Recognise oscillating failing and undefined iterations
Inspect the actual sequence and its domain instead of assuming every algebraic rearrangement leads toward a root.
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
- Generate iteration terms and solve a fixed-point equation.
- Check square-root and logarithmic input restrictions.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Does finding a fixed point of F guarantee every sequence xₙ₊₁ = F(xₙ) approaches it?
What you will learn
- Recognise a cycle, non-convergent behaviour or an undefined next iterate.
- Compare given rearrangements and reject invalid candidate values using the original equation.
An equivalent fixed-point equation does not ensure convergence
The equation x³ + x − 1 = 0 can be written x = 1 − x³. Yet the given iteration xₙ₊₁ = 1 − xₙ³ starting at x₀ = 0 gives 1, 0, 1, 0, … . It cycles forever between two distinct values, neither of which solves the original cubic. The root remains a fixed point of the rule, but this starting sequence does not approach it. The graph's closed stepping path shows the repeated values rather than an approach to the intersection with y = x.
A different given rearrangement of that same equation, xₙ₊₁ = 1/(1 + xₙ²) from x₀ = 0.5, approaches its root near 0.682327804. Thus equivalent equations can yield differently behaving numerical processes. Alternation alone is not failure: decreasing swings toward one value can converge. A non-shrinking two-cycle such as 0, 1, 0, 1 does not converge. This section expects you to recognise failure without requiring a general derivative-based convergence condition.
An undefined step cannot be used as a numerical approximation
For xₙ₊₁ = ln(2 − xₙ), choosing x₀ = 2.5 gives a negative log argument and no real first term. Do not report an error output as a root, replace the logarithm by an absolute value, or clamp the value: those choices change the supplied iteration. Use a permitted seed or another given rule. From the supplied alternative seed x₀ = 0, the terms remain real and approach a fixed point near 0.442854401 of eˣ + x − 2 = 0.
The golden-ratio equation x² − x − 1 = 0 can be rearranged for its positive root as x = √(x + 1). From seed 1 this sequence approaches approximately 1.618034. The alternative x = x² − 1 from the same seed gives 0, −1, 0, −1, … . Check those cycle values in the original polynomial: neither is zero. Do not assume increasing the iteration count can repair an exact persistent cycle, and do not choose a different root branch without checking what the problem asks for.
Pause and explain
What should happen when a given logarithmic iteration produces a negative log argument?
Worked example
For x³ + x − 1 = 0, examine the given iteration xₙ₊₁ = 1 − xₙ³ with x₀ = 0. Does it converge to a root?
Show the worked solution
- The first terms are x₁ = 1, x₂ = 0, x₃ = 1 and x₄ = 0.
- The same rule maps 0 to 1 and 1 to 0 forever, so this is an exact two-cycle.
- The original function values at 0 and 1 are −1 and 1; neither is a root, and the sequence has no single limit.
Answer The sequence cycles 0, 1, 0, 1, … and does not converge to a root.
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 = 2 − xₙ with x₀ = 0.8, give the first four new terms and explain whether the sequence converges. Find the original fixed point separately.
Give me a hint
Follow the exact alternating values and solve r = 2 − r independently.
Compare my reasoning
- The terms are 1.2, 0.8, 1.2 and 0.8.
- They form a persistent two-cycle, so this seed's sequence does not converge.
- The fixed point is r = 1, but neither cycle value equals it; the existence of that root did not make this sequence converge.
x₁ = 1.2, x₂ = 0.8, x₃ = 1.2, x₄ = 0.8; no convergence, though the fixed point is 1.
Look for these in your work
- Identify a repeated non-shrinking cycle.
- Distinguish the original fixed point from the actual sequence's values.
For the positive root of x² − x − 1 = 0, compare the given rules xₙ₊₁ = xₙ² − 1 and xₙ₊₁ = √(xₙ + 1), both starting at 1.
Give me a hint
Generate several terms and check the original equation at any repeating values.
Compare my reasoning
- The square-minus-one rule gives 0, −1, 0, −1, … ; the original polynomial is −1 at 0 and 1 at −1, so these are not roots.
- The square-root rule gives approximately 1.414214, 1.553774, 1.598053, … and approaches the positive root 1.618034.
- Keep the non-negative branch for the square-root rearrangement; algebraically related rules do not have identical convergence behaviour.
xₙ² − 1 cycles between 0 and −1 and fails; √(xₙ + 1) approaches the positive root approximately 1.618034.
Look for these in your work
- Use the same seed to compare the actual sequences.
- Check cycle values and the chosen branch against the original equation.
For xₙ₊₁ = ln(2 − xₙ), explain why seed 2.5 fails. Then use the supplied permitted seed 0 to approximate a fixed point to three decimal places, checking eˣ + x − 2 on (0.4428, 0.4429).
Give me a hint
An input must be below 2; the candidate must satisfy the original exponential equation.
Compare my reasoning
- Seed 2.5 gives log argument −0.5, so no real first term exists.
- From 0, the terms begin approximately 0.693147181, 0.267621819, 0.549495136, … and approach 0.442854401 while remaining in the real log domain.
- Original endpoint signs at 0.4428 and 0.4429 differ, and that whole bracket rounds to 0.443 to three decimal places.
Seed 2.5 is invalid; seed 0 approaches a valid root approximately 0.443 to three decimal places.
Look for these in your work
- Reject the undefined seed without altering the logarithm.
- Check the permitted sequence's domain and original-equation rounding evidence.
Common mistakes
- A valid algebraic fixed-point equation does not guarantee a convergent sequence from every seed.
- Distinguish shrinking oscillations from a persistent cycle.
- Reject undefined log or root steps without changing the given recurrence silently.
What can you explain now?
Does alternation across a candidate value always mean an iteration fails?
Compare with the explanation
No; shrinking alternation can converge to one limit.
Failure depends on whether the terms approach a single value, not solely whether consecutive steps switch sides.
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
Give a fixed-point rule with a persistent cycle and another with shrinking alternation. Explain why an undefined term or algebraic equivalence alone does not establish a numerical solution.
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.