Problem 30 - 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 \((1823-1891) . ]\) $$ \sum_{i=1}^{6} \sum_{j=1}^{7}\left(i^{2}-3 i+\delta_{i j}\right) $$

Step-by-Step Solution

Answer

The short answer for the given problem is: To evaluate the double summation, we first compute the inner sum for each \(i\), and then add up the results for the outer sum. Using the definition of \(\delta_{ij}\), we can rewrite the double summation as: $$ \sum_{i=1}^{6}\left(7\left(i^{2}-3 i\right)+1\right) $$ Computing the sum and simplifying the result, we get: $$ \sum_{i=1}^{6} \sum_{j=1}^{7}\left(i^{2}-3 i+\delta_{i j}\right)=-43 $$

Step 1: For each value of i from 1 to 6, we'll compute the sum: $$ \sum_{j=1}^{7}\left(i^{2}-3 i+\delta_{i j}\right) $$ Since \(\delta_{ij}\) is equal to 1 if i==j and 0 otherwise, we will separate the sum into two parts: $$ \sum_{j=1}^{7}\left(i^{2}-3 i\right)+\sum_{j=1}^{7}\delta_{i j} $$ The first part is the sum of a constant, therefore, $$ 7\left(i^{2}-3 i\right) $$ For the second part, we can use the properties of Kronecker's delta: $$ \sum_{j=1}^{7}\delta_{i j}=\delta_{i1}+\delta_{i2}+\delta_{i3}+\delta_{i4}+\delta_{i5}+\delta_{i6}+\delta_{i7} $$ Since \(\delta_{ij}\) is equal to 1 when i == j, there will be one element in this sum equal to 1 (when j == i), and all the others will be zero. So, this sum is equal to 1. Combining the two parts, the inner sum for each i becomes: $$ \sum_{j=1}^{7}\left(i^{2}-3 i+\delta_{i j}\right) = 7\left(i^{2}-3 i\right)+1 $$ #Step 2: Compute the outer sum (over i from 1 to 6)#

Now, we'll compute the outer sum by summing the expression we found in the previous step over i from 1 to 6: $$ \sum_{i=1}^{6}\left(7\left(i^{2}-3 i\right)+1\right) $$ To compute this sum, we can distribute the sum over the terms in the expression: $$ \sum_{i=1}^{6}7\left(i^{2}-3i\right)+\sum_{i=1}^{6}1 $$ Now compute the sums: $$ 7\sum_{i=1}^{6}i^{2}-7\sum_{i=1}^{6}3i+\sum_{i=1}^{6}1 $$ Separate the constant 7 and 3 from each sum: $$ 7\left(\sum_{i=1}^{6}i^{2}-3\sum_{i=1}^{6}i\right)+6 $$ Using the sum of squares formula and sum of numbers formula, we get: $$ 7\left(\frac{6(6+1)(2(6)+1)}{6}-3\left(\frac{6(6+1)}{2}\right)\right)+6 $$ Now, simplify and calculate the result: $$ 7\left(56-3(21)\right)+6 = 7(56-63)+6 = 7(-7)+6=-49+6=-43 $$ So the final result is: $$ \sum_{i=1}^{6} \sum_{j=1}^{7}\left(i^{2}-3 i+\delta_{i j}\right)=-43 $$

Step 2: Set up the problem

The short answer for the given problem is: To evaluate the double summation, we first compute the inner sum for each \(i\), and then add up the results for the outer sum. Using the definition of \(\delta_{ij}\), we can rewrite the double summation as:

Step 3: Simplify the expression

\sum_{i=1}^{6}\left(7\left(i^{2}-3 i\right)+1\right) Computing the sum and simplifying the result, we get:

Step 4: Arrive at the final answer

\sum_{i=1}^{6} \sum_{j=1}^{7}\left(i^{2}-3 i+\delta_{i j}\right)=-43

Key Concepts

Discrete Mathematics

Discrete Mathematics is an area of mathematics that is concerned with discrete rather than continuous structures. This field encompasses a wide range of topics including logic, set theory, combinatorics, graph theory, and algorithms. One important concept in discrete mathematics is the use of summation notation to represent the sum of a sequence of terms, which leads us to another fundamental tool, the Kronecker's delta.

Kronecker's delta, denoted as \(\delta_{ij}\), is a function of two variables that is 1 if the variables are equal and 0 otherwise. It’s widely used in various branches of mathematics, including linear algebra and functional analysis, to concisely represent the fact that certain terms in a sum or matrix are only included under specific conditions. Additionally, it serves a crucial role in the study of sequences and series, which are fundamental aspects of discrete math.

The use of Kronecker’s delta in discrete mathematics is instrumental due to its ability to handle different cases within mathematical summations and products. It can simplify complex expressions and is particularly useful in problems involving sums and identities.

Summation Notation

Summation notation is used to condense the summation of sequences of numbers into a compact form. The Greek capital letter sigma \(\Sigma\) represents the summation operator. Summation notation includes an expression for the terms to be summed, an index of summation that enumerates the terms, and the upper and lower bounds of the summation index.

For instance, in the given exercise, \(\sum_{i=1}^{6} \sum_{j=1}^{7}(i^2-3i+\delta_{ij})\) uses summation notation to sum terms over the indices i and j. The inner sum, \(\sum_{j=1}^{7}(i^2-3i+\delta_{ij})\), adds up terms for a fixed i and varies j from 1 to 7, while the outer sum, \(\sum_{i=1}^{6}\), takes the result of the inner sum and accumulates it from i equals 1 to 6.

Summation notation is extremely useful for representing series, especially in Discrete Mathematics and other sciences, because it provides a clear and concise way to represent the addition of terms that follow a particular pattern. Understanding how to manipulate and evaluate sums expressed in this notation is an essential skill for students.

Mathematical Induction

Mathematical Induction is a powerful proof technique used in mathematics, particularly in Discrete Mathematics, to show that a statement is true for all natural numbers. It is based on the principle that if a statement is true for the number 1 (base case), and if assuming the statement for a number k implies that the statement is true for the number k+1 (inductive step), then the statement is true for all natural numbers.

While not directly applied in this problem, mathematical induction can be used to derive formulas for summation expressions, such as the sum of the first n squares \(\sum_{i=1}^{n}i^2\) or the sum of the first n integers \(\sum_{i=1}^{n}i\), which can then be used to evaluate sums like the ones found in the exercise. This technique involves proving that the base case (n=1) is true and then assuming the formula holds for an arbitrary positive integer k to prove it also holds for k+1.

Mathematical induction is a cornerstone of many proofs and understanding its process is an essential aspect of a mathematician's toolkit. It allows for the handling of infinite cases by reducing them to just two critical steps - the base case and the inductive step.