Question

The function \(y(x)\) is known to be a quadratic function of \(x\). The following table gives the measured values and uncorrelated standard errors of \(y\) measured at various values of \(x\) (in which there is negligible error): $$ \begin{array}{lccccc} x & 1 & 2 & 3 & 4 & 5 \\ y(x) & 3.5 \pm 0.5 & 2.0 \pm 0.5 & 3.0 \pm 0.5 & 6.5 \pm 1.0 & 10.5 \pm 1.0 \end{array} $$ Construct the response matrix \(R\) using as basis functions \(1, x, x^{2} .\) Calculate the matrix \(\mathrm{R}^{\mathrm{T}} \mathrm{N}^{-1} \mathrm{R}\) and show that its inverse, the covariance matrix \(\mathrm{V}\), has the form $$ \mathrm{V}=\frac{1}{9184}\left(\begin{array}{ccc} 12592 & -9708 & 1580 \\ -9708 & 8413 & -1461 \\ 1580 & -1461 & 269 \end{array}\right). $$ Use this matrix to find the best values, and their uncertainties, for the coefficients of the quadratic form for \(y(x)\).

Short answer

Calculate the matrix R. Create the diagonal matrix N and its inverse. Show the form of the covariance matrix V and use these results to find the coefficients of the quadratic form.

Step-by-step solution

Define the Response Matrix R

Create the response matrix using the basis functions \(1, x, x^2\). This can be done by substituting each provided value of \(x\) into these basis functions to form the matrix \( R \). For example, when \(x = 1\), the values are \(1, 1, 1^2 = 1\): \[ R = \begin{pmatrix} 1 & 1 & 1^2 \ 1 & 2 & 2^2 \ 1 & 3 & 3^2 \ 1 & 4 & 4^2 \ 1 & 5 & 5^2 \end{pmatrix} = \begin{pmatrix} 1 & 1 & 1 \ 1 & 2 & 4 \ 1 & 3 & 9 \ 1 & 4 & 16 \ 1 & 5 & 25 \end{pmatrix} \]

Create the Diagonal Matrix N

We need the matrix \(N\), which is a diagonal matrix representing the variances (squared standard errors) of \(y(x)\). Based on the standard errors given in the problem, you form \(N\) as:\[ N = \begin{pmatrix} 0.5^2 & 0 & 0 & 0 & 0 \ 0 & 0.5^2 & 0 & 0 & 0 \ 0 & 0 & 0.5^2 & 0 & 0 \ 0 & 0 & 0 & 1.0^2 & 0 \ 0 & 0 & 0 & 0 & 1.0^2 \end{pmatrix} = \begin{pmatrix} 0.25 & 0 & 0 & 0 & 0 \ 0 & 0.25 & 0 & 0 & 0 \ 0 & 0 & 0.25 & 0 & 0 \ 0 & 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 0 & 1 \end{pmatrix} \]

Calculate R^T N^{-1} R

First, calculate \(N^{-1}\), the inverse of \(N\). For a diagonal matrix, the inverse is simply the reciprocal of each diagonal element:\[ N^{-1} = \begin{pmatrix} 4 & 0 & 0 & 0 & 0 \ 0 & 4 & 0 & 0 & 0 \ 0 & 0 & 4 & 0 & 0 \ 0 & 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 0 & 1 \end{pmatrix} \]Next, calculate \(R^T\) (the transpose of \(R\)) and then \(R^T N^{-1} R\):\[ R^T = \begin{pmatrix} 1 & 1 & 1 & 1 & 1 \ 1 & 2 & 3 & 4 & 5 \ 1 & 4 & 9 & 16 & 25 \end{pmatrix} \]The matrix multiplication gives us:\[ R^T N^{-1} R = \begin{pmatrix} 1 & 1 & 1 & 1 & 1 \ 1 & 2 & 3 & 4 & 5 \ 1 & 4 & 9 & 16 & 25 \end{pmatrix} \begin{pmatrix} 4 & 0 & 0 & 0 & 0 \ 0 & 4 & 0 & 0 & 0 \ 0 & 0 & 4 & 0 & 0 \ 0 & 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 & 1 & 1 & 1 \ 1 & 2 & 4 & 16 & 25 \end{pmatrix} \begin{pmatrix} 13 & -13.4 & 18 \ 8.2 & -13.4 & 6 \ 8 \ 8 \ VxT\end{pmatrix}\]

Calculate the Covariance Matrix V

The covariance matrix \(V\) is the inverse of \( R^T N^{-1} R \). The problem gives us the form of \(V\) and the constant. We recognize that:\[ V = \frac{1}{9184} \begin{pmatrix} 12592 & -9708 & 1580 \ -9708 & 8413 & -1461 \ 1580 & -1461 & 269 \end{pmatrix} \]

Find the Best Values of Coefficients

In order to find the coefficients of the quadratic function \( y(x) = a + bx + cx^2 \, use the normal equation \ f = (R^T N^{-1} R)^{-1} R^T N^{-1} y\). Calculate \(y\ \Vy\formula\backslash\fV\) using the know value:\N_1 \fb\fo}//ONTO_y\_{a,b,c}

Key concepts

Response Matrix

In mathematics, a response matrix helps to transform the input variables into a target function you wish to analyze. In the given problem, we are working with a quadratic function where the response matrix (R) has been generated using basis functions 1, x, and x². The matrix is constructed by substituting each provided value of x into these basis functions. The response matrix for the values of x = {1, 2, 3, 4, 5} is:

\[\begin{equation} R = \begin{pmatrix} 1 & 1 & 1 \ 1 & 2 & 4 \ 1 & 3 & 9 \ 1 & 4 & 16 \ 1 & 5 & 25 \end{pmatrix} \end{equation}\]
This matrix allows us to apply linear algebra methods to fitting the quadratic function to data, making it easier to estimate the coefficients a, b, and c of the function y(x) = a + bx + cx².

Covariance Matrix

The covariance matrix expresses the degree to which random variables change together. In this problem, it helps us understand how errors in our measurements propagate through our calculations of the quadratic function's coefficients.

The covariance matrix (V), given in the exercise, is:
\[\begin{equation} V = \frac{1}{9184} \begin{pmatrix} 12592 & -9708 & 1580 \ -9708 & 8413 & -1461 \ 1580 & -1461 & 269 \end{pmatrix} \end{equation}\]
We obtain this matrix through inverting \textcaron{R^T N^{-1} R}, which involves the variances from the response matrix and the standard errors of y(x). The covariance matrix provides important information for understanding the precision and uncertainty of the coefficients we calculate for the quadratic model.

Normal Equation

The normal equation provides a method for finding the best-fit line or curve to a set of data points. For a quadratic function, this is essential for calculating the coefficients that minimize the difference between the measured data and the quadratic model.

In this problem, the normal equation is given by:

\[\begin{equation} f = (R^T N^{-1} R)^{-1} R^T N^{-1} y \end{equation}\]
This equation combines the transpose of the response matrix, the inverse of our diagonal matrix N, and the measured data y. By solving this equation, we find the values of the coefficients a, b, and c of our quadratic function y(x) = a + bx + cx² that best fit the given data points.

Understanding and applying the normal equation allows us to derive these coefficients with their respective uncertainties, providing a clearer picture of our quadratic model's accuracy.