6 Charlie uses fixed point iteration to find a sequence of approximations to the root of the equation \(\sin ^ { - 1 } ( x ) - x ^ { 2 } + 1 = 0\). Charlie uses the iterative formula \(\mathrm { x } _ { \mathrm { n } + 1 } = \mathrm { g } \left( \mathrm { x } _ { \mathrm { n } } \right)\), where \(\mathrm { g } \left( \mathrm { x } _ { \mathrm { n } } \right) = \sin \left( \mathrm { x } _ { \mathrm { n } } ^ { 2 } - 1 \right)\).
Two sections of the associated spreadsheet output, showing \(x _ { 0 }\) to \(x _ { 6 }\) and \(x _ { 102 }\) to \(x _ { 108 }\), are shown in Fig. 6.1. \begin{table}[h] \end{table}

1. Use the information in Fig. 6.1 to find the value of the root as accurately as you can, justifying the precision quoted. The relaxed iteration \(\mathrm { x } _ { \mathrm { n } + 1 } = ( 1 - \lambda ) \mathrm { x } _ { \mathrm { n } } + \lambda \mathrm { g } \left( \mathrm { x } _ { \mathrm { n } } \right)\), with \(\lambda = 0.51\) and \(x _ { 0 } = 0\), is to be used to find the root of the equation \(\sin ^ { - 1 } ( x ) - x ^ { 2 } + 1 = 0\).
2. Complete the copy of Fig. 6.2 in the Printed Answer Booklet, giving the values of \(\mathrm { x } _ { \mathrm { r } }\) correct to 7 decimal places and the values in the difference column and ratio column correct to 3 significant figures. \begin{table}[h]

   \(r\)	\(\mathrm { x } _ { \mathrm { r } }\)	difference	ratio
   0	0
   1
   2
   3
   4		-0.000192
   5		\(- 1.99 \times 10 ^ { - 7 }\)	0.00103
   6		\(- 1.82 \times 10 ^ { - 10 }\)	0.000914

   \captionsetup{labelformat=empty} \caption{Fig. 6.2}
   \end{table}
3. Write down the value of the root correct to 7 decimal places.
4. Explain why extrapolation could not be used in this case to find an improved approximation using this sequence of iterates. In this case the method of relaxation has been used to speed up the convergence of an iterative scheme.
5. Name another application of the method of relaxation.