OCVX: TD Differentielle

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OCVX: TD Differentielle

But de la seance: comprendre les differentielles

Rappels

Definition premiere de la differentielle Une fonction \(\begin{aligned}f:\mathbb R^n&\to\mathbb R^n \\ x=(x_1,...,x_n)&\mapsto (f_1(x),...,f_n(x))\end{aligned}\) est differentiable en un point $x_0$ si on peut ecrire

\[f(x_0+h)=f(x_0)+\color{red}{d_{x_0}f(h)}+\Vert h\Vert\varepsilon(h)\]

$d_{x_0}f:h\mapsto d_{x_0}f(h)$ est une application lineaire: $d_{x_0}f(h_1+\lambda h_2)=d_{x_0}f(h_1)+\lambda d_{x_0}f(h_2)$

Notation: $d_{x_0}f$ / $df_{x_0} / Df(x_0)$

$1^{ere}$ maniere de calculer la differentielle

$1^{ere}$ maniere de calculer la differentielle en $x_0$ de $f$: ecrire et lineariser \(f(x_0+h)=f(x_0)+d_{x_0}f(h)+\underbrace{\Vert h\Vert\varepsilon(h)}_{\mathcal O_o(h)}\)

$2^{ere}$ maniere de calculer la differentielle

Si 1. \(\begin{aligned} f:\mathbb R&\to\mathbb R \\ d_{x_0}f:h&\mapsto hf'(x_0)\\ \end{aligned}\)

2. \(\begin{aligned} f:\mathbb R^n&\to\mathbb R \\ d_{x_0}f:h&\mapsto <\nabla_{x_0}f, h> = \nabla_xf^Th\\ \end{aligned}\\ \nabla_{x_0}f=\begin{pmatrix} \frac{\partial f}{\partial x_1}(x_0) \\ \vdots \\ \frac{\partial f}{\partial x_n}(x_0) \end{pmatrix}\)

3. \(\begin{aligned} f:\mathbb R^n&\to\mathbb R^m \\ d_{x_0}f:h&\mapsto Jac_{x_0}f\times h\\ \end{aligned}\\ Jac_{x_0}f=\text{matrice jacobienne}\\ [Jac_{x_0}f]_{ij} = \frac{\partial f_i}{\partial x_j}(x_0)\\ Jac_{x_0}f=\begin{bmatrix} \frac{\partial f_1}{\partial x_1}(x_0) & \dots & \frac{\partial f_1}{\partial x_n}(x_0)\\ \vdots & \ddots &\vdots\\ \frac{\partial f_n}{\partial x_1}(x_0) & \dots & \frac{\partial f_m}{\partial x_n}(x_0)\\ \end{bmatrix} = \begin{pmatrix}\nabla_{x_0}f_i^T \\ \vdots \\ \nabla_{x_0}f_m^T\end{pmatrix}\)

$f$ differentiable $\Rightarrow$ existence et continuite des $\frac{\partial f}{\partial x_i}$

Existence et continuite des $\frac{\partial f_i}{\partial x_j}$ $\Rightarrow$ $f$ differentiable

Exemple

\[f:(x,y)\mapsto \frac{xy}{x^2+y^2}\quad (x,y)\neq(0,0); 0\quad(x,y) = 0\\ f:(x,y)\mapsto \frac{xy}{\sqrt{x^2+y^2}}\quad (x,y)\neq(0,0); 0\quad(x,y) = 0\]

Fonction ou ca se passe mal

Differentielle de 2 fonctions

Rappel Pour des fonctions $f,g:\mathbb R\to\mathbb R$

• $(f+\lambda g)’=f’+\lambda g’$
• $(fg)’=f’g+g’f$
• $(f\circ g)’(x)=f’(g(x))\times g’(x)$

Pour la differentielle:

• $d_x(f+\lambda g)=d_xf+\lambda d_xg$
• $d_x(fg)=g(x)d_xf+f(x)d_xg$
• $d_x<f,g>=<d_xf,g(x)> + <f(x),d_xg>$ $\to$ pas ouf comme ecriture
  • $d_x<f,g>:f\mapsto <d_xf(h),g(x)> + <f(x),d_xg(h)>$
• $d_xg\circ f=d_{g(x)}f\circ d_xg$ $\to$ pas ouf comme ecriture
  • \[\begin{aligned} d_xg\circ f:h&\mapsto d_{g(x)}f\circ d_xg(h) \\ &=d_{g_x}f(d_xg(h)) \end{aligned}\]

Exercices

Exercice de cours

Differentielle de \(\begin{aligned}f:\mathbb R&\to\mathbb R \\ x&\mapsto \frac{\sin(x)}{x^2+1}\end{aligned}\) en tout point $x$

Solution

Rappel

\(\biggr(\frac{u(x)}{v(x)}\biggr)' = \frac{u'(x)v(x)-u(x)v'(x)}{v^2(x)}\)

Seconde methode:

Moyen memo technique

• si(mple) $\to$ $\cos$
• co(mplique) $\to$ $-\sin$

\[\begin{aligned} f'(x) & =\frac{\cos(x)(x^2+1)-\sin(x)2x}{(x^2+1)^2}\\ &=\frac{(x^2+1)\cos(x)-2x\sin(x)}{(x^2+1)^2} \end{aligned}\\ d_xf:h\mapsto hf'(x)\]

Premiere methode (mode bourrin): \(f(x+h)=\frac{\sin(x+h)}{(x+h)^2+1}=\frac{\sin(x)\cos(h)+\cos(x)\sin(h)}{x^2+2xh+h^2+1}\\ \underbrace{\frac{\sin(x)\cos(h)}{(x^2+1)(1+\frac{2x}{x^2+1}h+\frac{h^2}{x^2+1})}}_{\text{premier terme}} + \underbrace{\frac{\cos(x)\sin(h)}{(x^2+1)(1+\frac{2x}{x^2+1}h+\underbrace{\frac{h^2}{x^2+1}}_{\mathcal o(h)})}}_{\text{second terme}}\\ \begin{aligned} \text{Second terme }=&\frac{\cos(x)\sin(h)}{x^2+1}\underbrace{\frac{1}{(x^2+1)(1+\frac{2x}{x^2+1}h+o(h))}}_{1-\frac{2xh}{x^2+1}+o(h)}\quad\color{red}{\frac{1}{1+u}\sim 1-u+o(u)}\\ &\frac{\cos(x)\sin(h)}{x^2+1}\biggr(1-\frac{2xh}{x^2+1}+o(h)\biggr)\quad\color{red}{\sin(u)\sim u+o(u)}\\ &\frac{\cos(x)(h+o(h))}{x^2+1}\biggr(1-\frac{2xh}{x^2+1}+o(h)\biggr)\\ &= \frac{\cos(x)(h+o(h))}{x^2+1}-\underbrace{\frac{2xh}{x^2+1}\frac{\cos(x)(h+o(h))}{x^2+1}}_{o(h)}+o(h)\\ &= h\frac{\cos(x)}{x^2+1}+o(h) \end{aligned}\)

C’est que le second terme, on fait pas le premier parce qu’on a pas envie de crever.

Exercice 2-37

Solution

1.

\[\begin{aligned} f:\mathbb R^n&\to\mathbb R^m &A\in\mathcal M_{m,n}(\mathbb R)\\ x&\mapsto Ax+b &b\in\mathbb R^m \end{aligned}\\ f(x+h)=A(x+h)+b=\underbrace{Ax+b}_{f(x)} + \underbrace{Ah}_{\text{lineaire en }h}\\ \begin{aligned} d_xf:h&\mapsto Ah\\ d_xf(h)&=Jac_xf\times h \end{aligned} \biggr\} Jac_xf=A\]

2.

\[\begin{aligned} f:\mathbb R^n&\to\mathbb R \quad A\in\mathcal M_{n}(\mathbb R) \text{ symetrique}\\ x&\mapsto x^TAx \end{aligned}\\ \begin{aligned} f(x+h)&=(x+h)^TA(x+h)\\ &= \underbrace{x^TAx}_{f(x)} + \underbrace{x^TAh}_{\in\mathbb R} + \underbrace{h^TAx}_{\in\mathbb R} + \underbrace{h^TAh}_{= (h^TAx)^T=x^TA^Th=x^TAh}\\ &= f(x) + \underbrace{2x^TAh}_{d_xf(h)}+\underbrace{hTAh}_{o(h)} \end{aligned}\\ \begin{aligned} d_xf:h&\mapsto 2x^TAh\\ d_xf(h)&=2x^TAh = <\nabla_xf,h> = \nabla_xf^Th\\ &\to \nabla_xf^T=2x^TA\\ &\to \nabla_xf = 2A^Tx \end{aligned}\\\]

3.

\[\begin{aligned} f:\mathcal M_n(\mathbb R)&\to\mathbb R\\ X&\mapsto tr^2(X) \end{aligned}\\ \begin{aligned} f(X+H) &= tr^2(X+H) = (tr(X+H))^2 = (tr(X)+tr(H))^2\\ &= \underbrace{tr^2(X)}_{f(X)} + \underbrace{2tr(X)tr(H)}_{d_Xf(H)}+\underbrace{tr^2(H)}_{o(h)} \end{aligned}\\ \begin{aligned} d_Xf:H&\mapsto 2tr(X)tr(H)\\ d_Xf(H)&=\nabla_Xf^TH=2tr(X)tr(H) \end{aligned}\]

4.

\[\begin{aligned} f:\mathcal M_n(\mathbb R)&\to M_n(\mathbb R)\\ B&\mapsto tr(AB)B \end{aligned}\\ \begin{aligned} f(B+H)&= tr(A(B+H))(B+H) = tr(AB+AH)(B+H)\\ &= \underbrace{tr(AB)B}_{f(B)} + \underbrace{tr(AB)H + tr(AH)B}_{d_Bf(H)} + \underbrace{tr(AH)H}_{o(h)} \end{aligned}\\ \begin{aligned} d_Bf:H&\mapsto tr(AB)H + tr(AH)B\\ d_Bf(H)&=Jac_B(f)\times H \end{aligned}\]

Exercice 2-38

Solution

\[\begin{aligned} f:\mathbb R^n&\to\mathbb R^n &A\in\mathcal M_n(\mathbb R\\ X&\mapsto <\color{blue}{\underbrace{AX+b}_{f_1(X)}}, \color{red}{\underbrace{tr(A)X}_{f_2(X)}}> &b\in\mathbb R^n \end{aligned}\]

Rappel

\(d_x<f,g>:h\mapsto<d_xf(h),g(x)> + <f(x),d_xg(h)>\)

\[d_xf_1:h\mapsto \color{green}{Ah}\\ f_2(x+h) = tr(A)(x+h)=tr(A)x+\underbrace{tr(A)h}_{d_xf_2:h\mapsto \color{orange}{tr(A)h}}\]

Donc:

\[\begin{aligned} d_xf:h\mapsto d_x<f_1,f_2>(h)&= <\color{green}{d_xf_1(h)},\color{red}{f_2(x)}> + <\color{blue}{f_1(x)},\color{orange}{d_xf_2(h)}>\\ &= <Ah,tr(A)x> + <Ax+b,tr(A)h> \end{aligned}\\ d_xf:h\mapsto <Ah,tr(A)x> + <Ax+b,tr(A)h>\]

Exercice 2-39

Solution

1.

\[\begin{aligned} g:\mathbb R^n&\to \mathbb R\\ x&\mapsto \frac{1}{x^Tx+1} \end{aligned}\\\]

Rappel

\(d_xf\circ g = d_{g(x)}f\circ d_xg\\ d_xf\circ g(h) = d_{g(x)}f(d_xg(h))\)

\[g(x) = b\circ a(x)\\ \begin{aligned} a:\mathbb R^n&\to \mathbb R\\ x&\mapsto x^Tx+1\\ d_xa:h&\mapsto 2x^Th \end{aligned}\\ \begin{aligned} b:\mathbb R&\to \mathbb R\\ x&\mapsto \frac{1}{x}\\ d_xb:h&\mapsto hb'(x) = -\frac{1}{x^2}h \end{aligned}\\ d_xb(h) = -\frac{1}{x^2}h \quad d_xa(h)=2x^Th\\ \begin{aligned} d_xb\circ a(h)&=d_{a(x)}b(\underbrace{d_xa(h)}_{y})\\ &= d_{a(x)}b(y)\\ &= -\frac{1}{(a(x))^2}y\\ &= -\frac{1}{(x^Tx+1)^2}y\\ &= -\frac{1}{(x^Tx+1)^2}2x^Th\\ \to d_xg:h&\mapsto-\frac{2x^Th}{(x^tx+1)^2}\equiv \biggr(\frac{1}{u(x)}\biggr)' = -\frac{u'(x)}{u(x)} \end{aligned}\]

2.

\[\begin{aligned} f:\mathbb R^n&\to \mathbb R\\ x&\mapsto \cos^2(x^TAx) \end{aligned}\\ f(x) = b\circ a(x)\\ \begin{aligned} a:\mathbb R^n&\to \mathbb R\\ x&\mapsto x^TAx\\ d_xa:h&\mapsto 2x^TAh\\ b:\mathbb R&\to \mathbb R\\ x&\mapsto \cos^2(x)\\ d_xb:h&\mapsto hb'(x) = -2\cos(x)\sin(x)h = -sin(2x)h \end{aligned}\\ \begin{aligned} d_xf(h) = d_xb\circ a(h) = d_{a(x)}b(\underbrace{d_xa(h)}_{y}) = d_{a(x)}b(y) &= -\sin(2a(x))y\\ &= -\sin(2x^TAx)y\\ &= -\sin(2x^TAx)2x^TAh \end{aligned}\]

Exercice 3-42

On calcule un gradient ou une jacobienne ?

Solution

1. Gradient
2. Jacobienne
3. Jacobienne
4. Gradient
5. Gradient
6. Gradient

Exercice 3-43

$f:\mathbb R^3\to\mathbb R$ differentiable en tout point de $\mathbb R^3$

Soit \(\begin{aligned} g:\mathbb R^3&\to\mathbb R \\ (x,y,z)&\mapsto f(x-y,y-z,z-x) \end{aligned}\)

Montrer que

\[\frac{\partial g}{\partial x}(\alpha) + \frac{\partial g}{\partial y}(\alpha) + \frac{\partial g}{\partial z}(\alpha) = 0 \quad\forall \alpha=(a,b,c)\in\mathbb R^3\]

Solution

\[\begin{aligned} g(x,y,z) &=f(x-y,y-z,z-x)\\ &=f\circ u(x,y,z) \end{aligned}\]

avec \(\begin{aligned} u:\mathbb R^3&\to\mathbb R^3 \\ (x,y,z)&\mapsto (x-y,y-z,z-x) \end{aligned}\)

On vient de voir que

\[d_{\alpha}g:h\mapsto d_{\alpha} f\circ u(h)=\underbrace{d_{u(\alpha)}f(d_{\alpha}u(h))}_{Jac_{\alpha}g\times h=Jac_{u(\alpha)}f\times Jac_{\alpha}u\times h\mapsto Jac_{\alpha}g=Jac_{\underbrace{u(\alpha)}_{\beta\in\mathbb R^3=u(\alpha)}}f\times Jac_{\alpha}(u)}\\ u(x,y,z)=(\overbrace{x-y}^{u_1}, \overbrace{y-z}^{u_2}, \overbrace{z-x}^{u_3})\\ Jac_{(x,y,z)}u=\begin{bmatrix} \frac{\partial u_i}{\partial x_j} \end{bmatrix} = \begin{bmatrix} 1 &-1&0 \\ 0 & 1 &-1 \\ -1&0&1 \end{bmatrix}\\ \begin{aligned} Jac_{\beta}f&=\nabla_{\beta}f^T\\ &= (\frac{\partial f}{\partial x}(\beta), \frac{\partial f}{\partial y}(\beta), \frac{\partial f}{\partial z}(\beta))\\ Jac_{\alpha}g&=\nabla_{\alpha}g^T \\ &=(\frac{\partial g}{\partial x}(\alpha), \frac{\partial g}{\partial y}(\alpha), \frac{\partial g}{\partial z}(\alpha))\\ \nabla_{\alpha}g^T&=\nabla_{\beta}f^T \begin{bmatrix} 1 &-1 & 0\\ 0 & 1 &-1\\ -1 & 0 & 1 \end{bmatrix}\\ &= (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}) \begin{bmatrix} 1 &-1 & 0\\ 0 & 1 &-1\\ -1 & 0 & 1 \end{bmatrix}\\ &= (\frac{\partial f}{\partial x} - \frac{\partial f}{\partial z}, \frac{\partial f}{\partial y} - \frac{\partial f}{\partial x}, \frac{\partial f}{\partial z}- \frac{\partial f}{\partial y})\\ &= \frac{\partial g}{\partial x} + \frac{\partial g}{\partial y} + \frac{\partial g}{\partial z} = 0 \end{aligned}\]