GCSE · Iteration

Using an iterative formulaExam-style questions

How x_{n+1} = g(x_n) feeds each estimate into the next, the full-precision discipline (never round mid-chain), and what the settling sequence tells you about the equation's root.

Start practising Read the lesson

11 questions · GCSE (Higher) level · mark scheme and worked solution on each · no sign-up to try

DifficultyAll Level 4 Level 5

1
(a) Show that the equation $x^3 - 3x - 7 = 0$ can be written in the form $x = \sqrt[3]{3x + 7}$

(1 mark)
(b) Starting with $x_0 = 2$

use the iteration formula $x_{n+1} = \sqrt[3]{3x_n + 7}$ three times to find an estimate for a solution of $x^3 - 3x - 7 = 0$

(3 marks)
Level 4 · 4 marksTry it
2
$x_{n+1} = \sqrt[3]{7x_n + 8}, \qquad x_1 = 3$

Work out the values of $x_2$ and $x_3$

Give your answers correct to 3 decimal places.
Level 4 · 3 marksTry it
3
$x_{n+1} = \sqrt[3]{20 - x_n}, \qquad x_1 = 2.5$

Work out the values of $x_2$ and $x_3$

Give your answers correct to 4 decimal places.
Level 4 · 3 marksTry it
4
$x_{n+1} = \sqrt[3]{7x_n + 8}$

Given that $x_3 = 3.09009$ , work out the value of $x_4$

Give your answer correct to 3 decimal places.
Level 4 · 2 marksTry it
5
$x_{n+1} = \sqrt{3x_n + 7}, \qquad x_1 = 4$

(a) Work out the values of $x_2$ and $x_3$ , correct to 3 decimal places.

(b) Explain how you could find a more accurate solution of $x^2 - 3x - 7 = 0$
Level 4 · 3 marksTry it

3 more questions on using an iterative formula

Free with an account — just an email, no card. The whole bank, every mark scheme and worked solution.

Start free