Solve x=1+x/3 by starting at x₀=0 and substituting repeatedly. The iterates are 0,1,4/3,13/9. The exact fixed point is 3/2. Since g(x)−3/2=(x−3/2)/3, each error shrinks by exactly one third. This is a statement proved for every iterate, not just a pattern inferred from several decimal values. Write the recurrence for the error before calculating more terms.
Predict
Predict whether algebraically equivalent rearrangements give equally good iterations. The equation x²=2 can be written x=2/x, but applying that map twice returns the initial nonzero value. Unless the start is a root, it produces a two-cycle. The rearrangement x=(x+2/x)/2 behaves differently near the positive root. Equivalence of fixed-point equations does not imply equivalence of their iterative dynamics.
Experiment
Compare actual and certified errors in the linear example.
n
xₙ
Actual error
A priori bound
1
1
1/2
1/2
2
4/3
1/6
1/6
3
13/9
1/18
1/18
The bound here is qⁿ∣x₁−x₀∣/(1−q). Compute it and the latest-step posterior bound independently. In nonlinear problems the same formulas can overestimate the actual error.
Understand
Let g map a nonempty closed interval I into itself and satisfy |g(x)−g(y)|≤q|x−y| for 0≤q<1. Adjacent steps obey |xₙ₊₁−xₙ|≤qⁿ|x₁−x₀|. Summing a geometric tail shows the sequence is Cauchy. Completeness provides a limit r∈I, and continuity gives g(r)=r. Two fixed points r,s would satisfy |r−s|≤q|r−s|, forcing equality. Passing to the limit in the tail gives |xₙ−r|≤qⁿ|x₁−x₀|/(1−q). Restarting the same estimate from the latest observed step gives |xₙ−r|≤q|xₙ−xₙ₋₁|/(1−q). These estimates require a justified q, not one guessed from a short trace.
Apply
If g is differentiable and sup|g′|≤q<1 on an interval, the mean value theorem supplies the contraction bound. It does not prove g maps the interval into itself; verify that separately. If only |g′(r)|<1 is known and g′ is continuous near a fixed point, a sufficiently small neighborhood is attracting, but distant starts are not covered. A derivative magnitude greater than one leads to local repulsion under differentiability. Equality at one is inconclusive. For the averaging map above the derivative vanishes at √2, while the reciprocal map has magnitude one and a concrete two-cycle.
For g(x)=1+x/3, the interval [0,2] is invariant because its image is [1,5/3], contained inside it. The derivative bound is exactly 1/3 everywhere, so both theorem hypotheses can be verified without knowing the fixed point first. Repeat the check for a narrower proposed interval that fails to contain its image. This separates a legitimate convergence proof from circular reasoning that assumes all iterates are already close to the unknown solution.
Continue
A small latest step can be misleading when q is close to one, because q/(1−q) is large. Floating-point stagnation can also stop movement before accuracy is achieved. Report the invariant interval, contraction proof, bound and termination reason. The next lesson studies Newton’s map, which has a vanishing derivative at a simple root and can converge quadratically. That speed relies on stronger smoothness and a sufficiently close starting value; it does not remove the need for error evidence.
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Find the fixed point of g(x)=(x+9)/4.
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
For q=1/3 and latest step 1/10, compute q/(1−q) times the step.
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
State interval hypotheses guaranteeing convergence from every point in I.
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Prove the posterior error bound q|x_n−x_(n−1)|/(1−q).
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Compare x↦2/x and x↦(x+2/x)/2 near √2.
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Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Give a contraction whose fixed point is excluded by its domain.
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Reflect and continue
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