Recurrence models | Year 11

01 / LEARN

The idea, step by step

The order of events changes the rule

If a balance first earns 10% interest and then receives AUD 20, its update is $B_{n + 1} = 1.1 B_{n} + 20$ . A deposit before interest instead gives $1.1 (B_{n} + 20)$ . Define what each term measures and exactly when it is measured.

Generate a table using the complete update once per period. Carry unrounded values unless transaction-by-transaction rounding is explicitly required.

An equilibrium and its deviation

For $u_{n + 1} = r u_{n} + c$ , a constant equilibrium $L$ must satisfy $L = r L + c$ . When $r \neq 1$ , $L = c / (1 - r)$ . Subtracting $L$ shows that the deviations form a geometric sequence.

Deviations shrink if $∣ r ∣ < 1$ . If $r > 1$ , they grow unless the initial state is exactly $L$ . A negative mathematical balance may indicate that a physical model is no longer appropriate.

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

A common trap

Adding before or after multiplying gives the same update.

Instead: Usually $r (u + c) = r u + r c e r u + c$ . Match the order stated.

02 / WORKED EXAMPLES

See the thinking

Example 1

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.

$M_{n + 1} = 0.8 M_{n} + 6$ .
$M_{1} = 14, M_{2} = 17.2$ . Solve $L = 0.8 L + 6$ to get $L = 30$ .

M_{1} = 14

M_{2} = 17.2

g; equilibrium 30 g.

Example 2

Debt after repayments

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

$B_{n + 1} = 1.02 B_{n} - 100$ , $B_{0} = 1000$ .
$B_{1} = 920$ and $B_{2} = 1.02 (920) - 100 = 838.4$ .

920

and

838.40

Check your understanding

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

Show answer

$30$ .

$0.4 L = 12$ .

03 / WORKSHEET

Your turn

6 questions · 13 marks · Work at your own pace.

Worksheet PDF 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
$L = 8 / (1 - 0.75) = 32$ ; 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.
$u_{n} = 6 + 4 {(1 / 2)}^{n}$ .

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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.

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