Iterative Methods – OCR FSMQ Additional Maths

1 The Idea

NM2: use a simple iterative method to solve equations approximately. An iteration feeds each answer back into the same formula, so the estimates improve with every step — often very fast.

The method in two steps

1. Rearrange $\mathrm{f}(x) = 0$ into the form $x = \mathrm{g}(x)$
2. Iterate: $x_{n+1} = \mathrm{g}(x_n)$, starting from an estimate $x_1$

Why it works when it works. If the iterates settle on a value $L$, then $L = \mathrm{g}(L)$ — and rearranging that back gives $\mathrm{f}(L) = 0$. So the limit of the sequence is a root of the original equation.

Use the ANS key. Type your starting value and press $=$. Then type the formula with ANS wherever $x$ appears, and press $=$ repeatedly. Each press gives the next iterate. The specification expects a calculator with this facility.

Never round between steps. Feed the calculator's full-precision value back in each time. Rounding to 3 decimal places at every step can stop the sequence converging to the right answer at all.

2 Rearranging Into $x = \mathrm{g}(x)$

The same equation can be rearranged many ways, and the choice matters enormously — some rearrangements converge and others do not.

Worked Example 1 — Three rearrangements of one equation

Find three ways of writing $x^3 - 5x + 1 = 0$ in the form $x = \mathrm{g}(x)$.

①Isolate the $x$ term: $5x = x^3 + 1$, so $x = \dfrac{x^3+1}{5}$

②Isolate the $x^3$: $x^3 = 5x - 1$, so $x = \sqrt[3]{5x-1}$

③Divide through by $x$: $x^2 - 5 + \tfrac1x = 0$, so $x = \sqrt{5 - \tfrac1x}$

All three are algebraically equivalent, but they behave completely differently as iterations. Trying a rearrangement and finding it diverges is normal — switch to another.

Worked Example 2 — A converging iteration

Use $x_{n+1} = \dfrac{x_n^3+1}{5}$ with $x_1 = 0$ to find a root of $x^3 - 5x + 1 = 0$ to $4$ decimal places.

①$x_1 = 0$

②$x_2 = \dfrac{0+1}{5} = 0.2$

③$x_3 = \dfrac{0.008+1}{5} = 0.2016$

④$x_4 = \dfrac{0.0081936+1}{5} = 0.2016387$

⑤$x_5 = 0.2016397$

⑥$x_6 = 0.2016397$ — the value has stabilised.

⑦The root is $\mathbf{0.2016}$ to $4$ d.p.

⑧Verify: $\mathrm{f}(0.2016) = 0.00819 - 1.008 + 1 = 0.00019 \approx 0$ ✓

Always verify. Substituting your answer back into the original equation should give something very close to zero. It is the only way to be sure you converged to the right thing.

Worked Example 3 — A different root from a different start

The equation $x^3 - 5x + 1 = 0$ also has a root near $2$. Use $x_{n+1} = \sqrt[3]{5x_n - 1}$ with $x_1 = 2$.

①$x_2 = \sqrt[3]{9} = 2.080084$

②$x_3 = \sqrt[3]{9.400419} = 2.110486$

③$x_4 = \sqrt[3]{9.552428} = 2.121801$

④$x_5 = \sqrt[3]{9.609004} = 2.125981$

⑤Continuing, the iterates approach $2.128$.

⑥The root is $\mathbf{2.128}$ to $3$ d.p.

An iteration finds one root, not all of them. Which root you reach depends on both the rearrangement and the starting value. This cubic has three roots; each needs its own approach.

3 Convergence and Divergence

Behaviour	What you see
Converges	The iterates settle towards one value; successive differences shrink
Diverges	The values grow without limit, or produce a calculator error
Oscillates and converges	Alternates above and below the root, closing in each time
Oscillates and diverges	Alternates, but the swings get bigger

How to report the answer

Iterate until successive values agree to the required accuracy,
then round. Show at least the first four iterates as evidence.

4 When Iteration Fails (NM3)

Worked Example 4 — A diverging rearrangement

Try $x_{n+1} = x_n^3 - 4x_n$ … or rather, take the equation $x^3 - 5x + 1 = 0$ rearranged as $x = x^3 - 4x + 1$, with $x_1 = 0.2$.

①$x_1 = 0.2$

②$x_2 = 0.008 - 0.8 + 1 = 0.208$

③$x_3 = 0.008999 - 0.832 + 1 = 0.176999$

④$x_4 = 0.005545 - 0.707996 + 1 = 0.297549$

⑤$x_5 = 0.026344 - 1.190196 + 1 = -0.163854$

⑥$x_6 = -0.004399 + 0.655408 + 1 = 1.651018$

⑦The values are swinging further and further apart — the iteration diverges, even though it started right next to the root.

The same root, the same starting value, a different rearrangement — and now it fails. This is the central lesson of NM3.

The three ways iteration fails

1. The chosen rearrangement diverges from that root
2. The starting value is too far away
3. The formula becomes undefined — a negative square root, or division by zero

Worked Example 5 — Breaking down completely

Explain what happens using $x_{n+1} = \sqrt{x_n - 3}$ with $x_1 = 2$.

①$x_2 = \sqrt{2-3} = \sqrt{-1}$

②This is not a real number, so the calculator returns an error.

③The iteration fails immediately — not because it diverges, but because it is undefined.

④A starting value of $x_1 = 4$ or more would keep the expression under the root non-negative.

Check the domain before you start. An iteration involving a square root or a fraction needs a starting value that keeps it defined — and keeps every later iterate defined too.

What to do when an iteration fails

Try a different rearrangement, or a different starting value.
Use a change of sign first, to get a starting value close to the root.

5 Quick Reference

The form

Rearrange to $x = \mathrm{g}(x)$.

The iteration

$x_{n+1} = \mathrm{g}(x_n)$ with a starting value.

ANS key

One button press per step.

Full accuracy

Never round between steps.

Why it works

The limit satisfies $L = \mathrm{g}(L)$.

Stop when

Successive values agree to the accuracy wanted.

Verify

Substitute back into $\mathrm{f}(x) = 0$.

One root only

Which one depends on $\mathrm{g}$ and on $x_1$.

Can diverge

A different rearrangement may fail.

Can be undefined

Watch for negative square roots.

6 Practice Questions

Question 1

Rearrange $x^3 - 7x + 2 = 0$ into the form $x = \mathrm{g}(x)$ in two different ways.

▶ Show solution

$7x = x^3 + 2$, so $x = \dfrac{x^3+2}{7}$

$x^3 = 7x - 2$, so $x = \sqrt[3]{7x-2}$

Question 2

Use $x_{n+1} = \dfrac{x_n^3+2}{7}$ with $x_1 = 0$ to find $x_2$, $x_3$ and $x_4$.

▶ Show solution

$x_2 = \dfrac{0+2}{7} = 0.285714$

$x_3 = \dfrac{0.023324+2}{7} = 0.289046$

$x_4 = \dfrac{0.024148+2}{7} = 0.289164$

Question 3

Give the root found in Question 2 to $3$ decimal places.

▶ Show solution

The iterates are settling at $0.28917$, so the root is $\mathbf{0.289}$ to $3$ d.p.

Question 4

Use $x_{n+1} = \sqrt{\dfrac{10}{x_n + 1}}$ with $x_1 = 2$ to find $x_2$ and $x_3$.

▶ Show solution

$x_2 = \sqrt{10 \div 3} = \sqrt{3.333333} = 1.825742$

$x_3 = \sqrt{10 \div 2.825742} = \sqrt{3.538882} = 1.881195$

Question 5

An iteration gives $x_4 = 1.4142$ and $x_5 = 1.4142$. What can you conclude?

▶ Show solution

The iterates agree to $4$ decimal places, so the sequence has converged and the root is $1.4142$ to $4$ d.p.

(Strictly, agreement to $4$ d.p. is strong evidence rather than proof; a change of sign check on either side of $1.4142$ would confirm it.)

Question 6

Explain why $x_{n+1} = \sqrt{x_n - 10}$ with $x_1 = 3$ fails.

▶ Show solution

$x_2 = \sqrt{3 - 10} = \sqrt{-7}$, which is not a real number.

The iteration is undefined at the very first step, so it fails immediately. A starting value of at least $10$ would be needed to keep the expression under the root non-negative.

Question 7

Use $x_{n+1} = \sqrt[3]{2x_n + 3}$ with $x_1 = 2$ to find the root to $3$ d.p.

▶ Show solution

$x_2 = \sqrt[3]{7} = 1.912931$

$x_3 = \sqrt[3]{6.825862} = 1.896935$

$x_4 = \sqrt[3]{6.793871} = 1.893967$

$x_5 = \sqrt[3]{6.787934} = 1.893415$

$x_6 = \sqrt[3]{6.786831} = 1.893313$

Converging to $\mathbf{1.893}$ to $3$ d.p.

Check: this solves $x^3 - 2x - 3 = 0$; and $1.893^3 - 2(1.893) - 3 = 6.783 - 6.786 = -0.003 \approx 0$ ✓

Question 8

Show that the iteration $x_{n+1} = x_n^2 - 2$ with $x_1 = 3$ diverges.

▶ Show solution

$x_2 = 9 - 2 = 7$

$x_3 = 49 - 2 = 47$

$x_4 = 2209 - 2 = 2207$

$x_5 = 4\,870\,847$

The values grow explosively, so the iteration diverges.

(The equation $x = x^2 - 2$ does have roots, at $x = 2$ and $x = -1$, but this iteration runs away from both unless started between them.)

Question 9

Explain why two students using the same equation but different rearrangements might get different roots.

▶ Show solution

An iteration converges only to roots that are "attracting" for that particular function $\mathrm{g}$, and only from starting values close enough to them.

A different rearrangement gives a different $\mathrm{g}$, with a different pattern of which roots attract and which repel.

So one student's formula may converge to a root near $0$, while the other's converges to a root near $2$ — both correct roots of the same equation, reached by different routes. Neither student can claim to have found all the roots without further work.

Question 10

The equation $x^3 - 6x - 4 = 0$ is to be solved.

(a) Show there is a root between $2$ and $3$. (b) Show that the equation can be written as $x = \sqrt[3]{6x+4}$. (c) Use this iteration with $x_1 = 2.5$ to find the root to $3$ d.p. (d) Show that the rearrangement $x = \dfrac{x^3-4}{6}$ fails from the same starting value, and explain why.

▶ Show solution

(a) $\mathrm{f}(2) = 8 - 12 - 4 = -8 < 0$

$\mathrm{f}(3) = 27 - 18 - 4 = 5 > 0$

Sign change with $\mathrm{f}$ continuous, so a root lies in $(2, 3)$.

(b) $x^3 - 6x - 4 = 0 \;\Rightarrow\; x^3 = 6x + 4 \;\Rightarrow\; x = \sqrt[3]{6x+4}$ ✓

(c) $x_1 = 2.5$

$x_2 = \sqrt[3]{19} = 2.668402$

$x_3 = \sqrt[3]{20.010410} = 2.714888$

$x_4 = \sqrt[3]{20.289331} = 2.727444$

$x_5 = \sqrt[3]{20.364666} = 2.730816$

$x_6 = \sqrt[3]{20.384896} = 2.731720$

$x_7 = \sqrt[3]{20.390319} = 2.731962$

The iterates are converging, so the root is $\mathbf{2.731}$ to $3$ d.p.

Verify: $2.731^3 - 6(2.731) - 4 = 20.369 - 16.386 - 4 = -0.017 \approx 0$ ✓

(d) Using $x_{n+1} = \dfrac{x_n^3-4}{6}$ from $x_1 = 2.5$:

$x_2 = \dfrac{15.625-4}{6} = 1.937500$

$x_3 = \dfrac{7.273193-4}{6} = 0.545532$

$x_4 = \dfrac{0.162355-4}{6} = -0.639608$

$x_5 = \dfrac{-0.261625-4}{6} = -0.710277$

$x_6 = -0.726388$, $x_7 = -0.730545$ …

The sequence has moved away from the root near $2.731$ and is settling on a different root, near $-0.732$.

Why: the cubic $x^3-6x-4$ has three roots, at approximately $-2.000$, $-0.732$ and $2.731$. This rearrangement makes the root at $2.731$ repelling and the one at $-0.732$ attracting, so it converges to the wrong one. The first rearrangement has the opposite behaviour.

This is precisely the failure NM3 asks you to recognise: a perfectly valid rearrangement can converge to a root other than the one you wanted, or to none at all.

Iterative Methods (NM2, NM3) · OCR FSMQ Additional Maths · Created with MathJax