Algebra

Iteration

7 past-paper questions, easiest first. Try each one, then reveal the answer.

Higher only

Revision guide by The Full Marks

Iteration: what you need to know

Iteration is a way to get closer and closer to the solution of an equation you can't solve exactly. You feed each answer back into a formula, and the values settle down towards the root.

Key facts

An iteration formula looks like $x_{n+1} = \ldots x_n \ldots$ : put $x_0$ in to get $x_1$ , then $x_1$ to get $x_2$ , and so on.
A sign change of $f(x)$ between two values shows a root lies between them (for a continuous graph). For an equation like $f(x) = 7$ , show $f(x)$ goes from below 7 to above 7.
Rearranging the equation to $x = \ldots$ gives the iteration formula.
On a calculator, type the formula using ANS and press = repeatedly.
The values $x_1, x_2, x_3, \ldots$ are estimates of a solution of the original equation.
Real-life rules such as $P_{n+1} = 0.9P_n + 20$ work the same way: apply the rule once for each time step.

Showing a root lies in an interval

Substitute both ends of the interval into $f(x)$ .
Show one answer is negative and one is positive.
Write a conclusion: 'There is a sign change, so there is a solution between … and …'.

Using an iteration formula

Type the starting value and press =.
Type the formula with ANS in place of $x_n$ .
Press = once for each iteration and write each value down (at least 4 decimal places).

Worked examples

Example 1 Higher · grade 6

Use $x_{n+1} = 3 + \frac{2}{x_n}$ with $x_0 = 2$ to find $x_1$ , $x_2$ and $x_3$ .

Show the solution

$x_1 = 3 + \frac{2}{2} = 4$
$x_2 = 3 + \frac{2}{4} = 3.5$
$x_3 = 3 + \frac{2}{3.5} = 3.5714\ldots$

Example 2 Higher · grade 7

(a) Show that $x^3 - 4x - 2 = 0$ has a solution between 2 and 3. (b) Use $x_{n+1} = \sqrt[3]{4x_n + 2}$ with $x_0 = 2$ to find $x_3$ to 4 decimal places.

Show the solution

(a) $x = 2$ : $8 - 8 - 2 = -2$
$x = 3$ : $27 - 12 - 2 = 13$
There is a sign change, so there is a solution between 2 and 3.
(b) $x_1 = \sqrt[3]{10} = 2.1544\ldots$
$x_2 = 2.1979\ldots$
$x_3 = 2.2098$ (4 d.p.)

Example 3 Higher · grade 7

(a) Show that $x^3 - 2x^2 - 5 = 0$ can be rearranged to give $x = 2 + \frac{5}{x^2}$ . (b) Starting with $x_0 = 2.7$ , use $x_{n+1} = 2 + \frac{5}{x_n^2}$ to find $x_3$ to 4 decimal places. (c) What does your answer represent?

Show the solution

(a) $x^3 - 2x^2 - 5 = 0$
$x^3 = 2x^2 + 5$
Divide every term by $x^2$ : $x = 2 + \frac{5}{x^2}$
(b) $x_1 = 2.6858\ldots$
$x_2 = 2.6931\ldots$
$x_3 = 2.6894$ (4 d.p.)
(c) An estimate of a solution of $x^3 - 2x^2 - 5 = 0$ .

Common mistakes

Substituting values but not writing the conclusion about the sign change.
Rounding each value heavily before feeding it back in, so later values drift. Use ANS.
Doing one iteration too many or too few: $x_3$ means three steps from $x_0$ .

Exam tips

To show an equation rearranges to a formula, write every step of the rearranging. You can't just state it.

Questions people ask

What is iteration in GCSE Maths?

Repeatedly putting the answer back into a formula to get closer to the solution of an equation.

How do you do iteration on a calculator?

Enter the starting value and press =, then type the formula using ANS for $x_n$ . Each press of = gives the next value.

Exam questions: Iteration (7, easiest first)

Iteration3 marksGrade 6CalculatorHigher

June 2024 · Paper 3 Higher · Q17

Use an iterative formula to find the height of a bouncing ball after a given bounce.

Show answer

Answer	$1.331$ m (accept 1.33 or 1.3)	3 marks

Worked solution Checked against the mark scheme

$h_1 = 8$

$h_2 = 0.55 \times 8 = 4.4$

$h_3 = 0.55 \times 4.4 = 2.42$

$h_4 = 0.55 \times 2.42 = 1.331$ m

Watch out: Applying the multiplier four times ( $0.55^4 \times 8 = 0.732$ ) instead of three.

Iteration4 marksGrade 7CalculatorHigher

June 2023 · Paper 2 Higher · Q17

Rearrange a quartic equation into an iteration formula, then apply the formula three times from a starting value.

Show answer

(a)	$x^4 = x^2 + 5 \Rightarrow x = \sqrt[4]{x^2 + 5}$	1 mark
(b)	$1.669763...$	3 marks

Worked solution Checked against the mark scheme

(a) $x^4 - x^2 - 5 = 0 \Rightarrow x^4 = x^2 + 5 \Rightarrow x = \sqrt[4]{x^2 + 5}$

(b) $x_1 = \sqrt[4]{1.5^2 + 5} = 1.6409...$

$x_2 = \sqrt[4]{1.6409...^2 + 5} = 1.6653...$

$x_3 = \sqrt[4]{1.6653...^2 + 5} = 1.6697...$

Watch out: Using a square root instead of a fourth root on the calculator.

Iteration6 marksGrade 7CalculatorHigher

November 2023 · Paper 2 Higher · Q17

Show a cubic has a root between 1 and 2, rearrange it into an iterative form, and use the iteration twice.

Show answer

(a)	$f(1) = -3$ and $f(2) = 6$ ; sign change, so a root lies between 1 and 2	2 marks
(b)	$x^3 + 2x = 6 \Rightarrow x(x^2 + 2) = 6 \Rightarrow x = \frac{6}{x^2 + 2}$	1 mark
(c)	$1.4496$	3 marks

Worked solution Checked against the mark scheme

(a) $1^3 + 2(1) - 6 = -3$ and $2^3 + 2(2) - 6 = 6$ . The sign changes, so there is a solution between 1 and 2.

(b) $x^3 + 2x = 6$ , so $x(x^2 + 2) = 6$ , giving $x = \frac{6}{x^2 + 2}$ .

$x_2 = \frac{6}{1.46252...^2 + 2} = 1.44963... \approx 1.4496$

Watch out: In (a), working out the values but not stating that the sign changes.

Iteration4 marksGrade 7CalculatorHigher

November 2021 · Paper 2 Higher · Q16

Run an iteration formula three times from a starting value, then find the cubic equation it solves.

Show answer

(a)	$x_1 = 1.817$ , $x_2 = 1.853$ , $x_3 = 1.846$	3 marks
(b)	$a = 2$ , $b = -10$	1 mark

Worked solution Checked against the mark scheme

(a) $x_1 = \sqrt[3]{10 - 2 \times 2} = \sqrt[3]{6} = 1.817...$

$x_2 = \sqrt[3]{10 - 2 \times 1.817...} = 1.853...$

$x_3 = \sqrt[3]{10 - 2 \times 1.853...} = 1.846...$

(b) $x = \sqrt[3]{10 - 2x}$ gives $x^3 = 10 - 2x$ , so $x^3 + 2x - 10 = 0$ : $a = 2$ , $b = -10$ .

Watch out: Rounding each value heavily before putting it back in, so later values drift.

Iteration4 marksGrade 7CalculatorHigher

November 2019 · Paper 2 Higher · Q22

Apply a recurrence relation month by month to predict a rabbit population, then describe the long-term trend.

Show answer

(a)	163 (or 164)	3 marks
(b)	The number keeps falling, so eventually there will be no rabbits	1 mark

Worked solution Checked against the mark scheme

(a) April: $1.2 \times 200 - 50 = 190$

May: $1.2 \times 190 - 50 = 178$

June: $1.2 \times 178 - 50 = 163.6$ , so about 163 or 164 rabbits.

(b) The population drops each month, so over time it will keep decreasing until there are no rabbits left.

Watch out: Starting from March as n = 1 and doing only two steps, or one step too many.

Iteration6 marksGrade 7CalculatorHigher

June 2018 · Paper 3 Higher · Q18

Show a cubic equation has a root between 1 and 2, rearrange it into an iteration formula, and iterate three times from 2.

Show answer

(a)	$x = 1$ gives $2$ (below 7), $x = 2$ gives $10$ (above 7), so a solution lies between them	2 marks
(b)	$x^3 = 7 - x$ , so $x = \sqrt[3]{7 - x}$	1 mark
(c)	$1.74$	3 marks

Worked solution Checked against the mark scheme

(a) $1^3 + 1 = 2 < 7$ and $2^3 + 2 = 10 > 7$ . Since 7 is between 2 and 10, a solution lies between 1 and 2.

(b) $x^3 + x = 7 \Rightarrow x^3 = 7 - x \Rightarrow x = \sqrt[3]{7 - x}$

$x_2 = \sqrt[3]{5.29002} = 1.74242$

$x_3 = \sqrt[3]{5.25758} = 1.73885 \approx 1.74$

Watch out: In (a), substituting the values but not saying why that shows a solution.

Iteration6 marksGrade 7CalculatorHigher

Specimen · Paper 3 Higher · Q21

Rearrange a cubic equation into an iterative formula, carry out three iterations from a given start value, and say what the results mean.

Show answer

(a)	Divide $x^3 = 3x^2 + 3$ by $x^2$ to get $x = 3 + \frac{3}{x^2}$	2 marks
(b)	$x_1 = 3.2930$ , $x_2 = 3.2767$ , $x_3 = 3.2794$	3 marks
(c)	They are estimates (approximations) of a solution of $3x^2 - x^3 + 3 = 0$	1 mark

Worked solution Checked against the mark scheme

(a) $3x^2 - x^3 + 3 = 0 \Rightarrow x^3 = 3x^2 + 3$ . Divide every term by $x^2$ : $x = 3 + \frac{3}{x^2}$

(b) $x_1 = 3 + \frac{3}{3.2^2} = 3.29296...$

$x_2 = 3 + \frac{3}{3.29296...^2} = 3.27666...$

$x_3 = 3 + \frac{3}{3.27666...^2} = 3.27942...$

Watch out: Rounding each value before feeding it back in, which drifts the later answers.