/*! 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 12 Find the equilibria of the diffe... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the equilibria of the difference equation and classify them as stable or unstable. Use cobwebbing to find \(\lim _{t \rightarrow \infty} x_{t}\) for the given initial values. \(x_{t+1}=\frac{7 x_{t}^{2}}{x_{t}^{2}+10}, \quad x_{0}=1, \quad x_{0}=3\)

Short Answer

Expert verified
The stable equilibria are \(x=0\) and \(x=5\). Starting at \(x_0 = 1\) leads to \(\lim_{t\to\infty}x_t = 0\), and \(x_0 = 3\) leads to \(\lim_{t\to\infty}x_t = 5\).

Step by step solution

01

Find Equilibria

To find equilibria of the difference equation \(x_{t+1}=\frac{7 x_{t}^{2}}{x_{t}^{2}+10}\), set \(x_{t+1}=x_t\). Solve \(x = \frac{7x^2}{x^2 + 10}\) for \(x\).
02

Solve for Equilibria

Re-arrange the equation from Step 1 to \(x(x^2 + 10) = 7x^2\). Simplifying gives \(x^3 + 10x = 7x^2\), or, \(x^3 - 7x^2 + 10x = 0\). Factor this polynomial equation: \(x(x-5)(x-2) = 0\), so the equilibria points are \(x = 0, 2, 5\).
03

Classify Equilibria for Stability

To classify stability, determine the derivative \(f'(x)\) where \(f(x)=\frac{7x^2}{x^2+10}\), then evaluate it at each equilibrium point. If \(|f'(x)| < 1\), the equilibrium is stable; if \(|f'(x)| > 1\), it is unstable. Compute \(f'(x) = \frac{d}{dx}\left(\frac{7x^2}{x^2+10}\right)\) using the quotient rule.
04

Evaluate Stability

Calculate \(f'(x)\). Given \(f(x) = \frac{7x^2}{x^2+10}\), \(f'(x) = \frac{14x(x^2 + 10) - 14x^3}{(x^2 + 10)^2}\). Simplify, then evaluate at each equilibrium: - For \(x = 0\), \(f'(0)=0\), so stable.- For \(x = 2\), \(f'(2) = \frac{56}{49} \approx 1.14\), so unstable.- For \(x = 5\), \(f'(5) = \frac{35}{49} \approx 0.71\), so stable.
05

Cobwebbing for Convergence

Plot the function \(f(x) = \frac{7x^2}{x^2 + 10}\) and the line \(y=x\) to perform cobwebbing using initial values. Cobweb along the graph using \(x_0 = 1\) and \(x_0 = 3\). Observing the graphical pattern will show how \(x_t\) behaves as \(t\rightarrow \infty\).
06

Analyze Cobweb Plots

Using \(x_0 = 1\), cobwebbing approaches the stable point \(x = 0\). For \(x_0 = 3\), observe cobwebbing moving toward the stable point \(x = 5\). This confirms the stable equilibria for these initial conditions.

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Key Concepts

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

Stability Analysis of Equilibria
Stability analysis is crucial in understanding the behavior of equilibria in difference equations. To begin with, an equilibrium is a point where the state variable doesn't change over time. In our problem with the equation \(x_{t+1}=\frac{7 x_{t}^{2}}{x_{t}^{2}+10}\), equilibria are found where \(x_{t+1} = x_t\). This gives us a polynomial equation when rearranged.Once equilibria are determined, stability is analyzed by calculating the derivative \(f'(x)\) of the function \(f(x) = \frac{7x^2}{x^2+10}\). By evaluating this derivative at each equilibrium, we assess stability:
  • If \(|f'(x)| < 1\), the equilibrium is stable, indicating the system returns to equilibrium following minor perturbations.
  • If \(|f'(x)| > 1\), the system is unstable, meaning small disturbances drive the system away from equilibrium.
In this exercise, the equilibrium points are \(x = 0\) (stable), \(x = 2\) (unstable), and \(x = 5\) (stable). This analysis helps predict long-term trends.
Cobweb Plotting
Cobweb plotting is a graphical method used to visualize how a sequence behaves over iterations in a difference equation. It allows students to see the convergence or divergence of a sequence.To use cobweb plotting, the function \(f(x)\) is plotted alongside the line \(y = x\). For the sequence starting at \(x_0\), you draw vertical and horizontal lines to see how the sequence behaves:
  • Start at \(x_0\) on the x-axis.
  • Move vertically to the curve representing \(f(x)\).
  • Move horizontally to the line \(y = x\), and repeat.
For initial values \(x_0=1\) and \(x_0=3\), plotting revealed different paths. Starting at \(x_0=1\) settled down towards the stable equilibrium at \(x=0\). Whereas, \(x_0=3\) approached the stable point at \(x=5\). This visual method confirms analytical stability results, providing deeper insight into the behavior of difference equations.
Polynomial Factorization
Polynomial factorization can simplify complex equations, making equilibrium points easier to find. In our exercise, the polynomial \(x^3 - 7x^2 + 10x = 0\) was derived from the original problem setup.Factoring such a polynomial involves finding its roots:
  • First, factor out a common term, which in this case is \(x\), giving \(x(x^2 - 7x + 10) = 0\).
  • Then, further factor \(x^2 - 7x + 10\) into \((x-5)(x-2)\).
This factorization reveals the solutions or roots \(x = 0, 2, 5\). Finding these roots allows us to identify equilibria, which can then be analyzed for stability. Mastering polynomial factorization is a valuable tool in analyzing difference equations and other mathematical models.
Difference Equations in Biology
Difference equations play an essential role in modeling biological processes. They help understand population dynamics, the spread of diseases, and other systems that change over time.In this exercise, the equation \(x_{t+1}=\frac{7 x_{t}^{2}}{x_{t}^{2}+10}\) could represent a simplified biological system. For example, it might model population growth, where the population levels stabilize at certain values.Understanding equilibria and stability helps predict long-run behavior in these models. Stable equilibria correspond to steady states of a biological system—where a population neither grows nor shrinks. Conversely, unstable points might indicate population levels that are not viable.Applying these concepts enhances our ability to interpret the biological implications of mathematical models, offering insights into areas such as ecology and epidemiology. Tools like cobweb plotting further aid in visualizing these dynamics, making complex biological processes more intuitive.

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Most popular questions from this chapter

Find the most general antiderivative of the function.(Check your answer by differentiation.) \(u(r)=e^{-2 r}\)

$$ \begin{array}{c}{\text { The family of bell-shaped curves }} \\\ {y=\frac{1}{\sigma \sqrt{2 \pi}} e^{-(x-\mu)^{2} / / 2 \sigma^{2})}}\end{array} $$ $$ \begin{array}{l}{\text { occurs in probability and statistics, where it is called the }} \\ {\text { normal density function. The constant } \mu \text { is called the mean }} \\ {\text { and the positive constant } \sigma \text { is called the standard deviation. }}\end{array} $$ $$ \begin{array}{c}{\text { For simplicity, let's scale the function so as to remove the }} \\ {\text { factor } 1 /(\sigma \sqrt{2 \pi}) \text { and let's analyze the special case where }} \\ {\mu=0 . \text { So we study the function }} \\ {f(x)=e^{-x^{2} /\left(2 \sigma^{2}\right)}}\end{array} $$ $$ \begin{array}{l}{\text { (a) Find the asymptote, maximum value, and inflection }} \\ {\text { points of } f .} \\ {\text { (b) What role does } \sigma \text { play in the shape of the curve? }} \\ {\text { (c) Illustrate by graphing four members of this family on }} \\ {\text { the same screen. }}\end{array} $$

Find the most general antiderivative of the function.(Check your answer by differentiation.) \(f(x)=\frac{1+x-x^{2}}{x}\)

Nectar foraging by bumblebees Suppose that, instead of the specific nectar function in Example \(2,\) we have an arbitrary function \(N\) with \(N(0)=0, N(t) \geqslant 0, N^{\prime}(t)>0\) \(N^{\prime \prime}(t)<0,\) and arbitrary travel time \(T\) (a) Interpret the conditions on the function \(N .\) (b) Show that the optimal foraging time \(t\) satisfies the equation \(N^{\prime}(t)=\frac{N(t)}{t+T}\) (c) Show that, for any foraging time t satisfying the equation in part (b), the second derivative condition for a maximum value of the foraging function f in Example 2 is satisfied.

Crows and whelks Crows on the west coast of Canada feed on whelks by carrying them to heights of about 5 m and dropping them onto rocks (several times if necessary) to break open their shells. Two of the questions raised by the author of a study of this phenomenon were "Do crows drop whelks from the best height for breaking?" and "How energetically profitable is dropping of whelks?" The author constructed poles and dropped whelks from various heights. A model based on the study's data for the number of times a whelk needs to be dropped from a height h to be broken is \(n(h)=\frac{h+14.8}{h-1.2}\) where \(h\) is measured in meters. The energy expended by a crow in this activity is proportional to the height \(h\) and to the number \(n(h) :\) \(E=k h n(h)=\frac{k h(h+14.8)}{h-1.2}\) (a) What value of \(h\) minimizes the energy expended by the crows? (b) How does your answer to part (a) compare with the observed average dropping height of 5.3 m that is actually used by crows? Does the model support the existence of an optimal foraging strategy?

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