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Section3.2Dynamic Models Using Sequences

Overview

A dynamic model considers how quantities change in time. Sequences are often useful for such models. Many populations, including some plants and animals, reproduce on an annual cycle. It thus makes sense to census these populations on an annual basis so that the population is measured as a sequence. Financial models, such as paying off a loan or receiving amortized payments on a contract, involve interest accrual and periodic payments. In these cases, the balance of the loan or fund is a sequence relative to the number of periods. Even for quantities that do not change at such regular periods, we might take measurements at equal spacings for our own convenience. This also will result in a naturally observed sequence.

This section focuses on the formulation and interpretation of models using sequences. We often develop models by considering gain terms and loss terms. For example, in population growth, gains include births and immigration; losses include deaths and emigration. The development of a model involves creating formulas that compute or approximate the size of these terms based on the state of the system.

We will consider simple models that represent various rates of change. For a population model, we might consider the rate of births or the rate of deaths. For a financial model, we might consider the rate of interest or the rate of payment. Finding simple but meaningful models for different rates allows us to predict overall changes of the system. We will analyze the overall rate of change to understand the behavior of the model.

Populations are frequently modeled using sequences. Many population are adapted to reproduce on an annual cycle, so it makes sense that such populations might be censused on an annual basis. Even for populations that reproduce throughout the year, it might still make sense to measure the population at the same time to measure year-over-year growth or decline. Fast growing populations like bacteria or some species of insects might be measured on even shorter time scales, such as hourly (bacteria) or weekly (insects). Sequences are appropriate in these circumstances because we are interested in the population size at specific times rather than at all possible times.

There are many variables that determine how a population changes. Some of these are unpredictable. Unpredictability or randomness is called stochasticity. Populations are subject to environmental stochasticity and demographic stochasticity. Environmental effects might include temperature fluctuations or variation in rainfall. Demographic stochasticity includes the randomness in number of offspring (e.g., seeds or eggs) or randomness in mortality or the timing of development.

In spite of these random effects, it is often the case that the size of the population can be approximately predicted knowing the population of the previous year. Recursive equations for sequences provide a mathematical framework for modeling these sequences. If we use \(P\) for the name of our population sequence, then we attempt to find a function \(f\) that relates consecutive values of the population as a recursive equation

\begin{equation*} P_{n+1} = f(P_n). \end{equation*}

That function is called the projection function.

This section discusses strategies for creating projection functions and analyzing the behavior of the resulting sequences. Much of that discussion will relate to per capita growth rates.

Subsection3.2.1Population Models

Population sizes change because individuals are entering and leaving the population. Growth in the population includes births as well as immigration. Decline in the population includes deaths as well as emigration. The quantities measuring the number of births, deaths, and migration events per year are rates of change. For the state of our system, we include a variable representing the size of the population as well as a variable for each of the rates of change. In a more complex model, we might have variables for the number of individuals at different ages or stages of development.

For example, consider a population that only changes from births and deaths. Let \(P\) be the size of the population, let \(B\) be the annual birth rate, and let \(D\) be the annual death rate. These variables are each measured annually and can be considered as sequences. Our index variable \(t\) will measure the time in years. The population sequence will satisfy a recurrence relation

\begin{equation*} P_{t+1} - P_t = B_t - D_t. \end{equation*}

This equation simply states that the net change in the population, called the forward difference \(\Delta P_{t} = P_{t+1}-P_t\text{,}\) is equal to the number of births (a gain) minus the number of deaths (a loss). We usually consider a reference time \(t=0\) so that the first value in the sequence would actually be \(P_0\text{.}\)

The simplest model would be that the numbers of births and deaths are constant values every year. For such a model, the forward difference is also constant, \(\Delta P_t = \Delta P = B-D\text{.}\) The resulting recursive equation becomes

\begin{equation*} P_{t+1}= P_t + \Delta P\text{,} \end{equation*}

which we recognize as an arithmetic sequence with an increment \(\Delta P\text{.}\) Using Theorem 3.1.10, we know the explicit formula for this sequence is given by

\begin{equation*} P_t = P_0 + \Delta P \cdot t. \end{equation*}

Such a population either increases linearly (if \(B \gt D\)), decreases linearly (if \(B \lt D\)), or is constant (if \(B = D\)).

Example3.2.1

This example considers a dynamic graph for constant birth and death rates. There are sliders for the birth rate \(B\) and the death rate \(D\) and the initial population is also adjustable. The resulting population sequence automatically updates to visualize the result. Such a model gives an arithmetic (linear) sequence.

Figure3.2.2

Of course, it is not realistic to think that a population has the same number of births, regardless of how large the population is. Rather, we would expect that the population will see more births when the size of the population itself is larger. The simplest model for this would be that the number of births is proportional to the size of the population. That is, we expect that there is a parameter \(b\) so that \(B = b \cdot P\text{.}\) This parameter is the proportionality constant and is called the per capita birth rate. The phrase “per capita” literally means per head. If we rewrote the equation relating \(B\) and \(P\) as

\begin{equation*} b = \frac{B}{P}, \end{equation*}

we see that the model is really saying that the total number of births in a year divided by the population that year is always the same constant. In a similar way, we might expect the number of deaths to be proportional to the population size,

\begin{equation*} D = d \cdot P, \end{equation*}

where \(d\) is the per capita death rate.

Using constant per capita birth and death rates leads to a new model for the population. The recurrence relation is defined by

\begin{equation*} \Delta P_{t} = b \cdot P_t - d \cdot P_t = (b-d) P_t\text{.} \end{equation*}

The recursive equation becomes

\begin{equation*} P_{t+1} = P_t + (b-d) P_t = (1+b-d) P_t\text{.} \end{equation*}

We recognize this as the equation of a geometric sequence with the ratio \(\rho = 1+b-d\text{.}\) Again using Theorem 3.1.10, we know the explicit formula for this sequence is given by

\begin{equation*} P_t = P_0 \cdot (1+b-d)^{t}. \end{equation*}

This form of growth for the sequence is often called Malthusian growth.

Example3.2.3

This example considers a dynamic graph for constant per capita birth and death rates. There are sliders for the per capita birth rate \(b\) and the per capita death rate \(d\text{.}\) The initial population is also adjustable. The resulting population sequence automatically updates to visualize the result. Such a model gives a geometric (exponential) sequence.

Figure3.2.4

In many circumstances, we may not be as interested in the individual values of the per capita birth and death rates \(b\) and \(d\) as we are in their difference \(b-d\text{.}\) This quantity is called the net per capita growth rate and is frequently denoted by the symbol \(r=b-d\text{.}\) In that case, the explicit formula for the Malthusian growth model can be rewritten in the same form as compounded interest,

\begin{equation*} P_t = P_0 \cdot (1+r)^{t}. \end{equation*}

That is, we can interpret \(r\) as the decimal value corresponding to percent change in the population year-over-year.

Example3.2.5

Suppose a population of 2500 has 400 births and 250 deaths in the year. Compare the model for constant births and death rates with the model for constant per capita birth and death rates over the next five years.

Solution

The model for constant birth and death rates assumes that \(B=400\) and \(D=250\) are constants. The recursive equation for the population is then given by

\begin{equation*} P_{t+1} = P_t + 400 - 250 = P_t + 150. \end{equation*}

In this model, the population increases by a net number of 150 individuals per year with an explicit formula given by

\begin{equation*} P_t = 2500 + 150 t. \end{equation*}

The model for constant per capita birth and death rates assumes the ratios \(b = \frac{B}{P} = \frac{400}{2500} = 0.16\) and \(d = \frac{D}{P} = \frac{250}{2500} = 0.1\) are constants. The recursive equation for this model becomes

\begin{equation*} P_{t+1} = P_t + 0.16 P_t - 0.1 P_t = 1.06 P_t, \end{equation*}

with a corresponding explicit formula given by

\begin{equation*} P_t = 2500 \cdot 1.06^{t}. \end{equation*}

The table and figure below illustrate the growth of these two models. In the table for the Malthusian model (geometric growth), the model predicts non-integer values which I have shown to two decimal places. Of course, a population itself must be integer-valued. When working with mathematical models, we will leave the values exact until we are ready to interpret.

Year Linear Geometric
\(t\) \(P_t = 2500+150t\) \(P_t = 2500 \cdot 1.06^t\)
0 2500 2500
1 2650 2650
2 2800 2809
3 2950 2977.54
4 3100 3156.19
5 3250 3345.56

<<SVG image is unavailable, or your browser cannot render it>>

The arithmetic and geometric models agree at the initial value and after the first year. But from that point, the geometric model steadily grows faster than the arithmetic model. The geometric model grows each year by the same percentage. Since the population itself is getting larger, the increment of growth is going to be larger each year. The two models diverge from one another even more dramatically as time progresses.

<<SVG image is unavailable, or your browser cannot render it>>

Subsection3.2.2Other Models Using Sequences

The mathematical models introduced for sequences of populations can be applied and adapted to other situations. The ideas of per capita growth rates are mathematically the same as percentage growth or decay, such as appears in compounded interest investment problems. Any situation where a quantity increases or decreases by a fixed amount or by a fixed proportion or percentage will be modeled using a sequence defined in a similar way.

Example3.2.6Car Loan

Suppose you want to buy a car and obtain a loan for $10000 that includes an annual interest rate of 3%. A bank charges interest in a way that the annual percentage rate is divided equally into the months. The monthly rate of \(\frac{3}{12}\)% applies to the remaining balance of your loan. If you make a monthly payment of $250, find a model for your remaining loan balance.

Solution

Start by identifying the relevant variables. Our main concern is the outstanding balance on the loan. Let us use the variable \(B\) to represent our sequence. The initial balance on the loan is \(B_0 = 10000\text{.}\)

Next, we identify all sources to changes in the balance. A payment \(P\) on the loan reduces the loan balance. Interest \(I\) on the loan causes the loan balance to increase. If \(t\) represents the number of months since the loan began, then we have a recurrence relation describing how the loan changes,

\begin{equation*} \Delta B_{t} = B_{t+1}-B_t = -P_t + I_t\text{.} \end{equation*}

Solving for the new balance gives the recursive equation for the loan balance

\begin{equation*} B_{t+1} = B_t - P_t + I_t. \end{equation*}

For the car loan, the monthly payment is a constant, \(P_t = P = 250\text{.}\) The interest accrued each month is proportional to and depends on the current balance, \(I_t = 0.0025 B_t\text{.}\)

The model for our loan balance is given by the recursive equation and the initial value:

\begin{gather*} B_{t+1} = B_t - 250 + 0.0025 B_t, \\ B_0 = 10000. \end{gather*}

Our model is not arithmetic or geometric but a combination of the two. The projection function, \(f : B_t \mapsto B_{t+1}\text{,}\) is the linear function defined by

\begin{equation*} f(x) = x - 250 + 0.0025x = 1.0025x - 250. \end{equation*}

The values for the loan balance are plotted below.

<<SVG image is unavailable, or your browser cannot render it>>

How would you generate data for such a sequence? Computing the values by hand would be possible but quite tedious for more than just a few terms. (Yes, you need to know how to do that too.) Computers can do this efficiently and don't get tired. You just need to know how to instruct the computer what it should do.

Example3.2.7Using a Spreadsheet

One way to generate a sequence is with a spreadsheet. Common spreadsheet applications include Microsoft Excel, Apple Numbers, and Google Drive Sheets. A spreadsheet essentially starts as a giant blank table with rows and columns. The rows are numbered starting at 1 and the columns are labeled by letters. (After the first 26 columns, columns are labeled by pairs of letters.) Every cell in the table is identified by its column and row, called its address. So cell B4 would be the cell in the fourth row of the second column.

Spreadsheets are designed to perform calculations based on the values of other cells. Suppose that A1 contained the number 3 and A2 contained the number 5. If you were to type in cell A3 the formula =A1+A2 (including the equal sign), then A3 would show the value 8. If you were to change the values in either A1 or A2, the value in A3 would automatically be updated because A3 actually stores the formula not the value.

We can take advantage of these calculations to compute the values of sequences. For style, it usually is best to have some cells contain labels. In cell B1 we will enter the label Balance and then in B2 we will enter the initial loan amount 10000. (I am using the second column because I will later want to add a first column containing the number of months into the loan.) We want to continue computing values, but from this point on we want the spreadsheet to do the calculations for us.

We want the value in B3 to be based on the formula and the previous balance contained in B2. Since our recursive formula was

\begin{equation*} B_{t+1} = B_t -250 + 0.0025B_t, \end{equation*}

we will replace the variable \(B_t\) with the cell's reference. In B3, we type =B2-250+0.0025*B2 (include both the equal sign and the multiplication symbol). When you hit Enter, the value should update and show the value 9775 (the balance after one month).

To continue the process, we don't want to type in a formula for every cell. Spreadsheets have the ability to apply a formula repeatedly, and that is what we want to do. If you select the cell B3, you should see a small bold square in the bottom right corner. If you click and drag that corner down, you will generate a range of cells that all apply the same formula, except that they will refer to the cell directly above them instead of all referring to the value of B2 that was used for B3.

We'd like to plot our data. If we only plot a single column, then a spreadsheet labels the values with an index that starts with 1. We want our first index to be 0 corresponding to the original loan balance. So we need to create a column for index values, which we will put in column A. Put a header in A1 saying month. Then start our index values in the first two cells. In A2 type 0 and in A3 type 1. Once you have these two values, select both cells (click and drag) and again you will see the small bold square. When multiple cells are selected, if you drag the square, the spreadsheet will follow the pattern created by your cells, which in this case increase by 1.

Now that you have two columns, the first having the index and the second having the monthly balance, you can create a scatter plot. Select the two columns and then add a scatter plot. (This varies based on application.)

There are other tools than spreadsheets that are more efficient once you learn how to use them. Writing computer instructions in a scripting language is one of the most efficient approaches. A free online tool called Sage uses a scripting language based on the Python programming language. The online version of this text has interactive cells where you can try the scripts directly. Otherwise, you can use the following website and type the scripts: https://sagecell.sagemath.org/.

The following script will print out a table showing the same values as in the spreadsheet above. Python counts its range of values starting at 0. The pound/hashtag symbol marks a comment and is ignored by the computer.

Suppose we wanted the computer to create a figure as well. The following script will generate the values and plot a scatter plot. In Sage, that is called a list plot. To create a plot, the script needs to keep track of the history of all the previous calculations. We store the history of values in a list.

The computer calculations show that the last month in which there is a positive balance is month 42. That month, the remaining balance is $49.14, which we will pay off completely in month 43. The model continues to use the same rule, which is why it begins to predict a negative balance.

We can compute the total amount paid for this loan. For 42 months, we paid $250.00, followed by a final payment of $49.14. The total cost of the loan is

\begin{equation*} 42(250)+49.14 = 10549.14\text{.} \end{equation*}

Because the original cost of the car was $10000, we paid $549.14 in interest.

Another example that follows similar dynamics is in mixing solutions.

Example3.2.8Mixing Solutions

Suppose that you have 2 liters of salt water that initially has 200 grams salt. You pour out 0.5 liters from your bottle, replace it with a solution of pure water, and then shake well. This is repeated, making your bottle less and less salty. Use a sequence to describe the saltiness of the solution as a function of the number of dilutions.

Solution

Start by identifying the variables. We are interested in the amount of salt in the water. Use \(S\) as the variable representing the sequence of total salt (grams) in the water. The concentration \(C\) would be \(S/2\) (grams per liter). The initial value is \(S_0 = 200\text{.}\) Let \(n\) be the variable representing the number of dilutions performed, which we will use as our index for the sequence.

Next, identify what causes the change in the solution. Every dilution, a fraction of the solution is removed, \(\frac{0.5}{2}\text{,}\) along with all salt in that volume. Since the bottle is well-mixed, we have a fourth of the salt remaining taken out of the bottle. The replacement water is pure, so no new salt is added back in.

Based on our discussion, the recursive model for the salt includes only a single loss term:

\begin{gather*} S_{n+1} = S_n - 0.25 S_n = 0.75 S_n,\\ S_0 = 200. \end{gather*}

Thus our model is a simple geometric sequence. We have an explicit solution using Theorem 3.1.10:

\begin{equation*} S_n = 200 \cdot 0.75^n. \end{equation*}

When we have an explicit formula, Sage can generate a plot even more efficiently than described above. A script to use an explicit formula is given below.

Subsection3.2.3Nonlinear Projection Functions

Much more interesting (and surprising) dynamics occur when a sequence is defined by a nonlinear projection function. To motivate one example where this might occur, we return to the ideas of per capita growth for a population.

Recall that our earlier discussion used the idea that the net per capita growth rate was a constant and did not depend on the population size. That is, the number of births and deaths were simply proportional to the total population size. However, this is ultimately not physically possible. When a population gets too large, resources are limited and the population will eventually be unable to sustain such rapid growth. Either the per capita birth rate will decrease or the per capita death rate will increase (or both). Either way, the net per capita growth rate \(r=b-d\) will need to decrease as a function of population size.

Once we say that one variable decreases with respect to another variable, we can use a mathematical model to capture that idea. In this case, we want \(r\) to be a decreasing function of the population \(P\text{,}\) \(P \mapsto r\text{.}\) The simplest such model would be a linear function with a negative slope. If we had enough data, we could plot points \((P, r)\) and find a line of best fit. For now, we will use a parametrized model,

\begin{equation*} r(P) = r_0 - \alpha P, \end{equation*}

where the parameter \(r_0\) is called the intrinsic net per capita growth rate (because that is the growth rate for a very small population before resources are limited) and \(\alpha \gt 0\) is the magnitude of the negative slope. A better parametrization uses the formula for a line given both the intercepts, \((P,r) = (0,r_0)\) and \((P,r)=(K,0)\text{,}\) so that

\begin{equation*} r(P) = r_0\big(1-\frac{P}{K}\big). \end{equation*}

The value \(K\) is called the carrying capacity because for \(P \gt K\text{,}\) the growth rate will be negative (net decrease in population). That is, for any population greater than \(K\text{,}\) the available resources are inadequate to support such a population.

The population growth model is based on the model we just found. Recall that a population grows with a recursive model

\begin{equation*} P_{t+1} = P_t + r P_t, \end{equation*}

where \(r\) is the net per capita growth rate. Using the model given above for \(r=r_0(1-\frac{P}{K})\text{,}\) we construct a nonlinear model for the population sequence,

\begin{equation*} P_{t+1} = P_t + r_0\big(1-\frac{P_t}{K}\big) P_t = (1+r_0) P_t - \frac{r_0}{K} P_t^2. \end{equation*}

This model is called the discrete logistic model. The model only makes sense when the parameter \(r_0\) is in the interval \(r_0 \in (0,3)\text{.}\)

Different behaviors for the population arise, depending on the values of the parameter \(r_0\text{.}\) A Sage script is provided below that will generate plots of the population sequence for values of the parameters that you specify. In addition, a dynamic graph is given where you can adjust the parameters using sliders. The following graphs were generated by Sage using an initial value \(P_0=5\) and parameter values \(K=100\) (all plots) and \(r_0 = 0.2\text{,}\) \(r_0=1.8\text{,}\) \(r_0=2.2\) and \(r_0=2.6\text{.}\)

(a)\(r=0.2\)
(b)\(r=1.8\)
(c)\(r=2.2\)
(d)\(r=2.6\)
Figure3.2.9Logistic growth with \(K=100\) with \(P_0=5\) and varying values of \(r\text{.}\)
Figure3.2.10Dynamic graph of the discrete logistic model with variable \(r\) and \(P_0\text{.}\)

Subsection3.2.4Summary

  • Sequences can be used to model any quantities that are observed at regular intervals, with populations and financial balances as typical examples.

  • A common strategy for building a recurrence model is to add rates of gain and subtract rates of loss,

    \begin{equation*} \Delta x_{t} = x_{t+1}-x_t = +\text{Gains} - \text{Losses}\text{.} \end{equation*}
  • For a population, common gain rates include births and immigration; common loss rates include deaths and emigration.

  • A per capita rate is the ratio of the total rate to the population size. It represents the contribution toward the total rate for one individual. The total rate equals the per capita rate times the population size.

  • Finding models for individual rates or per capita rates in terms of the population size allows us to formulate a recursive equation for the population sequence. Simple examples are to assume constant rates or constant per capita rates. More complex models might fit models for density-dependent per capita rates.

  • We use computers to find values numerically for a sequence based on the recursive equation. This might be through a spreadsheet or through a scripting language like Python. These data allow us to create graphs.

Subsection3.2.5Exercises

1

A population of annual plants has all plants die every year. Before dying, each plant releases 20 seeds which will grow the following year.

  1. Find a recurrence equation for the population.
  2. If \(P_0=10\text{,}\) find \(P_1\) and \(P_2\) by hand.
  3. Find an explicit formula for the sequence \(P_t\text{.}\)
2

A population has constant per capita birth and death rates. When the population is \(P=1000\text{,}\) there are \(B=200\) births per year and \(D=250\) deaths per year. In addition, this population has a constant immigration rate of \(I=300\) individuals per year.

  1. Find a recurrence equation for the population.
  2. If \(P_0=1000\text{,}\) find \(P_1\) and \(P_2\) by hand.
  3. Use a computer to generate a plot of the sequence \((t,P_t)\) for \(t=0,\ldots,50\text{.}\) Describe the monotonicity and concavity of the sequence based on the plot.
3

You put $500 in a bank which pays 1% interest, compounded annually.

  1. Find a recurrence equation for the balance \(B\) of your account. What is the initial value, \(B_0\text{?}\)
  2. Compute \(B_1\) and \(B_2\) by hand.
  3. Find an explicit formula for the sequence \(B_t\text{.}\)
4

You inherit $50,000, which you immediately invest. Your investment fund guarantees an annual interest payment of 2%, compounded annually. You withdraw $2,000 each year to spend.

  1. Find a recurrence equation for the balance of your fund \(F\text{.}\) What is the initial value, \(F_0\text{?}\)
  2. Compute \(F_1\) and \(F_2\) by hand.
  3. Use a computer to generate a table and a plot of the sequence \((t,F_t)\) for \(t=0,\ldots,40\text{.}\) Describe the monotonicity and concavity of the sequence based on the plot.
  4. How long will the fund last? What was the total value of the inheritance?
5

You purchase a house with a home loan of $350,000 with an annual interest rate of 4%, which accrues monthly. You choose to make a monthly payment of $1,500.

  1. Find a recurrence equation for the balance of your loan \(L\text{.}\) What is the initial value, \(L_0\text{?}\)
  2. Compute \(L_1\) and \(L_2\) by hand.
  3. Use a computer to generate a table and a plot of the sequence \((t,L_t)\) for a long enough period to determine when the loan is completely paid.
  4. When will you pay off the house loan? What will have been your total cost? How much interest will you have paid?
  5. If you increase the monthly payments to $1,600, when will you pay off the loan? How much interest will you have paid?
6

A pond with 100,000 gallons of water has a stream flowing in and out at a rate of 5,000 gallons per day. One day, the stream flowing in is polluted with a chemical of 200 grams per gallon. Assuming that the pond mixes the water quickly, develop a model for the amount of chemical in the pond as a daily sequence.

  1. State your variables.
  2. What is your model for how much chemical enters the pond each day?
  3. What is your model for how much chemical leaves the pond each day?
  4. State your recurrence relation and the initial value for your sequence. Determine the resulting recursive equation.
  5. Find an explicit formula for your sequence. How much chemical is in the pond after 30 days?