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OCVX: TD Differentielle

But de la seance: comprendre les differentielles

Rappels

Definition premiere de la differentielle
Une fonction

f:RnRnx=(x1,...,xn)(f1(x),...,fn(x)) est differentiable en un point
x0
si on peut ecrire

f(x0+h)=f(x0)+\colorreddx0f(h)+hε(h)

dx0f:hdx0f(h) est une application lineaire:
dx0f(h1+λh2)=dx0f(h1)+λdx0f(h2)

Notation:

dx0f /
dfx0/Df(x0)

1ere
maniere de calculer la differentielle

1ere maniere de calculer la differentielle en
x0
de
f
: ecrire et lineariser
f(x0+h)=f(x0)+dx0f(h)+hε(h)Oo(h)

2ere
maniere de calculer la differentielle

Si
1.

f:RRdx0f:hhf(x0)

f:RnRdx0f:h↦<x0f,h>=xfThx0f=(fx1(x0)fxn(x0))

f:RnRmdx0f:hJacx0f×hJacx0f=matrice jacobienne[Jacx0f]ij=fixj(x0)Jacx0f=[f1x1(x0)f1xn(x0)fnx1(x0)fmxn(x0)]=(x0fiTx0fmT)

f differentiable
existence et continuite des
fxi

Existence et continuite des

fixj
f
differentiable

Exemple

f:(x,y)xyx2+y2(x,y)(0,0);0(x,y)=0f:(x,y)xyx2+y2(x,y)(0,0);0(x,y)=0

Fonction ou ca se passe mal

Differentielle de 2 fonctions

Rappel
Pour des fonctions

f,g:RR

  • (f+λg)=f+λg
  • (fg)=fg+gf
  • (fg)(x)=f(g(x))×g(x)

Pour la differentielle:

  • dx(f+λg)=dxf+λdxg
  • dx(fg)=g(x)dxf+f(x)dxg
  • dx<f,g>=<dxf,g(x)>+<f(x),dxg>
    pas ouf comme ecriture
    • dx<f,g>:f↦<dxf(h),g(x)>+<f(x),dxg(h)>
  • dxgf=dg(x)fdxg
    pas ouf comme ecriture
    • dxgf:hdg(x)fdxg(h)=dgxf(dxg(h))

Exercices

Exercice de cours

Differentielle de

f:RRxsin(x)x2+1 en tout point
x

Solution

Rappel

(u(x)v(x))=u(x)v(x)u(x)v(x)v2(x)

Seconde methode:

Moyen memo technique

  • si(mple)
    cos
  • co(mplique)
    sin

f(x)=cos(x)(x2+1)sin(x)2x(x2+1)2=(x2+1)cos(x)2xsin(x)(x2+1)2dxf:hhf(x)

Premiere methode (mode bourrin):

f(x+h)=sin(x+h)(x+h)2+1=sin(x)cos(h)+cos(x)sin(h)x2+2xh+h2+1sin(x)cos(h)(x2+1)(1+2xx2+1h+h2x2+1)premier terme+cos(x)sin(h)(x2+1)(1+2xx2+1h+h2x2+1o(h))second termeSecond terme =cos(x)sin(h)x2+11(x2+1)(1+2xx2+1h+o(h))12xhx2+1+o(h)\colorred11+u1u+o(u)cos(x)sin(h)x2+1(12xhx2+1+o(h))\colorredsin(u)u+o(u)cos(x)(h+o(h))x2+1(12xhx2+1+o(h))=cos(x)(h+o(h))x2+12xhx2+1cos(x)(h+o(h))x2+1o(h)+o(h)=hcos(x)x2+1+o(h)

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

:::

Exercice 2-37

Solution

f:RnRmAMm,n(R)xAx+bbRmf(x+h)=A(x+h)+b=Ax+bf(x)+Ahlineaire en hdxf:hAhdxf(h)=Jacxf×h}Jacxf=A

f:RnRAMn(R) symetriquexxTAxf(x+h)=(x+h)TA(x+h)=xTAxf(x)+xTAhR+hTAxR+hTAh=(hTAx)T=xTATh=xTAh=f(x)+2xTAhdxf(h)+hTAho(h)dxf:h2xTAhdxf(h)=2xTAh=<xf,h>=xfThxfT=2xTAxf=2ATx

f:Mn(R)RXtr2(X)f(X+H)=tr2(X+H)=(tr(X+H))2=(tr(X)+tr(H))2=tr2(X)f(X)+2tr(X)tr(H)dXf(H)+tr2(H)o(h)dXf:H2tr(X)tr(H)dXf(H)=XfTH=2tr(X)tr(H)

f:Mn(R)Mn(R)Btr(AB)Bf(B+H)=tr(A(B+H))(B+H)=tr(AB+AH)(B+H)=tr(AB)Bf(B)+tr(AB)H+tr(AH)BdBf(H)+tr(AH)Ho(h)dBf:Htr(AB)H+tr(AH)BdBf(H)=JacB(f)×H

Exercice 2-38

Solution

f:RnRnAMn(RX↦<\colorblueAX+bf1(X),\colorredtr(A)Xf2(X)>bRn

Rappel

dx<f,g>:h↦<dxf(h),g(x)>+<f(x),dxg(h)>

dxf1:h\colorgreenAhf2(x+h)=tr(A)(x+h)=tr(A)x+tr(A)hdxf2:h\colororangetr(A)h

Donc:

dxf:hdx<f1,f2>(h)=<\colorgreendxf1(h),\colorredf2(x)>+<\colorbluef1(x),\colororangedxf2(h)>=<Ah,tr(A)x>+<Ax+b,tr(A)h>dxf:h↦<Ah,tr(A)x>+<Ax+b,tr(A)h>

:::

Exercice 2-39

Solution

g:RnRx1xTx+1

Rappel

dxfg=dg(x)fdxgdxfg(h)=dg(x)f(dxg(h))

g(x)=ba(x)a:RnRxxTx+1dxa:h2xThb:RRx1xdxb:hhb(x)=1x2hdxb(h)=1x2hdxa(h)=2xThdxba(h)=da(x)b(dxa(h)y)=da(x)b(y)=1(a(x))2y=1(xTx+1)2y=1(xTx+1)22xThdxg:h2xTh(xtx+1)2(1u(x))=u(x)u(x)

f:RnRxcos2(xTAx)f(x)=ba(x)a:RnRxxTAxdxa:h2xTAhb:RRxcos2(x)dxb:hhb(x)=2cos(x)sin(x)h=sin(2x)hdxf(h)=dxba(h)=da(x)b(dxa(h)y)=da(x)b(y)=sin(2a(x))y=sin(2xTAx)y=sin(2xTAx)2xTAh

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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:R3R differentiable en tout point de
R3

Soit

g:R3R(x,y,z)f(xy,yz,zx)

Montrer que

gx(α)+gy(α)+gz(α)=0α=(a,b,c)R3

Solution

g(x,y,z)=f(xy,yz,zx)=fu(x,y,z)

avec

u:R3R3(x,y,z)(xy,yz,zx)

On vient de voir que

dαg:hdαfu(h)=du(α)f(dαu(h))Jacαg×h=Jacu(α)f×Jacαu×hJacαg=Jacu(α)βR3=u(α)f×Jacα(u)u(x,y,z)=(xyu1,yzu2,zxu3)Jac(x,y,z)u=[uixj]=[110011101]Jacβf=βfT=(fx(β),fy(β),fz(β))Jacαg=αgT=(gx(α),gy(α),gz(α))αgT=βfT[110011101]=(fx,fy,fz)[110011101]=(fxfz,fyfx,fzfy)=gx+gy+gz=0