An approximate solution of x^3 = x + 5 is found using the iterative formula x_n+1 = [3]x_n + 5 Starting with x_0 = 1.8, work out the values of x_1, x_2 and x_3. Give each value correct to 5 decimal places.

x1 = 1.89454, x2 = 1.90328, x3 = 1.90408

Is iteration on the Foundation paper?

No. Spec ref A20, finding approximate solutions to equations numerically using iteration, is Higher tier only on Edexcel 1MA1. Foundation candidates are never asked to use an iterative formula.

How many marks is iteration worth on Edexcel Higher?

It usually turns up as one question in two or three parts worth 4-5 marks altogether: one mark for showing the rearrangement, two or three for running the iteration, and sometimes one for interpreting the result. A sign change question showing a solution lies between two values is often a separate 2-mark item.

Can I use the ANS button on my calculator?

Yes, and it is much faster than retyping. Enter the starting value and press equals so it is stored, type the right-hand side of the iterative formula with ANS wherever x_n appears, then press equals once for each new estimate. Press it once per value and write each one down as it appears, because there is no way back if you overshoot.

What does it mean if the iteration does not converge?

The values move further apart instead of settling, or the formula asks for the square root of a negative number and the calculator gives an error. That rearrangement cannot find the solution from that starting value, even though the equation still has one. A different rearrangement of the same equation, or a starting value closer to the solution, will usually work.

Iteration - Edexcel GCSE Higher Maths

Typically 4-5 marks, almost always one multi-part question · Spec 1MA1

A cubic like $x^3 + 5x = 30$ has a solution, but no factorising or formula will find it. Iteration is the numerical way in: rearrange the equation so that one $x$ sits alone on the left, turn that rearrangement into $x_{n+1} = f(x_n)$ , feed in a starting value $x_0$ and keep feeding each output back in as the next input. The estimates $x_1$ , $x_2$ , $x_3$ close in on a solution. This is spec ref A20 on Edexcel's 1MA1 specification and it is Higher tier only, so a Foundation candidate never meets it.

Edexcel sets it as one question in three short parts, usually 4-5 marks in total. Part (a) is a show that: show that $x^3 + 5x = 30$ can be rearranged to give $x = \sqrt[3]{30 - 5x}$ , worth a single mark for the intermediate line of algebra. Part (b) gives you $x_0$ and asks for $x_1$ , $x_2$ and $x_3$ to a stated accuracy, or asks you to carry on and write the solution to 3 decimal places. Part (c) asks what those numbers estimate. The arithmetic is calculator work, so expect it on Paper 2 or Paper 3 rather than Paper 1, and learn the ANS key: type the starting value, press equals, then type the right-hand side using ANS in place of $x_n$ and press equals once per iteration.

Two habits earn the accuracy marks. Never round part way through, because the whole display of $x_1$ has to go into the calculation of $x_2$ , and a rounded $2.60$ instead of $2.596247$ will move the fourth decimal place of $x_3$ . And answer the question that was asked: if it says $x_3$ , stop at $x_3$ . Scripts that run the iteration to seven values, or that average $x_1$ , $x_2$ and $x_3$ and offer the mean, lose the accuracy mark with all the method marks already earned. The other half of the topic is the sign change: substitute the two ends of the interval into $f(x)$ written as $f(x) = 0$ , show one answer is negative and the other positive, then write the conclusion in words.

Worked example

(a) Show that the equation

x^3 + 5x = 30

can be rearranged to give

x = \sqrt[3]{30 - 5x}

(1 mark) (b) Starting with

x_0 = 2.5

, use the iterative formula

x_{n+1} = \sqrt[3]{30 - 5x_n}

to work out the values of

x_1

x_2

and

x_3

. Give each value to 4 decimal places. (3 marks) (c) Explain what the value of

x_3

is an estimate of. (1 mark)

[5 marks]

Subtract $5x$ from both sides to isolate the cube: $x^3 = 30 - 5x$ . Now take the cube root of both sides: $x = \sqrt[3]{30 - 5x}$ .
B1: the middle line $x^3 = 30 - 5x$ has to be on the page. A script that writes the given answer straight under the given equation shows nothing and scores zero.
Substitute $x_0 = 2.5$ : $x_1 = \sqrt[3]{30 - 5 \times 2.5} = \sqrt[3]{17.5} = 2.5962470...$ , so $x_1 = 2.5962$ to 4 decimal places.
M1: correct substitution of 2.5 into the right-hand side. Make sure the bar of the root sign covers the whole of $30 - 5x_n$ .
Put the full display back in, not the rounded 2.5962: $x_2 = \sqrt[3]{30 - 5 \times 2.5962470...} = 2.5722273...$ , so $x_2 = 2.5722$ .
A1: 2.5722. Using the rounded $x_1$ here gives 2.5722 as well this time, but on other iterations it shifts the last digit and loses the mark.
One more press of equals: $x_3 = \sqrt[3]{30 - 5 \times 2.5722273...} = 2.5782637...$ , so $x_3 = 2.5783$ . Stop here, because $x_3$ is what was asked for.
A1: 2.5783
$x_3$ is an estimate of a solution of the original equation $x^3 + 5x = 30$ .
B1: naming the equation being solved. Saying it is an estimate of $x$ on its own is not enough.

Practice questions

These are original questions in Edexcel 1MA1 Higher style. Iteration is calculator work, so use a scientific calculator and its ANS key, and keep every digit until the final rounding. Where a question states an accuracy, answer to exactly that accuracy: the marking accepts your unrounded value too, but the exam will not.

No account needed: the questions below are always free to check, plus 8 extra questions today. Your score so far updates as you go.

Question 1 Warm-up [1 mark]

A sequence is generated by the iterative formula $x_{n+1} = \sqrt{3x_n + 4}$ Starting with $x_0 = 7$ , work out the value of $x_1$ .

Your answer (fractions like 1/2 are fine)
Question 2 Warm-up [2 marks]

A sequence is generated by the iterative formula $x_{n+1} = \sqrt{2x_n + 6}$ Starting with $x_0 = 5$ , work out the values of $x_1$ and $x_2$ . Give each value correct to 3 decimal places.

x1 = x2 =
Question 3 Exam pace [3 marks]

The equation $x^3 - 5x + 1 = 0$ can be rearranged to give the iterative formula $x_{n+1} = \frac{(x_n)^3 + 1}{5}$ Starting with $x_0 = 0$ , work out the values of $x_1$ and $x_2$ . Give each value correct to 4 decimal places.

x1 = x2 =
Question 4 Exam pace [3 marks]

A reservoir loses 15% of its water each week, and 12000 litres are pumped in at the end of every week. The volume in litres at the start of week $n$ is modelled by $V_{n+1} = 0.85V_n + 12000$ The volume at the start of week 0 is 40000 litres. Work out the volume at the start of week 3. Give your answer in litres.

Your answer (fractions like 1/2 are fine)
Question 5 Stretch [3 marks]

An approximate solution of $x^3 = x + 5$ is found using the iterative formula $x_{n+1} = \sqrt[3]{x_n + 5}$ Starting with $x_0 = 1.8$ , work out the values of $x_1$ , $x_2$ and $x_3$ . Give each value correct to 5 decimal places.

x1 = x2 = x3 =
Question 6 Stretch [2 marks]

The iterative formula $x_{n+1} = \sqrt{6 - x_n}$ with $x_0 = 1$ produces the estimates 2.23607, 1.94009, 2.01492, 1.99627, 2.00093, ... These estimates converge to a solution of $x^2 + x - 6 = 0$ . Write down the exact value they converge to.

Your answer (fractions like 1/2 are fine)

Common mistakes examiners see

Rounding $x_1$ to the requested accuracy and then typing that rounded number back into the formula, so $x_2$ and $x_3$ drift in the last decimal place.

Round only when you write the value down. Keep the full number on the calculator by using ANS: enter $x_0$ , press equals, type the right-hand side with ANS where $x_n$ appears, then press equals once for each new value.
Misreading $x_{n+1}$ as one more than $x_n$ , so $x_0 = 2$ leads to writing $x_1 = 3$ before any substitution happens.

The subscript is a counter for position in the list, not a value. $x_{n+1}$ means the next estimate, produced by putting $x_n$ into the right-hand side.
Working out $x_1$ , $x_2$ and $x_3$ correctly and then averaging them, or carrying on to $x_6$ and offering that instead.

Write the three values on three separate lines, labelled, and stop at the one named in the question. The method marks are already yours by then; the accuracy mark is only for the value asked for.
In a show that part, writing the printed answer straight underneath the printed equation with no algebra between them.

Show the rearrangement one operation at a time. For $x^3 + 5x = 30$ that is $x^3 = 30 - 5x$ on its own line, then the cube root of both sides. One line of working is the whole mark.
For a sign change, substituting the two values into the iterative formula, or working out $f(1)$ and $f(2)$ and stopping without a sentence.

Rearrange to $f(x) = 0$ first, substitute both ends into that $f(x)$ , then write the conclusion: one value is negative, the other is positive, so $f(x)$ crosses zero and there is a solution between them.

Frequently asked questions

Is iteration on the Foundation paper?: No. Spec ref A20, finding approximate solutions to equations numerically using iteration, is Higher tier only on Edexcel 1MA1. Foundation candidates are never asked to use an iterative formula.
How many marks is iteration worth on Edexcel Higher?: It usually turns up as one question in two or three parts worth 4-5 marks altogether: one mark for showing the rearrangement, two or three for running the iteration, and sometimes one for interpreting the result. A sign change question showing a solution lies between two values is often a separate 2-mark item.
Can I use the ANS button on my calculator?: Yes, and it is much faster than retyping. Enter the starting value and press equals so it is stored, type the right-hand side of the iterative formula with ANS wherever $x_n$ appears, then press equals once for each new estimate. Press it once per value and write each one down as it appears, because there is no way back if you overshoot.
What does it mean if the iteration does not converge?: The values move further apart instead of settling, or the formula asks for the square root of a negative number and the calculator gives an error. That rearrangement cannot find the solution from that starting value, even though the equation still has one. A different rearrangement of the same equation, or a starting value closer to the solution, will usually work.

Iteration - Edexcel GCSE Higher Maths

Worked example

Practice questions

Common mistakes examiners see

Frequently asked questions

Related topics