Problem 24 - Discrete Mathematics with Applications

Question

Solve each LHRRWCC. $$\begin{aligned} &a_{n}=8 a_{n-1}-24 a_{n-2}+32 a_{n-3}-16 a_{n-4}, a_{0}=1, a_{1}=4, a_{2}=44\\\ &a_{3}=272 \end{aligned}$$

Step-by-Step Solution

Answer

The closed-form solution for the given Linear Homogeneous Recurrence Relation with Constant Coefficients is: \(a_n = (3 - n) - 4(2)^n + 3(4)^n\)

Step 1: Find the characteristic equation

To find the characteristic equation for the given recurrence relation, replace each term a_n, a_{n-1}, a_{n-2}, a_{n-3}, a_{n-4} with r^n, r^{n-1}, r^{n-2}, r^{n-3}, r^{n-4} respectively. The characteristic equation for the given recurrence relation will be: \(r^n=8r^{n-1}-24r^{n-2}+32r^{n-3}-16r^{n-4}\) Divide both sides by \(r^{n-4}\), the equation becomes: \(r^4=8r^3-24r^2+32r-16\)

Step 2: Solve the characteristic equation

To find the roots of the characteristic equation, we can rewrite it as: \(r^4 - 8r^3 + 24r^2 - 32r + 16 = 0\) Now, we will factor the equation using your favorite factoring method, in this case, we can directly observe that this can be factored as: \((r-1)(r-2)(r-4)(r-1) = 0\) This gives us roots r=1 (multiplicity 2), r=2, and r=4.

Step 3: Formulate the general solution

Using the roots and their multiplicities, we can formulate the general solution for the given recurrence relation: \(a_n = (c_1 + c_2n)r_1^n + c_3r_2^n + c_4r_3^n\) \(a_n = (c_1 + c_2n)(1)^n + c_3(2)^n + c_4(4)^n\)

Step 4: Solve for constants using initial conditions

Now, we will use the given initial conditions to solve for the constants c_1, c_2, c_3, and c_4. For \(n=0\), \(a_0=1\): \(1 = c_1 + 0 + c_3 + c_4\) For \(n=1\), \(a_1=4\): \(4 = (c_1 + c_2) + 2c_3 + 4c_4\) For \(n=2\), \(a_2=44\): \(44 = (c_1 + 2c_2) + 4c_3 + 16c_4\) For \(n=3\), \(a_3=272\): \(272 = (c_1 + 3c_2) + 8c_3 + 64c_4\) Solving this system of linear equations, we get: \(c_1 = 3, c_2 = -1, c_3 = -4, c_4 = 3\)

Step 5: Write the final solution

Using the values of the constants, we can now write the final closed-form solution for the recurrence relation: \(a_n = (3 - n)(1)^n - 4(2)^n + 3(4)^n\) Thus, the closed-form solution for the given Linear Homogeneous Recurrence Relation with Constant Coefficients is: \(a_n = (3 - n) - 4(2)^n + 3(4)^n\)

Key Concepts

Characteristic Equation

In the world of linear recurrence relations, the characteristic equation is a critical tool that aids in determining the behavior of the sequence solutions. To derive the characteristic equation, we substitute each term in the recurrence relation with powers of a common variable, typically denoted as \( r \). For the given recurrence relation:
\[ a_n = 8a_{n-1} - 24a_{n-2} + 32a_{n-3} - 16a_{n-4} \]
we replace \( a_n \) with \( r^n \), \( a_{n-1} \) with \( r^{n-1} \), and so on. After simplifying, we arrive at the characteristic polynomial:
\[ r^4 = 8r^3 - 24r^2 + 32r - 16 \]
which further simplifies to:
\[ r^4 - 8r^3 + 24r^2 - 32r + 16 = 0 \]
This equation's roots help us understand the form of the general solution.

General Solution

The general solution of a linear recurrence relation is constructed using the roots of its characteristic equation. These roots can have different behaviors: they could be distinct, real, complex, or have multiplicities. For the equation:
\[ (r-1)(r-2)(r-4)(r-1) = 0 \]
we find the roots: \( r = 1 \) with multiplicity 2, \( r = 2 \), and \( r = 4 \). The multiplicity of a root affects the general solution structure. Here, the general solution is expressed as:

• \( a_n = (c_1 + c_2 n)(1)^n + c_3(2)^n + c_4(4)^n \)

The terms \( c_1 + c_2 n \) reflect the root \( r = 1 \) repeating twice. The constants \( c_1, c_2, c_3, \) and \( c_4 \) are determined using initial conditions.

Initial Conditions

Initial conditions are essential for solving recurrence relations because they allow us to find specific values for the arbitrary constants in the general solution. They provide starting points for the sequence. For this particular problem, the initial conditions given are:

• \( a_0 = 1 \)
• \( a_1 = 4 \)
• \( a_2 = 44 \)
• \( a_3 = 272 \)

Using these initial values, we substitute back into the general solution equations to create a system of linear equations. Solving these equations will give us:

• \( c_1 = 3 \)
• \( c_2 = -1 \)
• \( c_3 = -4 \)
• \( c_4 = 3 \)

These defined constants allow us to specify the particular solution that meets these starting conditions.

Closed-form Solution

A closed-form solution is a neatly packaged formula that expresses the terms of a sequence without requiring previous terms. It's particularly useful because it provides a direct way to compute any term \( a_n \) in the sequence. For our recurrence relation, the closed-form solution, using our calculated constants, is:
\[ a_n = (3 - n) - 4(2)^n + 3(4)^n \]
In this solution:

• The term \((3 - n)\) correlates with the repeated root \( r = 1 \).
• The term \(- 4(2)^n\) corresponds to the root \( r = 2 \).
• The term \(+ 3(4)^n\) corresponds to the root \( r = 4 \).

This closed-form allows us to immediately compute values of \( a_n \) for any \( n \), providing a powerful tool to understand the sequence behavior across all indices.