Run given iterations with retained precision and original-equation checks
Generate the requested terms, retain calculator precision and support the final rounded root with evidence from the original continuous equation.
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
- Read iteration subscripts and recover the original fixed-point equation.
- Retain function precision and use a valid root bracket for rounding.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which value should be carried into the next calculator iteration?
What you will learn
- Use a given fixed-point rule or given rearrangement to approximate a root.
- Check domain validity, original-equation signs and the requested accuracy before reporting the result.
Use the formula as given and retain working digits
For x³ + x − 1 = 0, use the given rule xₙ₊₁ = 1/(1 + xₙ²) with x₀ = 0.5. The terms begin x₁ = 0.8, x₂ ≈ 0.609756098, x₃ ≈ 0.728967910, x₄ ≈ 0.652999725 and x₅ ≈ 0.701061373. They alternate across a limiting value near 0.682327804. The denominator is positive for every real input, so the rule does not encounter a reciprocal pole.
Keep the full stored term in the next calculation, even when a table displays only a few digits. Rounding each step changes the numerical sequence and can produce a false fixed value or cycle. Label a requested table accurately: if x₀ is the starting value, the first newly generated term is x₁. Use the number of terms, starting value and rearrangement that the question specifies; a different rule can behave differently.
Validate the approximation in the original equation
For f(x) = x³ + x − 1, f(0.6823) = −0.000066636233 and f(0.6824) = 0.000173044224. Continuity therefore brackets a root between these inputs. Both endpoints and the full interval round to 0.682 at three decimal places, supporting the final report. The iteration supplies approximations; the original sign bracket gives separate evidence for the rounding claim. The polynomial is increasing for all real x because 3x² + 1 > 0, so this root is unique.
Other supplied rules may involve square roots or logarithms. For xₙ₊₁ = √(3 − xₙ), every actual input must satisfy xₙ ≤ 3 and every output is non-negative. For xₙ₊₁ = ln(3 − xₙ), every actual input must satisfy xₙ < 3. Check those restrictions at each step and verify the final candidate in its original equation, since a squared rearrangement or a changed log domain can alter which solutions are admissible. These examples do not require the general convergence condition as a syllabus topic.
Pause and explain
Why check f(0.6823) and f(0.6824) in the original cubic?
Worked example
Use xₙ₊₁ = 1/(1 + xₙ²), x₀ = 0.5, to approximate the root of x³ + x − 1 = 0 to three decimal places, with an original-equation check.
Show the worked solution
- Generate successive terms 0.8, 0.609756098, 0.728967910, 0.652999725, … while retaining full stored precision.
- Continue toward approximately 0.682327804; every denominator remains positive.
- Original values at 0.6823 and 0.6824 have opposite signs, and the entire bracket rounds to 0.682 at three decimal places.
Answer Root approximately 0.682 to three decimal places, supported by the bracket (0.6823, 0.6824).
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 x³ − x − 2 = 0, use the given rule xₙ₊₁ = ∛(xₙ + 2), x₀ = 1.5. Find the first three new terms and a root to three decimal places, checking the bracket (1.5213, 1.5214).
Give me a hint
Use real cube roots and retain each unrounded term in the next step.
Compare my reasoning
- The first three terms are approximately 1.518294486, 1.520935263 and 1.521315698.
- Further terms approach 1.521379707; the real cube-root rearrangement recovers the original cubic.
- Original values at 1.5213 and 1.5214 are approximately −0.000473731403 and 0.000120620344. The continuous bracket rounds throughout to 1.521.
x₁ ≈ 1.518294486, x₂ ≈ 1.520935263, x₃ ≈ 1.521315698; root approximately 1.521 to three decimal places.
Look for these in your work
- Use the given seed and latest full-precision term.
- Recover the original equation and check signs within a single rounding band.
Use xₙ₊₁ = √(3 − xₙ), x₀ = 1, to approximate the positive root of x² + x − 3 = 0 to two decimal places. Check the square-root branch and bracket (1.3027, 1.3028).
Give me a hint
Keep the non-negative output restriction when relating the recurrence to the quadratic.
Compare my reasoning
- The sequence begins 1.414213562, 1.259280127, 1.319363435, … and approaches approximately 1.302775638.
- Every calculated input remains within the square-root domain; its fixed point is non-negative, so the negative quadratic root is not a valid square-root fixed point.
- The original values at 1.3027 and 1.3028 are −0.00027271 and 0.00008784. This bracket lies wholly in the two-decimal rounding band for 1.30.
Positive root approximately 1.30 to two decimal places; the non-negative square-root branch must be retained.
Look for these in your work
- Check the actual sequence's real domain and chosen positive branch.
- Support the requested two-decimal answer with original-equation signs.
For the supplied index R(t) = eᵗ + t on t ≥ 0 hours, use tₙ₊₁ = ln(3 − tₙ), t₀ = 1, to find when R reaches 3 to three decimal places. Check the original bracket (0.7920, 0.7921).
Give me a hint
The recurrence requires 3 − tₙ > 0; its fixed point must also satisfy eᵗ + t = 3.
Compare my reasoning
- The terms begin approximately 0.693147181, 0.835884180, 0.772011881, … and approach 0.792059968, with positive log arguments throughout the calculated sequence.
- For f(t) = eᵗ + t − 3, the endpoint values are approximately −0.000192371159 and 0.000128420643.
- Continuity and the common rounding band give t ≈ 0.792 hours; this is an approximate threshold in the supplied model.
Threshold time approximately 0.792 hours to three decimal places, with a valid logarithmic iteration and original-equation bracket.
Look for these in your work
- Use the given time recurrence within its positive log-argument domain.
- Verify the model equation and report the requested accuracy with time units.
Common mistakes
- Retain more digits internally than are displayed or finally requested.
- Check every iterate's square-root, reciprocal or logarithm domain.
- Final digit stability should be checked against the original equation and requested rounding.
What can you explain now?
For xₙ₊₁ = 1/(1 + xₙ²) with x₀ = 0.5, what are x₁ and x₂?
Compare with the explanation
x₁ = 0.8; x₂ = 25/41 ≈ 0.609756098.
The second step uses the newly obtained 0.8, not the original seed again.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Run a given recurrence without premature rounding. Explain how an original-domain check and a narrow sign-change bracket support the final root report.
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.