Recurrence models | Year 11

01 / LEARN

Understand the idea.

The order of events changes the rule

Define the state and the observation time before writing an update. If interest happens first and a fixed payment happens second, multiply first and then add or subtract. Reversing those events changes the amount on which interest is calculated.

Apply the whole update once per period and use the resulting state next time. Keep unrounded values through the calculation unless the question explicitly specifies rounding after each transaction. Also stop interpreting the recurrence when its physical assumptions cease to hold.

An equilibrium and its deviation

An equilibrium is a value unchanged by an update. For $u_{n + 1} = r u_{n} + c$ , solve $L = r L + c$ . When $r \neq 1$ , this gives $L = c / (1 - r)$ , which may or may not be a physically allowed state.

Subtracting $L$ gives $u_{n + 1} - L = r (u_{n} - L)$ , so the deviations form a geometric sequence. They shrink when $∣ r ∣ < 1$ and give $u_{n} = L + (u_{0} - L) r^{n}$ . A negative ratio causes alternating approaches; a ratio above one amplifies nonzero deviations.

u_{n} = L + (u_{0} - L) r^{n} (r \neq 1)

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A stable stock model

A tank retains 80% of its salt after each flush, then receives 6 g. Initially it has 10 g. Find two updates and its equilibrium.

Worked solution

1
Define the mass update.
$M_{0} = 10, M_{n + 1} = 0.8 M_{n} + 6$
Each flush retains eighty percent of the old salt before six grams are added.
2
Apply the first two updates.
$\begin{aligned} M_{1} & = 0.8 (10) + 6 = 14 g \\ M_{2} & = 0.8 (14) + 6 = 17.2 g \end{aligned}$
The second flush acts on the mass left after the first full update.
3
Solve for the unchanged mass.
$\begin{aligned} L & = 0.8 L + 6 \\ 0.2 L & = 6 \\ L & = 30 g \end{aligned}$
At equilibrium, retained mass plus the addition must equal the original mass.
4
Check the equilibrium and its stability.
$0.8 (30) + 6 = 30, ∣ 0.8 ∣ < 1$
Thirty grams is unchanged, and deviations from it shrink each cycle.

Answer

$M_{1} = 14$ g, $M_{2} = 17.2$ g; equilibrium 30 g.

EXAMPLE 02

Debt after repayments

A debt starts at AUD 1000, grows by 2% monthly and then has AUD 100 repaid. Find the first two balances.

Worked solution

1
Write the balance after a full monthly cycle.
$B_{0} = 1000, B_{n + 1} = 1.02 B_{n} - 100$
Interest is charged before the repayment is subtracted.
2
Find the first balance.
$B_{1} = 1.02 (1000) - 100 = 920$
The first month adds twenty dollars of interest before the payment.
3
Use that balance for the next month.
$B_{2} = 1.02 (920) - 100 = 838.40$
The second interest charge is based on the remaining debt of AUD 920.

Answer

$920$ and $838.40$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

Find the equilibrium of $u_{n + 1} = 0.6 u_{n} + 12$ .

Show answer & reasoning

$30$ .

An equilibrium satisfies $L = 0.6 L + 12$ . Thus $0.4 L = 12$ and $L = 30$ ; substituting gives $0.6 (30) + 12 = 30$ .

Keep practising

6 worksheet questions · 13 marks

Open worksheet Worked solutions

Read the questions online

$u_{0} = 4, u_{n + 1} = 3 u_{n} + 2$ . Find $u_{1}, u_{2}, u_{3}$ .
2 marks
A population starts at 200, keeps 90% yearly, then receives 50 newcomers. Find its rule and $P_{2}$ .
2 marks
Find the equilibrium and limiting value for $x_{n + 1} = 0.75 x_{n} + 8, x_{0} = 0$ .
2 marks
For $u_{n + 1} = 2 u_{n} - 6, u_{0} = 7$ , find the equilibrium and long-term behaviour.
2 marks
A machine adds 10 units then loses 20% of the new stock. Write a recurrence.
2 marks
Find an explicit rule for $u_{n + 1} = 0.5 u_{n} + 3, u_{0} = 10$ .
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

$u_{0} = 4, u_{n + 1} = 3 u_{n} + 2$ . Find $u_{1}, u_{2}, u_{3}$ .
2 marks
Apply multiplication then addition at each step.
$14, 44, 134$ .
A population starts at 200, keeps 90% yearly, then receives 50 newcomers. Find its rule and $P_{2}$ .
2 marks
$P_{1} = 230$ and $P_{2} = 257$ .
$P_{n + 1} = 0.9 P_{n} + 50$ , $P_{0} = 200$ ; $P_{2} = 257$ .
Find the equilibrium and limiting value for $x_{n + 1} = 0.75 x_{n} + 8, x_{0} = 0$ .
2 marks
$L = \frac{8}{1 - 0.75} = 32$ ; the deviation factor is 0.75.
Both equal 32.
For $u_{n + 1} = 2 u_{n} - 6, u_{0} = 7$ , find the equilibrium and long-term behaviour.
2 marks
$u_{n} = 6 + 2^{n}$ .
Equilibrium 6; the sequence increases without bound.
A machine adds 10 units then loses 20% of the new stock. Write a recurrence.
2 marks
The new units experience the loss too.
$S_{n + 1} = 0.8 (S_{n} + 10) = 0.8 S_{n} + 8$ .
Find an explicit rule for $u_{n + 1} = 0.5 u_{n} + 3, u_{0} = 10$ .
3 marks
The equilibrium is 6 and the initial deviation is 4.
$u_{n} = 6 + 4 {(\frac{1}{2})}^{n}$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics sequence and series content with introductory induction, regrouped into an original school-sized chapter.

Official curriculum source ↗

In this lesson

Understand the idea.

The order of events changes the rule

An equilibrium and its deviation

See the method in action.

A stable stock model

Define the mass update.

Apply the first two updates.

Solve for the unchanged mass.

Check the equilibrium and its stability.

Debt after repayments

Write the balance after a full monthly cycle.

Find the first balance.

Use that balance for the next month.

Try it yourself.

Check your understanding

Keep practising