Problem 41 - Discrete Mathematics with Applications

Question

Let \(r_{n}\) and \(s_{n}\) be two solutions of the recurrence relation \((5.8) .\) Prove that \(a_{n}=r_{n}+s_{n}\) is also a solution.

Step-by-Step Solution

Answer

In this exercise, we showed that \(a_n = r_n + s_n\), with \(r_n\) and \(s_n\) being solutions to the recurrence relation (5.8), is also a solution to the given recurrence relation. We proved this by demonstrating that \(a_n\) can be represented as a function of the sum of the previous terms of \(r_n\) and \(s_n\), satisfying the linear property of the given recurrence relation.

Step 1: Define the general form of a recurrence relation

A recurrence relation can be represented as: \[a_n = f(n, a_{n-1}, a_{n-2},..., a_{n-k})\] where \(a_n\) is the nth term of the sequence, \(f\) is a function, \(k\) is a positive integer, and \(a_{n-1}, a_{n-2},..., a_{n-k}\) are the previous terms in the sequence. We are given that \(r_n\) and \(s_n\) are solutions to the recurrence relation (5.8), so: \[r_n = f(n, r_{n-1}, r_{n-2},..., r_{n-k})\] \[s_n = f(n, s_{n-1}, s_{n-2},..., s_{n-k})\] We want to prove that \(a_n = r_n + s_n\) is also a solution.

Step 2: Show that \(a_n\) satisfies the conditions of the recurrence relation

Let's consider the sum of the equations for \(r_n\) and \(s_n\): \[a_n = r_n + s_n = f(n, r_{n-1}, r_{n-2},..., r_{n-k}) + f(n, s_{n-1}, s_{n-2},..., s_{n-k})\] Now, we can define the new sequence \(w_n\) as the sum of the two previous sequences: \[w_n = r_{n-1} + s_{n-1}\] We need to show that \(a_n\) can be represented as a function of \(w_n\): \[a_n = f(n, w_{n-1}, w_{n-2},..., w_{n-k})\] However, since \(r_n\) and \(s_n\) are solutions to the same recurrence relation, the function \(f\) satisfies the linear property, meaning that: \[f(n, r_{n-1} + s_{n-1}, r_{n-2} + s_{n-2},..., r_{n-k} + s_{n-k}) = f(n, r_{n-1}, r_{n-2},..., r_{n-k}) + f(n, s_{n-1}, s_{n-2},..., s_{n-k})\] Thus, we can write \(a_n\) as: \[a_n = f(n, r_{n-1} + s_{n-1}, r_{n-2} + s_{n-2},..., r_{n-k} + s_{n-k}) = f(n, w_{n-1}, w_{n-2},..., w_{n-k})\]

Step 3: Conclude that \(a_n\) is a solution to the given recurrence relation

We have demonstrated that \(a_n\), defined as the sum of \(r_n\) and \(s_n\), can be represented as a function of \(w_n\), which is defined as the sum of the previous terms of \(r_n\) and \(s_n\). Therefore, we have shown that \(a_n\) is also a solution to the same recurrence relation as \(r_n\) and \(s_n\), proving the statement in the exercise.

Key Concepts

Solutions of Recurrence Relations

When dealing with recurrence relations, the goal is to find the terms in a sequence based on previous terms. Imagine you have a sequence where each term can be determined by a specific rule. This rule is expressed in the form of a recurrence relation. In mathematics, a recurrence relation is an equation that recursively defines a sequence. Each term is a function of the preceding terms.

To solve recurrence relations, we look for solutions that satisfy these rules for every term in the sequence. When given a recurrence relation like \(a_n = f(n, a_{n-1}, a_{n-2}, \ldots, a_{n-k})\), a possible solution would involve finding such functions that work universally across the sequence.

For example, if you find solutions \ r_n \ and \ s_n \ for your recurrence, you might want to check if their sum \ a_n = r_n + s_n \ is also a solution. It would involve verifying that this new sequence fits into the original relation used by \ r_n \ and \ s_n \. When you're able to combine solutions like this, you're employing a powerful technique in addressing these mathematical mayhem of sequences.

Linear Recurrence

A linear recurrence is a specific type of recurrence relation where the next term in the sequence is a linear combination of previous terms. This means you will often see operations like addition or multiplication by constants being used in obtaining new terms from old ones.

Consider any relation of the form \(a_n = f(n, a_{n-1}, a_{n-2}, \ldots, a_{n-k})\) to be linear if for any sequences \ r_n \ and \ s_n \, as solutions, the property holds: \ f(n, r_{n-1} + s_{n-1}, \ldots) = f(n, r_{n-1}, \ldots) + f(n, s_{n-1}, \ldots)\. This linear property allows for simple additions or scaling of sequence elements without complication.

The beauty of linear recurrences is that it allows for straightforward combinations of known solutions to derive new ones, as seen when you sum two solutions \ r_n \ and \ s_n \ to get a new one \ a_n = r_n + s_n \. This simplicity in operations makes working with linear recurrences quite manageable compared to non-linear ones.

Sequence Functions

Sequence functions are essential in expressing how terms in a sequence relate to one another. For a given recurrence relation, the function \ f \ denotes the rule or the formula which binds terms of the sequence. These functions are key in dictifying how one term evolves from others.

When working with sequence functions, consider how changing parameters modifies the sequence. In any sequence where \(a_n = f(n, a_{n-1}, a_{n-2}, \ldots, a_{n-k})\), you can think of \ f \ as a machine that transforms previous states \(a_{n-1}, a_{n-2}, \ldots\) into a new state \ a_n \.

Especially in the context of linear recurrences, these functions often take on simpler forms like polynomials or simple arithmetic expressions. Knowing \ f \ well guarantees you can calculate future terms with ease and analyze the sequence comprehensively. This understanding is vital for predictive modeling wherever sequences are used, from computer science algorithms to forecasting models in economics.