Question

\(a_{n}-8 a_{n-1}+12 a_{n-2}=0\) for \(n \geq 2\) where \(a_{0}=3\) and \(a_{1}=-1\).

Short answer

The sequence is given by: \( a_n = -\frac{7}{4}(6^n) + \frac{19}{4}(2^n) \).

Step-by-step solution

Identify the Characteristics Equation

This is a linear homogeneous recurrence relation with constant coefficients. The general form is similar to that of a quadratic equation. The characteristic equation can be derived as follows:\[ r^2 - 8r + 12 = 0 \]

Solve the Characteristic Equation

To solve the characteristic equation \( r^2 - 8r + 12 = 0 \), we find the roots by factoring or using the quadratic formula. The equation factors as:\[ (r - 6)(r - 2) = 0 \]This gives us the roots: \( r_1 = 6 \) and \( r_2 = 2 \).

Write the General Solution

Since the roots are distinct, the general solution for the recurrence relation is:\[ a_n = A(6^n) + B(2^n) \]where \( A \) and \( B \) are constants to be determined.

Use Initial Conditions to Find Constants

We know that \( a_0 = 3 \) and \( a_1 = -1 \).For \( n = 0 \):\[ a_0 = A(6^0) + B(2^0) = A + B = 3 \]For \( n = 1 \):\[ a_1 = A(6^1) + B(2^1) = 6A + 2B = -1 \]

Solve the System of Equations

We have the following system of equations:1. \( A + B = 3 \)2. \( 6A + 2B = -1 \)First, solve the first equation for \( A \):\[ A = 3 - B \]Substitute \( A = 3 - B \) into the second equation:\[ 6(3 - B) + 2B = -1 \]\[ 18 - 6B + 2B = -1 \]\[ 18 - 4B = -1 \]\[ -4B = -19 \]\[ B = \frac{19}{4} \]Substitute \( B = \frac{19}{4} \) back into \( A = 3 - B \):\[ A = 3 - \frac{19}{4} \]\[ A = \frac{12}{4} - \frac{19}{4} = -\frac{7}{4} \]

Write the Final Solution

With these values for \( A \) and \( B \), the solution is:\[ a_n = -\frac{7}{4}(6^n) + \frac{19}{4}(2^n) \]This equation represents the sequence defined by the recurrence relation with the given initial conditions.

Key concepts

Characteristic Equation

When dealing with linear recurrence relations, one crucial step is formulating the characteristic equation. The characteristic equation is derived from the recurrence relation, which is a type of equation governing how each term in a sequence relates to its predecessors. In the exercise provided, the recurrence relation is linear and homogeneous. This means it can be represented in the general form:

• The relation is linear, as it involves linear combinations of previous terms.
• It is homogeneous, meaning all terms involve the sequence itself without any external additions or constant terms.

For our specific exercise, the characteristic equation becomes a quadratic:\[r^2 - 8r + 12 = 0\]This quadratic aspect allows us to solve for specific roots, which will help us determine the behavior of the sequence. Understanding how to transition from a recurrence relation to a characteristic equation is key to solving such problems.

Homogeneous Recurrence

A homogeneous recurrence relation is one where the terms depend solely on previous terms of the sequence. It does not include non-sequence terms. In the context of the exercise, the relation is given by:\(a_{n} - 8a_{n-1} + 12a_{n-2} = 0\)Key characteristics of a homogeneous recurrence include:

• All terms on one side and zero on the other.
• Involves a linear combination of terms of the sequence.

In simpler terms, a homogeneous recurrence establishes a self-contained system where all iterations and changes within the sequence naturally arise from preceding values. This makes the system predictable and deterministic once initial conditions are known. It's crucial to recognize the nature of homogeneity to properly apply mathematical tools to find solutions.

Initial Conditions

The initial conditions in problem-solving with recurrence relations are vital checks. They offer specific values for the initial terms of the sequence, allowing us to determine the constants in the general solution. In the example:

• Initial conditions are given as: \(a_0 = 3\) and \(a_1 = -1\).
• These are used to substitute into the general solution, found from the characteristic equation.

The general solution for our example was written as:\[a_n = A(6^n) + B(2^n)\]By substituting \(n = 0\) and \(n = 1\) along with the values of \(a_0\) and \(a_1\), we create equations to solve for the constants \(A\) and \(B\). Initial conditions transform a general solution into a complete and specific solution for a problem.

Linear Recurrence

Linear recurrence relations involve sequences in which terms are linearly dependent on earlier terms. These can be quite simple, yet they model complex patterns found in practical applications. The given relation:\(a_{n} - 8a_{n-1} + 12a_{n-2} = 0\)is linear because:

• Each term \(a_n\) is expressed as a linear combination of prior terms \(a_{n-1}\) and \(a_{n-2}\).
• Coefficients (here -8 and 12) are constants.

The simplicity of linear recurrence relations belies their power, which lies in their structured predictability and ease in deriving general patterns through characteristic equations. They underpin many sequences, making them a fundamental concept in understanding patterns and predictions in mathematical sequences.