$$\mathrm { f } ( x ) = 2 \sin \left( x ^ { 2 } \right) + x - 2 , \quad 0 \leqslant x < 2 \pi$$
- Show that \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) between \(x = 0.75\) and \(x = 0.85\)
The equation \(\mathrm { f } ( x ) = 0\) can be written as \(x = [ \arcsin ( 1 - 0.5 x ) ] ^ { \frac { 1 } { 2 } }\).
- Use the iterative formula
$$x _ { n + 1 } = \left[ \arcsin \left( 1 - 0.5 x _ { n } \right) \right] ^ { \frac { 1 } { 2 } } , \quad x _ { 0 } = 0.8$$
to find the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\), giving your answers to 5 decimal places.
- Show that \(\alpha = 0.80157\) is correct to 5 decimal places.