Problem 50 - Discrete Mathematics with Applications

Question

Consider the recurrence relation \(c_{n}=c_{[n / 2]}+c_{[L n+1 / 2]}+2,\) where \(c_{1}=0\) Compute \(c_{3}\) and \(c_{4}\)

Step-by-Step Solution

Answer

Using the given recurrence relation, we computed the values of \(c_3\) and \(c_4\). The results are \(c_3=4\) and \(c_4=6\).

Step 1: Evaluate \(c_{3}\) #

Step 1: Evaluate \(c_{[n / 2]}\) For \(n = 3\), we have \(c_{[3/2]} = c_{1}\). Since \(c_{1} = 0\), we get \(c_{[3/2]} = 0\). Step 2: Evaluate \(c_{[Ln+1 / 2]}\) For \(n = 3\), we have \(c_{[3+1 / 2]} = c_{2}\). As this is a new term, we calculate \(c_{2}\) using the recurrence relation. Step 1 of \(c_{2}\) calculation: \(c_{[2/2]} = c_{[1]} = c_1 = 0\) Step 2 of \(c_{2}\) calculation: \(c_{[2+1/2]} = c_{[3/2]} = c_{1} = 0\) So, \(c_{2} = 0 + 0 + 2 = 2\), Giving us, \(c_{[3+1 / 2]} = c_{2} = 2\). Step 3: Calculate \(c_{3}\) By adding the results from Step 1 and Step 2 along with the constant 2, we get: \(c_{3} = c_{[3/2]} + c_{[3+1 / 2]} + 2 = 0 + 2 + 2 = 4\).

Step 2: Evaluate \(c_{4}\) #

Step 1: Evaluate \(c_{[n / 2]}\) For \(n = 4\), we have \(c_{[4 / 2]} = c_{2}\). We have already calculated \(c_{2} = 2\), so we get \(c_{[4 / 2]} = 2\). Step 2: Evaluate \(c_{[Ln+1 / 2]}\) For \(n = 4\), we have \(c_{[4+1 / 2]} = c_{[5/2]} = c_{2}\). As we have already calculated \(c_{2} = 2\), so we get \(c_{[4+1 / 2]} = c_{2} = 2\). Step 3: Calculate \(c_{4}\) By adding the results from Step 1 and Step 2 along with the constant 2, we get: \(c_{4} = c_{[4/2]} + c_{[4+1 / 2]} + 2 = 2 + 2 + 2 = 6\). So, our final results are \(c_3=4\) and \(c_4=6\).

Key Concepts

Discrete Mathematics

Discrete mathematics is a branch of mathematics dealing with distinct and separate values or objects. Unlike continuous mathematics, which involves concepts that can change smoothly, discrete mathematics works with concepts that have clear, separate states.
In problems involving recurrence relations, such as the one presented in this exercise, discrete mathematics helps us analyze sequences and patterns using specific rules and assumptions.

In this given exercise, the sequence of numbers like \(c_1, c_2, c_3,\) and \(c_4\) is an example of a discrete set, as each term is distinct and derived using a formulaic or algorithmic approach. Discrete mathematics provides the framework and tools, such as algorithms, graphs, and logical statements, to solve these problems.

Integer Division

Integer division refers to a mathematical operation where two integers are divided to yield a quotient with no fractional part. The operation discards the remainder to provide an integer result instead of a floating-point number. It is fundamental in areas of discrete mathematics, especially when dealing with recurrence relations.

The exercise uses integer division in expressions like \([n/2]\) and \([Ln+1/2]\). In these terms, the brackets indicate that after dividing, the result should be rounded down to the nearest whole number if it is not already an integer. This concept is important for understanding how to calculate terms in a recurrence relation accurately.

• When \(n=3\), \([3/2]=1\) because dividing 3 by 2 yields 1.5, and taking the floor yields 1.
• Similarly, when \(n=4\), \([4/2]=2\), which divides evenly without rounding.

This operation is vital to ensure that calculations in the problem remain within distinct and structured integer values, a pillar in discrete mathematics.

Recursive Function

Recursive functions are functions that solve problems by calling themselves with simpler or smaller inputs. This method breaks down complex problems into manageable parts, allowing programmers or mathematicians to solve these problems step by step.
In the context of the recurrence relation \(c_n=c_{[n/2]}+c_{[Ln+1/2]}+2\), each term is defined in terms of earlier terms. It's a classic example of recursion, where you compute each value starting from known base cases.

For example, in the exercise:

• \(c_1 = 0\) is the base case which means it's directly known.
• \(c_2\) is computed using \(c_1\), and this process continues for \(c_3\) and \(c_4\).

This recursive approach helps in building towards a solution incrementally, from known values like \(c_1\). Recursive functions often use base cases, just like \(c_1\), to stop the infinite recursive calls and begin building solutions. This method is efficient for defining solutions in a structured manner.

Mathematical Induction

Mathematical induction is a powerful proof technique often used to prove statements or formulas about integers or sequences. While not directly applied in the solution of the given problem, understanding this concept allows for future validations of recurrence relations and their properties.
This technique consists of two main steps:

• Base Case: Verify that the statement holds for the initial value. Here, confirming that \(c_1 = 0\) satisfies the recurrence relation's foundation could be akin to this step.
• Inductive Step: Assume it holds for an arbitrary case \(n=k\), and then show it must hold for \(n=k+1\). For the problem here, assuming the formula works up to \(c_n\), you prove it extends to \(c_{n+1}\).

Inductive reasoning provides assurance about the correctness and the continuation of the recursive sequence or formula across all desired values of \(n\). Knowing induction helps understand and reinforce concepts in recursive relations.