Problem 25 - Discrete Mathematics with Applications

Question

Evaluate each sum, where \(\delta_{i j}\) is defined as follows. $$ \delta_{i j}=\left\\{\begin{array}{ll} 1 & \text { if } i=j \\ 0 & \text { otherwise } \end{array}\right. $$ $$\sum_{i=1}^{6} \sum_{j=1}^{5}\left(i^{2}-i\right)$$

Step-by-Step Solution

Answer

The short answer is: Evaluating the given sum $$\sum_{i=1}^{6} \sum_{j=1}^{5}\left(i^{2}-i\right)$$, we find that it is equal to \(\boxed{350}\).

Step 1: Understand Nested Summations

The two summations can be represented as: $$\sum_{i=1}^{6} \left(\sum_{j=1}^{5}\left(i^{2}-i\right)\right)$$. We need to first evaluate the expression inside the parenthesis for each value of \(i\) and \(j\) and sum up the results.

Step 2: Simplify the Inner Summation

Since \(j\) does not appear anywhere in the expression inside the inner summation, i.e., \((i^2 - i)\) is not a function of \(j\), we can simply multiply the value of \((i^2 - i)\) by the number of times it will be summed in the inner summation. $$\sum_{i=1}^{6} \left(5(i^{2}-i)\right)$$

Step 3: Simplify the Outer Summation

Now, evaluate the outer summation by substituting the values of \(i\) and summing the terms. $$5\left(\sum_{i=1}^{6}(i^2 - i)\right)$$ $$= 5\left((1^2 - 1) + (2^2 - 2) + (3^2 -3) + (4^2 - 4) + (5^2 - 5) + (6^2 - 6)\right)$$ $$= 5\left(0 + 2 + 6 + 12 + 20 + 30\right)$$ $$= 5\left(70\right)$$

Step 4: Calculate the Final Sum

Compute the final product. $$= 5 \cdot 70 = 350$$ Thus, the evaluated sum is \(\boxed{350}\).

Key Concepts

Summation

Summation is a fundamental operation in mathematics where you add a sequence of numbers or expressions. It's denoted by the sigma symbol, \( \Sigma \), and is a concise way to represent repeated addition of terms. In this exercise, we have a double summation to evaluate, which involves two layers of summation. Each term in the sequence is derived from a formula substituted with the index values.

• The inner summation deals with a fixed expression over a range of index values.
• The outer summation accumulates the results of the inner summation over its range.

In our specific problem, the summation goes over a range from 1 to 6 for \( i \) and 1 to 5 for \( j \). Evaluating each part carefully leads us to the final result.
Summations are a powerful tool to simplify expressions, especially when they involve repeated patterns or identical calculations over different indices.

Function Evaluation

Function evaluation involves calculating the value of a mathematical function given specific inputs. In discrete mathematics, these functions can often be algebraic expressions that require substitution of index or variable values.
For example, consider the expression \( i^2 - i \) used in our exercise, which is independent of \( j \), meaning it remains constant for each inner calculation loop.

• Substitute the index value (\( i \)) into the expression \( i^2 - i \).
• Calculate the result for each \( i \) value in the range.
• The resulting values are then used in the summation processes.

This step of evaluating the function simplifies the complexity by breaking it into manageable calculations, paving the way for easier summation.

Nested Summation

Nested summation refers to the scenario where one summation exists within another. It's a common situation in more complex mathematical problems and involves iterating through multiple levels of indices. In our exercise:

• The expression inside the outer summation \( \sum_{i=1}^{6} \) includes another summation over \( j \) from 1 to 5, \( \sum_{j=1}^{5}(i^2 - i) \).
• Since this expression doesn't involve \( j \), we can treat the inner summation as simply multiplying \( i^2 - i \) by 5.

Then, we carry this simplified result to the outer summation, summing over all \( i \).
By evaluating from the innermost layer outward, we leverage the opportunity to simplify nested processes significantly, reducing complexity step by step. This technique is crucial in calculus and algebra when working with large datasets or complex mathematical models. Nested summations can transform intricate calculations into a series of simpler steps, allowing for efficient and accurate problem-solving.