Math Insight

1. The one-dimensional analogue of this dynamical system is $z_{n+1}=a z_n$. Since this dynamical rule is multiplying by $a$ at each time step, the behavior of this system is dictated by the magnitude of parameter $a$. If $|a|<$
   , then $z_n$ approaches zero as $n$ increases, whereas if $|a|>$
   , $|z_n|$ approaches infinity as $n$ increases. We want to find a way to understand the two-dimensional system in a similar way, but we now have four parameters instead of one. What are the four parameters?
   (Enter the parameters separated by commas.)

   The starting point for analyzing the behavior of the two-dimensional system is re-writing it in a format similar to the one-dimensional system. We want to write it as a single equation, where the state variables at time $n+1$ are equal to one “parameter” times the state variables at time $n$. To do this, we use vectors and matrices. We start by writing the state variables as a vector. We will use two formats for writing vectors. Either $$(x_n, y_n)$$ or $$\begin{bmatrix} x_n\\ y_n \end{bmatrix}$$ will denote the vector of state variables. Typically, the first format will be used when we are discussing the vector in isolation. The second format will be used when a matrix is involved.

   Our objective is now to write the two-dimensional dynamical system as $$\begin{bmatrix} x_{n+1}\\ y_{n+1} \end{bmatrix}=A\begin{bmatrix} x_n\\ y_n \end{bmatrix}$$ In other words, we want to convert the two-dimensional dynamical system into a matrix equation.

2. We can use what we know about matrix-vector multiplication to determine what the matrix $A$ must be. When we multiply the matrix $A$ by the vector $(x_n, y_n)$, we want the first component to be the right-hand side of the first equation, $x_{n+1} = a x_n + b y_n$. Similarly, the second component should be the right-hand side of the second equation, $y_{n+1} = c x_n + d y_n$. Enter the appropriate parameters below to specify the components of $A$ and turn the two-dimensional system into a matrix equation.
   
   $\displaystyle\begin{bmatrix} x_{n+1}\\ y_{n+1} \end{bmatrix}=$

   $\displaystyle\begin{bmatrix} x_{n}\\ y_{n} \end{bmatrix}$

1. \begin{eqnarray*} u_{n+1} & = & 3u_n+7v_n\\ v_{n+1} & = & 4u_n-3v_n \end{eqnarray*}
   
   $\displaystyle\begin{bmatrix} u_{n+1}\\ v_{n+1} \end{bmatrix}=$

   $\displaystyle\begin{bmatrix} u_{n}\\ v_{n} \end{bmatrix}$
2. \begin{eqnarray*} x_{n+1} & = & 2x_n-3y_n+z_n\\ y_{n+1} & = & -x_n+2y_n+4z_n\\ z_{n+1} & = & -3x_n+y_n-2z_n \end{eqnarray*}
   
   $\displaystyle\begin{bmatrix} x_{n+1}\\ y_{n+1}\\z_{n+1} \end{bmatrix}=$

   $\displaystyle\begin{bmatrix} x_{n}\\ y_{n} \\z_n \end{bmatrix}$
3. \begin{eqnarray*} x_{n+1} & = & x_n+2y_n-3z_n+4w_n\\ y_{n+1} & = & 2x_n-y_n+2z_n-w_n\\ z_{n+1} & = & 3x_n+2y_n-5z_n+w_n\\ w_{n+1} & = & -x_n+3y_n+2z_n+5w_n \end{eqnarray*}
   
   $\displaystyle\begin{bmatrix} x_{n+1}\\ y_{n+1}\\z_{n+1}\\w_{n+1} \end{bmatrix}=$

   $\displaystyle\begin{bmatrix} x_{n}\\ y_{n} \\z_n\\w_n \end{bmatrix}$

Let's revisit the dynamical system from the first problem. above: \begin{align*} x_{n+1} &= 3.5 x_{n} - y_{n}, \qquad \text{for $n=0,1,2, \ldots$}\\ y_{n+1} &= x_{n} + y_{n} \end{align*} Rewrite this dynamical system in terms of a matrix-vector product:

Let's call this matrix $A$, so that we can write the dynamical system as \begin{gather*} \vc{x}_{n+1} = A\vc{x}_n, \qquad \text{for $n=0,1,2,\ldots$} \end{gather*} where \begin{align*} A = \begin{bmatrix}＿&＿\\＿&＿\end{bmatrix} \qquad \text{and} \qquad \vc{x}_n = \begin{bmatrix} x_{n}\\ y_{n} \end{bmatrix}. \end{align*}

The matrix equation $\vc{x}_{n+1} = A \vc{x}_n$ is shorthand for many equations, one for $n=0$, one for $n=1$, etc. The specific versions for $n=0$ and $n=1$ give the equations for $\vc{x}_1$ and $\vc{x}_2$:
$\vc{x}_1=$
, $\vc{x}_2 = $

As in the first problem, above, we'll use the initial condition $\vc{x}_0 = \left[\begin{matrix}1\\1\end{matrix}\right]$ to calculate $\vc{x}_1$ and $\vc{x}_2$. This time, since we've rewritten the dynamical system as a matrix-vector multiplication, we can easily ask the computer to do the work for us with the programing language R.

To enter the matrix $A$ in R, type the following command:

A=matrix(c(＿, ＿, ＿, ＿), 2, 2, byrow=TRUE)

Once you've entered that command in the console, you can just enter A in the console to view the matrix and make sure it turned out as you expected. R also outputs row and column headings like [,1] to show you the index corresponding to each row and column.

Create the initial condition vector by entering x0 = c(1,1) in the console. The next step is to tell R that we want $\vc{x}_1 = A \vc{x}_0$. Caution: we cannot use the command x1 = A*x0 in R. Instead, in R, A*x0 means to multiply the components of $A$ with the components of $\vc{x}_0$ (recycling the components of $\vc{x}_0$ since there are fewer components). The correct command for calculating $\vc{x}_1$ in R is

x1 = A %*% x0

The result is:
$\vc{x}_1 =$
, $\vc{x}_2 =$
, $\vc{x}_{10} =$

(If you are adventurous, you could try computing $\vc{x}_{10}$ using a for-loop, though it isn't that laborious to just type the required 10 equations, especially since you can hit the up-arrow in the R console to bring back the previous command, edit it, and run the new version.)

for(i in 1:10) {
   x = A %*% x
}

for(i in 0:9) {
  x[,i+2] = A %*% x[,i+1]
}