Teorema

Sea $f:\mathcal{U}\subseteq {\mathbb{R}}^2\to \mathbb{R}$ tal que existen las segundas derivadas parciales.

Si son continuas, entonces el polinomio de Taylor de 2° grado es el único $p(x,y)$ tal que

$$\underset{(h,k)\to (0,0)}{lim}{\displaystyle \frac{f(x_0+h,y_0+k)–p(x_0+h,y_0+k)}{\Vert (h,k){\Vert }^2}}=0$$

Observación: si NO son continuas la demostración no sería válida.

Demostración:

$p(x_0+h,y_0+k)=A{h}^2+Bhk+C{k}^2+Dh+Ek+F$ para algunas $A,B,C,D,E,F\in \mathbb{R}$ constantes.

$F=f(h,k)$ ya que $\underset{(h,k)\to (0,0)}{lim}f(x_0+h,y_0+k)–p(x_0+h,y_0+k)=f(x_0,y_0)–F=0$

Tenemos que $D={\displaystyle \frac{\partial f}{\partial x}}(x_0,y_0)$, y $E={\displaystyle \frac{\partial f}{\partial y}}(x_0,y_0)$ porque

$\underset{(h,k)\to (0,0)}{lim}{\displaystyle \frac{|f(x_0+h,y_0+k)–p(x_0+h,y_0+k)|}{\Vert (h,k)\Vert }}=0$

En particular, si tomamos el límite con puntos de la forma $(h,0)$ y $(0,k)$ entonces

$\underset{h\to 0}{lim}{\displaystyle \frac{|f(x_0+h,y_0)–p(x_0+h,y_0)|}{|(h,k)|}}=0$

$\underset{h\to 0}{lim}{\displaystyle \frac{|f(x_0+h,y_0)–f(x_0,y_0)–Dh–A{h}^2|}{|h|}}=0$

$⟺$

$\underset{h\to 0}{lim}{\displaystyle \frac{f(x_0+h,y_0)–f(x_0,y_0)}{h}}–D–{\cancel{A\underset{h\to 0}{lim}|h|}}^0=0$

$⟺$

$\underset{h\to 0}{lim}{\displaystyle \frac{f(x_0+h,y_0)–f(x_0,y_0)}{h}}=D$

$⟺$

$D={\displaystyle \frac{\partial f}{\partial x}}(x_0,y_0)$

Análogamente $E={\displaystyle \frac{\partial f}{\partial y}}(x_0,y_0)$

Ahora bien,

$$\underset{(h,k)\to (0,0)}{lim}{\displaystyle \frac{f(x_0+h,y_0+k)–p(x_0+h,y_0+k)}{\Vert (h,k){\Vert }^2}}=\underset{(h,k)\to (0,0)}{lim}{\displaystyle \frac{|f(x_0+h,y_0)–f(x_0,y_0)–{\displaystyle \frac{\partial f}{\partial x}}(x_0,y_0)h–{\displaystyle \frac{\partial f}{\partial y}}(x_0,y_0)k–A{h}^2–2Bhk–C{k}^2|}{\Vert (h,k){\Vert }^2}}$$

En particular, si tomamos puntos de la forma $(h,0)$

$\underset{h\to 0}{lim}{\displaystyle \frac{|f(x_0+h,y_0)–f(x_0,y_0)–{\displaystyle \frac{\partial f}{\partial x}}(x_0,y_0)h–A{h}^2|}{|h{|}^2}}=0$

$⟺$

$\underset{h\to 0}{lim}{\displaystyle \frac{f(x_0+h,y_0)–f(x_0,y_0)–{\displaystyle \frac{\partial f}{\partial x}}(x_0,y_0)h–A{h}^2}{h^2}}=0$

$⟺$

$\underset{h\to 0}{lim}{\displaystyle \frac{f(x_0+h,y_0)–f(x_0,y_0)–{\displaystyle \frac{\partial f}{\partial x}}(x_0,y_0)h}{h^2}}=A$

Aplicando la regla de L’Hôpital

$\underset{h\to 0}{lim}{\displaystyle \frac{f(x_0+h,y_0)–f(x_0,y_0)–{\displaystyle \frac{\partial f}{\partial x}}(x_0,y_0)h}{h^2}}=\underset{h\to 0}{lim}{\displaystyle \frac{{\displaystyle \frac{\partial f}{\partial x}}(x_0+h,y_0)–{\displaystyle \frac{\partial f}{\partial y}}(x_0,y_0)}{2h}}=\underset{h\to 0}{lim}\frac{G(h)}{2h}$ . . . (*)

Aplicaremos L’Hôpital por segunda vez a la expresión anterior (*), entonces tenemos

$\underset{h\to 0}{lim}{\displaystyle \frac{G^{\prime }(h)}{2}}$

Necesitamos que $\underset{h\to 0}{lim}G(h)=0$ , es decir ,

$\underset{h\to 0}{lim}{\displaystyle \frac{\partial f}{\partial x}}(x_0+h,y_0)–{\displaystyle \frac{\partial f}{\partial y}}(x_0,y_0)=0$

Esto lo garantizamos si ${\displaystyle \frac{\partial f}{\partial x}}(x,y)$ es continua en $(x_0,y_0)$.

Luego

$G^{\prime }(h)={\displaystyle \frac{{\partial }^{2}f}{\partial x^2}}(x_0+h,y_0)$ por continuidad, entonces cuando $h\to 0,G^{\prime }(0)={\displaystyle \frac{{\partial }^{2}f}{\partial x^2}}(x_0,y_0)$

Por lo tanto, $A={\displaystyle \frac{1}{2}}{\displaystyle \frac{{\partial }^{2}f}{\partial x^2}}(x_0,y_0)$

Análogamente usando trayectorias $x_0,y_0+k)$, cuando $k\to 0$ tenemos que $C={\displaystyle \frac{1}{2}}{\displaystyle \frac{{\partial }^{2}f}{\partial y^2}}(x_0,y_0)$

Reuniendo la información anterior , tenemos que

$\underset{(h,k)\to (0,0)}{lim}{\displaystyle \frac{|f(x_0+h,y_0+k)–f(x_0,y_0)–{\displaystyle \frac{\partial f}{\partial x}}(x_0,y_0)h–{\displaystyle \frac{\partial f}{\partial y}}(x_0,y_0)k–{\displaystyle \frac{1}{2}}{\displaystyle \frac{{\partial }^{2}f}{\partial x^2}}(x_0,y_0)h^2–2Bhk–{\displaystyle \frac{1}{2}}{\displaystyle \frac{{\partial }^{2}f}{\partial y^2}}(x_0,y_0)k^2|}{\Vert (h,k){\Vert }^2}}=0$

En particular cuando tomamos $h=k$

$\underset{(h,h)\to (0,0)}{lim}{\displaystyle \frac{|f(x_0+h,y_0+h)–f(x_0,y_0)–{\displaystyle \frac{\partial f}{\partial x}}(x_0,y_0)h–{\displaystyle \frac{\partial f}{\partial y}}(x_0,y_0)h–{\displaystyle \frac{1}{2}}{\displaystyle \frac{{\partial }^{2}f}{\partial x^2}}(x_0,y_0)h^2–2B{h}^2–{\displaystyle \frac{1}{2}}{\displaystyle \frac{{\partial }^{2}f}{\partial y^2}}(x_0,y_0)h^2|}{2h^2}}=0$

$⟺$

$\underset{h\to 0}{lim}{\displaystyle \frac{f(x_0+h,y_0+h)–f(x_0,y_0)–h({\displaystyle \frac{\partial f}{\partial x}}(x_0,y_0)–{\displaystyle \frac{\partial f}{\partial y}}(x_0,y_0))–{\displaystyle \frac{1}{2}}{h}^2({\displaystyle \frac{{\partial }^{2}f}{\partial x^2}}(x_0,y_0)–{\displaystyle \frac{{\partial }^{2}f}{\partial y^2}}(x_0,y_0))}{2h^2}}=2B$

$⟺$ Aplicando L’Hôpital

$\underset{h\to 0}{lim}{\displaystyle \frac{{\displaystyle \frac{\partial f}{\partial x}}(x_0+h,y_0+h).1–{\displaystyle \frac{\partial f}{\partial y}}(x_0,y_0).1–{\displaystyle \frac{\partial f}{\partial x}}(x_0,y_0)–{\displaystyle \frac{\partial f}{\partial y}}(x_0,y_0)h–{\displaystyle \frac{{\partial }^{2}f}{\partial y^2}}(x_0,y_0)h}{4h}}=B$

Aplicamos L’Hôpital nuevamente, entonces tenemos que:

$\underset{h\to 0}{lim}{\displaystyle \frac{{\displaystyle \frac{{\partial }^{2}f}{\partial x^2}}(x_0+h,y_0+h)+{\displaystyle \frac{{\partial }^{2}f}{\partial y\partial x}}(x_0+h,y_0+h)+{\displaystyle \frac{{\partial }^{2}f}{\partial x\partial y}}(x_0+h,y_0+h)+{\displaystyle \frac{{\partial }^{2}f}{\partial y^2}}(x_0+h,y_0+h)–{\displaystyle \frac{{\partial }^{2}f}{\partial x^2}}(x_0,y_0)–{\displaystyle \frac{{\partial }^{2}f}{\partial y^2}}(x_0,y_0)h}{4}}=B$

Si las segundas derivadas parciales son continuas, el último límite es igual a

${\displaystyle \frac{{\displaystyle \frac{{\partial }^{2}f}{\partial y\partial x}}(x_0,y_0)}{4}}=B$

y por tanto

$$B={\displaystyle \frac{1}{2}}{\displaystyle \frac{{\partial }^{2}f}{\partial x\partial y}}(x_0,y_0)$$

Decir que $p(x,y)$ es la mejor aproximación de 2° grado de $f(x,y)$ cerca del punto $(x_0,y_0)$, es decir que

$$\underset{(h,k)\to (0,0)}{lim}{\displaystyle \frac{|f(x_0+h,y_0+k)–p(x_0+h,y_0+k)|}{\Vert (h,k){\Vert }^2}}=0{}_{\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◼"\; src="https://s.w.org/images/core/emoji/14.0.0/svg/25fc.svg">}}$$

$$

Regresando al ejemplo de derivadas parciales NO continuas. (puedes hacer click en el siguiente enlace para acceder a esa entrada)

https://blog.nekomath.com/?p=101015&preview=true

Dada la función

$f(x,y)=\{\begin{array}{ll}{\displaystyle \frac{xy(x^2–y^2)}{x^2+y^2}}& \text{ si }(x,y)\ne (0,0)\\ \\ 0& \text{ si }(x,y)=(0,0)\end{array}$

https://www.geogebra.org/classic/cgm64kqa

¿Cuál será el polinomio de Taylor de 2° grado de $f$ alrededor del origen?

$f(0,0)=0$

${\displaystyle \frac{\partial f}{\partial x}}(0,0)=\underset{h\to 0}{lim}{\displaystyle \frac{f(h,0)–f(0,0)}{h}}=\underset{h\to 0}{lim}{\displaystyle \frac{0}{h}}=0$

Análogamente

${\displaystyle \frac{\partial f}{\partial y}}(0,0)=0$

${\displaystyle \frac{{\partial }^{2}f}{\partial x^2}}(0,0)=\underset{h\to 0}{lim}{\displaystyle \frac{{\displaystyle \frac{\partial f}{\partial x}}(h,0)–{\displaystyle \frac{\partial f}{\partial x}}(0,0)}{h}}={\displaystyle \frac{0}{h}}=0$

Análogamente

${\displaystyle \frac{{\partial }^{2}f}{\partial y^2}}(0,0)=0$

${\displaystyle \frac{{\partial }^{2}f}{\partial x\partial y}}(0,0)=1$

${\displaystyle \frac{{\partial }^{2}f}{\partial y\partial x}}(0,0)=-1$

Podemos examinar que pasa si tomamos un polinomio de la forma

$$p(x,y)=A{x}^2+Bxy+C{y}^2+Dx+Ey+F$$

$F=0$

$D=E=\mathrm{\nabla }f(0,0)=\overrightarrow{0}$ por lo tanto $D=0=E$

Si $\underset{(h,k)\to (0,0)}{lim}{\displaystyle \frac{|f(h,k)–p(h,k)|}{\Vert (h,k){\Vert }^2}}=0$, entonces

Para $k=0$

$\underset{h\to 0}{lim}{\displaystyle \frac{|f(h,0)–p(h,0)|}{h^2}}=\underset{h\to 0}{lim}{\displaystyle \frac{0–A{h}^2}{h}}=A=0$

Análogamente, para $h=0$ tenemos que $C=0$

luego

$\underset{(h,k)\to (0,0)}{lim}{\displaystyle \frac{|f(h,k)–2Bhk|}{\Vert (h,k){\Vert }^2}}=0$

En particular, para cuando $\underset{h\to 0}{lim}{\displaystyle \frac{f(h,h)–2B{h}^2}{2h^2}}=0$

$f(h,h)=0$

$\underset{h\to 0}{lim}{\displaystyle \frac{–\cancel{2}B\cancel{h^2}}{\cancel{2}\cancel{h^2}}}=\underset{h\to 0}{lim}–2B=0$

Por lo tanto $B=0$

$$

Sea $f(x,y)=(y–3x^2)(y–x^2)$

¿Cómo son las curvas de nivel de $f$?

¿Cómo es la gráfica de $f$?

«Curva» de nivel 0

$\{(x,y)\in {\mathbb{R}}^2|f(x,y)=0\}=f^{-1}(0)=\{(x,y)\in {\mathbb{R}}^2|(y–3x^2)(y–x^2)=0\}$

Entonces

$y–3x^2=0⟺y=3x^2$ o

$y–x^2=0⟺y=x^2$

Si $(x,y)$ cumple que $x^2<y<3x^2$ entonces

$0<y–x^2$ pero $y–3x^2<0$ por lo tanto $f(x,y)<0$

Si $(x,y)$ cumple que $y>3x^2$ entonces $f(x,y)>0$ ya que $y>x^2$