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Nov 2003 p3 q5
1897
(i) By sketching suitable graphs, show that the equation \(\sec x = 3 - x^2\) has exactly one root in the interval \(0 < x < \frac{1}{2}\pi\).
(ii) Show that, if a sequence of values given by the iterative formula \(x_{n+1} = \cos^{-1} \left( \frac{1}{3-x_n^2} \right)\) converges, then it converges to a root of the equation given in part (i).
(iii) Use this iterative formula, with initial value \(x_1 = 1\), to determine the root in the interval \(0 < x < \frac{1}{2}\pi\) correct to 2 decimal places, showing the result of each iteration.
Solution
(i) Sketch the graphs of \(y = \sec x\) and \(y = 3 - x^2\) for \(0 < x < \frac{1}{2}\pi\). The graph of \(y = \sec x\) has a vertical asymptote at \(x = \frac{1}{2}\pi\) and is positive in the interval. The graph of \(y = 3 - x^2\) is a downward-opening parabola with a positive y-intercept. The intersection of these graphs in the given interval shows exactly one root.
(ii) Assume \(\alpha = \cos^{-1} \left( \frac{1}{3-\alpha^2} \right)\). Rearrange to get \(\cos \alpha = \frac{1}{3-\alpha^2}\), which implies \(\sec \alpha = 3 - \alpha^2\). Thus, if the sequence converges, it converges to a root of the equation \(\sec x = 3 - x^2\).
(iii) Using the iterative formula \(x_{n+1} = \cos^{-1} \left( \frac{1}{3-x_n^2} \right)\) with \(x_1 = 1\):
