A20AlgebraHigherGrades 7–9

Iteration

There is no factorising trick for $x^{3} - 4 x - 1 = 0$ . Instead, you can find an approximate solution by making a guess and repeatedly improving it. Each improvement is an $iteration$ , and a calculator does the repetitive work in seconds.

40 min AQA GCSE Maths

The big picture

The method has three stages. First show a solution exists between two values by finding a change of sign. Next rearrange the equation into the form $x = g (x)$ . Finally use $x_{n + 1} = g (x_{n})$ , feeding each answer back in, until the values settle.

AQA questions often give the iterative formula and ask you to show where it came from, use it for a few steps, and then justify the accuracy of the final answer.

What you'll learn

Show a root lies in an interval using a change of sign
Rearrange an equation into the form $x = g (x)$
Use an iterative formula to find approximate solutions
Justify an answer to a given degree of accuracy

Locating a root

If a continuous graph goes from negative to positive (or the other way) between two $x$ -values, it must cross the $x$ -axis between them.

For $f (x) = x^{3} - 4 x - 1$ : $f (2) = 8 - 8 - 1 = - 1$ and $f (3) = 27 - 12 - 1 = 14$ .

f (2) = - 1 < 0

f (3) = 14 > 0

AnswerThere is a change of sign, so there is a solution between 2 and 3.

Your turnSame method, new numbers — try it before moving on.

For $f (x) = x^{3} - 5 x + 1$ , work out $f (2)$ .

Your answer

Tip — Write the conclusion in words. "Change of sign, so a root lies between…" is the mark.

Rearranging into x = g(x)

Isolate one $x$ term. From $x^{3} - 4 x - 1 = 0$ : $x^{3} = 4 x + 1$ , so $x = 3 4 x + 1$ .

Many rearrangements are possible; exam questions tell you which one to show.

x^{3} = 5 - 2 x

2Cube root both sides.

Answer

x = 3 5 - 2 x

as required.

Using the iterative formula

Turn the rearrangement into $x_{n + 1} = 3 4 x_{n} + 1$ and start from $x_{0} = 2$ .

On a calculator: type 2, press =, then type the formula using ANS and press = repeatedly.

x_{1} = 39 = 2.0801

x_{2} = 3 9.3203 = 2.1045

x_{3} = 2.1118

Answer

x_{1} = 2.0801

x_{2} = 2.1045

x_{3} = 2.1118

Your turnSame method, new numbers — try it before moving on.

Using $x_{n + 1} = 3 4 x_{n} + 1$ with $x_{0} = 2$ , find $x_{1}$ to 4 decimal places.

Your answer

To 4 decimal places.

Why this works

The values climb towards about 2.114 and change less each time. Each iteration is a better estimate of where the graph of $y = x^{3} - 4 x - 1$ crosses the $x$ -axis.

Justifying accuracy

To show a root is 2.115 to 3 decimal places, test the bounds 2.1145 and 2.1155.

$f (2.1145) = - 0.0038$ and $f (2.1155) = 0.0056$ . The sign changes, so the root lies between them and rounds to 2.115.

Tip — Always test the bounds of the rounded answer, not the rounded answer itself.

Think like an examiner

Show the substituted values when checking for a sign change.
Write each iteration, not just the final one.
Use unrounded calculator values in the next iteration.

Remember these

f (a) < 0 < f (b)

Sign change: a root lies between a and b.

x_{n + 1} = g (x_{n})

Iterative formula.

x = 3 4 x + 1

A rearrangement of x cubed minus 4x minus 1 = 0.

Watch out for these

Rounding each iterate before using it again.

Use ANS to carry the full value forward.

Saying

f (2) = - 1

is a solution.

A solution makes

f (x) = 0

; the sign change only locates it.

Testing 2.115 itself to justify 3 d.p.

Test 2.1145 and 2.1155.

Stating a sign change without a conclusion.

Say a root lies between the two values.

Stretch yourself

Show that $x^{3} + 2 x - 5 = 0$ has a root between 1.3 and 1.4. Starting with $x_{0} = 1$ , use $x_{n + 1} = 3 5 - 2 x_{n}$ to find $x_{4}$ to 4 decimal places.

Hint — Evaluate $x^{3} + 2 x - 5$ at 1.3 and 1.4 first.

Questions students ask

Do iterations always get closer to the root?

No — some rearrangements move away. GCSE questions give formulas that work.

Which starting value should I use?

Use the one given; otherwise a value from the sign-change interval.

How many iterations are enough?

Keep going until successive values agree to the accuracy required.

Key takeaways

A sign change shows a root lies in an interval.
Rearrange $f (x) = 0$ into $x = g (x)$ .
Iterate with $x_{n + 1} = g (x_{n})$ using the ANS key.
Justify accuracy by testing the bounds.

Exam-style questions

Practise it the way it is examined

3 original questions · 8 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 72 marks

Show that the equation $x^{3} - 6 x - 4 = 0$ has a solution between $x = 2$ and $x = 3$ .[2]

Question 2Grade 84 marks

(a)Using $x_{n + 1} = 3 6 x_{n} + 4$ with $x_{0} = 2.5$ , find $x_{1}$ , $x_{2}$ and $x_{3}$ to 4 decimal places.[2]
(b)Continue the iteration to find the solution correct to 2 decimal places.[2]

Question 3Grade 92 marks

Show that $x^{3} - 2 x^{2} - 5 = 0$ can be rearranged to give $x = 2 + \frac{5}{x ^{2}}$ .[2]

Unlimited practice

Skill drill

A new question every time, marked instantly, with a full worked solution. Aim for a streak of five, then move up a level.

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Real past-paper questions on Iteration, marked mark-by-mark. How you do feeds straight into your weak-topic list, so your revision keeps targeting what actually needs work.

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Iteration

Locating a root

Rearranging into x = g(x)

Using the iterative formula

Justifying accuracy

Practise it the way it is examined

Skill drill

Practise this topic

Related topics

Iteration

Locating a root

Rearranging into x = g(x)

Using the iterative formula

Justifying accuracy

Practise it the way it is examined

Skill drill

Practise this topic

Related topics