Introduccion

Diferencial de orden n

$$d^{n}f=\frac{{\partial }^{n}f}{\partial x^n}d{x}^n+\left(\begin{array}{c}n\\ 1\end{array}\right)\frac{{\partial }^{n-1}f}{\partial x^{n-1}\partial y}d{x}^{n-1}dy+\left(\begin{array}{c}n\\ 2\end{array}\right)\frac{{\partial }^{n-2}f}{\partial x^{n-2}\partial y^2}d{x}^{n-2}d{y}^2+\cdots +$$ $$\left(\begin{array}{c}n\\ k\end{array}\right)\frac{{\partial }^{n-k}f}{\partial x^{n-k}\partial y^k}d{x}^{n-k}d{y}^k+\cdots +\frac{{\partial }^{n}f}{\partial y^n}d{y}^n$$
que se puede escribir
$$d^{n}f=\sum _{j=0}^n\left(\begin{array}{c}n\\ j\end{array}\right)\frac{{\partial }^{n}f}{\partial x^{n-j}\partial y^j}d{x}^{n-j}d{y}^j$$

Ejercicio. Probar usando inducción
$$d^{n}f=\sum _{j=0}^n\left(\begin{array}{c}n\\ j\end{array}\right)\frac{{\partial }^{n}f}{\partial x^{n-j}\partial y^j}d{x}^{n-j}d{y}^j$$

Solución. Para n=1 se tiene
$$df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy$$
Suponemos valido para n

$$d^{n}f=\sum _{j=0}^n\left(\begin{array}{c}n\\ j\end{array}\right)\frac{{\partial }^{n}f}{\partial x^{n-j}\partial y^j}d{x}^{n-j}d{y}^j$$
Por demostrar que es valida para n+1
$$d^{n+1}f=d(d^{n}f)=\frac{\partial }{\partial x}\left(\sum _{j=0}^n\left(\begin{array}{c}n\\ j\end{array}\right)\frac{{\partial }^{n}f}{\partial x^{n-j}\partial y^j}d{x}^{n-j}d{y}^j\right)dx+\frac{\partial }{\partial y}\left(\sum _{j=0}^n\left(\begin{array}{c}n\\ j\end{array}\right)\frac{{\partial }^{n}f}{\partial x^{n-j}\partial y^j}d{x}^{n-j}d{y}^j\right)dy=$$

$$\sum _{j=0}^n\left(\begin{array}{c}n\\ j\end{array}\right)\frac{{\partial }^{n+1}f}{\partial x^{n+1-j}\partial y^j}d{x}^{n+1-j}d{y}^j+\sum _{j=0}^n\left(\begin{array}{c}n\\ j\end{array}\right)\frac{{\partial }^{n+1}f}{\partial x^{n-j}\partial y^{j+1}}d{x}^{n-j}d{y}^{j+1}=$$
$$\sum _{j=0}^n\left(\begin{array}{c}n\\ j\end{array}\right)\frac{{\partial }^{n+1}f}{\partial x^{n+1-j}\partial y^j}d{x}^{n+1-j}d{y}^j+\sum _{j=1}^{n+1}\left(\begin{array}{c}n\\ j-1\end{array}\right)\frac{{\partial }^{n+1}f}{\partial x^{n+1-j}\partial y^j}d{x}^{n+1-j}d{y}^j=$$

$$\frac{{\partial }^{n+1}f}{\partial x^{n+1}}d{x}^{n+1}+\sum _{j=1}^n\left(\begin{array}{c}n\\ j\end{array}\right)\frac{{\partial }^{n+1}f}{\partial x^{n+1-j}\partial y^j}d{x}^{n+1-j}d{y}^j+\sum _{j=1}^n\left(\begin{array}{c}n\\ j-1\end{array}\right)\frac{{\partial }^{n+1}f}{\partial x^{n+1-j}\partial y^j}d{x}^{n+1-j}d{y}^j+\frac{{\partial }^{n+1}f}{\partial y^{n+1}}d{y}^{n+1}=$$

$$\frac{{\partial }^{n+1}f}{\partial x^{n+1}}d{x}^{n+1}+\sum _{j=1}^n(\left(\begin{array}{c}n\\ j\end{array}\right)+\left(\begin{array}{c}n\\ j-1\end{array}\right))\frac{{\partial }^{n+1}f}{\partial x^{n+1-j}\partial y^j}d{x}^{n+1-j}d{y}^j+\frac{{\partial }^{n+1}f}{\partial y^{n+1}}d{y}^{n+1}=$$

$$\frac{{\partial }^{n+1}f}{\partial x^{n+1}}d{x}^{n+1}+\sum _{j=1}^n\left(\begin{array}{c}n+1\\ j\end{array}\right)\frac{{\partial }^{n+1}f}{\partial x^{n+1-j}\partial y^j}d{x}^{n+1-j}d{y}^j+\frac{{\partial }^{n+1}f}{\partial y^{n+1}}d{y}^{n+1}=\sum _{j=0}^{n+1}\left(\begin{array}{c}n\\ j\end{array}\right)\frac{{\partial }^{n}f}{\partial x^{n-j}\partial y^j}d{x}^{n-j}d{y}^j$$

La última fórmula puede expresarse simbólicamente por la ecuación
$$d^{n}f={(\frac{\partial }{\partial x}dx+\frac{\partial }{\partial y}dy)}^{n}f$$

donde primero debe desarrollarse le expresión de la derecha formalmente por medio del teorema del binomio y, a continuación deben sustituirse los términos
$$\frac{{\partial }^{n}f}{\partial x^n}d{x}^n,\frac{{\partial }^{n}f}{\partial x^{n-1}\partial y}d{x}^{n-1}dy,\cdots ,\frac{{\partial }^{n}f}{\partial y^n}d{y}^n$$
por los términos
$${\left(\frac{\partial }{\partial x}dx\right)}^{n}f,{\left(\frac{\partial }{\partial x}dx\right)}^{n-1}\left(\frac{\partial }{\partial y}dy\right)f,\cdots ,{\left(\frac{\partial }{\partial y}dy\right)}^{n}f$$

Teorema de Taylor para funciones $f:A\subset {\mathbb{R}}^2\to \mathbb{R}$}

Recordando el Teorema de Taylor para funciones $f:\mathbb{R}\to \mathbb{R}$

Teorema. Si $f(x)$ tiene n-ésima derivada continua en una vecindad de $x_0$, entonces en esa vecindad
$$f(x)=f(x_0)+\frac{1}{1!}{f}^{\prime }(x_0)(x-x_0)+\frac{1}{2!}f»(x_0)(x-x_0{)}^2+\frac{1}{3!}f{»}^{\prime }(x_0)(x-x_0{)}^3+\dots +\frac{1}{n!}{f}^n(x_0)(x-x_0{)}^n+R_n$$
donde
$$R_n=\frac{f^{n+1}(ϵ)}{(n+1)!}(x-x_0{)}^{n+1},dondeϵ\in (x_0,x)$$

Sea $f:A\subset {\mathbb{R}}^2\to \mathbb{R}$ y sea $F(t)=f(x_0+h_{1}t,y_0+h_{2}t)$ con $t\in [0,1]$, de esta manera f recorre el segmento de $[x_0,y_0]$ a $[x_0+h_{1}t,y_0+h_{2}t]$. Se tiene entonces que usando la regla de la cadena
$$F^{\prime }(t)=\frac{\partial f}{\partial x}(x_0+h_{1}t,y_0+h_{2}t)\cdot \frac{d(x_0+h_{1}t)}{dt}+\frac{\partial f}{\partial y}(x_0+h_{1}t,y_0+h_{2}t)\cdot \frac{d(y_0+h_{2}t)}{dt}=$$

$$\frac{\partial f}{\partial x}(x_0+h_{1}t,y_0+h_{2}t)\cdot h_1+\frac{\partial f}{\partial y}(x_0+h_{1}t,y_0+h_{2}t)\cdot h_2$$
Vamos ahora a calcular $F^{´´}(t)$

$$F^{´´}(t)=\frac{\partial }{\partial x}(\frac{\partial f}{\partial x}{h}_1+\frac{\partial f}{\partial y}{h}_2)h_1+\frac{\partial }{\partial y}(\frac{\partial f}{\partial x}{h}_1+\frac{\partial f}{\partial y}{h}_2)h_2=$$
$$\frac{{\partial }^{2}f}{\partial x^2}{h}_1^2+2\frac{{\partial }^{2}f}{\partial x\partial y}{h}_1{h}_2+\frac{{\partial }^{2}f}{\partial y^2}{h}_2^2$$

simbólicamente se puede escribir
$$F^{»}(t)={(\frac{\partial }{\partial x}\cdot h_1+\frac{\partial }{\partial y}\cdot h_2)}^{2}f$$
y en general

$$F^n(t)=\frac{{\partial }^{n}f}{\partial x^n}{h}_1^n+\left(\begin{array}{c}n\\ 1\end{array}\right)\frac{{\partial }^{n-1}f}{\partial x^{n-1}\partial y}{h}_1^{n-1}{h}_2+\left(\begin{array}{c}n\\ 2\end{array}\right)\frac{{\partial }^{n-2}f}{\partial x^{n-2}\partial y^2}{h}_1^{n-2}{h}_2^2+\cdots +\left(\begin{array}{c}n\\ k\end{array}\right)\frac{{\partial }^{n-k}f}{\partial x^{n-k}\partial y^k}{h}_1^{n-k}{h}_2^k+\cdots +\frac{{\partial }^{n}f}{\partial y^n}{h}_2^n$$

que simbólicamente se puede escribir
$$F^n=\sum _{j=0}^n\left(\begin{array}{c}n\\ j\end{array}\right)\frac{{\partial }^{n}f}{\partial x^{n-j}\partial y^j}{h}_1^{n-j}{h}_2^j={(\frac{\partial }{\partial x}\cdot h_1+\frac{\partial }{\partial y}\cdot h_2)}^{n}f$$

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 $t=0$, se tiene
$$F(t)=F(0)+\frac{1}{1!}{F}^{\prime }(0)t+\frac{1}{2!}{F}^{»}(0)t^2+\frac{1}{3!}F{»}^{\prime }(0)t^3+\dots ++\frac{1}{n!}{F}^{{}^n}(0)t^n+R_n$$
ahora bien con $t=1$
$$f(x_0+h_1,y_0+h_2)=f(x_0,y_0)+\frac{1}{1!}(\frac{\partial f}{\partial x}(x_0,y_0)\cdot h_1+\frac{\partial f}{\partial y}(x_0,y_0)\cdot h_2)+\frac{1}{2!}(\frac{{\partial }^{2}f}{\partial x^2}(x_0,y_0)h_1^2+2\frac{{\partial }^{2}f}{\partial x\partial y}(x_0,y_0)h_1{h}_2+\frac{{\partial }^{2}f}{\partial y^2}(x_0,y_0)h_2^2)$$
$$+\cdots +\frac{1}{n!}(\sum _{j=0}^{n+1}\left(\begin{array}{c}n\\ j\end{array}\right)\frac{{\partial }^{n}f}{\partial x^{n-j}\partial y^j}(x_0,y_0)h_1^{n-j}{h}_2^j)$$

$x=x_0+h_1$, $y_0+h_2=y$ por lo que $h_1=x-x_0$ y $h_2=y-y_0$ entonces

$$f(x,y)=f(x_0,y_0)+\frac{1}{1!}(\frac{\partial f}{\partial x}(x_0,y_0)\cdot (x-x_0)+\frac{\partial f}{\partial y}(x_0,y_0)\cdot (y-y_0))+$$

$$\frac{1}{2!}(\frac{{\partial }^{2}f}{\partial x^2}(x_0,y_0)(x-x_0{)}^2+2\frac{{\partial }^{2}f}{\partial x\partial y}(x_0,y_0)(x-x_0)(y-y_0)+\frac{{\partial }^{2}f}{\partial y^2}(x_0,y_0)(y-y_0{)}^2)+$$

$$\cdots +\frac{1}{n!}(\sum _{j=0}^{n+1}\left(\begin{array}{c}n\\ j\end{array}\right)\frac{{\partial }^{n}f}{\partial x^{n-j}\partial y^j}(x_0,y_0)(x-x_0{)}^{n-j}(y-y_0{)}^j)+R_n$$

donde
$$R_n=\frac{1}{n+1!}((x-x_0{)}^{n+1}\frac{{\partial }^{n+1}f}{\partial x^{n+1}}(\xi ,\eta )+\cdots +(y-y_0{)}^{n+1}\frac{{\partial }^{n+1}f}{\partial y^{n+1}}(\xi ,\eta ))$$ donde $\xi \in (x_0,x_0+h_1)$ y $\eta \in (y_0,y_0+h_2)$\En general el residuo $R_n$ se anula en un orden mayor que el término $d^{n}f$

Ejemplo. Desarrollar la fórmula de Taylor en $(x_0,y_0)=(0,0)$ con $n=3$ para la función $$f(x,y)=e^y\cos x$$

Solución. En este caso tenemos que
$$f(0,0)=e^0\cos (0)=1$$
Para la diferencial de orden 1
$$\frac{\partial f}{\partial x}(0,0)\Rightarrow \frac{\partial (e^y\cos (x))}{\partial x}(0,0)\Rightarrow -e^{y}sen\left(x\right)|(0,0)=0$$ $$\frac{\partial f}{\partial y}(0,0)\Rightarrow \frac{\partial (e^y\cos x)}{\partial y}(0,0)\Rightarrow -e^y\cos (x)|(0,0)=1$$
por lo tanto
$$\frac{1}{1!}(\frac{\partial f}{\partial x}(x_0,y_0)\cdot (x-x_0)+\frac{\partial f}{\partial y}(x_0,y_0)\cdot (y-y_0))=\frac{1}{1!}((0)(x)+(1)(y))=y$$
Para la diferencial de orden 2
$$\frac{{\partial }^{2}f}{\partial x^2}(x_0,y_0)\Rightarrow \frac{{\partial }^2(e^{y}cosx)}{\partial x^2}(0,0)\Rightarrow -e^y\cos x|(0,0)=-1$$ $$\frac{{\partial }^{2}f}{\partial y^2}(x0,y_0)\Rightarrow \frac{{\partial }^2(e^y\cos x)}{\partial y^2}(0,0)\Rightarrow e^y\cos x|(0,0)=1$$ $$\frac{{\partial }^{2}f}{\partial x\partial y}(x0,y_0)\Rightarrow \frac{{\partial }^2(e^y\cos x)}{\partial x\partial y}(0,0)\Rightarrow -e^{y}senx|(0,0)=0$$ Por lo tanto $$\frac{1}{2!}(\frac{{\partial }^{2}f}{\partial x^2}(x0,y_0)h_1^2+2\frac{{\partial }^{2}f}{\partial x\partial y}(x_0,y_0)h_1{h}_2+\frac{{\partial }^{2}f}{\partial y^2}(x_0,y_0)h_2^2)=\frac{1}{2!}((-1)x^2+2(0)xy+(1)y^2)$$
Para la diferencial de orden 3

$$\frac{{\partial }^{3}f}{\partial x^3}(x_0,y_0)\Rightarrow e^{y}senx{|}_{(0,0)}=0$$

$$\frac{{\partial }^{3}f}{\partial y^3}(x_0,y_0)\Rightarrow \frac{{\partial }^2(e^y\cos x)}{\partial y^3}(0,0)\Rightarrow e^y\cos x{|}_{(0,0)}=1$$

$$\frac{{\partial }^{3}f}{\partial x^2\partial y}(x_0,y_0)\Rightarrow \frac{{\partial }^2(e^y\cos x)}{\partial x^2\partial y}(0,0)\Rightarrow -e^y\cos x{|}_{(0,0)}=-1$$

$$\frac{{\partial }^{3}f}{\partial y^3}(x_0,y_0)\Rightarrow \frac{{\partial }^2(e^y\cos x)}{\partial y^3}(0,0)\Rightarrow e^y\cos x{|}_{(0,0)}=1$$

$$\frac{{\partial }^{3}f}{\partial x\partial y^2}(x_0,y_0)\Rightarrow \frac{{\partial }^2(e^y\cos x)}{\partial x\partial y^2}(0,0)\Rightarrow -e^{y}senx{|}_{(0,0)}=0$$

Por lo tanto
$$\frac{1}{3!}(\frac{{\partial }^{3}f}{\partial x^3}{h}_1^3+3\frac{{\partial }^{3}f}{\partial x^2\partial y}{h}_{}{1}^2{h}_2+3\frac{{\partial }^{3}f}{\partial x\partial y^2}{h}_1{h}_2^2+\frac{{\partial }^{3}f}{\partial y^3}{h}_{}{2}^3)=$$

$$\frac{1}{3!}((0)(x^3)+3(-1)x^{2}y+3(0)x{y}^2+(1)y^3)$$
Finalmente para el residuo se tiene

$$\frac{{\partial }^{4}f}{\partial x^4}(x_0,y_0)\Rightarrow \frac{{\partial }^4(e^y\cos (x))}{\partial y^3}(0,0)\Rightarrow e^y\cos x{|}_{(\xi ,\eta )}=e^{\eta }\cos \xi$$

$$\frac{{\partial }^{4}f}{\partial x^2\partial y^2}(x_0,y_0)\Rightarrow \frac{{\partial }^4(e^y\cos x)}{\partial x^2\partial y^2}(0,0)\Rightarrow -e^y\cos x{|}_{(\xi ,\eta )}=-e^{\eta }\cos \xi$$

$$\frac{{\partial }^{4}f}{\partial x\partial y^3}(x_0,y_0)\Rightarrow \frac{{\partial }^4(e^y\cos x)}{\partial x\partial y^3}(0,0)\Rightarrow -e^{y}senx{|}_{(\xi ,\eta )}=-e^{\eta }sen\xi$$

$$\frac{{\partial }^{4}f}{\partial y^4}(x_0,y_0)\Rightarrow \frac{{\partial }^4(e^y\cos x)}{\partial y^4}(0,0)\Rightarrow e^y\cos x{|}_{(\xi ,\eta )}=e^{\eta }\cos \xi$$

$$R_3=\frac{1}{4!}(\frac{{\partial }^{4}f}{\partial x^4}{h}_1^4+4\frac{{\partial }^{4}f}{\partial x^3\partial y}{h}_1^3{h}_2+6\frac{{\partial }^{4}f}{\partial x^2\partial y^2}{h}_1^2{h}_2^2+4\frac{{\partial }^{4}f}{\partial x\partial y^3}{h}_1{h}_2^3+\frac{{\partial }^{4}f}{\partial h_2^4}d{y}^4)$$

$$=\frac{1}{4!}(x^4{e}^{\eta }\cos \xi +4x^{3}y{e}^{\eta }senxi-6x^2{y}^2{e}^{\eta }\cos \xi -4x{y}^3{e}^{\eta }sen\xi +y^4{e}^{\eta }\cos \xi )$$

Por lo que nuestro desarrollo de Taylor nos queda
$$e^y\cos x=1+y+\frac{1}{2}(x^2-y^2)+\frac{1}{6}(x^3-3x{y}^2)+R_3$$
donde
$$R_3=\frac{1}{4!}(x^4{e}^{\eta }\cos \xi +4x^{3}y{e}^{\eta }senxi-6x^2{y}^2{e}^{\eta }\cos \xi -4x{y}^3{e}^{\eta }sen\xi +y^4{e}^{\eta }\cos \xi )$$
$\mathbf{\text{Ejercicio}}$ Use la fórmula de Taylor en
$$f(x,y)=\cos (x+y)$$
en el punto $(x_0,y_0)=(0,0)$ con $n=2$ para comprobar que
$$\underset{(x,y)\to (0,0)}{lim}\frac{1-\cos (x+y)}{(x^2+y^2{)}^2}=\frac{1}{2}$$

En este caso para
$$f(x,y)=\cos (x+y)$$
se tiene
$$f(0,0)=\cos (0+0)=1$$
Para la diferencial de orden 1
$$\frac{\partial f}{\partial x}(0,0)\Rightarrow \frac{\partial (\cos x+y)}{\partial x}(0,0)\Rightarrow -sen(x+y)|(0,0)=0$$ $$\frac{\partial f}{\partial y}(0,0)\Rightarrow \frac{\partial (\cos x+y)}{\partial y}(0,0)\Rightarrow -sen(x+y)|(0,0)=0$$
por lo tanto

$$\frac{1}{1!}(\frac{\partial f}{\partial x}(x_0,y_0)\cdot (x-x_0)+\frac{\partial f}{\partial y}(x_0,y_0)\cdot (y-y_0))=\frac{1}{1!}((0)(x)+(0)(y))=0$$

Para la diferencial de orden 2
$$\frac{{\partial }^{2}f}{\partial x^2}(x_0,y_0)\Rightarrow \frac{{\partial }^2(\cos x+y)}{\partial x^2}(0,0)\Rightarrow -\cos x+y|(0,0)=-1$$ $$\frac{{\partial }^{2}f}{\partial y^2}(x0,y_0)\Rightarrow \frac{{\partial }^2(\cos x+y)}{\partial y^2}(0,0)\Rightarrow -\cos x+y|(0,0)=-1$$ $$\frac{{\partial }^{2}f}{\partial x\partial y}(x0,y_0)\Rightarrow \frac{{\partial }^2(\cos x+y)}{\partial x\partial y}(0,0)\Rightarrow -\cos x+y{|}_{(0,0)}=-1$$
Por lo tanto

$$\frac{1}{2!}(\frac{{\partial }^{2}f}{\partial x^2}(x_0,y_0)h_1^2+2\frac{{\partial }^{2}f}{\partial x\partial y}(x_0,y_0)h_1{h}_2+\frac{{\partial }^{2}f}{\partial y^2}(x_0,y_0)h_2^2)=\frac{1}{2!}((-1)x^2-2xy+(-1)y^2)$$
Por lo que nuestro desarrollo de Taylor nos queda
$$\cos (x+y)=1-\frac{x^2}{2}-xy-\frac{y^2}{2}$$
De manera que

$$\underset{(x,y)\to (0,0)}{lim}\frac{1-\cos (x+y)}{(x^2+y^2{)}^2}=\underset{(x,y)\to (0,0)}{lim}\frac{1-(1-\frac{x^2}{2}-xy-\frac{y^2}{2})}{(x^2+y^2{)}^2}$$
$$=\underset{(x,y)\to (0,0)}{lim}\frac{1}{2}\frac{(x^2+y^2{)}^2}{(x^2+y^2{)}^2}=\frac{1}{2}$$

Mas adelante

Tarea Moral

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