R16 - General iterative processes

Some repeated changes combine a percentage with a fixed addition or removal. Form a rule for the next amount, use it repeatedly, undo its steps and explain whether its predictions settle towards a fixed value.

Modelling repeated percentage and fixed changes (Higher)

A tank starts with $20\,\mathrm{litres}$ . Each day it loses $20\%$ of its contents, then receives $10\,\mathrm{litres}$ . Will its contents keep growing without limit? First we need a rule that follows the events in the right order.

You already know

A $20\%$ decrease uses multiplier $0.8$ . In an iterative calculation, substitute the latest answer into the next use of the rule. A pure percentage model has no extra fixed addition or removal.

Write $u_0$ for the starting amount, $u_1$ for the amount after one complete step and $u_n$ for the amount after $n$ steps. The next amount is $u_{n+1}$ ; the subscripts count completed steps.

Iterative process

A process that repeatedly applies a rule, using the current value to calculate the next value. For example, $u_{n+1}=0.8u_n+10$ means multiply the current value by $0.8$ , then add $10$ .

A rule relating successive values is also called a recurrence. Its notation is not on the formula sheet: learn how to translate the stated steps rather than memorising a rule for every context.

Worked example 1: loss before addition

Question. A tank starts with $20\,\mathrm{litres}$ . Each day it loses $20\%$ , then receives $10\,\mathrm{litres}$ . Write a rule and find the amount after two days.

Plan. Apply the percentage loss to the current amount, add the water, then repeat.

\begin{aligned} u_0&=20 &&\text{starting amount}\\ u_{n+1}&=0.8u_n+10 &&\text{loss, then addition}\\ u_1&=0.8(20)+10=26 &&\text{first complete day}\\ u_2&=0.8(26)+10=30.8 &&\text{reuse the latest value} \end{aligned}

Why is $10$ outside the multiplication? The new water arrives after that day's loss.

Answer. $u_0=20$ , $u_{n+1}=0.8u_n+10$ ; after two days, $30.8\,\mathrm{litres}$ .

Check. On day two the loss is $0.2(26)=5.2$ ; $26-5.2+10=30.8$ .

Change only the event order: add the water before the percentage loss.

Worked example 2: addition before loss

Question. The same tank starts with $20\,\mathrm{litres}$ . Each day it receives $10\,\mathrm{litres}$ , then loses $20\%$ . Write a rule and find the amount after two days.

Plan. Add first inside brackets so that the loss applies to the entire new amount.

\begin{aligned} v_0&=20 &&\text{same starting amount}\\ v_{n+1}&=0.8(v_n+10) &&\text{addition, then loss}\\ v_1&=0.8(20+10)=24 &&\text{first complete day}\\ v_2&=0.8(24+10)=27.2 &&\text{repeat in the same order} \end{aligned}

Why the brackets? The added water now undergoes the same day's loss.

Answer. $v_0=20$ , $v_{n+1}=0.8(v_n+10)$ ; after two days, $27.2\,\mathrm{litres}$ .

Check. Expanding gives $v_{n+1}=0.8v_n+8$ , which adds $2$ fewer litres per update than the first rule.

Sometimes the order is known but a multiplier is missing. Use a known current value and its next value to form an equation. Do not assume the increase between those two values repeats as a fixed addition.

Worked example 3: recovering a multiplier

Question. An account follows $u_{n+1}=Ku_n-50$ . A balance of £200 becomes £170 after one update. Find $K$ .

Plan. Substitute the consecutive balances and undo the withdrawal.

\begin{aligned} 170&=200K-50 &&\text{substitute current and next}\\ 220&=200K &&\text{add }50\\ K&=1.1 &&\text{divide by }200 \end{aligned}

Why restore £50? The observed next balance is after the withdrawal, whereas $200K$ is before it.

Answer. $K=1.1$ , corresponding to $10\%$ growth before the withdrawal.

Check. $1.1(200)-50=170$ .

Change only the observed next balance to £190.

Worked example 4: a different observed update

Question. For $u_{n+1}=Ku_n-50$ , £200 becomes £190 after one update. Find $K$ .

Plan. Use the same equation with the changed next value.

\begin{aligned} 190&=200K-50 &&\text{use the new observation}\\ 240&=200K &&\text{restore the withdrawal}\\ K&=1.2 &&\text{divide by the current value} \end{aligned}

Why can a falling balance still have $K>1$ ? The withdrawal can exceed the interest added.

Answer. $K=1.2$ , a $20\%$ increase before the withdrawal.

Check. £40 interest is added, then £50 is removed: $200+40-50=190$ .

The same care with event order applies to any context.

Watch out

Wrong: $0.8(u_n+10)=0.8u_n+10$ . The multiplier must act on both terms in the brackets. Correct: $0.8(u_n+10)=0.8u_n+8$ . The two event orders therefore give different rules.

Change only the observed next balance once more.

Successive values and target crossings (Higher)

Use the complete rule on the latest value each time, and keep enough precision for later steps.

For $v_0=40$ , $v_{n+1}=0.85v_n+7$ , the next values are $41$ , $41.85$ , $42.5725$ and $43.186625$ . Rounding each new value to a whole number before feeding it back instead gives $41$ , $42$ , $43$ , $44$ . The fourth result would be $44$ , rather than the correctly rounded $43$ . Keep the stored answer and round only the requested final result, unless the model explicitly specifies rounding at every step.

Worked example 5: finding the first value below a target

Question. An account starts at £1000. Each year it loses $10\%$ of its current balance, then £100 is withdrawn. When does it first fall below £350?

Plan. Repeat $u_{n+1}=0.9u_n-100$ until consecutive balances bracket the target.

\begin{aligned} u_1&=0.9(1000)-100=800 &&\text{first year}\\ u_2&=0.9(800)-100=620 &&\text{second year}\\ u_3&=0.9(620)-100=458 &&\text{third year}\\ u_4&=0.9(458)-100=312.2 &&\text{fourth year} \end{aligned}

Why use the latest balance? Earlier withdrawals also reduce the base for later percentage changes.

Answer. Four complete years: £458 is above £350, while £312.20 is below.

Check. Each update reduces the positive balance, so the earlier values cannot already be below £350.

Change only the target from £350 to £500.

Worked example 6: a different target

Question. For the same account, when does the balance first fall below £500?

Plan. Compare the same successive balances with the new target.

\begin{aligned} u_2&=0.9(800)-100=620 &&\text{above }500\\ u_3&=0.9(620)-100=458 &&\text{below }500 \end{aligned}

Why show both values? They establish “first”, rather than merely showing a later value below the target.

Answer. Three complete years.

Check. The initial balance and the first two balances are £1000, £800 and £620; all exceed £500.

A contextual rule also needs a stopping condition. A withdrawal cannot continue as stated when the funds are insufficient, and a tank cannot contain negative water.

Watch out

Wrong: $u_4=1000(0.9)^4-100$ . This withdraws £100 only once. Correct: repeat the whole rule four times, giving £312.20. Nor can you subtract £400 after pure decay: each earlier withdrawal changes later losses.

Change only the target once again.

Reversing successive changes (Higher)

Undo the last operation first, working backwards one complete period at a time.

If a step multiplies by $m$ , then subtracts $w$ , its forward form is $u_{n+1}=mu_n-w$ . Restore $w$ before dividing by $m$ .

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Papers 1, 2 and 3

R16 - General iterative processes

Modelling repeated percentage and fixed changes (Higher)

Successive values and target crossings (Higher)

Reversing successive changes (Higher)

Continue this lesson.

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