Papers 2 and 3. Rearrange an equation into the form \(x = F(x)\), start with a rough value, and keep substituting each answer back into \(F\). When the sequence settles down, it has found a root; when it does not, a different rearrangement may succeed.
If the sequence given by \(x_{n+1} = F(x_n)\) converges to \(\alpha\), then \(\alpha = F(\alpha)\): the limit is a root of the equation \(x = F(x)\). Iterate until successive values agree to the accuracy required, then give the root to that accuracy.
Fig. 1: a cobweb diagram for x(n+1) = cos x(n) starting at x1 = 0.2: go up to the curve, across to y = x, up or down to the curve again; the path spirals in to the crossing point α ≈ 0.739, the root of x = cos x
Section 01
The notation and the fixed point.
An iterative formula \(x_{n+1} = F(x_n)\) builds a sequence \(x_1, x_2, x_3, \ldots\): each term is \(F\) of the one before. If the terms approach a limit \(\alpha\), then letting \(n \to \infty\) on both sides gives \(\alpha = F(\alpha)\). So the limit is where the curve \(y = F(x)\) meets the line \(y = x\), as in Fig. 1.
Which equation? Replace \(x_{n+1}\) and \(x_n\) by \(x\) and simplify: \(x_{n+1} = \sqrt[3]{x_n + 4}\) solves \(x = \sqrt[3]{x + 4}\), that is \(x^3 - x - 4 = 0\).
Exact limit: sometimes the equation can be solved exactly, giving \(\alpha\) as a surd or a root.
Many rearrangements: one equation can be rearranged into several forms \(x = F(x)\); some converge, some do not.
Worked example Find the exact limit
The sequence defined by \(x_{n+1} = \dfrac{2x_n}{3} + \dfrac{4}{3x_n^2}\), \(x_1 = 1.5\), converges to \(\alpha\). Find the exact value of \(\alpha\).
At the limit, \(\alpha = \dfrac{2\alpha}{3} + \dfrac{4}{3\alpha^2}\).
Show that the iteration \(x_{n+1} = \sqrt[3]{2x_n + 5}\) can be used to find a root of \(x^3 - 2x - 5 = 0\).
A limit \(\alpha\) satisfies \(\alpha = \sqrt[3]{2\alpha + 5}\).
Cube both sides: \(\alpha^3 = 2\alpha + 5\).
Rearrange: \(\alpha^3 - 2\alpha - 5 = 0\), so \(\alpha\) is a root of the equation.
Answer: The limit satisfies \(x^3 - 2x - 5 = 0\)
Your turn
Use the iteration in the first example, with \(x_1 = 1.5\), to find \(\alpha\) correct to 3 significant figures, and check against the exact value.
\(x_2 = 1.59259\), \(x_3 = 1.58742\), \(x_4 = 1.58740\). Both \(x_3\) and \(x_4\) round to 1.59, so \(\alpha = 1.59\) (3 s.f.). Check: \(\sqrt[3]{4} = 1.5874\).
Examiner trap
Keep \(x_{n+1}\) and \(x_n\) distinct while iterating, but set them BOTH equal to \(\alpha\) (or \(x\)) when finding the equation. Writing \(\alpha = \dfrac{2x}{3} + \ldots\) with a mixture of letters goes nowhere.
Section 02
Iterating to a required accuracy.
Use the calculator's ANS key: type the starting value, press =, then type \(F\) with ANS in place of \(x\) and press = repeatedly. Write down each iterate to at least one more decimal place than the answer needs.
Record iterates to 4 or 5 decimal places for an answer to 2 decimal places.
Stop when two successive iterates round to the same value at the required accuracy, and they have stopped drifting.
If asked, confirm with a sign change across the rounding bounds (see the previous lesson).
Worked example A cube root iteration
Use \(x_{n+1} = \sqrt[3]{2x_n + 5}\) with \(x_1 = 2\) to find the root of \(x^3 - 2x - 5 = 0\) correct to 2 decimal places. Show the result of each iteration to 4 decimal places.
\(x_7 = 2.2061\), \(x_8 = 2.2088\), \(x_9 = 2.2076\): the values zig-zag around the root and close in.
\(x_7\), \(x_8\) and \(x_9\) all round to 2.21.
Answer: \(2.21\) (2 d.p.)
Examiner trap
Do not round each iterate before feeding it back in: use the full calculator value (ANS). Rounding early can stall the sequence or change the final digit.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
The sequence \(x_{n+1} = F(x_n)\) converges to \(\alpha\). Which statement must be true?
Quick check+10 XP
The iteration \(x_{n+1} = \sqrt{x_n + 3}\) converges. Its limit is a root of which equation?
Quick check+10 XP
Given \(x_{n+1} = \dfrac{6}{x_n + 1}\) and \(x_1 = 1\), find \(x_3\).
Quick check+10 XP
Iterates are 2.2061, 2.2088, 2.2076. To what value has the iteration converged, correct to 2 decimal places?
Section 03
When iteration fails.
You do not need the condition for convergence, but you must recognise failure: the iterates may move away from the root, oscillate with growing size, or never settle. Then try a different rearrangement of the same equation.
Worked example Same equation, two rearrangements
The equation \(\ln x = 3 - x\) can also be rearranged as \(x = e^{3-x}\). Investigate \(x_{n+1} = e^{3 - x_n}\) with \(x_1 = 2.2\), close to the root 2.208.
The values jump from side to side of the root, further away each time.
Answer: The sequence diverges, even from a start very close to the root; \(x_{n+1} = 3 - \ln x_n\) works instead
Worked example Choosing a rearrangement that converges
The equation \(x = 2\cos x\) has a root near 1. Compare \(x_{n+1} = 2\cos x_n\) and \(x_{n+1} = \cos^{-1}\left(\dfrac{x_n}{2}\right)\), each with \(x_1 = 1\).
\(x_{n+1} = 2\cos x_n\): 1.0806, 0.9416, 1.1770, 0.7674, 1.4395, 0.2619: oscillating and growing, so it fails.
The second sequence settles: the last values all round to 1.03.
Answer: Only the inverse cosine form converges; the root is 1.03 (2 d.p.)
Examiner trap
A start close to the root does not guarantee convergence: the rearrangement decides. In an exam, if your values run away, recheck the arithmetic and calculator mode (radians!) before concluding that the iteration fails.
Stage 2 · Exam
Exam-style questions
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The sequence defined by \(x_{n+1} = \sqrt[3]{x_n + 4}\) converges to \(\alpha\). Which equation has \(\alpha\) as a root?
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The sequence \(x_{n+1} = \dfrac{x_n}{2} + \dfrac{3}{x_n}\), with \(x_1 = 2\), converges to \(\alpha\). Find the exact value of \(\alpha\).
Exam style+20 XP
Use the iteration \(x_{n+1} = \sqrt[3]{x_n + 4}\) with \(x_1 = 1.5\) to find the root correct to 3 significant figures.
Exam style+20 XP
For the equation \(\ln x = 3 - x\) (root 2.208), the iteration \(x_{n+1} = e^{3 - x_n}\) is used with \(x_1 = 2.2\). The next iterates are 2.2255, 2.1694, 2.2947, 2.0246, 2.6524. What should you conclude?
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