Problem 5 - Applied Discrete Structures

Question

Solve the following sets of recurrence relations and initial conditions: $$ S(k)-2 S(k-1)+S(k-2)=2, S(0)=25, S(1)=16 $$

Step-by-Step Solution

Answer

We need more steps to solve for exact sequence form and then use given initial conditions.

Step 1: Identifying the Type of Recurrence Relation

The given relation is a second-order linear homogeneous recurrence relation with constant coefficients, plus a constant, which means it includes both homogeneous and particular solutions. It can be represented as: \[ S(k) - 2S(k-1) + S(k-2) = 2. \]

Step 2: Finding the Homogeneous Solution

Start by solving the homogeneous equation:\[ S_h(k) - 2S_h(k-1) + S_h(k-2) = 0. \]Assume a solution of the form \( S_h(k) = r^k \). Substituting in gives the characteristic equation:\[ r^2 - 2r + 1 = 0. \]This can be factored as: \( (r-1)^2 = 0 \), yielding a repeated root \( r = 1 \). In this case, the homogeneous solution is:\[ S_h(k) = C_1 + C_2 k. \]

Step 3: Finding the Particular Solution

Assume a particular solution of the form \( S_p(k) = A \), since the non-homogeneous term is constant (2). Substitute \( S_p(k) = A \) into the non-homogeneous recurrence relation:\[ A - 2A + A = 2, \]which simplifies to \( 0 = 2 \). Since this is not solvable directly, the typical guess of a constant must be modified to match the form of a particular solution for the homogeneous part. Thus, assume \( S_p(k) = A k^2 \).

Key Concepts

Homogeneous Solution

When dealing with a recurrence relation that involves both homogeneous and non-homogeneous components, it's crucial first to solve for the homogeneous solution. This involves ignoring any constant terms or non-constant sequences on the right side of the equation. Here, we consider the equation: \[ S_h(k) - 2S_h(k-1) + S_h(k-2) = 0. \]Our task is to find a pattern or formula that satisfies this equation for all k.
To find this solution, assume a solution of the form \( S_h(k) = r^k \). This assumption helps us reduce the recurrence equation to a polynomial form, known as the characteristic equation, by substituting \( r^k \) into the equation and simplifying. Then, solve for the roots of this equation. For our scenario, this leads us to:

• A polynomial equation: \( r^2 - 2r + 1 = 0 \)
• Solutions that typically take the form of roots \( r_1, r_2 \)
• Which leads to the general solution based on these roots.

For repeated roots, the homogeneous solution becomes a linear combination of terms, such as \( C_1 + C_2 k \). These constants \( C_1 \) and \( C_2 \) will later be found using initial conditions.

Particular Solution

After determining the homogeneous solution, the next step involves finding a particular solution for the non-homogeneous recurrence relation. This solution complements the homogeneous one and accounts for any external 'forcing' effect (in this case, a constant 2 on the right side of the equation).
Initially, we assume a constant form \( S_p(k) = A \), where A is a constant, fits into the equation. However, substituting this into our equation:\[ A - 2A + A = 2 \]leads to a dead end (\( 0 = 2 \)), showing that the form is unsuitable. This is typical when the trial solution overlaps with the homogeneous form.

• To determine the correct particular solution, more investigation is required.
• Use a trial solution of a different form, sometimes \( A k^2 \) or \( A k \) might work.

For this equation, the correct modified form for a particular solution must be found by careful trial and error. In our example, the particular solution requires a different form to effectively account for non-homogeneity.

Characteristic Equation

The characteristic equation is a vital concept used to solve homogeneous linear recurrence relations. It arises when assuming a solution based on \( S_h(k) = r^k \), which transforms the recurrence relation into a polynomial equation involving powers of \( r \).
For the problem at hand, substituting into the homogeneous equation results in:\[ r^2 - 2r + 1 = 0. \]This characteristic polynomial reflects how the system behaves naturally, without external forces.

• One must factor this equation to find the roots.
• The roots represent the fundamental behavior of the sequence, referring to base solutions.

In our scenario, this equation factors to \((r-1)^2 = 0\), producing a repeated root, \(r = 1\). This indicates a specific type of behavior, where the solutions build naturally over steps, leading to a special form, \( C_1 + C_2 k \). This variety of solutions reflects the types of sequences the recurrence relation models.

Second-order Linear Equation

A second-order linear equation with constant coefficients in recurrence relations includes terms depending on the two previous values, such as:\[ S(k) - 2S(k-1) + S(k-2) = 2. \]This equation is termed 'second-order' because the highest backwards step is two periods before the current one. It is a backbone equation structure which shapes many dynamics, including physical, financial, or algorithmic processes.
Such equations encompass:

• A homogeneous part, which resolves without any added constants.
• A non-homogeneous part, which includes external forces or shifts.

Solving these equations involves breaking them into manageable parts: solving the homogeneous solution form first, followed by the specific particular solution. These solutions, when combined, help predict future behavior based on past values, specifically targeting situations like our example, where constants and previous terms combine to determine future sequence values.