5 You are given that
\(g ( x ) = \frac { \sqrt [ 3 ] { x ^ { x } + 25 } } { 2 }\).
Fig. 5.1 shows two values of \(x\) and the associated values of \(\mathrm { g } ( x )\).
\begin{table}[h]
| \(x\) | 1.45 | 1.55 |
| \(g ( x )\) | 1.49468 | 1.49949 |
\captionsetup{labelformat=empty}
\caption{Fig. 5.1}
\end{table}
- Use the central difference method to calculate an estimate of \(\mathrm { g } ^ { \prime } ( 1.5 )\), giving your answer correct to 3 decimal places.
The equation \(x ^ { x } - 8 x ^ { 3 } + 25 = 0\) has two roots, \(\alpha\) and \(\beta\), such that \(\alpha \approx 1.5\) and \(\beta \approx 4.4\).
- Obtain the iterative formula \(x _ { n + 1 } = g \left( x _ { n } \right) = \frac { \sqrt [ 3 ] { x _ { n } ^ { X _ { n } } + 25 } } { 2 }\).
- Use your answer to part (a) to explain why it is possible that the iterative formula \(x _ { n + 1 } = g \left( x _ { n } \right) = \frac { \sqrt [ 3 ] { x _ { n } ^ { X _ { n } } + 25 } } { 2 }\) may be used to find \(\alpha\).
- Starting with \(x _ { 0 } = 1.5\), use the iterative formula to find \(x _ { 1 } , x _ { 2 } , x _ { 3 } , x _ { 4 } , x _ { 5 }\), and \(x _ { 6 }\).
- Use your answer to part (d) to state the value of \(\alpha\) correct to 8 decimal places.
Starting with \(x _ { 0 } = 4.5\) the same iterative formula is used in an attempt to find \(\beta\). The results are shown in Fig. 5.2.
\begin{table}[h]
| \(n\) | \(x _ { n }\) |
| 0 | 4.5 |
| 1 | 4.81826433 |
| 2 | 6.27473453 |
| 3 | 23.2937196 |
| 4 | \(2.0654 \mathrm { E } + 10\) |
| 5 | \#NUM! |
\captionsetup{labelformat=empty}
\caption{Fig. 5.2}
- Explain why \#NUM! is displayed in the cell for \(x _ { 5 }\).
- On the diagram in the Printed Answer Booklet, starting with \(x _ { 0 } = 4.5\), illustrate how the iterative formula works to find \(x _ { 1 }\) and \(x _ { 2 }\).
- Determine what happens when the relaxed iteration \(x _ { n + 1 } = ( 1 - \lambda ) x _ { n } + \lambda g \left( x _ { n } \right)\) is used to try to find \(\beta\) with \(x _ { 0 } = 4.5\), in each of the following cases.
- \(\lambda = 0.5\)
- \(\lambda = - 0.4\)