Interpret iteration notation and verify an equivalent fixed-point equation
Read subscripts as sequence positions, use each latest value in the given rule and check that a possible limit solves the original equation.
Section 3.6 · Numerical solution of equations
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
- Substitute into an algebraic function and rearrange equations.
- Recognise square-root and denominator domain restrictions.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What does x₂ mean in iteration notation?
What you will learn
- Use x₀, x₁, … and xₙ₊₁ = F(xₙ) to generate a sequence of approximations.
- Relate a convergent sequence's fixed point to the original equation, preserving rearrangement restrictions.
A subscript labels a term, not a power or multiplier
The supplied starting value x₀ is the seed. In xₙ₊₁ = F(xₙ), evaluate F at the current term to get the next one: x₁ = F(x₀), x₂ = F(x₁), and so on. The subscript n counts steps; x₂ is not x squared and xₙ is not n times x. Use each newly computed value as the next input, rather than substitute the original seed repeatedly.
For xₙ₊₁ = √(xₙ + 1) and x₀ = 1, the terms begin x₁ = √2 ≈ 1.414214, x₂ = √(1 + √2) ≈ 1.553774 and x₃ ≈ 1.598053. They approach the positive value (1 + √5)/2 ≈ 1.618034 in this example. Keep the full stored calculator value for the next step; the displayed decimals are for reporting, not replacements for the internal input.
A valid limit must satisfy both the rule and the original equation
If the sequence converges to r and F is continuous there, taking the limit in xₙ₊₁ = F(xₙ) gives r = F(r). This is a fixed point, seen graphically where y = F(x) meets y = x. Convergent iterates may approach from one side or alternate across the limit. Merely finding a fixed point does not prove that a chosen starting value will approach it; the actual iteration can fail.
The rearrangement x = √(x + 1) implies x² − x − 1 = 0, but also requires x ≥ 0. Squaring gives an additional negative quadratic root (1 − √5)/2, which does not satisfy the original square-root fixed-point equation. For the requested positive root, the given recurrence is appropriate. When rearranging, retain lost denominators, chosen root branches and log arguments, then substitute the candidate back into the original equation. A fixed point of a changed or undefined rule is not enough.
Pause and explain
Does the negative root of r² − r − 1 = 0 satisfy r = √(r + 1)?
Worked example
For xₙ₊₁ = √(xₙ + 1), x₀ = 1, calculate the first three new terms and identify the original equation solved by a positive limit.
Show the worked solution
- x₁ = √2 ≈ 1.414214.
- Use the latest value: x₂ = √(1 + √2) ≈ 1.553774 and x₃ ≈ 1.598053.
- A continuous-rule limit satisfies r = √(r + 1), hence r² − r − 1 = 0 with r ≥ 0; its positive root is (1 + √5)/2.
Answer x₁ ≈ 1.414214, x₂ ≈ 1.553774, x₃ ≈ 1.598053; positive limit solves r² − r − 1 = 0 with r ≥ 0.
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 = (xₙ + 2)/3 with x₀ = 0, find x₁, x₂ and x₃ exactly, then solve the fixed-point equation.
Give me a hint
Use each latest fraction as the next input; solve r = (r + 2)/3 for the possible limit.
Compare my reasoning
- The first new term is x₁ = 2/3.
- Then x₂ = 8/9 and x₃ = 26/27.
- The fixed point satisfies 3r = r + 2, giving r = 1; these terms move toward that value.
x₁ = 2/3, x₂ = 8/9, x₃ = 26/27; fixed point r = 1.
Look for these in your work
- Advance one step using each latest term.
- Relate a possible sequence limit to its fixed-point equation.
A given rule for x³ + x − 1 = 0 is xₙ₊₁ = 1/(1 + xₙ²). Show the fixed-point equation is equivalent to the original equation and check its denominator.
Give me a hint
Multiply the fixed-point equation by 1 + r² only after confirming it cannot be zero for real r.
Compare my reasoning
- At a fixed point, r = 1/(1 + r²).
- For real r, 1 + r² > 0, so multiplication preserves the equation.
- The result r + r³ = 1 is exactly r³ + r − 1 = 0; the fixed point is positive because the reciprocal is positive.
Equivalent fixed-point equation r³ + r − 1 = 0; denominator 1 + r² is always positive for real r.
Look for these in your work
- Check the denominator before reversing the rearrangement.
- Recover the original equation and retain any implied sign information.
A learner squares r = √(r + 1) and keeps both roots (1 ± √5)/2. Which candidate is valid for the original fixed-point equation and why?
Give me a hint
The original right side is the non-negative square root.
Compare my reasoning
- Squaring gives r² − r − 1 = 0 with two algebraic candidates.
- The negative candidate cannot equal a non-negative square root.
- The positive candidate (1 + √5)/2 is non-negative and satisfies r² = r + 1, so it satisfies the original square-root equation.
Only (1 + √5)/2 is valid; the negative candidate fails the original non-negative branch.
Look for these in your work
- Retain the original branch restriction after squaring.
- Verify the accepted candidate in the unsquared equation.
Common mistakes
- Subscripts count sequence positions and are not exponents.
- Feed the latest unrounded term into the next step.
- A rearrangement must keep its branch and domain restrictions before a fixed point counts as an original solution.
What can you explain now?
In xₙ₊₁ = F(xₙ), which input should be used to calculate x₂?
Compare with the explanation
x₁, so x₂ = F(x₁).
The rule advances one sequence position using the latest term.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Explain what an iteration subscript means and how a continuous-rule limit becomes a fixed point. Give a rearrangement where checking the original branch removes an extra candidate.
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.