Finding the Minimum Sum of Squared Differences

The problem asks us to find the value of \(A\) that minimizes the expression \( \Sigma_{i=1}^{10}(x_i - A)^2 \) for the series of numbers \(x_i = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\). This expression represents the sum of the squared differences between each number in the series and a fixed value \(A\).

Understanding the Sum of Squared Differences

In statistics and data analysis, the sum of squared differences (also known as sum of squared errors or residuals) from a central value is a common measure. A fundamental property of the sum of squared differences is that it is minimized when the central value \(A\) is equal to the arithmetic mean (average) of the data points.

Let's look at the series provided:

• The series is \(x_i = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\).
• The number of terms in the series, \(n\), is 10.

Calculating the Mean of the Series

To find the value of \(A\) that minimizes the sum of squared differences, we need to calculate the mean of the series. The mean (\(\bar{x}\)) is calculated by summing all the values and dividing by the number of values.

The sum of the series is \( \Sigma_{i=1}^{10} x_i = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 \).

This is the sum of the first 10 natural numbers. The formula for the sum of the first \(n\) natural numbers is \( \frac{n(n+1)}{2} \).

Using this formula for \(n = 10\):

\( \text{Sum} = \frac{10(10+1)}{2} = \frac{10 \times 11}{2} = \frac{110}{2} = 55 \)

Now, we can calculate the mean:

\( \text{Mean} (\bar{x}) = \frac{\text{Sum}}{\text{Number of terms}} = \frac{55}{10} = 5.5 \)

Minimizing the Sum of Squared Differences with the Mean

As established, the sum of squared differences \( \Sigma_{i=1}^{10}(x_i - A)^2 \) is minimized when \(A\) is equal to the mean of the series. Our calculated mean is 5.5.

Therefore, the minimum possible value of \( \Sigma_{i=1}^{10}(x_i - A)^2 \) is attained when \(A = 5.5\).

Comparing with Options

Let's check the given options:

1. \(A = 6\)
2. \(A = 10\)
3. \(A = 5.5\)
4. \(A = 1\)

Our calculated mean is 5.5, which matches option 3.

Let's illustrate the concept with a small example series: 1, 2, 3. Mean is (1+2+3)/3 = 2.

If A=2 (mean): \((1-2)^2 + (2-2)^2 + (3-2)^2 = (-1)^2 + 0^2 + 1^2 = 1 + 0 + 1 = 2\)

If A=1: \((1-1)^2 + (2-1)^2 + (3-1)^2 = 0^2 + 1^2 + 2^2 = 0 + 1 + 4 = 5\)

If A=3: \((1-3)^2 + (2-3)^2 + (3-3)^2 = (-2)^2 + (-1)^2 + 0^2 = 4 + 1 + 0 = 5\)

The sum of squared differences is smallest when A is the mean (2 in this example). This principle holds true for any dataset, including the series 1 to 10.

Thus, for the series 1, 2, ..., 10, the minimum possible value of \( \Sigma_{i=1}^{10}(x_i - A)^2 \) is attained when \(A\) is equal to the mean, which is 5.5.

Revision Table: Minimizing Sum of Squares

Additional Information: Why Mean Minimizes Sum of Squared Differences

The property that the mean minimizes the sum of squared differences is a fundamental result in statistics and is related to the concept of variance. The variance of a dataset is defined as the average of the squared differences from the mean: \( \frac{\Sigma (x_i - \bar{x})^2}{n} \).

Mathematically, to find the value of \(A\) that minimizes \( f(A) = \Sigma (x_i - A)^2 \), we can use calculus. We take the derivative of \(f(A)\) with respect to \(A\) and set it to zero:

\( \frac{d}{dA} \Sigma (x_i - A)^2 = 0 \)

Using the sum rule and chain rule for derivatives:

\( \Sigma \frac{d}{dA} (x_i - A)^2 = 0 \)

\( \Sigma 2(x_i - A) \times (-1) = 0 \)

\( -2 \Sigma (x_i - A) = 0 \)

\( \Sigma (x_i - A) = 0 \)

\( \Sigma x_i - \Sigma A = 0 \)

\( \Sigma x_i - nA = 0 \)

\( nA = \Sigma x_i \)

\( A = \frac{\Sigma x_i}{n} \)

This shows that the value of \(A\) that minimizes the sum of squared differences is indeed the mean, \( \bar{x} = \frac{\Sigma x_i}{n} \). This principle is the basis for methods like linear regression (least squares method), where the goal is to minimize the sum of squared errors between the observed values and the values predicted by a model.