01 · First two segments
For g(x) = 0.5x + 1 and x₀ = 0, give the first vertical endpoint and the following point on y = x.
Hint
g(0) = 1.
Worked solution
Move from (0, 0) to (0, 1), then to (1, 1). The next input is 1.
Understand · explore · practise
Convergence and divergence of iteration
Understand convergent, divergent and oscillating fixed-point iterations using learner-controlled staircase and cobweb diagrams. Compare rearrangements, starting values and the local derivative test, with original worked practice.
Before you startFixed-point iteration; gradients
01 / Ask what repeated steps actually do
A fixed point can attract, repel or sit inside a cycle.
A sequence converges to α when its values approach α as the number of steps increases. Being able to solve α = g(α) only identifies a possible limit; it says nothing yet about your starting value.
Watch both the side of the fixed point and the distance from it. Alternating sides can occur in either convergence or divergence.
Same fixed point, different behaviourExplore
x₀ = 0; distance from 2 is 2.
The error halves at each step and keeps its sign: convergence without alternating sides.
Values: 0.
Start at (x₀, 0), go vertically to the green curve, then horizontally to y = x. Repeat those two moves. The plot rescales with equal scales on both axes; use the numerical distance to judge growth or shrinkage.
02 / Construct a staircase or cobweb carefully
The line y = x transfers an output into the next input.
Draw the iteration xₙ₊₁ = g(xₙ).Worked example
Mark (x₀, 0) on the horizontal axis
This records the starting input.
Move vertically to (x₀, g(x₀))
The curve gives output x₁.
Move horizontally to (x₁, x₁) on y = x
The output becomes the next input.
Move vertically to (x₁, g(x₁)), then horizontally to y = x again
Repeat at the learner’s pace.
A fixed point is where y = g(x) meets y = x
It is not usually where g crosses the horizontal axis.
Compare shrinking staircase and cobweb steps
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 · Fixed point versus root of g
For g(x) = 0.5x + 1, find its fixed point and its horizontal-axis intercept.
Hint
Solve g(x) = x and g(x) = 0 separately.
Worked solution
The fixed point is x = 2. The horizontal-axis intercept is x = −2. They answer different questions.
03 / Track the distance from the fixed point
For a linear rule, the error relationship is exact.
If g(x) = mx + c and α = mα + c, then eₙ₊₁ = m eₙ, where eₙ = xₙ − α.
All the model rules have fixed point α = 2.Worked example
Write g(x) = 2 + m(x − 2)
Then g(2) = 2 for any chosen m.
Subtract 2 from xₙ₊₁ = 2 + m(xₙ − 2)
This gives eₙ₊₁ = m eₙ.
Therefore eₙ = mⁿe₀
The size |m| controls error size and the sign of m controls alternating sides.
If x₀ = 2, every error is zero
Even a repelling fixed point stays fixed when started exactly there.
03 · Derive a general term
For xₙ₊₁ = 0.5xₙ + 1 and x₀ = 0, write xₙ explicitly.
Hint
The fixed point is 2 and e₀ = −2.
Worked solution
xₙ = 2 − 2(0.5)ⁿ. Since (0.5)ⁿ → 0, xₙ → 2.
04 / Distinguish a staircase from alternating convergence
Shrinking errors can keep or reverse their sign.
Compare m = 0.5 and m = −0.5 from x₀ = 0.Worked example
For g(x) = 0.5x + 1: 0, 1, 1.5, 1.75, …
The values approach 2 from below.
For g(x) = 3 − 0.5x: 0, 3, 1.5, 2.25, …
The values alternate around 2.
In both cases |eₙ₊₁| = 0.5|eₙ|
The distance halves at every step.
A crossing cobweb does not itself mean failure
Here its alternating jumps shrink.
04 · Alternating error
For g(x) = 3 − 0.5x and x₀ = 4, find x₁, x₂, x₃ and their errors relative to 2.
Hint
Apply the rule three times.
Worked solution
The values are 1, 2.5 and 1.75. Their signed errors are −1, 0.5 and −0.25: alternating and shrinking.
05 · Approach from above
For g(x) = 0.5x + 1 with x₀ = 4, does the sequence alternate around 2?
Hint
The initial error is positive and the multiplier is positive.
Worked solution
No. It gives 4, 3, 2.5, 2.25, … and approaches 2 from above.
05 / Identify growing errors
A fixed point may repel every nearby nonfixed start.
Use g(x) = 1.5x − 1, with x₀ = 0.Worked example
The values start 0, −1, −2.5, −4.75
They move away from the fixed point 2.
eₙ = −2(1.5)ⁿ
The error magnitude grows without bound.
Starting at 2 would stay at 2
This exception does not make nearby starts attractive.
For g(x) = 5 − 1.5x, the growing error also alternates
Oscillation and convergence are separate questions.
06 · Oscillating divergence
For xₙ₊₁ = 5 − 1.5xₙ and x₀ = 0, find x₁, x₂, x₃.
Hint
The fixed point is 2, but the multiplier has magnitude greater than 1.
Worked solution
x₁ = 5, x₂ = −2.5 and x₃ = 8.75. The signed errors 3, −4.5 and 6.75 grow in magnitude.
07 · Exact fixed-point start
For the same rule, what happens from x₀ = 2?
Hint
Evaluate g(2).
Worked solution
Every term is 2. An exact fixed-point start remains there even when the fixed point repels nearby starts.
06 / Recognise oscillation without convergence
A repeated cycle can keep the error size unchanged.
Use xₙ₊₁ = 4 − xₙ.Worked example
The fixed point solves x = 4 − x, so x = 2
There is one fixed point.
Starting at 0 gives 0, 4, 0, 4, …
The sequence has a two-cycle.
The error multiplier is −1
The sign reverses but its magnitude is unchanged.
Starting at 2 gives a constant sequence
Every other real start produces a nonconstant two-cycle.
08 · Another two-cycle
Starting at x₀ = 1.7 under g(x) = 4 − x, give the next three terms and decide whether the sequence converges.
Hint
Two applications return the starting value.
Worked solution
2.3, 1.7, 2.3. The sequence does not converge; it keeps cycling between two distinct values.
07 / Choose a rearrangement as well as a starting value
Algebraically equivalent equations can give very different iterations.
Compare two rules from x³ − x − 2 = 0.Worked example
g₁(x) = ∛(x + 2)
Near the positive root α ≈ 1.52138, g₁′(α) = 1/(3α²) ≈ 0.144.
g₂(x) = x³ − 2
At the same root, g₂′(α) = 3α² ≈ 6.944.
The first rule is locally attractive; the second is locally repelling
The following section explains the local derivative test.
From x₀ = 1, g₂ gives 1, −1, −3, −29, …
A valid rearrangement is not necessarily a useful numerical method.
09 · Compare the two local multipliers
At a nonzero root α of x³ − x − 2 = 0, show that g₁′(α)g₂′(α) = 1 for these two rules.
Hint
Use α³ = α + 2.
Worked solution
g₁′(α) = 1/[3(α + 2)^(2/3)] = 1/(3α²), and g₂′(α) = 3α², so their product is 1.
10 · Verify the divergent calculation
Calculate three new terms of xₙ₊₁ = xₙ³ − 2 from x₀ = 1.
Hint
Retain the signs of odd powers.
Worked solution
x₁ = −1, x₂ = −3 and x₃ = −29.
08 / Use the derivative test as a local guide
A slope test near a fixed point has conditions and limits.
For a continuously differentiable g near α: |g′(α)| < 1 gives local attraction; |g′(α)| > 1 gives local repulsion.
For a start sufficiently close to α, the error is approximately multiplied by g′(α). A positive derivative generally keeps the side of the error; a negative derivative reverses it. The size determines whether the local error shrinks or grows.
Local matters: a distant starting value may leave the domain or enter a different region. A sufficient interval guarantee is that g maps a closed interval into itself and |g′(x)| ≤ k < 1 throughout it; then iteration from that interval converges to its unique fixed point.
If |g′(α)| = 1, this simple derivative test is inconclusive. Higher-order behaviour can matter; do not automatically label all such cases divergent.
11 · Read a local derivative
A continuously differentiable iteration has a fixed point α with g′(α) = −0.7. What does the local test predict?
Hint
Separate the magnitude and sign.
Worked solution
Sufficiently nearby starts converge, generally alternating sides, because the multiplier is negative with magnitude less than 1. This does not certify arbitrary distant starts.
12 · An interval guarantee
Show that g(x) = 0.5x + 1 maps [0, 4] into itself and is a contraction there.
Hint
Find the image interval and derivative.
Worked solution
Its image is [1, 3], inside [0, 4], and |g′(x)| = 0.5 < 1. All starts in the interval converge to the unique fixed point 2.
09 / Treat starting values and displayed digits cautiously
The same rule can lead to different roots or undefined steps.
A nonlinear iteration can have more than one fixed point. Starting values may lie in different regions of attraction, and some may leave the real domain. A graph and a root bracket help choose a relevant start.
Two matching rounded values are not proof of a limit. A slowly moving or rounded sequence can look stable. Keep working precision and verify the final root independently using the original equation.
13 · An inconclusive boundary test
If g′(α) = 1, can the simple derivative test alone decide convergence?
Hint
The strict inequality is essential.
Worked solution
No. For example, near zero, g(x) = x − x³ attracts sufficiently small nonzero real starts, while g(x) = x + x³ repels them; both have g′(0) = 1.
10 / Explain the behaviour, not just the last number
Read the error size, the side and the chosen start.
- A fixed point solves g(x) = x.
- Vertical then horizontal moves construct the iteration diagram.
- Alternating sides can still converge if distances shrink.
- A cycle repeats without settling unless every term is the same.
- An exact repelling fixed point remains fixed.
- Compare rearrangements and check domains.
- Use derivative tests locally and check their hypotheses.
- Verify the final root and requested accuracy separately.
14 · Diagnose a linear iteration
For xₙ₊₁ = 0.8xₙ + 3, find the fixed point and describe the sequence from x₀ = 20.
Hint
Subtract the fixed point and track the error multiplier.
Worked solution
The fixed point is 15. The error is eₙ = 5(0.8)ⁿ, so the sequence decreases towards 15 from above, without alternating.
Section 1 of 10 · Ask what repeated steps actually do