Iterative Methods

GCSELevel 6-7AQAEdexcelOCR

Iterative Methods

Iterative methods or iterations is the idea of repeating a process over and over with the purpose of getting closer to an answer. In maths, iterative methods are often used when finding an exact answer is not so simple. There are 3 key skills involved with iterative method questions at GCSE level, which are shown below.

Make sure you are happy with the following topics before moving onto Iterative methods

Rearranging Formulas

Skill 1: Trial and Improvement

Trial and improvement is an iterative process whereby you try different solutions for an equation until you get the degree of accuracy that you want. This is easiest to see with an example.

Example: Use trial and improvement show that $textcolor{blue}{x}^3+6textcolor{blue}{x}=4$ has a solution between $0$ and $1$ .

Give your answer to $1$ dp.

Iterative Methods Example Table Too Big Too Small Trial And Improvement

First pick the largest number in the range ( $textcolor{blue}{x=1}$ ). Then, substitute this value into the equation.

$textcolor{blue}{x}^3 +6textcolor{blue}{x} = (textcolor{blue}{1})^3 +6(textcolor{blue}{1}) = 1 + 6 = 7$

We know $7>4$ (it’s too big), so we must try a smaller number. $textcolor{blue}{x=0}$

$textcolor{blue}{x}^3+6textcolor{blue}{x}=(textcolor{blue}{0})^3 + 6(textcolor{blue}{0})=0$

$0$ is too small, so we know the solution must be somewhere in between $0$ and $1$ .

We now repeat this process,

$begin{aligned}(textcolor{blue}{0.5})^3 + 6(textcolor{blue}{0.5})&=3.125.. , text{Too small}& (textcolor{blue}{0.7})^3 + 6(textcolor{blue}{0.7})&=4.543.., text{Too big}& (textcolor{blue}{0.6})^3 + 6(textcolor{blue}{0.6})&=3.816.., text{Too small}& (textcolor{blue}{0.65})^3 + 6(textcolor{blue}{0.65})&=4.17..,,,, text{Too big}&end{aligned}$

As we can now see, $0.6 =$ too small, but $0.65=$ too big. This means the actual solution must be between these two values.

We know that any number between $0.6$ and $0.65$ must round to $0.6$ to $1$ dp, so the solution must be $0.6$ to $1$ dp.

Skill 2: Using Iteration Machines

An iteration machine allows us to find an approximate solution to an equation we may not be able to solve any other way.

Example: Use the iterative formula shown to find the value of $x$ to $1$ dp

$textcolor{red}{x_{n+1}} = dfrac{4-textcolor{limegreen}{x_{n}}^3}{6}$

Use $x_1 = 1$ .

Step 1: Use $x_1$ as $textcolor{limegreen}{x_n}$ in the equation to give $textcolor{red}{x_{x+1}}$

$x_1 = 1 ,, text{gives } ,,, textcolor{red}{x_2} = dfrac{4-{textcolor{limegreen}{x_{1}}}^3}{6}=dfrac{4-(textcolor{limegreen}{1})^3}{6}=textcolor{red}{0.5}$

Step 2: Repeat the process using $x_2$ as $textcolor{limegreen}{x_n}$ to give $textcolor{red}{x_3}$

$x_2 = 0.5 ,,, text{ gives } ,,, textcolor{red}{x_3} = dfrac{4-{textcolor{limegreen}{x_{2}}}^3}{6}=dfrac{4-(textcolor{limegreen}{0.5})^3}{6}=textcolor{red}{0.6548333…}$

Step 3: Repeat the process to give $x_4$ , $x_5$ , ….

$x_3=0.6548333… ,,, text{ gives } ,,, textcolor{red}{x_4}=dfrac{4-{textcolor{limegreen}{x_{3}}}^2}{6}=dfrac{4-(textcolor{limegreen}{0.6548333…})^3}{6}=textcolor{red}{0.6217…}$

$x_4=0.6217… ,,, text{ gives } ,,, textcolor{red}{x_5}=dfrac{4-{textcolor{limegreen}{x_{4}}}^2}{6}=dfrac{4-(textcolor{limegreen}{0.6217…})^3}{6}=textcolor{red}{0.6266…}$

Step 4: Once two consecutive answers round to the same $1$ dp answer, we have our final answer.

$xcancel{0.6548333… ,,, text{rounds to} ,,, 0.7}$

$0.6217… ,,, text{rounds to} ,,, 0.6$

$0.6266… ,,, text{rounds to} ,,, 0.6$

Final answer $x = 0.6$ ( $1$ dp)

Note: When doing a question like this, it makes your life a lot easier if you use the ANS key on your calculator.

Skill 3: Creating the iteration formula

It is often necessary to form the iterative methods formula (e.g. $x_{n+1}=f(x_n))$ by rearranging an equation.

Example: Show that $x^3+x^2 =7$ can be rearranged to give to give $x_{n+1} = sqrt[3]{7-{x_{n}}^2}$

In order to do this we need to rearrange the equation to be in the form stated, then add in the notation required.

$begin{aligned} (-x^2) ,,,,,,,,, x^3+x^2 &=7 (sqrt[3]{}) ,,,,,,,,,,,,,,,,,,,,,,, x^3 &= 7-x^2 x &= sqrt[3]{7-x^2}end{aligned}$

Finally we add in the required notation to complete our answer.

$begin{aligned}x &= sqrt[3]{7-x^2} x_{n+1} &= sqrt[3]{7-{x_{n}}^2} end{aligned}$

Iterative Methods Example Questions

Question 1: Starting with $x=2$ , use trial and improvement to find a solution to the following equation to $1$ decimal place,

$2x^3-6x=1$

[3 marks]

We will form a table with one column of $x$ values, one column on the results of calculating $2x^3-6x$ , and one column stating if the answer is bigger or smaller than the desired $1$ .

So, if $1.8$ gives a result that is too small and $1.85$ gives a result that is too big, then the actual solution must be somewhere between these two values.

Given that any number between $1.8$ and $1.85$ must round to $1.8$ , the solution must be $1.8$ to $1$ decimal place.

Save your answers with

Gold Standard Education

Question 2: Starting with $x_1=2$ , use the iterative formula

$x_{n+1}=sqrt[3]{3x+9}$

to find a solution to $x^3-3x-9=0$ to $2$ decimal places.

[3 marks]

To find a solution we will use the recursive formula, until we get two consecutive terms which round to the same number to $2$ decimal places.

$begin{aligned} x_1&=2 x_2&=sqrt[3]{3(2)+9}=2.4662 x_3&=sqrt[3]{3(2.4662)+9}=2.54060 x_4&=sqrt[3]{3(2.54060 )+9}=2.55207 x_5&=sqrt[3]{3(2.55207 )+9}=2.55383end{aligned}$

These last two results both round to $2.55$ to $2$ dp, so the solution must be $2.55$ to $2$ decimal places.

Save your answers with

Gold Standard Education

Question 3: Starting with $x_1=1$ , use the iterative formula

$x_{n+1}=dfrac{-3}{{x_{n}}^2+5}$

to find a solution to $x^3+5x+3=0$ to $2$ decimal places.

[3 marks]

To find a solution we will use the recursive formula, until we get two consecutive terms which round to the same number to $2$ decimal places.

$begin{aligned}x_1&=1 x_2&=dfrac{-3}{(1)^2+5}=-0.5 x_3&=dfrac{-3}{(-0.5)^2+5}=-0.5714 x_4&=dfrac{-3}{(-0.5714 )^2+5}=-0.5632 x_5&= dfrac{-3}{(-0.5632 )^2+5}= -0.564end{aligned}$

These last two results both round to $-0.56$ to $2$ dp, so the solution must be $-0.56$ to $2$ decimal places.

Save your answers with

Gold Standard Education

Question 4: Starting with $x_1=4$ , use the iterative formula

$x_{n+1}=dfrac{3}{(x_n)^2}+3$

to find a solution to $3x^2-x^3+3=0$ to $3$ decimal places.

[3 marks]

To find a solution we will use the recursive formula, until we get two consecutive terms which round to the same number to $3$ decimal places.

$begin{aligned}x_1&=4 x_2&=dfrac{3}{(4)^2}+3=3.1875 x_3&=dfrac{3}{(3.1875)^2}+3=3.29527 x_4&=dfrac{3}{(3.29527)^2}+3=3.27627 x_5&=dfrac{3}{(3.27627)^2}+3=3.27949 x_6&=dfrac{3}{(3.27949)^2}+3=3.27894 x_7&=dfrac{3}{(3.27894)^2}+3=3.27903end{aligned}$

These last two results both round to $3.279$ to $3$ dp, so the solution must be $3.279$ to $3$ decimal places.

Save your answers with

Gold Standard Education

Question 5: Starting with $x_1=2$ , use the iterative formula

$x_{n+1}=sqrt[3]{6x+5}$

to find a solution to $x^3-6x-5=0$ to $2$ decimal places.

[3 marks]

To find a solution we will use the recursive formula, until we get two consecutive terms which round to the same number to $2$ decimal places.

$begin{aligned} x_1&=2 x_2&=sqrt[3]{6(2)+5}= 2.57128 x_3&=sqrt[3]{6(2.57128)+5}= 2.73363 x_4&=sqrt[3]{6(2.73363)+5}= 2.77641 x_5&=sqrt[3]{6(2.77641)+5}= 2.78746 x_6&=sqrt[3]{6(2.78746)+5}= 2.79030 x_7&=sqrt[3]{6(2.79030)+5}= 2.791036end{aligned}$

These last two results both round to $2.79$ to $2$ dp, so the solution must be $2.79$ to $2$ decimal places.

Save your answers with

Gold Standard Education

Specification Points Covered

Algebra – 20. find approximate solutions to equations numerically using iteration

Iterative Methods Worksheet and Example Questions

(NEW) Iterative Methods (Higher) Exam Style Questions - MME

NewOfficial MME

Exam Questions Mark Scheme

Revise

Try it yourself

Iterative Methods

Iterative Methods

Skill 1: Trial and Improvement

Skill 2: Using Iteration Machines

Skill 3: Creating the iteration formula

Iterative Methods Example Questions

Specification Points Covered

Iterative Methods Worksheet and Example Questions

(NEW) Iterative Methods (Higher) Exam Style Questions - MME

Iterative Methods Drill Questions

Iterative Methods

Related Topics

Rearranging Formulas Worksheets, Questions and Revision

Where next?

Previous GCSE Maths Topic

Graphical Inequalities

GCSE Maths Revision Home

Go back to the main GCSE Maths topic list

Next GCSE Maths Topic

Simultaneous Equations

A Level

GCSE

Key Stage 3

Key Stage 1 & 2

Functional Skills

Past Papers

A Level

GCSE

Other

Courses & Exams

By Type

Special Offers

For Teachers

Functional Skills

GCSE Exams

Tuition

A Level

GCSE

All

Revise

Try it yourself

Iterative Methods

Iterative Methods

Skill 1: Trial and Improvement

Skill 2: Using Iteration Machines

Skill 3: Creating the iteration formula

Iterative Methods Example Questions

Specification Points Covered

Iterative Methods Worksheet and Example Questions

(NEW) Iterative Methods (Higher) Exam Style Questions - MME

Iterative Methods Drill Questions

Iterative Methods

Related Topics

Rearranging Formulas Worksheets, Questions and Revision

Where next?

Previous GCSE Maths Topic

Graphical Inequalities

GCSE Maths Revision Home

Go back to the main GCSE Maths topic list

Next GCSE Maths Topic

Simultaneous Equations