Using iteration to solve equations - Teacher Slides.pptx

EDEXCEL GCSE MATHS · HIGHER

Using iteration to solve equations

Equations and graphs

Lesson 6 of 6

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EDEXCEL GCSE MATHS · HIGHER

Warm-up

Equations and graphs

Lesson 6 of 6

Answer each one, then check.

1

Work out 1³ − 1 − 1.

2

Work out 2³ − 2 − 1.

3

What is ∛8?

4

What does ANS do on a calculator?

5

What does "to 2 decimal places" mean?

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EDEXCEL GCSE MATHS · HIGHER

Warm-up - Answers

Equations and graphs

Lesson 6 of 6

1

Work out 1³ − 1 − 1.

−1

2

Work out 2³ − 2 − 1.

5

3

What is ∛8?

2

4

What does ANS do on a calculator?

Uses the previous answer

5

What does "to 2 decimal places" mean?

Rounded to two digits after the point

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EDEXCEL GCSE MATHS · HIGHER

Learning Objectives

Equations and graphs

Lesson 6 of 6

1

Show that a root lies between two values.

2

Use decimal search to narrow down a root.

3

Use an iteration formula x_(n+1) = f(x_n).

4

Rearrange an equation into an iteration formula.

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Equations and graphs

Lesson 6 of 6

CHANGE OF SIGN

If f(a) and f(b) have opposite signs, the graph crosses the x-axis between a and b, so there is a root in that interval.

This works when f is continuous, with no breaks in the curve.

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Iteration Staircase

Equations and graphs

Lesson 6 of 6

Each step feeds the last answer back in, moving towards the root.

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Change of Sign

Equations and graphs

Lesson 6 of 6

Show that x³ − x − 1 = 0 has a root between 1 and 2.

1

Write the function

f(x) = x³ − x − 1

2

f(1)

1 − 1 − 1 = −1

3

f(2)

8 − 2 − 1 = 5

4

Sign change

Negative to positive

ANSWER

f(1) < 0 and f(2) > 0, so there is a root between 1 and 2.

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Decimal Search

Equations and graphs

Lesson 6 of 6

Find the root of x³ − x − 1 = 0 to 2 decimal places.

1

f(1.3)

−0.103 (negative)

2

f(1.4)

0.344 (positive)

3

f(1.32)

−0.020 (negative)

4

f(1.33)

0.023 (positive)

5

Midpoint f(1.325)

0.001 (positive), so the root is below 1.325

ANSWER

The root is 1.32 to 2 decimal places.

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Iteration

Equations and graphs

Lesson 6 of 6

x_(n+1) = ∛(x_n + 1) with x₀ = 1. Work out x₁, x₂ and x₃.

1

x₁

∛2 = 1.2599

2

x₂

∛(1.2599 + 1) = 1.3123

3

x₃

∛(1.3123 + 1) = 1.3224

ANSWER

x₁ = 1.2599, x₂ = 1.3123, x₃ = 1.3224; the values are approaching the root 1.3247.

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Rearranging to an Iteration Formula

Equations and graphs

Lesson 6 of 6

Show that x³ − x − 1 = 0 can be rearranged to x = ∛(x + 1).

1

Add x + 1

x³ = x + 1

2

Cube root

x = ∛(x + 1)

ANSWER

x³ − x − 1 = 0 gives x³ = x + 1, so x = ∛(x + 1).

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Calculator Tips

Equations and graphs

Lesson 6 of 6

Use your calculator efficiently.

Type the starting value

Press =, then type the formula using ANS.

Keep pressing =

Each press gives the next term.

Do not round in between

Use the full value from ANS.

Compare

When two terms agree to the accuracy needed, stop.

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Key Terms

Equations and graphs

Lesson 6 of 6

Iteration

Repeating a calculation using the previous answer.

Iteration formula

A rule x_(n+1) = f(x_n).

Starting value

The first value, x₀.

Change of sign

f(a) and f(b) have opposite signs.

Root

A value of x where f(x) = 0.

Convergence

The values getting closer to a limit.

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Equations and graphs

Lesson 6 of 6

YOUR TASK

Close In on a Root

12 minutes

(a) Show that x³ + 2x − 7 = 0 has a root between 1 and 2. (b) Use decimal search to find the root to 1 decimal place.

1

Test the midpoint of the last interval.

2

Decide which way to round.

WHAT A GOOD ANSWER SHOWS

(a) f(1) = −4, f(2) = 5. (b) f(1.5) = −0.625, f(1.6) = 0.296, f(1.55) = −0.176; the root is between 1.55 and 1.6, so it is 1.6 to 1 d.p.

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Can I...?

Equations and graphs

Lesson 6 of 6

Evaluate f(a) and f(b).

Explain a change of sign.

Use decimal search.

Use a starting value.

Use ANS on a calculator.

Use an iteration formula.

Rearrange to an iteration formula.

Stop at the required accuracy.

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Summary & Exam Focus

Change of sign shows a root in an interval.

Decimal search narrows the interval.

Iteration: feed each answer back into the formula.

Stop when values agree to the required accuracy.

EXAM FOCUS

Show that x³ − x − 1 = 0 has a root between 1 and 2. (2 marks)

Show both values of f and state that the sign changes.

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Exam Practice: Using iteration to solve equations

Equations and graphs

Lesson 6 of 6

Answer all questions. Show your working. · 30 minutes

Question 1 · 2 marks · Show that

Show that the equation x³ − x − 1 = 0 has a root between 1 and 2.

Question 2 · 3 marks · Work out

x_(n+1) = ∛(x_n + 1) and x₀ = 1. Work out the values of x₁, x₂ and x₃.

Question 3 · 2 marks · Show that

Show that x³ − x − 1 = 0 can be rearranged to give x = ∛(x + 1).

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Exam Practice: Using iteration to solve equations (cont.)

Equations and graphs

Lesson 6 of 6

Question 4 · 4 marks · Find

The graph of y = x³ − x − 1 is shown. It crosses the x-axis between x = 1 and x = 2. Use a trial and improvement method to find this root…

Question 5 · 4 marks · Find

(a) Show that x³ + 2x − 7 = 0 has a root between 1 and 2. (b) Find this root correct to 1 decimal place.

Question 6 · 2 marks · Explain

Using x_(n+1) = ∛(x_n + 1) with x₀ = 1, the values are x₅ = 1.32463 and x₆ = 1.32470. Write down the root of x³ − x − 1 = 0 to 3 decimal…

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Question 1 · 2 marks · Show that

Equations and graphs

Lesson 6 of 6

“

Show that the equation x³ − x − 1 = 0 has a root between 1 and 2.

HOW TO ANSWER IT

Command word: Show that. Worth 2 marks, so plan before writing.

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Question 1 · mark scheme

Equations and graphs

Lesson 6 of 6

2 marks available. Award a mark for each point made.

f(1) = −1 and f(2) = 5

M1

Change of sign conclusion

C1

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Question 2 · 3 marks · Work out

Equations and graphs

Lesson 6 of 6

“

x_(n+1) = ∛(x_n + 1) and x₀ = 1. Work out the values of x₁, x₂ and x₃.

HOW TO ANSWER IT

Command word: Work out. Worth 3 marks, so plan before writing.

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Question 2 · mark scheme

Equations and graphs

Lesson 6 of 6

3 marks available. Award a mark for each point made.

x₁ = 1.26

B1

x₂ = 1.31

B1

x₃ = 1.32

B1

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Question 3 · 2 marks · Show that

Equations and graphs

Lesson 6 of 6

“

Show that x³ − x − 1 = 0 can be rearranged to give x = ∛(x + 1).

HOW TO ANSWER IT

Command word: Show that. Worth 2 marks, so plan before writing.

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EDEXCEL GCSE MATHS · HIGHER

Question 3 · mark scheme

Equations and graphs

Lesson 6 of 6

2 marks available. Award a mark for each point made.

x³ = x + 1

M1

Cube root both sides

A1

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Question 4 · 4 marks · Find

Equations and graphs

Lesson 6 of 6

The graph of y = x³ − x − 1 is shown. It crosses the x-axis between x = 1 and x = 2. Use a trial and improvement method to find this root correct to 1 decimal place. You must show all your working. (4 marks)

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Question 4 · mark scheme

Equations and graphs

Lesson 6 of 6

4 marks available. Award a mark for each point made.

Tests values in the interval

M1

f(1.3) < 0 and f(1.4) > 0

A1

Tests 1.35

M1

1.3

A1

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Question 5 · 4 marks · Find

Equations and graphs

Lesson 6 of 6

“

(a) Show that x³ + 2x − 7 = 0 has a root between 1 and 2. (b) Find this root correct to 1 decimal place.

HOW TO ANSWER IT

Command word: Find. Worth 4 marks, so plan before writing.

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EDEXCEL GCSE MATHS · HIGHER

Question 5 · mark scheme

Equations and graphs

Lesson 6 of 6

4 marks available. Award a mark for each point made.

f(1) = −4 and f(2) = 5

B1

Tests 1.5 or 1.6

M1

Tests 1.55

M1

1.6

A1

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Question 6 · 2 marks · Explain

Equations and graphs

Lesson 6 of 6

“

Using x_(n+1) = ∛(x_n + 1) with x₀ = 1, the values are x₅ = 1.32463 and x₆ = 1.32470. Write down the root of x³ − x − 1 = 0 to 3 decimal places and explain how you know.

HOW TO ANSWER IT

Command word: Explain. Worth 2 marks, so plan before writing.

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Question 6 · mark scheme

Equations and graphs

Lesson 6 of 6

2 marks available. Award a mark for each point made.

1.325

B1

Successive values agree to 3 d.p.

C1

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