/*! 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 9 Find the equilibria of $$x_{t+... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 9

Find the equilibria of $$x_{t+1}=4 x_{t}^{2}+x_{t}-1, \quad t=0,1,2, \ldots$$ and use the stability criterion for an equilibrium point to determine whether they are stable or unstable.

Short Answer

Expert verified
The equilibria are at \( x = \frac{1}{2} \) and \( x = -\frac{1}{2} \), both are unstable.

Step by step solution

01

Define Equilibrium Points

Equilibrium points occur where the system does not change in time, meaning \( x_{t+1} = x_t \). For the given equation \( x_{t+1} = 4x_t^2 + x_t - 1 \), setting \( x_{t+1} = x_t \) gives: \[ x_t = 4x_t^2 + x_t - 1. \] Simplifying it, we get the equation: \[ 4x_t^2 - 1 = 0. \]
02

Solve for Equilibrium Points

To find \( x_t \), solve the equation from Step 1: \[ 4x_t^2 - 1 = 0. \] Factor the left-hand side: \[ (2x_t - 1)(2x_t + 1) = 0. \] This gives solutions as \( 2x_t = 1 \) or \( 2x_t = -1 \). Therefore, the equilibrium points are \( x_t = \frac{1}{2} \) and \( x_t = -\frac{1}{2} \).
03

Determine Stability of Equilibrium Points

To determine stability, calculate the derivative \( f'(x) \) at each equilibrium point, where \( f(x) = 4x^2 + x - 1 \). The derivative is \( f'(x) = 8x + 1 \).
04

Calculate Stability at \( x = \frac{1}{2} \)

Substitute \( x = \frac{1}{2} \) into \( f'(x) \): \[ f'\left(\frac{1}{2}\right) = 8\left(\frac{1}{2}\right) + 1 = 4 + 1 = 5. \] Since \(|f'(x)| > 1\), the equilibrium at \( x = \frac{1}{2} \) is unstable.
05

Calculate Stability at \( x = -\frac{1}{2} \)

Substitute \( x = -\frac{1}{2} \) into \( f'(x) \): \[ f'\left(-\frac{1}{2}\right) = 8\left(-\frac{1}{2}\right) + 1 = -4 + 1 = -3. \] Since \(|f'(x)| > 1\), the equilibrium at \( x = -\frac{1}{2} \) is also unstable.

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 91影视!

Key Concepts

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

Understanding the Stability Criterion
In mathematics, the stability criterion is a method used to determine whether an equilibrium point, when perturbed, will return to its original state or not. In simpler terms, it checks if a system can balance itself if slightly adjusted.
  • Stable equilibrium points return to equilibrium after a disturbance.
  • Unstable equilibrium points tend to move away from equilibrium when disturbed.
When applied to difference equations, this criterion examines the behavior of a system over discrete points in time. For a system with an equilibrium point \( x^* \), and given a function \( f(x) \), the stability can often be assessed using the derivative \( f'(x) \).

The general rule is:
  • If \( |f'(x^*)| < 1 \), the system is stable at \( x^* \).
  • If \( |f'(x^*)| > 1 \), the system is unstable at \( x^* \).
  • If \( |f'(x^*)| = 1 \), the system's stability is inconclusive without further analysis.
Exploring Difference Equations
Difference equations are mathematical expressions that capture the relationship between a sequence of values. They are similar to differential equations but work over discrete, not continuous, points in time. The equation provided in the exercise, \( x_{t+1} = 4x_t^2 + x_t - 1 \), is an example of a first-order difference equation.

Key characteristics include:
  • A recurrence relation that connects different terms in the sequence.
  • The sequence \( x_t \) usually evolves step by step based on a function, in this case, \( f(x_t) = 4x_t^2 + x_t - 1 \).
  • Initial conditions (such as \( x_0 \)) are often needed to derive specific future values.
Equilibrium points in these equations occur where \( x_{t+1} = x_t \), meaning the system reaches a state of balance, and future values are the same as current ones. Solving for equilibrium gives critical insight into the behavior of these systems.
The Role of Derivative Calculation in Stability Analysis
Derivative calculation plays a pivotal role in our ability to determine the stability of an equilibrium point within a system. Given a function \( f(x) \) derived from a difference equation, its derivative \( f'(x) \) reveals how sensitive the system is to changes.
For the system defined by \( f(x) = 4x^2 + x - 1 \), the derivative is \( f'(x) = 8x + 1 \). Computing this derivative at the equilibrium points \( x = \frac{1}{2} \) and \( x = -\frac{1}{2} \) helps assess each point's stability:
  • Substitute \( x = \frac{1}{2} \) to find \( f'(\frac{1}{2}) = 5 \).
  • Substitute \( x = -\frac{1}{2} \) to find \( f'(-\frac{1}{2}) = -3 \).
Since both have \( |f'(x)| > 1 \), the equilibrium points are unstable.

By using derivatives, we not only identify the equilibrium points but also predict their long-term behavior.

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

Find the general antiderivative of the given function. $$ f(x)=\sec ^{2}(4 x) $$

Find the general antiderivative of the given function. $$ f(x)=x^{-7}+3 x^{5}+\sin (2 x) $$

Solve the initial-value problem. $$ \frac{d y}{d x}=3 x^{2}, \text { for } x \geq 0 \text { with } y(0)=1 $$

The dynamics of a population of fish is modeled using the Beverton-Holt model: $$N_{t+1}=\frac{2 N_{t}}{1+\frac{N_{t}}{100}}$$ (a) Calculate the first ten terms of the sequence when \(N_{0}=10\). (b) Calculate the first ten terms of the sequence when \(N_{0}=150\). (c) Find all equilibria of the system, and use the stability criterion to determine which of them (if any) are stable. (d) Explain why your answers from (a) and (b) are consistent with what you have determined about the equilibria of the system.

Find the equilibria of $$x_{t+1}=\frac{3}{5} x_{t}^{2}-\frac{2}{5}, \quad t=0,1,2, \ldots$$ and use the stability criterion for an equilibrium point to determine whether they are stable or unstable.
See all solutions

Recommended explanations on Biology Textbooks

Chemistry of Life

Read Explanation

Cell Communication

Read Explanation

Astrobiological Science

Read Explanation

Cell Cycle

Read Explanation

Reproduction

Read Explanation

Communicable Diseases

Read Explanation
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.