AS & A Level · Practice questions · Iterative methods

Round and round to a root.

Six original Cambridge-style questions on iterative formulae: which equation they solve, iterating to a required accuracy, exact limits and iterations that fail.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Rearrange, iterate, stop, check.

Rearrange the equation into \(x = F(x)\), or check that a given \(F\) comes from it.
Iterate with ANS, writing each value to one more decimal place than needed.
Stop when successive values agree to the required accuracy.
Exact limit: put \(x_{n+1} = x_n = \alpha\) and solve.
Failure: values that run away or oscillate more widely mean the rearrangement does not converge.

Recall

[4 marks]

(a) Show that the equation \(x^3 + 2x - 9 = 0\) can be written as \(x = \sqrt[3]{9 - 2x}\).

(b) Use the iteration \(x_{n+1} = \sqrt[3]{9 - 2x_n}\) with \(x_1 = 1.8\) to find the root correct to 2 decimal places, showing each iterate to 4 decimal places.

(a) \(x^3 = 9 - 2x\), so \(x = \sqrt[3]{9 - 2x}\). ✓
(b) \(x_2 = 1.7544\), \(x_3 = 1.7642\), \(x_4 = 1.7621\), \(x_5 = 1.7626\), \(x_6 = 1.7625\). ✓
\(x_5\) and \(x_6\) both round to 1.76, so the root is 1.76 (2 d.p.). ✓

Calculation

[3 marks]

Use the iteration \(x_{n+1} = \cos^{-1}\left(\dfrac{x_n}{2}\right)\) with \(x_1 = 1\) to find the root of \(x = 2\cos x\) correct to 2 decimal places.

Radians: \(x_2 = 1.0472\), \(x_3 = 1.0197\), \(x_4 = 1.0358\), \(x_5 = 1.0264\), \(x_6 = 1.0319\). ✓
\(x_7 = 1.0287\), \(x_8 = 1.0306\), \(x_9 = 1.0295\). ✓
\(x_7\), \(x_8\) and \(x_9\) all round to 1.03. Root: 1.03 (2 d.p.). ✓

Analysis

[5 marks]

The sequence defined by \(x_{n+1} = \dfrac{4x_n}{5} + \dfrac{10}{x_n^4}\), with \(x_1 = 2\), converges to \(\alpha\).

(a) Find the exact value of \(\alpha\).

(b) Use the iteration to find \(\alpha\) correct to 3 significant figures, and check your answer against part (a).

(a) \(\alpha = \dfrac{4\alpha}{5} + \dfrac{10}{\alpha^4} \Rightarrow \dfrac{\alpha}{5} = \dfrac{10}{\alpha^4} \Rightarrow \alpha^5 = 50\), so \(\alpha = \sqrt[5]{50}\). ✓
(b) \(x_2 = 2.2250\), \(x_3 = 2.1880\), \(x_4 = 2.1867\), \(x_5 = 2.1867\). ✓
\(\alpha = 2.19\) (3 s.f.). Check: \(50^{1/5} = 2.1867\). ✓

Proof

[4 marks]

Show that the iterative formula \(x_{n+1} = 3 - \ln x_n\) can be used to find the root of \(\ln x + x - 3 = 0\), and explain how a sketch of \(y = 3 - \ln x\) and \(y = x\) shows that this root is unique.

A limit \(\alpha\) satisfies \(\alpha = 3 - \ln\alpha\), which rearranges to \(\ln\alpha + \alpha - 3 = 0\). ✓
\(y = 3 - \ln x\) is decreasing for \(x > 0\) and \(y = x\) is increasing. ✓
A decreasing curve and an increasing line can cross at most once; they do cross (between 2 and 3), so the root is unique. ✓

Analysis

[4 marks]

The equation \(x^3 - 2x - 5 = 0\) is rearranged as \(x = \dfrac{x^3 - 5}{2}\).

(a) Using \(x_{n+1} = \dfrac{x_n^3 - 5}{2}\) with \(x_1 = 2\), find \(x_2\), \(x_3\) and \(x_4\).

(b) Comment on your results, given that the root is 2.09.

(a) \(x_2 = \dfrac{8 - 5}{2} = 1.5\); \(x_3 = \dfrac{3.375 - 5}{2} = -0.8125\); \(x_4 = \dfrac{-0.5364 - 5}{2} = -2.7682\). ✓
(b) The values move further and further from 2.09 (the next is about \(-13.1\)), so this rearrangement does not converge. ✓
A different rearrangement, such as \(x_{n+1} = \sqrt[3]{2x_n + 5}\), should be used. ✓

Analysis

[6 marks]

A chord of a circle subtends an angle \(\theta\) radians at the centre, and the minor segment it cuts off has area one quarter of the circle. It can be shown that \(\theta = \sin\theta + \dfrac{\pi}{2}\).

(a) Use the iteration \(\theta_{n+1} = \sin\theta_n + \dfrac{\pi}{2}\) with \(\theta_1 = 2.3\) to find \(\theta\) correct to 2 decimal places, showing each iterate to 4 decimal places.

(b) Hence find the length of the chord when the radius is 10 cm.

(a) \(\theta_2 = 2.3165\), \(\theta_3 = 2.3054\), \(\theta_4 = 2.3129\), \(\theta_5 = 2.3079\), \(\theta_6 = 2.3112\). ✓
From \(\theta_3\) on every value rounds to 2.31, and they zig-zag in towards it. \(\theta = 2.31\) (2 d.p.). ✓
(b) Chord \(= 2r\sin\dfrac{\theta}{2} = 20\sin(1.155)\), using the more accurate \(\theta = 2.3099\). ✓
\(= 20\times 0.9148 = 18.3\) cm (3 s.f.). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.