Expanding the Inner Summation

Oct 13, 2025 by sorusgentrification.com 113 views

Introduction

In algebra and precalculus, summation is a fundamental concept used to represent the sum of a series of numbers. A double summation, on the other hand, involves the sum of two or more summations. In this article, we will delve into the expansion of a double summation, specifically the expression $\sum_{i=1}^n{\sum_{j=1}^n{\left| x_i-x_j \right|}}$ . We will show that this expression can be simplified to a more manageable form, and provide a step-by-step analysis of the process.

The Double Summation

The double summation $\sum_{i=1}^n{\sum_{j=1}^n{\left| x_i-x_j \right|}}$ represents the sum of the absolute differences between each pair of numbers in a set of $n$ numbers. To expand this expression, we need to consider the absolute difference between each pair of numbers.

Let's consider the absolute difference between two numbers, $x_i$ and $x_j$ . The absolute difference between these two numbers is given by $\left| x_i-x_j \right|$ . This expression can be expanded as follows:

\left| x_i-x_j \right| = \begin{cases} x_i-x_j & \text{if } x_i \geq x_j \\ x_j-x_i & \text{if } x_i < x_j \end{cases}

Now, let's consider the double summation. We can rewrite the expression as follows:

\sum_{i=1}^n{\sum_{j=1}^n{\left| x_i-x_j \right|}} = \sum_{i=1}^n{\sum_{j=1}^n{\left( x_i-x_j \right)}}

Expanding the Inner Summation

The inner summation $\sum_{j=1}^n{\left( x_i-x_j \right)}$ represents the sum of the differences between $x_i$ and each of the other numbers in the set. We can expand this expression as follows:

\sum_{j=1}^n{\left( x_i-x_j \right)} = \sum_{j=1}^n{x_i} - \sum_{j=1}^n{x_j}

Since $x_i$ is a constant, we can pull it out of the summation:

\sum_{j=1}^n{\left( x_i-x_j \right)} = n x_i - \sum_{j=1}^n{x_j}

Expanding the Outer Summation

Now, let's consider the outer summation $\sum_{i=1}^n{\left( n x_i - \sum_{j=1}^n{x_j} \right)}$ . We can expand this expression as follows:

\sum_{i=1}^n{\left( n x_i - \sum_{j=1}^n{x_j} \right)} = n \sum_{i=1}^n{x_i} - \sum_{i=1}^n{\sum_{j=1}^n{x_j}}

Since the inner summation is independent of $i$ , we can pull it out of the summation:

\sum_{i=1}^n{\left( n x_i - \sum_{j=1}^n{x_j} \right)} = n \sum_{i=1}^n{x_i} - n \sum_{j=1}^n{x_j}

Simplifying the Expression

Now, let's simplify the expression by combining like terms:

\sum_{i=1}^n{\left( n x_i - \sum_{j=1}^n{x_j} \right)} = n \sum_{i=1}^n{x_i} - n \sum_{j=1}^n{x_j}

We can rewrite the expression as follows:

\sum_{i=1}^n{\left( n x_i - \sum_{j=1}^n{x_j} \right)} = n \left( \sum_{i=1}^n{x_i} - \sum_{j=1}^n{x_j} \right)

The Final Result

After simplifying the expression, we get:

\sum_{i=1}^n{\sum_{j=1}^n{\left| x_i-x_j \right|}} = 2 \left( \left( n-1 \right) x_n + \left( n-3 \right) x_{n-1} + \dots + \left(1-n\right) x_1 \right)

This is the final result of the expansion of the double summation.

Conclusion

In this article, we have shown that the double summation $\sum_{i=1}^n{\sum_{j=1}^n{\left| x_i-x_j \right|}}$ can be simplified to a more manageable form. We have expanded the inner summation, pulled out the constant term, and simplified the expression to get the final result. This result can be used to solve problems involving double summations and absolute differences.

References

[1] "Algebra and Pre-Calculus" by Michael Sullivan
[2] "Calculus" by Michael Spivak

Additional Resources

[1] Khan Academy: Double Summation
[2] MIT OpenCourseWare: Calculus

Q: What is the expansion of the double summation $\sum_{i=1}^n{\sum_{j=1}^n{\left| x_i-x_j \right|}}$ ?

A: The expansion of the double summation $\sum_{i=1}^n{\sum_{j=1}^n{\left| x_i-x_j \right|}}$ is given by:

\sum_{i=1}^n{\sum_{j=1}^n{\left| x_i-x_j \right|}} = 2 \left( \left( n-1 \right) x_n + \left( n-3 \right) x_{n-1} + \dots + \left(1-n\right) x_1 \right)

Q: How do I expand the inner summation $\sum_{j=1}^n{\left( x_i-x_j \right)}$ ?

A: To expand the inner summation $\sum_{j=1}^n{\left( x_i-x_j \right)}$ , you can rewrite it as:

\sum_{j=1}^n{\left( x_i-x_j \right)} = \sum_{j=1}^n{x_i} - \sum_{j=1}^n{x_j}

Since $x_i$ is a constant, you can pull it out of the summation:

\sum_{j=1}^n{\left( x_i-x_j \right)} = n x_i - \sum_{j=1}^n{x_j}

Q: How do I expand the outer summation $\sum_{i=1}^n{\left( n x_i - \sum_{j=1}^n{x_j} \right)}$ ?

A: To expand the outer summation $\sum_{i=1}^n{\left( n x_i - \sum_{j=1}^n{x_j} \right)}$ , you can rewrite it as:

\sum_{i=1}^n{\left( n x_i - \sum_{j=1}^n{x_j} \right)} = n \sum_{i=1}^n{x_i} - \sum_{i=1}^n{\sum_{j=1}^n{x_j}}

Since the inner summation is independent of $i$ , you can pull it out of the summation:

\sum_{i=1}^n{\left( n x_i - \sum_{j=1}^n{x_j} \right)} = n \sum_{i=1}^n{x_i} - n \sum_{j=1}^n{x_j}

Q: How do I simplify the expression $\sum_{i=1}^n{\left( n x_i - \sum_{j=1}^n{x_j} \right)}$ ?

A: To simplify the expression $\sum_{i=1}^n{\left( n x_i - \sum_{j=1}^n{x_j} \right)}$ , you can rewrite it as:

\sum_{i=1}^n{\left( n x_i - \sum_{j=1}^n{x_j} \right)} = n \left( \sum_{i=1}^n{x_i} - \sum_{j=1}^n{x_j} \right)

Q: What is the final result of the expansion of the double summation?

A: The final result of the expansion of the double summation is:

\sum_{i=1}^n{\sum_{j=1}^n{\left| x_i-x_j \right|}} = 2 \left( \left( n-1 \right) x_n + \left( n-3 \right) x_{n-1} + \dots + \left(1-n\right) x_1 \right)

Q: How do I use the final result to solve problems involving double summations and absolute differences?

A: You can use the final result to solve problems involving double summations and absolute differences by substituting the values of $x_i$ and $n$ into the expression. For example, if you have a set of numbers $\{x_1, x_2, \dots, x_n\}$ and you want to find the sum of the absolute differences between each pair of numbers, you can use the final result to calculate the sum.

Q: What are some common applications of the expansion of a double summation?

A: The expansion of a double summation has many common applications in mathematics, statistics, and computer science. Some examples include:

Calculating the sum of the absolute differences between each pair of numbers in a set
Finding the median or mode of a set of numbers
Calculating the variance or standard deviation of a set of numbers
Solving problems involving double summations and absolute differences in computer science and engineering

Q: How do I learn more about the expansion of a double summation?

A: You can learn more about the expansion of a double summation by studying algebra and precalculus, and by practicing problems involving double summations and absolute differences. You can also consult online resources, such as Khan Academy and MIT OpenCourseWare, for additional help and support.