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Fixed-point iteration and numerical roots — 9709 P3

Use a stated rearrangement and check its apparent convergence against a root bracket.

Notes and key methods

Rearrange \(\displaystyle f(x)=0\) into \(\displaystyle x=F(x)\), choose the stated starting value and compute \(\displaystyle x_{n}\)₊\(\displaystyle _{1}=F(x_{n})\), retaining guard digits. A different rearrangement of the same equation may fail to converge.

Agreement between successive rounded iterates is evidence to examine, not a universal error guarantee. Check the final rounded value against the original equation and, where possible, a sign-change bracket at the rounding boundaries.

Check the conditions

Three worked examples

Example 1 · Fixed-point iteration and accuracy

Solve \(x^{3} + x = 32\) using \(x_{n+1}=\sqrt[3]{32-x_n}\), with \(\displaystyle x_{0}=3. A\) calculator is appropriate; keep guard digits.

  1. Derive the stated iteration from the cubic.
  2. Calculate the first three iterates.
  3. Find the root to six decimal places and justify the rounding.
  4. Find the limiting root to at least six decimal places.
Show the reasoning

(a)

Isolate \(\displaystyle x^{3}\) and take its real cube root.

Answer: \(x=\sqrt[3]{32-x}\)

(b)

Use the recurrence in order, with guard digits retained.

Answer: \(3.07231682568585\), \(3.06976090454799\), \(3.06985131201532\)

(c)

The root is 3.069848. The original increasing cubic changes sign between 3.0698475 and 3.0698485, which are the half-unit rounding boundaries.

Answer: \(3.069848\)

(d)

Continue iterating and check a rounding bracket in the original cubic.

\(3.06984822347764\)

Answer: \(3.06984822347764\).

Example 2 · Fixed-point iteration and accuracy

Solve \(x^{3} + x = 71\) using \(x_{n+1}=\sqrt[3]{71-x_n}\), with \(\displaystyle x_{0}=4. A\) calculator is appropriate; keep guard digits.

  1. Derive the stated iteration from the cubic.
  2. Calculate the first three iterates.
  3. Find the root to six decimal places and justify the rounding.
  4. Find the limiting root to at least six decimal places.
Show the reasoning

(a)

Isolate \(\displaystyle x^{3}\) and take its real cube root.

Answer: \(x=\sqrt[3]{71-x}\)

(b)

Use the recurrence in order, with guard digits retained.

Answer: \(4.06154810044568\), \(4.06030403499317\), \(4.06032918871045\)

(c)

The root is 4.060329. The original increasing cubic changes sign between 4.0603285 and 4.0603295, which are the half-unit rounding boundaries.

Answer: \(4.060329\)

(d)

Continue iterating and check a rounding bracket in the original cubic.

\(4.06032869021048\)

Answer: \(4.06032869021048\).

Example 3 · Fixed-point iteration and accuracy

Solve \(x^{3} + x = 131\) using \(x_{n+1}=\sqrt[3]{131-x_n}\), with \(\displaystyle x_{0}=5. A\) calculator is appropriate; keep guard digits.

  1. Derive the stated iteration from the cubic.
  2. Calculate the first three iterates.
  3. Find the root to six decimal places and justify the rounding.
  4. Find the limiting root to at least six decimal places.
Show the reasoning

(a)

Isolate \(\displaystyle x^{3}\) and take its real cube root.

Answer: \(x=\sqrt[3]{131-x}\)

(b)

Use the recurrence in order, with guard digits retained.

Answer: \(5.01329793496458\), \(5.0131215623314\), \(5.01312390167098\)

(c)

The root is 5.013124. The original increasing cubic changes sign between 5.0131235 and 5.0131245, which are the half-unit rounding boundaries.

Answer: \(5.013124\)

(d)

Continue iterating and check a rounding bracket in the original cubic.

\(5.01312387104904\)

Answer: \(5.01312387104904\).

Common mistakes

Practise with answer checking

Original exam-style problems

Write complete solutions before opening the worked solution. The marking guidance rewards reasoning as well as answers.

Examination practice · 2020–2026

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