Mathematics 9709/23 — October/November 2016

Walkthrough

We start with the given initial value $x_1 = 2$ and substitute it into the iterative formula $x_{n+1} = \frac{4}{x_n^2} + \frac{2x_n}{3}$. Each step produces a new approximation. The problem asks us to give each result to 5 decimal places, which gives us enough precision to safely round the final answer to 3 decimal places. We keep iterating until two consecutive values agree to the required 3 decimal places. Here $x_5$ and $x_6$ both equal $2.28942$, which rounds to $2.289$. Since the sequence has clearly converged (two consecutive iterations give the same 5-decimal value), we can confidently state the root to 3 decimal places.

Key Takeaways

This question tests the core idea of an iterative formula: you repeatedly feed the previous output back into the formula to generate a sequence that converges to a root. It also emphasises the importance of maintaining sufficient precision during intermediate steps so the final rounded answer is reliable, and of explicitly showing enough iterations to justify the claimed accuracy.

Common Mistakes

• Rounding intermediate values to 3 decimal places too early, which can push the final answer off by one unit in the last digit.
• Making arithmetic slips when squaring or dividing, especially with recurring decimals.
• Stopping after only one or two iterations without showing that the value has stabilised, which loses the justification mark.

Things to Be Careful About

• Always give each iteration to 5 decimal places as instructed.
• The mark scheme requires you to justify accuracy to 3 dp, either by showing consecutive identical values or by demonstrating a sign change in the interval $(2.2885, 2.2895)$.
• Make sure you use the exact formula each time; do not round the formula's coefficients.

Walkthrough

For a converging iterative sequence, the limit $\alpha$ satisfies the fixed-point equation. This means that as $n$ becomes very large, both $x_n$ and $x_{n+1}$ become arbitrarily close to $\alpha$, so we can replace both with $\alpha$ in the formula. Doing this gives the equation $\alpha = \frac{4}{\alpha^2} + \frac{2\alpha}{3}$.

To solve it exactly, we first multiply every term by $\alpha^2$ to clear the fraction, obtaining $\alpha^3 = 4 + \frac{2\alpha^3}{3}$. Next, we multiply through by 3 to remove the remaining fraction, giving $3\alpha^3 = 12 + 2\alpha^3$. Subtracting $2\alpha^3$ from both sides yields $\alpha^3 = 12$. Finally, taking the cube root gives the exact value $\alpha = \sqrt[3]{12}$.

Key Takeaways

This part demonstrates the fundamental relationship between an iterative formula and the root it converges to: the limit satisfies the fixed-point equation $x = F(x)$. It also reinforces algebraic manipulation of equations involving fractions and indices, and how to express an exact irrational root.

Common Mistakes

• Forgetting to replace both $x_n$ and $x_{n+1}$ with $\alpha$ — a common error is leaving one as a variable.
• Making algebraic errors when multiplying by $\alpha^2$, such as forgetting to multiply the $\frac{2\alpha}{3}$ term correctly.
• Not simplifying the cubic equation fully, or incorrectly taking a square root instead of a cube root.

Things to Be Careful About

• The equation must be stated clearly before solving; the mark scheme awards a mark for the fixed-point equation itself.
• Ensure the final answer is given in exact form ($\sqrt[3]{12}$ or $12^{1/3}$), not as a decimal approximation.
• Double-check the algebra: multiplying by $\alpha^2$ and then by 3 must be applied to every term.