Question

Approximation \(\quad\) Consider the following approximations for a function \(f(x, y)\) centered at (0,0) Linear approximation: \(P_{1}(x, y)=f(0,0)+f_{x}(0,0) x+f_{y}(0,0) y\) Quadratic approximation: $$ \begin{aligned} P_{2}(x, y)=& f(0,0)+f_{x}(0,0) x+f_{y}(0,0) y+\\\ & \frac{1}{2} f_{x x}(0,0) x^{2}+f_{x y}(0,0) x y+\frac{1}{2} f_{y y}(0,0) y^{2} \end{aligned} $$ [Note that the linear approximation is the tangent plane to the surface at \((0,0, f(0,0)) .]\) (a) Find the linear approximation of \(f(x, y)=e^{(x-y)}\) centered at (0,0) (b) Find the quadratic approximation of \(f(x, y)=e^{(x-y)}\) centered at (0,0) . (c) If \(x=0\) in the quadratic approximation, you obtain the second-degree Taylor polynomial for what function? Answer the same question for \(y=0\). (d) Complete the table. $$ \begin{array}{|l|c|l|l|l|} \hline x & y & f(x, y) & P_{1}(x, y) & P_{2}(x, y) \\ \hline 0 & 0 & & & \\ \hline 0 & 0.1 & & & \\ \hline 0.2 & 0.1 & & & \\ \hline 0.2 & 0.5 & & & \\ \hline 1 & 0.5 & & & \\ \hline \end{array} $$ (e) Use a computer algebra system to graph the surfaces \(z=f(x, y), z=P_{1}(x, y),\) and \(z=P_{2}(x, y)\)

Short answer

The linear approximation of \(f(x, y) = e^{(x-y)}\) centered at (0,0) is \(P_1(x, y) = 1 + x - y\), and the quadratic approximation is \(P_2(x, y) = 1 + x - y + 1/2 * x^2 - x * y + 1/2* y^2\). If \(x=0\) in the quadratic approximation, it's a second-degree Taylor polynomial for \( e^{y}\). If \(y=0\), it's a second-degree Taylor polynomial for \( e^{x}\). The completed table consists of the values of the function \(f(x,y)\), linear approximation \(P_{1}(x,y)\), quadratic approximation \(P_{2}(x, y)\) for various x and y values.

Step-by-step solution

Computing the derivative

To compute the approximation, we'll start by computing the derivatives of \(f(x, y) = e^{(x-y)}\). The partial derivatives are \(f_x(0,0) = e^{(0-0)} = 1\) and \(f_y(0,0) = -e^{(0-0)} = -1\). The second partial derivatives also need to be computed for the quadratic approximation. We obtain: \(f_{xx}(0,0) = e^{(0-0)} = 1\), \(f_{yy}(0,0) = e^{(0-0)} = 1\), and \(f_{xy}(0,0) = -e^{(0-0)} = -1\).

Linear approximation

We substitute the derivatives into the formula for the linear approximation to get \(P_1(x, y) = e^{(0-0)} + 1 * x - 1 * y = 1 + x - y\).

Quadratic approximation

We substitute the derivatives into the formula for the quadratic approximation to get \(P_2(x, y) = 1 + x - y + 1/2 * x^2 - x * y + 1/2* y^2\).

Taylor polynomials

If \(x = 0\) in the quadratic approximation, we get \(P_{2}(0, y) = 1-y +1/2y^{2}\), which is a second-degree Taylor polynomial for \( e^{y}\). If \(y = 0\) in the quadratic approximation, we get \(P_{2}(x, 0) = 1+x +1/2x^{2}\), which is a second-degree Taylor polynomial for \( e^{x}\).

- Completing the table

Substitute the values of \(x\) and \(y\) into the function \(f(x,y) = e^{(x-y)}\), the linear approximation \(P_{1}(x,y) = 1 + x - y\), and the quadratic approximation \(P_{2}(x,y) = 1 + x - y + 1/2 * x^2 - x * y + 1/2* y^2\) to complete the respective columns in the table.