Chapter 5: Problem 3
Approximieren Sie die Abbildung \(\mathbf{f}: \mathbb{R}^{3} \rightarrow \mathbb{R}^{2}\) $$ \mathbf{f}(\mathbf{x})=\left(\begin{array}{c} x y \sin z \\ x+y^{3} z \end{array}\right) $$ durch eine lineare Abbildung um den Punkt \(\mathbf{x}_{0}=(0,0,0)^{T}\).
Short Answer
Expert verified
The linear approximation at \((0,0,0)\) is \(\begin{pmatrix} 0 \\ x \end{pmatrix}\).
Step by step solution
01
Understand the Problem
The exercise asks for the linear approximation of the given function \( \mathbf{f}(\mathbf{x}) \) around the point \( \mathbf{x}_0 = (0,0,0)^T \). Linear approximation involves finding the Jacobian matrix of \( \mathbf{f} \) at \( \mathbf{x}_0 \) and using it to approximate the function by a linear transformation.
02
Define the Function Components
The function \( \mathbf{f}(\mathbf{x}) \) is given as: \[\mathbf{f}(\mathbf{x}) = \begin{pmatrix} x y \sin z \ x + y^3 z \end{pmatrix} \]It's a vector-valued function from \( \mathbb{R}^3 \) to \( \mathbb{R}^2 \).
03
Compute the Partial Derivatives
Find the partial derivatives to form the Jacobian matrix. Compute the following:1. \( \frac{\partial f_1}{\partial x} = y \sin z \)2. \( \frac{\partial f_1}{\partial y} = x \sin z \)3. \( \frac{\partial f_1}{\partial z} = xy \cos z \)4. \( \frac{\partial f_2}{\partial x} = 1 \)5. \( \frac{\partial f_2}{\partial y} = 3y^2 z \)6. \( \frac{\partial f_2}{\partial z} = y^3 \)
04
Evaluate the Partial Derivatives at the Point
Evaluate each of these partial derivatives at \( \mathbf{x}_0 = (0,0,0)^T \):1. \( \frac{\partial f_1}{\partial x}(0,0,0) = 0 \)2. \( \frac{\partial f_1}{\partial y}(0,0,0) = 0 \)3. \( \frac{\partial f_1}{\partial z}(0,0,0) = 0 \)4. \( \frac{\partial f_2}{\partial x}(0,0,0) = 1 \)5. \( \frac{\partial f_2}{\partial y}(0,0,0) = 0 \)6. \( \frac{\partial f_2}{\partial z}(0,0,0) = 0 \)
05
Form the Jacobian Matrix
Using the evaluated partial derivatives, the Jacobian matrix \( J_{\mathbf{f}}(\mathbf{x}_0) \) is:\[J_{\mathbf{f}}(0,0,0) = \begin{pmatrix} 0 & 0 & 0 \ 1 & 0 & 0 \end{pmatrix}\]
06
Construct the Linear Approximation
The linear approximation \( \mathbf{L}(\mathbf{x}) \) of \( \mathbf{f}(\mathbf{x}) \) near \( \mathbf{x}_0 \) is given by:\[\mathbf{L}(\mathbf{x}) = \mathbf{f}(\mathbf{x}_0) + J_{\mathbf{f}}(\mathbf{x}_0) \cdot (\mathbf{x} - \mathbf{x}_0)\]Since \( \mathbf{f}(\mathbf{x}_0) = \begin{pmatrix} 0 \ 0 \end{pmatrix} \) and \( \mathbf{x}_0 = (0,0,0)^T \), this simplifies to:\[\mathbf{L}(\mathbf{x}) = \begin{pmatrix} 0 \ x \end{pmatrix}\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
When you're exploring the world of multi-variable calculus, understanding partial derivatives is crucial. Imagine you're working with a function that depends on several variables, like the function \( \mathbf{f}(\mathbf{x}) \) in our exercise, which has inputs \( x, y, \) and \( z \). A partial derivative helps you understand how the output of the function changes as you vary one variable, while keeping the others constant.
For example, in the exercise, the function \( f_1(x,y,z) = x y \sin z \) depends on all three variables. The partial derivative \( \frac{\partial f_1}{\partial x} = y \sin z \) tells us how \( f_1 \) changes with \( x \), holding \( y \) and \( z \) fixed.
Here are some helpful points to remember about partial derivatives:
For example, in the exercise, the function \( f_1(x,y,z) = x y \sin z \) depends on all three variables. The partial derivative \( \frac{\partial f_1}{\partial x} = y \sin z \) tells us how \( f_1 \) changes with \( x \), holding \( y \) and \( z \) fixed.
Here are some helpful points to remember about partial derivatives:
- Partial derivatives are like traditional derivatives, but consider multiple variables.
- Calculate one partial derivative for each variable the function depends on.
- They help form the building blocks for the Jacobian matrix which we'll discuss next.
Jacobian Matrix
The Jacobian matrix is like a treasure map for functions that have multiple inputs and outputs. It's a matrix built from the partial derivatives of a vector-valued function, showing how the function changes as each input variable changes. In our exercise, we needed the Jacobian matrix to approximate the function \( \mathbf{f}(\mathbf{x}) \) around the origin.
For \( \mathbf{f}(\mathbf{x}) \), the Jacobian matrix \( J_{\mathbf{f}} \) at a point \( \mathbf{x}_0 = (0,0,0)^T \) gives us a linear transformation that's the best approximation to \( \mathbf{f} \) near that point. Imagine it as a tool that transforms a small change in input to how the function will change.
Key aspects include:
For \( \mathbf{f}(\mathbf{x}) \), the Jacobian matrix \( J_{\mathbf{f}} \) at a point \( \mathbf{x}_0 = (0,0,0)^T \) gives us a linear transformation that's the best approximation to \( \mathbf{f} \) near that point. Imagine it as a tool that transforms a small change in input to how the function will change.
Key aspects include:
- The Jacobian matrix for a function \( \mathbf{f}(x, y, z) \) with outputs \( f_1, f_2 \) is constructed using the partial derivatives \( \frac{\partial f_i}{\partial x_j} \).
- This matrix encompasses all partial derivatives arranged in rows for each output component.
- In our solution, the Jacobian matrix is \( \begin{pmatrix} 0 & 0 & 0 \ 1 & 0 & 0 \end{pmatrix} \).
- Evaluating the Jacobian at a specific point, like \( (0,0,0)^T \), helps nail down linear approximations.
Vector-Valued Functions
Vector-valued functions are fascinating! Rather than spitting out just a single number, these functions output a vector. They map inputs from one space, say \( \mathbb{R}^3 \), to outputs in another space, like \( \mathbb{R}^2 \). This is exactly what we see with our exercise's function \( \mathbf{f}(\mathbf{x}) \).
Such functions are essential when handling systems that might involve multiple quantities, like direction and speed in physics. Here, \( \mathbf{f}(\mathbf{x}) \) outputs two components: \( f_1(x,y,z) \) and \( f_2(x,y,z) \).
A few things to keep in mind about vector-valued functions:
Such functions are essential when handling systems that might involve multiple quantities, like direction and speed in physics. Here, \( \mathbf{f}(\mathbf{x}) \) outputs two components: \( f_1(x,y,z) \) and \( f_2(x,y,z) \).
A few things to keep in mind about vector-valued functions:
- They often appear in physics and engineering, representing quantities like force or velocity.
- Understanding these functions requires understanding the role of each vector component.
- Linear approximation of such functions (using tools like the Jacobian matrix) makes them easier to work with in certain conditions.