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 \( c \) that minimizes the sum is \( c = \frac{1}{n} \sum_{i=1}^{n} x_i \).

Step 1: Understanding the Problem

We need to find a value of \( c \) such that it minimizes the sum \( \sum_{i=1}^{n}(x_i - c)^2 \). This is a classic optimization problem where we want to find the value of \( c \) that makes the squared differences between each \( x_i \) and \( c \) as small as possible.

Step 2: Function to Minimize

Consider the function \( f(c) = \sum_{i=1}^{n} (x_i - c)^2 \) which we want to minimize. This function represents the sum of the squared deviations of each \( x_i \) from \( c \).

Step 3: Taking the Derivative

Differentiate the function \( f(c) \) with respect to \( c \). This gives \[ f'(c) = \sum_{i=1}^{n} 2(x_i - c)(-1) = -2 \sum_{i=1}^{n} (x_i - c). \]

Step 4: Setting Derivative to Zero

To find the minimum, set the derivative of the function to zero: \[ -2 \sum_{i=1}^{n} (x_i - c) = 0. \] Simplifying, we get \[ \sum_{i=1}^{n} (x_i - c) = 0. \]

Step 5: Solve for c

The equation \( \sum_{i=1}^{n} (x_i - c) = 0 \) can be rewritten as \( \sum_{i=1}^{n} x_i = \sum_{i=1}^{n} c \). Thus, \( \sum_{i=1}^{n} x_i = nc \). Solving for \( c \) gives \[ c = \frac{1}{n} \sum_{i=1}^{n} x_i. \] This is the arithmetic mean of the \( x_i \) values.

Key Concepts

Derivatives

Derivatives are fundamental in understanding how functions change. They are crucial tools in calculus that help us determine where a function increases, decreases, or reaches extreme values like maximums or minimums.
When we discuss optimization problems, we often rely on derivatives to find points that minimize or maximize a function. In the given exercise, the function is defined as the sum of squared deviations:

• For a function, say, \( f(c) = \sum_{i=1}^{n} (x_i - c)^2 \), the derivative \( f'(c) \) tells us about the rate of change of \( f(c) \) with respect to \( c \).
• Taking the derivative, we apply the chain rule to differentiate terms involving \((x_i - c)^2\), leading to the expression \( f'(c) = -2 \sum_{i=1}^{n} (x_i - c)\).

By setting the derivative to zero, we identify where the function can have a minimum or maximum. Here, it ultimately helps us determine the optimal value of \( c \) that minimizes the total squared deviation.

Arithmetic Mean

The arithmetic mean is a measure of central tendency, often referred to as the average. It gives a single value that is representative of a set of values. In mathematical terms, for a series of numbers \( x_1, x_2, \ldots, x_n \), the arithmetic mean is calculated as:

• \( \text{Arithmetic Mean} = \frac{1}{n} \sum_{i=1}^{n} x_i \).

In the context of optimization and minimization, like in our exercise, the arithmetic mean plays a pivotal role.
When we aimed to minimize the function \( f(c) = \sum_{i=1}^{n} (x_i - c)^2 \), we arrived at the equation \( \sum_{i=1}^{n} x_i = nc \). Solving for \( c \), we found that \( c \) equates to the arithmetic mean of the values.
The arithmetic mean ensures that the square of the deviations of the numbers from this average is minimized. This makes it an optimal choice for such optimization problems.

Squared Deviations

Squared deviations refer to the square of the difference between each data point and a reference point, often the mean or another fixed value. This concept is widely used in statistics and optimization to measure dispersion.
In the problem \( \sum_{i=1}^{n}(x_i - c)^2 \), each term \((x_i - c)^2\) represents how far \( x_i \) is from \( c \), squared to ensure positivity and to penalize larger deviations more heavily.

• Squaring distances helps emphasize larger discrepancies, as larger variances will contribute more to the sum.
• Optimization of squared deviations helps in finding a value of \( c \) that centralizes \( x_i \) values, minimizing total spread or variance.

When the goal is to minimize squared deviations, as in the given exercise, the optimal value for \( c \) becomes the mean of the dataset because the arithmetic mean has the property of minimizing these squared deviations. It's a powerful method to find the point of balance or center of the data set.