Math Insight

Using cobwebbing as a graphical solution technique for discrete dynamical systems

Math 201, Spring 2017
Name:
ID #:
Due date:
Table/group #:
Group members:
Total points: 3
  1. For the dynamical system $x_{n+1} = f(x_n)$, where $f(x) = 2x-\frac{1}{3}x^2$, 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:
    Points:
    (Note: you won't be able to get full credit from this applet you get to the end of this question. Instructions for manipulating the applet are online in the “Need help” sections.)
    1. Find the equilibria using the above graph. $E =$

      (Separate multiple answers by commas)

    2. Let's start with the initial condition $x_0=1$. Estimate $x_1=f(x_0)$ from the graph of $f$ using the following procedure. Sketch a vertical line from the point on the horizontal axis where $x_n$ is 1, up to the graph of the function. The height of this point gives you $x_1=f(x_0)$. To read off the value of $x_1$, sketch a horizontal line from the point where you hit the graph, left to the vertical axis. The value of $x_{n+1}$ from the axis gives you $x_1$. When looking at the axis labels for this step, think of $n$ as being 0 so that the horizontal axis is $x_n=x_0$ and the vertical axis is $x_{n+1}=x_1$.

      The value of $x_1$ estimated from this graphical procedure is: $x_1 =$

      Also, use formula for $f(x)$ to calculate $x_1=f(x_0)=f($
      $) =$

      (Keep at least one digit to the right of the decimal.)

      In this first time step, is $x_n$ increasing or decreasing?

    3. Repeat the above procedure to estimate $x_2=f(x_1)$ using the graph. This time, rather than starting at the horizontal position of 1, start at the horizontal position corresponding to the value of $x_1$ estimated above. (When reading the values off the axes, think of $n=1$ for this step.)

      The value of $x_2$ estimated from this graphical procedure is: $x_2 =$

      Also, using the formula for $f(x)$ and the value for $x_1$, calculate
      $x_2=f(x_1)=f($
      $)=$

      In this second time step, is $x_n$ increasing or decreasing?

    4. Repeat this procedure one more time to estimate $x_3=f(x_2)$ using the graph, starting at the horizontal position corresponding the value of $x_2$ estimated above. (When reading the values off the axes, think of $n=2$ for this step.)

      The value of $x_2$ estimated from this graphical procedure is: $x_3 =$

      Also, using the formula for $f(x)$ and the value for $x_2$, calculate
      $x_3=f(x_2)=f($
      $)=$

      In this third time step, is $x_n$ increasing or decreasing?

  2. Continuing with the same quadratic dynamical system as the previous problem, we will repeat the same procedure, but this time take a shortcut. As before, the function $f(x) = 2x-\frac{1}{3}x^2$ along with the diagonal are plotted on a graph of $x_{n+1}$ versus $x_n$.
    Feedback from applet
    Initial condition:
    Points on diagonal:
    Points on function:
    1. In the previous problem, you drew horizontal and vertical lines all the way to the axes so that you could (1) read off the new value of the state variable (from the vertical $x_{n+1}$ axis), and then (2) start off using that new value in the next step (by matching the new value with the horizontal $x_n$ axis).

      We can take a shortcut by observing that the horizontal line from step one (at vertical position $x_1$) and the vertical line from step two (at horizontal position $x_1$) cross exactly at the diagonal line. (Online, you can see light gray lines that extend each line and show the intersection.) Exploiting this fact, we can, after moving vertically to the graph of $f$ to find the output value, draw a horizontal line to the diagonal rather than to the vertical axis. The point on the diagonal gives our next horizontal position, so we can simply move vertically to the graph of $f$ again to find the next value of $x$.

      This procedure works because moving horizontally to the diagonal makes your horizontal position match your vertical position, which was the output value of the previous step.

      Verify that this procedure works for calculating $x_1$, $x_2$, and $x_3$, starting with the initial condition $x_0=1$. You should get the same answers as you did in the previous problem. In fact, each time you move horizontally to the diagonal, you should reach the same vertical line that you used in the previous problem.

      If all went well, you should get the same estimates for the values of $x$:
      $x_1 =$
      , $x_2 = $
      , and $x_3=$
      .

    2. Continue this shortcut procedure, called cobwebbing, to calculate what happens to $x_n$ when, starting with the initial condition $x_0= 1$, you let the time point $n$ get larger and larger. What does $x_n$ approach as $n$ gets large?
    3. One thing cobwebbing is especially good for is determining trends and long term behavior. Cobweb starting with a different initial condition, $x_0=-0.1$. Use your result to answer these questions. Is $x$ increasing or decreasing?
      Although the plotted portion of the graph isn't enough to keep cobwebbing too long, knowing that the graph of $f$ is a parabola, you can infer that as $n$ gets large (going off toward $\infty$) the value of $x_n$ goes off toward
      .
      (Even online, sketch this cobwebbing by hand, as we don't have an applet.)

  3. For the dynamical system $p_{t+1}=f(p_t)$, a plot of the function $p_{t+1}=f(p_t)$ and the diagonal $p_{t+1}=p_t$ are shown below.
    Feedback from applet
    Initial condition:
    Points on diagonal:
    Points on function:
    1. What are the equilibria of the dynamical system? $E =$

      (All equilibria are integers. Even online, you'll have to estimate these by eye. Separate multiple answers by commas.)
    2. Cobweb starting with the initial condition $p_0 = 2.6$. As $t$ increases what value does $p_t$ approach? The
      $p = $
      . How is it moving toward that value?

      (When cobwebbing by hand, you'll have to cobweb carefully, making sure to draw horizontal and vertical lines. When cobwebbing using the applet, first move the step slider to 1, calculate $p_1$, then continue incrementing step to calculate the next values of $p_t$.)

    3. Cobweb starting with the initial condition $p_0 = -0.9$. At first, is $p_t$ increasing or decreasing?
      As $t$ get larger, what value does $p_t$ approach? The
      $p = $
      . How is it moving toward that value?

      (You'll need to cobweb this by hand, as we don't have an applet to help you.)

  4. For the dynamical system $x_{n+1}=f(x_n)$, where $f(x) = \frac{3}{2}x-\frac{1}{8}x^3$, the function $f$ along with the diagonal are plotted on a graph of $x_{n+1}$ versus $x_n$.
    Feedback from applet
    Points on diagonal:
    Points on function:
    1. Find the equilibria (they are all integers):
    2. Cobweb starting at the following initial conditions. For each initial condition, determine what value $x_n$ approaches as $n$ gets large.
      If $x_0=1$, $x_n$ approaches
      . If $x_0=-1$, $x_n$ approaches
      .
      If $x_0=3$, $x_n$ approaches
      . If $x_0=-3$, $x_n$ approaches
      .
      If $x_0=3.75$, $x_n$ approaches
      . If $x_0=-3.75$, $x_n$ approaches
      .
    3. When learning how to cobwebbing, folks often get confused when to move vertically and when to move horizontally. Here's one way to remember.

      When you graph a function $f(x)$ and you want to look up the value of, say $f(3)$ from the graph, you start at $x=3$ and move
      until you hit the graph. The height of the graph at that point is the value of $f(3)$. In other words, moving
      is the natural thing to do when looking up function values. When you are cobwebbing and moving vertically, you should always stop when you hit the
      .

      On the other hand, moving
      is the strange part. And, the new line that we plot with cobwebbing is the
      . If you associate those two strange additions together, you will remember that, during cobwebbing, when moving horizontally, you should stop when you hit the
      .

  5. The function $g$ is plotted below for the dynamical system $z_{t+1} = g(z_t)$.
    Feedback from applet
    Points on diagonal:
    Points on function:
    1. Find the equilibria of the dynamical system. $E=$
      (The equilibria are integers.)
    2. Use cobwebbing to solve the system with initial condition $z_0=2.8$. What happens to $z_t$ as $t$ gets large? $z_t$ tends toward
      .
    3. Use cobwebbing to solve the system with initial condition $z_0=3.2$. What happens to $z_t$ as $t$ gets large? $z_t$ tends toward
      . (You may have to extrapolate beyond the graph shown.)
    4. Use cobwebbing to solve the system with initial condition $z_0=-4$. What happens to $z_t$ as $t$ gets large? $z_t$ tends toward
      .

  6. Consider a linear dynamical system of the form $y_{n+1} = f(y_n)$ with $f(y)=ay$. We are going to look at different ranges of the slope $a$ and examine the resulting behavior of the dynamical system.
    1. Below is a graph of $f$ and the diagonal for the case when $a=0.4$.

      What is the equilibrium of this system? $E=$

      Cobweb with initial condition $y_0=4$. Do we get exponential growth or exponential decay?

      Also cobweb with initial condition $y_0=-4$. In both cases, describe the behavior of the solution.


      For what range of values of the slope $a$ should you get similar behavior?

      $< a <$

    2. Below is a graph of $f$ and the diagonal for the case when $a=1.5$.

      What is the equilibrium of this system? $E=$

      Cobweb with initial conditions $y_0=0.3$ Do we get exponential growth or exponential decay?

      Also cobweb with initial condition $y_0=-0.3$. In both cases, describe the behavior of the solution.


      For what range of values of the slope $a$ should you get similar behavior?
      $a > $

    3. Below is a graph of $f$ and the diagonal for the case when $a=-0.4$.

      What is the equilibrium of this system? $E=$

      Cobweb with initial conditions $y_0=4$ and $y_0=-4$. Describe the behavior of the solution.


      For what range of values of the slope $a$ should you get similar behavior?

      $< a <$

    4. Below is a graph of $f$ and the diagonal for the case when $a=-2$.

      What is the equilibrium of this system? $E=$

      Cobweb with initial conditions $y_0=0.3$ and $y_0=-0.3$. Describe the behavior of the solution.


      For what range of values of the slope $a$ should you get similar behavior?
      $a <$