Verify numerical accuracy without mistaking stable digits for proof
Combine a given iteration, original-equation evidence and rounding bands, recognising that small residuals and close successive terms need interpretation.
Exam scope and assessment notes
Pure Mathematics 2 (Paper 2, AS only) and Pure Mathematics 3 (Paper 3, A Level) · Mathematics 9709, 2026–2027. Paper 1 graph and algebra knowledge is assumed. The AS route using Papers 1 + 2 cannot be carried forward to A Level. Shared sections 2.6 and 3.6 include locating roots graphically or by sign changes, notation for convergent approximation sequences, and using a given iteration or rearrangement to a prescribed accuracy. An iteration may fail to converge. Knowledge of the general condition for convergence is not required here. Check the original equation, domains and requested rounding. Written practice is self-checked.
Before you begin
- Run given iterations and recognise failure or domain problems.
- Check original continuous sign brackets and decimal rounding bands.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Must a correctly rounded approximate root always give exactly zero in the original equation?
What you will learn
- Report a numerical root with its starting value, given recurrence and justified requested precision.
- Explain why close iterates or a small equation residual alone do not guarantee root accuracy.
A root report should show what the numbers establish
A useful numerical solution records the original equation, given recurrence and seed, enough terms to show the calculation, and the requested final precision. A convergent approximation is not normally an exact root after rounding. For the supplied rule xₙ₊₁ = (xₙ + 2/xₙ)/2, x₀ = 1, the terms 1.5, 1.416666667 and 1.414215686 approach the positive root of x² − 2 = 0. Its fixed-point equation recovers x² = 2 for x ≠ 0, and the positive iterates retain the requested branch.
The original sign bracket (1.4142, 1.4143) justifies 1.414 to three decimal places. Do not replace the given recurrence with a memorised unrelated method, and do not require the rounded value to produce exactly zero. For a cosine fixed-point equation, the standard numerical arguments use radians; a degree-mode calculation would generate a different sequence and solve a different equation.
Close values and small residuals need an independent accuracy check
Consider xₙ₊₁ = 0.9999xₙ from x₀ = 1, whose only fixed point is 0. The first values 1, 0.9999 and 0.99980001 all round to 1.000 at three decimal places, although they are far from the root 0. Agreement of displayed digits alone is therefore not a universal stopping proof. A small residual f(x) also depends on the equation's scale: f(x) = 10^(−9)(x − 100) has a tiny value at x = 0 despite that input being 100 units from its root.
For x³ + x − 1, a valid bracket (0.6823, 0.6824) establishes the three-decimal answer 0.682, but crosses the four-decimal rounding threshold 0.68235. Further original sign checks at 0.68232 and 0.68233 narrow the bracket to one four-decimal band, establishing 0.6823. Report only the precision supported by the evidence and retain working digits. A sketch, residual or pair of iterates can guide checks without automatically certifying every requested digit.
Pause and explain
Which bracket justifies the cubic root to four decimal places as 0.6823?
Worked example
Use the supplied rule xₙ₊₁ = (xₙ + 2/xₙ)/2, x₀ = 1, for the positive root of x² − 2 = 0. Report the first three new terms and justify a three-decimal answer.
Show the worked solution
- Generate x₁ = 1.5, x₂ = 17/12 ≈ 1.416666667 and x₃ = 577/408 ≈ 1.414215686, retaining full precision.
- The positive inputs avoid the rule's zero denominator; its fixed point gives x² = 2 on the non-zero branch.
- Original signs at 1.4142 and 1.4143 bracket a root inside one three-decimal rounding band, giving 1.414.
Answer x₁ = 1.5, x₂ = 17/12, x₃ = 577/408; positive root approximately 1.414 to three decimal places.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
For xₙ₊₁ = 0.9999xₙ, x₀ = 1, calculate x₁ and x₂, round them to three decimals and decide whether that proves the fixed point is near 1.
Give me a hint
Solve r = 0.9999r separately from examining two displayed terms.
Compare my reasoning
- The new terms are x₁ = 0.9999 and x₂ = 0.99980001.
- Both round to 1.000 at three decimal places.
- The fixed-point equation gives 0.0001r = 0, so r = 0. The displayed agreement does not establish closeness to the root.
x₁ = 0.9999, x₂ = 0.99980001; both display 1.000, but the fixed point is 0, so the accuracy claim fails.
Look for these in your work
- Generate and round the actual sequence values.
- Use the original fixed-point equation to expose the false stopping claim.
Use the given rule xₙ₊₁ = cos xₙ in radians, x₀ = 0.5, to approximate a solution of cos x = x to three decimal places. Check the original bracket (0.7390, 0.7391).
Give me a hint
Retain the full radian calculator values and evaluate f(x) = cos x − x at both bracket ends.
Compare my reasoning
- The sequence begins approximately 0.877582562, 0.639012494, 0.802685101, … and approaches 0.739085133.
- Original endpoint values are approximately 0.000142477295 and −0.000024881312, giving a continuous sign-change bracket.
- Every value in the bracket rounds to 0.739 to three decimal places; using degree arguments would solve a different fixed-point rule.
Root approximately 0.739 to three decimal places, using radian iteration and the original bracket.
Look for these in your work
- Use the given radian recurrence and seed with retained precision.
- Support the final rounding by original-equation signs.
For x³ + x − 1 = 0, the verified root bracket is (0.6823, 0.6824). State what it proves at three and four decimal places, then check (0.68232, 0.68233) to improve the report.
Give me a hint
Compare each interval with the relevant rounding threshold before increasing the claimed precision.
Compare my reasoning
- The first bracket lies wholly in the three-decimal band for 0.682, but crosses four-decimal boundary 0.68235.
- Original values at 0.68232 and 0.68233 are approximately −0.000018703417 and 0.000005263605, so continuity gives the finer root bracket.
- The finer bracket lies wholly in the four-decimal band for 0.6823; the supported report can now be improved.
First bracket proves 0.682 to three decimals but not a unique four-decimal value; finer signs justify 0.6823 to four decimals.
Look for these in your work
- Match each accuracy claim to its full rounding band.
- Use additional original-function evidence before reporting an extra digit.
Common mistakes
- Matching displayed digits can occur long before a sequence is close to its root.
- A small residual is not a universal root-error bound without information about the equation.
- Use the requested decimal places or significant figures and verify the entire supporting bracket.
What can you explain now?
Can two successive values both rounding to 1.000 alone prove a root rounds to 1.000?
Compare with the explanation
No.
A slowly changing sequence can display stable rounded digits while remaining far from its actual fixed point.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Explain how a complete numerical-root report links the recurrence, domain, original equation and requested precision. Give a counterexample to relying only on matching rounded iterates or a tiny residual.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.