Problem 3 - Applied Discrete Structures

Question

Solve the following sets of recurrence relations and initial conditions: $$ S(k)-0.25 S(k-1)=0, S(0)=6 $$

Step-by-Step Solution

Answer

The specific solution is \( S(k) = 6 (0.25)^k \).

Step 1: Identify the type of recurrence relation

The given recurrence relation is linear and homogeneous with constant coefficients. It can be expressed in the standard form: \[ S(k) - 0.25 S(k-1) = 0 \] where the constant coefficient of the previous term is 0.25.

Step 2: Find the characteristic equation

For homogeneous linear recurrence relations, we find the characteristic equation by replacing the recurrence relation with: \[ x^k - 0.25 x^{k-1} = 0 \] Simplifying, we find the characteristic equation is \[ x - 0.25 = 0 \] which gives the root \( x = 0.25 \).

Step 3: General solution of the recurrence relation

With the root \( x = 0.25 \), the general solution for the recurrence relation is given by: \[ S(k) = C (0.25)^k \] where \( C \) is a constant to be determined using the initial condition.

Step 4: Use the initial condition to find the constant

The initial condition provided is \( S(0) = 6 \). Substituting \( k = 0 \) into the general solution, we have: \[ S(0) = C (0.25)^0 = C \]Given that \( S(0) = 6 \), it follows that \( C = 6 \).

Step 5: Write the specific solution

Substitute the value of \( C \) back into the general solution: \[ S(k) = 6 (0.25)^k \] This is the specific solution satisfying the given initial condition.

Key Concepts

Linear Recurrence Relations

Linear recurrence relations are equations that define sequences where each term is a linear combination of its preceding terms. These relations are essential because they allow us to predict future values in a sequence based on previous ones. In our original exercise, the equation is

• \( S(k) = 0.25 S(k-1) \)

This represents a first-order linear recurrence relation because each term \( S(k) \) is expressed as a multiple (specifically 0.25) of the previous term \( S(k-1) \).
Linear recurrence relations can be either homogeneous or non-homogeneous. In our exercise, it is homogeneous because there are no additional terms (or constants) added during the formulation of the equation.

Homogeneous Equations

Homogeneous equations, particularly in the context of recurrence relations, are characterized by the absence of constant or external inputs. That means, the result of each step depends solely on the function itself and its previous values.
In a homogeneous recurrence relation like

• \( S(k) - 0.25 S(k-1) = 0 \)

there are no additional terms added. Each term arises from scaling the previous term by some constant.
Such relations are simpler to solve, since the solution involves finding a function that relies only on one or more of its own previous outputs, without external influences.

Characteristic Equation

The characteristic equation is a vital step in solving linear homogeneous recurrence relations. By finding it, we can determine the general behavior of the sequence.
To find the characteristic equation for our problem, we replace the sequence terms with powers of a variable, typically denoted as \( x \).
For

• \( S(k) - 0.25 S(k-1) = 0 \)

we substitute to get

• \( x^k - 0.25 x^{k-1} = 0 \)

which simplifies to the characteristic equation

• \( x - 0.25 = 0 \)

Solving this, we get the root \( x = 0.25 \), which indicates how the terms progress over time. A single root, as seen here, simplifies the process and shapes our general solution.

Initial Conditions

Initial conditions provide the starting point for a sequence, allowing us to find the specific solution from a general formula. For the given exercise, the initial condition is:

• \( S(0) = 6 \)

This tells us the first term of the sequence, which we use to calculate the constant \( C \) in the general solution formula:

• \( S(k) = C (0.25)^k \)

Plugging \( k = 0 \) into the formula gives us

• \( S(0) = C (0.25)^0 = C \)

So, \( C \) is directly 6. Initial conditions are crucial because they allow us to tailor the general solution to the specific needs of the problem, resulting in the specific solution

• \( S(k) = 6 (0.25)^k \)

which fully satisfies both the relational formula and given initial conditions.