3 The equation \(x ^ { 2 } - \cosh ( x - 2 ) = 0\) has two roots, \(\alpha\) and \(\beta\), such that \(\alpha < \beta\).
- Use the iterative formula
$$x _ { n + 1 } = g \left( x _ { n } \right) \text { where } g \left( x _ { n } \right) = \sqrt { \cosh \left( x _ { n } - 2 \right) } \text {, }$$
starting with \(x _ { 0 } = 1\), to find \(\alpha\) correct to \(\mathbf { 3 }\) decimal places.
The diagram shows the part of the graphs of \(\mathrm { y } = \mathrm { x }\) and \(\mathrm { y } = \mathrm { g } ( \mathrm { x } )\) for \(0 \leqslant x \leqslant 7\).
\includegraphics[max width=\textwidth, alt={}, center]{83a06341-74e9-4f47-9104-e8e0259e7dfa-3_760_657_753_246} - Explain why the iterative formula used to find \(\alpha\) cannot successfully be used to find \(\beta\), even if \(x _ { 0 }\) is very close to \(\beta\).
- Use the relaxed iteration
$$\mathrm { x } _ { \mathrm { n } + 1 } = ( 1 - \lambda ) \mathrm { x } _ { \mathrm { n } } + \lambda \mathrm { g } \left( \mathrm { x } _ { \mathrm { n } } \right) ,$$
with \(\lambda = - 0.21\) and \(x _ { 0 } = 6.4\), to find \(\beta\) correct to \(\mathbf { 3 }\) decimal places.
In part (c) the method of relaxation was used to convert a divergent sequence of approximations into a convergent sequence.
- State one other application of the method of relaxation.