Problem 22 - Discrete Mathematics with Applications

Question

Using induction, prove that each is a solution to the corresponding recurrence relation, where \(c\) is a constant and \(f(n)\) a function of \(n\). $$-a_{n}=c^{n} a_{0}+\frac{c^{n}-1}{c-1}, a_{n}=c a_{n-1}+1 \text { (assume } c \neq 1)$$

Step-by-Step Solution

Answer

To prove that the given function \(a_n = c^n a_0 + \frac{c^n - 1}{c - 1}\) is a solution to the recurrence relation \(a_n = ca_{n-1} + 1\), we use mathematical induction. First, we prove the base case for n = 0; then we assume the given function is true for n = k and show it also holds for n = k + 1. Following these steps, for n = 0, we find that the initial values match, proving the base case. Next, assuming the given function is true for n = k, i.e., \(a_k = c^k a_0 + \frac{c^k - 1}{c - 1}\), we use the given recurrence relation to calculate \(a_{k+1}\) and find that it matches the expression for the given function for n = k + 1. Through mathematical induction, we thus prove that the given function is indeed a solution to the recurrence relation.

Step 1: Base Case n = 0

Let's first check if the initial values match. Given: \(a_n = c^n a_0 + \frac{c^n - 1}{c - 1}\) For n = 0, we have: \(a_0 = c^0 a_0 + \frac{c^0 - 1}{c - 1}\) Now, \(c^0 = 1\). So, \(a_0 = a_0 + \frac{1 - 1}{c - 1}\) This simplifies to \(a_0 = a_0\), which is true. Thus, the base case holds.

Step 2: Induction Hypothesis

Now, let's assume the given function is true for n = k: \(a_k = c^k a_0 + \frac{c^k - 1}{c - 1}\)

Step 3: Prove It for n = k + 1

We need to show that the given function is also true for \(n = k + 1\), i.e., we need to prove the following: \(a_{k+1} = c^{k+1} a_0 + \frac{c^{k+1} - 1}{c - 1}\) Now, using the given recurrence relation, we have: \(a_{k+1} = c a_k + 1\) We can plug in our induction hypothesis (\(a_k = c^k a_0 + \frac{c^k - 1}{c - 1}\)): \(a_{k+1} = c\left(c^k a_0 + \frac{c^k - 1}{c - 1}\right) + 1\)

Step 4: Simplify the Expression

Let's simplify the expression for \(a_{k+1}\): \(a_{k+1} = c^{k+1} a_0 + \frac{c^{k+1} - c}{c - 1} + 1\) Now, we can combine the two fractions on the right: \(a_{k+1} = c^{k+1} a_0 + \frac{c^{k+1} - c + c - 1}{c - 1}\) This simplifies to: \(a_{k+1} = c^{k+1} a_0 + \frac{c^{k+1} - 1}{c - 1}\) This is exactly the expression we wanted to prove for \(n = k + 1\). Thus, the given function \(a_n\) is indeed a solution to the recurrence relation by mathematical induction.

Key Concepts

Recurrence Relations

Recurrence relations are mathematical formulas that define a sequence of values using preceding terms. They express each term in a series as a function of its predecessors, allowing us to understand complex sequences using simple equations. For example, the given recurrence relation is \( a_n = c a_{n-1} + 1 \), which suggests that each term \( a_n \) can be calculated from the previous term \( a_{n-1} \) using the constant \( c \). The solution to this recurrence relation was shown as \( a_n = c^n a_0 + \frac{c^n - 1}{c - 1} \), providing a direct formula to calculate any term in the sequence. These equations are powerful for modeling sequences in mathematics and computer science. Understanding how to manipulate them is crucial for solving complex problems.

Base Case

A base case is an essential part of mathematical induction. It serves as the foundation for proving that an inductive statement holds true for all natural numbers. In our example, the base case is tested for \( n = 0 \). We prove that the initial condition \( a_0 = c^0 a_0 + \frac{c^0 - 1}{c - 1} \) holds true. Simplifying this, we find \( a_0 = a_0 \), confirming that the initial step is valid. The importance of the base case cannot be overstated, as it serves as the starting point for the inductive process. Without a true base case, the induction cannot proceed.

Induction Hypothesis

The induction hypothesis is a key step in the proof of mathematical induction. It's an assumption that the given statement is true for an arbitrary natural number \( n = k \). In this exercise, the induction hypothesis assumed \( a_k = c^k a_0 + \frac{c^k - 1}{c - 1} \). This assumption is pivotal, as it's the logical bridge that allows us to prove something for one case and then generalize it for the next case.

• It provides a temporary assumption for a particular step \( n = k \).
• Allows proving the statement for \( n = k + 1 \) using this assumption.

Understanding how to properly state and use an induction hypothesis is crucial in facilitating the proof of statements that are believed to be always true.

Proof by Induction

Proof by induction is a fundamental technique in mathematics, used to demonstrate that a proposition is true for all natural numbers. It involves two key steps: the base case and the induction step. In our example, we began with the base case proving \( n = 0 \) and then moved on to the induction step. There, assuming the hypothesis \( a_k = c^k a_0 + \frac{c^k - 1}{c - 1} \) is true for \( n = k \), we show it must also be true for \( n = k + 1 \). By demonstrating that when a statement holds for any arbitrary number \( k \), it also holds for \( k + 1 \), we conclude it is valid for all \( n \).

• Start by validating the simplest case (base case).
• Assume it holds for an arbitrary step.
• Show this leads to it holding for the next step.

Proof by induction assures that what is proven true for some initial values will be true perpetually within defined parameters.