Iterative formula from rearrangement

A question is this type if and only if it involves using an iterative formula derived by rearranging an equation (not Newton-Raphson) to find a root, often with staircase/cobweb diagrams or convergence analysis.

3 questions

OCR C3 Q8
8. (i) Solve the equation $$\pi - 3 \cos ^ { - 1 } \theta = 0$$ (ii) Sketch on the same diagram the curves \(y = \cos ^ { - 1 } ( x - 1 ) , 0 \leq x \leq 2\) and \(y = \sqrt { x + 2 } , x \geq - 2\). Given that \(\alpha\) is the root of the equation $$\cos ^ { - 1 } ( x - 1 ) = \sqrt { x + 2 }$$ (iii) show that \(0 < \alpha < 1\),
(iv) use the iterative formula $$x _ { n + 1 } = 1 + \cos \sqrt { x _ { n } + 2 }$$ with \(x _ { 0 } = 1\) to find \(\alpha\) correct to 3 decimal places.
You should show the result of each iteration.
OCR FP2 2013 January Q8
8 It is required to solve the equation \(\ln ( x - 1 ) - x + 3 = 0\).
You are given that there are two roots, \(\alpha\) and \(\beta\), where \(1.1 < \alpha < 1.2\) and \(4.1 < \beta < 4.2\).
  1. The root \(\beta\) can be found using the iterative formula $$x _ { n + 1 } = \ln \left( x _ { n } - 1 \right) + 3$$ (a) Using this iterative formula with \(x _ { 1 } = 4.15\), find \(\beta\) correct to 3 decimal places. Show all your working.
    (b) Explain with the aid of a sketch why this iterative formula will not converge to \(\alpha\) whatever initial value is taken.
  2. (a) Show that the Newton-Raphson iterative formula for this equation can be written in the form $$x _ { n + 1 } = \frac { 3 - 2 x _ { n } - \left( x _ { n } - 1 \right) \ln \left( x _ { n } - 1 \right) } { 2 - x _ { n } }$$ (b) Use this formula with \(x _ { 1 } = 1.2\) to find \(\alpha\) correct to 3 decimal places.
OCR FP2 2012 June Q4
4 It is given that the equation \(x ^ { 4 } - 2 x - 1 = 0\) has only one positive root, \(\alpha\), and \(1.3 < \alpha < 1.5\).

  1. \includegraphics[max width=\textwidth, alt={}, center]{72a1330a-c6dc-4f3a-9b0e-333b099f4509-2_433_424_1119_817} The diagram shows a sketch of \(y = x\) and \(y = \sqrt [ 4 ] { 2 x + 1 }\) for \(x \geqslant 0\). Use the iteration \(x _ { n + 1 } = \sqrt [ 4 ] { 2 x _ { n } + 1 }\) with \(x _ { 1 } = 1.35\) to find \(x _ { 2 }\) and \(x _ { 3 }\), correct to 4 decimal places. On the copy of the diagram show how the iteration converges to \(\alpha\).
  2. For the same equation, the iteration \(x _ { n + 1 } = \frac { 1 } { 2 } \left( x _ { n } ^ { 4 } - 1 \right)\) with \(x _ { 1 } = 1.35\) gives \(x _ { 2 } = 1.1608\) and \(x _ { 3 } = 0.4077\), correct to 4 decimal places. Draw a sketch of \(y = x\) and \(y = \frac { 1 } { 2 } \left( x ^ { 4 } - 1 \right)\) for \(x \geqslant 0\), and show how this iteration does not converge to \(\alpha\).
  3. Find the positive root of the equation \(x ^ { 4 } - 2 x - 1 = 0\) by using the Newton-Raphson method with \(x _ { 1 } = 1.35\), giving the root correct to 4 decimal places.