OCVX: TD Differentielle

But de la seance: comprendre les differentielles

Rappels

Definition premiere de la differentielle
Une fonction

$$\begin{array}{rl}f:{\mathbb{R}}^n& \to {\mathbb{R}}^n\\ x=(x_1,...,x_n)& \mapsto (f_1(x),...,f_n(x))\end{array}$$ est differentiable en un point

$x_0$ si on peut ecrire

$$f(x_0+h)=f(x_0)+d_{x_0}f(h)+\Vert h\Vert \epsilon (h)$$

$d_{x_0}f:h\mapsto d_{x_0}f(h)$ 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:

$d_{x_0}f$ /

$d{f}_{x_0}/Df(x_0)$

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

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

$x_0$ de

$f$: ecrire et lineariser

$$f(x_0+h)=f(x_0)+d_{x_0}f(h)+\underset{{\mathcal{O}}_o(h)}{\underbrace{\Vert h\Vert \epsilon (h)}}$$

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

Si
1.

$$\begin{array}{rl}f:\mathbb{R}& \to \mathbb{R}\\ d_{x_0}f:h& \mapsto h{f}^{\prime }(x_0)\end{array}$$

$$\begin{gathered} \begin{array}{rl}f:{\mathbb{R}}^n& \to \mathbb{R}\\ d_{x_0}f:h& \mapsto <{\mathrm{\nabla }}_{x_0}f,h>={\mathrm{\nabla }}_x{f}^{T}h\end{array} \\ {\mathrm{\nabla }}_{x_0}f=\left(\begin{array}{c}\frac{\partial f}{\partial x_1}(x_0)\\ ⋮\\ \frac{\partial f}{\partial x_n}(x_0)\end{array}\right) \end{gathered}$$

$$\begin{gathered} \begin{array}{rl}f:{\mathbb{R}}^n& \to {\mathbb{R}}^m\\ d_{x_0}f:h& \mapsto Ja{c}_{x_0}f\times h\end{array} \\ Ja{c}_{x_0}f=\text{matrice jacobienne} \\ [Ja{c}_{x_0}f{]}_{ij}=\frac{\partial f_i}{\partial x_j}(x_0) \\ Ja{c}_{x_0}f=\left[\begin{array}{ccc}\frac{\partial f_1}{\partial x_1}(x_0)& \dots & \frac{\partial f_1}{\partial x_n}(x_0)\\ ⋮& \ddots & ⋮\\ \frac{\partial f_n}{\partial x_1}(x_0)& \dots & \frac{\partial f_m}{\partial x_n}(x_0)\end{array}\right]=\left(\begin{array}{c}{\mathrm{\nabla }}_{x_0}{f}_i^T\\ ⋮\\ {\mathrm{\nabla }}_{x_0}{f}_m^T\end{array}\right) \end{gathered}$$

$f$ 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

$$\begin{gathered} f:(x,y)\mapsto \frac{xy}{x^2+y^2}(x,y)\ne (0,0);0(x,y)=0 \\ f:(x,y)\mapsto \frac{xy}{\sqrt{x^2+y^2}}(x,y)\ne (0,0);0(x,y)=0 \end{gathered}$$

Fonction ou ca se passe mal

Differentielle de 2 fonctions

Rappel
Pour des fonctions

$f,g:\mathbb{R}\to \mathbb{R}$

• $(f+\lambda g{)}^{\prime }=f^{\prime }+\lambda g^{\prime }$
• $(fg{)}^{\prime }=f^{\prime }g+g^{\prime }f$
• $(f\circ g{)}^{\prime }(x)=f^{\prime }(g(x))\times g^{\prime }(x)$

Pour la differentielle:

• $d_x(f+\lambda g)=d_{x}f+\lambda d_{x}g$
• $d_x(fg)=g(x)d_{x}f+f(x)d_{x}g$
• $d_x<f,g>=<d_{x}f,g(x)>+<f(x),d_{x}g>$
  $\to$ pas ouf comme ecriture
  • $d_x<f,g>:f\mapsto <d_{x}f(h),g(x)>+<f(x),d_{x}g(h)>$
• $d_{x}g\circ f=d_{g(x)}f\circ d_{x}g$
  $\to$ pas ouf comme ecriture
  • $$\begin{array}{rl}{d}_{x}g\circ f:h& \mapsto d_{g(x)}f\circ d_{x}g(h)\\ & =d_{g_x}f(d_{x}g(h))\end{array}$$

Exercices

Exercice de cours

Differentielle de

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

$x$

Solution

Rappel

$$(\frac{u(x)}{v(x)}{)}^{\prime }=\frac{u^{\prime }(x)v(x)-u(x)v^{\prime }(x)}{v^2(x)}$$

Seconde methode:

Moyen memo technique

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

$$\begin{gathered} \begin{array}{rl}{f}^{\prime }(x)& =\frac{\cos (x)(x^2+1)-\sin (x)2x}{(x^2+1{)}^2}\\ & =\frac{(x^2+1)\cos (x)-2x\sin (x)}{(x^2+1{)}^2}\end{array} \\ {d}_{x}f:h\mapsto h{f}^{\prime }(x) \end{gathered}$$

Premiere methode (mode bourrin):

$$\begin{gathered} f(x+h)=\frac{\sin (x+h)}{(x+h{)}^2+1}=\frac{\sin (x)\cos (h)+\cos (x)\sin (h)}{x^2+2xh+h^2+1} \\ \underset{\text{premier terme}}{\underbrace{\frac{\sin (x)\cos (h)}{(x^2+1)(1+\frac{2x}{x^2+1}h+\frac{h^2}{x^2+1})}}}+\underset{\text{second terme}}{\underbrace{\frac{\cos (x)\sin (h)}{(x^2+1)(1+\frac{2x}{x^2+1}h+\underset{\mathcal{o}(h)}{\underbrace{\frac{h^2}{x^2+1}}})}}} \\ \begin{array}{rl}\text{Second terme }=& \frac{\cos (x)\sin (h)}{x^2+1}\underset{1-\frac{2xh}{x^2+1}+o(h)}{\underbrace{\frac{1}{(x^2+1)(1+\frac{2x}{x^2+1}h+o(h))}}}\frac{1}{1+u}\sim 1-u+o(u)\\ & \frac{\cos (x)\sin (h)}{x^2+1}(1-\frac{2xh}{x^2+1}+o(h))\sin (u)\sim u+o(u)\\ & \frac{\cos (x)(h+o(h))}{x^2+1}(1-\frac{2xh}{x^2+1}+o(h))\\ & =\frac{\cos (x)(h+o(h))}{x^2+1}-\underset{o(h)}{\underbrace{\frac{2xh}{x^2+1}\frac{\cos (x)(h+o(h))}{x^2+1}}}+o(h)\\ & =h\frac{\cos (x)}{x^2+1}+o(h)\end{array} \end{gathered}$$

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

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Exercice 2-37

Solution

$$\begin{gathered} \begin{array}{rlr}f:{\mathbb{R}}^n& \to {\mathbb{R}}^m& A\in {\mathcal{M}}_{m,n}(\mathbb{R})\\ x& \mapsto Ax+b& b\in {\mathbb{R}}^m\end{array} \\ f(x+h)=A(x+h)+b=\underset{f(x)}{\underbrace{Ax+b}}+\underset{\text{lineaire en }h}{\underbrace{Ah}} \\ \begin{array}{rl}{d}_{x}f:h& \mapsto Ah\\ d_{x}f(h)& =Ja{c}_{x}f\times h\end{array}\}Ja{c}_{x}f=A \end{gathered}$$

$$\begin{gathered} \begin{array}{rl}f:{\mathbb{R}}^n& \to \mathbb{R}A\in {\mathcal{M}}_n(\mathbb{R})\text{ symetrique}\\ x& \mapsto x^{T}Ax\end{array} \\ \begin{array}{rl}f(x+h)& =(x+h{)}^{T}A(x+h)\\ & =\underset{f(x)}{\underbrace{x^{T}Ax}}+\underset{\in \mathbb{R}}{\underbrace{x^{T}Ah}}+\underset{\in \mathbb{R}}{\underbrace{h^{T}Ax}}+\underset{=(h^{T}Ax{)}^T=x^T{A}^{T}h=x^{T}Ah}{\underbrace{h^{T}Ah}}\\ & =f(x)+\underset{d_{x}f(h)}{\underbrace{2x^{T}Ah}}+\underset{o(h)}{\underbrace{hTAh}}\end{array} \\ \begin{array}{rl}{d}_{x}f:h& \mapsto 2x^{T}Ah\\ d_{x}f(h)& =2x^{T}Ah=<{\mathrm{\nabla }}_{x}f,h>={\mathrm{\nabla }}_x{f}^{T}h\\ & \to {\mathrm{\nabla }}_x{f}^T=2x^{T}A\\ & \to {\mathrm{\nabla }}_{x}f=2A^{T}x\end{array} \end{gathered}$$

$$\begin{gathered} \begin{array}{rl}f:{\mathcal{M}}_n(\mathbb{R})& \to \mathbb{R}\\ X& \mapsto t{r}^2(X)\end{array} \\ \begin{array}{rl}f(X+H)& =t{r}^2(X+H)=(tr(X+H){)}^2=(tr(X)+tr(H){)}^2\\ & =\underset{f(X)}{\underbrace{t{r}^2(X)}}+\underset{d_{X}f(H)}{\underbrace{2tr(X)tr(H)}}+\underset{o(h)}{\underbrace{t{r}^2(H)}}\end{array} \\ \begin{array}{rl}{d}_{X}f:H& \mapsto 2tr(X)tr(H)\\ d_{X}f(H)& ={\mathrm{\nabla }}_X{f}^{T}H=2tr(X)tr(H)\end{array} \end{gathered}$$

$$\begin{gathered} \begin{array}{rl}f:{\mathcal{M}}_n(\mathbb{R})& \to M_n(\mathbb{R})\\ B& \mapsto tr(AB)B\end{array} \\ \begin{array}{rl}f(B+H)& =tr(A(B+H))(B+H)=tr(AB+AH)(B+H)\\ & =\underset{f(B)}{\underbrace{tr(AB)B}}+\underset{d_{B}f(H)}{\underbrace{tr(AB)H+tr(AH)B}}+\underset{o(h)}{\underbrace{tr(AH)H}}\end{array} \\ \begin{array}{rl}{d}_{B}f:H& \mapsto tr(AB)H+tr(AH)B\\ d_{B}f(H)& =Ja{c}_B(f)\times H\end{array} \end{gathered}$$

Exercice 2-38

Solution

$$\begin{array}{rlr}f:{\mathbb{R}}^n& \to {\mathbb{R}}^n& A\in {\mathcal{M}}_n(\mathbb{R}\\ X& \mapsto <\underset{f_1(X)}{\underbrace{AX+b}},\underset{f_2(X)}{\underbrace{tr(A)X}}>& b\in {\mathbb{R}}^n\end{array}$$

Rappel

$$d_x<f,g>:h\mapsto <d_{x}f(h),g(x)>+<f(x),d_{x}g(h)>$$

$$\begin{gathered} d_x{f}_1:h\mapsto Ah \\ {f}_2(x+h)=tr(A)(x+h)=tr(A)x+\underset{d_x{f}_2:h\mapsto tr(A)h}{\underbrace{tr(A)h}} \end{gathered}$$

Donc:

$$\begin{gathered} \begin{array}{rl}{d}_{x}f:h\mapsto d_x<f_1,f_2>(h)& =<d_x{f}_1(h),f_2(x)>+<f_1(x),d_x{f}_2(h)>\\ & =<Ah,tr(A)x>+<Ax+b,tr(A)h>\end{array} \\ {d}_{x}f:h\mapsto <Ah,tr(A)x>+<Ax+b,tr(A)h> \end{gathered}$$

:::

Exercice 2-39

Solution

$$\begin{array}{rl}g:{\mathbb{R}}^n& \to \mathbb{R}\\ x& \mapsto \frac{1}{x^{T}x+1}\end{array}$$

Rappel

$$\begin{gathered} d_{x}f\circ g=d_{g(x)}f\circ d_{x}g \\ {d}_{x}f\circ g(h)=d_{g(x)}f(d_{x}g(h)) \end{gathered}$$

$$\begin{gathered} g(x)=b\circ a(x) \\ \begin{array}{rl}a:{\mathbb{R}}^n& \to \mathbb{R}\\ x& \mapsto x^{T}x+1\\ d_{x}a:h& \mapsto 2x^{T}h\end{array} \\ \begin{array}{rl}b:\mathbb{R}& \to \mathbb{R}\\ x& \mapsto \frac{1}{x}\\ d_{x}b:h& \mapsto h{b}^{\prime }(x)=-\frac{1}{x^2}h\end{array} \\ {d}_{x}b(h)=-\frac{1}{x^2}h{d}_{x}a(h)=2x^{T}h \\ \begin{array}{rl}{d}_{x}b\circ a(h)& =d_{a(x)}b(\underset{y}{\underbrace{d_{x}a(h)}})\\ & =d_{a(x)}b(y)\\ & =-\frac{1}{(a(x){)}^2}y\\ & =-\frac{1}{(x^{T}x+1{)}^2}y\\ & =-\frac{1}{(x^{T}x+1{)}^2}2x^{T}h\\ \to d_{x}g:h& \mapsto -\frac{2x^{T}h}{(x^{t}x+1{)}^2}\equiv (\frac{1}{u(x)}{)}^{\prime }=-\frac{u^{\prime }(x)}{u(x)}\end{array} \end{gathered}$$

$$\begin{gathered} \begin{array}{rl}f:{\mathbb{R}}^n& \to \mathbb{R}\\ x& \mapsto {\cos }^2(x^{T}Ax)\end{array} \\ f(x)=b\circ a(x) \\ \begin{array}{rl}a:{\mathbb{R}}^n& \to \mathbb{R}\\ x& \mapsto x^{T}Ax\\ d_{x}a:h& \mapsto 2x^{T}Ah\\ b:\mathbb{R}& \to \mathbb{R}\\ x& \mapsto {\cos }^2(x)\\ d_{x}b:h& \mapsto h{b}^{\prime }(x)=-2\cos (x)\sin (x)h=-sin(2x)h\end{array} \\ \begin{array}{rl}{d}_{x}f(h)=d_{x}b\circ a(h)=d_{a(x)}b(\underset{y}{\underbrace{d_{x}a(h)}})=d_{a(x)}b(y)& =-\sin (2a(x))y\\ & =-\sin (2x^{T}Ax)y\\ & =-\sin (2x^{T}Ax)2x^{T}Ah\end{array} \end{gathered}$$

:::

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

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

Montrer que

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

Solution

$$\begin{array}{rl}g(x,y,z)& =f(x-y,y-z,z-x)\\ & =f\circ u(x,y,z)\end{array}$$

avec

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

On vient de voir que

$$\begin{gathered} d_{\alpha }g:h\mapsto d_{\alpha }f\circ u(h)=\underset{Ja{c}_{\alpha }g\times h=Ja{c}_{u(\alpha )}f\times Ja{c}_{\alpha }u\times h\mapsto Ja{c}_{\alpha }g=Ja{c}_{\underset{\beta \in {\mathbb{R}}^3=u(\alpha )}{\underbrace{u(\alpha )}}}f\times Ja{c}_{\alpha }(u)}{\underbrace{d_{u(\alpha )}f(d_{\alpha }u(h))}} \\ u(x,y,z)=(\stackrel{u_1}{\overbrace{x-y}},\stackrel{u_2}{\overbrace{y-z}},\stackrel{u_3}{\overbrace{z-x}}) \\ Ja{c}_{(x,y,z)}u=\left[\begin{array}{c}\frac{\partial u_i}{\partial x_j}\end{array}\right]=\left[\begin{array}{ccc}1& -1& 0\\ 0& 1& -1\\ -1& 0& 1\end{array}\right] \\ \begin{array}{rl}Ja{c}_{\beta }f& ={\mathrm{\nabla }}_{\beta }{f}^T\\ & =(\frac{\partial f}{\partial x}(\beta ),\frac{\partial f}{\partial y}(\beta ),\frac{\partial f}{\partial z}(\beta ))\\ Ja{c}_{\alpha }g& ={\mathrm{\nabla }}_{\alpha }{g}^T\\ & =(\frac{\partial g}{\partial x}(\alpha ),\frac{\partial g}{\partial y}(\alpha ),\frac{\partial g}{\partial z}(\alpha ))\\ {\mathrm{\nabla }}_{\alpha }{g}^T& ={\mathrm{\nabla }}_{\beta }{f}^T\left[\begin{array}{ccc}1& -1& 0\\ 0& 1& -1\\ -1& 0& 1\end{array}\right]\\ & =(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z})\left[\begin{array}{ccc}1& -1& 0\\ 0& 1& -1\\ -1& 0& 1\end{array}\right]\\ & =(\frac{\partial f}{\partial x}-\frac{\partial f}{\partial z},\frac{\partial f}{\partial y}-\frac{\partial f}{\partial x},\frac{\partial f}{\partial z}-\frac{\partial f}{\partial y})\\ & =\frac{\partial g}{\partial x}+\frac{\partial g}{\partial y}+\frac{\partial g}{\partial z}=0\end{array} \end{gathered}$$