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Use Taylor's formula for \(f(x, y)\) at the origin to find quadratic and cubic approximations of \(f\) near the origin. $$f(x, y)=x e^{y}$$

Short Answer

Expert verified
Quadratic: \( f(x, y) \approx x + xy \); Cubic: \( f(x, y) \approx x + xy + \frac{1}{2}xy^2 \)."

Step by step solution

01

Understand Taylor's formula

Taylor's formula for approximating a function around the origin includes terms derived from the function's derivatives evaluated at that point. For a function of two variables, the terms will include partial derivatives with respect to both variables.
02

Find the partial derivatives

First, find the partial derivatives of the function up to the third order.\1. \( f(x, y) = x e^{y} \)\2. First partial derivatives: \ \( f_x = e^{y} \), \( f_y = x e^{y} \)\3. Second partial derivatives: \ \( f_{xx} = 0 \), \( f_{yy} = x e^{y} \), \( f_{xy} = e^{y} \)\4. Third partial derivatives: \ \( f_{xxx} = 0 \), \( f_{yyy} = x e^{y} \), \( f_{xxy} = 0 \), \( f_{xyy} = e^{y} \), \( f_{yyx} = e^{y} \)
03

Evaluate derivatives at the origin

Substitute \(x = 0\) and \(y = 0\) into the partial derivatives:1. \( f(0,0) = 0 \)2. \( f_x(0,0) = e^{0} = 1 \)3. \( f_y(0,0) = 0 \)4. \( f_{xy}(0,0) = 1 \)5. All second and third derivatives evaluated at the origin are zero.
04

Construct the quadratic approximation

The quadratic approximation of \(f(x, y)\) about \((0,0)\) is given by:\[ f(x, y) \approx f(0,0) + f_x(0,0) x + f_y(0,0) y + \frac{1}{2}f_{xx}(0,0)x^2 + \frac{1}{2}f_{yy}(0,0)y^2 + f_{xy}(0,0)xy \]Substitute the values: \\[ f(x, y) \approx 0 + 1 \, x + 0 \, y + 0 \, x^2 + 0 \, y^2 + 1 \, xy \]Therefore, the quadratic approximation is \( f(x, y) \approx x + xy \).
05

Construct the cubic approximation

The cubic approximation of \(f(x, y)\) about \((0,0)\) is given by:\[ f(x, y) \approx f(x, y) \quad \text{(quadratic terms)} + \frac{1}{6} f_{xxx}(0,0)x^3 + \frac{1}{6} f_{yyy}(0,0)y^3 + \frac{1}{2} f_{xxy}(0,0)x^2y + \frac{1}{2} f_{xyy}(0,0)xy^2 \]Substitute the values (third degree terms vanish): \\[ f(x, y) \approx x + xy + 0 \, x^3 + 0 \, y^3 + 0 \, x^2y + \frac{1}{2} \, xy^2 \]Thus, the cubic approximation is \( f(x, y) \approx x + xy + \frac{1}{2}xy^2 \).
06

Conclusion

The quadratic approximation of \(f(x, y) = xe^y\) near the origin is \( f(x, y) \approx x + xy \) and the cubic approximation is \( f(x, y) \approx x + xy + \frac{1}{2}xy^2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Derivatives
Partial derivatives are like one-variable derivatives, but for functions with more than one variable. They show how a function changes as one variable changes, while the other variables remain constant. In the context of multivariable calculus, partial derivatives help us understand how a function behaves with respect to each of its inputs.

To illustrate, consider the function \( f(x, y) = x e^{y} \). Here, to find the partial derivative with respect to \( x \), you'll treat \( y \) as a constant, resulting in \( f_x = e^y \). Similarly, when finding the partial derivative with respect to \( y \), treat \( x \) as constant, yielding \( f_y = xe^y \).

Calculating higher-order partial derivatives involves finding derivatives of these derivatives. For example, the second-order partial derivatives of our function include \( f_{xx}, f_{yy}, \) and \( f_{xy} \), which describe how curvature or changes in slope occur involving each pair of variables.
Multivariable Calculus
Multivariable calculus extends the concepts of calculus to functions with more than one variable. Imagine you're studying a landscape to understand heights. Instead of a single slope in single-variable calculus, you now handle multiple directions, slopes, and curvatures.

In multivariable calculus, Taylor series come in handy to approximate functions around a point. They utilize partial derivatives to create polynomials that represent the function locally. These series are like a toolbox for breaking down complex surfaces into understandable bits, providing insight into how the function behaves close to a specified point.

When working with functions like \( f(x, y) = x e^{y} \), multivariable calculus helps by breaking down the problem into such derivatives and then constructing approximations like the quadratic approximation \( x + xy \) or even cubic approximations which include additional terms like \( \frac{1}{2}xy^2 \).
Function Approximation
Function approximation is a fundamental concept in calculus, especially when dealing with complex functions that are hard to work with directly. Taylor series is one powerful technique to achieve this. It helps you create a simpler polynomial that approximates the actual function near a specific point.

For instance, in the function \( f(x, y) = x e^y \), using Taylor's formula allows you to construct simpler versions of this function. The quadratic approximation \( f(x, y) = x + xy \) is a more manageable version that captures the essence of \( f(x, y) \) near the origin.

Function approximation is more than just a mathematical exercise—it is a tool that finds applications in physics, engineering, and economics, where exact models are often complex or unknown. By approximating, you can predict behavior, make calculations easier, and gain insights that lead to practical solutions.

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Most popular questions from this chapter

Find and sketch the domain for each function. $$f(x, y)=\frac{(x-1)(y+2)}{(y-x)\left(y-x^{3}\right)}$$

Find the specific function values. $$f(x, y)=\sin (x y)$$ a. \(f\left(2, \frac{\pi}{6}\right)\) b. \(f\left(-3, \frac{\pi}{12}\right)\) c. \(f\left(\pi, \frac{1}{4}\right)\) d. \(f\left(-\frac{\pi}{2},-7\right)\)

(A) find the function's domain, (b) find the function's range, (c) describe the function's level curves, (d) find the boundary of the function's domain, (e) determine if the domain is an open region, a closed region, or neither, and (f) decide if the domain is bounded or unbounded. $$f(x, y)=\sin ^{-1}(y-x)$$

By considering different paths of approach, show that the functions have no limit as \((x, y) \rightarrow(0,0)\). $$h(x, y)=\frac{x^{2} y}{x^{4}+y^{2}}$$

You will explore functions to identify their local extrema. Use a CAS to perform the following steps: a. Plot the function over the given rectangle. b. Plot some level curves in the rectangle. c. Calculate the function's first partial derivatives and use the CAS equation solver to find the critical points. How do the critical points relate to the level curves plotted in part (b)? Which critical points, if any, appear to give a saddle point? Give reasons for your answer. d. Calculate the function's second partial derivatives and find the discriminant \(f_{x x} f_{y y}-f_{x y}^{2}\). e. Using the max-min tests, classify the critical points found in part (c). Are your findings consistent with your discussion in part (c)? $$f(x, y)=x^{2}+y^{3}-3 x y, \quad-5 \leq x \leq 5, \quad-5 \leq y \leq 5$$

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