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June 2011 p33 q6
1889
(i) By sketching a suitable pair of graphs, show that the equation \(\cot x = 1 + x^2\), where \(x\) is in radians, has only one root in the interval \(0 < x < \frac{1}{2}\pi\).
(ii) Verify by calculation that this root lies between 0.5 and 0.8.
(iii) Use the iterative formula \(x_{n+1} = \arctan\left( \frac{1}{1 + x_n^2} \right)\) to determine this root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.
Solution
(i) Sketch the graphs of \(y = \cot x\) and \(y = 1 + x^2\) over the interval \(0 < x < \frac{1}{2}\pi\). The graph of \(\cot x\) decreases from infinity to 0, while \(y = 1 + x^2\) is a parabola opening upwards starting at 1. They intersect at only one point in this interval, indicating one root.
(ii) Calculate \(\cot 0.5 - (1 + 0.5^2)\) and \(\cot 0.8 - (1 + 0.8^2)\). The sign changes between these values, confirming a root exists between 0.5 and 0.8.
(iii) Use the iterative formula \(x_{n+1} = \arctan\left( \frac{1}{1 + x_n^2} \right)\) starting with \(x_0 = 0.5\). Iterate until the result stabilizes to 2 decimal places:
