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Find the least squares approximating function of the form \(r_{0}+r_{1} x^{2}+r_{2} \sin \frac{\pi x}{2}\) for each of the following sets of data pairs. a. (0,3),(1,0),(1,-1),(-1,2) b. \(\left(-1, \frac{1}{2}\right),(0,1),(2,5),(3,9)\)

Short Answer

Expert verified
Find coefficients using normal equations; then construct the least squares approximating function.

Step by step solution

01

Set Up Linear Equations

For each data point \(x, y\), write down the equation \r_{0} + r_{1} x^{2} + r_{2} \sin \frac{\pi x}{2} = y\. This results in a system of linear equations first for set (a) and then for set (b).
02

Formulate the Matrix

For the first data set (a), the matrix equation will be based on the linear equations from Step 1 and arranged as follows: \[\begin{bmatrix}1 & 0^2 & \sin 0 \1 & 1^2 & \sin \frac{\pi(1)}{2} \1 & 1^2 & \sin \frac{\pi(1)}{2} \1 & (-1)^2 & \sin \frac{\pi(-1)}{2} \\end{bmatrix}\begin{bmatrix} r_0 \ r_1 \ r_2 \end{bmatrix}= \begin{bmatrix} 3 \ 0 \ -1 \ 2 \end{bmatrix}\]Do the same for set (b).
03

Calculate the Normal Equations

Multiply the transpose of the matrix from Step 2 by itself and set it equals to the product of the transpose of the matrix and the results vector (right-hand side values). Solve the resulting normal equations to find \(r_0, r_1, r_2\) for both data sets.
04

Solve the Normal Equations

Use matrix algebra (or a calculator with matrix capabilities) to solve the system of equations from Step 3 and determine the coefficients \(r_0, r_1, r_2\) for both data sets.
05

Construct the Function

Using the coefficients \(r_0, r_1, r_2\) obtained from Step 4, construct the least squares approximating function for each data set. The function will be of the form \r_0 + r_1 x^2 + r_2 \sin \frac{\pi x}{2}\.

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Key Concepts

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

Understanding Matrix Equations
Matrix equations are crucial in many areas of mathematics and engineering. They simplify complex systems of equations into a more manageable form. When we describe our problem using a matrix equation, we can represent multiple linear equations as one system, using matrices and vectors.
How does it work?
Each equation, for a given data point, has a form like \(r_0 + r_1 x^2 + r_2 \sin \frac{\pi x}{2} = y\). Placed into matrix form, this becomes \(A\mathbf{r} = \mathbf{b}\), where:
  • \(A\) is the matrix comprising the coefficients of \(r_0\), \(r_1\), and \(r_2\) from each equation.
  • \(\mathbf{r}\) is the vector containing \(r_0\), \(r_1\), and \(r_2\). These are the values we're solving for.
  • \(\mathbf{b}\) is the vector of results, or target values, which the equations are intended to solve.
Writing down the matrix equation lets us use systematic methods to find \(r_0, r_1,\) and \(r_2\).
Exploring Normal Equations
Normal equations are an essential part of finding the least squares approximation. They help you solve the system \(A\mathbf{r} = \mathbf{b}\) when there is no exact solution because the equations are overdetermined (more equations than unknowns).
What's the process?
To form normal equations, multiply the transpose of matrix \(A\) by both sides of the matrix equation \(A\mathbf{r} = \mathbf{b}\), resulting in \(A^TA\mathbf{r} = A^T\mathbf{b}\).
  • \(A^T\) denotes the transpose of matrix \(A\). This step is crucial to balance the equation.
  • The multiplication reduces redundancy and ensures the resulting system is solvable using matrix algebra.
Solving the normal equations is then achievable through methods like numerical algorithms, depending on the matrix's complexity.
Demystifying Matrix Algebra
Matrix algebra is a set of mathematical methods used to perform operations on matrices. It is central to solving systems of linear equations, like those formed by our normal equations.
Key Operations:
  • Matrix Multiplication: The product of two matrices, \(A\) and \(B\), is formed by multiplying each row of \(A\) by each column of \(B\).
  • Transpose: Flipping a matrix over its diagonal. Rows become columns and vice-versa.
  • Inverse: The matrix \(A^{-1}\) is such that \(AA^{-1} = I\), where \(I\) is the identity matrix.
Using matrix algebra allows us to rearrange, simplify, and solve complex systems like those seen in least squares approximation. This understanding is pivotal in deriving the coefficients \(r_0, r_1,\) and \(r_2\).
Understanding Coefficient Determination
Finding the coefficients \(r_0, r_1,\) and \(r_2\) for a function to best fit given data points involves solving the normal equations derived from your data set. These coefficients are determined by minimizing the differences between your function's predictive values and the actual data points.
Why is it important?
  • By accurately determining the coefficients, you ensure your function best represents the data you've observed.
  • It allows for better predictions and interpretations of trends in data, essential in fields like statistics and data science.
  • It's not just about fitting the data; it's about reducing error, which is fundamental in creating trustworthy models.
Through solving the normal equations using matrix algebra, you obtain the best possible (least squares) approximating function that neatly fits your given data points.

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Most popular questions from this chapter

Find the least squares approximating line \(y=z_{0}+z_{1} x\) for each of the following sets of data points. $$ \begin{array}{l} \text { a. }(1,1),(3,2),(4,3),(6,4) \\ \text { b. }(2,4),(4,3),(7,2),(8,1) \end{array} $$ c. (-1,-1),(0,1),(1,2),(2,4),(3,6) $$ \text { d. }(-2,3),(-1,1),(0,0),(1,-2),(2,-4) $$

If \(A\) is \(m \times n\) of rank \(r,\) show that \(A\) can be factored as \(A=P Q\) where \(P\) is \(m \times r\) with \(r\) independent columns, and \(Q\) is \(r \times n\) with \(r\) independent rows. [Hint: Let \(U A V=\left[\begin{array}{cc}I_{r} & 0 \\ 0 & 0\end{array}\right]\) by Theorem 2.5.3, and write \(U^{-1}=\left[\begin{array}{ll}U_{1} & U_{2} \\ U_{3} & U_{4}\end{array}\right]\) and \(V^{-1}=\left[\begin{array}{ll}V_{1} & V_{2} \\\ V_{3} & V_{4}\end{array}\right]\) in block form, where \(U_{1}\) and \(V_{1}\) are \(\left.r \times r .\right]\)

If \(\left\\{\mathbf{x}_{1}, \mathbf{x}_{2}, \mathbf{x}_{3}, \ldots, \mathbf{x}_{k}\right\\}\) is independent, show \(\left\\{\mathbf{x}_{1}, \mathbf{x}_{1}+\mathbf{x}_{2}, \mathbf{x}_{1}+\mathbf{x}_{2}+\mathbf{x}_{3}, \ldots, \mathbf{x}_{1}+\mathbf{x}_{2}+\cdots+\mathbf{x}_{k}\right\\}\) is also independent.

We often write vectors in \(\mathbb{R}^{n}\) as rows. In each case determine if \(\mathbf{x}\) lies in \(U=\) \(\operatorname{span}\\{\mathbf{y}, \mathbf{z}\\} .\) If \(\mathbf{x}\) is in \(U,\) write it as a linear combination of \(\mathbf{y}\) and \(\mathbf{z}\); if \(\mathbf{x}\) is not in \(U\), show why not. a. \(\mathbf{x}=(2,-1,0,1), \mathbf{y}=(1,0,0,1),\) and \(\mathbf{z}=(0,1,0,1)\) b. \(\mathbf{x}=(1,2,15,11), \mathbf{y}=(2,-1,0,2),\) and \(\quad \mathbf{z}=(1,-1,-3,1)\) c. \(\mathbf{x}=(8,3,-13,20), \mathbf{y}=(2,1,-3,5),\) and \(\mathbf{z}=(-1,0,2,-3)\) d. \(\mathbf{x}=(2,5,8,3), \mathbf{y}=(2,-1,0,5),\) and \(\mathbf{z}=(-1,2,2,-3)\)

a. Show that \(\mathbf{x} \cdot \mathbf{y}=\frac{1}{4}\left[\|\mathbf{x}+\mathbf{y}\|^{2}-\|\mathbf{x}-\mathbf{y}\|^{2}\right]\) for all \(\mathbf{x}\), \(\mathbf{y}\) in \(\mathbb{R}^{n}\) b. Show that \(\|\mathbf{x}\|^{2}+\|\mathbf{y}\|^{2}=\frac{1}{2}\left[\|\mathbf{x}+\mathbf{y}\|^{2}+\|\mathbf{x}-\mathbf{y}\|^{2}\right]\) for all \(\mathbf{x}, \mathbf{y}\) in \(\mathbb{R}^{n}\)

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