Iteration at GCSE: Solving Equations You Cannot Solve Exactly

Iteration is one of those GCSE maths topics that looks far harder than it is. You are handed a formula full of subscripts, told to start with $x_0$, and expected to produce an answer to three decimal places. It feels like a different subject to the algebra you have been doing all year.

It is not. Iteration is a repeated calculation. You put a number in, you get a slightly better number out, and you feed that number back in. Do it enough times and the numbers stop changing, and the value they settle on is your answer.

This guide covers what an iterative formula means, how to rearrange an equation into one, how to run an iteration on your calculator without retyping every line, and the specific places students lose marks on these questions.

What an Iterative Formula Actually Says

An iterative formula looks something like this:

$$x_{n+1} = \frac{x_n^{3} + 1}{5}$$

The subscripts are the only new thing here. $x_0$ is your starting value, $x_1$ is the next one, $x_2$ the one after that. The $n$ is a placeholder for "whichever one you are on", so $x_{n+1}$ means "the next value" and $x_n$ means "the one you have got".

So the formula above says: take the value you have, cube it, add 1, divide by 5. That is your next value. Nothing more complicated than that.

Why bother? Because some equations cannot be solved with the methods you already have. You can factorise a quadratic, or use the quadratic formula, or complete the square. There is no equivalent method for $x^{3} - 5x + 1 = 0$ at GCSE. Iteration gets you a very accurate approximation instead, and in practice an answer correct to four decimal places is usually more than enough.

When the values settle on one number, we say the sequence converges. When they run away to enormous positive or negative numbers instead, it diverges. A diverging sequence tells you the rearrangement you chose does not work, not that you have made a mistake in the arithmetic.

Rearranging an Equation Into an Iterative Formula

Exam questions often ask you to show that an equation can be written as a particular iterative formula. This is pure algebra, and it is usually worth two marks.

Take $x^{3} + 2x = 9$. To turn it into a formula starting $x =$, isolate a single $x$ somewhere:

$$x^{3} = 9 - 2x$$

$$x = \sqrt[3]{9 - 2x}$$

Then replace the $x$ on the left with $x_{n+1}$ and the $x$ on the right with $x_n$:

$$x_{n+1} = \sqrt[3]{9 - 2x_n}$$

Here is the part that catches people out. The same equation can be rearranged more than one way, and the rearrangements are not equally useful. From $x^{3} + 2x = 9$ you could just as fairly write:

$$x = \frac{9 - x^{3}}{2} \quad \text{giving} \quad x_{n+1} = \frac{9 - x_n^{3}}{2}$$

Start both at $x_0 = 1$ and watch what happens. The first gives $1.913$, then $1.730$, then $1.769$, closing in on about $1.762$. The second gives $4$, then $-27.5$, then a number in the tens of thousands. Same equation, same starting value, and one of them falls apart completely.

You are not expected to predict which rearrangement converges at GCSE. You are expected to use the one the question gives you. If your values start growing out of control, check you have copied the formula correctly before you assume the maths has gone wrong.

Using Iteration on Your Calculator: A Worked Example

Here is a full question of the type you will meet in Paper 2 or Paper 3.

Show that $x^{3} - 5x + 1 = 0$ has a solution between $0$ and $1$. Using $x_{n+1} = \dfrac{x_n^{3} + 1}{5}$ with $x_0 = 0$, find the solution correct to 4 decimal places.

Step 1: show the sign change. Substitute both ends into $f(x) = x^{3} - 5x + 1$:

$$f(0) = 0 - 0 + 1 = 1$$

$$f(1) = 1 - 5 + 1 = -3$$

One value is positive and one is negative, so the graph crosses the $x$-axis somewhere between them. Write that sentence down. The mark is for the conclusion, not only the two numbers.

Step 2: run the iteration. Starting from $x_0 = 0$:

$$x_1 = \frac{0^{3} + 1}{5} = 0.2$$

$$x_2 = \frac{0.2^{3} + 1}{5} = 0.2016$$

$$x_3 = \frac{0.2016^{3} + 1}{5} = 0.2016387\ldots$$

$$x_4 = \frac{0.2016387^{3} + 1}{5} = 0.2016396\ldots$$

Step 3: decide when to stop. $x_3$ and $x_4$ agree to four decimal places, both rounding to $0.2016$. That is your answer: $x \approx 0.2016$.

There is a much faster way to do Step 2 on a scientific calculator. Type your starting value and press equals. Now type the formula using the Ans key wherever $x_n$ appears: open bracket, Ans, cube, plus 1, close bracket, divide by 5. Press equals, and every further press of equals gives you the next term. You never retype anything, and you never lose accuracy to rounding.

Common Mistakes in Iteration Questions

These are the places marks disappear, and most of them are about process rather than algebra.

Rounding partway through. If you round $x_1$ to $0.2$ and type that in by hand, then round $x_2$, the small errors build up and your final answer can be wrong in the fourth decimal place. Keep the full value in your calculator using the Ans key and only round at the very end.

Stopping too early. Two consecutive terms agreeing to three decimal places does not guarantee they agree to four. If the question asks for 4 d.p., carry on until two terms round to the same 4 d.p. value.

Giving the wrong term. Some questions ask for $x_1$, $x_2$ and $x_3$ and nothing else. Others ask for the solution. Read which one is wanted. Writing out the full sequence when a single value was asked for wastes time, and giving $x_3$ when the solution was wanted loses the mark.

Skipping the sign-change sentence. A sign change question needs three things: the value of $f$ at the lower bound, the value at the upper bound, and a written statement that the change of sign means there is a root between them. Two numbers with no conclusion is usually one mark out of two.

Writing terms to too few decimal places. When a question says "write down the first three iterates", you need enough digits to show the values are genuinely converging. Giving $0.2$, $0.2$, $0.2$ hides your working even if your calculator was right.

Practise now: Iteration and algebra questions on Bow Tie Maths. The app builds a Topic Radar from your answers so you always know where to focus next.

Where Iteration Sits in the GCSE Course

Iteration is a Higher tier topic, and it is usually pitched around grades 6 to 7. That makes it a sensible use of revision time if you are aiming for a 7 or above, because it is a self-contained method rather than a topic that needs years of fluency behind it. The algebra involved, cubing, cube roots and rearranging, is the same algebra you use elsewhere.

It also rewards calculator confidence more than almost any other topic. In my own lessons, the students who know the Ans key finish these questions in under a minute. The ones who retype every line take far longer and pick up rounding errors on the way. If you take one thing from this guide, make it that.

Summary

Iteration turns an equation you cannot solve exactly into a repeated calculation you can. Rearrange to get a single $x$ on the left, feed each value back in with the Ans key, and stop when two consecutive terms round to the same value at the accuracy you have been asked for. Show the sign change in words when the question asks for it.

If you want to put this into practice, try Bow Tie Maths. It generates questions on this topic at your level and tracks your progress over time.