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Nov 2016 p31 q6
1884
(i) By sketching a suitable pair of graphs, show that the equation \(\csc \frac{1}{2}x = \frac{1}{3}x + 1\) has one root in the interval \(0 < x \leq \pi\).
(ii) Show by calculation that this root lies between 1.4 and 1.6.
(iii) Show that, if a sequence of values in the interval \(0 < x \leq \pi\) given by the iterative formula \(x_{n+1} = 2 \sin^{-1} \left( \frac{3}{x_n + 3} \right)\) converges, then it converges to the root of the equation in part (i).
(iv) Use this iterative formula to calculate the root correct to 3 decimal places. Give the result of each iteration to 5 decimal places.
Solution
(i) Sketch the graphs of \(y = \csc \frac{1}{2}x\) and \(y = \frac{1}{3}x + 1\). The intersection of these graphs in the interval \(0 < x \leq \pi\) indicates a root.
(ii) Calculate the values of the expressions at \(x = 1.4\) and \(x = 1.6\). If there is a sign change, the root lies between these values.
(iii) Rearrange the iterative formula \(x_{n+1} = 2 \sin^{-1} \left( \frac{3}{x_n + 3} \right)\) to show it is equivalent to \(\csc \frac{1}{2}x = \frac{1}{3}x + 1\). This shows convergence to the root.
(iv) Use the iterative formula starting with an initial guess. Calculate successive values to 5 decimal places until they stabilize to 3 decimal places. The final answer is 1.471.