Problem 9 - Applied Discrete Structures

Question

Solve the following sets of recurrence relations and initial conditions: $$ S(k)-4 S(k-1)+4 S(k-2)=3 k+2^{k}, S(0)=1, S(1)=1 $$

Step-by-Step Solution

Answer

The solution is \( S(k) = 2^{k+2} + 2^k k - 4 + 3k \).

Step 1: Identify the Homogeneous Part

The given recurrence relation is \( S(k) - 4S(k-1) + 4S(k-2) = 3k + 2^k \). The homogeneous part of this is \( S(k) - 4S(k-1) + 4S(k-2) = 0 \). This will help us find the complementary solution.

Step 2: Solve the Characteristic Equation

To solve the homogeneous equation, we derive the characteristic equation: \( r^2 - 4r + 4 = 0 \). This can be factored as \( (r-2)^2 = 0 \), which has a repeated root \( r = 2 \).

Step 3: Find the Complementary Solution

The complementary solution \( S_c(k) \) for a repeated root is given by \( S_c(k) = (C_1 + C_2k)2^k \), where \( C_1 \) and \( C_2 \) are constants to be determined.

Step 4: Identify the Particular Solution

The non-homogeneous part of the equation is \( 3k + 2^k \). For \( 3k \), we guess a particular solution of the form \( A + Bk \) and for \( 2^k \), we guess \( Ck2^k \), since \( 2^k \) is already a part of the homogeneous solution.

Step 5: Formulate the Particular Solution

Combining guesses, our particular solution is \( S_p(k) = A + Bk + Ck2^k \). We substitute this back into the original recurrence relation to find \( A \), \( B \), and \( C \).

Step 6: Substitute Particular Solution into Recurrence

Substitute \( S_p(k) = A + Bk + Ck2^k \) into the left-hand side of the recurrence relation. This yields a system of equations by equating coefficients with the right-hand side \( 3k + 2^k \).

Step 7: Solve for Coefficients A, B, C

After substituting and collecting like terms, solve the resulting system of equations to find: \( A = -4 \), \( B = 3 \), \( C = -1 \).

Step 8: Construct the General Solution

The general solution is the sum of the complementary and particular solutions: \( S(k) = (C_1 + C_2k)2^k + (-4 + 3k - k2^k) \).

Step 9: Apply Initial Conditions

Use the initial conditions \( S(0) = 1 \) and \( S(1) = 1 \) to solve for \( C_1 \) and \( C_2 \). Substitute \( k = 0 \) and \( k = 1 \) into the general solution and set them to 1.

Step 10: Solve for Constants C1 and C2

Using \( S(0) = 1 \), we get \( C_1 - 4 = 1 \). Using \( S(1) = 1 \), we resolve \( C_1 + 2C_2 - 9 = 1 \). Solving yields \( C_1 = 5, \ C_2 = 2 \).

Step 11: Write the Final Solution

Substitute \( C_1 \) and \( C_2 \) back into the general solution: \( S(k) = (5 + 2k)2^k - 4 + 3k - k2^k \), which simplifies to \( S(k) = 2^{k+2} + 2^k k - 4 + 3k \).

Key Concepts

Homogeneous Equations

Homogeneous equations are fundamental when dealing with recurrence relations. A homogeneous equation is one where all the terms are dependent on the sequence itself, and there is no 'free' or independent term. In our example, the original recurrence relation is \( S(k) - 4 S(k-1) + 4 S(k-2) = 3k + 2^k \). The homogeneous part is derived by setting the right side to zero, resulting in: \[ S(k) - 4S(k-1) + 4S(k-2) = 0 \]. Discovering the homogeneous part is crucial because it helps us find what is known as the complementary solution, a core component of solving recurrence relations. By focusing on the homogeneous part, we are setting a foundation to explore step-by-step how individual terms in the sequence relate to their predecessors.

Characteristic Equation

To solve the homogeneous equation, we derive something known as the characteristic equation. This is a polynomial equation derived from the coefficients of the homogeneous recurrence relation. For the equation: \[ S(k) - 4S(k-1) + 4S(k-2) = 0 \] we substitute \( S(k) = r^k \), transforming it into a characteristic equation: \[ r^2 - 4r + 4 = 0 \].We then factor this polynomial, which in our case yields: \[ (r-2)^2 = 0 \]. This tells us that \( r = 2 \) is a repeated root. Understanding how to derive and solve the characteristic equation allows us to understand the nature of the sequence's growth or decline by revealing its fundamental "characteristics." Repeated roots indicate that the solution is more complex, moving us to include additional terms often involving multiples of the sequence variable \( k \) itself.

Complementary and Particular Solutions

The solution to a recurrence relation consists of two parts: the complementary and the particular solutions. For the homogeneous equation, with a repeated root of 2, the complementary solution is given by: \[ S_c(k) = (C_1 + C_2k)2^k \], where \( C_1 \) and \( C_2 \) are constants determined later through initial conditions.Next, the particular solution addresses the non-homogeneous part, \( 3k + 2^k \). We guess forms for the particular solution based on standard techniques:

• For \( 3k \), try \( A + Bk \)
• For \( 2^k \), attempt \( Ck2^k \)

This results in a particular solution: \[ S_p(k) = A + Bk + Ck2^k \].Combining them provides the general solution by adding the complementary and particular solutions. Finding the correct particular solution involves substituting back into the original relation and solving for \( A, B, \) and \( C \). These combined efforts yield a complete understanding of the sequence's dynamics.

Initial Conditions

Initial conditions are the specific values that determine the particular sequence described by a recurrence relation. They help in solving for the constants in our solutions, ensuring the sequence fits the problem’s requirements.In our problem, we have: \( S(0) = 1 \) and \( S(1) = 1 \). These values are used to solve for \( C_1 \) and \( C_2 \) in our complementary solution.By substituting \( k = 0 \) and \( k = 1 \) into the general form of our solution \( S(k) = (C_1 + C_2k)2^k + (-4 + 3k - k2^k) \), and setting the results equal to the given initial conditions, we are able to solve for these constants.For example:

• Using \( S(0) = 1 \), we derive \( C_1 - 4 = 1 \).
• From \( S(1) = 1 \), we obtain \( C_1 + 2C_2 - 9 = 1 \).

This yields \( C_1 = 5 \) and \( C_2 = 2 \), which are then substituted back into the solution for the complete final sequence. Initial conditions are thus crucial for aligning our theoretical solution with practical outcomes.