6 The equation \(0.5 \ln x - x ^ { 2 } + x + 1 = 0\) has two roots \(\alpha\) and \(\beta\), such that \(0 < \alpha < 1\) and \(1 < \beta < 2\).
- Use the Newton-Raphson method with \(x _ { 0 } = 1\) to obtain \(\beta\) correct to \(\mathbf { 6 }\) decimal places.
Fig. 6.1 shows part of the graph of \(y = 0.5 \ln x - x ^ { 2 } + x + 1\).
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{945883ad-c153-4c51-83d3-978e4c769ed5-06_1112_1156_529_354}
\captionsetup{labelformat=empty}
\caption{Fig. 6.1}
\end{figure} - On the copy of Fig. 6.1 in the Printed Answer Booklet, illustrate the Newton-Raphson method working to obtain \(x _ { 1 }\) from \(x _ { 0 } = 1\).
Beth is trying to find \(\alpha\) correct to 6 decimal places.
- Suggest a reason why she might choose the Newton-Raphson method instead of fixed point iteration.
Beth tries to find \(\alpha\) using the Newton-Raphson method with a starting value of \(x _ { 0 } = 0.5\). Her spreadsheet output is shown in Fig. 6.2.
\begin{table}[h]
| \(r\) | \(\mathrm { x } _ { \mathrm { r } }\) |
| 0 | 0.5 |
| 1 | - 0.40343 |
| 2 | \#NUM! |
\captionsetup{labelformat=empty}
\caption{Fig. 6.2}
- Explain how the display \#NUM! has arisen in the cell for \(x _ { 2 }\).
Beth decides to use the iterative formula
$$x _ { n + 1 } = g \left( x _ { n } \right) = \sqrt { 0.5 \ln \left( x _ { n } \right) + x _ { n } + 1 }$$
- Determine the outcome when Beth uses this formula with \(x _ { 0 } = 0.5\).
- 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 \(\lambda = - 0.041\) and \(x _ { 0 } = 0.5\) to obtain \(\alpha\) correct to \(\mathbf { 6 }\) decimal places.