## What you need to know

### Iterations

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 two main types of iterative method questions at GCSE level, which are shown below.

Having a good knowledge Algebra substitution and rearranging equations will help with iterations.

### 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.

### Recursive Formulae

A better way of finding solutions using iterative methods is to use a recursive formula. With a recursive formula, the difference is that you only choose the first input for the formula, and then after that, each number that the formula spits out is then fed back in as the input. There’s some notation we need to know, the first input is given by x_1, and then this is put into a recursive formula which is generally expressed like

x_{n+1} = f(x_n)

The idea is, if you substitute x_1 into the function on the right-hand side, then n=1, which means that n+1=2, so the result of the function will give us x_2. Then, you put x_2 into the function to give you x_3, then put that back into the function to get x_4, and so on.

Each of these values: x_1, x_2, x_3, x_4, are all estimates for the solution to some problem, and as the number increases, the values get closer to the actual solution. It is this most iterative method questions are testing at GCSE level.

### Example 1: Trial and Improvement

Use trial and improvement to find a solution to x^3+6x=4 to 1dp.

To do this, we will choose a number at random to start with (choose a small number first to make your life easier). Then, substitute this value into the equation in question, and see how whether the answer is above or below 4. If we pick x = 1, then we get

x^3 +6x = (1)^3 +6(1) = 1 + 6 = 7

7 is too big, so we must try a smaller number. If we try x=0, we get

x^3+6x=(0)^3 + 6(0)=0

0 is too small, so we know the solution must be somewhere in between 0 and 1. We now repeat this process, always choosing our next attempt to be some value within the range of values where we know the solution must lie – as seen in the table below

As we can now see, 0.6 gave an answer that was too small, but 0.65 gave an answer that was too big, so the actual solution must be between these two values. However, we know that any number between 0.6 and 0.65 must round to 0.6 to 1 decimal place, so the solution must be 0.6 to 1dp.

**Note:** A table like this is a common method of recording the trial and improvement results – it makes it easier for you to track your results and for your marker to see what you’ve done.

### Example 2: Iterative Methods

An iterative formula for finding a solution to an equation is given by

x_{n+1} = \dfrac{4-x_{n}^3}{6}

Taking x_1 = 1, find x_2, x_3, and x_4.

To find x_2, we substitute x_1 into the formula:

x_1 = 1\text{ gives }x_2 = \dfrac{4-{x_{1}}^3}{6}=\dfrac{4-(1)^3}{6}=0.5

Then, to find x_3, we put x_2=0.5 back into the formula (and then again for x_4):

x_2 = 0.5\text{ gives }x_3 = \dfrac{4-{x_{2}}^3}{6}=\dfrac{4-(0.5)^3}{6}=0.6548333...

x_3=0.6548333...\text{ gives }x_4=\dfrac{4-{x_{3}}^2}{6}=\dfrac{4-(0.6548333...)^3}{6}=0.6217...

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

### Example 3: Iterative Methods

Show that x^2-x-5=0 rearranges to x=\sqrt{5+x}. Starting with x_1=1, use the recursive formula x_{n+1}=\sqrt{5+x_n} to find a solution to x^2-x-5=0 to 2dp.

First, add 5 and x to both sides of the equation to get

x^2=5+x.

Then, square root both sides to get the desired

x=\sqrt{5+x}

Now, to find the solution to 2dp, all we need to do is (using the ANS key) repeat this iterative method enough times until the last two values we got are the same to 2dp.

x_2=\sqrt{5+x_1}=\sqrt{5+1}=2.4495

Then, x_3=2.7294, x_4=2.7802, x_5=2.7893, and x_6=2.7909. Both 2.7893 and 2.7909 round to 2.79 to 2dp, so the solution must be 2.79 to 2 decimal places.

### Example Questions

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

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.

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.

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}=2.55383\end{aligned}

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

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.

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}= -0.564\end{aligned}

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

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.

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=3.27903\end{aligned}

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

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.

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}= 2.791036\end{aligned}

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

### Worksheets and Exam Questions

#### Iterative Methods

Level 6-7### Videos

#### Iterative Methods Q1

GCSE MATHS#### Iterative Methods Q2

GCSE MATHS#### Iterative Methods Q3

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