Problem 50
Question
\mathrm{\\{} I n ~ P r o b l e m s ~ , ~ g r a p h ~ t h e ~ l i n e ~ \(\boldsymbol{N}_{t+1}=\boldsymbol{R} N_{t}\) in the \(\boldsymbol{N}_{t}-\boldsymbol{N}_{t+1}\) plane for the indicated value of \(R\) and locate the points \(\left(N_{t}, N_{t+1}\right), t=0,1\), and 2, for the given value of \(N_{0}\) $$ R=\frac{1}{4}, N_{0}=16 $$
Step-by-Step Solution
Verified Answer
Points: (16, 4), (4, 1), (1, 0.25).
1Step 1: Set up the relationship between \( N_t \) and \( N_{t+1} \)
The given equation is \( N_{t+1} = R N_t \). Since \( R = \frac{1}{4} \), we can rewrite this equation as \( N_{t+1} = \frac{1}{4} N_t \).
2Step 2: Calculate \( N_1 \) using \( N_0 \)
Given \( N_0 = 16 \), we use the equation from Step 1 to find \( N_1 \): \[ N_1 = \frac{1}{4} \times 16 = 4. \]
3Step 3: Calculate \( N_2 \) using \( N_1 \)
Now, use \( N_1 \) to calculate \( N_2 \): \[ N_2 = \frac{1}{4} \times 4 = 1. \]
4Step 4: Plot the points in the \( N_t - N_{t+1} \) plane
We need to plot the points corresponding to \( t = 0, 1, \) and \( 2 \):- At \( t = 0 \), the point is \( (16, 4) \).- At \( t = 1 \), the point is \( (4, 1) \).- At \( t = 2 \), the point is \( (1, 0.25) \).These points represent the graph of the line based on the given relationship.
Key Concepts
Difference EquationsStability AnalysisPhase Plane Analysis
Difference Equations
Difference equations are a fundamental concept in discrete dynamical systems. They are used to describe how something changes over discrete intervals, such as time steps, rather than continuously. In the exercise, we have a simple linear difference equation: \[ N_{t+1} = \frac{1}{4}N_t \]This equation tells us how to calculate the subsequent value of \( N_{t+1} \) given the current value \( N_t \), with a constant multiplier \( R = \frac{1}{4} \). Each iteration reflects how the system evolves step by step. For example, starting with an initial value \( N_0 = 16 \), you can predict \( N_1 \) and \( N_2 \) by following the rule of the equation, showing a clear linear and proportional decrease at each time step.
- At \( t = 0 \), \( N_1 = \frac{1}{4} \times 16 = 4 \).
- At \( t = 1 \), \( N_2 = \frac{1}{4} \times 4 = 1 \).
Stability Analysis
Stability analysis is a crucial part of understanding discrete dynamical systems. It helps predict the long-term behavior of the system described by a difference equation. For our example equation \( N_{t+1} = \frac{1}{4}N_t \), we consider the stability by examining the value of \( R \).If \( |R| < 1 \), the system tends towards a stable equilibrium, meaning the values will eventually approach zero as time progresses. Conversely, if \( |R| > 1 \), values can become unbounded, leading to instabilities or chaotic behavior. In this specific case, \( R = \frac{1}{4} \), which is less than 1, indicating that \( N_t \) will gradually approach zero, displaying a stable convergence. Over time, each successive \( N \) value becomes smaller, which is evident from our calculated steps:
- \( N_1 = 4 \)
- \( N_2 = 1 \)
- Progressing further would continue the trend towards zero
Phase Plane Analysis
Phase plane analysis involves visualizing the behavior of dynamical systems by plotting variables against each other. In the context of our exercise, it means plotting \( N_t \) against \( N_{t+1} \) over various time steps, thereby forming a trajectory that represents the system's evolution.For the equation \( N_{t+1} = \frac{1}{4}N_t \), the phase plane is defined by the points:
- \( (16, 4) \) when \( t = 0 \)
- \( (4, 1) \) when \( t = 1 \)
- \( (1, 0.25) \) when \( t = 2 \)
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