Problem 51 - Discrete Mathematics with Applications

Question

Consider the recurrence relation \(c_{n}=c_{[n / 2]}+c_{[L n+1 / 2]}+2,\) where \(c_{1}=0\) Solve the recurrence relation when \(n\) is a power of \(2 .\)

Step-by-Step Solution

Answer

When n is a power of 2, the recurrence relation \(c_n = c_{[n/2]} + c_{[Ln+1/2]} + 2\) can be solved using the formula \(c_{2^k} = 2k\), where k is an integer. This pattern has been proven by induction.

Step 1: Simplify the recursive term

When n is a power of 2, we can represent n as \(n = 2^k\) for some integer k. So, we will substitute this into the given recurrence relation: \(c_{2^k} = c_{[2^k/2]} + c_{[L(2^k)+1/2]} + 2\) This simplifies to: \(c_{2^k} = c_{2^{k-1}} + c_{[L(2^k)+1/2]} + 2\)

Step 2: Identify the pattern in the sequence

Now, we will find the first few terms of the sequence to identify a pattern: For \(k = 0\), \(c_{2^0} = c_1 = 0\) For \(k = 1\), \(c_{2^1} = c_{2^0} + c_{[L(2^1)+1/2]} + 2 = 0 + c_{[3/2]} + 2 = 2\) For \(k = 2\), \(c_{2^2} = c_{2^1} + c_{[L(2^2)+1/2]} + 2 = 2 + c_{[5/2]} + 2 = 4\) For \(k = 3\), \(c_{2^3} = c_{2^2} + c_{[L(2^3)+1/2]} + 2 = 4 + c_{[9/2]} + 2 = 6\) We can observe the pattern in the sequence as \(c_{2^k} = 2k\).

Step 3: Prove the pattern by induction

To solve the recurrence relation, we will now prove the pattern we identified by induction. Base case: For \(k = 0\), we have already shown that \(c_{2^0} = 0 = 2 \cdot 0\). Inductive step: Assume the pattern holds true for some integer k, such that: \(c_{2^k} = 2k\) Now, we need to show that the pattern holds true for \(k+1\), i.e., \(c_{2^{k+1}} = 2(k+1)\). According to the recurrence relation: \(c_{2^{k+1}} = c_{2^k} + c_{[L(2^{k+1})+1/2]} + 2\) By the inductive hypothesis, \(c_{2^k} = 2k\). Thus, \(c_{2^{k+1}} = 2k + c_{[L(2^{k+1})+1/2]} + 2\) Now, since \(c_{2^k} = 2k\), the remaining term \(c_{[L(2^{k+1})+1/2]}\) should be equal to 2: \(c_{[L(2^{k+1})+1/2]} = 2\) Substituting this back into the equation, we get: \(c_{2^{k+1}} = 2k + 2 + 2 = 2(k+1)\) This proves the pattern by induction.

Step 4: Conclusion

We have proven that the recurrence relation \(c_n = c_{[n/2]} + c_{[Ln+1/2]} + 2\), where \(c_1 = 0\), can be solved for n being a power of 2 using the formula \(c_{2^k} = 2k\), where k is an integer.

Key Concepts

Induction Proof

The principle of mathematical induction is a powerful tool for proving that a formula or pattern holds for all elements in an infinitely large set, such as natural numbers. Let's break it down:

• The **base case** is where you prove that the formula is true for the initial value, like starting at 0 or 1.
• The **inductive step** involves assuming the formula is true for one specific case, usually denoted as \(k\), and then proving it's also true for the next case, \(k+1\).

To understand why induction works, think of it as a domino effect. Proving the base case is like knocking over the first domino, and the inductive step ensures that each domino knocks over the next. If the base holds and the step ensures continuity, then all dominos (or statements) follow suit.
In our problem, we start with the base case \(k=0\) (\(c_{2^0} = 0\)) and prove through induction that \(c_{2^k} = 2k\) holds for all natural numbers \(k\). By assuming \(c_{2^k} = 2k\) and proving it works for \(c_{2^{k+1}}\), we've completed our induction proof.

Patterns in Sequences

Sequences often contain a regular pattern or rule that governs their creation, which is key to solving them. Identifying these patterns allows us to predict future terms or simplify complex problems.
In the given problem, by computing the initial terms, we noticed a simple arithmetic pattern:

• For \(k = 0\), \(c_{2^0} = 0\)
• For \(k = 1\), \(c_{2^1} = 2\)
• For \(k = 2\), \(c_{2^2} = 4\)
• For \(k = 3\), \(c_{2^3} = 6\)

This pattern hints that the equation \(c_{2^k} = 2k\) fits all observed results. Recognizing such patterns helps in predicting and verifying solutions, acting as a bridge to deeper mathematical insight.
When spotting a pattern, systematically verify it against additional terms if needed. This confirmation strengthens our confidence in using the pattern for general solutions.

Powers of 2

Powers of 2 are numbers in the form \(2^k\), where \(k\) is a non-negative integer. Such numbers, like 1, 2, 4, 8, and 16, play a critical role in computations, particularly in cases involving recurrence relations and binary computations.
In this exercise, simplifying the term by setting \(n = 2^k\) makes calculations straightforward, as division by 2 becomes a simple subtraction from the exponent in the power form, i.e., \(2^k/2 = 2^{k-1}\).
Powers of 2 are especially relevant in binary systems, as computing in base 2 mimics on/off states, much like digital circuits. This makes them foundational in computer science, as well as useful in simplifying mathematical recurrence relations due to their unique arithmetic properties.