f ( x ) = e x e 2 x + e f(x) = \frac{e^x}{e^{2x} + e} f ( x ) = e 2 x + e e x 1/a/ ( ∀ x ∈ R ) (\forall x\in\R) ( ∀ x ∈ R )
f ( 1 − x ) = ( e 1 − x ) e 2 x − 1 ( e 2 ( 1 − x ) + e ) e 2 x − 1 = e x e + e 2 x = f ( x ) \begin{align*}
f(1-x) &=\frac{(e^{1-x}){\color{red}e^{2x-1}}}{(e^{2(1-x)} + e){\color{red}e^{2x-1}}} \\
&=\frac{e^x}{e+e^{2x}}\\
&=f(x)
\end{align*} f ( 1 − x ) = ( e 2 ( 1 − x ) + e ) e 2 x − 1 ( e 1 − x ) e 2 x − 1 = e + e 2 x e x = f ( x ) car e 2 x − 1 ≠ 0 \text{car }e^{2x-1}\ne0 car e 2 x − 1 = 0
1/b/
Rappel g g g D g D_g D g ( C g ) (C_g) ( C g )
La droite d’équation x = a x=a x = a ( C g ) (C_g) ( C g )
∀ x ∈ D g : 2 a − x ∈ D g \forall x\in D_g ~:~ 2a-x\in D_g ∀ x ∈ D g : 2 a − x ∈ D g g ( 2 a − x ) = g ( x ) g(2a-x)=g(x) g ( 2 a − x ) = g ( x )
On a : f ( 2 × 1 2 − x ) = f ( x ) f(2\times\frac12-x)=f(x) f ( 2 × 2 1 − x ) = f ( x )
Donc la droite d’équation x = 1 2 x=\frac12 x = 2 1 ( Γ ) (\Gamma) ( Γ )
1/c/
lim x → − ∞ f ( x ) = lim x → − ∞ e x e 2 x + e = 0 \lim\limits_{x \to -\infty} f(x) = \lim\limits_{x \to -\infty} \frac{e^x}{e^{2x} + e}=0 x → − ∞ lim f ( x ) = x → − ∞ lim e 2 x + e e x = 0 car lim x → − ∞ e x = 0 \lim\limits_{x \to -\infty} e^x=0 x → − ∞ lim e x = 0
on pose t = 1 − x t=1-x t = 1 − x x = 1 − t x=1-t x = 1 − t
x → + ∞ ⟺ t → − ∞ x\to+\infty \iff t\to-\infty x → + ∞ ⟺ t → − ∞ lim x → + ∞ f ( x ) = lim t → − ∞ f ( 1 − t ) = lim t → − ∞ f ( t ) = 0 \begin{align*}
\lim\limits_{x \to +\infty} f(x) &= \lim\limits_{t \to -\infty} f(1-t)\\
&=\lim\limits_{t \to -\infty} f(t)=0
\end{align*} x → + ∞ lim f ( x ) = t → − ∞ lim f ( 1 − t ) = t → − ∞ lim f ( t ) = 0 1/d/
I.G : la droite d’équation y = 0 y=0 y = 0 ( Γ ) (\Gamma) ( Γ ) V − ∞ \mathcal{V}_{-\infty} V − ∞ V + ∞ \mathcal{V}_{+\infty} V + ∞
2/a/ la fonction f f f R \R R R \R R
( ∀ x ∈ R ) (\forall x\in\R) ( ∀ x ∈ R )
f ′ ( x ) = e x ( e 2 x + e ) − e x . 2 e 2 x ( e 2 x + e ) 2 = e x e 2 x + e × e 2 x + e − 2 e 2 x e 2 x + e = f ( x ) ( e − e 2 x ) e − 1 ( e + e 2 x ) e − 1 = f ( x ) 1 − e 2 x − 1 1 + e 2 x − 1 \begin{align*}
f'(x) &=\dfrac{e^x(e^{2x}+e)-e^x.2e^{2x}}{(e^{2x}+e)^2} \\
&=\frac{e^x}{e^{2x} + e} \times \frac{e^{2x}+e-2e^{2x}}{e^{2x} + e} \\
&=f(x)\frac{(e-e^{2x}){\color{red}e^{-1}}}{(e+e^{2x}){\color{red}e^{-1}}}\\
&=f(x)\dfrac{1-e^{2x-1}}{1+e^{2x-1}}
\end{align*} f ′ ( x ) = ( e 2 x + e ) 2 e x ( e 2 x + e ) − e x .2 e 2 x = e 2 x + e e x × e 2 x + e e 2 x + e − 2 e 2 x = f ( x ) ( e + e 2 x ) e − 1 ( e − e 2 x ) e − 1 = f ( x ) 1 + e 2 x − 1 1 − e 2 x − 1 2/b/
( ∀ x ∈ R ) (\forall x\in\R) ( ∀ x ∈ R ) f ( x ) 1 + e 2 x − 1 > 0 \dfrac{f(x)}{1+e^{2x-1}}>0 1 + e 2 x − 1 f ( x ) > 0
car : ( ∀ X ∈ R ) : e X > 0 \quad (\forall X\in\R)~:\quad e^X>0 ( ∀ X ∈ R ) : e X > 0
f ′ ( x ) = 0 ⟺ 1 − e 2 x − 1 = 0 ⟺ e 2 x − 1 = 1 = e 0 ⟺ 2 x − 1 = 0 ⟺ x = 1 2 \begin{align*}
f'(x)=0 &\iff 1-e^{2x-1}=0\\
&\iff e^{2x-1}=1=e^0 \\
&\iff 2x-1=0 \\
&\iff x=\frac12
\end{align*} f ′ ( x ) = 0 ⟺ 1 − e 2 x − 1 = 0 ⟺ e 2 x − 1 = 1 = e 0 ⟺ 2 x − 1 = 0 ⟺ x = 2 1 x ≥ 1 2 ⟹ 2 x − 1 ≥ 0 ⟹ e 2 x − 1 ≥ e 0 ⟹ 1 − e 2 x − 1 ≤ 0 ⟹ f ′ ( x ) ≤ 0 \begin{align*}
x\ge \frac12 &\implies 2x-1\ge0 \\
&\implies e^{2x-1}\ge e^0\\
&\implies 1- e^{2x-1}\le 0 \\
&\implies f'(x)\le0
\end{align*} x ≥ 2 1 ⟹ 2 x − 1 ≥ 0 ⟹ e 2 x − 1 ≥ e 0 ⟹ 1 − e 2 x − 1 ≤ 0 ⟹ f ′ ( x ) ≤ 0 car x ↦ e x x\mapsto e^x x ↦ e x ↗ \nearrow ↗
et
x ≤ 1 2 ⟹ f ′ ( x ) ≥ 0 \begin{align*}
x\le \frac12 &\implies f'(x)\ge0
\end{align*} x ≤ 2 1 ⟹ f ′ ( x ) ≥ 0 Alors la fonction f f f ↗ \nearrow ↗ ] − ∞ , 1 2 ] ]-\infty,\frac12] ] − ∞ , 2 1 ] ↘ \searrow ↘ [ 1 2 , + ∞ [ [\frac12,+\infty[ [ 2 1 , + ∞ [
f f f R \R R
f ( 1 2 ) f(\frac12) f ( 2 1 ) f f f R \R R
( ∀ x ∈ R ) : f ( x ) ≤ f ( 1 2 ) (\forall x\in\R)~:~f(x)\le f(\frac12) ( ∀ x ∈ R ) : f ( x ) ≤ f ( 2 1 ) f ( 1 2 ) = e 2 e f(\frac12)=\frac{\sqrt e}{2e} f ( 2 1 ) = 2 e e or e < e \sqrt e < e e < e e > 1 e>1 e > 1 e e < 1 \frac{\sqrt e}{e}<1 e e < 1
et donc :
( ∀ x ∈ R ) : f ( x ) < 1 2 (\forall x\in\R)~:~f(x)< \frac12 ( ∀ x ∈ R ) : f ( x ) < 2 1 et on sait que ( ∀ x ∈ R ) : e x > 0 (\forall x\in\R)~:~e^x>0 ( ∀ x ∈ R ) : e x > 0
donc ( ∀ x ∈ R ) : f ( x ) > 0 (\forall x\in\R)~:f(x)>0 ( ∀ x ∈ R ) : f ( x ) > 0
conclusion :
( ∀ x ∈ R ) : 0 < f ( x ) < 1 2 (\forall x\in\R)~:~0<f(x)< \frac12 ( ∀ x ∈ R ) : 0 < f ( x ) < 2 1 3/ \quad ( Γ ) (\Gamma) ( Γ )
4/a/ on a f ( 1 − x ) = f ( x ) f(1-x)=f(x) f ( 1 − x ) = f ( x )
posons t = 1 − x t=1-x t = 1 − x x = 1 − t x=1-t x = 1 − t d x = − d t dx=-dt d x = − d t
x = 0 ⟹ t = 1 x=0 \implies t=1 x = 0 ⟹ t = 1 x = 1 2 ⟹ t = 1 2 x=\frac12 \implies t=\frac12 x = 2 1 ⟹ t = 2 1
∫ 0 1 2 f ( x ) d x = ∫ 1 1 2 f ( 1 − t ) ( − d t ) = − ∫ 1 1 2 f ( t ) d t = ∫ 1 2 1 f ( x ) d x \begin{align*}
\int_0^{\frac{1}{2}} f(x) dx &= \int_1^{\frac{1}{2}} f(1-t)(-dt)
\\&=- \int_1^{\frac{1}{2}} f(t)dt \\
\\ &=\int^1_{\frac{1}{2}} f(x)dx
\end{align*} ∫ 0 2 1 f ( x ) d x = ∫ 1 2 1 f ( 1 − t ) ( − d t ) = − ∫ 1 2 1 f ( t ) d t = ∫ 2 1 1 f ( x ) d x 4/b/
∫ 0 1 f ( x ) d x = ∫ 0 1 2 f ( x ) d x + ∫ 1 2 1 f ( x ) d x = 2 ∫ 0 1 2 f ( x ) d x \begin{align*}
\int_0^{1} f(x) dx&=\int_0^{\frac{1}{2}} f(x) dx+\int^1_{\frac{1}{2}} f(x) dx \\
&=2\int_0^{\frac{1}{2}} f(x) dx
\end{align*} ∫ 0 1 f ( x ) d x = ∫ 0 2 1 f ( x ) d x + ∫ 2 1 1 f ( x ) d x = 2 ∫ 0 2 1 f ( x ) d x 5/a/
posons t = e x t=e^x t = e x x = ln t x=\ln t x = ln t d x = d t t dx=\dfrac{dt}{t} d x = t d t
x = 0 ⟹ t = 1 x=0 \implies t=1 x = 0 ⟹ t = 1 x = 1 2 ⟹ t = e x=\frac12 \implies t=\sqrt e x = 2 1 ⟹ t = e
∫ 0 1 2 f ( x ) d x = ∫ 1 e f ( ln t ) d t t = ∫ 1 e d t t 2 + e \begin{align*}
\int_0^{\frac{1}{2}} f(x) dx &=\int_1^{\sqrt e} f(\ln t) \dfrac{dt}{t}
\\&=\int_1^{\sqrt e} \frac{dt}{t^2+e}
\end{align*} ∫ 0 2 1 f ( x ) d x = ∫ 1 e f ( ln t ) t d t = ∫ 1 e t 2 + e d t car f ( ln t ) = t t 2 + e f(\ln t)=\dfrac{t}{t^2+e} f ( ln t ) = t 2 + e t
5/b/
∫ 1 e d t t 2 + e = ∫ 1 e 1 e ( t e ) ′ ( t e ) 2 + 1 d t = 1 e [ arctan ( t e ) ] 1 e = 1 e [ π 4 − arctan ( 1 e ) ] \begin{align*}
\int_1^{\sqrt e} \frac{dt}{t^2+e}&=
\int_1^{\sqrt e} \frac{1}{\sqrt e}\frac{\left(\frac{t}{\sqrt e}\right)'}{\left(\frac{t}{\sqrt e}\right)^2+1}dt \\
&=\frac{1}{\sqrt e}\left[ \arctan\left(\frac{t}{\sqrt e}\right)\right]_1^{\sqrt e} \\
\\&=\frac{1}{\sqrt e}\left[\frac\pi4-\arctan\left(\frac{1}{\sqrt e}\right)\right]
\end{align*} ∫ 1 e t 2 + e d t = ∫ 1 e e 1 ( e t ) 2 + 1 ( e t ) ′ d t = e 1 [ arctan ( e t ) ] 1 e = e 1 [ 4 π − arctan ( e 1 ) ] Or
arctan ( 1 e ) = π 2 − arctan ( e ) \arctan\left(\frac{1}{\sqrt e}\right)=\frac\pi2-\arctan\left(\sqrt e\right) arctan ( e 1 ) = 2 π − arctan ( e ) Alors
∫ 1 e d t t 2 + e = 1 e [ arctan ( e ) − π 4 ] \int_1^{\sqrt e} \frac{dt}{t^2+e}=\frac{1}{\sqrt{e}}\left[\arctan \left( \sqrt{e}\right) - \frac{\pi}{4} \right] ∫ 1 e t 2 + e d t = e 1 [ arctan ( e ) − 4 π ] 5/c
A = 2 ∫ 0 1 ∣ f ( x ) ∣ d x c m 2 \boxed{\mathcal{A}=2\int_0^{1} |f(x)|dx~cm^2}~ A = 2 ∫ 0 1 ∣ f ( x ) ∣ d x c m 2 ∥ i ⃗ ∥ . ∥ j ⃗ ∥ = 2 c m 2 \|\vec{i}\|.\|\vec{j}\|=2cm^2 ∥ i ∥.∥ j ∥ = 2 c m 2 d’aprés 1/b/ on a f ( x ) > 0 f(x)>0 f ( x ) > 0
∣ f ( x ) ∣ = f ( x ) pour tout x ∈ [ 0 , 1 ] |f(x)| = f(x) \quad \text{pour tout } x \in [0, 1] ∣ f ( x ) ∣ = f ( x ) pour tout x ∈ [ 0 , 1 ] ∫ 0 1 f ( x ) d x = 2 ∫ 0 1 2 f ( x ) d x = 2 e [ arctan ( e ) − π 4 ] \begin{align*}
\int_0^{1} f(x)dx &=2\int_0^{\frac{1}{2}} f(x) dx
\\&=\frac{2}{\sqrt{e}}\left[\arctan \left( \sqrt{e}\right) - \frac{\pi}{4} \right]
\end{align*} ∫ 0 1 f ( x ) d x = 2 ∫ 0 2 1 f ( x ) d x = e 2 [ arctan ( e ) − 4 π ] Alors
A = 4 e [ arctan ( e ) − π 4 ] c m 2 \mathcal{A}=\frac{4}{\sqrt{e}}\left[\arctan \left( \sqrt{e}\right) - \frac{\pi}{4} \right]~cm^2 A = e 4 [ arctan ( e ) − 4 π ] c m 2 Partie II :
( ∀ n ∈ N ) (\forall n \in \mathbb{N}) ( ∀ n ∈ N )
u 0 ∈ ] 0 ; 1 2 [ e t u n + 1 = f ( u n ) u_0 \in \left] 0; \frac{1}{2} \right[ \quad et \quad u_{n+1} = f(u_n) u 0 ∈ ] 0 ; 2 1 [ e t u n + 1 = f ( u n ) 1/ soit x ∈ R x\in\R x ∈ R
d’aprés la question I-2-2 on a :
f ′ ( x ) = f ( x ) 1 − e 2 x − 1 1 + e 2 x − 1 f'(x) = f(x) \dfrac{1 - e^{2x - 1}}{1 + e^{2x - 1}} f ′ ( x ) = f ( x ) 1 + e 2 x − 1 1 − e 2 x − 1 il est claire que :
f ( x ) 1 + e 2 x − 1 > 0 \dfrac{f(x)}{1 + e^{2x - 1}}>0 1 + e 2 x − 1 f ( x ) > 0 et on a :
∣ 1 − e 2 x − 1 ∣ ≤ ∣ 1 ∣ + ∣ e 2 x − 1 ∣ ⟹ ∣ 1 − e 2 x − 1 ∣ ≤ 1 + e 2 x − 1 ⟹ ∣ 1 − e 2 x − 1 ∣ 1 + e 2 x − 1 ≤ 1 ⟹ ∣ f ′ ( x ) ∣ ≤ f ( x ) \begin{align*}
&|1 - e^{2x - 1}|\le |1|+|e^{2x - 1}| \\~\\
&\implies|1 - e^{2x - 1}|\le 1+e^{2x - 1} \\~\\
&\implies\frac{|1 - e^{2x - 1}|}{1+e^{2x - 1}}\le 1 \\~\\
&\implies |f'(x)|\le f(x)
\end{align*} ∣1 − e 2 x − 1 ∣ ≤ ∣1∣ + ∣ e 2 x − 1 ∣ ⟹ ∣1 − e 2 x − 1 ∣ ≤ 1 + e 2 x − 1 ⟹ 1 + e 2 x − 1 ∣1 − e 2 x − 1 ∣ ≤ 1 ⟹ ∣ f ′ ( x ) ∣ ≤ f ( x ) 2/a Soit x ∈ [ 0 , 1 2 ] x\in\left[0,\frac12\right] x ∈ [ 0 , 2 1 ]
d’aprés la question I-2-a on a :
0 < f ( x ) < 1 2 0<f(x)<\frac12 0 < f ( x ) < 2 1 et on a d’aprés la question précédente : f ′ ( x ) ≤ f ( x ) f'(x)\le f(x) f ′ ( x ) ≤ f ( x )
donc
f ′ ( x ) < 1 2 f'(x)<\frac12 f ′ ( x ) < 2 1 montrons f ′ ( x ) ≥ 0 f'(x)\ge0 f ′ ( x ) ≥ 0 x ∈ [ 0 , 1 2 ] x\in\left[0,\frac12\right] x ∈ [ 0 , 2 1 ]
le signe de f ′ ( x ) f'(x) f ′ ( x ) 1 − e 2 x − 1 1-e^{2x-1} 1 − e 2 x − 1
x ∈ [ 0 , 1 2 ] ⟹ x ≤ 1 2 ⟹ 2 x − 1 ≤ 0 ⟹ e 2 x − 1 ≤ e 0 = 1 ⟹ 1 − e 2 x − 1 ≥ 0 \begin{align*}
x\in\left[0,\frac12\right] &\implies x \le \frac12 \\
&\implies 2x-1 \le 0 \\
&\implies e^{2x-1}\le e^{0}=1 \\
&\implies 1- e^{2x-1}\ge 0
\end{align*} x ∈ [ 0 , 2 1 ] ⟹ x ≤ 2 1 ⟹ 2 x − 1 ≤ 0 ⟹ e 2 x − 1 ≤ e 0 = 1 ⟹ 1 − e 2 x − 1 ≥ 0 et donc
( ∀ x ∈ [ 0 , 1 2 ] ) : 0 ≤ f ′ ( x ) < 1 2 (\forall x\in\left[0,\frac12\right])~:~ 0\le f'(x)<\frac12 ( ∀ x ∈ [ 0 , 2 1 ] ) : 0 ≤ f ′ ( x ) < 2 1 2/b/ la fonction g g g R \R R
( ∀ x ∈ R ) : g ′ ( x ) = f ′ ( x ) − 1 (\forall x\in\R)~:~g'(x)=f'(x)-1 ( ∀ x ∈ R ) : g ′ ( x ) = f ′ ( x ) − 1 D’aprés la question II-1 on a :
( ∀ x ∈ R ) : f ′ ( x ) ≤ f ( x ) (\forall x\in\R) : f'(x)\le f(x) ( ∀ x ∈ R ) : f ′ ( x ) ≤ f ( x ) et d’aprés la question I-2-b
( ∀ x ∈ R ) : f ( x ) ≤ 1 2 (\forall x\in\R) : f(x)\le \frac12 ( ∀ x ∈ R ) : f ( x ) ≤ 2 1 ⟹ ( ∀ x ∈ R ) : f ′ ( x ) ≤ 1 2 \implies (\forall x\in\R) : f'(x)\le \frac12 ⟹ ( ∀ x ∈ R ) : f ′ ( x ) ≤ 2 1 ⟹ ( ∀ x ∈ R ) : g ′ ( x ) ≤ 1 2 − 1 \implies (\forall x\in\R) : g'(x)\le \frac12-1 ⟹ ( ∀ x ∈ R ) : g ′ ( x ) ≤ 2 1 − 1 ⟹ ( ∀ x ∈ R ) : g ′ ( x ) ≤ − 1 2 < 0 \implies (\forall x\in\R) : g'(x)\le -\frac12<0 ⟹ ( ∀ x ∈ R ) : g ′ ( x ) ≤ − 2 1 < 0 Alors g g g ↘ \searrow ↘ R \R R
2/c/
on a :
g g g [ 0 , 1 2 ] [0,\frac12] [ 0 , 2 1 ]
g ( 1 2 ) = 1 2 e − 1 2 = 1 − e 2 e < 0 g(\frac12)=\frac{1}{2\sqrt e}-\frac12=\frac{1-\sqrt e}{2\sqrt e}<0 g ( 2 1 ) = 2 e 1 − 2 1 = 2 e 1 − e < 0
car e > 1 ⇒ 1 − e < 0 e>1 \Rightarrow 1-\sqrt e<0 e > 1 ⇒ 1 − e < 0
g ( 0 ) = 1 1 + e > 0 g(0)=\frac{1}{1+e}>0 g ( 0 ) = 1 + e 1 > 0
donc d’aprés le théorème des valeurs intermédiaires , l’équation g ( x ) = 0 g(x)=0 g ( x ) = 0 α ∈ ] 0 , 1 2 [ \alpha\in]0,\frac12[ α ∈ ] 0 , 2 1 [
et puisque g g g ↘ \searrow ↘ R \R R α \alpha α
g ( α ) = 0 ⟺ f ( α ) = α g(\alpha)=0 \iff f(\alpha)=\alpha g ( α ) = 0 ⟺ f ( α ) = α 3/a/ \quad par récurrence :
u 0 ∈ ] 0 , 1 2 [ u_0\in]0,\frac12[ u 0 ∈ ] 0 , 2 1 [ n = 0 n=0 n = 0
soit n ∈ N n\in\N n ∈ N u n ∈ ] 0 , 1 2 [ u_n\in]0,\frac12[ u n ∈ ] 0 , 2 1 [
on a d’aprés la question I-2-b
( ∀ x ∈ R ) ; 0 < f ( x ) < 1 2 (\forall x \in \mathbb{R}) \quad ; \quad 0 < f(x) < \frac{1}{2} ( ∀ x ∈ R ) ; 0 < f ( x ) < 2 1 or u n ∈ ] 0 , 1 2 [ u_n\in]0,\frac12[ u n ∈ ] 0 , 2 1 [ u n ∈ R u_n\in\R u n ∈ R
et donc 0 < f ( u n ) < 1 2 0 < f(u_n) < \frac{1}{2} 0 < f ( u n ) < 2 1
c’est à dire : 0 < u n + 1 < 1 2 \quad 0 < u_{n+1} < \frac{1}{2} 0 < u n + 1 < 2 1
D’aprés le principe de récurrence on a :
( ∀ n ∈ N ) : 0 < u n < 1 2 (\forall n\in\N)~:~0 < u_n < \frac{1}{2} ( ∀ n ∈ N ) : 0 < u n < 2 1 3/b/
On a :
∣ u n + 1 − α ∣ = ∣ f ( u n ) − f ( α ) ∣ |u_{n+1} - \alpha| = |f(u_n) - f(\alpha)| ∣ u n + 1 − α ∣ = ∣ f ( u n ) − f ( α ) ∣ On applique le théorème des accroissements finis à la fonction f f f R \mathbb{R} R u n u_n u n α \alpha α u n ∈ ] 0 ; 1 2 [ u_n \in \left] 0 ; \frac{1}{2} \right[ u n ∈ ] 0 ; 2 1 [
Donc, il existe c ∈ ] min ( u n , α ) ; max ( u n , α ) [ ⊂ ] 0 ; 1 2 [ c \in \left] \min(u_n,\alpha) ; \max(u_n,\alpha) \right[ \subset \left] 0 ; \frac{1}{2} \right[ c ∈ ] min ( u n , α ) ; max ( u n , α ) [ ⊂ ] 0 ; 2 1 [
f ( u n ) − f ( α ) = f ′ ( c ) ( u n − α ) ⇒ ∣ f ( u n ) − f ( α ) ∣ = ∣ f ′ ( c ) ∣ ⋅ ∣ u n − α ∣ \begin{align*}
f(u_n) - f(\alpha) = f'(c)(u_n - \alpha)
\quad \\
\Rightarrow
|f(u_n) - f(\alpha)| = |f'(c)| \cdot |u_n - \alpha|
\end{align*} f ( u n ) − f ( α ) = f ′ ( c ) ( u n − α ) ⇒ ∣ f ( u n ) − f ( α ) ∣ = ∣ f ′ ( c ) ∣ ⋅ ∣ u n − α ∣ Or d’après la question II.2-a) , on a ∀ x ∈ [ 0 ; 1 2 ] , 0 ≤ f ′ ( x ) < 1 2 \forall x \in \left[ 0 ; \frac{1}{2} \right],\quad 0 \leq f'(x) < \frac{1}{2} ∀ x ∈ [ 0 ; 2 1 ] , 0 ≤ f ′ ( x ) < 2 1
Donc :
∣ f ′ ( c ) ∣ < 1 2 ⇒ ∣ f ( u n ) − f ( α ) ∣ < 1 2 ∣ u n − α ∣ \begin{align*}
|&f'(c)| < \frac{1}{2} \quad \\
&\Rightarrow \quad |f(u_{n}) - f(\alpha)| < \frac{1}{2} |u_n - \alpha|
\end{align*} ∣ f ′ ( c ) ∣ < 2 1 ⇒ ∣ f ( u n ) − f ( α ) ∣ < 2 1 ∣ u n − α ∣ Et donc :
∀ n ∈ N , ∣ u n + 1 − α ∣ ≤ 1 2 ∣ u n − α ∣ \boxed{\forall n \in \mathbb{N},\quad |u_{n+1} - \alpha| \leq \frac{1}{2} |u_n - \alpha|} ∀ n ∈ N , ∣ u n + 1 − α ∣ ≤ 2 1 ∣ u n − α ∣ 3/c \quad par récurrence :
∀ n ∈ N , ∣ u n − α ∣ ≤ ( 1 2 ) n + 1 \forall n \in \mathbb{N}, \quad |u_n - \alpha| \leq \left( \frac{1}{2} \right)^{n+1} ∀ n ∈ N , ∣ u n − α ∣ ≤ ( 2 1 ) n + 1
Initialisation (pour n = 0 n = 0 n = 0
On sait que u 0 ∈ ] 0 ; 1 2 [ u_0 \in \left] 0 ; \frac{1}{2} \right[ u 0 ∈ ] 0 ; 2 1 [ α ∈ ] 0 ; 1 2 [ \alpha \in \left] 0 ; \frac{1}{2} \right[ α ∈ ] 0 ; 2 1 [ ∣ u 0 − α ∣ ≤ 1 2 |u_0 - \alpha| \leq \dfrac{1}{2} ∣ u 0 − α ∣ ≤ 2 1
Or :
( 1 2 ) 0 + 1 = 1 2 \left( \frac{1}{2} \right)^{0 + 1} = \frac{1}{2} ( 2 1 ) 0 + 1 = 2 1 Donc :
∣ u 0 − α ∣ ≤ ( 1 2 ) 1 |u_0 - \alpha| \leq \left( \frac{1}{2} \right)^{1} ∣ u 0 − α ∣ ≤ ( 2 1 ) 1
Hérédité : Supposons que pour un certain n ∈ N n \in \mathbb{N} n ∈ N
∣ u n − α ∣ ≤ ( 1 2 ) n + 1 |u_n - \alpha| \leq \left( \frac{1}{2} \right)^{n+1} ∣ u n − α ∣ ≤ ( 2 1 ) n + 1 Montrons que :
∣ u n + 1 − α ∣ ≤ ( 1 2 ) n + 2 |u_{n+1} - \alpha| \leq \left( \frac{1}{2} \right)^{n+2} ∣ u n + 1 − α ∣ ≤ ( 2 1 ) n + 2 D’après la question II.3-b , on a :
∣ u n + 1 − α ∣ ≤ 1 2 ∣ u n − α ∣ |u_{n+1} - \alpha| \leq \frac{1}{2} |u_n - \alpha| ∣ u n + 1 − α ∣ ≤ 2 1 ∣ u n − α ∣ Donc :
∣ u n + 1 − α ∣ ≤ 1 2 ⋅ ( 1 2 ) n + 1 = ( 1 2 ) n + 2 |u_{n+1} - \alpha| \leq \frac{1}{2} \cdot \left( \frac{1}{2} \right)^{n+1} = \left( \frac{1}{2} \right)^{n+2} ∣ u n + 1 − α ∣ ≤ 2 1 ⋅ ( 2 1 ) n + 1 = ( 2 1 ) n + 2
Conclusion : d’aprés le principe de récurrence , on a :
∀ n ∈ N , ∣ u n − α ∣ ≤ ( 1 2 ) n + 1 \boxed{\forall n \in \mathbb{N}, \quad |u_n - \alpha| \leq \left( \frac{1}{2} \right)^{n+1}} ∀ n ∈ N , ∣ u n − α ∣ ≤ ( 2 1 ) n + 1 3/d/ \quad
∀ n ∈ N , ∣ u n − α ∣ ≤ ( 1 2 ) n + 1 \forall n \in \mathbb{N}, \quad |u_n - \alpha| \leq \left( \frac{1}{2} \right)^{n+1} ∀ n ∈ N , ∣ u n − α ∣ ≤ ( 2 1 ) n + 1 et on a
lim n ( 1 2 ) n + 1 = 0 \lim_{n}\left(\frac12\right)^{n+1}=0 n lim ( 2 1 ) n + 1 = 0 car − 1 < 1 2 < 1 -1<\frac12<1 − 1 < 2 1 < 1
Alors
lim u n = α \boxed{
\lim u_n=\alpha
} lim u n = α ( u n ) (u_n) ( u n ) α \alpha α
Partie 3
( ∀ n ∈ N ∗ ) ; (\forall n \in \mathbb{N}^*) ; ( ∀ n ∈ N ∗ ) ;
S n = 1 n ( n + 1 ) ∑ k = 1 k = n k e k n + e − n − k n S_n = \frac{1}{n(n+1)} \sum_{k=1}^{k=n} \frac{k}{e^{\frac kn} + e^{-\frac{n-k}{n}}} S n = n ( n + 1 ) 1 k = 1 ∑ k = n e n k + e − n n − k k 1/a
S n = 1 n ( n + 1 ) ∑ k = 1 k = n k e k n + e − n − k n = 1 n + 1 ∑ k = 1 k = n k n e k n e k n ( e k n + e − n − k n ) = 1 n + 1 ∑ k = 1 k = n k n e k n ( e 2 k n + e ) = 1 n + 1 ∑ k = 1 k = n k n f ( k n ) \begin{align*}
S_n &= \frac{1}{n(n+1)} \sum_{k=1}^{k=n} \frac{k}{e^{\frac kn} + e^{-\frac{n-k}{n}}}\\
&=\frac{1}{n+1} \sum_{k=1}^{k=n} \frac{k}{n}\frac{\color{red}e^{\frac kn}}{{\color{red}e^{\frac kn}}(e^{\frac kn} + e^{-\frac{n-k}{n}})} \\
&=\frac{1}{n+1} \sum_{k=1}^{k=n} \frac{k}{n}\frac{e^{\frac kn}}{(e^{2\frac kn} + e)}\\
&=\frac{1}{n+1} \sum_{k=1}^{k=n} \frac{k}{n} f(\frac kn)
\end{align*} S n = n ( n + 1 ) 1 k = 1 ∑ k = n e n k + e − n n − k k = n + 1 1 k = 1 ∑ k = n n k e n k ( e n k + e − n n − k ) e n k = n + 1 1 k = 1 ∑ k = n n k ( e 2 n k + e ) e n k = n + 1 1 k = 1 ∑ k = n n k f ( n k ) 1/b
on pose t = 1 − x t=1-x t = 1 − x x = 1 − t x=1-t x = 1 − t d x = − d t dx=-dt d x = − d t
f ( 1 − t ) = f ( t ) f(1-t)=f(t) f ( 1 − t ) = f ( t ) x = 0 ⟹ t = 1 x=0\implies t=1 x = 0 ⟹ t = 1 x = 1 ⟹ t = 0 x=1\implies t=0 x = 1 ⟹ t = 0
∫ 0 1 x f ( x ) d x = ∫ 1 0 ( 1 − t ) f ( t ) ( − d t ) = ∫ 0 1 f ( t ) d t − ∫ 0 1 t f ( t ) d t \begin{align*}
\int_0^1 x f(x) dx &=\int_1^0 (1-t)f(t)(-dt) \\
&=\int_0^1f(t)dt - \int_0^1 tf(t)dt
\end{align*} ∫ 0 1 x f ( x ) d x = ∫ 1 0 ( 1 − t ) f ( t ) ( − d t ) = ∫ 0 1 f ( t ) d t − ∫ 0 1 t f ( t ) d t donc pour t = x t=x\quad t = x t t t variable muette )
∫ 0 1 x f ( x ) d x = 1 2 ∫ 0 1 f ( x ) d x \int_0^1 x f(x) dx=\frac12\int_0^1 f(x) dx ∫ 0 1 x f ( x ) d x = 2 1 ∫ 0 1 f ( x ) d x d’aprés la question I-4-b on a :
∫ 0 1 f ( x ) d x = 2 ∫ 0 1 2 f ( x ) d x \int_0^1 f(x) dx = 2 \int_0^{\frac{1}{2}} f(x) dx ∫ 0 1 f ( x ) d x = 2 ∫ 0 2 1 f ( x ) d x Alors
∫ 0 1 x f ( x ) d x = ∫ 0 1 2 f ( x ) d x \int_0^1 x f(x) dx= \int_0^{\frac{1}{2}} f(x) dx ∫ 0 1 x f ( x ) d x = ∫ 0 2 1 f ( x ) d x 2/
On a :
S n = 1 n + 1 ∑ k = 1 n k n f ( k n ) S_n = \frac{1}{n+1} \sum_{k=1}^{n} \frac{k}{n} f\left( \frac{k}{n} \right) S n = n + 1 1 k = 1 ∑ n n k f ( n k ) Posons la fonction :
φ ( x ) = x f ( x ) \varphi(x) = x f(x) φ ( x ) = x f ( x ) La fonction φ \varphi φ continue sur l’intervalle [ 0 , 1 ] [0,1] [ 0 , 1 ] f f f R \mathbb{R} R
S n = 1 n + 1 ∑ k = 1 n φ ( k n ) S_n = \frac{1}{n+1} \sum_{k=1}^{n} \varphi\left( \frac{k}{n} \right) S n = n + 1 1 k = 1 ∑ n φ ( n k ) D’après le théorème de convergence des sommes de Riemann , on a :
lim n → + ∞ 1 n ∑ k = 1 n φ ( k n ) = ∫ 0 1 φ ( x ) d x \lim_{n \to +\infty} \frac{1}{n} \sum_{k=1}^{n} \varphi\left( \frac{k}{n} \right) = \int_0^1 \varphi(x)\,dx n → + ∞ lim n 1 k = 1 ∑ n φ ( n k ) = ∫ 0 1 φ ( x ) d x Donc :
lim n S n = lim n n n + 1 1 n ∑ k = 1 n φ ( k n ) = ∫ 0 1 φ ( x ) d x \begin{align*}
\lim\limits_{n} S_n &= \lim\limits_{n} \frac{n}{n+1} \frac{1}{n} \sum_{k=1}^{n}\varphi\left( \frac{k}{n} \right) \\
&= \int_0^1\varphi(x)dx
\end{align*} n lim S n = n lim n + 1 n n 1 k = 1 ∑ n φ ( n k ) = ∫ 0 1 φ ( x ) d x car lim n n n + 1 = lim n n n = 1 \lim\limits_{n} \frac{n}{n+1} = \lim\limits_{n} \frac{n}{n}=1 n lim n + 1 n = n lim n n = 1
Or, d’après la question III-1-b , on a :
∫ 0 1 x f ( x ) d x = ∫ 0 1 2 f ( x ) d x = 1 e [ arctan ( e ) − π 4 ] \begin{align*}
\int_0^1 x f(x)\,dx &= \int_0^{\frac{1}{2}} f(x)\,dx
\\&= \frac{1}{\sqrt{e}} \left[ \arctan(\sqrt{e}) - \frac{\pi}{4} \right]
\end{align*} ∫ 0 1 x f ( x ) d x = ∫ 0 2 1 f ( x ) d x = e 1 [ arctan ( e ) − 4 π ] Conclusion :
la suite ( S n ) n ∈ N (S_n)_{n\in\N} ( S n ) n ∈ N
lim n → + ∞ S n = 1 e [ arctan ( e ) − π 4 ] \boxed{
\lim_{n \to +\infty} S_n = \frac{1}{\sqrt{e}} \left[ \arctan(\sqrt{e}) - \frac{\pi}{4} \right]
} n → + ∞ lim S n = e 1 [ arctan ( e ) − 4 π ]