Welcome, welcome, welcome to the first in our new series: Beyond’s Monthly Maths Mastery. Each month, our astute army of arithmetic Adonises will share their maths-teacher approved method for solving challenging maths problems. This month, we will look at the iterative method for solving nonlinear equations…strap yourselves in…

## What is an iterative formula?

An iterative formula is obtained by rearranging the equation you are trying to solve. This will usually be done for you, though sometimes you will need to verify that the original equations rearranges to give the correct formula.

Iteration means βthe process of doing something again and againβ, so to use an iterative formula, we substitute a starting value of *x* in to get a new value of *x* out. It’s a bit like using one of those penny arcade machines, hence the snazzy illustrations! We then do that over and over, until we get close to the actual answer (or to a required degree of accuracy such as 4 decimal places).

## How do you use an iterative formula?

Iterative formulae can look a little scary as they have something called subscripts in them. These look like *x*_{n }or *x*_{n+1}. We donβt need to be scared though; these equations are telling us that when we substitute a value of *x* in, we get the next value of *x* out.

For example,

Use the iterative formula with *x*_{1}=0.7 to find the value of *x*_{2} and *x*_{3}, giving your answers correct to 3 decimal places.

We are given the first value of *x* as 0.7, so we substitute this into the formula in place of *x*_{n}. Since we are substituting *x*_{1} into the formula, we know we are going to get *x*_{2} out.

This gives us a value of or 3.429 correct to 3 decimal places.

Next, we substitute this value back into the formula in place of *x*_{n}. Since we are substituting *x*_{2 }into the formula now, we are going to get *x*_{3}out. This is a great time to use the previous answer button on your calculator!

This gives us a value of or 2.292 correct to 3 decimal places.

## Iterative method for solving nonlinear equations: finding approximate solutions

The more we substitute values into the formula, the closer we get to the actual solution to the equation. We want to get to a stage where the value of *x*_{n }is equal to the value *x*_{n+1} to a given degree of accuracy. In other words, the value of x that goes into the formula matches the value of *x* we get out.

Use the iterative formula together with a starting value of *x*_{1} = -4 to find a solution to the equation , giving your answer correct to 3 decimal places.

Remember, this formula is telling us that when we substitute a value of *x* in, we get the next value of *x* out.

We are given a starting value of *x*_{1} = -4, so if we substitute this into the iterative formula, we will find the value of *x*_{2}.

This gives us *x*_{2} = -5.75.

We want an answer accurate to 3 decimal places, and if we compare *x*_{1} and *x*_{2}, we see they are not equal to this given degree of accuracy, so we continue.

Substitute the value for *x*_{2} into the formula to get a value of *x*_{3}. This is a great time to use the previous answer button on your calculator!

This gives us *x*_{3} = -5.522 (to 3 decimal places).

Comparing *x*_{2} and *x*_{3}, we see they are not equal to 3 decimal places, so we repeat the process. Letβs keep going until we find two numbers which are equal correct to 3 decimal places.

*x*_{4} = -5.543 (to 3 decimal places),

*x*_{5} = -5.541 (to 3 decimal places),

*x*_{6} = -5.541 (to 3 decimal places).

Notice that *x*_{5} is equal to *x*_{6.} This means we stop here, and we have answered the question.

A solution to the equation *x*^{2} + 5*x* β 3 = 0 is *x* = -5.541, correct to 3 decimal places.

Are you still with us? Shall we have a break? Meet back here in 5 with some biscuits…πͺ

## Finally, where does the iterative formula come from?

The iterative formula is found by rearranging the equation you are trying to solve.

For example,

Show that *x*^{2} β 5*x* β 8 = 0 can be rearranged to give the equation

Rearranging this formula looks different to our usual process for changing the subject. We would usually try to isolate *x* on one side of the equation, but here we see that we have an *x* on both sides. In fact, we have *x* as the denominator of a fraction which tells us we will need to divide through by *x* at some point.

Instead, letβs take the equation *x*^{2} β 5*x* β 8 = 0 and add 8 to both sides.

*x*^{2} β 5*x* = 8

Next, letβs add 5*x*.

*x*^{2} = 5*x* + 8

Now, we can see that we are starting to form an equation that looks like the one we want. Our last step is to divide by *x*.

We can cancel the *x* in our first fraction, leaving us with

, as required!

This equation could now be used as an iterative formula with a starting value of x to solve the quadratic equation we were originally given! Note that it can be easier to work backwards from the solution, remembering to present our work upside down.

If you’re struggling to picture all of this, our resident Maths Master has beautifully summarised in the video below:

So, there you have it! The iterative method for solving nonlinear equations laid bare. Why not share this post with your students or fellow maths aficionados? Until next month…happy mathsing!

Tackling iteration with your class? Why not try some of our resources below.

## GCSE Escape the Room: Iterations Exit Ticket

## Iterative Processes Worksheet

## Iteration to Find Approximate Solutions Lesson Pack

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