A derivada de ordem superior de uma função é simplesmente a derivada da derivada.f f f x i x_i x i x j x_j x j
∂ 2 f ∂ x 1 ∂ x j = ∂ ∂ x i ( ∂ f ∂ x j ) \frac{\partial^2f}{\partial x_1 \partial x_j} = \frac{\partial}{\partial x_i}\left(\frac{\partial f}{\partial x_j}\right) ∂ x 1 ∂ x j ∂ 2 f = ∂ x i ∂ ( ∂ x j ∂ f ) DEFINIÇÃO
f : R n → R f: \R^n \to \R f : R n → R C k C^k C k
∂ k f ∂ x 1 … ∂ x d \frac{\partial^k f}{\partial x_1 \dots \partial x_d} ∂ x 1 … ∂ x d ∂ k f (os índices podem-se repetir) existir e for função contínua.
Exemplo Seja f f f
f ( x , y ) = x 2 y + x y 3 = p 1 ( x , y ) 2 p 2 ( x , y ) + p 1 ( x , y ) p 2 ( x , y ) 3 f(x,y)=x^2y+xy^3=p_1(x,y)^2 p_2(x,y) + p_1(x,y)p_2(x,y)^3 f ( x , y ) = x 2 y + x y 3 = p 1 ( x , y ) 2 p 2 ( x , y ) + p 1 ( x , y ) p 2 ( x , y ) 3
∂ f ∂ x ( x , y ) = ∂ ∂ x ( x 2 y + x y 3 ) = y ⋅ 2 x + y 3 ⋅ 1 = 2 x y + y 3 \frac{\partial f}{\partial x}(x,y)=\frac{\partial}{\partial x}(x^2y+xy^3)=y\cdot 2x + y^3 \cdot 1 = 2xy+y^3 ∂ x ∂ f ( x , y ) = ∂ x ∂ ( x 2 y + x y 3 ) = y ⋅ 2 x + y 3 ⋅ 1 = 2 x y + y 3 ∂ f ∂ y ( x , y ) = ∂ ∂ y ( x 2 y + x y 3 ) = x 2 + 3 x y 2 \frac{\partial f}{\partial y}(x,y)=\frac{\partial}{\partial y}(x^2y+xy^3)=x^2+3xy^2 ∂ y ∂ f ( x , y ) = ∂ y ∂ ( x 2 y + x y 3 ) = x 2 + 3 x y 2
∂ 2 f ∂ x 2 ( x , y ) = ∂ ∂ x ( 2 x y + y 3 ) = 2 y \frac{\partial^2 f}{\partial x^2}(x,y)=\frac{\partial}{\partial x}(2xy+y^3)=2y ∂ x 2 ∂ 2 f ( x , y ) = ∂ x ∂ ( 2 x y + y 3 ) = 2 y ∂ 2 f ∂ y ∂ x ( x , y ) = ∂ ∂ y ( 2 x y + y 3 ) = 2 x + 3 y 2 \frac{\partial^2 f}{\partial y \partial x}(x,y)=\frac{\partial}{\partial y}(2xy+y^3)=2x+3y^2 ∂ y ∂ x ∂ 2 f ( x , y ) = ∂ y ∂ ( 2 x y + y 3 ) = 2 x + 3 y 2 ∂ 2 f ∂ y 2 ( x , y ) = ∂ ∂ x ( x 2 + 3 x y 2 ) = 6 x y \frac{\partial^2 f}{\partial y^2}(x,y)=\frac{\partial}{\partial x}(x^2+3xy^2)=6xy ∂ y 2 ∂ 2 f ( x , y ) = ∂ x ∂ ( x 2 + 3 x y 2 ) = 6 x y ∂ 2 f ∂ x ∂ y ( x , y ) = ∂ ∂ x ( x 2 + 3 x y 2 ) = 2 x + 3 y 2 \frac{\partial^2 f}{\partial x \partial y}(x,y)=\frac{\partial}{\partial x}(x^2+3xy^2)=2x+3y^2 ∂ x ∂ y ∂ 2 f ( x , y ) = ∂ x ∂ ( x 2 + 3 x y 2 ) = 2 x + 3 y 2
Este teorema mostra-nos que a ordem por que derivamos não é importante.
Se derivarmos uma função segundo x x x y y y y y y x x x
DEFINIÇÃO
Seja f ∈ C 2 f \in C^2 f ∈ C 2
∂ 2 f ∂ x i ∂ x j = ∂ 2 f ∂ x j ∂ x i \frac{\partial ^2 f}{\partial x_i \partial x_j} = \frac{\partial^2 f}{\partial x_j \partial x_i} ∂ x i ∂ x j ∂ 2 f = ∂ x j ∂ x i ∂ 2 f Recordar de CDI 1
Seja f : R → R f: \R \to \R f : R → R C k C^k C k
f ( x + h ) = f ( a ) + f ′ ( a ) h + 1 2 ! f ′ ′ ( a ) h 2 + ⋯ + 1 k ! f ( k ) ( a ) h k + R k ( a , h ) f(x+h)=f(a)+f'(a)h+\frac{1}{2!}f''(a)h^2 + \dots + \frac{1}{k!}f^{(k)}(a) h^k + R_k(a,h) f ( x + h ) = f ( a ) + f ′ ( a ) h + 2 ! 1 f ′′ ( a ) h 2 + ⋯ + k ! 1 f ( k ) ( a ) h k + R k ( a , h )
com lim h → 0 R k ( a , h ) h k = 0 \lim_{h\to 0} \frac{R_k(a,h)}{h^k}=0 lim h → 0 h k R k ( a , h ) = 0
Também tínhamos restos, por exemplo o resto de Lagrange:
R k ( a , h ) = f ( k + 1 ) ( a + ϵ h ) h k + 1 ( k + 1 ) ! , 0 < ϵ < 1 R_k(a,h)=\frac{f^{(k+1)}(a+\epsilon h) h^{k+1}}{(k+1)!}, 0 < \epsilon < 1 R k ( a , h ) = ( k + 1 )! f ( k + 1 ) ( a + ϵ h ) h k + 1 , 0 < ϵ < 1 DEFINIÇÃO
Seja f : R n → R f: \R^n \to \R f : R n → R f f f C n + 1 C^{n+1} C n + 1 a ∈ R n a \in \R^n a ∈ R n v ∈ R n v \in \R^n v ∈ R n g : R → R g: \R \to \R g : R → R C k + 1 C^{k+1} C k + 1 g ( t ) = f ( a + t v ) g(t) = f(a+tv) g ( t ) = f ( a + t v )
Então, a fórmula de taylor em g g g t = 0 t = 0 t = 0 k k k é representada pela expressão:
g ( t ) = g ( 0 ) + g ′ ( 0 ) t + 1 2 ! g ′ ′ ( 0 ) t 2 + ⋯ + g ( k ) ( 0 ) k ! t k + R k ( t ) g(t)=g(0)+g'(0)t+\frac{1}{2!}g''(0)t^2 + \dots + \frac{g^{(k)}(0)}{k!}t^k + R_k(t) g ( t ) = g ( 0 ) + g ′ ( 0 ) t + 2 ! 1 g ′′ ( 0 ) t 2 + ⋯ + k ! g ( k ) ( 0 ) t k + R k ( t ) Tentando agora obter os valores para as derivadas de g g g t = 0 t=0 t = 0
g ( 0 ) = f ( a + 0 ⋅ v ) = f ( 0 ) g(0)=f(a+0\cdot v)=f(0) g ( 0 ) = f ( a + 0 ⋅ v ) = f ( 0 ) g ′ ( 0 ) = ∇ f ( a + t v ) ⋅ v ∣ t = 0 = ∇ f ( a ) ⋅ v = ∑ i = 1 n ∂ f ∂ x i ( a ) v i \displaystyle g'(0) = \nabla f(a+tv)\cdot v \big|_{t=0} = \nabla f(a)\cdot v = \sum^n_{i=1} \frac{\partial f}{\partial x_i}(a)v_i g ′ ( 0 ) = ∇ f ( a + t v ) ⋅ v t = 0 = ∇ f ( a ) ⋅ v = i = 1 ∑ n ∂ x i ∂ f ( a ) v i g ′ ′ ( t ) = ∑ i , j = 1 n ∂ 2 f ∂ x j ∂ x y ( a + t v i ) v i v j \displaystyle g''(t)=\sum_{i,j=1}^n \frac{\partial^2 f}{\partial x_j \partial x_y}(a+tv_i)v_i v_j g ′′ ( t ) = i , j = 1 ∑ n ∂ x j ∂ x y ∂ 2 f ( a + t v i ) v i v j g ′ ′ ( 0 ) = ∑ i , j = 1 n ∂ 2 f ∂ x j ∂ x y ( a ) v i v j \displaystyle g''(0) = \sum_{i,j=1}^n \frac{\partial^2 f}{\partial x_j \partial x_y}(a)v_i v_j g ′′ ( 0 ) = i , j = 1 ∑ n ∂ x j ∂ x y ∂ 2 f ( a ) v i v j
Então, podemos obter a Fórmula de Taylor de f f f
f ( a + t v ) = f ( a ) + ∇ f ( a ) ⋅ t v + 1 2 ∑ i , j = 1 n ∂ 2 f ∂ x i ∂ x j ( a ) ⋅ ( t v i ) ⋅ ( t v j ) + R 2 ( t ) f(a+tv)=f(a)+\nabla f(a)\cdot tv + \frac{1}{2} \sum^n_{i,j=1} \frac{\partial^2 f}{\partial x_i \partial x_j} (a)\cdot (tv_i)\cdot(tv_j)+ R_2(t) f ( a + t v ) = f ( a ) + ∇ f ( a ) ⋅ t v + 2 1 i , j = 1 ∑ n ∂ x i ∂ x j ∂ 2 f ( a ) ⋅ ( t v i ) ⋅ ( t v j ) + R 2 ( t ) Se assumirmos h = t v h = tv h = t v ∣ ∣ v ∣ ∣ = 1 ||v||=1 ∣∣ v ∣∣ = 1
DEFINIÇÃO
Sejam f : R n → R f: \R^n \to \R f : R n → R C 3 C^3 C 3 h = t v h = tv h = t v ∣ ∣ v ∣ ∣ = 1 ||v||=1 ∣∣ v ∣∣ = 1 f f f
f ( a + h ) = f ( a ) + ∇ f ( a ) ⋅ h + 1 2 ⋅ ∑ i , j = 1 n ∂ 2 f ∂ x i ∂ x j ( a ) ⋅ h i ⋅ h j + R 2 ( t ) f(a+h)=f(a)+\nabla f(a)\cdot h + \frac{1}{2} \cdot \sum_{i,j=1}^n \frac{\partial^2 f}{\partial x_i \partial x_j} (a)\cdot h_i\cdot h_j+ R_2(t) f ( a + h ) = f ( a ) + ∇ f ( a ) ⋅ h + 2 1 ⋅ i , j = 1 ∑ n ∂ x i ∂ x j ∂ 2 f ( a ) ⋅ h i ⋅ h j + R 2 ( t ) com lim h → 0 ⃗ R 2 ( h ) ∣ ∣ h ∣ ∣ 2 = 0 ⃗ \displaystyle \lim_{h\to \vec 0}\frac{R_2(h)}{||h||^2}=\vec 0 h → 0 lim ∣∣ h ∣ ∣ 2 R 2 ( h ) = 0
Pode-se reescrever a expressão da Fórmula de Taylor de 2ª ordem da seguinte forma:
f ( a + h ) = f ( a ) + ∇ f ( a ) ⋅ h + 1 2 ( h 1 h 2 … h n ) ⋅ H f ( a ) ⋅ ( h 1 h 2 ⋮ h n ) + R 2 ( h ) f(a+h)=f(a)+\nabla f(a)\cdot h+\frac{1}{2}\begin{pmatrix}
h_{1} & h_{2} & \dotsc & h_{n}
\end{pmatrix} \cdot Hf(a)\cdot \begin{pmatrix}
h_{1}\\
h_{2}\\
\vdots \\
h_{n}
\end{pmatrix}
+ R_2(h) f ( a + h ) = f ( a ) + ∇ f ( a ) ⋅ h + 2 1 ( h 1 h 2 … h n ) ⋅ H f ( a ) ⋅ h 1 h 2 ⋮ h n + R 2 ( h ) ou ainda
f ( a + h ) = f ( a ) + ∇ f ( a ) ⋅ h + 1 2 h T H f ( a ) h + R 2 ( h ) f(a+h)=f(a)+\nabla f(a)\cdot h + \frac 12 h^T H_f(a)h + R_2(h) f ( a + h ) = f ( a ) + ∇ f ( a ) ⋅ h + 2 1 h T H f ( a ) h + R 2 ( h ) com lim h → 0 ⃗ R 2 ( h ) ∣ ∣ h ∣ ∣ 2 = 0 ⃗ \lim_{h \to \vec 0} \frac{R_2(h)}{||h||^2}=\vec 0 lim h → 0 ∣∣ h ∣ ∣ 2 R 2 ( h ) = 0 H f ( a ) Hf(a) H f ( a ) f f f a a a matriz simétrica pelo Teorema de Schwarz .
H f ( a ) = [ ∂ 2 f ∂ x 1 2 ∂ 2 f ∂ x 2 ∂ x 1 … ∂ 2 f ∂ x n ∂ x 1 ∂ 2 f ∂ x 1 ∂ x 2 ∂ 2 f ∂ x 2 2 … ∂ 2 f ∂ x 2 ∂ x 1 ⋮ ⋮ ⋱ ⋮ ∂ 2 f ∂ x 1 ∂ x n ∂ 2 f ∂ x 2 ∂ x n … ∂ 2 f ∂ x n 2 ] Hf( a) =\begin{bmatrix}
\frac{\partial ^{2} f}{\partial x^{2}_{1}} & \frac{\partial ^{2} f}{\partial x_{2} \partial x_{1}} & \dotsc & \frac{\partial ^{2} f}{\partial x_{n} \partial x_{1}}\\
\frac{\partial ^{2} f}{\partial x_{1} \partial x_{2}} & \frac{\partial ^{2} f}{\partial x^{2}_{2}} & \dotsc & \frac{\partial ^{2} f}{\partial x_{2} \partial x_{1}}\\
\vdots & \vdots & \ddots & \vdots \\
\frac{\partial ^{2} f}{\partial x_{1} \partial x_{n}} & \frac{\partial ^{2} f}{\partial x_{2} \partial x_{n}} & \dotsc & \frac{\partial ^{2} f}{\partial x^{2}_{n}}
\end{bmatrix} H f ( a ) = ∂ x 1 2 ∂ 2 f ∂ x 1 ∂ x 2 ∂ 2 f ⋮ ∂ x 1 ∂ x n ∂ 2 f ∂ x 2 ∂ x 1 ∂ 2 f ∂ x 2 2 ∂ 2 f ⋮ ∂ x 2 ∂ x n ∂ 2 f … … ⋱ … ∂ x n ∂ x 1 ∂ 2 f ∂ x 2 ∂ x 1 ∂ 2 f ⋮ ∂ x n 2 ∂ 2 f Exemplos de Polinómios de Taylor de 2ª ordem Escrever o polinómio de Taylor de 2ª ordem da função
f ( x , y ) = e x − e y f(x,y) = e^x - e^y f ( x , y ) = e x − e y a = ( 0 , 0 ) a=(0,0) a = ( 0 , 0 )
Começamos por ver se é diferenciável. Como
f ( x , y ) = e p 1 ( x , y ) − e p 2 ( x , y ) f(x,y)=e^{p_1(x,y)}-e^{p_2(x,y)} f ( x , y ) = e p 1 ( x , y ) − e p 2 ( x , y )
O primeiro passo é obter o gradiente de f f f ∇ f ( 0 , 0 ) \nabla f(0,0) ∇ f ( 0 , 0 )
∂ f ∂ x = ∂ ∂ x ( e x − e y ) = e x ∂ f ∂ x ( 0 , 0 ) = e 0 = 1 ∂ f ∂ y = ∂ ∂ y ( e x − e y ) = − e y ∂ f ∂ y ( 0 , 0 ) = − e 0 = − 1 \begin{darray}{l l}
\frac{\partial f}{\partial x}=\frac{\partial}{\partial x}(e^x-e^y)=e^x & \frac{\partial f}{\partial x}(0,0)=e^0=1\\
\\
\frac{\partial f}{\partial y}=\frac{\partial}{\partial y}(e^x-e^y)=-e^y & \frac{\partial f}{\partial y}(0,0)=-e^0 =-1
\end{darray} ∂ x ∂ f = ∂ x ∂ ( e x − e y ) = e x ∂ y ∂ f = ∂ y ∂ ( e x − e y ) = − e y ∂ x ∂ f ( 0 , 0 ) = e 0 = 1 ∂ y ∂ f ( 0 , 0 ) = − e 0 = − 1 Obtemos assim
∇ f ( 0 , 0 ) = ( 1 − 1 ) \nabla f(0,0) = \begin{pmatrix}
1 & -1
\end{pmatrix} ∇ f ( 0 , 0 ) = ( 1 − 1 ) De seguida, vamos calcular a Matriz Hessiana de f f f ( 0 , 0 ) (0,0) ( 0 , 0 )
∂ 2 f ∂ x 2 = ∂ ∂ x ( ∂ f ∂ x ) = ∂ ∂ x e x = e x ∂ 2 f ∂ x 2 ( 0 , 0 ) = e 0 = 1 ∂ 2 f ∂ y 2 = ∂ ∂ y ( ∂ f ∂ y ) = ∂ ∂ y ( − e y ) = − e y ∂ 2 f ∂ y 2 ( 0 , 0 ) = − e 0 = − 1 ∂ 2 f ∂ x ∂ y = ∂ f ∂ x ( ∂ f ∂ y ) = ∂ ∂ x ( − e ) y = 0 ∂ 2 f ∂ x ∂ y ( 0 , 0 ) = 0 \begin{darray}{l l}
\frac{\partial^2 f}{\partial x^2}=\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\right)=\frac{\partial }{\partial x}e^x=e^x
& \frac{\partial^2 f}{\partial x^2}(0,0)=e^0=1\\
\\
\frac{\partial^2 f}{\partial y^2}=\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial y}\right)=\frac{\partial }{\partial y}\left(-e^y\right)=-e^y
& \frac{\partial^2 f}{\partial y^2}(0,0)=-e^0=-1\\
\\
\frac{\partial^2 f}{\partial x \partial y}=\frac{\partial f}{\partial x}\left(\frac{\partial f}{\partial y}\right)=\frac{\partial}{\partial x}\left(-e\right)^y=0
& \frac{\partial^2 f}{\partial x \partial y}(0,0)=0
\end{darray} ∂ x 2 ∂ 2 f = ∂ x ∂ ( ∂ x ∂ f ) = ∂ x ∂ e x = e x ∂ y 2 ∂ 2 f = ∂ y ∂ ( ∂ y ∂ f ) = ∂ y ∂ ( − e y ) = − e y ∂ x ∂ y ∂ 2 f = ∂ x ∂ f ( ∂ y ∂ f ) = ∂ x ∂ ( − e ) y = 0 ∂ x 2 ∂ 2 f ( 0 , 0 ) = e 0 = 1 ∂ y 2 ∂ 2 f ( 0 , 0 ) = − e 0 = − 1 ∂ x ∂ y ∂ 2 f ( 0 , 0 ) = 0 Sendo que a Matriz Hessiana será:
H f ( 0 , 0 ) = ( 1 0 0 − 1 ) H_f(0,0)=\begin{pmatrix}
1 & 0\\
0 & -1
\end{pmatrix} H f ( 0 , 0 ) = ( 1 0 0 − 1 ) Obtemos assim o polinómio de Taylor de ordem 2 2 2 f f f ( 0 , 0 ) (0,0) ( 0 , 0 )
p 2 f ( x , y ) = f ( 0 , 0 ) + ∇ f ( 0 , 0 ) ⋅ ( x y ) + 1 2 ( x y ) H f ( 0 , 0 ) ( x y ) = 0 + ( 1 , − 1 ) ( x y ) + 1 2 ( x y ) ( 1 0 0 − 1 ) ( x y ) = x − y + 1 2 ( x − y ) ( x y ) = x − y + 1 2 ( x 2 − y 2 ) \begin{aligned}
p^{f}_{2} (x,y) & =f(0,0)+\nabla f(0,0)\cdot \begin{pmatrix}
x & y
\end{pmatrix} +\frac{1}{2}\begin{pmatrix}
x & y
\end{pmatrix} H_{f} (0,0)\begin{pmatrix}
x\\
y
\end{pmatrix}\\
& =0+(1,-1)\begin{pmatrix}
x & y
\end{pmatrix} +\frac{1}{2}\begin{pmatrix}
x & y
\end{pmatrix}\begin{pmatrix}
1 & 0\\
0 & -1
\end{pmatrix}\begin{pmatrix}
x\\
y
\end{pmatrix}\\
& =x-y+\frac{1}{2}\begin{pmatrix}
x & -y
\end{pmatrix}\begin{pmatrix}
x\\
y
\end{pmatrix}\\
& =x-y+\frac{1}{2} (x^{2} -y^{2} )
\end{aligned} p 2 f ( x , y ) = f ( 0 , 0 ) + ∇ f ( 0 , 0 ) ⋅ ( x y ) + 2 1 ( x y ) H f ( 0 , 0 ) ( x y ) = 0 + ( 1 , − 1 ) ( x y ) + 2 1 ( x y ) ( 1 0 0 − 1 ) ( x y ) = x − y + 2 1 ( x − y ) ( x y ) = x − y + 2 1 ( x 2 − y 2 ) O polinómio pedido é x − y + 1 2 ( x 2 − y 2 ) x-y+\frac 12 ( x^2-y^2) x − y + 2 1 ( x 2 − y 2 )
Tal como em CDI-I, sabemos que se um ponto é um extremo, então o gradiente nesse ponto vai ser nulo.
TEOREMA
Seja f : R n → R f: \R^n \to \R f : R n → R a ∈ R n a \in \R^n a ∈ R n
Então, ∇ f ( a ) = 0 ⃗ \nabla f(a) = \vec 0 ∇ f ( a ) = 0
Estes pontos, chamam-se pontos de estacionariedade ou pontos críticos .
DEFINIÇÃO
Seja f : R n → R f: \R^n \to \R f : R n → R a ∈ R n a \in \R^n a ∈ R n ∇ f ( a ) = 0 ⃗ \nabla f(a) = \vec 0 ∇ f ( a ) = 0
warning
É importante relembrar que nem todos os pontos ∇ f ( a ) = 0 ⃗ \nabla f(a) = \vec 0 ∇ f ( a ) = 0
Exemplos Seja f : R 2 → R f: \R^2 \to \R f : R 2 → R f ( x , y ) = x 2 + y 2 f(x,y)=x^2+y^2 f ( x , y ) = x 2 + y 2 f f f
( 0 , 0 ) = ∇ f ( x , y ) = ( 2 x , 2 y ) ⟹ x = 0 ∧ y = 0 (0,0) = \nabla f(x,y) = (2x, 2y) \implies x=0 \land y = 0 ( 0 , 0 ) = ∇ f ( x , y ) = ( 2 x , 2 y ) ⟹ x = 0 ∧ y = 0 Então, o único ponto crítico de f f f ( 0 , 0 ) (0,0) ( 0 , 0 )
Como f ( x , y ) = x 2 + y 2 ≥ 0 , ∀ ( x , y ) ∈ R 2 f(x,y)=x^2+y^2 \geq 0, \forall (x,y) \in \R^2 f ( x , y ) = x 2 + y 2 ≥ 0 , ∀ ( x , y ) ∈ R 2 f ( 0 , 0 ) = 0 f(0,0) = 0 f ( 0 , 0 ) = 0 f f f ( 0 , 0 ) (0,0) ( 0 , 0 )
Seja f : R 2 → R f: \R^2 \to \R f : R 2 → R f ( x , y ) = 1 1 + x 2 + y 2 f(x,y) = \frac{1}{1+x^2+y^2} f ( x , y ) = 1 + x 2 + y 2 1 f f f
∂ f ∂ x ( x , y ) = ∂ ∂ x 1 1 + x 2 + y 2 = ∂ ∂ x ( ( 1 + x 2 + y 2 ) − 1 ) = ( − 1 ) ( 1 + x 2 + y 2 ) − 1 − 1 ∂ ∂ x ( 1 + x 2 + y 2 ) = − 1 ( 1 + x 2 + y 2 ) 2 2 x = − 2 x ( 1 + x 2 + y 2 ) 2 \begin{aligned}
\frac{\partial f}{\partial x}(x,y) & = \frac{\partial }{\partial x}\frac{1}{1+x^2+y^2}\\
& =\frac{\partial}{\partial x} \left(\left(1+x^2+y^2\right)^{-1}\right)\\
& =(-1)(1+x^2+y^2)^{-1-1} \frac{\partial}{\partial x}\left(1+x^2+y^2\right)\\
& =\frac{-1}{\left(1+x^2+y^2\right)^2}2x\\
& =\frac{-2x}{\left(1+x^2+y^2\right)^2}
\end{aligned} ∂ x ∂ f ( x , y ) = ∂ x ∂ 1 + x 2 + y 2 1 = ∂ x ∂ ( ( 1 + x 2 + y 2 ) − 1 ) = ( − 1 ) ( 1 + x 2 + y 2 ) − 1 − 1 ∂ x ∂ ( 1 + x 2 + y 2 ) = ( 1 + x 2 + y 2 ) 2 − 1 2 x = ( 1 + x 2 + y 2 ) 2 − 2 x ∂ f ∂ y ( x , y ) = ∂ ∂ y 1 1 + x 2 + y 2 = ∂ ∂ y ( ( 1 + x 2 + y 2 ) − 1 ) = ( − 1 ) ( 1 + x 2 + y 2 ) − 1 − 1 ∂ ∂ y ( 1 + x 2 + y 2 ) = − 1 ( 1 + x 2 + y 2 ) 2 2 y = − 2 y ( 1 + x 2 + y 2 ) 2 \begin{aligned}
\frac{\partial f}{\partial y}(x,y) & = \frac{\partial }{\partial y}\frac{1}{1+x^2+y^2}\\
& =\frac{\partial}{\partial y} \left(\left(1+x^2+y^2\right)^{-1}\right)\\
& =(-1)(1+x^2+y^2)^{-1-1} \frac{\partial}{\partial y}\left(1+x^2+y^2\right)\\
& =\frac{-1}{\left(1+x^2+y^2\right)^2}2y\\
& =\frac{-2y}{\left(1+x^2+y^2\right)^2}
\end{aligned} ∂ y ∂ f ( x , y ) = ∂ y ∂ 1 + x 2 + y 2 1 = ∂ y ∂ ( ( 1 + x 2 + y 2 ) − 1 ) = ( − 1 ) ( 1 + x 2 + y 2 ) − 1 − 1 ∂ y ∂ ( 1 + x 2 + y 2 ) = ( 1 + x 2 + y 2 ) 2 − 1 2 y = ( 1 + x 2 + y 2 ) 2 − 2 y ( 0 , 0 ) = ∇ f ( x , y ) = ( − 2 x ( 1 + x 2 + y 2 ) 2 − 2 y ( 1 + x 2 + y 2 ) 2 ) ⟹ x = 0 ∧ y = 0 (0,0) = \nabla f(x,y) = \begin{pmatrix}
\frac{-2x}{\left(1+x^2+y^2\right)^2} &
\frac{-2y}{\left(1+x^2+y^2\right)^2}
\end{pmatrix}
\implies x = 0 \land y = 0 ( 0 , 0 ) = ∇ f ( x , y ) = ( ( 1 + x 2 + y 2 ) 2 − 2 x ( 1 + x 2 + y 2 ) 2 − 2 y ) ⟹ x = 0 ∧ y = 0 Portanto f f f ( x , y ) = ( 0 , 0 ) (x,y) = (0,0) ( x , y ) = ( 0 , 0 )
Pegando novamente na expressão da função, f ( x , y ) = 1 1 + x 2 + y 2 f(x,y)= \frac{1}{1+x^2+y^2} f ( x , y ) = 1 + x 2 + y 2 1
x 2 + y 2 ≥ 0 ⟹ 1 + x 2 + y 2 ≥ 1 ⟹ 1 1 ≥ 1 1 + x 2 + y 2 x^2+y^2\geq 0 \implies 1+x^2+y^2 \geq 1 \implies \frac 11 \geq \frac{1}{1+x^2+y^2} x 2 + y 2 ≥ 0 ⟹ 1 + x 2 + y 2 ≥ 1 ⟹ 1 1 ≥ 1 + x 2 + y 2 1 f ( x , y ) = 1 1 + x 2 + y 2 ≤ 1 , ∀ ( x , y ) ∈ R 2 f ( 0 , 0 ) = 1 1 + 0 2 + 0 2 = 1 \begin{darray}{l l}
f(x,y)=\frac{1}{1+x^2+y^2}\leq 1, \forall (x,y) \in \R^2
& f(0,0)= \frac{1}{1+0^2+0^2}=1
\end{darray} f ( x , y ) = 1 + x 2 + y 2 1 ≤ 1 , ∀ ( x , y ) ∈ R 2 f ( 0 , 0 ) = 1 + 0 2 + 0 2 1 = 1 Portanto ocorre máximo de f f f ( 0 , 0 ) (0,0) ( 0 , 0 )
Seja f : R 2 → R f: \R^2 \to \R f : R 2 → R f ( x , y ) = x y f(x,y) = xy f ( x , y ) = x y f f f
∂ f ∂ x = ∂ ∂ x ( x y ) = y ∂ x ∂ x = y ⋅ 1 = y ∂ f ∂ x = ∂ ∂ x ( x y ) = x \begin{darray}{l l}
\frac{\partial f}{\partial x}=\frac{\partial}{\partial x}(xy) = y\frac{\partial x}{\partial x} = y\cdot 1 = y\\
\frac{\partial f}{\partial x}=\frac{\partial}{\partial x}(xy) = x
\end{darray} ∂ x ∂ f = ∂ x ∂ ( x y ) = y ∂ x ∂ x = y ⋅ 1 = y ∂ x ∂ f = ∂ x ∂ ( x y ) = x Podemos agora calcular os pontos críticos:
( 0 , 0 ) = ∇ f ( x , y ) = ( y , x ) ⟹ y = 0 ∧ x = 0 (0,0) = \nabla f(x,y) = (y, x) \implies y = 0 \land x = 0 ( 0 , 0 ) = ∇ f ( x , y ) = ( y , x ) ⟹ y = 0 ∧ x = 0 Assim, o único ponto crítico de f f f ( x , y ) = ( 0 , 0 ) (x,y) = (0,0) ( x , y ) = ( 0 , 0 )
Será que ocorre extremo de f f f ( 0 , 0 ) (0,0) ( 0 , 0 )
Como é que esta função se comporta junto a ( 0 , 0 ) (0,0) ( 0 , 0 )
f ( x , y ) = x y f ( 0 , 0 ) = 0 \begin{darray}{l l}
f(x,y) = xy & f(0,0) = 0
\end{darray} f ( x , y ) = x y f ( 0 , 0 ) = 0
x y > 0 xy > 0 x y > 0 x x x y y y ( x > 0 ∧ y > 0 ) ∨ ( x < 0 ∧ y < 0 ) (x > 0 \land y > 0) \lor (x < 0 \land y < 0) ( x > 0 ∧ y > 0 ) ∨ ( x < 0 ∧ y < 0 ) x y < 0 xy < 0 x y < 0 x x x y y y ( x > 0 ∧ y < 0 ) ∨ ( x < 0 ∧ y > 0 ) (x > 0 \land y < 0) \lor (x < 0 \land y > 0) ( x > 0 ∧ y < 0 ) ∨ ( x < 0 ∧ y > 0 ) Então, temos ponto de sela em ( 0 , 0 ) (0,0) ( 0 , 0 )
Este é um exemplo de ponto crítico que não é extremo.
Seja f : R 2 → R f: \R^2 \to \R f : R 2 → R f ( x , y ) = x 2 − y 2 f(x,y) = x^2-y^2 f ( x , y ) = x 2 − y 2 f f f
( 0 , 0 ) = ∇ f ( x , y ) = ( 2 x , − 2 y ) ⟹ x = 0 ∧ y = 0 (0,0)=\nabla f(x,y) = (2x, -2y) \implies x=0 \land y=0 ( 0 , 0 ) = ∇ f ( x , y ) = ( 2 x , − 2 y ) ⟹ x = 0 ∧ y = 0 Portanto ponto crítico de f f f ( x , y ) = ( 0 , 0 ) (x,y)=(0,0) ( x , y ) = ( 0 , 0 )
Será que ocorre extremo de f f f ( 0 , 0 ) (0,0) ( 0 , 0 )
Se x = 0 x = 0 x = 0 f ( x , y ) = − y 2 ≤ 0 f(x,y) = -y^2 \leq 0 f ( x , y ) = − y 2 ≤ 0
Se y = 0 y = 0 y = 0 f ( x , y ) = x 2 ≥ 0 f(x,y) = x^2 \geq 0 f ( x , y ) = x 2 ≥ 0
Então, este ponto crítico não corresponde a extremo, pois temos ponto de sela em ( 0 , 0 ) (0,0) ( 0 , 0 )
Como a Matriz Hessiana é uma matriz simétrica, podemos usar as suas propriedades para
determinar os extremos de uma função.
Se A A A A A A u 1 , u 2 , … , u n u_1, u_2, \dots, u_n u 1 , u 2 , … , u n λ 1 , λ 2 , … , λ n \lambda_1, \lambda_2, \dots, \lambda_n λ 1 , λ 2 , … , λ n A u i = λ i u i Au_i = \lambda_i u_i A u i = λ i u i i = 1 , … , n i = 1, \dots, n i = 1 , … , n
Escrevendo h = a 1 u 1 + a 2 u 2 + ⋯ + a n u n h=a_1u_1+a_2u_2+\dots+a_nu_n h = a 1 u 1 + a 2 u 2 + ⋯ + a n u n a i ∈ R a_i \in \R a i ∈ R A A A
h T A h = h i ( A h ) = ( a 1 u 1 + a 2 u 2 + ⋯ + a n u n ) ⋅ ( a 1 λ 1 u 1 + a 2 λ 2 u 2 + ⋯ + a n λ n u n ) = λ 1 a 1 2 + λ 2 a 2 2 + ⋯ + λ n a n 2 \begin{aligned}
h^T A h &= h_i(Ah)\\
&=(a_1u_1+ a_2u_2+ \dots + a_nu_n) \cdot (a_1\lambda_1 u_1 + a_2 \lambda_2 u_2 + \dots + a_n \lambda_n u_n)\\
&= \lambda_1 a^2_1 + \lambda_2 a^2_2 + \dots + \lambda_n a^2_n
\end{aligned} h T A h = h i ( A h ) = ( a 1 u 1 + a 2 u 2 + ⋯ + a n u n ) ⋅ ( a 1 λ 1 u 1 + a 2 λ 2 u 2 + ⋯ + a n λ n u n ) = λ 1 a 1 2 + λ 2 a 2 2 + ⋯ + λ n a n 2 O sinal desta expressão é dado pelos λ i \lambda_i λ i
Se substituirmos este resultado na Fórmula de Taylor de 2ª Ordem em torno de um ponto crítico a ∈ R n a\in \R^n a ∈ R n ∇ f ( a ) = 0 ⃗ \nabla f (a) = \vec 0 ∇ f ( a ) = 0 f : R n → R f: \R^n \to \R f : R n → R
Então,
f ( a + h ) = f ( a ) + 0 + 1 2 h T H f ( a ) h + R 2 ( h ) f(a+h) = f(a) + 0 + \frac 12 h^T H_f(a) h + R_2(h) f ( a + h ) = f ( a ) + 0 + 2 1 h T H f ( a ) h + R 2 ( h ) com lim h → 0 ⃗ R 2 ( h ) ∣ ∣ h ∣ ∣ 2 = 0 ⃗ \lim_{h\to \vec 0} \frac{R_2(h)}{||h||^2} = \vec 0 lim h → 0 ∣∣ h ∣ ∣ 2 R 2 ( h ) = 0
Donde, pelas contas feitas acima, obtemos
f ( a + h ) − f ( a ) = 1 2 ( λ 1 a 1 2 + λ 2 a 2 2 + ⋯ + λ n a n 2 ) + R 2 ( h ) f(a+h) -f(a) = \frac 12 (\lambda_1 a^2_1+ \lambda_2 a^2_2 + \dots + \lambda_n a^2_n) + R_2(h) f ( a + h ) − f ( a ) = 2 1 ( λ 1 a 1 2 + λ 2 a 2 2 + ⋯ + λ n a n 2 ) + R 2 ( h ) Assim, como R 2 ( h ) R_2(h) R 2 ( h ) h → 0 ⃗ h\to \vec 0 h → 0 f ( a + h ) − f ( a ) f(a+h)-f(a) f ( a + h ) − f ( a )
f ( a + h ) − f ( a ) > 0 f(a+h)-f(a) > 0 f ( a + h ) − f ( a ) > 0 λ i > 0 \lambda_i > 0 λ i > 0 a a a ocorre mínimo f ( a + h ) − f ( a ) < 0 f(a+h)-f(a) < 0 f ( a + h ) − f ( a ) < 0 λ i < 0 \lambda_i < 0 λ i < 0 a a a ocorre máximo
TEOREMA
Seja f : R n → R f: \R^n \to \R f : R n → R a ∈ R a \in \R a ∈ R f f f ∇ f ( a ) = 0 ⃗ \nabla f(a) = \vec 0 ∇ f ( a ) = 0
Se H f ( a ) H_f(a) H f ( a )
definida positiva (cada valor próprio λ i > 0 \lambda_i > 0 λ i > 0 mínimo local em a a a
definida negativa (cada valor próprio λ i < 0 \lambda_i < 0 λ i < 0 máximo local em a a a
indefinida (valores próprios λ i > 0 \lambda_i >0 λ i > 0 λ j < 0 \lambda_j < 0 λ j < 0 ponto de sela a a a
semidefinida positiva (valores próprios λ i ≥ 0 \lambda_i \geq 0 λ i ≥ 0 mínimo ou ponto de sela em a a a
semidefinida negativa (valores próprios λ i ≤ 0 \lambda_i \leq 0 λ i ≤ 0 máximo ou ponto de sela em a a a
Exemplos Determine e clarifique os pontos de estacionariedade da função f : R 2 → R f: \R^2 \to \R f : R 2 → R
f ( x , y ) = − x 4 + 2 x 2 − y 2 + 3 f(x,y) = -x^4+2x^2-y^2+3 f ( x , y ) = − x 4 + 2 x 2 − y 2 + 3
Começamos por determinar o gradiente de f f f
∇ f ( x , y ) = ( − 4 x 3 + 4 x , − 2 y ) \nabla f(x,y) = (-4x^3+4x, -2y) ∇ f ( x , y ) = ( − 4 x 3 + 4 x , − 2 y )
E depois igualá-lo ao vetor nulo, para obtermos os pontos críticos de f f f
∇ f ( x , y ) = ( 0 , 0 ) ⇔ ( − 4 x 3 + 4 x , − 2 y ) = ( 0 , 0 ) ⟹ { 0 = − 4 x 3 + 2 x 0 = − 2 y ⇔ { 0 = 4 x ( 1 − x 2 ) y = 0 ⇔ { x = 0 ∨ x = ± 1 y = 0 \begin{array}{ c }
\nabla f( x,y) =( 0,0) \Leftrightarrow \left( -4x^{3} +4x,-2y\right) =( 0,0) \Longrightarrow \\
\begin{cases}
0=-4x^{3} +2x\\
0=-2y
\end{cases} \Leftrightarrow \begin{cases}
0=4x(1-x^{2} )\\
y=0
\end{cases} \Leftrightarrow \begin{cases}
x=0\lor x=\pm 1\\
y=0
\end{cases}
\end{array} ∇ f ( x , y ) = ( 0 , 0 ) ⇔ ( − 4 x 3 + 4 x , − 2 y ) = ( 0 , 0 ) ⟹ { 0 = − 4 x 3 + 2 x 0 = − 2 y ⇔ { 0 = 4 x ( 1 − x 2 ) y = 0 ⇔ { x = 0 ∨ x = ± 1 y = 0 Os pontos críticos de f f f ( 0 , 0 ) , ( 1 , 0 ) , ( − 1 , 0 ) (0,0), (1,0), (-1,0) ( 0 , 0 ) , ( 1 , 0 ) , ( − 1 , 0 )
∂ 2 f ∂ x 2 = ∂ ∂ x ( − 4 x 3 + 4 x ) = − 12 x 2 + 4 ∂ 2 f ∂ y 2 = ∂ ∂ y ( − 2 y ) = − 2 ∂ 2 f ∂ x ∂ y = ∂ ∂ x ( − 2 y ) = 0 \begin{darray}{l}
\frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x}(-4x^3+4x) = -12x^2+4\\
\\
\frac{\partial^2 f}{\partial y^2} = \frac{\partial}{\partial y}(-2y) = -2\\
\\
\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial}{\partial x}(-2y) = 0
\end{darray} ∂ x 2 ∂ 2 f = ∂ x ∂ ( − 4 x 3 + 4 x ) = − 12 x 2 + 4 ∂ y 2 ∂ 2 f = ∂ y ∂ ( − 2 y ) = − 2 ∂ x ∂ y ∂ 2 f = ∂ x ∂ ( − 2 y ) = 0
Em ( 0 , 0 ) (0,0) ( 0 , 0 )
H f ( 0 , 0 ) = [ 4 0 0 − 2 ] H_{f} (0,0)=\begin{bmatrix}
4 & 0\\
0 & -2
\end{bmatrix} H f ( 0 , 0 ) = [ 4 0 0 − 2 ] Como já está na forma diagonal, tem valores próprios 4 4 4 − 2 -2 − 2 f f f ( 0 , 0 ) (0,0) ( 0 , 0 )
Em ( 1 , 0 ) (1,0) ( 1 , 0 )
H f ( 1 , 0 ) = [ − 8 0 0 − 2 ] H_{f} (1,0)=\begin{bmatrix}
-8 & 0\\
0 & -2
\end{bmatrix} H f ( 1 , 0 ) = [ − 8 0 0 − 2 ] Como já está na forma diagonal, tem valores próprios − 8 -8 − 8 − 2 -2 − 2 f f f ( 1 , 0 ) (1,0) ( 1 , 0 )
Em ( − 1 , 0 ) (-1,0) ( − 1 , 0 )
H f ( − 1 , 0 ) = [ − 8 0 0 − 2 ] H_{f} (-1,0)=\begin{bmatrix}
-8 & 0\\
0 & -2
\end{bmatrix} H f ( − 1 , 0 ) = [ − 8 0 0 − 2 ] Como já está na forma diagonal, tem valores próprios − 8 -8 − 8 − 2 -2 − 2 f f f ( − 1 , 0 ) (-1,0) ( − 1 , 0 )
Para uma Matriz Hessiana 2 × 2 2\times2 2 × 2
Se f : R 2 → R f: \R^2 \to \R f : R 2 → R ∇ f ( x ) = 0 ⃗ \nabla f(x) = \vec 0 ∇ f ( x ) = 0
H f ( a ) = [ A B B C ] H_{f} (a)=\begin{bmatrix}
A & B\\
B & C
\end{bmatrix} H f ( a ) = [ A B B C ] com
A = ∂ 2 f ∂ x i 2 ( a ) B = ∂ 2 f ∂ x ∂ y ( a ) C = ∂ 2 f ∂ y 2 ( a ) A= \frac{\partial^2 f}{\partial x_i^2}(a) \quad
B=\frac{\partial^2 f}{\partial x\partial y}(a)\quad
C=\frac{\partial^2 f}{\partial y^2}(a) A = ∂ x i 2 ∂ 2 f ( a ) B = ∂ x ∂ y ∂ 2 f ( a ) C = ∂ y 2 ∂ 2 f ( a )
Determinante: det A = λ 1 × λ 2 × ⋯ × λ n \operatorname{det} A= \lambda_1 \times \lambda_2 \times \dots \times \lambda_n det A = λ 1 × λ 2 × ⋯ × λ n
Traço: tr A = λ 1 + λ 2 + ⋯ + λ n \operatorname{tr} A = \lambda_1 + \lambda_2 + \dots + \lambda_n tr A = λ 1 + λ 2 + ⋯ + λ n
Então, podemos concluir o seguinte :
Se det H f ( a ) < 0 \operatorname{det} H_f(a) < 0 det H f ( a ) < 0 f f f a a a
Se det H f ( a ) > 0 \operatorname{det} H_f(a) > 0 det H f ( a ) > 0 tr H f ( a ) > 0 \operatorname{tr} H_f(a) > 0 tr H f ( a ) > 0 f f f a a a
Se det H f ( a ) > 0 \operatorname{det} H_f(a) > 0 det H f ( a ) > 0 tr H f ( a ) < 0 \operatorname{tr} H_f(a) < 0 tr H f ( a ) < 0 f f f a a a
Exemplos Determinar e caracterizar os pontos críticos de f f f
f ( x , y ) = ( x − y ) 2 − x 4 − y 4 f(x,y) = (x-y)^2 -x^4-y^4 f ( x , y ) = ( x − y ) 2 − x 4 − y 4 ∂ f ∂ x = 2 ( x − y ) ⋅ 1 − 4 x 3 ∂ f ∂ y = 2 ( x − y ) ⋅ ( − 1 ) − 4 y 3 \begin{array}{ll}
\frac{\partial f}{\partial x} = 2(x-y)\cdot 1 - 4x^3 &
\frac{\partial f}{\partial y} = 2(x-y)\cdot (-1) - 4y^3
\end{array} ∂ x ∂ f = 2 ( x − y ) ⋅ 1 − 4 x 3 ∂ y ∂ f = 2 ( x − y ) ⋅ ( − 1 ) − 4 y 3 Igualando o gradiente de f f f
∇ f ( x , y ) = ( 0 , 0 ) ⇔ ( 2 ( x − y ) − 4 x 3 , − 2 ( x − y ) − 4 y 3 ) = ( 0 , 0 ) ⟹ { x − y = 2 x 3 x − y = − 2 y 3 ⇔ 2 x 3 = x − y = − 2 y 3 \begin{array}{ c }
\nabla f( x,y) =( 0,0) \Leftrightarrow \left( 2( x-y) -4x^{3} ,-2( x-y) -4y^{3}\right) =( 0,0) \Longrightarrow \\
\begin{cases}
x-y=2x^{3}\\
x-y=-2y^{3}
\end{cases} \Leftrightarrow 2x^{3} =x-y=-2y^{3}
\end{array} ∇ f ( x , y ) = ( 0 , 0 ) ⇔ ( 2 ( x − y ) − 4 x 3 , − 2 ( x − y ) − 4 y 3 ) = ( 0 , 0 ) ⟹ { x − y = 2 x 3 x − y = − 2 y 3 ⇔ 2 x 3 = x − y = − 2 y 3 Logo, os pontos críticos de f f f ( 0 , 0 ) , ( 1 , − 1 ) , ( − 1 , 1 ) (0,0), (1,-1), (-1, 1) ( 0 , 0 ) , ( 1 , − 1 ) , ( − 1 , 1 )
Determinando agora as derivadas de segunda ordem para determinar a Matriz Hessiana:
∂ 2 f ∂ x 2 = ∂ ∂ x ( 2 ( x − y ) − 4 x 3 ) = 2 − 12 x 2 ∂ 2 f ∂ y 2 = ∂ ∂ y ( − 2 ( x − y ) − 4 y 3 ) = 2 − 12 y 2 ∂ 2 f ∂ x ∂ y = ∂ ∂ x ( − 2 ( x − y ) − 4 y 3 ) = − 2 \begin{darray}{l}
\frac{\partial^2 f}{\partial x^2} = \frac{\partial }{\partial x} (2(x-y)-4x^3)= 2-12x^2\\
\\
\frac{\partial^2 f}{\partial y^2} = \frac{\partial }{\partial y} (-2(x-y)-4y^3)= 2-12y^2\\
\\
\frac{\partial^2 f}{\partial x\partial y} = \frac{\partial }{\partial x} (-2(x-y)-4y^3)= -2
\end{darray} ∂ x 2 ∂ 2 f = ∂ x ∂ ( 2 ( x − y ) − 4 x 3 ) = 2 − 12 x 2 ∂ y 2 ∂ 2 f = ∂ y ∂ ( − 2 ( x − y ) − 4 y 3 ) = 2 − 12 y 2 ∂ x ∂ y ∂ 2 f = ∂ x ∂ ( − 2 ( x − y ) − 4 y 3 ) = − 2 H f ( x , y ) = [ 2 − 12 x 2 − 2 − 2 2 − 12 y 2 ] H_{f} (x,y)=\begin{bmatrix}
2-12x^{2} & -2\\
-2 & 2-12y^{2}
\end{bmatrix} H f ( x , y ) = [ 2 − 12 x 2 − 2 − 2 2 − 12 y 2 ]
Em ( 1 , − 1 ) (1,-1) ( 1 , − 1 ) ( − 1 , 1 ) (-1,1) ( − 1 , 1 )
H f ( 1 , − 1 ) = H f ( − 1 , 1 ) = [ − 10 − 2 − 2 − 10 ] H_{f} (1,-1)=H_{f} (-1, 1)=\begin{bmatrix}
-10 & -2\\
-2 & -10
\end{bmatrix} H f ( 1 , − 1 ) = H f ( − 1 , 1 ) = [ − 10 − 2 − 2 − 10 ] Assim, como o determinante é 96 > 0 96 > 0 96 > 0 − 20 < 0 -20 < 0 − 20 < 0 ( 1 , − 1 ) (1,-1) ( 1 , − 1 ) ( − 1 , 1 ) (-1,1) ( − 1 , 1 )
Em ( 0 , 0 ) (0,0) ( 0 , 0 )
H f ( 0 , 0 ) = [ 2 − 2 − 2 2 ] H_{f} (0,0)=\begin{bmatrix}
2 & -2\\
-2 & 2
\end{bmatrix} H f ( 0 , 0 ) = [ 2 − 2 − 2 2 ] Assim, o determinante é 0 0 0 4 4 4
Como o determinante é 0 0 0 ( 0 , 0 ) (0,0) ( 0 , 0 )
Tentamos estudar a função em certas direções:
Em y = x y=x y = x f ( x , y ) = 0 2 − x 4 − x 4 = − 2 x 4 f(x,y) = 0^2 - x^4 - x^4= -2x^4 f ( x , y ) = 0 2 − x 4 − x 4 = − 2 x 4 máximo local segundo y = x y=x y = x
Em y = 0 y=0 y = 0 f ( x , y ) = x 2 − x 4 = x 2 ( 1 − x 2 ) f(x,y) = x^2 - x^4 = x^2(1-x^2) f ( x , y ) = x 2 − x 4 = x 2 ( 1 − x 2 ) mínimo local segundo y = 0 y=0 y = 0
Logo, o ponto ( 0 , 0 ) (0,0) ( 0 , 0 )
Identificar e classificar pontos críticos de f f f
f ( x , y ) = x y + 1 x + 8 y f(x,y) = xy + \frac 1x + \frac 8y f ( x , y ) = x y + x 1 + y 8 Começamos por igualar o gradiente de f f f
∇ f ( x , y ) = ( 0 , 0 ) ⇔ ( y − 1 x 2 , x − 8 y 2 ) = ( 0 , 0 ) ⟹ { y = 1 x 2 x = 8 y 2 ⟹ { y = 4 x = 1 2 \begin{array}{ c }
\nabla f( x,y) =( 0,0) \Leftrightarrow \left( y-\frac{1}{x^{2}} ,\ x-\frac{8}{y^{2}}\right) =( 0,0) \Longrightarrow \\
\begin{cases}
y=\frac{1}{x^{2}}\\
x=\frac{8}{y^{2}}
\end{cases} \Longrightarrow \begin{cases}
y=4\\
x=\frac{1}{2}
\end{cases}
\end{array} ∇ f ( x , y ) = ( 0 , 0 ) ⇔ ( y − x 2 1 , x − y 2 8 ) = ( 0 , 0 ) ⟹ { y = x 2 1 x = y 2 8 ⟹ { y = 4 x = 2 1 Logo, a função f f f ( 1 2 , 4 ) (\frac{1}{2}, 4) ( 2 1 , 4 )
H f ( x , y ) = [ 2 x 3 1 1 16 y 3 ] H_{f} (x,y)=\begin{bmatrix}
\frac{2}{x^{3}} & 1\\
1 & \frac{16}{y^{3}}
\end{bmatrix} H f ( x , y ) = [ x 3 2 1 1 y 3 16 ] Então, a Matriz Hessiana em ( 1 2 , 4 ) \left(\frac 12, 4\right) ( 2 1 , 4 )
H f ( 1 2 , 4 ) = [ 16 1 1 1 4 ] H_{f} \left(\frac 12, 4\right)=\begin{bmatrix}
16 & 1\\
1 & \frac 14
\end{bmatrix} H f ( 2 1 , 4 ) = [ 16 1 1 4 1 ] Assim, det = 3 > 0 \operatorname{det} = 3 > 0 det = 3 > 0 tr = 16 + 1 4 > 0 \operatorname{tr} = 16 + \frac 14 > 0 tr = 16 + 4 1 > 0
Logo, a função f f f ( 1 2 , 4 ) \left(\frac 12, 4\right) ( 2 1 , 4 )
Identificar e classificar pontos críticos de f f f
f ( x , y , z ) = x 2 + y 2 + z 2 + x y f (x,y,z) = x^2+ y^2+ z^2+ xy f ( x , y , z ) = x 2 + y 2 + z 2 + x y Começamos, novamente, por igualar o gradiente a zero:
∇ f ( x , y , z ) = ( 0 , 0 , 0 ) ⇔ ( 2 x + y , 2 y + x , 2 z ) = ( 0 , 0 ) ⟹ { 2 x + y = 0 2 y + x = 0 2 z = 0 ⟹ { x = 0 y = 0 z = 0 \begin{array}{ c }
\nabla f( x,y,z) =( 0,0,0) \Leftrightarrow ( 2x+y,2y+x,2z) =( 0,0) \Longrightarrow \\
\begin{cases}
2x+y=0\\
2y+x=0\\
2z=0
\end{cases} \Longrightarrow \begin{cases}
x=0\\
y=0\\
z=0
\end{cases}
\end{array} ∇ f ( x , y , z ) = ( 0 , 0 , 0 ) ⇔ ( 2 x + y , 2 y + x , 2 z ) = ( 0 , 0 ) ⟹ ⎩ ⎨ ⎧ 2 x + y = 0 2 y + x = 0 2 z = 0 ⟹ ⎩ ⎨ ⎧ x = 0 y = 0 z = 0 Então, o único ponto crítico de f f f ( 0 , 0 , 0 ) (0,0,0) ( 0 , 0 , 0 )
H f ( x , y , z ) = [ 2 1 0 1 2 0 0 0 2 ] → [ 3 2 0 0 1 2 0 0 0 2 ] → [ 3 2 0 0 0 2 0 0 0 2 ] H_{f}( x,y,z) =\begin{bmatrix}
2 & 1 & 0\\
1 & 2 & 0\\
0 & 0 & 2
\end{bmatrix}\rightarrow \begin{bmatrix}
\frac{3}{2} & 0 & 0\\
1 & 2 & 0\\
0 & 0 & 2
\end{bmatrix}\rightarrow \begin{bmatrix}
\frac{3}{2} & 0 & 0\\
0 & 2 & 0\\
0 & 0 & 2
\end{bmatrix} H f ( x , y , z ) = 2 1 0 1 2 0 0 0 2 → 2 3 1 0 0 2 0 0 0 2 → 2 3 0 0 0 2 0 0 0 2 Então, os valores próprios da matriz são 3 2 , 2 , 2 \frac 32, 2, 2 2 3 , 2 , 2 f f f ( 0 , 0 , 0 ) (0,0,0) ( 0 , 0 , 0 )
Slides: