Problem 37 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(x_{1}, x_{2}, \ldots, x_{n}\) be any real numbers. Find the value of \(c\) that minimizes \(\sum_{i=1}^{n}\left(x_{i}-c\right)^{2}\).

Step-by-Step Solution

Answer

The value of \( c \) is the average of \( x_i \): \( c = \frac{1}{n} \sum_{i=1}^{n}x_i \).

Step 1: Understand the Objective Function

The objective is to minimize the function \( f(c) = \sum_{i=1}^{n}(x_i - c)^2 \). This represents the sum of the squares of the differences between each number \( x_i \) and \( c \).

Step 2: Differentiate the Function

To find the critical points, we differentiate the function \( f(c) \) with respect to \( c \). The derivative is \( f'(c) = \sum_{i=1}^{n} 2(x_i - c) \cdot (-1) = -2\sum_{i=1}^{n}(x_i - c) \).

Step 3: Set the Derivative to Zero

Set the derivative found in Step 2 equal to zero to find the critical points: \(-2\sum_{i=1}^{n}(x_i - c) = 0\). This simplifies to \( \sum_{i=1}^{n}(x_i - c) = 0 \).

Step 4: Solve for \( c \)

From \( \sum_{i=1}^{n}(x_i - c) = 0 \), we obtain \( \sum_{i=1}^{n}x_i = n \cdot c \). Solving for \( c \) gives \( c = \frac{1}{n} \sum_{i=1}^{n}x_i \). This means \( c \) is the average of the \( x_i \) values.

Step 5: Confirm \( c \) is a Minimum

The second derivative is \( f''(c) = 2n \), which is positive. Hence, the function \( f(c) \) is convex, confirming that \( c = \frac{1}{n} \sum_{i=1}^{n}x_i \) is a minimum.

Key Concepts

Objective Function

In calculus optimization, the objective function is at the heart of finding extrema—either minima or maxima—of a certain problem. Here, the objective function is represented as:

• \( f(c) = \sum_{i=1}^{n}(x_i - c)^2 \)

This function calculates the sum of the squares of the differences between a set of real numbers \(x_i\) and a constant \(c\). The goal is to adjust \(c\) to make this sum as small as possible, effectively minimizing the function. This concept plays a crucial role because it encapsulates what you want to achieve in the optimization problem. By understanding the structure of your objective function, you can proceed to find its optimal solutions effectively. Often, such functions are designed to reflect real-world situations where minimizing cost, distance, or error is desirable.

Derivative

The derivative of a function gives us the rate at which the function's value is changing concerning its variable. In optimization problems, derivatives are used extensively to find critical points where these rates change direction.

To find the derivative of our objective function \( f(c) = \sum_{i=1}^{n}(x_i - c)^2 \), we employ the power rule and the chain rule:

• The derivative, denoted as \( f'(c) \), is calculated as: \(-2\sum_{i=1}^{n}(x_i - c)\).

This expression arises because differentiating \((x_i - c)^2\) with respect to \(c\) yields \(-2(x_i - c)\). Differentiation simplifies the problem by transforming it into a form where we can identify critical points more readily, thus directing us closer to the optimization solution.

Critical Points

Critical points are values of \(c\) where the derivative of our function equals zero. These points are significant because potential minima or maxima occur there. In our task:

• We set \(-2\sum_{i=1}^{n}(x_i - c) = 0\), simplifying to \(\sum_{i=1}^{n}(x_i - c) = 0\).

Solving this equation enables us to find \(c\) as:

• \(c = \frac{1}{n} \sum_{i=1}^{n}x_i\)

This result reveals that the critical point \(c\) coincides with the average of all \(x_i\) values. This simplification illustrates a central principle in statistics and optimization: the average is a natural balancing point of a data set, often leading to minimized squared differences.

Second Derivative Test

The second derivative test helps confirm the nature of a critical point found through the first derivative. Specifically, it tells us whether a critical point is a minimum or maximum by examining the concavity of the objective function at that point.

For our function \(f(c) = \sum_{i=1}^{n}(x_i - c)^2\), we compute the second derivative:

• \(f''(c) = 2n\)

Since \(f''(c)\) is positive, it indicates that the function is concave up, meaning the critical point we found is indeed a minimum point. This use of the second derivative shows us definitively that the chosen \(c\) minimizes the objective function, establishing robustness in our solution and reinforcing the importance of leveraging higher derivatives in optimization tasks.