(*) Sobre las derivadas de $g(t)=f(x_0+th,y_0+tk)$

Proposición:

$$g^{(n)}(t)=\sum _{i=0}^n\left(\begin{array}{c}n\\ i\end{array}\right){\displaystyle \frac{{\partial }^{n}f}{\partial x^i\partial y^{n-i}}}(x_0+th,y_0+tk)h^i{k}^{n-i}$$

Demostración (por inducción)

Base inductiva:

Para $n=1,2$ lo trabajamos en la entrada anterior. https://blog.nekomath.com/?p=101046&preview=true

Paso inductivo:

Supongamos la proposición válida para $n.$

$[$ por demostrar: que es válida para $n+1]$

Entonces

$\begin{array}{rl}{g}^{(n+1)}(t)& ={\displaystyle \frac{d}{dt}}\sum _{i=0}^n\left(\begin{array}{c}n\\ i\end{array}\right){\displaystyle \frac{{\partial }^{n}f}{\partial x^i\partial y^{n-i}}}(x_0+th,y_0+tk)h^i{k}^{n-i}\\ & =\sum _{i=0}^n\left(\begin{array}{c}n\\ i\end{array}\right)h^i{k}^{n-i}{\displaystyle \frac{d}{dt}}{\displaystyle \frac{{\partial }^{n}f}{\partial x^i\partial y^{n-i}}}(x_0+th,y_0+tk)\end{array}$

Podemos ver a la ${\displaystyle \frac{{\partial }^{n}f}{\partial x^i\partial y^{n-i}}}(x,y)=G(x,y)$

Entonces, derivando $G(x_0+th,y_0+tk)=G(\alpha (t))$ respecto a $t$ se tiene que

${\displaystyle \frac{d}{dt}}{\displaystyle \frac{{\partial }^{n}f}{\partial x^i\partial y^{n-i}}}(x_0+th,y_0+tk)={\displaystyle \frac{{\partial }^{n+1}f}{\partial x^{i+1}\partial y^{n-i}}}(x_0+th,y_0+tk)h+{\displaystyle \frac{{\partial }^{n+1}f}{\partial x^i\partial y^{n-i+1}}}(x_0+th,y_0+tk)k$

Entonces

$\begin{array}{rl}{g}^{(n+1)}(t)& =\sum _{i=0}^n\left(\begin{array}{c}n\\ i\end{array}\right)h^i{k}^{n-i}({\displaystyle \frac{{\partial }^{n+1}f}{\partial x^{i+1}\partial y^{n-i}}}(x_0+th,y_0+tk)h+{\displaystyle \frac{{\partial }^{n+1}f}{\partial x^i\partial y^{n-i+1}}}(x_0+th,y_0+tk)k)\end{array}$…(1)

$[$ por demostrar: $g^{(n+1)}(t)=\sum _{i=0}^{n+1}\left(\begin{array}{c}n+1\\ i\end{array}\right)h^i{k}^{n+1-i}{\displaystyle \frac{{\partial }^{n+1}f}{\partial x^i\partial y^{n+1-i}}}(x_0+th,y_0+tk)]$

De $(1)$ tenemos que

$g^{(n+1)}(t)=\sum _{i=0}^n\left(\begin{array}{c}n\\ i\end{array}\right)h^i{k}^{n-i}{\displaystyle \frac{{\partial }^{n+1}f}{\partial x^{i+1}\partial y^{n-i}}}+\sum _{i=0}^n\left(\begin{array}{c}n\\ i\end{array}\right)h^i{k}^{n+1-i}{\displaystyle \frac{{\partial }^{n+1}f}{\partial x^i\partial y^{n+1-i}}}$

haciendo $j=i+1$

$g^{(n+1)}(t)=\sum _{j=1}^{n+1}\left(\begin{array}{c}n\\ j-1\end{array}\right)h^i{k}^{n+1-i}{\displaystyle \frac{{\partial }^{n+1}f}{\partial x^j\partial y^{n+1-j}}}+\sum _{i=0}^n\left(\begin{array}{c}n\\ i\end{array}\right)h^i{k}^{n+1-i}{\displaystyle \frac{{\partial }^{n+1}f}{\partial x^i\partial y^{n+1-i}}}$

con $i=0$, $j=n+1$

$g^{(n+1)}(t)=\left(\begin{array}{c}n\\ 0\end{array}\right)h^0{k}^{n+1}{\displaystyle \frac{{\partial }^{n+1}f}{\partial y^{n+1}}}+\left(\begin{array}{c}n\\ n\end{array}\right)h^{n+1}{k}^0{\displaystyle \frac{{\partial }^{n+1}f}{\partial x^{n+1}}}+\sum _{i=1}^n[\left(\begin{array}{c}n\\ i-1\end{array}\right),+\left(\begin{array}{c}n\\ i\end{array}\right)]h^i{k}^{n+1-i}{\displaystyle \frac{{\partial }^{n+1}f}{\partial x^i\partial y^{n+1-i}}}$

entonces

$g^{(n+1)}(t)=\left(\begin{array}{c}n\\ 0\end{array}\right)h^0{k}^{n+1}{\displaystyle \frac{{\partial }^{n+1}f}{\partial y^{n+1}}}+\left(\begin{array}{c}n\\ n\end{array}\right)h^{n+1}{k}^0{\displaystyle \frac{{\partial }^{n+1}f}{\partial x^{n+1}}}+\sum _{i=1}^n\left(\begin{array}{c}n+1\\ i\end{array}\right)h^i{k}^{n+1-i}{\displaystyle \frac{{\partial }^{n+1}f}{\partial x^i\partial y^{n+1-i}}}$

$$

(*) Aplicamos el teorema de Taylor a la función $g(t)=f(x_0+th,y_0+tk)$ en $[0,1]$

Teorema de Taylor en una variable

$f(x)=f(a)+f^{\prime }(a)(x–a)+{\displaystyle \frac{{f^{\prime }}^{\prime }(a)}{2}}(x–a{)}^2+R_2(x)$

Fórmulas para $R_2(x)$

(1) Lagrange: $R_2(x)={\displaystyle \frac{f^{(3)}(t)}{3!}}(x–a{)}^3$ para algún $t\in (a,x)$.

(2) Integral: $R_2(x)=\underset{a}{\overset{x}{\int }}{\displaystyle \frac{f^{(3)}(t)}{2!}}(x–a{)}^{2}dt$

Si $[a,x]=[0,1]$ entonces $x–a=1$, por lo que

$g(1)=g(0)+g^{\prime }(0)+{\displaystyle \frac{{g^{\prime }}^{\prime }(0)}{2!}}+R_2(1)$

Si $(x_0,y_0)$ es un punto crítico, entonces

$f(x_0+h,y_0+k)=f(x_0,y_0)+\cancel{{\displaystyle \frac{\partial f}{\partial x}}(x_0,y_0)h}+\cancel{{\displaystyle \frac{\partial f}{\partial y}}(x_0,y_0)k}+{\displaystyle \frac{1}{2}}[{\displaystyle \frac{{\partial }^{2}f}{\partial x^2}}(x_0,y_0)h^2+2{\displaystyle \frac{{\partial }^{2}f}{\partial x\partial y}}(x_0,y_0)hk+{\displaystyle \frac{{\partial }^{2}f}{\partial y^2}}(x_0,y_0)k^2]+R_2(1)$

La expresión resaltada de color se conoce como $Formacuadr\acute{a}tica.$

Calculamos el error con cualquiera de las dos fórmulas vistas, de modo que:

(1) ${\displaystyle \frac{g^{(3)}(t)}{3!}}=\frac{1}{3!}({\displaystyle \frac{\partial f^{(3)}}{\partial x^3}}{h}^3+3{\displaystyle \frac{{\partial }^{3}f}{\partial x^2\partial y}}{h}^{2}k+3{\displaystyle \frac{{\partial }^{3}f}{\partial x\partial y^2}}h{k}^2+{\displaystyle \frac{\partial f^{(3)}}{\partial y^3}}{k}^3)$ en $t\in (0,1)$

(2) $\underset{0}{\overset{1}{\int }}{\displaystyle \frac{g^{(3)}(t)}{2}}(1–t{)}^{2}dt$

$$

(*) Un acercamiento a las formas cuadráticas.

Sea $H:{\mathbb{R}}^2\to \mathbb{R}$, donde

$H(x,y)=\left(\begin{array}{cc}x& y\end{array}\right)\left(\begin{array}{cc}a& b\\ b& c\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)$

$H(x,y)=a{x}^2+2bxy+c{y}^2$

Definición: $H$ es definida positiva ( o positivamente) si

$\{\begin{array}{l}H(x,y)>0\mathrm{\forall }(x,y)\ne (0,0)\\ \\ H(x,y)=0si(x,y)=(0,0)\end{array}$

Análogamente,

$H$ es definida negativa ( o negativamente) si

$H(x,y)<0\mathrm{\forall }(x,y)\ne (0,0)$

Además

$H$ es semi definida positiva, si $H(x,y)\ge 0\mathrm{\forall }(x,y)$

$H$ es semi definida negativa, si $H(x,y)\le 0\mathrm{\forall }(x,y)$

Las formas cuadráticas más fáciles de estudiar son las que tienen asociada una matriz diagonal $$\left(\begin{array}{cc}{\lambda }_1& 0\\ \\ 0& {\lambda }_2\end{array}\right)$$

Entonces

$$H(x,y)=\left(\begin{array}{cc}x& y\end{array}\right)\left(\begin{array}{cc}{\lambda }_1& 0\\ \\ 0& {\lambda }_2\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)$$

$$H(x,y)={\lambda }_1{x}^2+{\lambda }_2{y}^2$$

$H$ es definida positiva si $\{\begin{array}{l}{\lambda }_1>0\\ {\lambda }_2>0\end{array}$

$H$ es definida negativa si $\{\begin{array}{l}{\lambda }_1<0\\ {\lambda }_2<0\end{array}$

$H$ es semi definida positiva si $\{\begin{array}{l}{\lambda }_1\ge 0\\ {\lambda }_2\ge 0\end{array}$

$H$ es semi definida negativa si $\{\begin{array}{l}{\lambda }_1\le 0\\ {\lambda }_2\le 0\end{array}$

$H$ tiene un punto silla en el origen si ${\lambda }_1<0<{\lambda }_2$

$$

Veamos un ejemplo:

$H(x,y)=2xy$

Consideremos composiciones $H\circ T$ , con $T:{\mathbb{R}}^2\to {\mathbb{R}}^2$ lineal invertible.

$$\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)\to \left(\begin{array}{c}x\\ y\end{array}\right)$$

$T(x^{\prime },y^{\prime })=(x,y)$

$\left(\begin{array}{cc}a& b\\ b& c\end{array}\right)\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)=\left(\begin{array}{c}x\\ y\end{array}\right)$

$\{\begin{array}{l}x=a{x}^{\prime }+b{y}^{\prime }\\ y=c{x}^{\prime }+d{y}^{\prime }\end{array}$

$H(T(x^{\prime },y^{\prime }))=H(x,y)$

$\begin{array}{rl}H(T(x^{\prime },y^{\prime }))& =2(a{x}^{\prime }+b{y}^{\prime })(c{x}^{\prime }+d{y}^{\prime })\\ & =2ac(x^{\prime }{)}^2+2(ad+bc)x^{\prime }{y}^{\prime }+2bd(y^{\prime }{)}^2\end{array}$

Buscamos elegir $T$ tal que $ad+bc=0$

$H\circ T$ tiene asociada la matriz

$\left(\begin{array}{cc}2ac& ad+bc\\ ad+bc& 2bd\end{array}\right)$…(1)

Ahora bien, si $\overrightarrow{v}=\left(\begin{array}{c}x\\ y\end{array}\right)$ y $\overrightarrow{w}=\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)$

$A\overrightarrow{w}=\overrightarrow{v}$

$H(\overrightarrow{v})={\overrightarrow{v}}^{t}M\overrightarrow{v}$, donde $M=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$

Entonces

$\begin{array}{rl}H(\overrightarrow{v})& ={\overrightarrow{v}}^{t}MA\overrightarrow{w}\\ & =(A\overrightarrow{w}{)}^{t}MA\overrightarrow{w}\\ & ={\overrightarrow{w}}^t{A}^{t}MA\overrightarrow{w}\\ & =A^{t}MA\end{array}$

Comprobamos que (1) $=A^{t}MA$

$\begin{array}{rl}{A}^{t}MA& =\left(\begin{array}{cc}a& c\\ b& d\end{array}\right)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\\ \\ & =\left(\begin{array}{cc}a& c\\ b& d\end{array}\right)\left(\begin{array}{cc}c& d\\ a& b\end{array}\right)\\ \\ A^{t}MA& =\left(\begin{array}{cc}2ac& ad+bc\\ ad+bc& 2bd\end{array}\right)\end{array}$

Buscamos $T$ tal que $A^{t}MA$ sea diagonal.

Si $A=\left(\begin{array}{cc}\cos \theta & –\sin \theta \\ \\ \sin \theta & \cos \theta \end{array}\right)$ matriz de rotación de ejes.

$A^t=\left(\begin{array}{cc}\cos \theta & \sin \theta \\ \\ –\sin \theta & \cos \theta \end{array}\right)=\left(\begin{array}{cc}\cos (–\theta )& –\sin (–\theta )\\ \\ \sin (–\theta )& \cos (–\theta )\end{array}\right)=A^{-1}$

Luego $A^{t}MA=A^{-1}MA$ … (2)

Sabemos entonces que $A$ es inyectiva, invertible y su determinante $det(A)=1\ne 0$

Si $\overrightarrow{v_1}=\left(\begin{array}{c}\cos \theta \\ \\ \sin \theta \end{array}\right)$

$\overrightarrow{v_2}=\left(\begin{array}{c}–\sin \theta \\ \\ \cos \theta \end{array}\right)$

Entonces $A\overrightarrow{e_1}=\overrightarrow{v_1}$ y también $A\overrightarrow{e_2}=\overrightarrow{v_2}$, si además $\overrightarrow{v_1}$, $\overrightarrow{v_2}$ son eigenvectores ( o vectores propios) de $M$, entonces

$M\overrightarrow{v_1}={\lambda }_1\overrightarrow{v_1}$

$M\overrightarrow{v_2}={\lambda }_2\overrightarrow{v_2}$

$A^{-1}MA\overrightarrow{e_1}=A^{-1}M\overrightarrow{v_1}=A^{-1}{\lambda }_1\overrightarrow{v_1}={\lambda }_1{A}^{-1}\overrightarrow{v_1}={\lambda }_1\overrightarrow{e_1}$

Análogamente

$A^{-1}MA\overrightarrow{e_2}=A^{-1}M\overrightarrow{v_2}=A^{-1}{\lambda }_2\overrightarrow{v_2}={\lambda }_2{A}^{-1}\overrightarrow{v_2}={\lambda }_2\overrightarrow{e_2}$

Luego (2) es igual a $\left(\begin{array}{cc}{\lambda }_1& 0\\ \\ 0& {\lambda }_2\end{array}\right)$

En nuestro ejemplo:

$\overrightarrow{v_1}=\left(\begin{array}{c}{\displaystyle \frac{1}{\sqrt{2}}}\\ \\ {\displaystyle \frac{1}{\sqrt{2}}}\end{array}\right)$

$\overrightarrow{v_2}=\left(\begin{array}{c}{\displaystyle \frac{-1}{\sqrt{2}}}\\ \\ {\displaystyle \frac{1}{\sqrt{2}}}\end{array}\right)$

Además ${\lambda }_1=1$ y ${\lambda }_2=–1$

Luego $(x^{\prime }{)}^2–(y^{\prime }{)}^2=2xy=H(x,y)$