/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 50 \mathrm{\\{} I n ~ P r o b l e m... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

\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 \(N_{0}\) $$ R=\frac{1}{4}, N_{0}=16 $$

Short Answer

Expert verified
Points: (16, 4), (4, 1), (1, 0.25).

Step by step solution

01

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 \( N_{t+1} = \frac{1}{4} N_t \).
02

Calculate \( N_1 \) using \( N_0 \)

Given \( N_0 = 16 \), we use the equation from Step 1 to find \( N_1 \): \[ N_1 = \frac{1}{4} \times 16 = 4. \]
03

Calculate \( N_2 \) using \( N_1 \)

Now, use \( N_1 \) to calculate \( N_2 \): \[ N_2 = \frac{1}{4} \times 4 = 1. \]
04

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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 \( t = 1 \), \( N_2 = \frac{1}{4} \times 4 = 1 \).
Difference equations are particularly useful in modeling population dynamics, economics, and other areas where change occurs at distinct intervals.
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
This analysis provides insight into the system's predictability and reliability over an extended period.
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 \( t = 2 \)
These points plot a straightforward trajectory that visualizes the diminishing sequence of \( N_t \) values. The line formed by connecting these points would typically tilt downward, indicating decreasing values. This graphical representation helps students and analysts gain an intuitive understanding of how the discrete system behaves over time. By showing the interplay between the current and next states, phase plane analysis aids in predicting future behavior and verifying the results of our stability analysis.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In Problems 119-124, write each sum in expanded form. $$ \sum_{k=1}^{4} \sqrt{k} $$

Assume that the population growth is described by the Beverton-Holt recruitment curve with parameters \(R_{0}\) and a. Find the population sizes for \(t=1,2, \ldots, 5\) and find \(\lim _{t \rightarrow \infty} N_{t}\) for the given initial value \(N_{0} .\) \(R_{0}=2, a=0.1, N_{0}=2\)

Use the limit laws to determine \(\lim _{n \rightarrow \infty} a_{n}=a .\) $$ \lim _{n \rightarrow \infty}\left(\frac{n+1}{n^{2}-1}\right) $$

Hormone Implant You are studying an implanted contraceptive that releases hormone continuously into a patient's blood. The data in this question are from Nilsson et al. (1986). The device adds \(20 \mu \mathrm{g}\) of hormone to the blood each day. In the blood the hormone has first order elimination kinetics; \(4 \%\) of the hormone is eliminated each day. (a) Let the amount of hormone in the blood on day \(t\) be \(a_{t} .\) Write a word equation for the change in \(a_{t}\) over one day. (b) Put in mathematical formulas for each of the terms in your word equation from (a). (c) Assume that on day 0 no hormone is present in the patient's blood, in other words, \(a_{0}=0 .\) Use your equation from (b) to compute the amount of hormone in the blood on days \(1,2,3,4\), \(5,6 .\) (d) Over time the level of hormone in the blood converges to a limit. Find the value of this limit by looking for a fixed point of your recurrence relation in (b).

The sequence \(\left\\{a_{n} \mid\right.\) is recursively defined. Find all fixed points of \(\left[a_{n}\right\\} .\) $$ a_{n+1}=\frac{1}{3} a_{n}+\frac{4}{3} $$

See all solutions

Recommended explanations on Biology Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.