Exact value from iterative result

2 The sequence of values given by the iterative formula $$x _ { n + 1 } = \frac { x _ { n } \left( x _ { n } ^ { 3 } + 100 \right) } { 2 \left( x _ { n } ^ { 3 } + 25 \right) }$$ with initial value \(x _ { 1 } = 3.5\), converges to \(\alpha\).

1. Use this formula to calculate \(\alpha\) correct to 4 decimal places, showing the result of each iteration to 6 decimal places.
2. State an equation satisfied by \(\alpha\) and hence find the exact value of \(\alpha\).

6 It is given that \(\mathrm { f } ( x ) = 1 - \frac { 7 } { x ^ { 2 } }\).

1. Use the Newton-Raphson method, with a first approximation \(x _ { 1 } = 2.5\), to find the next approximations \(x _ { 2 }\) and \(x _ { 3 }\) to a root of \(\mathrm { f } ( x ) = 0\). Give the answers correct to 6 decimal places. [3]
2. The root of \(\mathrm { f } ( x ) = 0\) for which \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\) are approximations is denoted by \(\alpha\). Write down the exact value of \(\alpha\).
3. The error \(e _ { n }\) is defined by \(e _ { n } = \alpha - x _ { n }\). Find \(e _ { 1 } , e _ { 2 }\) and \(e _ { 3 }\), giving your answers correct to 5 decimal places. Verify that \(e _ { 3 } \approx \frac { e _ { 2 } ^ { 3 } } { e _ { 1 } ^ { 2 } }\).