Problem 4 - Calculus for the Life Sciences: A Modeling Approach Volume II

Question

Examine the behavior of \(x_{t}\) for \(t=2,3, \cdots, 20\) of the difference equation $$ \begin{aligned} x_{t+1} &=\alpha_{0} x_{t}+\alpha_{1} x_{t-1} \\ x_{0} &=0.1, \quad x_{1}=0.2 \end{aligned} $$ for a. \(\alpha_{0}=\frac{5}{6}, \quad \alpha_{1}=-\frac{1}{2}\). b. \(\alpha_{0}=\frac{3}{2}, \quad \alpha_{1}=-\frac{1}{2}\). c. \(\alpha_{0}=\frac{5}{2}, \quad \alpha_{1}=-1\).

Step-by-Step Solution

Answer

In part a, the sequence converges to zero; part b, it grows exponentially; part c, it oscillates.

Step 1: Equation Recurrence Setup

Given the recurrence relation \( x_{t+1} = \alpha_{0} x_{t} + \alpha_{1} x_{t-1} \), we start by substituting the initial conditions \( x_0 = 0.1 \) and \( x_1 = 0.2 \). We will calculate \( x_t \) for \( t = 2, 3, \, \ldots, 20 \) for each set of \( \alpha_0 \) and \( \alpha_1 \) values.

Step 2: Setup for Part a

For part a, substitute \( \alpha_0 = \frac{5}{6} \) and \( \alpha_1 = -\frac{1}{2} \) into the recurrence relation. The equation becomes:\[ x_{t+1} = \frac{5}{6} x_t - \frac{1}{2} x_{t-1} \]Using the initial conditions, calculate \( x_t \) from \( t = 2 \) to \( t = 20 \).

Step 3: Compute Values for Part a

Compute \( x_2 = \frac{5}{6}(0.2) - \frac{1}{2}(0.1) = 0.0833 \). Repeat this process for each \( t \) until 20 using the recurrence formula. Observe and note the trend or behavior of the sequence.

Step 4: Setup for Part b

For part b, substitute \( \alpha_0 = \frac{3}{2} \) and \( \alpha_1 = -\frac{1}{2} \). The equation becomes:\[ x_{t+1} = \frac{3}{2} x_t - \frac{1}{2} x_{t-1} \]Using the initial conditions, calculate \( x_t \) from \( t = 2 \) to \( t = 20 \).

Step 5: Compute Values for Part b

Compute \( x_2 = \frac{3}{2}(0.2) - \frac{1}{2}(0.1) = 0.25 \). Repeat this process for each \( t \) until 20 using the recurrence formula. Observe and note the trend or behavior of the sequence.

Step 6: Setup for Part c

For part c, substitute \( \alpha_0 = \frac{5}{2} \) and \( \alpha_1 = -1 \). The equation becomes:\[ x_{t+1} = \frac{5}{2} x_t - x_{t-1} \]Using the initial conditions, calculate \( x_t \) from \( t = 2 \) to \( t = 20 \).

Step 7: Compute Values for Part c

Compute \( x_2 = \frac{5}{2}(0.2) - 1(0.1) = 0.40 \). Repeat this process for each \( t \) until 20 using the recurrence formula. Observe and note the trend or behavior of the sequence.

Key Concepts

Recurrence Relations

Recurrence relations are equations that define sequences based on previous terms. They sound complex, but they're kind of like patterns in numbers. They help us calculate the next number in a sequence by using one or more of the numbers that came before. In mathematical terms, imagine you have a sequence of numbers: \( x_0, x_1, x_2, \ldots \). A recurrence relation gives us a formula to find \( x_{t+1} \), the next number, based on one or more preceding numbers such as \( x_t \) or \( x_{t-1} \). For our exercise, the difference equation is a second-order recurrence relation: \[ x_{t+1} = \alpha_0 x_t + \alpha_1 x_{t-1} \]Here, \( \alpha_0 \) and \( \alpha_1 \) are constants that dictate how much the two preceding terms influence the next term. It's like having a recipe with varying ingredients, where changing \( \alpha_0 \) and \( \alpha_1 \) slightly alters how the sequence progresses.

Initial Conditions

Initial conditions are the starting point of our sequence. Think of them as the first two stepping stones in a path that helps set the entire course. Without them, we wouldn't have a solid place to begin when utilizing a recurrence relation. For our problem, the initial conditions are given as \( x_0 = 0.1 \) and \( x_1 = 0.2 \). These values serve as the building blocks for all subsequent terms in our sequence. With these initial numbers, and our recurrence relation, we can calculate every future term. Without proper initial conditions, the whole sequence could not be fully determined.The choice of initial values can drastically influence the sequence's behavior, just like how starting at slightly different points can lead you to vastly different destinations on a map.

Sequence Behavior

Sequence behavior tells us about the trend or pattern in the numbers of a sequence. After we apply the recurrence relation and initial conditions, we can analyze how the sequence evolves. In our exercise, we examine different scenarios by modifying the values of \( \alpha_0 \) and \( \alpha_1 \), and observe how each affects sequence behavior:

• For \( \alpha_0=\frac{5}{6} \) and \( \alpha_1=-\frac{1}{2} \), the sequence tends to stabilize over time, converging toward a fixed limit.
• For \( \alpha_0=\frac{3}{2} \) and \( \alpha_1=-\frac{1}{2} \), we see that values increase, indicating exponential-like growth in the sequence.
• For \( \alpha_0=\frac{5}{2} \) and \( \alpha_1=-1 \), the sequence can exhibit oscillating behavior or even continue to grow larger uncontrollably.

Recognizing the behavior allows us to understand the long-term trends of sequences which can be crucial for applications in mathematical modeling and forecasting.