Solving linear discrete dynamical systems

Math 1241, Fall 2020
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Due date: Sept. 25, 2020, 11:59 p.m.
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Total points: 3

Consider the discrete dynamical system \begin{align*} y_{ n+1} &= 0.3 y_n \qquad \text{for $n = 0,1,2, \ldots$}\\ y_0 &= 15 \end{align*}
Calculate
$y_{1} =$
$\times y_0$
      $=$

$y_{2} =$
$\times y_1$
      $=$
$\times y_0$
      $=$

$y_{3} =$
$\times y_2$
      $ =$
$\times y_0$
      $=$

This means that to calculate $y_{3}$, you start with $y_0$ and multiply by
how many times?

Continuing the pattern $n$ times, to calculate $y_n$, you start with $y_0$ and multiply by
how many times?

Since repeated multiplication can be written as exponentiation, the formula for $y_n$ is:
$y_n =$
$\times y_0$
      $=$

(When answering online, you can use ^ for exponentiation, so enter $a^b$ as a^b.)

Calculate $y_{17}=$

(Keep at least 6 significant digits in your response. Online, you can enter a number like $1.2352 \times 10^{-5}$ as either 1.2352*10^-5 or 1.2352E-5.)
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Hint
This question all comes down to recognizing that the system, $y_{ n+1} = 0.3 y_n$, means that you multiply by $0.3$ to go from one time step to the next. To advance one time step (i.e., go from $x_0$ to $x_1$, or from $x_1$ to $x_2$, or from $x_7$ to $x_8$), you multiply by $0.3$ one time. Repeat the procedure to advance more than one time step.
Presumably, you will use your calculator to get these numbers. Be sure, though, to keep lots of digits. We require you keep 6 digits, but it's always a good idea to enter at least one more digit that we require to be one the safe side. So, if your calculator says something like 6.866455078125E-05 or 6.866455078125 × 10^-5, then go ahead and enter a number such as 6.8664551E-05 for your answer.

One last thing: don't round numbers in the middle of your calculations. Just keep all the digits your calculator has (for example by storing values in your calculator memory or recalling the previous answer and building on it). If you round numbers from the middle of your calculations to 6 digits, the end result might be accurate to only 5 digits, given the way rounding errors can accumulate.

Helpful pages
- Solving linear discrete dynamical systems
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Consider the discrete dynamical system \begin{align*} z_{ t+1} &= 2.1 z_t \qquad \text{for $t=0,1,2, \ldots$}\\ z_0 &= P \end{align*} where the parameter $P$ indicates the value of the initial condition. Calculate
$z_{1} =$

$z_{2} =$

$z_{3} =$

Continuing the pattern, calculate: $z_t =$

(When answering online, you can use ^ for exponentiation, so enter $a^b$ as a^b.)

Calculate $z_{33}=$

If we start with the tiny initial condition $P=7.4 \cdot 10^{-9}$, calculate $z_{47} =$

(Keep at least 6 significant digits in your response.)

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Hint
This problem is just like the previous, except that we keep the initial condition a variable $P$. Don't panic; if you start with $P$ and then multiply it by 2.1, what do you end up with? You have $2.1 P$. You can keep multiplying that value by 2.1 to proceed to the following step. Only for the last answer will we let you get to a simple number with no variable $P$ in it.
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Consider the dynamical system \begin{align*} x_{ t+1} &= c x_t \qquad \text{for $t = 0,1,2,\ldots$}\\ x_0 &= D. \end{align*}
1. Calculate:
  - $x_1 = $
    $\times x_0$
    $= $
  - $x_2 = $
    $\times x_1$
    $= $
  - $x_3 = $
    $\times x_2$
    $= $
  - . . . .
  - $x_t = $
  Need help?Hide help
  
  Hint
  We don't give you any number is this version, but really, it only makes it easier. Now, you will start getting powers of $c$ in your answer, and you don't need a calculator. You know that when, for example, you start with $D$ and multiply by $c$ eleven times, you get $c \times c \times c \times c \times c \times c \times c \times c \times c \times c \times D$, which is the same as $c^{11}D$.
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2. If $c=3$ and $D=2$, then
  - $x_t = $
  - $x_{10} = $
  - $x_{16} = $
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  Hint
  For the last two answers, you should get plain numbers. No rounding allowed here, just enter the numbers you get. If you happen to get big numbers, don't include any commas.
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Before you solve this dynamical system, \begin{align*} u_{n+1} &= 0.3 u_n \qquad \text{for $n=0,1,2,\ldots$}\\ u_0 &= 20,000,000. \end{align*} do you think that the solution should grow or shrink?

Solve the system and use the solution to calculate $u_5$, $u_{15}$, and $u_{25}$. Include a least 6 significant digits in your response.
$u_n =$

$u_5 =$

$u_{15} =$

$u_{25} =$

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Hint
The first question is reminding to always think about “sanity checks,” i.e., making sure your answer makes sense. Since the computer is telling you right away if you got the right answer, it's easy to forget about sanity checks. But they'll do you well in the context of an exam, where you don't get the immediate feedback from the computer.
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The above dynamical systems were written in function iteration form, given the value of the state variable as a function of the old value. Sometimes, though, we might have a dynamical system in difference form, where we are given the change in the state variables. In this case, we want to convert to function iteration form before solving.
Consider the dynamical system where the state variable $y$ decreases by 60% each time step: \begin{align*} y_{ t+1} - y_t &= -0.6 y_t \qquad \text{for $t=0,1,2\ldots$}\\ y_0 &= 19. \end{align*} Each time step, the change in $y$, i.e., $y_{ t+1} - y_t$, is -60% of the old value of $y$.
1. To solve the dynamical system, we don't want to view it as subtracting 60% at each time step. Instead, we want to view it as multiplying by some number at each time step. In other words, we want write the dynamical system in function iteration form, where the new $y_{ t+1}$ is set equal to some number times the old $y_t$.
  Converting to function iteration form is easy, we just need to add $y_t$ to both sides of the equation so we get $y_{ t+1}$ alone on the left side of the equation. The evolution rule for the dynamical system becomes
  
  $y_{ t+1} = -0.6y_t +$
  or, combining the terms on the right hand side, $y_{ t+1} = $
  . (Online, enter $y_t$ as y_t.)
  
  The function iteration form makes it clear that after each time step, the new value of $y$ is
  % of the previous value of $y$. In other words, at each time step, we multiply by
  .
  
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  Hint
  If you are losing 60% in each time step, how much do you have left after one time step? In each time step, you want to multiply by the fraction you have left (not the fraction you lose) because what you are keeping track of (i.e., the state variable) is the amount you have, not the amount you lost.
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2. The rest of the procedure for finding the solution should be familiar. Calculate:
  - $y_1 = $
    $\times y_0$
    $= $
  - $y_2 = $
    $\times y_1$
    $= $
  - $y_3 = $
    $\times y_2$
    $= $
  - . . . .
  - $y_t = $
Consider the dynamical system \begin{align*} y_{ t+1} - y_t &= k y_t \qquad \text{for $t=0,1,2,\ldots$} \\ y_0 &= L. \end{align*}
1. This dynamical system is written in
  form. To solve the system, we want to rewrite it in
  form. To convert it, we need to add
  to both sides of the equation. The result is: $y_{ t+1} = ky_t +$
  To write the result in a slightly more simple form, factor out the $y_t$ on the right hand side so that it is just some quantity times $y_t$. We end up with the following dynamical system in
  form: $y_{ t+1} = $
  
  $y_0 =$
  , where we repeated the initial condition for completeness.
  The result is a little uglier than the previous problems, but not bad.
  
  Need help?Hide help
  
  Hint
  If you have $ab +ac$ and we ask you to factor it, we want $a(b+c)$. Or, if you have $ab+a$, the factored form is $a(b+1)$. If you don't factor your answer, but it is otherwise correct, the computer will complain about your answer being mathematically equivalent to the correct answer but in the wrong form. We are being picky here. It will make the next part be easier if you write your answers as (something)y_t .
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2. Calculate:
  - $y_1 = ($
    $) \times y_0 = $
  - $y_2 = ($
    $) \times y_1 = $
  - $y_3 = ($
    $) \times y_2 = $
  - . . . .
  - $y_t = $
  Need help?Hide help
  
  Hint
  These answers are a little messy because they are going to look like (something)^(something else)L. But, now you should see why we insisted that your write $y_{t+1}$ as (something)y_t in the previous part.
  We'll massage the result to make it look prettier, below
  
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3. If $k=2$ and $L=4$, then
  - $y_t = $
  - $y_{10} = $
  - $y_{15} = $
  Need help?Hide help
  
  Hint
  Since $k=2$, do you multiply by 2 each time step or by something slightly larger than 2?
  These number are getting large since you are multiplying by a number larger than 1 each time step. We don't let you round, though. Don't enter any commas.
  
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4. The dynamical system would look a lot prettier in function iteration form if we could write is as \begin{align*} y_{ t+1} &= K y_t \qquad \text{for $t =0,1,2,\ldots$}\\ y_0 &= L. \end{align*} If we wanted to transform our system into this prettier form, what must we make $K$? (It must depend on our original parameter $k$.)
  
  $K = $
  
  With this definition of $K$, what is the solution to the dynamical system? (Your answer shouldn't have $k$ in it, only $K$.)
  
  $y_t = $
  
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  Hint
  Above, you had results that looked like (something)^(something else)L. In part b, that something depended on $k$. Now, we are going to let $K$ be that something, so $K$ should depend on $k$. Since we made $K$ be that something, we don't need parentheses any more, and the result should look prettier.
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5. If $K=3$ and $L=4$, then
  - $y_t = $
  - $y_{10} = $
  - $y_{15} = $
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  Hint
  These results should look familiar. When $k=2$ in part c, you were multiplying by what number in each time step?
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Consider the discrete dynamical system \begin{align*} m_{ t+1} - m_t&= 0.7 m_t \qquad \text{for $t=0,1,2,\ldots$}\\ m_0 &= c \end{align*}
Calculate
$m_{1} =$

$m_{2} =$

$m_{3} =$

Continuing the pattern, calculate: $m_t =$

If $c=3.2$, calculate $m_{70}=$

(Keep at least 6 significant digits in your response.)

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Hint
Is this dynamical system in function iteration or difference form? If it happens to be in difference form, then you should add what to both sides? This means that, in each time step, you multiply by what number?
In this case, you are adding $0.7$ times the previous value each time step. If you are adding 70% each time step, what percent do you have after one time step? (You started with 100% and added 70%, so you have....) It would appear that your numbers should be growing, so you should be multiplying by something larger than 1.

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In the first days of life, the cells in a human embryo divide into two cells approximately every day. Our goal is to write down a discrete dynamical system describing the growth of the embryo. After solving the model, we can use the model to predict the embryo growth. We can also evaluate the merits of the model.
1. Step 1: formulate the model
  
  The first step is to determine the state variable of the model. Given the above description, what is the most natural quantity to describe what quantity about the embryo is changing from day to day? The
  of the embryo
  
  Choose a variable to represent this quantity:
  . Choose a variable for time (either n or t):
  .
  
  With your choice of variables, $＿_{ ＿ }$ represents the
  of the embryo that are present
  days after conception.
  
  After fertilization (in day 0), the embryo consists of a single cell. (OK, it should be called a zygote until if divides into multiple cells, but you get the idea.) Therefor the initial condition is $＿_0=$
  
  Write the dynamical system both in difference form as
  $＿_{ ＿+1} - ＿_{ ＿ } =$
  
  $＿_0=$
  and in function iteration form as $＿_{ ＿+1} =$
  
  $＿_0=$
  
  The function iteration result is easy to understand because the number of cells in one day is
  times the number from the previous day. To understand the difference form result, you need to deduce that for any given day, the change in the number of cells from the previous day to that day is equal to
  times the number from the previous day.
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  Hint
  
  One tricky part of setting up the dynamical system is deciding whether it is easier to start with the difference form (such as $x_{t+1}-x_t =$ something) or the function iteration form (such as $x_{t+1}=$ something). If you choose to start with the difference form, then on the right hand side, you need to put the change in the number of cells. (And, if the number of cells is doubling, the change is not twice the previous number of cells.) If you choose to start with the function iteration form, then on the right hand side, you need to put an expression that gives the total number of cells after a day. (And, if the number of cells is doubling, the total number is indeed twice the previous number of cells.)
  
  Helpful pages
  
  Exponential growth and decay modeled by discrete dynamical systems
  
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2. Step 2: solve the model
  
  Solve the dynamical system to obtain an expression for the number of cells as a function of the number of days since fertilization.
  
  $_＿^{＿} = $
  
  Need help?Hide help
  
  Hint
  
  Since the increase in the number of cells is proportional to the number of the previous day (i.e., we have a linear dynamical system), we will get exponential growth. In each day, we multiply the number of cells by what number?
  
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3. Step 3: predict embryo growth and evaluate the merits of the model
  
  After one week, how many cells would there be in the embryo?
  After two weeks?
  After three weeks?
  These values seem
  , so we might
  that our model is capturing the dynamics of the embryo growth during these first weeks.
  
  What about the predictions of the model for the later stages of the pregnancy? If a pregnancy lasts 40 weeks, what does the model predict for the number of cells that the baby would have upon being born?
  
  How does this number compare to the number of atoms in the observable universe, which is estimated to be about $10^{80}$? The number is about
  times
  than the number of atoms in the observable universe.
  
  Do you believe the results of the model for the end of the pregnancy?
  What does the model assume about the rate of cell division? It assumes that the rate of division
  throughout the pregnancy. That assumption must be
  .
  
  What must be true of the rate of cell division? Can the cells continue to divide into two each day for the whole course of the pregnancy?
  
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  Hint
  
  Since you solved the system in the previous step, you just need to plug in the number of days of a pregnancy into the solution formula.
  
  If, for the end of the pregnancy, the number of cells is ridiculously large, then what can you conclude about the rate of cell division? It is at all possible that this model is correct, i.e., that the rate of cell division continues at a constant rate of one division per day for the whole pregnancy? If the model is clearly bad, then what must be happening to the rate of cell division?
  
  Hide help

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Solving linear discrete dynamical systems

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Math 1241, Fall 2020