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GCSE Maths · EdexcelChapter 25 · Functions and advanced graphs
GCSE Maths · Functions and Advanced Graphs · Higher · Lesson 25.10

GCSE Maths: Iteration and iterative processes (Higher)

Part of the GCSE Maths functions and advanced graphs chapter (Higher). Chapter overview · whole chapter on one page · Download worksheet (PDF) · Answers (PDF)

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25.10

Iteration and iterative processes

Edexcel 1MA1: A20, R16
Grade 8-9

Some equations cannot be solved with the methods you know. For example x3 − x − 1 = 0 cannot be factorised and there is no simple formula. But we can still find an answer as accurately as we like using iteration: repeating a calculation, using each answer as the input for the next, so that the numbers get closer and closer to the solution.

An iteration formula looks like xn+1 = g(xn). It says: to get the next term, put the current term into g. You are given a starting value x0. For example, with xn+1 = ∛(xn + 1) and x0 = 1: x1 = ∛(1 + 1) = 1.2599…, then x2 = ∛(1.2599 + 1) = 1.3122…, and so on. On a calculator type 1 =, then type ∛(Ans + 1) and press = repeatedly. The Ans button stores the full unrounded value, which keeps the working accurate.

Why it finds a solution. If the terms settle down to a value L, then xn and xn+1 are both about L, so L = g(L). So L solves the equation x = g(x). The iteration formula comes from rearranging the original equation into the form x = g(x). For x3 − x − 1 = 0 we can write x3 = x + 1 and so x = ∛(x + 1).

Where is the solution? To show that a solution lies between two numbers, put both into the expression. If one gives a negative answer and the other a positive answer (a change of sign), the graph crosses the x-axis between them. For f(x) = x3 − x − 1: f(1) = −1 and f(2) = 5, so there is a solution between 1 and 2.

xy0.511.520.511.52
A staircase (cobweb) diagram for xₙ₊₁ = ∛(xₙ + 1), starting at x₀ = 1. The blue curve is y = ∛(x + 1) and the grey line is y = x. Each step goes up to the curve and across to the line. The staircase closes in on where they cross: the solution of x = ∛(x + 1).

Giving the answer to a number of decimal places. Keep iterating until two (or three) successive terms agree to the required accuracy. To prove that the answer is correct to, say, 3 decimal places, substitute the two ends of the rounding interval (for 1.325 these are 1.3245 and 1.3255) into f(x): a change of sign proves the solution is between them.

Not every rearrangement works. The same equation can be rearranged in several ways. Rearranging x3 − x − 1 = 0 as x = x3 − 1 gives terms that fly away: starting at 1.5 we get 2.375, 12.4, 1905… (it diverges). In an exam you will always be told which formula to use.

xy0.511.522.530.511.522.53
The rearrangement xₙ₊₁ = xₙ³ − 1 starting from 1.5: the steps move away from the crossing point (the staircase spirals outwards), so the iteration diverges.

General iterative processes. Iteration also describes situations that repeat in steps, such as a population measured every year or the amount of a drug in the body measured every day. Write the rule as un+1 = aun + b: the next amount is the last amount multiplied by a factor, plus (or minus) a fixed amount added or removed. Compound interest is the simplest case (b = 0). If the terms settle on a limiting value L, then L = aL + b, so L = b ÷ (1 − a).

Key ideas

  • Iteration: xn+1 = g(xn). Start at x0 and repeat. Use the Ans button and keep all the digits.
  • The limit L of the terms satisfies L = g(L), which is a rearrangement of the original equation.
  • A change of sign of f(x) between two values shows a solution of f(x) = 0 lies between them.
  • For a solution to n d.p.: iterate until the terms agree to n d.p., then show a change of sign at the ends of the rounding interval.
  • Some rearrangements diverge. For un+1 = aun + b, the limit (if there is one) is b ÷ (1 − a).
Worked example 25.10.1
A sequence is given by xn+1 = ∛(xn + 1) with x0 = 1. Work out x1, x2, … x5, giving your answers to 6 decimal places.

Key words. An iteration repeats the same calculation, using each answer as the next input. x1, x2 ... are the terms.

Method. Substitute each value into the formula to get the next one, starting from x0.

  1. Write the iteration formula and the starting value. An iteration is a recipe that gives each term from the one before. The starting value is x0.
    xn+1 = ∛(xn + 1)
    x0 = 1
  2. Work out x1: put x0 into the right-hand side. Replace xn by the value you have, using brackets for negatives.
    x1 = ∛(1 + 1) = 1.259921
  3. Work out x2: use the answer for x1. Now x1 is the input. On a calculator type the value, press =, then type the formula using the Ans key in place of xn; each press of = gives the next term.
    x2 = ∛(1.259921 + 1) = 1.312294
  4. Keep pressing = to find the rest. Do not round during the calculation: keep all the digits in the calculator memory and only round the values you write down.
    nxn
    01.000000
    11.259921
    21.312294
    31.322354
    41.324269
    51.324633

Check. The terms are settling down towards 1.3246: each new term is closer to the last. ✓

Watch out. Use the calculator’s Ans button so you keep every digit. Rounding too early can give a wrong final digit.

Worked example 25.10.2
Show that the equation x3 − x − 1 = 0 has a solution between x = 1 and x = 2.

Method. Work out f(x) at each end. A change of sign shows the graph crosses the x-axis, so there is a solution in between.

  1. Substitute the lower value. Put the first value of x into the expression.
    f(1) = 13 − 1 − 1 = −1
  2. Substitute the upper value. Put the second value in.
    f(2) = 23 − 2 − 1 = 5
  3. Compare the signs. If the graph of a continuous function is below the x-axis at one value and above it at another, it must cross the axis between them. A crossing means f(x) = 0.
    f(1) is negative and f(2) is positive: there is a change of sign.
  4. State the conclusion.
    So the equation x3 − x − 1 = 0 has a solution between x = 1 and x = 2.

Check. The root is about 1.325, which lies between 1 and 2. ✓

Watch out. You must show both values and say there is a change of sign. Saying only “one is negative” earns no marks.

Worked example 25.10.3
Show that the equation x3 + 4x − 3 = 0 can be rearranged to give x = 3 ÷ (x2 + 4).

Method. To rearrange an equation into the form x = g(x), get an x (or x², x³) on its own on one side using inverse operations.

  1. Start with the equation.
    x3 + 4x − 3 = 0
  2. Move the plain number to the right. Subtract it from both sides.
    x3 + 4x = 3
  3. Factorise the left side. Take out x as a common factor.
    x(x2 + 4) = 3
  4. Divide both sides by the bracket. This leaves x on its own.
    x = 3x2 + 4

Check. At the solution (x ≈ 0.674) both sides agree: 0.674 and 0.674. ✓

Watch out. The same equation can be rearranged in several ways. Some rearrangements work for iteration and some do not (the numbers fly off), so always use the one the question gives.

Worked example 25.10.4
Use the iteration formula xn+1 = ∛(xn + 1) with x0 = 1.5 to find a solution of x3 − x − 1 = 0 correct to 3 decimal places. Show that your answer is correct to 3 decimal places.

Key words. The values converge when they get closer and closer to one number. A root (solution) of f(x) = 0 is a value of x that makes f(x) equal 0.

Method. Use the formula repeatedly until the terms agree to 3 d.p., then prove the answer by a change of sign.

  1. Write the iteration formula and the starting value. We repeat the same calculation, each time putting the last answer back in.
    xn+1 = ∛(xn + 1)
    x0 = 1.5
  2. Calculate the terms using the calculator’s Ans button. Type the starting value and press =. Then type the formula with Ans in place of xn and keep pressing = for each new term. Keep every digit; only round what you write down.
    nxn
    01.500000
    11.357209
    21.330861
    31.325884
    41.324939
    51.324760
    61.324726
    71.324719
  3. Stop when the terms agree to 3 decimal places. The last few terms all round to 1.325, so the solution is close to this value.
    x5, x6, x7 ≈ 1.3248, 1.3247, 1.3247: all round to 1.325
  4. Show the answer is correct to 3 d.p.: test the two ends of the rounding interval. Any number that rounds to 1.325 lies between 1.3245 and 1.3255. If f changes sign between these, the solution must lie in the interval.
    f(1.3245) = 1.32453 − 1.3245 − 1 = −0.00093
    f(1.3255) = 1.32553 − 1.3255 − 1 = 0.00334
  5. Conclude. There is a change of sign, so the solution is between the two values, and both round to the same number.
    x = 1.325 to 3 decimal places

Check. Put x = 1.325 into the original equation: 1.3253 − 1.325 − 1 = 0.0012, which is very close to 0. ✓

Watch out. The change-of-sign test must use the ends of the rounding interval (for example 1.3245 and 1.3255), not just the answer itself.

Worked example 25.10.5
A patient takes 20 mg of a drug at the start of each day. During each day 20% of the drug in the body is removed. The amount in the body just after each dose is un mg, with u0 = 0, and un+1 = 0.8un + 20. Work out u1, u2, u3 and u4.

Method. An iterative process gives each term from the previous one. Apply the rule repeatedly.

  1. Write the rule and the starting value. Each term is made from the one before.
    un+1 = 0.8un + 20,   u0 = 0
  2. Work out the terms one at a time. Put each term into the rule to get the next.
    u1 = 0.8 × 0 + 20 = 20
    u2 = 0.8 × 20 + 20 = 36
    u3 = 0.8 × 36 + 20 = 48.8
    u4 = 0.8 × 48.8 + 20 = 59.04
  3. Write the answers. List the terms asked for.
    20, 36, 48.8, 59.04

Check. Check the first step again: 0.8 × 0 + 20 = 20. ✓

Watch out. Use the previous term each time, not the starting value.

Watch out

Do not round during the iteration. If you round every term to 2 or 3 decimal places the errors build up and the final digit may be wrong. Use the Ans key. Also, do not show a root only with the answer: you must show a change of sign for the two ends of the rounding interval.

Exercise 25J  Iteration and iterative processes

Calculator allowed
1
FUse each iteration formula and starting value to work out x1, x2 and x3. Give your answers to 6 decimal places.
(a)xn+1 = ∛(xn + 1),   x0 = 1
(b)xn+1 = 3xn2 + 4,   x0 = 1
(c)xn+1 = xn35 + 15,   x0 = 0
(d)xn+1 = √(3xn + 2),   x0 = 3
2
FShow that each equation has a solution between the two given values.
(a)x3 − x − 1 = 0 between x = 1 and x = 2
(b)x3 + 4x − 3 = 0 between x = 0 and x = 1
(c)x3 − 5x + 1 = 0 between x = 0 and x = 1
(d)x2 − 3x − 2 = 0 between x = 3 and x = 4
3
RShow that each equation can be rearranged to give the iteration formula x = g(x) given.
(a)x3 + 2x − 7 = 0 can be rearranged to x = ∛(−2x + 7)
(b)x2 − 3x − 2 = 0 can be rearranged to x = √(3x + 2)
(c)x2 − 3x − 2 = 0 can be rearranged to x = 3x + 2x
(d)x3 − 5x + 1 = 0 can be rearranged to x = x3 + 15
4
PSUse the iteration formula xn+1 = ∛(xn + 1) with x0 = 1.5 to find a solution of x3 − x − 1 = 0 correct to 3 decimal places. Show that your answer is correct to 3 decimal places.
5
PSUse the iteration formula xn+1 = 3xn2 + 4 with x0 = 1 to find a solution of x3 + 4x − 3 = 0 correct to 3 decimal places. Show that your answer is correct to 3 decimal places.
6
PSUse the iteration formula xn+1 = √(3xn + 2) with x0 = 3 to find a solution of x2 − 3x − 2 = 0 correct to 3 decimal places. Show that your answer is correct to 3 decimal places.
7
PSUse the iteration formula xn+1 = xn35 + 15 with x0 = 0 to find a solution of x3 − 5x + 1 = 0 correct to 3 decimal places. Show that your answer is correct to 3 decimal places.
8
PSUse the iteration formula xn+1 = ∛(−2xn + 7) with x0 = 1.5 to find a solution of x3 + 2x − 7 = 0 correct to 2 decimal places. Show that your answer is correct to 2 decimal places.
9
RUsing xn+1 = xn3 − 1 with x0 = 1.5, work out x1, x2 and x3. Does this iteration find the solution of x3 − x − 1 = 0 near 1.32? Explain.
10
PSThe diagram shows the graph of y = √(3x + 2) (blue) and y = x (grey). A staircase has been drawn for the iteration xn+1 = √(3xn + 2) with x0 = 1. Does the iteration converge? Estimate the value it converges to.
xy1234512345
11
PSA drug is given in doses of 50 mg. Between doses 30% of the drug in the body is lost. The amount just after each dose is un mg, where un+1 = 0.7un + 50 and u0 = 0. Work out u1 to u4, and use L = 0.7L + 50 to find the limiting amount.
12
PSA loan of £5000 has interest of 1% added each month and a repayment of £200 is made at the end of each month. The balance after n months is Bn, where Bn+1 = 1.01Bn − 200 and B0 = 5000. Work out B1, B2 and B3. After how many months does the balance first fall below £4000?
13
PSA fish farm has Pn fish at the start of year n. Each year the population grows by 10% and then 50 fish are sold, so Pn+1 = 1.1Pn − 50. Work out P1, P2 and P3 when
(a)P0 = 600
(b)P0 = 400
14
RFor the fish farm in the last question, the amount of fish stays the same if Pn+1 = Pn = L. Find L and explain what happens if the farm starts with more than L fish and with fewer than L fish.
15
PSRavi puts £100 into an account. At the end of each year 3% interest is added and then he adds another £100. Write an iteration formula for the amount un after n years with u0 = 100, and find the amount after 3 years.
16
RThe iteration xn+1 = (xn + 7 ÷ xn) ÷ 2 with x0 = 3 is used to find √7. Work out x1, x2 and x3, and show why the limit L is √7.
17
Rxn+1 = 1 + 1 ÷ xn, x0 = 1. Work out x1 to x5. The terms settle on a limit L: show that L2 − L − 1 = 0 and find L using the quadratic formula.

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