Numerical Solutions Using Iteration Revision Notes | GCSE AQA Higher Maths

1. Why Iteration Exists

Some equations cannot be solved by factorising, completing the square, or any neat algebraic trick. Take $x^3 - x - 1 = 0$: there is no clean factorisation, and the quadratic formula does not apply. Yet the equation definitely has a solution somewhere. Iteration is the method that hunts it down numerically, step by step.

The core idea is surprisingly direct: start with a rough guess, feed it into a formula, get a better approximation, then repeat. Each cycle brings you closer to the true answer. This is what The higher-tier specification means by finding approximate solutions to equations numerically using iteration.

2. Suffix Notation and Recursive Formulae

You will always be given the iterative formula in the exam — your job is to apply it correctly. For example, the formula $x_{n+1} = \sqrt{x_n + 1}$ tells you: take your current approximation, add 1, then square root the result to get the next approximation.

The subscript $n$ is just a counter — it does not represent multiplication or a power. When $n = 1$, you compute $x_2 = f(x_1)$. When $n = 2$, you compute $x_3 = f(x_2)$. Each output becomes the next input.

3. Carrying Out an Iteration: Worked Example

Here is how the method looks in practice. The equation $x^3 - x - 1 = 0$ can be rearranged to give the iterative formula $x_{n+1} = \sqrt[3]{x_n + 1}$. Starting with $x_1 = 1$, the sequence converges to the root. This is a calculator-allowed exam question method — keep full calculator precision at every intermediate step and only round the final answer.

Worked example: Applying an Iterative Formula

Use the iterative formula $x_{n+1} = \sqrt[3]{x_n + 1}$ with $x_1 = 1$ to find $x_4$, giving your answer to 4 decimal places.

1. $x_1 = 1$ — The question gives us the first approximation. [Starting point — no mark yet]
2. $x_2 = \sqrt[3]{x_1 + 1} = \sqrt[3]{1 + 1} = \sqrt[3]{2} \approx 1.2599$ — Substitute $x_1 = 1$ into the formula. $\sqrt[3]{2}$ means the cube root of 2. [M1 for correct substitution]
3. $x_3 = \sqrt[3]{x_2 + 1} = \sqrt[3]{1.2599 + 1} = \sqrt[3]{2.2599} \approx 1.3122$ — Use the full unrounded value of $x_2$ from your calculator. Rounding mid-calculation causes drift. [M1 for correct process]
4. $x_4 = \sqrt[3]{x_3 + 1} = \sqrt[3]{1.3122 + 1} = \sqrt[3]{2.3122} \approx 1.3252$ — Again, use the full value of $x_3$. The sequence is converging — notice the values are getting closer together. [A1 for $x_4 \approx 1.3252$]
5. $x_4 \approx 1.3252$ — Round only at the final step. The true root of $x^3 - x - 1 = 0$ is approximately $1.3247$, so the sequence is closing in on it.

Answer: $x_4 \approx 1.3252$ (to 4 decimal places)

Notice how the values $1,\; 1.2599,\; 1.3122,\; 1.3252$ are getting closer together with each step. That narrowing gap is convergence — the hallmark of a working iterative formula.

The graph of $y = x^3 - x - 1$ crosses the $x$-axis once between $x = 1$ and $x = 2$. That crossing point is the root the iteration is homing in on. Spotting where a curve crosses the $x$-axis gives you a sensible starting value $x_1$.

4. Recognising Convergence and Knowing When to Stop

An exam question will usually ask you to find a root correct to a specified number of decimal places. The stopping rule is: keep iterating until two consecutive values agree when rounded to the required precision.

For example, if $x_5 = 1.3247$ and $x_6 = 1.3247$ both round to $1.3247$ to 4 decimal places, you can state the root as $1.3247$ to 4 d.p. Always write down enough iterations to show the convergence clearly — examiners award method marks for the working, not just the final value.

Worked example: Finding a Root to a Given Accuracy

The equation $x^3 + 2x - 5 = 0$ has a root near $x = 1$. Use the iterative formula $x_{n+1} = \frac{5 - x_n^3}{2}$ with $x_1 = 1$ to find the root correct to 2 decimal places.

1. $x_1 = 1$ — Given starting value.
2. $x_2 = \frac{5 - (1)^3}{2} = \frac{5 - 1}{2} = \frac{4}{2} = 2$ — $(1)^3 = 1 \times 1 \times 1 = 1$. Substituting gives $\frac{4}{2} = 2$. [M1 for correct substitution]
3. $x_3 = \frac{5 - (2)^3}{2} = \frac{5 - 8}{2} = \frac{-3}{2} = -1.5$ — $(2)^3 = 2 \times 2 \times 2 = 8$. The sequence has jumped to $-1.5$ — this formula is diverging from the root near $x = 1$.
4. Values: $1 \to 2 \to -1.5 \to \ldots$ (moving further apart) — This iterative formula does not converge for this starting value. A different rearrangement of the equation would be needed. This is an important exam trap.

Answer: This formula diverges. A convergent rearrangement such as $x_{n+1} = \sqrt[3]{5 - 2x_n}$ should be used instead.

Not every rearrangement of an equation produces a convergent iterative formula. If your sequence produces values that grow larger and larger (or oscillate wildly), the formula is diverging and a different rearrangement is needed. In the exam, you will always be given a formula that converges — but you may be asked to comment on whether a sequence is converging.

5. A Full Exam-Style Question Walkthrough

GCSE Higher exam questions on iteration typically ask you to show that a root lies in an interval, then apply a given formula to find it to a stated accuracy. Here is a complete example modelling both parts.

Worked example: Full Exam-Style Iteration Question

The equation $x^3 - 3x + 1 = 0$ has a root $\alpha$ in the interval $0 < \alpha < 1$. (a) Show that a root lies between $x = 0$ and $x = 1$. (b) Use the iterative formula $x_{n+1} = \frac{x_n^3 + 1}{3}$ with $x_1 = 0.5$ to find $x_3$, giving your answer to 4 decimal places.

1. $f(x) = x^3 - 3x + 1$ $f(0) = (0)^3 - 3(0) + 1 = 0 - 0 + 1 = 1$ $f(1) = (1)^3 - 3(1) + 1 = 1 - 3 + 1 = -1$ — $(0)^3 = 0$, $(1)^3 = 1 \times 1 \times 1 = 1$. [M1 for evaluating both endpoints]
2. $f(0) = 1 > 0$ and $f(1) = -1 < 0$ — There is a sign change between $x = 0$ and $x = 1$. Since $f$ is continuous, there must be a root in this interval. [A1 for sign-change conclusion]
3. $x_2 = \frac{(0.5)^3 + 1}{3} = \frac{0.125 + 1}{3} = \frac{1.125}{3} = 0.375$ — $(0.5)^3 = 0.5 \times 0.5 \times 0.5 = 0.125$. Substituting $x_1 = 0.5$ gives $x_2 = 0.375$. [M1 for correct substitution]
4. $x_3 = \frac{(0.375)^3 + 1}{3} = \frac{0.052734375 + 1}{3} = \frac{1.052734375}{3} \approx 0.3509$ — $(0.375)^3 = 0.375 \times 0.375 \times 0.375 = 0.140625 \times 0.375 = 0.052734375$. Keep full precision until the final rounding. [A1 for $x_3 \approx 0.3509$]
5. $x_3 \approx 0.3509$ (to 4 d.p.) — Round only at this final step. The sequence $0.5 \to 0.375 \to 0.3509$ is converging towards the root near $x \approx 0.347$.

Answer: (a) Sign change between $f(0) = 1 > 0$ and $f(1) = -1 < 0$ confirms a root in $(0, 1)$. (b) $x_3 \approx 0.3509$ (to 4 d.p.)

6. Practice Problems

The best way to consolidate iteration is to work through problems yourself before checking the solution. Try each one, writing out every step — examiners award marks for the working, not just the final value.

7. Exam Technique and Common Mistakes

Iteration questions are reliable mark-earners once you know the traps. The method itself is mechanical — the marks are lost through careless rounding, misreading the formula, or failing to show enough working.

Iteration is one of those topics where the method is entirely learnable — there are no surprises once you have practised the substitution process. Every mark on an iteration question is available to you through careful, systematic working using suffix notation and the recursive formula provided.