Feb/Mar 2019 p32 q2
1822
The sequence of values given by the iterative formula
\(x_{n+1} = \frac{2x_n^6 + 12x_n}{3x_n^5 + 8}\),
with initial value \(x_1 = 2\), converges to \(\alpha\).
(i) Use the formula to calculate \(\alpha\) correct to 4 decimal places. Give the result of each iteration to 6 decimal places. [3]
(ii) State an equation satisfied by \(\alpha\) and hence find the exact value of \(\alpha\). [2]
Solution
(i) Start with \(x_1 = 2\).
Calculate \(x_2 = \frac{2(2)^6 + 12(2)}{3(2)^5 + 8} = \frac{128 + 24}{96 + 8} = \frac{152}{104} = 1.461538\).
Calculate \(x_3 = \frac{2(1.461538)^6 + 12(1.461538)}{3(1.461538)^5 + 8} \approx 1.336405\).
Calculate \(x_4 = \frac{2(1.336405)^6 + 12(1.336405)}{3(1.336405)^5 + 8} \approx 1.320756\).
Calculate \(x_5 = \frac{2(1.320756)^6 + 12(1.320756)}{3(1.320756)^5 + 8} \approx 1.319545\).
Calculate \(x_6 = \frac{2(1.319545)^6 + 12(1.319545)}{3(1.319545)^5 + 8} \approx 1.319505\).
Calculate \(x_7 = \frac{2(1.319505)^6 + 12(1.319505)}{3(1.319505)^5 + 8} \approx 1.319505\).
Thus, \(\alpha \approx 1.3195\) to 4 decimal places.
(ii) The equation satisfied by \(\alpha\) is \(\alpha = \frac{2\alpha^6 + 12\alpha}{3\alpha^5 + 8}\).
Solving \(\alpha = \frac{2\alpha^6 + 12\alpha}{3\alpha^5 + 8}\) gives \(\alpha^5 = 4\).
Thus, \(\alpha = \sqrt[5]{4}\).
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