Problem 29 - Discrete Mathematics with Applications

Question

Evaluate each sum, where \(\delta_{i j}\) is defined as follows. $$ \delta_{i j}=\left\\{\begin{array}{ll}{1} & {\text { if } i=j} \\ {0} & {\text { otherwise }}\end{array}\right. $$ \(\left[\delta_{j} \text { is called Kronecker's delta, after the German mathematician Leopold }\right.\) Kronecker \((1823-1891) . ]\) $$ \sum_{i=1}^{3} \sum_{j=1}^{5}\left(2+3 \delta_{i j}\right) $$

Step-by-Step Solution

Answer

The short answer to the given question is: The evaluated sum of the double summation involving Kronecker's delta is 39. The expression simplifies as follows: $$ \sum_{i=1}^{3} \sum_{j=1}^{5}\left(2+3 \delta_{i j}\right) = 39 $$

Step 1: Understand the notation of the given expression

The given expression, \(\sum_{i=1}^{3} \sum_{j=1}^{5}\left(2+3 \delta_{i j}\right)\), is a double summation. We have two indices, \(i\) and \(j\). The index \(i\) will vary from 1 to 3, whereas the index \(j\) will vary from 1 to 5. In each term, we have the function \(2+3\delta_{ij}\).

Step 2: Evaluate the first summation

Let's start by evaluating the inner sum for each value of \(i\): $$ \begin{aligned} \sum_{j=1}^5(2+3\delta_{ij}) \end{aligned} $$ We will evaluate this expression separately for each value of \(i\) and sum the results afterward. For \(i=1\): $$ \begin{aligned} \sum_{j=1}^5(2+3\delta_{1j}) &= (2 + 3\delta_{11}) + (2 + 3\delta_{12}) + \cdots + (2 + 3\delta_{15}) \end{aligned} $$ Similarly, for \(i=2\) and \(i=3\): $$ \begin{aligned} \sum_{j=1}^5(2+3\delta_{2j}) &= (2 + 3\delta_{21}) + (2 + 3\delta_{22}) + \cdots + (2 + 3\delta_{25}) \\ \sum_{j=1}^5(2+3\delta_{3j}) &= (2 + 3\delta_{31}) + (2 + 3\delta_{32}) + \cdots + (2 + 3\delta_{35}) \end{aligned} $$ Now let's simplify these expressions using the properties of Kronecker's delta:

Step 3: Evaluate the second summation

Applying the properties of the Kronecker's delta, the expression simplifies as follows: For \(i=1\): $$ \sum_{j=1}^5(2+3\delta_{1j}) = (2+3\cdot1) + (2+3\cdot0) + \cdots + (2+3\cdot0) = 5+2+2+2+2 $$ For \(i=2\): $$ \sum_{j=1}^5(2+3\delta_{2j}) = (2+3\cdot0) + (2+3\cdot1) + \cdots + (2+3\cdot0) = 2+5+2+2+2 $$ For \(i=3\): $$ \sum_{j=1}^5(2+3\delta_{3j}) = (2+3\cdot0) + (2+3\cdot0) + \cdots + (2+3\cdot1) = 2+2+2+2+5 $$ We are now ready to sum these expressions and find the total sum.

Step 4: Simplify the resulting expression

Adding up the values from Steps 2 and 3, we get: $$ \sum_{i=1}^3 \sum_{j=1}^5 (2+3\delta_{ij}) = (5+2+2+2+2) + (2+5+2+2+2) + (2+2+2+2+5) = 3\cdot(5+4\cdot2) $$ Calculating the final value: $$ 3\cdot(5+4\cdot2) = 3\cdot(5+8) = 3\cdot(13) = 39 $$ Thus, the evaluated sum is 39.

Key Concepts

Understanding Double Summation

To understand double summation, imagine adding up values in a two-dimensional grid. Each cell in the grid represents a term in the sum, and we combine these terms in a systematic way.

Double summation involves summing over two indices, often indicated as \(i\) and \(j\). First, we fix \(i\) and sum over all values of \(j\); then, we move on to the next value of \(i\) and repeat the process. This continues until all terms have been included. This method is common in discrete mathematics, where we deal with well-defined separate values.

Visualize Double Summation

To better visualize, consider your two summations as directions on the grid: summing over \(j\) moves you horizontally across a row, while summing over \(i\) moves you vertically down a column. In our example, for each row indicated by \(i=1, 2, 3\), we sum across columns \(j=1\) to \(5\). Once done, we add up the results of each row to get the final answer.

The complexity of double summation calls for attention to detail to prevent errors and ensure that each term has been accounted for only once. This discipline is critical, especially in discrete mathematics, where counting and summation play a central role.

Discrete Mathematics: Core of Counting and Structure

Discrete mathematics is a branch of math that deals with discrete elements that can be counted. Unlike continuous mathematics, which involves quantities that can vary smoothly, discrete math works with distinct or separate values.

These ‘discrete’ elements form the foundation of modern computer science and informatics, where digital signals and binary choices reign supreme. It encompasses structures like graphs, sets, integers, and statements in logic. The precision of discrete math makes it key in proving algorithms and writing precise mathematical proofs.

Application in Summation

In the context of summation, discrete mathematics allows us to handle individual summation operations step by step. Kronecker's delta is a perfect example of a function that emerges from discrete concepts; it assigns a value of 1 or 0 based on the distinct comparison between two variables \(i\) and \(j\).

Mathematical Notation: Language of Precision

Mathematical notation is like a language that enables mathematicians to write complex ideas succinctly and precisely. It standardizes mathematical expressions, allowing anyone in the field to understand the concepts without ambiguity.

Notation ranges from simple symbols representing numbers or operations to more complex structures like Kronecker's delta, which uses subscript indices for its definition. Proper understanding of notation is absolutely crucial for computational accuracy, especially in expressions involving summations, integrals, and limits.

By using the compact and well-defined properties of notations like Kronecker's delta, mathematicians and students can significantly reduce the complexity and length of mathematical arguments, leading to clearer and more error-free computations. The proper use of mathematical notation not only serves to communicate but also aids in the analytical process, as seen in the execution of the steps in our example problem.