Numerical Methods and Iteration
Some equations do not have a convenient exact solution. Numerical methods use repeated calculations to narrow down or approximate a root.
Higher: this lesson covers locating roots by sign changes and decimal search, then using iterative formulae. AQA and Pearson Edexcel use iterative formulae; AQA expects suffix notation. OCR assesses systematic sign-change methods and says its questions do not involve iteration formulae.
You should already be comfortable substituting into expressions and rearranging formulae. The lessons on Rearranging Formulae and Functions are useful prerequisites.
What is a numerical solution?
A root of an equation is a value that makes the equation true. If we write an equation as f(x) = 0, roots are the x-values where the graph of y = f(x) crosses or touches the x-axis.
A numerical method does not necessarily give an exact surd or fraction. Instead, it produces an approximation accurate enough for the question.
Locating a root with a sign change
For a continuous function, if f(a) and f(b) have opposite signs, the graph must cross the x-axis somewhere between a and b.
Worked example — show a root lies between 1 and 2
Let f(x) = x³ + x − 5.
f(1) = 1 + 1 − 5 = −3, while f(2) = 8 + 2 − 5 = 5.
The sign changes from negative to positive, so a root lies between 1 and 2.
Decimal search
You can narrow the interval by testing closer values. Continue until the interval is small enough to justify the requested accuracy.
Worked example — narrow the same root
For f(x) = x³ + x − 5:
| x | f(x) |
|---|---|
| 1.50 | −0.125 |
| 1.51 | −0.047049 |
| 1.52 | 0.031808 |
The sign changes between 1.51 and 1.52, so a root lies in that interval. That interval alone does not justify 2 decimal places: it crosses the rounding boundary 1.515. Calculate f(1.515) = −0.007734125 and f(1.516) = 0.000156096. A root lies strictly between 1.515 and 1.516, and every value there rounds to 1.52 to 2 decimal places.
Do not round from a wide interval too early. Your interval must be tight enough to support the stated accuracy.
Iteration
Iteration starts from an initial value and repeatedly uses a rule such as . The output from one step becomes the input for the next.
Worked example
Solve x³ + x − 5 = 0 using the iteration xₙ₊₁ = ∛(5 − xₙ), starting with x₀ = 1.5. The table displays values to six decimal places; calculate each new iterate from the unrounded value.
| n | xₙ |
|---|---|
| 0 | 1.500000 |
| 1 | 1.518294 |
| 2 | 1.515644 |
| 3 | 1.516029 |
| 4 | 1.515973 |
| 5 | 1.515981 |
The values are settling near 1.51598, so the root is approximately x = 1.516 to 3 decimal places.
Using an iteration formula carefully
Write down the starting value, then calculate each new value from the previous one. Keep extra calculator digits during the process and round only the final answer.
A sequence of iterates may settle to a fixed value, oscillate, or move away. At GCSE, you may be given a suitable iteration formula or asked to use one obtained by rearranging an equation.
Not every rearrangement is equally useful for iteration. If the values do not settle, that form may not converge from the chosen starting value.
When can you stop?
A common practical stopping rule is to continue until successive iterates agree to the required number of decimal places.
For example, if xₙ = 2.43846 and xₙ₊₁ = 2.43849, both round to 2.438 to 3 decimal places. Depending on the question, one more iteration is a sensible check.
If the question asks you to show that a root lies in an interval, use substitution and a sign change instead of simply quoting an iterative value.
Higher — interval halving and contextual iteration
Worked example — Bisection keeps a sign change
For f(x) = x³ + x − 5, f(1) = −3 and f(2) = 5. Test the midpoint 1.5: f(1.5) = −0.125. Keep [1.5, 2], whose endpoints have opposite signs. Its midpoint is 1.75, and f(1.75) = 2.109375, so keep [1.5, 1.75]. Repeat to narrow the bracket. Continuity is essential: a jump across an undefined value does not guarantee a root.
Worked example — A repeated process in context
A tank starts with 100 litres. At the end of each day, 20% of the water is removed and then 10 litres is added. Let Vₙ be the volume after n complete days, so V₀ = 100.
Keep 80%, then add 10: Vₙ₊₁ = 0.8Vₙ + 10. Hence V₁ = 90, V₂ = 82 and V₃ = 75.6 litres. Use the latest volume each time.
Order matters. Adding 10 first and then removing 20% would give Vₙ₊₁ = 0.8(Vₙ + 10), with first value 88 litres. These describe different processes.
If the original process settles at L litres, then L = 0.8L + 10, so L = 50. In fact Vₙ₊₁ − 50 = 0.8(Vₙ − 50): the distance from 50 shrinks each day. This explains the limiting amount. Numerical percentage calculation remains in Number; this section models the repeated process.
Higher practice — For f(x) = x² − 2, start with [1, 2]. Test the midpoint and state the next bisection interval.
f(1) = −1 and f(2) = 2. The midpoint is 1.5; f(1.5) = 0.25.
Keep [1, 1.5], whose endpoint signs differ. Thus a root lies between 1 and 1.5.
Higher practice — A tank starts with 80 litres. Each day 5 litres is added, then 10% of the total is removed. Write the recurrence and find the first two daily volumes.
With V₀ = 80, Vₙ₊₁ = 0.9(Vₙ + 5), because the addition happens before the removal.
V₁ = 0.9 × 85 = 76.5 litres; V₂ = 0.9 × 81.5 = 73.35 litres.
At a fixed amount L, L = 0.9(L + 5), giving 0.1L = 4.5 and L = 45 litres.
Think Like a Mathematician
Why does iteration aim for a value that stops changing?
At a fixed point, the input and output are the same: x = g(x). That is exactly the rearranged equation we are trying to satisfy.
The repeated calculation is therefore searching for a value that reproduces itself under the iteration rule.
Common Mistakes
Rounding every line: this can move the sequence away from the correct value. Keep calculator accuracy until the final answer.
Using the original x each time: iteration always feeds the newest value into the next step.
Claiming a root from two positive values: a sign-change argument needs opposite signs at the ends of the interval.
Confusing an iterate with an exact answer: iteration normally gives an approximation.
Stopping too soon: check that the values are stable to the accuracy requested.
Practice — numerical methods
All questions are Higher. Use a calculator where appropriate and keep more digits than you finally report.
1. For f(x) = x³ − 4x − 1, calculate f(2) and f(3). Explain what the signs tell you.
f(2) = 8 − 8 − 1 = −1. f(3) = 27 − 12 − 1 = 14.
The signs are opposite, so a root lies between 2 and 3.
2. For f(x) = x³ + x − 5, use f(1.51) and f(1.52) to locate the root.
f(1.51) = −0.047049 and f(1.52) = 0.031808.
The sign changes, so 1.51 < x < 1.52.
3. The root of an equation lies between 3.274 and 3.276. Is that enough information to give the root to 2 decimal places?
The interval crosses the rounding midpoint 3.275. Values below 3.275 round to 3.27, while values at or above 3.275 round to 3.28.
More information is needed before giving a justified 2 d.p. answer.
4. Use xₙ₊₁ = √(7 − xₙ) with x₀ = 2. Find x₁ and x₂ to 4 decimal places.
x₁ = √5 ≈ 2.2361 (to 4 d.p.).
x₂ = √(7 − 2.2360679…) = √4.7639321… = 2.1826 to 4 d.p.
5. Using the same iteration, find x₃ to 4 decimal places.
Using the unrounded x₂ value, x₃ = √(7 − x₂) ≈ 2.1948 to 4 d.p.
6. Why should you use the full calculator value of x₂ rather than 2.1826 when finding x₃?
Rounding during the process introduces extra error. Using the stored calculator value keeps the iteration as accurate as possible.
Round only the final reported result.
7. The iteration values are 1.74281, 1.74294, 1.74291, 1.74292. Give a sensible approximation to 3 decimal places.
The values are settling near 1.7429, and they all round to 1.743 to 3 d.p.
8. For g(x) = x² − 3, calculate g(1.9) and g(2.0). What can you conclude?
g(1.9) = 3.61 − 3 = 0.61 and g(2.0) = 1.
Both are positive, so this pair of values does not establish a root between 1.9 and 2.0 by a sign-change argument.
9. Rearrange x³ + x − 5 = 0 into the iteration form used in the lesson.
x³ = 5 − x, so x = ∛(5 − x).
Therefore one possible iteration is xₙ₊₁ = ∛(5 − xₙ).
10. Starting with x₀ = 1.5 and xₙ₊₁ = ∛(5 − xₙ), find x₁ and x₂ to 5 decimal places.
x₁ = ∛3.5 ≈ 1.51829.
x₂ = ∛(5 − 1.518294…) ≈ 1.51564.
11. An iteration gives x₅ = 4.27641 and x₆ = 4.27639. Is 4.276 a reasonable answer to 3 decimal places?
Yes. Both values round to 4.276 to 3 d.p., so the sequence appears stable at that accuracy.
4.276 is a reasonable approximation, though another iteration is a useful check if requested.
12. A continuous function has f(2.31) = −0.004 and f(2.32) = 0.018. State an interval containing a root.
The signs are opposite, so 2.31 < x < 2.32.
Frequently Asked Questions
Is iteration the same as trial and improvement?
They are related but not identical. Decimal search tests chosen values to narrow an interval; iteration follows a repeated formula from one value to the next.
Why can an iteration fail to settle?
Some rearrangements or starting values do not produce a convergent sequence. At GCSE you should recognise this from the generated values rather than assume every formula will work.
Do I need to show every iterate?
Show enough working to make the repeated process clear, especially the starting value and the sequence of values requested by the question.
How do I justify a decimal answer?
Use the accuracy requested. A continuous function with opposite signs at interval endpoints guarantees at least one root inside; the entire interval must round to the reported value. Stable successive iterates support a practical estimate, but matching rounded values alone do not prove convergence or certify an error bound.