Question

Consider the recurrence $$ x_{k+2}=a x_{k+1}+b x_{k}+c $$ where \(c\) may not be zero. a. If \(a+b \neq 1\) show that \(p\) can be found such that, if we set \(y_{k}=x_{k}+p\), then \(y_{k+2}=a y_{k+1}+b y_{k}\). [Hence, the scquence \(x_{k}\) can be found provided \(y_{k}\) can be found by the methods of this section (or otherwise).] b. Use (a) to solve \(x_{k+2}=x_{k+1}+6 x_{k}+5\) where \(x_{0}=1\) and \(x_{1}=1\).

Short answer

Calculate \( p = \frac{5}{6} \), solve \( y_{k+2} = y_{k+1} + 6y_k \), and back-compute \( x_k = y_k - \frac{5}{6} \).

Step-by-step solution

Recurrence Transformation

We start by transforming the given recurrence relation. The problem is to show that for an appropriate choice of \( p \), the equation \( x_{k+2} = a x_{k+1} + b x_{k} + c \) can be rewritten in terms of a new sequence \( y_{k} = x_{k} + p \) which satisfies \( y_{k+2} = a y_{k+1} + b y_{k} \). To make these transformations, we substitute \( x_k = y_k - p \) in the recurrence relation:

Substituting and Reframing

Substitute \( x_k = y_k - p \) into the original recurrence:\[(y_{k+2}-p) = a(y_{k+1}-p) + b(y_k-p) + c\]Simplifying, we obtain:\[y_{k+2} - p = a y_{k+1} - ap + b y_k - bp + c\]Rearranging gives:\[y_{k+2} = a y_{k+1} + b y_k + (c - ap - bp + p)\]We want \( y_{k+2} = a y_{k+1} + b y_k \). Thus, set the constant term to zero:\[c - ap - bp + p = 0\]

Solving for p

Solve the equation from Step 2 for \( p \):\[p(a + b - 1) = c\]Since \( a + b eq 1 \), we can write:\[p = \frac{c}{a + b - 1}\]Now, \( y_k = x_k + \frac{c}{a + b - 1} \) will satisfy \( y_{k+2} = a y_{k+1} + b y_k \).

Solving the Example Sequence

Apply the result to the specific sequence \( x_{k+2} = x_{k+1} + 6 x_{k} + 5 \) with \( x_0 = 1 \) and \( x_1 = 1 \). Here, \( a = 1 \), \( b = 6 \) and \( c = 5 \).Compute \( p \):\[p = \frac{c}{a + b - 1} = \frac{5}{1 + 6 - 1} = \frac{5}{6}\]

Expressing in Terms of y_k

Set \( y_k = x_k + \frac{5}{6} \) so that:\[y_{k+2} = 1 y_{k+1} + 6 y_k\]This can be solved by methods discussed in previous sections or using characteristic equations for linear homogeneous recurrences.

Solving and Concluding

Find general solutions for \( y_k \) (depending on the roots of the characteristic equation) and back-substitute to find \( x_k = y_k - \frac{5}{6} \). Solve for initial conditions \( x_0 = 1 \) and \( x_1 = 1 \) to specify the sequence values.

Key concepts

Characteristic Equations

Characteristic equations play a crucial role in solving linear homogeneous recurrence relations. When faced with a recurrence of the form \( y_{k+2} = a y_{k+1} + b y_k \), we can find solutions by determining the characteristic equation. Typically, this equation is formed by replacing terms with powers of a variable \( r \), resulting in a characteristic equation:\[ r^2 = a r + b \]The solutions to the characteristic equation, known as characteristic roots, guide us in constructing the general solution for the sequence \( y_k \). The nature of the roots—whether they are real and distinct, real and repeated, or complex—will shape the form of this solution.

• **Real and Distinct Roots**: If the characteristic equation has two distinct real roots, say \( r_1 \) and \( r_2 \), then the general solution is \( y_k = C_1 r_1^k + C_2 r_2^k \), where \( C_1 \) and \( C_2 \) are constants determined by initial conditions.
• **Repeated Roots**: If there is a repeated root \( r \), the solution takes the form \( y_k = (C_1 + C_2 k) r^k \).
• **Complex Roots**: For complex roots \( r = \, e^{i\theta} \), the solution is expressed using sine and cosine, drawing from Euler's formula, leading to solutions like \( y_k = C_1 \cos(k\theta) + C_2 \sin(k\theta) \).

Finding and applying the characteristic equation is a cornerstone technique in handling complex recurrence relations, unlocking the ability to derive exact forms of sequence behaviors.

Linear Homogeneous Recurrences

Linear homogeneous recurrences are a particular type of relation in which each term is a linear combination of earlier terms, devoid of any additive constants. The classic form of such recurrences is \( y_{k+2} = a y_{k+1} + b y_k \). The absence of a constant makes the recurrence homogeneous.Describing a sequence fully using linear homogeneous recurrences often involves:

• **Analysis of Roots**: Determining the behavior of the sequence via characteristic equations and their roots.
• **Superposition Principle**: The principle that a linear combination of solutions to the homogenous equation is also a solution.
• **Initial Conditions**: Solving for constants using the initial terms of the sequence, crucial for tailoring the general solution to specific problems.

Linear homogeneous recurrences find their relevance in varied mathematical scenarios, from computer algorithms and numerical analysis to financial modeling and biological computations. The process of transforming a non-homogeneous recurrence into a homogeneous one (as seen in the original step by step solution) highlights the flexibility of these methods in wider mathematical problems.

Initial Conditions

Initial conditions provide the specific starting points of a recurrence sequence, crucial for deriving the particular solution from a general form. When solving recurrence relations, knowing the initial values, such as \( x_0 = 1 \) and \( x_1 = 1 \) in this exercise, is essential for calculating the constants in a closed-form solution.Every recurrence relation—whether linear, homogeneous, or non-homogeneous—relies on initial conditions to complete the narrative of a sequence. Typically, initial conditions give:

• **Specific Scenarios**: Adjusting the general solution to align with real-world or problem-specific data.
• **Insights into Evolution**: Enabling the tracking and forecasting of sequential behavior based on defined starting points.
• **Refinement of Solutions**: Helping solve unknown variables used in the general solution derived from the characteristic equation.

Understanding initial conditions helps create meaningful models of recurrence sequences, ensuring solutions are not only mathematically valid but also applicable to practical problems. In mathematics, just as in life, where you start can greatly influence where you end up, making initial conditions an integral part of recurrence analysis.