 # Iterative Methods Worksheets, Questions and Revision

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## What you need to know

Iteration 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.

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 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. Currently, this is between 0 and 1, but as we try more numbers – as seen in the table – the gap where we know the solution must be will get smaller and smaller as the result of the calculation gets closer to 4.

Recall that the point was to determine the solution to 1dp. 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.

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.

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 (it is not always used, but very often is): 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.

Example: A recursive 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 $x_4$):

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

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

When doing a question like this, it makes your life a lot easier if you use the ANS key on your calculator. To do this, type in the value of $x_1$ and press “=”. Then, type the recursive formula into your calculator, replacing $x_n$ with ANS – then, all you need to do to find the next step of the process every time is to press equals. This is a real time-saver.

This particular formula is actually trying to find the same solution as in the trial and improvement process (it finds its way towards the solution – which is actually 0.6258 to 4sf – more quickly and also requires less thinking – bonus!). This iterative formula was found by rearranging the equation we were trying to find a solution for. Sometimes you are asked to do the rearranging yourself.

Example: 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 process enough times until the last two values we got are the same to 2dp. Note: always write down more decimal places than you need for each iteration.

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

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.

Another way that an iterative process might be given to you is in the form of a flow-chart. The process is exactly the same as last two examples (including using the ANS key), it’s just set up differently. Try the third question below to see.

Iterative methods is another algebra topic that many students struggle to grasp and it is usually due to the notation involved. With the iteration questions and worksheets above teachers and tutors should find it easier to deliver iterations whilst helping students to understand the topic by providing them with many quality practice questions. For other GCSE Maths resources, similar to iterations, visit our comprehensive GCSE Maths revision page.

### Example Questions

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 1dp.

#### Is this a topic you struggle with? Get help now.

To find a solution to 2dp using this formula, we must use it to find $x_2, x_3,$ and so on until we get two consecutive terms which round to the same number to 2dp. So, we will put $x_1$ into the formula to get $x_2$, and then put that in the formula to get $x_3$ and so on until we get the desired accuracy of 2dp. Remember: use the ANS key!

$x_1=1,\text{ so }x_2=\dfrac{-3}{{x_{1}}^2+5}=\dfrac{-3}{(1)^2+5}=-0.5$

$\text{Then, }x_3=\dfrac{-3}{{x_{2}}^2+5}=\dfrac{-3}{(-0.5)^2+5}=-0.5714$

$x_4 = -0.5632$

$x_5 = -0.5642$

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

#### Is this a topic you struggle with? Get help now.

If you read through the flow chart, you’ll see that because the last step tells us to take each output and feed it back into the formula, this is precisely the same method as if we were given the recursive formula

$x_{n+1}=\dfrac{2{x_{n}}^3 + 12}{3{x_{n}}^2}$

and asked to find the solution to 3sf. That said, we’ll go through the flow-chart step-by-step regardless.

So, start with $x=3$. Then,

$\dfrac{2(3)^3+12}{3(3)^2} = 2.4444...$

This is not the same as $x$ when rounded to 3sf, so we make this our new $x$ and start again (using the ANS key!). Then, we get

$\dfrac{2(2.444...)^3+12}{3(2.444...)^2} = 2.299...$

This is not the same as $x$ when rounded to 3sf, so we make this our new $x$ and start again. Then, we get

$\dfrac{2(2.299...)^3+12}{3(2.299...)^2} = 2.289...$

This is not the same as $x$ when rounded to 3sf, so we make this our new $x$ and start again. Then, we get

$\dfrac{2(2.289...)^3+12}{3(2.289...)^2} = 2.289...$

This is the same as $x$ when rounded to 3sf, so 2.29 is our approximate solution to the equation $x^3-12=0$.

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