8.01h, 8.01i - Modelling with recurrence relations | OCR A-Level Further Mathematics A

Turn descriptions of births, deaths, arrivals and delayed effects into recurrence systems, and interpret what their solutions predict. In 8.01h you will build first-order models and use the integer-part function for individual counts; in 8.01i you will extend the modelling to second-order systems, including nonlinear rules and their limitations.

Births, deaths and the order of events

A model predicts that a population approaches $35$ , but keeping whole individuals at every update makes it settle at $31$ . How can the same growth rule produce those two forecasts? Rounding is itself part of a discrete model; you will investigate it after building the underlying rules.

Let $P_n$ be the population at a specified census time, with $n$ counting equal time intervals. Births and deaths must refer to a stated population and a stated interval. A fraction of the current population differs from a fixed number removed each year.

You already know

A recurrence system needs its initial value or values and the range of $n$ . Linear systems can be solved by complementary functions and particular solutions. A steady state remains unchanged by the update; solving its equation alone does not prove convergence.

Suppose births during an interval are modelled as $bP_n$ , deaths as $dP_n$ , and $I$ individuals arrive after those changes. Then the next census is

P_{n+1}=P_n+bP_n-dP_n+I=(1+b-d)P_n+I

This assumes both rates act on the starting population. If births or arrivals happen before the population used for deaths is counted, the formula must reflect that different timing.

A discrete population model

A discrete population model predicts counts at separate census times by an update rule. Rates such as $b$ and $d$ are fractions per interval; $I$ is a count per interval. A deterministic real-valued output can represent an approximate or average population, but cannot be an exact count of indivisible individuals. The birth-and-death update above is not in the formula booklet: construct it from the stated assumptions.

Write down what is counted at the start and what happens before the next count. That timeline prevents a percentage from being applied to the wrong total.

Worked example 1: arrivals after the population change

Question. A population starts at $100$ . During each year, births and deaths are modelled as $20\%$ and $30\%$ of the population at the start of the year. Then $20$ individuals are added. Form a recurrence system, solve it and find its long-term prediction.

Plan. Combine the two rates on the starting population, add the arrivals, then solve the linear system.

\begin{aligned} P_{n+1}&=(1+0.2-0.3)P_n+20=0.9P_n+20 && \text{census update} \\ P_0&=100,\qquad n\ge0 && \text{initial condition and range} \\ L&=0.9L+20\quad\Rightarrow\quad L=200 && \text{constant particular solution} \\ P_n&=C(0.9)^n+200 && \text{general solution} \\ C&=-100 && P_0=100 \end{aligned}

Why this step? The $20$ arrivals occur after the percentage changes, so they are not reduced by that year's net loss.

Answer. $P_n=200-100(0.9)^n$ , so the model predicts a population approaching $200$ .

Check. $P_1=110$ , $P_2=119$ ; both agree with direct updates, and the closed form proves convergence since $0.9^n\to0$ .

The next example changes only the timing of the arrivals. The rates now act on the enlarged starting population.

Worked example 2: arrivals before the population change

Question. Use the same initial population and rates, but add the $20$ individuals before applying that year's births and deaths. Form and solve the new system, and find its long-term prediction.

Plan. Put the arrivals inside the population to which the net factor $0.9$ is applied.

\begin{aligned} Q_{n+1}&=0.9(Q_n+20)=0.9Q_n+18 && \text{new event order} \\ Q_0&=100,\qquad n\ge0 && \text{same initial census} \\ L&=0.9L+18\quad\Rightarrow\quad L=180 && \text{particular solution} \\ Q_n&=C(0.9)^n+180,\qquad C=-80 && \text{apply }Q_0=100 \end{aligned}

Why this step? Those arriving now experience the same year's net reduction, so their contribution to the next census is $18$ , not $20$ .

Answer. $Q_n=180-80(0.9)^n$ , approaching $180$ .

Check. $Q_1=108$ , $Q_2=115.2$ . The different timing changes the limiting forecast by $20$ .

These model solutions are not in the formula booklet: derive them from the description. Constant rates and unchanged conditions are assumptions; crowding, varying food supply or a changing arrival policy may require a different update.

Watch out

Wrong: interpret a $30\%$ death rate as subtracting $30$ individuals. Right: subtract $0.3P_n$ when the rate applies to the population $P_n$ . A fixed removal and a percentage removal are different models.

State your population's census time before writing a formula; the same words can otherwise conceal different event orders.

Density dependence and competing steady states

A fixed percentage growth model can increase without bound. A density-dependent rate lets the change depend on how crowded the population already is.

For example, $P_{n+1}=1.5P_n-0.001P_n^2$ has increment $0.001P_n(500-P_n)$ : it is positive below $500$ and negative above $500$ , while positive populations are considered. A sign of the increment gives direction; a bound is still needed to prevent overshooting.

Density-dependent growth

A density-dependent model has a per-individual growth or survival contribution that changes with population size. A term such as $-kP_n^2$ , for $k>0$ , reduces growth more strongly at larger populations. Such a rule is nonlinear and need not have the linear closed forms from earlier lessons. Particular density-dependent formulae are not in the formula booklet: interpret the model supplied.

When there is more than one steady state, use the initial condition and the behaviour of the updates to select the one reached. A quadratic equation does not make both roots valid long-term predictions for the same initial population.

Worked example 3: growth towards a positive steady state

Question. A model is $P_0=450$ , $P_{n+1}=1.5P_n-0.001P_n^2$ . Show that the population increases towards $500$ .

Plan. Find the steady states, then show each update stays below $500$ while increasing.

For $f(P)=1.5P-0.001P^2$ ,

\begin{aligned} L=f(L)&\quad\Rightarrow\quad L(500-L)=0 && \text{steady states }0,500 \\ f(P)-P&=0.001P(500-P)>0 && 0<P<500 \\ 500-f(P)&=0.001(P-500)(P-1000)>0 && 0<P<500 \end{aligned}

Thus an update from $(0,500)$ stays in that interval and is larger than its input. Starting at $450$ , the sequence is increasing and bounded above, so it converges. Its limit is at least $450$ , excluding $0$ .

Why this step? The second inequality rules out an overshoot; the direction of the first increment alone would not prove the entire sequence increases.

Answer. The model converges monotonically to $500$ .

Check. $P_1=472.5$ , $P_2=485.49375$ , consistent with the proved bound and direction.

The next example changes only the policy: $40$ individuals are removed after each update. This creates two positive steady states.

Worked example 4: selecting between two positive roots

Question. The modified model is $Q_0=450$ , $Q_{n+1}=1.5Q_n-0.001Q_n^2-40$ . Show that the population decreases towards $400$ .

Plan. Find both candidate limits, then keep the sequence above the larger root while showing it decreases.

Writing $g(Q)=1.5Q-0.001Q^2-40$ ,

\begin{aligned} L=g(L)&\quad\Rightarrow\quad (L-100)(L-400)=0 && \text{candidate limits} \\ g(Q)-Q&=-0.001(Q-100)(Q-400)<0 && 400<Q\le450 \\ g(Q)-400&=-0.001(Q-400)(Q-1100)>0 && 400<Q\le450 \end{aligned}

Each update stays above $400$ and below its previous value, so the sequence converges to a limit at least $400$ . Of the candidates, only $400$ is possible.

Why this step? The lower bound selects the larger root; listing $100$ and $400$ would not decide the model's prediction.

Answer. The model decreases monotonically to $400$ .

Check. $Q_1=432.5$ , between $400$ and $450$ .

The same initial population now lies above the new equilibrium, so the direction reverses. Both limit arguments and factorisations are not in the formula booklet: construct them for the given model.

Watch out

Wrong: solve the fixed-point equation and choose whichever root looks realistic. Right: connect a root to the initial population using a justified interval and direction, or the convergence information supplied in the question.

Population rules may give negative outputs outside the region for which they were designed. That reveals a limitation of the model, rather than a meaningful negative number of organisms.

Integer-part updates for whole individuals

Keeping a whole-number population means changing the update itself, rather than rounding the final real-valued answer.

For non-negative $x$ , $\operatorname{INT}(x)$ is the integer part of $x$ : for example, $\operatorname{INT}(12.9)=12$ , while $\operatorname{INT}(12)=12$ . It rounds down for the non-negative quantities used here; it does not round to the nearest integer.

An integer-part update

For non-negative $x$ ,

\operatorname{INT}(x)=m\quad\Longleftrightarrow\quad m\le x<m+1

where $m$ is an integer. In an integer recurrence, use each rounded output as the input to the next update.
Not in the formula booklet: learn it.

Rounding may be appropriate for a rule that retains whole survivors, but it can bias a forecast downwards. The rounding convention is a modelling assumption, not a claim that rounding down is always the most accurate biological description.

Worked example 5: rounding changes the limiting count

Question. Compare $x_0=29,\ x_{n+1}=0.8x_n+7$ with $p_0=29,\ p_{n+1}=\operatorname{INT}(0.8p_n+7)$ . Find the limiting prediction of each.

Plan. Solve the real-valued linear model; iterate the integer model until a value repeats.

\begin{aligned} x_n&=35-6(0.8)^n\longrightarrow35 && \text{solve the linear system} \\ p_1&=\operatorname{INT}(30.2)=30 && \text{start with }29 \\ p_2&=\operatorname{INT}(31)=31 && \text{use the rounded }30 \\ p_3&=\operatorname{INT}(31.8)=31 && \text{use the rounded }31 \end{aligned}

Once $31$ maps to itself, every later integer-model value is $31$ .

Why this step? The second integer update uses $30$ , not the real-valued first output $30.2$ . Rounding after completing the real-valued iteration would describe a different sequence.

Answer. The real-valued model tends to $35$ ; the integer model becomes constant at $31$ .

Check. $0.8(31)+7=31.8$ , whose integer part is $31$ .

The next example changes only the initial count. Unlike the real-valued model, the integer model need not reach the same steady state.

Worked example 6: several integer steady states

Question. Start both models instead at $34$ . Find their limiting predictions and identify every non-negative integer steady state of the integer model.

Plan. Check the first integer update, then express the fixed-point condition using the defining inequalities for $\operatorname{INT}$ .

\begin{aligned} x_n&=35-(0.8)^n\longrightarrow35 && x_0=34 \\ p_1&=\operatorname{INT}(0.8(34)+7)=\operatorname{INT}(34.2)=34 && \text{fixed immediately} \\ m&\le0.8m+7<m+1 && \text{integer steady-state condition} \\ m&\le35,\qquad m>30 && \text{solve both inequalities} \end{aligned}

Why this step? An integer value remains fixed whenever the unrounded output lies anywhere from that value up to, but not including, the next integer.

Answer. The limits are $35$ and $34$ , respectively. The integer steady states are $31,32,33,34,35$ .

Check. At $m=35$ , the output is exactly $35$ ; at $m=30$ , the output is $31$ , so $30$ is not fixed.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account

Last Accessed

Pure Core

Statistics

Mechanics

Discrete Mathematics

Additional Pure Mathematics

8.01h, 8.01i - Modelling with recurrence relations

Births, deaths and the order of events

Density dependence and competing steady states

Integer-part updates for whole individuals

Continue this lesson.

Learn faster with Nova