6 The secant method is to be used to solve the equation \(x - \ln ( \cos x ) - 1 = 0\).
- Show that starting with \(x _ { 0 } = - 1\) and \(x _ { 1 } = 0\) leads to the method failing to find the root between \(x = 0\) and \(x = 1\).
The spreadsheet printout shows the application of the secant method starting with \(x _ { 0 } = 0\) and \(x _ { 1 } = 1\). Successive approximations to the root are in column E.
| A | B | C | D | E |
| 1 | \(x _ { n }\) | \(\mathrm { f } \left( x _ { n } \right)\) | \(x _ { n + 1 }\) | \(\mathrm { f } \left( x _ { n + 1 } \right)\) | \(x _ { n + 2 }\) |
| 2 | 0 | -1 | 1 | 0.6156265 | 0.6189549 |
| 3 | 1 | 0.6156265 | 0.6189549 | -0.175846 | 0.7036139 |
| 4 | 0.6189549 | -0.1758461 | 0.7036139 | -0.025245 | 0.7178053 |
| 5 | 0.7036139 | -0.0252451 | 0.7178053 | 0.0011619 | 0.7171808 |
| 6 | 0.7178053 | 0.0011619 | 0.7171808 | - 7.4 E -06 | 0.7171848 |
| 7 | 0.7171808 | -7.402E-06 | 0.7171848 | -2.16E-09 | 0.7171848 |
| 8 | 0.7171848 | -2.16E-09 | 0.7171848 | 3.997 E -15 | 0.7171848 |
- What feature of column B shows that this application of the secant method has been successful?
- Write down a suitable spreadsheet formula to obtain the value in cell E2.
Some analysis of convergence is carried out, and the following spreadsheet output is obtained.
| A | B | C | D | E | F | G | H |
| 1 | \(x _ { n }\) | \(\mathrm { f } \left( x _ { n } \right)\) | \(x _ { n + 1 }\) | \(\mathrm { f } \left( x _ { n + 1 } \right)\) | \(x _ { n + 2 }\) | | | |
| 2 | 0 | -1 | 1 | 0.6156265 | 0.6189549 | 0.084659 | 0.167629 | 1.980053 |
| 3 | 1 | 0.6156265 | 0.6189549 | -0.175846 | 0.7036139 | 0.0141913 | -0.044 | -3.10054 |
| 4 | 0.6189549 | -0.1758461 | 0.7036139 | -0.025245 | 0.7178053 | -0.0006244 | -0.00633 | 10.13727 |
| 5 | 0.7036139 | -0.0252451 | 0.7178053 | 0.0011619 | 0.7171808 | 3.953 E -06 | 0.000292 | 73.83899 |
| 6 | 0.7178053 | 0.0011619 | 0.7171808 | - 7.4 E -06 | 0.7171848 | 1.154 E -09 | -1.8E-06 | |
| 7 | 0.7171808 | -7.402E-06 | 0.7171848 | - 2.16 E -09 | 0.7171848 | - 2.109 E -15 | | |
| 8 | 0.7171848 | - 2.16 E -09 | 0.7171848 | 3.997 E -15 | 0.7171848 | | | |
The formula in cell F2 is =E3-E2. The formula in cell G2 is =F3/F2. The formula in cell H2 is =F3/(F2\^{}2). - (A) Explain the purpose of each of these three formulae.
(B) Explain the significance of the values in columns G and H in terms of the rate of convergence of the secant method. - Explain why the values in cells F6 and F7 are not 0 .
[Question 7 is printed overleaf.]