Problem 12
Question
Show that the characteristic equation of $$ \begin{aligned} x_{n+1} &=p \cdot x_{n}-q \cdot y_{n} \\ y_{n+1} &=x_{n} \end{aligned} $$ is Exercise 16.2 .12 Show that the characteristic equation of $$ \begin{array}{l} x_{n+1}=p \cdot x_{n}-q \cdot y_{n} \\ y_{n+1}=x_{n} \end{array} $$ is $$ x_{n+2}-p x_{n+1}+q x_{n}=0 $$
Step-by-Step Solution
Verified Answer
The characteristic equation is \( x_{n+2} - p x_{n+1} + q x_{n} = 0 \).
1Step 1: Write the given system of equations
The given system of equations is: \( x_{n+1} = p \cdot x_{n} - q \cdot y_{n} \) and \( y_{n+1} = x_{n} \). This represents a linear system in terms of \( x_n \) and \( y_n \).
2Step 2: Substitute for \( y_{n+1} \)
Using the second equation, substitute \( y_{n} = x_{n-1} \) into the first equation. We have \( x_{n+1} = p \cdot x_{n} - q \cdot x_{n-1} \).
3Step 3: Find an expression for \( x_{n+2} \)
Apply the transformation \( y_{n+1} = x_{n} \) to find \( x_{n+2} \): from the equation \( x_{n+1} = p \cdot x_{n} - q \cdot x_{n-1} \), set \( x_{n+2} = p \cdot x_{n+1} - q \cdot x_{n} \).
4Step 4: Substitute \( x_{n+1} \) into \( x_{n+2} \)
We know \( x_{n+1} = p \cdot x_{n} - q \cdot x_{n-1} \). Substitute this into the equation for \( x_{n+2} \):\[ x_{n+2} = p(p \cdot x_{n} - q \cdot x_{n-1}) - q \cdot x_{n} \].
5Step 5: Simplify the expression for \( x_{n+2} \)
Simplify the expression derived in Step 4 to obtain the homogeneous linear recurrence relation:\[ x_{n+2} = p^2 \cdot x_{n} - pq \cdot x_{n-1} - q \cdot x_{n} \].
6Step 6: Factor and rearrange terms
Combine terms to arrange the equation in the standard form:\[ x_{n+2} - p \cdot x_{n+1} + q \cdot x_{n} = 0 \].This is the characteristic equation of the given system.
Key Concepts
Linear Recurrence RelationHomogeneous EquationsSystem of Equations
Linear Recurrence Relation
A linear recurrence relation is a sequence of numbers where each term is a linear combination of previous terms. In simpler terms, the next term in the sequence depends linearly on earlier terms. For this exercise, we're looking at a second-order recurrence relation, which means that each term is determined by the two preceding terms. For instance, in the given characteristic equation \( x_{n+2} - p \cdot x_{n+1} + q \cdot x_{n} = 0 \), each term is expressed in terms of the two preceding terms, \( x_{n+1} \) and \( x_{n} \). The constants \( p \) and \( q \) act as coefficients, influencing how much each of these terms contribute to the next term in the sequence. Linear recurrence relations are pivotal in mathematical modeling because they provide a straightforward method to study systems' behaviors, predict future terms, and solve complex problems by breaking them down into simpler, recursive steps.
Homogeneous Equations
Homogeneous equations play a central role in understanding linear recurrence relations. A homogeneous equation, quite simply, is one where the function equals zero for every value of the variables. In the context of our exercise, it provides a neat framework where all terms on one side of the equation can completely define the behavior of the system without additional input.For the equation \( x_{n+2} - p \cdot x_{n+1} + q \cdot x_{n} = 0 \),it is classified as a homogeneous equation because all terms involve the sequence \( x \) itself, with no extraneous components. This structure ensures that solutions involve basic repeated patterns, leading to predictable geometric growth or decay, which is a fundamental aspect of solving and understanding these types of linear recurrence relations.
System of Equations
A system of equations is a collection of two or more equations with a common set of unknowns. They are incredibly useful in solving complex problems where multiple conditions must be satisfied simultaneously. In our context, the system consists of two linear equations:
- \( x_{n+1} = p \cdot x_{n} - q \cdot y_{n} \)
- \( y_{n+1} = x_{n} \)
Other exercises in this chapter
Problem 7
Determine whether the following systems are stable. a. \(\quad \begin{aligned} x_{t+1} &=0.48 x_{t}+0.2 y_{t} \\ y_{t+1} &=0.128 x_{t}+0.72 y_{t} \end{aligned}\
View solution Problem 11
In Equation 16.11 , we claim that for $$ x_{t}=C_{1} r_{1}^{t}+C_{2} \times t \times r_{1}^{t}, \quad \lim _{t \rightarrow \infty} A_{t}=0 \quad \text { if } \q
View solution Problem 17
Compute the characteristic values of $$ \begin{array}{l} \xi_{n+1}=0.7647 \xi_{n}+-0.04706 \eta_{n} \\ \eta_{n+1}=0.02157 \xi_{n}+0.7843 \eta_{n} \end{array} $$
View solution Problem 18
For each of the following systems, find an equilibrium point which has both coordinates positive, if there is one, and determine whether it is stable. $$ \begin
View solution