Ejercicio¶
Dada la función $f(x,y)=x^{3}+y^{3}$, calcular una aproximación de
- El gradiente $\nabla f(x,y)$
- La matriz Hessiana $\mathrm{Hess} f(x,y)$
para $(x_0,y_0)=(1,2)$ con $h_{x}=h_{y}=0.1$.
Gradiente $\nabla f(x,y)$¶
El Gradiente viene dado por $$ \nabla f=\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\right) $$
Y el gradiente exacto se calcularía calculando primero las derivadas parciales
$$\frac{\partial f}{\partial x} = 3x^2 \quad \frac{\partial f}{\partial y}=3y^2$$que para el valor $(x_0,y_0)=(1,2)$
$$\left(\frac{\partial f}{\partial x}\right)_{(1,2)} = 3(1)^2=3 \quad \left(\frac{\partial f}{\partial y}\right)_{(1,2)}=3(2)^2=12$$Y por lo tanto
$$\nabla f(x,y)_{(1,2)}=(3,12).$$Calculemos ahora una aproximación. La primera componente del vector será:
$$ \begin{align*} \frac{\partial f}{\partial x}\approx & \dfrac{f\left(x_{m}+h_{x},y_{n}\right)-f\left(x_{m}-h_{x},y_{n}\right)}{2h_{x}}=\dfrac{f\left(1+0.1,2\right)-f\left(1-0.1,2\right)}{2\left(0.1\right)}=\\[0.3cm] = & \dfrac{\left(\left(1+0.1\right)^{3}+\left(2\right)^{3}\right)-\left(\left(1-0.1\right)^{3}+\left(2\right)^{3}\right)}{2\left(0.1\right)}=\\[0.3cm] = & \dfrac{9.331-8.729}{0.2}=\dfrac{0.602}{0.2}=3.01\\ \end{align*} $$Y la segunda componente
$$ \begin{align*} \frac{\partial f}{\partial y}\approx & \dfrac{f\left(x_{m},y_{n}+h_{y}\right)- f\left(x_{m},y_{n}-h_{y}\right)}{2h_{y}}=\dfrac{f\left(1,2+0.1\right)-f\left(1,2-0.1\right)}{2\left(0.1\right)}=\\[0.3cm] = & \dfrac{\left(\left(1\right)^{3}+\left(2+0.1\right)^{3}\right)-\left(\left(1\right)^{3}+\left(2-0.1\right)^{3}\right)}{2\left(0.1\right)}=\\[0.3cm] = & \dfrac{10.261-7.859}{0.2}=\dfrac{2.401}{0.2}=12.01\\ \end{align*} $$Por lo tanto $$ \nabla f(1,2)\approx (3.01,12.01)$$
Matriz Hessiana $\mathrm{Hess} f(x,y)$¶
La matriz hessiana exacta para una función de dos variables viene dada por $$ H = \left( \begin{array}{cc} \dfrac{\partial^{2}f}{\partial x^{2}} & \dfrac{\partial^{2}f}{\partial x\partial y}\\ \dfrac{\partial^{2}f}{\partial y\partial x}& \dfrac{\partial^{2}f}{\partial y^{2}} \end{array} \right) $$
Como
$$\frac{\partial f}{\partial x} = 3x^2 \quad \frac{\partial f}{\partial y}=3y^2$$Se tiene que
$$ \frac{\partial^{2}f}{\partial x^{2}} = 6x \quad \frac{\partial^{2}f}{\partial y^{2}}=6y $$$$ \dfrac{\partial^{2}f}{\partial x\partial y} = 0 \quad \dfrac{\partial^{2}f}{\partial y\partial x} = 0 $$Para funciones suficientemente regulares, como es este caso, por el teorema de Schwarz, las derivadas cruzadas son iguales.
Para $(x_0,y_0)=(1,2)$
$$ \left(\frac{\partial^{2}f}{\partial x^{2}}\right)_{(1,2)} = 6(1)=6 \quad \left(\frac{\partial^{2}f}{\partial y^{2}}\right)_{(1,2)}=6(2)=12 \quad \left(\frac{\partial^{2}f}{\partial x\partial y}\right)_{(1,2)}=\left(\frac{\partial^{2}f}{\partial y\partial x}\right)_{(1,2)}=0 $$Y por lo tanto
$$ H = \left( \begin{array}{cc} 6 & 0 \\ 0 & 12 \end{array} \right) $$Calculemos ahora una aproximación. Tenemos que $$ \begin{align*} \frac{\partial^{2}f}{\partial x^{2}}& \approx \dfrac{f\left(x_{m}+h_{x},y_{n}\right)-2f\left(x_{m},y_{n}\right)+f\left(x_{m}-h_{x},y_{n}\right)}{h_{x}^{2}}=\\[0.3cm] &= \dfrac{f\left(1+0.1,2\right)-2f\left(1,2\right)+f\left(1-0.1,2\right)}{\left(0.1\right)^{2}}=\\[0.3cm] & = \dfrac{\left(\left(1+0.1\right)^{3}+\left(2\right)^{3}\right)-2\left(\left(1\right)^{3}+\left(2\right)^{3}\right)+\left(\left(1-0.1\right)^{3}+\left(2\right)^{3}\right)}{\left(0.1\right)^{3}}=\\[0.3cm] &= \dfrac{9.331-18+8.729}{0.01}=\dfrac{0.06}{0.01}=6\\ \end{align*} $$
Y $$ \begin{align*} \frac{\partial^{2}f}{\partial y^{2}} & \approx\dfrac{f\left(x_{m},y_{n}+h_{y}\right)-2f\left(x_{m},y_{n}\right)+f\left(x_{m},y_{n}-h_{y}\right)} {h_{y}^{2}}=\\[0.3cm] &=\dfrac{f\left(1,2+0.1\right)-2f\left(1,2\right)+f\left(1,2-0.1\right)}{\left(0.1\right)^{2}}=\\[0.3cm] & =\dfrac{\left(\left(1\right)^{3}+\left(2+0.1\right)^{3}\right)-2\left(\left(1\right)^{3}+\left(2\right)^{3}\right)+\left(\left(1\right)^{3}+\left(2-0.1\right)^{3}\right)}{\left(0.1\right)^{2}}=\\[0.3cm] & = \dfrac{10.261-18+7.859}{0.01}=\dfrac{0.12}{0.01}=12\\ \end{align*} $$
Calculemos solo una derivada cruzada puesto que son iguales. Usemos también fórmulas centradas
$$ \frac{\partial^{2}f}{\partial y\partial x}=\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right) = \frac{\partial f'_x}{\partial y} \approx \dfrac{f'_x\left(x_{m},y_{n}+h_{y}\right)-f'_x\left(x_{m},y_{n}-h_{y}\right)} {2h_{y}} $$Teniendo en cuenta que $$ f'_x\left(x_{m},y_{n}+h_{y}\right) \approx \dfrac{f\left(x_{m}+h_{x},y_{n}+h_{y}\right)-f\left(x_{m}-h_{x},y_{n}+h_{y}\right)}{2h_{x}} $$
y que $$ f'_x\left(x_{m},y_{n}-h_{y}\right) \approx \dfrac{f\left(x_{m}+h_{x},y_{n}-h_{y}\right)-f\left(x_{m}-h_{x},y_{n}-h_{y}\right)}{2h_{x}} $$
tenemos que $$ \frac{\partial^{2}f}{\partial y\partial x}\approx \frac{f\left(x_{m}+h_{x},y_{n}+h_{y}\right)-f\left(x_{m}-h_{x},y_{n}+h_{y}\right)-f\left(x_{m}+h_{x},y_{n}-h_{y}\right)+f\left(x_{m}-h_{x},y_{n}-h_{y}\right)}{4\,h_x\,h_y} $$
Y para el punto dado es
$$ \dfrac{f(1+0.1,2+0.1)-f(1-0.1,2+0.1)-f(1+0.1,2-0.1)+f(1-0.1,2-0.1)}{4(0.1)(0.1)}= $$tenemos
$$ =\dfrac{(1.1^3+2.1^3)-(0.9^3+2.1^3)-(1.1^3+1.9^3)+(0.9^3+1.9^3)}{4(0.1)^2}= $$Y operando con 3 decimales
$$ =\dfrac{10.592-9.990-8.190+7.588}{0.04}=\dfrac{0}{0.04}=0 $$Y por lo tanto, la matriz Hessiana aproximada es
$$ H \approx \left( \begin{array}{cc} 6 & 0 \\ 0 & 12 \end{array} \right) $$