How do you use the taylor series to find a quadratic approximation of #2x^2 - 3xy + x# at (1,1)?

Question

How do you use the taylor series to find a quadratic approximation of #2x^2 - 3xy + x#  at (1,1)?

Elijah Abram · Accepted Answer

# = x( 2x - 3y + 1 ) #

The 2-D Taylor Expansion about #(alpha, beta)# is:

# T(x,y)=f(alpha,beta)+(x-alpha )f_(x)(alpha,beta)+(y-beta)f_{y}(alpha,beta)+{1}/{2!} ( (x-alpha)^{2}f_(x x)(alpha,beta)+2(x-alpha)(y-beta)f_{xy}(alpha,beta)+(y-beta)^{2}f_{yy}(alpha,beta) )+\cdots #

We therefore need the following partial derivatives:

#f_x = 4x - 3y + 1, qquad f_x(1,1) = 2#

#f_y = - 3x, qquad f_y(1,1) = -3#

#f_(xy) = f_(yx) = -3, qquad f_(xy) (1,1) = -3#

#f_(x x) = 4, qquad f_(x x)(1,1) = 4#

#f_(yy) = 0, qquad f_(yy)(1,1) = 0#

This means that, working to only the quadratic terms:

# T(x,y) =0+2(x-1 )-3(y-1)+{1}/{2!} ( 4(x-1)^{2}+2(-3)(x-1)(y-1)+0 (y-1)^{2} ) #

# =2x- 2 -3y + 3 + 2x^2 - 4x + 2 - 3xy + 3x + 3y - 3 #

# = x( 2x - 3y + 1 ) #

Paisley Johnson · Answer

undefined To find a quadratic approximation of $2x^2 - 3xy + x$ at the point $(1,1)$ using Taylor series, follow these steps:

1. Find the first and second-order partial derivatives of the function with respect to $x$ and $y$.
2. Evaluate these derivatives at the point $(1,1)$.
3. Use the Taylor series expansion formula for a function of two variables up to the quadratic term.

The Taylor series expansion of a function $f(x,y)$ up to the quadratic term around the point $(x_0,y_0)$ is given by:

$$f(x,y) \approx f(x_0,y_0) + f_x(x_0,y_0)(x-x_0) + f_y(x_0,y_0)(y-y_0) + \frac{1}{2}[f_{xx}(x_0,y_0)(x-x_0)^2 + 2f_{xy}(x_0,y_0)(x-x_0)(y-y_0) + f_{yy}(x_0,y_0)(y-y_0)^2]$$

Now, let's proceed with finding the required derivatives:

$$f(x,y) = 2x^2 - 3xy + x$$

First-order partial derivatives:

$$f_x = 4x - 3y + 1$$
$$f_y = -3x$$

Second-order partial derivatives:

$$f_{xx} = 4$$
$$f_{xy} = -3$$
$$f_{yy} = 0$$

Now, evaluate these derivatives at the point $(1,1)$:

$$f(1,1) = 2(1)^2 - 3(1)(1) + 1 = 0$$
$$f_x(1,1) = 4(1) - 3(1) + 1 = 2$$
$$f_y(1,1) = -3(1) = -3$$
$$f_{xx}(1,1) = 4$$
$$f_{xy}(1,1) = -3$$
$$f_{yy}(1,1) = 0$$

Substitute these values into the Taylor series expansion formula:

$$f(x,y) \approx 0 + 2(x-1) - 3(y-1) + \frac{1}{2}[4(x-1)^2 - 6(x-1)(y-1)]$$

Simplify the expression to get the quadratic approximation:

$$f(x,y) \approx 2x - 3y + 2(x-1)^2 - 3(x-1)(y-1)$$

This is the quadratic approximation of $2x^2 - 3xy + x$ at the point $(1,1)$.