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June 2017 p33 q6
1824
The equation \(\cot x = 1 - x\) has one root in the interval \(0 < x < \pi\), denoted by \(\alpha\).
(i) Show by calculation that \(\alpha\) is greater than 2.5.
(ii) Show that, if a sequence of values in the interval \(0 < x < \pi\) given by the iterative formula \(x_{n+1} = \pi + \arctan \left( \frac{1}{1-x_n} \right)\) converges, then it converges to \(\alpha\).
(iii) Use this iterative formula to determine \(\alpha\) correct to 3 decimal places. Give the result of each iteration to 5 decimal places.
Solution
(i) Calculate \(\cot(2.5)\) and \(1 - 2.5 = -1.5\). Since \(\cot(2.5) > -1.5\), \(\alpha > 2.5\).
(ii) Assume \(x = \pi + \arctan \left( \frac{1}{1-x} \right)\). Rearrange to \(\cot x = 1 - x\), showing convergence to \(\alpha\).
(iii) Using the iterative formula:
1. \(x_1 = 3.00000\)
2. \(x_2 = 2.57732\)
3. \(x_3 = 2.57605\)
4. \(x_4 = 2.57600\)
5. \(x_5 = 2.57600\)
Thus, \(\alpha = 2.576\) correct to 3 decimal places.