$\textcolor{Red}{\textbf{Aproximación de Taylor para funciones $f:\mathbb{R}^{3}\rightarrow\mathbb{R}$}}$

El caso de la aproximación con $n=2$ nos queda
$$f(x,y)=f(x_{0},y_{0})+\textcolor{Blue}{\frac{1}{1!}\left(\frac{\partial f}{\partial x}(x_{0},y_{0})\cdot (x-x_{0})+\frac{\partial f}{\partial y}(x_{0},y_{0})\cdot (y-y_{0})\right)}+$$

$$\textcolor{Red}{\frac{1}{2!}\left(\frac{\partial^{2} f}{\partial x^{2}}(x_{0},y_{0})(x-x_{0})^{2}+2\frac{\partial^{2} f}{\partial x\partial y}(x_{0},y_{0})(x-x_{0})(y-y_{0})+\frac{\partial^{2} f}{\partial y^{2}}(x_{0},y_{0})(y-y_{0})^{2}\right)}+R_{2}$$
Donde la expresión azul se puede escribir

$$\textcolor{Blue}{\frac{1}{1!}\left(\frac{\partial f}{\partial x}(x_{0},y_{0})\cdot (x-x_{0})+\frac{\partial f}{\partial y}(x_{0},y_{0})\cdot (y-y_{0})\right)=\nabla f(x_{0},y_{0},z_{0})\cdot (h_{1},h_{2},h_{3})}$$
y la expresión en rojo

$$\textcolor{Red}{\frac{1}{2!}\left(\frac{\partial^{2}f}{\partial
x^{2}}{p}(x-x{0})^{2}+2\frac{\partial^{2}f}{\partial y \partial
x}{p}(x-x{0})(y-y_{0})+\frac{\partial^{2}f}{\partial
y^{2}}{p}(y-y{0})^{2}\right)}$$ Define una forma cuadratica que
podemos escribir

$$\textcolor{Red}{\frac{1}{2!}(x-x_{0}\quad y-y_{0})\left(\begin{array}{cc}
\frac{\partial^{2}f}{\partial x^{2}}&\frac{\partial^{2}f}{\partial y \partial x}\\
\frac{\partial^{2}f}{\partial x \partial y }&\frac{\partial^{2}f}{\partial y^{2}} \end{array}\right)\left(\begin{array}{c}
x-x_{0} \\y-y_{0} \end{array}\right)}$$

Por lo que el desarrollo de Taylor se puede escribir
$$f(x,y)=f(x_{0},y_{0})+\nabla f(x_{0},y_{0},z_{0})\cdot (h_{1},h_{2},h_{3})+\frac{1}{2!}(x-x_{0}\quad y-y_{0})\left(\begin{array}{cc}
\frac{\partial^{2}f}{\partial x^{2}}&\frac{\partial^{2}f}{\partial y \partial x} \\
\frac{\partial^{2}f}{\partial x \partial y }&\frac{\partial^{2}f}{\partial y^{2}} \end{array}\right)\left(\begin{array}{c}
x-x_{0} \\y-y_{0} \end{array}\right)$$

A la matriz

$$\left(\begin{array}{cc}
\frac{\partial^{2}f}{\partial x^{2}}&\frac{\partial^{2}f}{\partial y \partial x} \\
\frac{\partial^{2}f}{\partial x \partial y }&\frac{\partial^{2}f}{\partial y^{2}} \end{array}\right)$$

se le conoce como matriz Hessiana y se denota $H(x_{0},y_{0})$ por lo que el desarrollo de Taylor se puede escribir
$$f(x,y)=f(x_{0},y_{0})+\nabla f(x_{0},y_{0},z_{0})\cdot (h_{1},h_{2},h_{3})+\frac{1}{2!}(x-x_{0}\quad y-y_{0})(H(x_{0},y_{0}))\left(\begin{array}{c}
x-x_{0} \\y-y_{0} \end{array}\right)$$

$\textcolor{Red}{\textbf{Aproximación de Taylor para funciones $f:\mathbb{R}^{3}\rightarrow\mathbb{R}$}}$

Sea $f:A\subset\mathbb{R}^{3}\rightarrow\mathbb{R}$ y sea $F(t)=f(x_{0}+h_{1}t,y_{0}+h_{2}t,z_{0}+h_{3}t)$ con $t\in[0,1]$, de esta manera f recorre el segmento de $[x_{0},y_{0},z_{0}]$ a $[x_{0}+h_{1}t,y_{0}+h_{2}t,z_{0}+h_{3}t]$. Se tiene entonces que usando la regla de la cadena

$$F'(t)=\frac{\partial f}{\partial x}(x_{0}+h_{1}t,y_{0}+h_{2}t,z_{0}+h_{3}t)\cdot \frac{d(x_{0}+h_{1}t)}{dt}+\frac{\partial f}{\partial y}(x_{0}+h_{1}t,y_{0}+h_{2}t,z_{0}+h_{3}t)\cdot \frac{d(y_{0}+h_{2}t)}{dt}+$$

$$\frac{\partial f}{\partial z}(x_{0}+h_{1}t,y_{0}+h_{2}t,z_{0}+h_{3}t)\cdot \frac{d(z_{0}+h_{3}t)}{dt}=$$

$$\frac{\partial f}{\partial x}(x_{0}+h_{1}t,y_{0}+h_{2}t,z_{0}+h_{3})\cdot h_{1}+\frac{\partial f}{\partial y}(x_{0}+h_{1}t,y_{0}+h_{2}t,z_{0}+h_{3})\cdot h_{2}+\frac{\partial f}{\partial z}(x_{0}+h_{1}t,y_{0}+h_{2}t,z_{0}+h_{3})\cdot h_{3}$$

Vamos ahora a calcular $F^{´´}(t)$

$$F^{´´}(t)=\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}h_{1}+ \frac{\partial f}{\partial y}h_{2}+\frac{\partial f}{\partial z}h_{3}\right)h_{1}+\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}h_{1}+ \frac{\partial f}{\partial y}h_{2}+\frac{\partial f}{\partial z}h_{3}\right)h_{2}+\frac{\partial}{\partial z}\left(\frac{\partial f}{\partial x}h_{1}+ \frac{\partial f}{\partial y}h_{2}+\frac{\partial f}{\partial z}h_{3}\right)h_{3}=$$

$$\frac{\partial^{2}f}{\partial x^{2}}h_{1}^{2}+2\frac{\partial^{2}f}{\partial x\partial y}h_{1}h_{2}+\frac{\partial^{2}f}{\partial y^{2}}h_{2}^{2}+2\frac{\partial^{2}f}{\partial x\partial z}h_{3}h_{1}+2\frac{\partial^{2}f}{\partial y\partial z}h_{3}h_{2}+\frac{\partial^{2}f}{\partial z^{2}}h_{3}^{2}$$
Ahora bien si se aplica la fórmula de Taylor con la forma del residuo de Lagrange a la función $$F(t)=f(x_{0}+h_{1}t,y_{0}+h_{2}t)$$ y ponemos $t=0$, y $n=2$ se tiene

$$F(t)=F(0)+\frac{1}{1!}F'(0)t+\frac{1}{2!}F^{´´}(0)t^{2}+R_{2}$$

ahora bien con $t=1$, $x=x_{0}+h_{1}$, $y=y_{0}+h_{2}$, $z=z_{0}+h_{3}$

$$f(x,y)=f(x_{0},y_{0})+\textcolor{Blue}{\left(\frac{\partial f}{\partial
x}\right){p}(x-x_{0})+\left(\frac{\partial f}{\partial
y}\right){p}(y-y_{0})+\left(\frac{\partial f}{\partial
z}\right){p}(z-z_{0})}+$$

$$\textcolor{Red}{\frac{1}{2!}\left(\frac{\partial^{2}f}{\partial
x^{2}}{p}(x-x_{0})^{2}+2\frac{\partial^{2}f}{\partial x \partial
y}{p}(x-x_{0})(y-y_{0})+\frac{\partial^{2}f}{\partial
y^{2}}{p}(y-y_{0})^{2}+2\frac{\partial^{2}f}{\partial
x\partial z}{p}(z-z_{0})(x-x_{0})+2\frac{\partial^{2}f}{\partial
y\partial z}{p}(z-z_{0})(y-y_{0})\right)}$$

$$\textcolor{Red}{+\frac{\partial^{2}f}{\partial
z^{2}}{p}(z-z_{0})}+R_{2}$$
Donde la expresión en azul se puede escribir

$$\textcolor{Blue}{\left(\frac{\partial f}{\partial
x}\right){p}(x-x_{0})+\left(\frac{\partial f}{\partial
y}\right){p}(y-y_{0})+\left(\frac{\partial f}{\partial
z}\right){p}(z-z_{0})=\nabla f(x_{0},y_{0},z_{0})\cdot (h_{1},h_{2},h_{3})}$$

y la expresión en rojo
$$\textcolor{Red}{\frac{1}{2!}\left(\frac{\partial^{2}f}{\partial x^{2}}h_{1}^{2}+2\frac{\partial^{2}f}{\partial x\partial y}h_{1}h_{2}+\frac{\partial^{2}f}{\partial y^{2}}h_{2}^{2}+2\frac{\partial^{2}f}{\partial x\partial z}h_{3}h_{1}+2\frac{\partial^{2}f}{\partial y\partial z}h_{3}h_{2}+\frac{\partial^{2}f}{\partial z^{2}}h_{3}^{2}\right)}$$

se puede ver como producto de matrices

$$\frac{1}{2!}(h_{1}~h_{2}~h_{3})\left(\begin{matrix}\frac{\partial^{2}f}{\partial
x^{2}}&\frac{\partial^{2}f}{\partial y \partial x}&\frac{\partial^{2}f}{\partial z \partial x}\\ \frac{\partial^{2}f}{\partial x \partial y}&\frac{\partial^{2}f}{\partial
y^{2}}&\frac{\partial^{2}f}{\partial z \partial y}\\ \frac{\partial^{2}f}{\partial
x \partial z}&\frac{\partial^{2}f}{\partial y \partial z}&\frac{\partial^{2}f}{\partial
z^{2}}\end{matrix}\right)_{p}\left(\begin{matrix}h{1}\\h_{2}\\h_{3}\end{matrix}\right)$$

La matriz
$$\left(\begin{matrix}\frac{\partial^{2}f}{\partial
x^{2}}&\frac{\partial^{2}f}{\partial
y\partial x}&\frac{\partial^{2}f}{\partial
z\partial x}\\\frac{\partial^{2}f}{\partial
x \partial y}&\frac{\partial^{2}f}{\partial
y^{2}}&\frac{\partial^{2}f}{\partial
z \partial y}\\\frac{\partial^{2}f}{\partial
x \partial z}&\frac{\partial^{2}f}{\partial
y \partial z}&\frac{\partial^{2}f}{\partial
z^{2}}\end{matrix}\right)$$
se le conoce como matriz Hessiana y se le denota $H(x_{0},y_{0},z_{0})$, por lo que la aproximación de Taylor se puede escribir

$$f(x,y)=f(x_{0},y_{0})+\nabla f(x_{0},y_{0},z_{0})\cdot (h_{1},h_{2},h_{3})+\frac{1}{2!}(h_{1}~h_{2}~h_{3})H(x_{0},y_{0},z_{0})\left(\begin{matrix}h_{1}\\h_{2}\\h_{3}\end{matrix}\right)$$

$\textbf{Ejemplo}$ Considere la función $f(x,y)=e^{2x+3y}$
$f[(0,0)+(x,y)]=f(0,0) +\nabla f(0,0)\cdot(x,y)+\frac{1}{2}[xy]H(0,0)\left[\begin{array}{c} x\\y\end{array}\right]+r_2(x,y)$

donde $\displaystyle\lim _{(x,y)\rightarrow(0,0)} \displaystyle\frac{r(x,y)}{x^2+y^2}=0$

$\nabla f=\left(\displaystyle\frac{\partial f}{\partial x}, \displaystyle\frac{\partial f}{\partial y}\right)=(2e^{2x+3y},3e^{2x+3y})~~~~ \therefore \nabla f(0,0)=(2,3)$

$$
H(x,y)=\left[\begin{array}{cc}
\displaystyle\frac{\partial ^2f}{\partial x^2} & \displaystyle\frac{\partial ^2f}{\partial y\partial x}\\
\displaystyle\frac{\partial ^2f}{\partial x\partial y} & \displaystyle\frac{\partial ^2f}{\partial y^2}\end{array}\right]=
\left[\begin{array}{cc}
4e^{2x+3} & 6e^{2x+3y}\\
6e^{2x+3y} & 9e^{2x+3y}\end{array}\right] ~~~~ \therefore H(0,0)= \left(\begin{array}{cc} 4&6\\6&9\end{array}\right)
$$

Así

Así $f(x,y)=f(0,0)+(2,3)\cdot (x,y) +\frac{1}{2}[xy]\left(\begin{array}{cc} 4&6\\6&9 \end{array}\right)\left[\begin{array}{c} x\\y\end{array}\right]+r(x,y)$
$\therefore e^{2x+3y}=1+2x+3y+2x^2+6xy\frac{9}{2}y^2+r(x,y)$

$\textcolor{Red}{\textbf{Extremos Locales}}$

Entre las caracteristicas geometricas básicas de la gráficas de una función estan sus puntos extremos, en los cuales la función alcanza sus valores mayor y menor.

$\textbf{Definición 1.}$ Si $f:u\subset \mathbb{R}^n \rightarrow
\mathbb{R}$ es una función escalar, dado un punto $x_0 \in u$
se llama mínimo local de $f$ si existe una vecindad $v$ de $x_0$ tal que $\forall x \in v$ ,$f(x)>f(x_0)$. De manera analoga, $x_0 \in u$ es un máximo local si existe una vecindad $v$ de $x_0$ tal que $f(x)<f(x_0)$, $\forall \quad x \in v$. El punto $x_0 \in u$ es un extremo local o relativo, si es un mínimo local o máximo
local.

En la expresión del desarrollo de Taylor
$$f(x,y)=f(x_{0},y_{0})+\nabla f(x_{0},y_{0},z_{0})\cdot (h_{1},h_{2},h_{3})+\frac{1}{2!}(x-x_{0}\quad y-y_{0})(H(x_{0},y_{0}))\left(\begin{array}{c}
x-x_{0} \\y-y_{0} \end{array}\right)$$
Si consideramos los valores para los cuales

$$\nabla f(x_{0},y_{0},z_{0})=(0,0,0)$$
es decir los puntos críticos del gradiente entonces nuestra aproximación de Taylor nos queda

$$f(x,y)=f(x_{0},y_{0})+\frac{1}{2!}(x-x_{0}\quad y-y_{0})(H(x_{0},y_{0}))\left(\begin{array}{c}
x-x_{0} \\y-y_{0} \end{array}\right)$$
que se puede escribir
$$f(x,y)-f(x_{0},y_{0})=\frac{1}{2!}(x-x_{0}\quad y-y_{0})(H(x_{0},y_{0}))\left(\begin{array}{c}
x-x_{0} \\y-y_{0} \end{array}\right)$$

por lo que el signo del lado izquierdo $f(x,y)-f(x_{0},y_{0})$ dependerá del signo de la expresión
$$\frac{1}{2!}(x-x_{0}\quad y-y_{0})(H(x_{0},y_{0}))\left(\begin{array}{c}
x-x_{0} \\y-y_{0} \end{array}\right)$$

es decir dependerá del signo de la forma
$$\frac{1}{2!}(h_{1}~h_{2})\left(\begin{matrix}\frac{\partial^{2}f}{\partial
x^{2}}&\frac{\partial^{2}f}{\partial
y\partial x}\\\frac{\partial^{2}f}{\partial
x\partial y}&\frac{\partial^{2}f}{\partial
y^{2}}\end{matrix}\right)_{p}\left(\begin{matrix}h{1}\\h_{2}\\h_{3}\end{matrix}\right)$$

$\textbf{Teorema 1.}$ Sea $B=\left[\begin{array}{cc}
a & b \\
b & c \\
\end{array}
\right]$ y $H(h)=\frac{1}{2}[h_1,h_2]\left[
\begin{array}{cc}
a & b \\
b & c \\
\end{array}
\right]\left(
\begin{array}{c}
h_1 \\
h_2
\end{array}
\right)$ entonces $H(h)$ es definida positiva si y solo si $a>0$ y $ac-b^2>0$

$\small{Demostración.}$ Tenemos $$H(h)=\frac{1}{2}[h_1,h_2]\left[
\begin{array}{cc}
a h_1& bh_2 \\
b h_1& ch_2 \
\end{array}
\right]=\frac{1}{2}(ah_1^2+2bh_1h_2+ch_1^2)$$
si completamos el cuadrado
$$H(h)=\frac{1}{2}a\left(h_1+\frac{b}{a}h_2\right)^2+\frac{1}{2}\left(c-\frac{b^2}{a}\right)h_2^2$$
supongamos que $h$ es definida positiva. Haciendo
$h_2=0$ vemos que $a>0$. Haciendo $h_1=-\frac{b}{a}h_2$ $c-\frac{b^2}{a}>0$ ó $ac-b^2>0$. De manera analoga $H(h)$ es definida negativa si y solo si $a<0$ y $ac-b^2>0$. $\square$

Criterio del máximo y del mínimo para funciones de dos variables Sea $f(x,y)$ de clase
$C^3$ en un conjunto abierto $u$ de $\mathbb{R}^2$. Un punto $x_0,y_0$ es un mínimo local (Estricto) de $f$ si se cumple las siguientes tres condiciones:

$I)$ $\frac{\partial f}{\partial x}(x_0,y_0)=\frac{\partial f}{\partial y}(x_0,y_0)$
$II)$$\frac{\partial^2 f}{\partial x^2}(x_0,y_0)> 0$
$III)$ $\left(\frac{\partial^2 f}{\partial x^2}\right)\left(\frac{\partial^2 f}{\partial y^2}\right)-\left(\frac{\partial^2 f}{\partial x \partial y}\right)^2> 0$ en $(x_0,y_0)$ (Discriminante). Si en II) tenemos $<0$ en lugar de $>0$ sin cambiar III)
hay un máximo local.