4 It is given that the equation \(x ^ { 4 } - 2 x - 1 = 0\) has only one positive root, \(\alpha\), and \(1.3 < \alpha < 1.5\).
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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 \(\alpha\).- 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 \(\alpha\).
- 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.