7 Fig. 7.1 shows two values of \(x\) and the associated values of \(\mathrm { f } ( x )\).
\begin{table}[h]
| \(x\) | 3 | 3.5 |
| \(\mathrm { f } ( x )\) | 6.082763 | 4.596194 |
\captionsetup{labelformat=empty}
\caption{Fig. 7.1}
\end{table}
- Use the forward difference method to calculate an estimate of the gradient of \(\mathrm { f } ( x )\) at \(x = 3\), giving your answer correct to 4 decimal places.
Fig. 7.2 shows some spreadsheet output with additional values of \(x\) and the associated values of \(\mathrm { f } ( x )\).
\begin{table}[h]
| \(x\) | 3 | 3.00001 | 3.0001 | 3.001 | 3.01 | 3.1 |
| \(\mathrm { f } ( x )\) | 6.082763 | 6.08274 | 6.082541 | 6.08054 | 6.060454 | 5.848846 |
\captionsetup{labelformat=empty}
\caption{Fig. 7.2}
| \(h\) | 0.1 | 0.01 | 0.001 | 0.0001 | 0.00001 |
| estimate | -2.339165 | -2.230883 | -2.220532 | -2.219501 | -2.219398 |
| difference | 0.1082815 | 0.010352 | 0.0010307 | 0.000103 | |
| ratio | 0.095602 | 0.099567 | 0.0999568 | | |
\captionsetup{labelformat=empty}
\caption{Fig. 7.3}
- Explain whether or not Tommy is correct.
- Use extrapolation to determine the value of the gradient of \(\mathrm { f } ( x )\) at \(x = 3\) as accurately as possible, justifying the precision quoted.
- Calculate an estimate of the absolute error when \(\mathrm { f } ( 3 )\) is used as an approximation to \(\mathrm { f } ( 3.02 )\).