The shape of the lesson
By the end of the lesson, learners can
- 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.
- 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.
- 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.
- 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.
One picture every idea returns to
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.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Feed the answer back in.. Papers 2 and 3. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: 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. Model the worked example Find the exact limit on the board, one line at a time. 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. | Whole class, then pairs |
| 13 to 21 min | Teach and check: 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. Model the worked example A cube root iteration on the board, one line at a time. | Pairs |
| 21 to 30 min | Teach and check: 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. Model the worked example Same equation, two rearrangements on the board, one line at a time. | Think-Pair-Share |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 26 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in When iteration fails, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Think-Pair-Share
Pose the question, give thirty seconds of silent thinking, then a minute to compare with a partner before a few pairs share with the class. The silent minute is the part that cannot be cut.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| 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. |
| 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. |
| 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. |
Iterative methods: worksheet
6 questions, 26 marks. Written in the style of the papers; no past paper question is reproduced.
(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.
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.
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).
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.
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 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.
Answer key and marking notes
- (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.).
- 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.).
- (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: \(50^{1/5} = 2.1867\).
- 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.
- (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.
- (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.).
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
The sequence defined by \(x_{n+1} = \sqrt{5 - \dfrac{3}{x_n}}\) converges to \(\alpha\). Which equation has \(\alpha\) as a root?
- A\(x^3 - 5x + 3 = 0\)
- B\(x^3 + 5x - 3 = 0\)
- C\(x^2 - 5x + 3 = 0\)
- D\(x^3 - 5x - 3 = 0\)
The sequence defined by \(x_{n+1} = \dfrac{2x_n}{3} + \dfrac{6}{x_n^2}\) converges to \(\alpha\). Find the exact value of \(\alpha\).
- A\(\alpha = \sqrt[3]{18}\)
- B\(\alpha = \sqrt[3]{6}\)
- C\(\alpha = \sqrt[3]{2}\)
- D\(\alpha = \sqrt{18}\)
Use the iteration \(x_{n+1} = \sqrt[3]{x_n + 7}\) with \(x_1 = 2\) to find \(x_3\), giving your answer correct to 4 decimal places.
- A\(2.0862\)
- B\(2.0801\)
- C\(2.0867\)
- D\(3.1623\)
The iteration \(x_{n+1} = 2 + \ln x_n\) with \(x_1 = 3\) gives 3.0986, 3.1310, 3.1413, 3.1447, 3.1457, 3.1460, 3.1461, 3.1462. State the root correct to 2 decimal places.
- A\(3.15\)
- B\(3.14\)
- C\(3.1462\)
- D\(3.1\)
Use the iteration \(x_{n+1} = 1 + \dfrac{1}{2}\sin x_n\) with \(x_1 = 1\) to find the root of \(x = 1 + \dfrac{1}{2}\sin x\) correct to 2 decimal places (\(x\) in radians).
- A\(1.50\)
- B\(1.01\)
- C\(1.49\)
- D\(1.42\)
The sequence \(x_{n+1} = \ln(2x_n + 5)\), with \(x_1 = 2\), converges to \(\alpha\). Find \(\alpha\) correct to 2 decimal places, and state the equation that \(\alpha\) satisfies.
- A\(\alpha = 2.25\); \(e^{\alpha} = 2\alpha + 5\)
- B\(\alpha = 2.20\); \(e^{\alpha} = 2\alpha + 5\)
- C\(\alpha = 2.25\); \(\alpha = e^{2\alpha + 5}\)
- D\(\alpha = 2.25\); \(\ln\alpha = 2\alpha + 5\)
The equation \(x^3 + 2x - 5 = 0\) has a root near 1.3. Three rearrangements give the iterations below, each started at \(x_1 = 1.3\). Which one does NOT converge to the root?
- A\(x_{n+1} = \dfrac{5 - x_n^3}{2}\)
- B\(x_{n+1} = \sqrt[3]{5 - 2x_n}\)
- C\(x_{n+1} = \dfrac{5}{x_n^2 + 2}\)
- DAll three converge to the root.
The equation \(x^3 - 6x + 2 = 0\) has a root near 0.3. Which iteration is a rearrangement of this equation, and what is the root correct to 3 decimal places (start with \(x_1 = 0\))?
- A\(x_{n+1} = \dfrac{x_n^3 + 2}{6}\); root \(0.340\)
- B\(x_{n+1} = \dfrac{x_n^3 - 2}{6}\); root \(-0.340\)
- C\(x_{n+1} = \dfrac{x_n^3 + 2}{6}\); root \(0.333\)
- D\(x_{n+1} = \dfrac{x_n^3 + 2}{6}\); root \(0.339\)
Quiz key
\(\alpha^2 = 5 - \dfrac{3}{\alpha}\), so \(\alpha^3 = 5\alpha - 3\), i.e. \(\alpha^3 - 5\alpha + 3 = 0\). (From \(x_1 = 2\) the sequence goes \(1.871, 1.843, 1.836, \ldots\), and \(\alpha \approx 1.834\).) The tempting \(x^3 + 5x - 3 = 0\) moves both terms across with the wrong signs. \(x^2 - 5x + 3 = 0\) forgets to multiply \(\alpha^2\) by \(\alpha\), and \(x^3 - 5x - 3 = 0\) is a sign slip on the 3.
\(\dfrac{\alpha}{3} = \dfrac{6}{\alpha^2}\), so \(\alpha^3 = 18\) and \(\alpha = \sqrt[3]{18} \approx 2.62\). The tempting \(\sqrt[3]{6}\) ignores the \(\dfrac{2\alpha}{3}\) term and solves \(\alpha = \dfrac{6}{\alpha^2}\). \(\sqrt[3]{2}\) divides 6 by 3 instead of multiplying, and \(\sqrt{18}\) takes a square root where \(\alpha^3\) needs a cube root.
\(x_2 = \sqrt[3]{9} = 2.08008\ldots\); \(x_3 = \sqrt[3]{9.08008\ldots} = 2.08624\ldots = 2.0862\). The tempting 2.0801 is \(x_2\), one step too few. 2.0867 is the limit of the sequence, several steps further on, and 3.1623 uses square roots instead of cube roots (\(\sqrt{\sqrt{9} + 7} = \sqrt{10}\)).
The values settle at 3.146..., which rounds to 3.15 to 2 d.p. (The root of \(x - \ln x - 2 = 0\) is 3.14619.) The tempting 3.14 stops at \(x_5 = 3.1447\) before the sequence has settled, or truncates instead of rounding. 3.1462 is not to 2 decimal places, and 3.1 is to only 1.
\(x_2 = 1.42074\), \(x_3 = 1.49438\), \(x_4 = 1.49854\), \(x_5 = 1.49870\), \(x_6 = 1.49870\). The root is 1.4987, which is 1.50 to 2 d.p. The tempting 1.01 is what the sequence settles to in DEGREE mode. 1.49 truncates 1.4987 instead of rounding, and 1.42 is just \(x_2\).
\(x_2 = 2.19722\), \(x_3 = 2.24012\), \(x_4 = 2.24921\), \(x_5 = 2.25113\), \(x_6 = 2.25153\), \(x_7 = 2.25161\): \(\alpha = 2.25\). At the limit \(\alpha = \ln(2\alpha + 5)\), so \(e^{\alpha} = 2\alpha + 5\). The tempting 2.20 is \(x_2\), stopped too early. \(\alpha = e^{2\alpha + 5}\) undoes the ln on the wrong side, and \(\ln\alpha = 2\alpha + 5\) takes ln of \(\alpha\) instead of exponentiating.
\(x_{n+1} = \dfrac{5 - x_n^3}{2}\) gives \(1.3, 1.402, 1.124, 1.791, -0.371, 2.526, \ldots\), swinging further away each time: it fails. \(\sqrt[3]{5 - 2x_n}\) gives \(1.339, 1.324, 1.330, 1.328, \ldots\), converging to 1.328. \(\dfrac{5}{x_n^2 + 2}\) oscillates (\(1.355, 1.303, 1.352, \ldots\)) but the swings shrink, so it also converges to 1.328, only slowly. The tempting choice of \(\dfrac{5}{x_n^2 + 2}\) mistakes slow, oscillating convergence for failure. All three are correct rearrangements, but that alone does not guarantee convergence.
\(x^3 - 6x + 2 = 0 \iff 6x = x^3 + 2 \iff x = \dfrac{x^3 + 2}{6}\). From \(x_1 = 0\): \(0.33333, 0.33951, 0.33986, 0.33988, 0.33988\). The root is 0.33988, which is 0.340 to 3 d.p. The tempting \(\dfrac{x^3 - 2}{6}\) is a sign slip in the rearrangement and leads to \(-0.340\), a root of a different equation, \(x^3 - 6x - 2 = 0\). 0.333 stops at \(x_2\), and 0.339 truncates instead of rounding.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.