HR17 - Repeated change models | Eduqas GCSE Mathematics

Describe repeated fixed and proportional changes with an initial state, generate successive values, and compare their behaviour and assumptions. HR17 includes general iterative processes as well as growth and decay.

General iterative rules

Does multiplying by less than $1$ always make a quantity eventually approach zero? A repeated model can also add a fixed amount, so the whole rule matters.

State and update rule

A state is the value at a specified stage. An iterative rule calculates the next state from the current one. Here $x_0$ is the initial state and $x_n$ is the state after $n$ completed periods:

x_{n+1}=f(x_n).

Some questions start with $x_1$ instead; use the given indexing rather than assuming the first subscript is zero.

A fixed addition of $c$ gives $x_{n+1}=x_n+c$ . A repeated percentage change gives $x_{n+1}=rx_n$ , where $r$ is the multiplier. Combining proportional change with a fixed addition gives $x_{n+1}=rx_n+c$ , if the proportional change happens first.

For initial $100$ , adding $20$ each period gives $120,140,160$ ; multiplying by $1.2$ gives $120,144,172.8$ . The first updated value agrees, but the rules are different.

Worked example 1: growth followed by a fixed deposit

Question. An ideal savings model starts at £ $200$ . At each month end, the balance increases by $5\%$ , then £ $30$ is deposited. Write an initial condition and rule, and find the first three month-end balances.

Plan. Multiply the current balance by $1.05$ , then add $30$ . One update represents one month.

B_0=200,\qquad B_{n+1}=1.05B_n+30.

B_1=240,\qquad B_2=1.05(240)+30=282,\qquad B_3=1.05(282)+30=326.10.

Answer. The rule above gives £ $240$ , £ $282$ , £ $326.10$ .

Check. The first interest amount is $0.05\times200=$ £ $10$ , so $200+10+30=240$ . The second is £ $12$ , giving $240+12+30=282$ .

A multiplier below $1$ can describe removal of a fraction, with a separate fixed input afterwards.

Worked example 2: proportional loss followed by replenishment

Question. A tank initially contains $120$ litres. Each day $25\%$ of the current water is removed, then $18$ litres is added. Write the rule and find the next three states.

Plan. Removing $25\%$ leaves $75\%$ , so multiply by $0.75$ before adding $18$ .

V_0=120,\qquad V_{n+1}=0.75V_n+18.

V_1=108,\quad V_2=0.75(108)+18=99,\quad V_3=0.75(99)+18=92.25.

Answer. $108$ , $99$ , $92.25$ litres.

Check. The first day removes $30$ litres and adds $18$ , a net fall of $12$ . The second removes $27$ and adds $18$ , a net fall of $9$ .

Record an intermediate table so the output of one update becomes the next input. Keep unrounded values unless the question specifies rounding at each stage; real financial or counting processes may impose such rounding.

Watch out

$rx+c$ means “multiply, then add”. Adding first gives $r(x+c)$ , which also multiplies the new input by $r$ .

Build and use a fresh rule.

Long-term behaviour

Generating terms shows whether values rise, fall or appear to settle. A fixed point is a state that the rule maps back to itself. Check a proposed fixed point by substitution; a few rounded equal-looking terms alone do not prove exact equality.

Worked example 3: approaching a stable level

Question. Investigate $V_{n+1}=0.75V_n+18$ , starting at $120$ litres. Does the water approach zero?

Plan. Generate more values and test the candidate level where the daily removal matches the $18$ -litre input.

$n$	$0$	$1$	$2$	$3$	$4$	$5$
$V_n$ , litres	$120$	$108$	$99$	$92.25$	$87.1875$	$83.390625$

At $72$ litres, removal is $0.25\times72=18$ litres. Substitution gives $0.75(72)+18=72$ .

Answer. The values approach $72$ litres rather than zero. $72$ is a fixed point.

Check. The initial gap above $72$ is $48$ , then $36,27,20.25,\ldots$ : each gap is multiplied by $0.75$ . Starting at $0$ instead gives $18,31.5,41.625,\ldots$ , approaching the same level from below.

The fixed input changes the long-term level. Compare this with pure proportional changes.

Worked example 4: growth and decay without an input

Question. Two numerical models start at $100$ :

G_{n+1}=1.2G_n,\qquad D_{n+1}=0.8D_n.

Generate three updates and describe their longer-term behaviour if the rules continue unchanged.

Plan. Apply each multiplier repeatedly; distinguish absolute changes from percentage changes.

$n$	$0$	$1$	$2$	$3$
$G_n$	$100$	$120$	$144$	$172.8$
$D_n$	$100$	$80$	$64$	$51.2$

Answer. $G_n$ keeps growing under this unrestricted model. $D_n$ decreases towards zero but remains positive after every finite number of updates.

Check. Every growth ratio is $1.2$ ; every decay ratio is $0.8$ . The decreases are $20,16,12.8$ , not a fixed subtraction of $20$ .

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HR17 - Repeated change models

General iterative rules

Long-term behaviour

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