Problem 50 - Discrete Mathematics with Applications

Question

Consider the recurrence relation \(c_{n}=c_{\lfloor n / 2\rfloor}+c_{\lfloor(n+1) / 2 j}+2,\) where \(c_{1}=0\). Compute \(c_{3}\) and \(c_{4}\)

Step-by-Step Solution

Answer

We are given a recurrence relation \(c_n = c_{\lfloor n / 2 \rfloor} + c_{\lfloor(n+1) / 2 \rfloor} + 2,\) with base case \(c_1 = 0\). By calculating sequentially, we find that \(c_2 = 2\), \(c_3 = 4\), and \(c_4 = 6\). Thus, \(c_{3} = 4\) and \(c_{4} = 6\).

Step 1: Understand and write down given information

We are given a recurrence relation: \[c_n = c_{\lfloor n / 2 \rfloor} + c_{\lfloor(n+1) / 2 \rfloor} + 2,\] and the base case, for which \(c_1 = 0\). We need to compute \(c_3\) and \(c_4\).

Step 2: Compute \(c_2\)

To compute \(c_3\) and \(c_4\), we will first need to calculate \(c_2\). Using the given recurrence relation: \[c_2 = c_{\lfloor 2 / 2 \rfloor} + c_{\lfloor(2+1) / 2 \rfloor} + 2.\] Now, using the base case, we know that \(c_1 = 0\): \[c_2 = c_1 + c_1 + 2 = 0 + 0 + 2 = 2.\] So, \(c_2 = 2\).

Step 3: Compute \(c_3\)

Now we can use the calculated value of \(c_2\) to compute \(c_3\). Using the given recurrence relation: \[c_3 = c_{\lfloor 3 / 2 \rfloor} + c_{\lfloor(3+1) / 2 \rfloor} + 2.\] Plugging in the values of the calculated \(c_1\) and \(c_2\): \[c_3 = c_1 + c_2 + 2 = 0 + 2 + 2 = 4.\] Therefore, \(c_3 = 4\).

Step 4: Compute \(c_4\)

Lastly, we can use the calculated values of \(c_1\) and \(c_2\) to compute \(c_4\). Using the given recurrence relation: \[c_4 = c_{\lfloor 4 / 2 \rfloor} + c_{\lfloor(4+1) / 2 \rfloor} + 2.\] And plugging in the values of \(c_1\) and \(c_2\): \[c_4 = c_2 + c_2 + 2 = 2 + 2 + 2 = 6.\] Thus, \(c_4 = 6\).

Step 5: Conclusion

Using the given recurrence relation, we have determined that \(c_{3} = 4\) and \(c_{4} = 6\).

Key Concepts

Discrete Mathematics

Discrete mathematics deals with mathematical structures that are fundamentally discrete rather than continuous. This means it involves countable, distinct, and separate objects or events. The field is vast and includes topics such as logic, set theory, graph theory, and recursion. Recurrence relations, which we are focusing on here, are an essential part of discrete mathematics, as they describe sequences and help us find terms of these sequences by relying on previous terms.

• These relations provide a recursive method to define sequences.
• They express each term as a function of its predecessors.

In our exercise, we use a recurrence relation to find terms like \(c_3\) and \(c_4\). This example showcases how discrete mathematics enables us to build complex sequences from simple rules iteratively.

Base Case

The base case is a crucial component of any recursive definition, including recurrence relations. It serves as the starting point and is essential for ensuring that the recursive process can begin and terminate correctly. Without a base case, a recurrence would endlessly cycle without producing a tangible result.

• It initializes the sequence with known values.
• The base case prevents infinite recursion by providing a defined stopping condition.

In the problem above, the base case is defined as \(c_1 = 0\). This provides the initial value needed to compute subsequent terms like \(c_2\), \(c_3\), and \(c_4\) using the recurrence relation. Understanding the base case allows us to follow the sequence from a point of certainty.

Computation of Terms

The computation of terms in a sequence involves applying the recurrence relation iteratively, often starting from the base case. This method breaks down the process into simpler steps that are easier to manage:

• Start with the base case—here, \(c_1 = 0\).
• Use the recurrence relation to compute \(c_2 = 2\), \(c_3 = 4\), and \(c_4 = 6\).

This step-wise computation ensures we correctly apply the formula at each stage, using previously calculated terms. For instance, \(c_2\) was calculated using \(c_1\), while \(c_3\) depends on both \(c_1\) and \(c_2\). By systematically deriving terms in this manner, we can effectively understand and solve problems involving recurrence relations.

Floor Function

The floor function is a mathematical operation that takes a real number as input and gives the greatest integer less than or equal to that number as output. It is symbolized by \(\lfloor x \rfloor\) and is pivotal in the recurrence relations like the one in our exercise.

• Used to round down to the nearest whole number.
• Ensures integer results in calculations involving division.

In our problem, the floor function is utilized to determine terms such as \(c_{\lfloor n/2 \rfloor}\) and \(c_{\lfloor (n+1)/2 \rfloor}\). For instance, \(\lfloor 3/2 \rfloor\) results in \(1\), guiding us in the right direction when computing sequence values that demand integer inputs. Understanding how to apply the floor function ensures precise calculation and progression through the recurrence relation.