P9.2aIteration and staircase/cobweb diagrams

(a) Answer: \(x_{2} = 11.3979, \ x_{3} = 11.4334, \ x_{4} = 11.4365\)

• M1 for an attempt at \(x_2\) using the iterative formula (may be implied by a correct value)
• A1 for \(x_2 = 11.3979\) (awrt)
• A1 for \(x_3 = 11.4334\) and \(x_4 = 11.4365\) (awrt)

Worked solution: \(x_{2} = 9 + \ln 11 = 11.397895\ldots\)
\(x_{3} = 9 + \ln 11.397895\ldots = 11.433429\ldots\)
\(x_{4} = 9 + \ln 11.433429\ldots = 11.436541\ldots\)
So \(x_{2} = 11.3979, \ x_{3} = 11.4334, \ x_{4} = 11.4365\) (4 d.p.).

(b) Answer: \(\alpha = 11.437\)

• M1 for continuing the iteration until successive values agree to 3 decimal places
• A1 for \(\alpha = 11.437\)

Worked solution: Continuing the iteration: \(x_{5} = 11.4368, \ x_{6} = 11.4368\).
The values have settled to 3 decimal places, so \(\alpha = 11.437\).

(a) Answer: Shown

• M1 for \(x^{3} = 3x^{2} + 4\) and dividing by \(x^{2}\)
• A1* for \(x = 3 + \frac{4}{x^{2}}\) (cso)

Worked solution: \(x^{3} = 3x^{2} + 4\), and dividing by \(x^{2}\) (\(x \ne 0\)) gives \(x = 3 + \frac{4}{x^{2}}\).

(b) Answer: \(x_{2} = 3.2500, \ x_{3} = 3.3787, \ x_{4} = 3.3504\)

• M1 for an attempt at \(x_2\) using the iterative formula (may be implied by a correct value)
• A1 for \(x_2 = 3.2500\) (awrt)
• A1 for \(x_3 = 3.3787\) and \(x_4 = 3.3504\) (awrt)

Worked solution: \(x_{2} = 3 + \frac{4}{4^{2}} = 3.250000\ldots\)
\(x_{3} = 3 + \frac{4}{3.250000\ldots^{2}} = 3.378698\ldots\)
\(x_{4} = 3 + \frac{4}{3.378698\ldots^{2}} = 3.350398\ldots\)
So \(x_{2} = 3.2500, \ x_{3} = 3.3787, \ x_{4} = 3.3504\) (4 d.p.).

(c) Answer: Shown: sign change on \([3.3545, 3.3555]\)

• M1 for choosing the interval \([3.3545, 3.3555]\) (or a tighter interval) and evaluating \(x^{3} - 3x^{2} - 4\) at both ends
• A1 for both values correct (\(-0.0109\) and \(0.00271\)), a change of sign, continuity and the conclusion

Worked solution: Let \(f(x) = x^{3} - 3x^{2} - 4\). \(f(3.3545) = -0.0109 \lt 0\) and \(f(3.3555) = 0.00271 \gt 0\).
There is a change of sign and \(f\) is continuous, so \(3.3545 \lt \alpha \lt 3.3555\), and therefore \(\alpha = 3.355\) to 3 decimal places.

(a) Answer: (i) \(2x = x^3 - 5\); (ii) \(x^3 = 2x + 5\), take cube roots.

• B1 for (i): \(2x = x^3 - 5 \Rightarrow x = \frac{x^3 - 5}{2}\)
• B1 for (ii): \(x^3 = 2x + 5 \Rightarrow x = \sqrt[3]{2x + 5}\)

Worked solution: (i) \(x^3 - 2x - 5 = 0 \Rightarrow 2x = x^3 - 5 \Rightarrow x = \frac{x^3 - 5}{2}\).
(ii) \(x^3 = 2x + 5 \Rightarrow x = \sqrt[3]{2x + 5}\).

(b) Answer: (i) 1.5, −0.8125, −2.7682; (ii) 2.0801, 2.0924, 2.0942

• M1 for using either formula correctly at least twice
• A1 for (i): \(1.5, -0.8125, -2.7682\)
• A1 for (ii): \(2.0801, 2.0924, 2.0942\)

Worked solution: (i) \(x_2 = \frac{8 - 5}{2} = 1.5\), \(x_3 = \frac{3.375 - 5}{2} = -0.8125\), \(x_4 = \frac{-0.5364 - 5}{2} = -2.7682\).
(ii) \(x_2 = \sqrt[3]{9} = 2.0801\), \(x_3 = \sqrt[3]{9.1602} = 2.0924\), \(x_4 = 2.0942\).

(c) Answer: \(\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{x^3 - 5}{2}\right) = \frac{3x^2}{2} \approx 6.6\) near \(\alpha\) (greater than 1, diverges); \(\frac{\mathrm{d}}{\mathrm{d}x}(2x + 5)^{\frac{1}{3}} = \frac{2}{3}(2x + 5)^{-\frac{2}{3}} \approx 0.15\) (between −1 and 1, converges).

• M1 for the gradient of \(\frac{x^3 - 5}{2}\) at or near \(\alpha\): \(\frac{3x^2}{2} \approx 6.6\)
• M1 for the gradient of \(\sqrt[3]{2x + 5}\): \(\frac{2}{3}(2x + 5)^{-\frac{2}{3}} \approx 0.15\)
• A1 for concluding: an iteration \(x_{n + 1} = \mathrm{g}(x_n)\) converges when \(|\mathrm{g}'(\alpha)| \lt 1\) and moves away from \(\alpha\) when \(|\mathrm{g}'(\alpha)| \gt 1\)

Worked solution: For \(\mathrm{g}_1(x) = \frac{x^3 - 5}{2}\), \(\mathrm{g}_1'(x) = \frac{3x^2}{2}\), which is about 6.6 near \(\alpha\). As \(|\mathrm{g}_1'| \gt 1\), the distance from \(\alpha\) grows at each step, so (i) diverges.
For \(\mathrm{g}_2(x) = (2x + 5)^{\frac{1}{3}}\), \(\mathrm{g}_2'(x) = \frac{2}{3}(2x + 5)^{-\frac{2}{3}}\), about 0.15 near \(\alpha\). As \(|\mathrm{g}_2'| \lt 1\), the distance shrinks, so (ii) converges.

(d) Answer: A staircase, because the gradient of \(y = \sqrt[3]{2x + 5}\) near \(\alpha\) is positive (between 0 and 1); the iterates increase steadily towards \(\alpha\).

• B1 for staircase, because \(0 \lt \mathrm{g}'(\alpha) \lt 1\) (the iterates approach \(\alpha\) from one side)

Worked solution: The gradient of \(y = \sqrt[3]{2x + 5}\) near \(\alpha\) is positive (about 0.15), so the iterates approach \(\alpha\) from one side: a staircase diagram.