A-level · Numerical methods

Iterative methods (x = g(x))Exam-style questions

Solving f(x) = 0 by rearranging it into x = g(x) and iterating x_(n+1) = g(x_n) from a starting value — why some rearrangements converge and others diverge, and reading staircase and cobweb diagrams.

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20 questions · A-level level · mark scheme and worked solution on each · no sign-up to try

DifficultyAll Level 2 Level 3 Level 4 Level 5

1
$f(x) = x^3 - 6x + 2$ .
(a) Show $f(x) = 0$ has a root $\alpha$ in $[0, 1]$ .
(b) Using $x_{n+1} = \dfrac{x_n^3 + 2}{6}$ with $x_0 = 0$ , find $\alpha$ to 3 decimal places.
Level 4 · 5 marksTry it
2
$f(x) = x^{3} - 5x + 1$ .
(a) Show that $f(x) = 0$ has a root $\alpha$ in the interval $[0, 1]$ . (2 marks)
(b) Show that the equation can be rearranged as $x = \dfrac{x^{3} + 1}{5}$ . (1 mark)
(c) Using $x_{0} = 0.2$ , apply the iteration $x_{n+1} = \dfrac{x_{n}^{3} + 1}{5}$ to find $\alpha$ correct to 3 decimal places. (3 marks)
Level 4 · 6 marksTry it
3
$f(x) = x^3 + 3x - 7$ . The equation $f(x) = 0$ has a single real root $\alpha$ .
(a) Show that $\alpha$ lies between $1.4$ and $1.45$ . (2 marks)
(b) Show that the equation $f(x) = 0$ can be rearranged as $x = \sqrt[3]{7 - 3x}$ . (1 mark)
(c) Using the iterative formula $x_{n+1} = \sqrt[3]{7 - 3x_n}$ with $x_0 = 1.5$ , find $x_1$ , $x_2$ and $x_3$ , giving each to 4 decimal places. (3 marks)
(d) By choosing a suitable interval, show that $\alpha = 1.406$ correct to 3 decimal places. (2 marks)
Level 4 · 8 marksTry it
4
The equation

$2^x + 3x = 20$

has only one solution, which is denoted by $\alpha$
(a) Use a suitable change of sign to show that $3 < \alpha < 4$

(2 marks)
(b) Show that the equation $2^x + 3x = 20$ can be written as

$x = \frac{\ln(20 - 3x)}{\ln 2}$

(3 marks)
(c) Starting with $x_1 = 3$ , use the iterative formula

$x_{n+1} = \frac{\ln(20 - 3x_n)}{\ln 2}$

to find the values of $x_2$ , $x_3$ and $x_4$ , giving each one to three decimal places.

(2 marks)
Level 4 · 7 marksTry it
5
The function $\text{f}$ is defined by $\text{f}(x) = 4\ln x + 3 - x^2$ for $x > 0$ . The diagram shows the curve $y = \text{f}(x)$ , which crosses the $x$ -axis at two points. The smaller root of the equation $\text{f}(x) = 0$ is denoted by $\alpha$ .(a) Using the interval from $x = 0.25$ to $x = 3$ , a change of sign method gives no information about $\alpha$ . Explain why.

(1 mark)
(b) Show that the equation $\text{f}(x) = 0$ may be rearranged as $x = \text{e}^{\frac14(x^2 - 3)}$ .

(2 marks)
(c) Use the iterative formula $x_{n+1} = \text{e}^{\frac14(x_n^2 - 3)}$ , with $x_1 = 0.5$ , to find $\alpha$ correct to $4$ significant figures. You should show the result of each iteration.

(3 marks)
(d) The equation $\text{f}(x) = 0$ can also be written as $x = \text{F}(x)$ , where $\text{F}(x) = \sqrt{4\ln x + 3}$ .

Find $\text{F}'(0.5)$ , and hence explain why the iteration $x_{n+1} = \text{F}(x_n)$ will not converge to $\alpha$ .

(3 marks)
Level 4 · 9 marksTry it