Ch 6 · 2.6 / 3.6 Numerical Solution of Equations · A Level Mathematics 9709
Topic 6 of 9 in Pure Mathematics 2 and 3 — A Level Mathematics 9709 — 57 lessons.
Introduction
Most equations met in real problems, such as x^3 - 2x - 5 = 0 or ln x = 3 - x, cannot be solved exactly with algebra. Calculators and computers solve them by iteration: make a guess, improve it with a formula, and repeat until the answer stops changing. This content is identical for Paper 2 (section 2.6) and Paper 3 (section 3.6).
Key ideas — Locating a root by a change of sign
If f is continuous on [a, b] and f(a) and f(b) have opposite signs, then f(x) = 0 has at least one root between a and b. For f(x) = x^3 - 2x - 5: f(2) = -1 < 0 and f(3) = 16 > 0, so there is a root between 2 and 3. Always state the sign change and the conclusion.
🧑🏫 Teacher's explanation
Imagine you are walking on a hilly landscape. If you are at an altitude of -5 meters (below sea level) and you walk to a spot where you are at +5 meters (above sea level), you must have crossed sea level (0 meters) at some point along that path. This assumes the ground is continuous and doesn't have a sudden cliff or a hole (an asymptote).
📖 Definitions
🔍 Worked example — Easy
Show that the equation x^3 + 2x - 10 = 0 has a root between x=1 and x=3.
🔍 Worked example — Standard
Use a change of sign to locate the root of x^2 - 5 = 0 between consecutive integers.
🔍 Worked example — Exam-style
Show that x = cos x has a root in the interval [0, π/2].
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — Sequences that converge
An iteration produces a sequence x_1, x_2, x_3, dots. It converges to a root α if the terms get closer and closer to α: x_n → α as n → ∞.
🧑🏫 Teacher's explanation
Imagine you are playing a game of 'Hot or Cold' while walking toward a hidden object. Every time you take a step, someone tells you if you are getting closer or further away. If you keep moving based on their feedback, you will eventually stand right on top of the object. This 'walking' process is exactly what an iterative formula does. Instead of finding the exact answer in one giant leap, we take many tiny, calculated steps.
📖 Definitions
🔍 Worked example — Easy
Use the iterative formula x_n+1 = √(x_n + 5) to find the first three terms of the sequence, starting with x_1 = 1.
🔍 Worked example — Standard
Show that the iteration x_n+1 = x_n^2 - 2 starting with x_1 = 0 does not converge to a root of x^2 - x - 2 = 0.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
(a) Show that x_1 = 1 is a good starting value for finding a root. (b) Use this iteration to find x_3 correct to 3 decimal places. [4 marks]
Key ideas — Where an iterative formula comes from
Rearrange f(x) = 0 into the form x = F(x); then x_n+1 = F(x_n). If the sequence converges to α, then α = F(α), so α is a root of the original equation. For example, x^3 - 2x - 5 = 0 iff x^3 = 2x + 5 iff x = 3√(2x + 5), giving x_n+1 = 3√(2x_n + 5).
🧑🏫 Teacher's explanation
Imagine you are trying to find the exact center of a target, but you can only move by looking at where you currently are and adjusting your position based on a rule. This 'rule' is your iterative formula. If the rule is good, each step brings you closer to the bullseye. In mathematics, the 'bullseye' is the root of an equation, α, where f(α) = 0.
📖 Definitions
🔍 Worked example — Easy
Show that the equation x^2 - 5x + 2 = 0 can be written in the form x = F(x) and find the iterative formula x_n+1 = F(x_n).
🔍 Worked example — Standard
The sequence defined by x_n+1 = √(x_n + 6) converges to α. Find the equation that α satisfies.
🔍 Worked example — Exam-style
Given the iterative formula x_n+1 = 1/2(x_n + 5/x_n), find the equation that the limit α satisfies and solve for the positive root.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — Using the iteration to a given accuracy
Use the ANS key on your calculator. Write down each iterate to at least one more decimal place than the accuracy asked for, and stop when two successive iterates agree to the required accuracy. To confirm, check the sign change over the interval of rounding: e.g. for 2 d.p. answer 2.09, show f(2.085) and f(2.095) have opposite signs.
🧑🏫 Teacher's explanation
Imagine you are trying to find the exact center of a dark room by walking towards a wall. You take a step, check how far you are, and adjust your next step based on that information. You might never reach the wall perfectly, but you can get so close that, for all practical purposes, you are there. This is exactly what iteration does.
📖 Definitions
🔍 Worked example — Easy
Use the iteration x_n+1 = √(x_n + 5) to find the root of x^2 - x - 5 = 0, starting with x_0 = 3. Give your answer correct to 2 decimal places.
🔍 Worked example — Standard
Find the root of x^3 + 2x - 5 = 0 using the iteration x_n+1 = 5 - x_n^3/2x_n with x_0 = 1. Give your answer correct to 3 decimal places.
🔍 Worked example — Exam-style
Solve x = cos(x) using the iteration x_n+1 = cos(x_n) with x_0 = 0.5. Give your answer correct to 4 decimal places.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — When iteration fails
Different rearrangements of the same equation behave differently. Some converge (possibly slowly, or by oscillating either side of the root, a "cobweb"); others diverge, moving further away. You do not need the condition for convergence, but you should recognise failure when the iterates run away. Then try a different rearrangement.
🧑🏫 Teacher's explanation
Imagine you are trying to find the center of a target by throwing darts. If your first throw is slightly off, and your second throw is even further away, and your third is even further still, you are 'diverging'—you are moving away from the goal. In iteration, we use a formula x_n+1 = F(x_n) to guess the next value. If the formula is 'stable', each guess gets closer to the root, which we call 'convergence'.
📖 Definitions
🔍 Worked example — Easy
Show that the iteration x_n+1 = √(5 + x_n) with x_0 = 1 converges to a root of x^2 - x - 5 = 0. Calculate x_3 to 3 decimal places.
🔍 Worked example — Standard
Consider the equation x^3 - x - 1 = 0. An iterative formula is given by x_n+1 = 1 + x_n^2/x_n^2 (incorrectly rearranged). If x_0 = 1, show that the iteration fails to find the root near 1.32 by calculating x_1, x_2, x_3.
🔍 Worked example — Exam-style
An equation f(x) = 0 is solved using x_n+1 = 2x_n - 5. If x_0 = 4, show that the iteration diverges.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
(a) Calculate x_1, x_2, x_3 correct to 3 decimal places, starting with x_0 = 1. [M1, A1] (b) State whether the iteration is converging or diverging. [B1] [3 marks]
Worked examples
Use x_n+1 = 3√(2x_n + 5) with x_1 = 2 to find the root of x^3 - 2x - 5 = 0 to 3 d.p.
🔍 Worked example — Easy
Use the iterative formula x_n+1 = 3√(2x_n + 5) with x_1 = 2 to find the root of x^3 - 2x - 5 = 0 correct to 2 decimal places.
🔍 Worked example — Standard
Show that the iteration x_n+1 = x_n^3 - 5/2 with x_1 = 2 fails to converge to the root of x^3 - 2x - 5 = 0.
🔍 Worked example — Exam-style
An iteration x_n+1 = √(4 + sin x_n) converges to α. Find the equation that α is a root of and find α correct to 3 decimal places using x_1 = 2.
⚠️ Common errors
Common mistakes
Practice
Answers: (1) f(1) = -2 < 0, f(2) = 6 > 0; sign change and f continuous, so a root lies between 1 and 2. (2) 2.3069, 2.1641, 2.2280, 2.1989, 2.2120, 2.2061, 2.2088, 2.2076, dots (slowly converging); check g(2.205) < 0 < g(2.215) for g(x) = ln x + x - 3: α = 2.21. (3) x^3 = 4 - x; from x_1 = 1.5: 1.3572, 1.3826, 1.3781, 1.3789, 1.3788, so the root is 1.379. (4) α = 2α/3 + 4/α^2 ⇒ α/3 = 4/α^2 ⇒ α^3 = 12, α = 3√(12).