Esercizi sul polinomio di Taylor di ordine 2 per funzione di due variabili

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Esercizi sul polinomio di Taylor di ordine 2 per funzione di due variabili Esercizi sul polinomio di Taylor di ordine 2 per funzione di due variabili
Esercizi sul polinomio di Taylor di ordine 2 per funzione di due variabili

Versione italiana

Esercizi sul Polinomio di Taylor di Ordine 2 per Funzioni di Due Variabili

Concetti Principali

Definizione

Il polinomio di Taylor di ordine 2 di una funzione f(x, y)f(x,y)f(x, y) di due variabili attorno a un punto (a, b)(a,b)(a, b) è dato dalla seguente formula:

P(x, y) = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b) + \frac{1}{2}f_{xx}(a, b)(x - a)^2 + f_{xy}(a, b)(x - a)(y - b) + \frac{1}{2}f_{yy}(a, b)(y - b)^2
P(x,y)=f(a,b)+fx(a,b)(x−a)+fy(a,b)(y−b)+12fxx(a,b)(x−a)2+fxy(a,b)(x−a)(y−b)+12fyy(a,b)(y−b)2P(x, y) = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b) + \frac{1}{2}f_{xx}(a, b)(x - a)^2 + f_{xy}(a, b)(x - a)(y - b) + \frac{1}{2}f_{yy}(a, b)(y - b)^2

dove:

  • f_xfxf_x è la derivata parziale di fff rispetto a xxx
  • f_yfyf_y è la derivata parziale di fff rispetto a yyy
  • f_{xx}fxxf_{xx} è la derivata parziale seconda di fff rispetto a xxx
  • f_{yy}fyyf_{yy} è la derivata parziale seconda di fff rispetto a yyy
  • f_{xy}fxyf_{xy} è la derivata mista di fff

Derivate Parziali

Le derivate parziali sono calcolate come segue:

  • f_x(a, b) = \lim_{h \to 0} \frac{f(a + h, b) - f(a, b)}{h}fx(a,b)=lim⁡h→0f(a+h,b)−f(a,b)hf_x(a, b) = \lim_{h \to 0} \frac{f(a + h, b) - f(a, b)}{h}
  • f_y(a, b) = \lim_{k \to 0} \frac{f(a, b + k) - f(a, b)}{k}fy(a,b)=lim⁡k→0f(a,b+k)−f(a,b)kf_y(a, b) = \lim_{k \to 0} \frac{f(a, b + k) - f(a, b)}{k}
  • f_{xx}(a, b) = \frac{\partial^2 f}{\partial x^2}fxx(a,b)=∂2f∂x2f_{xx}(a, b) = \frac{\partial^2 f}{\partial x^2}
  • f_{yy}(a, b) = \frac{\partial^2 f}{\partial y^2}fyy(a,b)=∂2f∂y2f_{yy}(a, b) = \frac{\partial^2 f}{\partial y^2}
  • f_{xy}(a, b) = \frac{\partial^2 f}{\partial x \partial y}fxy(a,b)=∂2f∂x∂yf_{xy}(a, b) = \frac{\partial^2 f}{\partial x \partial y}

Esercizio

Sia f(x, y) = x^2 + y^2f(x,y)=x2+y2f(x, y) = x^2 + y^2. Calcola il polinomio di Taylor di ordine 2 attorno al punto (1, 1)(1,1)(1, 1).

Soluzione:

  1. Calcoliamo f(1, 1)f(1,1)f(1, 1):

    f(1, 1) = 1^2 + 1^2 = 2
    f(1,1)=12+12=2f(1, 1) = 1^2 + 1^2 = 2

  2. Calcoliamo le derivate parziali:

    • f_x = 2xfx=2xf_x = 2x e f_x(1, 1) = 2 \cdot 1 = 2fx(1,1)=2⋅1=2f_x(1, 1) = 2 \cdot 1 = 2
    • f_y = 2yfy=2yf_y = 2y e f_y(1, 1) = 2 \cdot 1 = 2fy(1,1)=2⋅1=2f_y(1, 1) = 2 \cdot 1 = 2
    • f_{xx} = 2fxx=2f_{xx} = 2 e f_{xx}(1, 1) = 2fxx(1,1)=2f_{xx}(1, 1) = 2
    • f_{yy} = 2fyy=2f_{yy} = 2 e f_{yy}(1, 1) = 2fyy(1,1)=2f_{yy}(1, 1) = 2
    • f_{xy} = 0fxy=0f_{xy} = 0 e f_{xy}(1, 1) = 0fxy(1,1)=0f_{xy}(1, 1) = 0

  3. Sostituiamo i valori nella formula del polinomio di Taylor:

    P(x, y) = 2 + 2(x - 1) + 2(y - 1) + \frac{1}{2}(2)(x - 1)^2 + 0 + \frac{1}{2}(2)(y - 1)^2
    P(x,y)=2+2(x−1)+2(y−1)+12(2)(x−1)2+0+12(2)(y−1)2P(x, y) = 2 + 2(x - 1) + 2(y - 1) + \frac{1}{2}(2)(x - 1)^2 + 0 + \frac{1}{2}(2)(y - 1)^2

    Semplificando:

    P(x, y) = 2 + 2(x - 1) + 2(y - 1) + (x - 1)^2 + (y - 1)^2
    P(x,y)=2+2(x−1)+2(y−1)+(x−1)2+(y−1)2P(x, y) = 2 + 2(x - 1) + 2(y - 1) + (x - 1)^2 + (y - 1)^2
    P(x, y) = 2 + 2x - 2 + 2y - 2 + (x - 1)^2 + (y - 1)^2
    P(x,y)=2+2x−2+2y−2+(x−1)2+(y−1)2P(x, y) = 2 + 2x - 2 + 2y - 2 + (x - 1)^2 + (y - 1)^2
    P(x, y) = (x - 1)^2 + (y - 1)^2 + 2x + 2y - 2
    P(x,y)=(x−1)2+(y−1)2+2x+2y−2P(x, y) = (x - 1)^2 + (y - 1)^2 + 2x + 2y - 2

English version

Exercises on Taylor Polynomial of Order 2 for Functions of Two Variables

Main Concepts

Definition

The Taylor polynomial of order 2 of a function f(x, y)f(x,y)f(x, y) of two variables around a point (a, b)(a,b)(a, b) is given by the following formula:

P(x, y) = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b) + \frac{1}{2}f_{xx}(a, b)(x - a)^2 + f_{xy}(a, b)(x - a)(y - b) + \frac{1}{2}f_{yy}(a, b)(y - b)^2
P(x,y)=f(a,b)+fx(a,b)(x−a)+fy(a,b)(y−b)+12fxx(a,b)(x−a)2+fxy(a,b)(x−a)(y−b)+12fyy(a,b)(y−b)2P(x, y) = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b) + \frac{1}{2}f_{xx}(a, b)(x - a)^2 + f_{xy}(a, b)(x - a)(y - b) + \frac{1}{2}f_{yy}(a, b)(y - b)^2

where:

  • f_xfxf_x is the partial derivative of fff with respect to xxx
  • f_yfyf_y is the partial derivative of fff with respect to a yyy
  • f_{xx}fxxf_{xx} is the second partial derivative of fff with respect to xxx
  • f_{yy}fyyf_{yy} is the second partial derivative of fff with respect to yyy
  • f_{xy}fxyf_{xy} is the mixed derivative of fff

Partial Derivatives

The partial derivatives are calculated as follows:

  • f_x(a, b) = \lim_{h \to 0} \frac{f(a + h, b) - f(a, b)}{h}fx(a,b)=lim⁡h→0f(a+h,b)−f(a,b)hf_x(a, b) = \lim_{h \to 0} \frac{f(a + h, b) - f(a, b)}{h}
  • f_y(a, b) = \lim_{k \to 0} \frac{f(a, b + k) - f(a, b)}{k}fy(a,b)=lim⁡k→0f(a,b+k)−f(a,b)kf_y(a, b) = \lim_{k \to 0} \frac{f(a, b + k) - f(a, b)}{k}
  • f_{xx}(a, b) = \frac{\partial^2 f}{\partial x^2}fxx(a,b)=∂2f∂x2f_{xx}(a, b) = \frac{\partial^2 f}{\partial x^2}
  • f_{yy}(a, b) = \frac{\partial^2 f}{\partial y^2}fyy(a,b)=∂2f∂y2f_{yy}(a, b) = \frac{\partial^2 f}{\partial y^2} - f_{xy}(a, b) = \frac{\partial^2 f}{\partial x \partial y}fxy(a,b)=∂2f∂x∂yf_{xy}(a, b) = \frac{\partial^2 f}{\partial x \partial y} ## Exercise Let f(x, y) = x^2 + y^2f(x,y)=x2+y2f(x, y) = x^2 + y^2. Compute the Taylor polynomial of order 2 around the point (1, 1)(1,1)(1, 1).

Solution:

  1. Let's calculate f(1, 1)f(1,1)f(1, 1):
f(1, 1) = 1^2 + 1^2 = 2
f(1,1)=12+12=2f(1, 1) = 1^2 + 1^2 = 2
  1. Let's calculate the partial derivatives:
  • f_x = 2xfx=2xf_x = 2x and f_x(1, 1) = 2 \cdot 1 = 2fx(1,1)=2⋅1=2f_x(1, 1) = 2 \cdot 1 = 2
  • f_y = 2yfy=2yf_y = 2y and f_y(1, 1) = 2 \cdot 1 = 2fy(1,1)=2⋅1=2f_y(1, 1) = 2 \cdot 1 = 2
  • f_{xx} = 2fxx=2f_{xx} = 2 and f_{xx}(1, 1) = 2fxx(1,1)=2f_{xx}(1, 1) = 2
  • f_{yy} = 2fyy=2f_{yy} = 2 and f_{yy}(1, 1) = 2fyy(1,1)=2f_{yy}(1, 1) = 2
  • f_{xy} = 0fxy=0f_{xy} = 0 and f_{xy}(1, 1) = 0fxy(1,1)=0f_{xy}(1, 1) = 0
  1. Let's substitute the values ​​in the Taylor polynomial formula:
P(x, y) = 2 + 2(x - 1) + 2(y - 1) + \frac{1}{2}(2)(x - 1)^2 + 0 + \frac{1}{2}(2)(y - 1)^2
P(x,y)=2+2(x−1)+2(y−1)+12(2)(x−1)2+0+12(2)(y−1)2P(x, y) = 2 + 2(x - 1) + 2(y - 1) + \frac{1}{2}(2)(x - 1)^2 + 0 + \frac{1}{2}(2)(y - 1)^2

Simplifying:

P(x, y) = 2 + 2(x - 1) + 2(y - 1) + (x - 1)^2 + (y 1)^ 2
P(x,y)=2+2(x−1)+2(y−1)+(x−1)2+(y1)2P(x, y) = 2 + 2(x - 1) + 2(y - 1) + (x - 1)^2 + (y 1)^ 2
P(x, y) = 2 + 2x - 2 + 2y - 2 + (x - 1)^2 + (y - 1)^2
P(x,y)=2+2x−2+2y−2+(x−1)2+(y−1)2P(x, y) = 2 + 2x - 2 + 2y - 2 + (x - 1)^2 + (y - 1)^2
P(x, y) = (x - 1)^2 + (y - 1)^2 + 2x + 2y - 2
P(x,y)=(x−1)2+(y−1)2+2x+2y−2P(x, y) = (x - 1)^2 + (y - 1)^2 + 2x + 2y - 2

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