OpenRevise

Maths · Functions, Sequences and Rates of Change

Viewing as

Student Teacher

Teacher view: every answer and mark scheme set out in full.

Iteration

Using iterative formulae, rearranging equations, and finding roots by a change of sign and trial and improvement.

  • Higher
  • 9 key terms
  • All boards

Learning Objectives

  1. 1Use an iterative formula \(x_{n+1} = f(x_n)\) to find the next terms from a starting value.
  2. 2Rearrange an equation into an iterative formula.
  3. 3Use a change of sign to show that a root lies between two values.
  4. 4Find a root to a given accuracy by trial and improvement, and find the limit of an iteration.

Solving equations by repeating a process

Many equations cannot be solved exactly by a simple method, but their solutions can be found as accurately as you like by repeating a calculation. Iteration starts with a guess, puts it into a formula, and uses the answer as the next guess. If the values settle down towards a single number, that number is a solution of the equation. Iteration is a Higher tier topic on every board. On a calculator paper, use the calculator to work with decimals: the Ans key feeds each answer back in, and you should keep the full display until the end.

Using a formula

\(x_{n+1} = 3 - \dfrac{2}{x_n}\) and \(x_0 = 4\). Work out \(x_1\) and \(x_2\).

Show the solutionHide the solution
  1. 1 First step \(x_1 = 3 - \dfrac{2}{4} = 3 - 0.5 = 2.5\).
  2. 2 Second step \(x_2 = 3 - \dfrac{2}{2.5} = 3 - 0.8 = 2.2\).
  3. 3 Pattern The values are getting closer to 2.
  4. 4 Next value \(x_3 = 3 - \dfrac{2}{2.2} = 2.0909\ldots\), still moving towards 2.

Answer\(x_1 = 2.5\) and \(x_2 = 2.2\)

Rearranging into an iterative formula

An iterative formula comes from rearranging an equation so that \(x\) is on one side.

  • Start

    \(x^2 - 3x + 2 = 0\).

  • Divide by x

    \(x - 3 + \dfrac{2}{x} = 0\).

  • Rearrange

    \(x = 3 - \dfrac{2}{x}\).

  • Iterative form

    \(x_{n+1} = 3 - \dfrac{2}{x_n}\).

A change of sign

A root of \(f(x) = 0\) lies between two values if \(f\) changes sign between them.

  • Method

    Work out \(f\) at both values. If one is negative and the other is positive, there is a root between them.

  • Condition

    The graph must be a continuous curve, with no gaps.

  • Narrowing down

    Try a value in the middle, and keep the interval where the sign changes.

  • To 1 decimal place

    Test the midpoint of the last interval, so that you know which way to round.

Trial and improvement

\(f(x) = x^3 + x - 3\). Show that \(f(x) = 0\) has a root between 1 and 2, and find this root to 1 decimal place.

Show the solutionHide the solution
  1. 1 Test 1 and 2 \(f(1) = 1 + 1 - 3 = -1\) and \(f(2) = 8 + 2 - 3 = 7\). The sign changes, so there is a root between 1 and 2.
  2. 2 Narrow down \(f(1.2) = 1.728 + 1.2 - 3 = -0.072\), and \(f(1.3) = 2.197 + 1.3 - 3 = 0.497\).
  3. 3 Interval The root is between 1.2 and 1.3.
  4. 4 Midpoint \(f(1.25) = 1.953125 + 1.25 - 3 = 0.203\ldots\) is positive, so the root is between 1.2 and 1.25, and rounds to 1.2.

AnswerThe root is 1.2 to 1 decimal place

Test yourself

  1. 1

    What does \(x_{n+1}\) mean in an iterative formula?

    Show answerHide answer

    The next value, found from \(x_n\).

  2. 2

    What is true at the limit of an iteration?

    Show answerHide answer

    \(x_{n+1} = x_n\).

  3. 3

    How do you show a root lies between two values?

    Show answerHide answer

    Show that \(f\) changes sign between them.

  4. 4

    Why do you test the midpoint when rounding?

    Show answerHide answer

    To see which side of the midpoint the root is, and so how to round.

  5. 5

    How do you find the limit of an iteration exactly?

    Show answerHide answer

    Replace \(x_{n+1}\) and \(x_n\) with \(a\) and solve.

Exam technique: iteration

Show every value, and write why you reach your conclusion.

  • Write each term

    Show \(x_1\), \(x_2\), and so on, so that errors can be spotted.

  • Limit equation

    Replace \(x_n\) and \(x_{n+1}\) with \(a\) and solve for \(a\).

  • Choose the right root

    The question will usually tell you the values are positive, or give a start value.

  • Conclusion

    Write that the sign changes, so there is a root between the two values.

Summary and exam focus

  • Iteration repeats a formula, using each answer as the next input.
  • The limit \(a\) satisfies \(a = f(a)\).
  • A change of sign shows a root between two values, for a continuous function.
  • Rearrange an equation to get an iterative formula.

Exam focus

\(x_{n+1} = 1 + \dfrac{6}{x_n}\) converges to a positive number \(a\). Show that \(a^2 - a - 6 = 0\), and find the value of \(a\). (4 marks) (4 marks)

At the limit \(a = 1 + \dfrac{6}{a}\), so \(a^2 = a + 6\) and \(a^2 - a - 6 = 0\). Factorise: \((a - 3)(a + 2) = 0\), and \(a\) is positive, so \(a = 3\).

Key terms

The words this lesson expects you to use. Each one is linked from the first place it appears above.

Iteration
A process that is repeated, using each output as the next input.
Iterative formula
A formula that gives the next value from the current one.
Limit
The value that the terms of an iteration approach.
Root
A solution of an equation \(f(x) = 0\).
Change of sign
The function is negative at one value and positive at another.
Continuous
A graph with no breaks or gaps.
Trial and improvement
Testing values to find a solution to a given accuracy.
Converge
To get closer and closer to a value.
Starting value
The first input of an iteration.

Questions and answers

15 questions set on this lesson, with the mark schemes and model answers open.

1. Exam question Calculate 2 marks Core

\(x_{n+1} = 4 - \dfrac{3}{x_n}\) and \(x_0 = 6\). Calculate \(x_1\) and \(x_2\). Give \(x_2\) as a fraction. [2 marks]

Mark scheme — 2 marks available

  • \(x_1 = 3.5\) — B1
  • \(x_2 = \dfrac{22}{7}\) — B1

Model answer

\(x_1 = 4 - \dfrac{3}{6} = 3.5\) and \(x_2 = 4 - \dfrac{3}{3.5} = 4 - \dfrac{6}{7} = \dfrac{22}{7}\).

2. Exam question Show that 2 marks Core

Show that the equation \(x^2 - 4x + 3 = 0\) can be rearranged to give \(x = 4 - \dfrac{3}{x}\). [2 marks]

Mark scheme — 2 marks available

  • Divides by \(x\), giving \(x - 4 + \dfrac{3}{x} = 0\) — M1
  • \(x = 4 - \dfrac{3}{x}\) — B1

Model answer

Divide every term by \(x\): \(x - 4 + \dfrac{3}{x} = 0\). Then \(x = 4 - \dfrac{3}{x}\).

3. Exam question Calculate 4 marks Core

\(x_{n+1} = 3 + \dfrac{10}{x_n}\). The values \(x_n\) tend to a positive limit \(a\). Show that \(a^2 - 3a - 10 = 0\), and calculate the value of \(a\). [4 marks]

Mark scheme — 4 marks available

  • \(a = 3 + \dfrac{10}{a}\) — M1
  • \(a^2 - 3a - 10 = 0\) — M1
  • \((a - 5)(a + 2) = 0\) — M1
  • \(5\) — A1

Model answer

At the limit, \(a = 3 + \dfrac{10}{a}\), so \(a^2 = 3a + 10\) and \(a^2 - 3a - 10 = 0\). Then \((a - 5)(a + 2) = 0\), and \(a\) is positive, so \(a = 5\).

4. Exam question Show that 4 marks Core

\(f(x) = x^3 - x - 1\) (a) Show that the equation \(f(x) = 0\) has a root between 1 and 2. [2 marks] (b) Find this root to 1 decimal place. Show your working. [2 marks]

Mark scheme — 4 marks available

  • (a) \(f(1) = -1\) and \(f(2) = 5\) — M1
  • (a) A change of sign, so there is a root — B1
  • (b) \(f(1.3) < 0\), \(f(1.4) > 0\) and a test of 1.35 — M1
  • (b) \(1.3\) — A1

Model answer

(a) \(f(1) = -1\) and \(f(2) = 5\). The sign changes, so there is a root between 1 and 2. (b) \(f(1.3) = -0.103\) and \(f(1.4) = 0.344\), and \(f(1.35) = 0.110\ldots\), which is positive, so the root is between 1.3 and 1.35, which is 1.3 to 1 decimal place.

5. Exam question Find 4 marks Stretch

Use trial and improvement to find the value of \(\sqrt{11}\) correct to 1 decimal place. Show all your working. [4 marks]

Mark scheme — 4 marks available

  • \(3.3^2 = 10.89\) and \(3.4^2 = 11.56\) — B1
  • Tests \(3.35\) — M1
  • \(3.35^2 = 11.2225\) — A1
  • \(3.3\) — A1

Model answer

\(3.3^2 = 10.89\) and \(3.4^2 = 11.56\), so \(\sqrt{11}\) is between 3.3 and 3.4. \(3.35^2 = 11.2225\), which is more than 11, so \(\sqrt{11}\) is below 3.35. So \(\sqrt{11} = 3.3\) to 1 decimal place.

6. Exam question Explain 2 marks Stretch

\(x_{n+1} = 4 - \dfrac{3}{x_n}\). Explain what happens to the values of \(x_n\) if \(x_0 = 1\). [2 marks]

Mark scheme — 2 marks available

  • \(x_1 = 4 - 3 = 1\) — M1
  • States that every term is 1, because the value does not change — B1

Model answer

\(x_1 = 4 - \dfrac{3}{1} = 1\), so every term is 1. The sequence stays at 1.

7. Multiple choice 1 mark Easier

\(x_{n+1} = 3 - \dfrac{2}{x_n}\) and \(x_0 = 4\). What is \(x_1\)?

  1. A 3.5
  2. B 2
  3. C 0.5
  4. D 2.5 Correct

Why: \(3 - \dfrac{2}{4} = 3 - 0.5 = 2.5\).

8. Multiple choice 1 mark Easier

What does \(x_0\) mean in an iteration?

  1. A The limit
  2. B The last value
  3. C The starting value Correct
  4. D The number of steps

Why: \(x_0\) is where the iteration starts.

9. Multiple choice 1 mark Easier

At the limit of an iteration, which statement is true?

  1. A \(x_{n+1} = 0\)
  2. B \(x_{n+1} = x_n\) Correct
  3. C \(x_n = 1\)
  4. D \(x_{n+1} = 2x_n\)

Why: At the limit the values stop changing.

10. Multiple choice 1 mark Core

\(x_{n+1} = 3 - \dfrac{2}{x_n}\) and \(x_1 = 2.5\). What is \(x_2\)?

  1. A 2.2 Correct
  2. B 2.8
  3. C 2.4
  4. D 1.8

Why: \(3 - \dfrac{2}{2.5} = 3 - 0.8 = 2.2\).

11. Multiple choice 1 mark Core

Which equation can be rearranged to \(x = 3 - \dfrac{2}{x}\)?

  1. A \(x^2 + 3x + 2 = 0\)
  2. B \(x^2 - 2x + 3 = 0\)
  3. C \(x^2 - 3x - 2 = 0\)
  4. D \(x^2 - 3x + 2 = 0\) Correct

Why: Divide \(x^2 - 3x + 2 = 0\) by \(x\) to get \(x - 3 + \dfrac{2}{x} = 0\).

12. Multiple choice 1 mark Core

\(f(x) = x^3 + x - 3\). What does \(f(1) = -1\) and \(f(2) = 7\) show?

  1. A The root is 1.5
  2. B There is no root
  3. C A root lies between 1 and 2 Correct
  4. D The root is 0

Why: The sign changes, so a root lies between them.

13. Multiple choice 1 mark Stretch

\(x_{n+1} = 1 + \dfrac{6}{x_n}\) converges to a positive number \(a\). Which equation does \(a\) satisfy?

  1. A \(a^2 + a - 6 = 0\)
  2. B \(a^2 - a - 6 = 0\) Correct
  3. C \(a^2 - 6a - 1 = 0\)
  4. D \(a^2 - a + 6 = 0\)

Why: \(a = 1 + \dfrac{6}{a}\), so \(a^2 = a + 6\).

14. Multiple choice 1 mark Stretch

What is the value of \(a\) in the previous question?

  1. A 3 Correct
  2. B \(-2\)
  3. C 6
  4. D 2

Why: \((a - 3)(a + 2) = 0\) and \(a\) is positive.

15. Multiple choice 1 mark Stretch

\(f(x) = x^2 - 7\). \(f(2.6) = -0.24\) and \(f(2.7) = 0.29\). Why is the root 2.6 to 1 decimal place, given \(f(2.65) = 0.0225\)?

  1. A The root is 2.7 because \(f(2.7)\) is positive
  2. B The root is exactly 2.65
  3. C It cannot be rounded
  4. D The root is below 2.65, so it rounds down to 2.6 Correct

Why: \(f(2.65) > 0\) means the root is between 2.6 and 2.65.