Numerical solution of equations

• Locate with both values and “continuous”; iterate without rounding; stop when two iterates agree when rounded; quote to exactly the accuracy asked; confirm with a sign change.
• Both values, both signs, “continuous”. Never round before feeding back. Agree when rounded, quote to that accuracy, confirm half a unit either side.

• 3.6.1 2.6.1 I can locate approximately a root of an equation, by means of graphical considerations and/or searching for a sign change — for example, find a pair of consecutive integers between which a root lies, stating both function values, their signs and that the function is continuous.
• 3.6.2 2.6.2 I can understand the idea of, and use the notation for, a sequence of approximations which converges to a root of an equation — write \(x_1, x_2, x_3, \ldots\) and \(x_{n+1} = F(x_n)\), and explain convergence as \(|x_n - \alpha| \to 0\).
• 3.6.3 2.6.3 I can understand how a given simple iterative formula of the form \(x_{n+1} = F(x_n)\) relates to the equation being solved, and use a given iteration, or an iteration based on a given rearrangement of an equation, to determine a root to a prescribed degree of accuracy — and recognise from the iterates when an iteration fails to converge.

Accuracy for this chapter. The question states the accuracy, so the three-significant-figure default gives way to it: “to 3 decimal places” and “to 3 significant figures” are different requests (\(1.564\) against \(1.56\)). Iterates are written to at least 4 d.p., and to at least 5 when the root is wanted to 4 d.p. Function values in a sign-change argument are written exactly or to at least 4 d.p., with their signs. The calculator is in radians for any trigonometric \(F\) or \(f\). An iterate is never retyped rounded: use ANS.

• “There is a sign change, so there is a root between 2 and 3.” Correct State both values and the word continuous. \(f(2) = -0.3069 < 0\), \(f(3) = 1.0986 > 0\); \(f\) is continuous and changes sign, so a root lies between 2 and 3. The values are the evidence and the continuity is the reason; the conclusion alone is an assertion.
• “\(x_2 = 1.26\), so \(x_3 = \sqrt[3]{5 - 2(1.26)}\).” Correct Never round an iterate before feeding it back. Keep the full calculator value (use the ANS key) and write at least 4 decimal places on the page. Rounded feedback adds an error at every step, and with it the last digit of the root can change.
• “\(x_8 = 1.3281\), \(x_9 = 1.3283\), so the root is \(1.3283\).” Correct Agree when rounded, then quote to that accuracy. To 3 d.p. both are \(1.328\), so you stop and the root is \(1.328\), written with three decimal places, no more. The two iterates are not equal; they agree when rounded.
• “\(f(-1) < 0\) and \(f(1) > 0\) for \(f(x) = \dfrac{1}{x}\), so there is a root between \(-1\) and \(1\).” Correct \(\dfrac{1}{x}\) has a break at \(x = 0\) and is never zero. The sign-change test needs a function that is continuous across the whole interval.
• “\(f(1)\) and \(f(2)\) are both positive, so there is no root between 1 and 2.” Correct No sign change proves nothing. \((x - 1.5)^2\) is positive at both ends and zero at \(1.5\); a curve that touches the axis, or crosses it twice, shows no sign change.
• “The iteration diverges, so the equation has no root.” Correct Divergence is a failure of that rearrangement, not of the equation. \(x_{n+1} = \dfrac{5 - x_n^3}{2}\) runs away from 1.3, yet \(x^3 + 2x - 5 = 0\) has the root \(1.328\), which another rearrangement finds.
• “Show that \(x = \sqrt[3]{5 - 2x}\): cube it, \(x^3 = 5 - 2x\), so \(x^3 + 2x - 5 = 0\), which is true.” Correct That starts from the result. Run it forwards from the given equation, or say that every step is reversible (\(\Leftrightarrow\) at each line).
• “\(x_{n+1} = \cos x_n\), \(x_1 = 1\): \(x_2 = 0.9998\).” Correct Degree mode. In radians \(\cos 1 = 0.5403\). Set radians before the first iterate of any trigonometric iteration.
• “\(f(2)\) is negative and \(f(3)\) is positive, so the root is 2.5.” Repair A sign change locates the root between 2 and 3; it says nothing about where in between. Iterate, or narrow the interval, to find it.
• “There is a sign change, so there is a root.” (No values written.) Repair Write \(f(2) = -0.3069 < 0\) and \(f(3) = 1.0986 > 0\), and the word continuous. The values are the evidence.
• “For \(f(x) = \dfrac{1}{x}\), \(f(-1) < 0\) and \(f(1) > 0\), so \(\dfrac{1}{x} = 0\) has a root between \(-1\) and \(1\).” Repair \(\dfrac{1}{x}\) is not continuous at \(x = 0\); the sign-change test needs a function with no break in the interval.
• “\(f(1)\) and \(f(2)\) are both positive, so there is no root between 1 and 2.” Repair No sign change proves nothing: \((x - 1.5)^2\) has a root there. Only a sketch, or a function known to be increasing or decreasing, settles the number of roots.
• “Show that \(x = \sqrt[3]{5 - 2x}\): \(x^3 = 5 - 2x\), so \(x^3 + 2x - 5 = 0\), which is true.” Repair That starts from the result. It is acceptable only if you say every step is reversible; better, run it forwards from the equation to the rearrangement, or write \(x^3 = 5 - 2x \Leftrightarrow x = \sqrt[3]{5 - 2x}\).
• “\(x_2 = 1.26\), so \(x_3 = \sqrt[3]{5 - 2.52}\).” Repair Keep every iterate at full calculator precision (ANS) and write at least four decimal places on the page. Rounding before feeding back puts an error into every later iterate.
• “\(x_4 = 1.3186\) and \(x_5 = 1.3319\), both about 1.33, so the root is 1.33 to 2 d.p.” Repair To 2 d.p. they are \(1.32\) and \(1.33\): they do not agree. Continue until two successive iterates agree when rounded to the required accuracy.
• “Root \(= 1.32827\) (correct to 3 decimal places).” Repair A root correct to 3 d.p. is written with three: \(1.328\).
• “Root \(= 1.328\), since \(x_8 = x_9\).” Repair \(x_8 = 1.3281\) and \(x_9 = 1.3283\) are not equal; they agree when rounded to 3 d.p. Say so, and confirm with the sign change at \(1.3275\) and \(1.3285\).
• “\(x_{n+1} = \cos x_n\), \(x_1 = 1\): \(x_2 = 0.9998\).” Repair The calculator was in degrees. \(x\) is a number, not an angle in degrees: set radians before the first iterate (\(\cos 1 = 0.5403\)).
• “The iteration diverges, so the equation has no root.” Repair Divergence is a failure of the rearrangement, not of the equation; the root is still there and another rearrangement finds it.
• “\(x = \dfrac{5 - x^3}{2}\) is not a rearrangement of \(x^3 + 2x - 5 = 0\), because it diverges.” Repair It is a correct rearrangement — its fixed point is the root — and a poor iteration. The two ideas are separate.
• Starting from \(x_1 = 1.5\) when the question gives \(x_1 = 1\). Repair A different start gives different iterates, and the reader cannot follow them. Use the value given.

• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.
• Interleave with the chapters that use this one. This chapter's equations come from other chapters: a stationary point with no exact solution in chapter 12 (re-answer mixed challenge question 2), an area equation in chapter 13 (worked example 5), and a time or value found from a model in chapter 16. Each time you meet an equation there that will not solve exactly, locate, iterate and stop as here. Recalling a method inside a new problem is worth more than another pass over this chapter on its own: later chapters use these methods without re-teaching them, and the syllabus says an individual examination question may involve ideas and methods from more than one section of the content for that paper, so nothing here is ever finished with.

• Cambridge International AS & A Level Mathematics 9709 has six components, and a candidate takes two of them for the AS Level and four for the A Level. This chapter is Pure Mathematics 3 content, examined in Paper 3 (and Paper 2 on the AS-only pure route). Paper 3 (Pure Mathematics 3) is compulsory for the A Level and is 30% of it. The Paper 2 (Pure Mathematics 2) outcomes this chapter also serves are marked in the syllabus map: Paper 2 is offered only in the AS-only Pure Mathematics route (Papers 1 and 2), where it is 40% of the AS Level, and that route cannot be carried forward to the A Level. Every paper is a written examination of compulsory structured questions, answered on the question paper, with MF19 (the list of formulae and statistical tables) supplied. Examinations are available in the June and November series, and in March in India.
• Across the whole qualification the assessment objectives are weighted AO1 55% (knowledge and understanding: concepts, terminology, notation and accurate manipulative technique) and AO2 45% (application and communication: choosing the procedure, combining techniques to solve problems, and presenting the work clearly and logically) at AS Level, and AO1 52%, AO2 48% at A Level. AS candidates are graded a–e; A Level candidates A*–E.
• A structured question on this section can run the whole routine in parts: sketch two graphs to show how many roots there are, or verify by calculation that a root lies between two values; show that the root satisfies a given rearrangement x = F(x); use the given iteration to determine the root to a stated accuracy. The equation itself can come from another section: a stationary point (chapter 12), an area (chapter 13), a modelling equation (chapter 16). The reasoning sits in the sign-change conclusion, the direction of the show-that, and the stopping decision.
• MF19 prints no numerical-methods content at all. The sign-change test and its continuity condition, the notation xn+1 = F(xn), the iteration procedure and the stopping rule must all be known. See the MF19 card.
• Here the question states the accuracy, so the three-significant-figure default gives way to it. Write every iterate to at least 4 decimal places (more when the root is wanted to 4 d.p.), each computed from the unrounded previous one; write the function values in a sign-change argument to at least 4 d.p. or exactly; quote the root to exactly the accuracy asked, no more. An iterate list with no formula, or a root with no iterates, is an unsupported calculator answer.
• Read the command word before you decide how much to write. This syllabus uses eleven: calculate, describe, determine, evaluate, explain, identify, justify, show (that), sketch, state and verify. Show that and verify give you the answer and mark the route to it, so every step must be visible and the argument must run forwards from what is given, never backwards from the result. Sketch means a simple freehand drawing showing the key features, taking care over proportions; it is not a plot. Determine means establish with certainty; justify means support a case with evidence or argument. Find, solve, express and hence are ordinary question wording; hence means the previous part is the intended route.