Problem 65 - Calculus Early Transcendentals

Question

If an unknown physical quantity \(x\) is measured \(n\) times, the measurements \(x_{1}, x_{2}, \ldots, x_{n}\) often vary because of uncontrollable factors such as temperature, atmospheric pressure, and so forth. Thus, a scientist is often faced with the problem of using \(n\) different observed measurements to obtain an estimate \(\bar{x}\) of an unknown quantity \(x .\) One method for making such an estimate is based on the least squares principle, which states that the estimate \(\bar{x}\) least squares principle, which states that the estimate \(\bar{x}\) should be chosen to minimize $$ s=\left(x_{1}-\bar{x}\right)^{2}+\left(x_{2}-\bar{x}\right)^{2}+\cdots+\left(x_{n}-\bar{x}\right)^{2} $$ which is the sum of the squares of the deviations between the estimate \(\bar{x}\) and the measured values. Show that the estimate resulting from the least squares principle is $$ \bar{x}=\frac{1}{n}\left(x_{1}+x_{2}+\cdots+x_{n}\right) $$ that is, \(\bar{x}\) is the arithmetic average of the observed values.

Step-by-Step Solution

Answer

\( \bar{x} = \frac{1}{n}(x_1 + x_2 + \cdots + x_n) \).

Step 1: Understand the Objective

The goal is to find the estimate \( \bar{x} \) that minimizes the sum of the squared deviations \( s = (x_1 - \bar{x})^2 + (x_2 - \bar{x})^2 + \cdots + (x_n - \bar{x})^2 \). This sum quantifies how much each measured value \( x_i \) deviates from the estimate \( \bar{x} \).

Step 2: Formulate the Function

We need to find an expression for \( s \), which is given by the formula \( s = \sum_{i=1}^{n} (x_i - \bar{x})^2 \). Our task is to find the value of \( \bar{x} \) that minimizes this expression.

Step 3: Take the Derivative of s

To find the minimum of the function \( s \), take the derivative with respect to \( \bar{x} \):\[ \frac{ds}{d\bar{x}} = \sum_{i=1}^{n} -2(x_i - \bar{x}) = -2 \sum_{i=1}^{n} (x_i - \bar{x}) \].

Step 4: Set Derivative to Zero

To find the critical points, set the derivative equal to zero:\[ -2 \sum_{i=1}^{n} (x_i - \bar{x}) = 0 \]This simplifies to \( \sum_{i=1}^{n} (x_i - \bar{x}) = 0 \).

Step 5: Solve for \( \bar{x} \)

Expanding \( \sum_{i=1}^{n} (x_i - \bar{x}) = \sum_{i=1}^{n} x_i - n\bar{x} = 0 \), we get:\[ \sum_{i=1}^{n} x_i = n \bar{x} \]Solving for \( \bar{x} \), we find:\[ \bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i \]Therefore, \( \bar{x} \) is the arithmetic average of the values.

Key Concepts

Arithmetic Mean

The arithmetic mean, often referred to simply as the mean or average, is a way of summarizing a set of numbers by taking their total and dividing it by the quantity of numbers in that set. This is a common method used to find an estimate or "central value" of a collection of data.
For example, if you have a set of measurements taken several times, compute the arithmetic mean to find your estimate.

• Add up all the measurements.
• Count how many measurements are in the set.
• Divide the total by the number of measurements.

Mathematically, if your measurements are denoted as \(x_1, x_2, \ldots, x_n\), the arithmetic mean \(\bar{x}\) is calculated using the formula:\[ \bar{x} = \frac{1}{n} (x_1 + x_2 + \cdots + x_n) \]This method provides a simple, yet powerful way to understand data and make estimations about unknown factors.

Sum of Squared Deviations

When trying to find the best estimate of an unknown quantity from multiple measurements, one common approach is the method of least squares. This involves minimizing the sum of squared deviations (or errors) between the estimated mean and each individual measurement.
The formula for the sum of squared deviations is:\[ s = (x_1 - \bar{x})^2 + (x_2 - \bar{x})^2 + \cdots + (x_n - \bar{x})^2 \]This calculation helps in understanding how far off each measurement is from the estimated mean. By squaring these differences, it ensures that positive and negative deviations do not cancel each other out.

• The squaring puts more weight on larger deviations.
• Minimizing this sum ensures the overall difference from the mean is minimized.
• This makes the least squares method very effective for finding the best estimate.

Thus, the sum of squared deviations is a crucial component helping to pinpoint the most accurate mean.

Derivatives in Optimization

Derivatives are a fundamental part of calculus and play a vital role in finding the point at which a function is minimized or maximized. In the context of least squares estimation, derivatives help us find the arithmetic mean that minimizes the sum of squared deviations.
To minimize the expression \( s = \sum_{i=1}^{n} (x_i - \bar{x})^2 \), take the derivative of \( s \) with respect to \( \bar{x} \).
The derivative is given by:\[ \frac{ds}{d\bar{x}} = -2 \sum_{i=1}^{n} (x_i - \bar{x}) \]Setting the derivative to zero helps us locate the critical points:

• \(-2 \sum_{i=1}^{n} (x_i - \bar{x}) = 0\)
• This simplifies to find our critical value of \( \bar{x} \).

Through this process, calculus verifies that the arithmetic mean really is the value that minimizes the total deviation, solving the optimization problem effectively.