01 · Form an iteration rule
Rearrange x³ − 2x − 4 = 0 using a cube root, and write the recurrence.
Hint
First isolate x³.
Worked solution
x³ = 2x + 4, so xₙ₊₁ = ∛(2xₙ + 4), using the real cube root.
Understand · explore · practise
Fixed-point iteration
Learn fixed-point iteration xₙ₊₁ = g(xₙ), rearrange equations, calculate labelled iterates and retain working precision. Explore real cube roots, square-root branches and logarithm domains with original practice.
Before you startLocating roots; functions; logarithms
01 / Turn an equation into a repeated calculation
A fixed point is a value left unchanged by g.
Rearrange f(x) = 0 into x = g(x), then calculate xₙ₊₁ = g(xₙ).
Choose a starting estimate x₀. Substitute it into g to obtain x₁, then substitute x₁ to obtain x₂. Each step uses the previous value as its input.
If a sequence converges to α and g is continuous there, passing to the limit gives α = g(α). Check that this fixed point satisfies the original equation and its domain.
One substitution at a timeExplore
x₀ = 1. No substitution has been performed.
The next value is ∛3 ≈ 1.44224957031.
The real cube root accepts every real input.
Original equation: x³ − x − 2 = 0.
Values shown: x₀ = 1.
Only your button presses advance the sequence. Values are stored without display rounding. Selecting a new rearrangement or start resets to n = 0. These iterates are estimates, not a proof of their limiting value.
02 / Make x the subject in a useful way
The same equation may allow several iteration rules.
Rearrange x³ − x − 2 = 0.Worked example
x³ = x + 2
Move the linear and constant terms.
x = ∛(x + 2)
Take the real cube root.
xₙ₊₁ = ∛(xₙ + 2)
The left side is the next value; the right side uses the current one.
Another algebraic form is x = x³ − 2
Algebraic equivalence does not mean that both iterations converge.
02 · Recognise a fixed point
Show that 2 is a fixed point of g(x) = ∛(2x + 4).
Hint
Evaluate g(2).
Worked solution
g(2) = ∛8 = 2. It also satisfies 2³ − 2(2) − 4 = 0.
03 / Label the start before counting steps
x₀ is the starting value, not the first result.
Use xₙ₊₁ = ∛(xₙ + 2), starting with x₀ = 1.Worked example
x₁ = ∛3 ≈ 1.44224957031
One substitution.
x₂ = ∛(x₁ + 2) ≈ 1.50989744933
Two substitutions; retain the full x₁ value.
x₃ ≈ 1.51972430499
Three substitutions.
Report the iterate requested, with rounding only at the end
The index tells you how many substitutions have been completed.
Follow three labelled substitutions
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 · Count three substitutions
For xₙ₊₁ = (xₙ + 6)/2 with x₀ = 2, find x₁, x₂ and x₃.
Hint
Each result becomes the next input.
Worked solution
x₁ = 4, x₂ = 5 and x₃ = 5.5. The start x₀ = 2 does not count as a substitution.
04 · Starting at x₁ instead
A question gives x₁ = 2 and the same rule xₙ₊₁ = (xₙ + 6)/2. How many substitutions produce x₄, and what is its value?
Hint
Follow the labels supplied in the question.
Worked solution
Three substitutions: x₂ = 4, x₃ = 5 and x₄ = 5.5.
04 / Carry the stored value into the next step
A displayed table can be rounded without rounding its calculation.
Keep the calculator’s previous answer or sufficient stored precision when evaluating g again. A table might display six decimal places while the calculation retains more digits internally.
Repeated digits suggest stability but do not prove that a sequence converges or that a rounded answer is correct. Use a sign bracket in the original function to verify a final accuracy claim.
05 · Verify the suggested root
Iteration suggests that the root of f(x) = x³ − x − 2 is 1.521 to 3 d.p. Which halfway boundaries should you test?
Hint
Use half a thousandth either side.
Worked solution
Test 1.5205 and 1.5215. f(1.5205) = −0.005225259875 and f(1.5215) = 0.000715063375. Continuity and the sign change put a root between these rounding boundaries; f′(x) = 3x² − 1 > 0 throughout this interval gives uniqueness.
05 / Use a real cube root for negative inputs
Odd roots preserve the sign of their argument.
∛(−a) = −∛a for a ≥ 0.
A real cube-root function accepts negative arguments. A general fractional-power key or programming expression may interpret a negative base differently. Use the real cube-root operation when the recurrence calls for it.
Taking a real cube root is reversible for every real input. It does not select only the positive roots of the original equation.
06 · Negative cube-root step
For xₙ₊₁ = ∛(xₙ + 2), find x₁ when x₀ = −10.
Hint
The argument is −8.
Worked solution
x₁ = ∛(−8) = −2. This is a real value.
06 / Separate positive and negative square-root branches
Taking √ selects a nonnegative output.
Rearrange x² − x − 2 = 0 using a square root.Worked example
x² = x + 2
Real square-root inputs need x + 2 ≥ 0.
x = √(x + 2) selects x ≥ 0
Its fixed point is 2.
x = −√(x + 2) selects x ≤ 0
Its fixed point is −1.
The original equation has roots −1 and 2
A single branch is not globally equivalent to the original equation.
07 · Test the lost root
Does −1 solve x = √(x + 2)? Does it solve x² − x − 2 = 0?
Hint
The square root of 1 is +1.
Worked solution
It does not solve the positive-branch equation because −1 ≠ 1. It does solve the original quadratic: 1 + 1 − 2 = 0.
08 · A negative-branch step
For xₙ₊₁ = −√(xₙ + 2), find x₁ from x₀ = 2. Can you find x₂?
Hint
Check the input at each step.
Worked solution
x₁ = −2, then x₂ = −√0 = 0. Both are real; the branch gives nonpositive outputs.
07 / Check every new input against the domain
An algebraic rearrangement can introduce restrictions.
Rearrange eˣ + x − 5 = 0.Worked example
eˣ = 5 − x
The right side must be positive.
x = ln(5 − x), requiring x < 5
The logarithm argument is strictly positive.
Starting at x₀ = 1 gives x₁ = ln4
This is a valid first step.
Starting at x₀ = 5 is undefined
ln0 is not a real finite number.
09 · A later domain failure
Use xₙ₊₁ = −√(xₙ + 2), starting at x₀ = 7. What happens?
Hint
Evaluate x₁ before checking the next input.
Worked solution
x₁ = −3 is real, but x₂ would require −√(−1), so the real iteration stops. A valid start does not guarantee all later steps are defined.
10 · A reciprocal rearrangement
Rearrange x² − 3x + 1 = 0 as x = 3 − 1/x. State its restriction and explain whether dividing by x loses a root of this original equation.
Hint
Test x = 0 in the original equation.
Worked solution
The rearrangement requires x ≠ 0. Zero gives 1, not 0, in the original equation, so no original root is lost by this division. Individual iterates must still avoid zero.
08 / Respect inverse-trigonometric branches
An inverse function returns a restricted range.
Consider sin x = x/2.Worked example
x = arcsin(x/2) selects one inverse-sine branch
Its input requires −2 ≤ x ≤ 2.
The principal arcsin output lies in [−π/2, π/2]
Use radians; other branches are omitted.
The original equation also has a positive root between π/2 and 2
For f(x) = sin x − x/2, f(π/2) = 1 − π/4 > 0 and f(2) = sin2 − 1 < 0.
That positive root cannot satisfy x = arcsin(x/2)
It lies outside the selected output range. Rearrangement using an inverse trig function needs a branch check.
11 · Audit a branch claim
For cos x = x²/4, does x = arccos(x²/4) retain every real solution?
Hint
The original equation is even, but principal arccos is nonnegative.
Worked solution
No. Any nonzero negative solution is omitted by this branch. Real arccos inputs require |x| ≤ 2 here, and the selected equation requires x ≥ 0. For a negative branch use x = −arccos(x²/4), then check the original equation.
09 / Check a proposed limit in the original equation
An equation for a possible limit is conditional on convergence.
From xₙ₊₁ = g(xₙ), it is tempting to set both sides equal to α immediately. This identifies possible fixed points; it does not establish that your chosen sequence reaches one.
The rearrangement and starting value matter. A sequence may diverge, oscillate without settling, or leave the real domain. Inspect the behaviour and verify the original equation before claiming a solution.
12 · A two-cycle
For xₙ₊₁ = 2 − xₙ with x₀ = 0, list four new values. The fixed-point equation gives α = 1. Does this sequence converge to 1?
Hint
Substitute repeatedly.
Worked solution
x₁ = 2, x₂ = 0, x₃ = 2, x₄ = 0. It does not converge. The existence of a fixed point does not make every start converge to it.
13 · Check the original equation
Squaring an equation produces a candidate root. What must you do before accepting a fixed point obtained from the squared equation?
Hint
Squaring need not be reversible.
Worked solution
Substitute the candidate into the original unsquared equation and check its domain and sign restrictions. Reject an extraneous candidate.
10 / Build a reliable iteration table
Method, labels and restrictions belong beside the values.
- State the original equation and the rearranged rule.
- Record the supplied starting index and value.
- Use the previous full value to calculate the next.
- Check real domains and selected square-root or inverse-trig branches.
- Stop and explain any undefined step.
- Distinguish a fixed point from demonstrated convergence.
- Verify final rounding with the original function.
14 · A complete short calculation
Use xₙ₊₁ = ln(5 − xₙ) with x₀ = 1 to find x₁ and x₂ to 4 d.p.
Hint
Keep ln4 unrounded when calculating x₂.
Worked solution
x₁ = ln4 ≈ 1.3863. Then x₂ = ln(5 − ln4) ≈ 1.2847. Both inputs are less than 5, so both logarithms are defined.
Section 1 of 10 · Turn an equation into a repeated calculation