Problem 23 - Discrete Mathematics with Applications

Question

Solve each LHRRWCC. $$\begin{array}{l}{a_{n}=9 a_{n-1}-30 a_{n-2}+44 a_{n-3}-24 a_{n-4}, a_{0}=5, a_{1}=12, a_{2}=38}{a_{3}=126}\end{array}$$

Step-by-Step Solution

Answer

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

Step 1: Find the characteristic equation

The characteristic equation is formed by replacing each \(a_n\) term in the LHRRWCC with its corresponding power of \(r\). The characteristic equation for our given LHRRWCC is: \[r^n - 9r^{n-1} + 30r^{n-2} - 44r^{n-3} + 24r^{n-4} = 0\]

Step 2: Solve for the roots of the characteristic equation

We can make our equation polynomial by factoring out \(r^{n-4}\): \[r^{n-4}(r^4 - 9r^3 + 30r^2 - 44r + 24) = 0\] Since the LHRRWCC is a fourth-order equation, we should expect four roots for our characteristic equation. We notice that the polynomial \(r^4 - 9r^3 + 30r^2 - 44r + 24\) can be factored to \((r-2)(r-3)(r-4)^2\). So, the roots are \(r_1 = 2, r_2 = 3,\) and \(r_3 = 4\) (with multiplicity 2).

Step 3: Form the general solution

Based on the roots found, the general solution will be of the form: \[a_n = C_1(2)^n + C_2(3)^n + C_3(4)^n + C_4n(4)^n\] Here, \(C_1\), \(C_2\), \(C_3\), and \(C_4\) are constants that we'll find next using the initial values.

Step 4: Apply the initial values to find the specific solution

We can use the initial values to find the values of our constants and get the specific solution for the general formula. For \(n = 0\): \[5 = a_0 = C_1(2)^0 + C_2(3)^0 + C_3(4)^0 + C_4(0)(4)^0\] \[5 = C_1 + C_2 + C_3\] For \(n = 1\): \[12 = a_1 = C_1(2)^1 + C_2(3)^1 + C_3(4)^1 + C_4(1)(4)^1\] \[12 = 2C_1 + 3C_2 + 4C_3 + 4C_4\] For \(n = 2\): \[38 = a_2 = C_1(2)^2 + C_2(3)^2 + C_3(4)^2 + C_4(2)(4)^1\] \[38 = 4C_1 + 9C_2 + 16C_3 + 16C_4\] For \(n = 3\): \[126 = a_3 = C_1(2)^3 + C_2(3)^3 + C_3(4)^3 + C_4(3)(4)^2\] \[126 = 8C_1 + 27C_2 + 64C_3 + 48C_4\] Now, solve this system of linear equations using any suitable method to find the constants \(C_1, C_2, C_3,\) and \(C_4\). After solving, we find that \(C_1 = 1, C_2 = 2, C_3 = 2,\) and \(C_4 = -1\).

Step 5: Final Answer

The specific solution for the given LHRRWCC is: \[a_n = (2)^n + 2(3)^n + 2(4)^n - n(4)^n\]

Key Concepts

Characteristic Equation

The characteristic equation bears great significance when solving linear homogeneous recurrence relations (LHRR). It functions as a bridge, translating the recurrence relation into an algebraic format that we can easily manage.

To construct a characteristic equation, we mimic the structure of the LHRR: each recurrence term corresponds to a power of an unknown value, usually denoted by 'r'. For the LHRR given by the relationship a_n = 9a_{n-1} - 30a_{n-2} + 44a_{n-3} - 24a_{n-4}, the characteristic equation takes the form r^4 - 9r^3 + 30r^2 - 44r + 24 = 0. Finding the roots of this polynomial will be pivotal in crafting the general solution to the original recurrence relation.

Roots of Characteristic Equation

Once we have the characteristic equation, the next crucial step is discovering its roots, as these will dictate the nature of the solution to the recurrence relation. A root of the characteristic equation is essentially a value for 'r' that satisfies the equation, setting it to zero.

In our problem, factoring yields (r-2)(r-3)(r-4)^2 = 0. This indicates that we have three distinct roots: 2, 3, and 4 (with the last one appearing twice, known as a root with multiplicity). Each root will contribute to the general solution of the recurrence relation, with the repeated root requiring a special treatment to account for its multiplicity.

General Solution of Recurrence Relation

The general solution to a recurrence relation is a formula that describes all possible sequences that satisfy the relation. It is built from the roots of the characteristic equation. If a root is simple (occurs once), it contributes a term like C_x(r_x)^n to the general solution. If a root has multiplicity, as is the case with the root 4 in our problem, it also requires additional terms involving 'n' to reflect its repeated nature.

Therefore, combining all the roots, the general solution assumes the shape a_n = C_1(2)^n + C_2(3)^n + C_3(4)^n + C_4n(4)^n. Here, C_1, C_2, C_3, and C_4 are constants determined by the initial conditions of the recurrence relation.

Initial Values of Recurrence Relation

Initial values play a critical role in determining the specific constants for our general solution. These are the specified terms at the beginning of the sequence that our recurrence relation must satisfy. In our exercise, we're given a_0 = 5, a_1 = 12, a_2 = 38, a_3 = 126.

By plugging these initial values into the general solution, we create a system of equations. Each equation corresponds to one of the initial terms, and solving this system will reveal the specific values for our constants C_1, C_2, C_3, and C_4.

Specific Solution of Recurrence Relation

The specific solution to a recurrence relation is the unique sequence that not only follows the given pattern but also fits the initial values that have been provided. This is what the exercise seeks: the exact formula to express each term of the sequence.

After using the initial values to solve for our constants, we integrate these into the general solution. Thus, the specific solution for our LHRR becomes a_n = 2^n + 2(3)^n + 2(4)^n - n(4)^n. This expresses every term of the sequence as a function of 'n' and is the direct answer to the exercise's query.