Math Insight

A graphical approach to finding equilibria of discrete dynamical systems

Math 201, Spring 2015
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  1. Let $f(x)=x^2-x$. Let's look at two different ways of finding values of $x$ where $f(x)$ does not change $x$, i.e., where $f(x)=x$. The first method is an analytic approach (meaning by manipulating equations) and the second is a graphical approach.
    1. Analytic approach. The equation we want to solve is $f(x)=x$, i.e., find values of $x$ that when plugged into the function $f(x)=x^2-x$, we get $x$ back out again. We want to find all values of $x$ where $x^2-x=x$.

      One thing we cannot do is divide both sides of the equation $x^2-x=x$ by $x$. Why can't we do that? We have to remember that we cannot divide both sides of the equation by
      and it is possible that $x$ might be equal to
      .

      Instead, we need to set the equation equal to zero by subtracting
      from both sides of the equation. The resulting equation is


      $=0$.
      Next, factor out the variable
      from the left hand side, obtaining:

      $=0$
      Since we have the product of two factors equal to zero, one of them must be zero, so we can conclude that

      $=0$ or
      $=0$.
      Therefore the solutions to $f(x)=x$ are $x=$
      . (Separate answers by commas.)
    2. Graphical approach. For the same function $f(x)=x^2-x$, let's explore a graphical way to solve $f(x)=x$. We can find a second way to solve $f(x)=x$ by plotting the function $y=f(x)=x^2-x$ on the same axes as the diagonal $y=x$. Where the two curves intersect, we know we found a value of $x$ where $f(x)=y=x$, i.e., where $f(x)=x$.

      On the below axes, graph the function $y=f(x)$. Using a different color (or different line thickness), also plot the diagonal line $y=x$. (Online, drag the blue points so the thick blue curve is the graph of $y=f(x)$ and drag the green points so the thin green line is the graph of $y=x$.)

      Feedback from applet
      function:
      intersections:
      line:

      Find the points where the diagonal and the graph of $f$ intersect and draw them on the graph. (Online, change the $n_i$ slider to indicate the number of intersections and drag the points that appear to the proper locations.) What are the coordinates of the points? If we label the points $A$ and $B$, the points are:

      $A =$
      and $B=$

      At those points $(x,y)$, both $y=f(x)$ and $y=x$. This means that $x=f(x)$, so the $x$-coordinates of those points are the solutions to our problem. The solutions to $f(x)=x$ are

      $x=$
      .

  2. Consider the dynamical system written in function iteration form \begin{align*} q_{t+1} &= f(q_t)\\ q_0 &= q_0, \end{align*} where $f(x)=x^2-x$. We will examine two methods for determining the equilibria: an analytical method and a graphical method. These two methods are exactly the same as the ones to solve the equation $f(x)=x$, since equilibria are the values of $q_t$ where $q_{t+1}=q_t$ (i.e., where $f(q_t)$ leaves $q_t$ unchanged). In fact, what you do for this problem should be eerily familiar.

    1. Analytic equilibria. Equilibria are values of $q_t$ where $f(q_t)=q_t$, i.e., where $$q_t^2 - q_t = q_t.$$ To stress that we are talking about equilibria, let's substitute $E$ for $q_t$, so the equation becomes

      $$E^2- E=E.$$
      Again, we can't divide both sides by $E$ because $E$ might be equal to
      . Instead, subtract
      from both sides of the equation, obtaining

      $=0$.
      Factor out the $E$ from the left hand size, obtaining:

      $=0$
      We can conclude that

      $=0$ or
      $=0$.
      Therefore, the equilibria are
      $E=$
      (separate answers by commas).
    2. Graphical equilibria. A second way to find equilibria is via a plot of $q_{t+1}$ versus $q_t$ (i.e., have $q_t$ on the $x$-axis and $q_{t+1}$ on the $y$-axis). On the below axes, graph the function $q_{t+1}=f(q_t)$. (Recall that the function is $f(x)=x^2-x$.) Using a different color (or different line thickness), also plot the diagonal line $q_{t+1}=q_t$. (Online, drag the blue points so the thick blue curve is the graph of $q_{t+1}=f(q_t)$ and drag the green points so the thin green line is the graph of $q_{t+1}=q_t$.) The graph should look familiar.

      Feedback from applet
      function:
      intersections:
      line:

      Find the points where the diagonal and the graph of $f$ intersect and draw them on the graph. (Online, change the $n_i$ slider to indicate the number of intersections and drag the points that appear to the proper locations.) What are the coordinates of the points? If we label the points $A$ and $B$, the points are:

      $A =$
      and $B=$

      At those points $(q_{t},q_{t+1})$, both $q_{t+1}=f(q_t)$ and $q_{t+1}=q_t$. This means that $q_{t+1}=f(q_t)$, so the value of $q_t$ atf those points are the equilibria $E$ for our dynamical system. The equilibria are

      $E=$
      .

  3. For the dynamical system \begin{align*} x_{n+1} &= f(x_n)\\ x_0 &= a, \end{align*} where $f(x) = x^3-8x$, the function $f$ along with the diagonal are plotted on a graph of $x_{n+1}$ versus $x_n$.
    Feedback from applet
    Equilibria:
    Number of equilibria:
    Point:

    1. Find the equilibria using the above graph. Label the points you used to determine the equilibria and give the resulting value of the equilibria. (Online, move the $n_e$ slider to indicate how many equilibria you found and drag the resulting points to the locations of the equilibria.)

      Remember the values of the equilibria are the first components of the points you found.
      The equilibria are: $E=$

      (Separate multiple values by commas.)

    2. To find the equilibria $x_n=x_{n+1}=E$ analytically, you must solve the equation (in terms of $E$)


      $=0$.
      You should be able to factor the left hand side of the equation into three factors. (Remember you can factor $a^2-b^2 = (a-b)(a+b)$.) Therefore, you get three equations for $E$. The equilibria are:
      $E=$

      (Separate multiple values by commas.)

    3. Starting with the initial condition $x_0 = -3$, calculate $x_1=$
      , $x_2=$
      , and $x_3 =$
      .
    4. If we start with the initial condition $x_0 = 1$, use the graph to approximate $x_1=f(x_0)$. Label the point on the graph that you used to estimate this value of $x_1$. (Online, check the “show point” box and move the resulting blue point to the location to estimate $x_1$.) $x_1=$

      Check the accuracy of your estimate by calculating $f(x_0)$ from the analytic expression for $f$.

  4. For the dynamical system \begin{align*} y_{t+1} &=g(y_t)\\ y_0 &= y_0, \end{align*} the function $g$ along with the diagonal are plotted on a graph of $y_{t+1}$ versus $y_t$.
    Feedback from applet
    Equilibria:
    Number of equilibria:
    Point:
    1. Find the equilibria using the above graph. Label the points you used to determine the equilibria and give the resulting values of the equilibria. (Online, move the $n_e$ slider to indicate how many equilibria you found and drag the resulting points to the locations of the equilibria.)

      The equilibria are: $E=$

    2. Starting with $y_0 = 1$, calculate $y_1 =$
      , $y_2 =$
      , and $y_3 =$
    3. If we start with $y_0 = -3$, use the graph to approximate $y_1=g(y_0)$. Label the point on the graph that you used to estimate this value of $y_1$. (Online, check the “show point” box and move the resulting blue point to the location to estimate $y_1$.)
      $y_1=$

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Math 201, Spring 2015