a. Rewrite the data points (2,1), (4,2), (8,6) and (9,6) with new x-coordinates in mean-deviation form. Let X be the associated design matrix. Why are the columns of X orthogonal? b. Write the normal equations for the data in part (a), and solve them to find the least-squares line, y =B₁ + B₁x², where x =x-5.75. The associated design matrix, X, is the entries in one column are all 1 and the entries in the other column sum to 0. (Simplify your answer. Use ascending order.) B. 1 -1.75 1 2.25 1 3.25 b. Write the normal equations for the data from part (a). Select the correct choice below and fill in the answer boxes within your choice. (Simplify your answer.) O A. Po P₁] = B1 Therefore, the columns of X are orthogonal because Bo B₁
a. Rewrite the data points (2,1), (4,2), (8,6) and (9,6) with new x-coordinates in mean-deviation form. Let X be the associated design matrix. Why are the columns of X orthogonal? b. Write the normal equations for the data in part (a), and solve them to find the least-squares line, y =B₁ + B₁x², where x =x-5.75. The associated design matrix, X, is the entries in one column are all 1 and the entries in the other column sum to 0. (Simplify your answer. Use ascending order.) B. 1 -1.75 1 2.25 1 3.25 b. Write the normal equations for the data from part (a). Select the correct choice below and fill in the answer boxes within your choice. (Simplify your answer.) O A. Po P₁] = B1 Therefore, the columns of X are orthogonal because Bo B₁
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![### Linear Algebra and Least-Squares Solutions
#### a. Data Point Transformation and Orthogonality of Design Matrix
We start with the data points \((2,1)\), \((4,2)\), \((8,6)\), and \((9,6)\). The x-coordinates are transformed into mean-deviation form with \(\bar{x} = 5.75\). Therefore, the new x-coordinates (\(x^*\)) are calculated as follows:
\[
x_{i}^* = x_{i} - \bar{x}
\]
This yields the transformed data points:
- \(x_1^* = 2 - 5.75 = -3.75\)
- \(x_2^* = 4 - 5.75 = -1.75\)
- \(x_3^* = 8 - 5.75 = 2.25\)
- \(x_4^* = 9 - 5.75 = 3.25\)
The associated design matrix, \(X\), which includes the transformed x-coordinates in mean-deviation form, is given by:
\[
X = \begin{pmatrix}
1 & -3.75 \\
1 & -1.75 \\
1 & 2.25 \\
1 & 3.25
\end{pmatrix}
\]
The columns of \(X\) are orthogonal because the entries in one column (the first column of ones) are all 1s, and the entries in the other column (transformed x-coordinates) sum to zero.
#### b. Normal Equations and Least-Squares Solution
We need to write the normal equations and solve them to find the least-squares line \(y = \beta_0 + \beta_1 x^*\). The normal equations are obtained from:
\[
X^T X \beta = X^T y
\]
Given the orthogonal nature of columns in \(X\), we proceed to solve for β0 and β1. After the computations, we seek the correct formulation from the provided options.
The two possible setups provided are:
- **Option A:**
\[
\begin{pmatrix}
\beta_0 \\
\beta_1
\end{pmatrix}
=
\begin{pmatrix}
\frac{23}{7} \\
\frac{6](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd5246d2a-5796-4784-a543-e596d3b5542c%2Fb6f50fa8-d32d-4ec4-a517-7437cf19fa5d%2F6wykjei_processed.png&w=3840&q=75)
Transcribed Image Text:### Linear Algebra and Least-Squares Solutions
#### a. Data Point Transformation and Orthogonality of Design Matrix
We start with the data points \((2,1)\), \((4,2)\), \((8,6)\), and \((9,6)\). The x-coordinates are transformed into mean-deviation form with \(\bar{x} = 5.75\). Therefore, the new x-coordinates (\(x^*\)) are calculated as follows:
\[
x_{i}^* = x_{i} - \bar{x}
\]
This yields the transformed data points:
- \(x_1^* = 2 - 5.75 = -3.75\)
- \(x_2^* = 4 - 5.75 = -1.75\)
- \(x_3^* = 8 - 5.75 = 2.25\)
- \(x_4^* = 9 - 5.75 = 3.25\)
The associated design matrix, \(X\), which includes the transformed x-coordinates in mean-deviation form, is given by:
\[
X = \begin{pmatrix}
1 & -3.75 \\
1 & -1.75 \\
1 & 2.25 \\
1 & 3.25
\end{pmatrix}
\]
The columns of \(X\) are orthogonal because the entries in one column (the first column of ones) are all 1s, and the entries in the other column (transformed x-coordinates) sum to zero.
#### b. Normal Equations and Least-Squares Solution
We need to write the normal equations and solve them to find the least-squares line \(y = \beta_0 + \beta_1 x^*\). The normal equations are obtained from:
\[
X^T X \beta = X^T y
\]
Given the orthogonal nature of columns in \(X\), we proceed to solve for β0 and β1. After the computations, we seek the correct formulation from the provided options.
The two possible setups provided are:
- **Option A:**
\[
\begin{pmatrix}
\beta_0 \\
\beta_1
\end{pmatrix}
=
\begin{pmatrix}
\frac{23}{7} \\
\frac{6
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