_{0})^{n}\right]}

.

On veut calculer

lim

x
→

x

0

(

f
(
x
)
−

∑

# k
0

n
−
1

(

f

(
k
)

(

x

0

)
×

(
x
−

x

0

)

k

k
!

)

(
x
−

x

0

)

n

)

{\displaystyle \lim\limits\_{x\to x\_{0}}\left( \frac{f(x) - \sum\limits\_{k=0}^{n-1}\left( f^{(k)}(x\_{0}) \times \frac{(x-x\_{0})^{k}}{k!}\right)}{(x-x\_{0})^{n}}\right)}

Soit

g
(
x
# )
f
(
x
)
−

∑

# k
0

n
−
1

(

f

(
k
)

(

x

0

)
×

(
x
−

x

0

)

k

k
!

)

{\displaystyle g(x) = f(x) - \sum\limits\_{k=0}^{n-1}\left( f^{(k)}(x\_{0}) \times \frac{(x-x\_{0})^{k}}{k!}\right)}

Et

h
(
x
# )
(
x
−

x

0

)

n

{\displaystyle h(x) = (x-x\_{0})^{n}}

### Première démonstration par récurrence: les dérivées de g

Le but de cette première démonstration par récurrence est de montrer que

g

(
n
−
1
)

(
x
# )

f

(
n
−
1
)

(
x
)
−

f

(
n
−
1
)

(

x

0

)

{\displaystyle g^{(n-1)}(x) = f^{(n-1)}(x) - f^{(n-1)}(x\_{0})}

Mais pour cela, on doit montrer par récurrence que

g

(
m
)

(
x
# )

f

(
m
)

(
x
)
−

∑

# k
0

n
−
1
−
m

(

f

(
k
m
)

(

x

0

)
×

(
x
−

x

0

)

k

k
!

)

{\displaystyle g^{(m)}(x) = f^{(m)}(x) - \sum\limits\_{k=0}^{n-1-m}\left(f^{(k+m)}(x\_{0}) \times \frac{(x-x\_{0})^{k}}{k!}\right)}

 et ce

∀
m
∈

N

{\displaystyle \forall m \in \mathbb{N}}

**Initialisation**

Pour

# m
1

{\displaystyle m = 1}

On veut donc montrer que

g
′

(
x
# )

f
′

(
x
)
−

∑

# k
0

n
−
2

(

f

(
k
1
)

(

x

0

)
×

(
x
−

x

0

)

k

k
!

)

{\displaystyle g'(x) = f'(x) - \sum\limits\_{k=0}^{n-2}\left(f^{(k+1)}(x\_{0}) \times \frac{(x-x\_{0})^{k}}{k!}\right)}

Or, on sait que

g
(
x
# )
f
(
x
)
−

∑

# k
0

n
−
1

(

f

(
k
)

(

x

0

)
×

(
x
−

x

0

)

k

k
!

)

{\displaystyle g(x) = f(x) - \sum\limits\_{k=0}^{n-1}\left( f^{(k)}(x\_{0}) \times \frac{(x-x\_{0})^{k}}{k!}\right)}

Donc

g
′

(
x
# )

f
′

(
x
)
−

∑

# k
0

n
−
1

(

d

[

f

(
k
)

(

x

0

)
×

(
x
−

x

0

)

k

k
!

]

d
x

)

{\displaystyle g'(x) = f'(x) - \sum\limits\_{k=0}^{n-1}\left( \frac{d\left[f^{(k)}(x\_{0}) \times \frac{(x-x\_{0})^{k}}{k!}\right]}{dx}\right)}

On s'intéresse à

d

[

f

(
k
)

(

x

0

)
×

(
x
−

x

0

)

k

k
!

]

d
x

{\displaystyle \frac{d\left[f^{(k)}(x\_{0}) \times \frac{(x-x\_{0})^{k}}{k!}\right]}{dx}}

* Si

# k
0

{\displaystyle k=0}

, alors la dérivée de ce terme de devient nulle, et donc on enlève le terme de toute la somme.

* Pour

k
≥
1

{\displaystyle k \ge 1}

, on a

d

[

f

(
k
)

(

x

0

)
×

(
x
−

x

0

)

k

k
!

]

d
x

=

f

(
k
)

(

x

0

)
×

(
x
−

x

0

)

k
−
1

(
k
−
1
)
!

{\displaystyle \frac{d\left[f^{(k)}(x\_{0}) \times \frac{(x-x\_{0})^{k}}{k!}\right]}{dx} = f^{(k)}(x\_{0}) \times \frac{(x-x\_{0})^{k-1}}{(k-1)!}}

On a donc

g
′

(
x
# )

f
′

(
x
)
−

∑

# k
1

n
−
1

(

f

(
k
)

(

x

0

)
×

(
x
−

x

0

)

k
−
1

(
k
−
1
)
!

)

{\displaystyle g'(x) = f'(x) - \sum\limits\_{k=1}^{n-1}\left( f^{(k)}(x\_{0}) \times \frac{(x-x\_{0})^{k-1}}{(k-1)!}\right)}

 (on part de

# k
1

{\displaystyle k = 1}

 car le terme pour

# k
0

{\displaystyle k=0}

 a disparu)

On pose

# p
k
−
1

{\displaystyle p = k-1}

Donc

g
′

(
x
# )

f
′

(
x
)
−

∑

# p
0

n
−
2

(

f

(
p
1
)

(

x

0

)
×

(
x
−

x

0

)

p

p
!

)

{\displaystyle g'(x) = f'(x) - \sum\limits\_{p=0}^{n-2}\left( f^{(p+1)}(x\_{0}) \times \frac{(x-x\_{0})^{p}}{p!}\right)}

Donc on a bien montré que

g
′

(
x
# )

f
′

(
x
)
−

∑

# k
0

n
−
2

(

f

(
k
1
)

(

x

0

)
×

(
x
−

x

0

)

k

k
!

)

{\displaystyle g'(x) = f'(x) - \sum\limits\_{k=0}^{n-2}\left(f^{(k+1)}(x\_{0}) \times \frac{(x-x\_{0})^{k}}{k!}\right)}

.

Donc c'est vrai pour

# m
1

{\displaystyle m = 1}

**Hérédité**

On va maintenant supposer que

g

(
m
)

(
x
# )

f

(
m
)

(
x
)
−

∑

# k
0

n
−
1
−
m

(

f

(
k
m
)

(

x

0

)
×

(
x
−

x

0

)

k

k
!

)

{\displaystyle g^{(m)}(x) = f^{(m)}(x) - \sum\limits\_{k=0}^{n-1-m}\left(f^{(k+m)}(x\_{0}) \times \frac{(x-x\_{0})^{k}}{k!}\right)}

 pour un seul

m

{\displaystyle m}

 quelconque.

On veut montrer que ça reste vrai pour

m
1

{\displaystyle m+1}

, c'est-à-dire que

g

(
m
1
)

(
x
# )

f

(
m
1
)

(
x
)
−

∑

# k
0

n
−
1
−
(
m
1
)

(

f

(
k
m
1
)

(

x

0

)
×

(
x
−

x

0

)

k

k
!

)

{\displaystyle g^{(m+1)}(x) = f^{(m+1)}(x) - \sum\limits\_{k=0}^{n-1-(m+1)}\left(f^{(k+m+1)}(x\_{0}) \times \frac{(x-x\_{0})^{k}}{k!}\right)}

Donc on a

g

(
m
)

(
x
# )

f

(
m
)

(
x
)
−

∑

# k
0

n
−
1
−
m

(

f

(
k
m
)

(

x

0

)
×

(
x
−

x

0

)

k

k
!

)

{\displaystyle g^{(m)}(x) = f^{(m)}(x) - \sum\limits\_{k=0}^{n-1-m}\left(f^{(k+m)}(x\_{0}) \times \frac{(x-x\_{0})^{k}}{k!}\right)}

Donc

g

(
m
1
)

=

f

(
m
1
)

(
x
)
−

∑

# k
0

n
−
1
−
m

(

d

[

f

(
k
m
)

(

x

0

)
×

(
x
−

x

0

)

k

k
!

]

d
x

)

{\displaystyle g^{(m+1)} = f^{(m+1)}(x) - \sum\limits\_{k=0}^{n-1-m}\left(\frac{d\left[f^{(k+m)}(x\_{0}) \times \frac{(x-x\_{0})^{k}}{k!}\right]}{dx}\right)}

On s'intéresse à

d

[

f

(
k
m
)

(

x

0

)
×

(
x
−

x

0

)

k

k
!

]

d
x

{\displaystyle \frac{d\left[f^{(k+m)}(x\_{0}) \times \frac{(x-x\_{0})^{k}}{k!}\right]}{dx}}

* Si

# k
0

{\displaystyle k=0}

, alors la dérivée de ce terme de devient nulle, et donc on enlève le terme de toute la somme.

* Pour

k
≥
1

{\displaystyle k \ge 1}

, on a

d

[

f

(
k
m
)

(

x

0

)
×

(
x
−

x

0

)

k

k
!

]

d
x

=

f

(
k
m
)

(

x

0

)
×

(
x
−

x

0

)

k
−
1

(
k
−
1
)
!

{\displaystyle \frac{d\left[f^{(k+m)}(x\_{0}) \times \frac{(x-x\_{0})^{k}}{k!}\right]}{dx} = f^{(k+m)}(x\_{0}) \times \frac{(x-x\_{0})^{k-1}}{(k-1)!}}

On a donc

g

(
m
1
)

=

f

(
m
1
)

(
x
)
−

∑

# k
1

n
−
1
−
m

(

f

(
k
m
)

(

x

0

)
×

(
x
−

x

0

)

k

k
!

)

{\displaystyle g^{(m+1)} = f^{(m+1)}(x) - \sum\limits\_{k=1}^{n-1-m}\left(f^{(k+m)}(x\_{0}) \times \frac{(x-x\_{0})^{k}}{k!}\right)}

On pose

# p
k
−
1

{\displaystyle p = k-1}

On a donc

g

(
m
1
)

=

f

(
m
1
)

(
x
)
−

∑

# p
0

n
−
1
−
(
m
1
)

(

f

(
p
m
1
)

(

x

0

)
×

(
x
−

x

0

)

p

p
!

)

{\displaystyle g^{(m+1)} = f^{(m+1)}(x) - \sum\limits\_{p=0}^{n-1-(m+1)}\left(f^{(p+m+1)}(x\_{0}) \times \frac{(x-x\_{0})^{p}}{p!}\right)}

On a donc bien montré que

g

(
m
1
)

(
x
# )

f

(
m
1
)

(
x
)
−

∑

# k
0

n
−
1
−
(
m
1
)

(

f

(
k
m
1
)

(

x

0

)
×

(
x
−

x

0

)

k

k
!

)

{\displaystyle g^{(m+1)}(x) = f^{(m+1)}(x) - \sum\limits\_{k=0}^{n-1-(m+1)}\left(f^{(k+m+1)}(x\_{0}) \times \frac{(x-x\_{0})^{k}}{k!}\right)}

**Conclusion**

On a

g

(
m
)

(
x
# )

f

(
m
)

(
x
)
−

∑

# k
0

n
−
1
−
m

(

f

(
k
m
)

(

x

0

)
×

(
x
−

x

0

)

k

k
!

)

{\displaystyle g^{(m)}(x) = f^{(m)}(x) - \sum\limits\_{k=0}^{n-1-m}\left(f^{(k+m)}(x\_{0}) \times \frac{(x-x\_{0})^{k}}{k!}\right)}

 et ce

∀
m
∈

N

{\displaystyle \forall m \in \mathbb{N}}

.

Le but de cette démonstration par récurrence est de montrer que

g

(
n
−
1
)

(
x
# )

f

(
n
−
1
)

(
x
)
−

f

(
n
−
1
)

(

x

0

)

{\displaystyle g^{(n-1)}(x) = f^{(n-1)}(x) - f^{(n-1)}(x\_{0})}

Ici donc,

# m
n
−
1

{\displaystyle m = n-1}

Donc d'après notre formule connue, on a

g

(
n
−
1
)

(
x
# )

f

(
n
−
1
)

(
x
)
−

∑

# k
0

n
−
1
−
(
n
−
1
)

(

f

(
k
(
n
−
1
)
)

(

x

0

)
×

(
x
−

x

0

)

k

k
!

)

{\displaystyle g^{(n-1)}(x) = f^{(n-1)}(x) - \sum\limits\_{k=0}^{n-1-(n-1)}\left(f^{(k+(n-1))}(x\_{0}) \times \frac{(x-x\_{0})^{k}}{k!}\right)}

On remarque que

k

{\displaystyle k}

 varie de

0

{\displaystyle 0}

 à

0

{\displaystyle 0}

: on retire sigma et on remplace

k

{\displaystyle k}

 par

0

{\displaystyle 0}

.

On obtient donc

g

(
n
−
1
)

(
x
# )

f

(
n
−
1
)

(
x
)
−

f

(
n
−
1
)

(

x

0

)

{\displaystyle g^{(n-1)}(x) = f^{(n-1)}(x) - f^{(n-1)}(x\_{0})}

### Deuxième démonstration par récurrence: les dérivées de h

On avait dit que

h
(
x
# )
(
x
−

x

0

)

n

{\displaystyle h(x) = (x-x\_{0})^{n}}

On veut montrer que

h

(
n
−
1
)

(
x
# )
n
!
(
x
−

x

0

)

{\displaystyle h^{(n-1)}(x) = n!(x-x\_{0})}

 et ce

∀
n
∈

N

{\displaystyle \forall n\in \N}

On va donc montrer par récurrence que pour obtenir la dérivée

m

{\displaystyle m}

-ième de

h

{\displaystyle h}

 (

m
<
n

{\displaystyle m < n}

), on utilise la formule

h

(
m
)

(
x
# )

n
!

(
n
−
m
)
!

(
x
−

x

0

)

n
−
m

{\displaystyle h^{(m)}(x) = \frac{n!}{(n-m)!}(x-x\_{0})^{n-m}}

**Initialisation**

Pour

# n
1

{\displaystyle n=1}

, on veut montrer que

h

(
1
)

(
x
# )

h
′

(
x
# )

n
!

(
n
−
1
)
!

(
x
−

x

0

)

n
−
1

=
n
(
x
−

x

0

)

n
−
1

{\displaystyle h^{(1)}(x) = h'(x) = \frac{n!}{(n-1)!}(x-x\_{0})^{n-1} = n(x-x\_{0})^{n-1}}

h
(
x
# )
(
x
−

x

0

)

n

=

e

n
×
l
n
(
x
−

x

0

)

=

e

u
(
x
)

{\displaystyle h(x) = (x-x\_{0})^{n} = e^{n \times ln(x-x\_{0})} = e^{u(x)}}

 avec

u
(
x
# )
n
×
l
n
(
x
−

x

0

)

{\displaystyle u(x) = n \times ln(x-x\_{0})}

Donc

h
′

(
x
# )

u
′

(
x
)
×

e

u
(
x
)

=

u
′

(
x
)
×
h
(
x
)

{\displaystyle h'(x) = u'(x) \times e^{u(x)} = u'(x) \times h(x)}

Or

u
′

(
x
# )

n

x
−

x

0

{\displaystyle u'(x) = \frac{n}{x-x\_{0}}}

Donc

h
′

(
x
# )

n

x
−

x

0

×
(
x
−

x

0

)

n

{\displaystyle h'(x) = \frac{n}{x-x\_{0}} \times (x-x\_{0})^{n}}

Donc

h
′

(
x
# )
n
×
(
x
−

x

0

)

n
−
1

{\displaystyle h'(x) = n\times(x-x\_{0})^{n-1}}

Donc c'est vrai pour

# n
1

{\displaystyle n=1}

.

**Hérédité**

On suppose que

h

(
m
)

(
x
# )

n
!

(
n
−
m
)
!

(
x
−

x

0

)

n
−
m

{\displaystyle h^{(m)}(x) = \frac{n!}{(n-m)!}(x-x\_{0})^{n-m}}

 est vrai pour un certain

m

{\displaystyle m}

 quelconque: on veut montrer que ça reste vrai pour

m
1

{\displaystyle m+1}

c'est-à-dire que

h

(
m
1
)

(
x
# )

n
!

(
n
−
m
−
1
)
!

(
x
−

x

0

)

n
−
m
−
1

{\displaystyle h^{(m+1)}(x) = \frac{n!}{(n-m-1)!}(x-x\_{0})^{n-m-1}}

On sait que

h

(
m
)

(
x
# )

n
!

(
n
−
m
)
!

(
x
−

x

0

)

n
−
m

=

n
!

(
n
−
m
)
!

×

e

(
n
−
m
)
×
l
n
(
x
−

x

0

)

=

n
!

(
n
−
m
)
!

×

e

u
(
x
)

{\displaystyle h^{(m)}(x) = \frac{n!}{(n-m)!}(x-x\_{0})^{n-m} = \frac{n!}{(n-m)!} \times e^{(n-m)\times ln(x-x\_{0})} = \frac{n!}{(n-m)!} \times e^{u(x)}}

avec

u
(
x
# )
(
n
−
m
)
×
l
n
(
x
−

x

0

)

{\displaystyle u(x) = (n-m)\times ln(x-x\_{0})}

 donc

u
′

(
x
# )

n
−
m

x
−

x

0

{\displaystyle u'(x) = \frac{n-m}{x-x\_{0}}}

Or

h

(
m
1
)

(
x
# )

n
!

(
n
−
m
)
!

×

u
′

(
x
)
×

e

u
(
x
)

=

n
!

(
n
−
m
)
!

×

u
′

(
x
)
×
(
x
−

x

0

)

n
−
m

{\displaystyle h^{(m+1)}(x) = \frac{n!}{(n-m)!} \times u'(x) \times e^{u(x)} = \frac{n!}{(n-m)!} \times u'(x) \times (x-x\_{0})^{n-m} }

Donc

h

(
m
1
)

(
x
# )

n
!

(
n
−
m
)
!

×

n
−
m

x
−

x

0

×
(
x
−

x

0

)

n
−
m

{\displaystyle h^{(m+1)}(x) = \frac{n!}{(n-m)!} \times \frac{n-m}{x-x\_{0}} \times (x-x\_{0})^{n-m}}

Donc

h

(
m
1
)

(
x
# )

n
!

(
n
−
m
−
1
)
!

×
(
x
−

x

0

)

n
−
m
−
1

{\displaystyle h^{(m+1)}(x) = \frac{n!}{(n-m-1)!} \times (x-x\_{0})^{n-m-1}}

Donc ça reste vrai pour

m
1

{\displaystyle m+1}

.

**Conclusion**

h

(
m
)

(
x
# )

n
!

(
n
−
m
)
!

(
x
−

x

0

)

n
−
m

{\displaystyle h^{(m)}(x) = \frac{n!}{(n-m)!}(x-x\_{0})^{n-m}}

Au final, on veut montrer que

h

(
n
−
1
)

(
x
# )
n
!
(
x
−

x

0

)

{\displaystyle h^{(n-1)}(x) = n!(x-x\_{0})}

Ici,

# m
n
−
1

{\displaystyle m = n-1}

, donc

h

(
m
)

(
x
# )

h

(
n
−
1
)

(
x
# )

n
!

(
n
−
(
n
−
1
)
)
!

(
x
−

x

0

)

n
−
(
n
−
1
)

=
n
!
(
x
−

x

0

)

{\displaystyle h^{(m)}(x) = h^{(n-1)}(x) =\frac{n!}{(n-(n-1))!}(x-x\_{0})^{n-(n-1)} = n!(x-x\_{0})}

### Fin de la démonstration

On a montré que

g

(
n
−
1
)

(
x
# )

f

(
n
−
1
)

(
x
)
−

f

(
n
−
1
)

(

x

0

)

{\displaystyle g^{(n-1)}(x) = f^{(n-1)}(x) - f^{(n-1)}(x\_{0})}

 et que

h

(
n
−
1
)

(
x
# )
n
!
(
x
−

x

0

)

{\displaystyle h^{(n-1)}(x) = n!(x-x\_{0})}

.

Donc

lim

x
→

x

0

(

g

(
n
−
1
)

(
x
)

h

(
n
−
1
)

(
x
)

)

=

lim

x
→

x

0

(

f

(
n
−
1
)

(
x
)
−

f

(
n
−
1
)

(

x

0

)

n
!
(
x
−

x

0

)

)

=

f

(
n
)

(

x

0

)

n
!

{\displaystyle \lim\limits\_{x\to x\_{0}}\left(\frac{g^{(n-1)}(x)}{h^{(n-1)}(x)}\right) = \lim\limits\_{x\to x\_{0}}\left(\frac{f^{(n-1)}(x) - f^{(n-1)}(x\_{0})}{n!(x-x\_{0})}\right) = \frac{f^{(n)}(x\_{0})}{n!}}

Donc

lim

x
→

x

0

(

f
(
x
)
−

∑

# k
0

n
−
1

[

f

(
k
)

(

x

0

)
×

(
x
−

x

0

)

k

k
!

]

(
x
−

x

0

)

n

)

=

lim

x
→

x

0

(

g
(
x
)

h
(
x
)

)

=

lim

x
→

x

0

(

g

(
n
−
1
)

(
x
)

h

(
n
−
1
)

(
x
)

)

=

f

(
n
)

(

x

0

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n
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{\displaystyle \lim\limits\_{x\to x\_{0}}\left( \frac{f(x) - \sum\limits\_{k=0}^{n-1}\left[ f^{(k)}(x\_{0}) \times \frac{(x-x\_{0})^{k}}{k!}\right]}{(x-x\_{0})^{n}}\right) = \lim\limits\_{x\to x\_{0}}\left(\frac{g(x)}{h(x)}\right) = \lim\limits\_{x\to x\_{0}}\left(\frac{g^{(n-1)}(x)}{h^{(n-1)}(x)}\right) = \frac{f^{(n)}(x\_{0})}{n!}}

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{\displaystyle \lim\limits\_{x\to x\_{0}}\left( \frac{f(x) - \sum\limits\_{k=0}^{n-1}\left[ f^{(k)}(x\_{0}) \times \frac{(x-x\_{0})^{k}}{k!}\right]}{(x-x\_{0})^{n}}\right) = \frac{f^{(n)}(x\_{0})}{n!}}

Donc

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# k
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{\displaystyle \lim\limits\_{x\to x\_{0}}\left( \frac{f(x) - \sum\limits\_{k=0}^{n-1}\left[ f^{(k)}(x\_{0}) \times \frac{(x-x\_{0})^{k}}{k!}\right]}{(x-x\_{0})^{n}} - \frac{f^{(n)}(x\_{0})}{n!} \right) = 0}

Donc

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f
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# k
0

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{\displaystyle \lim\limits\_{x\to x\_{0}}\left( \frac{f(x) - \sum\limits\_{k=0}^{n-1}\left[ f^{(k)}(x\_{0}) \times \frac{(x-x\_{0})^{k}}{k!}\right]}{(x-x\_{0})^{n}} - \frac{\frac{f^{(n)}(x\_{0})}{n!}(x-x\_{0})^{n}}{(x-x\_{0})^{n}} \right) = 0}

Donc

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x
→

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(

f
(
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# k
0

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{\displaystyle \lim\limits\_{x\to x\_{0}}\left( \frac{f(x) - \sum\limits\_{k=0}^{n-1}\left[ f^{(k)}(x\_{0}) \times \frac{(x-x\_{0})^{k}}{k!}\right] - \frac{f^{(n)}(x\_{0})}{n!}(x-x\_{0})^{n}}{(x-x\_{0})^{n}} \right) = 0}

Donc

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x
→

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(

f
(
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∑

# k
0

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1

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k
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n
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f

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0

{\displaystyle \lim\limits\_{x\to x\_{0}}\left( \frac{f(x) - \sum\limits\_{k=0}^{n-1}\left[ f^{(k)}(x\_{0}) \times \frac{(x-x\_{0})^{k}}{k!}\right] - \frac{(x-x\_{0})^{n}}{n!}f^{(n)}(x\_{0})}{(x-x\_{0})^{n}} \right) = 0}

Donc

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x
→

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(

f
(
x
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−

∑

# k
0

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[

f

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k
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x

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k
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{\displaystyle \lim\limits\_{x\to x\_{0}}\left( \frac{f(x) - \sum\limits\_{k=0}^{n}\left[ f^{(k)}(x\_{0}) \times \frac{(x-x\_{0})^{k}}{k!}\right]}{(x-x\_{0})^{n}} \right) = 0}

Donc

f
(
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# k
0

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f

(
k
)

(

x

0

)
×

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x
−

x

0

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k

k
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=
o

[

(
x
−

x

0

)

n

]

{\displaystyle f(x) - \sum\limits\_{k=0}^{n}\left[ f^{(k)}(x\_{0}) \times \frac{(x-x\_{0})^{k}}{k!}\right] = o\left[(x-x\_{0})^{n}\right]}

Donc

f
(
x

# k
0

n

[

f

(
k
)

(

x

0

)
×

(
x
−

x

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k

k
!

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o

[

(
x
−

x

0

)

n

]

{\displaystyle f(x) = \sum\limits\_{k=0}^{n}\left[ f^{(k)}(x\_{0}) \times \frac{(x-x\_{0})^{k}}{k!}\right] + o\left[(x-x\_{0})^{n}\right]}